Field Arithmetic [4 ed.] 9783031280191, 9783031280207

Table of contents :
Preface to the Fourth Edition
Main New Results
Detailed List of New Results
Problems of Field Arithmetic
Structural Changes
Typing Programs
Preface to the Third Edition
Preface to the Second Edition
Preface to the First Edition
Notation and Conventions
Contents
Chapter 1 Infinite Galois Theory and Profinite Groups
1.1 Inverse Limits
1.2 Profinite Groups
1.3 Infinite Galois Theory
1.4 The 𝒑-adic Integers and the Prüfer Group
1.5 The Absolute Galois Group of a Finite Field
Exercises
Notes
Chapter 2 Valuations
2.1 Valuations, Places, and Valuation Rings
2.2 Discrete Valuations
2.3 Extensions of Valuations and Places
2.4 Galois Extensions
2.5 Integral Extensions and Dedekind Domains
Exercises
Chapter 3 Linear Disjointness
3.1 Linear Disjointness of Fields
3.2 Purely Transcendental Extensions
3.3 Separable Extensions
3.4 Regular Extensions
3.5 Primary Extensions
3.6 The Imperfect Degree of a Field
3.7 Derivatives
Exercises
Chapter 4 Algebraic Function Fields of One Variable
4.1 Function Fields of One Variable
4.2 The Riemann–Roch Theorem
4.3 Holomorphy Rings
4.4 Extensions of Function Fields
4.5 Completions
4.6 The Different
4.7 Hyperelliptic Fields
4.8 Hyperelliptic Fields with a Rational Quadratic Subfield
Exercises
Notes
Chapter 5 The Riemann Hypothesis for Function Fields
5.1 Class Numbers
5.2 Zeta Functions
5.3 Zeta Functions under Constant Field Extensions
5.4 The Functional Equation
5.5 The Riemann Hypothesis and Degree 1 Prime Divisors
5.6 Reduction Steps
5.7 An Upper Bound
5.8 A Lower Bound
Exercises
Notes
Chapter 6 Plane Curves
6.1 Affine and Projective Plane Curves
6.2 Points and Prime Divisors
6.3 The Genus of a Plane Curve
6.4 Points on a Curve over a Finite Field
Exercises
Notes
Chapter 7 The Chebotarev Density Theorem
7.1 Decomposition Groups
7.2 The Artin Symbol over Global Fields
7.3 Dirichlet Density
7.4 Function Fields
7.5 Number Fields
Exercises
Notes
Chapter 8 Ultraproducts
8.1 First Order Predicate Calculus
8.2 Structures
8.3 Models
8.4 Elementary Substructures
8.5 Ultrafilters
8.6 Ultraproducts
8.7 Regular Ultrafilters
8.8 Regular Ultraproducts
8.9 Nonprincipal Ultraproducts of Finite Fields
Exercises
Notes
Chapter 9 Decision Procedures
9.1 Deduction Theory
9.2 Gödel’s Completeness Theorem
9.3 Primitive Recursive Functions
9.4 Primitive Recursive Relations
9.5 Recursive Functions
9.6 Recursive and Primitive Recursive Procedures
9.7 A Reduction Step in Decidability Procedures
Exercises
Notes
Chapter 10 Algebraically Closed Fields
10.1 Elimination of Quantifiers
10.2 A Quantifier Elimination Procedure
10.3 Effectiveness
10.4 Applications
Exercises
Notes
Chapter 11 Elements of Algebraic Geometry
11.1 Algebraic Sets
11.2 Varieties
11.3 Substitutions in Irreducible Polynomials
11.4 Rational Maps
11.5 Hyperplane Sections
11.6 Descent
11.7 Projective Varieties
11.8 About the Language of Algebraic Geometry
Exercises
Notes
Chapter 12 Pseudo Algebraically Closed Fields
12.1 PAC Fields
12.2 Reduction to Plane Curves
12.3 The PAC Property is an Elementary Statement
12.4 PAC Fields of Positive Characteristic
12.5 PAC Fields with Valuations
12.6 The Absolute Galois Group of a PAC Field
12.7 A non-PAC Field 𝑲 with 𝑲ins PAC
Exercises
Notes
Chapter 13 Hilbertian Fields
13.1 Hilbert Sets and Reduction Lemmas
13.2 Hilbert Sets under Separable Algebraic Extensions
13.3 Purely Inseparable Extensions
13.4 Imperfect Fields
Exercises
Notes
Chapter 14 The Classical Hilbertian Fields
14.1 Further Reduction
14.2 Function Fields Over Infinite Fields
14.3 Global Fields
14.4 Hilbertian Rings
14.5 Hilbertianity via Coverings
14.6 Non-Hilbertian 𝒈-Hilbertian Fields
Exercises
Notes
Chapter 15 The Diamond Theorem
15.1 Twisted Wreath Products
15.2 The Diamond Theorem
15.3 Weissauer’s Theorem
Exercises
Notes
Chapter 16 Nonstandard Structures
16.1 Higher Order Predicate Calculus
16.2 Enlargements
16.3 Concurrent Relations
16.4 The Existence of Enlargements
16.5 Examples
Exercises
Notes
Chapter 17 The Nonstandard Approach to Hilbert’s Irreducibility Theorem
17.1 Criteria for Hilbertianity
17.2 Arithmetical Primes Versus Functional Primes
17.3 Fields with the Product Formula
17.4 Generalized Krull Domains
17.5 Examples
Exercises
Notes
Chapter 18 Galois Groups over Hilbertian Fields
18.1 Galois Groups of Polynomials
18.2 Stable Polynomials
18.3 Regular Realization of Finite Abelian Groups
18.4 Split Embedding Problems with Abelian Kernels
18.5 Embedding Quadratic Extensions in Z/2𝒏Z-Extensions
18.6 Z𝒑-Extensions of Hilbertian Fields
18.7 Symmetric and Alternating Groups over Hilbertian Fields
18.8 GAR-Realizations
18.9 Embedding Problems over Hilbertian Fields
18.10 Regularity of Finite Groups over Complete Discrete-Valued Fields
Exercises
Notes
Chapter 19 Small Profinite Groups
19.1 Finitely Generated Profinite Groups
19.2 Abelian Extensions of Hilbertian Fields
Exercises
Notes
Chapter 20 Free Profinite Groups
20.1 The Rank of a Profinite Group
20.2 Profinite Completions of Groups
20.3 Formations of Finite Groups
20.4 Free pro-C Groups
20.5 Subgroups of Free Discrete Groups
20.6 Open Subgroups of Free Profinite Groups
20.7 An Embedding Property
Exercises
Notes
Chapter 21 The Haar Measure
21.1 The Haar Measure of a Profinite Group
21.2 Existence of the Haar Measure
21.3 Independence
21.4 Cartesian Product of Haar Measures
21.5 The Haar Measure of the Absolute Galois Group
21.6 The PAC Nullstellensatz
21.7 Baire’s Theorem
21.8 The Bottom Theorem
21.9 Triviality of a Group of Automorphisms
21.10 PAC Fields over Uncountable Hilbertian Fields
21.11 On the Stability of Fields
21.12 PAC Galois Extensions of Hilbertian Fields
21.13 Algebraic Groups
Exercises
Notes
Chapter 22 Effective Field Theory and Algebraic Geometry
22.1 Presented Rings and Fields
22.2 Extensions of Presented Fields
22.3 Galois Extensions of Presented Fields
22.4 The Algebraic and Separable Closures of Presented Fields
22.5 Constructive Algebraic Geometry
22.6 Presented Rings and Constructible Sets
22.7 Basic Normal Stratification
Exercises
Notes
Chapter 23 The Elementary Theory of 𝒆-Free PAC Fields
23.1 ℵ1-Saturated PAC Fields
23.2 The Elementary Equivalence Theorem of ℵ1-Saturated PAC
Fields
23.3 Elementary Equivalence of PAC Fields
23.4 On On 𝒆-Free PAC Fields
23.5 The Elementary Theory of Perfect 𝒆-Free PAC Fields
23.6 The Probable Truth of a Sentence
23.7 Change of Base Field
23.8 The Fields 𝑲sep(𝝈1, . . . , 𝝈𝒆)
23.9 The Transfer Theorem
23.10 The Elementary Theory of Finite Fields
Exercises
Notes
Chapter 24 Problems of Arithmetical Geometry
24.1 The Decomposition-Intersection Procedure
24.2 𝑪𝒊-Fields and Weakly 𝑪𝒊-Fields
24.3 Perfect PAC Fields which are 𝑪𝒊
24.4 The Existential Theory of PAC Fields
24.5 Kronecker Classes of Number Fields
24.6 Davenport’s Problem
24.7 On Permutation Groups
24.8 Schur’s Conjecture
24.9 The Generalized Carlitz Conjecture
Exercises
Notes
Chapter 25 Projective Groups and Frattini Covers
25.1 The Frattini Group of a Profinite Group
25.2 Cartesian Squares
25.3 On On C-Projective Groups
25.4 Projective Groups
25.5 Free Products of Finitely many Profinite Groups
25.6 Frattini Covers
25.7 The Universal Frattini Cover
25.8 Projective Pro- 𝒑-Groups
25.9 Supernatural Numbers
25.10 The Sylow Theorems
25.11 On Complements of Normal Subgroups
25.12 The Universal Frattini 𝒑-Cover
25.13 Examples of Universal Frattini 𝒑-covers
25.14 The Special Linear Group SL(2, Z𝒑)
25.15 The General Linear Group GL(2, Z𝒑)
25.16 Absolute Galois Groups
Exercises
Notes
Chapter 26 PAC Fields and Projective Absolute Galois Groups
26.1 Projective Groups as Absolute Galois Groups
26.2 Countably Generated Projective Groups
26.3 Perfect PAC Fields of Bounded Corank
26.4 Basic Elementary Statements
26.5 Reduction Steps
26.6 Application of Ultraproducts
Exercises
Notes
Chapter 27 Frobenius Fields
27.1 The Field Crossing Argument
27.2 The Beckmann–Black Problem
27.3 The Embedding Property and Maximal Frattini Covers
27.4 The Smallest Embedding Cover of a Profinite Group
27.5 A Decision Procedure
27.6 Examples
27.7 Non-projective Smallest Embedding Cover
27.8 A Theorem of Iwasawa
27.9 Free Profinite Groups of Countable Rank
27.10 Application of the Nielsen–Schreier Formula
Exercises
Notes
Chapter 28 Free Profinite Groups of Infinite Rank
28.1 Characterization of Free Profinite Groups by Embedding Problems
28.2 Applications of Theorem 28.1.7
28.3 The Pro-Completion of a Free Discrete Group
28.4 The Group Theoretic Diamond Theorem
28.5 The Melnikov Group of a Profinite Group
28.6 Homogeneous Pro-C Groups
28.7 The The 𝑺-rank of Closed Normal Subgroups
28.8 Closed Normal Subgroups with a Basis Element
28.9 Accessible Subgroups
Exercises
Notes
Chapter 29 Random Elements in Profinite Groups
29.1 Random Elements in a Free Profinite Group
29.2 Random Elements in Free pro-𝒑 Groups
29.3 Random Ẑ𝒏
29.4 The Golod–Shafarevich Inequality
29.5 On the Index of Normal Subgroups Generated by Random Elements
29.6 Freeness of Normal Subgroups Generated by Random Elements
Notes
Chapter 30 Omega-free PAC Fields
30.1 Model Companions
30.2 The Model Companion in an Augmented Theory of Fields
30.3 New Non-Classical Hilbertian Fields
30.4 An Abundance of 𝝎-Free PAC Fields
Notes
Chapter 31 Hilbertian Subfields of Galois Extensions
31.1 Small Extensions
31.2 Auxiliary Results
31.3 The Main Result
Notes
Chapter 32 Undecidability
32.1 Turing Machines
32.2 Computation of Functions by Turing Machines
32.3 Recursive Inseparability of Sets of Turing Machines
32.4 The Predicate Calculus
32.5 Undecidability in the Theory of Graphs
32.6 Assigning Graphs to Profinite Groups
32.7 The Graph Conditions
32.8 Assigning Profinite Groups to Graphs
32.9 Assigning Fields to Graphs
32.10 Interpretation of the Theory of Graphs in the Theory of Fields
Exercises
Notes
Chapter 33 Algebraically Closed Fields with Distinguished Automorphisms
33.1 The Base Field
33.2 Coding in PAC Fields with Monadic Quantifiers
33.3 The Theory of Almost all ⟨𝑲, 𝝈1, . . . , 𝝈𝒆⟩’s
33.4 The Probability of Truth Sentences
Chapter 34 Galois Stratification
34.1 The Artin Symbol
34.2 Conjugacy Domains under Projections
34.3 Normal Stratification
34.4 Elimination of One Variable
34.5 The Complete Elimination Procedure
34.6 Model-Theoretic Applications
34.7 A Limit of Theories
Exercises
Notes
Chapter 35 Galois Stratification over Finite Fields
35.1 The Elementary Theory of Frobenius Fields
35.2 The Elementary Theory of Finite Fields
35.3 Near Rationality of the Zeta Function of a Galois Formula
Exercises
Notes
Chapter 36 Problems of Field Arithmetic
36.1 Solved Problems
36.2 Open Problems
References
Index

Citation preview

Ergebnisse der Mathematik und ihrer Grenzgebiete, 3. Folge A Series of Modern Surveys in Mathematics  11

Michael D. Fried Moshe Jarden

Field Arithmetic Fourth Edition Revised by Moshe Jarden

Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics Volume 11

Series Editors L. Ambrosio, Pisa, Italy V. Baladi, Paris, France G.-M. Greuel, Kaiserslautern, Germany M. Gromov, Bures-sur-Yvette, France G. Huisken, Tübingen, Germany J. Jost, Leipzig, Germany J. Kollár, Princeton, USA G. Laumon, Orsay, France U. Tillmann, Oxford, UK D. B. Zagier, Bonn, Germany

Ergebnisse der Mathematik und ihrer Grenzgebiete, now in its third sequence, aims to provide reports, at a high level, on important topics of mathematical research. Each book is designed as a reliable reference covering a significant area of advanced mathematics, spelling out related open questions, and incorporating a comprehensive, up-to-date bibliography.

Michael D. Fried • Moshe Jarden

Field Arithmetic Fourth Edition

Michael D. Fried Emeritus Professor of Mathematics at the University of California Irvine Irvine, CA, USA

Moshe Jarden School of Mathematical Sciences University of Tel Aviv Fac. Exact Sciences Tel Aviv, Israel

ISSN 0071-1136 ISSN 2197-5655 (electronic) Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics ISBN 978-3-031-28020-7 (eBook) ISBN 978-3-031-28019-1 https://doi.org/10.1007/978-3-031-28020-7 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 1986, 2005, 2008, 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To those precious colleagues who can appreciate the goals of and connections to other areas. To those who acknowledge the depth of what we already know from the absorbed contribution of previous generations before we address our papers. To those who can transcend the hubris of today’s mathematical community.

‫הוּבים‬ ִ ֲ‫ִלנְ כ ָָדי הָ א‬ ‫ ְבּנֵי כְּ ֵמהָ ה וְאֶ ֶרז‬, ַ‫ ְתּכֶלֶ ת ו ְָר ִקיﬠ‬,‫ ָנוֶה‬,‫ָתּהֵ ל‬ ‫ ְבּנֵי הֶ ְמיַת ו ְִﬠדּוֹ‬,‫ יָם ו ְִﬠנְ בָּ ל‬,‫ ַמ ְﬠיָן‬,‫ר ֶֹתם‬ ‫עוּרי וַאֲ ִביטַ ל‬ ִ ‫ ָמטָ ר ְונ ִָתיב ְבּנֵי‬,‫לַ הַ ב‬ ‫ גֶּבַ ע וְאָ ָדר ְבּנֵי גָּאי וְאַ לּוֹנָה‬,‫ לוֹטָ ן‬,‫ַשחַ ק‬

Preface to the Fourth Edition

The fourth edition of “Field Arithmetic” improves the third edition (that appeared in 2008) in a few ways. It removes a lot of typos and mathematical inaccuracies that occur in the third edition. A great part of the improvements is due to Aharon Razon who carefully read the whole manuscript, pointed out mistakes in the text, and made suggestions toward their corrections. I thank him from all my heart for his immense help.

Main New Results The most important new result in this edition is an amazing generalization of Theorem 30.4.8. That theorem says that if 𝐾 is a countable Hilbertian field and 𝑒 is a positive integer, then for almost all 𝝈 ∈ Gal(𝐾) 𝑒 , the field 𝐾sep [𝝈] is Hilbertian. The generalization, due to Lior Bary-Soroker and Arno Fehm, says that if 𝐾 is as above and 𝑁 is a Galois extension of 𝐾, then for almost all 𝝈 ∈ Gal(𝑁/𝐾) 𝑒 the maximal Galois extension 𝑁 [𝝈] 𝐾 of 𝐾 in the fixed field 𝑁 (𝝈) of 𝝈 in 𝑁 is Hilbertian (Theorem 31.3.2). Corollary 29.6.8(a) is a profinite group theory analog of Theorem 30.4.8. It says that for an integer 𝑒 ≥ 2 and with 𝐹ˆ𝑒 being the free profinite group on 𝑒 generators, for almost all 𝑒-tuples 𝝈 ∈ 𝐹ˆ𝑒𝑒 the closed normal subgroup [𝝈] generated by the coordinates of 𝝈 is isomorphic to the free profinite group 𝐹ˆ𝜔 on ℵ0 generators. The proof of that corollary relies among results on the Golod–Shafarevich inequality for finite 𝑝-groups. Section 29.4 recapitulates a cohomological proof of that inequality due to Peter Roquette. Another new result of the current edition says that for a countable Hilbertian field 𝐾 and a positive integer 𝑒, the set 𝐴𝐾 ,𝑒 of all 𝝈 ∈ Gal(𝐾) 𝑒 such that 𝐾sep (𝝈) is PAC is a Baire space which is dense in Gal(𝐾) 𝑒 (Theorem 21.7.5). This sheds some more light on the set 𝐴𝐾 ,𝑒 which is known to have Haar measure 1 in Gal(𝐾) 𝑒 (Theorem 21.6.1). Section 25.16 has been added to give recent information about profinite groups that are not isomorphic to absolute Galois groups of fields. vii

viii

Preface to the Fourth Edition

Detailed List of New Results • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Example 2.3.11 has been generalized. The proof of Lemma 3.1.8 has been completed. The proof of Lemma 3.4.13 has been revised. Lemma 7.1.9 is new. Remark 12.6.7 is new. Remark 12.7.10 cites a negative answer of Arno Fehm to Problem 11.7.9 of the 3rd edition. Lemma 13.1.5 has been strengthened. Proposition 14.4.6 has been amended. Example 17.5.8 has been extended. 𝑛(𝑛−1) The sign (−1) 2 in formula (18.23) and in the proof of Proposition 18.7.8 𝑛(𝑛+1) has been corrected to (−1) 2 . Proposition 18.9.1 and its proof have been slightly improved. Proposition 18.9.3 has been slightly corrected. Section 21.7 is new. Section 21.9 that presents a result of Chatzidakis along with a result of BarySoroker (Remark 21.9.5) is new. Remark 21.13.3 has been updated. Lemma 23.6.3 is new. The end of proof of Proposition 24.7.7 has been changed. Part B of the proof of Proposition 24.7.8 has been slightly shortened. Lemma 25.2.5 has been amended. Part (d) of Example 25.2.7 has been fixed. Section 25.6 has been amended. A proof for the missing uniqueness of the universal Frattini cover due to Dan Haran has been added to Proposition 25.7.1. Example 25.8.15 has been revised. Section 25.16 is new. Theorem 26.1.1 has been strengthened. Remark 26.1.6 supplies a new alternative proof of Theorem 25.4.7. Remark 27.7.4 points out a solution of two open problems of [FrJ86]. Corollary 27.10.10 is new. Lemma 28.2.1 has been rephrased. The proof of Theorem 28.4.10 is new. The proof of Proposition 28.6.2 has been revised. Lemma 28.6.4 has been rephrased. Part B of the proof of Proposition 28.8.1 has been somewhat revised. Section 29.4 (Golod–Shafarevich inequality) is new. Chapter 31 is new. The proof of Part (d) of Proposition 33.1.1 has been adjusted. The proof of Theorem 33.3.1 has been slightly reorganized. Theorem 34.6.4 is new.

Preface to the Fourth Edition

ix

Problems of Field Arithmetic Section 36.1 lists 23 problems of the previous editions of “Field Arithmetic” which have been solved or partially solved. Section 36.2 lists 27 unsolved problems, some of them are old and others are new.

Structural Changes Three chapters of the third edition have been split, each of them into two chapters. One chapter has been added. Specifically, Chapter 2 of the third edition has been split into two chapters, “Valuations” and “Linear Disjointness”. Likewise, Chapter 13 of the third edition has been split into two chapters, “The Classical Hilbertian Fields” and “The Diamond Theorem”. Finally, Chapter 16 of the third edition has been split into two chapters, “Galois Groups over Hilbertian Fields” and “Small Profinite Groups”. Chapter 31, “Hilbertian Subfields of Galois Extensions”, is new. These changes have resulted in a renumbering of the chapters.

Typing Programs We started to work on the first edition of “Field Arithmetic” back in 1979 and used the then available typing programs “troff” and “eqn”. For the second and the third edition we applied “plain tex”, using “The TeX Book” of Donald E. Knuth. The current edition has been typed with “latex” using Springer’s “svmono” document class. This allows a reader of the electronic version of the book to jump to cited theorems and equations by clicking on the appropriate number. Tel Aviv, Israel, Autumn 2022

Moshe Jarden

Preface to the Third Edition

The third edition of “Field Arithmetic” improves the second edition in two ways. First it removes many typos and mathematical inaccuracies that occur in the second edition. In particular, it fills out a big gap in the References of the second edition, where unfortunately all references between “Gilmore and Robinson” and “Kantor and Lubotzky” are missing. Secondly, the third edition reports on five open problems of the second edition that were solved since that edition appeared in 2005. János Kollár solved Problem 2 by proving that if each projective plane curve defined over a field 𝐾 has a 𝐾-rational point, then 𝐾 is PAC. János Kollár also solved Problem 3 and proved that if 𝐾 is a PAC field, 𝑤 is a ˜ ˜ and 𝑉 is a variety defined over 𝐾, then 𝑉 (𝐾) is 𝑤-dense in 𝑉 ( 𝐾). valuation of 𝐾, János Kollár partially settled Problem 21. He proved that every PAC field of characteristic 0 is 𝐶1 . Problem 31 was affirmatively solved by Lior Bary-Soroker by establishing an analog of the diamond theorem for the finitely generated non-Abelian free profinite groups. Finally, Eric Rosen suggested to reorganize Corollary 28.5.3 of the second edition that led to an affirmative solution of Problem 33. Unfortunately, a full account of the first four solutions is out of the scope of the present volume. Much of the improvements made in the present edition are due to Arno Fehm and Dan Haran. I am really indebted to them for their contribution. Tel Aviv, Israel, Autumn 2007

Moshe Jarden

xi

Preface to the Second Edition

The first edition of “Field Arithmetic” appeared in 1986. At the end of that edition we gave a list of twenty-two open problems. It is remarkable that since then fifteen of them have been partially or fully solved. Parallel to this, Field Arithmetic has developed in many directions establishing itself as an independent branch of Algebra and Number Theory. Some of these developments have been documented in books. We mention here “Groups as Galois groups” [Vol96] on consequences of the Riemann existence theorem, “Inverse Galois Groups” [MaM99] with a comprehensive report on finite Galois groups over number fields, “Profinite groups” [RiZ00] including the cohomology of profinite groups, “Analytic pro-𝑝 Groups” [DSMS99] on closed subgroups of GL(𝑛, Z 𝑝 ), “Subgroup Growth” [LuS03] on counting the number of subgroups of finitely generating groups, and “Multi-Valued Fields” [Ers01] on the model theory of fields with several valuations. This led to an official recognition of Field Arithmetic by the Mathematical Reviews in the form of MSC number 12E30. The extent which Field Arithmetic has reached makes it impossible for us to report in one extended volume about all the exciting results which have been achieved. We have therefore made several choices which best suit the spirit of this book but do not extend beyond the scope of one volume. The new results and additional topics have made it necessary to reorganize and to enlarge the sections dealing with background material. Of course, we took the opportunity afforded by editing a second edition to correct flaws and mistakes which occurred in the first edition and to add more details to proofs wherever it seemed useful. We list the major changes and additions we made in the book: Chapter 2 has been reorganized. Sections 2.5–2.9 of the first edition, which survey the theory of algebraic function fields of one variable, were moved to Chapter 3. Sections 2.5–2.8 dealing with linear disjointness, regular extensions, and separability appeared in the first edition as Sections 9.1–9.3. A nice application of linear disjointness is Leptin’s construction (which preceded that of Waterhouse) of a Galois group isomorphic to a given profinite group (Proposition 2.6.12). In addition to the introductory material about the theory of algebraic function fields of one variable, Chapter 3 now includes a proof of the Riemann–Hurwitz formula and a discussion of hyperelliptic curves. xiii

xiv

Preface to the Second Edition

The proof of Theorem 4.9 of the first edition, estimating the number of zeros of an absolutely irreducible polynomial over a finite field, had a flaw. This has been fixed in the proof of Theorem 5.4.1. Likewise, the inequality given by [FrJ86, Prop. 5.16] is inaccurate. This inaccuracy is fixed in Proposition 6.4.8. We find it more convenient to use the language of algebraic sets as introduced in [Wei62] for model-theoretic applications. Section 10.8 translates the basic concepts of that language to the now more commonly used language of schemes. Theorem 10.14 of the first edition (due to Frey–Prestel) says that the Henselian hull of a PAC field 𝐾 is 𝐾sep . Proposition 11.5.3 (due to Prestel) strengthens this ˜ theorem. It says that 𝐾 is 𝑤-dense in 𝐾˜ for every valuation 𝑤 of 𝐾. What we called “a separably Hilbertian field” in the first edition, is now called “a Hilbertian field” (Section 12.1). This agrees with the common usage and seems more appropriate for applications. Section 13.5 gives an alternative definition of Hilbertianity via coverings leading to the notion of “𝑔-Hilbertianity”. This sets the stage for a generalization of a theorem of Zannier: Every global field has an infinite normal extension 𝑁 which is 𝑔-Hilbertian but not Hilbertian (Theorem 13.6.2). Moreover, there is a unique factorization subring 𝑅 of 𝑁 with infinitely many irreducible elements (Example 15.5.8). This answers negatively Problems 14.20 and 14.21 of the first edition. Chapter 13 includes now one of the major results of Field Arithmetic which we call “Haran’s diamond theorem”: Let 𝑀1 and 𝑀2 be Galois extensions of a Hilbertian field 𝐾 and 𝑀 a field between 𝐾 and 𝑀1 𝑀2 not contained in 𝑀1 nor in 𝑀2 . Then 𝑀 is Hilbertian (Theorem 13.8.3). In particular, if 𝑁 is a Galois extension of 𝐾, then 𝑁 is not the compositum of two Galois extensions of 𝐾 neither of which is contained in the other. This settles Problems 12.18 and 12.19 of the first edition. The immediate goal of Hilbert’s irreducibility theorem was to realize the groups 𝑆 𝑛 and 𝐴𝑛 as Galois groups over Q. Chapter 16 is dedicated to realizations of Galois groups over arbitrary Hilbertian fields. One of the most important of these results is due to Harbater (Proposition 16.12.1): Let 𝐾 be a complete valued field, 𝑡 an indeterminate, and 𝐺 a finite group. Then 𝐺 is regular over 𝐾, that is, 𝐾 (𝑡) has a finite Galois extension 𝐹, regular over 𝐾, with Gal(𝐹/𝐾 (𝑡))  𝐺. Unfortunately, none of the three proofs of this theorem fits into the scope of this book. Section 16.6 proves a theorem of Whaples: Let 𝐾 be a field and 𝑝 a prime number. Suppose Z/𝑝Z (resp. Z/4Z if 𝑝 = 2) occurs as a Galois group over 𝐾. Then Z 𝑝 is realizable over 𝐾. Section 16.7 generalizes a theorem of Hilbert: Let 𝐾 be a field and 𝑛 ≥ 2 an integer with char(𝐾) ∤ (𝑛 − 1)𝑛. Then 𝐴𝑛 is regular over 𝐾. One of the most far-reaching attempts to realize arbitrary finite Galois groups over Hilbertian fields uses Matzat’s notion of GAR realization of simple finite groups: Let 𝐾 be a Hilbertian field and 𝛼: 𝐺 → Gal(𝐿/𝐾) a finite embedding problem over 𝐾. Suppose every composition factor of Ker(𝛼) has a GAR realization over 𝐾. Then the embedding problem is solvable. This leads in particular to the realization of many finite groups over Q (Remark 16.9.5).

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Chapter 17 deals mainly with Melnikov’s formations C (i.e. sets consisting of all finite groups whose composition factors belong to a given set of finite simple groups). We prove that every free abstract group 𝐹 is residually-C. Thus, if the free pro-C group with a given rank 𝑚 exists, then the canonical injection of 𝐹 into 𝐹ˆ𝑚 (C) is injective (Proposition 17.5.11 – Ribes–Zalesskii). Konrad Neumann improved former results of Fried–Geyer–Jarden and proved that every field is stable (Theorem 18.9.3). This allows the construction of PAC Hilbertian Galois extensions of arbitrary countable Hilbertian fields (Theorems 18.10.2 and 18.10.3). We survey Neumann’s proof in Section 18.9. The full proof unfortunately falls outside the scope of this book. It seemed to be well known that the concept of absolute irreducibility of a variety is elementary. Unfortunately, we could find no solid proof for it in the literature. Proposition 19.5.9 fills in the gap by proving that result. Section 21.2 includes now the classical results about 𝐶𝑖 -fields and not only the corresponding results about weakly 𝐶𝑖 -fields as was the case in Section 19.2 of the first edition. Sections 21.8 gives a complete proof of Schur’s Conjecture: If 𝑓 (𝑋) is a polynomial with coefficients in a global field 𝐾 with char(𝐾) ∤ deg( 𝑓 ) and 𝑓 permutes 𝑂 𝐾 /𝔭 for infinitely many primes 𝔭 of 𝐾, then each composition factor of 𝑓 is linearly related over 𝐾 to a Dickson polynomial of a prime degree. Section 21.7 proves all lemmas about permutation groups which are used in the proof of Schur’s Conjecture (Theorem 21.8.13). This includes the classification of subgroups of AGL(1, F𝑙 ) (Lemma 21.7.2), and the theorems of Schur (Proposition 21.7.7) and Burnside (Proposition 21.7.8) about doubly transitive permutation groups. Section 21.9 contains the Fried–Cohen version of Lenstra’s proof of the generalized Carlitz conjecture: Let 𝑝 be a prime number, 𝑞 a power of 𝑝, and 𝑓 ∈ F𝑞 [𝑋] a polynomial of degree 𝑛 > 1 which is not a power of 𝑝. Suppose 𝑓 permutes infinitely many finite extensions of F𝑞 . Then gcd(𝑛, 𝑞 − 1) = 1. The universal Frattini 𝑝-cover of a finite group plays a central role in Fried’s theory of modular towers. Section 22.11 introduces the former concept and proves its basic properties. Corollary 22.13.4 shows then that PSL(2, Z 𝑝 ) is a 𝑝-Frattini cover of PSL(2, F 𝑝 ) although it is not the universal 𝑝-Frattini cover. Chapter 23 puts together material on PAC fields which appeared in Section 20.5 and Chapter 21 of the first edition. The Beckmann–Black problem is a refinement of the inverse problem of Galois Theory. Débes proved that the problem has an affirmative solution over PAC fields (Theorem 24.2.2). Chapter 25 substantially extends the study of free profinite groups 𝐹 of infinite rank which appeared in Section 24.4 of the first edition. Most of the material goes back to Melnikov. We characterize closed normal subgroups of 𝐹 by their 𝑆-ranks, and prove that a closed subgroup of 𝐹 is accessible if and only if it is homogeneous. The first part of Chapter 25 reproduces the group-theoretic version of Haran’s diamond theorem. Chapter 26 is completely new. It describes the properties of the closed subgroup ⟨x⟩ and the closed normal subgroup [x] generated by a random 𝑒-tuple x = (𝑥1 , . . . , 𝑥 𝑒 ) of elements of a finitely generated free profinite group 𝐹 of finite

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rank 𝑛 ≥ 2. For example, with probability 1, ⟨x⟩  𝐹ˆ𝑒 (Proposition 26.1.7). This solves Problem 16.16 of the first edition. In addition, with a positive probability, [x] has infinite rank and is isomorphic to 𝐹ˆ𝜔 (Theorem 26.4.5 and Corollary 26.5.7). The latter result is based on the Golod–Shafarevich inequality. Chapter 28 considers an infinite field 𝐾 which is finitely generated over its base field. It proves that for 𝑒 ≥ 2 the theory of all sentences 𝜃 which hold in almost all ˜ 𝜎1 , . . . , 𝜎𝑒 ⟩ with (𝜎1 , . . . , 𝜎𝑒 ) ∈ Gal(𝐾) 𝑒 is undecidable. Moreover, structures ⟨𝐾, ˜ 𝜎1 , . . . , 𝜎𝑒 ⟩ is in general a nonrational the probability that a sentence 𝜃 holds in ⟨𝐾, number. Perhaps the most significant achievement of Field Arithmetic since the first edition appeared is the solution of Problem 24.41 of that edition: The absolute Galois group of a countable PAC Hilbertian field is free of rank ℵ0 . It was originally proved in characteristic 0 with complex analysis by Fried–Völklein. Then it was proved in the general case by Pop using rigid geometry and by Haran–Jarden–Völklein using “algebraic patching”. The two latter methods also lead to the proof that Gal(𝐶 (𝑡)) is a free profinite group if 𝐶 is an arbitrary algebraically closed field (Harbater, Pop, Haran–Jarden). The method of Fried–Völklein led to the theory of modular towers of Fried. A remote goal in Galois theory is the classification of absolute Galois groups among all profinite groups. In this framework, one tries to construct new absolute Galois groups out of existing ones. For example, for all fields 𝐾1 , . . . , 𝐾𝑛 there exists a field 𝐾 with Gal(𝐾) isomorphic to the free product of Gal(𝐾1 ), . . . , Gal(𝐾𝑛 ) (Pop, Melnikov, Ershov, Koenigsmann). Generalization of this result to infinite families of closed subgroups generalize the concepts “projective groups” and “PRC fields” or “P𝑝C fields” to “relatively projective groups” and “pseudo closed fields” (Haran– Jarden–Pop). They generalize the classification of projective groups as those profinite groups appearing as absolute Galois groups of PAC fields. All of the exciting material mentioned in the preceding two paragraphs lie unfortunately outside the scope of this volume. It is my pleasure to thank colleagues and friends who critically read parts of the manuscript of the present edition of “Field Arithmetic”: Michael Bensimhoun, David Brink, Gregory Cherlin, Michael Fried, Wulf-Dieter Geyer, Peter Müller, Dan Haran, Wolfgang Herfort, Alexander Lubotzky, Nikolay Nikolov, Dan Segal, Aharon Razon, and Irene Zimmermann. Tel Aviv, Israel, Spring 2004

Moshe Jarden

Preface to the First Edition

Our topic is the use of algebraic tools — coming mainly from algebraic geometry, number theory, and the theory of profinite groups — in the study of the elementary properties of classes of fields, and related algorithmic problems. (We take the precise definition of “elementary” from first order logic.) This subject has its more distant roots in Tarski’s observation that, as a consequence of elimination theory, the full elementary theory of the class of all algebraically closed fields is decidable; this relies on the Euclid algorithm of finding the greatest common divisor of two polynomials in one variable over a field. In its first phase this line of thought led to similar results on real closed fields and 𝑝-adic fields. The subject took a new turn with the work of James Ax [Ax2] on the elementary theory of the class of finite fields, which represents a radical departure in terms of the algebraic methods used. The analysis is based entirely on three properties of a finite field 𝐾: (1a) 𝐾 is perfect. (1b) 𝐾 has a unique extension of each degree. (1c) There is an explicitly computable function 𝑞(𝑑, 𝑚) such that any absolutely irreducible variety 𝑉 defined over 𝐾 will have a 𝐾-rational point if |𝐾 | > 𝑞(dim(𝑉), deg(𝑉)). The validity of the third condition for finite fields is a consequence of Riemann’s hypothesis for curves over finite fields. Methods of logic, specifically ultraproducts, led Ax to consider this condition for infinite fields as well, in which case the lower bound afforded by the function 𝑞 is vacuous, and the condition becomes: (2) Every absolutely irreducible variety over 𝐾 has a 𝐾-rational point. Fields satisfying (2) are said to be pseudo algebraically closed, or PAC. The second condition may be interpreted as a description of the absolute Galois group Gal(𝐾) as a profinite group: Gal(𝐾) is the free profinite group on one generator. In Ax’s approach it was convenient to have an Abelian absolute Galois group, but a strong trend in later work has been the systematic analysis of situations involving progressively more general Galois groups. One of our central goals here is the presentation of the general theory of PAC fields in its modern form, and its connections xvii

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with other branches of algebra. From what we have said so far, some connections with algebraic geometry and profinite groups are visible; a number-theoretic connection will appear shortly. One important feature of PAC fields is that they occur in profusion in nature and are in fact typical in the following sense. Since the absolute Galois group Gal(Q) of the rationals is a compact topological group, it carries a canonical invariant probability measure, the Haar measure. We can therefore ask for the probability that ˜ ˜ will be of a sequence 𝝈 = (𝜎1 , ..., 𝜎𝑒 ) of automorphisms of Q the fixed field Q(𝝈) PAC; and we find that this occurs with probability 1. In addition, the absolute Galois ˜ is free on the 𝑒 generators 𝜎1 , ..., 𝜎𝑒 , again with probability 1. These group of Q(𝝈) facts are consequences of Hilbert’s irreducibility theorem for Q (Chapter 13), at least in the context of countable fields. We will develop other connections between the PAC property and Hilbertianity. There are also remarkable connections with number theory via the Chebotarev density theorem (Chapters 6, 13, 16, 20, 21, 31). For example, the probability that a ˜ coincides with the Dirichlet given elementary statement 𝜓 holds for the field Q(𝜎) density of the set of primes for which it holds for the field F 𝑝 , and this density is ˜ rational. Thus, the “probability 1” theory of the fixed fields Q(𝜎) coincides with the theory of “all sufficiently large” finite fields, which by Ax’s work is an algorithmically decidable theory. ˜ for 𝝈 of length Ax’s results extend to the “probability 1” theory of the fields Q(𝝈) 𝑒 > 1, by somewhat different methods (Chapter 20), although the connection with finite fields is lost. The elementary theories of such fields are largely determined by three properties: PAC, characteristic zero, and having an absolute Galois group which is free on 𝑒 generators. To determine the full elementary theory of one such ˜ field 𝐾, it is also necessary to describe the intersection 𝐾 ∩ Q. Although the absolute Galois group of a PAC field need not be free, it can be shown to be projective in a natural sense, and conversely any projective profinite group occurs as the Galois group of some PAC field. In extending the theory from PAC fields with free Galois group to the general (projective) case, certain obstacles arise: for example, the algorithmic results do not extend. There is nonetheless a quite general theory, which enables us to identify some broad classes of projective profinite groups for which the associated classes of profinite groups behave well, and also to pinpoint unruly behavior in other case. One approach to the algorithmic problems associated with PAC fields leads to the study of profinite groups 𝐺 with the embedding property (the terminology reflects a preoccupation with the corresponding fields): for each pair of continuous epimorphisms 𝜑: 𝐺 → 𝐴, 𝛼: 𝐵 → 𝐴, where 𝐵 is a finite quotient of 𝐺, we require that 𝜑 should factor through 𝛼. A perfect PAC field whose absolute Galois group is a group with the embedding property is called a Frobenius field. The elementary theory of all Frobenius fields can be computed quite explicitly. The algorithm has some relationship with elimination theory as used by Tarski. We associate to each elementary statement in the language of PAC fields a stratification of affine space into basic normal locally closed algebraic sets, each equipped with a Galois extension of its function field, and the given statement is reinterpreted as a statement about conjugacy classes of subgroups of the specified Galois groups.

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When the initial statement has no quantifiers this is a fairly trivial procedure, but addition of quantifiers corresponds to a special kind of “projection” of these Galois stratifications. This procedure has not yet been closely examined from the point of view of computational complexity. Like most procedures which operate by tracing through a series of projections, it is effective but hopelessly inefficient in its present form. It is not yet clear whether it is substantially less efficient than Tarski’s procedure for algebraically closed fields, nor whether, like that procedure, it can significantly reorganized and sped up. The Galois stratification algorithm relies on techniques of effective algebraic geometry, and also involves substantial algorithmic problems of a new type connected with the theory of profinite groups. Specifically, it is necessary to determine, given two collections 𝐴1 , ..., 𝐴𝑚 and 𝐵1 , ..., 𝐵𝑛 of finite groups, whether or not there is a projective group with the embedding property which has each 𝐴𝑖 as (continuous) image, but none of the groups 𝐵 𝑗 . The solution to this problem depends on recent work on projective covers (Chapter 22) and embedding covers (Chapter 24). Ultimately our decision problem reduces to the determination of the finite quotients of the projective cover of the embedding cover of a single finite group. The theory of projective covers leads also to the undecidability results alluded to earlier. A fairly natural encoding of graphs into profinite groups is lifted by this theory into the class of projective profinite groups, and then by looking at the corresponding PAC fields we see that their elementary theories encode algorithmically undecidable problems (the analogous results for graphs are well known). In the final chapter we return to our point of departure, the theory of finite fields. The zeta function of a Galois formula over a finite field is defined, and using a result of Dwork and Bombieri we show that some integral power of each such function is a product of an exponential and a rational function over Q. One of the goals of this book is to serve as a bridge between algebraists and logicians. For the algebraist there is a self contained introduction to the logic and model theory background for PAC fields (Chapter 7). Chapter 14 gives the “nonstandard” framework that suffices for Weissauer’s proof of Hilbert’s irreducibility theorem (Chapter 15), and Chapters 8 and 28 include basic recursion theory. On the other hand, for logicians with a basic algebraic background (e.g. Lang’s book “Algebra”) Chapter 4 has the Stepanov–Bombieri elementary proof of the Riemann hypotheses for curves, and Chapter 6 gives an elementary proof of the Chebotarev density theorem. Both groups of readers may find the extensive treatment of profinite groups (Chapters 1, 17, 18, 22, 24, 25 and 26) and of Hilbertian fields (Chapters 12, 13, 15, and 16) valuable. Although PAC fields arise over arithmetically rich fields, they themselves lack properties that we associate with the arithmetic, say, of the rationals. For example, a PAC field 𝐹 admits no orderings and all Henselizations of 𝐹 are separably closed (Section 11.5). Many PAC field results generalize to pseudo real closed (PRC) fields. A field 𝐹 is PRC if each absolutely irreducible variety defined over 𝐹 has an ˆ point in each real closure 𝐹ˆ 𝐹-rational point provided it has a nonsingular 𝐹-rational of 𝐹. Thus, a PRC field without orderings is PAC. This, and the development of the theory of pseudo 𝑝-adically closed P𝑝C fields are outside the scope of this book. We

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refer to [Pre81], to [Jar82c], [HaJ85], and to [HaJ86] for literature about PRC fields and to [HaJ88] for P𝑝C fields. Similarly, we give no account of the theories of real closed fields and 𝑝-adically closed fields that preceded the development of the theory of PAC fields. In particular, for Hilbert’s 17th problem and the Ax–Kochen–Ershov 𝑝-adic theory, we refer the reader to [Pre84], [AxK65a], [AxK65b], [AxK66] and [PrR84]. We are indebted to several colleagues who corrected errors in the process of critically reading the manuscript. In particular, Wulf-Dieter Geyer, Gregory L. Cherlin, and Dan Haran made crucial contributions. Gainesville, Florida Tel Aviv, Israel Summer 1986

Michael D. Fried Moshe Jarden

Notation and Conventions

• | 𝐴| = #𝐴 = the cardinality of a set 𝐴. • 𝐴 ⊆ 𝐵 means that 𝐴 is a subset of 𝐵, whereas 𝐴 ⊂ 𝐵 means that 𝐴 is a proper subset of 𝐵. • Given disjoint subsets 𝐴 and 𝐵 of a common set 𝐶, we write 𝐴 ∪· 𝐵 for the disjoint union of 𝐴 and 𝐵. Ð • · 𝑖 ∈𝐼 𝐵𝑖 is the disjoint union of sets 𝐵𝑖 , 𝑖 ∈ 𝐼. • We denote the trivial subgroup of a multiplicative group 𝐺 by 1𝐺 or simply by 1 if 𝐺 is clear from the context. • 𝑎 𝑥 = 𝑥 −1 𝑎𝑥, for elements 𝑎 and 𝑥 of a group 𝐺. • 𝐻 𝑥 = {ℎ 𝑥 | ℎ ∈ 𝐻}, for a subgroup 𝐻 of 𝐺. • Given subgroups 𝐴, 𝐵 of a group 𝐺, we use “𝐴 ≤ 𝐵” for “𝐴 is a subgroup of 𝐵” and “𝐴 < 𝐵” for “𝐴 is a proper subgroup of 𝐵”. • If 𝐺 is a subgroup of 𝐻, then 𝑁 𝐻 (𝐺) = {ℎ ∈ 𝐻 | 𝐺 ℎ = 𝐺} denotes the normalizer of 𝐺 in 𝐻. • Given an Abelian (additive) group 𝐴 and a positive integer 𝑛, we write Ð 𝐴𝑛 for the subgroup {𝑎 ∈ 𝐴 | 𝑛𝑎 = 0}. For a prime number 𝑝 we let 𝐴 𝑝∞ = ∞ 𝑖=1 𝐴 𝑝 𝑖 . • For a group 𝐵 that acts on a group 𝐴 from the right, we use 𝐵 ⋉ 𝐴 to denote the semidirect product of 𝐴 and 𝐵. • Bold face letters stand for 𝑛-tuples, e.g. x = (𝑥 1 , . . . , 𝑥 𝑛 ). • ord(𝑥) is the order of an element 𝑥 in a group 𝐺. • Whenever we speak about a homomorphism 𝛼: 𝐺 → 𝐻 between profinite groups, we tacitly assume that 𝛼 is continuous. • Z = the ring of rational integers. • Z 𝑝 = the ring of 𝑝-adic integers. Î • Zˆ = lim Z/𝑛Z = 𝑝 Z 𝑝 . ←− • Q = the field of rational numbers. • R = the field of real numbers. • C = the field of complex numbers. • F𝑞 = the field with 𝑞 elements. • 𝐾sep = the separable closure of a field 𝐾. • 𝐾ins = the maximal purely inseparable extension of a field 𝐾. • 𝐾˜ = the algebraic closure of a field 𝐾. xxi

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• Gal(𝐿/𝐾) = the Galois group of a Galois extension 𝐿/𝐾. ˜ • We call a polynomial 𝑓 ∈ 𝐾 [𝑋] separable if 𝑓 has no multiple root in 𝐾. • Gal( 𝑓 , 𝐾) = the Galois group of a separable polynomial 𝑓 ∈ 𝐾 [𝑋] over a field 𝐾 viewed as a permutation group of the roots of 𝑓 . • Gal(𝐾) = Gal(𝐾sep /𝐾) = the absolute Galois group of a field 𝐾. • irr(𝑥, 𝐾) = the monic irreducible polynomial of an algebraic element 𝑥 over a field 𝐾. • Whenever we form the compositum 𝐸 𝐹 of field extensions of a field 𝐾 we tacitly assume that 𝐸 and 𝐹 are contained in a common field. • In the context of fields, 𝜁 𝑛 stands for a primitive root of unity of order 𝑛. • Re(𝑧) is the real part of a complex number 𝑧. • 𝑅 × = the group of invertible elements of a ring 𝑅. • Quot(𝑅) = the quotient field of an integral domain 𝑅. • For a positive integer 𝑛 and an integer 𝑎 with gcd(𝑎, 𝑛) = 1, we use ord𝑛 𝑎 to denote the order of 𝑎 modulo 𝑛. Thus, ord𝑛 𝑎 is the minimal positive integer 𝑑 with 𝑎 𝑑 ≡ 1 mod 𝑛. • In the context of groups, 𝑆 𝑛 (resp. 𝐴𝑛 ) stands for the full permutation group (resp. alternating group) of {1, . . . , 𝑛}. • In the context of groups, 𝐶𝑛 stands for the cyclic multiplicative group of order 𝑛. Likewise we use Z/𝑛Z for the additive multiplicative group of order 𝑛.

Contents

Preface to the Fourth Edition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preface to the Third Edition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preface to the Second Edition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preface to the First Edition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notation and Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

vii xi xiii xvii xxi

1

Infinite Galois Theory and Profinite Groups . . . . . . . . . . . . . . . . . . . . . 1.1 Inverse Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Profinite Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Infinite Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 The 𝑝-adic Integers and the Prüfer Group . . . . . . . . . . . . . . . . . . . 1.5 The Absolute Galois Group of a Finite Field . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 4 10 13 16 17 19

2

Valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Valuations, Places, and Valuation Rings . . . . . . . . . . . . . . . . . . . . . 2.2 Discrete Valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Extensions of Valuations and Places . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Galois Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Integral Extensions and Dedekind Domains . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21 21 23 27 33 34 38

3

Linear Disjointness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Linear Disjointness of Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Purely Transcendental Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Separable Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Regular Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Primary Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 The Imperfect Degree of a Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41 41 47 48 49 54 55 60 62 xxiii

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4

Algebraic Function Fields of One Variable . . . . . . . . . . . . . . . . . . . . . . 4.1 Function Fields of One Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The Riemann–Roch Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Holomorphy Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Extensions of Function Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Completions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 The Different . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Hyperelliptic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Hyperelliptic Fields with a Rational Quadratic Subfield . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

63 63 65 67 69 72 77 80 83 85 86

5

The Riemann Hypothesis for Function Fields . . . . . . . . . . . . . . . . . . . . 5.1 Class Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Zeta Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Zeta Functions under Constant Field Extensions . . . . . . . . . . . . . . 5.4 The Functional Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 The Riemann Hypothesis and Degree 1 Prime Divisors . . . . . . . . 5.6 Reduction Steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 An Upper Bound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 A Lower Bound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87 87 89 91 92 94 95 96 98 101 103

6

Plane Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Affine and Projective Plane Curves . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Points and Prime Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 The Genus of a Plane Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Points on a Curve over a Finite Field . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

105 105 107 109 113 115 115

7

The Chebotarev Density Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Decomposition Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 The Artin Symbol over Global Fields . . . . . . . . . . . . . . . . . . . . . . . 7.3 Dirichlet Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Function Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Number Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

117 117 122 123 125 131 138 139

8

Ultraproducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 First Order Predicate Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Elementary Substructures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Ultrafilters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

141 141 143 144 146 147

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8.6 Ultraproducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 Regular Ultrafilters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8 Regular Ultraproducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.9 Nonprincipal Ultraproducts of Finite Fields . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

148 152 154 155 156 157

Decision Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Deduction Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Gödel’s Completeness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Primitive Recursive Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Primitive Recursive Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Recursive Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Recursive and Primitive Recursive Procedures . . . . . . . . . . . . . . . 9.7 A Reduction Step in Decidability Procedures . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

159 159 162 164 166 167 168 170 171 172

10 Algebraically Closed Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Elimination of Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 A Quantifier Elimination Procedure . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Effectiveness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

173 173 175 177 178 180 180

11 Elements of Algebraic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Algebraic Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Substitutions in Irreducible Polynomials . . . . . . . . . . . . . . . . . . . . 11.4 Rational Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 Hyperplane Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6 Descent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.7 Projective Varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.8 About the Language of Algebraic Geometry . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

183 183 186 188 189 191 194 196 198 200 202

12 Pseudo Algebraically Closed Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 PAC Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Reduction to Plane Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 The PAC Property is an Elementary Statement . . . . . . . . . . . . . . . 12.4 PAC Fields of Positive Characteristic . . . . . . . . . . . . . . . . . . . . . . . 12.5 PAC Fields with Valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.6 The Absolute Galois Group of a PAC Field . . . . . . . . . . . . . . . . . .

203 203 205 209 211 213 217

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12.7 A non-PAC Field 𝐾 with 𝐾ins PAC . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

221 226 227

13 Hilbertian Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 Hilbert Sets and Reduction Lemmas . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Hilbert Sets under Separable Algebraic Extensions . . . . . . . . . . . . 13.3 Purely Inseparable Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4 Imperfect Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

229 229 233 234 238 240 240

14 The Classical Hilbertian Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1 Further Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Function Fields Over Infinite Fields . . . . . . . . . . . . . . . . . . . . . . . . 14.3 Global Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 Hilbertian Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5 Hilbertianity via Coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.6 Non-Hilbertian 𝑔-Hilbertian Fields . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

243 243 247 249 253 256 260 263 265

15 The Diamond Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Twisted Wreath Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 The Diamond Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3 Weissauer’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

267 267 273 277 280 280

16 Nonstandard Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Higher Order Predicate Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Enlargements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3 Concurrent Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 The Existence of Enlargements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

281 281 282 284 286 288 289 290

17 The Nonstandard Approach to Hilbert’s Irreducibility Theorem . . . 17.1 Criteria for Hilbertianity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2 Arithmetical Primes Versus Functional Primes . . . . . . . . . . . . . . . 17.3 Fields with the Product Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.4 Generalized Krull Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

291 291 293 295 298 301 304 305

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18 Galois Groups over Hilbertian Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.1 Galois Groups of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2 Stable Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3 Regular Realization of Finite Abelian Groups . . . . . . . . . . . . . . . . 18.4 Split Embedding Problems with Abelian Kernels . . . . . . . . . . . . . 18.5 Embedding Quadratic Extensions in Z/2𝑛 Z-Extensions . . . . . . . . 18.6 Z 𝑝 -Extensions of Hilbertian Fields . . . . . . . . . . . . . . . . . . . . . . . . . 18.7 Symmetric and Alternating Groups over Hilbertian Fields . . . . . . 18.8 GAR-Realizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.9 Embedding Problems over Hilbertian Fields . . . . . . . . . . . . . . . . . 18.10 Regularity of Finite Groups over Complete Discrete-Valued Fields Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

307 307 310 314 319 322 324 331 336 342 344 346 346

19 Small Profinite Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.1 Finitely Generated Profinite Groups . . . . . . . . . . . . . . . . . . . . . . . . 19.2 Abelian Extensions of Hilbertian Fields . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

349 349 354 355 356

20 Free Profinite Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.1 The Rank of a Profinite Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2 Profinite Completions of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.3 Formations of Finite Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.4 Free pro-C Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.5 Subgroups of Free Discrete Groups . . . . . . . . . . . . . . . . . . . . . . . . 20.6 Open Subgroups of Free Profinite Groups . . . . . . . . . . . . . . . . . . . 20.7 An Embedding Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

357 357 360 364 366 370 377 380 381 382

21 The Haar Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1 The Haar Measure of a Profinite Group . . . . . . . . . . . . . . . . . . . . . 21.2 Existence of the Haar Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Cartesian Product of Haar Measures . . . . . . . . . . . . . . . . . . . . . . . . 21.5 The Haar Measure of the Absolute Galois Group . . . . . . . . . . . . . 21.6 The PAC Nullstellensatz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.7 Baire’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.8 The Bottom Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.9 Triviality of a Group of Automorphisms . . . . . . . . . . . . . . . . . . . . 21.10 PAC Fields over Uncountable Hilbertian Fields . . . . . . . . . . . . . . . 21.11 On the Stability of Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.12 PAC Galois Extensions of Hilbertian Fields . . . . . . . . . . . . . . . . . .

383 383 386 391 396 398 401 402 406 411 414 418 422

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21.13 Algebraic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

424 427 428

22 Effective Field Theory and Algebraic Geometry . . . . . . . . . . . . . . . . . 22.1 Presented Rings and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Extensions of Presented Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.3 Galois Extensions of Presented Fields . . . . . . . . . . . . . . . . . . . . . . 22.4 The Algebraic and Separable Closures of Presented Fields . . . . . 22.5 Constructive Algebraic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . 22.6 Presented Rings and Constructible Sets . . . . . . . . . . . . . . . . . . . . . 22.7 Basic Normal Stratification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

429 429 432 437 438 439 447 450 451 452

23 The Elementary Theory of 𝒆-Free PAC Fields . . . . . . . . . . . . . . . . . . . 23.1 ℵ1 -Saturated PAC Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.2 The Elementary Equivalence Theorem of ℵ1 -Saturated PAC Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.3 Elementary Equivalence of PAC Fields . . . . . . . . . . . . . . . . . . . . . 23.4 On 𝑒-Free PAC Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.5 The Elementary Theory of Perfect 𝑒-Free PAC Fields . . . . . . . . . 23.6 The Probable Truth of a Sentence . . . . . . . . . . . . . . . . . . . . . . . . . . 23.7 Change of Base Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.8 The Fields 𝐾sep (𝜎1 , . . . , 𝜎𝑒 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.9 The Transfer Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.10 The Elementary Theory of Finite Fields . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

455 455 456 459 462 464 465 468 470 472 474 477 478

24 Problems of Arithmetical Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.1 The Decomposition-Intersection Procedure . . . . . . . . . . . . . . . . . . 24.2 𝐶𝑖 -Fields and Weakly 𝐶𝑖 -Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.3 Perfect PAC Fields which are 𝐶𝑖 . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.4 The Existential Theory of PAC Fields . . . . . . . . . . . . . . . . . . . . . . . 24.5 Kronecker Classes of Number Fields . . . . . . . . . . . . . . . . . . . . . . . 24.6 Davenport’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.7 On Permutation Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.8 Schur’s Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.9 The Generalized Carlitz Conjecture . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

479 479 480 485 487 488 491 495 503 512 515 517

25 Projective Groups and Frattini Covers . . . . . . . . . . . . . . . . . . . . . . . . . . 25.1 The Frattini Group of a Profinite Group . . . . . . . . . . . . . . . . . . . . . 25.2 Cartesian Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.3 On C-Projective Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

519 519 521 525

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25.4 Projective Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.5 Free Products of Finitely many Profinite Groups . . . . . . . . . . . . . . 25.6 Frattini Covers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.7 The Universal Frattini Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.8 Projective Pro-𝑝-Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.9 Supernatural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.10 The Sylow Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.11 On Complements of Normal Subgroups . . . . . . . . . . . . . . . . . . . . . 25.12 The Universal Frattini 𝑝-Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.13 Examples of Universal Frattini 𝑝-covers . . . . . . . . . . . . . . . . . . . . . 25.14 The Special Linear Group SL(2, Z 𝑝 ) . . . . . . . . . . . . . . . . . . . . . . . 25.15 The General Linear Group GL(2, Z 𝑝 ) . . . . . . . . . . . . . . . . . . . . . . 25.16 Absolute Galois Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

528 530 531 536 537 543 545 547 551 555 557 560 562 563 566

26 PAC Fields and Projective Absolute Galois Groups . . . . . . . . . . . . . . . 26.1 Projective Groups as Absolute Galois Groups . . . . . . . . . . . . . . . . 26.2 Countably Generated Projective Groups . . . . . . . . . . . . . . . . . . . . . 26.3 Perfect PAC Fields of Bounded Corank . . . . . . . . . . . . . . . . . . . . . 26.4 Basic Elementary Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.5 Reduction Steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.6 Application of Ultraproducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

569 569 571 574 575 579 583 585 585

27 Frobenius Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.1 The Field Crossing Argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.2 The Beckmann–Black Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.3 The Embedding Property and Maximal Frattini Covers . . . . . . . . 27.4 The Smallest Embedding Cover of a Profinite Group . . . . . . . . . . 27.5 A Decision Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.7 Non-projective Smallest Embedding Cover . . . . . . . . . . . . . . . . . . 27.8 A Theorem of Iwasawa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.9 Free Profinite Groups of Countable Rank . . . . . . . . . . . . . . . . . . . . 27.10 Application of the Nielsen–Schreier Formula . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

587 587 590 592 594 599 600 604 605 607 610 615 617

28 Free Profinite Groups of Infinite Rank . . . . . . . . . . . . . . . . . . . . . . . . . . 28.1 Characterization of Free Profinite Groups by Embedding Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.2 Applications of Theorem 28.1.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.3 The Pro-C Completion of a Free Discrete Group . . . . . . . . . . . . . 28.4 The Group Theoretic Diamond Theorem . . . . . . . . . . . . . . . . . . . . 28.5 The Melnikov Group of a Profinite Group . . . . . . . . . . . . . . . . . . .

619 620 625 629 631 637

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28.6 Homogeneous Pro-C Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.7 The 𝑆-rank of Closed Normal Subgroups . . . . . . . . . . . . . . . . . . . . 28.8 Closed Normal Subgroups with a Basis Element . . . . . . . . . . . . . . 28.9 Accessible Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

639 644 647 650 657 657

29 Random Elements in Profinite Groups . . . . . . . . . . . . . . . . . . . . . . . . . . 29.1 Random Elements in a Free Profinite Group . . . . . . . . . . . . . . . . . 29.2 Random Elements in Free pro-𝑝 Groups . . . . . . . . . . . . . . . . . . . . 29.3 Random 𝑒-tuples in Zˆ 𝑛 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29.4 The Golod–Shafarevich Inequality . . . . . . . . . . . . . . . . . . . . . . . . . 29.5 On the Index of Normal Subgroups Generated by Random Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29.6 Freeness of Normal Subgroups Generated by Random Elements Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

659 659 664 666 670

30 Omega-free PAC Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.1 Model Companions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.2 The Model Companion in an Augmented Theory of Fields . . . . . 30.3 New Non-Classical Hilbertian Fields . . . . . . . . . . . . . . . . . . . . . . . 30.4 An Abundance of 𝜔-Free PAC Fields . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

687 687 691 696 698 701

31 Hilbertian Subfields of Galois Extensions . . . . . . . . . . . . . . . . . . . . . . . 31.1 Small Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 The Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

703 704 707 711 712

32 Undecidability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1 Turing Machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Computation of Functions by Turing Machines . . . . . . . . . . . . . . . 32.3 Recursive Inseparability of Sets of Turing Machines . . . . . . . . . . 32.4 The Predicate Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.5 Undecidability in the Theory of Graphs . . . . . . . . . . . . . . . . . . . . . 32.6 Assigning Graphs to Profinite Groups . . . . . . . . . . . . . . . . . . . . . . 32.7 The Graph Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.8 Assigning Profinite Groups to Graphs . . . . . . . . . . . . . . . . . . . . . . 32.9 Assigning Fields to Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.10 Interpretation of the Theory of Graphs in the Theory of Fields . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

713 713 714 718 720 723 728 729 731 734 735 737 738

33 Algebraically Closed Fields with Distinguished Automorphisms . . . 33.1 The Base Field 𝐾 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33.2 Coding in PAC Fields with Monadic Quantifiers . . . . . . . . . . . . . .

739 739 741

678 682 685

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˜ 𝜎1 , . . . , 𝜎𝑒 ⟩’s . . . . . . . . . . . . . . . . . The Theory of Almost all ⟨𝐾, The Probability of Truth Sentences . . . . . . . . . . . . . . . . . . . . . . . . .

744 746

34 Galois Stratification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34.1 The Artin Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34.2 Conjugacy Domains under Projections . . . . . . . . . . . . . . . . . . . . . . 34.3 Normal Stratification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34.4 Elimination of One Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34.5 The Complete Elimination Procedure . . . . . . . . . . . . . . . . . . . . . . . 34.6 Model-Theoretic Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34.7 A Limit of Theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

749 749 751 756 758 760 762 766 767 769

35 Galois Stratification over Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . 35.1 The Elementary Theory of Frobenius Fields . . . . . . . . . . . . . . . . . 35.2 The Elementary Theory of Finite Fields . . . . . . . . . . . . . . . . . . . . . 35.3 Near Rationality of the Zeta Function of a Galois Formula . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

771 771 775 779 787 789

36 Problems of Field Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36.2 Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

791 791 795

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

799

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

813

33.3 33.4

Chapter 1

Infinite Galois Theory and Profinite Groups

The usual Galois correspondence between subgroups of Galois groups of finite Galois extensions and intermediate fields is not valid for infinite Galois extensions. The Krull topology restores this correspondence for closed subgroups (Proposition 1.3.1). Since Galois groups are inverse limits of finite groups, they are profinite. Conversely, we define profinite groups, independently of Galois-theoretic properties. Each profinite group actually appears as a Galois group (Corollary 1.3.5). In particular, we study the procyclic groups Z 𝑝 and Zˆ and prove that every finite field has the latter as its absolute Galois group.

1.1 Inverse Limits Our interest in inverse limits comes from infinite Galois theory: Infinite Galois groups are inverse limits of finite Galois groups. As a preparation to the study of “profinite groups” we define in this section inverse limits of topological spaces and characterize inverse limits of finite topological spaces. Let 𝐼 be a set with a partial ordering ≤; that is, ≤ is a binary relation which is reflexive, transitive, and 𝑎 ≤ 𝑏 and 𝑏 ≤ 𝑎 imply 𝑎 = 𝑏. We call (𝐼, ≤) a directed partially ordered set if in addition (1.1) for all 𝑖, 𝑗 ∈ 𝐼 there exists a 𝑘 ∈ 𝐼 with 𝑖 ≤ 𝑘 and 𝑗 ≤ 𝑘. An inverse system (also called a projective system ) over a directed partially ordered set (𝐼, ≤) is a data (𝑆𝑖 , 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 where 𝑆𝑖 is a set and 𝜋 𝑗𝑖 : 𝑆 𝑗 → 𝑆𝑖 is a map for all 𝑖, 𝑗 ∈ 𝐼 with 𝑖 ≤ 𝑗 satisfying the following rules: (1.2a) 𝜋𝑖𝑖 is the identity map for each 𝑖 ∈ 𝐼. (1.2b) 𝜋 𝑘𝑖 = 𝜋 𝑗𝑖 ◦ 𝜋 𝑘 𝑗 if 𝑖 ≤ 𝑗 ≤ 𝑘.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_1

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2

1 Infinite Galois Theory and Profinite Groups

Î Let 𝑆 be the subset of the direct product 𝑖 ∈𝐼 𝑆𝑖 consisting of allÎelements 𝑠 = (𝑠𝑖 )𝑖 ∈𝐼 with 𝜋 𝑗𝑖 (𝑠 𝑗 ) = 𝑠𝑖 for all 𝑖 ≤ 𝑗. Note: 𝑆 may be empty. Let pr𝑖 : 𝑗 ∈𝐼 𝑆 𝑗 → 𝑆𝑖 be the projection on the 𝑖th coordinate. Denote the restriction of pr𝑖 to 𝑆 by 𝜋𝑖 . Then, 𝜋𝑖 = 𝜋 𝑗𝑖 ◦ 𝜋 𝑗 for every 𝑖 ≤ 𝑗. We say (𝑆, 𝜋𝑖 )𝑖 ∈𝐼 is the inverse limit (or projective limit ) of the family (𝑆𝑖 )𝑖 ∈𝐼 with respect to the maps 𝜋 𝑗𝑖 . Denote 𝑆 by lim 𝑆𝑖 . ←− Let (𝑆𝑖′, 𝜋 ′𝑗𝑖 )𝑖, 𝑗 ∈𝐼 be another inverse system over 𝐼. Suppose for each 𝑖 ∈ 𝐼 we are given a map 𝜃 𝑖 : 𝑆𝑖 → 𝑆𝑖′ with 𝜋 ′𝑗𝑖 ◦ 𝜃 𝑗 = 𝜃 𝑖 ◦ 𝜋 𝑗𝑖 for all 𝑖 ≤ 𝑗. (We say that the maps 𝜃 𝑖 , 𝑖 ∈ 𝐼, are compatible.) Then, there exists a unique map 𝜃: lim 𝑆𝑖 → lim 𝑆𝑖′ ←− ←− satisfying 𝜋𝑖′ ◦ 𝜃 = 𝜃 𝑖 ◦ 𝜋𝑖 for each 𝑖 ∈ 𝐼. Thus, 𝜃 maps 𝑠 = (𝑠𝑖 )𝑖 ∈𝐼 ∈ lim 𝑆𝑖 onto 𝜃 (𝑠) ←− with 𝜃 (𝑠)𝑖 = 𝜃 𝑖 (𝑠𝑖 ). Denote 𝜃 by lim 𝜃 𝑖 . ←− Similarly, let 𝑋 be a set and for each 𝑖 ∈ 𝐼 let 𝜃 𝑖 : 𝑋 → 𝑆𝑖 be a map satisfying 𝜋 𝑗𝑖 ◦ 𝜃 𝑗 = 𝜃 𝑖 whenever 𝑖 ≤ 𝑗. (Again, we say that the maps 𝜃 𝑖 , 𝑖 ∈ 𝐼, are compatible.) Then, there exists a unique map 𝜃: 𝑋 → lim 𝑆𝑖 with 𝜋𝑖 ◦ 𝜃 = 𝜃 𝑖 for each 𝑖 ∈ 𝐼. ←− When 𝑆𝑖 are topological spaces, we assume that the 𝜋 𝑗𝑖 are continuous. Î Then, we equip lim 𝑆𝑖 with the topology induced from the product topology of 𝑖 ∈𝐼 𝑆𝑖 (also ←− Î called the Tychonoff topology). Recall that the product topology on 𝑖 ∈𝐼 𝑆𝑖 has a Î base consisting of the sets 𝑖 ∈𝐼 𝑈𝑖 , with 𝑈𝑖 open in 𝑆𝑖 for each 𝑖 ∈ 𝐼, and 𝑈𝑖 = 𝑆𝑖 for all but finitely many 𝑖 ∈ 𝐼. Since pr𝑖 is continuous, so is 𝜋𝑖 , 𝑖 ∈ 𝐼. If 𝜃 𝑖 : 𝑆𝑖 → 𝑆𝑖′ are continuous, then 𝜃: lim 𝑆𝑖 → lim 𝑆𝑖′ is also continuous. ←− ←− Lemma 1.1.1 The collection of all subsets of 𝑆 = lim 𝑆𝑖 of the form 𝜋𝑖−1 (𝑈𝑖 ) with ←− 𝑈𝑖 open in 𝑆𝑖 is a base for the topology of 𝑆. Proof. Let Î𝑠 ∈ 𝑆. ByÎdefinition,
each basic open neighborhood of 𝑠 has the form 𝑉 =𝑆∩ 𝑗 ∈𝐽 𝑉 𝑗 × 𝑖 ∈𝐼 ∖ 𝐽 𝑆 𝑖 , where 𝐽 is a finite subset of 𝐼 and 𝑉 𝑗 is an open Ñ subset of 𝑆 𝑗 , 𝑗 ∈ 𝐽. Take 𝑘 ∈ 𝐼 with 𝑘 ≥ 𝑗 for all 𝑗 ∈ 𝐽. Then, 𝑈 𝑘 = 𝑗 ∈𝐽 𝜋 −1 𝑘 𝑗 (𝑉 𝑗 ) −1 is an open subset of 𝑆 𝑘 and 𝜋 𝑘 (𝑈 𝑘 ) is an open neighborhood of 𝑠 in 𝑉. Therefore, □ the collection 𝜋𝑖−1 (𝑈𝑖 )𝑖 ∈𝐼 is a base for the topology of 𝑆. Lemma 1.1.2 In the notation Î above, if each 𝑆𝑖 , 𝑖 ∈ 𝐼, is a Hausdorff space, then lim 𝑆𝑖 is a closed subset of 𝑆𝑖 . ←− Î Proof. Suppose 𝑠 = (𝑠𝑖 )𝑖 ∈𝐼 ∈ 𝑖 ∈𝐼 𝑆𝑖 does not belong to lim 𝑆𝑖 . Then, there are ←− 𝑈𝑖 and 𝑈𝑖′ of 𝑖, 𝑗 ∈ 𝐼 with 𝑖 ≤ 𝑗 and 𝜋 𝑗𝑖 (𝑠 𝑗 ) ≠ 𝑠𝑖 . Take open disjoint neighborhoods Î −1 ′ 𝑠𝑖 and 𝜋Î 𝑗𝑖 (𝑠 𝑗 ), respectively. Then, 𝑈𝑖 ×𝜋 𝑗𝑖 (𝑈𝑖 ) × 𝑘≠𝑖, 𝑗 𝑆 𝑘 is an open neighborhood of 𝑠 in 𝑖 ∈𝐼 𝑆𝑖 that does not intersect lim 𝑆𝑖 . □ ←− Î If, in addition, each 𝑆𝑖 is compact, then Tychonoff’s theorem implies that 𝑆𝑖 is also compact. Thus, lim 𝑆𝑖 with the induced topology is compact. ←− Lemma 1.1.3 The inverse limit 𝑆 of an inverse system of nonempty compact Hausdorff spaces 𝑆𝑖 , 𝑖 ∈ 𝐼, is a nonempty compact Hausdorff space. Proof. In view of Lemma 1.1.2, we only need to prove that 𝑆 is nonempty. Indeed, Ñ Î 𝑆 = 𝑘 ≥ 𝑗 𝑅 𝑘 𝑗 , where 𝑅 𝑘 𝑗 = {𝑠 ∈ 𝑖 ∈𝐼 𝑆 𝑖 | 𝜋 𝑘 𝑗 (𝑠 𝑘 ) = 𝑠 𝑗 }. The natural map Î pr 𝑘 × pr 𝑗 : 𝑖 ∈𝐼 𝑆𝑖 → 𝑆 𝑘 × 𝑆 𝑗 is continuous. The Hausdorff property of 𝑆 𝑗 implies

1.1 Inverse Limits

3

that 𝑇 = {(𝑠 𝑘 , 𝑠 𝑗 ) ∈ 𝑆 𝑘 × 𝑆 𝑗 | 𝜋 𝑘 𝑗 (𝑠 𝑘 ) = 𝑠 𝑗 } is a closed subset of 𝑆 𝑘 × 𝑆 𝑗 . Hence, Î Î 𝑅 𝑘 𝑗 = (pr 𝑘 × pr 𝑗 ) −1 (𝑇) is a closed subset of 𝑖 ∈𝐼 𝑆𝑖 . Since 𝑖 ∈𝐼 𝑆𝑖 is compact, we only need to show that the intersection of finitely many of the 𝑅 𝑘 𝑗 is nonempty. Indeed, let 𝑗 1 ≤ 𝑘 1 , . . . , 𝑗 𝑛 ≤ 𝑘 𝑛 be 𝑛 pairs in 𝐼. Choose 𝑙 ∈ 𝐼 with 𝑘 𝑖 ≤ 𝑙, 𝑖 = 1, . . . , 𝑛, and choose 𝑠𝑙 ∈ 𝑆𝑙 . Define 𝑠 𝑗𝑖 = 𝜋𝑙, 𝑗𝑖 (𝑠𝑙 ) and 𝑠 𝑘𝑖 = 𝜋𝑙,𝑘𝑖 (𝑠𝑙 ), for ∈ 𝐼 ∖{ 𝑗 1 , . . . , 𝑗 𝑛 , 𝑘 1 , . . . , 𝑘 𝑛 } let 𝑠𝑟 be an arbitrary element 𝑖 = 1, . . . , 𝑛. For each 𝑟 Ñ 𝑛 of 𝑆𝑟 . Then, 𝑠 = (𝑠𝑖 ) ∈ 𝑖=1 𝑅 𝑘𝑖 , 𝑗𝑖 . □ Unless stated otherwise, we consider each finite set 𝑋 as a topological space with the discrete topology. Thus, every subset of 𝑋 is open (hence, closed). In particular, 𝑋 is Hausdorff and compact. Corollary 1.1.4 The inverse limit of an inverse system of nonempty finite sets is nonempty. Corollary 1.1.5 Let (𝑆𝑖 , 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 and (𝑆𝑖′, 𝜋 ′𝑗𝑖 )𝑖, 𝑗 ∈𝐼 be inverse systems of compact Hausdorff spaces. Let 𝜃 𝑖 : 𝑆𝑖 → 𝑆𝑖′ be a compatible system of surjective continuous maps. Put 𝑆 = lim 𝑆𝑖 , 𝑆 ′ = lim 𝑆𝑖′, and 𝜃 = lim 𝜃 𝑖 . Then, 𝜃: 𝑆 → 𝑆 ′ is surjective. ←− ←− ←− Proof. Let 𝑠 ′ = (𝑠𝑖′)𝑖 ∈𝐼 be an element of 𝑆 ′. Then, (𝜃 𝑖−1 (𝑠𝑖′), 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 is an inverse system of nonempty compact Hausdorff spaces. By Lemma 1.1.3, the inverse limit of 𝜃 𝑖−1 (𝑠𝑖′) is nonempty. Each element in the inverse limit is mapped by 𝜃 onto 𝑠 ′. □ Corollary 1.1.6 Let 𝑋 be a compact space, (𝑆𝑖 , 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 an inverse system of Hausdorff spaces, and 𝜃 𝑖 : 𝑋 → 𝑆𝑖 a compatible system of continuous surjective maps. Put 𝜃 = lim 𝜃 𝑖 . Then, 𝜃: 𝑋 → lim 𝑆𝑖 is surjective. ←− ←− Proof. Consider 𝑠 = (𝑠𝑖 )𝑖 ∈𝐼 ∈ lim 𝑆𝑖 . Then, 𝜃 𝑖−1 (𝑠𝑖 ) is a closed nonempty subset ←− of 𝑋. For 𝑖 1 , . . . , 𝑖 𝑛 ∈ 𝐼 there is a 𝑗 ∈ 𝐼 with 𝑖 1 , . . . , 𝑖 𝑛 ≤ 𝑗. Then, 𝜃 −1 𝑗 (𝑠 𝑗 ) ⊆ −1 (𝑠 ), so 𝜃 −1 (𝑠 ) ∩ · · · ∩ 𝜃 −1 (𝑠 ) is nonempty. Since 𝑋 is 𝜃 ) ∩ · · · ∩ 𝜃 𝑖−1 (𝑠 𝑖1 𝑖𝑛 𝑖𝑛 𝑖1 𝑖1 𝑖𝑛 𝑖𝑛 1 compact, there is an 𝑥 ∈ 𝑋 which belongs to 𝜃 𝑖−1 (𝑠𝑖 ) for every 𝑖 ∈ 𝐼. It satisfies, □ 𝜃 (𝑥) = 𝑠. Thus, 𝜃 is surjective. A profinite space is an inverse limit of an inverse system of finite discrete spaces. Lemma 1.1.7 A compact Hausdorff space 𝑆 is profinite if and only if its topology has a base consisting of open-closed sets. Proof. Suppose first 𝑆 = lim 𝑆𝑖 is an inverse limit of finite discrete spaces 𝑆𝑖 . Let ←− 𝜋𝑖 : 𝑆 → 𝑆𝑖 be the projection on the 𝑖th coordinate, 𝑖 ∈ 𝐼. By Lemma 1.1.1, the sets 𝜋𝑖−1 (𝑈𝑖 ), where 𝑖 ∈ 𝐼 and 𝑈𝑖 is a subset of 𝑆𝑖 , form a base of the topology of 𝑆. Since 𝑆𝑖 is discrete, 𝑈𝑖 is open-closed. Hence, so is 𝜋𝑖−1 (𝑈𝑖 ). Now suppose the topology of 𝑆 has a base consisting of open-closed sets. An open-closed partition of 𝑆 is a finite set A of nonempty open-closed disjoint subsets of 𝑆 whose union is 𝑆. Denote the set of all open-closed partitions of 𝑆 by A. Let A, A′, B ∈ A. Write A ≤ B if for each 𝐵 ∈ B there is an 𝐴 ∈ A with 𝐵 ⊆ 𝐴. Then, 𝐴 is unique. Next note that C = {𝐴 ∩ 𝐴 ′ | 𝐴 ∈ A, 𝐴 ′ ∈ A′ } belongs to A and satisfies A, A′ ≤ C. Thus, (A, ≤) is a directed partially ordered set.

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When A ≤ B, define a map 𝜋B,A from B to A by 𝜋B,A (𝐵) = 𝐴, where 𝐴 is the unique element of A containing 𝐵. Equip each A ∈ A with the discrete topology. Then, (A, 𝜋B,A )A,B∈A is an inverse system of finite discrete spaces. Its limit 𝑆 ′ = lim A is a profinite space. We construct a homeomorphism of 𝑆 onto 𝑆 ′. ←− Let 𝑠 ∈ 𝑆 and A ∈ A. Define 𝜃 A (𝑠) to be the unique 𝐴 in A which contains 𝑠. Then, 𝜃 A : 𝑆 → A is a continuous surjective map. If A ≤ B, then 𝜋B,A ◦ 𝜃 B = 𝜃 A . Hence, by Corollary 1.1.6, there is a continuous surjective map 𝜃: 𝑆 → 𝑆 ′ satisfying 𝜋A ◦ 𝜃 = 𝜃 A for each A ∈ A. Suppose that 𝑠, 𝑠 ′ are distinct elements of 𝑆. Since 𝑆 is Hausdorff, there are disjoint open-closed subsets 𝐴 and 𝐴 ′ of 𝑆 with 𝑠 ∈ 𝐴 and 𝑠 ′ ∈ 𝐴 ′. Put 𝐴 ′′ = 𝑆 ∖ ( 𝐴 ∪ 𝐴 ′). Then, A = {𝐴, 𝐴 ′, 𝐴 ′′ } is an open-closed partition of 𝑆, 𝜃 A (𝑠) = 𝐴, and 𝜃 A (𝑠 ′) = 𝐴 ′. Thus, 𝜃 A (𝑠) ≠ 𝜃 A (𝑠 ′), so 𝜃 (𝑠) ≠ 𝜃 (𝑠 ′). Therefore, 𝜃 is bijective. Since 𝑆 is compact and 𝑆 ′ is Hausdorff, 𝜃 is a homeomorphism. Consequently, 𝑆 is a profinite space. □ Remark 1.1.8 (Totally disconnected spaces) Let 𝑆 be a compact Hausdorff space. Suppose that 𝑆 has a base for its topology consisting of open-closed sets. It is not difficult to see that 𝑆 is totally disconnected; that is each 𝑠 in 𝑆 is its own connected component. Conversely, if 𝑆 is totally disconnected, then the topology of 𝑆 has a base consisting of open-closed subsets [RiZ00, p. 11, Thm. 1.1.12]

1.2 Profinite Groups We survey here the basic properties of compact groups. In particular, this will apply to profinite groups. Topological Groups. A topological group is a group 𝐺 equipped with a topology in which the product (𝑥, 𝑦) ↦→ 𝑥𝑦 and the inverse map 𝑥 ↦→ 𝑥 −1 are continuous. It follows that for each 𝑎 ∈ 𝐺 the maps 𝑥 ↦→ 𝑎𝑥, 𝑥 ↦→ 𝑥𝑎, and 𝑥 ↦→ 𝑥 −1 are homeomorphisms. We always assume that 1 := {1} is a closed subset of 𝐺. Consequently, {𝑎} is a closed subset of 𝐺 for each 𝑎 ∈ 𝐺. It follows that 𝐺 is a Hausdorff space. Indeed, let 𝑎, 𝑏 be distinct elements of 𝐺. The identity 1 · 1−1 = 1 and the continuity of the group operations give open neighborhoods 𝑈 and 𝑊 of 1 with 𝑈𝑊 −1 ⊆ 𝐺 ∖{𝑎 −1 𝑏}. Thus, 𝑎𝑈 and 𝑏𝑊 are disjoint open neighborhoods of 𝑎 and 𝑏, respectively, as needed. In addition, each closed subgroup 𝐻 of 𝐺 of finite index is open. Indeed, there Ð are 𝑎 𝑖 ∈ 𝐺, 𝑖 ∈ 𝐼, with 𝐼 finite and 𝐺 = · 𝑖 ∈𝐼 𝑎 𝑖 𝐻. Each Ð of the sets 𝑎 𝑖 𝐻 is closed and there is a 𝑗 ∈ 𝐼 with 𝑎 𝑗 𝐻 = 𝐻. Therefore, 𝐻 = 𝐺 ∖ · 𝑖≠ 𝑗 𝑎 𝑖 𝐻 is open. Conversely, if 𝐺 is compact, then every open subgroup 𝐻 of 𝐺 is of finite index. Otherwise, 𝐺 would be a disjoint union of infinitely many cosets of 𝐻, each of which is open. Let 𝑁 be a closed normal subgroup of 𝐺 and 𝜋: 𝐺 → 𝐺/𝑁 the quotient map. Equip 𝐺/𝑁 with the quotient topology. Thus, a subset 𝑈¯ of 𝐺/𝑁 is open if and only ¯ is open. It follows that the group operations of 𝐺/𝑁 are continuous. In if 𝜋 −1 (𝑈) ¯ is closed. In particular, addition, a subset 𝐶¯ of 𝐺/𝑁 is closed if and only if 𝜋 −1 (𝐶)

1.2 Profinite Groups

5

since 𝑁 = 𝜋 −1 (1) is closed, 1 is a closed subset of 𝐺/𝑁. Consequently, 𝐺/𝑁 is a topological group and 𝜋: 𝐺 → 𝐺/𝑁 is a continuous map. Moreover, if 𝑈 is an open Ð subset of 𝐺, then 𝜋 −1 (𝜋(𝑈)) = 𝑛∈𝑁 𝑛𝑈, so 𝜋(𝑈) is open in 𝐺/𝑁. Therefore, 𝜋 is an open map. Suppose again that 𝐺 is compact. Let 𝜃 be a continuous homomorphism of 𝐺 into a topological group 𝐻. Since 𝐻 is Hausdorff, 𝜃 is a closed map [Bou89b, p. 87, Cor. I of Section I.9.4]. Suppose in addition, 𝜃 is surjective. Put 𝑁 = Ker(𝜃). Let ¯ 𝐺/𝑁 → 𝐻 the map induced by 𝜃. Then, 𝜃¯ 𝜋: 𝐺 → 𝐺/𝑁 be the quotient map and 𝜃: is an isomorphism of abstract groups. In addition, 𝜃¯ is a continuous bijective map of the compact group 𝐺/𝑁 onto the Hausdorff group 𝐻. Hence, 𝜃¯ is an isomorphism of topological groups and 𝜃 is an open map. This is the first isomorphism theorem for compact groups. Let 𝐻 be a closed subgroup and 𝑁 a closed normal subgroup of 𝐺. Then, 𝐻𝑁 = {ℎ𝑛 | ℎ ∈ 𝐻, 𝑛 ∈ 𝑁 } is the image of the compact group 𝐻 × 𝑁 under the continuous map (ℎ, 𝑛) ↦→ ℎ𝑛. Hence, 𝐻𝑁 is a closed subgroup of 𝐺. As in the preceding paragraph, the group-theoretic isomorphism 𝜃: 𝐻𝑁/𝑁 → 𝐻/𝐻 ∩ 𝑁 defined by 𝜃 (ℎ𝑁) = ℎ(𝐻 ∩ 𝑁), ℎ ∈ 𝐻, is a homeomorphism, so 𝜃 is an isomorphism of topological groups. Similarly, if 𝑀 and 𝑁 are closed normal subgroups of 𝐺 with 𝑁 ≤ 𝑀, then 𝑀/𝑁 is a closed normal subgroup of 𝐺/𝑁 and the map 𝐺/𝑀 → (𝐺/𝑁)/(𝑀/𝑁) given by 𝑔𝑀 ↦→ (𝑔𝑁) 𝑀/𝑁, 𝑔 ∈ 𝐺, is an isomorphism of topological groups. Here is one way to equip an abstract group 𝐺 with a topology. Let N be a family of normal subgroups of 𝐺, closed under finite intersections, such that the intersection of all 𝑁 ∈ N is 1. Take N to be a base for the open neighborhoods of 1. A base for the open neighborhoods of 𝑎 ∈ 𝐺 is the family N𝑎 = {𝑎𝑁 | 𝑁 ∈ N }. The union of the N𝑎 ’s is then a base for a group topology of 𝐺. For example, the identity 𝑥𝑁 · 𝑦𝑁 = 𝑥𝑦𝑁 for normal subgroups 𝑁 implies that multiplication is continuous. Let (𝐺 𝑖 , 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 be an inverse system of topological groups and continuous homomorphisms 𝜋 𝑗𝑖 : 𝐺 𝑗 → 𝐺 𝑖 , for each 𝑖, 𝑗 ∈ 𝐼 with 𝑗 ≥ 𝑖. Then, 𝐺 = lim 𝐺 𝑖 is a ←− topological group and the projections 𝜋𝑖 : 𝐺 → 𝐺 𝑖 are continuous homomorphisms. Let ⟨𝐺 𝑖′ , 𝜋 ′𝑗𝑖 ⟩𝑖, 𝑗 ∈𝐼 be another system of topological groups with 𝐺 ′ = lim 𝐺 𝑖′ . Sup←− pose that 𝜃 𝑖 : 𝐺 𝑖 → 𝐺 𝑖′ , 𝑖 ∈ 𝐼, is a compatible system of continuous homomorphisms. Then, the corresponding map 𝜃: 𝐺 → 𝐺 ′ is a continuous homomorphism. Profinite Groups. In this book, we primarily consider an inverse system of finite groups (𝐺 𝑖 , 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 , each equipped with the discrete topology. We call the inverse limit 𝐺 = lim 𝐺 𝑖 a profinite group. By Lemma 1.1.3, 𝐺 is a compact Hausdorff ←− group. By Lemma 1.1.1, the open-closed sets 𝜋𝑖−1 (𝑔𝑖 ), 𝑔𝑖 ∈ 𝐺 𝑖 , 𝑖 ∈ 𝐼, form a base for the topology on 𝐺. In particular, the open normal subgroups of 𝐺 form a base for the open neighborhoods of 1. Remark 1.2.1 (Basic rules) Here are some basic rules for profinite groups 𝐺 and 𝐻 which we use in this book without explicit reference: (a) A subgroup 𝐻 of 𝐺 is open if and only if 𝐻 is closed of finite index (by Subsection “Topological Groups”). Every open subset of 𝐺 is a union of cosets 𝑔𝑖 𝑁𝑖 with 𝑁𝑖 open normal and 𝑔𝑖 ∈ 𝐺.

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(b) If an abstract subgroup 𝐻 ′ of 𝐺 contains an open subgroup 𝐻 of 𝐺, then 𝐻 ′ is open (and closed). Indeed, by (a), 𝐻 ′ is the union of finitely many cosets of 𝐻. Each of them is an open subset of 𝐺. (c) The intersection of all open normal subgroups of 𝐺 is 1. Indeed, suppose that 𝑔 ∈ 𝐺 and 𝑔 lies in every open normal subgroup of 𝐺. If 𝑔 ≠ 1, then by the Hausdorff property of 𝐺, there exists an open normal subgroup 𝐻 of 𝐺 such that 1 ∉ 𝑔𝐻. Hence 𝑔 ∉ 𝐻, which is a contradiction. (d) Every profinite group is compact, Hausdorff, and has a base for its topology consisting of open-closed sets (Lemmas 1.1.3 and 1.1.7). (e) A subset 𝐶 of a profinite group is closed if and only if 𝐶 is compact (use (d)). (f) A subset 𝐵 of a profinite group is open-closed if it is a union of finitely many cosets 𝑔𝑖 𝑁 with 𝑁 open normal and 𝑔𝑖 ∈ 𝐺 (use (a) and the compactness of 𝐵). (g) Every homomorphism 𝜑: 𝐺 → 𝐻 is tacitly assumed to be continuous. In particular, 𝜑 maps compact subsets of 𝐺 onto compact subsets of 𝐻. Hence, 𝜑 maps closed subsets of 𝐺 onto closed subsets of 𝐻 (use (e)). (h) By the first isomorphism theorem for compact groups, every epimorphism 𝜑: 𝐺 → 𝐻 of profinite groups is an open map. In particular, 𝜑 maps open subgroups of 𝐺 onto open subgroups of 𝐻. (i) Every continuous map 𝑓 of a compact space 𝑋 into a Hausdorff space 𝑌 is closed, i.e. 𝑓 maps closed subsets of 𝑋 onto closed subsets of 𝑌 [Bou89b, p. 87, §9.4, Cor. 2]. In particular, every continuous map 𝐺 → 𝐻 is closed. In particular, since profinite groups are both Hausdorff and compact, each homomorphism 𝜑: 𝐺 → 𝐻 is closed. Note that, we say that a topological space 𝑋 is compact if every cover of 𝑋 by open subsets has a finite subcover, while [Bou89b] calls this property of 𝑋 quasi-compact.

We list below special properties of profinite groups not shared by all compact groups: Lemma 1.2.2 Let {𝐻𝑖 | 𝑖 ∈ 𝐼} be a directed family of closed subsets of a profinite group 𝐺; that is: Ñ (1.3) for every finite subset 𝐽 of 𝐼 there is an 𝑖 ∈ 𝐼 with 𝐻𝑖 ⊆ 𝑗 ∈𝐽 𝐻 𝑗 . Ñ Put 𝐻 = 𝑖 ∈𝐼 𝐻𝑖 . Then: (a) For every open subgroup 𝑈 of 𝐺 containing 𝐻, there is an 𝑖 ∈ 𝐼 with 𝐻𝑖 ⊆ 𝑈. (b) Suppose that each Ñ𝐻𝑖 is closed under the inverse operation. Then, Ñ 𝐾 of 𝐺. 𝑖 ∈𝐼 𝐾 𝐻𝑖 = 𝐾 𝐻 and 𝑖 ∈𝐼 𝐻𝑖 𝐾 = 𝐻𝐾 for every Ñ closed subset Ñ (c) Let 𝜑: 𝐺 → 𝐺¯ be an epimorphism. Then, 𝜑( 𝑖 ∈𝐼 𝐻𝑖 ) = 𝑖 ∈𝐼 𝜑(𝐻𝑖 ). (d) Let 𝐾 and 𝐾 ′ be closed subgroups of 𝐺. Then, the set 𝑆 = {𝜎 ∈ 𝐺 | 𝐾 𝜎 = 𝐾 ′ } is closed. (e) Let 𝐾 and 𝐾 ′ be closed subgroups of 𝐺 that contain 𝐻. Suppose that each 𝐻𝑖 is a normal subgroup of 𝐺 and 𝐾 𝐻𝑖 and 𝐾 ′ 𝐻𝑖 are conjugate. Then, 𝐾 and 𝐾 ′ are conjugate.

1.2 Profinite Groups

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(f) Let 𝐺 = lim 𝐺 𝑖 be an inverse limit of finite groups, 𝜋𝑖 : 𝐺 → 𝐺 𝑖 the quotient ←− maps, and 𝐾, 𝐾 ′ closed subgroups of 𝐺. Suppose that 𝜋𝑖 (𝐾) and 𝜋𝑖 (𝐾 ′) are conjugate in 𝐺 𝑖 for each 𝑖. Then, 𝐾 and 𝐾 ′ are conjugates. (g) Let 𝐾 and 𝐾 ′ be closed subgroups of 𝐺. Suppose that 𝐾 𝑁 = 𝐾 ′ 𝑁 for every open normal subgroup 𝑁 of 𝐺. Then 𝐾 = 𝐾 ′. Ñ Proof of (a). The set 𝐻𝑖 ∩ (𝐺 ∖ 𝑈) is closed for every 𝑖 ∈ 𝐼 and 𝑖 ∈𝐼 𝐻𝑖 ∩ (𝐺 ∖ 𝑈) = 𝐻 ∩ (𝐺 ∖ 𝑈) = ∅. Since 𝐺 is compact, there exists a finite subset 𝐽 of 𝐼 such that Ñ Ñ 𝐻 𝑗 ∩ (𝐺 ∖ 𝑈) = ∅, so 𝑗 ∈𝐽 𝐻 𝑗 ⊆ 𝑈. By (1.3), there exists an 𝑖 ∈ 𝐼 such that 𝑗 ∈𝐽 Ñ 𝐻𝑖 ⊆ 𝑗 ∈𝐽 𝐻 𝑗 . Hence, 𝐻𝑖 ⊆ 𝑈. Ñ Proof of (b). Let 𝑔 ∈ 𝑖 ∈𝐼 𝐾 𝐻𝑖 . Then, for each 𝑖 ∈ 𝐼 there are 𝑘 𝑖 ∈ 𝐾 and ℎ𝑖 ∈ 𝐻𝑖 with 𝑔 = 𝑘 𝑖 ℎ𝑖 . Hence, the closed subset 𝐻𝑖 ∩ 𝑔 −1 𝐾 of 𝐺 is nonempty. By (1.3), the intersection of finitely many of the sets 𝐻𝑖 ∩ 𝑔 −1 𝐾 is nonempty. Since 𝐺 is compact, Ñ there exists an ℎ ∈ 𝑖 ∈𝐼 𝐻𝑖 ∩ 𝑔 −1 𝐾. It satisfies ℎ ∈ 𝐻 and ℎ = 𝑔 −1 𝑘 for some 𝑘 ∈ 𝐾. Therefore, 𝑔 = 𝑘 ℎ−1 ∈ 𝐾 𝐻. Proof of (c). Let 𝐾 = Ker(𝜑). Then, 𝜑 induces a bijection between the set of closed ¯ Hence, by subgroups lying between 𝐾 and 𝐺 and the set of closed subgroups of 𝐺. Ñ Ñ Ñ (b), 𝜑(𝐻) = 𝜑(𝐾 𝐻) = 𝜑( 𝑖 ∈𝐼 𝐾 𝐻𝑖 ) = 𝑖 ∈𝐼 𝜑(𝐾 𝐻𝑖 ) = 𝑖 ∈𝐼 𝜑(𝐻𝑖 ). Proof of (d). Denote the set of all open normal subgroups of 𝐺 by N . For each 𝑁 ∈ N , let 𝑆 𝑁 be the inverse image under the quotient map 𝐺 → 𝐺/𝑁 of the finite set {𝑠 ∈ 𝐺/𝑁 | (𝐾 𝑁/𝑁) 𝑠 = 𝐾 ′ 𝑁/𝑁 }. Ñ Then, 𝑆 𝑁 is closed. If 𝑠 ∈ 𝑆 𝑁 for each 𝑁 ∈ N , then, by (b), 𝐾 𝑠 = 𝑁 ∈N 𝐾 𝑠 𝑁 = Ñ Ñ ′ ′ 𝑁 ∈N 𝑆 𝑁 . Therefore, 𝑆 is closed. 𝑁 ∈N 𝐾 𝑁 = 𝐾 . Hence, 𝑆 = Proof of (e). For each 𝑖 ∈ 𝐼, 𝑆𝑖 = {𝜎 ∈ 𝐺 | 𝐾 𝜎 𝐻𝑖 = 𝐾 ′ 𝐻𝑖 } is a nonempty subset Ñ of 𝐺. By (d), 𝑆𝑖 is closed. In theÑnotation of (1.3), we have 𝑆𝑖 ⊆ 𝑗 ∈𝐽 𝑆 𝑗 . Hence, by compactness, there is a 𝜎 ∈ 𝑖 ∈𝐼 𝑆𝑖 . Thus, 𝐾 𝜎 𝐻𝑖 = 𝐾 ′ 𝐻𝑖 for each 𝑖 ∈ 𝐼. We conclude from (b) that 𝐾 𝜎 = 𝐾 ′. Proof of (f). Apply (e) to 𝐻𝑖 = Ker(𝜋𝑖 ). Proof of (g). Since the intersection of all open normal subgroups 𝑁 of 𝐺 is 1, (g) □ follows from (b).

Lemma 1.2.3 Each closed subgroup 𝐻 of a profinite group 𝐺 is the intersection of open subgroups. Proof. Let N be the set of all open normal subgroups of 𝐺. By Lemma 1.2.2(b), Ñ 𝑁 □ 𝑁 ∈N 𝐻 = 𝐻. The following result gives a sufficient condition for a compact group to be profinite. We use it below to prove that the category of profinite groups is closed under various natural operations.

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1 Infinite Galois Theory and Profinite Groups

Lemma 1.2.4 Let 𝐺 be a compact group and {𝑁𝑖 | 𝑖 ∈ 𝐼} a directed family of closed normal subgroups of 𝐺 of finite index satisfying Ù 𝑁𝑖 = 1. (1.4) 𝑖 ∈𝐼

Then, 𝐺  lim 𝐺/𝑁𝑖 and 𝐺 is a profinite group. ←− Conversely, if 𝐺 is a profinite group, then 𝐺 = lim 𝐺/𝑁 with 𝑁 ranging over all ←− open normal subgroups of 𝐺. Proof. By definition, lim 𝐺/𝑁𝑖 is a profinite group. We only need to prove the ←− isomorphism. The quotient maps 𝐺 → 𝐺/𝑁𝑖 define a continuous embedding 𝜃 of 𝐺 into lim 𝐺/𝑁𝑖 . If (𝑔𝑖 𝑁𝑖 )𝑖 ∈𝐼 is an element of the latter group, then the closed subsets ←− 𝑔𝑖 𝑁𝑖 of 𝐺 have the finite intersection property, i.e. the intersection of every Ñ finite family of these sets is nonempty. Since 𝐺 is compact, there exists a 𝑔 ∈ 𝑖 ∈𝐼 𝑔𝑖 𝑁𝑖 . Then, 𝜃 (𝑔) = (𝑔𝑖 𝑁𝑖 )𝑖 ∈𝐼 . Thus, 𝜃 is bijective. Compactness of 𝐺 and the Hausdorff property of lim 𝐺/𝑁𝑖 imply that 𝜃 is an isomorphism of topological groups. ←− Conversely, suppose that 𝐺 is a profinite group. As mentioned at the beginning of the subsection “Profinite groups”, 𝐺 is compact and Hausdorff. In particular, 1 is a normal closed subgroup of 𝐺, so by Lemma 1.2.3, 1 is the intersection of all open subgroups 𝑁 of 𝐺. Therefore, 𝐺 = lim 𝐺/𝑁, as stated in the second part of ←− the lemma. □ Lemma 1.2.5 The following statements hold for each closed subgroup 𝐻 of a profinite group 𝐺: (a) The group 𝐻 is profinite. Moreover, for each open normal subgroup 𝑀 of 𝐻 there exists an open normal subgroup 𝑁 of 𝐺 with 𝐻 ∩ 𝑁 ≤ 𝑀. If 𝑀 is normal in 𝐺, then 𝑁 can be chosen such that 𝐻 ∩ 𝑁 = 𝑀. (b) The map 𝐺 0 ↦→ 𝐺 0 ∩ 𝐻 maps the set of all open subgroups of 𝐺 onto the set of all open subgroups of 𝐻. (c) Every homomorphism 𝜑0 of 𝐻 into a finite group 𝐴 extends to a homomorphism 𝜑: 𝐻 ′ → 𝐴, where 𝐻 ′ is an open subgroup of 𝐺 containing 𝐻. If 𝐻 is normal in 𝐺, then 𝐻 ′ can be chosen to be normal. Proof of (a). If 𝑁 is an open normal subgroup of 𝐺, then 𝐻 ∩ 𝑁 is an open normal subgroup of 𝐻. The family of all groups 𝐻 ∩ 𝑁 is directed and its intersection is trivial. Hence, by Lemma 1.2.4, 𝐻 = lim 𝐻/𝐻 ∩ 𝑁 is a profinite group. ←− Let 𝑀 be an open normal subgroup of 𝐻. By Lemma 1.2.2(a), 𝐻 ∩ 𝑁 ≤ 𝑀 for some open normal subgroup 𝑁 of 𝐺. If 𝑀 ⊳ 𝐺, then 𝑀 𝑁 ⊳ 𝐺 and 𝐻 ∩ 𝑀 𝑁 = 𝑀. Proof of (b). If 𝐺 0 is an open subgroup of 𝐺, then 𝐺 0 is closed and of finite index in 𝐺 (Remark 1.2.1(a)). Hence 𝐺 0 ∩ 𝐻 is a closed subgroup of 𝐻 of finite index. Again, by Remark 1.2.1(a), 𝐻0 is open in 𝐻. For the converse observe that the intersection of all conjugates of 𝐻0 in 𝐻 is an open normal subgroup of 𝐻. Hence, by the first statement of the lemma, there is an open normal subgroup 𝑁 of 𝐺 such that 𝑀 = 𝐻 ∩ 𝑁 ≤ 𝐻0 . Then, 𝐺 0 = 𝐻0 𝑁 is an open subgroup of 𝐺 and 𝐻 ∩ 𝐺 0 = 𝐻0 .

1.2 Profinite Groups

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Proof of (c). By (a), 𝐺 has an open normal subgroup 𝑁 with 𝐻 ∩ 𝑁 ≤ Ker(𝜑0 ). Put 𝐻 ′ = 𝐻𝑁 and define 𝜑: 𝐻 ′ → 𝐴 by 𝜑(ℎ𝑛) = 𝜑0 (ℎ). Then, 𝜑 is a well-defined homomorphism which extends 𝜑0 . □ Lemma 1.2.6 These statements hold: (a) If 𝑁 is a closed normal subgroup of a profinite group 𝐺, then 𝐺/𝑁 is profinite. Î (b) The direct product, 𝐺 = 𝑖 ∈𝐼 𝐺 𝑖 , of profinite groups is profinite. (c) Every inverse limit, 𝐺 = lim 𝐺 𝑖 , of profinite groups is profinite. ←− Proof of (a). A closed normal group 𝑁 is an intersection of open subgroups (Lemma 1.2.3) and therefore of open normal subgroups 𝑁𝑖 , 𝑖 ∈ 𝐼, of 𝐺. Now apply Lemma 1.2.4 to 𝐺/𝑁 and {𝑁𝑖 /𝑁 | 𝑖 ∈ 𝐼}. Proof of (b). Consider a finite subset 𝐽 of 𝐼. For each 𝑗 ∈ 𝐽 let 𝑁 𝑗 be an open normal Î Î subgroup of 𝐺 𝑗 . Then, 𝑁 = 𝑗 ∈𝐽 𝑁 𝑗 × 𝑖 ∈𝐼 ∖ 𝐽 𝐺 𝑖 is an open normal subgroup of 𝐺. The family of all these normal subgroups is directed and its intersection is 1. By Lemma 1.2.4, 𝐺 = lim 𝐺/𝑁 is a profinite group. ←− Î Proof of (c). Observe that 𝐺 is a closed subgroup of the direct product 𝑖 ∈𝐼 𝐺 𝑖 □ (Lemma 1.1.2). By (b), this product is profinite. Now apply Lemma 1.2.5(a). Lemma 1.2.7 Each epimorphism 𝜑: 𝐺 → 𝐴 of profinite groups has a continuous set-theoretic section 𝜑 ′: 𝐴 → 𝐺. That is, 𝜑 ′ is a continuous map satisfying 𝜑 ◦ 𝜑 ′ = id 𝐴. Proof. Let 𝐾 = Ker(𝜑). We split the proof into two parts. Part A: 𝐾 is finite. Then, 1 is an open normal subgroup of 𝐾, so Lemma 1.2.5(a) gives an open subgroup 𝐻 of 𝐺 with 𝐾 ∩ 𝐻 = 1. Thus, 𝜑 maps 𝐻 bijectively onto 𝐵 = 𝜑(𝐻). As both 𝐻 and 𝐵 are Hausdorff and compact, 𝜑| 𝐻 is a homeomorphism. Set 𝛽 = (𝜑| 𝐻 ) −1 . Next let ( 𝐴 : 𝐵) = 𝑛, choose 𝑎 1 , . . . , 𝑎 𝑛 ∈ 𝐴 and 𝑔1 , . . . , 𝑔𝑛 ∈ 𝐺 with 𝐴 = Ð𝑛 · 𝑖=1 𝑎 𝑖 𝐵 and 𝜑(𝑔𝑖 ) = 𝑎 𝑖 , 𝑖 = 1, . . . , 𝑛. Define 𝜑 ′: 𝐴 → 𝐺 by 𝜑 ′ (𝑎 𝑖 𝑏) = 𝑔𝑖 𝛽(𝑏), 𝑏 ∈ 𝐵. Then, 𝜑 ′ is a continuous set-theoretic section to 𝜑. Part B: 𝐾 is arbitrary. Denote the set of all closed normal subgroups of 𝐺 which are contained in 𝐾 by L. For each 𝐿 ∈ L, let 𝜑 𝐿 : 𝐺/𝐿 → 𝐴 be the epimorphism induced by 𝜑. For each 𝐿 ′ ∈ L with 𝐿 ′ ≤ 𝐿 let 𝜑 𝐿′ ,𝐿 : 𝐺/𝐿 ′ → 𝐺/𝐿 be the quotient map. Let Φ′ be the set of all pairs (𝐿, 𝜑 ′𝐿 ) where 𝐿 ∈ L and 𝜑 ′𝐿 : 𝐴 → 𝐺/𝐿 is a settheoretic section to 𝜑 𝐿 . Since 𝜑 𝐾 : 𝐺/𝐾 → 𝐴 is an isomorphism, it has an inverse ′ . Hence, (𝐾, 𝜑 ′ ) is in Φ ′ . 𝜑𝐾 𝐾 Define a partial ordering on Φ′ as follows: (𝐿 ′, 𝜑 ′𝐿′ ) ≤ (𝐿, 𝜑 ′𝐿 ) if 𝐿 ′ ≤ 𝐿 and 𝜑 𝐿′ ,𝐿 ◦ 𝜑 ′𝐿′ = 𝜑 ′𝐿 . Suppose that Φ′′ = {(𝐿 𝑖 , 𝜑𝑖′ ) | 𝑖 ∈ 𝐼} is a descending chain in Ñ Φ′; that is, every two elements of Φ′′ are comparable. Put 𝐿 ′′ = 𝑖 ∈𝐼 𝐿 𝑖 . Then, by Lemmas 1.2.6(a) and 1.2.4, 𝐺/𝐿 ′′ = lim 𝐺/𝐿 𝑖 . Thus, the compatible maps ←− 𝜑𝑖′ : 𝐴 → 𝐺/𝐿 𝑖 give a section 𝜑 ′𝐿′′ : 𝐴 → 𝐺/𝐿 ′′ to 𝜑 𝐿′′ . Thus, (𝐿 ′′, 𝜑 ′𝐿′′ ) is a lower bound for Φ′′.

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1 Infinite Galois Theory and Profinite Groups

Zorn’s lemma gives a minimal element (𝐿, 𝜑 ′𝐿 ) of Φ′. Assume by contradiction that 𝐿 ≠ 1. Then, 𝐿 has a proper open subgroup 𝐿 ′ which is normal in 𝐺. Consider the epimorphism 𝜑 𝐿′ ,𝐿 : 𝐺/𝐿 ′ → 𝐺/𝐿 with the finite kernel 𝐿/𝐿 ′. Part A gives a set-theoretic section 𝜑 ′𝐿,𝐿′ to 𝜑 𝐿′ ,𝐿 . Set 𝜑 ′𝐿′ = 𝜑 ′𝐿,𝐿′ ◦ 𝜑 ′𝐿 . Then, (𝐿 ′, 𝜑 ′𝐿′ ) is an element of Φ′ which is smaller than (𝐿, 𝜑 ′𝐿 ). This contradiction to the minimality of (𝐿, 𝜑 ′𝐿 ) proves that 𝐿 = 1. Set 𝜑 ′ = 𝜑 ′𝐿 . Then, 𝜑 ′ is a continuous set-theoretic section to 𝜑. □

1.3 Infinite Galois Theory Let 𝑁 be a Galois extension of a field 𝐾. The Galois group Gal(𝑁/𝐾) associated with 𝑁/𝐾 consists of all automorphisms of 𝑁 that fix each element of 𝐾. If 𝑁/𝐾 is a finite extension and 𝐻1 , 𝐻2 are subgroups of Gal(𝑁/𝐾) with the same fixed fields in 𝑁, then 𝐻1 = 𝐻2 . This is not the case any more if 𝑁/𝐾 is infinite. Consider for example the case where 𝐾 = F 𝑝 for some prime number 𝑝 and Ð∞ F 𝑝𝑖 . Let 𝜑 be the Frobenius automorphism of 𝑁/F 𝑝 . It is defined by 𝑁 = 𝑖=1 the rule 𝜑𝑥 = 𝑥 𝑝 for each 𝑥 ∈ 𝑁. Let 𝐺 0 be the discrete subgroup of Gal(𝑁/F 𝑝 ) generated by 𝜑. It is a countable group and F 𝑝 is its fixed field in 𝑁. On the other hand, each element of Gal(F 𝑝2𝑖 /F 𝑝 ) has exactly two extensions to F 𝑝2𝑖+1 . Hence, there are 2ℵ0 sequences (𝜎1 , 𝜎2 , 𝜎3 , . . .) with 𝜎𝑖 ∈ Gal(𝐹 𝑝2𝑖 /F 𝑝 ) such that the restriction of 𝜎𝑖+1 to F 𝑝2𝑖 is 𝜎𝑖 for 𝑖 = 1, 2, 3, . . . . Each such sequence defines a unique 𝜎 ∈ Gal(𝑁/F 𝑝 ) whose restriction to F 𝑝2𝑖 is 𝜎𝑖 , 𝑖 = 1, 2, 3, . . . . It follows that the cardinality of Gal(𝑁/F 𝑝 ) is 2ℵ0 . In particular, Gal(𝑁/F 𝑝 ) is different from 𝐺 0 but has the same fixed field, namely F 𝑝 . Galois groups as profinite groups. The Galois correspondence is restored for closed subgroups of Gal(𝑁/𝐾) in the “Krull topology”, which we now introduce. Denote the set of all intermediate fields 𝐾 ⊆ 𝐿 ⊆ 𝑁, with 𝐿/𝐾 finite and Galois, by L. To each 𝐿 ∈ L associate the (finite) Galois group Gal(𝐿/𝐾). If 𝐿 ′ ∈ L and 𝐿 ⊆ 𝐿 ′, then res 𝐿 : Gal(𝐿 ′/𝐾) → Gal(𝐿/𝐾) is an epimorphism. Consider the inverse limit lim Gal(𝐿/𝐾), with 𝐿 ranging over L. Every 𝜎 ∈ Gal(𝑁/𝐾) defines ←− a unique element (res 𝐿 𝜎) 𝐿 ∈L of lim Gal(𝐿/𝐾). Conversely, every (𝜎𝐿 ) 𝐿 ∈L ∈ ←− lim Gal(𝐿/𝐾) defines a unique 𝜎 ∈ Gal(𝑁/𝐾) with res 𝐿 𝜎 = 𝜎𝐿 for each 𝐿 ∈ L. ←− Thus, 𝜎 ↦→ (res 𝐿 𝜎) 𝐿 ∈L is an isomorphism Gal(𝑁/𝐾)  lim Gal(𝐿/𝐾) of abstract ←− groups. This isomorphism induces a topology on Gal(𝑁/𝐾) through the topology on lim Gal(𝐿/𝐾): the Krull topology. Thus, under the Krull topology, Gal(𝑁/𝐾) ←− becomes a profinite group and the family N = {Gal(𝑁/𝐿) | 𝐿 ∈ L} is a base for the open neighborhoods of 1. If 𝑁/𝐾 is a finite extension, then the Krull topology is discrete. Suppose that 𝐿 is a finite extension of 𝐾 contained in 𝑁. Then, its Galois closure 𝐿ˆ is the smallest Galois extension of 𝐾 that contains 𝐿. It is finite over 𝐾 and is ˆ to see that contained in 𝑁. Write Gal(𝑁/𝐿) as a union of right cosets of Gal(𝑁/ 𝐿) Gal(𝑁/𝐿) is an open closed subgroup of Gal(𝑁/𝐾).

1.3 Infinite Galois Theory

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Suppose that 𝐿 is an arbitrary extension of 𝐾 in 𝑁. Then, 𝐿 is the union of a Ñ family {𝐿 𝑖 | 𝑖 ∈ 𝐼} of finite extensions of 𝐾. Hence, Gal(𝑁/𝐿) = 𝑖 ∈𝐼 Gal(𝑁/𝐿 𝑖 ). Therefore, Gal(𝑁/𝐿) is a closed subgroup of Gal(𝑁/𝐾). If 𝑆 is a set of automorphisms of 𝑁, then 𝑁 (𝑆) = {𝑥 ∈ 𝑁 | 𝜎𝑥 = 𝑥 for every 𝜎 ∈ 𝑆} is the fixed field of 𝑆 in 𝑁. 𝑁 (𝑆) is also the fixed field in 𝑁 of the closed subgroup ⟨𝑆⟩ of Gal(𝑁/𝐾) generated by 𝑆. If 𝑆 = {𝜎1 , . . . , 𝜎𝑒 } is a finite set, replace 𝑁 (𝑆) by 𝑁 (𝜎1 , . . . , 𝜎𝑒 ). Let 𝑀 be a Galois extension of 𝐾 in 𝑁. Denote the homomorphism from Gal(𝑁/𝐾) into Gal(𝑀/𝐾) that maps 𝜎 ∈ Gal(𝑁/𝐾) onto its restriction res 𝑀 𝜎 to 𝑀 by res 𝑀 (or res). It is a continuous surjective map. Proposition 1.3.1 Let 𝑁 be a Galois extension of a field 𝐾. Then, 𝐿 ↦→ Gal(𝑁/𝐿) is a bijection from the family of fields 𝐿 lying between 𝐾 and 𝑁 onto the family of closed subgroups of 𝐺 = Gal(𝑁/𝐾). The inverse map is 𝐻 ↦→ 𝑁 (𝐻). Proof. Consider a field extension 𝐿 of 𝐾 in 𝑁. Put 𝐻 = Gal(𝑁/𝐿). Then, 𝐿 ⊆ 𝑁 (𝐻). Each 𝑥 ∈ 𝑁 (𝐻) is contained in a finite Galois extension 𝑀 ⊆ 𝑁 of 𝐿. Since the map res: Gal(𝑁/𝐿) → Gal(𝑀/𝐿) is surjective, 𝜎𝑥 = 𝑥 for every 𝜎 ∈ Gal(𝑀/𝐿). By finite Galois theory, 𝑥 ∈ 𝐿. Hence, 𝑁 (Gal(𝑁/𝐿)) = 𝐿. Conversely, let 𝐻 be a closed subgroup of 𝐺 and put 𝐿 = 𝑁 (𝐻). Then, 𝐻 ≤ Gal(𝑁/𝐿). Consider 𝜎 ∈ Gal(𝑁/𝐿). In order to prove 𝜎 ∈ 𝐻 it suffices to show that 𝜎 is in the closure of 𝐻. Indeed, let 𝑀 ⊆ 𝑁 be a finite Galois extension of 𝐾. Then, 𝑀 ∩ 𝐿 = 𝑀 (res 𝑀 𝐻). Finite Galois theory shows that res 𝑀 𝜎 ∈ Gal(𝑀/𝑀 ∩ 𝐿) = res 𝑀 𝐻. Hence, 𝐻 ∩ 𝜎Gal(𝑁/𝑀) is nonempty. It follows that Gal(𝑁/𝑁 (𝐻)) = 𝐻. □ Basic rules of Galois theory. As in finite Galois theory [Lan97, pp. 192–199], Proposition 1.3.1 gives the following rules for the Galois correspondence: (1.5a) 𝐿 1 ⊆ 𝐿 2 ⇐⇒ Gal(𝑁/𝐿 2 ) ≤ Gal(𝑁/𝐿 1 ). (1.5b) 𝐻1 ≤ 𝐻2 ⇐⇒ 𝑁 (𝐻2 ) ⊆ 𝑁 (𝐻1 ). (1.5c) 𝑁 (𝐻1 ) ∩ 𝑁 (𝐻2 ) = 𝑁 (⟨𝐻1 , 𝐻2 ⟩), where ⟨𝐻1 , 𝐻2 ⟩ is the closed subgroup of 𝐺 generated by the closed subgroups 𝐻1 and 𝐻2 . (1.5d) Gal(𝑁/𝐿 1 ∩ 𝐿 2 ) = ⟨Gal(𝑁/𝐿 1 ), Gal(𝑁/𝐿 2 )⟩. (1.5e) Gal(𝑁/𝐿 1 𝐿 2 ) = Gal(𝑁/𝐿 1 ) ∩ Gal(𝑁/𝐿 2 ). (1.5f) 𝑁 (𝐻1 ∩ 𝐻2 ) = 𝑁 (𝐻1 )𝑁 (𝐻2 ). Since Galois groups are compact, their images under restriction are closed. As in the finite case this produces other theorems of infinite Galois theory. (1.6a) 𝑁 (𝜎𝐻𝜎 −1 ) = 𝜎𝑁 (𝐻) and (1.6b) Gal(𝑁/𝜎𝐿) = 𝜎Gal(𝑁/𝐿)𝜎 −1 , for every 𝜎 ∈ 𝐺. (1.6c) A closed subgroup 𝐻 of 𝐺 is normal if and only if 𝐿 = 𝑁 (𝐻) is a Galois extension of 𝐾. (1.6d) If 𝐿 is a Galois extension of 𝐾 and 𝐿 ⊆ 𝑁, then res: Gal(𝑁/𝐾) → Gal(𝐿/𝐾) is a continuous open epimorphism with kernel Gal(𝑁/𝐿) and Gal(𝐿/𝐾)  Gal(𝑁/𝐾)/Gal(𝑁/𝐿).

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(1.6e) res: Gal(𝐿 𝑀/𝑀) → Gal(𝐿/𝐿 ∩ 𝑀) is an isomorphism for every Galois extension 𝐿 of 𝐾 and every extension 𝑀 of 𝐾. (1.6f) If in (1.6e), 𝑀 is also a Galois extension of 𝐾, then 𝜎 ↦→ (res 𝐿 𝜎, res 𝑀 𝜎) is an isomorphism Gal(𝐿 𝑀/𝐿 ∩ 𝑀)  Gal(𝐿/𝐿 ∩ 𝑀) × Gal(𝑀/𝐿 ∩ 𝑀) and Gal(𝐿 𝑀/𝐾)  {(𝜎, 𝜏) ∈ Gal(𝐿/𝐾) × Gal(𝑀/𝐾) | res 𝐿∩𝑀 𝜎=res 𝐿∩𝑀 𝜏}. In both cases we use the product topology on products of groups. The first isomorphism of (1.6f) is a special case of the second one. To prove the second isomorphism, note first that the map 𝜎 ↦→ (res 𝐿 𝜎, res 𝑀 𝜎) is a continuous injective map of the left-hand side onto the right-hand side. Hence, it suffices to prove surjectivity. Thus, consider 𝜌 ∈ Gal(𝐿/𝐾) and 𝜏 ∈ Gal(𝑀/𝐾) with res 𝐿∩𝑀 𝜌 = res 𝐿∩𝑀 𝜏. Extend 𝜌 to an automorphism 𝜌1 of 𝐿 𝑀 and let 𝜌0 = res 𝑀 𝜌1 . Then, 𝜌0−1 𝜏 ∈ Gal(𝑀/𝐿 ∩ 𝑀). By (1.6e), there is a 𝜆 ∈ Gal(𝐿 𝑀/𝐿) with res 𝑀 𝜆 = 𝜌0−1 𝜏. The element 𝜎 = 𝜌1 𝜆 of Gal(𝐿 𝑀/𝐾) satisfies res 𝐿 𝜎 = 𝜌 and res 𝑀 𝜎 = 𝜏, as desired. These statements are useful when 𝑁 is the separable closure 𝐾sep of 𝐾. Denote Gal(𝐾sep /𝐾) by Gal(𝐾), the absolute Galois group of 𝐾. Profinite groups as Galois groups. Next we show that profinite groups are Galois groups. Lemma 1.3.2 Suppose that a profinite group 𝐺 acts faithfully as automorphisms of a field 𝐹. Suppose that for each 𝑥 ∈ 𝐹, the stabilizer 𝑆(𝑥) = {𝜎 ∈ 𝐺 | 𝜎𝑥 = 𝑥}, is an open subgroup of 𝐺. Then, 𝐹 is a Galois extension of the fixed field 𝐾 = 𝐹 (𝐺) and 𝐺 = Gal(𝐹/𝐾). Proof. If 𝐺 is a finite group, this is a result of Artin [Lan97, p. 264]. In general the group 𝐻 = 𝑆(𝑥1 ) ∩ · · · ∩ 𝑆(𝑥 𝑛 ) is open in 𝐺 for every 𝑥1 , . . . , 𝑥 𝑛 ∈ 𝐹. Therefore, so is the intersection 𝑁 of all the conjugates of 𝐻 (Exercise 4). The finite quotient group 𝐺/𝑁 acts faithfully on the field 𝐿 = 𝐾 (𝐺𝑥1 , . . . , 𝐺𝑥 𝑛 ). It has 𝐾 as its fixed field. Thus, 𝐿 is a finite Galois extension of 𝐾 and 𝐺/𝑁  Gal(𝐿/𝐾). The field 𝐹 is the union of all above 𝐿 and 1 is the intersection of all the above 𝑁. Hence, 𝐹 is a Galois extension of 𝐾 and Gal(𝐹/𝐾)  lim Gal(𝐿/𝐾)  lim 𝐺/𝑁  ←− ←− 𝐺. □ Proposition 1.3.3 Let 𝐿/𝐾 be a Galois extension and 𝛼: 𝐺 → Gal(𝐿/𝐾) an epimorphism of profinite groups. Then, there is a Galois extension 𝐹/𝐸 and an isomorphism 𝜑: Gal(𝐹/𝐸) → 𝐺 such that 𝐹 is a purely transcendental extension of 𝐿, 𝐿 ∩𝐸 = 𝐾, and 𝛼 ◦ 𝜑 = res 𝐿 . Proof (Waterhouse). Let 𝑋 be the disjoint union of all quotient groups 𝐺/𝑁, where 𝑁 ranges over all open normal subgroups of 𝐺. Consider the elements of 𝑋 as independent over 𝐿 and set 𝐹 = 𝐿(𝑋). Define an action of each 𝜎 in 𝐺 on 𝐹 by

1.4 The 𝑝-adic Integers and the Prüfer Group

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𝜎(𝜏𝑁) = 𝜎𝜏𝑁, for 𝜏𝑁 ∈ 𝑋, and 𝜎(𝑎) = 𝛼(𝜎) (𝑎) for 𝑎 ∈ 𝐿. This action of 𝐺 is faithful. We have 𝑆(𝜏𝑁) = 𝑁 and 𝑆(𝑎) = 𝛼−1 (Gal(𝐿/𝐾 (𝑎))). Any 𝑢 ∈ 𝐹 is a rational function with integral coefficients in 𝑎 1 , . . . , 𝑎 𝑚 ∈ 𝐿 and 𝑥 1 , . . . , 𝑥 𝑛 ∈ 𝑋. The stabilizer 𝑆(𝑢) of 𝑢 contains the open subgroup 𝑆(𝑎 1 ) ∩ · · · ∩ 𝑆(𝑎 𝑚 ) ∩ 𝑆(𝑥 1 ) ∩ · · · ∩ 𝑆(𝑥 𝑛 ). Hence, 𝑆(𝑢) is an open subgroup. Let 𝐸 be the fixed field of 𝐺 in 𝐹. By Lemma 1.3.2, 𝐺 = Gal(𝐹/𝐸) and the conclusion of the proposition follows from the definitions. □ Remark 1.3.4 In the notation of Proposition 1.3.3 and its proof, if 𝑋 is finite, then 𝐸 can be chosen to be finitely generated over 𝐾. If the cardinality 𝑚 of the number of open subgroups of 𝐺 is infinite, then 𝐸 can be chosen to have transcendence degree ≤ 𝑚 over 𝐾.

Corollary 1.3.5 (Leptin) Every profinite group is isomorphic to a Galois group of some Galois extension.

1.4 The 𝒑-adic Integers and the Prüfer Group The first examples of profinite groups are the group Z 𝑝 of 𝑝-adic numbers and the Prüfer group Zˆ = lim Z/𝑛Z. ←− The 𝑝-adic group Z 𝑝 . Let 𝑝 be a rational prime. Consider the quotient rings Z/𝑝 𝑖 Z with their canonical homomorphisms Z/𝑝 𝑗 Z → Z/𝑝 𝑖 Z given for 𝑗 ≥ 𝑖 by 𝑥 + 𝑝 𝑗 Z ↦→ 𝑥 + 𝑝 𝑖 Z. The inverse limit Z 𝑝 = lim Z/𝑝 𝑖 Z is the ring of 𝑝-adic integers. ←− It is a profinite ring that becomes a profinite group if one ignores the multiplication of Z 𝑝 . Each 𝑥 ∈ Z 𝑝 is a sequence (𝑥𝑖 + 𝑝 𝑖 Z)𝑖 ∈N where 𝑥𝑖 ∈ Z and 𝑥 𝑗 ≡ 𝑥𝑖 mod 𝑝 𝑖 Z for 𝑗 ≥ 𝑖. Each integer 𝑚 ≥ 0 corresponds to a basic neighborhood of 𝑥 consisting of all elements 𝑦 = (𝑦 𝑖 + 𝑝 𝑖 Z)𝑖 ∈N with 𝑦 𝑚 ≡ 𝑥 𝑚 mod 𝑝 𝑚 Z. The map 𝑎 ↦→ (𝑎 + 𝑝 𝑖 Z)𝑖 ∈N is an embedding of Z into Z 𝑝 . Identify Z with its image in Z 𝑝 . The sequence (𝑥𝑖 )𝑖 ∈N converges to 𝑥 = (𝑥𝑖 + 𝑝 𝑖 Z)𝑖 ∈N in the 𝑝-adic topology. Hence, Z is dense in Z 𝑝 . Yet, Z is not equal to Z 𝑝 . For example, if 𝑝 ≠ 2, Í𝑛−1 𝑖 Í𝑛−1 𝑖 𝑛 then ( 𝑖=0 𝑝 + 𝑝 𝑛 Z) 𝑛∈N belongs to Z 𝑝 but not to Z. For 𝑝 = 2, ( 𝑖=0 4 + 2 Z) 𝑛∈N belongs to Z2 but not to Z (see also Exercise 15). Lemma 1.4.1 The ring Z 𝑝 has the following properties: (a) An element 𝑥 = (𝑥𝑖 + 𝑝 𝑖 Z)𝑖 ∈N is invertible if and only if 𝑝 ∤ 𝑥 1 . (b) Z 𝑝 is an integral domain. Proof of (a). Suppose that 𝑥 ′ = (𝑥𝑖′ + 𝑝 𝑖 Z)𝑖 ∈N is an inverse of 𝑥. Then, 𝑥1′ 𝑥1 ≡ 1 mod 𝑝, hence 𝑝 ∤ 𝑥1 . Conversely, suppose 𝑝 ∤ 𝑥1 . Then, for each 𝑖, 𝑝 ∤ 𝑥 𝑖 . Hence, there exists an 𝑥 𝑖′ ∈ Z which is unique modulo 𝑝 𝑖 with 𝑥𝑖′𝑥𝑖 ≡ 1 mod 𝑝 𝑖 . Thus, 𝑥 ′ = (𝑥𝑖′ + 𝑝 𝑖 Z)𝑖 ∈N is in Z 𝑝 and 𝑥 ′𝑥 = 1.

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1 Infinite Galois Theory and Profinite Groups

Proof of (b). Let 𝑥 = (𝑥𝑖 + 𝑝 𝑖 )𝑖 ∈N and 𝑦 = (𝑦 𝑖 + 𝑝 𝑖 )𝑖 ∈N be nonzero elements of Z 𝑝 . Then, there are 𝑚, 𝑛 ∈ N with 𝑥 𝑚 ̸≡ 0 mod 𝑝 𝑚 and 𝑦 𝑛 ̸≡ 0 mod 𝑝 𝑛 . Hence, 𝑥 𝑚+𝑛 𝑦 𝑚+𝑛 ̸≡ 0 mod 𝑝 𝑚+𝑛 . Therefore, 𝑥𝑦 ≠ 0. Consequently, Z 𝑝 is an integral □ domain. The quotient field of Z 𝑝 is called the field of 𝑝-adic numbers. It is denoted by Q𝑝. Lemma 1.4.2 The group Z 𝑝 satisfies the following conditions: (a) For each 𝑖, 𝑝 𝑖 Z 𝑝 is the kernel of the projection 𝜋𝑖 : Z 𝑝 → Z/𝑝 𝑖 Z. Thus, 𝑝 𝑖 Z 𝑝 is an open subgroup of Z 𝑝 of index 𝑝 𝑖 . (b) If 𝐻 is a subgroup of Z 𝑝 of a finite index, then 𝐻 = 𝑝 𝑖 Z 𝑝 for some 𝑖 ∈ N. (c) 0 is the only closed subgroup of Z 𝑝 of infinite index. (d) 𝑝Z 𝑝 is the unique closed maximal subgroup of Z 𝑝 . (e) All nonzero closed subgroups of Z 𝑝 are isomorphic to Z 𝑝 . Proof of (a). Suppose that 𝑥 = (𝑥 𝑗 + 𝑝 𝑗 Z) 𝑗 ∈N belongs to Ker(𝜋𝑖 ). Then, 𝑥 𝑖 ≡ 0 mod 𝑝 𝑖 . Hence, 𝑥 𝑗 ≡ 0 mod 𝑝 𝑖 for each 𝑗 ≥ 𝑖. Write 𝑥 𝑗 = 𝑝 𝑖 𝑦 𝑗 for 𝑗 ≥ 𝑖 and 𝑦 𝑗 = 𝑦 𝑖 for 𝑗 < 𝑖. Let 𝑧 = (𝑦 𝑗+𝑖 + 𝑝 𝑗 Z) 𝑗 ∈N . Then, 𝑧 ∈ Z 𝑝 and 𝑝 𝑖 𝑧 = 𝑥. Indeed, 𝑝 𝑖 𝑧 𝑗 = 𝑝 𝑖 𝑦 𝑗+𝑖 = 𝑥 𝑗+𝑖 ≡ 𝑥 𝑗 mod 𝑝 𝑗 for every positive integer 𝑗. Proof of (b). Conversely, let 𝐻 be a subgroup of Z 𝑝 of index 𝑛. Suppose that 𝑛 = 𝑘 𝑝 𝑖 with 𝑝 ∤ 𝑘. By Lemma 1.4.1(a), 𝑘 is invertible in the ring Z 𝑝 , so 𝑛Z 𝑝 = 𝑝 𝑖 Z 𝑝 . Thus, 𝑝 𝑖 Z 𝑝 = 𝑛Z 𝑝 ≤ 𝐻. It follows that 𝑝 𝑖 = (Z 𝑝 : 𝑝 𝑖 Z 𝑝 ) ≥ (Z 𝑝 : 𝐻) = 𝑘 𝑝 𝑖 . Therefore, 𝑘 = 1 and 𝐻 = 𝑝 𝑖 Z 𝑝 . Proof of (c). A closed subgroup 𝐽 of Z 𝑝 of infinite index is the intersection of infinitely many open groups (Lemma 1.2.3). Hence, by (a), all subgroups 𝑝 𝑖 Z 𝑝 contain 𝐽. Consequently 𝐽 = 0. Proof of (d). By (a), (b), and (c), 𝑝Z 𝑝 is the unique closed maximal subgroup of Z𝑝. Proof of (e). By Lemma 1.4.1(b), the map 𝑥 ↦→ 𝑝 𝑖 𝑥 is an isomorphism of Z 𝑝 onto 𝑝𝑖 Z 𝑝 . □ Every element 𝑥 = (𝑥𝑖 + 𝑝 𝑖 Z)𝑖 ∈N has a unique representation as a formal power Í𝑛−1 Í 𝑛 𝑖 𝑖 series ∞ 𝑖=0 𝑎 𝑖 𝑝 , with 0 ≤ 𝑎 𝑖 < 𝑝 for all 𝑖. Indeed, 𝑥 𝑛 ≡ 𝑖=0 𝑎 𝑖 𝑝 mod 𝑝 , for every 𝑛 ∈ N. Lemma 1.4.3 Let 𝛼: Z 𝑝 → Z/𝑝 𝑛 Z be an epimorphism with 𝑛 ≥ 1 and 𝐻 a closed subgroup of Z 𝑝 . Suppose that 𝛼(𝐻) = Z/𝑝 𝑛 Z. Then, 𝐻 = Z 𝑝 . Proof. By Lemma 1.4.2(a), Ker(𝛼) = 𝑝 𝑛 Z 𝑝 . Thus, by assumption, 𝐻 + 𝑝 𝑛 Z 𝑝 = Z 𝑝 . Assume that 𝐻 ≠ Z 𝑝 . Then, by Lemma 1.4.2(d), 𝐻 ≤ 𝑝Z 𝑝 . Therefore, Z 𝑝 = 𝐻 + 𝑝 𝑛 Z 𝑝 ≤ 𝑝Z 𝑝 < Z 𝑝 . It follows from this contradiction that 𝐻 = Z 𝑝 . □ In the terminology of Section 25.6, Lemma 1.4.3 says that 𝛼: Z 𝑝 → Z/𝑝 𝑛 Z is a Frattini cover.

1.4 The 𝑝-adic Integers and the Prüfer Group

15

The Prüfer group. For each 𝑛 ∈ N consider the quotient group Z/𝑛Z and the canonical homomorphisms Z/𝑛Z → Z/𝑚Z defined for 𝑚|𝑛 by 𝑥 + 𝑛Z ↦→ 𝑥 + 𝑚Z. The inverse limit Zˆ = lim Z/𝑛Z is the Prüfer group. Like with Z 𝑝 , embed Z as a ←− dense subgroup of Zˆ by 𝑥 ↦→ (𝑥 + 𝑛Z) 𝑛∈N . Thus, Zˆ is the closure of the subgroup ˆ Also, the subgroups 𝑛Z generated by 1. Write Zˆ = ⟨1⟩ and say that 1 generates Z. of Z form a base for the neighborhoods of 0 in the induced topology. ˆ Lemma 1.4.4 For each 𝑛 ∈ N, 𝑛Zˆ is an open subgroup of Zˆ of index 𝑛 and 𝑛Zˆ  Z. ˆ If 𝐻 is a subgroup of Zˆ of index 𝑛, then 𝐻 = 𝑛Z. Proof. Suppose that 𝑥 = (𝑥 𝑘 + 𝑘Z) 𝑘 ∈Z lies in the kernel 𝑍 𝑛 of the projection Zˆ → Z/𝑛Z. Then, 𝑥 𝑛 ≡ 0 mod 𝑛. Hence, for each 𝑟 ∈ N we have 𝑥𝑟 𝑛 ≡ 𝑥 𝑛 ≡ 0 mod 𝑛, so 𝑥𝑟 𝑛 = 𝑛𝑦 𝑟 𝑛 for some 𝑦 𝑟 𝑛 ∈ Z. Let 𝑧 = (𝑦 𝑟 𝑛 + 𝑟Z)𝑟 ∈Z . If 𝑟 ′ is a multiple of 𝑟, ˆ Moreover, then 𝑛𝑦 𝑟 ′ 𝑛 ≡ 𝑛𝑦 𝑟 𝑛 mod 𝑟𝑛, so 𝑦 𝑟 ′ 𝑛 ≡ 𝑦 𝑟 𝑛 mod 𝑟. Therefore, 𝑧 ∈ Z. ˆ On the other hand, 𝑥𝑟 ≡ 𝑥𝑟 𝑛 ≡ 𝑛𝑦 𝑟 𝑛 ≡ 𝑛𝑧𝑟 mod 𝑟. Hence, 𝑥 = 𝑛𝑧, so 𝑍 𝑛 ≤ 𝑛Z. by definition, 𝑛Zˆ ≤ 𝑍 𝑛 , so 𝑛Zˆ = 𝑍 𝑛 . Consequently, 𝑛Zˆ is an open subgroup of Zˆ of index 𝑛. ˆ Indeed, if 𝑛𝑥 = 0, Next note that the map 𝑥 ↦→ 𝑛𝑥 is an isomorphism of Zˆ onto 𝑛Z. then 𝑛𝑥𝑟 𝑛 ≡ 0 mod 𝑟𝑛, so 𝑥𝑟 ≡ 𝑥𝑟 𝑛 ≡ 0 mod 𝑟 for each 𝑟 ∈ N. Hence, 𝑥 = 0. Finally, if 𝐻 is a subgroup of Zˆ of index 𝑛, then 𝑛Zˆ is contained in 𝐻 and has the ˆ □ same index. Therefore, 𝐻 = 𝑛Z. The following result relates Zˆ to the groups Z 𝑝 . Î Lemma 1.4.5 The group Zˆ is topologically isomorphic to the direct product Z 𝑝 where 𝑝 ranges over all prime numbers. Î 𝑘𝑝 Proof. Let 𝑛 = 𝑝 be the decomposition of a positive integer 𝑛 into a product of prime powers. The Chinese remainder theorem gives a canonical isomorphism Î Î Î 𝑘𝑝 Z/𝑝 𝑘 𝑝 Z → Z/𝑛Z. Combine this Îwith the projection Z 𝑝 → Z/𝑝 Z to obtain Z 𝑝 → Z/𝑛Z. The maps a compatible a continuous epimorphism 𝑓𝑛 : Î 𝑓𝑛 form ˆ Since Î Z 𝑝 is Z 𝑝 → Z. system, so they give a continuous homomorphism 𝑓 : ˆ Moreover, Z embeds compact and ZˆÎis Hausdorff, Im( 𝑓 ) is a closed subgroup of Z. diagonally in Z 𝑝 and 𝑓 (𝑚) = 𝑚 for each 𝑚 ∈ Z. Thus, Z ⊆ Im( 𝑓 ). Since Z is ˆ we have Im( 𝑓 ) = Z, ˆ so 𝑓 is surjective. The kernel of 𝑓 is Ñ Ker( 𝑓𝑛 ) = 0. dense in Z, Hence, 𝑓 is also injective. The compactness and Hausdorff properties imply that 𝑓 is a topological isomorphism. □ As a consequence of Lemma 1.4.5, Z 𝑝 is both a closed subgroup and a quotient ˆ for each prime 𝑝. of Z, Procyclic groups. A profinite group 𝐺 is procyclic if 𝐺 is the closure of an abstract group generated by one element 𝜎. Lemma 1.4.6 Let 𝑚 and 𝑛 be positive integers such that 𝑚|𝑛 and let 𝑥 be a positive integer such that 𝑥 + 𝑚Z generates Z/𝑚Z. Then, there exists a 𝑦 ∈ Z such that 𝑦 + 𝑛Z generates Z/𝑛Z and 𝑦 + 𝑚Z = 𝑥 + 𝑚Z.

16

1 Infinite Galois Theory and Profinite Groups

Proof. Write 𝑚 = 𝑝 1𝑘1 · · · 𝑝 𝑟𝑘𝑟 and 𝑛 = 𝑝 𝑙11 · · · 𝑝 𝑟𝑙𝑟 with integers 𝑘 1 , . . . , 𝑘 𝑟 ≥ 0 and 𝑙 1 , . . . , 𝑙𝑟 ≥ 1 such that 𝑘 𝑖 ≤ 𝑙𝑖 for 𝑖 = 1, . . . , 𝑟. Then, there exists a 𝑦 ∈ Z such that 𝑦 ≡ 𝑥 mod 𝑝 𝑙𝑖𝑖 if 𝑘 𝑖 ≥ 1 and 𝑦 ≡ 1 mod 𝑝 𝑙𝑖𝑖 if 𝑘 𝑖 = 0. In particular, 𝑦 ≡ 𝑥 mod 𝑚. Since 𝑥 + 𝑚Z generates Z/𝑚Z, we have that 𝑥 is relatively prime to 𝑚. Hence, 𝑦 is relatively prime to 𝑛. Therefore, 𝑦 + 𝑛Z generates Z/𝑛Z, as desired. □ Remark 1.4.7 (Group-theoretic section) In other words, Lemma 1.4.6 says that if 𝜑: 𝐺 → 𝐻 is an epimorphism of finite cyclic groups and 𝑥 is a generator of 𝐻, then 𝐺 has a generator 𝑦 with 𝜑(𝑦) = 𝑥. More generally, let 𝜑: 𝐺 → 𝐻 be an epimorphism of procyclic groups and let ℎ be a generator of 𝐻. For each open normal subgroup 𝑁 of 𝐺, 𝑀 = 𝜑(𝑁) is an open normal subgroup of 𝐻 and 𝜑 induces an epimorphism 𝜑 𝑁 : 𝐺/𝑁 → 𝐻/𝑀 of finite cyclic groups. Moreover, 𝐻/𝑀 = ⟨ℎ𝑀⟩. Since 𝐺/𝑁 is a finite cyclic group, Lemma 1.4.6 yields 𝑔 ∈ 𝐺 with ⟨𝑔𝑁⟩ = 𝐺/𝑁 and 𝜑 𝑁 (𝑔𝑁) = ℎ𝑀. If 𝑁 ′ is an open normal subgroup of 𝐺 which is contained in 𝑁, 𝑀 ′ = 𝜑(𝑁 ′), and 𝑔 ′ is an element of 𝐺 with ⟨𝑔 ′ 𝑁 ′⟩ = 𝐺/𝑁 ′ and 𝜑 𝑁 ′ (𝑔 ′ 𝑁 ′) = ℎ𝑀 ′, then ⟨𝑔 ′ 𝑁⟩ = 𝐺/𝑁 and 𝜑 𝑁 (𝑔 ′ 𝑁) = ℎ𝑀. It follows from Corollary 1.1.4 that 𝐺 has a generator 𝑔 such that 𝜑(𝑔) = ℎ. Lemma 20.7.3 generalizes this statement to arbitrary finitely generated profinite groups. Corollary 1.4.8 Let 𝜑: Zˆ → 𝐴 and 𝛼: 𝐵 → 𝐴 be epimorphisms of procyclic groups. Then, there exists an epimorphism 𝛾: Zˆ → 𝐵 such that 𝛼 ◦ 𝛾 = 𝜑. Proof. By assumption, 𝑎 := 𝜑(1) generates 𝐴. Use Remark 1.4.7 to choose 𝑏 ∈ 𝐵 that generates 𝐵 with 𝛼(𝑏) = 𝑎. Then, the map 1 ↦→ 𝑏 extends to an epimorphism 𝛾: Zˆ → 𝐵 with 𝛼 ◦ 𝛾 = 𝜑, as desired. □ Proposition 20.7.4 generalizes Corollary 1.4.8 to arbitrary “finitely generated free profinite groups”.

1.5 The Absolute Galois Group of a Finite Field For every prime power 𝑞 there exists a field F𝑞 (unique up to isomorphism) with 𝑞 elements. It is characterized within its algebraic closure F˜ 𝑞 by F𝑞 = {𝑥 ∈ F˜ 𝑞 | 𝑥 𝑞 = 𝑥}. The field F𝑞 has, for each 𝑛 ∈ N, exactly one extension, F𝑞 𝑛 , of degree 𝑛. It is Galois with a cyclic group generated by the Frobenius automorphism 𝜋𝑞,𝑛 defined by 𝑎 is an isomorphism of Z/𝑛Z onto 𝜋𝑞,𝑛 (𝑥) = 𝑥 𝑞 for 𝑥 ∈ F𝑞 𝑛 . The map 𝑎 + 𝑛Z ↦→ 𝜋𝑞,𝑛 Gal(F𝑞 𝑛 /F𝑞 ). If 𝑚|𝑛, there is a canonical commutative diagram / Z/𝑚Z

Z/𝑛Z  Gal(F𝑞 𝑛 /F𝑞 )

res

 / Gal(F𝑞 𝑚 /F𝑞 ).

1.5 The Absolute Galois Group of a Finite Field

17

Take the inverse limits to obtain an isomorphism Zˆ  Gal(F𝑞 ) mapping the identity element 1 of Zˆ to the Frobenius automorphism 𝜋𝑞 , defined on all of F˜ 𝑞 by 𝜋𝑞 (𝑥) = 𝑥𝑞 . Ð∞ F𝑞𝑙𝑖 . Then, the projection Zˆ ↦→ Z𝑙 Let 𝑙 be a prime number and F𝑞(𝑙) = 𝑖=1 corresponds to res: Gal(F𝑞 ) → Gal(F𝑞(𝑙) /F𝑞 ). By Lemma 1.4.5, Gal(F𝑞 )  Î Gal(F𝑞(𝑙) /F𝑞 ). On the other hand, let 𝑁𝑙 be the fixed field of Z𝑙 in F˜ 𝑞 . Then, Gal(𝑁𝑙 ) = Z𝑙 , F𝑞(𝑙) 𝑁𝑙 = F˜ 𝑞 , and F𝑞(𝑙) ∩ 𝑁𝑙 = F𝑞 . It follows that Ö Z𝑙′ × Z𝑙 . Gal(F𝑞 ) = Gal(F𝑞(𝑙) ) × Gal(𝑁𝑙 )  𝑙′ ≠𝑙

Exercises 1. Let (𝑆𝑖 , 𝜋 𝑗𝑖 ) be an inverse system of finite sets with all 𝜋 𝑗𝑖 surjective and let 𝑆 = lim 𝑆𝑖 . Use Lemma 1.1.3 to prove that all maps 𝜋𝑖 : 𝑆 → 𝑆𝑖 determined by the ←− 𝜋 𝑗𝑖 ’s are surjective. 2. Let 𝐻1 , . . . , 𝐻𝑟 be closed subgroups of a profinite group 𝐺. Prove that Ù (𝐻1 𝑁 ∩ · · · ∩ 𝐻𝑟 𝑁), 𝐻 1 ∩ · · · ∩ 𝐻𝑟 = 𝑁

where 𝑁 ranges over all open normal subgroups of 𝐺. Hint: Use Lemma 1.2.2(b). 3. Suppose 𝐻 is a closed subgroup of a profinite group 𝐺. Prove: If 𝐻𝑁/𝑁 = 𝐺/𝑁 for every open normal subgroup 𝑁 of 𝐺, then 𝐻 = 𝐺. 4. Let 𝐻 be an open subgroup of index 𝑛 of a profinite group 𝐺. Denote the intersection of all conjugates of 𝐻 in 𝐺 by 𝑁. Note: Multiplication of 𝐺 on the left cosets of 𝐻 induces a homomorphism of 𝐺 into the symmetric group 𝑆 𝑛 with kernel 𝑁. Conclude that 𝐺/𝑁 is isomorphic to a subgroup of 𝑆 𝑛 and (𝐺 : 𝑁) ≤ 𝑛!. 5. Let 𝐺 be a compact group and 𝐻 an open subgroup. Suppose 𝐻 is profinite. Prove: 𝐺 is profinite. Hint: Use Lemma 1.2.4. 6. Let 𝑆 be a set of prime numbers. Consider the profinite group Zˆ 𝑆 = lim Z/𝑛Z, ←− with 𝑛 running over all positive integers with prime factors in 𝑆. (a) Prove: The finite homomorphic images of Zˆ 𝑆 are exactly the groups Z/𝑛Z, where the prime factors of 𝑛 belong to 𝑆. (b) Embed Z in Zˆ 𝑆 and determine the topology on Z induced by that of Zˆ 𝑆 . (c) Prove that Zˆ 𝑆 is procyclic. Î (d) Follow the proof of Lemma 1.4.5 to prove that Zˆ 𝑆  𝑝 ∈𝑆 Z 𝑝 . 7. Let 𝐺 be a procyclic group. Use that 𝐺Îis a homomorphic image of Zˆ to show there exists a set 𝑆 of primes with 𝐺 = 𝑝 ∈𝑆 𝐺 𝑝 , where for each 𝑝 ∈ 𝑆 either 𝐺 𝑝 ÎZ/𝑝 𝑖 𝑝 Z for some 𝑖 𝑝 ∈ N or 𝐺 𝑝  Z 𝑝 . In particular, if 𝐺 is torsion free, then 𝐺  𝑝 ∈𝑆 Z 𝑝 . ˆ Prove that every finite quotient of 𝐺 is a cyclic 8. Let 𝐺 be a closed subgroup of Z. group. Conclude that Î 𝐺 is procyclic and therefore that there exists a set 𝑆 of prime numbers with 𝐺  𝑝 ∈𝑆 Z 𝑝 .

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1 Infinite Galois Theory and Profinite Groups

9. Let 𝐺 be a profinite group. Prove that each of the following statements is equivalent ˆ to 𝐺  Z. (a) 𝐺 has exactly one open subgroup of each index 𝑛. ˆ (b) 𝐺 is procyclic and there is an epimorphism 𝜋: 𝐺 → Z. 10. Let 𝐺 be a procyclic group. Use Corollary 1.4.8 to prove that each epimorphism 𝜋: 𝐺 → Zˆ is an isomorphism. 11. Let 𝐺 be a profinite group with at most one open subgroup of every index 𝑛. (a) Prove that every open subgroup is normal. (b) Observe that (a) holds for every finite homomorphic image 𝐺¯ of 𝐺. Conclude that 𝐺¯ is nilpotent. (c) Let 𝑃 be a finite 𝑝-group with the above properties. Prove that every element 𝑥 of 𝑃 of maximal order generates 𝑃. (d) Conclude that 𝐺 is a procyclic group. 12. Define powers in a profinite group 𝐺 with exponents in Zˆ in the following way: ˆ Then, there exists a sequence {𝜈1 , 𝜈2 , 𝜈3 , . . .} of elements of Let 𝑔 ∈ 𝐺 and 𝜈 ∈ Z. Z that converges to 𝜈. By compactness there is a subsequence of {𝑔 𝜈1 , 𝑔 𝜈2 , 𝑔 𝜈3 , . . .} that converges to an element ℎ of 𝐺. (a) Prove ℎ does not depend on the sequence {𝜈1 , 𝜈2 , 𝜈3 , . . .}. So we may denote ℎ by 𝑔 𝜈 . Hint: If 𝑁 is a normal subgroup of 𝐺 of index 𝑛, then 𝑥 𝑛 ∈ 𝑁 for every 𝑥 ∈ 𝐺. (b) Prove the usual rules for the power operations. For example,

𝑔 𝜇 𝑔 𝜈 = 𝑔 𝜇+𝜈 ,

(𝑔 𝜇 ) 𝜈 = 𝑔 𝜇𝜈 ,

and

𝑔 𝜈 ℎ 𝜈 = (𝑔ℎ) 𝜈 if 𝑔ℎ = ℎ𝑔.

(c) Prove that the map (𝑔, 𝜈) ↦→ 𝑔 𝜈 of 𝐺 × Zˆ into 𝐺 is continuous. 13. Prove the following statements about the ring Z 𝑝 : (a) Every closed subgroup of Z 𝑝 is an ideal of Z 𝑝 . (b) 𝑝Z 𝑝 is the unique maximal ideal of Z 𝑝 ; observe that Z 𝑝 /𝑝Z 𝑝  F 𝑝 . (c) Deduce: 𝛼 ∈ Z 𝑝 is a unit (i.e. invertible in this ring) if and only if 𝛼 is congruent modulo 𝑝Z 𝑝 to one of the numbers 1, 2, . . . , 𝑝 − 1. 14. Define multiplication in the additive group Zˆ in a manner analogous to the definition of multiplication in Z 𝑝 . Prove that this makes Zˆ a commutative topological ring with zero divisors. Î (a) Prove that the isomorphism of additive groups Zˆ  Z 𝑝 established in Lemma 1.4.5 is an isomorphism of rings. (b) Prove that every closed subgroup of Zˆ is also an ideal. 15. Use the power series representation of the elements of Z 𝑝 (Section 1.4) to show |Z 𝑝 | = 2ℵ0 . Conclude that Q 𝑝 has elements that are transcendental over Q.

1.5 The Absolute Galois Group of a Finite Field

19

√ 16. For a prime number 𝑝, let 𝐾 𝑝 be Q(𝜁 𝑝 ) if 𝑝 ≠ 2 and Q( −1) if 𝑝 = 2. Also, let Ð 𝐿𝑝 = ∞ 𝑖=1 Q(𝜁 𝑝 𝑖 ). (a) Prove that if 𝐾 ′ is a field such that 𝐾 𝑝 ⊆ 𝐾 ′ ⊂ 𝐿 𝑝 , then Gal(𝐿 𝑝 /𝐾 ′)  Z 𝑝 and [𝐾 ′ : Q] < ∞. (b) Prove that Gal(𝐿 𝑝 /Q) is isomorphic to Z 𝑝 ×Z/( 𝑝−1)Z if 𝑝 ≠ 2 and to Z2 ×Z/2Z if 𝑝 = 2.

Notes More about topological groups can be found in [Pon66] and also in [HeR63]. A detailed exposition on the Galois theory of finite extensions appears in [Lan97, Chapter VI, Section 1]. For finite fields see [Lan97, Chapter V, Section 5]. Leptin’s proof of Corollary 1.3.5 uses linear disjointness of fields [Lep95]. We reproduce it in Proposition 3.4.12. The proofs of Lemma 1.3.2 and Proposition 1.3.3 appear in [Wat74]. This chapter overlaps with [Rbs70, Chapter 1].

Chapter 2

Valuations

This chapter introduces the basic elements of the theory of valuations, especially discrete valuations, and of Dedekind domains. These sections are primarily a survey. We prove that an overring of a Dedekind domain is again a Dedekind domain (Proposition 2.5.7).

2.1 Valuations, Places, and Valuation Rings The literature treats the arithmetic theory of fields through three intimately connected classes of objects: valuations, places, and valuation rings. We briefly review the basic definitions. Call an Abelian (additive) group Γ with a binary relation < an ordered group if the following statements hold for all 𝛼, 𝛽, 𝛾 ∈ Γ. (2.1a) Either 𝛼 < 𝛽, or 𝛼 = 𝛽, or 𝛽 < 𝛼. (2.1b) If 𝛼 < 𝛽 and 𝛽 < 𝛾, then 𝛼 < 𝛾. (2.1c) If 𝛼 < 𝛽, then 𝛼 + 𝛾 < 𝛽 + 𝛾. Some examples of ordered groups are the additive groups Z, R, and Z ⊕ Z with the order (𝑚, 𝑛) < (𝑚 ′, 𝑛 ′) if either 𝑚 < 𝑚 ′ or 𝑚 = 𝑚 ′ and 𝑛 < 𝑛 ′ (the lexicographic order). A valuation 𝑣 of a field 𝐹 is a map of 𝐹 into a set Γ ∪ {∞}, where Γ is an ordered group, with these properties: (2.2a) 𝑣(𝑎𝑏) = 𝑣(𝑎) + 𝑣(𝑏).
(2.2b) 𝑣(𝑎 + 𝑏) ≥ min 𝑣(𝑎), 𝑣(𝑏) . (2.2c) 𝑣(𝑎) = ∞ if and only if 𝑎 = 0. (2.2d) There exists an 𝑎 ∈ 𝐹 × with 𝑣(𝑎) ≠ 0. By definition, the symbol ∞ satisfies these rules: (2.3a) ∞ + ∞ = 𝛼 + ∞ = ∞ + 𝛼 = ∞; and (2.3b) 𝛼 < ∞ for each 𝛼 ∈ Γ.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_2

21

22

2 Valuations

Conditions (2.2a)–(2.2d) imply several more properties of 𝑣: (2.4a) 𝑣(1) = 0, 𝑣(−𝑎) = 𝑣(𝑎). (2.4b) If 𝑣(𝑎) < 𝑣(𝑏), then 𝑣(𝑎 + 𝑏) = 𝑣(𝑎) (use the identity 𝑎 = (𝑎 + 𝑏) − 𝑏 and (2.2b)). Í𝑛 𝑎 𝑖 = 0, then there exist 𝑖 ≠ 𝑗 such that 𝑣(𝑎 𝑖 ) = 𝑣(𝑎 𝑗 ) and 𝑣(𝑎 𝑖 ) = (2.4c) If 𝑖=1 min(𝑣(𝑎 1 ), . . . , 𝑣(𝑎 𝑛 )) (use (2.2b) and (2.4b)). We refer to the pair (𝐹, 𝑣) as a valued field. The subgroup Γ𝑣 = 𝑣(𝐹 × ) of Γ is the value group of 𝑣. The set 𝑂 𝑣 = {𝑎 ∈ 𝐹 | 𝑣(𝑎) ≥ 0} is the valuation ring of 𝑣 whose group of units is 𝑂 ×𝑣 = {𝑢 ∈ 𝐹 | 𝑣(𝑢) = 0}. It follows that 𝔪𝑣 = {𝑎 ∈ 𝐹 | 𝑣(𝑎) > 0} is the unique maximal ideal of 𝑂 𝑣 . Refer to the residue field 𝐹¯𝑣 = 𝑂 𝑣 /𝔪𝑣 as the residue field of 𝐹 at 𝑣. Likewise, whenever there is no ambiguity, we denote the coset 𝑎 + 𝔪𝑣 by 𝑎¯ and call it the residue of 𝑎 at 𝑣. Two valuations 𝑣 1 , 𝑣 2 of a field 𝐹 with value groups Γ1 , Γ2 are equivalent if there exists an isomorphism 𝑓 : Γ1 → Γ2 with 𝑣 2 = 𝑓 ◦ 𝑣 1 . Starting from Section 2.2, we abuse our language and say that 𝑣 1 and 𝑣 2 are distinct if they are inequivalent. A place of a field 𝐹 is a map 𝜑 of 𝐹 into a set 𝑀 ∪ {∞}, where 𝑀 is a field, with these properties: (2.5a) 𝜑(𝑎 + 𝑏) = 𝜑(𝑎) + 𝜑(𝑏). (2.5b) 𝜑(𝑎𝑏) = 𝜑(𝑎)𝜑(𝑏). (2.5c) There exist 𝑎, 𝑏 ∈ 𝐹 with 𝜑(𝑎) = ∞ and 𝜑(𝑏) ≠ 0, ∞. By definition the symbol ∞ satisfies the following rules: (2.6a) 𝑥 + ∞ = ∞ + 𝑥 = ∞ for each 𝑥 ∈ 𝑀. (2.6b) 𝑥 · ∞ = ∞ · 𝑥 = ∞ · ∞ = ∞ for each 𝑥 ∈ 𝑀 × . (2.6c) Neither ∞ + ∞ nor 0 · ∞ are defined. It is understood that (2.5a) and (2.5b) hold whenever the right-hand side is defined. These conditions imply that 𝜑(1) = 1, 𝜑(0) = 0 and 𝜑(𝑥 −1 ) = 𝜑(𝑥) −1 . In particular, if 𝑥 ≠ 0, then 𝜑(𝑥) = 0 if and only if 𝜑(𝑥 −1 ) = ∞. We call an element 𝑥 ∈ 𝐹 with 𝜑(𝑥) ≠ ∞ finite at 𝜑, and say that 𝜑 is finite at 𝑥. The subring of all elements finite at 𝜑, 𝑂 𝜑 = {𝑎 ∈ 𝐹 | 𝜑(𝑎) ≠ ∞},

(2.7)

is the valuation ring of 𝜑. It has a unique maximal ideal 𝔪 𝜑 = {𝑎 ∈ 𝐹 | 𝜑(𝑎) = 0}.

(2.8)

The quotient ring 𝑂 𝜑 /𝔪 𝜑 is a field which is canonically isomorphic to the residue field 𝐹¯𝜑 = {𝜑(𝑎) | 𝑎 ∈ 𝑂 𝜑 } of 𝐹 at 𝜑. The latter is a subfield of 𝑀. Call 𝜑 a 𝐾-place if 𝐾 is a subfield of 𝐹 and 𝜑(𝑎) = 𝑎 for each 𝑎 ∈ 𝐾. Two places 𝜑1 and 𝜑2 of a field 𝐹 with residue fields 𝑀1 and 𝑀2 are equivalent if there exists an isomorphism 𝜆: 𝑀1 → 𝑀2 with 𝜑2 = 𝜆 ◦ 𝜑1 . A valuation ring of a field 𝐹 is a proper subring 𝑂 of 𝐹 such that if 𝑥 ∈ 𝐹 × , then 𝑥 ∈ 𝑂 or 𝑥 −1 ∈ 𝑂. The subset 𝔪 = {𝑥 ∈ 𝑂 | 𝑥 −1 ∉ 𝑂} is the unique maximal ideal of 𝑂 (Exercise 1). The map 𝜑: 𝐹 → 𝑂/𝔪 ∪ {∞} which maps 𝑥 ∈ 𝑂 onto its residue class modulo 𝔪 and maps 𝑥 ∈ 𝐹 ∖ 𝑂 onto ∞ is a place of 𝐹 with valuation ring 𝑂.

2.2 Discrete Valuations

23

Denote the units of 𝑂 by 𝑈 = {𝑥 ∈ 𝑂 | 𝑥 −1 ∈ 𝑂}. Then, 𝐹 × /𝑈 is a multiplicative group ordered by the rule 𝑥𝑈 ≤ 𝑦𝑈 ⇐⇒ 𝑦𝑥 −1 ∈ 𝑂. The map 𝑥 ↦→ 𝑥𝑈 defines a valuation of 𝐹 with 𝑂 being its valuation ring. These definitions give a bijective correspondence between the valuation classes, the place classes, and the valuation rings of a field 𝐹 which is compatible with the three notions of maximal ideals and residue fields. An isomorphism 𝜎: 𝐹 → 𝐹 ′ of fields induces a bijective map of the valuations and places of 𝐹 onto those of 𝐹 ′ according to the following rule: If 𝑣 is a valuation of 𝐹, then 𝜎(𝑣) is defined by 𝜎(𝑣) (𝑥) = 𝑣(𝜎 −1 𝑥) for every 𝑥 ∈ 𝐹 ′. If 𝜑 is a place of 𝐹, then 𝜎(𝜑) (𝑥) = 𝜑(𝜎 −1 𝑥). In particular, 𝜎 induces an isomorphism 𝐹¯𝜑  𝐹¯ ′ 𝜎 ( 𝜑)

of residue fields. It is also clear that if 𝜑 corresponds to 𝑣, then 𝜎(𝜑) corresponds to 𝜎(𝑣). A valuation 𝑣 of a field 𝐹 is real (or of rank 1) if Γ𝑣 is isomorphic to a subgroup of R. Real valuations satisfy the so-called weak approximation theorem, a generalization of the Chinese remainder theorem [CaF67, p. 48, Lemma]: Proposition 2.1.1 Consider the following objects: inequivalent real valuations 𝑣 1 , . . . , 𝑣 𝑛 of a field 𝐹, elements 𝑥1 , . . . , 𝑥 𝑛 of 𝐹, and real numbers 𝛾1 , . . . , 𝛾𝑛 . Then, there exists an 𝑥 ∈ 𝐹 with 𝑣 𝑖 (𝑥 − 𝑥𝑖 ) ≥ 𝛾𝑖 , 𝑖 = 1, . . . , 𝑛. An analog of the weak approximation theorem is the Chinese remainder theorem from number theory that extends to arbitrary commutative rings. Proposition 2.1.2 Let 𝑅 be a commutative ring. (a) Suppose that 𝑃1 , . . . , 𝑃𝑟 are ideals of 𝑅 with 𝑃𝑖 + 𝑃 𝑗 = 𝑅 for all distinct integers 𝑖, 𝑗 between 1 and 𝑟. Let 𝑎 1 , . . . , 𝑎𝑟 ∈ 𝑅. Then, there exists an 𝑥 ∈ 𝑅 such that 𝑥 ≡ 𝑎 𝑖 mod 𝑃𝑖 for 𝑖 = 1, . . . , 𝑟. (b) Suppose that 𝑃1 , . . . , 𝑃𝑟 are distinct maximal ideals of 𝑅. Let 𝑎 1 , . . . , 𝑎𝑟 , 𝑏 1 , . . . , 𝑏𝑟 ∈ 𝑅 with 𝑎 𝑖 ∉ 𝑃𝑖 for 𝑖 = 1, . . . , 𝑟. Then, there exists an 𝑥 ∈ 𝑅 such that 𝑎 𝑖 𝑥 + 𝑏 𝑖 ≡ 0 mod 𝑃𝑖 for 𝑖 = 1, . . . , 𝑟. Proof. We refer to [Lan97, p. 94, Thm. 2.1] for statement (a) and prove statement (b). For each 𝑖 between 1 and 𝑟 we have 𝑅𝑎 𝑖 + 𝑃𝑖 = 𝑅. Hence, there exists an 𝑎 𝑖′ ∈ 𝑅 with 𝑎 𝑖′ 𝑎 𝑖 ≡ 1 mod 𝑃𝑖 . By assumption, 𝑃𝑖 + 𝑃 𝑗 = 𝑅 if 𝑖 ≠ 𝑗. Hence, by (a), there exists an 𝑥 ∈ 𝑅 with 𝑥 ≡ −𝑎 𝑖′ 𝑏 𝑖 mod 𝑃𝑖 , hence 𝑎 𝑖 𝑥 + 𝑏 𝑖 ≡ 0 mod 𝑃𝑖 for 𝑖 = 1, . . . , 𝑟, □ as desired.

2.2 Discrete Valuations A valuation 𝑣 of a field 𝐹 is discrete if 𝑣(𝐹 × )  Z. In this case we normalize 𝑣 by replacing it with an equivalent valuation such that 𝑣(𝐹 × ) = Z. Each element 𝜋 ∈ 𝐹 with 𝑣(𝜋) = 1 is a prime element of 𝑂 𝑣 . In particular, discrete valuations are real, so they satisfy the weak approximation theorem 2.1.1.

24

2 Valuations

Prime elements of a unique factorization domain 𝑅 produce discrete valuations of 𝐹 = Quot(𝑅). If 𝑝 is a prime element of 𝑅, then every element 𝑥 of 𝐹 × has a unique representation as 𝑥 = 𝑢 𝑝 𝑚 , where 𝑢 is relatively prime to 𝑝 and 𝑚 ∈ Z. Define ord 𝑝 (𝑥) to be 𝑚. Then, ord 𝑝 is a discrete valuation of 𝐹. Suppose that 𝑝 ′ is another prime element of 𝑅. Then, ord 𝑝′ is equivalent to ord 𝑝 if and only if 𝑝 ′ 𝑅 = 𝑝𝑅, that is if 𝑝 ′ = 𝑢 𝑝 with 𝑢 ∈ 𝑅 × . Example 2.2.1 Basic examples of discrete valuations. (a) The ring of integers Z is a unique factorization domain. For each prime number 𝑝 the residue field of Q at ord 𝑝 is F 𝑝 . When 𝑝 ranges over all prime numbers, ord 𝑝 ranges over all valuations of Q (Exercise 3). (b) Let 𝑅 = 𝐾 [t] be the ring of polynomials in 𝑛 indeterminates t = (𝑡1 , . . . , 𝑡 𝑛 ) over a field 𝐾. Then, 𝑅 is a unique factorization domain [Lan97, p. 183, Cor. 2.4 with 𝐴 = 𝐾]. The prime elements of 𝑅 are the irreducible polynomials 𝑝 over 𝐾. Units of 𝑅 are the elements 𝑢 of 𝐾 × , so ord 𝑝 (𝑢) = 0 and we say ord 𝑝 is trivial on 𝐾. Note that if 𝑝 ′ = 𝑢 𝑝 with 𝑢 ∈ 𝐾 × , then ord 𝑝′ = ord 𝑝 . (c) The case where 𝑛 = 1 and 𝑡 = 𝑡1 deserves special attention. In this case, in addition to the valuations ord 𝑝 , with irreducible elements 𝑝 of 𝐾 [𝑡], we have a discrete valuation that we denote by ord∞ . It is defined for a quotient 𝑔𝑓 of elements of 𝐾 [𝑡] by the formula ord∞ ( 𝑔𝑓 ) = deg(𝑔) − deg( 𝑓 ). In particular ord∞ is trivial on 𝐾. In this case, the set of ord 𝑝 ’s and ord∞ give all valuations of 𝐾 (𝑡) that are trivial on 𝐾. Thus, all valuations of 𝐾 (𝑡) which are trivial on 𝐾 are discrete (Exercise 4). The residue field of 𝐾 (𝑡) at ord 𝑝 is isomorphic to the field 𝐾 (𝑎), where 𝑎 is a root of 𝑝. An arbitrary irreducible polynomial 𝑝 may have several roots 𝑎 in the algebraic closure 𝐾˜ of 𝐾. Each of them defines a place 𝜑 𝑎 : 𝐾 (𝑡) → 𝐾˜ ∪ {∞} by 𝜑 𝑎 (𝑡) = 𝑎 and 𝜑 𝑎 (𝑐) = 𝑐 for each 𝑐 ∈ 𝐾. These places are equivalent. If 𝑝(𝑡) = 𝑡 − 𝑎, then 𝜑 𝑎 is the unique place of 𝐾 (𝑡) corresponding to ord 𝑝 . Similarly, there is a unique place 𝜑∞ corresponding to ord∞ . It is defined by 𝜑∞ (𝑡) = ∞. We may view each 𝑓 (𝑡) ∈ 𝐾 (𝑡) as a function from 𝐾 ∪ {∞} into itself: 𝑓 (𝑎) = (𝑡) 𝜑 𝑎 ( 𝑓 (𝑡)). Explicitly, write 𝑓 (𝑡) = 𝑔ℎ(𝑡) with 𝑔, ℎ ∈ 𝐾 [𝑋] and gcd(𝑔, ℎ) = 1. Let 𝑎 ∈ 𝐾. Then, 𝑓 (𝑎) = let 𝑢 =

𝑡 −1

𝑔 (𝑎) ℎ (𝑎)

if ℎ(𝑎) ≠ 0 and 𝑓 (𝑎) = ∞ if ℎ(𝑎) = 0. To compute 𝑓 (∞)

and write 𝑓 (𝑡) =

𝑔1 (𝑢) ℎ1 (𝑢)

with 𝑔1 , ℎ1 ∈ 𝐾 [𝑋] and gcd(𝑔1 , ℎ1 ) = 1. Then,

𝑔1 (0) ℎ1 (0)

𝑓 (∞) = if ℎ1 (0) ≠ 0 and 𝑓 (∞) = ∞ if ℎ1 (0) = 0. Suppose for example 𝑓 (𝑡) ∈ 𝐾 [𝑡] and 𝑓 ≠ 0. Then, 𝑓 maps 𝐾 into itself and 𝑎 𝑓 (∞) = ∞. Now suppose 𝑓 (𝑡) = 𝑎𝑡+𝑏 𝑐𝑡+𝑑 with 𝑎𝑑 − 𝑏𝑐 ≠ 0 and 𝑐 ≠ 0, then 𝑓 (∞) = 𝑐 . When 𝐾 is algebraically closed, each irreducible polynomial is linear. Hence, each valuation of 𝐾 (𝑡) which is trivial over 𝐾 is either ord𝑡−𝑎 for some 𝑎 ∈ 𝐾 or 𝑣 ∞ .

More examples of discrete valuations arise through extensions of the basic examples (Section 2.3).

2.2 Discrete Valuations

25

Proposition 2.2.2 Every discrete valuation ring 𝑅 is a principal ideal domain. Proof. Let 𝑣 be the valuation of 𝐾 = Quot(𝑅) with 𝑂 𝑣 = 𝑅 and 𝑣(𝐾 × ) = Z. Choose a prime element 𝑝 of 𝑅. Now consider a nonzero ideal 𝔞 of 𝑅. Then, the minimal integer 𝑚 with 𝑝 𝑚 ∈ 𝔞 is positive. It satisfies, 𝔞 = 𝑝 𝑚 𝑅. □ As a consequence of Proposition 2.2.2, finitely generated modules over 𝑅 have a simple structure. Proposition 2.2.3 Let 𝑅 be a discrete valuation ring, 𝑝 a prime element of 𝑅, 𝐾 = Quot(𝑅), and 𝑀 a finitely generated 𝑅-module. Put 𝐾¯ = 𝑅/𝑝𝑅. Let 𝑟 = dim𝐾 𝑀 ⊗𝑅 𝐾, 𝑛 = dim𝐾¯ 𝑀/𝑝𝑀, and 𝑚 = 𝑛 − 𝑟. Then, 𝑚 ≥ 0 and there is a unique 𝑚-tuple of positive integers (𝑘 1 , 𝑘 2 , . . . , 𝑘 𝑚 ) with 𝑘 1 ≤ 𝑘 2 ≤ · · · ≤ 𝑘 𝑚 and 𝑀  𝑅/𝑝 𝑘𝑚 𝑅 ⊕· · ·⊕ 𝑅/𝑝 𝑘1 𝑅 ⊕ 𝑅𝑟 . Moreover, 𝑟 is the maximal number of elements of 𝑀 which are linearly independent over 𝑅 and 𝑛 is the minimal number of generators of 𝑀. Proof. By Proposition 2.2.2, 𝑅 is a principal ideal domain, so 𝑀 = 𝑀tor ⊕ 𝑁, where 𝑀tor = {𝑥 ∈ 𝑀 | 𝑟𝑥 = 0 for some 𝑟 ∈ 𝑅, 𝑟 ≠ 0} is the 𝑅-torsion part of 𝑀 and 𝑁 is a free 𝑅-module [Lan97, p. 147, Thm. 7.3]. Both 𝑀tor and 𝑁 are finitely generated [Lan97, p. 147, Cor. 7.2]. In particular, 𝑁  𝑅 𝑠 for some integer 𝑠 ≥ 0. Suppose that 𝑥 ∈ 𝑀tor and 𝑎𝑥 = 0 with 𝑎 ∈ 𝑅, 𝑎 ≠ 0. Then, 𝑥 ⊗ 1 = 𝑎𝑥 ⊗ 𝑎1 = 0. Hence, 𝑀tor ⊗𝑅 𝐾 = 0 and 𝑀 ⊗𝑅 𝐾  𝐾 𝑠 . Therefore, 𝑠 = 𝑟. By [Lan97, p. 151, Thm. 7.7], 𝑀tor  𝑅/𝑞 𝑚′ 𝑅 ⊕ · · · ⊕ 𝑅/𝑞 1 𝑅 where 𝑞 1 , . . . , 𝑞 𝑚′ are elements of 𝑅 which are neither zero nor units and 𝑞 𝑖 |𝑞 𝑖+1 , 𝑖 = 1, . . . , 𝑚 ′ − 1. Multiplying each 𝑞 𝑖 by a unit, we may assume that 𝑞 𝑖 = 𝑝 𝑘𝑖 with 𝑘 𝑖 an integer and 1 ≤ 𝑘 1 ≤ 𝑘 2 ≤ · · · ≤ 𝑘 𝑚′ . Moreover, the above cited theorem assures that 𝑅𝑞 1 , . . . , 𝑅𝑞 𝑚′ are uniquely determined by the above conditions. Hence, 𝑘 1 , . . . , 𝑘 𝑚′ are also uniquely determined. Combining the first two paragraphs gives: 𝑀  𝑅/𝑝 𝑘𝑚′ 𝑅 ⊕ · · · ⊕ 𝑅/𝑝 𝑘1 𝑅 ⊕ 𝑅𝑟 . ′ ′ Hence, 𝑀/𝑝𝑀 = (𝑅/𝑝𝑅) 𝑚 +𝑟  𝐾¯ 𝑚 +𝑟 , so 𝑛 = 𝑚 ′ + 𝑟 and 𝑚 ′ = 𝑚. Now recall that elements 𝑣 1 , . . . , 𝑣 𝑠 of 𝑀 are linearly independent over 𝑅 if Í𝑠 𝑖=1 𝑎 𝑖 𝑣 𝑖 = 0 with 𝑎 1 , . . . , 𝑎 𝑠 ∈ 𝑅 implies that 𝑎 1 = · · · = 𝑎 𝑠 = 0. Alternatively, 𝑣 1 ⊗ 1, . . . , 𝑣 𝑠 ⊗ 1 are linearly independent over 𝐾. Thus, 𝑟 is the maximal number of 𝑅-linearly independent elements of 𝑀. Finally, by Nakayama’s lemma [Lan97, p. 425, Lemma 4.3], 𝑛 is the minimal number of generators of 𝑀. □

Definition 2.2.4 Let 𝑅 be an integral domain with quotient field 𝐹. An overring of 𝑅 is a ring 𝑅 ⊆ 𝑅 ′ ⊂ 𝐹. It is said to be proper if 𝑅 ≠ 𝑅 ′. Lemma 2.2.5 A discrete valuation ring 𝑂 has no proper overrings. Proof. Let 𝑅 be an overring of 𝑂. Assume there exists an 𝑥 ∈ 𝑅 ∖ 𝑂. Then, 𝑥 −1 is a nonunit of 𝑂. Choose a prime element 𝑝 for 𝑂. Then, 𝑥 = 𝑢 𝑝 −𝑚 for some 𝑢 ∈ 𝑂 × and a positive integer 𝑚. Hence, 𝑝 −1 = 𝑢 −1 𝑝 𝑚−1 𝑥 ∈ 𝑅. Therefore, 𝑢 ′ 𝑝 𝑘 ∈ 𝑅 for all 𝑢 ′ ∈ 𝑂 × and 𝑘 ∈ Z. We conclude that 𝑅 = Quot(𝑂). □

26

2 Valuations

Composita of places attached to discrete valuations of rational function fields of one variable give rise to useful places of rational function fields of several variables. Remark 2.2.6 (Composition of places) Suppose that 𝜓 is a place of a field 𝐾 with residue field 𝐿 and 𝜑 is a place of 𝐿 with residue field 𝑀. Then, 𝜓 −1 (𝑂 𝜑 ) is a valuation ring of 𝐾 with maximal ideal 𝜓 −1 (𝔪 𝜑 ) and residue field 𝜓 −1 (𝑂 𝜑 )/𝜓 −1 (𝔪 𝜑 )  𝑂 𝜑 /𝔪 𝜑  𝑀. Define a map 𝜑 ◦ 𝜓: 𝐾 → 𝑀 ∪ {∞} as follows: 𝜑 ◦ 𝜓(𝑥) = 𝜑(𝜓(𝑥)) if 𝜓(𝑥) ≠ ∞ and 𝜑 ◦ 𝜓(𝑥) = ∞ if 𝜓(𝑥) = ∞. Then, 𝜑 ◦ 𝜓 is a homomorphism on 𝜓 −1 (𝑂 𝜑 ) and {𝑥 ∈ 𝐾 | 𝜑 ◦ 𝜓(𝑥) = ∞} = 𝐾 ∖ 𝜓 −1 (𝑂 𝜑 ). Therefore, 𝜑 ◦ 𝜓 is a place of 𝐾, called the compositum of 𝜓 and 𝜑, 𝑂 𝜑◦𝜓 = 𝜓 −1 (𝑂 𝜑 ), and 𝔪 𝜑◦𝜓 = 𝜓 −1 (𝔪 𝜑 ). / 𝐿 ∪ {∞}

𝜑

/ 𝑀 ∪ {∞}

𝜓

/𝐿

𝜑

/ 𝑀 ∪ {∞}

𝜓

/ 𝑂𝜑

𝜑

/𝑀

𝜓

/ 𝔪𝜑

𝜑

/0

𝜓

/0

𝜑

/0

𝜓

𝐾 𝑂𝜓 𝑂 𝜑◦𝜓 𝔪 𝜑◦𝜓 𝔪𝜓

In addition, 𝐿 = 𝐾¯ 𝜓 and 𝑀 = 𝐿¯ 𝜑 = 𝐾¯ 𝜑◦𝜓 .

˜ 𝑡1 , . . . , 𝑡𝑟 indeterminates, Lemma 2.2.7 Let 𝐾 be a field, 𝑎 1 , . . . , 𝑎𝑟 elements of 𝐾, and 𝐿 a finite extension of 𝐾. Then, there exists a 𝐾-place 𝜑: 𝐾 (t) → 𝐾 (a) ∪ {∞} such that 𝜑(𝑡 𝑖 ) = 𝑎 𝑖 , 𝑖 = 1, . . . , 𝑟. Moreover, every extension of 𝜑 to an 𝐿-place of 𝐿 (t) maps 𝐿(t) onto 𝐿 (a) ∪ {∞}. Proof. For each 𝑖 there is a 𝐾 (𝑎 1 , . . . , 𝑎 𝑖−1 , 𝑡𝑖+1 , . . . , 𝑡𝑟 )-place 𝜑𝑖 : 𝐾 (𝑎 1 , . . . , 𝑎 𝑖−1 , 𝑡𝑖 , 𝑡𝑖+1 . . . , 𝑡𝑟 ) → 𝐾 (𝑎 1 , . . . , 𝑎 𝑖−1 , 𝑎 𝑖 , 𝑡𝑖+1 , . . . , 𝑡𝑟 ) with 𝜑𝑖 (𝑡 𝑖 ) = 𝑎 𝑖 (Example 2.2.1). The compositum 𝜑 = 𝜑𝑟 ◦ · · · ◦ 𝜑1 is a 𝐾-place of 𝐾 (𝑡 1 , . . . , 𝑡𝑟 ) with residue field 𝐾 (𝑎 1 , . . . , 𝑎𝑟 ) and 𝜑(𝑡 𝑖 ) = 𝑎 𝑖 , 𝑖 = 1, . . . , 𝑟. Let now 𝜑 be an extension of 𝜑 to an 𝐿-place of 𝐿(t). Choose a basis 𝑏 1 , . . . , 𝑏 𝑛 for 𝐿/𝐾. Then, 𝑏 1 , . . . , 𝑏 𝑛 is also a basis for 𝐿 (t)/𝐾 (t). Hence, each 𝑓 ∈ 𝐿 (t) has Í𝑛 a presentation 𝑓 = 𝑖=1 𝑏 𝑖 𝑓𝑖 with 𝑓𝑖 ∈ 𝐾 (t). Assume without loss that 𝑓𝑓1𝑖 is finite Í𝑛 𝑓𝑖 under 𝜑 for 𝑖 = 1, . . . , 𝑛. Then, 𝑓 = 𝑓1 𝑖=1 𝑓1 𝑏 𝑖 and 𝜑( 𝑓 ) ∈ 𝐿(𝑎 1 , . . . , 𝑎 𝑟 ) ∪ {∞}. Thus, 𝜑(𝐿 (t)) = 𝐿 (a) ∪ {∞}. □

2.3 Extensions of Valuations and Places

27

2.3 Extensions of Valuations and Places The examples of Section 2.2 and the following extension results give a handle on describing valuations of function fields in one variable. Proposition 2.3.1 (Chevalley, [Lan64], p. 8, Thm. 1) Let 𝜑0 be a homomorphism of an integral domain 𝑅 into an algebraically closed field 𝑀 and let 𝐹 be a field containing 𝑅. Then, 𝜑0 extends either to an embedding 𝜑 of 𝐹 into 𝑀 or to a place 𝜑 of 𝐹 into 𝑀 ∪ {∞}. When 𝐹 is algebraic over 𝑅, the proposition has a more precise form: Let 𝑓 ∈ 𝑅[𝑋] be an irreducible polynomial over 𝐸 = Quot(𝑅) and 𝑓¯ ∈ 𝑀 [𝑋] the result of applying 𝜑0 to the coefficients of 𝑓 . Suppose that 𝑓¯ is not identically zero. Assume 𝑥 and 𝑥¯ are roots of 𝑓 and 𝑓¯ in 𝐸˜ and 𝑀, respectively. Then, 𝜑0 extends to a place 𝜑 of 𝐸 (𝑥) into 𝑀 ∪ {∞} with 𝜑(𝑥) = 𝑥¯ [Lan64, p. 10, Thm. 2]. Moreover, if 𝜑0 is injective, so is 𝜑 [Lan64, p. 8, Prop. 2]. In particular, suppose 𝑣 is a valuation of a field 𝐸 and 𝐹 is an extension of 𝐸. Then, 𝑣 extends to a valuation 𝑤 0 of 𝐹. Each valuation 𝑤 of 𝐹 which is equivalent to 𝑤 0 lies over 𝑣. Thus, 𝑤 lies over 𝑣 if and only if 𝑂 𝑣 ⊆ 𝑂 𝑤 and 𝔪𝑣 = 𝔪𝑤 ∩ 𝑂 𝑣 . We call 𝑒 𝑤/𝑣 = (𝑤(𝐹 × ) : 𝑤(𝐸 × )) the ramification index of 𝑤 over 𝑣 (and also over 𝐸). The field degree [𝐹 : 𝐸] bounds 𝑒 𝑤/𝑣 (Exercise 5). Similarly, 𝐸¯ 𝑣 embeds in 𝐹¯𝑤 to give the inequality 𝑓𝑤/𝑣 = [ 𝐹¯𝑤 : 𝐸¯ 𝑣 ] ≤ [𝐹 : 𝐸] (Exercise 7). Both the ramification index and the residue field degree are multiplicative. Thus, if (𝐹 ′, 𝑤 ′) is an extension of (𝐹, 𝑤), then 𝑒 𝑤′ /𝑣 = 𝑒 𝑤′ /𝑤 𝑒 𝑤/𝑣 and 𝑓𝑤′ /𝑣 = 𝑓𝑤′ /𝑤 𝑓𝑤/𝑣 . If [𝐹 : 𝐸] < ∞, then the number of valuations of 𝐹 that lie over 𝑣 is finite (a consequence of Proposition 2.3.2). Proposition 2.3.2 Let 𝐹/𝐸 be a finite extension of fields and 𝑣 a valuation of 𝐸. Let 𝑤 1 , . . . , 𝑤 𝑔 be all inequivalent extensions of 𝑣 to 𝐹. Then, 𝑔 ∑︁

𝑒 𝑤𝑖 /𝑣 𝑓𝑤𝑖 /𝑣 ≤ [𝐹 : 𝐸]

(2.9)

𝑖=1

(see [Bou89a, p. 420, Thm. 1] or [EnP05, p. 75, Thm. 3.3.4]). If, in addition, 𝑣 is discrete and 𝐹/𝐸 is separable, then each 𝑤 𝑖 is discrete and (see [Bou89a, p. 425, Cor. 1] or [EnP05, p. 76, Thm. 3.3.5]) 𝑔 ∑︁

𝑒 𝑤𝑖 /𝑣 𝑓𝑤𝑖 /𝑣 = [𝐹 : 𝐸].

(2.10)

𝑖=1

Suppose that (𝐹, 𝑤)/(𝐸, 𝑣) is an extension of discrete valued fields. In particular, 𝑤(𝑎) = 𝑣(𝑎) for each 𝑎 ∈ 𝐸. By definition, 𝑒 𝑤/𝑣 = (𝑤(𝐹 × ) : 𝑣(𝐸 × )). However, as in Section 2.2, it is customary to replace 𝑣 and 𝑤 by equivalent valuations with 𝑣(𝐸 × ) = 𝑤(𝐹 × ) = Z. The new valuations satisfy 𝑤(𝑎) = 𝑒 𝑤/𝑣 𝑣(𝑎)

for each

𝑎 ∈ 𝐸.

28

2 Valuations

Whenever we speak about an extension of discrete valuations, we mean they are normalized and satisfy the latter relation. Suppose that 𝐹 is a finite Galois extension of 𝐸 with a Galois group 𝐺. Let 𝑤 be a discrete valuation of 𝐹 and let 𝜎 ∈ 𝐺. Then, 𝜎(𝑤) is a valuation of 𝐹 (Section 2.1), both 𝑤 and 𝜎(𝑤) lie over the same valuation 𝑣 of 𝐸, and 𝑒 𝑤/𝑣 = 𝑒 𝜎 (𝑤)/𝑣

and

𝑓𝑤/𝑣 = 𝑓 𝜎 (𝑤)/𝑣 .

𝑤′

Conversely, suppose 𝑤 and are two discrete valuations of 𝐹 over the same valuation 𝑣 of 𝐸. Then, there exists a 𝜎 ∈ 𝐺 such that 𝜎(𝑤) = 𝑤 ′ (Exercise 8). Thus, if 𝑤 1 , . . . , 𝑤 𝑔 are all distinct valuations of 𝐹 that lie over 𝑣, then they all have the same residue degree 𝑓 and ramification index 𝑒 over 𝑣. In this case formula (2.10) simplifies to 𝑒 𝑓 𝑔 = [𝐹 : 𝐸]. (2.11) The subgroups 𝐷 𝑤 = 𝐷 𝑤/𝑣 = {𝜎 ∈ 𝐺 | 𝜎𝑂 𝑤 = 𝑂 𝑤 }, 𝐼 𝑤 = 𝐼 𝑤/𝑣 = {𝜎 ∈ 𝐺 | 𝑤(𝑥 − 𝜎𝑥) > 0 for all 𝑥 ∈ 𝑂 𝑤 } are the decomposition group and the inertia group, respectively, of 𝑤 over 𝐸. Obviously 𝐼 𝑤 ⊳ 𝐷 𝑤 . If 𝐹¯𝑤 /𝐸¯ 𝑣 is separable, then, by [Ser68b, p. 33, Cor.], |𝐼 𝑤 | = 𝑒 𝑤/𝑣

and

|𝐷 𝑤 | = 𝑒 𝑤/𝑣 𝑓𝑤/𝑣 .

(2.12)

If 𝜑 𝑤 is a place that corresponds to 𝑤, then 𝐷 𝑤 = {𝜎 ∈ 𝐺 | ∀𝑥 ∈ 𝐹 : 𝜑 𝑤 (𝑥) ≠ ∞ ⇐⇒ 𝜑 𝑤 (𝜎𝑥) ≠ ∞}, 𝐼 𝑤 = {𝜎 ∈ 𝐺 | 𝜑 𝑤 (𝑥 − 𝜎𝑥) = 0 for all 𝑥 ∈ 𝐹 with 𝜑 𝑤 (𝑥) ≠ ∞}.

(2.13) (2.14)

Section 3.3 generalizes the notion of separable algebraic extension of fields to arbitrary extensions of fields. In particular, purely transcendental extensions of fields are separable. We use this notion in the following definition. Suppose that (𝐹, 𝑤)/(𝐸, 𝑣) is an arbitrary extension of valued fields. We say that 𝑤 is unramified (resp. tamely ramified) over 𝑣 (or also over 𝐸) if 𝐹¯𝑤 /𝐸¯ 𝑣 is a separable extension and 𝑒 𝑤/𝑣 = 1 (resp. char( 𝐸¯ 𝑣 ) ∤ 𝑒 𝑤/𝑣 ). We say that 𝑣 is unramified (resp. tamely ramified) in 𝐹 if each extension of 𝑣 to 𝐹 is unramified (resp. tamely ramified) over 𝑣. Example 2.3.3 Purely transcendental extensions. Let (𝐸, 𝑣) be a valued field. Consider a transcendental element 𝑡 over 𝐸. Extend 𝑣 to a valuation 𝑣 ′ of 𝐸 (𝑡) as follows. First define 𝑣 ′ on 𝐸 [𝑡] by the following rule: 𝑣′

𝑚 ∑︁



𝑎 𝑖 𝑡 𝑖 = min 𝑣(𝑎 0 ), . . . , 𝑣(𝑎 𝑚 )

(2.15)

𝑖=0

for 𝑎 0 , . . . , 𝑎 𝑚 ∈ 𝐸. The same argument used to prove Gauss’ lemma proves that 𝑣 ′ ( 𝑓 𝑔) = 𝑣 ′ ( 𝑓 ) + 𝑣 ′ (𝑔) for all 𝑓 , 𝑔 ∈ 𝐸 [𝑡]. Í𝑚 Í 𝑎 𝑖 𝑡 𝑖 and 𝑔(𝑡) = 𝑛𝑗=0 𝑏 𝑗 𝑡 𝑗 . Let 𝑟 be the minimal integer Indeed, let 𝑓 (𝑡) = 𝑖=0
with 𝑣(𝑎𝑟 ) = min 𝑣(𝑎
0 ), . . . , 𝑣(𝑎 𝑚 ) and let 𝑠 be the minimal integer with 𝑣(𝑏 𝑠 ) = min 𝑣(𝑏 0 ), . . . , 𝑣(𝑏 𝑛 ) . If 𝑖 + 𝑗 = 𝑟 + 𝑠 and (𝑖, 𝑗) ≠ (𝑟, 𝑠), then either 𝑖 < 𝑟 or 𝑗 < 𝑠. In both cases 𝑣(𝑎𝑟 ) + 𝑣(𝑏 𝑠 ) < 𝑣(𝑎 𝑖 ) + 𝑣(𝑏 𝑗 ). Hence

2.3 Extensions of Valuations and Places

𝑣′

𝑚 ∑︁

29

𝑛 ∑︁

𝑎𝑖 𝑡 𝑖 + 𝑣 ′ 𝑏 𝑗 𝑡 𝑗 = 𝑣(𝑎𝑟 ) + 𝑣(𝑏 𝑠 ) 𝑗=0

𝑖=0

∑︁

= min

𝑣(𝑎 𝑖 𝑏 𝑗 ) | 𝑘 = 0, . . . , 𝑚 + 𝑛




𝑖+ 𝑗=𝑘

= 𝑣′

𝑚 ∑︁ 𝑖=0

𝑎𝑖 𝑡 𝑖 ·

𝑛 ∑︁


𝑏 𝑗𝑡 𝑗 ,

𝑗=0

as claimed. We extend 𝑣 ′ to 𝐸 (𝑡) by the rule 𝑣 ′ ( 𝑔𝑓 ) = 𝑣 ′ ( 𝑓 ) − 𝑣 ′ (𝑔). Then, we prove
𝑣 ′ (𝑢 1 + 𝑢 2 ) ≥ min 𝑣 ′ (𝑢 1 ), 𝑣 ′ (𝑢 2 ) first for 𝑢 1 , 𝑢 2 ∈ 𝐸 [𝑡] and then for 𝑢 1 , 𝑢 2 ∈ 𝐸 (𝑡). ′ Thus, 𝑣 is a valuation of 𝐸 (𝑡). ÍNote that the residue 𝑡¯ of 𝑡 at 𝑣 ′ is transcen𝑛 ¯𝑖 dental over 𝐸¯ 𝑣 . Indeed,
suppose Í𝑛 𝑖=0𝑖
𝑎¯ 𝑖 𝑡 = 0 for some 𝑎 0 , . . . , 𝑎 𝑛 ∈ 𝑂 𝑣 . Then, ′ min 𝑣(𝑎 0 ), . . . , 𝑣(𝑎 𝑛 ) = 𝑣 𝑖=0 𝑎 𝑖 𝑡 > 0. Hence, 𝑎¯ 𝑖 = 0, 𝑖 = 0, . . . , 𝑛. ¯ It follows that 𝐸 (𝑡) 𝑣′ = 𝐸 𝑣 ( 𝑡¯) is a rational function field over 𝐸¯ 𝑣 . By definition, Γ𝑣′ = Γ𝑣 . In particular, if 𝑣 is discrete, then so is 𝑣 ′ and 𝑒 𝑣′ /𝑣 = 1. Suppose that 𝑣 ′′ is another extension of 𝑣 to 𝐸 (𝑡) with the residue of 𝑡 at 𝑣 ′′ transcendental over 𝐸¯ 𝑣 . We show that 𝑣 ′′ = 𝑣 ′. Indeed, for 𝑎 0 , . . . , 𝑎 𝑛 ∈
𝐸, not all zero, choose 𝑗 between 0 and 𝑛 with 𝑣(𝑎 𝑗 ) = min 𝑣(𝑎 0 ), . . . , 𝑣(𝑎 𝑛 ) . Then, Í𝑛 ¯𝑖 𝑖=0 𝑎 𝑖 /𝑎 𝑗 𝑡 ≠ 0. Therefore, 𝑣 ′′

𝑛 ∑︁

𝑛 ∑︁

𝑎 𝑖 𝑡 𝑖 = 𝑣(𝑎 𝑗 ) + 𝑣 ′′ (𝑎 𝑖 /𝑎 𝑗 )𝑡 𝑖

𝑖=0

𝑖=0 𝑛 ∑︁

= min 𝑣(𝑎 0 ), . . . , 𝑣(𝑎 𝑛 ) = 𝑣 ′ 𝑎𝑖 𝑡 𝑖 , 𝑖=0

as claimed. Lemma 2.3.4 Let 𝑣 be a discrete valuation of a field 𝐸, ℎ ∈ 𝑂 𝑣 [𝑋] a monic ˜ and 𝐹 = 𝐸 (𝑥). Suppose irreducible polynomial of degree 𝑛, 𝑥 a root of ℎ(𝑋) in 𝐸, ¯ that the residue polynomial ℎ(𝑋) is separable. Then, 𝑣 is unramified in 𝐹. Î ¯ Proof. By assumption, ℎ(𝑋) = 𝑟𝑖=1 ℎ𝑖 (𝑋), where ℎ𝑖 ∈ 𝐸¯ 𝑣 [𝑋] are distinct monic irreducible polynomials. For each 𝑖 between 1 and 𝑟 choose a root 𝑎 𝑖 of ℎ𝑖 (𝑋) in ( 𝐸¯ 𝑣 )sep . Use Proposition 2.3.1 to extend the residue map 𝑂 𝑣 → 𝐸¯ 𝑣 to a place 𝜑𝑖 of 𝐹 with 𝜑𝑖 (𝑥) = 𝑎 𝑖 . Denote the corresponding valuation by 𝑤 𝑖 . Then, 𝐸¯ 𝑣 (𝑎 𝑖 ) ⊆ 𝐹¯𝑤𝑖 , so [ 𝐸¯ 𝑣 (𝑎 𝑖 ) : 𝐸¯ 𝑣 ] ≤ 𝑓𝑤𝑖 /𝑣 ≤ 𝑒 𝑤𝑖 /𝑣 𝑓𝑤𝑖 /𝑣 . Since ℎ𝑖 (𝑋) and ℎ 𝑗 (𝑋) have no common root for 𝑖 ≠ 𝑗 for 𝑖 = 1, . . . , 𝑟, the valuations 𝑤 1 , . . . , 𝑤 𝑟 are mutually inequivalent extensions of 𝑣. Label any further extensions of 𝑣 to valuations of 𝐹 as 𝑤 𝑟+1 , . . . , 𝑤 𝑔 . By 2.9, 𝑔 𝑟 𝑟 ∑︁ ∑︁ ∑︁ ¯ ¯ 𝑛= deg(ℎ𝑖 ) = [ 𝐸 𝑣 (𝑎 𝑖 ) : 𝐸 𝑣 ] ≤ 𝑒 𝑤𝑖 /𝑣 𝑓𝑤𝑖 /𝑣 ≤ 𝑛. 𝑖=1

𝑖=1

𝑖=1

Hence, 𝑒 𝑤𝑖 /𝑣 = 1 and 𝐸¯ 𝑣 (𝑎 𝑖 ) = 𝐹¯𝑤𝑖 for 𝑖 = 1, . . . , 𝑟. Moreover, 𝑤 1 , . . . , 𝑤 𝑟 are all extensions of 𝑣 to 𝐹 and each of them is unramified over 𝐸. Therefore, 𝑣 is unramified in 𝐹. □

30

2 Valuations

The converse of Lemma 2.3.4 requires 𝐸¯ 𝑣 to be infinite. Lemma 2.3.5 Let 𝑣 be a discrete valuation of a field 𝐸. Let 𝐹 be a separable extension of 𝐸 of degree 𝑛. Suppose that 𝑣 is unramified in 𝐹 and 𝐸¯ 𝑣 is an infinite field. Then, 𝐹/𝐸 has a primitive element 𝑥 with irr(𝑥, 𝐸) ∈ 𝑂 𝑣 [𝑋] and the residue of irr(𝑥, 𝐸) at 𝑣 is a separable polynomial. Proof. Let 𝑤 1 , . . . , 𝑤 𝑔 be all extensions of 𝑣 to 𝐹. By 2.10, [𝐹 : 𝐸] = Í 𝑔 ¯ ¯ ¯ ¯ 𝑖=1 [ 𝐹𝑤𝑖 : 𝐸 𝑣 ]. Moreover, for each 𝑖 the extension 𝐹𝑤𝑖 / 𝐸 𝑣 is finite and separable. Hence, we may choose 𝑐 𝑖 in 𝐹 with 𝑤 𝑖 (𝑐 𝑖 ) = 0 and the residue 𝑐¯𝑖 of 𝑐 𝑖 at 𝑤 𝑖 is a primitive element of 𝐹¯𝑤𝑖 /𝐸¯ 𝑣 . Let ℎ𝑖 = irr( 𝑐¯𝑖 , 𝐸¯ 𝑣 ). Since 𝐸¯ 𝑣 is infinite, we may choose 𝑐 1 , . . . , 𝑐 𝑔 such that 𝑐¯1 , . . . , 𝑐¯𝑔 are mutually nonconjugate over 𝐸¯ 𝑣 . Thus, ℎ1 , . . . , ℎ𝑔 are relatively prime. Use Proposition 2.1.1 to find 𝑥 ∈ 𝐹 with 𝑤 𝑖 (𝑥 − 𝑐 𝑖 ) > 0, 𝑖 = 1, . . . , 𝑔. Then, 𝑤 𝑖 (𝑥) = 0, 𝑖 = 1, . . . , 𝑔. Extend each 𝑤 𝑖 to the Galois closure 𝐹ˆ of 𝐹/𝐸. Then, all ˆ 𝐸-conjugates of 𝑥 have nonnegative values under each extended valuation of 𝑣 to 𝐹. Hence, the elementary symmetric polynomials in the 𝐸-conjugates of 𝑥 belong to 𝑂 𝑣 . Therefore, 𝑓 (𝑋) = irr(𝑥, 𝐸) ∈ 𝑂 𝑣 [𝑋]. ¯ ¯ Let 𝑓¯ be the residue of 𝑓 at 𝑣. By construction, Î𝑔 𝑓 ( 𝑐¯𝑖 ) = 0, therefore ℎ𝑖 | 𝑓 , 𝑖 = 1, . . . , 𝑔. Since ℎ1 , . . . , ℎ𝑔 are relatively prime, 𝑖=1 ℎ𝑖 | 𝑓¯. Hence, [𝐹 : 𝐸] =

𝑔 ∑︁ 𝑖=1

[ 𝐹¯𝑤𝑖 : 𝐸¯ 𝑣 ] =

𝑔 ∑︁

deg(ℎ𝑖 )

𝑖=1

≤ deg( 𝑓¯) = deg( 𝑓 ) = [𝐸 (𝑥) : 𝐸] ≤ [𝐹 : 𝐸]. Consequently, 𝐸 (𝑥) = 𝐹, as desired.

□

Example 4.5.4 shows that the assumption on 𝐸¯ 𝑣 to be infinite is necessary for Lemma 2.3.5 to hold. The next lemma says that an arbitrary change of the base field preserves unramified discrete valuations. Lemma 2.3.6 Let (𝐸, 𝑣) be a discrete valued field. Consider a separable algebraic extension 𝐹 of 𝐸 and a discrete valued field (𝐸 1 , 𝑣 1 ) which extends (𝐸, 𝑣). Suppose that 𝑣 is unramified in 𝐹. Then, 𝑣 1 is unramified in 𝐹𝐸 1 . Proof. Suppose without loss that [𝐹 : 𝐸] < ∞. Let 𝐹1 = 𝐹𝐸 1 . Suppose first that 𝐸¯ 𝑣 is infinite. Choose 𝑥 as in Lemma 2.3.5 and let 𝑓 (𝑋) = irr(𝑥, 𝐸). Then, 𝐹 = 𝐸 (𝑥) and 𝑓¯(𝑋) is separable. Hence, 𝐹1 = 𝐸 1 (𝑥) and the residue 𝑓¯1 (𝑋) of 𝑓1 (𝑋) := irr(𝑥, 𝐸 1 ) at 𝑣 1 divides 𝑓¯(𝑋), so 𝑓¯1 (𝑋) is separable. By Lemma 2.3.4, 𝑣 1 is unramified in 𝐹1 . In the case where 𝐸¯ 𝑣 is finite we consider an extension 𝑤 1 of 𝑣 1 to a valuation of 𝐹1 . Denote the restriction of 𝑤 1 to 𝐹 by 𝑤. By assumption 𝑒 𝑤/𝑣 = 1.

(2.16)

Let 𝑡 be transcendental over 𝐹1 . Example 2.3.3 extends 𝑣 (resp. 𝑤, 𝑣 1 , 𝑤 1 ) in a canonical way to a discrete valuation 𝑣 ′ (resp. 𝑤 ′, 𝑣 1′ , 𝑤 1′ ) of 𝐸 (𝑡) (resp. 𝐹 (𝑡), 𝐸 1 (𝑡), 𝐹1 (𝑡)). Further, 𝑒 𝑣′ /𝑣 = 1 (resp. 𝑒 𝑤′ /𝑤 = 1, 𝑒 𝑣1′ /𝑣1 = 1, 𝑒 𝑤1′ /𝑤1 = 1)

(2.17)

2.3 Extensions of Valuations and Places

31

and 𝐸 (𝑡) 𝑣′ = 𝐸¯ 𝑣 ( 𝑡¯) (resp. 𝐹 (𝑡) 𝑤′ = 𝐹¯𝑤 ( 𝑡¯), 𝐸 1 (𝑡) 𝑣′ = 𝐸¯ 1,𝑣1 ( 𝑡¯), 𝐹1 (𝑡) 𝑤′ = 𝐹¯1,𝑤1 ( 𝑡¯)), 1 1 where 𝑡¯ is transcendental over 𝐹¯1,𝑤1 . Moreover, 𝑤 1′ extends 𝑤 ′ and 𝑣 1′ extends 𝑣 ′ giving this diagram:

(𝐹, 𝑤)

(𝐸, 𝑣)

(𝐹 (𝑡), 𝑤 ′) ♣ ♣ ♣♣♣

(𝐹1 (𝑡), 𝑤 1′ ) ♥ ♥ ♥♥♥ (𝐹1 , 𝑤 1 )

(𝐸 (𝑡), 𝑣 ′) ♣ ♣ ♣♣♣

(𝐸 1 (𝑡), 𝑣 1′ ) ♥ ♥ ♥♥♥

(𝐸 1 , 𝑣 1 )

We claim 𝑣 ′ is unramified in 𝐹 (𝑡). Indeed, 𝐹 (𝑡) 𝑤′ = 𝐹¯𝑤 · 𝐸¯ 𝑣 ( 𝑡¯) is a separable extension of 𝐸 (𝑡) 𝑣′ . Also, (2.17)

𝑒 𝑤′ /𝑣′ = 𝑒 𝑤′ /𝑣′ 𝑒 𝑣′ /𝑣 = 𝑒 𝑤′ /𝑣 = 𝑒 𝑤′ /𝑤 𝑒 𝑤/𝑣

(2.17),(2.16)

=

1.

Hence, 𝑤 ′ is unramified over 𝑣 ′. If 𝑢 ∗ is an arbitrary extension of 𝑣 ′ to 𝐹 (𝑡) and 𝑢 is its restriction to 𝐹, then the residue of 𝑡 at 𝑢 ∗ is 𝑡¯, which is transcendental over 𝐹¯𝑢 . Thus, by uniqueness of the construction in Example 2.3.3, 𝑢 ∗ = 𝑢 ′, where 𝑢 ′ is the canonical extension of 𝑢 to 𝐹 (𝑡). By the above, 𝑢 ∗ is unramified over 𝑣 ′. Since 𝐸 (𝑡) 𝑣′ is infinite, the first paragraph of the proof implies that 𝑣 1′ is unramified in 𝐹1 (𝑡). Thus, 𝐹¯1,𝑤1 ( 𝑡¯)/𝐸¯ 1,𝑣1 ( 𝑡¯) is a separable extension and 𝑒 𝑤1′ /𝑣1′ = 1. Therefore, 𝐹¯1,𝑤1 /𝐸¯ 1,𝑣1 is a separable extension and (2.17)

(2.17)

𝑒 𝑤1 /𝑣1 = 𝑒 𝑤1′ /𝑤1 𝑒 𝑤1 /𝑣1 = 𝑒 𝑤1′ /𝑣1 = 𝑒 𝑤1′ /𝑣1′ 𝑒 𝑣1′ /𝑣1 = 1. Consequently, 𝑣 1 is unramified in 𝐹1 .

□

Combine the multiplicativity of the ramification index and the residue field degree with Lemma 2.3.6 to prove: Corollary 2.3.7 Let (𝐸, 𝑣) ⊆ (𝐸 ′, 𝑣 ′) ⊆ (𝐸 ′′, 𝑣 ′′) be a tower of discrete valued fields. The following hold: (a) 𝑣 ′′/𝑣 is unramified if and only if 𝑣 ′′/𝑣 ′ and 𝑣 ′/𝑣 are unramified. (b) 𝑣 is unramified in 𝐸 ′′ if and only if 𝑣 is unramified in 𝐸 ′ and each extension of 𝑣 to 𝐸 ′ is unramified in 𝐸 ′′. (c) Let 𝐹1 and 𝐹2 be field extensions of 𝐸 which are contained in a common field. Suppose that 𝐹1 /𝐸 is separable algebraic and 𝑣 is unramified in 𝐹1 and in 𝐹2 . Then, 𝑣 is unramified in 𝐹1 𝐹2 . Example 2.3.8 (Radical extensions) Let (𝐸, 𝑣) be a discrete valued field and 𝑛 a positive integer with char( 𝐸¯ 𝑣 ) ∤ 𝑛. Consider an extension 𝐹 = 𝐸 (𝑥) of degree 𝑛 of 𝐸, where 𝑥 𝑛 = 𝑎 is in 𝐸. In particular, 𝐹/𝐸 is a separable extension. Let 𝑤 be an extension of 𝑣 to a valuation of 𝐹 and let 𝑒 = 𝑒 𝑤/𝑣 . Assume that both 𝑣 and 𝑤 are normalized. Then, 𝑛𝑤(𝑥) = 𝑒𝑣(𝑎) and 𝑒 ≤ 𝑛. (2.18)

32

2 Valuations

There are three cases to consider:
Case A: gcd 𝑛, 𝑣(𝑎) = 1. By (2.18), 𝑛|𝑒, so 𝑛 = 𝑒. By (2.10), 𝑤 is the unique extension of 𝑣 to 𝐹. Therefore, 𝑣 totally ramifies in 𝐹. Case B: 𝑛 ∤ 𝑣(𝑎). By (2.18), 𝑒 ≠ 1. Hence, 𝑤 ramifies over 𝐸. Case C: 𝑛|𝑣(𝑎). Choose 𝜋 ∈ 𝐸 with 𝑣(𝜋) = 1. Write 𝑎 = 𝑏𝜋 𝑘𝑛 with 𝑘 ∈ Z and 𝑏 ∈ 𝐸 such that 𝑣(𝑏) = 0. Then, 𝑦 = 𝑥𝜋 −𝑘 satisfies 𝑦 𝑛 = 𝑏 and 𝐹 = 𝐸 (𝑦). Moreover, 𝑌 𝑛 − 𝑏¯ is separable over 𝐸¯ 𝑣 , because char( 𝐸¯ 𝑣 ) ∤ 𝑛. Therefore, by Lemma 2.3.4, 𝑣 is unramified in 𝐹. Example 2.3.9 (Artin–Schreier Extensions) Let (𝐸, 𝑣) be a discrete-valued field of positive characteristic 𝑝. An Artin–Schreier extension 𝐹 of degree 𝑝 has the form 𝐸 (𝑥) where 𝑥 𝑝 − 𝑥 = 𝑎 with 𝑎 ∈ 𝐸. We consider two cases: Case A: 𝑣(𝑎) < 0 and 𝑝 ∤ 𝑣(𝑎). Let 𝑤 be an extension of 𝑣 to 𝐹. Then, 𝑤(𝑥) must be negative and 𝑤(𝑥 𝑝 ) < 𝑤(𝑥). Hence, 𝑝𝑤(𝑥) = 𝑒𝑣(𝑎), where 𝑒 = 𝑒 𝑤/𝑣 . Hence, 𝑝 = 𝑒 and 𝑤(𝑥) = 𝑣(𝑎). Thus, 𝑣 totally ramifies in 𝐹. Case B: 𝑣(𝑎) ≥ 0. Then, 𝑋 𝑝 − 𝑋 − 𝑎¯ is a separable polynomial. By Lemma 2.3.4, 𝑣 is unramified in 𝐹. Î 𝑝−1 In particular, if 𝑣(𝑎) > 0, then 𝑋 𝑝 −𝑋 = 𝑖=0 (𝑋 −𝑖) in 𝐸¯ 𝑣 . Hence, by Proposition 2.3.2, 𝑣 has exactly 𝑝 extensions to 𝐹. Label
them 𝑣 0 , . . . , 𝑣 𝑝−1 with 𝑣 𝑖 (𝑥 − 𝑖) > 0, 𝑖 = 0, . . . , 𝑝 −1. Since 𝑣 𝑖 (𝑥 −𝑖) < 𝑣 𝑖 (𝑥 −𝑖) 𝑝 , we conclude from (𝑥 −𝑖) 𝑝 − (𝑥 −𝑖) = 𝑎 that 𝑣 𝑖 (𝑥 − 𝑖) = 𝑣(𝑎). Lemma 2.3.10 (Eisenstein’s Criterion) Let 𝑅 be a unique factorization domain, 𝑝 a prime element of 𝑅, and 𝑓 (𝑋) = 𝑎 𝑛 𝑋 𝑛 + 𝑎 𝑛−1 𝑋 𝑛−1 + · · · + 𝑎 0 a polynomial with coefficients 𝑎 𝑖 ∈ 𝑅. Then, each of the following conditions suffices for 𝑓 to be irreducible over 𝐾 = Quot(𝑅): (a) 𝑝 ∤ 𝑎 𝑛 , 𝑝 divides 𝑎 0 , . . . , 𝑎 𝑛−1 , and 𝑝 2 ∤ 𝑎 0 . In this case we call 𝑓 (𝑋) an Eisenstein polynomial. (b) 𝑝 ∤ 𝑎 0 , 𝑝 divides 𝑎 1 , . . . , 𝑎 𝑛 , and 𝑝 2 ∤ 𝑎 𝑛 . Proof of (a). See [Lan97, p. 183, Thm. 3.1]. Proof of (b). By (a), the polynomial 𝑋 𝑛 𝑓 (𝑋 −1 ) = 𝑎 𝑛 + 𝑎 𝑛−1 𝑋 + · · · + 𝑎 0 𝑋 𝑛 is irreducible over 𝐾. Therefore, 𝑓 (𝑋) is irreducible. □ Example 2.3.11 (Ramification at infinity) Let (𝐸, 𝑣) be a normalized discrete valued field, 𝑡 an element of 𝐸 with 𝑣(𝑡) = −1, and 𝑓 (𝑋) = 𝑎 𝑛 𝑋 𝑛 + · · · + 𝑎 0 a polynomial in 𝐸 [𝑋] with 𝑣(𝑎 𝑖 ) = 0 for 𝑖 = 0, . . . , 𝑛. Choose 𝑥 ∈ 𝐸˜ with 𝑓 (𝑥) = 𝑡 and extend 𝑣 to a normalized discrete valuation 𝑤 of 𝐸 (𝑥). By Lemma 2.3.10(b), applied to the ring 𝑂 𝑣 , the polynomial 𝑡 −1 𝑓 (𝑋) − 1 is irreducible in 𝐸 [𝑋], hence so is 𝑓 (𝑋) − 𝑡, so [𝐸 (𝑥) : 𝐸] = 𝑛. The relation 𝑎 𝑛 𝑥 𝑛 + · · · + 𝑎 1 𝑥 + 𝑎 0 = 𝑡 implies that 𝑤(𝑥) < 0. Hence, 𝑛𝑤(𝑥) = 𝑤( 𝑓 (𝑥)) = 𝑤(𝑡) = 𝑒 𝑤/𝑣 𝑣(𝑡) = −𝑒 𝑤/𝑣 . (2.19) By (2.9), 𝑒 𝑤/𝑣 ≤ 𝑛. Hence, by (2.19), 𝑒 𝑤/𝑣 = 𝑛 = [𝐸 (𝑥) : 𝐸]. We conclude from (2.9) that 𝑣 totally ramifies in 𝐸 (𝑥).

2.4 Galois Extensions

33

In particular, if 𝐸 = 𝐾 (𝑡) is the field of rational functions in 𝑡 over a field 𝐾, 𝑣 is the 𝐾-valuation of 𝐸/𝐾 with 𝑣(𝑡) = −1, and 𝑓 (𝑥) = 𝑡, then 𝑣 totally ramifies in 𝐾 (𝑥).

2.4 Galois Extensions Let (𝐿, 𝑤)/(𝐾, 𝑣) be a finite Galois extension of discrete valued fields. By Section 2.3, {𝜎𝑤 | 𝜎 ∈ Gal(𝐿/𝐾)} is the set of all extensions of 𝑣 to 𝐿. Let 𝐿 0 be the decomposition field of 𝑤 over 𝐾. Denote the restriction of 𝑤 to 𝐿 0 by 𝑤 0 . Let 𝜎1 , . . . , 𝜎𝑚 , with 𝜎1 = 1, be representatives for the left cosets Ð 𝑚 of Gal(𝐿/𝐾)−1modulo Ð𝑚 Gal(𝐿/𝐿 0 ), that is Gal(𝐿/𝐾) = · 𝑖=1 𝜎𝑖 Gal(𝐿/𝐿 0 ) = · 𝑖=1 Gal(𝐿/𝐿 0 )𝜎𝑖 . Then, −1 | 𝜎1−1 | 𝐿0 , . . . , 𝜎𝑚 𝐿0 are the distinct 𝐾-embeddings of 𝐿 0 into 𝐾sep . We prove that (𝐿 0 , 𝑤 0 ) is an immediate extension of (𝐾, 𝑣). This means that the residue field and the value group of (𝐿 0 , 𝑤 0 ) coincide with those of (𝐾, 𝑣). Lemma 2.4.1 In the above notation, (a) 𝜎1 𝑤, . . . , 𝜎𝑚 𝑤 are the distinct extensions of 𝑣 to 𝐿, and (b) the valuation 𝑤 is the unique (up to equivalence) extension of 𝑤 0 to 𝐿 0 . Proof of (a). If 𝑖, 𝑗 ∈ {1, . . . , 𝑚} satisfy 𝜎𝑖 𝑤 = 𝜎 𝑗 𝑤, then 𝜎 −1 𝑗 𝜎𝑖 𝑂 𝑤 = 𝑂 𝑤 , so −1 𝜎 𝑗 𝜎𝑖 ∈ 𝐷 𝑤 = Gal(𝐿/𝐿 0 ). Hence, 𝜎𝑖 Gal(𝐿/𝐿 0 ) = 𝜎 𝑗 Gal(𝐿/𝐿 0 ) and therefore 𝑖 = 𝑗. Proof of (b). If 𝑤 ′ is an extension of 𝑤 0 to 𝐿, then there exists a 𝜎 ∈ Gal(𝐿/𝐿 0 ) such that 𝜎𝑂 𝑤 = 𝑂 𝑤′ . But since Gal(𝐿/𝐿 0 ) is the decomposition group of 𝑤 over 𝐾, we have 𝜎𝑂 𝑤 = 𝑂 𝑤 . Hence, 𝑂 𝑤′ = 𝑂 𝑤 , so 𝑤 and 𝑤 ′ are equivalent, as claimed. □ Proposition 2.4.2 In the above notation, (𝐿 0 , 𝑤 0 ) is an immediate extension of (𝐾, 𝑣). Proof (Ax [Ax71, Prop. 14]). Let 𝑤 0 , 𝑣 1 , . . . , 𝑣 𝑘 be the inequivalent extensions of 𝑣 to 𝐿 0 . By Lemma 2.4.1, the restriction of 𝜎 𝑗 𝑤 to 𝐿 0 belongs to {𝑣 1 , . . . , 𝑣 𝑘 } for 𝑗 = 2, . . . , 𝑚. The rest of the proof breaks up into two parts. Part A: The residue field of 𝐿 0 with respect to 𝑤 0 is 𝐾¯ 𝑣 . Indeed, let 𝑧 ∈ 𝐿 0 be an element with 𝑤 0 (𝑧) ≥ 0. By the weak approximation theorem (Proposition 2.1.1) there exists a 𝑦 ∈ 𝐿 0 such that 𝑤 0 (𝑦 − 𝑧) > 0 and 𝑣 𝑖 (𝑦) > 0 for 𝑖 = 1, . . . , 𝑘.

(2.20)

−1 𝑦 of 𝑦 belongs to 𝐾. For each 2 ≤ 𝑗 ≤ 𝑚 Then, the trace 𝑎 = 𝑦 + 𝜎2−1 𝑦 + · · · + 𝜎𝑚 Lemma 2.4.1(b) yields an 𝑖 ∈ {1, . . . , 𝑘 } with 𝜎 𝑗 𝑤| 𝐿0 = 𝑣 𝑖 , so by (2.20), 𝑤(𝜎 −1 𝑗 𝑦) = −1 𝑦) > 0. Therefore, (𝜎 𝑗 𝑤) (𝑦) = 𝑣 𝑖 (𝑦) > 0. Hence, 𝑤 0 (𝑎 − 𝑦) = 𝑤(𝜎2−1 𝑦 + · · · + 𝜎𝑚 by (2.20), 𝑤 0 (𝑎 − 𝑧) > 0. Conclude that 𝐿¯ 0,𝑤0 = 𝐾¯ 𝑣 .

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Part B: The value group of 𝐿 0 with respect to 𝑤 0 is Γ𝑣 . We have to find for each 𝑥 ∈ 𝐿 0× an element 𝑑 ∈ 𝐾 × such that 𝑤 0 (𝑥) = 𝑤 0 (𝑑). Apply (2.20) with 𝑧 = 1 to find 𝑠 ∈ 𝐿 0× such that 𝑤 0 (𝑠) = 0 and 𝑣 𝑖 (𝑠) > 0 for 𝑖 = 1, . . . , 𝑘. Then, 𝑤(𝑠) = 0 and −1 2 −1 −1 3 𝑤(𝜎 −1 𝑗 𝑠) > 0 for 𝑗 = 2, . . . , 𝑚. Hence, 𝑤(𝜎 𝑗 (𝑠𝑥)), 𝑤(𝜎 𝑗 (𝑠 𝑥)), 𝑤(𝜎 𝑗 (𝑠 𝑥)), . . . 𝑛 is an infinite sequence of elements of Γ𝑤 for 𝑗 = 2, . . . , 𝑚. Since 𝑤(𝑠 𝑥) = 𝑛𝑤 0 (𝑠) + 𝑤 0 (𝑥) = 𝑤 0 (𝑥) is independent of 𝑛, there exists a positive integer 𝑛 such that −1 (𝑠 𝑛 𝑥)). Replace 𝑥 by 𝑠 𝑛 𝑥, if necessary, to assume 𝑤(𝑠 𝑛 𝑥) ≠ 𝑤(𝜎2−1 (𝑠 𝑛 𝑥)), . . . , 𝑤(𝜎𝑚 −1 −1 that 𝑤(𝑥) ≠ 𝑤(𝜎2 𝑥), . . . , 𝑤(𝜎𝑚 𝑥). Under this assumption reenumerate 𝜎2 , . . . , 𝜎𝑚 if necessary to find an integer 𝑙 between 1 and 𝑚 such that −1 −1 𝑤(𝜎2−1 𝑥), . . . , 𝑤(𝜎𝑙−1 𝑥) < 𝑤(𝑥) < 𝑤(𝜎𝑙+1 𝑥). 𝑥), . . . , 𝑤(𝜎𝑚

(2.21)

(Note that 𝑙 = 1 means that only the right inequalities exist, while 𝑙 = 𝑚 means that only the left inequalities exist.) Then consider the fundamental symmetric polynomial of degree 𝑙 − 1: ∑︁ 𝜏2−1 𝑥 · · · 𝜏𝑙−1 𝑥, 𝑏 = 𝜎2−1 𝑥 · · · 𝜎𝑙−1 𝑥 + where {𝜏2 , . . . , 𝜏𝑙 } ranges over all subsets of {𝜎1 , 𝜎2 , . . . , 𝜎𝑚 } of cardinality 𝑙 − 1 which are different from {𝜎2 , . . . , 𝜎𝑙 }. Thus, there exists an 𝑖 such that 𝜏𝑖 is not in {𝜎2 , . . . , 𝜎𝑙 }. Hence, by (2.21), 𝑤(𝜎2−1 𝑥 · · · 𝜎𝑙−1 𝑥) < 𝑤(𝜏2−1 𝑥 · · · 𝜏𝑙−1 𝜏). It follows that 𝑤(𝑏) = 𝑤(𝜎2−1 𝑥 · · · 𝜎𝑙−1 𝑥). Likewise consider the fundamental symmetric polynomial of degree 𝑙: ∑︁ (𝜏1−1 𝑥) · · · (𝜏𝑙−1 𝑥), 𝑐 = 𝑥(𝜎2−1 𝑥) · · · (𝜎𝑙−1 𝑥) + where {𝜏1 , . . . , 𝜏𝑙 } ranges over all subsets of {𝜎1 , . . . , 𝜎𝑚 } of cardinality 𝑙 which are different from {𝜎1 , . . . , 𝜎𝑙 }. As in the preceding paragraph, (2.21) implies that 𝑤(𝑐) = 𝑤(𝑥(𝜎2−1 𝑥) · · · (𝜎𝑙−1 𝑥)). Both 𝑏 and 𝑐 belong to 𝐾. Hence, 𝑑 = 𝑐/𝑏 is an element of 𝐾 for which 𝑤(𝑑) = 𝑤(𝑥), as desired. □

2.5 Integral Extensions and Dedekind Domains Integral extensions of Z in number fields are Dedekind domains. Although they are in general not unique factorization domain, their ideals uniquely factor as products of prime ideals. In this section we survey the concepts of integral extensions of rings and of Dedekind domains and prove that every overring of a Dedekind domain is again a Dedekind domain. Let 𝐹 be a field containing an integral domain 𝑅. An element 𝑥 ∈ 𝐹 is integral over 𝑅 if it satisfies an equation of the form 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 + · · · + 𝑎 0 = 0 with 𝑎 1 , . . . , 𝑎 𝑛 ∈ 𝑅. The set of all elements of 𝐹 which are integral over 𝑅 form a ring (e.g. by Proposition 2.5.1 below), the integral closure of 𝑅 in 𝐹. Call 𝑅 integrally closed if 𝑅 coincides with its integral closure in Quot(𝑅). For example, every valuation ring 𝑂 of 𝐹 is integrally closed. Indeed, assume by contradiction that

2.5 Integral Extensions and Dedekind Domains

35

𝑥 ∈ 𝐹 ∖ 𝑂 and 𝑥 is integral over 𝑂. Then, 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 + · · · + 𝑎 0 = 0 for some 𝑎 0 , . . . , 𝑎 𝑛−1 ∈ 𝑂. Hence, 𝑥 −1 is in the maximal ideal 𝔪 of 𝑂 and 1 + 𝑎 𝑛−1 𝑥 −1 + · · · + 𝑎 0 𝑥 −𝑛 = 0. Thus, 1 ∈ 𝔪, a contradiction. Proposition 2.5.1 ([Lan64], p. 12, Prop. 4) An element 𝑥 of 𝐹 is integral over 𝑅 if and only if every place of 𝐹 finite on 𝑅 is finite at 𝑥. Thus, the integral closure of 𝑅 in 𝐹 is the intersection of all valuation rings of 𝐹 which contain 𝑅. In particular, every valuation ring of 𝐹 is integrally closed. Suppose that 𝜑 is a place of a field 𝐹 and 𝐾 is a subfield of 𝐹. We say that 𝜑 is trivial on 𝐾, or also that 𝜑 is a place of 𝐹/𝐾, if 𝜑(𝑥) ≠ ∞ for all 𝑥 ∈ 𝐾. Then, 𝜑(𝑦) ≠ 0 for all 𝑦 ∈ 𝐾 × . Thus, 𝜑 maps 𝐾 isomorphically onto 𝜑(𝐾). Lemma 2.5.2 Let 𝐾 ⊆ 𝐿 ⊆ 𝐹 be a tower of fields and 𝜑 a place of 𝐹. Suppose that 𝜑 is trivial on 𝐾 and 𝐿 is algebraic over 𝐾. Then, 𝜑 is trivial on 𝐿. Proof. Each 𝑥 ∈ 𝐿 is integral over 𝐾, so by Proposition 2.5.1 𝜑(𝑥) ≠ ∞. Thus, 𝜑 is also trivial on 𝐿. □ Let 𝑆 be a subring of 𝐹 containing 𝑅. Call 𝑆 integral over 𝑅 if every element of 𝑆 is integral over 𝑅. If 𝑆 = 𝑅[𝑥 1 , . . . , 𝑥 𝑚 ] and 𝑆 is integral over 𝑅, then 𝑆 is a finitely generated 𝑅-module. Indeed, every element of 𝑆 is a linear combination 𝛼𝑚 , where 0 ≤ 𝛼𝑖 < with coefficients in 𝑅
of the set of monomials 𝑥1𝛼1 𝑥2𝛼2 · · · 𝑥 𝑚 deg irr(𝑥𝑖 , Quot(𝑅)) . Propositions 2.3.1 and 2.5.1 give the following: Proposition 2.5.3 Let 𝑅 ⊆ 𝑆 be integral domains with 𝑆 finitely generated as an 𝑅-algebra. Suppose that 𝑆 is integral over 𝑅. Then, the following hold: (a) 𝑆 is finitely generated as an 𝑅-module. (b) Let 𝜑: 𝑅 → 𝑀 be a homomorphism into an algebraically closed field 𝑀. Then, the set of all homomorphisms 𝜓: 𝑆 → 𝑀 that extend 𝜑 is finite and nonempty. Suppose that 𝑅1 ⊆ 𝑅2 ⊆ 𝑅3 are integral domains. Proposition 2.5.1 implies that 𝑅3 is integral over 𝑅1 if and only if 𝑅2 is integral over 𝑅1 and 𝑅3 is integral over 𝑅2 . Call an integral domain 𝑅 Noetherian if every ideal of 𝑅 is finitely generated. For example, since a discrete valuation ring 𝑂 is a principal ideal domain, it is integrally closed and Noetherian. If 𝑅 is an integral domain and 𝔭 is a prime ideal of 𝑅, then 𝑎 | 𝑎 ∈ 𝑅 and 𝑏 ∈ 𝑅 ∖ 𝔭 𝑅𝔭 = 𝑏 is the local ring of 𝑅 at 𝔭. It has a unique maximal ideal, 𝔭𝑅𝔭 . If 𝑅 is a Noetherian domain, then 𝑅𝔭 is also Noetherian. If 𝑅 is integrally closed, then so is 𝑅𝔭 . Ñ Lemma 2.5.4 Suppose that 𝑅 is an integral domain. Then, Ñ 𝑅 = 𝑅𝔪 , where 𝔪 ranges over all maximal ideals of 𝑅. More generally, 𝔞 = 𝔞𝑅𝔪 for each ideal 𝔞 of 𝑅. Proof. Suppose that 𝑥 belongs to each 𝔞𝑅𝔪 . For each 𝔪, 𝑥 = 𝑎 𝔪 /𝑏 𝔪 , with 𝑎 𝔪 ∈ 𝔞 and 𝑏 𝔪 ∈ 𝑅 ∖ 𝔪. Denote the ideal generated by all the 𝑏 𝔪 ’s by 𝔟. If 𝔟 ≠ 𝑅, then 𝔟 is contained in a maximal ideal 𝔪. Hence, 𝑏 𝔪 ∈ 𝔪, a contradiction. Hence, 𝔟 = 𝑅. In Í particular, 1 = 𝔪 ∈𝑀 𝑏 𝔪 𝑐 𝔪 where Í 𝑀 is a finite set Í of maximal ideals, and 𝑐 𝔪 ∈ 𝑅 for each 𝔪 ∈ 𝑀. Therefore 𝑥 = 𝔪 ∈𝑀 𝑥𝑏 𝔪 𝑐 𝔪 = 𝔪 ∈𝑀 𝑎 𝔪 𝑐 𝔪 ∈ 𝔞. □

36

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Let 𝑅 be an integral domain with the quotient field 𝐹. A nonzero 𝑅-submodule 𝔞 of 𝐹 is said to be a fractional ideal of 𝑅 if there exists a nonzero 𝑥 ∈ 𝑅 with 𝑥𝔞 ⊆ 𝑅. In particular, every ideal of 𝑅 is a fractional ideal. Define the product, 𝔞𝔟, of two fractional ideals 𝔞 and 𝔟 to be the 𝑅-submodule generated by the products 𝑎𝑏, with 𝑎 ∈ 𝔞 and 𝑏 ∈ 𝔟. Define the inverse of a fractional ideal 𝔞 as 𝔞 −1 = {𝑥 ∈ 𝐹 | 𝑥𝔞 ⊆ 𝑅}. If 𝑎 ∈ 𝔞, then 𝑎𝔞 −1 ⊆ 𝑅. Therefore, both 𝔞𝔟 and 𝔞 −1 are fractional ideals. Proposition 2.5.5 ([CaF67], p. 6, Prop. 1) The following conditions on an integral domain 𝑅 are equivalent: (a) 𝑅 is Noetherian, integrally closed, and its nonzero prime ideals are maximal. (b) 𝑅 is Noetherian and the local ring, 𝑅𝔭 , of every nonzero prime ideal 𝔭 is a discrete valuation ring. (c) Every fractional ideal 𝔞 is invertible (i.e. 𝔞𝔞 −1 = 𝑅). When these conditions hold, 𝑅 is called a Dedekind domain. By Proposition 2.5.5, the set of all fractional ideals of a Dedekind domain 𝑅 forms an Abelian group, with 𝑅 being the unit. One proves that this group is free and the maximal ideals of 𝑅 are free generators of this group. Thus, every ideal 𝔞 of 𝑅 has a unique presentation 𝔞 = 𝔭1𝑚1 𝔭2𝑚2 · · · 𝔭𝑟𝑚𝑟 , as the product of powers of maximal ideals with positive exponents [CaF67, p. 8, Prop. 2]. Every principal ideal domain is a Dedekind domain. Thus, Z and 𝐾 [𝑥], where 𝑥 is a transcendental element over a field 𝐾, are Dedekind domains. By the same reason, every discrete valuation ring is a Dedekind domain. In the notation of Proposition 2.5.5(b), 𝑅𝔭 is the valuation ring of a discrete valuation 𝑣𝔭 of 𝐾 = Quot(𝑅). The corresponding place 𝜑𝔭 is finite on 𝑅. Conversely, if 𝜑 is such a place, then 𝔭 = {𝑥 ∈ 𝑅 | 𝜑(𝑥) = 0} is a nonzero prime ideal of 𝑅. Since 𝑅𝔭 ⊆ 𝑂 𝜑 , Lemma 2.2.5 implies that 𝑅𝔭 = 𝑂 𝜑 . This establishes a bijection between the nonzero prime ideals of 𝑅 and the equivalence classes of places of 𝐾 finite on 𝑅. Proposition 2.5.6 ([CaF67], p. 13, Prop. 1) Let 𝑆 be the integral closure of a Dedekind domain 𝑅 in a finite algebraic extension of Quot(𝑅). Then, 𝑆 is also a Dedekind domain. Let 𝔭 be a prime ideal of 𝑅. Then, 𝔭𝑆 = 𝔓1𝑒1 𝔓2𝑒2 · · · 𝔓𝑟𝑒𝑟 , where 𝔓1 , 𝔓2 , . . . , 𝔓𝑟 are the distinct prime ideals of 𝑆 that lie over 𝔭; that is, 𝔓𝑖 ∩ 𝑅 = 𝔭, 𝑖 = 1, . . . , 𝑟. For each 𝑖 we have 𝔭𝑆𝔓𝑖 = 𝔓𝑖𝑒𝑖 𝑆𝔓𝑖 . Hence, 𝑒 𝑖 is the ramification index of ord𝔓𝑖 over ord𝔭 . We say that 𝔓𝑖 is unramified over 𝐾 if ord𝔓𝑖 /ord𝔭 is unramified; that is, 𝑒 𝑖 = 1 and 𝑆/𝔓𝑖 is a separable extension of 𝑅/𝔭. The prime ideal 𝔭 is unramified in 𝐿 if each 𝔓𝑖 is unramified over 𝐾. By Proposition 2.5.6, the integral closure of Z in a finite extension 𝐿 of Q is a Dedekind domain, 𝑂 𝐿 , called the ring of integers of 𝐿. Proposition 2.5.7 (Noether–Grell) Every overring 𝑅 ′ of a Dedekind domain 𝑅 is a Dedekind domain. Proof. We show that 𝑅 ′ satisfies Condition (b) of Proposition 2.5.5. Part A: An injective map. If 𝔭 ′ is a nonzero prime ideal of 𝑅 ′, then 𝔭 = 𝑅 ∩ 𝔭 ′ is a nonzero prime ideal of 𝑅. Indeed, for 0 ≠ 𝑥 ∈ 𝔭 ′, write 𝑥 = 𝑏𝑎 , where 𝑎, 𝑏 ∈ 𝑅. Thus, 0 ≠ 𝑎 = 𝑏𝑥 ∈ 𝑅 ∩ 𝔭 ′ = 𝔭. Since 𝑅𝔭 ⊆ 𝑅𝔭′ ′ and 𝑅𝔭 is a discrete valuation ring, Lemma 2.2.5 implies that 𝑅𝔭 = 𝑅𝔭′ ′ . Hence,

2.5 Integral Extensions and Dedekind Domains

𝔭𝑅𝔭 = 𝔭 ′ 𝑅𝔭′ ′ .

37

(2.22)

In addition, 𝔭 ′ 𝑅𝔭′ ′ ∩ 𝑅 ′ = 𝔭 ′. Therefore, the map 𝔭 ′ ↦→ 𝑅 ∩𝔭 ′ from the set of nonzero prime ideals of 𝑅 ′ into the set of nonzero prime ideals of 𝑅 is injective. Part B: A finiteness condition. Let 𝑥 be a nonzero element of 𝑅 ′, 𝔭 ′ a prime ideal of 𝑅 ′ which contains 𝑥, and 𝔭 = 𝑅 ∩𝔭 ′. Then, 𝑅𝔭 = 𝑅𝔭′ ′ . Hence, ord𝔭 (𝑥) > 0. But this relation holds only for the finitely many prime ideals of 𝑅 that appear with positive exponents in the factorization of the fractional ideal 𝑥𝑅. Hence, by Part A, 𝑥 belongs to only finitely many prime ideals of 𝑅 ′. Part C: The ring 𝑅 ′ is Noetherian. Let 𝔞 be a nonzero ideal of 𝑅 ′. Choose a nonzero element 𝑥 ∈ 𝔞 and denote the finite set of prime ideals of 𝑅 ′ that contain 𝑥 by 𝑃. For each 𝔭 ∈ 𝑃 the local ring 𝑅𝔭′ is a discrete valuation domain. Hence, there exists an 𝑎𝔭 ∈ 𝔞 such that 𝔞𝑅𝔭′ = 𝑎𝔭 𝑅𝔭′ . Denote the ideal of 𝑅 ′ generated by 𝑥 and by all 𝑎𝔭 , for 𝔭 ∈ 𝑃, by 𝔞0 . It is contained in 𝔞. To show that 𝔞 is finitely generated, we need only to prove that 𝔞 ⊆ 𝔞0 . Indeed, consider a prime ideal 𝔮 of 𝑅 ′ not in 𝑃. Then, 𝑥 ∉ 𝔮, so 𝔞0 ̸ ⊆ 𝔮. Ñ Hence, 𝔞0 𝑅𝔮′ = 𝑅𝔮′ . It follows from Lemma 2.5.4 that 𝔞0 = 𝔭∈𝑃 𝔞0 𝑅𝔭′ . Therefore, Ñ Ñ Ñ ′ ′ ′ □ 𝔞 ⊆ 𝔭∈𝑃 𝔞𝑅𝔭 = 𝔭∈𝑃 𝑎𝔭 𝑅𝔭 ⊆ 𝔭∈𝑃 𝔞0 𝑅𝔭 = 𝔞0 , as desired. Lemma 2.5.8 Let (𝐸, 𝑣) be a discrete valued field, 𝐹1 , 𝐹2 , 𝐹 finite separable extensions of 𝐸 with 𝐹 = 𝐹1 𝐹2 , and 𝑤 an extension of 𝑣 to 𝐹. Suppose that 𝑣 is unramified in 𝐹1 . Then, the residue fields with respect to 𝑤 satisfy 𝐹¯ = 𝐹¯1 𝐹¯2 . Proof. Choose a finite Galois extension 𝑁 of 𝐸 which contains 𝐹 and an extension 𝑤 ′ of 𝑤 to 𝑁. Denote the decomposition groups of 𝑤 ′ over 𝐸, 𝐹1 , 𝐹2 , 𝐹 by 𝐷 𝐸 , 𝐷 𝐹1 , 𝐷 𝐹2 , 𝐷 𝐹 , respectively. Let 𝐸 ′, 𝐹1′, 𝐹2′, 𝐹 ′ be the fixed fields in 𝑁 of 𝐷 𝐸 , 𝐷 𝐹1 , 𝐷 𝐹2 , 𝐷 𝐹 , respectively. Let 𝑣 ′ = 𝑤 ′ | 𝐸 ′ . Since all valuations of 𝑁 lying over 𝑣 ′ are conjugate over 𝐸 ′, the definition of 𝐸 ′ as the fixed field of 𝐷 𝐸 implies that 𝑤 ′ is the unique extension of 𝑣 ′ to 𝑁. Also, 𝐷 𝐹1 = Gal(𝑁/𝐹1 ) ∩ 𝐷 𝐸 , so 𝐹1 𝐸 ′ = 𝐹1′. By assumption, 𝑣 is unramified in 𝐹1 , hence by Lemma 2.3.6, 𝑣 ′ is unramified in 𝐹1′. (𝑁, 𝑤 ′) 𝐹′ ✈ ❍❍❍❍ ✈ ❍❍ ✈✈ ❍ ✈✈ ✈ ′ 𝐹1 ● (𝐹, 𝑤)● 𝐹′ ●● ✇✇ ✇ 2 ● ✇ ●● ✇✇●● ✇✇✇ ● ′ ′✇✇ ● 𝐹1 ❍ (𝐸 , 𝑣 ) 𝐹2 ❍❍ ✈✈ ❍❍ ✈ ✈ ❍ ✈✈ (𝐸, 𝑣) Finally, by Proposition 2.4.2, the residue fields of 𝐸, 𝐹1 , 𝐹2 , 𝐹 at 𝑤 coincide with the residue fields of 𝐸 ′, 𝐹1′, 𝐹2′, 𝐹 ′ at 𝑤 ′, respectively.

38

2 Valuations

We may therefore replace 𝐸, 𝐹1 , 𝐹2 , 𝐹, respectively, by 𝐸 ′, 𝐹1′, 𝐹2′, 𝐹 ′, if necessary, to assume that 𝑤| 𝐹1 is the unique extension of 𝑣 to 𝐹1 . Having made this assumption, set 𝑤 𝑖 = 𝑤| 𝐹𝑖 , 𝑖 = 1, 2. By Proposition 2.5.1, 𝑂 𝑤1 is the integral closure of 𝑂 𝑣 in 𝐹1 . ¯ where Since 𝑣 is unramified in 𝐹1 , Proposition 2.3.2 implies [𝐹1 : 𝐸] = [ 𝐹¯1 : 𝐸], the bar denotes reduction modulo 𝑤. Choose 𝑥 ∈ 𝑂 𝑤1 such that 𝑥¯ is a primitive element for the separable extension ¯ Let 𝑓 = irr(𝑥, 𝐸) and 𝑝 = irr( 𝑥, ¯ Then, 𝑓 ∈ 𝑂 𝑣 [𝑋] and 𝑓 (𝑥) = 0. Hence, 𝐹¯1 /𝐸. ¯ 𝐸). 𝑓¯( 𝑥) ¯ = 0 and 𝑝| 𝑓¯. Therefore, ¯ = [𝐹1 : 𝐸]. ¯ = [ 𝐹¯1 : 𝐸] [𝐹1 : 𝐸] ≥ deg( 𝑓 ) ≥ deg( 𝑝) ¯ Consequently, 𝑝 = 𝑓¯, 𝐹1 = 𝐸 (𝑥), and 𝐹¯1 = 𝐸¯ ( 𝑥). By Lemma 2.3.6, 𝑤 2 is unramified in 𝐹. Thus, we may apply the result of the preceding paragraph to 𝐹/𝐹2 and conclude that 𝐹¯ = 𝐹¯2 ( 𝑥). ¯ Consequently, ¯ 𝐹¯1 𝐹¯2 = 𝐸¯ ( 𝑥) ¯ 𝐹¯2 = 𝐹¯2 ( 𝑥) □ ¯ = 𝐹. Exercise 9 shows that Lemma 2.5.8 is false for arbitrary real valuations. The following result is a special case of Proposition 2.1.2(b). Proposition 2.5.9 Let 𝑃1 , . . . , 𝑃𝑟 be distinct prime ideals of a Dedekind domain 𝑅. Let 𝑎 1 , . . . , 𝑎𝑟 , 𝑏 1 , . . . , 𝑏𝑟 ∈ 𝑅 with 𝑎 𝑖 ∉ 𝑃𝑖 for 𝑖 = 1, . . . , 𝑟. Then, there exists an 𝑥 ∈ 𝑅 such that 𝑎 𝑖 𝑥 + 𝑏 𝑖 ≡ 0 mod 𝑃𝑖 for 𝑖 = 1, . . . , 𝑟.

Exercises 1. Let 𝑂 be a valuation ring of a field 𝐹 and consider the subset 𝔪 = {𝑥 ∈ 𝑂 | 𝑥 −1 ∉ 𝑂}. Show that if 𝑥 ∈ 𝔪 and 𝑎 ∈ 𝑂, then 𝑎𝑥 ∈ 𝔪. Prove that 𝔪 is closed under addition. Hint: Use the identity 𝑥 + 𝑦 = (1 + 𝑥𝑦 −1 )𝑦 for 𝑦 ≠ 0. Show that 𝔪 is the unique maximal ideal of 𝑂. 2. Use Exercise 1 to prove that every valuation ring is integrally closed. 3. Let 𝑣 be a valuation of Q. Observe that 𝑣(𝑛) ≥ 𝑣(1) = 0, for each 𝑛 ∈ N. Hence, there exists a smallest 𝑝 ∈ N such that 𝑣( 𝑝) > 0. Prove that 𝑝 is a prime element of 𝑂 𝑣 and 𝑣 is equivalent to ord 𝑝 . Hint: If a positive integer 𝑚 is relatively prime to 𝑝, then there exist 𝑥, 𝑦 ∈ Z such that 𝑥 𝑝 + 𝑦𝑚 = 1. 4. Let 𝑣 be a valuation of the rational function field 𝐹 = 𝐾 (𝑡) which is trivial on 𝐾. Suppose that there exists a 𝑝 ∈ 𝐾 [𝑡] with 𝑣( 𝑝) > 0. Now suppose that 𝑝 has smallest degree with this property. Show that 𝑣 is equivalent to ord 𝑝 . Otherwise, there exists an 𝑓 ∈ 𝐾 [𝑡] such that 𝑣( 𝑓 (𝑡)) < 0. Conclude that 𝑣(𝑡) < 0, and that 𝑣 is equivalent to ord∞ . 5. Let 𝐹/𝐸 be a field extension, 𝑤 a valuation of 𝐹, and 𝑥 1 , . . . , 𝑥 𝑒 elements of 𝐹 such that 𝑤(𝑥 1 ), . . . , 𝑤(𝑥 𝑒 ) represent distinct classes of 𝑤(𝐹 × ) modulo 𝑤(𝐸 × ). Show that 𝑥1 , . . . , 𝑥 𝑒 are linearly independent over 𝐸. Thus, (𝑤(𝐹 × ) : 𝑤(𝐸 × )) ≤ [𝐹 : 𝐸]. Hint: Use (2.4b).

2.5 Integral Extensions and Dedekind Domains

39

6. Let Δ be an ordered group containing Z as a subgroup of index 𝑒. Show that there exists no positive element 𝛿 ∈ Δ such that 𝑒𝛿 < 1. Conclude that Δ contains a smallest positive element and hence that Δ  Z. Combine this with Exercise 5 to prove that if the restriction of 𝑤 to 𝐸 is discrete, then 𝑤 is discrete. 7. In the notation of Exercise 5, let 𝑣 be the restriction of 𝑤 to 𝐸. Let 𝑦 1 , . . . , 𝑦 𝑓 be elements of 𝐹 with 𝑤(𝑦 1 ), . . . , 𝑤(𝑦 𝑓 ) ≥ 0 with residue classes 𝑦¯ 1 , . . . , 𝑦¯ 𝑓 linearly independent over 𝐸¯ 𝑣 . Show that 𝑦 1 , . . . , 𝑦 𝑓 are linearly independent over 𝐸. Conclude that [ 𝐹¯𝑤 : 𝐸¯ 𝑣 ] ≤ [𝐹 : 𝐸]. Hint:
If 𝑎 1 , . . . , 𝑎
𝑓 ∈ 𝐹 are not all zero, then there exists a 𝑗, 1 ≤ 𝑗 ≤ 𝑓 , such 𝑎 that 𝑣 𝑎𝑎1𝑗 , . . . , 𝑣 𝑎 𝑓𝑗 ≥ 0. 8. Let 𝑣 be a discrete valuation of a field 𝐾 and let 𝑤 be an extension of 𝑣 to a finite Galois extension 𝐿 of 𝐾. Assume that 𝑤 ′ is also an extension of 𝑣 to 𝐿 such that 𝑤 ′ ≠ 𝜎(𝑤) for all 𝜎 ∈ Gal(𝐿/𝐾). Combine Exercise 7 with Proposition 2.1.1 to produce 𝑥 ∈ 𝐿 such that 𝑤 ′ (𝑥) > 0 and 𝑤(𝜎𝑥 − 1) > 0 for all 𝜎 ∈ Gal(𝐿/𝐾). With 𝑦 = 𝑁 𝐿/𝐾 (𝑥), conclude that the former condition gives 𝑣(𝑦) > 0, while the latter implies 𝑣(𝑦 − 1) > 0. Use this contradiction to prove that Gal(𝐿/𝐾) acts transitively on the extensions of 𝑣 to 𝐿. 9. (Geyer) The following example shows that Lemma 2.5.8 is false for arbitrary real √ valuations. Consider the field Q2 of 2-adic numbers. Show that the field 𝐾 = 𝑛 Q2 ( 2 | 𝑛 ∈ N) is a totally ramified extension of Q2 with value group Q.√Hence, each √ extension of 𝐾 is unramified. Prove that the residue √ field of both 𝐾 ( 3) and 𝐾 ( −1) is F2 . However, their compositum contains 𝐾 ( −3) and therefore has F4 as its residue field.

Chapter 3

Linear Disjointness

This chapter centers around the notion of linear disjointness of fields. We use this notion to define separable, regular, and primary extensions of fields. In particular, we prove that an extension 𝐹/𝐾 with a 𝐾-rational place is regular. Section 3.7 gives a useful criterion for separability with derivatives. We use the notion of regularity in Chapter 11 to characterize “varieties” over a field 𝐾 and in Chapter 12 to characterize “PAC fields”. It turns out that linear disjointness of finite field extensions of a field 𝐾 is intimately connected with “independence of subsets of Gal(𝐾)” with respect to the Haar measure of Gal(𝐾) (Section 21.3). This in turn leads in Chapter 22 to interesting properties of “large algebraic extensions” of 𝐾.

3.1 Linear Disjointness of Fields Central to field theory is the concept “linear disjointness of fields”, an analog of linear independence of vectors. We repeat the convention made in “Notation and Convention” that whenever we form the compositum of fields, we tacitly assume they are contained in a common field. Lemma 3.1.1 Let 𝐸 and 𝐹 be extensions of a field 𝐾. The following conditions are equivalent: (a) Each 𝑚-tuple (𝑥 1 , . . . , 𝑥 𝑚 ) of elements of 𝐸 which is linearly independent over 𝐾 is also linearly independent over 𝐹. (b) Each 𝑛-tuple (𝑦 1 , . . . , 𝑦 𝑛 ) of elements of 𝐹 which is linearly independent over 𝐾 is also linearly independent over 𝐸. Proof. It suffices to prove that (a) implies (b). Let 𝑦 1 , . . . , 𝑦 𝑛 be elements of 𝐹 that are linearly independent over 𝐾. Suppose that 𝑎 1 , . . . , 𝑎 𝑛 ∈ 𝐸 satisfy 𝑎 1 𝑦 1 + · · · + Í 𝑎 𝑛 𝑦 𝑛 = 0. Let {𝑥 𝑗 | 𝑗 ∈ 𝐽} be a linear basis for 𝐸 over 𝐾 and write 𝑎 𝑖 = 𝑗 ∈𝐽 𝑎 𝑖 𝑗 𝑥 𝑗 with 𝑎 𝑖 𝑗 elements of 𝐾, only finitely many different from 0. Then, © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_3

41

42

3 Linear Disjointness 𝑛 ∑︁ ∑︁ 𝑗 ∈𝐽


𝑎 𝑖 𝑗 𝑦 𝑖 𝑥 𝑗 = 0.

𝑖=1

Í By (a), {𝑥 𝑗 | 𝑗 ∈ 𝐽} is linearly independent over 𝐹. Hence, 𝑎 𝑖 𝑗 𝑦 𝑖 = 0 for every 𝑗. By our assumption on the 𝑦 𝑖 ’s, 𝑎 𝑖 𝑗 = 0 for every 𝑖 and 𝑗, so 𝑎 𝑖 = 0, 𝑖 = 1, . . . , 𝑚. □ Thus, 𝑦 1 , . . . , 𝑦 𝑚 are linearly independent over 𝐸. This proves (b). With 𝐸 and 𝐹 field extensions of a field 𝐾, refer to 𝐸 and 𝐹 as linearly disjoint over 𝐾 if (a) (or (b)) of Lemma 3.1.1 holds. Alternatively, we say that 𝐸 is linearly disjoint from 𝐹 over 𝐾. It follows that 𝐸 and 𝐹 are linearly disjoint over 𝐾 if and only if 𝐸 0 and 𝐹0 are linearly disjoint over 𝐾 for all finitely generated extensions 𝐸 0 and 𝐹0 of 𝐾 with 𝐸 0 ⊆ 𝐸 and 𝐹0 ⊆ 𝐹. Corollary 3.1.2 Let 𝐸 and 𝐹 be extensions of a field 𝐾 such that [𝐸 : 𝐾] < ∞. Then, 𝐸 and 𝐹 are linearly disjoint over 𝐾 if and only if [𝐸 : 𝐾] = [𝐸 𝐹 : 𝐹]. If in addition [𝐹 : 𝐾] < ∞, this is equivalent to [𝐸 𝐹 : 𝐾] = [𝐸 : 𝐾] [𝐹 : 𝐾]. Proof. The Í assumption that [𝐸 : 𝐾] < ∞ implies that each element of 𝐸 𝐹 can be written as 𝑟𝑖=1 𝑒 𝑖 𝑓𝑖 with 𝑒 1 , . . . , 𝑒𝑟 ∈ 𝐸 and 𝑓1 , . . . , 𝑓𝑟 ∈ 𝐹. It follows that if 𝐸 and 𝐹 are linearly disjoint over 𝐾 and 𝑤 1 , . . . , 𝑤 𝑛 is a basis for 𝐸/𝐾, then 𝑤 1 , . . . , 𝑤 𝑛 is also a basis for 𝐸 𝐹 over 𝐹. Hence, [𝐸 𝐹 : 𝐹] = 𝑛 = [𝐸 : 𝐾]. Conversely, suppose that [𝐸 : 𝐾] = [𝐸 𝐹 : 𝐹] and let 𝑥 1 , . . . , 𝑥 𝑚 ∈ 𝐸 be linearly independent over 𝐾. Extend {𝑥1 , . . . , 𝑥 𝑚 } to a basis {𝑥1 , . . . , 𝑥 𝑛 } of 𝐸/𝐾. Since {𝑥1 , . . . , 𝑥 𝑛 } generates 𝐸 𝐹 over 𝐹 and 𝑛 = [𝐸 𝐹 : 𝐹], {𝑥1 , . . . , 𝑥 𝑛 } is a basis of 𝐸 𝐹/𝐹. In particular, 𝑥1 , . . . , 𝑥 𝑚 are linearly independent over 𝐹. □ Let 𝐸/𝐾 be a finite Galois extension. Then, the condition 𝐸 ∩ 𝐹 = 𝐾 is equivalent to “res: Gal(𝐸 𝐹/𝐹) → Gal(𝐸/𝐾) is an isomorphism” and also to “res: Gal(𝐸 𝐹/𝐹) → Gal(𝐸/𝐾) is surjective”. Hence, by Corollary 3.1.2, 𝐸 ∩ 𝐹 = 𝐾 if and only if 𝐸 and 𝐹 are linearly disjoint over 𝐾. For arbitrary extensions this condition is clearly necessary, but not sufficient. Let 𝐿 be a degree 𝑛 > 1 extension of 𝐾 for which 𝐿 ′ is conjugate to 𝐿 over 𝐾 and to Corollary 3.1.2, 𝐿 and 𝐿 ′ ∩ 𝐿 = 𝐾. Then, [𝐿𝐿 ′ : 𝐾] ≤ 𝑛(𝑛 − 1). Thus, according √3 𝐿 ′ are√not linearly disjoint over 𝐾. For example, Q( 2) is not linearly disjoint from 3 Q(𝜁3 2) over Q although their intersection is Q. Lemma 3.1.3 (Tower Property) Let 𝐾 ⊆ 𝐸 and 𝐾 ⊆ 𝐿 ⊆ 𝐹 be four fields. Then, 𝐸 is linearly disjoint from 𝐹 over 𝐾 if and only if 𝐸 is linearly disjoint from 𝐿 over 𝐾 and 𝐸 𝐿 is linearly disjoint from 𝐹 over 𝐿. Proof. The only nontrivial part is to show that if 𝐸 and 𝐹 are linearly disjoint over 𝐾, then 𝐸 𝐿 and 𝐹 are linearly disjoint over 𝐿. Apply Lemma 3.1.1. Suppose that 𝑦 1 , . . . , 𝑦 𝑚 are elements of 𝐹 which are linearly Í𝑚 independent over 𝐿, but 𝑎 1 , . . . , 𝑎 𝑚 are elements of 𝐸 𝐿 such that 𝑖=1 𝑎 𝑖 𝑦 𝑖 = 0. Clear Í denominators to assume that 𝑎 𝑖 ∈ 𝐿 [𝐸], so that 𝑎 𝑖 = 𝑎 𝑖 𝑗 𝑥 𝑗 with 𝑎 𝑖 𝑗 ∈ 𝐿, where {𝑥 𝑗 | 𝑗 ∈ 𝐽} is a linear basis for 𝐸 over 𝐾. Then,

3.1 Linear Disjointness of Fields

∑︁ ∑︁ 𝑗

43


𝑎𝑖 𝑗 𝑦𝑖 𝑥 𝑗 =

∑︁ ∑︁ 𝑖

𝑖


𝑎𝑖 𝑗 𝑥 𝑗 𝑦𝑖 =

𝑗

∑︁

𝑎 𝑖 𝑦 𝑖 = 0.

𝑖

By assumption, the 𝑥 𝑗 are linearly independent over 𝐹. Hence, 𝑎 𝑖 𝑗 = 0 for all 𝑖 and 𝑗. Consequently, 𝑎 𝑖 = 0 for 𝑖 = 1, . . . , 𝑚.

Í

𝑗

𝑎 𝑖 𝑗 𝑦 𝑖 = 0, so □

Corollary 3.1.4 Suppose that in the diagram of fields 𝐸

𝑀

𝐾

𝐿

𝐸 𝐹= 𝑀 𝐹

𝐹

𝐸 is linearly disjoint from 𝐹 over 𝐾 and 𝑀 is linearly disjoint from 𝐹 over 𝐿. Then, 𝑀 = 𝐸 𝐿. Proof. Consider the commutative diagram of fields 𝐹

𝐹𝐸 𝐿

𝐹𝑀

𝐿

𝐸𝐿

𝑀

Since 𝑀 is linearly disjoint from 𝐹 over 𝐿, we have by the tower property that 𝑀 is linearly disjoint from 𝐹𝐸 𝐿 over 𝐸 𝐿. If 𝐸 𝐿 ⊂ 𝑀, there exists an 𝑥 ∈ 𝑀 ∖ 𝐸 𝐿, so 1, 𝑥 are linearly independent over 𝐸 𝐿. Hence, 1, 𝑥 are linearly independent over 𝐹𝐸 𝐿. Therefore, 𝑥 ∈ 𝐹 𝑀 ∖ 𝐹𝐸 𝐿, a contradiction. It follows that 𝐸 𝐿 = 𝑀, as asserted. □ Lemma 3.1.5 Let 𝐿 be a separable algebraic extension of a field 𝐾 and let 𝑀 be a purely inseparable extension of 𝐾. Then, 𝐿 and 𝑀 are linearly disjoint over 𝐾. Proof. Let 𝐿ˆ be the Galois closure of 𝐿/𝐾. Then, 𝐿ˆ ∩ 𝑀 = 𝐾. Hence, 𝐿ˆ and 𝑀 are linearly disjoint over 𝐾. Therefore, by Lemma 3.1.3, 𝐿 and 𝑀 are linearly disjoint over 𝐾. □ Let 𝐸 1 , . . . , 𝐸 𝑛 be 𝑛 extensions of a field 𝐾. We say that 𝐸 1 , . . . , 𝐸 𝑛 are linearly disjoint over 𝐾 if 𝐸 1 · · · 𝐸 𝑚−1 and 𝐸 𝑚 are linearly disjoint over 𝐾 for 𝑚 = 2, . . . , 𝑛. Induction on 𝑛 shows that this is the case if and only if the following condition holds: If 𝑤 𝑖, 𝑗𝑖 , 𝑗 𝑖 ∈ 𝐽𝑖 , are elements of 𝐸 𝑖 which are linearly independent over 𝐾, Î𝑛 𝑖 = 1, . . . , 𝑛, then 𝑖=1 𝑤 𝑖, 𝑗𝑖 , ( 𝑗1 , . . . , 𝑗 𝑛 ) ∈ 𝐽1 × · · · × 𝐽𝑛 , are linearly independent over 𝐾. It follows that if 𝐸 1 , . . . , 𝐸 𝑛 are linearly disjoint over 𝐾, then 𝐸 𝜋 (1) , . . . , 𝐸 𝜋 (𝑛) are linearly disjoint over 𝐾 for every permutation 𝜋 of {1, . . . , 𝑛}. Lemma 3.1.6 In the above notation, (a) 𝐸 1 , . . . , 𝐸 𝑛 are linearly disjoint over 𝐾 if and only if (b) the canonical homomorphism of 𝐸 1 ⊗𝐾 · · · ⊗𝐾 𝐸 𝑛 into 𝐸 1 · · · 𝐸 𝑛 that maps 𝑥1 ⊗ · · · ⊗ 𝑥 𝑛 onto 𝑥 1 · · · 𝑥 𝑛 is injective.

44

3 Linear Disjointness

Proof. We prove only the direction we need in this book, namely that (a) implies (b). Again, induction on 𝑛 shows that it suffices to consider only the case 𝑛 = 2. Thus, we set 𝐸 = 𝐸 1 , 𝐹 = 𝐸 2 and assume that 𝐸, 𝐹 are linearly disjoint over 𝐾. In order to prove that the canonical map 𝐸 ⊗𝐾 𝐹 → 𝐸 𝐹 is injective, it suffices to prove that the map is injective on 𝑈 ⊗𝐾 𝑉 for every finitely generated 𝐾-subspace 𝑈 of 𝐸 and every finitely generated 𝐾-subspace 𝑉 of 𝐹. Indeed, let 𝑢 1 , . . . , 𝑢 𝑙 and 𝑣 1 , . . . , 𝑣 𝑚 be 𝐾-basis of 𝑈 and 𝑉, respectively. In particular, 𝑢 1 , . . . , 𝑢 𝑙 are linearly independent over 𝐾. Hence, by the linear disjointness of 𝐸 and 𝐹 over 𝐾, 𝑢 1 , . . . , 𝑢 𝑙 are also linearly independent over 𝐹. Therefore, the 𝑢 𝑖 𝑣 𝑗 are linearly independent over 𝐾. In addition, the 𝑢 𝑖 𝑣 𝑗 ’s generate the 𝐾-space 𝑈𝑉, so they form a basis for that space. Thus, dim(𝑈𝑉) = 𝑙𝑚 = dim(𝑈 ⊗𝐾 𝑉). Since the canonical map is surjective, we conclude that it is also injective, as claimed. □ The application of tensor products makes the following lemma an easy observation. Lemma 3.1.7 Let 𝐸 1 , . . . , 𝐸 𝑛 (resp. 𝐹1 , . . . , 𝐹𝑛 ) be linearly disjoint field extensions of 𝐾 (resp. 𝐿). For each 𝑖 between 1 and 𝑛 let 𝜑𝑖 : 𝐸 𝑖 → 𝐹𝑖 ∪ {∞} be either a place or an embedding. Suppose that 𝜑1 , . . . , 𝜑 𝑛 coincide on 𝐾 and 𝜑𝑖 (𝐾) = 𝐿, 𝑖 = 1, . . . , 𝑛. Let 𝐸 = 𝐸 1 · · · 𝐸 𝑛 and 𝐹 = 𝐹1 · · · 𝐹𝑛 . Then, there exists a place 𝜑: 𝐸 → 𝐹˜ ∪ {∞} that extends each of the 𝜑𝑖 ’s. If each 𝜑𝑖 is an isomorphism of 𝐸 𝑖 onto 𝐹𝑖 , then 𝜑 is an isomorphism of 𝐸 onto 𝐹. Proof. Let 𝑂 𝑖 be the valuation ring of 𝜑𝑖 if 𝜑𝑖 is a place and 𝐸 𝑖 if 𝜑𝑖 is an isomorphism. By assumption and by Lemma 3.1.6, the map 𝑥 1 · · · 𝑥 𝑛 → 𝑥1 ⊗ · · · ⊗ 𝑥 𝑛 is an isomorphism 𝑂 1 · · · 𝑂 𝑛  𝑂 1 ⊗𝐾 · · · ⊗𝐾 𝑂 𝑛 of rings. Hence, there exists a ring homomorphism 𝜑0 : 𝑂 1 · · · 𝑂 𝑛 → 𝐹 such that 𝜑0 (𝑥) = 𝜑𝑖 (𝑥) for each 𝑥 ∈ 𝑂 𝑖 , 𝑖 = 1, . . . , 𝑛. Extend 𝜑0 to a place 𝜑: 𝐸 → 𝐹˜ ∪ {∞} (Proposition 2.3.1). If 𝑥 ∈ 𝐸 𝑖 ∖ 𝑂 𝑖 , then 𝜑(𝑥 −1 ) = 𝜑𝑖 (𝑥 −1 ) = 0, so 𝜑(𝑥) = 𝜑𝑖 (𝑥) = ∞. We conclude that 𝜑 coincides with 𝜑𝑖 on 𝐸 𝑖 . □ Finally, define a family {𝐸 𝑖 | 𝑖 ∈ 𝐼} of field extensions of 𝐾 to be linearly disjoint over 𝐾 if every finite subfamily is linearly disjoint over 𝐾. It follows from the discussion preceding Lemma 3.1.6 that a sequence (𝐸 1 , 𝐸 2 , 𝐸 3 , . . .) of field extensions of 𝐾 is linearly disjoint over 𝐾 if 𝐸 𝑛 is linearly disjoint from 𝐸 1 · · · 𝐸 𝑛−1 over 𝐾 for 𝑛 = 2, 3, 4, . . . . Then, 𝐸 𝜋 (1) , 𝐸 𝜋 (2) , 𝐸 𝜋 (3) , . . . are linearly disjoint over 𝐾 for every permutation 𝜋 of N. Lemma 3.1.8 Let L = {𝐿 𝑖 | 𝑖 ∈ 𝐼} be a family of Galois extensions of a field 𝐾. (a) If the family L is linearly disjoint over 𝐾, then Ö Ö 𝐿 𝑖 )/𝐾)  Gal(𝐿 𝑖 /𝐾). (3.1) Gal(( 𝑖 ∈𝐼

𝑖 ∈𝐼

(b) If 𝐼 is finite and (3.1) holds, then the family L is linearly disjoint over 𝐾. Î Î Proof of (a). Since 𝑖 ∈𝐼 Gal(𝐿 𝑖 /𝐾)  lim 𝑖 ∈𝐼0 Gal(𝐿 𝑖 /𝐾), where 𝐼0 ranges ←− over all finite Î subsets of 𝐼, we Î may assume that 𝐼 is finite. In this case, the embedding Gal(( 𝑖 ∈𝐼 𝐿 𝑖 )/𝐾) → 𝑖 ∈𝐼 Gal(𝐿 𝑖 /𝐾) given by 𝜎 ↦→ (𝜎| 𝐿𝑖 )𝑖 ∈𝐼 is surjective (Lemma (3.1.7)). Therefore, the embedding is an isomorphism.

3.1 Linear Disjointness of Fields

45

Proof of (b). First consider Galois extensions 𝐿 and 𝐿 ′ of 𝐾 with Gal(𝐿𝐿 ′/𝐾)  Gal(𝐿/𝐾) × Gal(𝐿 ′/𝐾). Without loss assume that [𝐿 : 𝐾], [𝐿 ′ : 𝐾] < ∞. Then, [𝐿𝐿 ′ : 𝐾] = [𝐿 : 𝐾] [𝐿 ′ : 𝐾], so by Corollary 3.1.2, 𝐿 and 𝐿 ′ are linearly disjoint □ over 𝐾. The general case follows now by induction on the cardinality of 𝐼. Example 3.1.9 Part (b) of Lemma 3.1.8 need not hold if 𝐼 is an infinite family. Indeed, let 𝐾 = Q(𝑡 1 , 𝑡2 , 𝑡3 , . . .) for a sequence 𝑡 1 , 𝑡2 , 𝑡3 , . . . of indeterminates and √ set 𝐿 𝑖 = Q( 𝑡 𝑖 ) for 𝑖 = 1, 2, 3, . . . . Then, the 𝐿 1 , 𝐿 2 , 𝐿 3 , . . . are linearly disjoint over 𝐾. However, the sequence 𝐿 0 , 𝐿 1 , 𝐿 2 , 𝐿 3 , . . . for which 𝐿 0 = 𝐿 1 is not linearly Î Î∞ disjoint over 𝐾. Nevertheless, ∞ 𝑖=0 𝐿 𝑖 = 𝑖=1 𝐿 𝑖 (where here the product symbol stands for combination of fields). Therefore, ∞ ∞ ∞ ∞ Ö Ö Ö Ö Gal( 𝐿 𝑖 /𝐾) = Gal( 𝐿 𝑖 /𝐾)  Gal(𝐿 𝑖 /𝐾)  Gal(𝐿 𝑖 /𝐾𝑖 ), 𝑖=0

𝑖=1

𝑖=1

𝑖=0

where the first isomorphism follows from Lemma 3.1.8(a) and the second isomorphism maps Gal(𝐿 𝑖 /𝐾) onto Gal(𝐿 𝑖−1 /𝐾) for 𝑖 = 1, 2, 3, . . . . Lemma 3.1.10 Let 𝐾 be a field, 𝐾1 , 𝐾2 , 𝐾3 , . . . a linearly disjoint sequence of extensions of 𝐾, and 𝐿 a finite separable extension of 𝐾. Then, there exists a positive integer 𝑛 such that 𝐿, 𝐾𝑛 , 𝐾𝑛+1 , 𝐾𝑛+2 , . . . are linearly disjoint over 𝐾. Proof. Replace 𝐿 by its Galois closure over 𝐾, if necessary, to assume that 𝐿 is Galois over 𝐾. Assume by contradiction that for each positive integer 𝑛 the field 𝐿 is not linearly disjoint from 𝐾𝑛 𝐾𝑛+1 𝐾𝑛+2 · · · over 𝐾. Then, 𝐿 𝑛 = 𝐿 ∩ 𝐾𝑛 𝐾𝑛+1 𝐾𝑛+2 · · · is a proper extension of 𝐾. Since 𝐿 has only finitely many subfields that contain 𝐾 and since 𝐿 𝑛 ⊇ 𝐿 𝑛+1 ⊇ 𝐿 𝑛+2 ⊇ · · · , there is an 𝑚 such that 𝐿 𝑛 = 𝐿 𝑚 for all 𝑛 ≥ 𝑚. Since 𝐿 is a finite extension of 𝐿 𝑚 , there is an 𝑛 > 𝑚 with 𝐿 𝑚 = 𝐿 ∩ 𝐾𝑚 𝐾𝑚+1 𝐾𝑚+2 · · · = 𝐿 ∩ 𝐾𝑚 · · · 𝐾𝑛−1 ⊆ 𝐾𝑚 · · · 𝐾𝑛−1 . Similarly, there exists an 𝑟 > 𝑛 with 𝐿 𝑚 = 𝐿 𝑛 ⊆ 𝐾𝑛 · · · 𝐾𝑟−1 . By assumption, 𝐾𝑚 · · · 𝐾𝑛−1 and 𝐾𝑛 · · · 𝐾𝑟−1 are linearly disjoint over 𝐾. In particular, their intersection is 𝐾. Therefore, 𝐿 𝑚 = 𝐾. This contradiction proves that there exists an 𝑛 such that 𝐿, 𝐾𝑛 , 𝐾𝑛+1 , 𝐾𝑛+2 , . . . are linearly disjoint over 𝐾. □ Lemma 3.1.11 Let 𝑣 be a discrete valuation of a field 𝐾 and 𝐿, 𝑀 finite extensions of 𝐾. Suppose that 𝑣 is unramified in 𝐿 but totally ramified in 𝑀. Then, 𝐿 and 𝑀 are linearly disjoint over 𝐾. proof. Let 𝑣 ′ be an extension of 𝑣 to 𝑀, 𝑤 ′ an extension of 𝑣 ′ to 𝐿 𝑀, and 𝑤 the restriction of 𝑤 ′ to 𝐿. By assumption, 𝑒 𝑤/𝑣 = 1 and 𝑒 𝑣′ /𝑣 = [𝑀 : 𝐾].

(3.2)

Using the multiplicativity of the ramification indices (paragraph preceding Proposition 2.3.2), we have (3.2)

[𝐿 𝑀 : 𝐿] ≥ 𝑒 𝑤′ /𝑤 = 𝑒 𝑤′ /𝑤 𝑒 𝑤/𝑣 = 𝑒 𝑤′ /𝑣 = 𝑒 𝑤′ /𝑣′ 𝑒 𝑣′ /𝑣 (3.2)

= 𝑒 𝑤′ /𝑣′ [𝑀 : 𝐾] ≥ 𝑒 𝑤′ /𝑣′ [𝐿 𝑀 : 𝐿].

(3.3)

46

3 Linear Disjointness

It follows that 1 ≥ 𝑒 𝑤′ /𝑣′ , so 𝑒 𝑤′ /𝑣′ = 1. We conclude from (3.3) that [𝑀 : 𝐾] ≥ [𝐿 𝑀 : 𝐿] ≥ 𝑒 𝑤′ /𝑤 = [𝑀 : 𝐾], so [𝑀 : 𝐾] = [𝐿 𝑀 : 𝐿], hence 𝐿, 𝑀 are linearly disjoint over 𝐾, as claimed.

□

Example 3.1.12 (Roots of unity) For each 𝑛 consider the Galois extension Q(𝜁 𝑛 ) of Q obtained by adjoining a primitive root of unity of order 𝑛. It is well known that 𝜑(𝑛) = [Q(𝜁 𝑛 ) : Q] is the number of integers between 1 and 𝑛 which are relatively prime to 𝑛 [Lan97, p. 278, Thm. 3.1]. If 𝑚 is relatively prime to 𝑛, then 𝜑(𝑚𝑛) = 𝜑(𝑚)𝜑(𝑛) [LeV58, p. 28, Thm. 3-7]. In addition, Q(𝜁 𝑚 , 𝜁 𝑛 ) = Q(𝜁 𝑚𝑛 ) [Lan97, p. 278]. Hence, [Q(𝜁 𝑚 , 𝜁 𝑛 ) : Q] = [Q(𝜁 𝑚𝑛 ) : Q] = 𝜑(𝑚𝑛) = 𝜑(𝑚)𝜑(𝑛) = [Q(𝜁 𝑚 ) : Q] [Q(𝜁 𝑛 ) : Q]. It follows from Corollary 3.1.2 that Q(𝜁 𝑚 ) and Q(𝜁 𝑛 ) are linearly disjoint over Q. Here is an application of linear disjointness to integral closures of domains. Lemma 3.1.13 Let 𝐾 be a field, 𝐿 a separable algebraic extension of 𝐾, and 𝑅 an integrally closed integral domain containing 𝐾. Let 𝐸 = Quot(𝑅), 𝐹 = 𝐸 𝐿, and 𝑆 the integral closure of 𝑅 in 𝐹. Suppose that 𝐸 and 𝐿 are linearly disjoint over 𝐾. Then, 𝑆 = 𝑅𝐿  𝑅 ⊗𝐾 𝐿. Proof. Assume without loss that 𝐿/𝐾 is finite. Choose a basis 𝑤 1 , . . . , 𝑤 𝑛 for 𝐿/𝐾. Let 𝜎1 , . . . , 𝜎𝑛 be the distinct 𝐾-embeddings of 𝐿 into 𝐾sep . Then, det(𝜎𝑖 𝑤 𝑗 ) ≠ 0 [Lan97, p. 286, consequence of Cor. 5.4]. Each element of 𝐿 is integral over 𝐾, hence over 𝑅, so 𝑅𝐿 ⊆ 𝑆. Conversely, let 𝑥 ∈ 𝑆. By the linear disjointness, 𝑤 1 , . . . , 𝑤 𝑛 form a basis for 𝐹/𝐸. Hence, Í 𝑥 = 𝑛𝑗=1 𝑒 𝑗 𝑤 𝑗 with 𝑒 𝑗 ∈ 𝐸, 𝑗 = 1, . . . , 𝑛. Also, Í each 𝜎𝑖 extends to an 𝐸-embedding of 𝐹 into 𝐸 sep (Lemma 3.1.7). Thus, 𝜎𝑖 𝑥 = 𝑛𝑗=1 𝑒 𝑗 𝜎𝑖 𝑤 𝑗 , 𝑖 = 1, . . . , 𝑛. Apply Kramer’s law to present each 𝑒 𝑘 as a polynomial in 𝜎𝑖 𝑥, 𝜎𝑖 𝑤 𝑗 , with 𝑖, 𝑗 = 1, . . . , 𝑛, divided by det(𝜎𝑖 𝑤 𝑗 ). Since det(𝜎𝑖 𝑤 𝑗 ) −1 belongs to the Galois closure of 𝐿 over 𝐾, it is integral over 𝐾, hence also over 𝑅. Thus, 𝑒 𝑘 is an element of 𝐸 which is integral over 𝑅. Since 𝑅 is integrally closed, 𝑒 𝑘 ∈ 𝑅, 𝑘 = 1, . . . , 𝑛. Consequently, 𝑥 ∈ 𝑅𝐿, as needed. Finally, the canonical embedding, 𝐸 ⊗𝐾 𝐿 → 𝐹 (Lemma 3.1.6(b)), maps 𝑅 ⊗𝐾 𝐿 onto 𝑅𝐿 = 𝑆. □ We generalize the tower property to families of field extensions: Lemma 3.1.14 Let 𝐾 be a field and 𝐼 a set. For each 𝑖 ∈ 𝐼 let 𝐹𝑖 /𝐸 𝑖 be a field extension with 𝐾 ⊆ 𝐸 𝑖 . Suppose that {𝐹𝑖 | 𝑖 ∈ 𝐼} is linearly disjoint over 𝐾. Denote the compositum of all 𝐸 𝑖 ’s by 𝐸. Then, the set {𝐹𝑖 𝐸 | 𝑖 ∈ 𝐼} is linearly disjoint over 𝐸. Moreover, for each 𝑖 ∈ 𝐼, the field 𝐹𝑖 is linearly disjoint from 𝐸 over 𝐸 𝑖 . Proof. It suffices to consider the case where 𝐼 = {1, 2, . . . , 𝑛}. By induction assume that 𝐹𝑖 𝐸 1 · · · 𝐸 𝑛−1 , 𝑖 = 1, . . . , 𝑛 − 1, are linearly disjoint over 𝐸 1 · · · 𝐸 𝑛 . By assumption, 𝐹1 · · · 𝐹𝑛−1 is linearly disjoint from 𝐹𝑛 over 𝐾. Hence, by the tower property, 𝐹1 · · · 𝐹𝑛−1 is linearly disjoint from 𝐸 over 𝐸 1 · · · 𝐸 𝑛−1 , so 𝐹𝑖 𝐸, 𝑖 = 1, . . . , 𝑛 − 1, are linearly disjoint over 𝐸.

3.2 Purely Transcendental Extensions

47

𝐹1 · · · 𝐹𝑛−1

𝐹1 · · · 𝐹𝑛−1 𝐸

𝐹𝑖 𝐸 1 · · · 𝐸 𝑛−1

𝐹𝑖 𝐸

𝐸 1 · · · 𝐸 𝑛−1

𝐸

𝐸 𝐹𝑛

𝐾

𝐸𝑛

𝐹𝑛

Moreover, 𝐹1 · · · 𝐹𝑛−1 𝐸 is linearly disjoint from 𝐸 𝐹𝑛 over 𝐸. Consequently, 𝐸 is linearly disjoint from 𝐹𝑛 over 𝐸 𝑛 and 𝐹𝑖 𝐸, 𝑖 = 1, . . . , 𝑛 are linearly disjoint over 𝐸, as claimed. □

3.2 Purely Transcendental Extensions Let 𝐹/𝐾 be an extension of fields and consider a subset 𝑇 of 𝐹. We say that 𝑇 is algebraically independent over 𝐾 if 𝑓 (𝑡1 , . . . , 𝑡 𝑛 ) ≠ 0 for all 𝑡1 , . . . , 𝑡 𝑛 ∈ 𝑇 and for each nonzero 𝑓 ∈ 𝐾 [𝑋1 , . . . , 𝑋𝑛 ]. In this case, we say that 𝐾 (𝑇)/𝐾 is a purely transcendental extension. If in addition 𝐹/𝐾 (𝑇) is an algebraic extension, then 𝑇 is a transcendence basis of 𝐹/𝐾. The cardinality of 𝑇 depends only on 𝐹/𝐾. It is the transcendence degree of 𝐹/𝐾. We denote it by trans.deg(𝐹/𝐾). For example, trans.deg(𝐹/𝐾) = 0 if and only if 𝐹/𝐾 is an algebraic extension. If 𝑆 is a subset of 𝐹 such that 𝐹/𝐾 (𝑆) is algebraic, then 𝑆 contains a transcendence basis for 𝐹/𝐾 [Lan97, p. 356, Thm. 1.1]. In particular, if 𝐹/𝐾 is finitely generated, ˜ then trans.deg(𝐹/𝐾) < ∞. The converse is false. For example, trans.deg( Q/Q) =0 ˜ although Q/Q is not finitely generated. If 𝑇0 is a subset of 𝐹 which is algebraically independent over 𝐾, choose a transcendence basis 𝑇1 for 𝐹/𝐾 (𝑇0 ). Then, 𝑇0 ∩ 𝑇1 = ∅ and 𝑇0 ∪ 𝑇1 is a transcendence basis for 𝐹/𝐾. This argument also gives the additivity of the transcendence degree for a tower 𝐾 ⊆ 𝐸 ⊆ 𝐹 of fields: trans.deg(𝐹/𝐾) = trans.deg(𝐸/𝐾) + trans.deg(𝐹/𝐸).

(3.4)

Lemma 3.2.1 Let 𝑡1 , . . . , 𝑡 𝑛 be algebraically independent elements over a field 𝐾 and set t = (𝑡1 , . . . , 𝑡 𝑛 ). Then, (a) 𝑡1 , . . . , 𝑡 𝑛 are algebraically independent over 𝐾˜ and (b) 𝐾 (t) is linearly disjoint from 𝐾˜ over 𝐾. ˜ Proof of (a). Since trans.deg( 𝐾/𝐾) = 0, an application of (3.4) implies that ˜ = 𝑛. Another application of (3.4) implies that 𝑡1 , . . . , 𝑡 𝑛 are trans.deg( 𝐾˜ (t)/𝐾) ˜ algebraically independent over 𝐾. Proof of (b). Let 𝑓1 , . . . , 𝑓𝑚 be elements of 𝐾 (t) which are linearly dependent ˜ Thus, there are 𝑐˜1 , . . . , 𝑐˜𝑚 ∈ 𝐾˜ not all zero with Í𝑚 𝑐˜𝑖 𝑓𝑖 = 0. Clearing over 𝐾. 𝑖=0 Í 𝑗 𝑗 denominators, we may assume that all 𝑓𝑖 ∈ 𝐾 [t]. Write 𝑓𝑖 (t) = j 𝑎 𝑖j 𝑡 11 · · · 𝑡 𝑛𝑛 . Then,
𝑗1 Í Í𝑚 Í𝑚 Í𝑚 𝑗𝑛 j 𝑖=1 𝑐˜𝑖 𝑎 𝑖j 𝑡 1 · · · 𝑡 𝑛 = 𝑖=1 𝑐˜𝑖 𝑓𝑖 (t) = 0. Hence, by (a), 𝑖=1 𝑐˜𝑖 𝑎 𝑖j = 0 for all j.

48

3 Linear Disjointness

Í𝑚 𝑋𝑖 𝑎 𝑖j = 0 with coefficients Thus, the homogeneous linear system of equations 𝑖=1 𝑎 𝑖j ∈ 𝐾 has a nonzero solution in 𝐾˜ 𝑚 . Therefore, it has a nonzero solution in 𝐾 𝑚 . In Í𝑚 other words, there are 𝑐 1 , . . . , 𝑐 𝑚 ∈ 𝐾 not all zero with 𝑖=1 𝑐 𝑖 𝑎 𝑖j = 0 for all j. They Í𝑚 satisfy 𝑖=1 𝑐 𝑖 𝑓𝑖 = 0. Hence, 𝑓1 , . . . , 𝑓𝑚 are linearly dependent over 𝐾. It follows that 𝐾 (t) and 𝐾˜ are linearly disjoint over 𝐾, as claimed. □

3.3 Separable Extensions We generalize the notion of “separable algebraic extension” to arbitrary field extensions. Let 𝐾 be a field of positive characteristic 𝑝. The field generated over 𝐾 by the 𝑝th roots of all elements of 𝐾 is denoted 𝐾 1/ 𝑝 . We denote the maximal purely ∞ inseparable extension of 𝐾 by 𝐾ins (or 𝐾 1/ 𝑝 ). Let 𝐹 be a finitely generated extension of 𝐾. A collection 𝑡 1 , . . . , 𝑡𝑟 ∈ 𝐹 of elements algebraically independent over 𝐾 is a separating transcendence basis if 𝐹/𝐾 (𝑡1 , . . . , 𝑡𝑟 ) is a finite separable extension.

Lemma 3.3.1 An extension 𝐹 of a field 𝐾 is separable if it satisfies one of the following equivalent conditions: (a) 𝐹 is linearly disjoint from 𝐾ins over 𝐾. (b) 𝐹 is linearly disjoint from 𝐾 1/ 𝑝 over 𝐾. (c) Every finitely generated extension 𝐸 of 𝐾 which is contained in 𝐹 has a separating transcendence basis. Moreover, a separating transcendence basis can be selected from a given set of generators for 𝐹/𝐾. Proof. The implication “(a) ⇒ (b)” is an immediate consequence of the tower property (Lemma 3.1.3). Suppose that (c) holds. Let 𝐸 be a finitely generated extension of 𝐾 in 𝐹 and let t = (𝑡1 , . . . , 𝑡𝑟 ) be a separating transcendence basis for 𝐸/𝐾. By Lemma 3.2.1(b), ˜ hence also from 𝐾ins . Since 𝐸/𝐾 (t) is 𝐾 (t) is linearly disjoint over 𝐾 from 𝐾, separable, 𝐸 is linearly disjoint from 𝐾ins (t) over 𝐾 (t). It follows from Lemma 3.1.3 that 𝐸 and 𝐾ins are linearly disjoint over 𝐾. Therefore, 𝐹 is linearly disjoint from 𝐾ins over 𝐾, so (a) holds. For “(b) ⇒ (c)” see [Lan64, p. 54]. Lemma 22.2.4 gives a constructive proof. □ In particular, every separable algebraic extension satisfies conditions (a), (b), and (c) of Lemma 3.3.1. Now apply the rules of linear disjointness to conclude the following result.

3.4 Regular Extensions

49

Corollary 3.3.2 The following rules hold: (a) If 𝐸/𝐾 and 𝐹/𝐸 are separable extensions, then 𝐹/𝐾 is also separable. (b) If 𝐹/𝐾 is a separable extension, then 𝐸/𝐾 is separable for every field 𝐾 ⊆ 𝐸 ⊆ 𝐹. (c) Every extension of a perfect field is separable. (d) If 𝐸/𝐾 is a purely inseparable extension and 𝐹/𝐾 is a separable extension, then 𝐸 and 𝐹 are linearly disjoint over 𝐾. Example 3.3.3 (A separable tower does not imply separable steps.) Consider the tower of fields F 𝑝 ⊂ F 𝑝 (𝑡 𝑝 ) ⊂ F 𝑝 (𝑡), where 𝑡 is transcendental over F 𝑝 . The extension F 𝑝 (𝑡)/F 𝑝 is separable, but F 𝑝 (𝑡)/F 𝑝 (𝑡 𝑝 ) is not. Example 3.3.4 (Roquette) It is well known for a tower 𝐾 ⊆ 𝐸 ⊆ 𝐹 of algebraic extensions of fields that 𝐹/𝐾 is separable if and only if 𝐸/𝐾 and 𝐹/𝐸 are separable. Moreover, if 𝐸/𝐾 and 𝐸 ′/𝐾 are separable, then so is 𝐸 𝐸 ′/𝐾 [Lan97, p. 241, Thm. 4.5]. Example 3.3.3 shows that the first statement is not true for non-algebraic separable extensions. Similarly, the second statement is in general false for non-algebraic separable extension. Indeed, choose a purely inseparable element 𝑎 of degree greater than 1 over a field 𝐾. Let 𝑥 be a transcendental element over 𝐾 and set 𝑦 = 𝑎 + 𝑥. Then, 𝑦 is transcendental over 𝐾, so both 𝐾 (𝑥) and 𝐾 (𝑦) are separable extensions over 𝐾. However, 𝐾 ⊂ 𝐾 (𝑎) ⊆ 𝐾 (𝑥, 𝑦), so by Corollary 3.3.2(b), 𝐾 (𝑥, 𝑦) = 𝐾 (𝑥)𝐾 (𝑦) is not a separable extension of 𝐾.

3.4 Regular Extensions Finitely generated regular extensions characterize absolutely irreducible varieties (Section [REF] 10.2). Lemma 3.4.1 A field extension 𝐹/𝐾 is regular if it satisfies one of the following equivalent conditions: (a) 𝐹/𝐾 is separable and 𝐾 is algebraically closed in 𝐹. (b) 𝐹 is linearly disjoint from 𝐾˜ over 𝐾. Proof. The implication “(b) ⇒ (a)” is immediate from the definition. To prove “(a) ⇒ (b)”, it suffices to assume that 𝐹/𝐾 is finitely generated. Then, 𝐹/𝐾 has a separating transcendence basis, 𝑡 1 , . . . , 𝑡𝑟 . That basis is also a separating transcendence basis for the extension 𝐹𝐾sep /𝐾sep . Since 𝐾˜ = (𝐾sep )ins , Lemma 3.3.1 implies that 𝐹𝐾sep is linearly disjoint from 𝐾˜ over 𝐾sep . Also, 𝐾sep /𝐾 is a Galois extension and 𝐹 ∩𝐾sep = 𝐾. Hence, 𝐹 is linearly disjoint from 𝐾sep over 𝐾. Therefore, by Lemma 3.1.3, 𝐹 is linearly disjoint from 𝐾˜ over 𝐾. □ Statements (a), (b), and (c) of the following result follow from Lemma 3.4.1 and the tower property of linear disjointness.

50

3 Linear Disjointness

Corollary 3.4.2 (a) If 𝐸/𝐾 and 𝐹/𝐸 are regular extensions, then 𝐹/𝐾 is regular. (b) If 𝐹/𝐾 is a regular extension, then 𝐸/𝐾 is regular for every field 𝐸 lying between 𝐾 and 𝐹. (c) Every extension of an algebraically closed field is regular. (d) Let 𝑚 be a cardinal number Ð and 𝐾 𝛼 , 𝛼 ≤ 𝑚, an ascending transfinite sequence of fields such that 𝐾 𝛾 = 𝛼 0. ˜ Let 𝑥1 , . . . , 𝑥 𝑛 be elements of 𝐹 that are Í𝑛linearly dependent over 𝐾. Thus, there ˜ exist 𝑎 1 , . . . , 𝑎 𝑛 ∈ 𝐾, not all zero, such that Í𝑖=1 𝑎 𝑖 𝑥 𝑖 = 0. Choose a power 𝑞 of 𝑝 such 𝑛 𝑎 𝑖𝑞 𝑥 𝑖𝑞 = 0 shows that 𝑥 1𝑞 , . . . , 𝑥 𝑛𝑞 are that 𝑎 1𝑞 , . . . , 𝑎 𝑞𝑛 ∈ 𝐾sep . Then, the relation 𝑖=1 linearly dependent over 𝐾sep . Since 𝐹/𝐾 is primary, 𝑥1𝑞 , . . . , 𝑥 𝑛𝑞Íare linearly dependent 𝑛 over 𝐾. Thus, there exist 𝑏 1 , . . . , 𝑏 𝑛 ∈ 𝐾, not all zero, with 𝑖=1 𝑏 𝑖 𝑥 𝑖𝑞 = 0. Taking Í𝑛 1/𝑞 𝑏 𝑖 𝑥 𝑖 = 0, that is 𝑥1 , . . . , 𝑥 𝑛 are linearly dependent over 𝐾ins . 𝑞th roots, we get 𝑖=1 Since 𝐹/𝐾 is separable, 𝑥 1 , . . . , 𝑥 𝑛 are linearly dependent over 𝐾. This proves that 𝐹/𝐾 is regular. □ Lemma 3.5.4 (a) Let 𝐸 be a primary extension of a field 𝐾 which is algebraically independent from an extension 𝐹 of 𝐾. Then, 𝐸 𝐹 is a primary extension of 𝐹. (b) If two primary extensions 𝐸 and 𝐹 of 𝐾 are algebraically independent, then 𝐸 𝐹/𝐾 is primary. Proof. Assertion (b) follows from (a) and from Corollary 3.5.2(a). To prove (a), choose a transcendence basis 𝑇 for 𝐸/𝐾 and let 𝑀 be the maximal separable extension of 𝐾 (𝑇) in 𝐸. Then, 𝑀 is a separable and primary extension of 𝐾. Hence, by Lemma 3.5.3, it is regular. Also, 𝑀 is algebraically independent from 𝐹sep over 𝐾. By Lemma 3.4.7, 𝑀 𝐹 is linearly disjoint from 𝐹sep over 𝐹. Since 𝐸 𝐹 is a purely inseparable extension of 𝑀 𝐹, it is linearly disjoint from 𝑀 𝐹sep . It follows that 𝐸 𝐹 is linearly disjoint from 𝐹sep over 𝐹; that is, 𝐸 𝐹 is a primary extension of 𝐹. □

3.6 The Imperfect Degree of a Field We classify fields of positive characteristic by their imperfect degree and characterize those fields for which every finite extension has a primitive element as fields of imperfect degree 1. Let 𝐹 be a field of positive characteristic 𝑝. Consider a subfield 𝐹0 of 𝐹 that contains the field 𝐹 𝑝 of all 𝑝th powers in 𝐹. Observe that for 𝑥1 , . . . , 𝑥 𝑛 ∈ 𝐹, the set of monomials 0 ≤ 𝑖1 , . . . , 𝑖 𝑛 ≤ 𝑝 − 1, (3.6) 𝑥1𝑖1 · · · 𝑥 𝑛𝑖𝑛 , generates 𝐹0 (x) as a vector space over 𝐹0 . Hence, [𝐹0 (x) : 𝐹0 ] ≤ 𝑝 𝑛 . If [𝐹0 (x) : 𝐹0 ] = 𝑝 𝑛 , then 𝑥1 , . . . , 𝑥 𝑛 are said to be 𝑝-independent over 𝐹0 . Equivalently, each of the fields 𝐹0 (𝑥1 ), . . . , 𝐹0 (𝑥 𝑛 ) has degree 𝑝 over 𝐹0 and they are linearly

56

3 Linear Disjointness

disjoint over 𝐹0 . This means that the set of monomials (3.6) is linearly independent over 𝐹0 . A subset 𝐵 of 𝐹 is 𝑝-independent over 𝐹0 if every finite subset of 𝐵 is 𝑝-independent over 𝐹0 . If in addition 𝐹0 (𝐵) = 𝐹, then 𝐵 is said to be a 𝑝-basis for 𝐹 over 𝐹0 . As in the theory of vector spaces, each maximal 𝑝-independent subset of 𝐹 over 𝐹0 is a 𝑝-basis for 𝐹 over 𝐹0 . If 𝑥 1 , . . . , 𝑥 𝑛 ∈ 𝐹 are 𝑝-independent over 𝐹 𝑝 , we call them 𝑝-independent elements of 𝐹. The 𝑝-power 𝑝 𝑛 = [𝐹 : 𝐹 𝑝 ] is then the imperfect degree of 𝐹, 𝑛 is the imperfect exponent of 𝐹, and we say that 𝐹 is 𝑛-imperfect. Thus, a perfect field has imperfect exponent 0. Both quantities are infinite if [𝐹 : 𝐹 𝑝 ] = ∞. In this case 𝐹 is ∞-imperfect. The following result is an analog of the Steinitz exchange lemma for finitely generated vector spaces over fields. Lemma 3.6.1 (Exchange Principle) Let 𝐹0 be a subfield of 𝐹 which contains 𝐹 𝑝 . (a) Let 𝑥 1 , . . . , 𝑥 𝑚 , 𝑦 1 , . . . , 𝑦 𝑛 ∈ 𝐹 be such that 𝑥1 , . . . , 𝑥 𝑚 are 𝑝-independent over 𝐹0 and 𝑥1 , . . . , 𝑥 𝑚 ∈ 𝐹0 (𝑦 1 , . . . , 𝑦 𝑛 ). Then, 𝑚 ≤ 𝑛, and there is a reordering of 𝑦 1 , . . . , 𝑦 𝑛 so that 𝑦 1 , . . . , 𝑦 𝑚 ∈ 𝐹0 (𝑥1 , . . . , 𝑥 𝑚 , 𝑦 𝑚+1 , . . . , 𝑦 𝑛 ). (b) Every subset of 𝐹 which is 𝑝-independent over 𝐹0 extends to a 𝑝-basis for 𝐹 over 𝐹0 . Proof. We use induction on 𝑚. Assume the lemma is true for 𝑚 = 𝑘. Thus, after a reordering of 𝑦 1 , . . . , 𝑦 𝑛 , we have 𝐹0 (𝑥1 , . . . , 𝑥 𝑘 , 𝑦 𝑘+1 , . . . , 𝑦 𝑛 ) = 𝐹0 (𝑦 1 , . . . , 𝑦 𝑛 ), so 𝑥 𝑘+1 ∈ 𝐹1 := 𝐹0 (𝑥 1 , . . . , 𝑥 𝑘 , 𝑦 𝑘+1 , . . . , 𝑦 𝑛 ). Then, [𝐹1 : 𝐹0 ] ≤ 𝑝 𝑛 and there exists an 𝑙 between 𝑘 + 1 and 𝑛 such that 𝑦 𝑙 ∈ 𝐹0 (𝑥1 , . . . , 𝑥 𝑘+1 , 𝑦 𝑘+1 , . . . , 𝑦 𝑙−1 ), because otherwise [𝐹1 : 𝐹0 ] ≥ 𝑝 𝑛+1 , a contradiction. Thus, 𝑦 𝑙 can be exchanged for 𝑥 𝑘+1 . This proves the first part of the lemma for 𝑚 = 𝑘 + 1. For the last part start from a subset 𝐴 of 𝐾 which is 𝑝-independent over 𝐹0 . Use Zorn’s lemma to prove the existence of a maximal subset 𝐵 of 𝐹 which contains 𝐴 □ and which is 𝑝-independent over 𝐹0 . Then, 𝐵 is a 𝑝-basis of 𝐹 over 𝐹0 . Lemma 3.6.2 Suppose that 𝐹 is a finitely generated extension of transcendence degree 𝑛 of a perfect field 𝐾 of positive characteristic 𝑝. Then, the imperfect exponent of 𝐹 is 𝑛. Proof. Choose a separating transcendence basis 𝑡1 , . . . , 𝑡 𝑛 for 𝐹/𝐾. Then, 𝐾 (t) 𝑝 = 𝐾 (t 𝑝 ) and 𝑡1 , . . . , 𝑡 𝑛 is a 𝑝-basis for 𝐾 (t)/𝐾 (t 𝑝 ); that is, [𝐾 (t) : 𝐾 (t 𝑝 )] = 𝑝 𝑛 . Since 𝐾 (t) is a purely inseparable extension of 𝐾 (t 𝑝 ) and 𝐹 𝑝 is a separable extension of 𝐾 (t 𝑝 ), these extensions of 𝐾 (t 𝑝 ) are linearly disjoint. Also, 𝐹 is both a separable extension and a purely inseparable extension of 𝐾 (t)𝐹 𝑝 . Hence, 𝐹 = 𝐾 (t)𝐹 𝑝 . □ Consequently, [𝐹 : 𝐹 𝑝 ] = [𝐾 (t) : 𝐾 (t 𝑝 )] = 𝑝 𝑛 , as claimed. Lemma 3.6.3 Let 𝐵 be a subset of 𝐹 which is 𝑝-independent over 𝐹 𝑝 and let 𝐹 ′ be a separable extension of 𝐹. Then, 𝐵 is 𝑝-independent over (𝐹 ′) 𝑝 . If, in addition, 𝐹 ′ is separable algebraic over 𝐹, then the imperfect degree of 𝐹 ′ is equal to that of 𝐹.

3.6 The Imperfect Degree of a Field

57

Proof. Assume without loss that 𝐵 consists of 𝑛 elements. Then, [(𝐹 ′) 𝑝 (𝐵) : (𝐹 ′) 𝑝 ] = [𝐹 𝑝 (𝐵) : 𝐹 𝑝 ] = 𝑝 𝑛 . Hence, 𝐵 is 𝑝-independent over (𝐹 ′) 𝑝 . Suppose now that 𝐹 ′/𝐹 is separably algebraic. Then, 𝐹 ′ is both separable and purely inseparable over 𝐹 (𝐹 ′) 𝑝 , so, 𝐹 ′ = 𝐹 (𝐹 ′) 𝑝 . Hence, [𝐹 ′ : (𝐹 ′) 𝑝 ] = [𝐹 : 𝐹 𝑝 ]. Therefore, the imperfect degree of 𝐹 ′ is equal to that of 𝐹.

□

Lemma 3.6.4 Let 𝐾 be a field of positive characteristic 𝑝, let 𝑎, 𝑏 1 , . . . , 𝑏 𝑚 be 𝑝independent elements of 𝐾, and let 𝑥1 , . . . , 𝑥 𝑚 be algebraically independent over 𝐾. Suppose that 𝑦 1 , . . . , 𝑦 𝑚 satisfy 𝑎𝑥𝑖𝑝 + 𝑏 𝑖 𝑦 𝑖𝑝 = 1,

𝑖 = 1, . . . , 𝑚.

(3.7)

Then, 𝐾 is algebraically closed in 𝐾𝑚 := 𝐾 (x, y). Proof. We use induction on 𝑚. Part A: 𝑚 = 1. Let 𝑥 = 𝑥 1 , 𝑦 = 𝑦 1 , and 𝑏 = 𝑏 1 and assume that 𝑢 is a nonzero element of 𝐾1 = 𝐾 (𝑥, 𝑦) which is algebraic over 𝐾. Then, 𝑢 is also algebraic over (3.7)

𝐾 (𝑎 1/ 𝑝 , 𝑏 1/ 𝑝 ). But, 𝐾 (𝑥, 𝑦, 𝑎 1/ 𝑝 , 𝑏 1/ 𝑝 ) = 𝐾 (𝑥, 𝑎 1/ 𝑝 , 𝑏 1/ 𝑝 ) is a purely transcendental extension of 𝐾 (𝑎 1/ 𝑝 , 𝑏 1/ 𝑝 ). Hence, 𝑢 ∈ 𝐾 (𝑎 1/ 𝑝 , 𝑏 1/ 𝑝 ) and therefore 𝑢 𝑝 ∈ 𝐾. Since 𝑢 ∈ 𝐾 (𝑥, 𝑦) and [𝐾 (𝑥, 𝑦) : 𝐾 (𝑥)] ≤ 𝑝, we may write 𝑢=

ℎ0 (𝑥) ℎ1 (𝑥) ℎ 𝑘 (𝑥) 𝑘 + 𝑦 +···+ 𝑦 ℎ(𝑥) ℎ(𝑥) ℎ(𝑥)

(3.8)

with 𝑘 ≤ 𝑝 − 1, ℎ(𝑥), ℎ0 (𝑥), . . . , ℎ 𝑘 (𝑥) ∈ 𝐾 [𝑥] and ℎ(𝑥), ℎ 𝑘 (𝑥) ≠ 0. With no loss we may assume that 𝑥 does not divide the greatest common divisor of ℎ(𝑥), ℎ0 (𝑥), . . . , ℎ 𝑘 (𝑥). Raise (3.8) to the 𝑝th power, multiply it by ℎ(𝑥) 𝑝 and substitute 𝑦 𝑝 = (1 − 𝑎𝑥 𝑝 )𝑏 −1 to obtain: (ℎ(𝑥)𝑢) 𝑝 = ℎ0 (𝑥) 𝑝 + ℎ1 (𝑥) 𝑝 (1 − 𝑎𝑥 𝑝 )𝑏 −1 + · · · + ℎ 𝑘 (𝑥) 𝑝 (1 − 𝑎𝑥 𝑝 ) 𝑘 𝑏 −𝑘 .

(3.9)

If ℎ(0) = 0, then the substitution 𝑥 = 0 in (3.9) gives 0 = ℎ0 (0) 𝑝 + ℎ1 (0) 𝑝 𝑏 −1 + · · · + ℎ 𝑘 (0) 𝑝 𝑏 −𝑘 . Therefore, ℎ0 (0) = ℎ1 (0) = · · · = ℎ 𝑘 (0) = 0, contrary to assumption. Thus, we may assume that ℎ(0) ≠ 0. Then, the substitution 𝑥 = 0 in (3.9) shows that 𝑢 ∈ 𝐾 (𝑏 1/ 𝑝 ). Similarly, 𝑢 ∈ 𝐾 (𝑎 1/ 𝑝 ). Since 𝑎 and 𝑏 are 𝑝-independent in 𝐾, 𝑢 ∈ 𝐾 (𝑎 1/ 𝑝 ) ∩ 𝐾 (𝑏 1/ 𝑝 ) = 𝐾. Thus, 𝐾 is algebraically closed in 𝐾 (𝑥, 𝑦). Part B: Induction. Assume that the lemma is true for 𝑚−1. Then, 𝐾 is algebraically closed in 𝐾𝑚−1 = 𝐾 (𝑥1 , . . . , 𝑥 𝑚−1 , 𝑦 1 , . . . , 𝑦 𝑚−1 ). If we prove that 𝑎 and 𝑏 𝑚 are 𝑝independent in 𝐾𝑚−1 , then with 𝐾𝑚−1 replacing 𝐾 in Part A, 𝐾𝑚−1 is algebraically closed in 𝐾𝑚 , so 𝐾 is algebraically closed in 𝐾𝑚 . Since 𝑥1 , . . . , 𝑥 𝑚 are algebraically independent over 𝐾, the field 𝑝 1/ 𝑝 𝐾 (𝑎 1/ 𝑝 , 𝑏 1/ 1 , . . . , 𝑏 𝑚 ) is linearly disjoint from 𝐸 𝑚−1 = 𝐾 (𝑥 1 , . . . , 𝑥 𝑚−1 ) over 𝐾. Thus, 𝑝 1/ 𝑝 𝑚+1 [𝐸 𝑚−1 (𝑎 1/ 𝑝 , 𝑏 1/ . (3.10) 1 , . . . , 𝑏 𝑚 ) : 𝐸 𝑚−1 ] = 𝑝

58

3 Linear Disjointness

Also, from (3.7), 𝐾𝑚−1 = 𝐸 𝑚−1 (𝑦 1 , . . . , 𝑦 𝑚−1 )

and

𝑝 𝑝 1/ 𝑝 1/ 𝑝 𝐾𝑚−1 (𝑎 1/ 𝑝 , 𝑏 1/ , 𝑏 1 , . . . , 𝑏 1/ 𝑚 ). 𝑚 ) = 𝐸 𝑚−1 (𝑎

Thus, [𝐾𝑚−1 : 𝐸 𝑚−1 ] ≤ 𝑝 𝑚−1

and

𝑝 2 [𝐾𝑚−1 (𝑎 1/ 𝑝 , 𝑏 1/ 𝑚 ) : 𝐾 𝑚−1 ] ≤ 𝑝 .

(3.11)

Combine (3.10) and (3.11) to conclude that (3.11) consists of equalities. In particular, □ 𝑎 and 𝑏 𝑚 are 𝑝-independent in 𝐾𝑚−1 . Lemma 3.6.5 The following conditions on a field 𝐾 are equivalent: (a) The imperfect exponent of 𝐾 is at most 1. (b) Every finite extension of 𝐾 has a primitive element. (c) If 𝐾 is algebraically closed in a field extension 𝐹, then 𝐹 is regular over 𝐾. Proof. If 𝐾 is perfect, then (a), (b), and (c) are true. Therefore, we may assume char(𝐾) = 𝑝 > 0 and 𝐾 is imperfect. Proof of “(a) =⇒ (b)”. By assumption, [𝐾 1/ 𝑝 : 𝐾] = [𝐾 : 𝐾 𝑝 ] = 𝑝. Hence, 𝐾1 = 𝐾 1/ 𝑝 is the unique purely inseparable extension of 𝐾 of degree 𝑝. Moreover, 𝑛 𝐾1 = 𝐾 (𝑎 1/ 𝑝 ) for some 𝑎 ∈ 𝐾, so 𝐾𝑛 = 𝐾 (𝑎 1/ 𝑝 ) is a purely inseparable extension of 𝐾 of degree 𝑝 𝑛 . Assume that for each 𝑚 ≤ 𝑛, 𝐾𝑚 is the unique purely inseparable extension of 𝐾 of degree 𝑝 𝑚 . Let 𝐿 be a purely inseparable extension of 𝐾 of degree 𝑝 𝑛+1 . If we prove that 𝐿 = 𝐾𝑛+1 , then we may conclude by induction that each finite purely inseparable extension of 𝐾 has a primitive element. To this end choose 𝑥 ∈ 𝐿 ∖ 𝐾𝑛 . Let 𝑚 be the smallest positive integer with 𝑚 𝑥 𝑝 ∈ 𝐾. Then, 𝐾 (𝑥) is a purely inseparable extension of 𝐾 of degree 𝑝 𝑚 . If 𝑚 ≤ 𝑛, then by the induction hypothesis 𝐾 (𝑥) = 𝐾𝑚 ⊆ 𝐾𝑛 , so 𝑥 ∈ 𝐾𝑛 . This contradiction proves that 𝑚 = 𝑛 + 1 and 𝐿 = 𝐾 (𝑥). The same argument implies that 𝑥 𝑝 ∈ 𝐾𝑛 . Hence, with 𝑞 = 𝑝 𝑛 , we have 𝑥 𝑝 = Í𝑞−1 𝑖/ 𝑝 𝑛 for some 𝑐 , . . . , 𝑐 0 𝑞−1 ∈ 𝐾. Therefore, 𝑖=0 𝑐 𝑖 𝑎 𝑥=

𝑞−1 ∑︁

𝑝 𝑖/ 𝑝 𝑐1/ 𝑖 𝑎

𝑛+1

∈ 𝐾1 (𝑎 1/ 𝑝

𝑛+1

) = 𝐾𝑛+1 .

𝑖=0

It follows that 𝐿 ⊆ 𝐾𝑛+1 . As both fields have degree 𝑝 𝑛+1 over 𝐾, they coincide, as desired. Now let 𝐸 be a finite extension of 𝐾. Denote the maximal separable extension of 𝐾 in 𝐸 by 𝐸 0 . By the primitive element theorem, 𝐸 0 = 𝐾 (𝑥). Since 𝐸 0 is both separable and purely inseparable over 𝐾 𝐸 0𝑝 we have 𝐸 0 = 𝐾 𝐸 0𝑝 . Therefore, [𝐸 0 : 𝐸 0𝑝 ] = [𝐾 : 𝐾 𝑝 ] = 𝑝. Apply the first part of the proof to 𝐸 0 and conclude that 𝐸 = 𝐸 0 (𝑦), for some element 𝑦. Thus, 𝐸 = 𝐾 (𝑥, 𝑦) with 𝑥 separable over 𝐾. By [Wae91b, §6.10], 𝐸/𝐾 has a primitive element Proof of “(b) =⇒ (c)”. Let 𝐾 (𝑥) be a finite extension of 𝐾 and let 𝑓 = irr(𝑥, 𝐾). If 𝐾 is algebraically closed in 𝐹, then 𝑓 remains irreducible over 𝐹. Otherwise, its factors would have coefficients algebraic over 𝐾 and in 𝐹, and therefore in 𝐾. Thus, 𝐹 is linearly disjoint from 𝐾 (𝑥) over 𝐾. Hence, (b) implies that 𝐹 is regular over 𝐾.

3.6 The Imperfect Degree of a Field

59

Proof of “(c) =⇒ (a)”. Assume 𝑎 and 𝑏 are 𝑝-independent elements of 𝐾. Then, [𝐾 (𝑎 1/ 𝑝 , 𝑏 1/ 𝑝 ) : 𝐾] = 𝑝 2 . Let 𝑥 and 𝑦 be transcendental elements over 𝐾 with 𝑎𝑥 𝑝 + 𝑏𝑦 𝑝 = 1. Put 𝐹 = 𝐾 (𝑥, 𝑦). By Lemma 3.6.4, 𝐾 is algebraically closed in 𝐹. Hence, by (c), 𝐹 is regular over 𝐾. Therefore, [𝐹 (𝑎 1/ 𝑝 , 𝑏 1/ 𝑝 ) : 𝐹] = [𝐾 (𝑎 1/ 𝑝 , 𝑏 1/ 𝑝 ) : 𝐾] = 𝑝 2 . On the other hand, 𝐹 (𝑎 1/ 𝑝 ) = 𝐹 (𝑏 1/ 𝑝 ), so [𝐹 (𝑎 1/ 𝑝 , 𝑏 1/ 𝑝 ) : 𝐹] ≤ 𝑝. This contradiction proves that the imperfect exponent of 𝐾 is at most 1. □ Remark 3.6.6 (Relative algebraic closedness does not imply regularity) Let 𝐾 be a field of positive characteristic 𝑝. Suppose that 𝐾 has 𝑝-independent elements 𝑎, 𝑏 (e.g. 𝐾 = F 𝑝 (𝑡, 𝑢) where 𝑡, 𝑢 are algebraically independent over F 𝑝 ). Let 𝑥, 𝑦 be transcendental elements over 𝐾 with 𝑎𝑥 𝑝 + 𝑏𝑦 𝑝 = 1. Put 𝐹 = 𝐾 (𝑥, 𝑦). The proof of “(c) =⇒ (a)” of Lemma 3.6.5 then shows that 𝐾 is algebraically closed in 𝐹 but 𝐹 is not linearly disjoint from 𝐾 1/ 𝑝 over 𝐾. Thus, 𝐹 is not a separable extension of 𝐾. A fortiori, 𝐹/𝐾 is not regular. We present another version of Lemma 3.6.4 in which all of the 𝑥𝑖 ’s coincide. However, we prove it only for 𝑛 = 2 and 𝑛 = 3. The best way to prove the general case is to use differentials, which we do not develop in this book. However, the case 𝑛 = 2 suffices for the proof that the theory 𝑇𝑅 introduced in Example 30.2.4 does not have the “amalgamation property”. Lemma 3.6.7 Let 𝐸 be a field of positive characteristic 𝑝, let 𝑛 ≥ 2 be an integer, let 𝑎 1 , . . . , 𝑎 𝑛 be 𝑝-independent elements of 𝐸, and let 𝑥 be an indeterminate. For 𝑖 = 𝑝 1/ 𝑝 2, . . . , 𝑛 let 𝑦 𝑖 = 𝑎 1/ and let 𝐿 = 𝐸 (𝑥, 𝑦 2 , . . . , 𝑦 𝑛 ). Then, 𝐸 is algebraically 𝑖−1 𝑥 + 𝑎 𝑖 closed in 𝐿. Moreover, for 𝑖 = 2, . . . , 𝑛 we have [𝐸 (𝑥, 𝑦 2 , . . . , 𝑦 𝑖 ) : 𝐸 (𝑥)] = 𝑝 𝑖−1 . Proof. As mentioned above, we prove the lemma only for 𝑛 = 3 (hence, also for 𝑛 = 2). Let 𝑎, 𝑏, 𝑐 be 𝑝-independent elements of 𝐸 and set 𝑦 = 𝑎 1/ 𝑝 𝑥 + 𝑏 1/ 𝑝 and 𝑧 = 𝑏 1/ 𝑝 𝑥 + 𝑐1/ 𝑝 . 𝐸′

𝐸 (𝑎 1/ 𝑝 , 𝑏 1/ 𝑝 , 𝑐1/ 𝑝 )

We set = and 𝐿 = 𝐸 (𝑥, 𝑦, 𝑧). Then, 𝐸 ′ (𝑥). Let 𝑢 be an element of 𝐿 which is algebraic over 𝐸.

𝐿𝑝

(3.12) ⊆ 𝐸 (𝑥) and 𝐿𝐸 ′ =

Claim A: 𝑢 ∈ 𝐸 (𝑏 1/ 𝑝 , 𝑐1/ 𝑝 ). Indeed, since 𝐿 𝑝 ⊆ 𝐿 (𝑥), we may write ∑︁ ℎ𝑖 𝑗 (𝑥) 𝑖 𝑗 𝑢= 𝑦𝑧 , ℎ(𝑥) 1≤𝑖, 𝑗 ≤ 𝑝−1

(3.13)

where ℎ, ℎ𝑖 𝑗 ∈ 𝐸 [𝑋] are polynomials such that ℎ(𝑥) ≠ 0 and 0 is not a simultaneous zero of ℎ and all of the ℎ𝑖 𝑗 ’s. By definition, 𝑦, 𝑧 ∈ 𝐸 ′ (𝑥), hence 𝑢 ∈ 𝐸 ′ (𝑥). Since 𝑢 is algebraic over 𝐸, it is also algebraic over 𝐸 ′, so 𝑢 ∈ 𝐸 ′. The specialization 𝑥 ↦→ 0 extends to an 𝐸 ′-rational place 𝜑 of 𝐸 ′ (𝑥). By (3.12), 𝜑(𝑦) = 𝑏 1/ 𝑝 and 𝜑(𝑧) = 𝑐1/ 𝑝 . If ℎ(0) = 0, then we multiply (3.13) by ℎ(𝑥) and apply 𝜑 on the resulting equality Í to find that 0 = 1≤𝑖, 𝑗 ≤ 𝑝−1 ℎ𝑖 𝑗 (0)𝑏 𝑖/ 𝑝 𝑐 𝑗/ 𝑝 . Since 𝑏, 𝑐 are 𝑝-independent elements of 𝐸, we have ℎ𝑖 𝑗 (0) = 0 for all 0 ≤ 𝑖, 𝑗 ≤ 𝑝 −1. This contradiction to the assumption we have made on the polynomials ℎ and ℎ𝑖 𝑗 implies that ℎ(0) ≠ 0.

60

3 Linear Disjointness

Applying 𝜑 on (3.13) again, we now find that ∑︁ ℎ𝑖 𝑗 (0) 𝑖/ 𝑝 𝑗/ 𝑝 𝑢= 𝑏 𝑐 ∈ 𝐸 (𝑏 1/ 𝑝 , 𝑐1/ 𝑝 ), ℎ(0) 1≤𝑖, 𝑗 ≤ 𝑝−1 as claimed. Claim B: 𝑢 ∈ 𝐸 (𝑏 1/ 𝑝 , 𝑎 1/ 𝑝 ). Indeed, we may rewrite (3.12) in the form 𝑧 1 𝑦 1 = 𝑐1/ 𝑝 + 𝑏 1/ 𝑝 and = 𝑏 1/ 𝑝 + 𝑎 1/ 𝑝 𝑥 𝑥 𝑥 𝑥 and note that 𝐸 ( 1𝑥 ) = 𝐸 (𝑥) and 𝐸 1𝑥 , 𝑥𝑧 , 𝑦𝑥 ) = 𝐿. By Claim A, applied to 1𝑥 , 𝑐, 𝑏, 𝑎 rather than to 𝑥, 𝑎, 𝑏, 𝑐, we have 𝑢 ∈ 𝐸 (𝑏 1/ 𝑝 , 𝑎 1/ 𝑝 ), as claimed. Claim C: 𝑢 ∈ 𝐸. The assumption on 𝑎, 𝑏, 𝑐 implies that the fields 𝐸 (𝑎 1/ 𝑝 ), 𝐸 (𝑏 1/ 𝑝 ), and 𝐸 (𝑐1/ 𝑝 ) are linearly disjoint over 𝐸. It follows from Claim A and Claim B that 𝑢 ∈ 𝐸 (𝑏 1/ 𝑝 ). In addition, [𝐸 (𝑏 1/ 𝑝 ) : 𝐸] = 𝑝. Hence, if 𝑢 ∉ 𝐸, then 𝐸 (𝑢) = 𝐸 (𝑏 1/ 𝑝 ). Therefore, 𝑏 1/ 𝑝 ∈ 𝐸 (𝑢). It follows from (3.12) that also 1 𝑎 1/ 𝑝 , 𝑐1/ 𝑝 ∈ 𝐸 (𝑢). Therefore, 𝐸 ′ = 𝐸 (𝑢) = 𝐸 (𝑏 𝑝 ), so [𝐸 ′ : 𝐸] = 𝑝. However, since 𝑎, 𝑏, 𝑐 are 𝑝-independent elements of 𝐸, we have [𝐸 ′ : 𝐸] = 𝑝 3 . We conclude from this contradiction that 𝑢 ∈ 𝐸, as claimed. In addition, [𝐸 ′ (𝑥) : 𝐸 (𝑥)] = 𝑝 3 .

(3.14)

Finally, (3.12)

𝐸 (𝑥, 𝑦, 𝑧, 𝑎 1/ 𝑝 ) = 𝐸 (𝑥, 𝑎 1/ 𝑝 , 𝑏 1/ 𝑝 , 𝑐1/ 𝑝 ) = 𝐸 ′ (𝑥). It follows from (3.14) that [𝐸 (𝑥, 𝑦) : 𝐸 (𝑥)] = 𝑝 and [𝐸 (𝑥, 𝑦, 𝑧) : 𝐸 (𝑥, 𝑦)] = 𝑝, as claimed. □

3.7 Derivatives We develop a criterion for a finitely generated field extension of positive characteristic 𝑝 to be separable in terms of derivatives.. Definition 3.7.1 A map 𝐷: 𝐹 → 𝐹 is called a derivation of a field 𝐹 if 𝐷 (𝑥 + 𝑦) = 𝐷 (𝑥) + 𝐷 (𝑦) and 𝐷 (𝑥𝑦) = 𝐷 (𝑥)𝑦 + 𝑥𝐷 (𝑦) for all 𝑥, 𝑦 ∈ 𝐹. If 𝐷 vanishes on a subfield 𝐾 of 𝐹, then 𝐷 is a derivation of 𝐹 over 𝐾 (or a 𝐾-derivation). Let 𝐹 (𝑥) be a field extension of 𝐹 and 𝑓 ∈ 𝐹 [𝑋]. Suppose that 𝐷 extends to 𝐹 (𝑥). Then, 𝐷 satisfies the classical chain rule: 𝐷 ( 𝑓 (𝑥)) = (𝐷 𝑓 ) (𝑥) + 𝑓 ′ (𝑥)𝐷 (𝑥),

(3.15)

where 𝐷 𝑓 is the polynomial obtained by applying 𝐷 to the coefficients of 𝑓 and 𝑓 ′ is the usual derivative of 𝑓 . There are three cases:

3.7 Derivatives

61

Case 1: 𝑥 is separably algebraic over 𝐹. Then, with 𝑓 = irr(𝑥, 𝐹), 𝑓 ′ (𝑥) ≠ 0. By (3.15), 0 = (𝐷 𝑓 ) (𝑥) + 𝑓 ′ (𝑥)𝐷 (𝑥). Thus, 𝐷 extends uniquely to 𝐹 (𝑥). Case 2: 𝑥 is transcendental. Then, 𝐷 extends to 𝐹 (𝑥) by rule (3.15) and 𝐷 (𝑥) may be chosen arbitrarily. 𝑚

Case 3: 𝑥 satisfies 𝑥 𝑝 = 𝑎 ∈ 𝐹, for some 𝑚. Then, 𝐷 extends to 𝐹 (𝑥) if and only if 𝐷 (𝑎) = 0. In this case 𝐷 (𝑥) may be chosen arbitrarily. Lemma 3.7.2 A necessary and sufficient condition for a finitely generated extension 𝐹/𝐾 to be separably algebraic is that 0 is the only 𝐾-derivation of 𝐹. Proof. Necessity follows from Case 1. Now suppose that 𝐹/𝐾 is not separably algebraic. Then, we may write 𝐹 = 𝐾 (𝑥1 , . . . , 𝑥 𝑛 ) such that 𝑥𝑖 is transcendental over 𝐾 (𝑥 1 , . . . , 𝑥 𝑖−1 ) for 𝑖 = 1, . . . , 𝑘, 𝑥𝑖 is separably algebraic over 𝐾 (𝑥 1 , . . . , 𝑥 𝑖−1 ) for 𝑖 = 𝑘 + 1, . . . , 𝑙, and 𝑥𝑖 is purely inseparable over 𝐾 (𝑥1 , . . . , 𝑥 𝑖−1 ) for 𝑖 = 𝑙 + 1, . . . , 𝑛. Moreover, either 𝑛 > 𝑙 or 𝑛 = 𝑙 and 𝑘 > 0. If 𝑛 > 𝑙, then Case 1 allows us to extend the zero derivation of 𝐾 (𝑥1 , . . . , 𝑥 𝑛−1 ) to a nonzero derivation of 𝐹. If 𝑛 = 𝑙 and 𝑘 > 0, then by Case 2, the zero derivation of 𝐾 (𝑥1 , . . . , 𝑥 𝑘−1 ) extends to a nonzero derivation 𝐷 of 𝐾 (𝑥1 , . . . , 𝑥 𝑘 ). Applying Case 3 several times, we may then extend 𝐷 to a derivation □ of 𝐹 in more than one way. Lemma 3.7.3 Let 𝐹/𝐾 be a finitely generated extension of positive characteristic 𝑝 and transcendence degree 𝑛. Then, 𝐹/𝐾 is separable if and only if [𝐹 : 𝐾 𝐹 𝑝 ] = 𝑝 𝑛 . In this case 𝑡1 , . . . , 𝑡 𝑛 form a 𝑝-basis for 𝐹 over 𝐾 𝐹 𝑝 if and only if they form a separating transcendence basis for 𝐹/𝐾. Proof. Suppose first that [𝐹 : 𝐾 𝐹 𝑝 ] = 𝑝 𝑛 . Let 𝑡 1 , . . . , 𝑡 𝑛 be a 𝑝-basis for 𝐹/𝐾 𝐹 𝑝 . Every derivation 𝐷 of 𝐹 vanishes on 𝐹 𝑝 . If 𝐷 vanishes on 𝐾 (t), it vanishes on 𝐹 = 𝐾 (t) · 𝐹 𝑝 . By Lemma 3.7.2, 𝐹/𝐾 (t) is separably algebraic and 𝑡 1 , . . . , 𝑡 𝑛 is a separating transcendence basis for 𝐹/𝐾. Conversely, suppose that 𝐹/𝐾 is separable. Let 𝑡1 , . . . , 𝑡 𝑛 be a separating transcendence basis for 𝐹/𝐾. The extension 𝐹/𝐾 (t) · 𝐾 𝐹 𝑝 is both separable and purely inseparable. Hence, 𝐹 = 𝐾 (t) · 𝐾 𝐹 𝑝 . Since 𝐹 𝑝 /𝐾 (t) 𝑝 is separably algebraic and since 𝐾 (t 𝑝 )𝐹 𝑝 = 𝐾 𝐹 𝑝 , we conclude that 𝐾 𝐹 𝑝 /𝐾 (t 𝑝 ) is separably algebraic. 𝐾 (t)

𝐹

𝐾 (t 𝑝 )

𝐾𝐹 𝑝

𝐾 (t) 𝑝

𝐹𝑝

Therefore, 𝐾 𝐹 𝑝 is linearly disjoint from 𝐾 (t) over 𝐾 (t 𝑝 ), and [𝐹 : 𝐾 𝐹 𝑝 ] = [𝐾 (t) : 𝐾 (t 𝑝 )] = 𝑝 𝑛 . Moreover, t is a 𝑝-basis for 𝐹/𝐾 𝐹 𝑝 . □

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3 Linear Disjointness

Corollary 3.7.4 Let 𝐹/𝐾 be a finitely generated separable extension of positive characteristic 𝑝 and let 𝑡 ∈ 𝐹. (a) If there exists a derivation 𝐷 of 𝐹/𝐾 such that 𝐷 (𝑡) ≠ 0, then 𝐹 is a separable extension of 𝐾 (𝑡). (b) If 𝑡 is transcendental over 𝐾 and 𝐹/𝐾 (𝑡) is separable, then there exists a derivation 𝐷 of 𝐹/𝐾 such that 𝐷 (𝑡) ≠ 0. Proof of (a). By assumption, 𝑡 ∉ 𝐾 𝐹 𝑝 . Let 𝑛 = trans.deg(𝐹/𝐾). By Lemma 3.7.3, [𝐹 : 𝐾 𝐹 𝑝 ] = 𝑝 𝑛 . Hence, 𝑡 can be extended to a 𝑝-basis 𝑡, 𝑡2 , . . . , 𝑡 𝑛 for 𝐹/𝐾 𝐹 𝑝 . Again, by Lemma 3.7.3, 𝑡, 𝑡2 , . . . , 𝑡 𝑛 is a separating transcendence basis for 𝐹/𝐾. Therefore, 𝐹 is a separable extension of 𝐾 (𝑡). Proof of (b). Let 𝑡 2 , . . . , 𝑡 𝑛 be a separating transcendence basis for 𝐹/𝐾 (𝑡). By Case 2, there exists a derivation 𝐷 0 of 𝐾 (𝑡, 𝑡2 , . . . , 𝑡 𝑛 )/𝐾 such that 𝐷 0 (𝑡) = 1, 𝐷 0 (𝑡2 ) = 0, □ . . . , 𝐷 0 (𝑡 𝑛 ) = 0. By Case 1, 𝐷 0 extends to a derivation 𝐷 of 𝐹/𝐾.

Exercises 1. Let 𝐿, 𝐾1 , . . . , 𝐾𝑛 be extensions of a field 𝐾. Let 𝐿 𝑖 = 𝐾𝑖 𝐿, 𝑖 = 1, . . . , 𝑛. Suppose that 𝐾𝑖 is linearly disjoint from 𝐿 over 𝐾 for 𝑖 = 1, . . . , 𝑛 and 𝐿 1 , . . . , 𝐿 𝑛 are linearly disjoint over 𝐿. Prove that 𝐾1 , . . . , 𝐾𝑛 are linearly disjoint over 𝐾. 2. Let 𝐸 be a regular extension of a perfect field 𝐾 and let 𝐹 be a purely inseparable extension of 𝐸. Prove that 𝐹/𝐾 is a regular extension. 3. Let 𝐾 be a field algebraically closed in an extension 𝐹. Prove that 𝐾 (𝑥) is linearly ˜ disjoint from 𝐹 for every 𝑥 ∈ 𝐾. Hint: Check the irreducibility of irr(𝑥, 𝐾) over 𝐹. 4. Prove that a field extension 𝐹/𝐾 is primary if and only if 𝐹𝐾ins ∩ 𝐾˜ = 𝐾ins . Use this criterion to give another proof to Corollary 3.5.2(a). 5. Let 𝐹/𝐾 be a finitely generated field extension of characteristic 𝑝 > 0 and of 𝑛 transcendence degree 1. Prove that for each positive integer 𝑛, 𝐾 𝐹 𝑝 is the unique subfield 𝐸 of 𝐹 which contains 𝐾 such that 𝐹/𝐸 is a purely inseparable extension of degree 𝑝 𝑛 . 6. Let 𝐾 ⊆ 𝐿 ⊆ 𝐹 and 𝐾 ⊆ 𝐸 be fields. Suppose that a field 𝐸 ′ contains 𝐸 𝐿, is contained in 𝐸 𝐹, and is linearly disjoint from 𝐹 over 𝐿. Use Lemma 3.1.3 to prove that 𝐸 ′ = 𝐸 𝐿.

Chapter 4

Algebraic Function Fields of One Variable

Sections 4.1–4.4 survey the theory of functions of one variable; the Riemann–Roch Theorem; properties of holomorphy rings of function fields; and extensions of the field of constants. Sections 4.5–4.6 include a proof of the Riemann–Hurwitz formula. The rest of the chapter applies these concepts and results to hyperelliptic curves.

4.1 Function Fields of One Variable Call a field extension 𝐹/𝐾 an algebraic function field of one variable (briefly, function field of one variable or just function field) if these conditions hold: (4.1a) The transcendence degree of 𝐹/𝐾 is 1. (4.1b) 𝐹/𝐾 is finitely generated and regular. In this case there exists a 𝑡 ∈ 𝐹, transcendental over 𝐾, with 𝐹/𝐾 (𝑡) a finite separable extension. All valuations of 𝐾 (𝑡) trivial on 𝐾 are discrete (Example 2.2.1), so their extensions to 𝐹 are also discrete (Proposition 2.3.2). Also, since the residue fields of the valuations of 𝐾 (𝑡) are finite extensions of 𝐾, so are the residue fields of the valuations of 𝐹. We define a prime divisor of 𝐹/𝐾 as an equivalence class of 𝐾-places of 𝐹. For 𝔭 a prime divisor of 𝐹/𝐾, choose a place 𝜑𝔭 in 𝔭. Then, 𝜑𝔭 fixes the elements of 𝐾 and maps 𝐹 into 𝐾˜ ∪ {∞}. Denote its residue field by 𝐹¯𝔭 . As mentioned above, 𝐹¯𝔭 is a finite extension of 𝐾 of degree deg(𝔭) = [ 𝐹¯𝔭 : 𝐾] which we call the degree of 𝔭. Also, choose a valuation ord𝔭 corresponding to 𝔭 and normalize it so that ord𝔭 (𝐹 × ) = Z. Each element 𝜋 of 𝐹 with ord𝔭 (𝜋) = 1 is a local parameter of 𝐹 at 𝔭. Denote the free Abelian group that the prime divisors of 𝐹/𝐾 generate by Div(𝐹/𝐾). Each element 𝔞 of Div(𝐹/𝐾) is a divisor of 𝐹/𝐾. It has the form Í 𝔞 = 𝛼𝔭𝔭, where 𝔭 runs over the prime divisors of 𝐹/𝐾, the 𝛼𝔭 are integers and all but finitely many of them are zero. Define a homomorphism ord𝔭 : Div(𝐹/𝐾) → Z by ord𝔭 (𝔞) = 𝛼𝔭 . © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_4

63

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4 Algebraic Function Fields of One Variable

The symbol ord𝔭 appears for two distinct functions. We show these uses are compatible as follows. Introduce the divisor of a nonzero element 𝑥 of 𝐹 as div(𝑥) = Í ord𝔭 (𝑥)𝔭. Since ord𝔭 (𝑥) = 0 for all but finitely many 𝔭 the right-hand side is well defined. If 𝑥 is a constant (i.e. 𝑥 ∈ 𝐾), then div(𝑥) = 0. If 𝑥 ∈ 𝐹 ∖ 𝐾, then the place of 𝐾 (𝑥) taking 𝑥 to 0 (resp. to ∞) has finitely many extensions to the field 𝐹 (Proposition 2.3.2). Equivalence classes of these extensions are the zeros (resp. poles) of 𝑥. Thus, div(𝑥) is not zero if 𝑥 is not a constant. Define the divisor of zeros and the divisor of poles of 𝑥 as follows: ∑︁ ∑︁ ord𝔭 (𝑥)𝔭, div0 (𝑥) = ord𝔭 (𝑥)𝔭. div∞ (𝑥) = − ord𝔭 ( 𝑥) 0

Then, div(𝑥) = div0 (𝑥) − div∞ (𝑥). In particular, ord𝔭 (𝑥) (using ord𝔭 as a valuation) is the same as ord𝔭 applied to the divisor div(𝑥). Í ord𝔭 (𝔞) deg(𝔭). Then, Define the degree deg(𝔞) of a divisor 𝔞 to be deg : Div(𝐹/𝐾) → Z is a homomorphism. Since deg(div0 (𝑥)) = deg(div∞ (𝑥)) = [𝐹 : 𝐾 (𝑥)]

(4.2)

for 𝑥 ∈ 𝐹 ∖ 𝐾 ([Deu73, p. 27] or Proposition 2.3.2), deg(div(𝑥)) = 0. To each divisor 𝔞 attach a vector space L (𝔞) over 𝐾: L (𝔞) = {𝑥 ∈ 𝐹 × | div(𝑥) + 𝔞 ≥ 0} ∪ 0. The phrase ‘𝔟 ≥ 0’ for a divisor 𝔟 means that ord𝔭 (𝔟) ≥ 0 for every prime divisor 𝔭. In this case we abuse our language and say that 𝔟 is a positive divisor. The 𝐾-vector space L (𝔞) is finite-dimensional [Deu73, p. 23] (Note: Deuring uses the notation 𝐿 (𝔞) for L (−𝔞).) Denote the nonnegative integer dim𝐾 L (𝔞) by dim(𝔞). Extend the notation 0 ≤ 𝔟 to a partial ordering on Div(𝐹/𝐾) by writing 𝔞 ≤ 𝔟 if ord𝔭 (𝔞) ≤ ord𝔭 (𝔟) for every prime divisor 𝔭 of 𝐹/𝐾. This implies L (𝔞) ⊆ L (𝔟). In this case, L (𝔞) ⊂ L (𝔟) is equivalent to dim(𝔞) < dim(𝔟). For divisors 𝔞1 , . . . , 𝔞 𝑛 of 𝐹/𝐾, write ∑︁
min ord𝔭 (𝔞1 ), . . . , ord𝔭 (𝑎 𝑛 ) 𝔭. min(𝔞1 , . . . , 𝔞 𝑛 ) = 𝔭

This is the maximal divisor of 𝐹/𝐾 less than or equal to 𝔞𝑖 for 𝑖 = 1, . . . , 𝑛. Similarly, define max(𝔞1 , . . . , 𝔞 𝑛 ). Call a divisor of the form div(𝑥), where 𝑥 ∈ 𝐹 × , a principal divisor. Since div(𝑥𝑦) = div(𝑥) + div(𝑦), the set of principal divisors of 𝐹/𝐾 is a subgroup of Div(𝐹/𝐾). The quotient group is called the group of divisor classes of 𝐹/𝐾. Denote it by C. Every element of C is a class of divisors. Call two divisors linearly equivalent if they differ by a principal divisor. Two linearly equivalent divisors have the same degree and the same dimension. This defines the degree and the dimension of a class of divisors. and 𝑥 1 , . . . , 𝑥 𝑚 ∈ 𝐹 × . Put 𝔞 = Lemma 4.1.1 Let 𝐹/𝐾 be a function field Ð Í ∞ 𝑚 𝑘=1 L (𝑘𝔞) is the integral closure 𝑖=1 div∞ (𝑥 𝑖 ) and 𝑅 = 𝐾 [𝑥 1 , . . . , 𝑥 𝑚 ]. Then, of 𝑅 in 𝐹.

4.2 The Riemann–Roch Theorem

65

Proof. Denote the integral closure of 𝑅 in 𝐹 by 𝑆 and the set of poles of 𝑥 1 , . . . , 𝑥 𝑚 by 𝑃. First consider a nonzero 𝑧 ∈ L (𝑘𝔞). Then, div(𝑧) + 𝑘𝔞 ≥ 0. Hence, ord𝔭 (𝑧) ≥ 0 for each prime divisor 𝔭 of 𝐹/𝐾 which is not in 𝑃; that is for each 𝔭 with 𝜑𝔭 finite on 𝑅. By Proposition 2.5.1, 𝑧 ∈ 𝑆. Conversely, let 𝑧 ∈ 𝑆. By Proposition 2.5.1, ord𝔭 (𝑧) ≥ 0 for each 𝔭 ∉ 𝑃. For each 𝔭 ∈ 𝑃 the set 𝐼𝔭 = {1 ≤ Í 𝑖 ≤ 𝑚 | ord𝔭 (𝑥 𝑖 ) < 0} is not empty. Choose a positive integer 𝑘 with ord𝔭 (𝑧) − 𝑘 𝑖 ∈𝐼𝔭 ord𝔭 (𝑥𝑖 ) ≥ 0 for all 𝔭 ∈ 𝑃. Then, 𝑧 ∈ L (𝑘𝔞). □

4.2 The Riemann–Roch Theorem The degree of a divisor 𝔞 of a function field 𝐹/𝐾 can often be read from its definition. It is more difficult to compute the dimension of 𝔞 directly from the definition. The Riemann–Roch theorem allows us in many cases to compute dim(𝔞) from deg(𝔞) and a constant of 𝐹/𝐾 called the “genus”. In this section we present the Riemann–Roch theorem in three forms and explain how to apply it in the computation of dim(𝔞). Repartition. Let 𝐹/𝐾 be a function field. A repartition of 𝐹/𝐾 is a function 𝛼 from prime divisors of 𝐹/𝐾 into 𝐹 (denote the image of 𝔭 by 𝛼𝔭 ) with ord𝔭 (𝛼𝔭 ) ≥ 0 for all but finitely many 𝔭. Denote the set of all repartitions of 𝐹/𝐾 by A. The following definitions turn A into an 𝐹-algebra: For 𝛼, 𝛽 ∈ A and 𝑥 ∈ 𝐹, (𝛼 + 𝛽)𝔭 = 𝛼𝔭 + 𝛽𝔭 , (𝛼𝛽)𝔭 = 𝛼𝔭 𝛽𝔭

and

(𝑥𝛼)𝔭 = 𝑥 · 𝛼𝔭 .

Extend ord𝔭 to A by ord𝔭 (𝛼) = ord𝔭 (𝛼𝔭 ). If 𝔞 is a divisor of 𝐹/𝐾, then Λ(𝔞) = {𝛼 ∈ A | ord𝔭 (𝛼) + ord𝔭 (𝔞) ≥ 0 for every 𝔭} is a vector space over 𝐾. Identify 𝐹 as a subset of A by the diagonal mapping. Then, Λ(𝔞) contains L (𝔞) and A/(Λ(𝔞) + 𝐹) is a finite-dimensional 𝐾-vector space. Moreover, there exists a nonnegative integer 𝑔, called the genus of 𝐹/𝐾, which is independent of 𝔞 such that dim(𝔞) − deg(𝔞) − 1 + 𝑔 = dim𝐾

A . Λ(𝔞) + 𝐹

(4.3)

This is the first form of the Riemann–Roch theorem [Deu73, p. 34]. Differential. The second form of the theorem interprets the right-hand side of (4.3) as the dimension of a space of differentials. A differential (also called a Weil differential) of the function field 𝐹/𝐾 is a 𝐾-linear map of A into 𝐾 that vanishes on some subspace of the form Λ(𝔞) + 𝐹, where 𝔞 is a divisor of 𝐹/𝐾. The set of all differentials that vanish on Λ(𝔞) + 𝐹, denoted Ω(𝔞), is the dual space of A/(Λ(𝔞) + 𝐹). We denote its dimension by 𝛿(𝔞) and rewrite (4.3) as dim(𝔞) = deg(𝔞) + 1 − 𝑔 + 𝛿(𝔞).

(4.4)

The third form of the theorem identifies 𝛿(𝔞) with the dimension of a divisor of 𝐹/𝐾 related to 𝔞. To each nonzero differential 𝜔 of 𝐹/𝐾 there corresponds a unique divisor of 𝐹/𝐾, denoted div(𝜔), with div(𝜔) ≥ 𝔞 if and only if 𝜔 ∈ Ω(𝔞) [Deu73, p. 39]. We call div(𝜔) the divisor of 𝜔.

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Theorem 4.2.1 Let 𝜔 be a nonzero differential of an algebraic function field 𝐹/𝐾 of one variable of genus 𝑔. Put 𝔴 = div(𝜔). Then, for every divisor 𝔞 of 𝐹/𝐾 we have dim(𝔞) = deg(𝔞) + 1 − 𝑔 + dim(𝔴 − 𝔞).

(4.5)

Proof. For each 𝑥 ∈ 𝐹 × define a 𝐾-linear map 𝑥𝜔: A → 𝐾 by (𝑥𝜔) (𝛼) = 𝜔(𝑥𝛼). If 𝜔 vanishes on Λ(𝔞), then 𝑥𝜔 vanishes on Λ(div(𝑥) + 𝔞). Hence, 𝑥𝜔 is a differential and div(𝑥𝜔) = div(𝑥) + div(𝜔). The map 𝑥 ↦→ 𝑥𝜔 gives an isomorphism of L (div(𝜔) − 𝔞) onto Ω(𝔞). Hence, dim(div(𝜔) − 𝔞) = dim(Ω(𝔞)) = 𝛿(𝔞). Thus, (4.5) is a reformulation of (4.4). □ If 𝜔 ′ is another nonzero differential, then there exists an 𝑥 ∈ 𝐹 × with 𝜔 ′ = 𝑥𝜔. Thus, div(𝜔 ′) = div(𝑥) + div(𝜔) [Deu73, p. 37] and the divisor of a differential is determined up to a principal divisor. The class of these divisors is the canonical class of 𝐹/𝐾. Each of its elements is a canonical divisor. A few additional observations help us apply the third form of the Riemann-Roch theorem in (4.5): Lemma 4.2.2 The following hold for each function field 𝐹/𝐾 of genus 𝑔: (a) dim(0) = 1, and if 𝑥 ∈ 𝐹 × , then deg(div(𝑥)) = 0 and dim(div(𝑥)) = 1. (b) If 𝔴 is a canonical divisor, then dim(𝔴) = 𝑔 and deg(𝔴) = 2𝑔 − 2. Hence, dim𝐾 (A/Λ(0) + 𝐹) = dim(Ω(0)) = dim(L (𝔴)) = 𝑔. (c) deg(𝔞) < 0 implies dim(𝔞) = 0. (d) deg(𝔞) > 2𝑔 − 2 implies 𝛿(𝔞) = 0 and dim(𝔞) = deg(𝔞) + 1 − 𝑔 . Proof of (a). L (0) = 𝐾 because each 𝑥 ∈ 𝐹 ∖ 𝐾 has a pole. Similarly, L (div(𝑥)) = 𝐾𝑥 −1 . Proof of (b). Take 𝔞 = 0 and then 𝔞 = 𝔴 in (4.5). Proof of (c). Use that deg(div(𝑥)) = 0. Proof of (d). By (b), deg(𝔴 − 𝔞) < 0. Hence, by (c), 𝛿(𝔞) = dim(𝔴 − 𝔞) = 0. Thus, (4.5) simplifies to dim(𝔞) = deg(𝔞) + 1 − 𝑔. □ Here is our first application of Lemma 4.2.2(d): Lemma 4.2.3 Let 𝐹/𝐾 be a function field of genus 𝑔, 𝔭 a prime divisor of 𝐹/𝐾, and 𝑛 a positive integer satisfying (𝑛 − 1) deg(𝔭) > 2𝑔 − 2. Then, there exists an 𝑥 ∈ 𝐹 × with div∞ (𝑥) = 𝑛𝔭. Proof. By Lemma 4.2.2(d), dim((𝑛 − 1)𝔭) = (𝑛 − 1) deg(𝔭) + 1 − 𝑔 and dim(𝑛𝔭) = 𝑛 deg(𝔭) + 1 − 𝑔. Hence, dim((𝑛 − 1)𝔭) < dim(𝑛𝔭), so L ((𝑛 − 1)𝔭) ⊂ L (𝑛𝔭). If 𝑥 ∈ L (𝑛𝔭) ∖ L ((𝑛 − 1)𝔭), then div0 (𝑥) + 𝑛𝔭 ≥ div∞ (𝑥). Since div0 (𝑥) and div∞ (𝑥) have no common component, we have 𝑛𝔭 ≥ div∞ (𝑥). Similarly, (𝑛 − 1)𝔭 ̸ ≥ div∞ (𝑥). □ Therefore, div∞ (𝑥) = 𝑛𝔭. Example 4.2.4 (Rational function field) Let 𝐹 = 𝐾 (𝑡), where 𝑡 is a transcendental element over 𝐾. Denote the prime divisor corresponding to the valuation ord∞ (Example 2.2.1(c)) by 𝔭∞ . We determine the linear space L (𝑛𝔭∞ ), where 𝑛 is a positive integer.

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If 𝑢 ∈ L (𝑛𝔭∞ ), then ord𝔭 (𝑢) ≥ 0 for every prime divisor 𝔭 ≠ 𝔭∞ . This means 𝑢 ∈ 𝐾 [𝑡]. Also ord𝔭∞ (𝑢) ≥ −𝑛, so the degree of 𝑢 as a polynomial in 𝑡 is bounded by 𝑛. Thus, L (𝑛𝔭∞ ) = {𝑢 ∈ 𝐾 [𝑡] | deg(𝑢) ≤ 𝑛} and dim(𝑛𝔭∞ ) = 𝑛 + 1. For 𝑛 > 2𝑔 − 2, Lemma 4.2.2(d) gives 𝑛 + 1 = 𝑛 − 𝑔 + 1. Hence, 𝑔 = 0. Conversely, let 𝐹/𝐾 be a function field of one variable of genus 0. Assume that 𝐹/𝐾 has a prime divisor 𝔭 of degree 1 (e.g. if 𝐾 is algebraically closed). Lemma 4.2.3 gives 𝑡 ∈ 𝐹 × with div∞ (𝑡) = 𝔭. By Section 4.1, [𝐹 : 𝐾 (𝑡)] = deg(div∞ (𝑡)) = 1. Consequently, 𝐹 = 𝐾 (𝑡). Í𝑛 𝑎 𝑖 𝑡 𝑖 with 𝑎 𝑖 ∈ 𝐾 and 𝑎 𝑛 ≠ 0. For each Consider now a polynomial 𝑓 (𝑡) = 𝑖=0 𝑖 𝑛 𝑖 < 𝑛 we have ord∞ (𝑎 𝑛 𝑡 ) < ord∞ (𝑎 𝑖 𝑡 ), so ord∞ ( 𝑓 (𝑡)) = −𝑛. For 𝔭 ≠ 𝔭∞ we have ord𝔭 (𝑡) ≥ 0, hence ord𝔭 ( 𝑓 (𝑡)) ≥ 0. Thus, div∞ ( 𝑓 (𝑡)) = deg( 𝑓 )𝔭∞ . (𝑡) Suppose that 𝑔(𝑡) ∈ 𝐾 [𝑡] is relatively prime to 𝑓 (𝑡). Put 𝑢 = 𝑔𝑓 (𝑡) . Then, 𝑡 is a root of the polynomial ℎ(𝑢, 𝑇) = 𝑢 · 𝑔(𝑇) − 𝑓 (𝑇). Since ℎ(𝑢, 𝑇) is linear in 𝑢 and gcd( 𝑓 , 𝑔) = 1, ℎ(𝑢, 𝑇) is irreducible over 𝐾 (𝑢), and ℎ(𝑢, 𝑡) = 0. In addition, deg𝑇 (ℎ) = max(deg( 𝑓 ), deg(𝑔)). Therefore,
[𝐾 (𝑡) : 𝐾 (𝑢)] = max deg( 𝑓 ), deg(𝑔) . (4.6) In particular, suppose 𝐾 (𝑢) = 𝐾 (𝑡). Then, 𝑓 (𝑡) and 𝑔(𝑡) are linear and relatively prime. This means, 𝑢 = 𝑎𝑡+𝑏 𝑐𝑡+𝑑 with 𝑎, 𝑏, 𝑐, 𝑑 ∈ 𝐾 and 𝑎𝑑 − 𝑏𝑐 ≠ 0.

4.3 Holomorphy Rings Let 𝐹/𝐾 be a function field of genus 𝑔. Denote the set of prime divisors of 𝐹/𝐾 by R. For each 𝔭 ∈ R let 𝑂𝔭 = {𝑥 ∈ 𝐹 | ord𝔭 (𝑥) ≥ 0} be the corresponding valuation Ñ ring. To every subset 𝑆 of R we attach the holomorphy ring 𝑂 𝑆 = 𝔭∈𝑆 𝑂𝔭 . By definition, 𝐾 ⊆ 𝑂 𝑆 . If 𝑆 is empty, then, by definition, 𝑂 𝑆 = 𝐹. If 𝑆 = R, then the elements of 𝑂 𝑆 have no poles. They are therefore constants. Thus, 𝑂 R = 𝐾. The case where 𝑆 is a nonempty proper subset of R requires a strengthening of the weak approximation theorem (Proposition 2.1.1). Proposition 4.3.1 (Strong Approximation Theorem) Let 𝑆 be a finite subset of R. Consider 𝔮 ∈ R ∖ 𝑆 and let 𝑆 ′ = 𝑆 ∪ {𝔮}. Suppose that for each 𝔭 ∈ 𝑆 we have an 𝑥𝔭 ∈ 𝐹 and a positive integer 𝑚𝔭 . Then, there exists an 𝑥 ∈ 𝐹 with ord𝔭 (𝑥 − 𝑥𝔭 ) = 𝑚𝔭 for each 𝔭 ∈ 𝑆 and ord𝔭 (𝑥) ≥ 0 for each 𝔭 ∈ R ∖ 𝑆 ′ .

(4.7)

Moreover, if 𝑚 is an integer with 𝑚 · deg(𝔮) > 2𝑔 − 2 +

∑︁

(𝑚𝔭 + 1) deg(𝔭),

𝔭∈𝑆

then 𝑥 can be chosen such that, in addition to (4.7), it satisfies ord𝔮 (𝑥) ≥ −𝑚.

(4.8)

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Proof. Let 𝑚 be a positive integer satisfying (4.8). Consider the divisor 𝔞 = 𝑚𝔮 − Í 𝔭∈𝑆 𝑚𝔭𝔭. Then, deg(𝔞) > 2𝑔 − 2. By Lemma 4.2.2(d), 𝛿(𝔞) = 0, so A = 𝐹 + Λ(𝔞). Define 𝜉 ∈ A by 𝜉𝔭 = 𝑥𝔭 for 𝔭 ∈ 𝑆 and 𝜉𝔭 = 0 for 𝔭 ∈ R ∖ 𝑆. Then, there exists a 𝑦 ∈ 𝐹 such that 𝑦 − 𝜉 ∈ Λ(𝔞). Thus, ord𝔭 (𝑦 − 𝑥𝔭 ) ≥ 𝑚𝔭 for 𝔭 ∈ 𝑆, ord𝔮 (𝑦) ≥ −𝑚, (4.9) and ord𝔭 (𝑦) ≥ 0 for 𝔭 ∈ R ∖ 𝑆 ′ . Í Now consider the divisor 𝔟 = 𝑚𝔮 − 𝔭∈𝑆 (𝑚𝔭 + 1)𝔭. For each 𝔭 ∈ 𝑆 we have deg(𝔟 + 𝔭) > deg(𝔟) > 2𝑔 − 2 (by (4.8)). By Lemma 4.2.2(d), dim(𝔟 + 𝔭) = deg(𝔟 + 𝔭) + 1 − 𝑔 > deg(𝔟) + 1 − 𝑔 = dim(𝔟). Hence, L (𝔟) ⊂ L (𝔟 + 𝔭). Choose 𝑧𝔭 ∈ L (𝔟 + 𝔭) ∖ L (𝔟). Then, ord𝔭 (𝑧𝔭 ) = 𝑚𝔭 . Also, ord𝔭′ (𝑧𝔭 ) ≥ 𝑚𝔭′ + 1 if 𝔭 ′ ∈ 𝑆 ∖{𝔭}, ord𝔮 (𝑧𝔭 ) ≥ −𝑚, and ord𝔭′ (𝑧𝔭 ) ≥ 0 for 𝔭 ′ ∈ R ∖ 𝑆 ′. Í Let 𝑃 = {𝔭 ∈ 𝑆 | ord𝔭 (𝑦 − 𝑥𝔭 ) > 𝑚𝔭 } and let 𝑄 = 𝑆 ∖ 𝑃. Then, 𝑧 = 𝔭∈𝑃 𝑧𝔭 has the following property: ord𝔭 (𝑧) = 𝑚𝔭 if 𝔭 ∈ 𝑃, ord𝔭 (𝑧) ≥ 𝑚𝔭 + 1 if 𝔭 ∈ 𝑄, ord𝔮 (𝑧) ≥ −𝑚, and ord𝔭 (𝑧) ≥ 0 for 𝔭 ∈ R ∖ 𝑆 ′ .

(4.10)

Combine (4.9) and (4.10) to see that 𝑥 = 𝑧 + 𝑦 satisfies ord𝔭 (𝑥 − 𝑥𝔭 ) = 𝑚𝔭 , for 𝔭 ∈ 𝑆, □ ord𝔮 (𝑥) ≥ −𝑚, and ord𝔭 (𝑥) ≥ 0 for 𝔭 ∈ R ∖ 𝑆 ′. If 𝔭 belongs to a subset 𝑆 of R, then 𝑂 𝑆 ⊆ 𝑂𝔭 . Also, 𝑃 = {𝑥 ∈ 𝑂 𝑆 | ord𝔭 (𝑥) > 0} is a prime ideal of 𝑂 𝑆 , the center of 𝔭 at 𝑂 𝑆 . Denote the local ring of 𝑂 𝑆 at 𝑃 by 𝑂 𝑆,𝑃 . Proposition 4.3.2 (Holomorphy Ring Theorem) Let 𝑆 be a nonempty proper subset of R. Then, 𝑆 has these properties: (a) Quot(𝑂 𝑆 ) = 𝐹. (b) If 𝔭 ∈ 𝑆 and 𝑃 is the center of 𝔭 at 𝑂 𝑆 , then 𝑂𝔭 = 𝑂 𝑆,𝑃 . (c) If 𝔮 ∈ R ∖ 𝑆, then 𝑂 𝑆 ̸ ⊆ 𝑂𝔮 . (d) Every nonzero prime ideal of 𝑂 𝑆 is the center of a prime 𝔭 ∈ 𝑆. (e) Distinct primes in 𝑆 have distinct centers at 𝑂 𝑆 , and the center of each 𝔭 ∈ 𝑆 is a maximal ideal of 𝑂 𝑆 . (f) 𝑂 𝑆 is a Dedekind domain. Proof of (a). Consider 𝑧 ∈ 𝐹 ∖ 𝐾. Since there are only finitely many 𝔭 ∈ 𝑆 with ord𝔭 (𝑧) < 0 and since R ∖ 𝑆 is nonempty, Proposition 4.3.1 gives 𝑦 ∈ 𝐹 such that the following holds for each 𝔭 ∈ 𝑆: If ord𝔭 (𝑧) < 0, then ord𝔭 (𝑦 − 𝑧−1 ) = ord𝔭 (𝑧−1 ) + 1, so ord𝔭 (𝑦) = −ord𝔭 (𝑧); while if ord𝔭 (𝑧) ≥ 0, then ord𝔭 (𝑦) ≥ 0. Let 𝑥 = 𝑦𝑧. Then, both 𝑥 and 𝑦 belong to 𝑂 𝑆 . If 𝑧 ∈ 𝑂 𝑆 , then 𝑧 ∈ Quot(𝑂 𝑆 ). If 𝑧 ∉ 𝑂 𝑆 , then there is a 𝔭 ∈ 𝑆 with ord𝔭 (𝑧) < 0. Hence, ord𝔭 (𝑦) = ord𝔭 (𝑧−1 ) ≠ ∞, so 𝑦 ≠ 0. Therefore, 𝑧 = 𝑥𝑦 −1 ∈ Quot(𝑂 𝑆 ). Proof of (b). Let 𝑧 ∈ 𝑂𝔭 ∖ 𝐾. As in the proof of (a), there exists a 𝑦 ∈ 𝐹 such that ord𝔭 (𝑦) = 0, ord𝔭′ (𝑦) = −ord𝔭′ (𝑧) if 𝔭 ′ ∈ 𝑆 ∖{𝔭} and ord𝔭′ (𝑧) < 0, while ord𝔭′ (𝑦) ≥ 0 if 𝔭 ′ ∈ 𝑆 ∖{𝔭} and ord𝔭′ (𝑧) ≥ 0. Therefore, 𝑥 = 𝑦𝑧 is in 𝑂 𝑆 and 𝑧 ∈ 𝑂 𝑆, 𝑃 . Since the inclusion 𝑂 𝑆, 𝑃 ⊆ 𝑂𝔭 is clear, 𝑂𝔭 = 𝑂 𝑆,𝑃 .

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69

Proof of (c). If 𝑆 ∪ {𝔮} = R and 𝑂 𝑆 ⊆ 𝑂𝔮 , then 𝑂 𝑆 = 𝑂 R = 𝐾, a contradiction to (a). Therefore, assume that 𝑆 ∪ {𝔮} is a proper subset of R. By Proposition 4.3.1, there exists an 𝑥 ∈ 𝐹 such that ord𝔮 (𝑥) = −1 and ord𝔭 (𝑥) ≥ 0 for each 𝔭 ∈ 𝑆. This element belongs to 𝑂 𝑆 but not to 𝑂𝔮 . Proof of (d). Let 𝑃 be a nonzero prime ideal of 𝑂 𝑆 . Proposition 2.3.1 extends the quotient map 𝑂 𝑆 → 𝑂 𝑆 /𝑃 to a place 𝜑 of 𝐹 trivial on 𝐾. Let 𝔭 be the prime divisor of 𝐹/𝐾 which is defined by 𝜑. Then, 𝑂 𝑆 ⊆ 𝑂𝔭 and 𝑃 = {𝑥 ∈ 𝑂 𝑆 | ord𝔭 (𝑥) > 0}. By (c), 𝔭 ∈ 𝑆. Proof of (e). Let 𝔭 and 𝔭 ′ be two distinct prime divisors in 𝑆. By the strong approximation theorem (Proposition 4.3.1), there exists an 𝑥 ∈ 𝑂 𝑆 with ord𝔭 (𝑥) > 0 and ord𝔭′ (𝑥) = 0. This means that the center 𝑃 of 𝔭 is not contained in the center of 𝔭 ′. The maximality of 𝑃 now follows from (d). Proof of (f). Let 𝔮 ∈ R ∖ 𝑆, let 𝑆 ′ = R ∖{𝔮}, and choose 𝔭 ′ ∈ 𝑆 ′. Then, 𝑆 ⊆ 𝑆 ′, so 𝑂 𝑆′ ⊆ 𝑂 𝑆 . Hence, by Proposition 2.5.7, it suffices to prove that 𝑂 𝑆′ is a Dedekind domain. Indeed, the strong approximation theorem gives 𝑥 ∈ 𝑂 𝑆′ with ord𝔭′ (𝑥) > 0. Since 𝑥 must have a pole, it must beÑ𝔮. Thus, 𝐾 [𝑥] ⊆ 𝑂𝔭 if and only if 𝔭 ∈ 𝑆 ′. Therefore, by Proposition 2.5.1, 𝑂 𝑆′ = 𝔭∈𝑆′ 𝑂𝔭 is the integral closure of 𝐾 [𝑥] in 𝐹. Since 𝐾 [𝑥] is a Dedekind domain, 𝑂 𝑆′ is also a Dedekind domain (Proposition 2.5.6). □ Corollary 4.3.3 The following conditions are equivalent for a nonempty subset 𝑆 of R: (a) 𝑆 = R; (b) 𝑂 𝑆 = 𝐾; and (c) 𝑂 𝑆 is a field. Proof. Every nonconstant element of 𝐹 has a pole. Therefore, (a) implies (b). The implication “(b) ⇒ (c)” is trivial. To prove that (c) implies (a), note that 𝑂 𝑆 ≠ 𝐹, since 𝑆 is nonempty and for each 𝔭 ∈ 𝑆 there exists an 𝑥 ∈ 𝐹 with ord(𝑥) < 0. If 𝑆 ≠ R, then, by Proposition 4.3.2(a), the quotient field of 𝑂 𝑆 is 𝐹. Hence, 𝑂 𝑆 is not □ a field. The following converse to Proposition 4.3.2 is useful: Proposition 4.3.4 Let 𝐹/𝐾 be a function field and 𝑅 a proper subring of 𝐹 containing 𝐾. Suppose that 𝑅 is integrally closed and Quot(𝑅) = 𝐹. Then, there is a nonempty proper subset 𝑆 of R with 𝑅 = 𝑂 𝑆 . Thus, 𝑅 is a Dedekind domain. Proof. Let 𝑆 be the set of all prime divisors of 𝐹/𝐾 which are finite on 𝑅. By Proposition 2.5.1, 𝑂 𝑆 = 𝑅. As 𝑅 ⊂ 𝐹, 𝑆 is nonempty. By Corollary 4.3.3, 𝑆 ≠ R. □ Hence, by Proposition 4.3.2(f), 𝑅 is a Dedekind domain.

4.4 Extensions of Function Fields Let 𝐸/𝐾 and 𝐹/𝐿 be algebraic function fields of one variable. We say that 𝐹/𝐿 is an extension of 𝐸/𝐾 if 𝐸 ⊆ 𝐹, 𝐾 ⊆ 𝐿, and 𝐿 ∩ 𝐸 = 𝐾. We call 𝐹/𝐿 a constant field extension of 𝐸/𝐾 if 𝐸 is linearly disjoint from 𝐿 over 𝐾 and 𝐹 = 𝐸 𝐿. Thus,

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in this case [𝐹 : 𝐸] = [𝐿 : 𝐾]. Recall that 𝐸/𝐾 is a regular extension (by (4.1b)). Hence, the linear disjointness of 𝐸 and 𝐿 over 𝐾 is automatic if 𝐿 is an algebraic extension of 𝐾. Likewise, 𝐸 is linearly disjoint from 𝐿 over 𝐾 if 𝐿 = 𝐾 (𝑥1 , . . . , 𝑥 𝑛 ) and 𝑥1 , . . . , 𝑥 𝑛 are algebraically independent over 𝐸 (Lemma 3.4.7). Example 4.4.1 Let 𝐸/𝐾 be an algebraic function field of one variable, let 𝐹 be a finite separable extension of 𝐸, and let 𝐿 be the algebraic closure of 𝐾 in 𝐹. By Lemma 3.4.5, 𝐹/𝐿 is a finitely generated regular extension of transcendence degree 1. Thus, 𝐹/𝐿 is an algebraic function field of one variable. Since 𝐸 is linearly disjoint from 𝐾˜ over 𝐾, we have 𝐿 ∩ 𝐸 = 𝐾. Hence, 𝐹/𝐿 is an extension of 𝐸/𝐾. Let 𝐹/𝐿 be an extension of 𝐸/𝐾, 𝔭 a prime divisor of 𝐸/𝐾, and 𝔓 a prime divisor of 𝐹/𝐿 that lies over 𝔭, that is ord𝔓 lies over ord𝔭 . Denote the ramification index of ord𝔓 over ord𝔭 by 𝑒𝔓/𝔭 . Since both ord𝔓 and ord𝔭 are discrete and normalized, ord𝔓 (𝑥) = 𝑒𝔓/𝔭 ord𝔭 (𝑥) for each 𝑥 ∈ 𝐸. Refer to 𝔓 as unramified over 𝐸 if ord𝔓 is unramified over 𝐸. If 𝐹 is separable algebraic over 𝐸, then only finitely many prime divisors of 𝐹/𝐿 are ramified over 𝐸 [Deu73, p. 111]. Over every prime divisor of 𝐸/𝐾 there lie only finitely many prime divisors of 𝐹/𝐿 [Deu73, p. 96]. Use this result to embed the group of divisors Div(𝐸/𝐾) of 𝐸/𝐾 into Div(𝐹/𝐿) as follows: For 𝔭 a prime divisor of 𝐸/𝐾 and 𝔓1 , . . . , 𝔓𝑑 the prime Í𝑑 divisors of 𝐹/𝐿 lying over 𝔭, map 𝔭 to the divisor 𝑖=1 𝑒𝔓𝑖 /𝔭 𝔓𝑖 of 𝐹/𝐿. Extend this map to Div(𝐸/𝐾) by linearity. The principal divisor of 𝑥 (in Div(𝐸/𝐾)) maps to the principal divisor of 𝑥 in Div(𝐹/𝐿), so there is no ambiguity in using div(𝑥) for that divisor. In particular, for every divisor 𝔞 of 𝐸/𝐾 we have L 𝐸 (𝔞) = 𝐸 ∩ L 𝐹 (𝔞). Suppose that [𝐹 : 𝐸] < ∞. If 𝐹/𝐸 is separable, apply (2.10) to conclude: 𝑑 ∑︁

𝑒𝔓𝑖 /𝔭 [ 𝐹¯𝔓𝑖 : 𝐸¯𝔭 ] = [𝐹 : 𝐸].

(4.11)

𝑖=1

Even if 𝐹/𝐸 is not separable, the following argument shows that (4.11) still holds: By Lemma 4.2.3, there
are an integer 𝑚 and 𝑥 ∈ 𝐸 with
div∞ (𝑥) = 𝑚𝔭. Apply the rules deg𝐹 div∞ (𝑥) = [𝐹 : 𝐿 (𝑥)] and deg𝐸 div∞ (𝑥) = [𝐸 : 𝐾 (𝑥)] to conclude that (4.11) is true in general [Deu73, p. 97, Thm.]. Next consider a function field 𝐹/𝐾 and let 𝐸 be a proper extension of 𝐾 contained in 𝐹. Since 𝐹 is linearly disjoint from 𝐾˜ over 𝐾, so is 𝐸. In particular, there is a
transcendental element 𝑥 in 𝐸. Then, [𝐹 : 𝐾 (𝑥)] = deg𝐹 div∞ (𝑥) < ∞. Thus, [𝐸 : 𝐾 (𝑥)] < ∞, 𝐸/𝐾 is a function field of one variable, and [𝐹 : 𝐸] < ∞. Lemma 4.4.2 Let 𝐸/𝐾 and 𝐹/𝐾 be algebraic function fields of one variable with 𝐸 ⊆ 𝐹. Then, deg𝐹 (𝔞) = [𝐹 : 𝐸] deg𝐸 (𝔞) for each 𝔞 ∈ Div(𝐸/𝐾). Proof. Assume by linearity that 𝔞 = 𝔭 is a prime divisor. Let 𝔓1 , . . . , 𝔓𝑑 be the prime divisors of 𝐹/𝐾 which lie over 𝔭. Then, 𝐾 ⊆ 𝐸¯𝔭 ⊆ 𝐹¯𝔓𝑖 , so deg𝐹 𝔓𝑖 = 𝑓𝔓𝑖 /𝔭 deg𝐸 (𝔭), 𝑖 = 1, . . . , 𝑑. By (4.11), deg𝐹 (𝔭) =

𝑑 ∑︁

𝑒𝔓𝑖 /𝔭 deg𝐹 (𝔓𝑖 )

𝑖=1

=

𝑑 ∑︁ 𝑖=1

𝑒𝔓𝑖 /𝔭 𝑓𝔓𝑖 /𝔭 deg𝐸 (𝔭) = [𝐹 : 𝐸] deg𝐸 (𝔭).□

4.4 Extensions of Function Fields

71

Proposition 4.4.3 Let 𝐹/𝐿 be a constant field extension of an algebraic function field 𝐸/𝐾 of one variable with 𝐿/𝐾 separable. Then: (a) For a divisor 𝔞 of 𝐸/𝐾, deg𝐸 (𝔞) = deg𝐹 (𝔞), dim𝐸 (𝔞) = dim𝐹 (𝔞), and L 𝐸 (𝔞)𝐿 = L 𝐹 (𝔞). (b) genus(𝐸/𝐾) = genus(𝐹/𝐿). (c) If 𝔭 and 𝔓 are respective prime divisors of 𝐸/𝐾 and 𝐹/𝐿, with 𝔓 lying over 𝔭, then 𝐹¯𝔓 = 𝐿 𝐸¯𝔭 . Moreover, if 𝐿/𝐾 is algebraic, then 𝔓 is unramified over 𝔭. (d) Let 𝑅 be an integrally closed subring of 𝐸 containing 𝐾. Suppose that 𝐿/𝐾 is algebraic. Then, 𝑅𝐿 is the integral closure of 𝑅 in 𝐹. (e) Let 𝔭 be a prime divisor of 𝐸/𝐾 of degree 1. Then, there is a unique prime divisor 𝔓 of 𝐹/𝐿 lying over 𝔭 and deg(𝔓) = 1. (f) Let 𝑥1 , . . . , 𝑥 𝑚 be elements of 𝐸. Denote the integral closure of 𝐾 [x] in 𝐸 by 𝑆. Then, 𝑆𝐿 is the integral closure of 𝐿 [x] in 𝐹. Proof of (a). [Deu73, p. 126, Thm.] shows deg𝐸 (𝔞) = deg𝐹 (𝔞). We show dim𝐸 (𝔞) = dim𝐹 (𝔞). Let 𝐵 be a basis of L 𝐸 (𝔞) over 𝐾. Then, 𝐵 is contained in L 𝐹 (𝔞) and is linearly independent over 𝐿. Therefore, dim𝐸 (𝔞) ≤ dim𝐹 (𝔞). Conversely, let 𝐵 ′ be a basis of L 𝐹 (𝔞). Then, 𝐵 ′ is contained in 𝐸 𝐿 0 for some finitely generated extension 𝐿 0 /𝐾 of 𝐿/𝐾. By [Deu73, p. 132, Thm. 2], dim𝐸 𝐿0 (𝔞) = dim𝐸 (𝔞). Hence, dim𝐹 (𝔞) = |𝐵 ′ | ≤ dim𝐸 𝐿0 (𝔞) = dim𝐸 (𝔞). Therefore, dim𝐸 (𝔞) = dim𝐹 (𝔞). It follows that L 𝐸 (𝔞)𝐿 ⊆ L 𝐹 (𝔞) have the same dimension over 𝐿, so L 𝐸 (𝔞)𝐿 = L 𝐹 (𝔞). Proof of (b). Let 𝑔 𝐸 = genus(𝐸/𝐾) and 𝑔𝐹 = genus(𝐹/𝐿). Choose a divisor 𝔞 of 𝐸/𝐾 with deg𝐸 (𝔞) > max(2𝑔 𝐸 − 2, 2𝑔𝐹 − 2). By Lemma 4.2.2(d), dim𝐸 (𝔞) = deg𝐸 (𝔞) + 1 − 𝑔 𝐸 and dim𝐹 (𝔞) = deg𝐹 (𝔞) + 1 − 𝑔𝐹 . Therefore, by (a), 𝑔 𝐸 = 𝑔𝐹 . Proof of (c). The general case reduces to the case where 𝐿/𝐾 is finitely generated. In this case [Deu73, p. 128, Thm.] proves that 𝐹¯𝔓 = 𝐿 𝐸¯ 𝑝 . Now assume that 𝐿/𝐾 is algebraic, choose a primitive element 𝑐 of 𝐿/𝐾, and let 𝑓 = irr(𝑐, 𝐾). By the preceding paragraph, 𝐹¯𝔓 = 𝐸¯𝔭 (𝑐). The reduction of 𝑓 modulo 𝔓 is 𝑓 itself. Since 𝐿/𝐾 is separable, 𝑓 is separable. Hence, by Lemma 2.3.4, 𝔓/𝔭 is unramified. Proof of (d). Assume without loss that [𝐿 : 𝐾] < ∞. Choose a basis 𝑤 1 , . . . , 𝑤 𝑛 of 𝐿/𝐾. Let 𝜎1 , . . . , 𝜎𝑛 be the 𝐾-embeddings of 𝐿 into 𝐾sep . Since 𝐸 is linearly disjoint from 𝐿 over 𝐾, each 𝜎𝑖 extends uniquely to an 𝐸-embedding of 𝐹 into 𝐸 sep . Í Now consider 𝑧 ∈ 𝐹 which Í is integral over 𝑅. Write 𝑧 = 𝑛𝑗=1 𝑎 𝑗 𝑤 𝑗 with 𝑎 1 , . . . , 𝑎 𝑛 ∈ 𝐸. Then, 𝜎𝑖 𝑧 = 𝑛𝑗=1 𝑎 𝑗 𝜎𝑖 𝑤 𝑗 , 𝑖 = 1, . . . , 𝑛. By Cramer’s rule, 𝑏

𝑏 Δ

𝑎 𝑗 = Δ𝑗 = Δ𝑗2 , where 𝑏 𝑗 is in the ring generated over 𝑅 by the 𝜎𝑘 𝑧’s and the 𝜎𝑘 𝑤 𝑙 ’s, and Δ = det(𝜎𝑘 𝑤 𝑙 )1≤𝑘,𝑙 ≤𝑛 ∈ 𝐾 × . By assumption, each 𝑏 𝑗 is integral over 𝑅. Since 𝐿/𝐾 is separable, Δ2 ∈ 𝐾 × [Lan97, p. 286, Cor. 5.4]. Hence, each 𝑎 𝑗 is integral over 𝑅. Since 𝑅 is integrally closed, 𝑎 𝑗 ∈ 𝑅, 𝑗 = 1, . . . , 𝑛. Consequently, 𝑧 ∈ 𝑅𝐿. Proof of (e). To prove (e), it suffices to consider two cases. In one case, 𝐿 = 𝐾 (𝑢) with 𝑢 purely transcendental over 𝐾. [Deu73, pp. 128–129] handles this case. The other case is when 𝐿/𝐾 is separable and finite. Let in this case 𝔓1 , . . . , 𝔓𝑑 be the

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4 Algebraic Function Fields of One Variable

prime divisors of 𝐹/𝐿 lying over 𝔭. By (c), 𝑒𝔓𝑖 /𝔭 = 1, 𝑖 = 1, . . . , 𝑚. By assumption, deg(𝔭) = 1, so 𝐸¯𝔭 = 𝐾. Hence, by (c), 𝐹¯𝔓𝑖 = 𝐿, 𝑖 = 1, . . . , 𝑑. Since 𝐹/𝐿 is a constant field extension of 𝐸/𝐾, we have [𝐹 : 𝐸] = [𝐿 : 𝐾]. Therefore, by (4.11), 𝑑 = 1 and 𝐹¯𝔓1 = 𝐿. Thus, deg(𝔓1 ) = 1. Proof of (f). Put 𝔞 =

Í𝑚

𝑖=1 div∞,𝐸 (𝑥 𝑖 ).

𝑆=

By Lemma 4.1.1,

∞ Ø

L 𝐸 (𝑘𝔞).

𝑘=1

Ð∞ Ð By (a), 𝑆𝐿 = ∞ 𝑘=1 L 𝐹 (𝑘𝔞). Hence, again by Lemma 4.1.1, 𝑆𝐿 is 𝑘=1 L 𝐸 (𝑘𝔞)𝐿 = the integral closure of 𝐿 [x] in 𝐹. □

4.5 Completions The completion of a function field 𝐹/𝐾 at a prime divisor 𝔭 gives a powerful tool to investigate the behavior of 𝐹 at 𝔭. For example, it allows us to determine the decomposition of 𝔭 into prime divisors in finite extensions of 𝐹 (Proposition 4.5.3). We also use completions to define the ‘different’ of an extension. This notion plays a central role in the Riemann–Hurwitz genus formula, to be introduced in the next section. Let 𝑣 be a rank-1 valuation of a field 𝐹. Then, 𝑣 induces a topology on 𝐹. Two elements 𝑥, 𝑦 of 𝐹 are ‘close’ in this topology if 𝑣(𝑥 − 𝑦) is ‘large’. A sequence ∞ of elements of 𝐹 is a Cauchy sequence if for every integer 𝑚 there exists {𝑥𝑖 }𝑖=1 a 𝑘 such that 𝑖, 𝑗 ≥ 𝑘 implies 𝑣(𝑥 𝑖 − 𝑥 𝑗 ) ≥ 𝑚. If every Cauchy sequence converges, 𝐹 is complete. Each 𝐹 embeds as a dense subfield in a complete field 𝐹ˆ𝑣 with a valuation 𝑣 extending the valuation of 𝐹 [BoS66, Chap. 1, Sec. 4.1]. In particular, 𝐹ˆ𝑣 has the same residue field and value group at 𝑣 as 𝐹. We call ( 𝐹ˆ𝑣 , 𝑣) (or also just 𝐹ˆ𝑣 ) the completion of (𝐹, 𝑣). We also say that 𝐹ˆ𝑣 is the completion of 𝐹 at 𝑣. The completion ( 𝐹ˆ𝑣 , 𝑣) of (𝐹, 𝑣) is unique up to an 𝐹-automorphism. Example 4.5.1 (Q 𝑝 and 𝐾 ((𝑡))) The completion of Q at the 𝑝-adic valuation ord 𝑝 of Q×𝑝 has a (Example 2.2.1(a)) is the field Q 𝑝 of 𝑝-adic numbers. Every element 𝑥 Í 𝑛 unique presentation as a convergent (in the ord 𝑝 -topology) power series ∞ 𝑛=𝑚 𝑎 𝑛 𝑝 with 𝑚 = ord 𝑝 (𝑥), 𝑎 𝑛 ∈ Z, 0 ≤ 𝑎 𝑛 ≤ 𝑝 − 1, and 𝑎 𝑚 ≠ 0. The valuation ring Z 𝑝 of Q 𝑝 consists of all 𝑥 ∈ Q 𝑝 with 𝑚 ≥ 0. Next consider a field 𝐾 and a transcendental element 𝑡 over 𝐾. Let 𝑣 be the unique valuation of 𝐾 (𝑡) with 𝑣(𝑡) = 1 (Example 2.2.1(b)). The completion of 𝐾 (𝑡) at 𝑣 is the field 𝐾 ((𝑡)) of formal power series in 𝑡 with coefficients in 𝐾. Each nonzero Í 𝑛 element 𝑓 of 𝐾 ((𝑡)) has a unique presentation 𝑓 = ∞ 𝑛=𝑚 𝑎 𝑛 𝑡 with 𝑚 = 𝑣( 𝑓 ) and 𝑎 𝑛 ∈ 𝐾. The valuation ring of 𝐾 ((𝑡)) is the ring 𝐾 [[𝑡]] of formal power series in 𝑡 with coefficients in 𝐾. The residue field of 𝐾 ((𝑡)) under 𝑣 is 𝐾. By Corollary 3.4.9(b), 𝐾 ((𝑡)) is a regular extension of 𝐾.

4.5 Completions

73

Consider now a finite extension 𝐿 of 𝐾 with a basis 𝑤 1 , . . . , 𝑤 𝑑 . Then, 𝑤 1 , . . . , 𝑤 𝑑 Í are linearly independent over 𝐾 ((𝑡)). Let 𝑥 = ∞ 𝑎 𝑡 𝑛 with 𝑎 𝑛 ∈ 𝐿 be an Í𝑑 𝑛=𝑚 𝑛 element of 𝐿((𝑡)). For each 𝑛 write 𝑎 𝑛 = 𝑖=1 𝑎 𝑛𝑖 𝑤 𝑖 with 𝑎 𝑛𝑖 ∈ 𝐾. Then, Í𝑑 Í∞ 𝑥 = 𝑖=1 ( 𝑛=𝑚 𝑎 𝑛𝑖 𝑡 𝑛 )𝑤 𝑖 . Therefore, 𝐿((𝑡)) = 𝐾 ((𝑡))𝐿 and 𝑤 1 , . . . , 𝑤 𝑑 form a basis for 𝐿((𝑡))/𝐾 ((𝑡)). The next result contains various versions of Hensel’s lemma: Proposition 4.5.2 Let (𝐹, 𝑣) be a complete discrete valued field. (a) Let 𝑓 ∈ 𝑂 𝑣 [𝑋] and 𝑎 ∈ 𝑂 𝑣 with 𝑣( 𝑓 (𝑎)) > 2𝑣( 𝑓 ′ (𝑎)). Then, there is a unique 𝑥 ∈ 𝐹 with 𝑓 (𝑥) = 0 and 𝑣(𝑥 −𝑎) ≥ 𝑣( 𝑓 (𝑎)) −𝑣( 𝑓 ′ (𝑎)) [CaF67, p. 83, Lemma]. (b) Let 𝑓 ∈ 𝑂 𝑣 [𝑋] be a monic polynomial. Denote reduction at 𝑣 by a bar. Suppose that 𝑓¯(𝑋) = 𝜁 (𝑋)𝜂(𝑋) with 𝜁, 𝜂 ∈ 𝐹¯𝑣 [𝑋] monic and relatively prime. Then, there are monic polynomials 𝑔, ℎ ∈ 𝑂 𝑣 [𝑋] with 𝑔¯ = 𝜁, ℎ¯ = 𝜂, and 𝑓 (𝑋) = 𝑔(𝑋)ℎ(𝑋) [ZaS60, p. 279, Thm. 17]. (c) Let 𝐹 ′ be a finite algebraic extension of 𝐹. Then, 𝑣 has a unique extension 𝑣 ′ to 𝐹 ′ [CaF67, p. 56, Thm.] and 𝐹 ′ is complete under 𝑣 ′ [CaF67, p. 57, Cor. 2]. Here is a global application of completions: Proposition 4.5.3 Let (𝐸, 𝑣) be a discrete valued field. Denote its completion by ˆ 𝑣ˆ ). Consider a finite separable extension 𝐹 of 𝐸. Let 𝑧 be a primitive element ( 𝐸, for 𝐹/𝐸 which is integral over 𝑂 𝑣 . Put ℎ = irr(𝑧, 𝐸). Let ℎ = ℎ1 · · · ℎ𝑟 be the ˆ For each 𝑖 let decomposition of ℎ into a product of irreducible polynomials over 𝐸. ˆ ˆ ˆ 𝑧 𝑖 be a root of ℎ𝑖 in 𝐸 𝑠 . Denote the unique extension of 𝑣ˆ to 𝐹𝑖 = 𝐸 (𝑧𝑖 ) by 𝑣ˆ 𝑖 . The following hold: (a) The map 𝑧 ↦→ 𝑧𝑖 extends to an 𝐸-embedding of 𝐹 into 𝐹ˆ𝑖 . (b) 𝐹ˆ𝑖 is the completion of 𝐹 at the restriction 𝑣 𝑖 of 𝑣ˆ 𝑖 to 𝐹. (c) The valuations 𝑣 1 , . . . , 𝑣 𝑟 are mutually nonequivalent. Every extension of 𝑣 to 𝐹 coincides with one of the 𝑣 𝑖 ’s. 𝑒 with ℎ ∈ 𝐸¯ [𝑋] irreducible and 𝑒 ∈ N. (d) 𝑒 𝑣ˆ𝑖 /𝑣ˆ = 𝑒 𝑣𝑖 /𝑣 , 𝑓𝑣ˆ𝑖 /𝑣ˆ = 𝑓𝑣𝑖 /𝑣 , and ℎ¯ 𝑖 = ℎ𝑖0 𝑖0 ˆ (e) The map 𝑔(𝑧)É↦→ 𝑔(𝑧1 ), . . . , 𝑔(𝑧𝑟 )) for 𝑔 ∈ 𝐸ˆ [𝑋] is an 𝐸-isomorphism of 𝑟 ˆ ˆ 𝐸 ⊗𝐸 𝐹 onto 𝐹 . 𝑖 𝑖=1 (f) Each 𝑧 in 𝐹 satisfies 𝑟 ∑︁ trace𝐹/𝐸 𝑧 = trace𝐹ˆ𝑖 /𝐸ˆ 𝑧 𝑖 and 𝑖=1

norm𝐹/𝐸 𝑧 =

𝑟 Ö

norm𝐹ˆ𝑖 /𝐸ˆ 𝑧 𝑖 .

𝑖=1

Proof. See [CaF67, p. 57, Thm.]. The last statement of (d) follows from Proposition 4.5.2(b). □ Example 4.5.4 (Dedekind) We apply Proposition 4.5.3 to show the necessity of the assumption “𝐸¯ 𝑣 is infinite” in Lemma 2.3.5. Consider the polynomial 𝑓 (𝑋) = 𝑋 3 − 𝑋 2 −2𝑋 −8. Observe that 𝑓 (𝑋) has no root ˜ modulo 3, hence no root in Q. It is therefore irreducible. Let 𝑧 be a root of 𝑓 (𝑋) in Q ′ and 𝐹 = Q(𝑧). Then, [𝐹 : Q] = 3. Next observe that ord2 ( 𝑓 (0)) = 3, ord2 ( 𝑓 (0)) =

74

4 Algebraic Function Fields of One Variable

1, ord2 ( 𝑓 (1)) = 1, ord2 ( 𝑓 ′ (1)) = 0, ord2 ( 𝑓 (2)) = 3, and ord2 ( 𝑓 ′ (2)) = 1. Hence, by Proposition 4.5.2, 𝑓 has three roots 𝑥1 , 𝑥2 , 𝑥3 in Q2 with ord2 (𝑥1 ) ≥ 2, ord2 (𝑥2 −1) ≥ 1, and ord2 (𝑥 3 −2) ≥ 2. Thus, 𝑓 (𝑋) = (𝑋 −𝑥1 ) (𝑋 −𝑥 2 ) (𝑋 −𝑥3 ) is a decomposition of 𝑓 (𝑋) over Q2 into three distinct irreducible polynomials. By Proposition 4.5.3, ord2 has three distinct extensions to 𝐹: 𝑤, 𝑤 ′, and 𝑤 ′′. By Proposition 2.3.2, 𝐹¯𝑤 = F2 . Now assume that 𝐹/Q has a primitive element 𝑧 with 𝑔 = irr(𝑧, Q) ∈ 𝑂 ord2 [𝑋] such that 𝑔¯ is separable. Then, deg(𝑔) = 3. By the preceding paragraph, 𝐹¯𝑤 = F2 . Hence, all three distinct roots of 𝑔¯ belong to F2 . But, F2 has only two elements, so 𝑧 does not exist. Lemma 4.5.5 Let (𝐹, 𝑣) be a discrete valued field and 𝐼 a nonzero 𝑂 𝑣 submodule of 𝐹 which is not 𝐹. Then, 𝐼 is a fractional ideal of 𝑂 𝑣 and there exists an 𝑚 ∈ Z with 𝐼 = 𝔪𝑣−𝑚 . If 𝐹 is a finite extension of a field 𝐸, and 𝑂 𝑣 is the unique valuation ring of 𝐹 lying over 𝑂 𝑣 ∩ 𝐸 (e.g. (𝐸, 𝑣) is complete), then trace𝐹/𝐸 𝐼 is a fractional ideal of 𝑂 𝑣 ∩ 𝐸. Proof. Choose 𝜋 ∈ 𝐹 with 𝑣(𝜋) = 1. Let 𝑥 be a nonzero element of 𝐼. If 𝑥 ′ ∈ 𝐹 ′ and 𝑣(𝑥 ′) ≥ 𝑣(𝑥), then 𝑥 ′ = 𝑥𝑥 𝑥 ∈ 𝐼. Since 𝐼 ≠ 𝐹, this implies that 𝑣(𝐼) is bounded from below. Hence, −𝑚 = inf(𝑣(𝑥) | 𝑥 ∈ 𝐼) is an integer. Therefore, 𝐼 = 𝜋 −𝑚 𝑂 𝑣 . Now suppose that 𝑂 𝑣 is the unique valuation ring of 𝐹 over 𝑂 = 𝑂 𝑣 ∩ 𝐸. By Proposition 2.5.1, 𝑂 𝑣 is the integral closure of 𝑂 in 𝐹. Let 𝑎 be an element of 𝑂 with 𝑣(𝑎) sufficiently large. By the preceding paragraph, 𝑎𝐼 ⊆ 𝑂 𝑣 . Then, 𝑎 ·trace𝐹/𝐸 (𝐼) = trace𝐹/𝐸 (𝑎𝐼) ⊆ trace𝐹/𝐸 (𝑂 𝑣 ) ⊆ 𝑂. Thus, trace𝐹/𝐸 (𝐼) is a fractional ideal of 𝑂. □ Using completions, we generalize the notions of repartition and differential (Section 4.2) of a function field. Completions of 𝐹. Let 𝔭 be a prime divisor of 𝐹/𝐾 and let 𝑂𝔭 = {𝑥 ∈ 𝐹 | ord𝔭 (𝑥) ≥ 0} be the corresponding discrete valuation ring. Suppose that the residue field 𝐹¯𝔭 of 𝐹 at 𝔭 is separable over 𝐾. By Hensel’s lemma, 𝐹¯𝔭 embeds into 𝐹ˆ𝔭 . Indeed, choose 𝑎 ∈ 𝑂𝔭 with 𝐹¯𝔭 = 𝐾 ( 𝑎). ¯ Put 𝑓 = irr( 𝑎, ¯ 𝐾). Then, ord𝔭 ( 𝑓 (𝑎)) > 0 and ord𝔭 ( 𝑓 ′ (𝑎)) = 0. By Proposition 4.5.2(a), there is an 𝑥 ′ ∈ 𝐹ˆ𝔭 with 𝑓 (𝑥 ′) = 0. The map 𝑎¯ ↦→ 𝑥 ′ extends to a 𝐾-embedding of 𝐹¯𝔭 into 𝐹ˆ𝔭 . Let 𝜋 be an element of the completion 𝐹ˆ𝔭 of 𝐹 with respect to 𝔭 such that ord𝔭 (𝜋) = 1. Then, 𝐹ˆ𝔭 is isomorphic to the field 𝐹¯𝔭 ((𝜋)) of formal power series in 𝜋 over 𝐹¯𝔭 . Indeed, if 𝛼 ∈ 𝐹ˆ𝔭× with ord𝔭 (𝛼) = 𝑚, then there exists a unique 𝑎 𝑚 ∈ 𝐹¯𝔭 with ord𝔭 (𝛼𝜋 −𝑚 − 𝑎 𝑚 ) > 0, so ord𝔭 (𝛼 − 𝑎 𝑚 𝜋 𝑚 ) > 𝑚.ÍBy induction, there exists for 𝑛 eachÍ 𝑛 ≥ 𝑚 a unique element 𝑎 𝑛 ∈ 𝐹¯𝔭 with ord𝔭 (𝛼 − 𝑖=𝑚 𝑎 𝑖 𝜋 𝑖 ) > 𝑛. It follows that ∞ 𝑖 𝛼 = 𝑖=𝑚 𝑎 𝑖 𝜋 [Che51, p. 46]. Î Adeles. Consider the direct product 𝐹ˆ𝔭 with 𝔭 ranging over all prime divisors Î ˆ of 𝐹/𝐾. An 𝛼 ∈ 𝐹𝔭 is an adele if ord𝔭 (𝛼𝔭 ) ≥ 0 for all but finitely many 𝔭. In ˆ (or by particular, each repartition of 𝐹/𝐾 is an adele. Denote the set of by A Î adeles ˆ A𝐹 , if 𝐹 is not clear from the context). It is an 𝐹-subalgebra of 𝐹ˆ𝔭 which contains ˆ as the algebra A of repartitions. For each prime divisor 𝔭 of 𝐹/𝐾 embed 𝐹ˆ𝔭 in A ′ follows. Identify 𝑥 ∈ 𝐹ˆ𝔭 with the adele 𝜉 having 𝜉𝔭 = 𝑥 and 𝜉𝔭′ = 0 if 𝔭 ≠ 𝔭. For each 𝔞 ∈ Div(𝐹/𝐾) consider the 𝐾-vector space ˆ | ord𝔭 (𝛼) + ord𝔭 (𝔞) ≥ 0 for each 𝔭}, ˆ Λ(𝔞) = {𝛼 ∈ A

4.5 Completions

75

ˆ = Λ(𝔞) and where we have abbreviated ord𝔭 (𝛼𝔭 ) by ord𝔭 (𝛼). Then, A ∩ Λ(𝔞) ˆ Indeed, let 𝛼ˆ ∈ A. ˆ For each 𝔭 choose 𝛼𝔭 ∈ 𝐹 with ord𝔭 (𝛼𝔭 − 𝛼ˆ 𝔭 ) ≥ ˆ A + Λ(𝔞) = A. ˆ and 𝛼ˆ = 𝛼 − (𝛼 − 𝛼) ˆ ∈ −ord𝔭 (𝔞). Then, 𝛼 = (𝛼𝔭 ) belongs to A, 𝛼 − 𝛼ˆ is in Λ(𝔞), ˆ A + Λ(𝔞). Therefore, ˆ ˆ ˆ (4.12) + 𝐹) = A. + 𝐹) = Λ(𝔞) + 𝐹 and A + ( Λ(𝔞) A ∩ ( Λ(𝔞)

ˆ Λ(𝔞) ˆ + 𝐹  A/ Λ(𝔞) + 𝐹 . Thus, in the notation of Section 4.2, Hence, A/
ˆ Λ(𝔞) ˆ (4.13) + 𝐹) = 𝛿(𝔞) = dim(𝔴 − 𝔞), dim𝐾 A/( where 𝔴 is a canonical divisor of 𝐹/𝐾. Differentials. Recall that a differential of 𝐹/𝐾 is a 𝐾-linear map, 𝜔: A → 𝐾, which vanishes on a subspace of the form Λ(𝔞) + 𝐹. By (4.12), we can extend 𝜔 ˆ → 𝐾 which vanishes on Λ(𝔞) ˆ uniquely to a 𝐾-linear map 𝜔: ˆ A + 𝐹. So, from now ˆ → 𝐾 which vanishes on Λ(𝔞) ˆ ˆ A on, a differential of 𝐹/𝐾 is a 𝐾-linear map 𝜔: +𝐹 for some 𝔞 ∈ Div(𝐹/𝐾). The restriction 𝜔 of 𝜔ˆ to A is a differential in the old sense. The divisor of 𝜔ˆ is the maximum of all 𝔞 ∈ Div(𝐹/𝐾) such that 𝜔ˆ vanishes on ˆ ˆ = div(𝜔). ˆ By the above, div( 𝜔) Λ(𝔞). Denote it by div( 𝜔). ˆ ˆ + 𝐹 by Ω(𝔞). Denote the 𝐾-vector space of all differentials that vanish on Λ(𝔞) ˆ The natural map 𝜔 ↦→ 𝜔ˆ is an isomorphism of Ω(𝔞) onto Ω(𝔞). In particular, these spaces have the same dimension 𝛿(𝔞). Traces. Suppose that 𝐹/𝐸 is a finite separable extension of function fields of one ˆ 𝐸 . Given a prime divisor 𝔓 of 𝐹/𝐾, denote its restriction to variable over 𝐾. Let 𝛼 ∈ A ˆ ˆ 𝐸 by 𝔭 and let É𝛼𝔓 = 𝛼𝔭 . This identifies A𝐸 as a subalgebra of A𝐹 . The isomorphisms ˆ ˆ 𝐸𝔭 ⊗ 𝐸 𝐹  𝔓|𝔭 𝐹𝔓 of Proposition 4.5.3(e) combine to an isomorphism ˆ 𝐸 ⊗𝐸 𝐹  A ˆ𝐹 A

(4.14)

([CaF67, p. 64] or [Art67, p. 244, Thm. 2]). In addition, define a trace function ˆ𝐹 → A ˆ 𝐸: trace𝐹/𝐸 : A ∑︁ trace𝐹/𝐸 (𝛽)𝔭 = trace𝐹ˆ𝔓 /𝐸ˆ𝔭 (𝛽𝔓 ). 𝔓|𝔭

Proposition 4.5.6 Let 𝐹/𝐸 be a finite separable extension of function fields of one ˆ 𝐹: variable over a field 𝐾. The following hold for all 𝛼, 𝛽 ∈ A (a) trace𝐹/𝐸 (𝛼 + 𝛽) = trace𝐹/𝐸 (𝛼) + trace𝐹/𝐸 (𝛽). ˆ 𝐸. (b) trace𝐹/𝐸 (𝛼𝛽) = 𝛼 · trace𝐹/𝐸 (𝛽) if 𝛼 ∈ A (c) The trace of an element 𝑥 of 𝐹 coincides with the trace of 𝑥 as an adele. (d) Let 𝔓 be a prime divisor of 𝐹/𝐾 and 𝔭 = 𝔓| 𝐸 . Then, the trace of an element 𝑥 of 𝐹ˆ𝔓 to 𝐸ˆ𝔭 coincides with the trace of 𝑥 as an adele. ˆ 𝐹 such that trace𝐹/𝐸 (𝛼) ≠ 0. (e) There exists an 𝛼 ∈ A Proof. Statements (a) and (b) follow from the corresponding properties of the trace function on fields. Statement (c) follows from Proposition 4.5.3(f). Statement (d) follows from the definition. Finally, use (c) and the corresponding fact for the trace of fields [Lan97, p. 286, Thm. 5.2], to prove (e). □

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4 Algebraic Function Fields of One Variable

Remark 4.5.7 (Complementary modules) (a) Let (𝐸, 𝑣) be a complete discrete valued field. Denote the valuation ring of (𝐸, 𝑣) by 𝑂 𝐸 . Let 𝐹 be a finite separable extension of 𝐸. Denote the unique extension of 𝑣 to 𝐹 by 𝑣. Choose a generator 𝜋 𝐹 of the maximal ideal of 𝑂 𝐹 . The complementary module of 𝑂 𝐹 over 𝐸 is ′ 𝑂 𝐹/𝐸 = {𝑥 ∈ 𝐹 | trace𝐹/𝐸 (𝑥𝑂 𝐹 ) ⊆ 𝑂 𝐸 }.

It is a fractional ideal of 𝑂 𝐹 which contains 𝑂 𝐹 [Lan70, p. 58, Cor.]. By Lemma −𝑑 ′ = 𝜋 𝐹 𝐹/𝐸 𝑂 𝐹 for some nonnegative integer 𝑑 𝐹/𝐸 . Call 𝑑 𝐹/𝐸 the different 4.5.5, 𝑂 𝐹/𝐸 exponent of 𝐹/𝐸. It is known [Lan70, p. 62, Prop. 8] that 𝑑 𝐹/𝐸 > 0 if and only if 𝐹/𝐸 is ramified. Moreover, 𝑑 𝐹/𝐸 ≥ 𝑒 𝐹/𝐸 − 1. If the residue field extension is separable, then equality holds if and only if 𝐹/𝐸 is tamely ramified [Ser68b, p. 67, Prop. 13]. (b) Let 𝐹/𝐸 be a finite separable extension of function fields of one variable over a field 𝐾. Consider prime divisors 𝔭 and 𝔓 of 𝐸/𝐾 and 𝐹/𝐾, respectively, with 𝔓 lying over 𝔭. Choose completions 𝐸ˆ𝔭 and 𝐹ˆ𝔓 with 𝐸ˆ𝔭 ⊆ 𝐹ˆ𝔓 . Denote the valuation rings of 𝐸ˆ𝔭 and 𝐹ˆ𝔓 , by 𝑂ˆ 𝔭 and 𝑂ˆ 𝔓 , respectively. We occasionally write 𝑑 𝐹ˆ𝔓 /𝐸ˆ𝔭 as 𝑑𝔓/𝔭 or 𝑑𝔓/𝐸 and call 𝑑𝔓/𝔭 the different exponent of 𝔓 over 𝔭 (or over 𝐸). By (a), 𝑑𝔓/𝔭 > 0 if and only if 𝔓/𝔭 is ramified. This happens for only finitely many 𝔓’s. Lemma 4.5.8 Let 𝐸 ⊆ 𝐹 ⊆ 𝐹 ′ be function fields of one variable over a field 𝐾 with 𝐹 ′/𝐸 separable. Consider prime divisors 𝔭, 𝔓, 𝔓′ of 𝐸/𝐾, 𝐹/𝐾, 𝐹 ′/𝐾, respectively, with 𝔓 lying over 𝔭 and 𝔓′ lying over 𝔓. Then, 𝑑𝔓′ /𝔭 = 𝑒𝔓′ /𝔓 𝑑𝔓/𝔭 + 𝑑𝔓′ /𝔓 . In particular, if 𝔓′/𝔓 is unramified, then 𝑑𝔓′ /𝔭 = 𝑑𝔓/𝔭 . If 𝔓/𝔭 is unramified, then 𝑑𝔓′ /𝔭 = 𝑑𝔓′ /𝔓 . Proof. Suppose that the formula holds. If 𝔓′/𝔓 is unramified, then 𝑑𝔓′ /𝔓 = 0 and 𝑒𝔓′ /𝔓 = 1. Hence, 𝑑𝔓′ /𝔭 = 𝑑𝔓/𝔭 . If 𝔓/𝔭 is unramified, then 𝑑𝔓/𝔭 = 0, so 𝑑𝔓′ /𝔭 = 𝑑𝔓′ /𝔓 . To prove the formula, assume without loss that 𝐸, 𝐹, and 𝐹 ′ are complete with respective valuation rings 𝑂 𝐸 , 𝑂 𝐹 , and 𝑂 𝐹 ′ . Choose prime elements 𝜋 𝐸 , 𝜋 𝐹 , and 𝜋 𝐹 ′ for the respective maximal ideals. Put 𝑑 = 𝑑 𝐹/𝐸 , 𝑑 ′ = 𝑑 𝐹 ′ /𝐹 , and 𝑒 ′ = 𝑒 𝐹 ′ /𝐹 . We have to prove that 𝑒 ′ 𝑑 + 𝑑 ′ = 𝑑 𝐹 ′ /𝐸 . ′ First note that 𝜋 𝐹𝑒 ′ 𝑂 𝐹 ′ = 𝜋 𝐹 𝑂 𝐹 ′ . Hence, ′

′

′

𝑑−𝑑 −𝑑 trace𝐹 ′ /𝐸 (𝜋 −𝑒 𝑂 𝐹 ′ ) = trace𝐹/𝐸 (trace𝐹 ′ /𝐹 (𝜋 −𝑑 𝐹′ 𝐹 𝜋 𝐹 ′ 𝑂 𝐹 ′ )) ′

−𝑑 = trace𝐹/𝐸 (𝜋 −𝑑 𝐹 trace𝐹 ′ /𝐹 (𝜋 𝐹 ′ 𝑂 𝐹 ′ ))

⊆ trace𝐹/𝐸 (𝜋 −𝑑 𝐹 𝑂𝐹 ) ⊆ 𝑂𝐸 . Thus,

′

′

𝑑−𝑑 𝜋 −𝑒 𝑂 𝐹 ′ ⊆ 𝑂 𝐹′ ′ /𝐸 . 𝐹′

(4.15)

′ By definition, 𝜋 −𝑑−1 ∉ 𝑂 𝐹/𝐸 . Hence, trace𝐹/𝐸 (𝜋 −𝑑−1 𝑂 𝐹 ) ̸⊆ 𝑂 𝐸 . By the second 𝐹 𝐹 part of Lemma 4.5.5, trace𝐹/𝐸 (𝜋 −𝑑−1 𝑂 ) is a fractional ideal of 𝑂 𝐸 . Hence, by the 𝐹 𝐹 −𝑑−1 −𝑚 first part of Lemma 4.5.5, trace𝐹/𝐸 (𝜋 𝐹 𝑂 𝐹 ) = 𝜋 𝐸 𝑂 𝐸 for some positive integer 𝑚. Therefore,

4.6 The Different

77 −𝑑−1 𝑂 𝐹 ). 𝜋 −1 𝐸 𝑂 𝐸 ⊆ trace𝐹/𝐸 (𝜋 𝐹

Similarly,

(4.16)

′

−𝑑 −1 𝜋 −1 𝑂 𝐹 ′ ). 𝐹 𝑂 𝐹 ⊆ trace𝐹 ′ /𝐹 (𝜋 𝐹 ′

Assume that 𝜋

−𝑒′ 𝑑−𝑑′ 𝐹′

(4.17)

𝑂 𝐹 ′ ≠ 𝑂 𝐹′ ′ /𝐸 . Then,

(4.15)

′

′

𝑑−𝑑 −1 𝑂 𝐸 ⊇ trace𝐹 ′ /𝐸 (𝜋 −𝑒 𝑂 𝐹′ ) 𝐹′ ′

−𝑑 −1 = trace𝐹/𝐸 (𝜋 −𝑑 𝑂 𝐹 ′ )) 𝐹 trace𝐹 ′ /𝐹 (𝜋 𝐹 ′ (4.17)

(4.16)

𝑂 𝐹 ) ⊇ 𝜋 −1 ⊇ trace𝐹/𝐸 (𝜋 −𝑑−1 𝐸 𝑂𝐸, 𝐹 ′

′

−𝑑

′

𝑑−𝑑 𝑂 𝐹 ′ = 𝑂 𝐹′ ′ /𝐸 = 𝜋 𝐹 ′ 𝐹 /𝐸 𝑂 𝐹 ′ . Consewhich is a contradiction. Therefore, 𝜋 −𝑒 𝐹′ □ quently, 𝑒 ′ 𝑑 + 𝑑 ′ = 𝑑 𝐹 ′ /𝐸 , as claimed.

4.6 The Different The Riemann–Hurwitz genus formula enables us to compute the genus of a function field of one variable 𝐹 over a field 𝐾 from the genus of a function subfield 𝐸 in terms of the ‘different’ of the extension 𝐹/𝐸. This formula is in particular useful when the genus of 𝐸 is known, e.g. when 𝐸 is the field of rational functions over 𝐾. Let 𝐹/𝐸 be a finite separable extension of function fields of one variable over a field 𝐾. The different of 𝐹/𝐸 is a divisor of 𝐹/𝐾: ∑︁ 𝑑𝔓/𝐸 𝔓, Diff(𝐹/𝐸) = where 𝔓 ranges over all prime divisors of 𝐹/𝐾 and 𝑑𝔓/𝐸 is the different exponent of 𝔓/𝐸. By Remark 4.5.7(a), 𝑑𝔓/𝐸 ≥ 0 for all 𝔓. Moreover, 𝑑𝔓/𝐸 > 0 if and only if 𝔓 is ramified over 𝐸. Since only finitely many 𝔓 ramify over 𝐸, Diff(𝐹/𝐸) is well defined, Diff(𝐹/𝐸) ≥ 0, and deg(Diff(𝐹/𝐸)) ≥ 0. Theorem 4.6.1 (Riemann–Hurwitz Genus Formula) Let 𝐹/𝐸 be a finite separable extension of function fields of one variable over a field 𝐾. Put 𝑔 𝐸 = genus(𝐸/𝐾) and 𝑔𝐹 = genus(𝐹/𝐾). Then, 2𝑔𝐹 − 2 = [𝐹 : 𝐸] (2𝑔 𝐸 − 2) + deg(Diff(𝐹/𝐸)).

(4.18)

Proof. Let 𝔭𝑖 , 𝑖 ∈ 𝐼, be the prime divisors of 𝐸/𝐾. Choose a nonzero differential 𝜔 Í of 𝐸/𝐾. Write div(𝜔) = 𝑖 ∈𝐼 𝑘 𝑖 𝔭𝑖 with 𝑘 𝑖 ∈ Z and 𝑘 𝑖 = 0 for all but finitely many 𝑖 ∈ 𝐼. By its definition in Subsection “Differential” of Section 4.5, ˆ 𝐸 ( Í𝑖 ∈𝐼 𝑘 𝑖 𝔭𝑖 ) but not on (4.19) 𝜔 vanishes on 𝐸 and on Λ Í ˆ 𝐸 ((𝑘 𝑟 + 1)𝔭𝑟 + 𝑖≠𝑟 𝑘 𝑖 𝔭𝑖 ) for any 𝑟 ∈ 𝐼. Λ ˆ 𝐹 → 𝐾: Define a map Ω: A Ω(𝛼) = 𝜔(trace𝐹/𝐸 (𝛼)). It is 𝐾-linear (Proposition 4.5.6(a),(b)) and vanishes on 𝐹 (by (4.19)).

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4 Algebraic Function Fields of One Variable

Í For each 𝑖 ∈ 𝐼 let 𝔭𝑖 = 𝑗 ∈𝐽𝑖 𝑒 𝑖 𝑗 𝔓𝑖 𝑗 , where the 𝔓𝑖 𝑗 are the distinct prime divisors of 𝐹/𝐾 lying over 𝔭𝑖 and the 𝑒 𝑖 𝑗 are the corresponding ramification indices. We simplify the notation and for all 𝑖 ∈ 𝐼 and 𝑗 ∈ 𝐽𝑖 let 𝐸 𝑖 = 𝐸ˆ𝔭𝑖 , 𝑂 𝑖 = 𝑂ˆ 𝔭𝑖 , 𝑣 𝑖 = ord𝔭𝑖 , 𝐹𝑖 𝑗 = 𝐹ˆ𝔓𝑖 𝑗 , 𝑂 𝑖 𝑗 = 𝑂ˆ 𝔓𝑖 𝑗 , 𝑣 𝑖 𝑗 = ord𝔓𝑖 𝑗 , and 𝑑𝑖 𝑗 = 𝑣 𝑖 𝑗 (Diff(𝐹/𝐸)) = −𝑣 𝑖 𝑗 (𝑂 𝑖′ 𝑗 ),

(4.20)

where 𝑂 𝑖′ 𝑗 is the complementary module of 𝑂 𝑖 𝑗 with respect to trace𝐹/𝐸 (Remark ˆ 𝐸 and 𝛽 ∈ A ˆ 𝐹 let 𝛼𝑖 = 𝛼𝔭𝑖 and 𝛽𝑖 𝑗 = 𝛽𝔓 . We prove that 4.5.7). Also, for all 𝛼 ∈ A 𝑖𝑗 Í ˆ 𝐹 ( 𝑖, 𝑗 (𝑒 𝑖 𝑗 𝑘 𝑖 + 𝑑𝑖 𝑗 )𝔓𝑖 𝑗 ) and (4.21a) Ω vanishes on Λ (4.21b) for each 𝑟 ∈ 𝐼 and each 𝑠 Í∈ 𝐽𝑟 the differential Ω does
not vanish on ˆ 𝐹 (𝑒𝑟 𝑠 𝑘 𝑟 + 𝑑𝑟 𝑠 + 1)𝔓𝑟 𝑠 + (𝑖, 𝑗)≠(𝑟 ,𝑠) (𝑒 𝑖 𝑗 𝑘 𝑖 + 𝑑𝑖 𝑗 )𝔓𝑖 𝑗 . Λ By (4.20), this will imply that ∑︁ div𝐹 (Ω) = (𝑒 𝑖 𝑗 𝑘 𝑖 + 𝑑𝑖 𝑗 )𝔓𝑖 𝑗 = div𝐸 (𝜔) + Diff(𝐹/𝐸). 𝑖, 𝑗

The formulas deg(div𝐹 (Ω)) = 2𝑔𝐹 − 2 and deg(div𝐸 (𝜔)) = [𝐹 : 𝐸] (2𝑔 𝐸 − 2) (Lemma 4.2.2(b) and Lemma 4.4.2) will give (4.18). For all 𝑖, 𝑗 choose 𝜋𝑖 ∈ 𝐸 and 𝜋𝑖 𝑗 ∈ 𝐹 with 𝑣 𝑖 (𝜋𝑖 ) = 1 and 𝑣 𝑖 𝑗 (𝜋𝑖 𝑗 ) = 1. ˆ 𝐹 ( Í𝑖 𝑗 (𝑒 𝑖 𝑗 𝑘 𝑖 + 𝑑𝑖 𝑗 )𝔓𝑖 𝑗 ). By definition, 𝑣 𝑖 𝑗 (𝛼) ≥ Proof of (4.21a): Consider 𝛼 ∈ Λ −𝑒 𝑖 𝑗 𝑘 𝑖 − 𝑑𝑖 𝑗 . Hence, 𝑣 𝑖 𝑗 (𝜋𝑖𝑘𝑖 𝛼) ≥ −𝑑𝑖 𝑗 . By (4.20), (𝜋𝑖𝑘𝑖 𝛼)𝑖 𝑗 ∈ 𝑂 𝑖′ 𝑗 , so trace𝐹𝑖 𝑗 /𝐸𝑖 (𝜋𝑖𝑘𝑖 𝛼)𝑖 𝑗 ∈ 𝑂 𝑖 . Thus, (trace𝐹/𝐸 (𝜋𝑖𝑘𝑖 𝛼))𝑖 =

∑︁

trace𝐹𝑖 𝑗 /𝐸𝑖 (𝜋𝑖𝑘𝑖 𝛼) ∈ 𝑂 𝑖 ,

𝑗 ∈𝐽𝑖

hence 𝑣 𝑖 (trace𝐹/𝐸 (𝛼)) + 𝑘 𝑖 ≥ 0 for each 𝑖 ∈ 𝐼. Therefore, trace𝐹/𝐸 (𝛼) ∈ ˆ 𝐸 ( Í𝑖 ∈𝐼 𝑘 𝑖 𝔭𝑖 ). Consequently, by (4.19), Ω(𝛼) = 𝜔(trace𝐹/𝐸 (𝛼)) = 0. Λ Proof of (4.21b): Assume that there exist 𝑟 ∈ 𝐼 and 𝑠 ∈ 𝐽𝑟 such that Ω vanishes on ∑︁
ˆ 𝐹 (𝑒𝑟 𝑠 𝑘 𝑟 + 𝑑𝑟 𝑠 + 1)𝔓𝑟 𝑠 + (𝑒 𝑖 𝑗 𝑘 𝑖 + 𝑑𝑖 𝑗 )𝔓𝑖 𝑗 . 𝑉 =Λ (𝑖, 𝑗)≠(𝑟 ,𝑠)

By Lemma 4.5.5, there is an 𝑚 ∈ Z with 𝑟 𝑠 −1 𝐼𝑟 𝑠 = trace𝐹𝑟 𝑠 /𝐸𝑟 (𝜋𝑟−𝑘𝑟 𝜋𝑟−𝑑 𝑂 𝑟 𝑠 ) = 𝜋𝑟−𝑚 𝑂 𝑟 . 𝑠

We distinguish between two cases:
ˆ 𝐸 (𝑘 𝑟 + 1)𝔭𝑟 + Í𝑖≠𝑟 𝑘 𝑖 𝔭𝑖 . Then, 𝑣 𝑟 (𝛽) ≥ Case A: 𝑚 ≥ 𝑘 𝑟 + 1. Consider 𝛽 ∈ Λ ˆ 𝐸 ( Í𝑖 ∈𝐼 𝑘 𝑖 𝔭𝑖 ), so 𝜔(𝛽) = 0 (by (4.19)). −𝑘 𝑟 − 1. If 𝑣 𝑟 (𝛽) ≥ −𝑘 𝑟 , then 𝛽 ∈ Λ ˆ𝐸 Otherwise, 𝑣 𝑟 (𝛽) = −𝑘 𝑟 − 1. In this case write 𝛽 = 𝛼 + 𝛾, where 𝛼, 𝛾 ∈ A satisfy 𝛼𝑟 = (1 − 𝜋𝑟 ) 𝛽𝑟 , 𝛾𝑟 = 𝜋𝑟 𝛽𝑟 , 𝛼𝑖 = 0, and 𝛾𝑖 = 𝛽𝑖 for 𝑖 ≠ 𝑟. Then, 𝛼𝑟 ∈ 𝑟 𝑠 −1 𝑂 𝑟 𝑠 with trace𝐹𝑟 𝑠 /𝐸𝑟 𝛿𝑟 𝑠 = 𝜋𝑟−𝑘𝑟 −1 𝑂 𝑟 ⊆ 𝜋𝑟−𝑚 𝑂 𝑟 = 𝐼𝑟 𝑠 , so there is a 𝛿𝑟 𝑠 ∈ 𝜋𝑟−𝑘𝑟 𝜋𝑟−𝑑 𝑠 𝛼𝑟 . For (𝑖, 𝑗) ≠ (𝑟, 𝑠) let 𝛿𝑖 𝑗 = 0. Then, 𝛿 := (𝛿𝑖 𝑗 )𝑖, 𝑗 ∈ 𝑉, 𝛼 = trace𝐹/𝐸 𝛿, and ˆ 𝐸 ( Í𝑖 ∈𝐼 𝑘 𝑖 𝔭𝑖 ), so 𝜔(𝛾) = 0 (by 𝜔(𝛼) = 𝜔(trace(𝛿)) = Ω(𝛿) = 0. Also, 𝛾 ∈ Λ
ˆ 𝐸 (𝑘 𝑟 + 1)𝔭𝑟 + Í𝑖≠𝑟 𝑘 𝑖 𝔭𝑖 , (4.19)). It follows that 𝜔(𝛽) = 0. Thus, 𝜔 vanishes on Λ in contradiction to (4.19).

4.6 The Different

79

Case B: 𝑚 ≤ 𝑘 𝑟 . Then, 𝑟 𝑠 −1 trace𝐹𝑟 𝑠 /𝐸𝑟 (𝜋𝑟−𝑘𝑟 𝜋𝑟−𝑑 𝑂 𝑟 𝑠 ) = 𝜋𝑟−𝑚 𝑂 𝑟 ⊆ 𝜋𝑟−𝑘𝑟 𝑂 𝑟 . 𝑠 𝑟 𝑠 −1 𝑟 𝑠 −1 Hence, trace𝐹𝑟 𝑠 /𝐸𝑟 (𝜋𝑟−𝑑 𝑂 𝑟 𝑠 ) ⊆ 𝑂 𝑟 . Therefore, 𝜋𝑟−𝑑 𝑂 𝑟 𝑠 ⊆ 𝑂 𝑟′ 𝑠 , so 𝑠 𝑠 ′ 𝑣 𝑟 𝑠 (𝑂 𝑟 𝑠 ) ≤ −𝑑𝑟 𝑠 − 1. This contradiction to (4.20) completes the proof of Case B and the proof of the whole theorem. □

Remark 4.6.2 (Applications of the Riemann–Hurwitz formula) Let 𝐹/𝐸 be a finite separable extension of function fields of one variable of a field 𝐾. We say that 𝐹/𝐸 is unramified (resp. tamely ramified) if each prime divisor of 𝐹/𝐾 is unramified (resp. tamely ramified) over 𝐸. Let 𝑔 𝐸 = genus(𝐸/𝐾) and 𝑔𝐹 = genus(𝐹/𝐾). Suppose that [𝐹 : 𝐸] ≥ 2. (a) Comparison of genera: We have mentioned at the beginning of this section that deg(Diff(𝐹/𝐸)) ≥ 0. Hence, by (4.18), 𝑔𝐹 ≥ 𝑔 𝐸 . Both 𝑔 𝐸 and 𝑔𝐹 have the same value if and only if 𝑔
𝐸 = 1 and 𝐹/𝐸 is unramified, or 𝑔 𝐸 = 0 and deg(Diff(𝐹/𝐸)) = 2 [𝐹 : 𝐸] − 1 . In particular, if 𝐹 = 𝐾 (𝑡), then 𝑔𝐹 = 0 (Example 4.2.4). Hence, 𝑔 𝐸 = 0. Each prime divisor of 𝐹/𝐾 of degree 1 induces a prime divisor of 𝐸/𝐾 of degree 1. We conclude from Example 4.2.4 that 𝐸 = 𝐾 (𝑢) is also a rational function field. This is Lüroth’s theorem. This theorem actually holds for arbitrary algebraic extensions 𝐾 (𝑡)/𝐸 and not only for separable extensions and may be proved by elementary arguments on polynomials [Wae91a, p. 218]. (b) An analog of a theorem of Minkowski: Suppose that 𝐹/𝐸 is unramified. In this case the Riemann–Hurwitz formula simplifies to 𝑔𝐹 − 1 = [𝐹 : 𝐸] (𝑔 𝐸 − 1). Hence, 𝑔 𝐸 > 0. In other words, a function field 𝐸/𝐾 of genus 0 has no proper finite unramified extension 𝐹 which is regular over 𝐾. In particular, 𝐾 (𝑡) has no finite proper separable unramified extension 𝐹 which is regular over 𝐾. This is an analog of a theorem of Minkowski saying that Q has no proper unramified extensions [Jan73, p. 57, Cor. 11.11]. (c) The Riemann–Hurwitz formula for tamely ramified extensions: Suppose that 𝐹/𝐸 is tamely ramified. By Remark 4.5.7(a), the Riemann–Hurwitz formula simplifies to ∑︁ ∑︁
𝑒𝔓/𝔭 − 1 deg(𝔓). 2𝑔𝐹 − 2 = [𝐹 : 𝐸] (2𝑔 𝐸 − 2) + 𝔭 𝔓|𝔭

(d) An analog of Minkowski’s theorem in the tamely ramified case: Suppose that 𝐾 is algebraically closed and 𝑔 𝐸 = 0. Then, 𝐸 = 𝐾 (𝑡) (Example 4.2.4) and the degree of each prime divisor is 1. Suppose that 𝐹/𝐸 is a proper tamely ramified extension. Then, 𝐸 has at least two prime divisors that ramify in 𝐹. Indeed, assume that 𝐸 has only one prime divisor 𝔭 that ramifies in 𝐹. Let 𝔓1 , . . . , 𝔓𝑟 be the prime divisors of 𝔭 in 𝐹. Then, −2 ≤ 2𝑔𝐹 − 2 = −2[𝐹 : 𝐸] +

𝑟 ∑︁

𝑒𝔓𝑖 /𝔭 − 1




𝑖=1

= −2[𝐹 : 𝐸] + [𝐹 : 𝐸] − 𝑟 = −[𝐹 : 𝐸] − 𝑟.

80

4 Algebraic Function Fields of One Variable

Hence, 3 ≤ [𝐹 : 𝐸] + 𝑟 ≤ 2, a contradiction. In particular, if 𝐾 is algebraically closed and char(𝐾) = 0, then every proper extension of 𝐾 (𝑡) is ramified over at least two prime divisors. (e) Generation of Galois groups by inertia groups: Let 𝐾 be an algebraically closed field, 𝐸 = 𝐾 (𝑡), and 𝐹 a finite Galois extension of 𝐸. Denote the prime divisors of 𝐹/𝐾 which ramify over 𝐸 by 𝔓1 , . . . , 𝔓𝑟 . For each 𝑖 let 𝐷 𝑖 be the decomposition group of 𝔓𝑖 over 𝐸. Since 𝐾 is algebraically closed, 𝐷 𝑖 is also the inertia group of 𝔓𝑖 over 𝐸. Let 𝐸 𝑖 be the fixed field of 𝐷 𝑖 in 𝐹. Then, 𝔓𝑖 | 𝐸𝑖 is unramified over 𝐸. Hence, 𝐸 0 = 𝐸 1 ∩ · · · ∩ 𝐸𝑟 is unramified over 𝐸. By (b), 𝐸 0 = 𝐸. In other words, the inertia groups of the prime divisors of 𝐹/𝐾 which ramify over 𝐸 generate Gal(𝐹/𝐸). (f) Quasi-𝑝 groups: Let 𝐾 be an algebraically closed field of positive characteristic 𝑝. Consider the rational field 𝐸 = 𝐾 (𝑡), a prime divisor 𝔭 of 𝐸/𝐾, and a finite Galois extension 𝐹 of 𝐸 which is ramified only over 𝔭. Denote the fixed field in 𝐹 of all 𝑝-Sylow subgroups of Gal(𝐹/𝐸) by 𝐸 𝑝 . Then, 𝐸 𝑝 /𝐸 is a Galois extension of degree relatively prime to 𝑝. Hence, 𝐸 𝑝 /𝐸 is tamely ramified. The only prime divisor of 𝐸/𝐾 which is possibly ramified in 𝐸 𝑝 is 𝔭. It follows from (d) that 𝐸 𝑝 = 𝐸. In other words, Gal(𝐹/𝐸) is generated by its 𝑝-Sylow subgroups. One says that Gal(𝐹/𝐸) is quasi-𝑝. (g) Abhyankar’s conjecture: Let 𝐾 be an algebraically closed field of positive characteristic 𝑝. Let 𝐸 = 𝐾 (𝑡) and 𝔭 a prime divisor of 𝐸/𝐾. In 1957, Abhyankar conjectured that for each finite quasi-𝑝 group 𝐺 there exists a finite Galois extension 𝐹 of 𝐸 which is unramified outside 𝔭 and Gal(𝐹/𝐸)  𝐺 [Abh57]. Serre proved the conjecture for solvable 𝐺 in 1990 [Ser90]. Raynaud treated all other cases in [Ray94]. (h) The generalized Abhyankar conjecture: Let again 𝐾 be an algebraically closed field of positive characteristic 𝑝. Put 𝐸 = 𝐾 (𝑡). Consider a set 𝑆 = {𝔭1 , . . . , 𝔭𝑟 } of prime divisors of 𝐸/𝐾 and a finite Galois extension 𝐹 of 𝐸 which is unramified over 𝐸 outside 𝑆. Let 𝐺 = Gal(𝐹/𝐸). Denote the subgroup of 𝐺 generated by all 𝑝-Sylow subgroups of 𝐺 by 𝐺 ( 𝑝). Let 𝐸 𝑝 be the fixed field of 𝐺 ( 𝑝) in 𝐹. Then, 𝐸 𝑝 is a Galois extension of 𝐸 which is tamely ramified over 𝐸 and unramified outside 𝑆. By [Gro71, XIII, Cor. 2.12], Gal(𝐸 𝑝 /𝐸) is generated by 𝑟 − 1 elements. This led Abhyankar to conjecture that for every finite group 𝐺 such that 𝐺/𝐺 ( 𝑝) is generated by 𝑟 − 1 elements there is a Galois extension 𝐹 of 𝐸 which is unramified outside 𝑆 and Gal(𝐹/𝐸)  𝐺. Harbater proved this conjecture in [Hrb94] by reducing it to the special case 𝑟 = 1 proved by Raynaud.

4.7 Hyperelliptic Fields We apply the concepts and results of this chapter to the study of a special kind of algebraic function field of one variable which we now introduce. A function field 𝐹/𝐾 is hyperelliptic if its genus is at least 2 and if it is a quadratic extension of a function field 𝐸/𝐾 of genus 0. We then say 𝐸 is a quadratic subfield of 𝐹. It turns out that 𝐸 is uniquely determined. Moreover, we will be able to identify 𝐸 from the arithmetic of 𝐹.

4.7 Hyperelliptic Fields

81

Lemma 4.7.1 Let 𝐹/𝐾 be a function field of genus 𝑔. Let 𝔞, 𝔟 be divisors of 𝐹/𝐾 such that 𝔞, 𝔟 ≥ 0 and let 𝔴 be a canonical divisor of 𝐹/𝐾. (a) If L (𝔟 − 𝔞) = L (𝔟) and dim(𝔟) ≥ 1, then dim(𝔞) = 1. (b) If 𝑔 ≥ 1 and 𝔞 > 0, then dim(𝔴 − 𝔞) < dim(𝔴). (c) If 𝑔 ≥ 1 and 𝑥1 , . . . , 𝑥 𝑔 is a basis for L (𝔴), then
−𝔴 = min div(𝑥 1 ), . . . , div(𝑥 𝑔 ) . Proof of (a). Since 𝔞 ≥ 0, we have 𝐾 ⊆ L (𝔞). Conversely, let 𝑥 ∈ L (𝔞). Then, div(𝑥) + 𝔞 ≥ 0. Consider 𝑦 ∈ L (𝔟). By assumption, 𝑦 ∈ L (𝔟 − 𝔞), hence div(𝑦) + 𝔟 − 𝔞 ≥ 0. Therefore, div(𝑥𝑦) + 𝔟 ≥ 0, so 𝑥𝑦 ∈ L (𝔟). Apply this result to a basis 𝑦 1 , . . . , 𝑦 𝑛 of L (𝔟). Find 𝑎 𝑖 𝑗 ∈ 𝐾 such that 𝑥𝑦 𝑖 = Í𝑛 𝑗=1 𝑎 𝑖 𝑗 𝑦 𝑗 , 𝑖 = 1, . . . , 𝑛. Hence, det(𝑥𝐼 − 𝐴) = 0, where 𝐴 = (𝑎 𝑖 𝑗 )1≤𝑖, 𝑗 ≤𝑛 . Therefore, 𝑥 satisfies a monic equation with coefficients in 𝐾. Since 𝐾 is algebraically closed in 𝐹, we have 𝑥 ∈ 𝐾. Consequently, dim(𝔞) = 1. Proof of (b). Assume that dim(𝔴 − 𝔞) = dim(𝔴). Then, L (𝔴 − 𝔞) = L (𝔴). By Lemma 4.2.2(b), dim(𝔴 − 𝔞) = dim(𝔴) = 𝑔. So, by (a), dim(𝔞) = 1. By Theorem 4.2.1, dim(𝔞) = deg(𝔞) + 1 − 𝑔 + dim(𝔴 − 𝔞). Hence, deg(𝔞) = 0, which is a contradiction. Therefore, dim(𝔴 − 𝔞) < dim(𝔴).
Proof of (c). Set 𝔪 = min div(𝑥1 ), . . . , div(𝑥 𝑔 ) . Since div(𝑥𝑖 ) + 𝔴 ≥ 0 we have 𝔪 ≥ −𝔴. If 𝔪 > −𝔴, then there exists a prime divisor 𝔭 of 𝐹/𝐾 with 𝔪 − 𝔭 ≥ −𝔴. Hence, div(𝑥𝑖 ) +𝔴 −𝔭 ≥ 0, so 𝑥𝑖 ∈ L (𝔴 −𝔭) for 𝑖 = 1, . . . , 𝑔. Therefore, L (𝔴 −𝔭) = L (𝔴), contradicting (b). Consequently, 𝔪 = −𝔴. □ Proposition 4.7.2 Let 𝐹/𝐾 be a function field of genus 𝑔 ≥ 2. Consider a canonical 𝑥
divisor 𝔴 of 𝐹/𝐾. Let 𝑥1 , . . . , 𝑥 𝑔 be a basis of L (𝔴). If 𝐸 = 𝐾 𝑥𝑥12 , . . . , 𝑥𝑔1 is a proper subfield of 𝐹, then genus(𝐸/𝐾) = 0 and [𝐹 : 𝐸] = 2. Thus, 𝐹/𝐾 is a hyperelliptic field. 𝑥

Proof. Let 𝔴 ′ = div(𝑥1 ) + 𝔴. Then, 𝔴 ′ is also a canonical divisor and 1, 𝑥𝑥21 , . . . , 𝑥𝑔1 is a basis of L (𝔴 ′). Replace 𝔴 by 𝔴 ′, if necessary, to assume that 𝑥 1 = 1 and 𝐸 = 𝐾 (𝑥2 , . . . , 𝑥 𝑔 ). Since 𝑔 ≥ 2, the element 𝑥 2 is transcendental over 𝐾. Hence, 𝐸 is also an algebraic function field of one variable over 𝐾 and 𝑑 = [𝐹 : 𝐸] < ∞ (paragraph preceding Lemma 4.4.2). Set 𝑔 𝐸 = genus(𝐸/𝐾) and let 𝔴𝐸 be a
canonical divisor of 𝐸/𝐾. By Lemma 4.7.1(c), 𝔴 = − min 0, div(𝑥2 ), . . . , div(𝑥 𝑔 ) . In particular, 𝔴 ≥ 0. By Lemma 4.2.2(b), deg(𝔴) = 2𝑔 − 2 ≥ 2. Therefore, 𝔴 > 0.

(4.22)

Observe that div(𝑥𝑖 ) ∈ Div(𝐸/𝐾), 𝑖 = 2, . . . , 𝑔, so 𝔴 ∈ Div(𝐸/𝐾). We may therefore apply Riemann–Roch to 𝐸/𝐾 and 𝔴:

dim L 𝐸 (𝔴) = deg𝐸 (𝔴) + 1 − 𝑔 𝐸 + dim L 𝐸 (𝔴𝐸 − 𝔴) . (4.23) Since 1, 𝑥1 , . . . , 𝑥 𝑔 are in 𝐸 and generate L 𝐹 (𝔴), we have L 𝐸 (𝔴) = L 𝐹 (𝔴). Hence,
by Lemma 4.2.2(b), dim L 𝐸 (𝔴) = 𝑔. Applying Lemma 4.2.2(b) again and Lemma 4.4.2, we have 2𝑔 − 2 = deg𝐹 (𝔴) = 𝑑 · deg𝐸 (𝔴). Substituting this into (4.23) gives

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2𝑔 − 2 + 1 − 𝑔 𝐸 + dim L 𝐸 (𝔴𝐸 − 𝔴) , 𝑑 which may be rewritten as
𝑑 dim L 𝐸 (𝔴𝐸 − 𝔴) − 𝑔 𝐸 ) = (𝑔 − 1) (𝑑 − 2).

𝑔=

(4.24)

(4.25)

By assumption, both 𝑔 and 𝑑 are at least 2. Hence, the right-hand side of (4.25) is at least 0. On the other hand, by Lemma 4.2.2(b), dim L 𝐸 (𝔴𝐸 − 𝔴)


(4.22)
≤ dim L 𝐸 (𝔴𝐸 ) = 𝑔 𝐸 .

(4.26)

Hence, both sides of (4.25) are 0. Since 𝑔 ≥ 2, this gives 𝑑 = 2. Finally, by (4.24), 𝑔 𝐸 = dim(L 𝐸 (𝔴𝐸 − 𝔴)). Hence, by (4.26), dim(L 𝐸 (𝔴𝐸 − 𝔴)) = dim(L 𝐸 (𝔴𝐸 )). We conclude from Lemma 4.7.1(b) that 𝑔 𝐸 = 0.

□

Lemma 4.7.3 Let 𝐹/𝐾 be a function field of genus 𝑔 ≥ 1, 𝔴 a canonical divisor of 𝐹/𝐾, and Í𝑔 𝑥1 , . . . , 𝑥 𝑔 a basis of L 𝐹 (𝔴). Let 𝐸/𝐾 be a subfield of genus 0 of 𝐹/𝐾. Then, 𝑖=1 𝐸𝑥𝑖 ≠ 𝐹. ˆ 𝐸 (0)𝑥𝑖 ⊆ Λ ˆ 𝐹 (𝔴). Indeed, if 𝛼 ∈ Λ ˆ 𝐸 (0), then for each Proof. For each 𝑖 we have Λ prime divisor 𝔭 of 𝐸/𝐾 we have ord𝔭 (𝛼) ≥ 0. Since 𝑥 𝑖 ∈ L 𝐹 (𝔴), we have div(𝑥 𝑖 ) + 𝔴 ≥ 0. Hence, for each prime divisor 𝔭 of 𝐹/𝐾 we have ord𝔭 (𝛼𝑥 𝑖 ) + ord𝔭 (𝔴) ≥ 0, ˆ 𝐹 (𝔴). This proves our claim. It follows from the claim that so 𝛼𝑥 𝑖 ∈ Λ 𝑔 ∑︁


ˆ 𝐸 (0) + 𝐸 𝑥 𝑖 ⊆ Λ ˆ 𝐹 (𝔴) + 𝐹. Λ

(4.27)

𝑖=1

Choose a canonical divisor 𝔴𝐸 of 𝐸/𝐾. By (4.13),
ˆ 𝐸 /( Λ ˆ 𝐸 (0) + 𝐸) = dim(𝔴𝐸 ) = genus(𝐸/𝐾) = 0. dim𝐾 A Hence, ˆ𝐸 =Λ ˆ 𝐸 (0) + 𝐸 . A

(4.28)

Another application of (4.13) yields
ˆ 𝐹 /( Λ ˆ 𝐹 (𝔴) + 𝐹) = dim(0) = 1. dim𝐾 A ˆ 𝐹 , so ˆ 𝐹 (𝔴) + 𝐹 ⊂ A Hence, Λ 𝑔 ∑︁ 𝑖=1

ˆ 𝐸 𝑥 𝑖 (4.28) = A

𝑔 ∑︁

(4.27)

ˆ 𝐹. ˆ 𝐸 (0) + 𝐸)𝑥𝑖 ⊆ Λ ˆ 𝐹 (𝔴) + 𝐹 ⊂ A (Λ

(4.29)

𝑖=1

Í𝑔 Í𝑔 ˆ ˆ If 𝑖=1 𝐸𝑥 Í𝑔𝑖 = 𝐹, then by (4.14), A𝐹 = 𝑖=1 A𝐸 𝑥 𝑖 , contradicting (4.29). Consequently, 𝑖=1 𝐸𝑥𝑖 ≠ 𝐹. □ Proposition 4.7.4 Let 𝐹/𝐾 be a hyperelliptic function field, 𝔴 a canonical divisor 𝑥
of 𝐹/𝐾, and 𝑥1 , . . . , 𝑥 𝑔 a basis of L 𝐹 (𝔴). Then, 𝐸 = 𝐾 𝑥𝑥21 , . . . , 𝑥𝑔1 is the only quadratic subfield of 𝐹.

4.8 Hyperelliptic Fields with a Rational Quadratic Subfield

83

Proof. By definition, 𝐹 has quadratic subfields of genus 0. Let 𝐸 ′ be one of them. Í𝑔 By Lemma 4.7.3, 𝑖=1 𝐸 ′𝑥 𝑖 ≠ 𝐹. Since [𝐹 : 𝐸 ′] = 2, this implies that 𝑥𝑖 ∈ 𝐸 ′, 𝑖 = 1, . . . , 𝑔. It follows that 𝐸 ⊆ 𝐸 ′. In particular, 𝐸 ⊂ 𝐹. By Proposition 4.7.2, □ [𝐹 : 𝐸] = 2. Comparing degrees, we have 𝐸 ′ = 𝐸.

4.8 Hyperelliptic Fields with a Rational Quadratic Subfield A hyperelliptic field with a rational quadratic subfield is generated by two generators which satisfy an equation of a special type. This situation arises, for example, when the hyperelliptic field has a prime divisor of degree 1 (third paragraph of Example 4.2.4). Proposition 4.8.1 Let 𝐹/𝐾 be a hyperelliptic field of genus 𝑔. Suppose that the quadratic subfield of 𝐹 is 𝐾 (𝑥) with 𝑥 indeterminate. Then, 𝐹 = 𝐾 (𝑥, 𝑦), where 𝑦 satisfies a relation 𝑦 2 + ℎ1 (𝑥)𝑦 + ℎ2 (𝑥) = 0 with ℎ1 , ℎ2 ∈ 𝐾 [𝑋], deg(ℎ1 ) ≤ 𝑔 + 1, and deg(ℎ2 ) ≤ 2𝑔 + 2. If char(𝐾) ≠ 2, we may choose 𝑦 with 𝑦 2 = 𝑓 (𝑥), where 𝑓 ∈ 𝐾 [𝑋] is a polynomial which is not a square in 𝐾˜ (𝑋) and deg( 𝑓 (𝑥)) ≤ 2𝑔 + 2. Proof. Let 𝔭∞ be the pole of 𝑥 in 𝐾 (𝑥)/𝐾. By Lemma 4.4.2, deg𝐹 (𝑛𝔭∞ ) = 2𝑛 for each positive integer 𝑛. If 𝑛 ≥ 𝑔, then 2𝑛 > 2𝑔 − 2. By Lemma 4.2.2(d), dim(L 𝐹 (𝑛𝔭∞ )) = 2𝑛 + 1 − 𝑔. In particular, for 𝑛 = 𝑔 the elements 1, 𝑥, . . . , 𝑥 𝑔 form a basis for L 𝐹 (𝑔𝔭∞ ).
For 𝑛 = 𝑔 +1 we have dim L 𝐹 ((𝑔 +1)𝔭∞ ) = 𝑔 +3. Hence,
two more elements are needed to complete 1, 𝑥, . . . , 𝑥 𝑔 to a basis of L 𝐹 (𝑔 + 1)𝔭∞ . We take one of them as 𝑥 𝑔+1 and denote the other one by 𝑦. By Riemann–Roch, dim(L 𝐾 ( 𝑥) ((𝑔 + 1)𝔭∞ )) =
in 𝐾 (𝑥), then 𝑦 is is . If (𝑔 + L a basis of 1)𝔭 𝑔 + 2. So, 1, 𝑥, . . . , 𝑥 𝑔+1 form 𝑦 ∞ 𝐾 ( 𝑥)
also in L 𝐾 ( 𝑥) (𝑔 + 1)𝔭∞ . This implies that 1, 𝑥, . . . , 𝑥 𝑔+1 , 𝑦 are linearly dependent that 𝑦 ∉ 𝐾 (𝑥), so 𝐹 = 𝐾 (𝑥, 𝑦). over 𝐾. We conclude from this contradiction
2𝑔+2 , Next note that dim L 𝐹 ((2𝑔 +2)𝔭∞ ) = 3𝑔 +5.
All 3𝑔 +6 elements 1, 𝑥, . . . , 𝑥 2 𝑔+1 𝑦, 𝑦𝑥, . . . , 𝑦𝑥 , 𝑦 belong to L 𝐹 (2𝑔 + 2)𝔭∞ . Therefore, there are 𝑎 𝑖 , 𝑏 𝑗 , 𝑐 ∈ 𝐾, Í2𝑔+2 Í𝑔+1 not all zero, with 𝑖=0 𝑎 𝑖 𝑥 𝑖 + 𝑗=0 𝑏 𝑗 𝑦𝑥 𝑗 + 𝑐𝑦 2 = 0. Since 𝑦 ∉ 𝐾 (𝑥), we have 𝑐 ≠ 0. Í2𝑔+2 Í𝑔+1 Let ℎ1 (𝑋) = 𝑗=0 𝑐−1 𝑏 𝑗 𝑋 𝑗 , ℎ2 (𝑋) = 𝑖=0 𝑐−1 𝑎 𝑖 𝑋 𝑖 . Then, 𝑦 2 + ℎ1 (𝑥)𝑦 + ℎ2 (𝑥) = 0.

(4.30)

If char(𝐾) ≠ 2, replace 𝑦 by 2𝑦 + ℎ1 (𝑥), if necessary, to assume that (4.30) has the form 𝑦 2 = 𝑓 (𝑥). Here 𝑓 ∈ 𝐾 [𝑋] has degree at most 2𝑔 + 2. Finally, 𝑓 factors in 𝐾 [𝑋] as 𝑓 = ℎ2 ℎ1′ · · · ℎ 𝑠′ with ℎ, ℎ1′ , . . . , ℎ 𝑠′ irreducible in 𝐾 [𝑋] such that ℎ𝑖′ ≠ 𝑐ℎ ′𝑗 if 1 ≤ 𝑖 < 𝑗 ≤ 𝑠 and 𝑐 ∈ 𝐾. Let ℎ ′ = ℎ1′ · · · ℎ 𝑠′ . Since char(𝐾) ≠ 2, ℎ ′ is not a square in 𝐾˜ [𝑋]. Replace 𝑦 by 𝑦ℎ(𝑥) −1 to assume that 𝑓 is □ not a square in 𝐾˜ (𝑋). Our first task is to compute the genus of the hyperelliptic field in characteristic ≠ 2 from deg( 𝑓 ).

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Proposition 4.8.2 Let 𝐾 be a field of characteristic ≠ 2 and 𝑓 ∈ 𝐾 [𝑋] a polynomial of degree 𝑑 ≥ 1 with no multiple roots. Let 𝐹 = 𝐾 (𝑥, 𝑦), with 𝑥 transcendental over 𝐾 and 𝑦 2 = 𝑓 (𝑥). Then, 𝐹/𝐾 is an algebraic function field of one variable of genus 𝑑−1 2 if 𝑑 is odd 𝑑−2 and of genus 2 if 𝑑 is even. In particular, if 𝑑 ≥ 5, then 𝐹/𝐾 is hyperelliptic. Proof. Since 𝑓 (𝑥) is not a square in 𝐾˜ (𝑥), [𝐾 (𝑥, 𝑦) : 𝐾 (𝑥)] = [ 𝐾˜ (𝑥, 𝑦) : 𝐾˜ (𝑥)] = 2. Hence, 𝐾 (𝑥, 𝑦) is linearly disjoint from 𝐾˜ (𝑥) over 𝐾 (𝑥). Since 𝑥 is transcendental over 𝐾, 𝐾 (𝑥) is linearly disjoint from 𝐾˜ over 𝐾. Hence, by the tower property of linear disjointness (Lemma 3.1.3), 𝐹 is linearly disjoint from 𝐾˜ over 𝐾. It follows that 𝐹/𝐾 is an algebraic function field of one variable. Since char(𝐾) ≠ 2, 𝐹/𝐾 (𝑥) is tamely ramified. Since the genus of 𝐾 (𝑥) is 0, the Riemann–Hurwitz formula reduces to ∑︁ ∑︁ 2𝑔 − 2 = −4 + (𝑒𝔓/𝔭 − 1) deg(𝔓) (4.31) 𝔭 𝔓|𝔭

(Remark 4.6.2(c)). If 𝔭 is a prime divisor of 𝐾 (𝑥) which ramifies in 𝐹, then 𝔭 has only one extension 𝔓 to 𝐹, the ramification index of 𝔓 is 2, and its residue degree is 1. Hence, (4.31) simplifies to ∑︁ 2𝑔 = −2 + deg(𝔭), (4.32) where 𝔭 ranges over all prime divisors of 𝐾 (𝑥)/𝐾 which ramify in 𝐹. Let 𝑓 (𝑥) = 𝑝 1 (𝑥) · · · 𝑝 𝑟 (𝑥) be the decomposition of 𝑓 (𝑥) into a product of distinct irreducible polynomials in 𝐾 [𝑥]. To each 𝑝 𝑖 there corresponds a prime divisor 𝔭𝑖 of 𝐾 (𝑥)/𝐾 of degree deg( 𝑝 𝑖 ) and a valuation 𝑣 𝑖 = ord𝔭𝑖 such that 𝑣 𝑖 ( 𝑓 (𝑥)) = 𝑣 𝑖 ( 𝑝 𝑖 (𝑥)) = 1 (Example 2.2.1(b)). By Example 2.3.8, each 𝔭𝑖 ramifies in 𝐹. In addition, let 𝑣 ∞ be the valuation of 𝐾 (𝑥)/𝐾 with 𝑣 ∞ (𝑥) = −1 and let 𝔭∞ be the corresponding prime divisor; its degree is 1. Since 𝑣 ∞ ( 𝑓 (𝑥)) = −𝑑, the prime divisor 𝔭∞ is ramified in 𝐹 if 𝑑 is odd and unramified if 𝑑 is even (Example 2.3.8). All other prime divisors of 𝐾 (𝑥)Íare unramified in 𝐹. The sum of the degrees of the ramified prime divisors is 𝛿 + 𝑟𝑖=1 deg( 𝑝 𝑖 ) = 𝛿 + 𝑑, where 𝛿 = 1 if 𝑑 is odd and 𝑑−2 𝛿 = 0 if 𝑑 is even. It follows from (4.32) that 𝑔 = 𝑑−1 2 if 𝑑 is odd and 𝑔 = 2 if 𝑑 is even. □ In characteristic 2 we compute the genus of only a special type of a hyperelliptic field. Lemma 4.8.3 Let (𝐸, 𝑣) be a valued field of characteristic 2 and 𝑡 an element of 𝐸 with 𝑣(𝑡) = 1. Consider an Artin-Schreier extension 𝐹 = 𝐸 (𝑥) with 𝑥 2 + 𝑥 = 1𝑡 . Denote the unique extension of 𝑣 to 𝐹 by 𝑤. Let 𝑦 = 𝑡𝑥. ′ = 𝑦 −2 𝑂 . Then, 𝑤(𝑦) = 1, 𝑂 𝑤 = 𝑂 𝑣 [𝑦] is the integral closure of 𝑂 𝑣 in 𝐹, and 𝑂 𝑤 𝑤 Proof. By Example 2.3.11, 𝑣 has a unique extension 𝑤 to 𝐹. It totally ramifies over 𝐸 and satisfies 𝑤(𝑥) = −1, 𝑤(𝑡) = 2, and 𝑤(𝑦) = 1. In particular, 𝑂 𝑤 is the integral closure of 𝑂 𝑣 in 𝐹 (Proposition 2.5.1). Since 𝐸¯ 𝑣 = 𝐹¯𝑤 , each 𝑧 ∈ 𝑂 𝑤 can be written as 𝑧 = 𝑎 + 𝑏𝑦 with 𝑎 ∈ 𝑂 𝑣 and 𝑏 ∈ 𝑂 𝑤 . Hence, 𝑂 𝑤 = 𝑂 𝑣 [𝑦] [Lan70, p. 26, Prop. 23].

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85

′ = {𝑧 ∈ 𝐹 | trace To compute 𝑂 𝑤 𝐹/𝐸 (𝑧𝑂 𝑤 ) ⊆ 𝑂 𝑣 } observe that 𝑔(𝑋) = ′ = 𝑡 −1 𝑂 = 𝑦 −2 𝑂 [Lan70, irr(𝑦, 𝐸) = 𝑌 2 + 𝑡𝑌 + 𝑡 and 𝑔 ′ (𝑋) = 𝑡. Therefore, 𝑂 𝑤 𝑤 𝑤 ′ directly. □ p. 59, Cor.]. Of course, one may also compute 𝑂 𝑤

Proposition 4.8.4 Let 𝐾 be a field of characteristic 2 and 𝑎 1 , . . . , 𝑎 𝑑 distinct elements 𝑥) of 𝐾. Let 𝐹 = 𝐾 (𝑥, 𝑦) with 𝑥 transcendental over 𝐾 and 𝑦 2 + 𝑦 = ( 𝑥−𝑎1ℎ( ) ··· ( 𝑥−𝑎𝑑 ) , where ℎ ∈ 𝐾 [𝑋] with deg(ℎ) ≤ 𝑑 and ℎ(𝑎 𝑖 ) ≠ 0 for 𝑖 = 1, . . . , 𝑑. Then, 𝐹/𝐾 is an algebraic function field of one variable of genus 𝑑 − 1. In particular, if 𝑑 ≥ 3, then 𝐹/𝐾 is hyperelliptic. Proof. For each 𝑖 between 1 and 𝑑 let 𝑣 𝑖 be the discrete normalized valuation of 𝐾 (𝑥)/𝐾 with 𝑣 𝑖 (𝑥 − 𝑎 𝑖 ) = 1. Let 𝑡 = (𝑥 − 𝑎 1 ) · · · (𝑥 − 𝑎 𝑑 ). Then, 𝑣 𝑖 (𝑡) = 1. ˜ If 𝑦 is in 𝐾˜ (𝑥), then 2𝑣˜ 𝑖 (𝑦) = −1, which is a Extend 𝑣 𝑖 to a valuation 𝑣˜ 𝑖 of 𝐾˜ (𝑥)/𝐾. contradiction. Hence, 𝑦 is not in 𝐾˜ (𝑥), so [𝐾 (𝑥, 𝑦) : 𝐾 (𝑥)] = [ 𝐾˜ (𝑥, 𝑦) : 𝐾˜ (𝑥)] = 2. It follows that 𝐹 is linearly disjoint from 𝐾˜ over 𝐾, so 𝐹/𝐾 is an algebraic function field of one variable. By Example 2.3.9, 𝑣 1 , . . . , 𝑣 𝑑 are the only valuations of 𝐾 (𝑥)/𝐾 that ramify in 𝐹. Hence, only 𝑣 1 , . . . , 𝑣 𝑑 contribute to the different of 𝐹/𝐾 (𝑥). Let 𝑤 𝑖 be the unique š extension of 𝑣 𝑖 to 𝐹. Then, 𝐹ˆ𝑤𝑖 = 𝐾 (𝑥) 𝑣𝑖 (𝑦). By Lemma 4.8.3, the contribution of 𝑤 𝑖
to deg Diff(𝐹/𝐾 (𝑥)) is 2. Hence, deg(Diff(𝐹/𝐾 (𝑥))) = 2𝑑. Let 𝑔 = genus(𝐹/𝐾). By Riemann–Hurwitz (Theorem 4.6.1), 2𝑔 − 2 = −4 + 2𝑑. Hence, 𝑔 = 𝑑 − 1, as claimed. □

Exercises In the following exercises 𝐹 is a function field of one variable over a field 𝐾 and 𝑡 is a transcendental element over 𝐾. 1. Let 𝔞 be a divisor of 𝐹/𝐾. Note: If 𝔞 ≥ 0 and deg(𝔞) = 0, then 𝔞 = 0. (a) Prove that if deg(𝔞) = 0 and 𝔞 is not a principal divisor, then dim(𝔞) = 0. (b) Let 𝑔 be the genus of 𝐹. Suppose that 𝔞 is a noncanonical divisor with deg(𝔞) = 2𝑔 − 2. Show that dim(𝔞) = 𝑔 − 1. Hint: Use Riemann–Roch and (a). 2. Let 𝔞 and 𝔟 be divisors of 𝐹/𝐾 with 𝔟 ≥ 0. Use Riemann–Roch to prove that (a) dim(𝔞) ≤ dim(𝔞 + 𝔟) ≤ dim(𝔞) + deg(𝔟); and (b) dim(𝔞) ≤ max(0, deg(𝔞) + 1). Hint: Write 𝔞 as the difference of its “positive” and “negative” parts. (c) Conclude that if dim(𝔞) > 2𝑔 − 1, then dim(𝔞) = deg(𝔞) − 𝑔 + 1. 3. Let 𝔭1 , . . . , 𝔭𝑛 be Í𝑛distinct prime divisors of 𝐹/𝐾 and let 𝑚 1 , . . . , 𝑚 𝑛 be positive integers such that 𝑖=1 𝑚 𝑖 deg(𝔭𝑖 ) > 2𝑔 − 1. If 𝐾 is infinite, prove that there exists an Í𝑛 Í𝑛 𝑥 ∈ 𝐹 × such that div∞ (𝑥) = 𝑖=1 𝑚 𝑖 𝔭𝑖 . In particular, [𝐹 : 𝐾 (𝑥)] = 𝑖=1 𝑚 𝑖 deg(𝔭𝑖 ). Í𝑛 Hint: Consider 𝔞 = 𝑖=1 𝑚 𝑖 𝔭𝑖 and 𝔞 𝑗 = 𝔞 − 𝔭 𝑗 for 𝑗 = 1, . . . , 𝑛. For each 𝑗, 1 ≤ 𝑗 ≤ 𝑛, distinguish between the cases deg(𝔞 𝑗 ) ≤ 2𝑔 − 2 and deg(𝔞 𝑗 ) > 2𝑔 − 2, and prove that L (𝔞 𝑗 ) ⊂ L (𝔞).

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4. Suppose that 𝐹 = 𝐾 (𝑡). Prove, in the notation of Example 4.2.4, that −2𝔭∞ is a canonical divisor of 𝐹/𝐾. 5. Suppose that the genus of 𝐹/𝐾 is 0. Prove that every divisor 𝔞 with deg(𝔞) = 0 is principal. Hint: Compute dim(𝔞) and apply Exercise 1(a). 6. In the notation of Proposition 4.3.2 let 𝔭1 , . . . , 𝔭𝑟 ∈ 𝑆, let 𝑘 1 , . . . Î , 𝑘 𝑟 ∈ N, and let 𝑛 𝑃𝑖 be the center of 𝔭𝑖 at 𝑂Í𝑆 , 𝑖 = 1, . . . , 𝑟. Consider the ideal 𝐴 = 𝑖=1 𝑃𝑖𝑘𝑖 of 𝑂 𝑆 . 𝑟 Prove that the divisor 𝔞 = 𝑖=1 𝑘 𝑖 𝔭𝑖 of 𝐹/𝐾 satisfies deg(𝔞) = (𝑂 𝑆 : 𝐴). Î Hint: Prove that 𝑂 𝑆 /𝐴  𝑟𝑖=1 𝑂 𝑆 /𝑃𝑖𝑘𝑖 and that 𝑂 𝑆 /𝑃𝑖  𝑃𝑖𝑘 /𝑃𝑖𝑘+1 , as groups, for each nonnegative integer 𝑘. 7. Prove that every algebraic function field 𝐹/𝐾 of genus 2 is hyperelliptic with a rational quadratic subfield. Hint: Choose a positive canonical divisor 𝔴 of 𝐹/𝐾. Then, choose a nonconstant 𝑥 in L (𝔴). 8. Let 𝐾 be a field of characteristic that does not divide 𝑛. Consider the function field 𝐹 = 𝐾 (𝑥, 𝑦) over 𝐾 with 𝑥, 𝑦 satisfying 𝑥 𝑛 +𝑦 𝑛 = 1. Prove that genus(𝐹) = (𝑛−1)2(𝑛−2) . Hint: Use Example 2.3.8 and the Riemann–Hurwitz formula (Remark 4.6.2(c)).

Notes In addition to [Che51], one may find a proof of the Riemann–Roch theorem in [Lan64, Chapter 10, Section 2], [Deu73, Section 15], and [Sti93, Section I.5]. The content of Section 4.7 is borrowed from [Art67]. One may also find it in [Sti93, Section VI.2] along with computations of genera of various function fields using the Riemann–Hurwitz genus formula. However, in contrast to our exposition, [Sti93] assumes the fields of constants to be perfect.

Chapter 5

The Riemann Hypothesis for Function Fields

In this chapter 𝐾 is a finite field of characteristic 𝑝 with 𝑞 elements. Let 𝐹 be an algebraic function field of one variable over 𝐾 and 𝑔 the genus of 𝐹/𝐾. Denote the group of divisors and the of divisor classes of 𝐹/𝐾 by D and C, respectively. Í group −𝑠 defines the classical Riemann zeta function of a The series 𝜁 (𝑠) = ∞ 𝑛=1 𝑛 complex variable. It converges absolutely for Re(𝑠) > 1. Hence, 𝜁 (𝑠) is an analytic function in this domain. The series diverges for 𝑠 = 1. The function, however, can be analytically continued to a meromorphic function on the whole 𝑠-plane. The 1 production of this requires two stages: an analytic continuation of 𝜁 (𝑠) + 1−𝑠 to the half plane Re(𝑠) > 0 by a rearrangement of the series via Abel summation; and (𝑠) then, (the difficult part) a demonstration that 𝜁 𝜁(1−𝑠) is the product of the classical Gamma function and an elementary function in the domain 0 < Re(𝑠) < 1. The expression that results from the last stage is called the functional equation for 𝜁 (𝑠). The resulting analytic continuation of 𝜁 (𝑠) yields a function with a simple pole at 𝑠 = 1 with residue 1 and zeros at the points −2, −4, −6, −8, . . . . There are no other zeros in the domains Re(𝑠) ≥ 1 and Re(𝑠) ≤ 0 [Tit51, p. 30]. The classical Riemann hypothesis is still unproven. It states that the only zeros of 𝜁 (𝑠) in the strip 0 < Re(𝑠) < 1 lie on the line Re(𝑠) = 12 ; its applications are legion. There is an analog for 𝐹/𝐾 of the Riemann zeta function (Section 5.2). It satisfies a functional equation (Proposition 5.4.1). Our main goal is the proof of an analog to the Riemann hypothesis (Theorem 5.5.1). In Chapter 6 we extract from this an explicit estimate for the number of points on any curve over a finite field.

5.1 Class Numbers The assumption that 𝐾 is finite results in the finiteness of other sets connected to 𝐹/𝐾. For example, 𝐹/𝐾 has only finitely many ideal classes of degree 0 and only finitely many positive divisors of a given degree 𝑛. The main result of this section (Lemma 5.1.4) computes the latter number in terms of the former.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_5

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5 The Riemann Hypothesis for Function Fields

Let 𝔭 be a prime divisor of 𝐹/𝐾. Its residue field 𝐹¯𝔭 is a finite extension of 𝐾 of degree deg(𝔭). Thus, 𝐹¯𝔭 is a finite field whose order we indicate by 𝑁𝔭 = 𝑞 deg(𝔭) , the norm of 𝔭. Extend the definition of the norm to arbitrary divisors by the formula 𝑁𝔞 = 𝑞 deg(𝔞) . Then, 𝑁 (𝔞 + 𝔟) = 𝑁𝔞 · 𝑁𝔟. Definition 5.1.1 Denote the number of divisor classes of 𝐹/𝐾 of degree zero by ℎ; the class number of 𝐹/𝐾. If 𝑔 = 0, then every divisor of degree 0 of 𝐹/𝐾 is principal (Exercise 5 of Chapter 4). Therefore, ℎ = 1 if 𝐹 is 𝐾 (𝑥) or any genus 0 function field. Lemma 5.1.2 Only finitely many positive divisors of 𝐹/𝐾 have degree equal to a given integer 𝑚. In addition, the class number of 𝐹/𝐾 is finite. Proof. Let 𝑥 ∈ 𝐹 be a transcendental element over 𝐾 and let 𝐸 = 𝐾 (𝑥). Denote the collection of all prime divisors of 𝐸/𝐾 of degree ≤ 𝑚 by 𝑆0 . Each element of 𝑆0 , except possibly 𝔭∞ , corresponds to a monic irreducible polynomial in 𝐾 [𝑥] of degree ≤ 𝑚. Hence, 𝑆0 is a finite set. Only finitely many prime divisors of 𝐹/𝐾 lie over a given prime divisor 𝔭0 of 𝐸/𝐾. Each of them has degree at least as large as deg(𝔭0 ). Thus, there are only finitely many prime divisors of 𝐹/𝐾 of degree ≤ 𝑚. Therefore, the set A 𝑚 of all positive divisors of 𝐹/𝐾 of degree 𝑚 is finite. For the second part of the lemma choose a positive divisor 𝔪 of degree 𝑚 ≥ 𝑔. Denote the set of divisor classes of degree 𝑚 by C𝑚 . For each 𝔟 ∈ C𝑚 Riemann-Roch (Theorem 4.2.1) implies that dim(𝔟) ≥ 𝑚 + 1 − 𝑔 ≥ 1. Hence, there is an 𝑥 ∈ 𝐹 × with div(𝑥) + 𝔟 ≥ 0, so the class of 𝔟 contains a positive divisor. It follows that the map A 𝑚 → C𝑚 mapping each 𝔞 ∈ A 𝑚 onto its class is surjective. Therefore, by the preceding paragraph, C𝑚 is finite. Finally, the map 𝔞 ↦→ 𝔞 + 𝔪 induces a bijective map of C0 onto C𝑚 . Consequently, C0 is a finite group. □ Lemma 5.1.3 The number of positive divisors in a given class of divisors 𝐶 of 𝐹/𝐾 dim(𝐶) is 𝑞 𝑞−1 −1 . Proof. If 𝔞 ≥ 0, then div(𝑎) + 𝔞 ≥ 0 for each 𝑎 ∈ 𝐾 × , so 𝐾 ⊆ L (𝔞). Hence, dim(𝔞) ≥ 1. This gives the formula if dim(𝐶) = 0. Suppose that 𝑛 := dim(𝐶) > 0. Let 𝔠 be a divisor in 𝐶. The number of positive divisors in 𝐶 is equal to the number of principal divisors div(𝑥) with 𝑥 ∈ L (𝔠). Let 𝑥Í1 , . . . , 𝑥 𝑛 be a basis for L (𝔠) over 𝐾. The number of elements of 𝐹 × of the form 𝑛 𝑛 ′ 𝑖=1 𝑎 𝑖 𝑥 𝑖 , with 𝑎 1 , . . . , 𝑎 𝑛 ∈ 𝐾, is equal to 𝑞 − 1. Since div(𝑥) = div(𝑥 ) if and × ′ only if there exists an 𝑎 ∈ 𝐾 such that 𝑥 = 𝑎𝑥, the formula follows. □ Denote the greatest common divisor of the degrees of the divisors of 𝐹/𝐾 by 𝛿. Eventually we prove that 𝛿 = 1. In the meantime notice that a positive integer 𝑛 is a multiple of 𝛿 if and only if there exists a divisor of degree 𝑛. Indeed, there are divisors 𝔞1 , . . . , 𝔞𝑟 with 𝛿 = gcd(deg(𝔞 1 ), . . . , deg(𝔞𝑟 )). For each multiple Í Í 𝑛 of 𝛿 there are 𝑎 1 , . . . , 𝑎𝑟 ∈ Z with 𝑟𝑖=1 𝑎 𝑖 deg(𝔞𝑖 ) = 𝑛. The divisor 𝔞 = 𝑟𝑖=1 𝑎 𝑖 𝔞𝑖 satisfies deg(𝔞) = 𝑛. In particular, since the degree of the canonical divisor is 2𝑔 − 2 (Lemma 4.2.2(b)), 𝛿 divides 2𝑔 − 2. (5.1)

5.2 Zeta Functions

89

Lemma 5.1.4 Let 𝐴𝑛 be the number of positive divisors of 𝐹/𝐾 of degree 𝑛. If 𝑛 ≥ 0 𝑛+1−𝑔 is a multiple of 𝛿 larger than 2𝑔 − 2, then 𝐴𝑛 = ℎ 𝑞 𝑞−1 −1 . Proof. Let 𝑛 be a multiple of 𝛿 with 𝑛 > 2𝑔 − 2. Choose a divisor 𝔠 of degree 𝑛. 𝑛+1−𝑔 By Riemann-Roch, dim(𝔠) = 𝑛 + 1 − 𝑔. By Lemma 5.1.3, there are 𝑞 𝑞−1 −1 positive divisors in the class of 𝔠. Moreover, the map 𝔞 ↦→ 𝔞 + 𝔠 defines a bijection of the set of divisor classes of degree 0 with the set of divisor classes of degree 𝑚. Hence, there are ℎ divisor 𝑛+1−𝑔 classes of degree 𝑛. It follows that 𝐴𝑛 = ℎ 𝑞 𝑞−1 −1 . □

5.2 Zeta Functions The zeta function of a function field over 𝐾 is a rational function with coefficients in Q and with simple poles at several points including 1. We define the zeta function of the function field 𝐹/𝐾 to be the Dirichlet series ∑︁ 𝜁 (𝑠) = 𝜁 𝐹/𝐾 (𝑠) = (𝑁𝔞) −𝑠 (5.2) 𝔞 ≥0

where 𝔞 runs over the positive divisors of 𝐹/𝐾. Check the domain of convergence of the series 5.2 using the substitution 𝑡 = 𝑞 −𝑠 and the identity 𝑁𝔞 = 𝑞 deg(𝔞) . We obtain a power series for 𝜁 (𝑠) in terms of 𝑡: ∑︁ ∑︁ ∑︁ (5.2) 𝑍 (𝑡) := 𝑡 deg(𝔞) = 𝑞 −𝑠·deg(𝔞) = (𝑁𝔞) −𝑠 = 𝜁 (𝑠) (5.3) 𝔞 ≥0

𝔞 ≥0

𝔞 ≥0

and write 𝑍 (𝑡) =

∑︁

𝑡 deg(𝔞) =

𝔞 ≥0

∞ ∑︁

𝐴𝑛 𝑡 𝑛 ,

(5.4)

𝑛=0

where 𝐴𝑛 is the number of positive divisors of degree 𝑛. By Lemma 5.1.4 and by (5.1), 2𝑔−2 ∞ ∑︁ ∑︁ 𝑞 𝑚 𝛿+1−𝑔 − 1 𝑚 𝛿 𝑍 (𝑡) = 𝐴𝑛 𝑡 𝑛 + ℎ 𝑡 𝑞−1 𝑛=0 𝑚=𝑑 where 𝑑 = and

2𝑔−2+ 𝛿 . 𝛿

The right-hand side converges for |𝑡| < 𝑞 −1 (i.e. for Re(𝑠) > 1)

𝑍 (𝑡) = Φ(𝑡) +

ℎ𝑞 𝑔−1+ 𝛿 𝑡 2𝑔−2+ 𝛿 ℎ 𝑡 2𝑔−2+ 𝛿 · − · , 𝑞−1 1 − (𝑞𝑡) 𝛿 𝑞 − 1 1 − 𝑡 𝛿

where Φ(𝑡) =

2𝑔−2 ∑︁

𝐴𝑛 𝑡 𝑛

𝑛=0

is a polynomial of degree ≤ 2𝑔 − 2. We summarize:

(5.5)

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5 The Riemann Hypothesis for Function Fields

Proposition 5.2.1 The power series 𝑍 (𝑡) in (5.4) converges in the circle |𝑡| < 𝑞 −1 . Formula (5.5) continues 𝑍 (𝑡) to a meromorphic function on the whole plane. The only poles of 𝑍 (𝑡) occur for values of 𝑡 with 𝑡 𝛿 = 1 or 𝑡 𝛿 = 𝑞 − 𝛿 , and they are simple. The Dirichlet series for 𝜁 (𝑠) in (5.2) converges in the right half plane Re(𝑠) > 1. The substitution 𝑡 = 𝑞 −𝑠 in (5.5) continues 𝜁 (𝑠) to a meromorphic function in the whole plane. Like the Riemann zeta function, 𝜁 𝐹/𝐾 (𝑠) has a multiplicative presentation. Let 𝑎 1 , 𝑎 2 , 𝑎 3 , . . . be a sequenceÎof complex numbers of absolute value less than 1. We ∞ 1 say that the infinite product 𝑖=1 1−𝑎𝑖 converges (resp. absolutely converges) if the limit 𝑛 𝑛 Ö Ö 1 1
lim resp. lim 𝑛→∞ 𝑛→∞ 1 − 𝑎𝑖 1 − |𝑎 𝑖 | 𝑖=1 𝑖=1 converges to a nonzero complex number. Proposition 5.2.2 If Re(𝑠) > 1 and |𝑡| < 𝑞 −1 , then Ö Ö 1 1 𝑍 (𝑡) = 𝜁 (𝑠) = = , −𝑠 deg(𝔭) 1 − (𝑁𝔭) 𝔭 𝔭 1−𝑡

(5.6)

where 𝔭 runs over the prime divisors of 𝐹/𝐾. The product converges absolutely. Therefore, it is independent of the order of the factors. In particular, if Re(𝑠) > 1, then 𝜁 (𝑠) ≠ 0. Proof. The prime divisors are free generators of the group of divisors. Thus, for every positive integer 𝑚, if Re(𝑠) > 1, then ∞ Ö ∑︁ ∑︁ ∑︁ ′ 1 −𝑠𝑘 −𝑠 = (𝑁𝔭) = (𝑁𝔞) + (𝑁𝔞) −𝑠 , 1 − (𝑁𝔭) −𝑠 𝑁𝔭 ≤𝑚 𝑘=0 𝔞≥0 𝔞≥0 𝑁𝔭 ≤𝑚

Ö

𝑁𝔞≤𝑚

𝑁𝔞>𝑚

where the prime in the second sum means that 𝔞 runs over all positive divisors with norm exceeding 𝑚 whose prime divisors 𝔭 satisfy 𝑁𝔭 ≤ 𝑚. It follows that Ö ∑︁ 1 |(𝑁𝔞) −𝑠 |. 𝜁 (𝑠) − ≤ −𝑠 1 − (𝑁𝔭) 𝔞≥0 𝑁𝔭 ≤𝑚 𝑁𝔞>𝑚

The right-hand side is the tail of a convergent series (for Re(𝑠) > 1), hence it converges to zero as 𝑚 tends to infinity. Now we prove that 𝜁 (𝑠) ≠ 0 when Re(𝑠) > 1. Indeed, consider a positive integer 𝑚. Then, Ö Ö 1 1 − (𝑁𝔭) −𝑠 = 𝜁 (𝑠) −𝑠 1 − (𝑁𝔭) 𝑁𝔭 ≤𝑚 𝑁𝔭>𝑚 ∑︁′′ ∑︁ = 1 + (𝑁𝔞) −𝑠 ≥ 1 − (𝑁𝔞) Re(𝑠) > 0, 𝔞>0 𝑁𝔞>𝑚

(where the double primes mean summation over all positive divisors 𝔞 with only prime divisors 𝔭 satisfying 𝑁𝔭 > 𝑚) if 𝑚 is large enough. Therefore, 𝜁 (𝑠) ≠ 0.

5.3 Zeta Functions under Constant Field Extensions

91

Î In particular, 𝔭 1−( 𝑁𝔭)1 −Re(𝑠) converges. Since, |𝑁𝔭−𝑠 | = 𝑁𝔭−Re(𝑠) , this means Î 1 that 𝔭 1−( 𝑁𝔭) −𝑠 absolutely converges. By Exercise 3(c), the value of the product is independent of the order of the factors. □

5.3 Zeta Functions under Constant Field Extensions The analytic properties of the zeta function of a function field 𝐹/𝐾 proved in Proposition 5.2.2 result in the conclusion that 𝛿 = 1 (Corollary 5.3.3). Denote the unique extension of 𝐾 of degree 𝑟 by 𝐾𝑟 . Then, 𝐹𝑟 = 𝐹𝐾𝑟 is a function field of one variable over 𝐾𝑟 . Use 𝑟 as a subscript to denote the “extension” of objects of 𝐹 to 𝐹𝑟 . For example, if 𝔞 is a divisor of 𝐹/𝐾, then deg𝑟 𝔞 denotes the degree of 𝔞 as a divisor of 𝐹𝑟 /𝐾𝑟 . We have already noted that deg𝑟 𝔞 = deg(𝔞), dim𝑟 𝔞 = dim(𝔞), and 𝑔𝑟 = 𝑔 (Proposition 4.4.3). Lemma 5.3.1 Let 𝔭 be a prime divisor of 𝐹/𝐾. Then, 𝔭 decomposes in 𝐹𝑟 as 𝔭 = 𝔓1 + 𝔓2 + · · · + 𝔓𝑑 , where 𝔓1 , 𝔓2 , . . . , 𝔓𝑑 are distinct prime divisors of 𝐹𝑟 /𝐾𝑟 , deg(𝔓𝑖 ) = 𝑟 −1 · lcm(𝑟, deg(𝔭)), and 𝑑 = gcd(𝑟, deg(𝔭)). Proof. Put 𝑚 = deg(𝔭). Since 𝔭 is unramified in 𝐹𝑟 (Proposition 4.4.3(c)), 𝔓1 , . . . , 𝔓𝑑 are distinct. Moreover, (𝐹𝑟 )𝔓𝑖 = 𝐾𝑟 𝐹¯𝔭 . Hence, [(𝐹𝑟 )𝔓𝑖 : 𝐾] = lcm(𝑟, 𝑚), and thus deg(𝔓𝑖 ) = [(𝐹𝑟 )𝔓𝑖 : 𝐾𝑟 ] = 𝑟 −1 · lcm(𝑟, 𝑚), 𝑖 = 1, . . . , 𝑑. Also, by Propositions 2.3.2 and 4.4.3, 𝑟 = [𝐾𝑟 : 𝐾] = [𝐹𝑟 : 𝐹] = 𝑑 · [ 𝐹¯𝑟 ,𝔓𝑖 : 𝐹¯𝔭 ] = 𝑑 · lcm(𝑟, 𝑚)𝑚 −1 . Therefore, 𝑑 = gcd(𝑟, 𝑚).

□

Proposition 5.3.2 For every complex number 𝑡, Ö 𝑍𝑟 (𝑡 𝑟 ) = 𝑍 (𝜉𝑡),

(5.7)

𝜉 𝑟 =1

where 𝜉 runs over the 𝑟th roots of unity. Proof. Since both sides of (5.7) are rational functions of 𝑡 (by (5.5)), it suffices to prove (5.7) for |𝑡| < 𝑞 −𝑟 . First apply the product formula (5.6) for 𝑍𝑟 rather than for 𝑍. Then, by Lemma 5.3.1, we have: ÖÖ Ö 𝑍𝑟 (𝑡 𝑟 ) −1 = (1 − 𝑡 𝑟 ·deg(𝔓) ) = (1 − 𝑡 lcm(𝑟 ,deg(𝔭)) ) gcd(𝑟 ,deg(𝔭)) (5.8) 𝔭

𝔭

𝔓|𝔭

Ö 𝜉 𝑟 =1

𝑍 (𝜉𝑡) −1 =

ÖÖ 𝔭

(1 − (𝜉𝑡) deg(𝔭) ).

(5.9)

𝜉 𝑟 =1

Thus, (5.7) follows if we show equality of the corresponding factors on the right-hand sides of (5.8) and (5.9). Indeed, for a fixed 𝔭 let 𝑚 = deg(𝔭) and 𝑑 = gcd(𝑟, 𝑚). We must show that Ö (1 − 𝑡 𝑟 𝑚/𝑑 ) 𝑑 = (1 − (𝜉𝑡) 𝑚 ). (5.10) 𝜉 𝑟 =1

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5 The Riemann Hypothesis for Function Fields

Substitute 𝑡 𝑚 = 𝑥 −1 in (5.10) to rewrite it as Ö (𝑥 − 𝜉 𝑚 ). (𝑥 𝑟/𝑑 − 1) 𝑑 =

(5.11)

𝜉 𝑟 =1

Both monic polynomials in (5.11) have each (𝑟/𝑑)th root of unity as a zero of multiplicity 𝑑. Indeed, if 𝜁𝑟 is a primitive root of unity of order 𝑟, then 𝜁𝑟𝑚 is a primitive root of unity of order 𝑟/𝑑 and each power of 𝜁𝑟𝑚 appears 𝑑 times among 1, 𝜁𝑟𝑚 , 𝜁𝑟2𝑚 , . . . , 𝜁𝑟(𝑟−1)𝑚 . Therefore, the polynomials are equal. □ Corollary 5.3.3 (F. K. Schmidt) 𝛿 = 1. Í 𝑚 𝛿 . Hence, if 𝜉 𝛿 = 1, then 𝑍 (𝜉𝑡) = 𝑍 (𝑡). From Proof. By (5.4), 𝑍 (𝑡) = ∞ 𝑚=0 𝐴𝑚 𝛿 𝑡 𝛿 𝛿 (5.7), 𝑍 𝛿 (𝑡 ) = 𝑍 (𝑡) . However, by (5.7) and (5.5), 𝑍 𝛿 (𝑡 𝛿 ) has a simple pole at □ 𝑡 = 1, while 𝑍 (𝑡) 𝛿 has a pole of order 𝛿 at 𝑡 = 1. Consequently, 𝛿 = 1. Corollary 5.3.4 For every integer 𝑛 there are exactly ℎ divisor classes of 𝐹/𝐾 of degree 𝑛. Proof. By Corollary 5.3.3 and by the paragraph preceding Lemma 5.1.4, 𝐹/𝐾 has a divisor 𝔠 with deg(𝔠) = 𝑛. The map 𝔞 → 𝔞 + 𝔠 induces a bijection of the set of divisor classes of 𝐹/𝐾 of degree 0 onto the set of divisor classes of degree 𝑛. Hence, the number of elements in the latter set equals the number of elements in the former set, namely ℎ. □ Remark 5.3.5 Consider the case where 𝑔 = 0. By the observation preceding Lemma 5.1.2, ℎ = 1. Since by Corollary 5.3.3, 𝛿 = 1, each integer is a multiple of 𝛿. Hence, 𝑛+1 by Lemma 5.1.4, 𝐴𝑛 = 𝑞 𝑞−1−1 for each 𝑛 ≥ 0. Therefore, by (5.4), 𝑍 (𝑡) =

∞ ∑︁ 𝑞 𝑛+1 − 1 𝑛 1 𝑡 = . 𝑞 − 1 (1 − 𝑡) (1 − 𝑞𝑡) 𝑛=0

□

5.4 The Functional Equation Like the Riemann zeta functions, 𝑍 (𝑡) satisfies a functional equation relating its 1 values in 𝑡 and 𝑞𝑡 . The main tool in the proof is the Riemann-Roch theorem. Proposition 5.4.1 𝑍 (𝑡) satisfies the functional equation   1 √ √ . ( 𝑞𝑡) 1−𝑔 𝑍 (𝑡) = ( 𝑞𝑡) 𝑔−1 𝑍 𝑞𝑡 Proof. If 𝑔 = 0, the result follows from the explicit presentation 𝑍 (𝑡) = (1−𝑡) 1(1−𝑞𝑡) in Remark 5.3.5. Therefore, assume 𝑔 > 0. The basic idea is to split 𝑍 (𝑡) into the sum of a polynomial 𝑃(𝑡) and an infinite series 𝑄(𝑡), each of which satisfies the same functional equation in the statement of the proposition.

5.4 The Functional Equation

93

Apply Lemmas 5.1.3 and 5.1.4 with 𝛿 = 1 (Corollary 5.3.3). In addition use Corollary 5.3.4 to obtain: 𝑍 (𝑡) =

∑︁

𝑡

deg(𝔞)

2𝑔−2 ∑︁

=

𝔞 ≥0

=

deg(𝐶)=0

2𝑔−2 ∑︁ deg(𝐶)=0

=

∑︁

h 1 𝑞−1

𝑡 deg(𝔞) +

𝔞≥0 𝔞∈𝐶

∞ ∑︁

𝐴𝑛 𝑡 𝑛

𝑛=2𝑔−1

∞ ∑︁ 𝑞 dim(𝐶) − 1 deg(𝐶) 𝑞 𝑛+1−𝑔 − 1 𝑛 𝑡 + ℎ 𝑡 𝑞−1 𝑞−1 𝑛=2𝑔−1 2𝑔−2 ∑︁

𝑞 dim(𝐶) 𝑡 deg(𝐶)

i

deg(𝐶)=0

+

∞ ∞ h ℎ ∑︁ ℎ ∑︁ 𝑛 i 𝑞 𝑛+1−𝑔 𝑡 𝑛 − 𝑡 𝑞 − 1 𝑛=2𝑔−1 𝑞 − 1 𝑛=0

= 𝑃(𝑡) + 𝑄(𝑡), where 𝑃(𝑡) (resp. 𝑄(𝑡)) is the expression in the first (resp. second) brackets. First we analyze 𝑃(𝑡). The Riemann-Roch theorem relates dim(𝐶) to dim(𝑊 −𝐶), where 𝑊 is the canonical class. Recall that deg(𝑊) = 2𝑔−2 (Lemma 4.2.2(b)). Hence dim(𝐶) −

1 1 deg(𝐶) = deg(𝐶) + 1 − 𝑔 + dim(𝑊 − 𝐶) 2 2 1 = dim(𝑊 − 𝐶) − deg(𝑊 − 𝐶). 2

As 𝐶 varies over all divisor classes of degree between 0 and 2𝑔 − 2 so does 𝑊 − 𝐶. Hence √

( 𝑞𝑡)

2−2𝑔

√ ( 𝑞𝑡) 2−2𝑔 𝑃(𝑡) = 𝑞−1 =

=

=

1 𝑞−1 1 𝑞−1 1 𝑞−1

2𝑔−2 ∑︁

𝑞 dim(𝐶) 𝑡 deg(𝐶)

deg(𝐶)=0

2𝑔−2 ∑︁

1 √ 𝑞 dim(𝐶)− 2 deg(𝐶) ( 𝑞𝑡) 2−2𝑔+deg(𝐶)

deg(𝐶)=0 2𝑔−2 ∑︁

1 √ 𝑞 dim(𝑊−𝐶)− 2 deg(𝑊−𝐶) ( 𝑞𝑡) − deg(𝑊−𝐶)

deg(𝐶)=0 2𝑔−2 ∑︁ deg(𝐶 ′ )=0

′

′

𝑞 dim(𝐶 ) (𝑞𝑡) − deg(𝐶 ) = 𝑃

  1 . 𝑞𝑡

Now evaluate the geometric series involved in the expression for 𝑄(𝑡): 𝑄(𝑡) =

∞ ∞ ℎ h ∑︁ 𝑛+1−𝑔 𝑛 ∑︁ 𝑛 i ℎ h 𝑞 𝑔 𝑡 2𝑔−1 1 i 𝑞 𝑡 − 𝑡 = − . 𝑞 − 1 𝑛=2𝑔−1 𝑞 − 1 1 − 𝑞𝑡 1−𝑡 𝑛=0

94

5 The Riemann Hypothesis for Function Fields

A direct computation shows that   1 √ ( 𝑞𝑡) 2−2𝑔 𝑄(𝑡) = 𝑄 . 𝑞𝑡 This completes the proof of the proposition.

□

5.5 The Riemann Hypothesis and Degree 1 Prime Divisors We reformulate here the Riemann hypothesis for a function field 𝐹 of one variable over a finite field and draw an estimate for the number of prime divisors of 𝐹 of degree 1. Rewrite formula (5.5) with 𝛿 = 1 (Corollary 5.3.3) as 𝑍 (𝑡) = Φ(𝑡) +

ℎ𝑞 𝑔 𝑡 2𝑔−1 ℎ 𝑡 2𝑔−1 · − · 𝑞 − 1 1 − 𝑞𝑡 𝑞 − 1 1 − 𝑡

where Φ(𝑡) is a polynomial of degree ≤ 2𝑔 − 2. Hence 𝑍 (𝑡) =

𝐿 (𝑡) (1 − 𝑡) (1 − 𝑞𝑡)

(5.12)

with 𝐿(𝑡) = 𝑎 0 + 𝑎 1 𝑡 + · · · + 𝑎 2𝑔 𝑡 2𝑔 a polynomial with rational coefficients. We determine some of these coefficients: First: 𝑎 0 = 𝐿(0) = 𝑍 (0) = 𝐴0 = 1, since the zero divisor is the only positive divisor of degree 0. Second: 𝐴1 is equal to the number of prime divisors of 𝐹/𝐾 of degree 1. Write 𝑁 = 𝐴1 , so that 𝐿(𝑡) = (1 − 𝑡) (1 − 𝑞𝑡)

∞ ∑︁

𝐴𝑛 𝑡 𝑛 ≡ 1 + (𝑁 − (𝑞 + 1))𝑡 mod 𝑡 2 .

𝑛=0

Therefore, 𝑎 1 = 𝑁 − (𝑞 + 1).

(5.13)

Now let 𝑥 ∈ 𝐹 be transcendental over 𝐾 and write 𝐹0 = 𝐾 (𝑥). The Zeta function of 𝐹0 /𝐾 is 𝑍0 (𝑡) = (1−𝑡) 1(1−𝑞𝑡) (Remark 5.3.5). By Proposition 5.4.1
  1 𝑞 𝑔−1 𝑡 2𝑔−2 𝑍 𝑞𝑡 𝑍 (𝑡) 1 𝑔 2𝑔 𝐿 (𝑡) = = = 𝑞 𝑡 𝐿 . (5.14)
1 −1 −2 𝑍0 (𝑡) 𝑞𝑡 𝑞 𝑡 𝑍0 𝑞𝑡 This functional equation for 𝐿(𝑡), written explicitly, has the form 2𝑔 ∑︁ 𝑖=0

Equivalently, 𝑎 𝑖 =

𝑞 𝑖−𝑔 𝑎

2𝑔−𝑖 .

𝑎 2𝑔 = 𝑞 𝑔

𝑎𝑖 𝑡 𝑖 =

2𝑔 ∑︁

𝑎 2𝑔−𝑖 𝑞 𝑖−𝑔 𝑡 𝑖 .

𝑖=0

In particular, deduce for 𝑖 = 0 and 𝑖 = 1 that
and 𝑎 2𝑔−1 = 𝑞 𝑔−1 𝑁 − (𝑞 + 1) .

These formulas imply that deg(𝐿(𝑡)) = 2𝑔. Decompose 𝐿(𝑡) over C as

(5.15)

5.6 Reduction Steps

95

𝐿(𝑡) =

2𝑔 Ö

(1 − 𝜔𝑖 𝑡)

(5.16)

𝑖=1

where the 𝜔𝑖−1 ’s are the zeros of 𝐿 (𝑡). Formulas (5.13), (5.15), and (5.16) give 𝑔

𝑞 =

2𝑔 Ö

𝜔𝑖

and

𝑁 − (𝑞 + 1) = −

𝑖=1

2𝑔 ∑︁

𝜔𝑖 .

(5.17)

𝑖=1

Moreover, the functional equation (5.14) for 𝐿 (𝑡) implies that     1 𝜔𝑖 𝐿 = 0 if and only if 𝐿 = 0. 𝜔𝑖 𝑞 √ √ √ √ Rename 𝜔1 , . . . , 𝜔2𝑔 as 𝜔1 , 𝜔1′ , . . . , 𝜔 𝑓 , 𝜔 ′𝑓 , 𝑞, . . . , 𝑞, − 𝑞, . . . , − 𝑞 with √ √ 𝑓 ≤ 𝑔 such that 𝜔𝑖 𝜔𝑖′ = 𝑞, 𝑖 = 1, . . . , 𝑓 , and 𝑞 (resp. − 𝑞) appear 𝑘 (resp. 𝑙) times. Then, 2 𝑓 + 𝑘 + 𝑙 = 2𝑔 shows that if 𝑘 is odd, then 𝑙 is odd. In this case (5.17) gives 𝑞 𝑔 = 𝑞 𝑓 𝑞 𝑘/2 (−1) 𝑙 𝑞 𝑙/2 = −𝑞 𝑔 , a contradiction. Hence, both 𝑘 and 𝑙 are even and we may take 𝑓 = 𝑔. Thus, 𝐿(𝑡) =

𝑔 Ö

(1 − 𝜔𝑖 𝑡) (1 − 𝜔𝑖′𝑡),

𝑖=1

𝜔𝑖 𝜔𝑖′

with = 𝑞 for 𝑖 = 1, . . . , 𝑔. Here is a reformulation of the Riemann hypothesis for the function field 𝐹/𝐾. Sections 5.6–5.8 complete the proof. Theorem 5.5.1 (a) The zeros of the function 𝜁 𝐹/𝐾 (𝑠) lie on the line Re(𝑠) = 12 . 1 (b) The zeros of the function 𝑍 𝐹/𝐾 (𝑡) lie on the circle |𝑡| = 𝑞 − 2 . (c) If the 𝜔𝑖 are the inverses of the zeros of the polynomial 𝐿 𝐹/𝐾 (𝑡), then √ |𝜔𝑖 | = 𝑞, 𝑖 = 1, 2, . . . , 2𝑔. By (5.3), statements (a) and (b) are equivalent. Also, (c) is equivalent to (b) since the poles of 𝑍 (𝑡) are 𝑡 = 1 and 𝑡 = 𝑞 −1 (by (5.12)). Theorem 5.5.1 with (5.17) provide an estimate on the number of prime divisors of degree 1: Theorem 5.5.2 Let 𝐹 be a function field of one variable over a finite field 𝐾 of 𝑞 elements. Denote the genus of 𝐹/𝐾 by 𝑔 and let 𝑁 be the number of prime divisors √ of 𝐹/𝐾 of degree 1. Then, |𝑁 − (𝑞 + 1)| ≤ 2𝑔 𝑞.

5.6 Reduction Steps Theorem 5.5.2 is a consequence of the Riemann hypothesis. As a first step, this section shows that an appropriate version of Theorem 5.5.2 implies the Riemann hypothesis. As in Section 5.3, denote the unique extension of 𝐾 of degree 𝑟 by 𝐾𝑟 . Lemma 5.6.1 The Riemann hypothesis holds for the function field 𝐹/𝐾 if and only if it holds for the function field 𝐹𝑟 /𝐾𝑟 .

96

5 The Riemann Hypothesis for Function Fields

Proof. We have: (5.14)

𝐿 𝑟 (𝑡 𝑟 ) =

𝑍𝑟 (𝑡 𝑟 ) (5.7) Ö 𝑍 (𝜉𝑡) (5.12) Ö = = 𝐿(𝜉𝑡) 𝑍0,𝑟 (𝑡 𝑟 ) 𝑍 (𝜉𝑡) 𝜉 𝑟 =1 0 𝜉 𝑟 =1 (5.16)

=

2𝑔 ÖÖ 𝜉 𝑟 =1

(1 − 𝜔𝑖 𝜉𝑡) =

𝑖=1

2𝑔 Ö

(1 − 𝜔𝑟𝑖 𝑡 𝑟 ).

𝑖=1

Î2𝑔

Hence, 𝐿 𝑟 (𝑡) = 𝑖=1 (1 − 𝜔𝑟𝑖 𝑡). Thus, (5.18) 𝜔𝑟1 , . . . , 𝜔𝑟2𝑔 are the inverses of the zeros of 𝐿 𝑟 . √ √ Since |𝜔𝑖 | = 𝑞 if and only if |𝜔𝑟𝑖 | = 𝑞 𝑟 , the lemma follows.

□

Denote the number of prime divisors of 𝐹𝑟 /𝐾𝑟 of degree 1 by 𝑁𝑟 . Lemma 5.6.2 Let 𝐹 be a function field of one variable over a field 𝐾 of 𝑞 elements. If there exists a constant 𝑐 such that |𝑁𝑟 − (𝑞 𝑟 + 1)| ≤ 𝑐𝑞 𝑟/2 for every positive integer 𝑟, then the Riemann hypothesis holds for 𝐹/𝐾. Proof. Apply the differential operator 𝐷 = −𝑡 Î2𝑔 𝐿(𝑡) = 𝑖=1 (1 − 𝜔𝑖 𝑡): 𝐷 (𝐿 (𝑡)) =

2𝑔 ∑︁ 𝑖=1

∞

𝑑 log 𝑑𝑡

to both sides of the formula

2𝑔

∑︁  ∑︁  𝜔𝑖 𝑡 = 𝜔𝑟 𝑡 𝑟 . 1 − 𝜔𝑖 𝑡 𝑟=1 𝑖=1 𝑖

(5.19)

Í2𝑔 Combine (5.18) with (5.17) to obtain − 𝑖=1 𝜔𝑟𝑖 = 𝑁𝑟 −(𝑞 𝑟 +1). The hypothesis of the Í2𝑔 𝑟 lemma thus implies | 𝑖=1 𝜔𝑖 | ≤ 𝑐𝑞 𝑟/2 . Therefore, the radius of convergence 𝑅 of the 1 −1 right-hand side of (5.19) satisfies 𝑅 ≥ 𝑞 − 2 . But (5.19) implies that 𝜔−1 1 , . . . , 𝜔2𝑔 are √ −1 the only singularities of 𝐷 (𝐿(𝑡)). Hence, 𝑅 = min1≤𝑖 ≤2𝑔 |𝜔𝑖 |. Therefore, |𝜔𝑖 | ≤ 𝑞 Î 2𝑔 for 𝑖 = 1, . . . , 2𝑔. This together with the equality 𝑞 𝑔 = 𝑖=1 𝜔𝑖 (5.17) implies that √ |𝜔𝑖 | = 𝑞 for 𝑖 = 1, . . . , 2𝑔, which is (c) of Theorem 5.5.1. □

5.7 An Upper Bound Assume, by extension of constants if necessary, that 𝐾 and 𝐹 satisfy these conditions: (5.20a) 𝑞 = 𝑎 2 is a square; (5.20b) 𝑞 > (𝑔 + 1) 4 ; and (5.20c) 𝐹 has a prime divisor 𝔬 of degree 1. By Lemma 5.6.1, a proof of the Riemann hypothesis under these conditions suffices for the general case. We prove a result that implies, as a special case, the inequality √ 𝑁 − (𝑞 + 1) < (2𝑔 + 1) 𝑞. (5.21) Let 𝜎 be an automorphism of 𝐹 over 𝐾. It induces a permutation of the prime divisors of 𝐹/𝐾. If 𝔭 is a prime divisor of 𝐹/𝐾, the 𝔭 𝜎 is the prime divisor corresponding to the place

5.7 An Upper Bound

97

𝜑𝔭𝜎 (𝑥) = 𝜑𝔭 (𝜎𝑥) for 𝑥 ∈ 𝐹 (where, as in Section 2.1, 𝜑𝔭𝜎 = 𝜎 −1 𝜑𝔭 ). Also, recall that the map 𝑥 ↦→ 𝑥 𝑞 is an automorphism of 𝐹˜ over 𝐾. Hence, the formula 𝜑𝔭𝑞 (𝑥) = 𝜑𝔭 (𝑥) 𝑞 defines a place 𝜑𝔭𝑞 of 𝐹/𝐾. Although 𝜑𝔭𝑞 is equivalent to 𝜑𝔭 , it is convenient to use 𝜑𝔭𝑞 , because 𝜑𝔭 = 𝜑𝔭𝑞 if and only if deg(𝔭) = 1. The remainder of this section investigates the expression ∑︁ 𝑁 ( 𝜎) = (5.22) deg(𝔭) 𝑞

𝜑𝔭𝜎 =𝜑𝔭

in order to show that

√ 𝑁 ( 𝜎) − (𝑞 + 1) < (2𝑔 + 1) 𝑞.

(5.23)

If 𝜎 is the identity automorphism, then 𝑁 ( 𝜎) = 𝑁 and (5.23) becomes (5.21). With the notation 𝑚 = 𝑎 − 1, 𝑛 = 𝑎 + 2𝑔, and 𝑟 = 𝑚 + 𝑎𝑛, rewrite (5.23) as 𝑁 ( 𝜎) − 1 ≤ 𝑟.

(5.24)

With 𝔬 as in (5.20c), consider the ascending sequence of 𝐾-vector spaces L (𝔬) ⊆ L (2𝔬) ⊆ L (3𝔬) ⊆ · · · . By Exercise 2 of Chapter 4, dim(L (𝑖𝔬)) − dim(L ((𝑖 − 1)𝔬)) ≤ 1.

(5.25)

With 𝑘 a positive integer, let 𝐼 𝑘 be the set of 𝑖, with 1 ≤ 𝑖 ≤ 𝑘, for which equality holds in (5.25). For each 𝑖 ∈ 𝐼 𝑘 , choose 𝑢 𝑖 ∈ L (𝑖𝔬) ∖ L ((𝑖 − 1)𝔬). Then, div∞ (𝑢 𝑖 ) = 𝑖𝔬 and the system (𝑢 𝑖 )𝑖 ∈𝐼𝑘 is a basis for L (𝑘𝔬). In particular, this holds for 𝑘 = 𝑚. Since 𝑎 2 = 𝑞, 𝑎 is a power of char(𝐾). Thus, the set L (𝑛𝔬) 𝑎 = {𝑦 𝑎 | 𝑦 ∈ L (𝑛𝔬)}, consisting of elements in the field 𝐹 𝑎 , is a 𝐾-vector space with basis (𝑢 𝑎𝑗 ) 𝑗 ∈𝐼𝑛 having the same dimension as L (𝑛𝔬). Therefore, n ∑︁ o L= 𝑢 𝑖 𝑦 𝑖𝑎 | 𝑦 𝑖 ∈ L (𝑛𝔬) 𝑖 ∈𝐼𝑚

is a 𝐾-vector space generated by the set 𝑈 = (𝑢 𝑖 𝑢 𝑎𝑗 )𝑖 ∈𝐼𝑚 ,

𝑗 ∈𝐼𝑛 .

Lemma 5.7.1 The set 𝑈 is linearly independent over 𝐾. Proof. It suffices to prove that the system (𝑢 𝑖 )𝑖 ∈𝐼𝑚 is linearly independent over the field 𝐹 𝑎 . Indeed, assume that there exist 𝑦 𝑖 ∈ 𝐹, 𝑖 ∈ 𝐼𝑚 , not all zero, such that Í 𝑎 𝑎 𝑖 ∈𝐼𝑚 𝑢 𝑖 𝑦 𝑖 = 0. Then, by (2.4c), there exist distinct 𝑖, 𝑗 ∈ 𝐼 𝑚 such that 𝑣 𝔬 (𝑢 𝑖 𝑦 𝑖 ) = 𝑎 𝑣𝔬 (𝑢 𝑗 𝑦 𝑗 ), 𝑦 𝑖 ≠ 0, 𝑦 𝑗 ≠ 0. Thus, −𝑖 + 𝑎𝑣𝔬 (𝑦 𝑖 ) = − 𝑗 + 𝑎𝑣𝔬 (𝑦 𝑗 ), so 𝑖 ≡ 𝑗 mod 𝑎. Since 1 ≤ 𝑖, 𝑗 ≤ 𝑚 < 𝑎, this is a contradiction. □ By Lemma 5.7.1, dim(L) = dim(L (𝑚𝔬)) · dim(L (𝑛𝔬)). Apply Riemann-Roch to the right-hand side terms: √ dim(L) ≥ (𝑚 + 1 − 𝑔) (𝑛 + 1 − 𝑔) = 𝑞 + 𝑞 − 𝑔(𝑔 + 1). (5.26) Now consider the 𝐾-vector space n ∑︁ o L′ = (𝜎 −1 𝑢 𝑖 ) 𝑎 𝑦 𝑖 | 𝑦 𝑖 ∈ L (𝑛𝔬) . 𝑖 ∈𝐼𝑚

98

5 The Riemann Hypothesis for Function Fields

Check that L ′ ⊆ L (𝑚𝑎𝔬 𝜎 + 𝑛𝔬) and deg(𝑚𝑎𝔬 𝜎 + 𝑛𝔬) = 𝑞 + 2𝑔 > 2𝑔 − 2. By Riemann-Roch, dim(L ′) ≤ dim(𝑚𝑎𝔬 𝜎 + 𝑛𝔬) = deg(𝑚𝑎𝔬 𝜎 + 𝑛𝔬) + 1 − 𝑔 = 𝑞 + 𝑔 + 1. (5.27) √ By (5.20b), 𝑞 − 𝑔(𝑔 + 1) > 𝑔 + 1. Thus, the right-hand side of (5.26) is greater than the right-hand side of (5.27). Therefore, dim(L ′) < dim(L).

(5.28)

𝜎∗

Finally, use Lemma 5.7.1 to define a 𝐾-linear map from L into  ∑︁  ∑︁ 𝜎∗ 𝑢 𝑖 𝑦 𝑖𝑎 = (𝜎 −1 𝑢 𝑖 ) 𝑎 𝑦 𝑖 . 𝑖 ∈𝐼𝑚

By (5.28), the kernel of all zero, with

𝜎∗

L′

by

𝑖 ∈𝐼𝑚

is nontrivial. Hence, there exist 𝑦 𝑖 ∈ L (𝑛𝔬), 𝑖 ∈ 𝐼𝑚 , not ∑︁ (𝜎 −1 𝑢 𝑖 ) 𝑎 𝑦 𝑖 = 0. (5.29) 𝑖 ∈𝐼𝑚

In particular, 𝑢 =

Í

𝑖 ∈𝐼𝑚

𝑢 𝑖 𝑦 𝑖𝑎

is a nonzero element of L (𝑟𝔬). Thus, div∞ (𝑢) ≤ 𝑟𝔬.

(5.30)

If 𝔭 is a prime divisor of 𝐹/𝐾 and 𝔭 ≠ 𝔬, then 𝜑𝔭 (𝑦 𝑖 ) ≠ ∞ and 𝜑𝔭 (𝑢 𝑖 ) ≠ ∞. If in addition 𝜑𝔭𝜎 = 𝜑𝔭𝑞 , then (5.29) implies ∑︁ ∑︁ 𝜑𝔭 (𝑢) = 𝜑𝔭 (𝑢 𝑖 )𝜑𝔭 (𝑦 𝑖 ) 𝑎 = 𝜑𝔭 (𝜎 −1 𝑢 𝑖 ) 𝑞 𝜑𝔭 (𝑦 𝑖 ) 𝑎 𝑖 ∈𝐼𝑚

= 𝜑𝔭

𝑖 ∈𝐼𝑚

 ∑︁

(𝜎 −1 𝑢 𝑖 ) 𝑎 𝑦 𝑖

𝑎

= 0.

𝑖 ∈𝐼𝑚

Hence, 𝔭 occurs in the divisor of zeros of 𝑢. In other words, ∑︁ 𝔭 ≤ div0 (𝑢). 𝔭≠𝔬 𝔭 𝜎 =𝔭 𝑞

Thus, (5.30)

𝑁 ( 𝜎) − 1 ≤ deg(div0 (𝑢)) = deg(div∞ (𝑢)) ≤ 𝑟 and the proof of (5.24) is complete.

5.8 A Lower Bound In the notation of Section 5.7, we establish a lower bound inequality of the form √ 𝑁 ( 𝜎) − (𝑞 + 1) > 𝑐 ′ 𝑞 where 𝑐 ′ is an explicit constant depending only on 𝐹. This will complete the construction of Lemma 5.6.2 toward the proof of the Riemann hypothesis for 𝐹/𝐾. To this end we rewrite the symbol 𝑁 ( 𝜎) introduced in (5.22) as 𝑁 ( 𝜎) (𝐹) and use that symbol also for extensions of 𝐹.

5.8 A Lower Bound

99

Lemma 5.8.1 Let 𝐹 be a function field of one variable over a field 𝐾 with 𝑞 elements. Let 𝐹 ′ be a finite Galois extension of 𝐹 with a Galois group 𝐺 such that 𝐾 is also ′ algebraically closed in 𝐹 ′ and consider Í an automorphism 𝜎 of 𝐹 that leaves 𝐹 invariant. Then, 𝑁 ( 𝜎) (𝐹) = [𝐹 ′ : 𝐹] −1 𝜏 ∈𝐺 𝑁 ( 𝜎 𝜏) (𝐹 ′). Proof. Let 𝔭 ′ be a prime divisor of 𝐹 ′/𝐾 and let 𝔭 be its restriction to 𝐹. Suppose that 𝜑𝔭𝜎 = 𝜑𝔭𝑞 . Then, there exists a 𝜏0 ∈ 𝐺 with 𝜑𝔭𝜎′ 𝜏0 = 𝜑𝔭𝑞′ [Lan97, p. 343, Cor. 2.6]. Claim: #{𝜏 ∈ 𝐺 | 𝜑𝔭𝜎′ 𝜏 = 𝜑𝔭𝑞′ } is the ramification index 𝑒𝔭 of 𝔭 ′ over 𝔭. Indeed, replacing 𝜎 by 𝜎𝜏0 we may assume that 𝜑𝔭𝜎′ = 𝜑𝔭𝑞′ . Then, for each 𝑥 ∈ 𝐹 ′ and every 𝜏 ∈ 𝐺, 𝜑𝔭𝜎′ 𝜏 (𝑥) = 𝜑𝔭𝜎′ (𝜏𝑥) = (𝜑𝔭′ (𝜏𝑥)) 𝑞 = (𝜑𝔭𝜏′ (𝑥)) 𝑞 = (𝜑𝔭𝜏′ ) 𝑞 (𝑥). Thus, 𝜑𝔭𝜎′ 𝜏 = (𝜑𝔭𝜏′ ) 𝑞 . Therefore, 𝜑𝔭𝜎′ 𝜏 = 𝜑𝔭𝑞′ ⇐⇒ (𝜑𝔭𝑞′ ) 𝜏 = 𝜑𝔭𝑞′ ⇐⇒ (𝜑𝔭𝜏′ ) 𝑞 = 𝜑𝔭𝑞′ ⇐⇒ 𝜑𝔭𝜏′ = 𝜑𝔭′ ⇐⇒ 𝜏 ∈ 𝐼𝔭′ /𝔭 , where the latter double implication follows from (2.14). It follows that #{𝜏 ∈ 𝐺 | 𝜑𝔭𝜎′ 𝜏 = 𝜑𝔭𝑞′ } = 𝑒𝔭 , as claimed. Put 𝑓𝔭 = [ 𝐹¯𝔭′′ : 𝐹¯𝔭 ] and denote the number of prime divisors of 𝐹 ′ lying over 𝔭 by 𝑔𝔭 . Then: ∑︁

𝑁 ( 𝜎 𝜏) (𝐹 ′) =

𝜏 ∈𝐺

=

∑︁

∑︁

𝜏 ∈𝐺

𝑞 𝜑𝔭𝜎′ 𝜏 =𝜑𝔭′

∑︁

deg(𝔭 ′) =

∑︁ ∑︁ 𝑞 𝜑𝔭𝜎 =𝜑𝔭

𝑒𝔭 deg(𝔭 ′)

𝔭′ |𝔭

𝑒𝔭 𝑓𝔭 𝑔𝔭 deg(𝔭) 𝑞

𝜑𝔭𝜎 =𝜑𝔭 (2.11)

= [𝐹 ′ : 𝐹]

∑︁

deg(𝔭) = [𝐹 ′ : 𝐹]𝑁 ( 𝜎) (𝐹).□

𝑞 𝜑𝔭𝜎 =𝜑𝔭

Let 𝐹 be a function field of one variable over a field 𝐾 of 𝑞 elements and let 𝜎 be an automorphism of 𝐹 over 𝐾 of finite order. Denote the fixed field of 𝜎 in 𝐹 by 𝐸. Then, 𝐹 is a finite Galois extension of 𝐸. As a finite field, 𝐾 is perfect. Hence, there exists an 𝑥 ∈ 𝐸, transcendental over 𝐾, such that 𝐸 is a finite separable extension of 𝐾 (𝑥). Let 𝐹ˆ be the Galois closure of 𝐹/𝐾 (𝑥) and let 𝐾ˆ be the algebraic closure of ˆ Then, 𝐹, ˆ as well as 𝐹 𝐾, ˆ are function fields of one variable over 𝐾ˆ and 𝜎 𝐾 in 𝐹. extends to an automorphism of 𝐹ˆ over 𝐾ˆ (𝑥). After an additional finite extension of ˆ 𝐾, ˆ 𝐾ˆ (if necessary), assume that Conditions (5.20a), (5.20b), and (5.20c) hold for 𝐹/ ˆ 𝐾ˆ (use Proposition 4.4.3). and therefore for 𝐹 𝐾/ Starting with a given 𝐹 we have extended the field of constants so as to assume these conditions: (5.31a) 𝐹/𝐾 has a separating transcendence element 𝑥; the field 𝐹 has a finite ˆ which is Galois over 𝐾 (𝑥), and 𝐾 is algebraically closed in extension 𝐹, ˆ 𝐹; ˆ 𝐾; ˆ and (5.31b) 𝑞 is a square larger than ( 𝑔ˆ + 1) 4 where 𝑔ˆ is the genus of 𝐹/ ˆ ˆ (5.31c) 𝐹/𝐾 has a prime divisor of degree 1.

100

5 The Riemann Hypothesis for Function Fields

Lemma 5.8.2 Under these conditions 𝑁 ( 𝜎) (𝐹) − (𝑞 + 1) ≥ −

𝑛−𝑚 √ (2𝑔ˆ + 1) 𝑞, 𝑚

(5.32)

where 𝑚 = [ 𝐹ˆ : 𝐹] and 𝑛 = [ 𝐹ˆ : 𝐾 (𝑥)]. ˆ ˆ (𝑥)). From Lemma 5.8.1 Proof. Let 𝐻 = Gal( 𝐹/𝐹) and 𝐺 = Gal( 𝐹/𝐾 𝑁 ( 𝜎) (𝐹) =

1 ∑︁ ( 𝜎 𝜏) ˆ 1 ∑︁ ( 𝜃) ˆ 𝑁 ( 𝐹) and 𝑞 + 1 = 𝑁 (𝐾 (𝑥)) = 𝑁 ( 𝐹). 𝑚 𝜏 ∈𝐻 𝑛 𝜃 ∈𝐺

(5.33)

Then: ∑︁

ˆ = 𝑁 ( 𝜃) ( 𝐹)

∑︁ 𝜏 ∈𝐻

𝜃 ∈𝐺

(5.23)

≤

∑︁

ˆ + 𝑁 ( 𝜎 𝜏) ( 𝐹)

𝜃 ∈𝐺 ∖ 𝜎𝐻

∑︁

ˆ + 𝑁 ( 𝜎 𝜏) ( 𝐹)

𝜏 ∈𝐻

=

∑︁

ˆ 𝑁 ( 𝜃) ( 𝐹)

𝑁

∑︁

√ (𝑞 + 1 + (2𝑔ˆ + 1) 𝑞)

𝜃 ∈𝐺 ∖ 𝜎𝐻

( 𝜎 𝜏)

ˆ + (𝑛 − 𝑚) (𝑞 + 1 + (2𝑔ˆ + 1) √𝑞). ( 𝐹)

𝜏 ∈𝐻

From the second half of (5.33), ∑︁ ˆ ≥ 𝑛(𝑞 + 1) − (𝑛 − 𝑚) (𝑞 + 1 + (2𝑔ˆ + 1) √𝑞) 𝑁 ( 𝜎 𝜏) ( 𝐹) 𝜏 ∈𝐻

√ = 𝑚(𝑞 + 1) − (𝑛 − 𝑚) (2𝑔ˆ + 1) 𝑞.

Thus, the first half of (5.33) implies that 𝑁 ( 𝜎) (𝐹) ≥ (𝑞 + 1) − as claimed.

𝑛−𝑚 √ (2𝑔ˆ + 1) 𝑞, 𝑚 □

Note that Conditions (5.31a), (5.31b), and (5.31c), as well as the numbers 𝑔, ˆ 𝑚, 𝑛, are independent of extension of the field of constants. We may therefore combine Lemma 5.8.2 with the results of Section 5.7 to conclude: Proposition 5.8.3 Let 𝐹 be a function field of one variable over a finite field 𝐾 and let 𝜎 be an automorphism of 𝐹 over 𝐾 of finite order. Then, 𝐾 has a finite extension 𝐾 ′ with 𝑞 ′ elements and there exists a positive constant 𝑐 such that for every positive integer 𝑟 we have |𝑁 ( 𝜎) (𝐹𝑟′ ) − ((𝑞 ′) 𝑟 − 1)| ≤ 𝑐(𝑞 ′) 𝑟/2 , where 𝐾𝑟′ is the unique extension of 𝐾 ′ of degree 𝑟, 𝐹𝑟′ = 𝐹 ′ 𝐾𝑟′ , and 𝜎 extends to an automorphism, also denoted by 𝜎, of 𝐹𝑟′ over 𝐾𝑟′ . In particular, Proposition 5.8.3 is valid in the case 𝜎 = 1. By Lemma 5.6.2, the Riemann hypothesis is true for the function field 𝐹 ′/𝐾 ′. It follows from Lemma 5.6.1 that it is also true for 𝐹/𝐾.

5.8 A Lower Bound

101

Exercises 1. For real-valued functions 𝑓 , 𝑔 write 𝑓 (𝑥) = 𝑂 (𝑔(𝑥)) as 𝑥 → 𝑎 if there exists a positive constant 𝑐 such that | 𝑓 (𝑥)| ≤ 𝑐|𝑔(𝑥)| for all values of 𝑥 in a neighborhood of 𝑎. Let 𝐹 be a function field of one variable over a field 𝐾 of 𝑞 elements. Denote the set of prime divisors of 𝐹/𝐾 of degree 𝑟 by 𝑃𝑟 (𝐹/𝐾). Follow these instructions to prove that 1 |𝑃𝑟 (𝐹/𝐾)| = 𝑞 𝑟 + 𝑂 (𝑞 𝑟/2 ). 𝑟 (a) Use Theorem 5.5.2 to prove that |𝑃1 (𝐹𝑟 /𝐾𝑟 )| = 𝑞 𝑟 + 𝑂 (𝑞 𝑟/2 ). (b) Observe that if 𝔓 ∈ 𝑃1 (𝐹𝑟 /𝐾𝑟 ) and if 𝔭 is the prime divisor of 𝐹/𝐾 that lies below 𝔓, then deg(𝔭)|𝑟. (c) Deduce from Lemma 5.3.1 that if 𝑑|𝑟, then over each 𝔭 ∈ 𝑃 𝑑 (𝐹/𝐾) there lie exactly 𝑑 elements of 𝑃1 (𝐹𝑟 /𝐾𝑟 ). Thus, ∑︁ |𝑃1 (𝐹𝑟 /𝐾𝑟 )| = 𝑑|𝑃 𝑑 (𝐹/𝐾)|. 𝑑 |𝑟

(d) Use the estimates |𝑃 𝑑 (𝐹/𝐾)| = and 𝑑 ≤ 𝑟2 for proper divisors 𝑑 of 𝑟 and 𝑟 𝑟 2 the inequality 𝑞 + 𝑞 + · · · + 𝑞 2 ≤ 2𝑞 2 to complete the proof. 2. Let 𝐹 be a function field of one variable over a field 𝐾 of 𝑞 elements and genus 𝑔. Let 𝜎 be an automorphism of 𝐹 over 𝐾 of finite order and let 𝑁 ( 𝜎) be as in Section √ 5.7. Prove that |𝑁 ( 𝜎) − (𝑞 + 1)| ≤ 2𝑔 𝑞. ˜ 𝐾˜ by 𝜎𝑥 ˜ Hint: Extend 𝜎 to an automorphism 𝜎 ˜ of 𝐹 𝐾/ ˜ = 𝑥 𝑞 for each 𝑥 ∈ 𝐾. ˜ Then, let 𝐸 be the fixed field of 𝜎 ˜ in 𝐹 𝐾. Show that 𝐸 is a function field of one ˜ variable over 𝐾 and that 𝐸 𝐾˜ = 𝐹 𝐾. 3. This exercise establishes basic facts about infinite products which lie behind the multiplicative presentation of the zeta functions. Consider a sequence {𝑧 𝑖 }∞ 𝑖=1 of nonzero complex numbers. If the sequence of the Î𝑛 partial productsÎ 𝑖=1 𝑧𝑖 converges to a nonzero complex number 𝑧, weÎsay that the 𝑛 infinite product ∞ and that 𝑧 is its value. We say that 𝑖=1 (1 + 𝑧𝑖 ) 𝑖=1 𝑧 𝑖 converges Î∞ absolutely converges if 𝑖=1 (1 + |𝑧𝑖 |) converges. The logarithm function makes a connection between the theory of infinite products and the theory of infinite series: 𝑂 (𝑞 𝑑 )

Proposition ([Kno28], Î p. 434): Let 𝑎 𝑖 ≠ −1, 𝑖 = 1, 2, 3, . . . be complex Í∞ numbers. Then, the product ∞ 𝑖=1 (1 + 𝑎 𝑖 ) converges if, and only if, the series 𝑖=1 log(1 + 𝑎 𝑖 ) whose terms are the principal values of log(1 + 𝑎 𝑖 ) converges. If 𝑙 is the sum of this Î 𝑙 series, then ∞ 𝑖=1 (1 + 𝑎 𝑖 ) = e . (a) Prove that if |𝑧| ≤ 12 , then |𝑧| ≤ 2| log(1 + 𝑧)| ≤ 2|𝑧|. (b) Suppose that 0 < |𝑎 𝑖 | ≤ 12 for 𝑖 = 1, 2, 3, . . . and consider the following series: ∞ ∑︁ 𝑖=1

| log(1 + 𝑎 𝑖 )|,

∞ ∑︁ 𝑖=1

|𝑎 𝑖 |,

∞ ∑︁ 𝑖=1

log(1 + |𝑎 𝑖 |).

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5 The Riemann Hypothesis for Function Fields

Use (a) to prove that the convergence of each one of these series implies the convergence of the two others. Î −1 converges. (c) Suppose that 0 < |𝑎 𝑖 | ≤ 21 for 𝑖 = 1, 2, 3, . . . and ∞ 𝑖=1 (1 − |𝑎 𝑖 |) Î∞ −1 Then, for each permutation 𝜋 of N, the product 𝑖=1 (1 − 𝑎 𝜋 (𝑖) ) converges and its value is independent of 𝜋. 4. Let 𝐹 be a function field of one variable over a field 𝐾 with 𝑞 elements. Consider 𝜎 ∈ Aut(𝐹/𝐾). As in Section 5.7, define 𝑁 ( 𝜎) to be the sum of all deg(𝔭), where 𝔭 ranges over all prime divisors of 𝐹/𝐾 for which 𝜑𝔭𝜎 = 𝜑𝔭𝑞 . Extend 𝜎 to an ˜ Use Lemma 5.3.1 for automorphism of 𝐹 𝐾˜ by the formula 𝜎𝑥 ˜ = 𝑥 𝑞 for each 𝑥 ∈ 𝐾. 𝑟 a multiple of deg(𝔭) to prove that 𝑁 ( 𝜎) is also the number of prime divisors 𝔓 of ˜ 𝐾˜ for which 𝜑 𝜎˜ = 𝜑𝑞 . 𝐹 𝐾/ 𝔓 𝔓 5. Let 𝐹 be a function field of one variable over a field with 𝑞 elements and of genus 𝑔. Let 𝜎 be an element of Aut(𝐹/𝐾) of order 𝑛. Define 𝑁 ( 𝜎) as in Exercise 4. Prove that √ |𝑁 ( 𝜎) − (𝑞 + 1)| ≤ 2𝑔 𝑞. Hint: Let 𝜎 ˜ be as in Exercise 4. Denote the fixed field of 𝜎 ˜ in 𝐹 𝐾˜ by 𝐹 ′. Prove that 𝐹 ′ is also the fixed field of 𝜎 ˜ in 𝐹𝐾𝑟 for each multiple 𝑟 of 𝑛. Deduce that 𝐹 ′ ˜ Use Exercise 4 to is a function field of one variable over 𝐾 such that 𝐹 𝐾˜ = 𝐹 ′ 𝐾. prove that 𝑁 ( 𝜎) is the number of prime divisors of 𝐹 ′/𝐾 of degree 1. Then, apply Theorem 5.5.2. 6. Let 𝑞 be a power of an odd prime. Prove that 𝑎𝑥 2 + 𝑏𝑦 2 = 𝑐 has a solution (𝑥, 𝑦) ∈ F2𝑞 if 𝑎, 𝑏, 𝑐 ∈ F×𝑞 . Let (𝑥0 , 𝑦 0 ) be one of these solutions. Use the substitution 𝑥 = 𝑥 0 + 𝑡𝑧, 𝑦 = 𝑦 0 + 𝑡 and solve for 𝑡 in terms of 𝑧 to show that the function field of 𝑎𝑋 2 + 𝑏𝑌 2 − 𝑐 = 0 is isomorphic to F𝑞 (𝑧). Conclude that the function field has exactly 𝑞 + 1 degree 1 places. Hint: Assume 𝑎 = 1 and observe that 𝑐 − 𝑏𝑦 2 takes on (𝑞 + 1)/2 values in F𝑞 . On the other hand there are 𝑞+1 2 squares in F𝑞 . 7. Let 𝑞 be a prime power ≡ 1 mod 3 and let 𝛼 be a generator of F×𝑞 . Follow these instructions to prove that 𝛼𝑋 3 + 𝛼2𝑌 3 = 1 has a solution in F𝑞 : Let 𝐴𝑖 = {𝛼𝑖 𝑥 3 | 𝑥 ∈ F×𝑞 }, 𝑖 = 0, 1, 2 and note that 𝛼𝐴0 = 𝐴1 , 𝛼𝐴1 = 𝐴2 , and 𝛼𝐴2 = 𝐴0 . Let 𝐴𝑖 +𝐴 𝑗 = {𝑥+𝑦 | 𝑥 ∈ 𝐴𝑖 ∧𝑦 ∈ 𝐴 𝑗 }. Now assume that 𝐴1 +𝐴2 ⊆ 𝐴1 ∪𝐴2 to obtain a contradiction. For this use the notation 𝐴1 + 𝐴2 = 𝑚 1 𝐴1 ∪ 𝑚 2 𝐴2 to indicate that for each 𝑎 𝑖 ∈ 𝐴𝑖 there are 𝑚 𝑖 pairs (𝑥, 𝑦) ∈ 𝐴1 × 𝐴2 such that 𝑥 + 𝑦 = 𝑎 𝑖 (independent of 𝑎 𝑖 ∈ 𝐴𝑖 ), 𝑖 = 1, 2. Multiply this “equality” by 𝛼 (resp. 𝛼2 ) to compute 𝐴2 + 𝐴0 (resp. 𝐴0 + 𝐴1 ). For 𝑎 0 ∈ 𝐴0 , use this to compute the number of triples (𝑥, 𝑦, 𝑧) ∈ 𝐴0 × 𝐴1 × 𝐴2 such that 𝑥 + 𝑦 + 𝑧 = 𝑎 0 in two ways: first compute ( 𝐴0 + 𝐴1 ) + 𝐴2 and then 𝐴0 + ( 𝐴1 + 𝐴2 ). The resulting expressions for 𝑚 1 and 𝑚 2 will lead to the contradiction 𝑚 1 = 𝑚 2 = 0. The attractiveness of the Riemann hypothesis for curves over finite fields has resulted in extensive lists of problems in a number of books treating the combinatorics of finite fields. For the sake of completeness we reference two such problem sources: [IrR72, Chap. 8, pp. 105–107; Chap. 11, pp. 169–171] and [LiN83, see notes, Chap. 6, pp. 339–346]. In those chapters that use the Riemann hypothesis, our material and problems will tend to concentrate on the connections between the

5.8 A Lower Bound

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Riemann hypothesis and arithmetic properties of fields that are special to this book (e.g. an explicit form of Hilbert’s irreducibility theorem for global fields that follows from the Riemann hypothesis – Theorem 14.3.6).

Notes An extensive survey of the literature giving estimates on the number of points on an affine variety 𝑉 appears in [LiN83, pp. 317–339]. Although a considerable literature on the Riemann hypothesis for curves over finite fields (Theorem 5.5.1) existed long before the two proofs given by Weil [Wei48], subsequent concerns included two sophisticated — and interrelated — developments. Both of Weil’s proofs employed elements of the theory of algebraic geometry outside the domain of algebraic curves. Indeed, Weil’s far reaching generalization of Theorem 5.5.1 was suggested to him by the latter proof. This generalization, now called the Riemann hypothesis for nonsingular projective varieties over finite fields, was proved by Deligne in [Del74]. However, since Weil’s theorem had so many applications to apparently elementary results about finite fields, many practitioners were anxious for a more accessible proof. Stepanov [Ste69] was the first to make serious progress on an elementary proof (e.g. not applying the theory of algebraic surfaces) of Weil’s result. He introduced elements of diophantine approximation to the problem in the case of hyperelliptic curves, in a style suggested by the Thue–Siegel–Roth theorem. To do this he constructed an auxiliary nonvanishing function on the hyperelliptic curve with a “lot” of prescribed zeros of “high” multiplicity. Eventually [Ste72] realized Theorem 5.5.1 for all hyperelliptic curves over finite fields. Continuing with the purely diophantine approximation approach, Stepanov [Ste77], for prime fields (and some extra conditions on the equations) and W. M. Schmidt [[Sch73] and [Sch73]] were able to prove Theorem 5.5.1 for all curves. Indeed, Schmidt’s method were even applicable to prove results like those of Deligne even for some complete intersections. Our proof of Theorem 5.5.1, however, follows [Bom74]. In this proof the Riemann hypotheses replaces diophantine approximation to give the construction of the auxiliary functions that appear in Stepanov’s proof (the function 𝑢 at the end of Section 5.7). Voloch [Vol85] has an elementary proof of the Riemann hypothesis for function fields that sometimes gives a better bound than Weil’s estimate (see also [StV86]).

Chapter 6

Plane Curves

The estimate on the number of prime divisors of degree 1 of a function field 𝐹 over F𝑞 (Theorem 5.5.2) leads in this chapter to an estimate on the number 𝑁 of 𝐾-rational zeros of an absolutely irreducible polynomial 𝑓 ∈ F𝑞 [𝑋, 𝑌 ]. We prove √ (Theorem 6.4.1) that |𝑁 + (𝑞 + 1)| ≤ (𝑑 − 1) (𝑑 − 2) 𝑞 + 𝑑, where 𝑑 = deg( 𝑓 ).

6.1 Affine and Projective Plane Curves Let 𝐾 be a field and let Ω be an algebraically closed extension of 𝐾 of infinite transcendence degree over 𝐾. We denote by A2 the affine plane: all pairs (𝑥, 𝑦) ∈ Ω2 . We denote by P2 the projective plane: all nonzero triples x = (𝑥 0 , 𝑥1 , 𝑥2 ) ∈ Ω3 modulo the equivalence relation x ∼ x′ if and only if x′ = 𝑐x for some 𝑐 ∈ Ω× . We denote the equivalence class of (𝑥0 , 𝑥1 , 𝑥2 ) by (𝑥0 :𝑥1 :𝑥2 ). Embed the affine plane A2 in P2 by the map (𝑥, 𝑦) → (1:𝑥:𝑦). With this understood, the points of A2 are then referred to as the finite points on P2 , whereas the points of the form (0:𝑥 1 :𝑥2 ) are the points at infinity on P2 . An affine plane curve defined over K is a set Γ = {(𝑥, 𝑦) ∈ A2 | 𝑓 (𝑥, 𝑦) = 0},

(6.1)

where 𝑓 ∈ 𝐾 [𝑋, 𝑌 ] is a nonconstant absolutely irreducible polynomial. Write it in the form 𝑓 (𝑋, 𝑌 ) = 𝑓 𝑑 (𝑋, 𝑌 ) + 𝑓 𝑑−1 (𝑋, 𝑌 ) + · · · + 𝑓0 (𝑋, 𝑌 ), (6.2) where 𝑓 𝑘 (𝑋, 𝑌 ) is a homogeneous polynomial of degree 𝑘, for 𝑘 = 0, . . . , 𝑑, and 𝑓 𝑑 (𝑋, 𝑌 ) ≠ 0. Then, 𝑑 is the degree of Γ. Attach to 𝑓 the homogeneous polynomial 𝑓 ∗ (𝑋0 , 𝑋1 , 𝑋2 ) of degree 𝑑: 𝑓 ∗ (𝑋0 , 𝑋1 , 𝑋2 ) = 𝑓 𝑑 (𝑋1 , 𝑋2 ) + 𝑋0 𝑓 𝑑−1 (𝑋1 , 𝑋2 ) + · · · + 𝑋0𝑑 𝑓0 (𝑋1 , 𝑋2 )

(6.3)

and let Γ∗ = {(𝑥0 :𝑥1 :𝑥2 ) ∈ P2 | 𝑓 ∗ (𝑥0 , 𝑥1 , 𝑥2 ) = 0}. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_6

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Then, Γ∗ is the projective plane curve corresponding to Γ. It is also called the projective completion of Γ. We have Γ = Γ∗ ∩ A2 , and Γ∗ ∖ Γ is a finite set corresponding to the points of 𝑓 𝑑 (𝑋1 , 𝑋2 ) = 0 in P1 . The infinite points of Γ∗ are sometimes referred to as the points at infinity on Γ. If 𝑓 (𝑋, 𝑌 ) = 𝑎+𝑏𝑋 +𝑐𝑌 with 𝑎, 𝑏, 𝑐 ∈ 𝐾 and 𝑏 ≠ 0 or 𝑐 ≠ 0 (i.e. deg( 𝑓 ) = 1), then Γ is a line. The points on the corresponding projective line satisfy 𝑎𝑋0 +𝑏𝑋1 +𝑐𝑋2 = 0. The line at infinity is given by 𝑋0 = 0. Call a point (𝑥, 𝑦) of an affine curve Γ generic if trans.deg𝐾 𝐾 (𝑥, 𝑦) = 1. Because 𝑓 is absolutely irreducible, the field 𝐹 = 𝐾 (𝑥, 𝑦) is a regular extension of 𝐾. Thus, 𝐹 is a function field of one variable which we call the function field of Γ over 𝐾. Up to 𝐾-isomorphism it is independent of the choice of the generic point. The map (𝑋, 𝑌 ) ↦→ (𝑥, 𝑦) extends to a 𝐾-epimorphism of rings 𝐾 [𝑋, 𝑌 ] → 𝐾 [𝑥, 𝑦] with 𝑓 (𝑋, 𝑌 )𝐾 [𝑋, 𝑌 ] as the kernel (Gauss’ lemma). Thus, a polynomial 𝑔 ∈ 𝐾 [𝑋, 𝑌 ] vanishes on Γ if and only if 𝑔(𝑥, 𝑦) = 0, or, equivalently, 𝑔 is a multiple of 𝑓 . Define the genus of Γ (and of Γ∗ ) to be the genus of 𝐹. The coordinate ring of Γ over 𝐾 is 𝑅 = 𝐾 [𝑥, 𝑦]. There is a bijective correspondence between the set of all maximal ideals 𝔪 of 𝑅 with quotient field 𝑅/𝔪 = 𝐾 and the set Γ(𝐾) of all 𝐾-rational points of Γ. If (𝑎, 𝑏) ∈ Γ(𝐾), then the corresponding maximal ideal of 𝑅 is 𝔪 = {𝑔(𝑥, 𝑦) ∈ 𝑅 | 𝑔(𝑎, 𝑏) = 0}. If 𝔭 is a prime ideal of 𝑅 and 𝔭 ⊆ 𝔪, then the transcendence degree of the quotient field of 𝑅/𝔭 is either 0 or 1. In the latter case 𝔭 = 0, and in the former case 𝑅/𝔭 is already a field so that 𝔭 = 𝔪. The local ring of Γ at a = (𝑎, 𝑏) over 𝐾 is the local ring of 𝑅 at 𝔪: n 𝑔(𝑥, 𝑦) o 𝑔(𝑥, 𝑦), ℎ(𝑥, 𝑦) ∈ 𝑅 and ℎ(𝑎, 𝑏) ≠ 0 . 𝑂 Γ, (𝑎,𝑏),𝐾 = 𝑅𝔪 = ℎ(𝑥, 𝑦) The unique nonzero prime ideal of 𝑅𝔪 is generated by the elements of 𝔪. As a local ring of a Noetherian domain, 𝑅𝔪 is itself a Noetherian domain. Similarly, a point (𝑥0 :𝑥 1 :𝑥2 ) of Γ∗ with 𝑥0 ≠ 0 is said to be generic over 𝐾, if 𝑥1 𝑥2
𝑥0 , 𝑥0 is a generic point of Γ. Define the local ring of a point a = (𝑎 0 :𝑎 1 :𝑎 2 ) of Γ∗ over 𝐾 as n 𝑔(𝑥 , 𝑥 , 𝑥 ) 𝑔, ℎ are homogeneous polynomials of o 0 1 2 𝑂 Γ∗ ,a,𝐾 = . ℎ(𝑥0 , 𝑥1 , 𝑥2 ) the same degree and ℎ(𝑎 0 , 𝑎 1 , 𝑎 2 ) ≠ 0 If 𝑎 0 ≠
0, then this ring coincides with the local ring of the corresponding point 𝑎1 𝑎2 𝑎0 , 𝑎0 of Γ. If 𝑎 1 ≠ 0, then regard a as a point on the affine curve Γ1 , defined by the equation 𝑓 ∗ (𝑋0 , 1, 𝑋2 ) = 0. Then, the projective completion of Γ1 is also Γ∗ , but the line 𝑋1 = 0 is taken as “the line at infinity”. This shows that the local ring of a projective plane curve at each point is equal to the local ring of some affine curve at a point. It is therefore a Noetherian domain. In any case, call a 𝐾-rational point of Γ (or of Γ∗ ) 𝐾-normal if its local ring over 𝐾 is integrally closed. Two affine plane curves Γ1 and Γ2 are 𝐾- isomorphic if their coordinate rings are 𝐾-isomorphic. In this case generators of the coordinate ring of Γ2 (resp. Γ1 ) can be expressed as polynomials in the generators of the coordinate ring of Γ1 (resp. Γ2 ), and the composition of these two polynomial maps is the identity when applied to generators of the coordinate ring of Γ1 (resp. Γ2 ).

6.2 Points and Prime Divisors

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Two projective plane curves Γ1∗ and Γ2∗ are 𝐾-isomorphic if (6.4a) for each x ∈ Γ1∗ there exist homogeneous polynomials 𝑔0 , 𝑔1 , 𝑔2 in 𝐾 [𝑋0 , 𝑋1 , 𝑋2 ] of the same degree such that 𝑦 𝑖 = 𝑔𝑖 (x) ≠ 0 for at least one 𝑖 with 0 ≤ 𝑖 ≤ 2 we have y ∈ Γ2∗ , and there exist homogeneous polynomials ℎ0 , ℎ1 , ℎ2 in 𝐾 [𝑋0 , 𝑋1 , 𝑋2 ] of the same degree such that ℎ𝑖 (y) = 𝑥𝑖 , 𝑖 = 0, 1, 2; and (6.4b) the same condition with the roles of Γ1∗ and Γ2∗ exchanged. For example, if 𝑔0 , 𝑔1 , 𝑔2 are linear polynomials with a nonsingular coefficient matrix, then (𝑔0 , 𝑔1 , 𝑔2 ) is called a nonsingular homogeneous linear transformation. The function fields of two isomorphic plane curves are 𝐾-isomorphic. So are the local rings of corresponding points. It follows that the genera of isomorphic plane curves are the same; and if p1 and p2 are corresponding points of the curves, then p1 is 𝐾-normal if and only if p2 is 𝐾-normal. In particular, both curves have the same number of nonnormal points. If Γ is a plane curve defined over a field 𝐾 and 𝐿 is an algebraic extension of 𝐾, then Γ is also defined over 𝐿. The function field 𝐹𝐿 of Γ over 𝐿 is the extension of 𝐹 by the field of constants 𝐿. By Proposition 4.4.3(b) the genus of Γ remains unchanged if 𝐿 is separable over 𝐾.

6.2 Points and Prime Divisors As in Section 6.1, let Γ be an affine plane curve of degree 𝑑 defined by an absolutely irreducible equation 𝑓 (𝑋, 𝑌 ) = 0 over a field 𝐾. We establish a bijective correspondence between the 𝐾-rational points of 𝑓 and the prime divisors of the function field of 𝑓 of degree 1 (Lemma 6.2.2). We prove that a 𝐾-rational point of Γ is normal if and only if it is simple (Lemma 6.2.3). Let (𝑥, 𝑦) be a generic point of Γ over 𝐾. Denote the coordinate ring and the function field, respectively, of Γ over 𝐾 by 𝑅 = 𝐾 [𝑥, 𝑦] and 𝐹 = 𝐾 (𝑥, 𝑦). Consider a 𝐾-rational point (𝑎, 𝑏) of Γ. Then, the map (𝑥, 𝑦) ↦→ (𝑎, 𝑏) uniquely extends to a 𝐾-homomorphism 𝜑 of the local ring 𝑂 (𝑎,𝑏) into 𝐾. The homomorphism 𝜑 extends ˜ further (not necessarily uniquely) to a 𝐾-valued place 𝜑 ′ of 𝐹. Call (𝑎, 𝑏) the center of the corresponding prime divisor 𝔭 of 𝐹/𝐾. Then, 𝔭 lies over the unique prime divisor 𝔭0 of 𝐾 (𝑥)/𝐾 determined by the map 𝑥 ↦→ 𝑎. Since [𝐹 : 𝐾 (𝑥)] ≤ 𝑑, there exist at most 𝑑 prime divisors 𝔭 of 𝐹/𝐾 with the point (𝑎, 𝑏) as a center on Γ. Since each point of Γ∗ is a finite point of some affine representative of Γ∗ , this holds for each 𝐾-rational point of the projective completion Γ∗ of Γ. If a point p of Γ∗ (𝐾) is 𝐾-normal, then its local ring 𝑂 p is a Noetherian integrally closed domain. By Section 6.1, each nonzero prime ideal of 𝑂 p is maximal. Hence, by Proposition 2.5.5, 𝑂 p is a discrete valuation ring. Therefore, there exists a unique prime divisor 𝔭 of 𝐹/𝐾 with p as a center on Γ. The degree of 𝔭 is 1. Conversely, consider a prime divisor 𝔭 of 𝐹/𝐾 of degree 1. Suppose that 𝜑𝔭 is finite on 𝑅. Then, (𝑎, 𝑏) = 𝜑𝔭 (𝑥, 𝑦) is a 𝐾-rational point of Γ and it is the center of 𝔭 on Γ.

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6 Plane Curves

The following result, referred to as Noether’s Normalization Theorem, shows that the number of prime divisors 𝔭 of 𝐹/𝐾 which are not finite on 𝑅 does not exceed 𝑑. Proposition 6.2.1 Let 𝐾 [𝑥1 , . . . , 𝑥 𝑛 ] be a finitely generated integral domain over a field 𝐾 with a quotient field 𝐹. If the transcendence degree of 𝐹 over 𝐾 is 𝑟, then there exist elements 𝑡1 , . . . , 𝑡𝑟 in 𝐾 [x] such that 𝐾 [x] = 𝐾 [𝑡1 , . . . , 𝑡𝑟 , 𝑥1 , . . . , 𝑥 𝑛−𝑟 ] (after rearranging the 𝑥𝑖 ’s) and 𝐾 [x] is integral over 𝐾 [t] [Lan64, p. 22]. If 𝐾 is an infinite field, then 𝑡1 , . . . , 𝑡𝑟 can be chosen to be linear combinations of 𝑥 1 , . . . , 𝑥𝑟 with coefficients in 𝐾 [ZaS58, p. 266]. Lemma 22.5.1 gives a constructive version of Noether’s normalization theorem, where 𝑡 1 , . . . , 𝑡𝑟 are chosen to be an invertible linear combinations of 𝑥1 , . . . , 𝑥𝑟 with coefficients in 𝐾. Return now to the plane curve Γ. Noether’s normalization theorem, in its linear form, allows us to replace 𝑥 by a linear combination 𝑡 of 𝑥 and 𝑦 such that 𝑅 is integral over 𝐾 [𝑡] and [𝐹 : 𝐾 (𝑡)] ≤ 𝑑. If 𝐾 is finite, replace 𝐾 by a suitable finite extension to achieve the linear dependence of 𝑡 on 𝑥 and 𝑦. If a place 𝜑𝔭 of 𝐹 is not finite on 𝑅, then it is infinite at 𝑡 (Proposition 2.5.1). Hence, 𝔭 lies over the infinite prime divisor 𝔭∞ of 𝐾 (𝑡)/𝐾. There are at most 𝑑 prime divisors of 𝐹/𝐾 that lie over 𝔭∞ . This proves our contention. Now we summarize. Lemma 6.2.2 Let Γ be an affine plane curve of degree 𝑑 defined over a field 𝐾. Denote the coordinate ring and the function field, respectively, of Γ over 𝐾 by 𝑅 and 𝐹. Then: (a) For each 𝐾-normal point p ∈ Γ(𝐾) there exists exactly one prime divisor 𝔭 of 𝐹/𝐾 with center p on Γ; the degree of 𝔭 is 1. (b) There are at most 𝑑 prime divisors of 𝐹/𝐾 whose centers on Γ are a given point p. (c) If a prime divisor 𝔭 of 𝐹/𝐾 is of degree 1 and if 𝜑𝔭 is finite on 𝑅, then its center is in Γ(𝐾). (d) There are at most 𝑑 prime divisors of 𝐹/𝐾 which are not finite on 𝑅. Finally, we point out that the 𝐾-normal points and simple points on Γ are the 𝜕𝑓 𝜕𝑓 same. Here, a point (𝑎, 𝑏) on Γ is simple if 𝜕𝑋 (𝑎, 𝑏) ≠ 0 or 𝜕𝑌 (𝑎, 𝑏) ≠ 0. Lemma 6.2.3 A 𝐾-rational point (𝑎, 𝑏) of Γ is normal if and only if it is simple. 𝜕𝑓 Moreover, if 𝜕𝑋 (𝑎, 𝑏) ≠ 0, then 𝐹 has a discrete normalized valuation 𝑣 with 𝑣(𝑦 − 𝑏) = 1. Proof. Suppose first that (𝑎, 𝑏) is normal. Then, its local ring 𝑅 = 𝑂 (𝑎,𝑏) over 𝐾 is a discrete valuation ring. Hence, 𝑅(𝑥 − 𝑎) ⊆ 𝑅(𝑦 − 𝑏) or 𝑅(𝑦 − 𝑏) ⊆ 𝑅(𝑥 − 𝑎). Suppose for example that 𝑅(𝑥 −𝑎) ⊆ 𝑅(𝑦 −𝑏). Then, there exist 𝑔, ℎ, 𝑞 ∈ 𝐾 [𝑋, 𝑌 ] such that ℎ(𝑎, 𝑏) ≠ 0 and ℎ(𝑋, 𝑌 ) (𝑋 − 𝑎) = 𝑔(𝑋, 𝑌 ) (𝑌 − 𝑏) + 𝑞(𝑋, 𝑌 ) 𝑓 (𝑋, 𝑌 ). Apply

𝜕 𝜕𝑋

to both sides and substitute (𝑎, 𝑏) for (𝑋, 𝑌 ): ℎ(𝑎, 𝑏) = 𝑞(𝑎, 𝑏)

𝜕𝑓 (𝑎, 𝑏). 𝜕𝑋

6.3 The Genus of a Plane Curve

109

𝜕𝑓 Hence, 𝜕𝑋 (𝑎, 𝑏) ≠ 0 and therefore (𝑎, 𝑏) is simple. 𝜕𝑓 Conversely, suppose that 𝜕𝑋 (𝑎, 𝑏) ≠ 0. We show that 𝑦 − 𝑏 generates the maximal ideal 𝔪 of 𝑅. Indeed,

𝑓 (𝑋, 𝑌 ) =

𝜕𝑓 𝜕𝑓 (𝑎, 𝑏) (𝑋 − 𝑎) + (𝑎, 𝑏) (𝑌 − 𝑏) + higher terms, 𝜕𝑋 𝜕𝑌

and 0 = 𝑓 (𝑥, 𝑦) =

𝜕𝑓 𝜕𝑓 (𝑎, 𝑏) (𝑥 − 𝑎) + (𝑎, 𝑏) (𝑦 − 𝑏) + (𝑥 − 𝑎)𝑢 + (𝑦 − 𝑏)𝑣, 𝜕𝑋 𝜕𝑌

with 𝑢, 𝑣 ∈ 𝔪. Hence, 𝜕𝑓  𝜕 𝑓  0= (𝑎, 𝑏) + 𝑢 (𝑥 − 𝑎) + (𝑎, 𝑏) + 𝑣 (𝑦 − 𝑏). 𝜕𝑋 𝜕𝑌 𝜕𝑓 But, since 𝜕𝑋 (𝑎, 𝑏) is a nonzero constant and 𝑢 ∈ 𝔪, the coefficient of 𝑥 − 𝑎 is a unit of 𝑅. Therefore, 𝑥 − 𝑎 ∈ 𝑅(𝑦 − 𝑏), so − 𝑎, 𝑦 − 𝑏) = 𝑅(𝑦 − 𝑏). Ñ𝔪 = 𝑅(𝑥 𝑛 Since 𝑅 is Noetherian, the ideal 𝔞 = ∞ 𝑛=1 𝔪 is a finitely generated 𝑅-module. Since 𝔪𝔞 = 𝔞, Nakayama’s Lemma [Lan64, p. 195, Prop. 1] implies that 𝔞 = 0 (alternatively, use Krull’s intersection theorem [Eis95, p. 152]). Hence, each 𝑧 ∈ 𝑅 has a unique representation 𝑧 = 𝑤(𝑦 − 𝑏) 𝑛 , with 𝑤 a unit of 𝑅 and 𝑛 ≥ 0. Thus, 𝑅 is a discrete valuation ring. Therefore, 𝑅 is integrally closed (Exercise 2 of Chapter 2). Let 𝑣 be the normalized discrete valuation of 𝐹 corresponding to 𝑅. Since 𝑦 − 𝑏 generates 𝔪, we have 𝑣(𝑦 − 𝑏) = 1. □

For a projective curve Γ∗ defined by 𝑓 ∗ (𝑋0 , 𝑋1 , 𝑋2 ) = 0 as in (6.3), the point 𝜕𝑓 (𝑎 0 :𝑎 1 :𝑎 2 ) is simple if 𝜕𝑋 (𝑎 0 , 𝑎 1 , 𝑎 2 ) ≠ 0, for some 𝑖, 𝑖 = 0, 1 or 2. Otherwise, 𝑖 (𝑎 0 :𝑎 1 :𝑎 2 ) is singular.

6.3 The Genus of a Plane Curve Here we bound the genus and the number of nonnormal points of a plane curve by a function of its degree. We first prove a finiteness result for the coordinate ring of a plane curve. Lemma 6.3.1 Let 𝑅 = 𝐾 [𝑥, 𝑦] be an integral domain with quotient field 𝐹 of transcendence degree 1 over 𝐾. Let 𝑆 be the integral closure of 𝑅 in 𝐹. Then, 𝑆/𝑅 is a finitely generated 𝐾-vector space. Proof. It is well known that 𝑆 is a finitely generated 𝑅-module ([Lan64, p. 120, Thm. 2] or [ZaS58, p. 267]). Hence, there exists a nonzero element 𝑧 ∈ 𝑅 such that 𝑧𝑆 ⊆ 𝑅. Thus, 𝑆/𝑅  𝑧𝑆/𝑧𝑅 ⊆ 𝑅/𝑧𝑅. It suffices to prove that dim𝐾 𝑅/𝑧𝑅 < ∞. By Noether’s normalization theorem (Proposition 6.2.1), we may assume without loss that 𝑅 is integral over 𝐾 [𝑥]. The element 𝑧 satisfies an equation of the form 𝑧 𝑚 + 𝑎(𝑥)𝑧 𝑚−1 + · · · + 𝑔(𝑥) = 0, with 0 ≠ 𝑔(𝑥) ∈ 𝐾 [𝑥] of degree, say, 𝑘. Since 𝑔(𝑥) belongs to 𝑧𝑅, every power of 𝑥 is a linear combination modulo 𝑧𝑅 of 1, 𝑥, . . . , 𝑥 𝑘−1 with coefficients in 𝐾.

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If 𝑑 is the degree of a monic equation for 𝑦 over 𝐾 [𝑥], then each element of 𝑅 can be written as ℎ(𝑥, 𝑦), with ℎ ∈ 𝐾 [𝑋, 𝑌 ] of degree at most 𝑑 − 1 in 𝑌 . Our lemma follows. □ In the notation of the proof of Lemma 6.3.1, 𝑆 is the integral closure in 𝐹 of the Dedekind domain 𝐾 [𝑥]. Hence, 𝑆 is a Dedekind domain and therefore every nonzero prime ideal of 𝑆 is maximal. Hence, by [AtM69, p. 61, Cor. 5.8], every nonzero prime ideal of 𝑅 is maximal. Consider a nonzero prime ideal 𝔭 of 𝑅. Let 𝑅𝔭 be the local ring of 𝑅 at 𝔭. Then, the integral closure of 𝑅𝔭 in 𝐹 is 𝑆𝔭 = 𝑏𝑠 | 𝑠 ∈ 𝑆 and 𝑏 ∈ 𝑅 ∖ 𝔭 . Lemma 6.3.2 Í (a) The quotient 𝑆/𝑅 is isomorphic to the direct sum 𝑆𝔭 /𝑅𝔭 , where 𝔭 runs over the nonzero prime ideals of 𝑅. (b) Let 𝔭1 , . . . , 𝔭𝑘 be the prime ideals of 𝑆 that lie over a prime ideal 𝔭 of 𝑅. Then, dim𝐾 𝑆𝔭 /𝑅𝔭 ≥ 𝑘 − 1. Proof of (a). Let 𝔠 = {𝑐 ∈ 𝑅 | 𝑐𝑆 ⊆ 𝑅} be the conductor of 𝑆 over 𝑅. It is a nonzero ideal of both 𝑅 and 𝑆. By Lemma 6.3.1, 𝑆 is a finitely generated 𝑅-module. Hence, by [ZaS58, p. 269, Cor.], for each nonzero prime ideal 𝔭 of 𝑅, the local ring 𝑅𝔭 is integrally closed and therefore equal to 𝑆𝔭 , if and only if 𝔠 ̸ ⊆ 𝔭. Î 𝑘 Î𝑙 (𝑖) 𝑒𝑖 𝑗 Let 𝔠 = 𝑖=1 𝑗=1 𝔭𝑖 𝑗 be a factorization of 𝔠 into the product of prime ideal powers of 𝑆 where 𝔭𝑖1 , . . . , 𝔭𝑖,𝑙 (𝑖) all lie over the same prime ideal 𝔭𝑖 of 𝑅, 𝑖 = 1, . . . , 𝑘. If 𝔭 is a nonzero prime ideal of 𝑅 not in the set {𝔭1 , . . . , 𝔭𝑘 }, then 𝑅𝔭 = 𝑆𝔭 . Hence, with 𝑅𝑖 = 𝑅𝔭𝑖 and 𝑆𝑖 = 𝑆𝔭𝑖 , 𝑖 = 1, . . . , 𝑘, the preceding paragraph and Ñ Lemma 2.5.4 gives 𝑆 ∩ 𝑅1 ∩ · · · ∩ 𝑅 𝑘 = 𝑅𝔭 = 𝑅, where 𝔭 runs over the nonzero prime ideals of 𝑅. Therefore, the Í 𝑘map 𝑠 + 𝑅 ↦→ (𝑠 + 𝑅1 , . . . , 𝑠 + 𝑅 𝑘 ), for 𝑠 ∈ 𝑆, is an injective homomorphism into 𝑖=1 𝑆𝑖 /𝑅𝑖 . The proof of (a) is complete if we show that the map is surjective. Let 𝑠1 , . . . , 𝑠 𝑘 be elements of 𝑆 and let 𝑎 1 , . . . , 𝑎 𝑘 ∈ 𝑅 with 𝑎 𝑖 ∉ 𝔭𝑖 . The ideals Î (𝑖) 𝑒𝑖, 𝑗 𝔮𝑖 = 𝑙𝑗=1 𝔭𝑖 𝑗 of 𝑆 are pairwise relatively prime, 𝑖 = 1, . . . , 𝑘. Hence, by Proposition 2.5.9, there exist 𝑠 ∈ 𝑆, 𝑞 𝑖 ∈ 𝔮𝑖 , and 𝑎 𝑖′ ∈ 𝑅 ∖ 𝔭𝑖 such that 𝑠𝑖 𝑞𝑖 = ′ 𝑖 = 1, . . . , 𝑘. 𝑎𝑖 𝑎𝑖 Î For each 𝑖 choose 𝑏 𝑖 ∈ (𝑅 ∖ 𝔭𝑖 ) ∩ 𝑗≠𝑖 𝔮 𝑗 . Then, 𝑐 𝑖 = 𝑏 𝑖 𝑞 𝑖 ∈ 𝔠 ⊆ 𝑅 and 𝑠 = Thus, 𝑠 + 𝑅𝑖 = 𝑎𝑠𝑖𝑖 + 𝑅𝑖 for 𝑖 = 1, . . . , 𝑘. 𝑠−

𝑠𝑖 𝑎𝑖

+ 𝑏𝑐𝑖 𝑎𝑖 ′ . 𝑖

Proof of (b). Let 𝔭1 , . . . , 𝔭𝑘 be the prime ideals of 𝑆 that lie over a prime ideal 𝔭 of 𝑅. For 𝑖 = 1, . . . , 𝑘 − 1 use the weak approximation theorem to find 𝑠𝑖 ∈ 𝑆 such that 𝑠𝑖 ≡ 1 mod 𝔭𝑖 and 𝑠𝑖 ≡ 0 mod 𝔭 𝑗 , for 𝑗 = 1, . . . , 𝑘 and 𝑗 ≠ 𝑖. We need only to show that 𝑠1 , . . . , 𝑠 𝑘−1 are 𝐾-linearly independent modulo 𝑅𝔭 . Indeed, Í 𝑘−1 if 𝑟 = 𝑖=1 𝑎 𝑠 ∈ 𝑅𝔭 with 𝑎 𝑖 ∈ 𝐾 (and 𝑠𝑖 ∈ 𝔭𝑘 ), then 𝑟 belongs to 𝔭𝑅𝔭 . Since Í 𝑘−1 𝑖 𝑖 𝑎 𝑗 ≡ 𝑖=1 𝑎 𝑖 𝑠𝑖 ≡ 0 mod 𝔭𝑘 , this gives 𝑎 𝑗 = 0, 𝑗 = 1, . . . , 𝑘 − 1. Consequently, dim𝐾 𝑆𝔭 /𝑅𝔭 ≥ 𝑘 − 1. □

6.3 The Genus of a Plane Curve

111

Lemma 6.3.3 Let Γ be an affine plane curve defined over a field 𝐾 with a generic point x = (𝑥1 , 𝑥2 ) and let a ∈ Γ(𝐾). Consider a separable algebraic extension 𝐿 ′ ) be the local ring of Γ at a over 𝐾 (resp. its integral of 𝐾 and let 𝑂 a,𝐾 (resp. 𝑂 a,𝐾 closure). Then, (a) 𝑂 a,𝐿 ∩ 𝐾 (x) = 𝑂 a,𝐾 ; and ′ /𝑂 ′ (b) dim𝐾 𝑂 a,𝐾 a,𝐾 ≤ dim 𝐿 𝑂 a,𝐿 /𝑂 a,𝐿 . Proof. Without loss assume that 𝐿/𝐾 is finite. Let 𝑤 1 , . . . , 𝑤 𝑛 be a linear basis for 𝐿/𝐾. Since Γ is defined by an absolutely irreducible polynomial, 𝐾 (x)/𝐾 is a regular extension of 𝐾 (Section 6.1). Hence, 𝐾 (x) and 𝐿 are linearly disjoint over 𝐾. Therefore, 𝑤 1 , . . . , 𝑤 𝑛 is also a linear basis for 𝐿(x)/𝐾 (x). Consider 𝑢 ∈ 𝑂 a,𝐿 ∩ 𝐾 (x). Then, 𝑢 = 𝑓 (x)/𝑔(x) with 𝑓 (x), 𝑔(x) ∈ 𝐿 [x] and Í Í = 𝑔𝑖 (x)𝑤 𝑖 , with 𝑓𝑖 (x), 𝑔𝑖 (x) ∈ 𝑔(a) ≠ 0. Write 𝑓 (x) = Í 𝑓𝑖 (x)𝑤 𝑖 and 𝑔(x) Í 𝑓𝑖 (x)𝑤 𝑖 deduce that 𝑢𝑔𝑖 (x) = 𝑓𝑖 (x), 𝐾 [x], 𝑖 = 1, . . . , 𝑛. From 𝑢𝑔𝑖 (x)𝑤 𝑖 = 𝑖 = 1, . . . , 𝑛. Since there is an 𝑖 with 𝑔𝑖 (a) ≠ 0, we have 𝑢 ∈ 𝑂 a,𝐾 . This proves (a). ′ To prove (b), consider 𝑢 1 , . . . , 𝑢 𝑚 ∈ 𝑂 a,𝐾 linearly dependent over 𝐿 modulo 𝑂 a,𝐿 . We show that they are also linearly dependent over 𝐾 modulo 𝑂 a,𝐾 . Indeed, Í there exist 𝑏 1 , . . . , 𝑏 𝑚 ∈ 𝐿 not all zero and 𝑣 ∈ 𝑂 a,𝐿 such that 𝑏 𝑖 𝑢 𝑖 = 𝑣. Write Í𝑚 Í𝑛 𝑏 𝑖 = 𝑗=1 𝑎 𝑖 𝑗 𝑤 𝑗 with 𝑎 𝑖 𝑗 ∈ 𝐾 and let 𝑣 𝑗 = 𝑖=1 𝑎 𝑖 𝑗 𝑢 𝑖 . Then, 𝑣 𝑗 ∈ 𝐾 (x) and Í 𝑣 𝑗𝑤 𝑗 = closure of 𝐾 (x) Í𝑣. For each𝜎𝐾 (x)-embedding 𝜎 of 𝐿 (x) into the algebraic 𝜎 ) ≠ 0 [Lan97, p. 286, we have 𝑣 𝑗 𝑤 𝜎 det(𝑤 𝑗 𝑗 = 𝑣 . Since 𝐿 (x)/𝐾 (x) is separable, Í 𝜎 Cor. 5.4]. Apply Cramer’s rule to solve the system 𝑣 𝑗 𝑤 𝜎 𝑗 = 𝑣 for all 𝜎, and write 𝜎 𝑣 𝑗 as a linear combination of the 𝑣 ’s with coefficients in the Galois closure 𝐿˜ of 𝐿/𝐾. Since a is 𝐾-rational, each 𝑣 𝜎 is in 𝑂 a, 𝐿˜ . Thus, 𝑣 𝑗 ∈ 𝑂 a, 𝐿˜ ∩ 𝐾 (x). Hence, by (a), 𝑣 𝑗 ∈ 𝑂 a,𝐾 . Therefore, the 𝑢 𝑖 ’s are linearly dependent over 𝐾 modulo 𝑂 a,𝐾 . □ Proposition 6.3.4 Let 𝐾 be an algebraically closed field and Γ∗ a projective plane curve of degree 𝑑 and genus 𝑔 defined over 𝐾. Then, ∑︁ 1 𝑔 = (𝑑 − 1) (𝑑 − 2) − dim𝐾 𝑂 p′ /𝑂 p , (6.5) 2 where p runs over the points of Γ∗ (𝐾), 𝑂 p denotes the local ring of Γ∗ at p, and 𝑂 p′ is the integral closure of 𝑂 p in the function field of Γ∗ over 𝐾. Proof. The five parts of the proof draw information from the form of Γ∗ after a change of variables that puts 𝑑 distinct points at infinity. Part A: A linear transformation. Choose a line 𝐿 in the projective plane over 𝐾 passing through no singular point of Γ∗ and not tangent to Γ∗ . Then, 𝐿 cuts Γ∗ in 𝑑 distinct simple points p1 , . . . , p𝑑 of Γ∗ [Sei68, p. 37, Cor. 6.7]. By Lemma 6.2.3, they are normal. After a suitable linear homogeneous transformation [Sei68, Chapter 5] we may assume that 𝐿 is the line at infinity and that the infinite points on the 𝑋axis and the 𝑌 -axis, (0:1:0) and (0:0:1) do not belong to the set {p1 , . . . , p𝑑 }. Such a transformation does not change the genus or the degree of the curve. Moreover, corresponding points have the same local rings. View Γ∗ as the projective completion of an affine curve Γ defined by an equation 𝑓 (𝑋, 𝑌 ) = 0, as given by (6.1). Let (𝑥, 𝑦) be a generic point of Γ over 𝐾, let 𝑅 = 𝐾 [𝑥, 𝑦] be the coordinate ring of Γ, and let 𝐹 be its function field.

112

6 Plane Curves

Part B: The divisor at ∞. By Part A, p𝑖 = (0:𝑎 𝑖 :1), with 𝑎 𝑖 ∈ 𝐾 × for 𝑖 = 1, . . . , 𝑑, and 𝑎 1 , . . . , 𝑎 𝑑 distinct. In the notation of (6.2), obtain the factorization 𝑓 𝑑 (𝑋, 𝑌 ) = 𝑐

𝑑 Ö

(𝑋 − 𝑎 𝑖𝑌 ), with 0 ≠ 𝑐 ∈ 𝐾.

(6.6)

𝑖=1

Therefore, 𝑓 (𝑥, 𝑌 ) is an irreducible polynomial of degree 𝑑 over 𝐾 [𝑥]. Thus, 𝑅 is integral over 𝐾 [𝑥] and each element of 𝑅 can be uniquely expressed as a polynomial ℎ(𝑥, 𝑦) with coefficients in 𝐾 [𝑥] such that deg𝑌 (ℎ) ≤ 𝑑 − 1. Similarly, 𝑅 is integral over 𝐾 [𝑦]. Over the infinite prime divisor 𝔭∞ of 𝐾 (𝑥) there lie exactly 𝑑 distinct prime divisors 𝔭1 , . . . , 𝔭𝑑 of 𝐹/𝐾, with 𝔭𝑖 being the unique prime divisor with center p𝑖 (Lemma 6.2.2(b)). In particular, 𝔭∞ is unramified in 𝐹 (Proposition 2.3.2) and we may therefore normalize ord𝔭𝑖 such that ord𝔭𝑖 (𝑥)
= −1. Since 𝑅 is integral over 𝐾 [𝑥], this implies that ord𝔭𝑖 (𝑦) < 0, so ord𝔭𝑖 𝑦𝑥 ≥ 0. By (6.2) and (6.6),

−1
Î𝑑 𝑥 𝑥 𝑐 𝑖=1 + · · · + 𝑓0 𝑦𝑥 , 1 𝑦 −𝑑 = 0. Hence, 𝑦𝑥 has residue 𝑎 𝑖 𝑦 − 𝑎 𝑖 + 𝑓 𝑑−1 𝑦 , 1 𝑦 at 𝔭𝑖 , so ord𝔭𝑖 (𝑦) = −1. Let 𝔬 = 𝔭1 + · · · + 𝔭𝑑 . Then, the pole divisors div∞ (𝑥) and div∞ (𝑦) of 𝑥 and 𝑦 in 𝐹 are both 𝔬. Part C: The integral Ðclosure of 𝑅 in 𝐹. Let 𝑆 be the integral closure of 𝑅 in 𝐹. By Lemma 4.1.1, 𝑆 = ∞ 𝑛=1 𝑆 𝑛 , where 𝑆 𝑛 = L (𝑛𝔬). Let 𝑅 𝑛 = 𝑅 ∩ 𝑆 𝑛 , 𝑛 = 1, 2, . . . . By Lemma 6.3.1, dim𝐾 𝑆/𝑅 < ∞. Hence, for 𝑛 sufficiently large, 𝑅 + 𝑆 𝑛 = 𝑅 + 𝑆 𝑛+1 = 𝑅 + 𝑆 𝑛+2 = . . . . Therefore, 𝑆 = 𝑅 + 𝑆 𝑛 , so 𝑆 𝑛 /𝑅𝑛  𝑆/𝑅. For 𝑛 > 2𝑔 − 2, Lemma 4.2.2(d) implies that dim𝐾 𝑆 𝑛 = 𝑛𝑑 + 1 − 𝑔. It follows that 𝑛𝑑 + 1 − 𝑔 = dim𝐾 𝑅𝑛 + dim𝐾 𝑆/𝑅. Part D:

(6.7)

For 𝑛 > 𝑑, we have 𝑅𝑛 = {ℎ(𝑥, 𝑦) ∈ 𝑅 | deg(ℎ) ≤ 𝑛 and deg𝑌 (ℎ) ≤ 𝑑 − 1}.

(6.8)

Indeed, let 𝑘 ≥ 0 be an integer and ℎ 𝑘 ∈ 𝐾 [𝑋, 𝑌 ] a homogeneous polynomial of degree 𝑘 with deg𝑌 (ℎ 𝑘 ) ≤ 𝑑 − 1. If 𝑖 + 𝑗 = 𝑘, then div∞ (𝑥 𝑖 𝑦 𝑗 ) = 𝑘𝔬 (Part B). Hence, ℎ 𝑘 (𝑥, 𝑦) ∈ 𝑆 𝑘 . We prove that ℎ 𝑘 (𝑥, 𝑦) ∉ 𝑆 𝑘−1 : Since 𝐾 is algebraically closed, we may factor ℎ 𝑘 (𝑥, 𝑦) as 𝑑−1 Ö ℎ 𝑘 (𝑥, 𝑦) = (𝑥 − 𝑏 𝑗 𝑦) · 𝑎𝑥 𝑘−𝑑+1 𝑗=1

with 𝑏 𝑗 ∈ 𝐾 and 𝑎 ∈

𝐾 ×.

Then,

ord𝔭𝑖 (ℎ 𝑘 (𝑥, 𝑦)) =

𝑑−1 ∑︁

ord𝔭𝑖 (𝑥 − 𝑏 𝑗 𝑦) − (𝑘 − 𝑑 + 1).

𝑗=1

If ℎ 𝑘 (𝑥, 𝑦) ∈ 𝑆 𝑘−1 , then for each 𝑖 we have 𝑑−1 ∑︁ 𝑗=1

ord𝔭𝑖 (𝑥 − 𝑏 𝑗 𝑦) − (𝑘 − 𝑑 + 1) + 𝑘 − 1 ≥ 0.

6.4 Points on a Curve over a Finite Field

113

Í Hence, 𝑑−1 𝑗=1 ord𝔭𝑖 (𝑥 − 𝑏 𝑗 𝑦) ≥ −𝑑 + 2, so there exists 1 ≤ 𝑗 ≤ 𝑑 − 1 with ord𝔭𝑖 (𝑥 − 𝑏 𝑗 𝑦) ≥ 0. This implies that ord𝔭𝑖 𝑦𝑥 − 𝑏 𝑗 ) ≥ 1. Therefore, 𝑏 𝑗 = 𝑎 𝑖 . Since there are 𝑑 𝑎 𝑖 ’s and at most 𝑑 − 1 𝑏 𝑗 ’s, this implies that two of the 𝑎 𝑖 ’s are equal, a contradiction. To complete Part D, write each ℎ(𝑥, 𝑦) in 𝐾 [𝑥, 𝑦] with deg𝑌 (ℎ) ≤ 𝑑 − 1 as Í ℎ(𝑥, 𝑦) = 𝑚 ℎ 𝑘=1 𝑘 (𝑥, 𝑦) with ℎ 𝑘 as above and ℎ 𝑚 (𝑥, 𝑦) ≠ 0. If 𝑚 ≤ 𝑛, then ℎ 𝑘 (𝑥, 𝑦) ∈ 𝑆 𝑘 ⊆ 𝑆 𝑛 , 𝑘 = 0, . . . , 𝑚, and therefore ℎ(𝑥, 𝑦) ∈ 𝑅𝑛 . If 𝑚 > 𝑛, then ℎ(𝑥, 𝑦) ∈ 𝑆 𝑚 ∖ 𝑆 𝑚−1 and therefore ℎ(𝑥, 𝑦) ∉ 𝑅𝑛 . Ð 𝑖 𝑗 Part E: Computation of the genus. By Part D, the set 𝑑−1 𝑗=0 {𝑥 𝑦 | 𝑖 = 0, . . . , 𝑛 − 𝑗 } is a basis of 𝑅𝑛 . Hence dim𝐾 𝑅𝑛 =

𝑑−1 ∑︁

(𝑛 − 𝑗 + 1) = 𝑛𝑑 + 1 −

𝑗=0

1 (𝑑 − 1) (𝑑 − 2). 2

Substitute this into (6.7) to conclude that 1 𝑔 = (𝑑 − 1) (𝑑 − 2) − dim𝐾 𝑆/𝑅. (6.9) 2 É By Lemma 6.3.2(a), 𝑆/𝑅 = 𝑆 𝑃 /𝑅 𝑃 , where 𝑃 ranges over all nonzero prime ideals of 𝑅. Since 𝐾 is algebraically closed, there is a bijective correspondence between the nonzero prime 𝑅 and the finite points p of Γ∗ and we have 𝑅 𝑃 = 𝑂 p . É ideals 𝑃 of ′ Hence, 𝑆/𝑅 = p∈Γ∗ (𝐾) 𝑂 p /𝑂 p . Substituting this expression for 𝑆/𝑅 into 6.9, we get (6.5). □ Corollary 6.3.5 Let Γ∗ be a projective plane curve of degree 𝑑 defined over a perfect field 𝐾. Then, genus(Γ∗ ) ≤ 12 (𝑑 − 1) (𝑑 − 2). Proof. The genus of Γ∗ does not change by going from 𝐾 to its algebraic closure (Proposition 4.4.3(b)). Now apply Proposition 6.3.4. □ Corollary 6.3.6 Let Γ∗ be a projective plane curve of degree 𝑑 defined over an algebraically closed field 𝐾. Then, genus(Γ∗ ) = 12 (𝑑 − 1) (𝑑 − 2) if and only if Γ∗ is smooth; that is, all 𝐾-rational points of Γ∗ are simple. Proof. The condition “Γ∗ is smooth” is equivalent to 𝑂 p = 𝑂 p′ for all 𝐾-rational points p of Γ∗ . Now apply Proposition 6.3.4. □

6.4 Points on a Curve over a Finite Field Using the formula for the genus of an absolutely irreducible curve Γ over F𝑞 (Proposition 6.3.4), we translate the estimate of the number of prime divisors of degree 1 of the function field of Γ to an estimate on |Γ(F𝑞 )| which involves only 𝑞 and the degree of Γ. Theorem 6.4.1 Let 𝑓 ∈ F𝑞 [𝑋, 𝑌 ] be an absolutely irreducible polynomial of degree 𝑑. Denote the affine curve defined by the equation 𝑓 (𝑋, 𝑌 ) = 0 by Γ. Then, √ √ 𝑞 + 1 − (𝑑 − 1) (𝑑 − 2) 𝑞 − 𝑑 ≤ |Γ(F𝑞 )| ≤ 𝑞 + 1 + (𝑑 − 1) (𝑑 − 2) 𝑞.

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Proof. Put 𝐾 = F𝑞 . Denote the function field of Γ over 𝐾 by 𝐹, and denote the number of prime divisors of degree 1 of 𝐹/𝐾 by 𝑁. By Theorem 5.5.2, √ √ 𝑞 + 1 − 2𝑔 𝑞 ≤ 𝑁 ≤ 𝑞 + 1 + 2𝑔 𝑞 , (6.10) where 𝑔 = genus(𝐹/𝐾). Let Γ∗ be the projective completion of Γ. For each p ∈ Γ∗ (𝐾) let 𝑘 (p) (resp. 𝑘 1 (p)) be the number of prime divisors (resp. prime divisors of degree 1) of 𝐹/𝐾 with center at p. Then, ∑︁ ∑︁ 𝑁 − |Γ∗ (𝐾)| = (𝑘 1 (p) − 1) ≤ (𝑘 (p) − 1). (6.11) p∈Γ∗ (𝐾)

p∈Γ∗ (𝐾)

Let 𝛿=

∑︁

′ dim𝐾 𝑂 p,𝐾 /𝑂 p,𝐾 and 𝛿˜ =

p∈Γ∗ (𝐾)

∑︁

′ dim𝐾˜ 𝑂 p, /𝑂 p, 𝐾˜ . 𝐾˜

(6.12)

p∈Γ∗ (𝐾)

′ /𝑂 Consider p ∈ Γ∗ (𝐾). If p is 𝐾-normal, then dim𝐾 𝑂 p,𝐾 p,𝐾 = 0 and 𝑘 1 (p) = ′ /𝑂 𝑘 (p) = 1 (Lemma 6.2.2(a)). If p is not 𝐾-normal, then dim𝐾 𝑂 p,𝐾 p,𝐾 ≥ 1. Hence, by (6.12) and by Lemma 6.3.3(b), ∑︁ ˜ (1 − 𝑘 1 (p)) ≤ 𝛿 ≤ 𝛿. p∈Γ∗ (𝐾)

We conclude that (6.11)

∑︁

|Γ(𝐾)| ≤ |Γ∗ (𝐾)| = 𝑁 +

(6.10) √ (1 − 𝑘 1 (p)) ≤ 𝑞 + 1 + 2𝑔 𝑞 + 𝛿˜

p∈Γ∗ (𝐾)

√ √ = 𝑞 + 1 + (𝑑 − 1) (𝑑 − 2) 𝑞 − 2𝛿˜ 𝑞 + 𝛿˜ √ ≤ 𝑞 + 1 + (𝑑 − 1) (𝑑 − 2) 𝑞.

(6.5)

Next we give a lower bound for |Γ(𝐾)|. Note first that Γ∗ has at most 𝑑 points at infinity [Sei68, p. 37, Cor. 6.7]. Hence, |Γ(𝐾)| ≥ |Γ∗ (𝐾)| − 𝑑. Therefore, ∑︁ (6.11) |Γ(𝐾)| ≥ |Γ∗ (𝐾)| − 𝑑 ≥ 𝑁 − (𝑘 (p) − 1) − 𝑑 p∈Γ∗ (𝐾) 6.3.2(𝑏)

≥

𝑁−

∑︁

′ dim𝐾 𝑂 𝐾 ,p /𝑂 𝐾 ,p − 𝑑 = 𝑁 − 𝛿 − 𝑑

p∈Γ∗ (𝐾)

√ ≥ 𝑞 + 1 − 2𝑔 𝑞 − 𝛿˜ − 𝑑

(6.10)

√ √ = 𝑞 + 1 − (𝑑 − 1) (𝑑 − 2) 𝑞 + 2𝛿˜ 𝑞 − 𝛿˜ − 𝑑 √ ≥ 𝑞 + 1 − (𝑑 − 1) (𝑑 − 2) 𝑞 − 𝑑.

(6.5)

This completes the proof of the theorem.

□

Corollary 6.4.2 The curve Γ of Theorem 6.4.1 satisfies the following conditions: (a) For each 𝑚 there exists a 𝑞 0 such that |Γ(F𝑞 )| ≥ 𝑚 for all 𝑞 ≥ 𝑞 0 . (b) If 𝑞 > (𝑑 − 1) 4 , then Γ(𝐾) is not empty. √ Proof of (b). By Theorem 6.4.1, |Γ(F𝑞 )| ≥ 𝑞 + 1 − (𝑑 − 1) (𝑑 − 2) 𝑞 − 𝑑 =
√ √ √ 𝑞 𝑞 − (𝑑 − 1) (𝑑 − 2) − (𝑑 − 1) > 𝑞 − (𝑑 − 1) > 0. □

6.4 Points on a Curve over a Finite Field

115

Exercises 1. Use Proposition 6.3.4 to prove that any projective plane curve of degree 2 is smooth. 2. Let Γ∗ be a projective plane curve of degree 3. Use Proposition 6.3.4 to show that either Γ∗ is smooth, in which case it has genus 1, or Γ∗ has exactly one singular point, in which case it has genus 0 and therefore its function field is rational (Example 4.2.4). 3 2 2 3. Consider the affine plane curve Γ defined by the equation √ 𝑌 − 2𝑋 − 𝑋 = 0 over a field 𝐾 of characteristic ≠ 2 that does not contain 2. Note that (0, 0) is a singular point of Γ. Let (𝑥, 𝑦) be a generic point of Γ over 𝐾. Show that the map ˜ (𝑥, 𝑦) → (0, 0) does not extend to a 𝐾-rational (rather than 𝐾-rational) place of 𝐾 (𝑥, 𝑦) (i.e. the singular point of Γ is not the center of a 𝐾-rational place of the function field of Γ). Hint: Consider the element 𝑦𝑥 of 𝐾 (𝑥, 𝑦).

4. Prove directly that the function field of the curve 𝑌 2 − 2𝑋 2 − 𝑋 3 = 0 is rational. 5. Consider the projective plane curve Γ∗ defined for 𝑑 ≥ 2 by 𝑋0 𝑋1𝑑−1 − 𝑋0 𝑋2𝑑−1 − 𝑋2𝑑 = 0 over an algebraically closed field 𝐾. (a) Use Lemma 6.2.3 to show that the only nonnormal point of Γ∗ is (1:0:0). (b) Let (𝑥, 𝑦) be a generic point of the affine part Γ of Γ∗ defined by 𝑋 𝑑−1 − 𝑌 𝑑−1 − 𝑌 𝑑 = 0 and let 𝑧 = 𝑦𝑥 . Use Lemma 2.5.4 to conclude that 𝐾 [𝑦, 𝑧] is the integral closure of 𝐾 [𝑥, 𝑦] in 𝐾 (𝑥, 𝑦). Conclude that the genus of Γ is 0. 𝑦,𝑧) (c) Prove that 𝑆0 = { 𝑔𝑓 ((𝑥,𝑦) | 𝑓 ∈ 𝐾 [𝑌 , 𝑍], 𝑔 ∈ 𝐾 [𝑋, 𝑌 ], 𝑔(0, 0) ≠ 0} is the integral closure of the local ring 𝑅0 of Γ at (0, 0). (d) Use Proposition 6.3.4 to conclude that the set {𝑥 𝑖 𝑧 𝑗 | 𝑗 = 1, . . . , 𝑑 − 2; 𝑖 = 0, . . . , 𝑗 − 1} gives a basis for 𝑆0 /𝑅0 over 𝐾. 6. Count the number of points on the projective plane curve 𝑋03 + 𝑋13 + 𝑋23 = 0 over the finite field F𝑞 , where 𝑞 is a prime power such that gcd(𝑞 − 1, 3) = 1. Hint: The map 𝑥 ↦→ 𝑥 3 from F×𝑞 into itself is bijective. 7. Let 𝑓 ∈ F𝑞 [𝑋] be a polynomial of degree 𝑑 such that 𝑔(𝑋, 𝑌 ) = 𝑓 (𝑌𝑌)−−𝑋𝑓 (𝑋) is absolutely irreducible. Suppose that either 𝑞 is not a power of 2 or 𝑑 ≥ 3. Use Theorem 6.4.1 to prove that if 𝑞 > (𝑑 − 1) 4 , then there exist distinct 𝑥, 𝑦 ∈ F𝑞 such that 𝑓 (𝑦) = 𝑓 (𝑥). Hint: Observe that 𝑔(𝑋, 𝑋), the derivative of 𝑓 (𝑋), has at most 𝑑 − 1 zeros.

Notes The proof of Proposition 6.3.4 is an elaboration on [Sam66, p. 52, Thm. 2]. Denote the maximum number of F𝑞 -points on a curve of genus 𝑔 which is defined over F𝑞 by 𝑁𝑞 (𝑔). For fixed 𝑞, put 𝐴(𝑞) = lim sup 𝑔1 𝑁𝑞 (𝑔). Weil’s estimate 𝑔→∞ √ 𝑁𝑞 (𝑔) ≤ 𝑞 + 1 + 2𝑔 𝑞 (Theorem 5.5.2 and the proof of Theorem 6.4.1) implies

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that 𝐴(𝑞) ≤ 2𝑞 2 . [Ser83] improves Weil’s estimate (via an interpretation of the Frobenius automorphism as an endomorphism on Jacobians) to give the bound √ 𝑁𝑞 (𝑔) ≤ 𝑞 + 1 + 𝑔[2 𝑞] (where [x] is the greatest integer not exceeding 𝑥). Thus, √ √ 𝐴(𝑞) ≤ [2 𝑞]. But [DrV83] obtains the much improved estimate 𝐴(𝑞) ≤ 𝑞 − 1. When 𝑞 is a square [Iha81] and [TVZ82] have shown this bound to be exact. As for a lower estimate, [Ser83] proves the existence of 𝑐 > 0 such that 𝐴(𝑞) ≥ 𝑐 log(𝑞) 8 (e.g. 𝐴(2) ≥ 39 ). The exact lower and upper bound for 𝐴(𝑞) for general 𝑞 have yet to be found. The case 𝑞 = 2 has applications to coding theory, as first noted by [Gop81] (or [LiN83, p. 530] for a survey of work in this direction).

Chapter 7

The Chebotarev Density Theorem

The major connection between the theory of finite fields and the arithmetic of number fields and function fields is the Chebotarev density theorem. Explicit decision procedures and transfer principles of Chapters 23 and 35 depend on the theorem or some analogs. In the function field case our proof, using the Riemann hypothesis for curves, is complete and elementary. In particular, we make no use of the theory of analytic functions. The number field case, however, uses an asymptotic formula for the number of ideals in an ideal class, and only simple properties of analytic functions. In particular, we do not use Artin’s reciprocity law (or any equivalent formulation of class field theory). This proof is close to Chebotarev’s original field crossing argument, which gave a proof of a piece of Artin’s reciprocity law for cyclotomic extensions.

7.1 Decomposition Groups Let 𝑅 be an integrally closed domain with quotient field 𝐾. Consider a finite Galois extension 𝐿 of 𝐾 with Galois group 𝐺 and denote the integral closure of 𝑅 in 𝐿 by 𝑆. Suppose that 𝔭 is a prime ideal of 𝑅. By Chevalley’s theorem (Proposition 2.3.1), there exists a prime ideal 𝔓 of 𝑆 lying over 𝔭 (i.e. 𝔭 = 𝑅 ∩ 𝔓). Denote the quotient ¯ The set of all 𝜎 ∈ 𝐺 satisfying fields of 𝑅/𝔭 and 𝑆/𝔓, respectively, by 𝐾¯ and by 𝐿. 𝜎(𝔓) = 𝔓 is a group 𝐷 𝔓 called the decomposition group of 𝔓 over 𝐾. Its fixed field in 𝐿 is the decomposition field of 𝔓 over 𝐾. ¯ Each 𝜎 ∈ 𝐷 𝔓 induces For 𝑥 ∈ 𝑆 denote the equivalence class of 𝑥 modulo 𝔓 by 𝑥. ¯ 𝑥¯ = 𝜎𝑥 for each 𝑥 ∈ 𝑆. The map ¯ of 𝐿¯ over 𝐾¯ satisfying 𝜎 a unique automorphism 𝜎 ¯ 𝐾). ¯ Its kernel is the inertia group 𝜎 ↦→ 𝜎 ¯ is a homomorphism of 𝐷 𝔓 into Aut( 𝐿/ 𝐼𝔓 of 𝔓 over 𝐾, 𝐼𝔓 = {𝜎 ∈ 𝐺 | 𝜎𝑥 ∈ 𝑥 + 𝔓 for every 𝑥 ∈ 𝑆}. The fixed field of 𝐼𝔓 in 𝐿 is the inertia field of 𝔓 over 𝐾. If 𝜎 ∈ 𝐺, then 𝜎𝑆 = 𝑆 and 𝜎𝔓 is another prime ideal of 𝑆 that lies over 𝔭. In this case 𝐷 𝜎𝔓 = 𝜎 · 𝐷 𝔓 · 𝜎 −1 and 𝐼 𝜎𝔓 = 𝜎 · 𝐼𝔓 · 𝜎 −1 . Conversely, any two prime ideals of 𝑆 lying over the same prime ideal of 𝑅 are conjugate over 𝐾 [Lan97, p. 340]. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_7

117

118

7 The Chebotarev Density Theorem

In the proof of Lemma 7.1.1 we use the expression “to localize 𝑅 and 𝑆 at 𝔭”. This means that we replace 𝑅 by 𝑅𝔭 , 𝔭 by 𝔭𝑅𝔭 , 𝑆 by 𝑆𝔭 = { 𝑎𝑠 | 𝑠 ∈ 𝑆 and 𝑎 ∈ 𝑅 ∖ 𝔭}, and 𝔓 by 𝔓𝑆𝔭 . The local ring 𝑅𝔭 is integrally closed, 𝔭𝑅𝔭 is its maximal ideal, and 𝐾¯ = 𝑅𝔭 /𝔭𝑅𝔭 . In addition, 𝑆𝔭 is the integral closure of 𝑅𝔭 in 𝐿 [Lan97, p. 338, Prop. 1.8 and 1.9], and 𝔓𝑆𝔭 is a prime ideal that lies over 𝔭𝑅𝔭 . Thus, 𝑆𝔭 /𝔓𝑆𝔭 is a ¯ Hence, 𝑆𝔭 /𝔓𝑆𝔭 = 𝐿¯ is a field [Lan97, domain which is integral over the field 𝐾. p. 339, Prop. 1.11] and 𝔓𝑆𝔭 is a maximal ideal. Moreover, 𝐷 𝔓𝑆𝔭 = 𝐷 𝔓 and 𝐼𝔓𝑆𝔭 = 𝐼𝔓 . Lemma 7.1.1 ¯ 𝐾¯ is normal, and the map 𝜎 ↦→ 𝜎 ¯ 𝐾) ¯ (a) The field extension 𝐿/ ¯ from 𝐷 𝔓 into Aut( 𝐿/ is surjective. ¯ 𝐾¯ is separable and [ 𝐿¯ : 𝐾] ¯ = [𝐿 : 𝐾]. Then, 𝐿/ ¯ 𝐾¯ is Galois, (b) Suppose that 𝐿/ 𝐷 𝔓 = Gal(𝐿/𝐾), and the map 𝜎 ↦→ 𝜎 ¯ is an isomorphism of Gal(𝐿/𝐾) onto ¯ 𝐾). ¯ Gal( 𝐿/ Proof of (a). Denote the decomposition field of 𝔓 by 𝐿 0 , let 𝑆0 = 𝑆 ∩ 𝐿 0 , and let 𝔓0 = 𝔓 ∩ 𝐿 0 . We prove that 𝑆0 /𝔓0 = 𝑅/𝔭. Consider 𝑥 ∈ 𝑆0 . We need only to find 𝑎 ∈ 𝑅 such that 𝑥 ≡ 𝑎 mod 𝔓0 . For each 𝜎 ∈ 𝐺 ∖ 𝐷 𝔓 we have 𝜎 −1 𝔓 ≠ 𝔓. If 𝜎 −1 𝔓 ∩ 𝐿 0 = 𝔓0 , then there exists a 𝜏 ∈ Gal(𝐿/𝐿 0 ) = 𝐷 𝔓 such that 𝜏𝜎 −1 𝔓 = 𝔓. Therefore, 𝜎 ∈ 𝐷 𝔓 , a contradiction. Thus, 𝜎 −1 𝔓∩ 𝐿 0 ≠ 𝔓0 . Localize at 𝔭, to assume that 𝔭 is a maximal ideal of 𝑅. Then, 𝔓0 and 𝜎 −1 𝔓 ∩ 𝐿 0 are maximal ideals of 𝑆0 . Hence, 𝔓0 + (𝜎 −1 𝔓 ∩ 𝐿 0 ) = 𝑆0 . By the Chinese remainder theorem [Lan97, p. 94], there exists a 𝑦 ∈ 𝑆0 with 𝑦 ≡ 𝑥 mod 𝔓0 and 𝑦 ≡ 1 mod 𝜎 −1 𝔓 ∩ 𝐿 0 for every 𝜎 ∈ 𝐺 ∖ 𝐷 𝔓 . Thus, 𝑦 ≡ 𝑥 mod 𝔓 and 𝜎𝑦 ≡ 1 mod 𝔓 for every 𝜎 ∈ 𝐺 ∖ 𝐷 𝔓 . Since 𝐿 0 /𝐾 is a separable extension, the element 𝑎 = norm 𝐿0 /𝐾 (𝑦) of 𝑅 is a product of 𝑦 and elements 𝜎𝑦 with 𝜎 running over nonidentity coset representatives of 𝐷 𝔓 in 𝐺. Consequently, 𝑎 ≡ 𝑥 mod 𝔓0 , as desired. To continue, localize 𝑆0 and 𝑆 at 𝔓0 to assume 𝐾¯ = 𝑆0 /𝔓0 and 𝐿¯ =Î𝑆/𝔓. Write 𝑚 each element of 𝐿¯ as 𝑥, ¯ with 𝑥 ∈ 𝑆, and let 𝑓 = irr(𝑥, 𝐿 0 ). Then, 𝑓 (𝑋) = 𝑖=1 (𝑋−𝑥𝑖 ), where 𝑥 𝑖 ∈ 𝑆, 𝑖 = 1, . . . , 𝑚, are the conjugates of 𝑥 over 𝐿 0 . The coefficients of 𝑓 (𝑋) belong to 𝐿 0 ∩𝑆 = 𝑆0 . Hence, 𝑓¯(𝑋) ∈ 𝐾¯ [𝑋]. In addition, 𝑥¯ is a root of the polynomial Î𝑚 ¯ Hence, 𝐿¯ is a normal extension of 𝐾. ¯ Moreover, 𝑓¯(𝑋) = 𝑖=1 (𝑋 − 𝑥¯𝑖 ) with roots in 𝐿. ¯ ≤ deg( 𝑓 ) ≤ [𝐿 : 𝐿 0 ]. [ 𝐾¯ ( 𝑥) ¯ : 𝐾] Separable extensions have primitive generators [Lan97, p. 243]. Hence, the max¯ ≤ [𝐿 : 𝐿 0 ] and imal subfield 𝐸 of 𝐿¯ which is separable over 𝐾¯ satisfies [𝐸 : 𝐾] ¯ 𝐾) ¯  Gal(𝐸/𝐾), ¯ 𝐸 = 𝐾¯ ( 𝑥) ¯ for some 𝑥 ∈ 𝑆. In the above notation, if 𝜏 ∈ Aut( 𝐿/ we have 𝜏 𝑥¯ = 𝑥¯ 𝑗 for some 𝑗, 1 ≤ 𝑗 ≤ 𝑚. The map 𝑥 ↦→ 𝑥 𝑗 extends to a field automorphism 𝜎 ∈ Gal(𝐿/𝐿 0 ) with 𝜎 ¯ = 𝜏. Consequently, the map 𝜎 ↦→ 𝜎 ¯ from ¯ 𝐾) ¯ is surjective. 𝐷 𝔓 into Aut( 𝐿/ ¯ 𝐾¯ is separable by assumption and normal by (a), Proof of (b). The extension 𝐿/ ¯ = |Gal( 𝐿/ ¯ 𝐾)| ¯ ≤ hence it is Galois. Moreover, by assumption, [𝐿 : 𝐾] = [ 𝐿¯ : 𝐾] ¯ 𝐾)|. ¯ Therefore, 𝐷 𝔓 = |𝐷 𝔓 | ≤ [𝐿 : 𝐾]. Hence, |𝐷 𝔓 | = |Gal(𝐿/𝐾)| = |Gal( 𝐿/ ¯ 𝐾) ¯ is an isomorphism. Gal(𝐿/𝐾) and the map Gal(𝐿/𝐾) → Gal( 𝐿/ □ The discriminant of 𝑓 ∈Î𝑅[𝑋], disc( 𝑓 ), gives information about ramification. 𝑛 Assume that 𝑓 is monic and 𝑖=1 (𝑋 − 𝑥𝑖 ) is the factorization of 𝑓 into linear factors. Then

7.1 Decomposition Groups

119

disc( 𝑓 ) = (−1)

𝑛(𝑛−1) 2

Ö

(𝑥 𝑖 − 𝑥 𝑗 ) = (−1)

𝑖≠ 𝑗

𝑛(𝑛−1) 2

𝑛 Ö

𝑓 ′ (𝑥 𝑗 ).

(7.1)

𝑗=1

This is an element of 𝑅, and disc( 𝑓 ) ≠ 0 if and only if the 𝑥𝑖 ’s are distinct. Assume that 𝑓 is irreducible. Then, (7.1) implies that disc( 𝑓 ) = (−1)

𝑛(𝑛−1) 2

norm𝐾 ( 𝑥1 )/𝐾 ( 𝑓 ′ (𝑥1 )).

We call norm𝐾 ( 𝑥1 )/𝐾 ( 𝑓 ′ (𝑥1 )) the discriminant of 𝑥1 over 𝐾. Lemma 7.1.2 Let 𝑅 be an integrally closed domain with quotient field 𝐾. Let 𝑆 be the integral closure of 𝑅 in a finite separable extension 𝐿 of 𝐾. Assume that 𝐿 = 𝐾 (𝑧) with 𝑧 integral over 𝑅 and let 𝑓 = irr(𝑧, 𝐾). Suppose that 𝑑 = disc( 𝑓 ) is a unit of 𝑅. Then, 𝑆 = 𝑅[𝑧]. Proof. Let 𝑛 = [𝐿 : 𝐾]. Assume that 𝑦 = 𝑎 0 +𝑎 1 𝑧+· · ·+𝑎 𝑛−1 𝑧 𝑛−1 , with 𝑎 𝑖 ∈ 𝐾, is an element of 𝑆. We must prove that 𝑎 𝑖 ∈ 𝑅, 𝑖 = 0, . . . , 𝑛 − 1. To this end, let 𝜎1 , . . . , 𝜎𝑛 ˜ Then, each of 𝜎𝑖 𝑦 = Í𝑛−1 𝑎 𝑗 𝜎𝑖 𝑧 𝑗 , 𝑖 = be the isomorphisms of 𝐿 over 𝐾 into 𝐾. 𝑗=0 1, . . . , 𝑛, is integral over 𝑅. 𝑏 Δ𝑏 Solve for the 𝑎 𝑗 ’s by Cramer’s rule. This gives 𝑎 𝑗 = Δ𝑗 = Δ2𝑗 , 𝑗 = 0, . . . , 𝑛 − 1, where Δ = det(𝜎𝑖 𝑧 𝑗 ) with 𝑏 𝑗 integral over 𝑅, 𝑗 = 0, . . . , 𝑛 − 1. But Δ is a Vandermonde determinant: Ö Δ2 = ± (𝜎𝑖 𝑧 − 𝜎 𝑗 𝑧) = ±norm 𝐿/𝐾 𝑓 ′ (𝑧) = ±𝑑. 𝑖≠ 𝑗

Since 𝑑 is a unit of 𝑅, 𝑎 𝑗 is integral over 𝑅, 𝑗 = 0, . . . , 𝑛 − 1. Since 𝑅 is integrally closed, all 𝑎 𝑗 are in 𝑅. □ Definition 7.1.3 (Ring covers) As in the preceding lemmas, consider two integrally closed integral domains 𝑅 ⊆ 𝑆 with 𝐾 ⊆ 𝐿 their respective quotient fields such that 𝐿/𝐾 is finite and separable. Suppose that 𝑆 = 𝑅[𝑧], where 𝑧 is integral over 𝑅 and the discriminant of 𝑧 over 𝐾 is a unit of 𝑅. In this set up we say that 𝑆/𝑅 is a ring cover and 𝐿/𝐾 is the corresponding field cover. In this case, Lemma 7.1.2 implies that 𝑆 is the integral closure of 𝑅 in 𝐿. Call the element 𝑧 a primitive element for the cover. If in addition 𝐿/𝐾 is Galois, call 𝑆/𝑅 a Galois ring cover for 𝐿/𝐾. Remark 7.1.4 Every ring cover is “standard étale” [Ray70, p. 19].

□

Remark 7.1.5 (Creating ring covers) Let 𝐾 = 𝐾0 (𝑥1 , . . . , 𝑥 𝑛 ) be a finitely generated extension of a field 𝐾0 and let 𝐿 be a finite separable extension of 𝐾. The subring 𝑅 = 𝐾0 [𝑥1 , . . . , 𝑥 𝑛 ] of 𝐾 is not necessarily integrally closed. But there exists a nonzero 𝑦 ∈ 𝑅 with 𝑅 ′ = 𝐾0 [𝑥1 , . . . , 𝑥 𝑛 , 𝑦 −1 ] integrally closed ([Lan64, p. 120, Thm. 2]; a constructive proof of this fact appears in the proof of Lemma 22.7.2. Suppose that 𝑧 is a primitive generator for 𝐿/𝐾, 𝑓 ∈ 𝑅[𝑍] is an irreducible polynomial over 𝐾, and 𝑓 (𝑧) = 0. Multiply 𝑦 by the product of the leading coefficient and the discriminant of 𝑓 . Then, 𝑆 ′ = 𝑅 ′ [𝑧] is a ring cover of 𝑅 ′ with 𝑧 a primitive element.

120

7 The Chebotarev Density Theorem

Remark 7.1.6 (Decomposition groups of places) Suppose that 𝐿/𝐾 is a finite Galois extension and 𝜑 is a place of 𝐿 with a valuation ring 𝑂. Then, 𝐿¯ = 𝜑(𝑂) is the residue field of 𝐿 under 𝜑. Also, 𝑅 = 𝑂 ∩ 𝐾 is the valuation ring of the restriction of 𝜑 to 𝐾 and 𝐾¯ = 𝜑(𝑅) is its residue field. By Proposition 2.5.1, 𝑂 contains the integral closure 𝑆 of 𝑅 in 𝐿. Let 𝔪 be the maximal ideal of 𝑂, 𝔓 = 𝑆 ∩ 𝔪, and 𝔭 = 𝔓 ∩ 𝑅. Then, 𝑂 = 𝑆𝔓 [Lan64, p. 18, Thm. 4] and 𝔓 is maximal [Lan97, p. 339, ¯ We call 𝐷 𝜑 = 𝐷 𝔓 and 𝐼 𝜑 = 𝐼𝔓 the Prop. 1.11]. Hence, 𝐿¯ = 𝑆/𝔓 and 𝑅/𝔭  𝐾. decomposition group and inertia group, respectively, of 𝜑 over 𝐾. The fixed fields of 𝐷 𝜑 and 𝐼 𝜑 in 𝐿 are the decomposition field and the inertia field, respectively, of ¯ 𝐾¯ is a normal extension and the map 𝜎 ↦→ 𝜎 𝜑 over 𝐾. By Lemma 7.1.1, 𝐿/ ¯ from ¯ is surjective. ¯ 𝐾) 𝐷 𝔓 into Aut( 𝐿/ ¯ 𝐾¯ is separable and 𝐼 𝜑 = 1. (This holds if 𝑆/𝑅 is a ring cover.) Suppose now that 𝐿/ ¯ under the map ¯ 𝐾¯ is Galois and 𝐷 𝜑 is isomorphic to Gal( 𝐿/ ¯ 𝐾) By Lemma 7.1.1, 𝐿/ 𝜎 ↦→ 𝜎. ¯ Denote the decomposition field of 𝜑 over 𝐾 by 𝐿 0 . For each 𝑥 ∈ 𝑂 ∩ 𝐿 0 and ¯ Therefore, 𝜑(𝐿 0 ) = 𝐾¯ ∪ {∞}. each 𝜎 ∈ 𝐷 𝜑 we have 𝜎 ¯ 𝑥¯ = 𝜎𝑥 = 𝑥. ¯ Hence, 𝑥¯ ∈ 𝐾. Let 𝔓0 = 𝔓∩ 𝐿 0 . Then, each prime ideal of 𝑆 that lies over 𝔓0 is conjugate to 𝔓 by an element of 𝐷 𝔓 . Hence, 𝔓 is the only prime ideal of 𝑆 lying over 𝔓0 . By Proposition ¯ = [𝐿 : 𝐿 0 ], we have 2.3.2, 𝑒(𝔓/𝔓0 ) [ 𝐿¯ : 𝐿¯ 0 ] ≤ [𝐿 : 𝐿 0 ]. Since [ 𝐿¯ : 𝐿¯ 0 ] = [ 𝐿¯ : 𝐾] 𝑒(𝔓/𝔓0 ) = 1. In particular, if 𝑂 is discrete, 𝔭 is unramified in 𝐿.

Remark 7.1.7 (Ring covers under change of base ring) Consider an integrally closed domain 𝑅 with quotient field 𝐾. Let 𝐿 be a finite separable extension of 𝐾, 𝑆 the integral closure of 𝑅 in 𝐿,Î𝑧 an element of 𝑆 with 𝐿 = 𝐾 (𝑧), and 𝑓 = irr(𝑧, 𝐾). Then, norm 𝐿/𝐾 ( 𝑓 ′ (𝑧)) = 𝑓 ′ (𝑧) 𝜎 , where 𝜎 ranges over all 𝐾-embeddings of 𝐿 ′ 𝜎 into 𝐾sep . Each 𝑓 (𝑧) is integral over 𝑅. Hence, norm 𝐿/𝐾 ( 𝑓 ′ (𝑧)) is a unit of 𝑅 if and only if 𝑓 ′ (𝑧) is a unit of 𝑆. Suppose that 𝑓 ′ (𝑧) is a unit of 𝑆. By Lemma 7.1.2, 𝑆 = 𝑅[𝑧] and 𝑧 is a primitive element of the ring cover 𝑆/𝑅. Let 𝜑 be a homomorphism of 𝑅 into an integrally ¯ Put 𝐾¯ = Quot( 𝑅). ¯ Then, 𝜑 extends to a homomorphism 𝜓 of closed domain 𝑅. ¯ ¯ 𝑧¯], 𝑆 into the algebraic closure of 𝐾 (Proposition 2.3.1). Put 𝑧¯ = 𝜓(𝑧), 𝑆¯ = 𝑅[ ¯ 𝑓¯ = 𝜑( 𝑓 ), and 𝑔 = irr( 𝑧¯, 𝐾). ¯𝐿 = Quot( 𝑆), ¯ Then, 𝑓¯( 𝑧¯) = 0, 𝑓¯′ ( 𝑧¯) is a unit of ¯ 𝑧¯], and there is a monic polynomial ℎ ∈ 𝐾¯ [𝑋] with 𝑓¯(𝑋) = 𝑔(𝑋)ℎ(𝑋). The 𝑅[ coefficients of ℎ are polynomials in the roots of 𝑓¯. Since the latter are integral ¯ we have ℎ ∈ 𝑅[𝑋]. ¯ over 𝑅, Deduce from 𝑔 ′ ( 𝑧¯)ℎ( 𝑧¯) = 𝑓¯′ ( 𝑧¯) that 𝑔 ′ ( 𝑧¯) is a unit of ¯ 𝑅¯ is a ring cover with 𝑧¯ as a primitive element. In ¯ 𝑅[ 𝑧¯]. Hence, by Lemma 7.1.2, 𝑆/ ¯ ¯ Since 𝑓 is monic and 𝐿¯ = 𝐾¯ ( 𝑧¯), we particular, 𝑆 is the integral closure of 𝑅¯ in 𝐿. ¯ ¯ ¯ ¯ then 𝑆¯ = 𝐿. ¯ have [ 𝐿 : 𝐾] = deg(𝑔) ≤ deg( 𝑓 ) = deg( 𝑓 ) = [𝐿 : 𝐾]. Also, if 𝑅¯ = 𝐾, ¯ ¯ If in addition, 𝐿/𝐾 is Galois, then so is 𝐿/𝐾 (Lemma 7.1.8(b)). As an example, let 𝐿/𝐾 be a finite separable extension, 𝑧 a primitive element for 𝐿/𝐾 and 𝑓 = irr(𝑧, 𝐾). Then, 𝑓 ′ (𝑧) ≠ 0. Hence, 𝐿 = 𝐾 [𝑧] is a ring cover of 𝐾. Therefore, 𝑅[𝑧]/𝑅 is a ring cover whenever 𝑅 is an integrally closed ring containing 𝐾. For another example suppose in the notation of the first two paragraphs that 𝑅 is a valuation ring and Ker(𝜑) is the maximal ideal of 𝑅. Then, 𝑅¯ = 𝐾¯ is a field ¯ Hence, 𝑆¯ = 𝐿¯ is a field. In addition, the and 𝑆¯ = 𝐾¯ [ 𝑧¯] is an integral extension of 𝐾.

7.1 Decomposition Groups

121

local ring of 𝑆 at Ker(𝜓) is a valuation ring lying over 𝑅 [Lan64, p. 18, Thm. 4.7]. ¯ Hence, 𝜑 and 𝜓 extend uniquely to places of 𝐾 and 𝐿 with residue fields 𝐾¯ and 𝐿, respectively. The next result is another consequence of Lemma 7.1.2 which may be applied to covers. Lemma 7.1.8 (a) Let 𝑅 be an integral domain with quotient field 𝐾, 𝐿 a finite Galois extension of 𝐾, and 𝑆 the integral closure of 𝑅 in 𝐿. Consider a monic polynomial 𝑓 ∈ 𝑅[𝑋] having all of its roots in 𝐿, a prime ideal 𝔭 of 𝑅, and a prime ideal 𝔓 of 𝑆 lying over 𝔭. Assume that disc( 𝑓 ) ∉ 𝔭 and denote reduction modulo 𝔓 by a bar. Also ¯ 𝜎). let 𝜎 ∈ 𝐷 𝔓 and 𝐹 a field containing 𝐾¯ such that 𝐿¯ ∩ 𝐹 = 𝐿( ¯ Then, the number of the roots of 𝑓 in 𝐿(𝜎) is equal to the number of the roots of 𝑓¯ in 𝐹. (b) Suppose that 𝐿 = 𝐾 (𝑧) with 𝑧 integral over 𝑅, 𝑓 = irr(𝑧, 𝐾), and disc( 𝑓 ) ∉ 𝔭. ¯ 𝐾¯ is Galois, 𝐼𝔓 = 1, and 𝔭 is unramified in 𝐿. Moreover, the map Then, 𝐿/ ¯ 𝐾). ¯ 𝜎 ↦→ 𝜎 ¯ given by 𝜎 ¯ 𝑥¯ = 𝜎𝑥 for 𝑥 ∈ 𝑆 is an isomorphism of 𝐷 𝔓 onto Gal( 𝐿/ Proof of (a). The roots of 𝑓¯ are distinct, because disc( 𝑓¯) ≠ 0. In addition deg( 𝑓 ) = deg( 𝑓¯). Thus, 𝑥 ↦→ 𝑥¯ maps the roots of 𝑓 bijectively onto the roots of 𝑓¯. For 𝑥 a root of 𝑓 , 𝜎𝑥 = 𝑥 if and only if 𝜎 ¯ 𝑥¯ = 𝑥. ¯ Moreover, since all of the roots of 𝑓¯ belong to ¯𝐿, we have 𝑥¯ ∈ 𝐿¯ ( 𝜎) ¯ if and only if 𝑥¯ ∈ 𝐹. Consequently, the number of the roots of 𝑓 in 𝐿(𝜎) is equal to the number of the roots of 𝑓¯ in 𝐹. Proof of (b). Replace 𝑅 by 𝑅𝔭 , if necessary, to assume that 𝑅 is a local ring and 𝔭 is its maximal ideal. By Lemma 7.1.2, 𝑆 = 𝑅[𝑧] is the integral closure of 𝑅 in 𝐿. ¯ Under the assumptions of (b), 𝐿¯ = 𝐾¯ ( 𝑧¯). From (a), 𝐿¯ is a separable extension of 𝐾. ¯ Also, if 𝜎 Hence, by Lemma 7.1.1(a), 𝐿¯ is a Galois extension of 𝐾. ¯ = 1 for some 𝜎 ∈ 𝐷 𝔓 , then 𝜎 ¯ 𝑧¯ = 𝑧¯. Hence, 𝜎𝑧 = 𝑧, so 𝜎 = 1. Hence, 𝐼𝔓 = 1. By (2.12), 𝔭 is unramified in 𝐿. Moreover, by Lemma 7.1.1(b), the map 𝜎 ↦→ 𝜎 ¯ is an epimorphism of 𝐷 𝔓 onto ¯ 𝐾). ¯ Since its kernel 𝐼𝔓 is trivial, that map is an isomorphism. Gal( 𝐿/ □ Finally, we consider towers of ring covers. Lemma 7.1.9 Let 𝑅 be an integrally closed domain and let 𝐾 = Quot(𝑅). Let 𝑅[𝑦] be a ring cover of 𝑅 with primitive element 𝑦 and let 𝑅[𝑧] be a ring cover of 𝑅 with primitive element 𝑧 such that 𝐾 (𝑦) ⊆ 𝐾 (𝑧). Then, 𝑦 ∈ 𝑅[𝑧] and 𝑅[𝑧]/𝑅[𝑦] is a ring cover with primitive element 𝑧. Moreover, if 𝜓 is a place of 𝐾 (𝑧) which is finite on 𝑅 and we set 𝑦¯ = 𝜓(𝑦) and 𝑧¯ = 𝜓(𝑧), then 𝐾¯ ( 𝑦¯ ) is the residue field of 𝜓| 𝐾 ( 𝑦) and 𝐾¯ ( 𝑧¯) is the residue field ¯ ≤ [𝐾 (𝑦) : 𝐾], [ 𝐾¯ ( 𝑧¯) : 𝐾¯ [≤ [𝐾 (𝑧) : 𝐾], and of 𝜓. In particular, [ 𝐾¯ ( 𝑦¯ ) : 𝐾] [ 𝐾¯ ( 𝑧¯) : 𝐾¯ ( 𝑦¯ )] ≤ [𝐾 (𝑧) : 𝐾 (𝑦)]. Proof. By Lemma 7.1.2, 𝑅[𝑦] is the integral closure of 𝑅 in 𝐾 (𝑦) and 𝑅[𝑧] is the integral closure of 𝑅 in 𝐾 (𝑧). Hence, 𝑅[𝑧] is the integral closure of 𝑅[𝑦] in 𝑅(𝑧), so 𝑅[𝑧] ∩ 𝐾 (𝑦) = 𝑅[𝑦]. Let 𝑍 be the set of zeros of irr(𝑧, 𝐾) in 𝐾 (𝑦)sep and let 𝑍0 be the set of zeros of irr(𝑧, 𝐾 (𝑦)) in 𝐾 (𝑦)sep . Then, irr(𝑧, 𝐾 (𝑦)) divides irr(𝑧, 𝐾), so 𝑍0 ⊆ 𝑍. Moreover, by (7.1), disc(irr(𝑧, 𝐾)) is the product of ±1 by the product of

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7 The Chebotarev Density Theorem

all differences 𝑧 − 𝑧 ′ where 𝑧 and 𝑧 ′ ranges over the distinct elements of 𝑍. A similar description applies to irr(𝑧, 𝐾 (𝑦)), where now 𝑍 is replaced by 𝑍0 . In particular, disc(irr(𝑧, 𝐾 (𝑦))) is an element of 𝑅[𝑦] that divides disc(irr(𝑧, 𝐾)) in the latter ring. In addition, disc(irr(𝑧, 𝐾)) is a unit of 𝑅, hence also a unit of 𝑅[𝑦]. It follows that disc(irr(𝑧, 𝐾 (𝑦))) is a unit of 𝐾 [𝑦], so 𝐾 [𝑧]/𝐾 [𝑦] is a ring cover. Now suppose that 𝜓 is a place of 𝐾 (𝑧) which is finite on 𝑅. Then, 𝑅 is contained in the valuation ring 𝑅 ′ of 𝜓| 𝐾 . By Remark 7.1.7, 𝑅 ′ [𝑦]/𝑅 ′, 𝑅 ′ [𝑧]/𝑅 ′, and 𝑅 ′ [𝑧]/𝑅 ′ [𝑦] are ring covers. Moreover, in the notation of the lemma, the residue field of 𝜓| 𝐾¯ ( 𝑦) is 𝐾¯ ( 𝑦¯ ) and the residue field of 𝜓 is 𝐾¯ ( 𝑧¯). The inequalities at the end of the lemma follow therefore from Proposition 2.3.2. □

7.2 The Artin Symbol over Global Fields The Artin symbol over number fields is a generalization of the Legendre symbol for quadratic residues. Here we define the Artin symbol over global fields and state some of its basic properties. Let 𝑅 be a Dedekind domain with quotient field 𝐾. Consider a finite separable extension 𝐿 of 𝐾. Let 𝑆 be the integral closure of 𝑅 in 𝐿. Take 𝑧 ∈ 𝑆 with 𝐿 = 𝐾 (𝑧). If 𝑓 = irr(𝑧, 𝐾), then 𝑑 = disc( 𝑓 ) ∈ 𝑅 and 𝑑 ≠ 0. Consider 𝑅1 = 𝑅[𝑑 −1 ] and 𝑆1 = 𝑆[𝑑 −1 ]. Then, 𝑑 is a unit in 𝑅1 and 𝑆1 = 𝑅1 [𝑧] (Lemma 7.1.2). Thus, adjoining 𝑑 −1 gives a ring cover 𝑆1 /𝑅1 for 𝐿/𝐾. Maximal ideals 𝔓 for which 𝔓𝑆1 = 𝑆1 are exactly those containing 𝑑. For all others 𝑆/𝔓  𝑆1 /𝔓𝑆1 and 𝑆/𝔓 = (𝑅/𝔭) [ 𝑧¯]. If in addition 𝐿/𝐾 is a Galois extension, then extending 𝔓 to 𝑆1 leaves the decomposition group and the inertia group unchanged. Hence, if 𝔓 does not contain 𝑑, then 𝔓𝑆1 , and therefore also 𝔓, is unramified over 𝐾 (Lemma 7.1.8(b)). Thus, a prime ideal 𝔭 of 𝑅 not containing 𝑑 does not ramify in 𝐿. Since only finitely many prime ideals of 𝑅 contain 𝑑, only finitely many prime ideals ramify in 𝐿. Denote the greatest common divisor of all the principal ideals 𝑓 ′ (𝑧)𝑆 with 𝑧 ∈ 𝑆 by Diff(𝑆/𝑅) and call it the different of 𝑆 over 𝑅. Then, (Diff(𝑆/𝑅)) −1 = {𝑥 ∈ 𝐿 | trace 𝐿/𝐾 (𝑥𝑆) ⊆ 𝑅} [Lan70, p. 60, just before Prop. 6]. A prime ideal 𝔓 of 𝑆 ramifies over 𝐾 if and only if it divides Diff(𝑆/𝑅) [Lan70, p. 62, Prop. 8]. Hence, the prime divisors of the discriminant 𝐷 𝑆/𝑅 = norm 𝐿/𝐾 Diff(𝑆/𝑅) of 𝑆 over 𝑅 (an ideal of 𝑅), are exactly those primes that ramify in 𝐿. Call 𝐾 a global field if 𝐾 is either a finite extension of Q (𝐾 is a number field) or 𝐾 is a function field of one variable over a finite field. In the number field case denote the integral closure of Z in 𝐾 by 𝑂 𝐾 . In the function field case 𝐾 is a finite separable extension of F 𝑝 (𝑡), where 𝑝 = char(𝐾) and 𝑡 is transcendental over F 𝑝 . Denote the integral closure of F 𝑝 [𝑡] in 𝐾 by 𝑂 𝐾 . With the understanding it depends on 𝑡, call 𝑂 𝐾 the ring of integers of 𝐾. The local ring of 𝑂 𝐾 at a prime ideal 𝔭 is a valuation ring. Denote its residue class field by 𝐾¯ 𝔭 . Note that 𝐾¯ 𝔭 is a finite field. We call 𝑁𝔭 = | 𝐾¯ 𝔭 | the absolute norm of 𝔭.

7.3 Dirichlet Density

123

Let 𝐿 be a finite Galois extension of 𝐾. Suppose that 𝔭 is unramified in 𝐿. If 𝔓 is a prime ideal of 𝑂 𝐿 over 𝔭, then reduction modulo 𝔓 gives a canonical isomorphism of ¯ (Lemma 7.1.8(b)). The latter group the decomposition group 𝐷 𝔓 and Gal( 𝐿¯ 𝔓 /𝐾𝔭) is cyclic. It contains a canonical generator Frob, the Frobenius automorphism. It acts on 𝐿¯ 𝔓 by this rule: Frob(𝑥) = 𝑥 𝑁𝔭

for all 𝑥 ∈ 𝐿¯ 𝔓 .

(7.2)

Call the element of 𝐷 𝔓 that corresponds to Frob the Frobenius automorphism at   𝔓 and denote it by 𝐿/𝐾 𝔓 . It is uniquely determined in Gal(𝐿/𝐾) by the condition   𝐿/𝐾 𝑥 ≡ 𝑥 𝑁𝔭 mod 𝔓 for all 𝑥 ∈ 𝑂 𝐿 . (7.3) 𝔓 If 𝐾 ⊆ 𝐾 ′ ⊆ 𝐿 and 𝐾 ′/𝐾 is a Galois extension, this immediately implies    ′  𝐿/𝐾 𝐾 /𝐾 res𝐾 ′ = . (7.4) 𝔓 𝔓 ∩ 𝐾′    𝐿/𝐾  −1 If 𝜎 ∈ Gal(𝐿/𝐾), then 𝐿/𝐾 𝜎𝔓 = 𝜎 𝔓 𝜎 . Therefore, as 𝔓 ranges over prime ideals of 𝑂 𝐿 lying over 𝔭, the Frobenius automorphism ranges over some conjugacy class in Gal(𝐿/𝐾) that depends on 𝔭. This conjugacy class is the Artin symbol, 𝐿/𝐾
𝔭 . It is tacit in this symbol that 𝔭 is unramified in 𝐿. If 𝐿/𝐾 is Abelian, then     𝐿/𝐾
𝐿/𝐾
is one element, 𝐿/𝐾 for 𝐿/𝐾 𝔭 𝔓 . In this case write 𝔭 𝔓 . In defining the Frobenius automorphism and the Artin symbol, replacing 𝑂 𝐾 by the local ring 𝑂 𝐾 ,𝔭 does not change these objects. If 𝐾 is a function field over F𝑞 , the local rings 𝑂 𝐾 ,𝔭 bijectively correspond to prime divisors 𝔭 ′ of 𝐾/F𝑞 finite at 𝑡.

′ We use 𝐿/𝐾 as a substitute for 𝐿/𝐾 𝔭′ 𝔭 . Since each prime divisor 𝔭 of 𝐾/F𝑞 is either
finite at 𝑡 or at 𝑡 −1 , the symbol 𝐿/𝐾 is well defined if 𝔭 ′ is unramified in 𝐿. 𝔭′ Example 7.2.1 (Quadratic extensions of Q) Let 𝑎√be a non-square integer and 𝑝 an odd prime number not dividing 𝑎. Put 𝐿 = Q( 𝑎). Then, 𝑝 is unramified in 𝐿 (Example 2.3.8, Case C). Let 𝔭 be a prime divisor of 𝐿 lying over 𝑝. By Euler’s criterion,     𝑝−1 √ √ 𝑝 𝐿/Q √ 𝑎 √ 2 𝑎 ≡ ( 𝑎) = 𝑎 𝑎≡ 𝑎 mod 𝔭 𝑝 𝑝
√ [LeV58, p. 46, Thm. 3-24]. Thus, the Frobenius symbol 𝐿/Q acts on 𝑎 as the 𝑝
Legendre symbol 𝑎𝑝 .

7.3 Dirichlet Density For 𝐾 a global field denote the set of all prime ideals of 𝑂 𝐾 by 𝑃(𝐾). If 𝐴 is a subset of 𝑃(𝐾), then the Dirichlet density, 𝛿( 𝐴), of 𝐴 is the limit Í −𝑠 𝔭∈ 𝐴 (𝑁𝔭) 𝛿( 𝐴) = lim+ Í , −𝑠 𝑠→1 𝔭∈𝑃 (𝐾) (𝑁𝔭)

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7 The Chebotarev Density Theorem

if it exists. The Dirichlet density is a quantitative measure on subsets of 𝑃(𝐾). We apply it to test if specific subsets are infinite. Clearly 𝛿( 𝐴) is a real number between 0 and 1. For example, 𝛿(𝑃(𝐾)) = 1. If 𝐾 is a number field, then ∑︁ (𝑁𝔭) −𝑠 = ∞ lim+ 𝑠→1

𝔭∈𝑃 (𝐾)

[Lan70, p. 162]. Relation (7.19) implies that the same holds if 𝐾 is a function field. Í Hence, in both cases, 𝛿( 𝐴) = 0 if 𝔭∈ 𝐴 (𝑁𝔭) −1 is finite. In particular, 𝛿( 𝐴) = 0 if 𝐴 is finite. If 𝐴 and 𝐵 are disjoint subsets of 𝑃(𝐾) having a density, then 𝛿( 𝐴 ∪ 𝐵) = 𝛿( 𝐴) + 𝛿(𝐵). Here is the main result: Theorem 7.3.1 (Chebotarev Density Theorem) Let 𝐿/𝐾 be a finite Galois extension of global fields and let C be a conjugacy class in Gal(𝐿/𝐾). Then, the Dirichlet 
|C| density of 𝔭 ∈ 𝑃(𝐾) | 𝐿/𝐾 = C exists and is equal to [𝐿:𝐾 𝔭 ]. Section 7.4 proves Theorem 7.3.1 for function fields and Section 7.5 proves the theorem for number fields. A non-obvious special case is Dirichlet’s theorem showing the arithmetic progression {𝑎, 𝑎 + 𝑛, 𝑎 + 2𝑛, . . .} has infinitely many primes when gcd(𝑎, 𝑛) = 1. Corollary 7.3.2 (Dirichlet) Suppose that 𝑎 and 𝑛 are relatively prime positive inte1 gers. Then, the Dirichlet density of {𝑝 ∈ 𝑃(Q) | 𝑝 ≡ 𝑎 mod 𝑛} is 𝜑 (𝑛) , where 𝜑(𝑛) is the Euler totient function. Proof. Denote a primitive 𝑛th root of unity by 𝜁 𝑛 and let 𝐿 = Q(𝜁 𝑛 ). Then, Gal(𝐿/𝐾) is isomorphic to (Z/𝑛Z) × . If 𝜎 ∈ Gal(𝐿/𝐾) and 𝜎𝜁 𝑛 = 𝜁 𝑛𝑎 , then this isomorphism maps 𝜎 to 𝑎 mod 𝑛. Also, for 𝑎, 𝑏 relatively prime to 𝑛, we have 𝜁 𝑛𝑎 ≡ 𝜁 𝑛𝑏 mod 𝑝 if and only if 𝜁 𝑛𝑎 = 𝜁 𝑛𝑏 . Thus, for 𝑝 ∤ 𝑛, 𝑝 ≡ 𝑎 mod 𝑛 if and only if 𝐿/Q
𝑎 □ 𝑝 (𝜁 𝑛 ) = 𝜁 𝑛 . Now apply Theorem 7.3.1. Example 7.3.3 Let 𝑓 ∈ Z[𝑋] be a monic polynomial. Write 𝑓 (𝑋) =

𝑟 Ö

𝑓𝑖 (𝑋)

𝑖=1

with 𝑓1 (𝑋), . . . , 𝑓𝑟 (𝑋) monic and irreducible and let 𝑑 be the product of the discriminants of 𝑓1 , . . . , 𝑓𝑟 (see (7.1)). Consider the following hypotheses: (7.5) 𝑓 (𝑋) ≡ 0 mod 𝑝 has a solution for all but finitely many primes 𝑝. (7.6) 𝑓 (𝑋) ≡ 0 mod 𝑝 has a solution for all primes 𝑝 ∤ 𝑑. Let 𝐿 be the splitting field of 𝑓 over Q. According to Theorem 7.3.1, each element   of Gal(𝐿/Q) has the form 𝐿/Q for infinitely many prime ideals 𝔭 of 𝑂 𝐿 . Therefore, 𝔭 Lemma 7.1.8(a) implies that each of the statements (7.5) and (7.6) is equivalent to the following: (7.7) Each 𝜎 ∈ Gal(𝐿/Q) fixes a root of 𝑓 (𝑋). In particular, (7.5)and (7.6) are equivalent.

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125

Example 7.3.4 Let 𝐾 be a number field and 𝐵 the set of all prime ideals of 𝑂 𝐾 whose Í Í 1 absolute norm is not a prime number. Then, 𝔭∈𝐵 𝑁𝔭 ≤ 𝑝 [𝐾:Q] < ∞. Hence, 𝑝2 𝛿(𝐵) = 0. Suppose that 𝐿 is a finite Galois extension of 𝐾 and C a conjugacy class in Gal(𝐿/𝐾). Then, in view of the preceding paragraph, the Chebotarev density theorem
gives infinitely many prime ideals 𝔭 of 𝑂 𝐾 such that 𝐿/𝐾 = C and 𝑁𝔭 is a prime 𝔭 number.

7.4 Function Fields This section contains the proof of the Chebotarev density theorem in the function field case. Apart from elementary algebraic manipulations it depends only on the Riemann hypothesis for curves. To fix notation, let 𝑞 be a power of a prime number. Consider a function field 𝐾 over F𝑞 , a finite Galois extension 𝐿 of 𝐾, and a conjugacy class C of Gal(𝐿/𝐾) with 𝑐 elements. Let F𝑞 𝑛 be the algebraic closure of F𝑞 in 𝐿 and fix a separating transcendence element 𝑡 for 𝐾/F𝑞 . Denote the Frobenius element of F𝑞 𝑛 /F𝑞 by Frob𝑞 . As in Section 7.2, let 𝑂 𝐾 be the integral closure of F𝑞 [𝑡] in 𝐿 and let 𝑃(𝐾) be the set of all nonzero prime ideals of 𝑂 𝐾 . Denote the set of prime divisors of 𝐾/F𝑞 by
P(𝐾). Identify 𝑃(𝐾) with a cofinite subset of P(𝐾). Thus, 𝐶 = {𝔭 ∈ P(𝐾) | 𝐿/𝐾 = 𝔭 𝐿/𝐾
C} and {𝔭 ∈ 𝑃(𝐾) | 𝔭 = C} differ by finitely many elements. It suffices therefore 𝑐 to prove that 𝛿(𝐶) = [𝐿:𝐾 ]. In addition to 𝑛 = [F𝑞 𝑛 : F𝑞 ], two more degrees enter the proof: 𝑑 = [𝐾 : F𝑞 (𝑡)] and 𝑚 = [𝐿 : 𝐾F𝑞 𝑛 ] as in the following diagram. 𝑛

𝐾

𝐾 · F𝑞 𝑛

𝑚

𝐿

𝑑

F𝑞 (𝑡)

𝑛

F𝑞 𝑛 (𝑡)

Fix the following notation: P′ (𝐾) = {𝔭 ∈ P(𝐾) | 𝔭 is unramified over F𝑞 (𝑡) and in 𝐿}. P 𝑘 (𝐾) = {𝔭 ∈ P(𝐾) | deg(𝔭) = 𝑘 }. P′𝑘 (𝐾) = {𝔭 ∈ P′ (𝐾) | deg(𝔭) = 𝑘 }.
𝐶 𝑘 (𝐿/𝐾, C) = {𝔭 ∈ P′𝑘 (𝐾) | 𝐿/𝐾 = C}. 𝔭 If 𝔓 ∈ P(𝐿), then 𝔓| 𝐾 is the prime of 𝐾 that lies under 𝔓.   𝐷 𝑘 (𝐿/𝐾, 𝜏) = {𝔓 ∈ P(𝐿) | 𝔓| 𝐾 ∈ P 𝑘 (𝐾) and 𝐿/𝐾 = 𝜏}, 𝔓 for 𝜏 ∈ Gal(𝐿/𝐾). Ð 𝐶′ = ∞ 𝑘=1 𝐶 𝑘 (𝐿/𝐾, C). 𝑔𝐾 = the genus of 𝐾.

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7 The Chebotarev Density Theorem

The sets 𝐶 ′ and 𝐶 differ by only finitely many elements. Hence, they have the same Dirichlet density. To compute this density, we compute the cardinality of each finite set 𝐶 𝑘 (𝐿/𝐾, C). This is also of independent interest, especially when 𝑘 = 1. Lemma 7.4.1 Let 𝑘 be a positive integer, 𝔭 ∈ 𝐶 𝑘 (𝐿/𝐾, C), and 𝜏 ∈ C. (a) There are exactly [𝐿 : 𝐾]/ord(𝜏) primes of P(𝐿) over 𝔭. (b) If 𝐶 𝑘 is a subset of 𝐶 𝑘 (𝐿/𝐾, C) and 𝐷 𝑘 (𝜏) = {𝔓 ∈ 𝐷 𝑘 (𝐿/𝐾, 𝜏) | 𝔓| 𝐾 ∈ 𝐶 𝑘 }, then |𝐶 𝑘 | = |C| · ord(𝜏) · |𝐷 𝑘 (𝜏)| · [𝐿 : 𝐾] −1 . Proof of (a). Suppose that 𝔓 ∈ P(𝐿) lies over 𝔭. Then, by (2.10), [𝐿 : 𝐾] = 𝑒𝔓/𝔭 𝑓𝔓/𝔭 𝑔𝔭 ,   where 𝑓𝔓/𝔭 is the order of the decomposition group ⟨ 𝐿/𝐾 𝔓 ⟩ and 𝑔𝔭 is the number of   primes of 𝐿 that lie over 𝔭. In our case 𝑒𝔓/𝔭 = 1 and 𝜏 is conjugate to 𝐿/𝐾 𝔓 . Thus, 𝑓𝔓/𝔭 = ord(𝜏) and (a) holds. Proof of (b). For each 𝜎 ∈ Gal(𝐿/𝐾), 𝐷 𝑘 (𝜎𝜏𝜎 −1 ) = 𝜎𝐷 𝑘 (𝜏). If 𝜏 ′ ∈ 𝐶 and Ð 𝜏 ′ ≠ 𝜏, then 𝐷 𝑘 (𝜏 ′) and 𝐷 𝑘 (𝜏) are disjoint. Thus, · 𝜏 ∈ C 𝐷 𝑘 (𝜏) is the set of primes in P(𝐿) lying over the elements of 𝐶 𝑘 . By (a), |𝐶 𝑘 | · [𝐿 : 𝐾] ∑︁ = |𝐷 𝑘 (𝜏)| = |C| · |𝐷 𝑘 (𝜏)|, ord(𝜏) 𝜏∈C and the formula follows.

□

Lemma 7.4.2 Let 𝐾 ⊆ 𝐾 ′ ⊆ 𝐿 and 𝜏 ∈ Gal(𝐿/𝐾 ′). Let F𝑞𝑟 be the algebraic closure of F𝑞 in 𝐾 ′. Suppose that 𝑟 |𝑘. Then, 𝐷 𝑘 (𝐿/𝐾, 𝜏) = 𝐷 𝑘/𝑟 (𝐿/𝐾 ′, 𝜏) ∩ {𝔓 ∈ P(𝐿) | deg(𝔓| 𝐾 ) = 𝑘 }. Proof. Let 𝔓 ∈ P(𝐿). Suppose that 𝔭 := 𝔓| 𝐾 ∈ P 𝑘 (𝐾) and let 𝔭 ′ = 𝔓| 𝐾 ′ ∈ P(𝐾 ′). Then, 𝑁𝔭 = | 𝐾¯ 𝔭 | = 𝑞 𝑘 and 𝑁𝔭 ′ = |𝐾 ′𝔭′ | = 𝑞 𝑟𝑙 , where 𝑙 = deg(𝔭 ′) = [𝐾 ′𝔭′ : F𝑞𝑟 ]. By (7.3),   𝑘 𝐿/𝐾 = 𝜏 ⇐⇒ 𝜏𝑥 ≡ 𝑥 𝑞 mod 𝔓 for every 𝑥 ∈ 𝑂 𝐿 ; and (7.8) 𝔓   𝑟𝑙 𝐿/𝐾 ′ = 𝜏 ⇐⇒ 𝜏𝑥 ≡ 𝑥 𝑞 mod 𝔓 for every 𝑥 ∈ 𝑂 𝐿 . (7.9) 𝔓   Thus, it suffices to show that 𝐿/𝐾 = 𝜏 implies 𝑟𝑙 = 𝑘. 𝔓 Since 𝜏 ∈ Gal(𝐿/𝐾 ′), (7.8) implies 𝑥 ≡ 𝑥 𝑞 mod 𝔓 for every 𝑥 ∈ 𝑂 𝐾 ′ . Hence, ⊆ F𝑞 𝑘 . On the other hand, 𝐾 ′𝔭′ ⊇ 𝐾¯ 𝔭 = F𝑞 𝑘 . Therefore, F𝑞𝑟𝑙 = 𝐾 ′𝔭′ = F𝑞 𝑘 . Consequently, 𝑟𝑙 = 𝑘. □ 𝑘

𝐾 ′𝔭′

Corollary 7.4.3 With the hypotheses of Lemma 7.4.2, let C and C ′ be the re′ spective conjugacy classes of 𝜏 in Gal(𝐿/𝐾) and in Gal(𝐿/𝐾 ′) and 𝐶 𝑘/𝑟 = 𝑘 ′ ′ ′ ′ ′ 𝐶 𝑘/𝑟 (𝐿/𝐾 , C ) ∖{𝔭 ∈ 𝑃(𝐾 ) | deg(𝔭 | 𝐾 ) ≤ 2 }. Then, |𝐶 𝑘 (𝐿/𝐾, C)| =

′ | |C||C𝑘/𝑟

|C ′ | [𝐾 ′ : 𝐾]

.

7.4 Function Fields

127

Proof. Let 𝑙 = 𝑟𝑘 . Then, 𝐷 ′𝑘 (𝜏) := 𝐷 𝑙 (𝐿/𝐾 ′, 𝜏) ∩ {𝔓 ∈ P(𝐿) | deg(𝔓| 𝐾 ) = 𝑘 } is the set of primes in 𝐷 𝑙 (𝐿/𝐾 ′, 𝜏) lying over 𝐶 𝑘′′ := 𝐶𝑙 (𝐿/𝐾 ′, C ′) ∩ {𝔭 ′ ∈ P(𝐾 ′) | deg(𝔭 ′ | 𝐾 ) = 𝑘 }. 𝐷 ′𝑘 (𝜏) is also the set of primes in 𝐷 𝑙 (𝐿/𝐾, 𝜏) over 𝐶 𝑘 (𝐿/𝐾, C). By Lemma 7.4.2, 𝐷 ′𝑘 (𝜏) = 𝐷 𝑘 (𝐿/𝐾, 𝜏). Applying Lemma 7.4.1 twice, first for the case where 𝐶 𝑘 = 𝐶 𝑘 (𝐿/𝐾, C) and then for the case where 𝐶 𝑘 = 𝐶 𝑘′′ (and for 𝐿/𝐾 ′ rather than for 𝐿/𝐾), gives a chain of equalities: [𝐿 : 𝐾] [𝐿 : 𝐾 ′] |𝐶 𝑘 (𝐿/𝐾, C)| = |𝐷 𝑘 (𝐿/𝐾, 𝜏)| = |𝐷 ′𝑘 (𝜏)| = ′ |𝐶 ′′ |. |C| · ord(𝜏) |C | · ord(𝜏) 𝑘 Thus, it suffices to show that 𝐶 𝑘′′ = 𝐶𝑙′. Indeed, if 𝔭 ′ ∈ P(𝐾 ′) is of degree 𝑙 and 𝔭 = 𝔭 ′ | 𝐾 , then F𝑞 ⊆ 𝐾¯ 𝔭 ⊆ 𝐾 ′𝔭′ = F𝑞𝑟𝑙 = F𝑞 𝑘 . Hence, deg(𝔭)|𝑘. Thus, either deg(𝔭) = 𝑘 or deg(𝔭) ≤ 𝑘2 . Therefore, 𝐶 𝑘′′ = 𝐶𝑙′. □ Lemma 7.4.4 Let 𝑘 be a positive integer such that resF𝑞 𝑛 (𝜏) = resF𝑞 𝑛 (Frob𝑞𝑘 ) 𝑛′

(7.10)

𝐿′

for every 𝜏 ∈ C. Let be a multiple of 𝑛 and = 𝐿F𝑞 𝑛′ . Then, 𝐿 ′/𝐾 is a Galois extension, F𝑞 𝑛′ is the algebraic closure of F𝑞 in 𝐿 ′, and 𝑔 𝐿 = 𝑔 𝐿′ (Proposition 4.4.3). Moreover, for each 𝜏 ∈ C there exists a unique 𝜏 ′ ∈ Gal(𝐿 ′/𝐾) with res 𝐿 𝜏 ′ = 𝜏 and resF𝑞 𝑛′ (𝜏 ′) = resF𝑞 𝑛′ (Frob𝑞𝑘 ). Furthermore: (a) ord(𝜏 ′) = lcm(ord(𝜏), [F𝑞 𝑛′ : F𝑞 𝑛′ ∩ F𝑞 𝑘 ]); (b) C ′ = {𝜏 ′ | 𝜏 ∈ C} is a conjugacy class in Gal(𝐿 ′/𝐾); and (c) 𝐶 𝑘 (𝐿 ′/𝐾, C) = 𝐶 𝑘 (𝐿/𝐾, C ′). Proof. Note that 𝐿/F𝑞 𝑛 is a regular extension, because F𝑞 𝑛 is perfect and algebraically closed in 𝐿. Given 𝜏 ∈ C, the existence of 𝜏 ′ follows from (7.10), because 𝐾F𝑞 𝑛′ ∩ 𝐿 = 𝐾F𝑞 𝑛 . Uniqueness follows from 𝐿 ′ = F𝑞 𝑛′ 𝐿. To prove (a), note that

ord(𝜏 ′) = lcm ord(res 𝐿 𝜏 ′), ord(resF𝑞 𝑛′ 𝜏 ′) = lcm ord(𝜏), [F𝑞 𝑛′ : F𝑞 𝑛′ ∩ F𝑞 𝑘 ] . Since F𝑞 𝑛′ /F𝑞 is an Abelian extension, assertion (b) follows from the uniqueness of 𝜏 ′.   To prove (c), we show that 𝔓 ∈ P(𝐿 ′) and 𝔭 = 𝔓| 𝐾 ∈ P 𝑘 (𝐾) imply 𝐿/𝐾 𝔓| 𝐿 =  ′    𝑞 𝑘 mod 𝔓| for each 𝑥 ∈ 𝑂 . 𝜏 ⇐⇒ 𝐿 𝔓/𝐾 = 𝜏 ′. Indeed, if 𝐿/𝐾 𝐿 𝐿 𝔓| 𝐿 = 𝜏, then 𝜏𝑥 ≡ 𝑥 𝑘

By definition, Frob𝑞𝑘 𝑥 = 𝑥 𝑞 for each 𝑥 ∈ F𝑞 𝑛′ . Since 𝑂 𝐿′ = F𝑞 𝑛 𝑂 𝐿 (Proposition  ′  𝑘 4.4.3(d)), 𝜏 ′𝑥 ≡ 𝑥 𝑞 mod 𝔓 for each 𝑥 ∈ 𝑂 𝐿′ . Hence, 𝐿 𝔓/𝐾 = 𝜏 ′. The converse is a special case of (7.4). □ Corollary 7.4.5 If 𝐿 = F𝑞 𝑛 𝐾 and 𝜏 ∈ Gal(𝐿/𝐾) satisfies (7.10), then 𝐶 𝑘 (𝐾/𝐾, 1) = 𝐶 𝑘 (𝐿/𝐾, {𝜏}).

128

7 The Chebotarev Density Theorem

Now we give estimates for the key sets. Lemma 7.4.6 Suppose that 𝐿 = 𝐾F𝑞 𝑛 , C = {𝜏}, and 𝜏| F𝑞 𝑛 = Frob𝑞 . Then, √ |#𝐶1 (𝐿/𝐾, C) − 𝑞| < 2(𝑔 𝐿 𝑞 + 𝑔 𝐿 + 𝑑).

(7.11)

Proof. First note that 𝐶1 (𝐿/𝐾, C) = P1′ (𝐾) and each 𝔭 ∈ P(𝐾) is unramified in 𝐿. Thus, P1 (𝐾) ∖ 𝐶1 (𝐿/𝐾, C) consists exactly of all prime divisors of Diff(𝐾/F𝑞 (𝑡)). By Riemann-Hurwitz (Theorem 4.6.1), deg(Diff(𝐾/F𝑞 (𝑡))) = 2(𝑔𝐾 + 𝑑 − 1). By Theorem 5.5.2, √ |#P1 (𝐾) − (𝑞 + 1)| ≤ 2𝑔𝐾 𝑞. √ Hence, |#𝐶1 (𝐿/𝐾, C) − 𝑞| ≤ 2𝑔𝐾 𝑞 + 1 + 2(𝑔𝐾 + 𝑑 − 1). Since 𝑔𝐾 = 𝑔 𝐿 , this proves (7.11). □ Denote the situation when 𝑎 divides 𝑏 and 𝑎 < 𝑏 by 𝑎 pd 𝑏. Lemma 7.4.7 Let 𝐾 ′ be a degree 𝑘𝑚 extension of 𝐾 containing F𝑞 𝑘 . Then, #{𝔮 ∈ P(𝐾 ′) | deg(𝔮| 𝐾 ) pd 𝑘 } ≤ 2𝑚(𝑞 𝑘/2 + (2𝑔𝐾 + 1)𝑞 𝑘/4 ).

(7.12)

Proof. If 𝑗 |𝑘, then F𝑞 𝑗 ⊆ F𝑞 𝑘 . If in addition, 𝔭 ∈ P 𝑗 (𝐾), then by Lemma 5.3.1, 𝔭 decomposes in 𝐾F𝑞 𝑗 into 𝑗 prime divisors of degree 1. Each has exactly one extension to 𝐾F𝑞 𝑘 . The latter decomposes in 𝐾 ′ into at most 𝑚 prime divisors. Hence, #{𝔮 ∈ P(𝐾 ′) | deg(𝔮| 𝐾 ) pd 𝑘 } =

∑︁

{𝔮 ∈ P(𝐾 ′) | deg(𝔮| 𝐾 ) = 𝑗 }

𝑗 |𝑘 𝑗≤𝑘/2

≤𝑚

∑︁

{𝔮 ∈ P(𝐾F𝑞 𝑘 ) | deg(𝔮| 𝐾 ) = 𝑗 }

𝑗 |𝑘 𝑗≤𝑘/2

=𝑚

∑︁

(7.13) {𝔮 ∈ P(𝐾F𝑞 𝑗 ) | deg(𝔮| 𝐾 ) = 𝑗 }

𝑗 |𝑘 𝑗≤𝑘/2

≤𝑚

∑︁

|P1 (𝐾F𝑞 𝑗 )|.

𝑗 |𝑘 𝑗≤𝑘/2

By Theorem 5.5.2, |P1 (𝐾F𝑞 𝑗 )| ≤ 𝑞 𝑗 + 2𝑔𝐾 𝑞 𝑗/2 + 1. Now verify the inequalities Í 𝑗 𝑘/2 and 𝑘 ≤ 2𝑞 𝑘/4 and use them to deduce (7.12) from (7.13). □ 𝑗 ≤𝑘/2 𝑞 ≤ 2𝑞 2 We now come to the key result of this section, Proposition 7.4.8. It is the main result from which the Chebotarev density theorem follows. It is also the main ingredient in an arithmetic proof of Hilbert irreducibility theorem (Lemma 14.3.3). A variant of it (Lemma 35.2.1) is crucial to the Galois stratification procedure for the elementary theory of finite fields. Recall: F𝑞 𝑛 is the algebraic closure of F𝑞 in 𝐿.

7.4 Function Fields

129

Proposition 7.4.8 Let 𝑎 be a positive integer with resF𝑞 𝑛 𝜏 = resF𝑞 𝑛 Frob𝑞𝑎 for each 𝜏 ∈ C. Let 𝑘 be a positive integer. If 𝑘 ̸≡ 𝑎 mod 𝑛, then 𝐶 𝑘 (𝐿/𝐾, C) is empty. If 𝑘 ≡ 𝑎 mod 𝑛, then 𝑐 𝑘 𝑞 #𝐶 𝑘 (𝐿/𝐾, C) − 𝑘𝑚 (7.14)  2𝑐  < (𝑚 + 𝑔 𝐿 )𝑞 𝑘/2 + 𝑚(2𝑔𝐾 + 1)𝑞 𝑘/4 + 𝑔 𝐿 + 𝑑𝑚 . 𝑘𝑚 Proof. Suppose that 𝐶 𝑘 (𝐿/𝐾, C) ≠ ∅ and consider 𝔓 ∈ P(𝐿) that lies over 𝔭 ∈   𝐶 𝑘 (𝐿/𝐾, C). Then, 𝐿/𝐾 ∈ C, so 𝔓   𝐿/𝐾 resF𝑞 𝑛 (Frob𝑞𝑎 ) = resF𝑞 𝑛 = resF𝑞 𝑛 (Frob𝑞𝑘 ). 𝔓 Hence, 𝑘 ≡ 𝑎 mod 𝑛. Conversely, suppose that 𝑘 ≡ 𝑎 mod 𝑛. Let 𝜏 ∈ C and 𝑛 ′ = 𝑛𝑘 · ord(𝜏). Extend 𝐿 to 𝐿 ′ = 𝐿F𝑞 𝑛′ . Since 𝐿/F𝑞 𝑛 is a regular extension, we have [𝐿 ′ : 𝐾F𝑞 𝑛′ ] = [𝐿 : 𝐾F𝑞 𝑛 ] = 𝑚. Since 𝑘 ≡ 𝑎 mod 𝑛, there exists a 𝜏 ′ ∈ Gal(𝐿 ′/𝐾) with 𝜏 ′ | 𝐿 = 𝜏 and 𝜏 ′ | F𝑞 𝑛′ = Frob𝑞𝑘 . By Lemma 7.4.4(a), ord(𝜏 ′) = lcm(ord(𝜏), [F𝑞 𝑛′ : F𝑞 𝑘 ]) = lcm(ord(𝜏), 𝑛 · ord(𝜏)) = 𝑛 · ord(𝜏). Denote the conjugacy class of 𝜏 ′ in Gal(𝐿 ′/𝐾) by C ′. Denote the fixed field of 𝜏 ′ in 𝐿 ′ by 𝐾 ′. Then, 𝐾 ′ ∩ F𝑞 𝑛′ = 𝐾 ′ ∩ F˜ 𝑞 = F𝑞 𝑘 and ′ 𝐾 F𝑞 𝑛′ = 𝐿 ′. 𝐾′ 𝑚

𝐾 𝑑

𝐾F𝑞 𝑘

ord( 𝜏 ′ )

𝐿′ 𝑚

𝐾F𝑞 𝑛′

𝑑

F𝑞 (𝑡)

F𝑞 𝑘 (𝑡)

F𝑞 𝑛′ (𝑡)

F𝑞

F𝑞 𝑘

F𝑞 𝑛′

Thus, [𝐾 ′ : 𝐾F𝑞 𝑘 ] = [𝐿 ′ : 𝐾F𝑞 𝑛′ ] = [𝐿 : 𝐾F𝑞 𝑛 ] = 𝑚. Hence, [𝐾 ′ : F𝑞 𝑘 (𝑡)] = 𝑑𝑚. Applying Lemma 7.4.3 with 𝐿 ′, C ′, {𝜏 ′ }, 𝑘 replacing 𝐿, C, C ′, 𝑟 we conclude that 𝑐 |𝐶 𝑘 (𝐿 ′/𝐾, C ′)| = |𝐶1 (𝐿 ′/𝐾 ′, {𝜏 ′ }) ∖{𝔮 ∈ P(𝐾 ′) | deg(𝔮| 𝐾 ) pd 𝑘 }|. [𝐾 ′ : 𝐾] Since [𝐾 ′ : 𝐾] = 𝑘𝑚, Lemma 7.4.7 implies #𝐶 𝑘 (𝐿 ′/𝐾, C ′)− 𝑐 #𝐶1 (𝐿 ′/𝐾 ′, {𝜏 ′ }) 𝑘𝑚 𝑐 ≤ #{𝔮 ∈ P(𝐾 ′) | deg(𝔮| 𝐾 ) pd 𝑘 } (7.15) 𝑘𝑚 𝑐 ≤ · 2𝑚(𝑞 𝑘/2 + (2𝑔𝐾 + 1)𝑞 𝑘/4 ). 𝑘𝑚

130

7 The Chebotarev Density Theorem

By Lemma 7.4.6, now with 𝐾 ′, 𝐿 ′, 𝑛 ′, 𝜏 ′, 𝑞 𝑘 replacing 𝐾, 𝐿, 𝑛, 𝜏, 𝑞, |#𝐶1 (𝐿 ′/𝐾 ′, {𝜏 ′ }) − 𝑞 𝑘 | < 2(𝑔 𝐿′ 𝑞 𝑘/2 + 𝑔 𝐿′ + 𝑑𝑚).

(7.16)

Now we combine (7.15)and (7.16) using the equalities 𝑔 𝐿 = 𝑔 𝐿′ and 𝐶 𝑘 (𝐿/𝐾, C) = 𝐶 𝑘 (𝐿 ′/𝐾, C ′) (see Lemma 7.4.4(c)) to prove (7.14): 𝑐 𝑘 𝑐 𝑘 |#𝐶 𝑘 (𝐿/𝐾, C) − 𝑞 | = |#𝐶 𝑘 (𝐿 ′/𝐾, C ′) − 𝑞 | 𝑘𝑚 𝑘𝑚 𝑐 𝑐 ≤ #𝐶 𝑘 (𝐿 ′/𝐾, C ′) − #𝐶1 (𝐿 ′/𝐾 ′, {𝜏 ′ }) + #𝐶1 (𝐿 ′/𝐾 ′, {𝜏 ′ }) − 𝑞 𝑘 𝑘𝑚 𝑘𝑚
2𝑐
𝑐 ≤ · 2𝑚 𝑞 𝑘/2 + (2𝑔𝐾 + 1) 𝑘/4 + 𝑔 𝐿 𝑞 𝑘/2 + 𝑔 𝐿 + 𝑑𝑚 𝑘𝑚 𝑘𝑚
2𝑐 = (𝑚 + 𝑔 𝐿 )𝑞 𝑘/2 + (2𝑔𝐾 + 1)𝑞 𝑘/4 + 𝑔 𝐿 + 𝑑𝑚 . □ 𝑘𝑚 We deduce the function field case of Theorem 7.3.1 by summing over 𝑘 in the conclusion of Proposition 7.4.8. We use the big 𝑂 notation, i.e. for real valued functions 𝑓 (𝑥) and 𝑔(𝑥) write 𝑓 (𝑥) = 𝑂 (𝑔(𝑥)),

𝑥→𝑎

to mean there exists a constant 𝑐 with | 𝑓 (𝑥)| ≤ 𝑐|𝑔(𝑥)| for all 𝑥 close to 𝑎. In particular, if 𝑔(𝑥) = 1, then 𝑓 (𝑥) is bounded near 𝑎. Lemma 7.4.9 Let 𝑎 and 𝑛 be positive integers. Then, ∞ ∑︁ 𝑥 𝑎+ 𝑗𝑛 1 = − log(1 − 𝑥) + 𝑂 (1), 𝑎 + 𝑗𝑛 𝑛 𝑗=0

𝑥 → 1− .

(7.17)

Proof. If 𝜁 ≠ 1 is an 𝑛th root of unity, then (1 + 𝜁 + · · · + 𝜁 𝑛−1 ) (1 − 𝜁) = 1 − 𝜁 𝑛 = 0, so 1 + 𝜁 + · · · + 𝜁 𝑛−1 = 0. Using the expansion of log(1 − 𝑦) for real values 𝑦 in the neighborhood of 1 (e.g. [Kno28, p. 212, 120(a)]), we get −

𝑛−1 ∞ 𝑛−1 1 ∑︁ 1 ∑︁ 𝑥 𝑘 ∑︁ 𝑖 (𝑘−𝑎) log(1 − 𝜁 𝑖 𝑥)𝜁 −𝑖𝑎 = 𝜁 = 𝑛 𝑖=0 𝑛 𝑘=1 𝑘 𝑖=0

∑︁ 𝑘≡𝑎 mod

𝑥𝑘 . 𝑘 𝑛

Since, for 1 ≤ 𝑖 ≤ 𝑛 − 1, log(1 − 𝜁 𝑖 𝑥) is bounded in the neighborhood of 1, the result follows. □ Lemma 7.4.10 Suppose that 0 < 𝑎 ≤ 𝑛 is an integer with resF𝑞 𝑛 𝜏 = resF𝑞 𝑛 Frob𝑞𝑎 for each 𝜏 ∈ C. Then, ∑︁ 𝑐 (𝑁𝔭) −𝑠 = − log(1 − 𝑞 1−𝑠 ) + 𝑂 (1), 𝑠 → 1+ . (7.18) [𝐿 : 𝐾] 𝔭∈𝐶

7.5 Number Fields

131

Ð Proof. The set 𝐶 ′ = ∞ 𝑘=1 𝐶 𝑘 (𝐿/𝐾, C) differs from 𝐶 by only finitely many elements. We apply Proposition 7.4.8 and Lemma 7.4.9 for 𝑥 = 𝑞 1−𝑠 to compute: ∑︁

(𝑁𝔭) −𝑠 =

𝔭∈𝐶

∞ ∑︁

∑︁

(𝑁𝔭) −𝑠

𝑗=0 𝔭∈𝐶𝑎+ 𝑗𝑛 (𝐿/𝐾 , C) ∞ (7.14) ∑︁

=

=

𝑐 𝑚

𝑗=0 ∞ ∑︁ 𝑗=0



 1 𝑐 𝑞 𝑎+ 𝑗𝑛 + 𝑂 (𝑞 2 (𝑎+ 𝑗𝑛) ) 𝑞 −(𝑎+ 𝑗𝑛)𝑠 𝑚(𝑎 + 𝑗𝑛) ∞

∑︁ 1
1 𝑞 (1−𝑠) (𝑎+ 𝑗𝑛) + 𝑂 𝑞 ( 2 −𝑠) 𝑎 𝑞 ( 2 −𝑠) 𝑗𝑛 𝑎 + 𝑗𝑛 𝑗=0

 𝑞 ( 12 −𝑠) 𝑎  𝑐 1−𝑠 log(1 − 𝑞 ) + 𝑂 (1) + 𝑂 1 𝑚𝑛 1 − 𝑞 ( 2 −𝑠)𝑛 𝑐 =− log(1 − 𝑞 1−𝑠 ) + 𝑂 (1), 𝑠 → 1+ .□ [𝐿 : 𝐾]

(7.17)

= −

When 𝐿 = 𝐾, Lemma 7.4.10 simplifies to ∑︁ 𝑁𝔭−𝑠 = − log(1 − 𝑞 1−𝑠 ) + 𝑂 (1),

𝑠 → 1+ .

(7.19)

𝔭∈𝑃 (𝐸)

Dividing (7.18) by (7.19) and taking the limit as 𝑠 → 1+ gives the Dirichlet density of 𝐶: Í −𝑠 𝑐 𝔭∈𝐶 (𝑁𝔭) 𝛿(𝐶) = lim+ Í = . −𝑠 𝑠→1 [𝐿 : 𝐾] 𝔭∈𝑃 (𝐸) (𝑁𝔭) This concludes the proof of the Chebotarev density theorem for function fields.

7.5 Number Fields Let 𝐿/𝐾 be a finite Galois extension of number fields. The proof of the Chebotarev density theorem for 𝐿/𝐾 splits into eight parts. It uses the asymptotic formula (7.21) for counting ideals with bounded norm in a given class (which we quote without proof). We say that 𝐿/𝐾 is cyclotomic if 𝐿 ⊆ 𝐾 (𝜁) with 𝜁 a root of 1. The case when 𝐿/𝐾 is cyclotomic produces the general case from an easy reduction to 𝐿/𝐾 cyclic. Part A: Ideals with a bounded norm in a given class. Let 𝔠 be a nonzero ideal of 𝑂 𝐾 . Denote the group of all fractional ideals of 𝐾 relatively prime to 𝔠 by 𝐽 (𝔠). Let 𝑃(𝔠) be the subgroup of all principal fractional ideals 𝑥𝑂 𝐾 , where 𝑥 satisfies the following conditions. (7.20a) If 𝔭 is a prime ideal of 𝑂 𝐾 that divides 𝔠, then 𝑥 lies in the local ring 𝑂𝔭 of 𝑂 𝐾 at 𝔭 and 𝑥 ≡ 1 mod 𝔠𝑂𝔭 . (7.20b) 𝑥 is totally positive: 𝜎𝑥 is positive for each embedding 𝜎: 𝐾 → R. The factor group 𝐺 (𝔠) = 𝐽 (𝔠)/𝑃(𝔠) is finite [Lan70, p. 127, Thm. 1]. Denote the order of 𝐺 (𝔠) by ℎ𝔠 .

132

7 The Chebotarev Density Theorem

Extend the absolute norm 𝑁𝔭 of prime ideals multiplicatively to all fractional ideals. Consider a class K of 𝐽 (𝔠) modulo 𝑃(𝔠). Denote the number of ideals 𝔞 ∈ K (of 𝑂 𝐾 ) with 𝑁𝔞 ≤ 𝑛 by 𝑗 (K, 𝑛). The key asymptotic formula is: −1

𝑗 (K, 𝑛) = 𝜌𝔠 𝑛 + 𝑂 (𝑛1−[𝐾:Q] ),

𝑛 → ∞,

(7.21)

where 𝜌𝔠 is a positive constant dependent on 𝔠 and 𝐾 but not on K [Lan70, p. 132, Thm. 3]. Part B: Abelian characters. A character of a finite Abelian group 𝐺 is a homomorphism 𝜒: 𝐺 → C× . Define multiplication of characters by ( 𝜒1 𝜒2 ) (𝜎) = 𝜒1 (𝜎) 𝜒2 (𝜎). The set 𝐺ˆ of characters of 𝐺 forms a group isomorphic to 𝐺. Here are the standard character formulas. Í (7.22a) 𝜎 ∈𝐺 𝜒1 (𝜎 −1 ) 𝜒2 (𝜎) = |𝐺 | if 𝜒1 = 𝜒2 and 0 otherwise. Í (7.22b) 𝜒 ∈𝐺ˆ 𝜒(𝜎 −1 ) 𝜒(𝜏) = |𝐺 | if 𝜎 = 𝜏 and 0 otherwise. Î (7.22c) 𝜒 ∈𝐺ˆ (1 − 𝜒(𝜎) 𝑋) = (1 − 𝑋 𝑓 ) |𝐺 |/ 𝑓 if 𝑓 = ord(𝜎). Formulas (7.22a)and (7.22b) are known as the orthogonality relations [Jan73, p.127, Prop. 3.3]. Formula (7.22c) follows in the case where 𝐺 = ⟨𝜎⟩ by observing that the zeros of both sides are the roots of 1 of order 𝑓 and from the relation | 𝐺ˆ | = |𝐺 |. The general case follows from the cyclic case and the canonical isomorphism ⊥ , where ⟨𝜎⟩ ⊥ = { 𝜒 ∈ 𝐺 d  𝐺/⟨𝜎⟩ ˆ ˆ | 𝜒(𝜎) = 1}. ⟨𝜎⟩ Part C: 𝐿-series. For a given ideal 𝔠 of 𝑂 𝐾 and a character 𝜒 of 𝐺 (𝔠) consider the Dirichlet series ∑︁ 𝜒(𝔞) 𝐿𝔠 (𝑠, 𝜒) = , Re(𝑠) > 1, (7.23) (𝑁𝔞) 𝑠 gcd(𝔞,𝔠)=1

where 𝔞 ranges over all ideals of 𝑂 𝐾 relatively prime to 𝔠, and 𝜒(𝔞) = 𝜒(K) if K is the class of 𝔞. Call 𝐿𝔠 (𝑠, 𝜒) an 𝐿-series. The function 𝜒(𝔞) is multiplicative on 𝐽 (𝔠). Therefore, 𝐿𝔠 (𝑠, 𝜒) satisfies the Euler identity Ö 𝜒(𝔭)  −1 𝐿𝔠 (𝑠, 𝜒) = 1− for Re(𝑠) > 1 (7.24) (𝑁𝔭) 𝑠 𝔭∤𝔠 [Jan73, p. 128, Thm. 4.2]. We quote the following result of complex analysis: Lemma 7.5.1 ([Lan70], p. 158) Let {𝑎 𝑖 }∞ 𝑖=1 be a sequence of complex numbers, for which there is a 0 ≤ 𝜎 < 1 and a complex number 𝜌 with 𝑛 ∑︁

𝑎 𝑖 = 𝜌𝑛 + 𝑂 (𝑛 𝜎 )

as

𝑛 → ∞.

𝑖=1

Í −𝑠 for Re(𝑠) > 1 analytically continues to Re(𝑠) > 𝜎, except Then, 𝑓 (𝑠) = ∞ 𝑛=1 𝑎 𝑛 𝑛 for a simple pole with residue 𝜌 at 𝑠 = 1. Lemma 7.5.2 The function 𝐿𝔠 (𝑠, 𝜒) has an analytic continuation to the half plane 1 Re(𝑠) > 1 − [𝐾:Q] . If 𝜒 = 1, then it has a simple pole at 𝑠 = 1 with residue ℎ𝔠 𝜌𝔠 . If 𝜒 ≠ 1, then 𝐿𝔠 (𝑠, 𝜒) is analytic in the entire half plane.

7.5 Number Fields

133

Proof. We use (7.21) to substitute for 𝑗 (K, 𝑡) and use the orthogonality relation (7.22a) for the finite group 𝐺 (𝔠), to conclude that ∑︁ (𝔞,𝔠) =1 𝑁𝔞≤𝑛

∑︁

∑︁

K ∈𝐺 (𝔠)

𝔞∈K 𝑁𝔞≤𝑛

𝜒(𝔞) =

(7.22a)

=

(

(7.21)

𝜒(𝔞) =

∑︁

−1

𝜒(K) 𝜌𝔠 𝑛 + 𝑂 (𝑛1−[𝐾:Q] )




K ∈𝐺 (𝔠) −1

ℎ𝔠 𝜌𝔠 𝑛 + 𝑂 (𝑛1−[𝐾:Q] ) −1 𝑂 (𝑛1−[𝐾:Q] )

if 𝜒 = 1 if 𝜒 ≠ 1,

𝑛 → ∞.

Thus, our lemma is a special case of Lemma 7.5.1.

□

Part D: Special case of Artin’s reciprocity law. This law is the central result of class field theory. Consider a finite Abelian extension 𝐿/𝐾 of number fields. Let 𝔠 be an ideal of 𝑂 𝐾 divisible by all prime ideals ramifying in 𝐿 (we say that 𝔠 is admissible).
If a prime ideal 𝔭 does not divide 𝔠, then 𝐿/𝐾 defines a unique element of Gal(𝐿/𝐾). 𝔭 𝐿/𝐾
The map 𝔭 ↦→ 𝔭 extends to a homomorphism 𝜔𝔠 : 𝐽 (𝔠) → Gal(𝐿/𝐾) called the reciprocity map. When referring to the extension 𝐿/𝐾 we will denote 𝜔𝔠 by 𝜔 𝐿/𝐾 ,𝔠 . Let 𝐿 ′ be any Abelian extension of 𝐾 containing 𝐿. Suppose that each prime ideal of 𝑂 𝐾 ramified in 𝐿 ′ divides 𝔠. Then, res 𝐿 (𝜔 𝐿′ /𝐾 ,𝔠 (𝔭)) = 𝜔 𝐿/𝐾 ,𝔠 (𝔭) for each prime 𝔭 ∈ 𝐽 (𝔠) (see (7.4)). So, res 𝐿 (𝜔 𝐿′ /𝐾 ,𝔠 (𝔞)) = 𝜔 𝐿/𝐾 ,𝔠 (𝔞)

for each 𝔞 ∈ 𝐽 (𝔠).

(7.25)

Class field theory proves that 𝜔𝔠 is surjective [Lan70, p. 199, Thm. 1]. In order to describe its kernel, let Norm = Norm 𝐿/𝐾 be the norm map of fractional ideals of 𝑂 𝐿 onto fractional ideals of 𝑂 𝐾 . If 𝔄 is an ideal of 𝑂 𝐿 , then Norm(𝔄) is the ideal of 𝑂 𝐾 generated by norm 𝐿/𝐾 (𝑎) for all 𝑎 ∈ 𝔄. If 𝔅 is another ideal, then Norm(𝔄𝔅) = Norm(𝔄)Norm(𝔅), so Norm extends multiplicatively to the group of all fractional ideals of 𝑂 𝐿 . If 𝔓 is a prime ideal of 𝑂 𝐿 , 𝔭 = 𝔓 ∩ 𝑂 𝐾 , and 𝑓 = 𝑓𝔓/𝔭 , then Norm(𝔓) = 𝔭 𝑓 . Finally, Norm 𝐿/Q and the absolute norm of ideals relate to each other by the formula Norm 𝐿/Q (𝔄) = 𝑁 (𝔄)Z [Jan73, pp. 35–37]. Artin’s reciprocity law gives an admissible ideal 𝔠 of 𝑂 𝐾 such that 𝜔𝔠 : 𝐽 (𝔠) → Gal(𝐿/𝐾) is surjective and Ker(𝜔𝔠 ) = Norm(𝔠)𝑃(𝔠) [Lan70, p. 205, Thm. 3]. Here we prove one part of Artin’s reciprocity law for cyclotomic extensions of 𝐾. Lemma 7.5.3 Let 𝜁 be a primitive 𝑚th root of 1, 𝐿 a subfield of 𝐾 (𝜁) containing 𝐾, and 𝔠 an ideal of 𝑂 𝐾 divisible by 𝑚. Then, 𝑃(𝔠) ⊆ Ker(𝜔𝔠 ). Proof. Each prime ideal of 𝑂 𝐾 which ramifies in 𝐾 (𝜁) divides 𝑚 [Gol71, p. 98, Cor. 6-2-4]. This defines 𝜔𝔠 = 𝜔 𝐿/𝐾 ,𝔠 . Use (7.25) to assume that 𝐿 = 𝐾 (𝜁). Consider the natural embedding 𝑖: Gal(𝐿/𝐾) → (Z/𝑚Z) × determined by 𝜎(𝜁) = 𝜁 𝑖 ( 𝜎) , 𝜎 ∈ Gal(𝐿/𝐾). If 𝔭 ∈ 𝐽 (𝔠) is prime and 𝔓 is a prime ideal of 𝑂 𝐿 lying over 𝔭, then by (7.3), 𝜔𝔠 (𝔭) (𝜁) ≡ 𝜁 𝑁𝔭 mod 𝔓. Since reduction modulo 𝔓 is

134

7 The Chebotarev Density Theorem

injective on {1, 𝜁, . . . , 𝜁 𝑚−1 } [Gol71, p. 97, Prop. 6-2-2], 𝜔𝔠 (𝔭) (𝜁) = 𝜁 𝑁𝔭 . Hence, 𝑖(𝜔𝔠 (𝔭)) ≡ 𝑁𝔭 mod 𝑚. Therefore, 𝑖(𝜔𝔠 (𝔞)) ≡ 𝑁𝔞 mod 𝑚 for each 𝔞 ∈ 𝐽 (𝔠).

(7.26)

Now let 𝑥𝑂 𝐾 ∈ 𝑃(𝔠), with 𝑥 ∈ 𝐾 × . Since 𝑥 is totally positive (by (7.20b)), 𝑁 𝐾/Q (𝑥) is a positive rational number congruent by (7.20a) to 1 mod 𝑚. It generates the same fractional Z-ideal as 𝑁 (𝑥𝑂 𝐾 ) [Jan73, p. 37, 4th paragraph]. Since 𝑁 (𝑥𝑂 𝐾 ) is also a positive rational number, 𝑁 (𝑥𝑂 𝐾 ) = 𝑁 𝐾/Q (𝑥) ≡ 1 mod 𝑚. We conclude from (7.26) that 𝜔𝔠 (𝑥𝑂 𝐾 ) = 1. □ Part E: Cyclotomic 𝐿-series and the Dedekind zeta function. Let 𝐿 and 𝔠 be as in Lemma 7.5.3. Then, 𝜔𝔠 induces a homomorphism 𝜔¯ 𝔠 from 𝐺 (𝔠) := 𝐽 (𝔠)/𝑃(𝔠) onto a subgroup 𝐺 of Gal(𝐿/𝐾). (In Corollary 7.5.5(c) we prove that 𝐺 = Gal(𝐿/𝐾).) ˆ 𝜒 ◦ 𝜔¯ 𝔠 is a character of 𝐺 (𝔠). Thus, for each 𝜒 ∈ 𝐺, In the notation of Part C, the 𝐿-series 𝐿𝔠 (𝑠, 1) of the trivial character is the Dedekind zeta function of 𝐾 with respect to 𝔠: ∑︁ Ö 1 1  −1 𝜁𝔠 (𝑠, 𝐾) := = 1 − , Re(𝑠) > 1. (𝑁𝔞) 𝑠 𝔭∤𝔠 (𝑁𝔭) 𝑠 gcd(𝔞,𝔠)=1

Lemma 7.5.4 In the above notation let 𝑛 = (Gal(𝐿/𝐾) : 𝐺) and let ℭ = 𝔠𝑂 𝐿 . Then, Ö 𝜁ℭ (𝑠, 𝐿) = 𝐿𝔠 (𝑠, 𝜒 ◦ 𝜔¯ 𝔠 ) 𝑛 . (7.27) 𝜒 ∈𝐺ˆ

Proof. Let 𝔭 be a prime ideal of 𝑂 𝐾 with 𝔭 ∤ 𝔠. Suppose that 𝔭 factors in 𝐿 into a product of 𝑔 prime ideals 𝔓, each of degree 𝑓 . Since 𝔭 is unramified in 𝐿, we have 𝑓 𝑔 = [𝐿 : 𝐾] = 𝑛|𝐺 | (7.28) 𝐿/𝐾
(Proposition 2.3.2) and 𝜔𝔠 (𝔭) = 𝔭 is of order 𝑓 . By (7.22c) and the relation 𝑁𝔓 = 𝑁𝔭 𝑓 , Ö 𝜒 ◦ 𝜔¯ 𝔠 (𝔭)  𝑛  1  𝑛 |𝐺 |/ 𝑓 1− = 1 − (𝑁𝔭) 𝑠 (𝑁𝔭) 𝑠 𝑓 ˆ 𝜒 ∈𝐺

(7.28)

=



1−

1 𝑔 Ö  1  = 1− . 𝑠 (𝑁𝔓) (𝑁𝔓) 𝑠 𝔓|𝔭

Applying the product over all 𝔭 ∤ 𝔠, we conclude (7.27).

□

Corollary 7.5.5 Let 𝜒1 be a nontrivial character of 𝐺. Then: (a) 𝐿𝔠 (1, 𝜒1 ◦ 𝜔¯ 𝔠 ) ≠ 0; (b) log 𝜁𝔠 (𝑠, 𝐾) = − log(𝑠 − 1) + 𝑂 (1), 𝑠 → 1+ ; and (c) 𝐺 = Gal(𝐿/𝐾). Proof. Since 𝜔¯ 𝔠 : 𝐺 (𝔠) → 𝐺 is surjective, 𝜒 ◦ 𝜔¯ 𝔠 is a nontrivial character of 𝐺 (𝔠) for each nontrivial character of 𝐺. By Lemma 7.5.2, 𝜁𝔠 (𝑠, 𝐾) has a simple pole at 𝑠 = 1 (so (b) holds). All other factors of the right-hand side of (7.27) are regular at

7.5 Number Fields

135

𝑠 = 1. Assume that 𝐿𝔠 (𝑠, 𝜒1 ◦ 𝜔¯ 𝔠 ) has a zero at 𝑠 = 1, then the zero of 𝐿𝔠 (𝑠, 𝜒 ◦ 𝜔¯ 𝔠 ) 𝑛 in (7.27) at 𝑠 = 1 cancels the pole of 𝜁𝔠 (𝑠, 𝐾) 𝑛 . Hence, 𝜁ℭ (𝑠, 𝐿) is analytic at 𝑠 = 1, contradicting Lemma 7.5.1. This proves (a). Finally, by the preceding paragraph, the left-hand side of (7.27) has a pole of order 1 at 𝑠 = 1. By (a), the right-hand side has pole of order 𝑛. Therefore, (Gal(𝐿/𝐾) : 𝐺) = 𝑛 = 1, so 𝐺 = Gal(𝐿/𝐾). □ Lemma 7.5.6 If 𝜒 is a character of 𝐺, then ∑︁ 𝜒 ◦ 𝜔¯ 𝔠 (𝔭) log 𝐿𝔠 (𝑠, 𝜒 ◦ 𝜔¯ 𝔠 ) = + 𝑂 (1), (𝑁𝔭) 𝑠 𝔭∤𝔠

𝑠 → 1+ .

Proof. We apply the Euler identity (7.24) for Re(𝑠) > 1 to obtain: ∞  ∑︁ 𝜒 ◦ 𝜔¯ 𝔠 (𝔭)  ∑︁ ∑︁ 𝜒 ◦ 𝜔¯ 𝔠 (𝔭) 𝑘 log 𝐿𝔠 (𝑠, 𝜒 ◦ 𝜔¯ 𝔠 ) = − log 1 − = . (𝑁𝔭) 𝑠 𝑘 · (𝑁𝔭) 𝑘𝑠 𝔭∤𝔠 𝔭∤𝔠 𝑘=1 Next let 𝜎 = Re(𝑠). Then, ∞ ∞ ∑︁ ∑︁ | 𝜒 ◦ 𝜔¯ 𝔠 (𝔭)| 𝑘 ∑︁ ∑︁ ∑︁ 1 ≤ 𝑘𝑠 𝑓𝔭/ 𝑝 𝑘 𝜎 𝑘 (𝑁𝔭) 𝑝 𝑝 𝔭 | 𝑝 𝑘=2 𝔭∤𝔠 𝑘=2 ∞  ∑︁ ∑︁ ∑︁ 1  1 1 = [𝐾 : Q] 𝑝𝑘 𝜎 𝑝 2𝜎 1 − 𝑝 −𝜎 𝑝 𝑘=2 𝑝 ∑︁ 1 ≤ 2[𝐾 : Q] 1 with Í −𝑠 1 𝜀 𝔭∈𝐶 𝜎 (𝑁𝔭) Í > − , −𝑠 [𝐿 : 𝐾] [𝐿 : 𝐾] 𝔭∈𝑃 (𝐾) (𝑁𝔭)

1 < 𝑠 < 𝑠0 .

(7.33)

As 𝜎 ranges on Gal(𝐿/𝐾), the sum of the left-hand side of (7.33) is 1. Hence, by (7.33) applied to all 𝜎 ′ ∈ Gal(𝐿/𝐾) with 𝜎 ′ ≠ 𝜎, Í −𝑠 1 [𝐿 : 𝐾] − 1 𝔭∈𝐶 𝜎 (𝑁𝔭) Í < +𝜀 , 1 < 𝑠 < 𝑠0 . (7.34) −𝑠 (𝑁𝔭) [𝐿 : 𝐾] [𝐿 : 𝐾] 𝔭∈𝑃 (𝐾) It follows from (7.33) and (7.34) that each 𝐶 𝜎 has Dirichlet density [𝐿 : 𝐾] −1 . This proves the Chebotarev density theorem for Abelian extensions. Part H: Reduction to the cyclic case. We now reduce the Chebotarev density theorem for an arbitrary finite Galois extension 𝐿/𝐾 of number fields to the case when 𝐿/𝐾 is cyclic. Let C be a conjugacy class in 𝐺 := Gal(𝐿/𝐾) and let    𝐿/𝐾 𝐶 = 𝔭 ∈ 𝑃(𝐾) | =C . 𝔭 Choose 𝜏 ∈ C and let 𝐾 ′ = 𝐿 (𝜏) be its fixed field. Denote the set of primes 𝔮 ∈ 𝑃(𝐾 ′) which are unramified over 𝐾, have a relative degree 1 over 𝐾, and satisfy 𝐿/𝐾 ′
= {𝜏} by 𝐷 ′. If 𝔮 ∈ 𝐷 ′ and 𝔭 = 𝔮 ∩ 𝐾, then 𝑁𝔭 = 𝑁𝔮. Hence, if 𝔓 ∈ 𝑃(𝐿) 𝔮    𝐿/𝐾 ′  lies over 𝔮, then 𝐿/𝐾 = 𝔓 = 𝜏, so 𝔭 ∈ 𝐶. Moreover, since ord(𝜏) = [𝐿 : 𝐾 ′], 𝔓 𝔓 is the unique element of 𝑃(𝐿) over 𝔮. Conversely, if 𝔭 ∈ 𝐶, then there exists a   prime 𝔓 ∈ 𝑃(𝐿) that lies over 𝔭 with 𝐿/𝐾 = 𝜏. Hence, 𝔮 := 𝔓 ∩ 𝐾 ′ belongs to 𝐷 ′ 𝔓 and lies over 𝔭.

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Let 𝐶𝐺 (𝜏) be the centralizer of 𝜏 in 𝐺. The map 𝜎 ↦→ 𝜎𝜏𝜎 −1 from 𝐺 onto C has fibers whose order is |𝐶𝐺 (𝜏)|, so |𝐺 | = |C| · |C𝐺 (𝜏)|. All primes 𝔓′ ∈ 𝑃(𝐿) over 𝔭   that satisfy 𝐿/𝐾 = 𝜏 are conjugate to 𝔓 by some element 𝜎 ∈ 𝐶𝐺 (𝜏). Hence, their 𝔓′ 𝐺 ( 𝜏) | number is |𝐶|𝐷 = [𝐿:𝐾|𝐺′ ]|· | C | . 𝔓| This is therefore the number of 𝔮 ∈ 𝐷 ′ that lie over 𝔭. Hence, for 𝑠 > 1 ∑︁ 1 |C|[𝐿 : 𝐾 ′] ∑︁ 1 = . (7.35) 𝑠 (𝑁𝔭) [𝐿 : 𝐾] 𝔮 ∈𝐷′ (𝑁𝔮) 𝑠 𝔭∈𝐶  ′
The set 𝐷 ′ differs from 𝐷 = 𝔮 ∈ 𝑃(𝐾 ′) | 𝐿/𝐾 = {𝜏} by only primes of relative 𝔮 degree at least 2 over 𝐾. For these primes we have ∑︁ ∑︁ 1 1 ′ ≤ [𝐾 : Q] < ∞. (𝑁𝔮) 𝑠 𝑞2 𝑞 deg(𝔮) ≥2

(Note: This elimination of the set of absolute degree 2 primes won’t work in the function field case, for there are only finitely many primes of degree 1.) Hence, by Part G for cyclic extensions ∑︁ 1 ∑︁ 1 1 = + 𝑂 (1) = − log(𝑠 − 1) + 𝑂 (1), 𝑠 → 1+ . 𝑠 𝑠 ′] (𝑁𝔮) (𝑁𝔮) [𝐿 : 𝐾 ′ 𝔮 ∈𝐷 𝔮 ∈𝐷 Combining this with (7.35) gives ∑︁ 1 |C| =− log(𝑠 − 1) + 𝑂 (1), 𝑠 (𝑁𝔭) [𝐿 : 𝐾] 𝔭∈𝐶 Finally, by (7.29), 𝛿(𝐶) =

|C| [𝐿:𝐾 ] ,

as stated.

𝑠 → 1+ . □

Exercises 1. Consider the ring 𝑅 = Z[𝜁 𝑝 ] [𝑥], where 𝑝 is a prime number and 𝑥 is an indeterminate. Let 𝐾 be the quotient field of 𝑅 and 𝐿 = 𝐾 (𝑥 1/ 𝑝 ). Give an example, in the ¯ 𝐾¯ is not Galois. setup of Lemma 7.1.1, where the residue field extension 𝐿/ 2. Let 𝑆/𝑅 be a ring cover. Consider prime ideals 𝔭 of 𝑅 and 𝔓 of 𝑆 with 𝔓 ∩ 𝑅 = 𝔭. Prove that 𝔭𝑆𝔓 = 𝔓𝑆𝔓 (this, together with the separability of 𝑆𝔓 /𝔓𝑆𝔓 over 𝑅𝔭 /𝔭𝑅𝔭 means that 𝑆/𝑅 is an unramified extension of rings). Hint: Assume without loss that 𝑅 and 𝑆 are local rings with maximal ideals 𝔭 and 𝔓, respectively. Let 𝑧 be a primitive element for 𝑆/𝑅 and 𝑓 = irr(𝑧, Quot(𝑅)). Denote the reduction modulo 𝔭𝑆 by a bar. Observe that 𝑆¯ is a local ring with maximal ¯ Prove that 𝑆¯  𝑅[𝑋]/ ¯ ¯ ideal 𝔓. 𝑓¯(𝑋) 𝑅[𝑋] and the right-hand side is a direct sum of fields corresponding to the irreducible factors of 𝑓¯(𝑋). Conclude that there is only one such factor. 3. Let 𝐿 1 , 𝐿 2 be finite Galois extensions of a global field 𝐾 and let 𝐿 = 𝐿 1 𝐿 2 . Consider a prime ideal 𝔓 of 𝑂 𝐿 unramified over 𝐾 and put 𝔓𝑖 = 𝔓 ∩ 𝐿 𝑖 , 𝑖 = 1, 2.    𝐿𝑖 /𝐾  Prove that 𝐿/𝐾 𝔓 is the unique element of Gal(𝐿/𝐾) whose restriction to 𝐿 𝑖 is 𝔓𝑖 , 𝑖 = 1, 2.

7.5 Number Fields

139

4. Show that if 𝑓 (𝑋) ∈ Z[𝑋] is irreducible, then there exists infinitely many primes 𝑝 for which 𝑓 (𝑋) ≡ 0 mod 𝑝 has no solution. Hint: Use the equivalence of (7.5) and (7.7) and Lemma 14.3.2. 5. Let 𝐿/𝐾 be a finite Galois extension of global fields. Denote the set of prime ideals 𝔭 of 𝑂 𝐾 that split completely in 𝐿 (i.e. 𝔭𝑂 𝐿 = 𝔓1 · · · 𝔓𝑛 , where 𝑛 = [𝐿 : 𝐾]) by Splt(𝐿/𝐾). (a) Show that a prime 𝔭 ∈ 𝑃(𝐾), unramified in 𝐿, belongs to Splt(𝐿/𝐾) if and
only if 𝐿/𝐾 = 1. 𝔭 (b) (Bauer) Suppose that 𝐿 and 𝐿 ′ are finite Galois extensions of a global field 𝐾 such that Splt(𝐿/𝐾) and Splt(𝐿 ′/𝐾) differ by a finite set. Prove that 𝐿 = 𝐿 ′. Hint: Apply the Chebotarev density theorem to the field 𝐿𝐿 ′. 6. Let 𝐾 and 𝐿 be number fields with equal zeta functions (Section 7.5, Part E). For each positive integer 𝑛 prove that the number of ideals of 𝑂 𝐾 with absolute norm 𝑛 is the number of ideals of 𝑂 𝐿 with absolute norm 𝑛. Apply this to any prime 𝑝 unramified in 𝐾 𝐿. Prove that 𝑝 splits completely in 𝐾 if and only if 𝑝 splits completely in 𝐿. Conclude from Exercise 5 that if 𝐾 and 𝐿 are Galois over Q, then 𝐾=𝐿 7. Let 𝐾 be a global field. Denote the set of all prime ideals 𝔭 of 𝑂 𝐾 whose absolute degree is at least 2 by 𝑃 ′ (𝐾). That is, 𝑁𝔭 = 𝑝 𝑑 where 𝑑 ≥ 2. When 𝐾 is a number field prove that the Dirichlet density as well as the natural density of 𝑃 ′ (𝐾) is 0. If, however, 𝐾 is a function field, show that almost all primes of 𝑃(𝐾) belong to 𝑃 ′ (𝐾). Thus, 𝛿(𝑃 ′ (𝐾)) = 1.

Notes Frobenius [Fro96] conjectured what we now call the Chebotarev density theorem for finite Galois extension 𝐿/𝐾 of number fields. His result replaced the conjugacy class appearing in the conjecture by the union of all conjugates of 𝜎 𝑖 , where 𝜎 is a given element of Gal(𝐿/𝐾) and 𝑖 ranges over all integers relatively prime to ord(𝜎). A fine account of the Frobenius density theorem appears in [Jan73]. Chebotarev [Tsc26] used cyclotomic fields to prove the Frobenius conjecture via a more difficult version of the field crossing argument of Part G of Section 7.5. Artin [Art24] introduced his 𝐿-series; then he proved his reciprocity law and applied it to reprove the conjecture [Art27]. Our proof is a mixture of both methods, with the addition of Deuring’s reduction to the cyclic case [Deu34]. It was elaborated for this book by Haran. Deuring’s reduction was reproduced in [MaC68]. For the function field case note that Reichardt proved Proposition 7.4.8 when 𝑎 = 𝑘 = 1 and 𝐾 is algebraically closed in 𝐹, (i.e. 𝑚 = 1) [Rei36]. The restriction 𝑚 = 1 does not appear explicitly in [Rei36]. Without it, however, the result as well as its proof would be false. It is also interesting to note that [Rei36] appeared before Weil proved the Riemann hypothesis for curves. Thus, Reichardt’s (analytic) proof uses only that the maximum of the real parts of the zeros of the Zeta function is less than 1.

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7 The Chebotarev Density Theorem

Serre [Ser65] gives a unified approach to the number field and function field case. He considers a scheme 𝑋 of finite type over Z, takes an étale Galois covering of 𝑋 and attaches an 𝐿-series 𝐿 (𝑋, 𝜒; 𝑠) to the cover. He says that an induction on dim(𝑋) shows if 𝜒 ≠ 1, then 𝐿(𝑋, 𝜒; 𝑠) is holomorphic and ≠ 0 at the point 𝑠 = dim(𝑋). This implies the Chebotarev density theorem by the classical Dirichlet argument (e.g. as in Part F of Section 7.5). Serre’s program for the function field case appears in [Fri80]. An early version of our proof for the function field case appears in [Jar82a]. The proof of [FrJ86, Prop. 5.16] applies [FrJ86, Lemma 5.14] in a faulty way. Indeed, 𝑑 on [FrJ86, p. 63, line -3] should be replaced by 𝑚𝑑. This version of Field Arithmetic follows [GeJ98, Appendix] and corrects this mistake, also improving the estimate of [FrJ86, Prop. 5.16]. An extensive account of [Ser65] can be found in Holschbach’s Diplomarbeit [Hol04]. There are several effective versions of the Chebotarev density theorem. One of the most valuable for the problems in this book is [LMO79, p. 416, Theorem]. This isolates the contribution of the absolute discriminant 𝑑 𝐿 of a number field 𝐿 over Q to the error term. It proves that there is an effectively computable constant 𝐴 with the following property: For each Galois extension 𝐿/𝐾 of Gal(𝐿/𝐾) of number fields and each conjugacy class C of Gal(𝐿/𝐾) there is a prime 𝔭 of 𝐾, unramified in
𝐿 with 𝐿/𝐾 = C, 𝑝 = 𝑁 𝐿/Q𝔭 is a rational prime and 𝑝 ≤ 2𝑑 𝐿𝐴. This result is 𝔭 independent of the generalized Riemann hypothesis.

Chapter 8

Ultraproducts

We develop the basic concepts of logic and model theory required for applications to field theory. These include the Skolem–Löwenheim theorem, Łoš’ theorem and an ℵ1 -saturation property for ultraproducts. Finally, we apply regular ultraproducts of families of models to the theory of finite fields.

8.1 First Order Predicate Calculus There is no general test to decide whether a given polynomial 𝑓 (𝑋1 , . . . , 𝑋𝑛 ) with integral coefficients has a zero in Z𝑛 ; this is the negative solution to Hilbert’s 10th problem. A partial decision test must satisfy two criteria: it must be conclusive for a significant body of polynomials; and it must be effective in concrete situations. The simplest, and most famous, such test is the congruence test, whereby we test the congruence 𝑓 (𝑋1 , . . . , 𝑋𝑛 ) ≡ 0 mod 𝑝 for solutions for all primes 𝑝. Regard the coefficients of 𝑓 as elements of F 𝑝 to see that the above congruence is equivalent to solving the equation 𝑓 (𝑋1 , . . . , 𝑋𝑛 ) = 0 in F 𝑝 . If 𝑓 (𝑋1 , . . . , 𝑋𝑛 ) = 0 has no solution in F 𝑝 for one 𝑝, then 𝑓 (𝑋1 , . . . , 𝑋𝑛 ) = 0 has no solution in Z. This chapter develops the language and techniques for the formulation of analogs of the diophantine problem and of the corresponding congruence test over general rings and fields. We start with the introduction of a first order language, the concept of a theory in the first order language, and a model for this theory. A language (more precisely, first order language) consists of letters, rules for combining letters into meaningful words, and, finally, an interpretation of the meaningful words. The language we now describe depends on functions 𝜇 and 𝜈 from sets 𝐼 and 𝐽 to N and on a set 𝐾. It is denoted L (𝜇, 𝜈, 𝐾). Here are its letters: (8.1a) (8.1b) (8.1c) (8.1d) (8.1e) (8.1f)

Countably many variable symbols: 𝑋1 , 𝑋2 , 𝑋3 , . . . ; constant symbols 𝑐 𝑘 , one for each 𝑘 ∈ 𝐾; a 𝜇(𝑖)-ary relation symbol, 𝑅𝑖 , one for each 𝑖 ∈ 𝐼; the equality symbol =; a 𝜈( 𝑗)-ary function symbol, 𝐹 𝑗 , one for each 𝑗 ∈ 𝐽; the negation symbol ¬, and the disjunction symbol ∨;

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_8

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(8.1g) the existential symbol ∃; and (8.1h) parentheses ( ) and brackets [ ]. A finite sequence of letters of L (𝜇, 𝜈, 𝐾) is a string. Among the strings of L (𝜇, 𝜈, 𝐾) the (meaningful) words include, “terms”, “formulas” and “sentences”; which we now define. The collection of terms of L (𝜇, 𝜈, 𝐾) is the smallest collection of strings that contains all of the following: (8.2a) all the variable symbols 𝑋𝑖 ; (8.2b) all the constant symbols 𝑐 𝑘 ; and (8.2c) all the strings 𝐹 𝑗 (𝑡 1 , . . . , 𝑡 𝜈 ( 𝑗) ) where 𝑗 ∈ 𝐽 and (𝑡 1 , . . . , 𝑡 𝜈 ( 𝑗) ) is a 𝜈( 𝑗)tuple of previously defined terms. We list the atomic formulas: (8.3a) 𝑡 = 𝑡 ′ for each pair of terms (𝑡, 𝑡 ′); and (8.3b) 𝑅𝑖 (𝑡 1 , . . . , 𝑡 𝜇 (𝑖) ), for all 𝑖 ∈ 𝐼 and all 𝜇(𝑖)-tuples (𝑡1 , . . . , 𝑡 𝜇 (𝑖) ) of terms. The set of formulas is the smallest collection of strings containing all atomic formulas and satisfying the following: (8.4a) ¬[𝜑] is a formula, if 𝜑 is a formula; (8.4b) 𝜑1 ∨ 𝜑2 is a formula if 𝜑1 and 𝜑2 are formulas; and (8.4c) (∃𝑋𝑙 ) [𝜑] is a formula, if 𝜑 is a formula and 𝑙 ∈ N. This definition allows us to prove a property of formulas or to make definitions depending on formulas by an induction on structure. We first prove the property for atomic formulas. Then, assuming the validity of the formula for 𝜑, 𝜑1 , and 𝜑2 we prove it for ¬[𝜑], 𝜑1 ∨ 𝜑2 , and (∃𝑋𝑙 ) [𝜑]. As a first example we define the notion of free occurrence of a variable in a formula by an induction on structure: Any occurrence of 𝑋 in an atomic formula 𝜑 is free. If an occurrence of 𝑋 in a formula 𝜑 is free, and 𝜓 is an arbitrary formula, then this occurrence is free in ¬[𝜑], 𝜑 ∨ 𝜓 and (∃𝑌 ) [𝜑], for 𝑌 distinct from 𝑋. Any occurrence of 𝑋 which is not free is bounded. Any variable 𝑋 which has a free occurrence in a formula 𝜑 is said to be a free variable of 𝜑. Frequently we write 𝜑(𝑋1 , . . . , 𝑋𝑛 ) (or 𝑡 (𝑋1 , . . . , 𝑋𝑛 )) to indicate that 𝑋1 , . . . , 𝑋𝑛 include all the free variables of 𝜑 (or 𝑡). Example 8.1.1 In the formula  1  2 3 (∃𝑋) 𝑋 = 𝑌 ∨ (∃𝑋) [ 𝑋 = 𝑐] ∨ ¬𝑅( 𝑋 , 𝑌 ) occurrences 1 and 2 are bounded, while occurrence 3 is free; both occurrences of 𝑌 are free. Hence, 𝑋 and 𝑌 are free variables of the formula. A formula without free variables is a sentence. Some abbreviations simplify this language: (8.5a) 𝜑 ∧ 𝜓 for ¬[¬𝜑 ∨ ¬𝜓] (∧ is the conjunction symbol); (8.5b) 𝜑 → 𝜓 for ¬𝜑 ∨ 𝜓 (→ is the implication symbol); (8.5c) 𝜑 ↔ 𝜓 for [𝜑 → 𝜓] ∧ [𝜓 → 𝜑] (↔ is the double implication symbol);

8.2 Structures

143

(8.5d) (∀𝑋𝑙 ) [𝜑] for ¬(∃𝑋𝑙 ) [¬𝜑] (∀ is the universal quantifier); Ó𝑛 (8.5e) 𝑖=1 𝜑𝑖 for 𝜑1 ∧ 𝜑2 ∧ · · · ∧ 𝜑 𝑛 ; and Ô𝑛 (8.5f) 𝑖=1 𝜑𝑖 for 𝜑1 ∨ 𝜑2 ∨ · · · ∨ 𝜑 𝑛 .

8.2 Structures The sentences of a first order language are interpreted in “structures” of this language. In each of these structures, they are either true or false. A structure for the language L (𝜇, 𝜈, 𝐾) is a system A = ⟨𝐴, 𝑅¯𝑖 , 𝐹¯ 𝑗 , 𝑐¯𝑘 ⟩𝑖 ∈𝐼, 𝑗 ∈𝐽 , 𝑘 ∈𝐾 where 𝐴 is a nonvoid set, called the domain of A, 𝑅¯𝑖 is a 𝜇(𝑖)-ary relation of 𝐴 (i.e. a subset of 𝐴 𝜇 (𝑖) ), 𝐹¯ 𝑗 : 𝐴 𝜈 ( 𝑗) → 𝐴 is a 𝜈( 𝑗)-ary function on 𝐴, and 𝑐¯𝑘 is an element of 𝐴, called a constant. Sometimes we use the same letter for the logical symbol and its interpretation in the structure. Also, for well-known binary relations and binary functions we write the relation and function symbols as usual, between the argument (e.g. “𝑎 ≤ 𝑏” for “𝑎 less than or equal to 𝑏”). Occasionally we add to L = L (𝜇, 𝜈, 𝐾) a new constant symbol
𝑎˜ for each 𝑎 ∈ 𝐴. This gives an extended structure L ( 𝐴) = L 𝜇, 𝜈, 𝐾 ∪· { 𝑎˜ | 𝑎 ∈ 𝐴} . A substitution into 𝐴 is a function 𝑓 (𝑋𝑖 ) = 𝑥𝑖 , from the set of variables into 𝐴. The following recursive rules extend this uniquely to a function from the set of terms into 𝐴: (8.6a) 𝑓 (𝑐 𝑘 ) = 𝑐¯𝑘 , and (8.6b) 𝑓 (𝐹 𝑗 (𝑡 1 , . . . , 𝑡 𝜈 ( 𝑗) )) = 𝐹¯ 𝑗 ( 𝑓 (𝑡1 ), . . . , 𝑓 (𝑡 𝜈 ( 𝑗) )), where 𝑡1 , . . . , 𝑡 𝜈 ( 𝑗) , are terms for which 𝑓 has already been defined. Define the truth value of a formula 𝜑 under a substitution 𝑓 (either “true” or “false”) by induction on structure: (8.7a) 𝑡 = 𝑡 ′ is true if 𝑓 (𝑡) = 𝑓 (𝑡 ′); and (8.7b) 𝑅𝑖 (𝑡1 , . . . , 𝑡 𝜇 (𝑖) ) is true if ( 𝑓 (𝑡1 ), . . . , 𝑓 (𝑡 𝜇 (𝑖) )) ∈ 𝑅¯𝑖 . Continue by assuming that the truth values of 𝜑, 𝜑1 , and 𝜑2 have been defined for all possible substitutions. Then, (8.8a) ¬𝜑 is true if 𝜑 is false; (8.8b) 𝜑1 ∨ 𝜑2 is true if 𝜑1 is true or if 𝜑2 is true (so if both 𝜑1 and 𝜑2 are true, then 𝜑1 ∨ 𝜑2 is also true); and (8.8c) (∃𝑋𝑙 ) [𝜑] is true if there exists an 𝑥 in 𝐴 such that 𝜑 is true under the substitution 𝑔 defined by: 𝑔(𝑋𝑙 ) = 𝑥 and 𝑔(𝑋𝑚 ) = 𝑓 (𝑋𝑚 ) if 𝑚 ≠ 𝑙. The truth values of the additional logical symbols introduced above are as follows: (8.9a) 𝜑1 ∧ 𝜑2 is true if both 𝜑1 and 𝜑2 are true; (8.9b) 𝜑 → 𝜓 is true if the truth of 𝜑 implies the truth of 𝜓 (i.e. either 𝜑 is false, or both 𝜑 and 𝜓 are true); (8.9c) 𝜑 ↔ 𝜓 is true if both 𝜑 and 𝜓 are true or both 𝜑 and 𝜓 are false; and

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(8.9d) (∀𝑋𝑙 ) [𝜑] is true if for each 𝑥 in 𝐴, 𝜑 is true under the substitution 𝑔 defined by 𝑔(𝑋𝑙 ) = 𝑥 and 𝑔(𝑋𝑚 ) = 𝑓 (𝑋𝑚 ) if 𝑚 ≠ 𝑙. By an easy induction on structure one observes that the truth value of a formula 𝜑(𝑋1 , . . . , 𝑋𝑛 ) under a substitution 𝑓 depends only on 𝑓 (𝑋1 ) = 𝑥 1 , . . . , 𝑓 (𝑋𝑛 ) = 𝑥 𝑛 . If 𝜑 is true under 𝑓 , write A |= 𝜑(𝑥1 , . . . , 𝑥 𝑛 ). In particular, for 𝜑 a sentence, the truth value of 𝜑 is independent of 𝑓 . It is either true in A, or false in A. In the former case write A |= 𝜑, and in the latter, A ̸ |= 𝜑

8.3 Models Models of a first order language L are generalizations of groups, rings, fields, and ordered sets. They are structures where a given set of sentences, called axioms, is true. A theory in a first order language L = L (𝜇, 𝜈, 𝐾) is a set of sentences 𝑇 of L. A structure A for L is called a model of 𝑇 if A |= 𝜃 for every 𝜃 ∈ 𝑇. In this case write A |= 𝑇. If 𝑇 ′ is another theory in L for which every model of 𝑇 ′ is also a model of 𝑇, write 𝑇 ′ |= 𝑇. If Π is a theory of L and 𝑇 is the theory of all sentences 𝜃 of L such that Π |= 𝜃, then Π is said to be a set of axioms for 𝑇. Denote the class of all models of a theory 𝑇 by Mod(𝑇). If A is a structure for a language L, then Th(A, L) (or Th(A) if L is known from the context) is the set of all sentences of L which are true in A. Example 8.3.1 (The theory of fields) Denote the first order language that contains the two binary functions symbols + (addition) and · (multiplication), and two constant symbols 0 and 1 by L (ring). For each integral domain 𝑅 let L (ring, 𝑅) be the language L (ring) extended by all elements of 𝑅 as constant symbols. Denote the set of usual axioms for the theory of fields by Π: (∀𝑋) (∀𝑌 )(∀𝑍) [(𝑋 + 𝑌 ) + 𝑍 = 𝑋 + (𝑌 + 𝑍)]; (∀𝑋)(∀𝑌 ) [𝑋 + 𝑌 = 𝑌 + 𝑋]; (∀𝑋) [𝑋 + 0 = 𝑋]; (∀𝑋)(∃𝑌 ) [𝑋 + 𝑌 = 0]; (∀𝑋)(∀𝑌 )(∀𝑍) [(𝑋𝑌 )𝑍 = 𝑋 (𝑌 𝑍)]; (∀𝑋) (∀𝑌 ) [𝑋𝑌 = 𝑌 𝑋]; (∀𝑋) [1 · 𝑋 = 𝑋]; (∀𝑋) [𝑋 ≠ 0 → (∃𝑌 ) [𝑋𝑌 = 1]]; 1 ≠ 0; and (∀𝑋)(∀𝑌 ) (∀𝑍) [𝑋 (𝑌 + 𝑍) = 𝑋𝑌 + 𝑋 𝑍]. Every model of Π is a field. Extend Π by all equalities — the positive diagram of 𝑅— (8.10) 𝑎 1 + 𝑏 1 = 𝑐 1 and 𝑎 2 𝑏 2 = 𝑐 2 , for 𝑎 𝑖 , 𝑏 𝑖 , 𝑐 𝑖 ∈ 𝑅 that are true in 𝑅. Denote the set obtained by Π(𝑅). A model of Π(𝑅) is a field that contains a subset 𝑅¯ = { 𝑎¯ | 𝑎 ∈ 𝑅} whose elements satisfy the equalities

8.3 Models

145

𝑎¯ 1 + 𝑏¯ 1 = 𝑐¯1

and

𝑎¯ 2 𝑏¯ 2 = 𝑐¯2

whenever the corresponding equalities of (8.10) are true in 𝑅. That is, 𝑅¯ is a homomorphic image of 𝑅. If 𝑅 = 𝐾 is a field, then 𝐾¯ is an isomorphic copy of 𝐾. Thus, a model of Π(𝐾) is (up to an isomorphism) a field containing 𝐾. Example 8.3.2 (Irreducible Polynomials) Let 𝑅 be an integral domain. An elementary statement about models of Π(𝑅) is a mathematical statement that applies to each member of Mod(Π(𝑅)) and for which there exists a sentence 𝜃 of L (ring, 𝑅) which is true in any given model 𝐹 if and only if the statement is true. Consider, for example, a polynomial 𝑓 (𝑋1 , . . . , 𝑋𝑛 ) of degree 𝑑 with coefficients in 𝑅. Then “ 𝑓 (X) is irreducible” is an elementary statement about models of Π(𝑅). Indeed, it is equivalent to the conjunction of the statements “there exist no polynomials 𝑔, ℎ of degree 𝑑1 and 𝑑2 respectively such that 𝑓 (X) = 𝑔(X)ℎ(X),” where (𝑑1 , 𝑑2 ) runs over all pairs of positive integers with 𝑑1 + 𝑑2 = 𝑑. Rewrite the phrase “there exists no polynomial 𝑔(𝑋1 , . . . , 𝑋𝑛 ) of degree 𝑑1 ” as “¬(∃𝑢 1 ) · · · (∃𝑢 𝑘 )”, where 𝑢 1 , . . . , 𝑢 𝑘 are variables for the coefficients of 𝑔(X). A system of equalities between corresponding coefficients on both sides of “=” replaces “ 𝑓 (X) = 𝑔(X)ℎ(X).” Similarly, we may consider a polynomial ∑︁ 𝑓 (u, 𝑋1 , . . . , 𝑋𝑛 ) = 𝑢 i 𝑋1𝑖1 · · · 𝑋𝑛𝑖𝑛 with intermediate coefficients 𝑢 i . The same argument as above gives a formula 𝜑(u) in L (ring) such that for each field 𝐾 and all tuples a with entries in 𝐾, the polynomial 𝑓 (a, X) is irreducible in 𝐾 [X] if and only if 𝜑(a) is true in 𝐾. Two structures A = ⟨𝐴, 𝑅𝑖 , 𝐹 𝑗 , 𝑐 𝑘 ⟩ and B = ⟨𝐵, 𝑆𝑖 , 𝐺 𝑗 , 𝑑 𝑘 ⟩ of the language L = L (𝜇, 𝜈, 𝐾) are isomorphic if there exists a bijective function 𝑓 : 𝐴 → 𝐵 such that (8.11a) (𝑎 1 , . . . , 𝑎 𝜇 (𝑖) ) ∈ 𝑅𝑖 ⇐⇒ ( 𝑓 (𝑎 1 ), . . . , 𝑓 (𝑎 𝜇 (𝑖) )) ∈ 𝑆𝑖 , for each 𝑖 ∈ 𝐼; (8.11b) 𝑓 (𝐹 𝑗 (𝑎 1 , . . . , 𝑎 𝜈 ( 𝑗) )) = 𝐺 𝑗 ( 𝑓 (𝑎 1 ), . . . , 𝑓 (𝑎 𝜈 ( 𝑗) )), for each 𝑗 ∈ 𝐽; and (8.11c) 𝑓 (𝑐 𝑘 ) = 𝑑 𝑘 , for each 𝑘 ∈ 𝐾. In this case write A  B. The structures A and B are elementarily equivalent if A |= 𝜃 ⇐⇒ B |= 𝜃 for every sentence 𝜃 of L. If this is the case, we write A ≡ B. Clearly, if A  B, then A ≡ B. But we will have many examples that show the converse is false. Two fields 𝐿 and 𝐿 ′ that contain a field 𝐾 are isomorphic as models of Π(𝐾) if and only if there exists a field isomorphism of 𝐿 onto 𝐿 ′ that fixes every element of 𝐾: write 𝐿 𝐾 𝐿 ′. However, if they are elementarily equivalent as models of Π(𝐾), we write 𝐿 ≡𝐾 𝐿 ′. Call A a substructure of B (A ⊆ B, and B is an extension of A) if 𝐴 ⊆ 𝐵, 𝑅𝑖 = 𝐴 𝜇 (𝑖) ∩ 𝑆𝑖 for each 𝑖 ∈ 𝐼, 𝐹 𝑗 (𝑎 1 , . . . , 𝑎 𝜈 ( 𝑗) ) = 𝐺 𝑗 (𝑎 1 , . . . , 𝑎 𝜈 ( 𝑗) ) for each 𝑗 ∈ 𝐽 and all 𝑎 1 , . . . , 𝑎 𝜈 ( 𝑗) ∈ 𝐴; and 𝑐 𝑘 = 𝑑 𝑘 for each 𝑘 ∈ 𝐾.

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146

More generally, an embedding of A into B is an injective map 𝑓 : 𝐴 → 𝐵 that satisfies Conditions (8.11a), (8.11b), and (8.11c). Note that an arbitrary map 𝑓 : 𝐴 → 𝐵 is an embedding of A into B if and only if for each quantifier-free formula 𝜑(𝑋1 , . . . , 𝑋𝑛 ) of L and for all 𝑎 1 , . . . , 𝑎 𝑛 ∈ 𝐴, the condition A |= 𝜑(𝑎 1 , . . . , 𝑎 𝑛 ) implies B |= 𝜑( 𝑓 (𝑎 1 ), . . . , 𝑓 (𝑎 𝑛 )). Indeed, an application of the latter condition to the formula 𝑋1 ≠ 𝑋2 implies that 𝑓 is injective. Suppose now that A ⊆ B. We say A is existentially closed in B if for each quantifier-free formula 𝜑(𝑋1 , . . . , 𝑋𝑛 ) and for all 𝑏 1 , . . . , 𝑏 𝑛 ∈ 𝐵 with B |= 𝜑(𝑏 1 , . . . , 𝑏 𝑛 ) there exist 𝑎 1 , . . . , 𝑎 𝑛 ∈ 𝐴 with A |= 𝜑(𝑎 1 , . . . , 𝑎 𝑛 ). Call A an elementary substructure of B and B an elementary extension of A (in symbols A ≺ B) if A ⊆ B and if for each formula 𝜑(𝑋1 , . . . , 𝑋𝑛 ) of L and for every 𝑎 1 , . . . , 𝑎 𝑛 in 𝐴, the truth of 𝜑(𝑎 1 , . . . , 𝑎 𝑛 ) in A is equivalent to its truth in B. It follows, in particular, that a sentence 𝜃 of L is true in A if and only if it is true in B, (i.e. A ≡ B). The converse is false (Example 8.3.3). However, if A ⊆ B, then “A ≡ B as models of L ( 𝐴)” is equivalent to “A ≺ B as models of L”. Transitivity of elementarily equivalence follows immediately: A ≺ B, B ≺ C implies A ≺ C. In addition, A ⊆ B ≺ C and A ≺ C imply A ≺ B. Example 8.3.3 (Elementary subfields) If a field 𝐾 is an elementary subfield of a field 𝐹, then 𝐹 is a regular extension of 𝐾. In other words, 𝐾 is algebraically closed in 𝐹 and 𝐹/𝐾 is separable (Lemma 3.4.1). First of all let 𝑥 ∈ 𝐾˜ ∩ 𝐹 and 𝑓 = irr(𝑥, 𝐾). Then, the sentence (∃𝑋) [ 𝑓 (𝑋) = 0] holds in 𝐹, so also in 𝐾. Therefore, deg( 𝑓 ) = 1, hence 𝑥 ∈ 𝐾. Consequently, 𝐾˜ ∩ 𝐹 = 𝐾. Now assume that 𝑝 := char(𝐾) > 0. Consider 𝑢 1 , . . . , 𝑢 𝑛 ∈ 𝐾 and 𝑏 1 , . . . , 𝑏 𝑛 ∈ 𝐹 Í𝑛 𝑝 Í𝑛 𝑝 with 𝑖=1 𝑏 𝑖 𝑢 1/ 𝑏 𝑖 𝑢 𝑖 = 0. Hence, = 0. Then, 𝑖=1 𝑖 (∃𝑋1 ) · · · (∃𝑋𝑛 )

𝑛 ∑︁

𝑋𝑖𝑝 𝑢 𝑖

𝑖=1

is 𝐹, so also in 𝐾. In other words, there exist 𝑎 1 , . . . , 𝑎 𝑛 ∈ 𝐾 with Í𝑛true in 1/ 𝑝 𝑎 = 0. Therefore, 𝐹 is linearly disjoint from 𝐾 1/ 𝑝 over 𝐾. Consequently, 𝑢 𝑖 𝑖=1 𝑖 𝐹/𝐾 is separable. For example, let 𝑥 be an indeterminate. Then, Q(𝑥 2 )  Q(𝑥). Hence, Q(𝑥 2 ) ≡ Q(𝑥). But Q(𝑥) is a proper algebraic extension of Q(𝑥 2 ). Therefore, Q(𝑥 2 ) is not an elementary subfield of Q(𝑥).

8.4 Elementary Substructures We develop criteria for one structure to be an elementary substructure of another. Let 𝑚 be a cardinal number. Consider a transfinite sequence {A 𝛼 | 𝛼 < 𝑚} of structures for a language L = L (𝜇, 𝜈, 𝐾) with A 𝛼 = ⟨𝐴 𝛼 , 𝑅 𝛼𝑖 Ð , 𝐹𝛼 𝑗 , 𝑐 𝛼𝑘 ⟩. Suppose that A 𝛼 ⊆ A 𝛽 for each 𝛼 ≤ 𝛽 < 𝑚 and define the union 𝛼 0 and 𝑣(𝑎) < 0. Then, there is an 𝑥 ∈ 𝐾 𝑣 with 𝑥 𝑝 − 𝑥 − 𝑎 = 0. Hence, 𝑝1 𝑣(𝑎) = 𝑣(𝑥) ∈ 𝑣(𝐾 × ). Consequently, 𝑣(𝐾 × ) is divisible. □ Corollary 12.5.6 Let 𝐾 be a PAC field which is not separably closed. Then, 𝐾 is Henselian with respect to no valuation. Denote the maximal Abelian, nilpotent, and solvable extensions of Q by Qab , Qnil , and Qsolv , respectively. Corollary 12.5.7 ([Fre73]) The fields Qab and Qnil are not PAC fields. Proof. By Corollary 12.2.5, we have only to prove the statement for Qnil . For each 𝑝, Q 𝑝,alg is Henselian. Assume Qnil is PAC. Then, Qnil Q 𝑝,alg is PAC (Corollary ˜ Hence, Gal(Q 𝑝,alg )  12.2.5) and Henselian. By Corollary 12.5.5, Qnil Q 𝑝,alg = Q. Gal(Qnil ∩ Qnil Q 𝑝,alg ) is pronilpotent. That is, the Galois group of every irreducible separable polynomial over Q 𝑝,alg is nilpotent. In particular, this is true for 𝑝 = 5. However, 𝑋 3 + 5 is irreducible over Q5 (e.g. by Eisenstein’s criterion) and its discriminant −27 · 52 [Lan97, p. 270] is not a square in Q5,alg (e.g. −27 is a quadratic non-residue modulo 5). Hence, 𝑋 3 + 5 is irreducible over Q5,alg and its discriminant is not a square in Q5,alg . Therefore, Gal(𝑋 3 +5, Q5,alg ) is 𝑆3 , which is not nilpotent. We conclude from this contradiction to the preceding paragraph that Qnil is not a PAC field. □ ˜ for each prime number 𝑝. In contrast to Qab and Qnil , we have Qsolv Q 𝑝,alg = Q ˜ Proposition 12.5.8 The Henselian closure of each valuation of Qsolv is Q. Proof. Let 𝑣 be a valuation of Qsolv . Replacing 𝑣 by an equivalent valuation, we may assume that 𝑣| Q = 𝑣 𝑝 for some prime number 𝑝. Thus, Qsolv can be embedded ˜ 𝑝 such that 𝑣 is the restriction to Qsolv of the unique extension of 𝑣 𝑝 to Q ˜ 𝑝 . Let in Q ˜ 𝑀 = Qsolv Q 𝑝,alg . Then, 𝑀 is a Henselian closure of 𝑀 at 𝑣. Assume 𝑀 ≠ Q.

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12 Pseudo Algebraically Closed Fields

Since Gal(Q 𝑝 ) is prosolvable [CaF67, p. 31, Cor. 1], so is Gal(𝑀). Since 𝑀 √ 𝑛 contains all roots of unity, there exist 𝑏 ∈ 𝑀 and√𝑛 ∈ N such√that 𝑏 ∉ 𝑀. If 𝑎 ∈ 𝑀 is 𝑛 𝑛 sufficiently 𝑣 𝑝 -close to 𝑏, then by Krasner, 𝑀 ( 𝑏) ⊆ 𝑀 ( 𝑎) [Jar91, Í𝑚Lemma 12.1]. Choose 𝑤 1 , . . . , 𝑤 𝑚 ∈ Qsolv and 𝑏 1 , . . . , 𝑏 𝑚 ∈ Q 𝑝,alg such that 𝑏 = 𝑖=1 𝑏 𝑖 𝑤 𝑖 . Since Q is 𝑣 𝑝 -dense in Q , we may choose 𝑎 , . . . , 𝑎 which are 𝑣 -close to 𝑏1 , . . . , 𝑏 𝑝 . 𝑝 1 𝑝 𝑝 √ Í𝑚 Then, 𝑎 = 𝑖=1 𝑎 𝑖 𝑤 𝑖 is 𝑣-close to 𝑏 and lies in Qsolv . It follows that 𝑛 𝑎 ∈ Qsolv ⊆ 𝑀, √ 𝑛 ˜ so 𝑏 ∈ 𝑀. We conclude from this contradiction that 𝑀 = Q. □ Thus, Corollary 12.5.5 fails to solve the following problem. Problem 12.5.9 (a) Is Qsolv a PAC field? (b) Does there exist an infinite non-PAC field 𝐾 of a finite transcendence degree over its prime field such that 𝐾 is not formally real and all of its Henselian closures are separably closed? Remark 12.5.10 (On Problem 12.5.9(a)) Problem 12.5.9(a) leads to other problems with respectable classical connections. Here is one example: By Theorem 12.2.3, we have only to check if each absolutely irreducible curve, 𝑓 (𝑋, 𝑌 ) = 0, with 𝑓 ∈ Q[𝑋, 𝑌 ], has a point with coordinates in Qsolv . Let 𝐸 = Q(𝑥, 𝑦) be the function field of this curve. Suppose that there exists a 𝑡 ∈ 𝐸 such that 𝐸/Q(𝑡) is an extension ˆ is solvable over Q(𝑡). Then, each specialization 𝑡 → 𝑡0 whose Galois closure, 𝐸, such that 𝑡0 ∈ Q and 𝑥 and 𝑦 are integral over the corresponding local ring extends to a specialization (𝑥, 𝑦) → (𝑥 0 , 𝑦 0 ) with 𝑥0 , 𝑦 0 ∈ Qsolv (Lemma 7.1.1(a)). Since there are infinitely many such specializations, 𝑓 (𝑋, 𝑌 ) = 0 certainly has infinitely many Qsolv -rational points. This idea fails, however, if the function field 𝐹 = C(𝑥, 𝑦) ˆ over C for a “general” curve, 𝑓 (𝑋, 𝑌 ) = 0, has no subfield C(𝑡) with Gal( 𝐹/C(𝑡)) solvable. The proper subfields of a “general” curve of genus 𝑔 > 1 are of genus 0 [Fri77, p. 26-27]. Therefore, with no loss, assume that there are no proper fields between ˆ C(𝑡) and 𝐹; that is, Gal( 𝐹/C(𝑡)) is a primitive solvable group. A theorem of Galois implies that [𝐹 : C(𝑡)] = 𝑝 𝑟 for some prime 𝑝 [Bur55, p. 202]. Combining [Fri77, p. 26] and [Rit22] one may prove that for 𝐹 “general” of genus suitably large (> 6) ˆ that if 𝑟 = 1 then it is impossible for Gal( 𝐹/C(𝑡)) to be solvable. But higher values of 𝑟 have not yet been excluded.

Remark 12.5.11 (On Problem 12.5.9(b)) Theorem D of [GeJ01] gives for each characteristic 𝑝 (including 𝑝 = 0) an example of an infinite non-PAC field 𝐾 of characteristic 𝑝 which is not formally real and all of its Henselian closures of 𝐾 are separably closed. This gives an affirmative answer to Problem 11.5.1(b) of [FrJ86]. The proof is based on results and ideas of [Efr01]. It uses Galois cohomology, valuations of higher rank, and the Jacobian varieties of curves. Problem 12.5.9(b) is a reformulation of the older problem.

12.6 The Absolute Galois Group of a PAC Field

217

12.6 The Absolute Galois Group of a PAC Field We show that the absolute Galois groups of PAC fields are “projective” (Theorem 12.6.2). All decidability and undecidability results on PAC fields depend on this result. Lemma 12.6.1 Let 𝐿/𝐾 be a finite Galois extension, 𝐵 a finite group, and 𝛼: 𝐵 → Gal(𝐿/𝐾) an epimorphism. Then, there exists a finite Galois extension 𝐹/𝐸 with Gal(𝐹/𝐸) = 𝐵 such that 𝐸 is a regular finitely generated extension of 𝐾, 𝐹 is a purely transcendental extension of 𝐿 with trans.deg(𝐹/𝐿) = |𝐵|, and 𝛼 = res𝐹/𝐿 . Proof. Let {𝑦 𝛽 | 𝛽 ∈ 𝐵} be a set of indeterminates of cardinality |𝐵|. Put 𝐹 = ′ ′ ′ ′ 𝐿 (𝑦 𝛽 | 𝛽 ∈ 𝐵). Define an action of 𝐵 on 𝐹 by (𝑦 𝛽 ) 𝛽 = 𝑦 𝛽𝛽 and 𝑎 𝛽 = 𝑎 𝛼(𝛽 ) for ′ 𝛽, 𝛽 ∈ 𝐵 and 𝑎 ∈ 𝐿. Then, let 𝐸 be the fixed field of 𝐵 in 𝐹. By Galois theory, 𝐹 is a Galois extension of 𝐸 with Galois group 𝐵 [Lan97, p. 264, Thm. 1.8] and 𝛼(𝛽) = res𝐹/𝐿 (𝛽) for each 𝛽 ∈ 𝐵. Moreover, 𝐹 is a finitely generated extension of 𝐾. Hence, by Lemma 11.5.1, 𝐸 is a finitely generated extension of 𝐾. By construction, 𝐸 ∩ 𝐿 = 𝐾 and 𝐹 is a purely transcendental extension of 𝐿, so 𝐹 is linearly disjoint from 𝐾˜ over 𝐿. Hence, 𝐸 𝐿 is linearly disjoint from 𝐾˜ over 𝐿. It follows from Lemma 3.1.3 that 𝐸 is linearly disjoint from 𝐾˜ over 𝐾; that is 𝐸 is a regular extension of 𝐾. □ Let 𝐸 be a finitely generated extension of a field 𝐾 and 𝐹 a finite Galois extension of 𝐸. By Remark 7.1.5, there is a Galois ring cover 𝑆/𝑅 for 𝐹/𝐸. Thus, 𝑅 = 𝐾 [𝑥1 , . . . , 𝑥 𝑚 ] is an integrally closed domain with quotient field 𝐸; 𝑆 is the integral closure of 𝑅 in 𝐹, and 𝑆 = 𝑅[𝑧] where, if 𝑓 = irr(𝑧, 𝐸), then 𝑓 ′ (𝑧) is a unit of 𝑆 (Definition 7.1.3). Every homomorphism 𝜑0 of 𝑅 onto a field 𝐸¯ extends to a ¯ The map 𝜑 induces an homomorphism 𝜑 of 𝑆 onto a Galois extension 𝐹¯ of 𝐸.
¯ → Gal(𝐹/𝐸) such that 𝜑 𝜑∗ (𝜎) (𝑥) = 𝜎(𝜑(𝑥)) for each ¯ 𝐸) embedding 𝜑∗ : Gal( 𝐹/ ¯ and 𝑥 ∈ 𝑆 (Lemma 7.1.8(b)). ¯ 𝐸) 𝜎 ∈ Gal( 𝐹/ Theorem 12.6.2 ([Ax68], p. 269) Let 𝐾 be a PAC field, 𝐴 and 𝐵 finite groups, and 𝜌: Gal(𝐾) → 𝐴 and 𝛼: 𝐵 → 𝐴 epimorphisms. Then, there exists a homomorphism 𝛾: Gal(𝐾) → 𝐵 such that 𝜌 = 𝛼 ◦ 𝛾 (i.e. Gal(𝐾) is projective). Proof (Haran). Denote the fixed field of Ker(𝜌) in 𝐾sep by 𝐿. Then, 𝐿 is a finite Galois extension of 𝐾 and 𝜌 defines an isomorphism Gal(𝐿/𝐾) → 𝐴. Thus, we may identify 𝐴 with Gal(𝐿/𝐾) and 𝜌 with the restriction map. Let 𝐸 and 𝐹 be as in Lemma 12.6.1. Let 𝑆/𝑅 be a Galois ring cover for 𝐹/𝐸. Since 𝐾 is a PAC field, there exists a 𝐾-homomorphism 𝜑0 : 𝑅 → 𝐾 (Proposition 12.1.3). Let 𝜑 be an extension of 𝜑0 to 𝑆 which is the identity on 𝐿. Then, 𝑀 = 𝜑(𝑆) is a Galois extension of 𝐾 which contains 𝐿 and 𝜑 induces an embedding 𝜑∗ : Gal(𝑀/𝐾) → Gal(𝐹/𝐸) such that res𝐹/𝐿 ◦ 𝜑∗ = res 𝑀/𝐿 . Compose 𝜑∗ with the map res 𝑀 : Gal(𝐾) → Gal(𝑀/𝐾) to obtain the desired homomorphism □ 𝛾: Gal(𝐾) → 𝐵 with 𝜌 = 𝛼 ◦ 𝛾. There are non-PAC fields 𝐾 with Gal(𝐾) projective (e.g. 𝐾 is finite or 𝐾 = C(𝑡)). On the other hand, if 𝐺 is a projective group, then there exists some PAC field 𝐾 such that 𝐺  Gal(𝐾) (Corollary 26.1.2).

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The projectivity of the absolute Galois group of a field 𝐾 is closely related to the vanishing of the Brauer group Br(𝐾) of 𝐾, although it is not equivalent to it. We survey the concept of the Brauer group and prove that Br(𝐾) = 0 if 𝐾 is PAC. A central simple 𝐾-algebra is a 𝐾-algebra 𝐴 whose center is 𝐾 and which has no two sided ideals except 0 and 𝐴. In particular, if 𝐷 is a division ring with center 𝐾, then the ring 𝑀𝑛 (𝐷) of all 𝑛 × 𝑛 matrices with entries in 𝐷 is a central simple 𝐾-algebra for each positive integer 𝑛 [Hup67, p. 472]. Conversely, if 𝐴 is a finite-dimensional central simple 𝐾-algebra, then, by a theorem of Wedderburn, there exists a unique division ring 𝐷 with center 𝐾 and a positive integer 𝑛 such that 𝐴 𝐾 𝑀𝑛 (𝐷) [Hup67, p. 472]. Suppose that 𝐴 ′ is another finite-dimensional central simple 𝐾-algebra. Then, 𝐴 ′ is equivalent to 𝐴 if there exists a positive integer 𝑛 ′ such that 𝐴 ′  𝑀𝑛′ (𝐷). In particular, 𝐷 is equivalent to 𝐴. We denote the equivalence class of 𝐴 by [ 𝐴] and let Br(𝐾) be the set of all equivalence classes of finite-dimensional central simple 𝐾-algebras. The tensor product of two finite-dimensional central simple 𝐾-algebras is again a finite-dimensional central simple 𝐾-algebra [Wei67, p. 166]. Moreover, the tensor product respects the equivalence relation between finite-dimensional central simple 𝐾-algebras. Hence, [ 𝐴] · [𝐵] = [ 𝐴 ⊗𝐾 𝐵] is an associative multiplication rule on Br(𝐾). Since 𝐴 ⊗𝐾 𝐵  𝐵 ⊗𝐾 𝐴, multiplication in Br(𝐾) is commutative. Further, the equivalence class [𝐾] is a unit in Br(𝐾), because 𝐴 ⊗𝐾 𝐾  𝐴. Finally, let 𝐴o be the opposite algebra of 𝐴. It consists of all elements 𝑎 o , with 𝑎 ∈ 𝐴. Addition and multiplication are defined by the rules 𝑎 o + 𝑏 o = (𝑎 + 𝑏) o and 𝑎 o 𝑏 o = (𝑏𝑎) o . One proves that 𝐴 ⊗𝐾 𝐴o  𝑀𝑛 (𝐾), where 𝑛 = dim𝐾 ( 𝐴). Thus, [ 𝐴o ] = [ 𝐴] −1 . Therefore, Br(𝐾) is an Abelian group. For each field extension 𝐿 of 𝐾 the map 𝐴 ↦→ 𝐴 ⊗𝐾 𝐿 induces a homomorphism res 𝐿/𝐾 : Br(𝐾) → Br(𝐿). The kernel of res 𝐿/𝐾 consists of all [ 𝐴] such that 𝐴 ⊗𝐾 𝐿  𝐿 𝑀𝑛 (𝐿) for some positive integer 𝑛. If 𝐴 satisfies the latter relation, then 𝐴 is said to split over 𝐿. If 𝐿 ′ contains 𝐿, then 𝐴 also splits over 𝐿 ′. It is known that each 𝐴 splits over 𝐾sep [Wei67, p. 167]. Thus, Br(𝐾sep ) is trivial. Remark 12.6.3 (The reduced norm of a central simple algebra) Let 𝐴 be a finitedimensional central simple algebra 𝐴 over a field 𝐾. Choose a 𝐾sep -isomorphism 𝛼: 𝐴 ⊗𝐾 𝐾sep → 𝑀𝑛 (𝐾sep ) for some positive integer 𝑛. In particular, dim𝐾 𝐴 = dim𝐾sep (𝑀𝑛 (𝐾sep )) = 𝑛2 . Let {𝑒 𝑖 𝑗 | 1 ≤ 𝑖, 𝑗 ≤ 𝑛} be a basis of 𝐴 over 𝐾. Then, e𝑖 𝑗 = 𝛼(𝑒 𝑖 𝑗 ⊗ 1), 1 ≤ 𝑖, 𝑗 ≤ 𝑛, form a basis of 𝑀𝑛 (𝐾sep ) over 𝐾sep . Each 𝑎 ∈ 𝐴 has a Í unique presentation as 𝑎 = 𝑖,𝑛 𝑗=1 𝑎 𝑖 𝑗 𝑒 𝑖 𝑗 with 𝑎 𝑖 𝑗 ∈ 𝐾. The matrix a = (𝑎 𝑖 𝑗 )1≤𝑖, 𝑗 ≤𝑛 satisfies 𝑛 ∑︁
𝛼(𝑎 ⊗ 1) = 𝑎 𝑖 𝑗 e𝑖 𝑗 = 𝜆 𝑘𝑙 (a) 1≤𝑘,𝑙 ≤𝑛 , 𝑖, 𝑗=1 2 where 𝜆 𝑘𝑙 are linear forms over Í𝑛 𝐾sep in the 𝑛 variables 𝑋𝑖 𝑗 . Indeed, if e𝑖 𝑗 = (𝜀𝑖 𝑗,𝑘𝑙 )1≤𝑘,𝑙 ≤𝑛 , then 𝜆 𝑘𝑙 (X) = 𝑖 𝑗=1 𝜀𝑖 𝑗,𝑘𝑙 𝑋𝑖 𝑗 , where X = (𝑋𝑖 𝑗 )1≤𝑖, 𝑗 ≤𝑛 . The reduced norm of 𝑎 is defined by

red.norm(𝑎) = det(𝛼(𝑎 ⊗ 1)).

(12.15)

12.6 The Absolute Galois Group of a PAC Field

219

If 𝛼 ′: 𝐴 ⊗𝐾 𝐾sep → 𝑀𝑛 (𝐾sep ) is another 𝐾sep -isomorphism, then 𝛼 ′ ◦ 𝛼−1 is a 𝐾sep automorphism of 𝑀𝑛 (𝐾sep ). By Skolem–Noether [Wei67, p. 166, Prop. 4], 𝛼 ′ ◦ 𝛼−1 is a conjugation by an invertible matrix of 𝑀𝑛 (𝐾sep ). Hence, det(𝛼 ′ (𝑎 ⊗ 1)) = det(𝛼(𝑎 ⊗ 1)). Thus, red.norm(𝑎) is independent of the particular choice of 𝛼. Moreover, each 𝜎 ∈ Gal(𝐾) fixes red.norm(𝑎). Therefore, red.norm(𝑎) ∈ 𝐾. Indeed, 𝜎 induces a 𝐾-automorphism 1⊗𝜎 −1 of 𝐴⊗𝐾 𝐾sep and a 𝐾-automorphism 𝜎𝑛 of 𝑀𝑛 (𝐾sep ). Then, 𝛼 ′ = 𝜎𝑛 ◦ 𝛼 ◦ 1 ⊗ 𝜎 −1 : 𝐴 ⊗𝐾 𝐾sep → 𝑀𝑛 (𝐾sep )
′ is a 𝐾sep -isomorphism
satisfying 𝛼 (𝑎 ⊗ 1)
= 𝜎𝑛 𝛼(𝑎 ⊗ 1) for each
𝑎 ∈ 𝐾. It follows
that 𝜎 red.norm(𝑎) = 𝜎 det(𝛼(𝑎 ⊗ 1)) = det 𝜎𝑛 (𝛼(𝑎 ⊗ 1)) = det 𝛼 ′ (𝑎 ⊗ 1) = red.norm(𝑎), as claimed.
Now let 𝑝(X) = det 𝜆 𝑘𝑙 (X) . It is a homogeneous polynomial of degree 𝑛 over 𝐾sep such that 𝑝(a) ∈ 𝐾 for each a ∈ 𝑀𝑛 (𝐾). It follows that the coefficients of 𝑝 belong to 𝐾 (Exercise 9). Next observe that the linear forms 𝜆 𝑘𝑙 are linearly independent over 𝐾sep because the 𝑛2 × 𝑛2 matrix (e𝑖 𝑗 )1≤𝑖, 𝑗 ≤𝑛 is nonsingular. We may therefore form a change of variables 𝑌𝑘𝑙 = 𝜆 𝑘𝑙 (X). It maps 𝑝(X) onto det(Y), which is an absolutely irreducible polynomial [Boc47, p. 176, Thm. 1]. Hence, 𝑝(X) is also absolutely irreducible. We call 𝑝(X) the reduced form of 𝐴. Theorem 12.6.4 Let 𝐾 be a PAC field. Then, its Brauer group Br(𝐾) is trivial. Proof ([Ax68], p. 269). Assume Br(𝐾) is nontrivial. Then, there exists a division ring 𝐷 with center 𝐾 such that dim𝐾 (𝐷) = 𝑛2 and 𝑛 > 1. Let 𝑝(X) be the associated reduced form. Since 𝑝(X) is an absolutely irreducible polynomial (Remark 12.6.3), 𝑝(X) has a nontrivial zero a ∈ 𝑀𝑛 (𝐾) (Proposition 12.1.1). In the notation of Remark Í 12.6.3 (with 𝐴 = 𝐷), let 𝑎 = 𝑖,𝑛 𝑗=1 𝑎 𝑖 𝑗 𝑒 𝑖 𝑗 . By (12.15), red.norm(𝑎) = 𝑝(a) = 0. On the other hand, 𝑎 is a nonzero element of 𝐷, hence invertible. Therefore, 𝛼(𝑎 ⊗ 1) is a regular matrix, so red.norm(𝑎) = det(𝛼(𝑎 ⊗ 1)) ≠ 0. This contradiction proves that Br(𝐾) is trivial. □ Remark 12.6.5 (Varieties of Severi–Brauer) An alternative proof of Theorem 12.6.4 uses varieties of Severi–Brauer. They are varieties 𝑉 which are defined over a field 𝐾 and are isomorphic over 𝐾sep to P𝑛 for some positive integer 𝑛. There is a bijective correspondence between 𝐾-isomorphism classes of varieties 𝑉 of Severi–Brauer and equivalence classes of finite-dimensional central simple 𝐾-algebras 𝐴. If 𝑉 has a 𝐾-rational point, then 𝐴 splits over 𝐾 [Jac96, p. 113]. In particular, if 𝐾 is PAC, this implies that Br(𝐾) = 0. The connection between the projectivity of the absolute Galois group of a field 𝐾 and its Brauer group is based on the canonical isomorphism × 𝐻 2 (Gal(𝐾), 𝐾sep )  Br(𝐾)

(12.16)

([Deu35, p. 56, Satz 1] or [Ser68b, §X5]). Here we assume that the reader is familiar with Galois cohomology, e.g. as presented in [Rbs70] or in [Ser97]. In particular, it follows from (12.16) that

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(12.17) every element of Br(𝐾) has a finite order [Rbs70, p. 138, Cor. 6.7]. For each prime number 𝑝 and a profinite group 𝐺 the notation cd 𝑝 (𝐺) stands for the 𝑝th cohomological dimension of 𝐺. It is the smallest positive integer 𝑛 such that 𝐻 𝑞 (𝐺, 𝐴) 𝑝∞ = 0 for each torsion 𝐺-module 𝐴 and every 𝑞 ≥ 𝑛 + 1. Finally, cd(𝐺) = sup 𝑝 (cd 𝑝 (𝐺)) is the cohomological dimension of 𝐺. Proposition 12.6.6 The following conditions on a field 𝐾 are equivalent: (a) Gal(𝐾) is projective. (b) cd(Gal(𝐾)) ≤ 1. (c) For each prime number 𝑝 ≠ char(𝐾) and for each finite separable extension 𝐿 of 𝐾, Br(𝐿) 𝑝∞ is trivial. Proof of “(a) ⇐⇒ (b)”. Let 𝑝 be a prime number. By [Rbs70, p. 211, Prop. 3.1], cd 𝑝 (Gal(𝐾)) ≤ 1 if and only if for every finite Galois extension 𝐿 of 𝐾 and for every short exact sequence 𝛼

0 −→ (Z/𝑝Z) 𝑚 −→ 𝐵 −→ Gal(𝐿/𝐾) −→ 1 there exists a homomorphism 𝛽: Gal(𝐾) → 𝐵 such that 𝛼 ◦ 𝛽 = res𝐾sep /𝐿 . By a theorem of Gruenberg (Corollary 25.4.3), the latter condition holds for all 𝑝 if and only if Gal(𝐾) is projective. Consequently, (a) and (b) are equivalent. Proof of “(b) ⇐⇒ (c)”. Let 𝑝 be again a prime number. First suppose that 𝑝 ≠ char(𝐾). Then, cd 𝑝 (Gal(𝐾)) ≤ 1 if and only if Br(𝐿) 𝑝∞ is trivial for every finite separable extension 𝐿 of 𝐾 [Rbs70, p. 261, Cor. 3.7]. If 𝑝 = char(𝐾), then cd 𝑝 (Gal(𝐿)) ≤ 1 for every field 𝐿 of characteristic 𝑝 [Rbs70, p. 256, Thm. 3.3]. Since the Brauer group of each field is torsion (by (12.17)), this establishes the equivalence of (b) and (c). □ Remark 12.6.7 The proof of the equivalence of (a) and (b) in Proposition 12.6.6 remains valid for each profinite group 𝐺 replacing Gal(𝐾). Thus, 𝐺 is projective if and only if cd(𝐺) ≤ 1. Proposition 12.6.8 ([Rbs70], p. 264, Prop. 3.10) The following conditions on a field 𝐾 are equivalent. (a) Br(𝐿) is trivial for every finite separable extension 𝐿 of 𝐾. (b) The norm map, norm: 𝑁 × → 𝐿 × , is surjective for every finite separable extension 𝐿 of 𝐾 and for every finite Galois extension 𝑁 of 𝐿. We summarize consequences of the previous results for PAC fields: Corollary 12.6.9 The following statements hold for every PAC field 𝐾: (a) Gal(𝐾) is projective. (b) Br(𝐾) is trivial. (c) cd(Gal(𝐾)) ≤ 1. (d) The map norm: 𝑁 × → 𝐾 × is surjective for each finite Galois extension 𝑁 of 𝐾. Proof. Let 𝐿 be a finite separable extension of 𝐾. By Corollary 12.2.5, 𝐿 is PAC. Hence, by Theorem 12.6.4, Br(𝐿) is trivial. Therefore, cd(Gal(𝐾)) ≤ 1 (Proposition 12.6.6) and the norm map 𝑁 × → 𝐾 × is surjective for each finite Galois extension 𝑁 of 𝐾 (Proposition 12.6.8). □

12.7 A non-PAC Field 𝐾 with 𝐾ins PAC

221

Example 12.6.10 (Geyer) We construct an example of a field 𝐾 with Gal(𝐾) projective, a finite Galois extension 𝐾 ′ of 𝐾, and an element 𝑢 of 𝐾 which is not a norm of an element of 𝐾 ′. By Proposition 12.6.8, 𝐾 has a finite separable extension 𝐿 such that Br(𝐿) ≠ 0. This will show that it is impossible to omit the condition “𝑝 ≠ char(𝐾)” in Condition (c) of Proposition 12.6.6. We start from a transcendental element 𝑢 over F2 and let 𝐾0 = F2 (𝑢)sep . Then, choose a transcendental element 𝑡 over 𝐾0 and let 𝐾 = 𝐾0 (𝑡). By Tsen’s Theorem, Gal(𝐾) is projective [Jar11, p. 185, Prop. 9.4.6(b)]. Consider the Artin–Schreier extension 𝐾 ′ = 𝐾 (𝑥) of 𝐾 with 𝑥 2 + 𝑥 + 𝑡 = 0. Each element 𝑦 of 𝐾 ′ has the form 𝑦 = 𝑣 + 𝑤𝑥 with 𝑣, 𝑤 ∈ 𝐾 and norm𝐾 ′ /𝐾 (𝑦) = (𝑣 + 𝑤𝑥) (𝑣 + 𝑤(1 + 𝑥)) = 𝑓 (𝑡) 𝑣 2 + 𝑣𝑤 + 𝑤 2 𝑡. Write 𝑣 = ℎ(𝑡) and 𝑤 = 𝑔ℎ (𝑡) (𝑡) , where 𝑓 , 𝑔, ℎ ∈ 𝐾0 [𝑡] and ℎ ≠ 0. Let 𝑎 (resp. 𝑏, 𝑐) be the leading coefficient of 𝑓 (resp. 𝑔, ℎ). If norm𝐾 ′ /𝐾 𝑦 = 𝑢, then 𝑓 (𝑡) 2 + 𝑓 (𝑡)𝑔(𝑡) + 𝑔(𝑡) 2 𝑡 = ℎ(𝑡) 2 𝑢. Compare the leading coefficients of both sides of this equality. If deg( 𝑓 ) > deg(𝑔), then 𝑎 2 = 𝑐2 𝑢. If deg( 𝑓 ) ≤ deg(𝑔), then 𝑏 2 = 𝑐2 𝑢. In both cases we find that 𝑢 is a square in F2 (𝑢)sep , which is not the case. This contradiction proves that 𝑢 is not a norm of an element of 𝐾 ′.

12.7 A non-PAC Field 𝑲 with 𝑲ins PAC Let 𝐿/𝐾 be a purely inseparable extension of fields. If 𝐾 is PAC, then so is 𝐿 (Corollary 12.2.5). Problem 12.4 of [GeJ89] asks whether the converse is true. An example of Hrushovski shows that this is not the case. The main ingredient of this example is the analog of the Mordell conjecture for function fields: Proposition 12.7.1 (Grauert–Manin [Sam66], pp. 107 and 118) Let 𝐾 be a finitely generated regular extension of a field 𝐾0 and 𝐶 a nonconstant curve over 𝐾/𝐾0 . Suppose that the genus of 𝐶 over 𝐾 𝐾˜ 0 is at least 2. Then, 𝐶 (𝐾) is a finite set. Here we say that 𝐶 is a nonconstant curve over 𝐾/𝐾0 if 𝐶 is defined over 𝐾 and if 𝐶 is not birationally equivalent over 𝐾˜ to a curve 𝐶0 defined over 𝐾˜ 0 . Lemma 12.7.2 Let 𝐹 = 𝐾 (𝑥1 , . . . , 𝑥 𝑛 ) be a finitely generated extension of a field 𝐾 of positive characteristic 𝑝. Suppose that 𝐾 is algebraically closed in 𝐹. Then, ∞ Ù 𝑘 𝑘 𝐾 (𝑥 1𝑝 , . . . , 𝑥 𝑛𝑝 ) = 𝐾. (12.18) 𝑘=1

Proof. Denote the left-hand side of (12.18) by 𝐹0 . First suppose that 𝐾 is perfect. Thus, 𝐾 𝑝 = 𝐾, so 𝐹0 = 𝐹0𝑝 is also perfect. In addition 𝐹0 , as a subfield of 𝐹, is finitely generated over 𝐾 (Lemma 11.5.1). Assume that 𝐹0 is transcendental over 𝐾. Choose a transcendence basis 𝑡 1 , . . . , 𝑡𝑟 with 𝑟 ≥ 1. Then, 𝐹0 has a finite degree over 𝑚 𝐸 = 𝐾 (𝑡 1 , . . . , 𝑡𝑟 ). On the other hand, since 𝐹0 is perfect, 𝐸 (𝑡11/ 𝑝 ) is contained in 𝐹0 and has degree 𝑝 𝑚 over 𝐸 for each positive integer 𝑚. This contradiction proves that 𝐹0 is algebraic over 𝐾. Since 𝐾 is algebraically closed in 𝐹, we conclude that 𝐹0 = 𝐾.

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12 Pseudo Algebraically Closed Fields

In the general case 𝐾ins is a perfect field. Hence, by the preceding paragraph, 𝐹0 ⊆ 𝐹 ∩

∞ Ù

𝑘

𝑘

𝐾ins (𝑥 1𝑝 , . . . , 𝑥 𝑛𝑝 ) = 𝐹 ∩ 𝐾ins = 𝐾.

𝑘=1

Therefore, 𝐹0 = 𝐾.

□

Lemma 12.7.3 Let 𝐾 be a finitely generated regular transcendental extension of a field 𝐾0 of positive characteristic 𝑝. Let 𝐶 be a curve which is defined over 𝐾 and whose genus over 𝐾 𝐾˜ 0 is at least 2. Let 𝐹 be a finitely generated regular extension of 𝐾. Suppose that 𝐶 is a nonconstant curve over 𝐹/𝐾0 . Then, 𝐾 has a finitely generated extension 𝐸 which is contained in 𝐹 such that 𝐹/𝐸 is a finite purely inseparable extension and 𝐶 (𝐾) = 𝐶 (𝐸). 𝑘

𝑘

Proof. Let 𝐹 = 𝐾 (𝑥1 , . . . , 𝑥 𝑛 ) and for each 𝑘 write 𝐹𝑘 = 𝐾 (𝑥1𝑝 , . . . , 𝑥 𝑛𝑝 ). By Lemma 12.7.2, the intersection of all 𝐹𝑘 is 𝐾. Since 𝐹 𝐾˜ 0 /𝐾 𝐾˜ 0 is a regular extension, the genus of 𝐶 over 𝐹 𝐾˜ 0 is the same as the genus of 𝐶 over 𝐾 𝐾˜ 0 (Proposition 4.4.3(b)), so at least 2. By Proposition 12.7.1, 𝐶 (𝐹) is a finite set. Hence, there exists a positive integer 𝑘 such that 𝐶 (𝐹𝑘 ) = 𝐶 (𝐾), so 𝐸 = 𝐹𝑘 satisfies the assertion of the lemma. □ Remark 12.7.4 (On Möbius
transformations) Let 𝐾 be a field and 𝑥 an indeterminate. To each matrix 𝐴 = 𝑎𝑐 𝑑𝑏 in GL(2, 𝐾) we associate a Möbius transformation 𝜏𝐴 (also called a linear fractional transformation). It is the 𝐾-isomorphism of 𝐾 (𝑥) into 𝐾 (𝑥) defined by the following rule: 𝑎𝑥 + 𝑏 . (12.19) 𝑐𝑥 + 𝑑 If 𝐵 is another matrix in GL(2, 𝐾), then 𝜏𝐴 ◦ 𝜏𝐵 = 𝜏𝐵 𝐴. If 𝐼 is the unit matrix, then 𝜏𝐼 is the identity map of 𝐾 (𝑥). In particular, 𝜏𝐴−1 ◦ 𝜏𝐴 = id, so 𝜏𝐴 is an automorphism of 𝐾 (𝑥)/𝐾. If 𝑘 ∈ 𝐾 × , then 𝜏𝑘𝐼 = id. Conversely, if 𝜏𝐴 = id, then, by (12.19), 𝑐(𝑥 ′) 2 + (𝑑 − 𝑎)𝑥 ′ − 𝑏 = 0 for all 𝑥 ′ ∈ 𝐾. Hence, 𝑐 = 𝑏 = 0 and 𝑑 = 𝑎, so 𝐴 = 𝑎𝐼. Therefore, the kernel of the map 𝐴 ↦→ 𝜏𝐴 consists of the group of scalar matrices. If 𝜏 is an
arbitrary element of Aut 𝐾 (𝑥)/𝐾 , then 𝐾 (𝑥) = 𝐾 (𝜏(𝑥)). Hence, by Example 4.2.4, there exists an 𝐴 ∈ GL(2, 𝐾) such that 𝜏(𝑥) = 𝜏𝐴 (𝑥). Thus, the map 𝐴 ↦→ 𝜏𝐴 defines an isomorphism
PGL(2, 𝐾)  Aut 𝐾 (𝑥)/𝐾 . (12.20) 𝜏𝐴 (𝑥) =

Substituting elements of 𝐾 ∪ {∞} into (12.19), we may also view 𝜏𝐴 as a bijective 𝑎 map of 𝐾 ∪ {∞} onto itself. For example, if 𝑐 ≠ 0, then 𝑎·∞+𝑏 𝑐·∞+𝑑 = 𝑐 . Note that ′ since the pairs (𝑎, 𝑏) and (𝑐, 𝑑) are linearly independent over 𝐾, no 𝑥 ∈ 𝐾 satisfies both 𝑎𝑥 ′ + 𝑏 = 0 and 𝑐𝑥 ′ + 𝑑 = 0. Hence, the value of 𝜏𝐴 (𝑥 ′) is well defined. The arithmetic with ∞ becomes clearer if we substitute 𝑥 = 𝑥𝑥10 into (12.19) and view 𝜏𝐴 as a bijective map of P1 (𝐾) onto itself: 𝜏𝐴 (𝑥0 :𝑥1 ) = (𝑐𝑥1 + 𝑑𝑥0 : 𝑎𝑥1 + 𝑏𝑥 0 ).

□

(12.21)

The map 𝑥 ′ ↦→ 𝜏𝐴 (𝑥 ′) of 𝐾 ∪ {∞} onto itself determines 𝜏𝐴. This is one of the consequences of the next lemma.

12.7 A non-PAC Field 𝐾 with 𝐾ins PAC

223

Lemma 12.7.5 Let 𝐾/𝐾0 be an extension of fields. Let (𝑥1 , 𝑥2 , 𝑥3 ) and (𝑦 1 , 𝑦 2 , 𝑦 3 ) be triples of distinct elements of 𝐾 ∪{∞}. Then, there is a unique Möbius transformation 𝜏 over 𝐾 such that 𝜏(𝑥𝑖 ) = 𝑦 𝑖 , 𝑖 = 1, 2, 3. Moreover, 𝜏 can be presented as 𝜏𝐴, where the entries of 𝐴 belong to the field 𝐾0 (𝑥1 , 𝑥2 , 𝑥3 , 𝑦 1 , 𝑦 2 , 𝑦 3 ). Proof of uniqueness. If 𝜏𝑖 (𝑥 𝑗 ) = 𝑦 𝑗 for 𝑖 = 1, 2 and 𝑗 = 1, 2, 3, then 𝜏 = 𝜏1−1 𝜏2 satisfies 𝜏(𝑥 𝑗 ) = 𝑥 𝑗 for 𝑗 = 1, 2, 3. Suppose first that none of the 𝑥 𝑗 is ∞ and 𝜏 = 𝜏𝐴
with 𝐴 = 𝑎𝑐 𝑑𝑏 . Then, 𝑐𝑥 2𝑗 + (𝑑 − 𝑎)𝑥 𝑗 − 𝑏 = 0 for 𝑗 = 1, 2, 3. Hence, 𝑏 = 𝑐 = 0 and 𝑎 = 𝑑. Now assume that 𝑥 1 = ∞; that is, 𝑥 1 is (0:1) in homogeneous coordinates. By (12.21), 𝑐 = 0. Then, we may assume that 𝑑 = 1 and conclude from 𝑎𝑥 𝑗 + 𝑏 = 𝑥 𝑗 , 𝑗 = 2, 3, that 𝑎 = 1 and 𝑏 = 0. In both cases 𝜏 = id and 𝜏1 = 𝜏2 . 1 ′ Proof of existence. The Möbius transformation 𝜏(𝑥) = 𝑥−𝑥 ′ maps the element 𝑥 1 of 𝐾 onto ∞. Likewise, 𝜏(𝑥) = 𝑥 exchanges 0 and ∞. Hence, we may assume that 𝑥 1 = ∞ and 𝑦 1 = ∞. Since 𝑥2 ≠ 𝑥 3 , 𝑦2 − 𝑦3 𝑥2 𝑦 3 − 𝑥3 𝑦 2 𝜏(𝑥) = 𝑥+ 𝑥2 − 𝑥3 𝑥2 − 𝑥3 maps 𝑥𝑖 onto 𝑦 𝑖 , 𝑖 = 1, 2, 3, as desired. □

Remark 12.7.6 (Conservation of branch points) Let 𝐾 be an algebraically closed field, 𝑥 an indeterminate, and 𝐹 a finite separable extension of 𝐾 (𝑥). Then, 𝐹/𝐾 is a function field of one variable. For each 𝑎 ∈ 𝐾 ∪ {∞} let 𝜑 𝑎 : 𝐾 (𝑥) → 𝐾 ∪ {∞} be the 𝐾-place of 𝐾 (𝑥) with 𝜑 𝑎 (𝑥) = 𝑎. Denote the corresponding prime divisor of 𝐾 (𝑥)/𝐾 by 𝔭𝑎 . We say that 𝑎 is a branch point of 𝐹/𝐾 (𝑥) (with respect to 𝑥) if 𝔭𝑎 ramifies in 𝐹. There are only finitely many prime divisors of 𝐾 (𝑥)/𝐾 which ramify in 𝐹 (Section 4.4), so 𝐹/𝐾 (𝑥) has only finitely many branch points. If 𝜏 is a Möbius transformation of 𝐾 (𝑥) and 𝜏(𝑥) = 𝑥 ′, then 𝜏 maps the set of branch points of 𝐹/𝐾 (𝑥) with respect to 𝑥 onto the set of branch points of 𝐹/𝐾 (𝑥) with respect to 𝑥 ′. By Lemma 12.7.5, there exists a Möbius transformation 𝜏 of 𝐾 (𝑥) mapping ∞ onto a finite nonbranch point of 𝐹/𝐾 (𝑥). Replacing 𝑥 by 𝜏(𝑥), if necessary, we may assume that ∞ is not a branch point. Let 𝑆 = 𝐾 [𝑦 1 , . . . , 𝑦 𝑛 ] be the integral closure of 𝐾 [𝑥] in 𝐹. Then, 𝑆 is a Dedekind domain (Proposition 2.5.6). Let 𝐶 be the curve generated in A𝑛 over 𝐾 by y. The local ring 𝑂 𝐶,b of 𝐶 at each b ∈ 𝐶 (𝐾) is the local ring of 𝑆 at the kernel of the 𝐾-homomorphism mapping y onto b. This kernel is a maximal ideal of 𝑆. Hence, 𝑂 𝐶,b is a discrete valuation ring. The corresponding prime divisor 𝔮b of 𝐹/𝐾 is uniquely determined by b. Conversely, each prime divisor 𝔮 of 𝐹/𝐾 with 𝜑𝔮 finite at 𝑥 is also finite on 𝑆 and 𝜑𝔮 (y) ∈ 𝐶 (𝐾). Thus, the map b ↦→ 𝔮b is a bijective correspondence between 𝐶 (𝐾) and the set of prime divisors of 𝐹/𝐾 which are finite at 𝑥. Choose a polynomial 𝑓 ∈ 𝐾 [𝑌1 , . . . , 𝑌𝑛 ] such that 𝑥 = 𝑓 (y). Then, 𝑓 defines a morphism 𝜋: 𝐶 → A1 by 𝜋(b) = 𝑓 (b) for each b ∈ 𝐶 (𝐾). The prime divisor 𝔮b of 𝐹/𝐾 lies over 𝔭𝑎 if and only if 𝜋(b) = 𝑎. Now let 𝔭 be a prime divisor of 𝐾 (𝑥)/𝐾 which is finite at 𝑥. Denote the prime divisors of 𝐹/𝐾 lying over 𝔭 by 𝔮1 , . . . , 𝔮𝑟 and theÍcorresponding ramification indices by 𝑒 1 , . . . , 𝑒𝑟 . Since 𝐾 is algebraically closed, 𝑟𝑖=1 𝑒 𝑖 = [𝐹 : 𝐾 (𝑥)] (Proposition

224

12 Pseudo Algebraically Closed Fields

2.3.2). Hence, 𝔭 is ramified in 𝐹 if and only if 𝑟 < [𝐹 : 𝐾 (𝑥)]. Therefore, by the preceding two paragraphs, an element 𝑎 of 𝐾 is a branch point of 𝐹/𝐾 (𝑥) if and only if there are less than [𝐹 : 𝐾 (𝑥)] points of 𝐶 (𝐾) lying over 𝑎. This implies that “𝑎 is a branch point of 𝐹/𝐾 (𝑥) with respect to 𝑥” is an elementary statement on 𝐾. Next consider an algebraically closed subfield 𝐾0 of 𝐾. Suppose that there is a function field 𝐹0 over 𝐾0 which contains 𝑥, 𝑦 1 , . . . , 𝑦 𝑛 such that 𝐹0 𝐾 = 𝐹 and 𝑓 has coefficients in 𝐾0 . Since 𝑥 is transcendental over 𝐾 and 𝐹0 is algebraic over 𝐾0 (𝑥), the fields 𝐹0 and 𝐾 are algebraically independent over 𝐾0 . Since 𝐾0 is algebraically closed, 𝐹0 is regular over 𝐾0 . Hence, by Lemma 3.4.7, 𝐹0 is linearly disjoint from 𝐾 over 𝐾0 . Therefore, by Lemma 3.1.3, 𝐹0 is linearly disjoint from 𝐾 (𝑥) over 𝐾0 (𝑥). Since 𝐹0 𝐾 = 𝐹, we have [𝐹0 : 𝐾0 (𝑥)] = [𝐹 : 𝐾 (𝑥)]. Finally, since 𝑦 1 , . . . , 𝑦 𝑛 ∈ 𝐹0 , the curve 𝐶 is defined over 𝐾0 . By the two preceding paragraphs, “𝑎 is a branch point with respect to 𝑥” is an elementary statement in the language L (ring, 𝐾0 ). Let 𝑎 1 , . . . , 𝑎 𝑚 be all branch points of 𝐹0 /𝐾0 (𝑥). Then, “𝑎 is a branch point with respect to 𝑥 if and only if 𝑎 = 𝑎 𝑖 for some 𝑖 between 1 and 𝑚” is an elementary statement which holds over 𝐾0 . Since 𝐾0 is an elementary subfield of 𝐾 (Corollary 10.3.2), the same statement holds over 𝐾. It follows that each branch point of 𝐹/𝐾 (𝑥) with respect to 𝑥 belongs to 𝐾0 . Remark 12.7.7 (Construction of nonconstant curves) Let 𝐾0 be a field. Choose five distinct elements 𝑡 1 , . . . , 𝑡 5 in some regular extension of 𝐾0 such that 𝑡 4 ∉ 𝐾˜ 0 (𝑡 1 , 𝑡2 , 𝑡3 ).

(12.22)

Consider a regular field extension 𝐾 of 𝐾0 containing 𝑡1 , 𝑡2 , 𝑡3 , 𝑡4 , 𝑡5 . Put 𝑓 (𝑋) = Î5 (𝑋 − 𝑡 𝑖 ) and define a curve 𝐶 over 𝐾 by the equation 𝑌 2 = 𝑓 (𝑋) if char(𝐾) ≠ 2 𝑖=1 1 and 𝑌 2 +𝑌 = 𝑓 (𝑋) if char(𝐾) = 2. By Propositions 4.8.2 and 4.8.4, 𝐶 is a hyperelliptic curve. More precisely, the genus 𝑔 of 𝐶 is 2 if char(𝐾) ≠ 2 and 4 if char(𝐾) = 2. ˜ The genus of 𝐶 does not change over 𝐾. Claim: 𝐶 is a nonconstant curve over 𝐾/𝐾0 . Proof. Choose a generic point (𝑥, 𝑦) for 𝐶 over 𝐾 and let 𝐹 = 𝐾˜ (𝑥, 𝑦). Then, 𝐾˜ (𝑥) is a quadratic subfield of 𝐹. Assume that 𝐶 is a constant curve over 𝐾/𝐾0 . Then, there exists a function field of one variable 𝐹0 over 𝐾˜ 0 such that 𝐹0 𝐾˜ = 𝐹. Since 𝐹0 /𝐾˜ 0 is a regular extension, 𝐹0 is linearly disjoint from 𝐾˜ over 𝐾˜ 0 (Lemma 3.4.7). ˜ (Proposition 4.4.3(b)). Denote In particular, 𝑔 = genus(𝐹0 /𝐾˜ 0 ) = genus(𝐹/𝐾) ˜ the canonical divisor of 𝐹0 /𝐾˜ 0 by 𝔴0 and denote the image of 𝔴0 in Div(𝐹/𝐾) by 𝔴. By Proposition 4.4.3 and Lemma 4.2.2(b), dim(𝔴) = dim(𝔴0 ) = 𝑔 and deg(𝔴) = deg(𝔴0 ) = 2𝑔 − 2. Hence, 𝔴 is the canonical divisor of 𝐹/𝐾˜ (Exercise 1(b) of Chapter 4). Choose a basis 𝑧1 , 𝑧2 , . . . , 𝑧 𝑔 for L (𝔴0 ) over 𝐾˜ 0 . Then, by the linear disjointness, ˜ By Proposition 4.7.4, 𝐾˜ ( 𝑧2 , . . . , 𝑧𝑔 ) 𝑧 1 , 𝑧2 , . . . , 𝑧 𝑔 form a basis for L (𝔴) over 𝐾. 𝑧1 𝑧1 𝑧 is the unique quadratic subfield of 𝐹, so 𝐾˜ ( 𝑧𝑧21 , . . . , 𝑧𝑔1 ) = 𝐾˜ (𝑥). In particu𝑧 ˜ Hence, 𝐾˜ 0 ( 𝑧2 , . . . , 𝑧𝑔 ) lar, 𝐾˜ ( 𝑧𝑧21 , . . . , 𝑧𝑔1 ) is a function field of genus 0 over 𝐾. 𝑧1 𝑧1 ˜ By Example 4.2.4, there is an 𝑥 0 with is a function field of genus 0 over 𝐾. 𝑧 𝐾˜ 0 ( 𝑧𝑧21 , . . . , 𝑧𝑔1 ) = 𝐾˜ 0 (𝑥0 ). It satisfies 𝐾˜ (𝑥0 ) = 𝐾˜ (𝑥). By Lemma 12.7.5, there exists a Möbius transformation 𝜏 over 𝐾˜ such that 𝜏(𝑥) = 𝑥 0 .

225

12.7 A non-PAC Field 𝐾 with 𝐾ins PAC

˜ For each 𝑖 between 1 and 5 let 𝜑𝑖 : 𝐾˜ (𝑥) → 𝐾˜ ∪ {∞} be the 𝐾-place with 𝜑𝑖 (𝑥) = 𝑡 𝑖 . By Examples 2.3.8 and 2.3.9, 𝑡 𝑖 is a branch point of 𝐹/𝐾˜ (𝑥) with respect to 𝑥. Put 𝑎 𝑖 = 𝜏(𝑡𝑖 ). Then, 𝜑𝑖 (𝑥0 ) = 𝜑𝑖 (𝜏(𝑥)) = 𝜏(𝜑𝑖 (𝑥)) = 𝜏(𝑡𝑖 ) = 𝑎 𝑖 . Thus, 𝑎 𝑖 is a branch point of 𝐹/𝐾˜ (𝑥0 ) with respect to 𝑥0 . It follows from Remark 12.7.6 that 𝑎 𝑖 ∈ 𝐾˜ 0 . By Lemma 12.7.5, 𝜏 is already defined over 𝐾˜ 0 (𝑡1 , 𝑡2 , 𝑡3 ). Hence, 𝑡 4 = 𝜏 −1 (𝑎 4 ) ∈ 𝐾˜ 0 (𝑡1 , 𝑡2 , 𝑡3 ). This contradiction to assumption (12.22) proves that 𝐶 is a nonconstant curve over 𝐾/𝐾0 .

Theorem 12.7.8 ([Hru94], Cor. 5) For each prime number 𝑝 there exists a countable non-PAC field 𝐸 of characteristic 𝑝 such that 𝐸 ins is PAC. Proof. Choose algebraically independent elements 𝑡1 , 𝑡2 , 𝑡3 , 𝑡4 , 𝑡5 over F 𝑝 . Let 𝐶 be the curve defined over 𝐾 := F 𝑝 (𝑡 1 , 𝑡2 , 𝑡3 , 𝑡4 , 𝑡5 ) in Remark 12.7.7. Then, 𝐶 is a nonconstant curve over 𝐹/F 𝑝 for every regular field extension 𝐹 of F 𝑝 that contains 𝐾. By Proposition 12.7.1, 𝐶 (𝐾) is a finite set. By induction we construct two ascending towers of fields 𝐾 = 𝐸 1 ⊆ 𝐸 2 ⊆ 𝐸 3 ⊆ · · · and 𝐹1 ⊆ 𝐹2 ⊆ 𝐹3 ⊆ · · · and for each positive integer 𝑚 we enumerate the varieties which are defined over 𝐸 𝑚 in a sequence, 𝑉𝑚1 , 𝑉𝑚2 , 𝑉𝑚3 , . . . such that (12.23a) 𝐸 𝑚 and 𝐹𝑚 are finitely generated regular extensions of 𝐾, (12.23b) 𝐹𝑚 is a finite purely inseparable extension of 𝐸 𝑚 , (12.23c) 𝐶 (𝐸 𝑚 ) = 𝐶 (𝐾), and (12.23d) 𝑉𝑖 𝑗 (𝐹𝑚 ) ≠ ∅ for 𝑖, 𝑗 = 1, . . . , 𝑚 − 1. Indeed, suppose 𝐸 1 , . . . , 𝐸 𝑚−1 , 𝐹1 , . . . , 𝐹𝑚−1 , and 𝑉𝑖 𝑗 for 𝑖 < 𝑚 and all 𝑗 have been defined such that they satisfy (12.23a)–(12.23d). Let 𝑉 be the direct product of 𝑉𝑖 𝑗 for 𝑖, 𝑗 = 1, . . . , 𝑚 − 1. It is a variety defined over 𝐸 𝑚−1 . Let x be a generic point ′ = 𝐸 of 𝑉 over 𝐸 𝑚−1 . Then, 𝐸 𝑚 𝑚−1 (x) is a finitely generated regular extension of ′ (instead 𝐾, 𝐹) and 𝐸 𝑚−1 and therefore also of 𝐾. Apply Lemma 12.7.3 to 𝐸 𝑚−1 , 𝐸 𝑚 ′ ′ /𝐸 is a construct an extension 𝐸 𝑚 of 𝐸 𝑚−1 which is contained in 𝐸 𝑚 such that 𝐸 𝑚 𝑚 finite purely inseparable extension and 𝐶 (𝐸 𝑚 ) = 𝐶 (𝐸 𝑚−1 ). By (12.23c) for 𝑚 − 1 we ′ is a regular extension of 𝐸 have 𝐶 (𝐸 𝑚 ) = 𝐶 (𝐾). Since 𝐸 𝑚 𝑚−1 , it is linearly disjoint ′ 𝐹 from 𝐹𝑚−1 𝐾˜ over 𝐸 𝑚−1 . Hence, 𝐹𝑚 = 𝐸 𝑚 is linearly disjoint from 𝐹𝑚−1 𝐾˜ over 𝑚−1 𝐹𝑚−1 . ′ 𝐸𝑚 𝐹𝑚 𝐸𝑚 𝐸 𝑚−1 𝐾

𝐹𝑚−1

𝐹𝑚−1 𝐾˜ 𝐾˜

Since 𝐹𝑚−1 is linearly disjoint from 𝐾˜ over 𝐾, 𝐹𝑚 is linearly disjoint from 𝐾˜ over 𝐾. Thus, 𝐹𝑚 is a regular extension of 𝐾. By construction, 𝐹𝑚 is a finite purely inseparable extension of 𝐸 𝑚 and 𝑉𝑖 𝑗 (𝐹𝑚 ) ≠ ∅ for 𝑖, 𝑗 = 1 . . . , 𝑚 − 1.

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Ð Ð∞ Let 𝐸 = ∞ 𝑚=1 𝐸 𝑚 and 𝐹 = 𝑚=1 𝐹𝑚 . Then, 𝐸 and 𝐹 are countable regular extensions of 𝐾 and 𝐹 is purely inseparable over 𝐸. Hence, in order to prove that 𝐸 ins is PAC it suffices, by Theorem 12.2.3, to prove that each absolutely irreducible variety 𝑉 defined over 𝐸 has an 𝐹-rational point. Indeed, if 𝑉 is such a variety, then 𝑉 = 𝑉𝑖 𝑗 for some 𝑖 and 𝑗. Let 𝑚 = max{𝑖, 𝑗 } +1. By (12.23d), 𝑉 has an 𝐹𝑚 -rational point, which is, of course, an 𝐹-rational point. Finally, each point of 𝐶 (𝐸) belongs to 𝐶 (𝐸 𝑚 ) for some 𝑚 and therefore, by (12.23c), to 𝐶 (𝐾). Thus, 𝐶 (𝐸) = 𝐶 (𝐾) is a finite set. By Proposition 12.1.1, 𝐸 is not PAC.

(12.24) □

Remark 12.7.9 We call a field 𝐾 ample if 𝑉 (𝐾) is Zariski-dense in 𝑉 for every variety 𝑉 defined over 𝐾 with a simple point a ∈ 𝑉 (𝐾) [Jar11, p. 68, Def. 5.3.2]. In particular, every PAC field is ample (Proposition 12.1.1). The family of ample fields is much larger than the family of PAC fields. In addition to PAC fields, it contains all Henselian fields [Jar11, p. 74, Example 5.6.2], all fields 𝐾0 ((𝑋1 , . . . , 𝑋𝑟 )) of formal power series fields with 𝑟 ≥ 1 over arbitrary fields 𝐾0 [Jar11, p. 79, Remark 5.7.8], and all fields 𝐹 with the property that the degree of every finite extension 𝐹 ′ of 𝐹 is 𝑝 𝑟 for a fixed prime number 𝑝 and arbitrary non-negative integer 𝑟 [Jar11, p. 83, Thm. 5.8.3]. In the notation of the proof of Theorem 12.7.8, there exists an 𝑚 such that 𝐶 (𝐸 𝑚 ) contains a generic point of 𝐶 over 𝐾. Thus, 𝐶 (𝐸 𝑚 ), hence also 𝐶 (𝐸), contains a simple point of 𝐶. By (12.24), 𝐶 (𝐸) is finite. Therefore, 𝐸 is even not ample. Remark 12.7.10 Arno Fehm proved that if 𝐾 is a field of positive characteristic and 𝐹 is a finitely generated transcendental extension of 𝐾, then 𝐹ins is not PAC [Jar11, p. 103, Prop. 6.1.10]. Thus, it is impossible to strengthen the condition on 𝐸 in Theorem 12.7.8 to be finitely generated over F 𝑝 . This gives a negative answer to [FrJ08, p. 217, Problem 11.7.9].

Exercises 1. Let 𝑉 be an absolutely irreducible variety which is defined over a PAC field 𝐾. Suppose that 𝐴 and 𝐵 are Zariski 𝐾-closed subsets of 𝑉 such that 𝑉 (𝐾) = 𝐴(𝐾) ∪ 𝐵(𝐾). Use Proposition 12.1.1 to prove that 𝑉 = 𝐴 or 𝑉 = 𝐵. Conclude that 𝑉 (𝐾) is connected in the Zariski-topology. 2. By the Artin–Schreier theorem, the only torsion that occurs in the absolute Galois group of a field 𝐾 comes from real closed fields. Prove that if 𝐾 is PAC, then Gal(𝐾) is torsion free.

12.7 A non-PAC Field 𝐾 with 𝐾ins PAC

227

3. A more elementary characterization of PAC fields, just as useful as Theorem 12.2.3, can be derived from Lemma 11.4.1. Prove that a field 𝐾 is PAC if and only if for each nonconstant absolutely irreducible polynomial 𝑓 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑟 , 𝑋], separable in 𝑋, and for every nonzero polynomial 𝑔 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑟 ], there exist 𝑎 1 , . . . , 𝑎𝑟 , 𝑏 ∈ 𝐾 such that 𝑓 (a, 𝑏) = 0 and 𝑔(a) ≠ 0. 4. Prove the following assertion without using descent: If 𝐾 is a PAC field of characteristic 𝑝 > 0, then 𝐾ins is also a PAC field. 5. Let Γ be the plane curve defined over F3 by the equation 𝑋04 + 𝑋14 − 𝑋24 − 𝑋02 𝑋22 + 𝑋13 𝑋2 = 0. Prove that Γ is absolutely irreducible and has no F3 -rational point. Hint: Show that Γ is smooth. 6. (a) Let 𝑓 ∈ 𝐿 [𝑋1 , . . . , 𝑋𝑛 ] be a nonconstant homogeneous polynomial with at least one absolutely irreducible factor of multiplicity 1. Prove that if 𝑎 ∈ 𝐿 × , then 𝑓 (𝑋1 , . . . , 𝑋𝑛 ) − 𝑎 is absolutely irreducible. (b) Let 𝑀/𝐿 be a finite Galois extension with basis 𝑤 1 , . . . , 𝑤 𝑛 . Describe the norm map norm 𝑀/𝐿 : 𝑀 → 𝐿 as a homogeneous polynomial of degree 𝑛 in 𝑋1 , . . . , 𝑋𝑛 . (c) Combine (a) and (b) to show that if 𝐿 is a finite separable extension of a PAC field 𝐾, then norm 𝑀/𝐿 is surjective. Thus, show directly property (b) of Proposition 12.6.8. 7. This exercise suggests an alternative way to achieve some consequences of Corollary 12.5.5. (a) Use Eisenstein’s criterion to prove that the polynomial 𝑓 (𝑋, 𝑌 ) = (𝑋 𝑞 − 𝑋) (𝑌 𝑞 − 𝑌 ) + 1 defined over the field F𝑞 (𝑞 a prime power) is absolutely irreducible and has no F𝑞 -rational zeros. Deduce that a PAC field is infinite. (b) (Ax) Let 𝐾 be a field with a valuation 𝑣 whose residue class field has 𝑞 elements. Let 𝜋 ∈ 𝐾 with 𝑣(𝜋) > 0 and show that (𝑋 𝑞 − 𝑋 − 1) (𝑌 𝑞 − 𝑌 − 1) − 𝜋 = 0 defines an absolutely irreducible variety with no 𝐾-rational point. Thus, show directly that 𝐾 is not a PAC field. Conclude, in particular, that fields finitely generated over their prime fields are not PAC fields. 8. Use the idea included in the proof of Corollary 12.5.7 to show that Gal(Q 𝑝 ) is pronilpotent for no prime 𝑝. 9. Let 𝐾 be a field, 𝐿 an infinite extension of 𝐾, and 𝑓 ∈ 𝐿 [𝑋1 , . . . , 𝑋𝑛 ]. Prove that if 𝑓 (a) ∈ 𝐾 for each a ∈ 𝐿 𝑛 , then the coefficients of 𝑓 belong to 𝐾. Hint: For 𝑛 = 1 use Cramer’s rule to compute the coefficients of 𝑓 . Then, continue by induction on 𝑛.

Notes Apparently PAC fields appear for the first time in J. Ax’s papers [Ax67] and [Ax68]. Ax observes that the Riemann hypothesis for curves implies that nonprincipal ultraproducts of finite fields are PAC fields. From this he establishes a recursive decision procedure for the theory of finite fields. Following a suggestion of the second author,

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Frey introduces in [Fre73] the name PAC for fields over which each variety has rational points. He also proves a place-theoretic predecessor to Lemma 12.2.1. Remark 12.2.10, due to János Kollár, settles Problem 11.2.10 of the second edition. The notion of “ample fields” is introduced by Florian Pop in [Pop96, Thm. A]. Our proof of Proposition 12.4.1 appears to be simpler than Tamagawa’s. In [Fre73], Frey uses the theory of homogeneous spaces attached to elliptic curves and Tate’s theory of bad reduction in order to prove Corollary 12.5.5 for real valuations. In a private communication, Prestel showed the authors an elementary direct proof of the corollary for arbitrary valuations. A variation of that proof serves also for Proposition 12.5.3.

Chapter 13

Hilbertian Fields

David Hilbert proved his celebrated irreducibility theorem during his attempt to solve a central problem of Galois theory: Is every finite group realizable over Q? He proved that a general specialization of the coefficients of the general polynomial of degree 𝑛 to elements of Q gives a polynomial whose Galois group is 𝑆 𝑛 . Further, if 𝑓 ∈ Q[𝑇1 , . . . , 𝑇𝑟 , 𝑋] is an irreducible polynomial, then there exist 𝑎 1 , . . . , 𝑎𝑟 ∈ Q such that 𝑓 (a, 𝑋) remains irreducible. This result is now known as Hilbert’s irreducibility theorem. Since then, many more finite groups have been realized over Q. Most of those have been realized via Hilbert’s theorem. This has brought the theorem to the center of the theory of fields. Various alternative proofs of the irreducibility theorem apply to other fields (including all infinite finitely generated fields). We call them Hilbertian fields. We give several reductions of the irreducibility theorem, and one especially valuable result (Corollary 13.2.3): Let 𝐾 be a Hilbertian field, 𝐿 a finite separable extension of 𝐾, and 𝑓 ∈ 𝐿 [𝑇1 , . . . , 𝑇𝑟 , 𝑋] an irreducible polynomial. Then, there exist 𝑎 1 , . . . , 𝑎𝑟 ∈ 𝐾 such that 𝑓 (a, 𝑋) is irreducible in 𝐿 [𝑋]. Chapters 14, 15, and 17 give a proof of the Hilbertian property for most known Hilbertian fields. This lays the foundations of a subject central to the book, the model theory of PAC fields.

13.1 Hilbert Sets and Reduction Lemmas Consider a field 𝐾 and two sets 𝑇1 , . . . , 𝑇𝑟 and 𝑋1 , . . . , 𝑋𝑛 of variables. Let 𝑓1 (T, X), . . . , 𝑓𝑚 (T, X) be polynomials in 𝑋1 , . . . , 𝑋𝑛 with coefficients in 𝐾 (T). Assume these are irreducible in the ring 𝐾 (T) [X]. For 𝑔 ∈ 𝐾 [T] a nonzero polynomial, denote the set of all r-tuples (𝑎 1 , . . . , 𝑎𝑟 ) ∈ 𝐾 𝑟 with 𝑔(a) ≠ 0 and 𝑓1 (a, X), . . . , 𝑓𝑚 (a, X) defined and irreducible in 𝐾 [X] by 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ; 𝑔). Call 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ; 𝑔) a Hilbert subset of 𝐾 𝑟 . If in addition 𝑛 = 1 and each 𝑓𝑖 is sep-

229 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_13

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arable in 𝑋, call 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ; 𝑔) a separable Hilbert subset of 𝐾 𝑟 . A Hilbert set (resp. separable Hilbert set) of 𝐾 is a Hilbert subset (resp. separable Hilbert subset) of 𝐾 𝑟 for some positive integer 𝑟. The intersection of finitely many Hilbert subsets of 𝐾 𝑟 is again a Hilbert subset of 𝐾 𝑟 . Hence, if each Hilbert subset of 𝐾 𝑟 is nonempty, then each Hilbert subset of 𝐾 𝑟 is Zariski 𝐾-dense in 𝐾 𝑟 . By Lemma 11.2.5, each Hilbert subset of 𝐾 𝑟 is Zariski dense in 𝐾 𝑟 . The same holds for separable Hilbert sets. We say that 𝐾 is Hilbertian if each separable Hilbert set of 𝐾 is nonempty. In particular, a Hilbertian field must be infinite. The next lemmas reduce the infiniteness of arbitrary Hilbert sets (resp. separable Hilbert sets) to the infiniteness of special Hilbert sets (resp. separable Hilbert sets). Lemma 13.1.1 Each Hilbert subset (resp. separable Hilbert subset) 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ; 𝑔) of 𝐾 𝑟 contains a Hilbert subset (resp. separable Hilbert subset) 𝐻𝐾 ( 𝑓1′, . . . , 𝑓𝑚′ ; 𝑔 ′) of 𝐾 𝑟 with 𝑓𝑖′ irreducible in 𝐾 [T, X] and 𝑓𝑖′ ∉ 𝐾 [T], 𝑖 = 1, . . . , 𝑚. Proof. By assumption, 𝑓𝑖 is irreducible in 𝐾 (T) [X]. Hence, at least one 𝑋 𝑗 occurs in 𝑓𝑖 . Multiply 𝑓𝑖 (T, X) by a polynomial, 𝑔𝑖 (T), to ensure that its coefficients lie in 𝐾 [T]. Then, divide the resulting polynomial by the greatest common divisor, 𝑑𝑖 (T), of its coefficients. Since 𝐾 [T, X] has unique factorization, we obtain an irreducible polynomial 𝑓𝑖′ ∈ 𝐾 [T, X]. Now put 𝑔 ′ = 𝑔 · 𝑔1 𝑑1 . . . 𝑔𝑚 𝑑 𝑚 and conclude that 𝐻𝐾 ( 𝑓1′, . . . , 𝑓𝑚′ ; 𝑔 ′) ⊆ 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ; 𝑔). Finally, suppose that 𝑛 = 1 and each □ 𝑓𝑖 (T, 𝑋) is separable in 𝑋. Then, so is each 𝑓𝑖′ (T, 𝑋). Lemma 13.1.2 Suppose that every Hilbert set 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ; 𝑔) of 𝐾 with 𝑓𝑖 irreducible in 𝐾 [𝑇, 𝑋1 , . . . , 𝑋𝑛 ], 𝑖 = 1, . . . , 𝑚, is nonempty. Then, every Hilbert set of 𝐾 is nonempty. Proof. Start with irreducible polynomials 𝑓𝑖 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑟 , 𝑋1 , . . . , 𝑋𝑛 ], 𝑖 = 1, . . . , 𝑚, and a nonzero polynomial 𝑔 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑟 ]. By assumption, there exists an 𝑎 1 ∈ 𝐾 with 𝑓𝑖 (𝑎 1 , 𝑇2 , . . . , 𝑇𝑟 , X) irreducible, 𝑖 = 1, . . . , 𝑚, and 𝑔(𝑎 1 , 𝑇2 , . . . , 𝑇𝑟 ) ≠ 0. Repeat this procedure 𝑟 times to find 𝑎 1 , . . . , 𝑎𝑟 ∈ 𝐾 with 𝑓𝑖 (a, X) irreducible and 𝑔(a) ≠ 0, 𝑖 = 1, . . . , 𝑚. By Lemma 13.1.1, every Hilbert set of 𝐾 is nonempty. □ Lemma 13.1.3 Every Hilbert subset of 𝐾 contains a Hilbert set of the form 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ; 𝑔), where 𝑓𝑖 is an irreducible polynomial in 𝐾 [𝑇, 𝑋] with deg𝑋 ( 𝑓𝑖 ) ≥ 1, 𝑖 = 1, . . . , 𝑚. Proof. Let 𝑓 ∈ 𝐾 [𝑇, 𝑋1 , . . . , 𝑋𝑛 ] be an irreducible polynomial with 𝑓 ∉ 𝐾 [𝑇] and 0 ≠ 𝑔0 ∈ 𝐾 [𝑇]. By Lemma 13.1.1, it suffices to find irreducible polynomials ℎ1 , . . . , ℎ𝑟 ∈ 𝐾 [𝑇, 𝑌 ] ∖ 𝐾 [𝑇] and 0 ≠ 𝑔 ∈ 𝐾 [𝑇] with 𝐻𝐾 (ℎ1 , . . . , ℎ𝑟 ; 𝑔) ⊆ 𝐻𝐾 ( 𝑓 ; 𝑔0 ). Indeed, choose 𝑑 > max1≤ 𝑗 ≤𝑛 deg𝑋 𝑗 ( 𝑓 ). Apply the Kronecker substitution 𝑆 𝑑 : 𝑆 𝐾 [𝑇 ] (𝑛, 𝑑) → 𝑆 𝐾 [𝑇 ] (1, 𝑑 𝑛 ) on 𝑓 (Section 12.3). Consider the factorization of 𝑆 𝑑 ( 𝑓 ) into irreducible factors of 𝐾 [𝑇, 𝑌 ]: Ö ℎ𝑖 (𝑇, 𝑌 ). 𝑆 𝑑 ( 𝑓 ) (𝑇, 𝑌 ) = (13.1) 𝑖 ∈𝐼

13.1 Hilbert Sets and Reduction Lemmas

231

The polynomials 𝑆 𝑑 ( 𝑓 ) (𝑇, 𝑌 ) and 𝑓 (𝑇, 𝑋1 , . . . , 𝑋𝑛 ) have the same coefficients in 𝐾 [𝑇]. Since 𝑓 is irreducible in 𝐾 [𝑇, 𝑋1 , . . . , 𝑋𝑛 ], the greatest common divisor of its coefficients in 𝐾 [𝑇] is 1, so none of the ℎ𝑖 is in 𝐾 [𝑇]. partition ofÎ𝐼. The exponent to which 𝑌 appears in Let 𝐼 = 𝐽 ∪· 𝐽 ′ be a nontrivial Î each of the polynomials 𝑖 ∈𝐽 ℎ𝑖 (𝑇, 𝑌 ) and 𝑖 ∈𝐽 ′ ℎ𝑖 (𝑇, 𝑌 ) does not exceed 𝑑 𝑛 − 1. Since 𝑆 𝑑 is bijective on 𝑆 𝐾 [𝑇 ] (𝑛, 𝑑) (Section 12.3), there exist polynomials 𝑝 𝐽 , 𝑝 𝐽 ′ ∈ 𝐾 [𝑇, X] with deg𝑋 𝑗 ( 𝑝 𝐽 ), deg𝑋 𝑗 ( 𝑝 𝐽 ′ ) < 𝑑, 𝑗 = 1, . . . , 𝑛, and Ö Ö ℎ𝑖 (𝑇, 𝑌 ) and 𝑆 𝑑 ( 𝑝 𝐽 ′ ) (𝑇, 𝑌 ) = 𝑆 𝑑 ( 𝑝 𝐽 ) (𝑇, 𝑌 ) = ℎ𝑖 (𝑇, 𝑌 ). (13.2) 𝑖 ∈𝐽 ′

𝑖 ∈𝐽

Note: The product 𝑝 𝐽 (𝑇, X) 𝑝 𝐽 ′ (𝑇, X) contains a monomial of the form 𝑔 𝐽 (𝑇) 𝑋1𝜈1 · · · 𝑋𝑛𝜈𝑛 in which at least one of the 𝜈 𝑗 exceeds 𝑑 − 1. Otherwise, the relation 𝑆 𝑑 ( 𝑓 ) = 𝑆 𝑑 ( 𝑝 𝐽 ) · 𝑆 𝑑 ( 𝑝 𝐽 ′ ) = 𝑆 𝑑 ( 𝑝 𝐽 𝑝 𝐽 ′ ) would imply a nontrivial factorization of the irreducible polynomial 𝑓 into 𝑝 𝐽 𝑝 𝐽 ′ . Let 𝑔 be the product of 𝑔0 with all 𝑔 𝐽 (one for each nontrivial partition). Let 𝑎 be an element of 𝐾 with each ℎ𝑖 (𝑎, 𝑌 ) irreducible in 𝐾 [𝑌 ] and 𝑔(𝑎) ≠ 0. We show 𝑓 (𝑎, X) is also irreducible. Indeed, assume 𝑓 (𝑎, X) = 𝑞(X)𝑞 ′ (X) is a nontrivial factorization of 𝑓 (𝑎, X) in 𝐾 [X], then (13.1) implies Ö ℎ𝑖 (𝑎, 𝑌 ) = 𝑆 𝑑 ( 𝑓 ) (𝑎, 𝑌 ) = 𝑆 𝑑 (𝑞) (𝑌 ) · 𝑆 𝑑 (𝑞 ′) (𝑌 ). 𝑖 ∈𝐼

Hence, (13.2) implies a nontrivial partition 𝐼 = 𝐽 ∪· 𝐽 ′ of 𝐼 with Ö 𝑆 𝑑 (𝑞) (𝑌 ) = ℎ𝑖 (𝑎, 𝑌 ) = 𝑆 𝑑 ( 𝑝 𝐽 ) (𝑎, 𝑌 ) and 𝑖 ∈𝐽 ′

𝑆 𝑑 (𝑞 ) (𝑌 ) =

Ö

ℎ𝑖 (𝑎, 𝑌 ) = 𝑆 𝑑 ( 𝑝 𝐽 ′ ) (𝑎, 𝑌 ).

𝑖 ∈𝐽 ′

Hence, 𝑞(X) = 𝑝 𝐽 (𝑎, X) and 𝑞 ′ (X) = 𝑝 𝐽 ′ (𝑎, X). Thus, 𝑝 𝐽 (𝑎, X) 𝑝 𝐽 ′ (𝑎, X) = 𝑓 (𝑎, X). The left-hand side contains the monomial 𝑔 𝐽 (𝑎) 𝑋1𝜈1 · · · 𝑋𝑛𝜈𝑛 in which at least one 𝜈 𝑗 exceeds 𝑑 − 1, while 𝑓 (𝑎, X) contains no such monomial. Therefore, □ 𝑓 (𝑎, X) is irreducible. Lemma 13.1.4 Let 𝐻 := 𝐻𝐾 (𝑔1 , . . . , 𝑔𝑚 ; ℎ) be a Hilbert subset of 𝐾 𝑟 with 𝑔𝑖 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑟 , 𝑋] irreducible and deg𝑋 (𝑔𝑖 ) ≥ 1, 𝑖 = 1, . . . , 𝑚. Then, 𝐻 contains a Hilbert set of the form 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ) in which 𝑓𝑖 is a monic irreducible polynomial in 𝐾 [T, 𝑋] of degree at least 2 in 𝑋, 𝑖 = 1, . . . , 𝑚. In addition, if 𝑔1 , . . . , 𝑔𝑚 are separable in 𝑋, then so are 𝑓1 , . . . , 𝑓𝑚 . If 𝑔1 , . . . , 𝑔𝑚 are absolutely irreducible, then so are 𝑓1 , . . . , 𝑓𝑚 . Moreover, suppose a ∈ 𝐾 𝑟 and none of the polynomials 𝑓𝑖 (a, 𝑋) has a root in 𝐾. Then, none of the polynomials 𝑔𝑖 (a, 𝑋) with deg𝑋 (𝑔𝑖 ) ≥ 2 has a root in 𝐾. Proof. Let 𝑐 𝑖 be the leading coefficient of 𝑔𝑖 as a polynomial in 𝑋, 𝑛𝑖 = deg𝑋 (𝑔𝑖 ), and 𝑞 = ℎ𝑐 1 · · · 𝑐 𝑚 . Choose a prime number 𝑙 ≠ char(𝐾) with 𝑞 not an 𝑙th power in 𝐾˜ [T]. By assumption, 𝑛𝑖 ≥ 1. If 𝑚 = 0 or 𝑛𝑖 = 1, let 𝑓𝑖 (T, 𝑋) = 𝑋 𝑙 − 𝑞(T). If 𝑛𝑖 ≥ 2, let 𝑓𝑖 (T, 𝑋) = 𝑞(T) 𝑛𝑖 𝑐 𝑖 (T) −1 𝑔𝑖 (T, 𝑞(T) −1 𝑋).

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Then, 𝑓𝑖 (T, 𝑋) = 𝑋 𝑛𝑖 + 𝑏 𝑖,𝑛𝑖 −1 (T) 𝑋 𝑛𝑖 −1 + 𝑞(T)

𝑛∑︁ 𝑖 −2

𝑏 𝑖 𝑗 (T) 𝑋 𝑗 ,

𝑗=0

with 𝑏 𝑖 𝑗 ∈ 𝐾 [T]. In each case 𝑓𝑖 (T, 𝑋) is monic in 𝑋 and irreducible in 𝐾 [T, 𝑋] (see [Lan97, p. 297, Lemma 9.1] for the case 𝑚 = 0 or 𝑛𝑖 = 1). Note that if 𝑔1 , . . . , 𝑔𝑚 ˜ we conclude that if 𝑔𝑖 are separable in 𝑋, then so are 𝑓1 , . . . , 𝑓𝑚 . Replacing 𝐾 by 𝐾, is absolutely irreducible, then so is 𝑓𝑖 , for 𝑖 = 1, . . . , 𝑚. We prove that 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ) ⊆ 𝐻. Indeed, consider a ∈ 𝐾 𝑟 with 𝑓1 (a, 𝑋), . . . , 𝑓𝑚 (a, 𝑋) irreducible in 𝐾 [X]. Consider an 𝑖 between 1 and 𝑚. Suppose first that 𝑛𝑖 = 1. Then, 𝑋 𝑙 − 𝑞(a) is irreducible. Hence, 𝑞(a) ≠ 0. Therefore, ℎ(a) ≠ 0, 𝑐 𝑖 (a) ≠ 0, and 𝑔𝑖 (a, 𝑋) = 𝑐 𝑖 (a) 𝑋 + 𝑏 𝑖 for some 𝑏 𝑖 ∈ 𝐾. Thus, 𝑔𝑖 (a, 𝑋) is irreducible. Now suppose that 𝑛𝑖 ≥ 2. Then, 𝑞(a) ≠ 0 (otherwise 𝑓𝑖 (a, 𝑋) = 𝑋 𝑛𝑖 + 𝑏 𝑖,𝑛𝑖 −1 (a) 𝑋 𝑛𝑖 −1 is reducible) and 𝑔𝑖 (a, 𝑞(a) −1 𝑋) = 𝑞(a) −𝑛𝑖 𝑐 𝑖 (a) 𝑓𝑖 (a, 𝑋) is irreducible. Hence, so is 𝑔𝑖 (a, 𝑋). Similarly, if 𝑛𝑖 ≥ 2 and 𝑓𝑖 (a, 𝑋) has no root in 𝐾, □ neither has 𝑔𝑖 (a, 𝑋). Lemma 13.1.5 Suppose that each Hilbert set of the form 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ) with 𝑓𝑖 ∈ 𝐾 [𝑇, 𝑋] monic and of degree at least 2 in 𝑋 is nonempty. Then, every Hilbert set of 𝐾 is nonempty. Proof. By Lemma 13.1.4 for 𝑟 = 1, each Hilbert subset of 𝐾 of the form 𝐻𝐾 (𝑔1 , . . . , 𝑔𝑚 ; ℎ) with 𝑔𝑖 ∈ 𝐾 [𝑇, 𝑋] irreducible and deg𝑋 (𝑔𝑖 ) ≥ 1, 𝑖 = 1, . . . , 𝑚, is nonempty. Hence, by Lemma 13.1.3, each Hilbert subset of 𝐾 is nonempty. Consequently, by Lemma 13.1.2, every Hilbert set of 𝐾 is nonempty. □ The corresponding result for separable Hilbert sets is also true. We postpone its proof until Section 14.2. Here we show that separable Hilbert subsets of 𝐾 𝑟 contain especially simple separable Hilbert subsets of 𝐾 𝑟 . Lemma 13.1.6 Let 𝐻 be a separable Hilbert subset of 𝐾 𝑟 . Then, there exists an irreducible polynomial 𝑓 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑟 , 𝑋], separable, monic, and of degree at least 2 in 𝑋 with 𝐻𝐾 ( 𝑓 ) ⊆ 𝐻. Proof. Use Lemma 13.1.1 to assume that 𝐻 = 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ; 𝑔) with 𝑓1 , . . . , 𝑓𝑚 ∈ 𝐾 [T, 𝑋] ∖ 𝐾 [T] irreducible and separable in 𝑋 and 0 ≠ 𝑔 ∈ 𝐾 [T]. For each 𝑖, 1 ≤ 𝑖 ≤ 𝑚, let 𝑥𝑖 be a root of 𝑓𝑖 (T, 𝑋) in 𝐾 (T)sep . Then, the finite separable extension 𝐾 (T, 𝑥1 , . . . , 𝑥 𝑚 ) of 𝐾 (T) is contained in a separable extension 𝐾 (T, 𝑦) of 𝐾 (T) of degree at least 2. Assume without loss that 𝑦 is integral over 𝐾 [T] and let ℎ = irr(𝑦, 𝐾 (T)). Now apply Remark 7.1.5 to find a nonzero multiple 𝑔 ′ ∈ 𝐾 [T] of 𝑔 such that with the notation 𝑅 = 𝐾 [T, 𝑔 ′ (T) −1 ], 𝑅𝑖 = 𝐾 [T, 𝑔 ′ (T) −1 , 𝑥𝑖 ], 𝑖 = 1, . . . , 𝑚, and 𝑆 = 𝐾 [T, 𝑔 ′ (T) −1 , 𝑦], the extensions 𝑅𝑖 /𝑅, 𝑖 = 1, . . . , 𝑚, and 𝑆/𝑅 are ring covers (Definition 7.1.3). In particular, 𝑅𝑖 and 𝑆 are the integral closures of 𝑅 in 𝐾 (T, 𝑥𝑖 ) and 𝐾 (T, 𝑦), respectively. Thus, by Lemma 7.1.9, 𝑅𝑖 ⊆ 𝑆 and 𝑆/𝑅𝑖 is also a ring cover, 𝑖 = 1, . . . , 𝑚. We prove that 𝐻𝐾 (ℎ; 𝑔 ′) ⊆ 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ; 𝑔).

13.2 Hilbert Sets under Separable Algebraic Extensions

233

Indeed, if a ∈ 𝐻𝐾 (ℎ; 𝑔 ′), then ℎ(a, 𝑋) is irreducible and 𝑔 ′ (a) ≠ 0. Example 3.4.10 gives a 𝐾-place 𝜑 of 𝐾 (T) with residue field 𝐾 such that 𝜑(T) = a. In particular, 𝜑(𝑔 ′ (T)) = 𝑔 ′ (a) ≠ 0, so 𝜑 is finite on 𝑅. We extend 𝜑 to a place 𝜓 of 𝐾 (T, 𝑦). Then, 𝜓 is finite on each 𝑅𝑖 and on 𝑆. We set 𝜑(𝑦) = 𝑐 and 𝑏 𝑖 = 𝜑(𝑥 𝑖 ). Then, [𝐾 (𝑐) : 𝐾] = [𝐾 (T, 𝑦) : 𝐾 (T)]. (13.3) Moreover, by Lemma 7.1.9, [𝐾 (𝑏 𝑖 ) : 𝐾] ≤ [𝐾 (T, 𝑥𝑖 ) : 𝐾 (T)] and [𝐾 (𝑐) : 𝐾 (𝑏 𝑖 )] ≤ [𝐾 (T, 𝑦) : 𝐾 (T, 𝑥𝑖 )]. Multiply the terms in these two inequalities and apply (13.3) to conclude that they are equalities. Therefore, 𝑓1 (a, 𝑋), . . . , 𝑓𝑚 (a, 𝑋) are irreducible over 𝐾. By Lemma 13.1.4, there exists an irreducible polynomial 𝑓 ∈ 𝐾 [T, 𝑋], separable, □ monic, and of degree at least 2 in 𝑋, with 𝐻𝐾 ( 𝑓 ) ⊆ 𝐻𝐾 (ℎ; 𝑔 ′).

13.2 Hilbert Sets under Separable Algebraic Extensions Let 𝐿/𝐾 be a finite separable extension. We prove that every Hilbert subset of 𝐿 𝑟 contains a Hilbert subset of 𝐾 𝑟 . Lemma 13.2.1 Let 𝐿 be a separable extension of degree 𝑑 of an infinite field 𝐾 and 𝜎0 , . . . , 𝜎𝑑−1 distinct representatives of the cosets of Gal(𝐿) in Gal(𝐾). Suppose that 𝑓 ∈ 𝐿 [𝑋1 , . . . , 𝑋𝑛 ] is a nonconstant polynomial. Then, there exist 𝑐 1 , . . . , 𝑐 𝑛 ∈ 𝐿 with 𝑓 (𝑋1 + 𝑐 1 , . . . , 𝑋𝑛 + 𝑐 𝑛 ) 𝜎𝑖 , 𝑖 = 0, . . . , 𝑑 − 1, pairwise relatively prime in 𝐾˜ [𝑋1 , . . . , 𝑋𝑛 ]. Proof. Let 𝜃 be a primitive element for 𝐿/𝐾. Choose algebraically independent elements 𝑡 𝑖𝑘 , 𝑖 = 0, . . . , 𝑑 − 1 and 𝑘 = 1, . . . , 𝑛, over 𝐾. For every 𝑖 and 𝑘 consider 𝑢 𝑖𝑘 =

𝑑−1 ∑︁

(𝜃 𝜎𝑖 ) 𝑗 𝑡 𝑗 𝑘 .

𝑗=0

. . , u𝑑−1 ). The linear transformation t → Write u𝑖 = (𝑢 𝑖1 , . . . , 𝑢 𝑖𝑛 ) and u = (u0 , .Î u has as determinant the 𝑛th power of 𝑖< 𝑗 (𝜃 𝜎 𝑗 − 𝜃 𝜎𝑖 ) ≠ 0. Thus, the 𝑡 𝑖𝑘 are ˜ so the 𝑢 𝑖𝑘 are algebraically linear combinations of the 𝑢 𝑖𝑘 with coefficients in 𝐾; independent over 𝐾. ˜ factors. Since 𝑓 𝜇 is nonconstant Write 𝑓 = 𝑓1 · · · 𝑓𝑚 , a product of 𝐾-irreducible 𝜎 and the 𝑢 𝑖𝑘 are algebraically independent, 𝑓 𝜇𝜎𝑖 (u𝑖 ) ≠ 𝑓 𝜈 𝑗 (u 𝑗 ) for 𝑖 ≠ 𝑗 and each 𝜇 and 𝜈, 1 ≤ 𝜇, 𝜈 ≤ 𝑚. Therefore, ÖÖ 𝜎 ℎ(t) = ( 𝑓 𝜇𝜎𝑖 (u𝑖 ) − 𝑓 𝜈 𝑗 (u 𝑗 )) ≠ 0, 𝑖< 𝑗 𝜇,𝜈

so there exist 𝑎 𝑖𝑘 ∈ 𝐾 with ℎ(a) ≠ 0. Let 𝑐𝑘 =

𝑑−1 ∑︁ 𝑗=0

𝑎 𝑗𝑘𝜃 𝑗,

𝑘 = 1, . . . , 𝑛.

234

13 Hilbertian Fields

˜ t → a maps 𝑢 𝑖𝑘 to 𝑐 𝑘𝜎𝑖 . Hence 𝑓 𝜇 (c) 𝜎𝑖 ≠ 𝑓 𝜈 (c) 𝜎 𝑗 for 𝑖 ≠ 𝑗. The 𝐾-specialization Therefore, 𝑓 𝜇 (X + c) 𝜎𝑖 ≠ 𝑓 𝜈 (X + c) 𝜎 𝑗 for all 𝑖, 𝑗, 𝜇, 𝜈 with 0 ≤ 𝑖 < 𝑗 ≤ 𝑑 − 1 and 1 ≤ 𝜇, 𝜈 ≤ 𝑚. Since 𝑓1 (X + c) 𝜎𝑖 , . . . , 𝑓𝑚 (X + c) 𝜎𝑖 are exactly the irreducible factors of 𝑓 (X + c) 𝜎𝑖 in 𝐾˜ [X], 𝑓 (X + c) 𝜎0 , . . . , 𝑓 (X + c) 𝜎𝑑−1 are relatively prime in pairs. □ Lemma 13.2.2 Let 𝐿 be a finite separable extension of a field 𝐾 and 𝑓 ∈ 𝐿 (𝑇1 , . . . , 𝑇𝑟 ) [𝑋1 , . . . , 𝑋𝑛 ] irreducible. Then, there exists an irreducible 𝑝 ∈ 𝐾 (T) [X] with 𝐻𝐾 ( 𝑝) ⊆ 𝐻 𝐿 ( 𝑓 ). If 𝑛 = 1 and 𝑓 is separable in 𝑋, then 𝑝 is separable in 𝑋. Proof. Let 𝑆 be a set of representatives of the right cosets of Gal(𝐿) in Gal(𝐾). First consider the case where the 𝑓 𝜎 , with 𝜎 ∈ 𝑆, are pairwise relatively prime in Î  (T) [X]. Let 𝑝 = 𝜎 ∈𝑆 𝑓 𝜎 . Since Gal(𝐾) fixes the coefficients of 𝑝, we have 𝐾 𝑝 ∈ 𝐾 (T) [X]. Moreover, if 𝑞 is an irreducible factor of 𝑝 in 𝐾 (T) [X], then one of the 𝑓 𝜎 ’s (and therefore all) divides 𝑞 in 𝐾˜ (T) [X]. Since the 𝑓 𝜎 are pairwise relatively prime, 𝑝 divides 𝑞. Hence, 𝑝 is irreducible in 𝐾 (T) [X]. In addition, if 𝑛 = 1 and 𝑓 is separable in 𝑋, then 𝑝 is separable in 𝑋. Let 𝑎 1 , . . . , 𝑎𝑟 be elements of 𝐾 such that 𝑝(a, X) is defined and irreducible in 𝐾 [X]. If Î 𝑓 (a, X) = 𝑔(X)ℎ(X) were a nontrivial decomposition in 𝐿 [X], then Î 𝑝(a, X) = ( 𝜎 ∈𝑆 𝑔 𝜎 (X)) ( 𝜎 ∈𝑆 ℎ 𝜎 (X)) would be a nontrivial decomposition in 𝐾 [X], which is a contradiction. Therefore, 𝑓 (a, X) is irreducible in 𝐿 [X] and 𝐻𝐾 ( 𝑝) ⊆ 𝐻 𝐿 ( 𝑓 ). In the general case apply Lemma 13.2.1, for 𝐾 (T) and 𝐿(T) rather than to 𝐾 and 𝐿, to find 𝑐 1 , . . . , 𝑐 𝑛 ∈ 𝐿 (T) with 𝑓 (T, X + c) 𝜎 pairwise relatively prime in  𝐾 (T) [X], 𝜎 running over 𝑆. Let 𝑔(T, X) = 𝑓 (T, X + c). Irreducibility of 𝑔(a, X) is equivalent to that of 𝑓 (a, X) for each a ∈ 𝐾 𝑟 such that both polynomials are defined. □ Thus, the lemma follows from the first part of the proof. Corollary 13.2.3 Let 𝐿 be a finite separable extension of a field 𝐾. Then, every Hilbert subset (resp. separable Hilbert subset) of 𝐿 𝑟 contains a Hilbert subset (resp. separable Hilbert subset) of 𝐾 𝑟 . In particular, if 𝐾 is Hilbertian, so is 𝐿. Remark 13.2.4 The converse of Corollary 13.2.3 is false: Remark 15.3.5 gives a field 𝐾 with an empty Hilbert set and a finite separable extension 𝐿 whose Hilbert sets are all nonempty.

13.3 Purely Inseparable Extensions Corollary 13.2.3 is false if 𝐿/𝐾 is inseparable. For example, let 𝐾 = F 𝑝 (𝑡), with 𝑡 transcendental over F 𝑝 , and let 𝐿 = 𝐾 (𝑡 1/ 𝑝 ). Then, 𝑋 𝑝 − 𝑇 is irreducible over 𝐿. Yet every 𝑎 ∈ 𝐾 is a 𝑝th power in 𝐿, so 𝑋 𝑝 − 𝑎 is reducible over 𝐿. Hence, 𝐻 𝐿 (𝑋 𝑝 − 𝑇) contains no elements of 𝐾. Still, 𝑋 𝑝 − 𝑡 1/ 𝑝 is irreducible over 𝐿, so 𝐻 𝐿 (𝑋 𝑝 − 𝑇) is nonempty. The following results generalize this observation. They assert: if every Hilbert set of 𝐾 is nonempty, then every Hilbert set of 𝐿 is nonempty.

13.3 Purely Inseparable Extensions

235

First we list simple properties of purely inseparable extensions. To this end we say that two polynomials 𝑓 , 𝑔 ∈ 𝐿 [X] with X = (𝑋1 , . . . , 𝑋𝑟 ) are equivalent and write 𝑓 ∼ 𝑔 if there exists a 𝑐 ∈ 𝐿 × such that 𝑓 = 𝑐𝑔. In this case, 𝑓 is irreducible in 𝐿 [X] if and only if 𝑔 is irreducible in 𝐿 [X]. Note that 𝑓1 ∼ 𝑔1 and 𝑓2 ∼ 𝑔2 imply that 𝑓1 𝑓2 ∼ 𝑔1 𝑔2 . Also, factorization of polynomials in 𝐿 [X] is unique up to equivalence [Lan97, p. 183, Cor. 2.4]. If 𝑓 and 𝑔 are equivalent polynomials in 𝐿 [𝑇, X], then 𝑓 (𝑎, X) and 𝑔(𝑎, X) are equivalent in 𝐿 [X] for all 𝑎 ∈ 𝐿. In addition, if 𝑓1 , . . . , 𝑓𝑟 , 𝑓1′, . . . , 𝑓𝑟′ are irreducible polynomials in 𝐿 [T, 𝑋] and 𝑔, 𝑔 ′ ∈ 𝐿 [T] are nonzero polynomials such that 𝑓𝑖 ∼ 𝑓𝑖′ for 𝑖 = 1, . . . , 𝑟 and 𝑔 ∼ 𝑔 ′, then 𝐻 𝐿 ( 𝑓1 , . . . , 𝑓𝑟 ; 𝑔) = 𝐻 𝐿 ( 𝑓1′, . . . , 𝑓𝑟′, 𝑔 ′). Lemma 13.3.1 Let 𝐿/𝐾 be a purely inseparable extension of fields of characteristic 𝑝. (a) For each irreducible 𝑔 ∈ 𝐾 [X] there exist an irreducible 𝑓 ∈ 𝐿 [X] and 𝑒 ≥ 0 𝑒 with 𝑔 ∼ 𝑓 𝑝 . 𝑒 𝑒 (b) Consider 𝑓 ∈ 𝐿 [X]. Suppose that 𝐿 𝑝 ⊆ 𝐾 and 𝑔 ∼ 𝑓 𝑝 is irreducible in 𝐾 [X]. Then, 𝑓 is irreducible in 𝐿 [X]. (c) Let 𝑔 be an irreducible polynomial in 𝐾 [X] and let 𝑓 ∈ 𝐿 [X] be a polynomial 𝑒 which is not equivalent to a 𝑝th power such that 𝑔 ∼ 𝑓 𝑝 is irreducible in 𝐾 [X]. Then, 𝑓 is irreducible in 𝐿 [X]. 𝑒 (d) If 𝑓 ∈ 𝐿 [X] is irreducible and 𝑒 is the least nonnegative integer with 𝑔 := 𝑓 𝑝 ∈ 𝐾 [X], then 𝑔 is irreducible in 𝐾 [X]. Proof of (a). Let 𝑔 = 𝑓1 · · · 𝑓𝑚 be a factorization of 𝑔 into irreducible factors in 𝑘 𝑘 𝐿 [X]. Choose 𝑘 ≥ 0 such that 𝑓𝑖𝑝 ∈ 𝐾 [X], 𝑖 = 1, . . . , 𝑚. The relation 𝑔 𝑝 = 𝑘

𝑘

𝑘

𝑓1𝑝 · · · 𝑓𝑚𝑝 and the unique factorization in 𝐾 [X] imply that 𝑓1𝑝 ∼ 𝑔𝑟 for some positive integer 𝑟. Unique factorization over 𝐿 [X] then gives a positive integer 𝑠 𝑘 𝑒 with 𝑔 ∼ 𝑓1𝑠 . Then, 𝑓1𝑝 ∼ 𝑓1𝑟 𝑠 implies that 𝑠 = 𝑝 𝑒 , so 𝑔 ∼ 𝑓1𝑝 for some positive integer 𝑒. Proof of (b). Let 𝑓1 · · · 𝑓𝑚 be a factorization of 𝑓 into irreducible factors in 𝐿 [X]. 𝑒 𝑒 Then, up to equivalence, 𝑓1𝑝 · · · 𝑓𝑚𝑝 is a factorization of 𝑔 in 𝐾 [X]. Since 𝑔 is irreducible over 𝐾, we have 𝑚 = 1. Therefore, 𝑓 is irreducible over 𝐿. 𝑒′

Proof of (c). By (a), 𝑔 ∼ ℎ 𝑝 for some irreducible ℎ ∈ 𝐿 [X] and 𝑒 ′ ≥ 0. Hence, 𝑒′ 𝑒 𝑒′ −𝑒 ℎ 𝑝 ∼ 𝑓 𝑝 , so 𝑒 ′ ≥ 𝑒. Therefore, ℎ 𝑝 ∼ 𝑓 . The assumption on 𝑓 implies that 𝑒 ′ = 𝑒 and 𝑓 ∼ ℎ. Thus, 𝑓 is irreducible in 𝐿 [X]. Proof of (d). Let 𝑔1 · · · 𝑔𝑚 be a factorization of 𝑔 into irreducible factors in 𝐾 [X]. 𝑒𝑖 By (a), 𝑔𝑖 ∼ 𝑓𝑖𝑝 for some irreducible polynomial 𝑓𝑖 in 𝐿 [X] and 𝑒 𝑖 ≥ 0. Hence, 𝑒𝑚 𝑒1 𝑒 𝑓 𝑝 ∼ 𝑓1𝑝 · · · 𝑓𝑚𝑝 . Therefore, 𝑓𝑖 ∼ 𝑓 for 𝑖 = 1, . . . , 𝑚 and 𝑝 𝑒1 + · · · + 𝑝 𝑒𝑚 = 𝑝 𝑒 . 𝑒 In addition, 𝑓 𝑝 𝑖 = 𝑔𝑖 ∈ 𝐾 [𝑋], so by assumption, 𝑒 𝑖 ≥ 𝑒. It follows that 𝑚 = 1 and 𝑔 is irreducible in 𝐾 [X]. □

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Lemma 13.3.2 Let 𝐾 be a field with char(𝐾) = 𝑝 > 0, 𝐿 a purely inseparable extension of 𝐾, and 𝑓 an irreducible polynomial in 𝐿 [𝑇1 , . . . , 𝑇𝑟 , 𝑋] with deg𝑋 𝑓 ≥ 1. (a) If 𝑓 is separable in 𝑋, then 𝐻 𝐿 ( 𝑓 ) contains a separable Hilbert subset of 𝐾 𝑟 . (b) If all occurrences of 𝑇1 , . . . , 𝑇𝑟 , 𝑋 in 𝑓 are powers of 𝑝, then 𝐻 𝐿 ( 𝑓 ) contains a Hilbert subset of 𝐾 𝑟 . 𝑒

Proof. Let 𝑒 be the least nonnegative integer with 𝑔 := 𝑓 𝑝 ∈ 𝐾 [T, 𝑋]. By Lemma 13.3.1(d), 𝑔 is irreducible in 𝐾 [T, 𝑋]. Distinguish between two cases to find a Hilbert subset of 𝐾 𝑟 contained in 𝐻 𝐿 ( 𝑓 ). Case A: f is separable with respect to X. Consider a ∈ 𝐾 𝑟 with 𝑔(a, 𝑋) irreducible in 𝐾 [𝑋] and 𝑓 (a, 𝑋) separable. Then, 𝑓 (a, 𝑋) is not equivalent to a 𝑝th power in 𝐿 [𝑋]. By Lemma 13.3.1(c), 𝑓 (a, 𝑋) is irreducible in 𝐿 [𝑋]. This settles (a). Case B: All occurrences of 𝑇1 , . . . , 𝑇𝑟 , 𝑋 in 𝑓 are powers of 𝑝. Since 𝑓 is irreducible in 𝐿 [T, 𝑋], our assumption implies that 𝑓 is not a 𝑝th power in 𝐿 [T, 𝑋], so at least one coefficient 𝑐(T) of 𝑓 (T, 𝑋) is not a 𝑝th power in 𝐿 [T]. In particular, 𝐿 is an infinite field. Hence, 𝐾 is also infinite. Claim: There is a nonempty Zariski 𝐿-open subset 𝑈 of A𝑟 such that 𝑐(a) ∉ 𝐿 𝑝 for all a ∈ 𝑈 (𝐿). To prove the claim, let 𝑑1 = deg𝑇1 (𝑐(T))/𝑝, . . . , 𝑑𝑟 = deg𝑇𝑟 (𝑐(T))/𝑝, Í write 𝑐(T) = 𝑐 i𝑇1𝑖1 𝑝 · · · 𝑇𝑟𝑖𝑟 𝑝 with i = (𝑖 1 , . . . , 𝑖𝑟 ), 0 ≤ 𝑖1 ≤ 𝑑1 , . . . , 0 ≤ 𝑖𝑟 ≤ 𝑑𝑟 , and 𝑐 i ∈ 𝐿. Let 𝑑 = (𝑑1 + 1) · · · (𝑑𝑟 + 1). Assume that there are 𝑑 𝑟-tuples (𝑏 𝑘1 , . . . , 𝑏 𝑘𝑟 )

(13.4) 𝑗′

𝑗′

≤ 𝑟 such that with 0 ≤ 𝑘 𝑗 ≤ 𝑑1 and 𝑏 𝑘 𝑗 ≠ 𝑏 𝑘 𝑗′ if 𝑗 ≠ and 1 ≤ 𝑗, 𝑐(𝑏 𝑘1 , . . . , 𝑏 𝑘𝑟 ) = 𝜆 𝑝 with 𝜆 ∈ 𝐿. Thus, there is a 𝜆 𝑘1 ,...,𝑘𝑟 ∈ 𝐿 with ∑︁ (13.5) 𝑐 i 𝑏 𝑖𝑘11𝑝 · · · 𝑏 𝑖𝑘𝑟𝑟𝑝 = 𝜆 𝑘𝑝1 ,...,𝑘𝑟 . i

Consider (13.5) as a system of 𝑑 linear equations in the 𝑐 i ’s. The coefficient matrix of (13.5) is a Kronecker product of 𝑟 Vandermonde matrices. The determinant of this matrix is 𝑟 Ö 𝑟 Ö Ö ′ ′ 𝑖 (𝑏 𝑘 𝑗′ − 𝑏 𝑘 𝑗 ) 𝑑 𝑗 𝑝 , det(𝑏 𝑘𝑗𝑗 ) 𝑑 𝑗 𝑝 = 𝐷 = det(𝑏 𝑖𝑘11𝑝 · · · 𝑏 𝑖𝑘𝑟𝑟𝑝 ) = 𝑗=1

𝑗=1 𝑗< 𝑗 ′

where 𝑑 ′𝑗 = (𝑑 𝑗 + 1) −1 (𝑑1 + 1) · · · (𝑑𝑟 + 1), 𝑗 = 1, . . . , 𝑟. (See [Bou74, p. 534, (3)] for the determinant of the Kronecker product of two matrices, from which the determinant of the Kronecker product of 𝑟 matrices can be derived.) Thus, 𝐷 is a 𝑝th power of a nonzero element of 𝐿. Applying Cramer’s rule to the system (13.5), we conclude that 𝑐 i ∈ 𝐿 𝑝 for each i. This contradiction to the assumption we made on 𝑐(T) proves that there are at most 𝑑 − 1 𝑟-tuples (13.4). Let 𝐵 be the set of all these 𝑟-tuples. Define, 𝑈 = {(𝑥1 , . . . , 𝑥𝑟 ) ∈ A𝑟 | (𝑥1 , . . . , 𝑥𝑟 ) ∉ 𝐵 and 𝑥 𝑗 ≠ 𝑥 𝑗 ′ if 𝑗 ≠ 𝑗 ′ }. Then, 𝑈 is a Zariski 𝐿-open subset of A𝑟 satisfying our claim. Now we may use the claim to choose a ∈ 𝐾 𝑟 with 𝑐(a) ∉ 𝐿 𝑝 (therefore 𝑓 (a, 𝑋) is not a 𝑝th power in 𝐿 [𝑋]) and 𝑔(a, 𝑋) irreducible in 𝐾 [𝑋]. By Lemma 13.3.1(c), □ 𝑓 (a, 𝑋) is irreducible in 𝐿 [𝑋]. This settles (b).

13.3 Purely Inseparable Extensions

237

Proposition 13.3.3 Let 𝐿/𝐾 be an algebraic extension of fields of a finite separable degree. Then, each separable Hilbert subset of 𝐿 𝑟 contains a separable Hilbert subset of 𝐾 𝑟 . Thus, if 𝐾 is Hilbertian, then so is 𝐿. In particular, this is the case if [𝐿 : 𝐾] < ∞. Proof. By assumption, 𝐿 is a purely inseparable extension of a finite separable extension of 𝐾. Apply Corollary 13.2.3 to assume that 𝐿/𝐾 is purely inseparable. Let 𝐻 be a separable Hilbert subset of 𝐿 𝑟 . By Lemma 13.1.6, 𝐻 contains 𝐻 𝐿 ( 𝑓 ) with irreducible 𝑓 ∈ 𝐿 [T, 𝑋] separable in 𝑋. From Lemma 13.3.2(a), 𝐻 𝐿 ( 𝑓 ) contains a separable Hilbert subset of 𝐾 𝑟 . □ Example 13.3.4 The separable degree of 𝐾ins /𝐾 for a field 𝐾 is 1. Hence, by Proposi𝑟 contains a separable Hilbert subset tion 13.3.3, each separable Hilbert subset of 𝐾ins 𝑟 of 𝐾 . In particular, if 𝐾 is Hilbertian, so is 𝐾ins . On the other hand, 𝐾sep is not Hilbertian. For example, if char(𝐾) ≠ 2 (resp. char(𝐾) ≠ 3), then 𝑋 2 − 𝑇 (resp. 𝑋 3 − 𝑇) is irreducible in 𝐾sep [𝑇, 𝑋], but there is no 𝑎 ∈ 𝐾sep such that 𝑋 2 − 𝑎 (resp. 𝑋 3 − 𝑎) is irreducible in 𝐾sep [𝑋]. Lemma 13.3.5 Let 𝐾 be a field of positive characteristic 𝑝, 𝐿 an extension of 𝐾 𝑟 with 𝐿 𝑝 ⊆ 𝐾 for some 𝑟 ≥ 0, and 𝑓1 , . . . , 𝑓𝑚 irreducible polynomials in 𝐿 [𝑇, 𝑋]. Suppose that deg𝑋 𝑓𝑖 ≥ 1 and 𝑓𝑖 is separable in 𝑇. Suppose in addition that each Hilbert subset of 𝐾 is nonempty. Then, there exists a 𝛾 ∈ 𝐿 such that 𝐻 𝐿 ( 𝑓𝑖 (𝑇 +𝛾, 𝑋)) contains a Hilbert subset of 𝐾, 𝑖 = 1, . . . , 𝑚. Proof. Let 𝑖 be between 1 and 𝑚. At least one of the coefficients 𝑐 𝑖 (𝑇) of 𝑓𝑖 (𝑇, 𝑋) has nonzero derivative 𝑐 𝑖′ (𝑇). Choose a nonzero coefficient 𝛾𝑖 of 𝑐 𝑖′ (𝑇). Let 𝑒 be the least nonnegative integer with 𝐿 𝑝 ⊆ 𝐾. Then, 𝐿 𝑝 ̸ ⊆ 𝐾, so 𝐿 𝑝 ̸ ⊆ 𝛾𝑖− 𝑝 𝐾, Ð𝑚 − 𝑝 𝑒−1 𝑒−1 𝑖 = 1, . . . , 𝑚. Since 𝐾 is infinite, 𝐿 𝑝 ̸ ⊆ 𝑖=1 𝛾𝑖 𝐾. Hence, there exists a 𝜆 ∈ 𝐿 𝑒−1 𝑒−1 𝑝 ′ independent of 𝑖 with (𝛾𝑖 𝜆) ∉ 𝐾. This implies (𝑐 𝑖 (𝑇)𝜆) 𝑝 ∉ 𝐾 [𝑇]. The Taylor expansion 𝑐 𝑖 (𝑇 + 𝜆𝑍) = 𝑐 𝑖 (𝑇) + 𝑐 𝑖′ (𝑇) · 𝜆𝑍 + · · · shows that 𝑒−1 𝑒 𝑐 𝑖 (𝑇 + 𝜆𝑍) 𝑝 ∉ 𝐾 [𝑇, 𝑍]. Hence, ℎ𝑖 (𝑇, 𝑍) = 𝑐 𝑖 (𝑇 + 𝜆𝑍) 𝑝 is not a 𝑝th power of a 𝑝 polynomial belonging to 𝐾 [𝑇, 𝑍]. Therefore, 𝑋 − ℎ𝑖 (𝑇, 𝑍) is irreducible over 𝐾. By assumption, there exists a 𝑏 ∈ 𝐾, independent of 𝑖, with 𝑋 𝑝 −ℎ𝑖 (𝑇, 𝑏) irreducible over 𝑒 𝐾. Thus, 𝑒 is the least nonnegative integer for which 𝑐 𝑖 (𝑇 + 𝜆𝑏) 𝑝 is in 𝐾 [𝑇]. Hence, 𝑒 𝑒 is the least nonnegative integer for which 𝑔𝑖 (𝑇, 𝑋) := 𝑓𝑖 (𝑇 + 𝜆𝑏, 𝑋) 𝑝 ∈ 𝐾 [𝑇, 𝑋]. By Lemma 13.3.1(d), 𝑔𝑖 (𝑇, 𝑋) is irreducible in 𝐾 [𝑇, 𝑋]. If for 𝑎 ∈ 𝐾, the polynomial 𝑔𝑖 (𝑎, 𝑋) is irreducible in 𝐾 [𝑋], then from Lemma 13.3.1(b), 𝑓𝑖 (𝑎 + 𝜆𝑏, 𝑋) is irreducible in 𝐿 [𝑋]. Thus, 𝐻𝐾 (𝑔𝑖 ) ⊆ 𝐻 𝐿 ( 𝑓𝑖 (𝑇 + 𝜆𝑏, 𝑋)). □ 𝑒

𝑒−1

𝑒−1

𝑒−1

Proposition 13.3.6 Let 𝐾 be a field with all Hilbert subsets nonempty and 𝐿 an algebraic extension of 𝐾. Then, in each of the following cases all Hilbert sets of 𝐿 are nonempty. 𝑟 (a) 𝑝 := char(𝐾) > 0 and 𝐿 𝑝 ⊆ 𝐾 for some 𝑟 ≥ 0. (b) 𝐿 is a finite extension of 𝐾.

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13 Hilbertian Fields

Proof of (a). Consider a Hilbert subset 𝐻 𝐿 ( 𝑓1 , . . . , 𝑓𝑛 ) of 𝐿 with 𝑓𝑖 ∈ 𝐿 [𝑇, 𝑋] irreducible and deg𝑋 𝑓𝑖 ≥ 2, 𝑖 = 1, . . . , 𝑛. Reorder the 𝑓𝑖 ’s, if necessary, to assume the following: For 𝑖 = 1, . . . , 𝑚, 𝑓𝑖 is separable in 𝑋 or inseparable in both 𝑇 and 𝑋. For 𝑖 = 𝑚 + 1, . . . , 𝑛, 𝑓𝑖 is separable in 𝑇. Lemma 13.3.5 gives 𝛾 in 𝐿 such that 𝐻 𝐿 ( 𝑓𝑖 (𝑇 + 𝛾, 𝑋)) contains a Hilbert subset 𝐻𝑖 of 𝐾. For each 𝑖 between 1 and 𝑚, the polynomial 𝑓𝑖 (𝑇 + 𝛾, 𝑋) is either separable in 𝑋 or inseparable in both 𝑇 and 𝑋. By Lemma 13.3.2, 𝐻 𝐿 ( 𝑓𝑖 (𝑇 + 𝛾, 𝑋)) contains a Hilbert subset 𝐻𝑖 of 𝐾, so 𝐻 := 𝐻1 ∩ · · · ∩ 𝐻𝑚 is a Hilbert subset of 𝐾 which is contained in 𝐻 𝐿 ( 𝑓1 , . . . , 𝑓𝑛 ). By Lemma 13.1.5, every Hilbert set of 𝐿 is nonempty. Proof of (b). Combine (a) with Lemma 13.2.2.

□

13.4 Imperfect Fields Suppose that char(𝐾) = 𝑝 ≠ 0 and every Hilbert set of 𝐾 is nonempty. Then, there exists an 𝑎 ∈ 𝐾 with 𝑋 𝑝 − 𝑎 irreducible. Thus, 𝐾 is imperfect. We show that, conversely, every Hilbert set of an imperfect Hilbertian field is nonempty. Lemma 13.4.1 Let 𝑝 := char(𝐾) > 0 and let 𝑓 ∈ 𝐾 [𝑋] ∖ 𝐾 𝑝 [𝑋] be irreducible 𝑚 and monic. Then, 𝑓 (𝑋 𝑝 ) is irreducible in 𝐾 [𝑋] for each 𝑚 ≥ 0. Proof. By induction it suffices to prove the lemma for 𝑚 = 1. If 𝑥 is a root of 𝑓 (𝑋 𝑝 ), then 𝑦 := 𝑥 𝑝 is a root of 𝑓 (𝑋). We show that [𝐾 (𝑥) : 𝐾 (𝑦)] = 𝑝. If not, then 𝐾 (𝑥) = 𝐾 (𝑦) and irr(𝑥, 𝐾) = 𝑋 𝑛 + 𝑏 𝑛−1 𝑋 𝑛−1 + · · · + 𝑏 0 , where 𝑛 = [𝐾 (𝑥) : 𝐾] = [𝐾 (𝑦) : 𝐾] = deg( 𝑓 ). Therefore, 𝑦 is a root of 𝑔(𝑋) = 𝑝 𝑋 𝑛 + 𝑏 𝑛−1 𝑋 𝑛−1 + · · · + 𝑏 0𝑝 . Hence, 𝑓 = 𝑔 and 𝑓 ∈ 𝐾 𝑝 [𝑋], contrary to our hypotheses. It follows that [𝐾 (𝑥) : 𝐾] = 𝑝 · deg( 𝑓 ) = deg( 𝑓 (𝑋 𝑝 )). Consequently, 𝑓 (𝑋 𝑝 ) is irreducible. □ Lemma 13.4.2 Let 𝐾 be an imperfect field of characteristic 𝑝. Suppose that ℎ ∈ 𝐾 [𝑇] is not a 𝑝th power. Then: (a) The additive group 𝐾 has infinitely many congruence classes modulo 𝐾 𝑝 . (b) All 𝑎 ∈ 𝐾 with ℎ(𝑇 + 𝑎) ∈ 𝐾 𝑝 [𝑇] are congruent modulo 𝐾 𝑝 . (c) For each 𝑏 ∈ 𝐾 with ℎ(𝑇 + 𝑏) ∉ 𝐾 𝑝 [𝑇] only finitely many 𝑐 ∈ 𝐾 satisfy ℎ(𝑐 𝑝 + 𝑏) ∈ 𝐾 𝑝 . Proof of (a). Since 𝐾 is imperfect, 𝐾/𝐾 𝑝 is a nonzero vector space over the infinite field 𝐾 𝑝 . Hence, 𝐾/𝐾 𝑝 is infinite. Proof of (b). Let 𝑎, 𝑏 ∈ 𝐾 with ℎ(𝑇 + 𝑎), ℎ(𝑇 + 𝑏) ∈ 𝐾 𝑝 [𝑇]. Since ℎ(𝑇 + 𝑎) is not a 𝑝th power, ℎ(𝑇 + 𝑎) = 𝑔(𝑇 𝑝 ) + 𝑓 (𝑇) where 𝑔, 𝑓 ∈ 𝐾 𝑝 [𝑇] but 𝑝 ∤ deg( 𝑓 ). Then, with 𝑐 = 𝑏 − 𝑎, we have ℎ(𝑇 + 𝑏) = 𝑔(𝑇 𝑝 + 𝑐 𝑝 ) + 𝑓 (𝑇 + 𝑐). Let 𝑓 (𝑇) = 𝑝 𝑚 𝑝 𝑑𝑚 𝑇 + 𝑑 𝑚−1 𝑇 𝑚−1 + · · · + 𝑑0𝑝 with 𝑑 𝑚 ≠ 0. Then, 𝑝 ∤ 𝑚 and the coefficient of 𝑇 𝑚−1 𝑝 𝑝 in 𝑓 (𝑇 + 𝑐) is 𝑚𝑐𝑑 𝑚 . Since ℎ(𝑇 + 𝑏) and 𝑔(𝑇 𝑝 + 𝑐 𝑝 ) are in 𝐾 𝑝 [𝑇], this + 𝑑 𝑚−1 𝑝 element belongs to 𝐾 . Finally, 𝑚 𝑝 = 𝑚 ≠ 0. Therefore, 𝑐 is a 𝑝th power.

13.4 Imperfect Fields

239

Proof of (c). Write ℎ(𝑇 + 𝑏) ∉ 𝐾 𝑝 [𝑇] in the form ℎ(𝑇 + 𝑏) = 𝑓1 (𝑇) + 𝑓2 (𝑇)𝑢 2 + · · · + 𝑓𝑟 (𝑇)𝑢𝑟 , where 𝑟 ≥ 2, 𝑢 2 , . . . , 𝑢𝑟 ∈ 𝐾, 1, 𝑢 2 , . . . , 𝑢𝑟 linearly independent over 𝐾 𝑝 , 𝑓1 , . . . , 𝑓𝑟 ∈ 𝐾 𝑝 [𝑇], and 𝑓2 , . . . , 𝑓𝑟 ≠ 0. If 𝑐 ∈ 𝐾 and ℎ(𝑐 𝑝 + 𝑏) ∈ 𝐾 𝑝 , then 𝑓1 (𝑐 𝑝 ) + 𝑓2 (𝑐 𝑝 )𝑢 2 + · · · + 𝑓𝑟 (𝑐 𝑝 )𝑢𝑟 ∈ 𝐾 𝑝 . Hence, 𝑓1 (𝑐 𝑝 ) = 𝑓2 (𝑐 𝑝 ) = · · · = 𝑓𝑟 (𝑐 𝑝 ) = 0. The number of such 𝑐’s is finite. □ Proposition 13.4.3 (Uchida) Let 𝐾 be a field satisfying these conditions: (a) 𝐾 is Hilbertian, i.e. every separable Hilbert subset of 𝐾 is nonempty. (b) If char(𝐾) > 0, then 𝐾 is imperfect. Then, every Hilbert set of 𝐾 is nonempty. Proof. By Lemma 13.1.5, it suffices to prove that each Hilbert set of the form 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ) with 𝑓1 , . . . , 𝑓𝑚 ∈ 𝐾 [𝑇, 𝑋] irreducible, monic, and of degree at least 2 in 𝑋, is nonempty. If char(𝐾) = 0, this statement follows from (a). So, we assume that 𝑝 := char(𝐾) > 0, 𝑓1 , . . . , 𝑓𝑙 are not separable in 𝑋, and 𝑓𝑙+1 , . . . , 𝑓𝑚 are separable in 𝑋. Application of Lemma 13.4.2: For each 𝑖, 1 ≤ 𝑖 ≤ 𝑙, 𝑓𝑖 is inseparable in 𝑋, so there exists an irreducible polynomial 𝑔𝑖 ∈ 𝐾 [𝑇, 𝑋] which is separable and monic in 𝑋, and there exists a power 𝑞 𝑖 of 𝑝 different from 1 with 𝑓𝑖 (𝑇, 𝑋) = 𝑔𝑖 (𝑇, 𝑋 𝑞𝑖 ). In particular, 𝑔𝑖 (𝑇, 𝑋) has a coefficient 𝑔𝑖, 𝑗 (𝑖) ∈ 𝐾 [𝑇] which is not a 𝑝th power. Choose 𝛼𝑖 ∈ 𝐾 with 𝑔𝑖, 𝑗 (𝑖) (𝑇 + 𝛼𝑖 ) ∈ 𝐾 𝑝 [𝑇] if there exists any, otherwise let Ð 𝛼𝑖 = 0. By Lemma 13.4.2(a), there exists a 𝛽 ∈ 𝐾 ∖ 𝑙𝑖=1 (𝛼𝑖 + 𝐾 𝑝 ). By Lemma Ð 13.4.2(c), the set 𝐶 := 𝑙𝑖=1 {𝛾 ∈ 13.4.2(b), 𝑔𝑖, 𝑗 (𝑖) (𝑇 +𝛽) ∉ 𝐾 𝑝 [𝑇]. Hence, by Lemma Î 𝐾 | 𝑔𝑖, 𝑗 (𝑖) (𝛾 𝑝 + 𝛽) ∈ 𝐾 𝑝 } is finite. Let 𝑐(𝑇) = 𝛾 ∈𝐶 (𝑇 − 𝛾). For each 𝑖 between 𝑙 + 1 and 𝑚, we set 𝑔𝑖 = 𝑓𝑖 . Thus, for each 𝑖 between 1 and 𝑚 the polynomial 𝑔𝑖 is irreducible in 𝐾 [𝑇, 𝑋], and is separable and monic in 𝑋. Claim: For each 𝑖 between 1 and 𝑚 the polynomial 𝑔𝑖 (𝑇 𝑝 + 𝛽, 𝑋) is irreducible in 𝐾 [𝑇, 𝑋]. Indeed, consider the irreducible polynomial ℎ(𝑇, 𝑋) := 𝑔𝑖 (𝑇 + 𝛽, 𝑋) of 𝐾 [𝑇, 𝑋]. Since 𝑔𝑖 (𝑇, 𝑋) is separable and monic in 𝑋, so is ℎ(𝑇, 𝑋). We write ℎ(𝑇, 𝑋) = ℎ0 (𝑋) + ℎ1 (𝑋)𝑇 + · · · + ℎ𝑟 (𝑋)𝑇 𝑟 with ℎ0 , . . . , ℎ𝑟 ∈ 𝐾 [𝑋] and ℎ𝑟 ≠ 0. If 𝑟 = 0, then ℎ(𝑇 𝑝 , 𝑋) = ℎ(𝑇, 𝑋) is irreducible. Hence, we may assume that 𝑟 ≥ 1. In this case (13.6a) ℎ 𝑗 ∉ 𝐾 [𝑋] 𝑝 for at least one 𝑗 between 0 and 𝑟, and (13.6b) ℎ0 , ℎ1 , . . . , ℎ𝑟 have no common irreducible divisor in 𝐾 [𝑋]. Assume toward contradiction that ℎ𝑟 1(𝑋) ℎ(𝑇, 𝑋) ∈ 𝐾 (𝑋) 𝑝 [𝑇]. Then, for each 𝑗 between 0 and 𝑟 − 1 there exist relatively prime polynomials 𝑏 𝑗 , 𝑐 𝑗 ∈ 𝐾 [𝑋] such 𝑝

that

ℎ𝑗 ℎ𝑟

=

𝑏𝑗 𝑝 𝑐𝑗

, so ℎ 𝑗 𝑐 𝑝𝑗 = ℎ𝑟 𝑏 𝑝𝑗 .

𝑏 𝑝𝑗

𝑐 𝑝𝑗

𝑐 𝑝𝑗 |ℎ𝑟 .

(13.7)

are also relatively prime, Let 𝑑 = lcm(𝑐 0 , . . . , 𝑐𝑟−1 ). Then,
𝑝 𝑝 𝑝 = lcm(𝑐 0 , . . . , 𝑐𝑟−1 ) divides ℎ𝑟 . By (13.7), ℎ 𝑗 = 𝑑ℎ𝑟𝑝 𝑐𝑑𝑗 𝑏 𝑝𝑗 . In addition ℎ𝑟 = ℎ𝑟 𝑝 ℎ𝑟 × 𝑑 𝑝 𝑑 . Hence, 𝜂 = 𝑑 𝑝 is a common factor of ℎ0 , . . . , ℎ𝑟 . By (13.6b), 𝜂 ∈ 𝐾 . Thus,

Since 𝑑𝑝

and

240

13 Hilbertian Fields

 ℎ 𝑗 (𝑋) = 𝜂 ·

𝑑 (𝑋) 𝑐 𝑗 (𝑋)

𝑝 𝑏 𝑗 (𝑋) 𝑝 for 𝑗 = 0, . . . , 𝑟 − 1 and

(13.8)

ℎ𝑟 (𝑋) = 𝜂 · 𝑑 (𝑋) 𝑝 . Now observe that for each 1 ≤ 𝑗 ≤ 𝑟, 𝑇 𝑗 divides the leading coefficient of ℎ 𝑗 (𝑋)𝑇 𝑗 as a polynomial in 𝑋. Since ℎ(𝑇, 𝑋) is monic in 𝑋, this implies that the leading coefficient of ℎ(𝑇, 𝑋), as a polynomial in 𝑋, is the leading coefficient of ℎ0 (𝑋). Thus, that coefficient is 1. Hence, by (13.8) with 𝑗 = 0, 𝜂 ∈ (𝐾 × ) 𝑝 . It follows from (13.8) that ℎ 𝑗 ∈ 𝐾 [𝑋] 𝑝 for each 𝑗 between 0 and 𝑟. This contradicts (13.6a). Application of Lemma 13.4.1:

It follows from the latter contradiction that

1 ℎ(𝑇, 𝑋) ∈ 𝐾 (𝑋) [𝑇] ∖ 𝐾 (𝑋) 𝑝 [𝑇]. ℎ𝑟 (𝑋) In addition, the monic polynomial ℎ𝑟 1(𝑋) ℎ(𝑇, 𝑋) is irreducible in 𝐾 (𝑋) [𝑇]. By Lemma 13.4.1, applied to the field 𝐾 (𝑋) rather than to 𝐾 and to the variable 𝑇 rather than to 𝑋, the polynomial ℎ𝑟 1(𝑋) ℎ(𝑇 𝑝 , 𝑋) is irreducible in 𝐾 (𝑋) [𝑇]. Since the coefficients ℎ0 , ℎ1 , . . . , ℎ𝑟 have no irreducible common divisor in 𝐾 [𝑋], we conclude from [Lan97, p. 182, Thm. 2.3] that ℎ(𝑇 𝑝 , 𝑋) is irreducible in 𝐾 [𝑇, 𝑋], as claimed. End of proof: By (a), there exists an 𝛼 ∈ 𝐾 such that 𝑐(𝛼) ≠ 0 and 𝑔𝑖 (𝛼 𝑝 + 𝛽, 𝑋) is separable and irreducible in 𝐾 [𝑋], 𝑖 = 1, . . . , 𝑚. Thus, 𝑓𝑖 (𝛼 𝑝 + 𝛽, 𝑋) is irreducible for 𝑖 = 𝑙 + 1, . . . , 𝑚. Now consider 𝑖 between 1 and 𝑙. Since 𝑐(𝛼) ≠ 0, we have 𝛼 ∉ 𝐶, so 𝑔𝑖, 𝑗 (𝑖) (𝛼 𝑝 + 𝛽) ∉ 𝐾 𝑝 . Hence, 𝑔𝑖 (𝛼 𝑝 + 𝛽, 𝑋) ∉ 𝐾 𝑝 [𝑋]. By Lemma 13.4.1, 𝑓𝑖 (𝛼 𝑝 + 𝛽, 𝑋) = 𝑔𝑖 (𝛼 𝑝 + 𝛽, 𝑋 𝑞𝑖 ) is irreducible over 𝐾. Consequently, 𝛼 𝑝 + 𝛽 ∈ 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ), as desired. □

Exercises 1. Let 𝐾 be a Hilbertian field and let 𝑓 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑛 , 𝑋] be a separable polynomial in 𝑋 such that for every t ∈ 𝐾 𝑛 there exists an 𝑥 ∈ 𝐾 with 𝑓 (t, 𝑥) = 0. Prove that 𝑓 (T, 𝑋) has a factor of degree 1 in 𝑋. 2. LetÐ {𝐾 𝛼 | 𝛼 < 𝑚} be a transfinite ascending tower of Hilbertian fields such that 𝐾 𝛾 = 𝛼 0, then 𝐾 is imperfect. Hence, by Uchida (Proposition 13.4.3), every Hilbert set of 𝐾 is nonempty and 𝐾 is Hilbertian. Lemmas 13.1.6 and 14.1.2 reduce the proof of the first statement to the following one: For an absolutely irreducible polynomial 𝑓 ∈ 𝐾 [𝑋, 𝑌 ], monic and separable in 𝑌 , with deg𝑌 ( 𝑓 ) ≥ 2, there exists a nonempty Zariski 𝐾0 -open subset 𝑈 of A2 such that 𝑓 (𝑎 + 𝑏𝑡, 𝑌 ) has no zeros in 𝐾 for each (𝑎, 𝑏) ∈ 𝑈 (𝐾0 ). If necessary, multiply 𝑓 by a suitable element of 𝐾 and make a linear change in the variable 𝑌 to assume that 𝑓 (𝑋, 𝑌 ) = 𝑔(𝑡, 𝑋, 𝑌 ) ∈ 𝐾0 [𝑡, 𝑋, 𝑌 ] is an absolutely irre𝜕𝑔 ducible polynomial and 𝜕𝑌 ≠ 0. Then, 𝑔(𝑡, 𝑍0 + 𝑍1 𝑡, 𝑌 ) is an irreducible polynomial in 𝐿˜ [𝑡, 𝑌 ], where 𝐿 = 𝐾0 (𝑍0 , 𝑍1 ) (Proposition 11.5.4). By Proposition 10.4.3, there exists a nonzero polynomial ℎ ∈ 𝐾0 [𝑍0 , 𝑍1 ] such that 𝑔(𝑡, 𝑎 + 𝑏𝑡, 𝑌 ) is irreducible in 𝐾˜ 0 [𝑡, 𝑌 ] for every 𝑎, 𝑏 ∈ 𝐾0 such that ℎ(𝑎, 𝑏) ≠ 0. In particular, 𝑓 (𝑎 + 𝑏𝑡, 𝑌 ) has no zeros in 𝐾 if ℎ(𝑎, 𝑏) ≠ 0. □ Proposition 14.2.1 implies an analog of Corollary 13.1.5 for separable Hilbert sets. Lemma 14.2.2 Let 𝑅 be an integral domain with quotient field 𝐾. Suppose that each separable Hilbert subset of 𝐾 of the form 𝐻𝐾 ( 𝑓 ) with irreducible 𝑓 ∈ 𝐾 [𝑇, 𝑋], separable, monic, and of degree at least 2 in 𝑋, contains an element of 𝑅. Then, every separable Hilbert subset 𝐻 of 𝐾 𝑟 contains a point in 𝑅𝑟 . Proof. By Lemma 13.1.6, it suffices to consider a separable irreducible polynomial 𝑓 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑟 , 𝑋], monic and of degree at least 2 in 𝑋, and to prove that 𝑅𝑟 ∩ 𝐻𝐾 ( 𝑓 ) ≠ ∅. The case 𝑟 = 1 is covered by the assumption of the lemma. Suppose that 𝑟 ≥ 2 and the statement holds for 𝑟 − 1. The assumption of the lemma implies that 𝐾 is infinite. Let 𝐾 ′ = 𝐾 (𝑇1 , . . . , 𝑇𝑟−2 ), 𝑡 = 𝑇𝑟−1 , and regard 𝑓 as a polynomial in 𝐾 ′ (𝑡) [𝑇𝑟 , 𝑋]. By Proposition 14.2.1, there exists a nonempty Zariski 𝐾 ′-open subset 𝑈 of A2 such that {𝑎 + 𝑏𝑡 | (𝑎, 𝑏) ∈ 𝑈 (𝐾 ′)} ⊆ 𝐻𝐾 ′ (𝑡) ( 𝑓 ). Since 𝐾 is infinite, so is 𝑅. Hence, we may choose 𝑎, 𝑏 ∈ 𝑅 such that 𝑎 + 𝑏𝑡 ∈ 𝐻𝐾 ′ (𝑡) ( 𝑓 ). Thus, 𝑓 (𝑇1 , . . . , 𝑇𝑟−1 , 𝑎+𝑏𝑇𝑟−1 , 𝑋) is irreducible and separable in 𝐾 (𝑇1 , . . . , 𝑇𝑟−1 ) [𝑋]. The induction hypothesis gives 𝑎 1 , . . . , 𝑎𝑟−1 ∈ 𝑅 such that 𝑓 (𝑎 1 , . . . , 𝑎𝑟−1 , 𝑎 + 𝑏𝑎𝑟−1 , 𝑋) is irreducible and separable in 𝐾 [𝑋]. Let 𝑎𝑟 = 𝑎 + 𝑏𝑎𝑟−1 . Then, 𝑎𝑟 ∈ 𝑅 and 𝑓 (𝑎 1 , . . . , 𝑎𝑟 , 𝑋) is irreducible in 𝐾 [𝑋]. □ The following result is a consequence of Lemma 14.1.2 and Lemma 14.2.2. Lemma 14.2.3 Suppose that 𝐻𝐾′ (ℎ1 , . . . , ℎ 𝑚 ) ≠ ∅ for all absolutely irreducible polynomials ℎ1 , . . . , ℎ 𝑚 ∈ 𝐾 [𝑇, 𝑋], monic and separable in 𝑋 with deg𝑋 (ℎ𝑖 ) ≥ 2. Then, 𝐾 is Hilbertian. An immediate corollary of Lemma 14.1.4 and Lemma 14.2.2 is the following criterion for Hilbertianity.

14.3 Global Fields

249

Corollary 14.2.4 (Haran’s criterion for Hilbertianity) Let 𝐾 be an infinite field. Suppose that for every absolutely irreducible polynomial 𝑔 ∈ 𝐾 [𝑇, 𝑋] and every finite Galois extension 𝐿 of 𝐾 such that 𝑔 is monic of degree ≥ 2 as a polynomial in 𝑋, and Galois over 𝐿(𝑇), there exists an 𝑎 ∈ 𝐾 such that 𝑔(𝑎, 𝑋) is irreducible over 𝐿. Then, 𝐾 is Hilbertian. Remark 14.2.5 Following a question posed in [DeH99, p. 284], Bary-Soroker gives the following criterion for a field 𝐾 to be Hilbertian: For each absolutely irreducible polynomial 𝑓 ∈ 𝐾 [𝑇, 𝑋], separable in 𝑋, and every nonzero polynomial 𝑝 ∈ 𝐾 [𝑇], there exists an 𝑎 ∈ 𝐾 with 𝑝(𝑎) ≠ 0 and 𝑓 (𝑎, 𝑋) irreducible over 𝐾 [BaS08b, Theorem 1].

14.3 Global Fields For 𝐾 a global field it is easy to choose the ℎ𝑖 ’s in Lemma 14.1.2 with coefficients in the ring of integers 𝑂 𝐾 . If, in particular, 𝐾 is a number field, 𝑟 = 1, and each of the curves defined by ℎ𝑖 = 0 has positive genus, a celebrated theorem of Siegel implies that each of the ℎ𝑖 ’s has only finitely many zeros (𝑎, 𝑏) ∈ 𝑂 2𝐾 ([Lan62, p. 121] or [RoR75]). In this case 𝐻𝐾′ (ℎ1 , . . . , ℎ 𝑚 ) is clearly infinite. If, however, the curve ℎ𝑖 = 0 is of genus zero, we may use Riemann–Hurwitz to replace ℎ𝑖 (𝑇, 𝑋) by 𝑔𝑖 (𝑇, 𝑋) = ℎ𝑖 (𝑚(𝑇), 𝑋) for some 𝑚(𝑇) ∈ 𝑂 𝐾 [𝑇], so that 𝑔𝑖 (𝑇, 𝑋) = 0 has positive genus. Thus, Siegel’s theorem gives Hilbert’s theorem effortlessly. Of course, Siegel’s theorem, a deep result in arithmetic, applies only to number fields (more generally, to fields which are finitely generated over Q [Lan62, p. 127, Thm. 4]). Besides, its power masks subtle connections between the irreducibility theorem and other arithmetic results. Our approach to the Hilbert irreducibility theorem for global fields is based on the Chebotarev density theorem for function fields over finite fields. More accurately, we use a special case of Proposition 7.4.8. As a bonus we will prove that each Hilbert set over Q contains arithmetic progressions. We start the proof with a weak consequence of Bauer’s theorem (Exercise 5(b) of Chapter 7 or [Neu99, p. 548, Prop. 13.9]). In keeping with our elementary treatment we use only Euclid’s argument for proving the infinitude of primes: Lemma 14.3.1 Let 𝐿/𝐾 be a finite separable extension of global fields. Then, there exist infinitely many primes 𝔭 of 𝐾 such that 𝐿¯ 𝔓 = 𝐾¯ 𝔭 for every prime 𝔓 of 𝐿 lying over 𝔭. Proof. Assume, without loss, that 𝐾 = Q if char(𝐾) = 0 and 𝐾 = F 𝑝 (𝑡) if char(𝐾) = 𝑝. In particular, 𝑂 𝐾 is a principal ideal domain with only finitely many units. Replace 𝐿 by the Galois hull of 𝐿/𝐾 to assume that 𝐿/𝐾 is Galois. Consider a primitive element 𝑧 ∈ 𝑂 𝐿 for the extension 𝐿/𝐾. Then, 𝑓 (𝑋) = irr(𝑧, 𝐾) = 𝑋 𝑛 + 𝑐 𝑛−1 𝑋 𝑛−1 + · · · + 𝑐 0 with 𝑐 0 , . . . , 𝑐 𝑛−1 ∈ 𝑂 𝐾 and 𝑐 0 ≠ 0. Moreover, by (7.1), 𝑑 = disc( 𝑓 ) ∈ 𝑂 𝐾 . Suppose that 𝔭1 , . . . , 𝔭𝑘 are prime ideals of 𝑂 𝐾 satisfying the conclusion of the lemma. For each 𝑖 between 1 and 𝑘 choose a nonzero

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𝜋𝑖 ∈ 𝔭𝑖 and let 𝜋 = 𝑑 · 𝜋1 · · · 𝜋 𝑘 . Since 𝑐 0 ≠ 0 and 𝑂 𝐾 has only finitely many units, there exists a positive integer 𝑚 such that 𝑚 𝑛−1 𝑚𝑛 𝑐−1 + 𝑐 𝑛−1 𝑐 0𝑛−2 𝜋 𝑚(𝑛−1) + · · · + 1 0 𝑓 (𝑐 0 𝜋 ) = 𝑐 0 𝜋

is a nonunit of 𝑂 𝐾 . Therefore, this element has a prime divisor 𝔭 = 𝔭𝑘+1 different from 𝔭1 , . . . , 𝔭𝑘 and relatively prime to 𝑑. Let 𝔓 be a prime ideal of 𝑂 𝐿 lying over 𝔭𝑘+1 . Denote the reduction modulo 𝔓 by a bar. Then, 𝑓¯(𝑋) has a root in 𝐾¯ 𝔭 , namely 𝑐 0 𝜋 𝑚 modulo 𝔭, which we may assume to be 𝑧¯. By Remark 7.1.7, 𝐿¯ 𝔓 = 𝐾¯ 𝔭 ( 𝑧¯) = 𝐾¯ 𝔭 . □ Lemma Ð 14.3.2 For 𝐻 a proper subgroup of a finite group 𝐺, the set 𝐺 ∖ 𝜎 ∈𝐺 𝜎 −1 𝐻𝜎 is nonempty. For 𝑓 ∈ 𝐾 [𝑋] an irreducible separable polynomial with roots 𝑥1 , . . . , 𝑥 𝑛 and splitting field 𝑁 = 𝐾 (𝑥1 , . . . , 𝑥 𝑛 ), there exists a 𝜎 ∈ Gal(𝑁/𝐾) such that 𝜎𝑥𝑖 ≠ 𝑥𝑖 , for each 𝑖 = 1, . . . , 𝑛. Proof. Let 𝐻1 , . . . , 𝐻𝑚 be the distinct subgroups of 𝐺 conjugate to 𝐻. Since 𝐻 ≠ 𝐺, we may assume that 𝐻 is not normal in 𝐺, so 𝑚 ≥ 2. Then, 𝑚 is equal to the index of the normalizer of 𝐻. Thus, 𝑚 ≤ (𝐺 : 𝐻). The intersection 𝐻1 ∩ · · · ∩ 𝐻𝑚 contains Ð𝑚 Í𝑚 1, so | 𝑖=1 |𝐻𝑖 | = 𝑚 · |𝐻| ≤ |𝐺 |. This proves the first statement of the 𝐻𝑖 | < 𝑖=1 lemma. With Gal(𝑁/𝐾) = 𝐺 and Gal(𝑁/𝐾 (𝑥 1 )) = 𝐻 the conjugates of 𝐻 are Gal(𝑁/𝐾 (𝑥1 )), . . .Ð , Gal(𝑁/𝐾 (𝑥 𝑛 )). The last statement of the lemma follows by □ choosing 𝜎 ∈ 𝐺 ∖ 𝜏 ∈𝐺 𝜏 −1 𝐻𝜏. Lemma 14.3.3 Let 𝑞 be a prime power, 𝐾¯ = F𝑞 , 𝑡 an indeterminate, and 𝐸¯ = 𝐾¯ (𝑡). ¯ 𝑧) = 0 with 𝑔¯ ∈ 𝐾¯ [𝑇, 𝑍] an Consider a Galois extension 𝐹¯ = 𝐾¯ (𝑡, 𝑧), where 𝑔(𝑡, ¯ Let 𝑑 = deg( 𝑔), ¯ irreducible polynomial. Suppose that 𝐾¯ is algebraically closed in 𝐹. ¯ Then, the number 𝑁 of ¯ 𝐸). 𝑚 = deg 𝑍 ( 𝑔), ¯ and C be a conjugacy class in Gal( 𝐹/ ¯ with Artin symbol equal to C satisfies ¯ 𝐾¯ of degree 1, unramified in 𝐹, primes of 𝐸/ 𝑁 − |C| 𝑞 < 10𝑑 2 |C| √𝑞. (14.4) 𝑚 ¯ Since 𝐾¯ is algebraically closed in 𝐹, ¯ 𝑔¯ is absolutely Proof. Write 𝑔𝐹¯ = genus( 𝐹). irreducible (Corollary 11.2.2). Hence, 𝑔(𝑇, ¯ 𝑍) = 0 defines a curve of degree 𝑑 over ¯ Recalling that 𝐸¯ has genus 0, Proposition 7.4.8, with 𝑘 = 1, gives the inequality 𝐾. h i 𝑁 − |C| 𝑞 < 2|C| (𝑚 + 𝑔 ¯ )𝑞 12 + 𝑚𝑞 14 + 𝑔 ¯ + 𝑚 . 𝐹 𝐹 𝑚 𝑚 Combining this with the inequality 𝑔𝐹¯ ≤ 12 (𝑑−1) (𝑑−2) for the genus of 𝐹¯ (Corollary 6.3.5) we obtain (14.4). □ An absolute value of a field 𝐾 is a map | |: 𝐾 → R which satisfies the following conditions for all 𝑥, 𝑦 ∈ 𝐾: (14.5a) |𝑥| ≥ 0 and |𝑥| = 0 if and only if 𝑥 = 0. (14.5b) |𝑥𝑦| = |𝑥||𝑦|. (14.5c) |𝑥 + 𝑦| ≤ |𝑥| + |𝑦|.

14.3 Global Fields

251

(14.5d) There is an 𝑎 ∈ 𝐾 × with |𝑎| ≠ 1. If, instead of (14.5c), | | satisfies the stronger condition (14.6) |𝑥 + 𝑦| ≤ max(|𝑥|, |𝑦|), it is ultrametric (or non-archimedean), otherwise, it is metric (or archimedean). An absolute value | | ′ is equivalent to | | if there exists a 𝑐 > 0 such that |𝑥| ′ = |𝑥| 𝑐 for all 𝑥 ∈ 𝐾. Remark 14.3.4 If | | is ultrametric, then 𝑂 | | = {𝑥 ∈ 𝐾 | |𝑥| ≤ 1} is a valuation ring of 𝐾. In particular, |𝑛| ≤ 1 for each positive integer 𝑛. Conversely, suppose that |𝑛| ≤ 1 for each of 𝐾 with |𝑧| ≤ 1. integer 𝑛. Let 𝑧 be an element 1/𝑘 Í 𝑘 positive 𝑘
𝑖 𝑘 |1 + 𝑧| + 1) . Letting 𝑘 go to ≤ 𝑧 Then, |1 + 𝑧| 𝑘 = 𝑖=0 + Hence, 1. (𝑘 ≤ 𝑖 infinity, we find that |1 + 𝑧| ≤ 1. Next let 𝑥, 𝑦 ∈ 𝐾 with |𝑥| ≤ |𝑦| and 𝑦 ≠ 0. Then, |𝑥 + 𝑦| = |1 + 𝑦𝑥 ||𝑦| ≤ |𝑦| = max(|𝑥|, |𝑦|). Consequently, | | is ultrametric. In particular, suppose that char(𝐾) = 𝑝 > 0. Then, 𝑛 𝑝−1 = 1 for each positive integer 𝑛 not divisible by 𝑝. Hence, |𝑛| = 1. Therefore, | | is ultrametric. If 𝐾 is a global field, the map | | → 𝑂 | | is a bijection between the equivalence classes of ultrametric absolute values of 𝐾 and the valuation rings of 𝐾. In particular, if 𝐾 is a function field of one variable over a finite field 𝐾0 , then the equivalence classes of absolute values of 𝐾 bijectively correspond to the prime divisors of 𝐾/𝐾0 . If 𝐾 is a number field, then the equivalence classes of ultrametric absolute values bijectively correspond to the prime ideals of 𝑂 𝐾 . In addition, each embedding 𝜎: 𝐾 → C defines a metric absolute value | | 𝜎 of 𝐾: |𝑥| 𝜎 = |𝜎𝑥|, where | | is here the usual absolute value of C. Each metric absolute value of 𝐾 is equivalent to some | | 𝜎 and there are at most [𝐾 : Q] such equivalence classes [CaF67, p. 57, Thm.]. In each case we call an equivalence class of an absolute value a prime of 𝐾. For each prime 𝔭 of 𝐾 we choose an absolute value | |𝔭 which represents it. The set of all primes of a global field 𝐾 satisfies the strong approximation theorem: Let 𝔭0 , 𝔭1 , . . . , 𝔭𝑛 be distinct primes of 𝐾. For each 𝑖 between 1 and 𝑛 consider an element 𝑎 𝑖 of 𝐾 and let 𝜀 > 0. Then, there exists an 𝑥 ∈ 𝐾 such that |𝑥 − 𝑎 𝑖 |𝔭𝑖 < 𝜀, 𝑖 = 1, . . . , 𝑛, and |𝑥|𝔭 ≤ 1 for each prime 𝔭 not in {𝔭0 , 𝔭1 , . . . , 𝔭𝑛 } [CaF67, p. 67]. In the function field case, this theorem is also a consequence of the strong approximation theorem (Proposition 4.3.1). Generalizations of the following lemma appear as ingredients in the Galois stratification procedure of Chapter 34. In this lemma we write for a global field 𝐾, for an ultrametric prime 𝔭 of 𝐾, and for elements 𝑎, 𝑏 ∈ 𝑂 𝐾 ,𝔭 that 𝑎 ≡ 𝑏 mod 𝔭 if |𝑎 − 𝑏|𝔭 < 1. Lemma 14.3.5 Let 𝐾 be a global field and 𝑓 ∈ 𝐾 [𝑇, 𝑋] an absolutely irreducible polynomial which is separable in 𝑋 with deg𝑋 ( 𝑓 ) ≥ 2. Then, 𝐾 has infinitely many ultrametric primes 𝔭 for which there is an 𝑎𝔭 ∈ 𝑂 𝐾 with this property: if 𝑎 ∈ 𝑂 𝐾 satisfies 𝑎 ≡ 𝑎𝔭 mod 𝔭, then 𝑓 (𝑎, 𝑏) ≠ 0 for every 𝑏 ∈ 𝐾. Proof. We assume without loss that 𝑓 ∈ 𝑂 𝐾 [𝑇, 𝑋]. Let 𝐸 = 𝐾 (𝑡), with 𝑡 an indeterminate, and let 𝑥1 , . . . , 𝑥 𝑛 be the roots of 𝑓 (𝑡, 𝑋) in 𝐸 sep . Denote the algebraic closure of 𝐾 in the Galois extension 𝐹 = 𝐸 (𝑥1 , . . . , 𝑥 𝑛 ) of 𝐸 by 𝐿. Then, 𝐹 is a regular extension of 𝐿; write it as 𝐹 = 𝐿 (𝑡, 𝑧) with 𝑧 integral over 𝑂 𝐿 [𝑡]. Thus, there

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is an absolutely irreducible polynomial 𝑔 ∈ 𝐿 [𝑇, 𝑋] with 𝑔(𝑡, 𝑋) = irr(𝑧, 𝐿(𝑡)). By Lemma 14.3.1, the set of ultrametric primes 𝔭 of 𝐾 for which 𝐿¯ 𝔓 = 𝐾¯ 𝔭 for each prime 𝔓 of 𝐿 over 𝔭 is infinite; call it 𝐴1 . The rest of the proof derives the result from the case where 𝐿 replaces 𝐾 and 𝑔 replaces 𝑓 . Part A: Geometrically exceptional primes. Let 𝐴2 be the restriction to 𝐾 of the set of ultrametric primes 𝔓 of 𝐿 for which both 𝑓 mod 𝔓 and 𝑔 mod 𝔓 are defined, separable in 𝑋, and absolutely irreducible. By Bertini–Noether (Proposition 10.4.3), 𝐴2 is a cofinite set. Express 𝑧 as a polynomial in 𝑥1 , . . . , 𝑥 𝑛 (with coefficients that are ratios of elements of 𝑂 𝐾 [𝑡]) and express each 𝑥𝑖 as a polynomial in 𝑧 (with coefficients that are ratios of elements of 𝑂 𝐿 [𝑡]). Denote the cofinite set of the ultrametric primes of 𝐾 all of their prime divisors in 𝐿 divide none of the denominators of these coefficients by 𝐴3 . 𝜕𝑔 Finally, consider 𝑑 (𝑔; 𝑡) = norm𝐹/𝐿 (𝑡) ( 𝜕𝑋 (𝑡, 𝑧)) ∈ 𝑂 𝐿 [𝑡]. Denote the cofinite set consisting of the restriction to 𝐾 of the ultrametric primes 𝔓 of 𝐿 for which the degree of 𝑑 (𝑔; 𝑡) modulo 𝔓 is equal to the degree of 𝑑 (𝑔; 𝑡) by 𝐴4 . Then, the set 𝐴 = 𝐴1 ∩ 𝐴2 ∩ 𝐴3 ∩ 𝐴4 is infinite. Part B: Reduction modulo primes of 𝐴. For 𝔭 ∈ 𝐴 and 𝔓 a prime of 𝐿 over 𝔭, extend the residue map 𝑂 𝐿 → 𝐿¯ 𝔓 = 𝐾¯ 𝔭 to a place 𝜑 of 𝐹 such that 𝑡¯ = 𝜑(𝑡) is a transcendental element over 𝐾¯ 𝔭 . Denote 𝜑(𝑥) by 𝑥¯ for every 𝑥 ∈ 𝐹 finite under 𝜑. Let 𝐸¯ = 𝐾¯ 𝔭 ( 𝑡¯) and 𝐹¯ = 𝐸¯ ( 𝑧¯) = 𝐸¯ ( 𝑥¯1 , . . . , 𝑥¯ 𝑛 ), where 𝑥¯1 , . . . , 𝑥¯ 𝑛 are the distinct roots of 𝑓¯( 𝑡¯, 𝑋). ¯ Hence, by Lemma By the choice of 𝔭, the polynomial 𝑓¯( 𝑡¯, 𝑋) is irreducible over 𝐸. ¯ 𝐸) ¯ such that 𝜎 𝑥¯𝑖 ≠ 𝑥¯𝑖 , 𝑖 = 1, . . . , 𝑛. Denote the 14.3.2, there exists a 𝜎 ∈ Gal( 𝐹/ ¯ 𝐸) ¯ by Con(𝜎). Next note that since 𝑔¯ is absolutely conjugacy class of 𝜎 in Gal( 𝐹/ ¯ In addition, the irreducible and 𝑔( ¯ 𝑡¯, 𝑧¯) = 0, the field 𝐾¯ 𝔭 is algebraically closed in 𝐹. 𝜕𝑔¯ ¯ polynomial 𝑑 ( 𝑔; ¯ 𝑡¯) = 𝑁 𝐹/ ¯ 𝐸¯ ( 𝜕𝑋 ( 𝑡¯, 𝑧¯)) with coefficients in 𝐾𝔭 has the same degree as 𝑑 (𝑔; 𝑡). If 𝑎¯ ∈ 𝐾¯ 𝔭 is not a root of 𝑑 ( 𝑔; ¯ 𝑡¯), then the prime 𝔭𝑎¯ corresponding to the specialization 𝑡¯ → 𝑎¯ of 𝐸¯ is unramified in 𝐹¯ (Lemma 7.1.8(b)). The number of primes of degree 1 of 𝐸¯ which ramify in 𝐹¯ is therefore bounded by 1 + deg(𝑑 (𝑔; 𝑡)). Let 𝑓𝑛 ∈ 𝑂 𝐾 [𝑇] be the leading coefficient of 𝑓 as a polynomial in 𝑋. Let 𝐴 ′ be the set of all 𝔭 ∈ 𝐴 finite at 𝑡 for which there exists an 𝑎¯𝔭 ∈ 𝐾¯ 𝔭 with 𝑓¯𝑛 ( 𝑎¯𝔭 )𝑑 ( 𝑓¯; 𝑎¯𝔭 ) ≠ 0 ¯ 𝐸¯
such that the Artin symbol 𝐹/ 𝔭𝑎¯ (here 𝑎¯ = 𝑎¯𝔭 ) corresponding to the prime 𝔭𝑎¯ equals Con(𝜎). By Lemma 14.3.3, 𝐴 ′ is cofinite in 𝐴. For each 𝔭 ∈ 𝐴 ′ there exists a prime 𝑄 of 𝐹¯ lying over 𝔭𝑎¯ with these properties: for 𝑞 = | 𝐾¯ 𝔭 | (14.7a) 𝜎𝑥 ≡ 𝑥 𝑞 mod 𝑄 for Î every 𝑥 ∈ 𝐹¯ integral with respect to 𝑄; and, since 𝑑 ( 𝑓¯; 𝑡¯) = 𝑖≠ 𝑗 ( 𝑥¯𝑖 − 𝑥¯ 𝑗 ), (14.7b) 𝑥¯𝑖 ̸≡ 𝑥¯ 𝑗 mod 𝑄 for every 𝑖 ≠ 𝑗. Part C: Finding 𝑎𝔭 . For each 𝑖 between 1 and 𝑛 let 𝑗 be an integer with 𝜎 𝑥¯𝑖 = 𝑥¯ 𝑗 . By the choice of 𝜎 in Part B, 𝑖 ≠ 𝑗. Hence, by (14.7a) and (14.7b), 𝑥¯𝑖 ̸≡ 𝑥¯𝑖𝑞 mod 𝑄. That is, the polynomial 𝑓¯( 𝑎¯𝔭 , 𝑋) has no roots in 𝐾¯ 𝔭 .

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Let 𝑎𝔭 be an element of 𝑂 𝐾 which is mapped to 𝑎¯𝔭 by 𝜑. If 𝑎 belongs to the localization 𝑂 𝐾 ,𝔭 of 𝑂 𝐾 and satisfies 𝑎 ≡ 𝑎𝔭 mod 𝔭, then 𝑎¯ = 𝑎¯𝔭 , and therefore, by the preceding paragraph, 𝑓 (𝑎, 𝑏) ≠ 0 for each 𝑏 ∈ 𝑂 𝐾 ,𝔭 . Moreover, since 𝑓¯𝑛 ( 𝑎¯𝔭 ) ≠ 0, the element 𝑓𝑛 (𝑎) is a unit of 𝑂 𝐾 ,𝔭 . Since the latter ring is integrally closed (because it is a localization of an integrally closed domain), 𝑓 (𝑎, 𝑋) has no zero in 𝐾. □ Let 𝐾 be a global field. An arithmetic progression of 𝑂 𝐾 is a set of the form 𝑎 + 𝔞, where 𝑎 ∈ 𝑂 𝐾 and 𝔞 is an ideal of 𝑂 𝐾 . Theorem 14.3.6 Let 𝐾 be a global field and 𝐻 a separable Hilbert subset of 𝐾. Then: (a) 𝐻 contains an arithmetic progression of 𝑂 𝐾 . (b) Let 𝔮0 , 𝔮1 , . . . , 𝔮𝑛 be distinct primes of 𝐾, 𝑏 1 , . . . , 𝑏 𝑛 elements of 𝐾, and 𝜀 > 0. Then, there exists an 𝑥 ∈ 𝐻 with |𝑥 − 𝑏 𝑖 |𝔮𝑖 < 𝜀 for 𝑖 = 1, . . . , 𝑛 and |𝑥|𝔮 ≤ 1 for each prime of 𝐾 not in {𝔮0 , 𝔮1 , . . . , 𝔮𝑛 }. (c) The intersection of 𝐻 with every arithmetical progression of 𝑂 𝐾 is nonempty. Proof. By Lemma 14.1.2 there exist absolutely irreducible polynomials ℎ1 , . . . , ℎ 𝑚 ∈ 𝑂 𝐾 [𝑇, 𝑋], separable in 𝑋, with deg𝑋 (ℎ𝑖 ) ≥ 2, 𝑖 = 1, . . . , 𝑚, and 𝐻𝐾′ (ℎ1 , . . . , ℎ 𝑚 ) ⊆ 𝐻. Apply Lemma 14.3.5 to find distinct prime ideals 𝔭1 , . . . , 𝔭𝑚 of 𝑂 𝐾 and elements 𝑎 1 , . . . , 𝑎 𝑚 ∈ 𝑂 𝐾 such that 𝑥 ∈ 𝐻𝐾′ (ℎ𝑖 ) for each 𝑥 ∈ 𝐾 with |𝑥 − 𝑎 𝑖 |𝔭𝑖 < 1, 𝑖 = 1, . . . , 𝑚. Proposition 2.1.1 produces an element 𝑎 ∈ 𝑂 𝐾 with 𝑎 + 𝔭𝑖 = 𝑎 𝑖 + 𝔭𝑖 for 𝑖 = 1, . . . , 𝑚. Thus, with 𝔞 = 𝔭1 · · · 𝔭𝑚 , we have 𝑎 + 𝔞 ⊆ 𝐻. This proves (a). For the proof of (b) choose 𝔭1 , . . . , 𝔭𝑚 above not in {𝔮0 , 𝔮1 , . . . , 𝔮𝑛 }. Then, use the strong approximation theorem for global fields [CaF67, p. 67] to find 𝑥 ∈ 𝐾 with |𝑥 − 𝑎 𝑖 |𝔭𝑖 < 1, 𝑖 = 1, . . . , 𝑚, |𝑥 − 𝑏 𝑗 |𝔮 𝑗 < 𝜀, 𝑗 = 1, . . . , 𝑛, and |𝑥|𝔮 ≤ 1 for each prime 𝔮 not in {𝔭1 , . . . , 𝔭𝑚 , 𝔮0 , . . . , 𝔮𝑛 }. Then, 𝑥 ∈ 𝐻 and |𝑥|𝔭𝑖 ≤ 1 for 𝑖 = 1, . . . , 𝑚, as desired. Each arithmetical progression in 𝑂 𝐾 has the form 𝐴 = {𝑥 ∈ 𝑂 𝐾 | 𝑣𝔮𝑖 (𝑥 − 𝑎) ≥ 𝑘 𝑖 , 𝑖 = 1, . . . , 𝑟 } for some 𝑎 ∈ 𝑂 𝐾 , ultrametric primes 𝔮1 , . . . , 𝔮𝑟 which are finite on 𝑂 𝐾 , and positive integers 𝑘 1 , . . . , 𝑘 𝑟 . If 𝐾 is a number field, choose 𝔮0 to be a metric prime. If 𝐾 is a function field of one variable over a finite field, choose 𝔮0 to be a prime which is not finite on 𝑂 𝐾 . Statement (b) gives 𝑥 ∈ 𝐻 with 𝑣𝔮𝑖 (𝑥 − 𝑎) ≥ 𝑘 𝑖 , 𝑖 = 1, . . . , 𝑟, and 𝑣𝔮 (𝑥) ≥ 0 for all ultrametric primes 𝔮 of 𝐾 which are finite on 𝑂 𝐾 . Thus, 𝑥 ∈ 𝑂 𝐾 . □ So, 𝑥 ∈ 𝐴 ∩ 𝐻. This proves (c).

14.4 Hilbertian Rings We call an integral domain 𝑅 with quotient field 𝐾 Hilbertian if every separable Hilbert subset of 𝐾 𝑟 contains elements all of whose coordinates are in 𝑅. In this case, each overring of 𝑅 is Hilbertian. In particular, 𝐾 is Hilbertian.

254

14 The Classical Hilbertian Fields

Proposition 14.4.1 Let 𝑅 be an integral domain with quotient field 𝐾. Suppose that either 𝑅 is finitely generated over Z or 𝑅 is finitely generated over a field 𝐾0 and 𝐾/𝐾0 is transcendental. Then, 𝑅 is Hilbertian. Proof. Let 𝑅0 be either Z or 𝐾0 . In the former case let 𝐾0 = Q. By assumption, 𝑅 = 𝑅0 [𝑢 1 , . . . , 𝑢 𝑛 ]. Assume without loss that 𝑢 1 , . . . , 𝑢 𝑚 with 𝑚 ≤ 𝑛 form a transcendence basis for 𝐾/𝐾0 . By Proposition 13.3.3, every separable Hilbert subset of 𝐾 𝑟 contains a separable Hilbert subset of 𝐾0 (𝑢 1 , . . . , 𝑢 𝑚 ) 𝑟 . We may therefore assume that 𝑢 1 , . . . , 𝑢 𝑛 are algebraically independent over 𝐾0 . Consider an irreducible polynomial 𝑓 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑟 , 𝑋] which is separable in 𝑋. We have to prove that 𝐻𝐾 ( 𝑓 ) ∩ 𝑅𝑟 ≠ ∅ (Lemma 13.1.6). There are several cases to consider. Case 1: 𝑟 = 1 and 𝑛 = 0. Then, 𝑅 = 𝑅0 = Z. By Theorem 14.3.6, there exists an 𝑎 ∈ 𝑅 such that 𝑓 (𝑎, 𝑋) is irreducible. Case 2: 𝑟 = 1, 𝑛 = 1, and 𝑅0 = 𝐾0 is finite. Since 𝐾0 is finite, 𝐾 is global and 𝑅 = 𝑂 𝐾 . Thus, Theorem 14.3.6 gives 𝑎 ∈ 𝑅 with 𝑓 (𝑎, 𝑋) irreducible. Case 3: 𝑟 = 1, 𝑛 = 1, and 𝑅0 is infinite. Proposition 14.2.1 gives a nonempty Zariski 𝐾0 -open subset 𝑈 of A2 such that 𝑓 (𝑎 + 𝑏𝑢 1 , 𝑋) is irreducible for all (𝑎, 𝑏) ∈ 𝑈 (𝐾0 ). Since 𝑅0 is infinite, we may choose (𝑎, 𝑏) in 𝑈 (𝑅0 ). Case 4: 𝑟 = 1 and 𝑛 ≥ 2. Then, 𝑅0 [𝑢 1 , . . . , 𝑢 𝑛−1 ] is an infinite ring. Hence, by Proposition 14.2.1, that ring has elements 𝑎, 𝑏 such that 𝑓 (𝑎 + 𝑏𝑢 𝑛 , 𝑋) is irreducible. Case 5: 𝑟 ≥ 2. Consider 𝑓 as a polynomial in 𝑇𝑟 , 𝑋 with coefficients in the infinite ring 𝐾 [𝑇1 , . . . , 𝑇𝑟−1 ]. As such, 𝑓 is irreducible. Replace 𝑅0 and 𝐾0 in Cases 3 and 4 by 𝑅0 [u, 𝑇1 , . . . , 𝑇𝑟−1 ] and 𝐾0 (u, 𝑇1 , . . . , 𝑇𝑟−1 ) to find 𝑔 ∈ 𝑅0 [u, 𝑇1 , . . . , 𝑇𝑟−1 ] such that 𝑓 (𝑇1 , . . . , 𝑇𝑟−1 , 𝑔(𝑇1 , . . . , 𝑇𝑟−1 ), 𝑋) is irreducible. Now use an induction hypothesis on 𝑟 to find 𝑎 1 , . . . , 𝑎𝑟−1 ∈ 𝑅 with 𝑓 (𝑎 1 , . . . , 𝑎𝑟−1 , 𝑔(𝑎 1 , . . . , 𝑎𝑟−1 ), 𝑋) irreducible. Then, 𝑎𝑟 = 𝑔(𝑎 1 , . . . , 𝑎𝑟−1 ) is in 𝑅 and 𝑓 (a, 𝑋) is irreducible. □ The quotient fields of the rings mentioned in Proposition 14.4.1 are the classical Hilbertian fields. Theorem 14.4.2 Suppose that 𝐾 is a global field or a finitely generated transcendental extension of an arbitrary field 𝐾0 . Then, 𝐾 is Hilbertian. Moreover, each Hilbert set of 𝐾 is nonempty. Proof. As a consequence of Proposition 14.4.1, 𝐾 is Hilbertian. To prove the second statement, assume char(𝐾) = 𝑝 > 0. Then, 𝐾 is imperfect. Hence, by Uchida □ (Proposition 13.4.3), every Hilbertian set of 𝐾 is nonempty. The proof of Proposition 14.4.1 gives another useful version of Theorem 14.4.2: Proposition 14.4.3 Let 𝐾 be a finitely generated separable extension of a field 𝐾0 with 𝑚 = trans.deg(𝐾/𝐾0 ) > 0. Let 𝑟 ≤ 𝑚 and let 𝐻 be a separable Hilbert subset of 𝐾 𝑟 . Then, 𝐻 contains a point (𝑢 1 , . . . , 𝑢𝑟 ) with 𝑢 1 , . . . , 𝑢𝑟 algebraically independent over 𝐾0 .

14.4 Hilbertian Rings

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Proof. Choose a separating transcendence basis 𝑡 1 , . . . , 𝑡 𝑚 for 𝐾/𝐾0 . Then, 𝐾/𝐾0 (t) is a finite separable extension. By Corollary 13.2.3, 𝐻 contains a separable Hilbert subset of 𝐾0 (t) 𝑟 . We may therefore assume 𝐾 = 𝐾0 (t). By Lemma 13.1.6, we may assume 𝐻 = 𝐻𝐾 ( 𝑓 ) with an irreducible 𝑓 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑟 , 𝑋], monic and separable in 𝑋. For 𝑚 = 1 and 𝐾0 finite, Theorem 14.3.6 gives 𝑢 ∈ 𝐻 transcendental over 𝐾0 . For 𝑚 = 1 and 𝐾0 infinite, Proposition 14.2.1 does the job. Assume 𝑚 ≥ 2. Then, 𝐾 ′ = 𝐾0 (𝑡 2 , . . . , 𝑡 𝑚 , 𝑇2 , . . . , 𝑇𝑟 ) is an infinite field, 𝑡 1 is transcendental over 𝐾 ′, and 𝐾 ′ (𝑡1 ) = 𝐾 (𝑇2 , . . . , 𝑇𝑟 ). Consider 𝑓 as an irreducible polynomial in 𝑇1 , 𝑋 over 𝐾 ′ (𝑡1 ). Proposition 14.2.1 gives nonzero 𝑔, ℎ ∈ 𝐾0 [𝑡 2 , . . . , 𝑡 𝑚 , 𝑇2 , . . . , 𝑇𝑟 ] such that
𝑓1 (𝑇2 , . . . , 𝑇𝑟 , 𝑋) = 𝑓 𝑔(𝑇2 , . . . , 𝑇𝑟 ) + ℎ(𝑇2 , . . . , 𝑇𝑟 )𝑡 1 , 𝑇2 , . . . , 𝑇𝑟 , 𝑋 is irreducible over 𝐾. Let 𝐾1 = 𝐾0 (𝑡1 ). Apply the induction hypothesis to 𝐾1 and 𝑓1 instead of to 𝐾 and 𝑓 and find 𝑢 2 , . . . , 𝑢𝑟 in 𝐾 algebraically independent over 𝐾1 such that 𝑓 (𝑔(𝑢 2 , . . . , 𝑢𝑟 ) + ℎ(𝑢 2 , . . . , 𝑢𝑟 )𝑡1 , 𝑢 2 , . . . , 𝑢𝑟 , 𝑋) is irreducible over 𝐾. Put 𝑢 1 = 𝑔(𝑢 2 , . . . , 𝑢𝑟 ) + ℎ(𝑢 2 , . . . , 𝑢𝑟 )𝑡1 . Then, 𝑓 (u, 𝑋) is irreducible over 𝐾 and 𝐾0 (𝑢 1 , 𝑢 2 , . . . , 𝑢𝑟 ) = 𝐾0 (𝑡1 , 𝑢 2 , . . . , 𝑢𝑟 ). Hence, 𝑢 1 , . . . , 𝑢𝑟 are algebraically inde□ pendent over 𝐾0 , as desired. Remark 14.4.4 (More Hilbertian rings) Exercise 4 proves that each valuation ring of a Hilbertian field is Hilbertian. Geyer extends this result to several valuations of rank 1. He considers a Hilbertian field 𝐾, nonequivalent absolute values | | 1 , . . . , | | 𝑛 of 𝐾, and a separable Hilbert subset 𝐻 of 𝐾 𝑟 . Let a1 , . . . , a𝑛 be tuples in 𝐾 𝑟 and let 𝜀 > 0. Then, there exists an x ∈ 𝐻 such that |x − a𝑖 | 𝑖 < 𝜀, 𝑖 = 1, . . . , 𝑛 [Gey78, Lemma 3.4]. In particular, if 𝑣 1 , . . . , 𝑣 𝑛 are valuations of rank 1, then their holomorphy ring 𝑅 = {𝑎 ∈ 𝐾 | 𝑣 𝑖 (𝑎) ≥ 0, 𝑖 = 1, . . . , 𝑛} is Hilbertian. Proposition 19.7 of [Jar91] generalizes Geyer’s result. Here one considers valuations 𝑣 1 , . . . , 𝑣 𝑚 and orderings 𝑗 0 for 𝑗 = 1, . . . , 𝑛. Then, there exists an x ∈ 𝐻 with 𝑣 𝑖 (x − a𝑖 ) > 𝛼𝑖 , 𝑖 = 1, . . . , 𝑚, and −𝑐 𝑗 < 𝑗 x − b 𝑗 < 𝑗 𝑐 𝑗 , 𝑗 = 1, . . . , 𝑛. Moreover, the holomorphy ring of arbitrary finitely many valuations of 𝐾 is Hilbertian [Jar91, Proposition 19.6]. In contrast, the holomorphy ring of infinitely many valuations of a field 𝐾 may not be Hilbertian even if 𝐾 is Hilbertian. For example, if 𝐾0 is a finite field, the holomorphy ring of all valuations of 𝐾0 (𝑡) is 𝐾0 . The latter field is not Hilbertian, although 𝐾0 (𝑡) is Hilbertian. More interesting examples can be found in Example 17.5.3 and Example 17.5.6. The following lemma improves Exercise 2 of Section 13: Lemma 14.4.5 Let 𝑚 be a cardinal number and {𝐾 𝛼 | 𝛼 < 𝑚} a transfinite sequence of fields. Suppose that for each 𝛼 < Ð 𝑚 the field 𝐾 𝛼+1 is a proper finitely generated regular extension of 𝐾 𝛼 . Then, 𝐾 = 𝛼 max(char(𝐾), 2𝑔 − 2 + 2𝑙).

(14.9)

Extend 𝐾, if necessary, to assume 𝜁𝑙 , 𝜁𝑞 ∈ 𝐾. Then, use the Hilbertianity of 𝐾 √ (Theorem 14.4.2) to choose 𝑢 ∈ 𝐾 which is not a 𝑞th power. Thus, 𝑞 𝑢 ∉ 𝐾ins . By Eisenstein’s criterion (Lemma 2.3.10), 𝑓 (𝑋, 𝑌 ) = 𝑌 𝑙 − 𝑋 𝑙𝑞 + 𝑢 is an absolutely irreducible polynomial. Now define an ascending √𝑙 sequence of Galois extensions 𝐾𝑛 of 𝐾 inductively: 𝐾0 = 𝐾 and 𝐾𝑛+1 = 𝐾𝑛 ( 𝑎 𝑙𝑞 − 𝑢 | 𝑎 ∈ 𝐾𝑛 ). Then, 𝐾𝑛 /𝐾 is a pro-𝑙-extension. Indeed, by [Lan97, p. 297, Thm. 9.1], 𝐾𝑛 /𝐾 is Galois and [𝐿 : 𝐾] is an 𝑙-power for each finite subextension 𝐿/𝐾 of 𝐾𝑛 /𝐾.

14.6 Non-Hilbertian 𝑔-Hilbertian Fields

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Ð∞ Ð Let 𝑀′ (𝐾) = ′ 𝑛=0 𝐾𝑛 . This is an infinite pro-𝑙 extension of 𝐾. Let 𝑁 = 𝑁 (𝐾) = 𝑀 (𝐾 ) with 𝐾 ranging over all finite purely inseparable extensions of 𝐾. Then, 𝑁 is a perfect field which is infinite and normal over 𝐾. For each 𝑎 ∈ 𝑁 there exists a 𝑏 ∈ 𝑁 such that 𝑏 𝑙 − 𝑎 𝑙𝑞 + 𝑢 = 0. Hence, 𝑁 is non-Hilbertian. We prove however that 𝑁 is 𝑔-Hilbertian. Note: We may replace 𝐾 in the proof by any finite extension 𝐿 in 𝑁. Indeed, then 𝑀 (𝐿) is contained in 𝑀 (𝐾 ′) for some finite purely inseparable extension 𝐾 ′ of 𝐾. Hence, 𝑁 = 𝑁 (𝐿). Claim A: Suppose that √ 𝑀 is a finite Galois extension of 𝐾 in 𝑁. Then, [𝑀 : 𝐾] is a power of 𝑙, and 𝑀 ( 𝑞 𝑢) is a Galois extension of 𝑀 of degree 𝑞. Indeed, there is a finite purely inseparable extension 𝐾 ′ of 𝐾 with 𝑀 ⊆ 𝑀 (𝐾 ′), so [𝑀 : 𝐾] = [𝑀𝐾 ′ : 𝐾 ′] is√a power of 𝑙. By the choice of 𝑢 and √ again, by [Lan97, p. 297, Thm. 9.1], 𝐾 ( 𝑞 𝑢)/𝐾 is Galois of degree 𝑞, so 𝑀 ( 𝑞 𝑢)/𝑀 is a Galois extension of degree 𝑞. Let 𝑃 be the set of all ultrametric prime divisors 𝔭 of 𝐾 with 𝑣𝔭 (𝑢) = 0, 𝑣𝔭 (𝑙) = 0, √ 𝐾 ( 𝑞 𝑢)/𝐾
𝑣𝔭 (𝑞) = 0, and ≠ 1. 𝔭 Claim B: Each 𝔭 in 𝑃 is unramified in 𝑀 (𝐾). Indeed, it suffices to consider a finite extension 𝐿 of 𝐾 in 𝑀 (𝐾) in which 𝔭 is unramified, to take a prime √𝑙 divisor 𝔮 of 𝐿 over 𝔭 and an element 𝑎 ∈ 𝐿, and to prove that 𝔮 is unramified in 𝐿( 𝑎 𝑙𝑞 − 𝑢). By Case C of Example 2.3.8, it suffices to prove that 𝑙 divides 𝑣𝔮 (𝑎 𝑙𝑞 − 𝑢). Suppose first that 𝑣𝔮 (𝑎) < 0. Then, 𝑣𝔮 (𝑎 𝑙𝑞 − 𝑢) = 𝑣𝔮 (𝑎 𝑙𝑞 ) = 𝑙𝑞𝑣𝔮 (𝑎). Now suppose that 𝑣𝔮 (𝑎) ≥ 0. Denote reduction modulo 𝔮 by a bar. The assumptions on 𝔮 √ √ √ imply that 𝑂𝔭 [ 𝑞 𝑢]/𝑂𝔭 is a ring cover. By Remark 7.1.7, 𝐾¯ ( 𝑞 𝑢) ¯ = 𝐾 ( 𝑞 𝑢). Since √ √ 𝑞
𝐾 ( 𝑢)/𝐾 ¯ = 𝑞. Next, let 𝑀 be the Galois closure ≠ 1, this implies that [ 𝐾¯ ( 𝑞 𝑢) ¯ : 𝐾] 𝔭 ¯ is a power of 𝑙. Therefore, of 𝐿/𝐾. Then, 𝑀 ⊆ 𝑀 (𝐾). Hence, by Claim A, [ 𝑀¯ : 𝐾] √ 𝑞 ¯ ¯ ¯ ¯ ¯ In [ 𝐿 : 𝐾] is also a power of 𝑙. Thus, 𝐾 ( 𝑢) ¯ ̸ ⊆ 𝐿, so 𝑢¯ is not a 𝑞th power in 𝐿. 𝑙𝑞 𝑙𝑞 particular, 𝑎¯ ≠ 𝑢. ¯ Consequently, 𝑣𝔮 (𝑎 − 𝑢) = 0. Claim C: Let 𝐿 be a finite extension of 𝐾 in 𝑀 (𝐾). Consider nonconstant polynomials ℎ1 , . . . , ℎ𝑟 ∈ 𝐿 [𝑋]. Suppose that ℎ𝑖 has an irreducible 𝐿-factor 𝑓𝑖 which is separable, of multiplicity not divisible by 𝑙; suppose further that the degree of each irreducible 𝐿-factor√︁of ℎ𝑖 is less than 𝑞, 𝑖 = 1, . . . , 𝑟. Then, 𝑂 𝐾 has an arithmetical progression 𝐴 with 𝑙 ℎ𝑖 (𝑎) ∉ 𝑁 for all 𝑎 ∈ 𝐴 and 𝑖 = 1, . . . , 𝑟. Indeed, write ℎ𝑖 = 𝑓𝑖𝑘 𝑔𝑖 with 𝑓𝑖 , 𝑔𝑖 ∈ 𝐿 [𝑋], 𝑓𝑖 irreducible, 𝑙 ∤ 𝑘, and 𝑔𝑖 relatively ′ ′ ′ prime to 𝑓𝑖 . Choose 𝑘 ′, 𝑙 ′ ∈ Z with 𝑘 ′ ≥ 1 and 𝑘 𝑘 ′ + 𝑙𝑙 ′ = 1. Then, ℎ𝑖𝑘 = 𝑓𝑖𝑘𝑘 𝑔𝑖𝑘 = √︁ √︁ ′ ′ 𝑓𝑖 𝑔𝑖𝑘 𝑓𝑖−𝑙𝑙 . For every 𝑎 ∈ 𝐾, 𝑙 ℎ𝑖 (𝑎) ∉ 𝑁 if and only if 𝑙 𝑓𝑖 (𝑎)𝑔𝑖 (𝑎) 𝑘′ ∉ 𝑁. Each ′ ′ root of 𝑓𝑖 in 𝐾sep is a simple root of 𝑓𝑖 𝑔𝑖𝑘 . Therefore, we may replace ℎ𝑖 by 𝑓𝑖 𝑔𝑖𝑘 , if necessary, to assume ℎ𝑖 has a simple root 𝑎 𝑖 in 𝐾sep . Now extend 𝐿,√if necessary, to assume 𝐿 is Galois over 𝐾 with Galois group 𝐺. By Claim A, 𝐿( 𝑞 𝑢)/𝐿 is a Galois extension of degree 𝑞. Let 𝑀 be the splitting Î Î field of 𝑟𝑖=1 𝜎 ∈𝐺 ℎ𝑖𝜎 (𝑋) over 𝐿. Since the degree of each irreducible 𝐿-factor of ℎ1 , . . . , ℎ𝑟 is less than 𝑞, 𝑀 is a finite Galois extension of 𝐾 whose degree is divisible only by prime numbers smaller than 𝑞. Hence, 𝑀 is a proper subfield of

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√ √ 𝑀 ( 𝑞 𝑢). Denote the set of all ultrametric primes√𝔭 of 𝐾 which unramify in 𝑀 ( 𝑞 𝑢) 𝑞
√ 𝑀 ( 𝑢)/𝐾 with 𝑣𝔭 (𝑢) = 0, 𝑣𝔭 (𝑙) = 0, 𝑣𝔭 (𝑞) = 0, and ⊆ Gal(𝑀 ( 𝑞 𝑢)/𝑀) ∖{1} by 𝔭 𝑃0 . Then, 𝑃0 is a subset of 𝑃. By Chebotarev (Theorem 7.3.1), 𝑃0 is infinite. Omit finitely many elements from 𝑃0 to assume that for each 𝔭 ∈ 𝑃0 and for 𝑖 = 1, . . . , 𝑟 all nonzero coefficients of the ℎ𝑖 are 𝔮-units and 𝑎 𝑖 is a simple root of ℎ𝑖 modulo 𝔮 ′, for each prime 𝔮 of 𝐿 over 𝔭 and each prime 𝔮 ′ of 𝑀 over 𝔮. Now choose distinct primes 𝔭1 , . . . , 𝔭𝑟 in 𝑃0 . Consider 𝑖 between 1 and 𝑟. Let 𝔭 = 𝔭𝑖 . Choose a prime 𝔮 = 𝔮𝑖 of 𝐿 over 𝔭, a ¯ so prime 𝔮 ′ of 𝑀 over 𝔮, and denote reduction modulo 𝔮 ′ by a bar. Then, 𝑀¯ = 𝐾, ¯ℎ𝑖 is a polynomial with coefficients in 𝐾¯ which decomposes into linear factors. In ¯ ℎ¯ 𝑖 ( 𝑎¯ 𝑖 ) = 0, and ℎ¯ ′ ( 𝑎¯ 𝑖 ) ≠ 0, where ℎ ′ is the derivative of ℎ𝑖 . particular, 𝑎¯ 𝑖 ∈ 𝐾, 𝑖 𝑖 Choose 𝑏 𝑖 ∈ 𝐾 with 𝑣𝔮′ (𝑏 𝑖 − 𝑎 𝑖 ) > 0. Then, 𝑣𝔮 (ℎ𝑖 (𝑏 𝑖 )) ≥ 1 and 𝑣𝔮 (ℎ𝑖′ (𝑏 𝑖 )) = 0. If 𝑣 𝑞 (ℎ𝑖 (𝑏 𝑖 )) ≥ 2, choose 𝜋𝑖 ∈ 𝐾 with 𝑣𝔭 (𝜋𝑖 ) =
1. Since 𝔭 is unramified in 𝑀, 𝑣𝔮 (𝜋𝑖 ) = 1. Then, 𝑣𝔮 ℎ𝑖 (𝑏 𝑖 + 𝜋𝑖 ) − ℎ𝑖 (𝑏 𝑖 ) − ℎ𝑖′ (𝑏 𝑖 )𝜋𝑖 ≥ 2. Hence, 𝑣𝔮 (ℎ𝑖 (𝑏 𝑖 + 𝜋𝑖 )) = 1. Thus, replacing 𝑏 𝑖 by 𝑏 𝑖 + 𝜋𝑖 , if necessary, we may assume that in any case 𝑣𝔮 (ℎ𝑖 (𝑏 𝑖 )) = 1. The weak approximation theorem (Proposition 2.1.1) gives 𝑏, 𝜋 ∈ 𝐾 with 𝑣𝔭𝑖 (𝑏 − 𝑏 𝑖 ) ≥ 2 and 𝑣𝔭𝑖 (𝜋 − 𝜋𝑖 ) ≥ 2, 𝑖 = 1, . . . , 𝑟. Let 𝐴 = 𝑏 + 𝜋 2 𝑂 𝐾 . Then, for each prime 𝔮𝑖 of 𝐿 over 𝔭𝑖 and all 𝑎 ∈ 𝐴, we have 𝑣𝔮𝑖 (ℎ𝑖 (𝑎)) = 1. Consider an arbitrary finite purely inseparable extension 𝐾 ′ of 𝐾. Denote the unique extension of 𝔭𝑖 to 𝐾 ′ by 𝔭𝑖′ and let 𝔮𝑖′ be a prime of 𝐿𝐾 ′ over 𝔭𝑖′. Then, 𝑣𝔮𝑖′ (ℎ𝑖 (𝑎)) = 1 if char(𝐾) = 0 and 𝑣𝔮𝑖′ (ℎ𝑖 (𝑎)) is a power of char(𝐾) if char(𝐾) > 0. In each case, 𝑙 ∤ 𝑣𝔮𝑖′ (ℎ𝑖 (𝑎)). By Claim B, applied to 𝐾 ′ instead of to 𝐾, 𝔭𝑖′ is unramified in 𝑀 (𝐾 ′). Hence, 𝑙 ∤ 𝑣 𝑖 (ℎ𝑖 (𝑎)) for each extension 𝑣 𝑖 of 𝑣𝔭𝑖′ to 𝑀 (𝐾 ′), √︁ √︁ so 𝑙 ℎ𝑖 (𝑎) ∉ 𝑀 (𝐾 ′). Consequently, 𝑙 ℎ𝑖 (𝑎) ∉ 𝑁, as claimed. Claim D: Let 𝐿 be a finite extension of Î 𝐾 and 𝐶 a curve of genus at most 𝑔 over 𝐿. Suppose that 𝐶 is defined by 𝑌 𝑙 = 𝑐 𝑠𝑗=1 𝑓 𝑗 (𝑋) 𝑚 𝑗 , where 𝑐 ∈ 𝐾 × , 𝑓1 , . . . , 𝑓𝑠 are distinct monic separable irreducible polynomials in 𝐿 [𝑋], and 𝑙 ∤ 𝑚 𝑗 , 𝑗 = 1, . . . , 𝑠. Then, the 𝑓 𝑗 ’s can be chosen such that deg( 𝑓 𝑗 ) < 𝑞, 𝑗 = 1, . . . , 𝑠. Indeed, replace 𝑚 𝑗 by its residue modulo 𝑙, if necessary, to assume 1 ≤ 𝑚 𝑗 ≤ 𝑙 −1. Î Choose a transcendental element 𝑥 over 𝐿 and 𝑦 ∈ 𝐿 (𝑥)sep with 𝑦 𝑙 = 𝑐 𝑠𝑗=1 𝑓 𝑗 (𝑥) 𝑚 𝑗 . Then, 𝐹 = 𝐿 (𝑥, 𝑦) is the function field of 𝐶 over 𝐿. Denote the prime divisor of 𝐿(𝑥)/𝐿 corresponding to 𝑓 𝑗 by 𝑃 𝑗 (Sections 4.1 and 2.2). Then, Î𝑠deg(𝑃 𝑗 )𝑚=𝑖 deg( 𝑓 𝑗 ), 𝑣 𝑃 𝑗 ( 𝑓 𝑗 (𝑥)) = 1, and 𝑣 𝑃 𝑗 ( 𝑓𝑖 (𝑥)) = 0 for 𝑗 ≠ 𝑖. Hence, 𝑣 𝑃 𝑗 (𝑐 𝑖=1 𝑓𝑖 (𝑥) ) = 𝑚 𝑗 is not divisible by 𝑙. By Example 2.3.8(Case A), 𝑃 𝑗 totally ramifies in 𝐹. Thus, 𝐹 has a unique prime divisor 𝑄 𝑗 lying over 𝑃 𝑗 , 𝑒 𝑄 𝑗 /𝑃 𝑗 = 𝑙, and deg(𝑄 𝑗 ) = deg(𝑃 𝑗 ) = deg( 𝑓 𝑗 ). Since 𝑙 ≠ char(𝐿), the ramification is tame. Hence, by Riemann–Hurwitz (Remark 4.6.2(c)), 2𝑔 − 2 ≥ 2 · genus(𝐹) − 2 ≥ −2𝑙 + (𝑙 − 1) deg( 𝑓 𝑗 ). Therefore, by (14.9), deg( 𝑓 𝑗 ) ≤ 2𝑔 − 2 + 2𝑙 < 𝑞, 𝑗 = 1, . . . , 𝑠, as claimed. Claim E: 𝑁 is 𝑔-Hilbertian. Ð𝑚 Indeed, assume toward contradiction that 𝑁 = 𝑖=1 𝜑𝑖 (𝐶𝑖 (𝑁)) with 𝐶𝑖 a curve over 𝑁 of genus at most 𝑔 and 𝜑𝑖 : 𝐶𝑖 → A1 a dominant separable rational map over 𝑁 of degree at least 2. Replace 𝐾 by a finite extension in 𝑁, if necessary, to assume that 𝐶𝑖 and 𝜑𝑖 are defined over 𝐾 and the genus of 𝐶𝑖 over 𝐾 is the same as the genus of 𝐶𝑖 over 𝑁. Choose a transcendental element 𝑥 over 𝐾 and a generic point

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z𝑖 for 𝐶𝑖 over 𝐾 with 𝜑𝑖 (z𝑖 ) = 𝑥, 𝑖 = 1, . . . , 𝑚. Then, 𝐸 𝑖 = 𝐾 (z𝑖 ) is a finite separable extension of 𝐾 (𝑥) of degree at least 2. Denote the Galois closure of 𝐸 𝑖 over 𝐾 (𝑥) by 𝐹𝑖 . Next, list the cyclic extensions of 𝐾 (𝑥) of degree 𝑙 which are contained in one of the fields 𝐸 𝑖 as 𝐷 1 , . . . , 𝐷 𝑟 . For each 𝑗 between 1 and 𝑟 choose a primitive element 𝑦 𝑗 for 𝐷 𝑗 /𝐾 (𝑥) such that 𝑦 𝑙𝑗 = ℎ 𝑗 (𝑥) with ℎ 𝑗 ∈ 𝐾 [𝑋]. Then, ℎ 𝑗 (𝑥) is not an 𝑙th power in 𝐾 (𝑥). Since each 𝐸 𝑖 is a regular extension of 𝐾, so is 𝐷 𝑗 /𝐾. Hence, deg(ℎ 𝑗 ) ≥ 1. Since 𝑁 is perfect, we may again replace 𝐾 by a finite purely inseparable extension Î𝑠 𝑗 𝑚 𝑗 𝑘 in 𝑁 to assume that ℎ 𝑗 = 𝑐 𝑗 𝑘=1 ℎ 𝑗 𝑘 such that 𝑐 𝑗 ∈ 𝐾 × , ℎ 𝑗 𝑘 are distinct monic separable irreducible polynomials of positive degrees, and 𝑚 𝑗 𝑘 are positive integers. If 𝑙 |𝑚 𝑗 𝑘 for some 𝑗, 𝑘, replace 𝑦 𝑗 by 𝑦 𝑗 ℎ 𝑗 𝑘 (𝑥) −𝑚 𝑗 𝑘 /𝑙 . Thus, assume without loss, 𝑙 ∤ 𝑚 𝑗 𝑘 for all 𝑗 and 𝑘. By Claim D, we may assume that deg(ℎ 𝑗 𝑘 ) < 𝑞. Hence, by Claim C, there √︁ exists an arithmetical progression 𝐴 of 𝑂 𝐾 such that 𝑙 ℎ 𝑗 (𝑎) ∉ 𝑁 for each 𝑎 ∈ 𝐴, Î𝑚 𝑗 = 1, . . . , 𝑟. Let 𝐹 = 𝑖=1 𝐹𝑖 . Let 𝐻 be the set of all 𝑎 ∈ 𝐾 such that each 𝐾-place 𝜓: 𝐹 → 𝐾˜ ∪ {∞} with 𝜓(𝑥) = 𝑎 maps 𝐹𝑖 onto 𝐹¯𝑖 ∪ {∞} and 𝐸 𝑖 onto 𝐸¯ 𝑖 ∪ {∞} with these properties: (14.10a) 𝐹¯𝑖 /𝐾 is a Galois extension and Gal( 𝐹¯𝑖 /𝐾)  Gal(𝐹𝑖 /𝐾 (𝑥)). (14.10b) 𝜓 is finite at z𝑖 and c𝑖 = 𝜓(z𝑖 ) satisfies 𝐸¯ 𝑖 = 𝐾 (c𝑖 ) and [𝐾 (c𝑖 ) : 𝐾] = [𝐸 𝑖 : 𝐾 (𝑥)]. By Lemma 14.1.1, 𝐻 contains a separable Hilbert subset. Moreover, Theorem 14.3.6(c) gives 𝑎 ∈ 𝐴 ∩ 𝐻. Since 𝜑𝑖 : 𝐶𝑖 → A1 is dominant over 𝑁, we have 𝑎 = 𝜑𝑖 (c) with c ∈ 𝐶𝑖 (𝑁) for some 𝑖 between 1 and 𝑚. Extend the 𝐾-specialization (𝑥, z𝑖 ) → (𝑎, c) to a place 𝜓: 𝐹 → 𝐾˜ ∪ {∞}. Then, (14.10a) and (14.10b) are true with c𝑖 = c. Since 𝐹𝑖 is the Galois closure of 𝐸 𝑖 /𝐾 (x), 𝐹¯𝑖 is the Galois closure of 𝐾 (c)/𝐾. Since 𝐾 (c) ⊆ 𝑁 and 𝑁/𝐾 is normal, 𝐹¯𝑖 ⊆ 𝑁. By Claim A, Gal( 𝐹¯𝑖 /𝐾) is an 𝑙-group. By (14.10b) and the first paragraph of the current claim, [ 𝐸¯ 𝑖 : 𝐾] = [𝐸 𝑖 : 𝐾 (𝑥)] > 1. Thus, Gal( 𝐹¯𝑖 /𝐸¯ 𝑖 ) is a proper subgroup of Gal( 𝐹¯𝑖 /𝐾). Hence, Gal( 𝐹¯𝑖 /𝐸¯ 𝑖 ) is contained in a normal subgroup of index 𝑙 of Gal( 𝐹¯𝑖 /𝐾) [Hal59, p. 45, Cor. 4.2.2]. By (14.10a), Gal(𝐹𝑖 /𝐸 𝑖 ) is contained in a normal subgroup of Gal(𝐹𝑖 /𝐾 (𝑥)) of index 𝑙, so 𝐾 (𝑥) has a cyclic extension of degree 𝑙 in 𝐸 𝑖 . It is 𝐷 𝑗 for some 𝑗 between 1 and 𝑟. Then, 𝑏 = 𝜓(𝑦 𝑗 ) satisfies 𝑏 𝑙 = 𝜓(𝑦 𝑙𝑗 ) = 𝜓(ℎ 𝑗 (𝑥)) = ℎ 𝑗 (𝑎) and 𝑏 ∈ 𝐸¯ 𝑖 ⊆ 𝑁. This contradicts the choice of 𝑎 in 𝐴. Ð𝑚 Thus, our initial assumption 𝑁 = 𝑖=1 𝜑𝑖 (𝐶𝑖 (𝑁)) is false, so 𝑁 is 𝑔-Hilbertian. □

Exercises 1. [Fri85, §2] This exercise shows that it is not always possible to take 𝑚 = 1 in Lemma 14.1.2. Consider an irreducible polynomial 𝑓 ∈ Q[𝑇, 𝑋], monic in 𝑋, and 𝑔1 , 𝑔2 ∈ Q[𝑌 ], polynomials of positive degree for which 𝑓 (𝑔𝑖 (𝑌 ), 𝑋) is reducible, 𝑖 = 1, 2. In addition, suppose that there exists an irreducible ℎ ∈ Q[𝑇, 𝑋] with

14 The Classical Hilbertian Fields

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deg𝑋 (ℎ) > 1 such that 𝐻Q′ (ℎ) = {𝑎 ∈ Q | ℎ(𝑎, 𝑋) has no zero in Q} ⊆ 𝐻Q ( 𝑓 ). (a) Use Exercise 1 of Chapter 13 to conclude that ℎ(𝑔𝑖 (𝑌 ), 𝑋) has a factor of degree 1 in 𝑋, say 𝑋 − 𝑚 𝑖 (𝑌 ), where 𝑚 𝑖 ∈ Q(𝑌 ), 𝑖 = 1, 2. (b) Use field theory to interpret (a): Let 𝑡 be an indeterminate, 𝑦 𝑖 a zero of 𝑔𝑖 (𝑌 ) − 𝑡, and 𝑥𝑖 = 𝑚 𝑖 (𝑦 𝑖 ), 𝑖 = 1, 2. Note that 𝑥1 and 𝑥 2 are both zeros of ℎ(𝑡, 𝑋), so that they are conjugate over Q(𝑡). Conclude that [Q(𝑦 1 ) ∩ Q(𝑦 2′ ) : Q(𝑡)] > 1 for some conjugate 𝑦 2′ of 𝑦 2 over Q(𝑡). (c) Consider the case 𝑓 (𝑇, 𝑋) = 𝑋 4 + 2𝑋 2 − 𝑇 and 𝑔1 (𝑌 ) = 𝑌 4 + 2𝑌 2 , 𝑔2 (𝑌 ) = −4𝑌 4 − 4𝑌 2 − 1. Prove that the splitting field of 𝑔𝑖 (𝑌 ) − 𝑡 over Q(𝑡) has the dihedral group of order 8 as its Galois group over Q(𝑡). Observe that since Q(𝑦 1 )/Q(𝑡) is a nonnormal extension of degree 4, Q(𝑦 21 )/Q(𝑡) is its only quadratic subextension. Prove that the prime of Q(𝑡) corresponding to the specialization 𝑡 → 0 is unramified in Q(𝑦 21 ) but ramified in Q(𝑦 22 ). Conclude that there is no irreducible ℎ ∈ Q[𝑇, 𝑋] with deg𝑋 (ℎ) > 1 and 𝐻Q′ (ℎ) ⊆ 𝐻Q ( 𝑓 ). 2. Let 𝑓1 (𝑇, 𝑋), . . . , 𝑓𝑚 (𝑇, 𝑋), with deg𝑋 ( 𝑓𝑖 ) > 1, 𝑖 = 1, . . . , 𝑚, be absolutely irreducible polynomials, separable in 𝑋, with coefficients in a global field 𝐾. Let 𝑡 be transcendental over 𝐾 and denote the splitting fields of 𝑓1 (𝑡, 𝑋), . . . , 𝑓𝑚 (𝑡, 𝑋), respectively, over 𝐸 = 𝐾 (𝑡) by 𝐹1 , . . . , 𝐹𝑚 . Assume 𝐹1 , . . . , 𝐹𝑚 are linearly disjoint over 𝐸 and let 𝐹 = 𝐹1 · · · 𝐹𝑚 . Observe that there is a 𝜎 ∈ Gal(𝐹/𝐸) that fixes none of the roots of 𝑓1 (𝑡, 𝑋) · · · 𝑓𝑚 (𝑡, 𝑋) and improve Lemma 14.3.5. Show that there exists a prime ideal 𝔭 of 𝑂 𝐾 and an element 𝑎𝔭 ∈ 𝑂 𝐾 such that for 𝑎 ∈ 𝑂 𝐾 , if 𝑎 ≡ 𝑎𝔭 mod 𝔭, then 𝑓𝑖 (𝑎, 𝑋) has no zeros in 𝐾, 𝑖 = 1, . . . , 𝑚. 3. Consider the three absolutely irreducible polynomials 𝑓1 (𝑇, 𝑋) = 𝑋 2 − 𝑇, 𝑓2 (𝑇, 𝑋) = 𝑋 2 − (𝑇 + 1), 𝑓3 (𝑇, 𝑋) = 𝑋 2 − 𝑇 (𝑇 + 1) and let 𝐻 = 𝐻Q ( 𝑓1 , 𝑓2 , 𝑓3 ). Choose a prime number 𝑝 and an integer 𝑎 such that both 𝑎 and 𝑎 + 1 are quadratic nonresidues modulo 𝑝. Prove that 𝑎 + 𝑝Z ⊆ 𝐻, even though 𝐹1 , 𝐹2 , and 𝐹3 (in the notation of Exercise 2) are not linearly disjoint over 𝐸. 4. Let 𝐾 be a Hilbertian field with valuation 𝑣. Prove that every Hilbert subset 𝐻 of 𝐾 𝑟 is 𝑣-dense in 𝐾 𝑟 . Hint: Let 𝐻 = 𝐻𝐾 ( 𝑓1 , . . . , 𝑓𝑚 ), where 𝑓1 , . . . , 𝑓𝑚 are irreducible polynomials in 𝐾 (𝑇1 , . . . , 𝑇𝑟 ) [𝑋1 , . . . , 𝑋𝑛 ]. For (𝑎 1 , . . . , 𝑎𝑟 ) ∈ 𝐾 𝑟 and 𝛾 = 𝑣(𝑐) an element of the value group, each polynomial in the set { 𝑓𝑖 (𝑎 1 + 𝑐𝑇1𝜀1 , . . . , 𝑎𝑟 + 𝑐𝑇𝑟𝜀𝑟 , X) | 1 ≤ 𝑖 ≤ 𝑚 and 𝜀1 , . . . , 𝜀𝑟 ∈ {±1}} is irreducible in 𝐾 (T) [X]. Substitute elements 𝑡1 , . . . , 𝑡𝑟 ∈ 𝐾 for 𝑇1 , . . . , 𝑇𝑟 so that these polynomials remain irreducible in 𝐾 [X]. Thus, find (𝑏 1 , . . . , 𝑏𝑟 ) ∈ 𝐻 such that 𝑣(𝑏 𝑖 − 𝑎 𝑖 ) ≥ 𝛾, 𝑖 = 1, . . . , 𝑟.

14.6 Non-Hilbertian 𝑔-Hilbertian Fields

265

Notes Hilbert [Hil92, p. 280] proves Lemma 14.1.1(b) for number fields. Corollary 14.2.4 reproduces [Har99a, Lemma 2.4]. Our proof of Proposition 14.2.1 is a version of Inaba’s proof [Ina44]. Theorem 14.3.6(a) for number fields appears in [Eic39]. We follow [Fri74a]. Lang [Lan62, p. 152] reproduces Franz’s power series expansion proof [Fra31]. Theorem 14.6.2 for 𝑔 = 0 and 𝐾 a number field appears in [CoZ98]. Generalization to arbitrary 𝑔 and arbitrary countable Hilbertian field 𝐾 of characteristic 0 appears in [FrJ98]. The proof uses a deep group-theoretic result due to Guralnick, Thompson, et al. and the Riemann existence theorem. Thus, one cannot generalize the proof to positive characteristic. An elementary proof of the theorem for arbitrary 𝑔 but still for 𝐾 a number field can be found in [Zan98]. Our proof generalizes that of Zannier. We replace the polynomial 𝑌 2 − 𝑋 2𝑞 + 2 which Zannier uses by the polynomial 𝑌 𝑙 − 𝑋 𝑙𝑞 + 𝑢 in order to make the proof works for each global field 𝐾. Proposition 14.4.1 says that the ring of integers 𝑂 𝐾 of a number field 𝐾 is Hilbertian. Thus, 𝑂 𝐾 ∩ 𝐻𝐾 ( 𝑓 ) is an infinite set for each irreducible polynomial 𝑓 ∈ 𝐾 [𝑇, 𝑋]. One may further ask when 𝑂 𝐾 ∖ 𝐻𝐾 ( 𝑓 ) is finite. This is not always the case. For example, 𝑂 𝐾 ∖ 𝐻𝐾 (𝑋 2 − 𝑇) is the infinite set of all squares in 𝑂 𝐾 . But, for 𝐾 = Q, Müller proves that the set Z ∖ 𝐻Q ( 𝑓 ) is finite in each of the following cases: deg𝑋 ( 𝑓 ) is a prime number and the curve 𝑓 (𝑇, 𝑋) = 0 has positive genus or Gal( 𝑓 (𝑇, 𝑋), Q(𝑇))  𝑆 𝑛 for some positive integer 𝑛 ≠ 5 [Mul97, Thm. 1.2]. The proof uses Siegel’s theorem about integral points on algebraic curves and classical results about finite groups.

Chapter 15

The Diamond Theorem

Each finite proper separable extension of a Galois extension of a Hilbertian field 𝐾 is Hilbertian. This is a theorem of Weissauer. Moreover, if 𝐿 1 and 𝐿 2 are Galois extensions of 𝐾 and neither of them contains the other, then 𝐿 1 𝐿 2 is Hilbertian. The diamond theorem says even more: each extension 𝑀 of 𝐾 in 𝐿 1 𝐿 2 which is contained in neither 𝐿 1 nor in 𝐿 2 is Hilbertian (Theorem 15.2.3). An essential tool in the proof is the twisted wreath product (Section 15.1).

15.1 Twisted Wreath Products Given finite groups 𝐴 and 𝐺, there are several ways of constructing a new group out of them. We describe here two of them: the semidirect product and the wreath product. Definition 15.1.1 (Semidirect products) Consider profinite groups 𝐴 and 𝐺. Suppose that 𝐺 acts on 𝐴 continuously from the right. That is, there is a continuous map 𝐺 × 𝐴 → 𝐴 mapping (𝜎, 𝑎) onto 𝑎 𝜎 and satisfying these rules: (𝑎𝑏) 𝜎 = 𝑎 𝜎 𝑏 𝜎 , 𝑎 1 = 𝑎, and (𝑎 𝜎 ) 𝜏 = 𝑎 𝜎 𝜏 for 𝑎, 𝑏 ∈ 𝐴 and 𝜎, 𝜏 ∈ 𝐺. The semidirect product 𝐺 ⋉ 𝐴 consists of all pairs (𝜎, 𝑎) ∈ 𝐺 × 𝐴 with the product rule (𝜎, 𝑎) (𝜏, 𝑏) = (𝜎𝜏, 𝑎 𝜏 𝑏). This makes 𝐺 ⋉ 𝐴 a profinite group with unit element (1, 1) and the inverse operation −1 (𝜎, 𝑎) −1 = (𝜎 −1 , (𝑎 −1 ) 𝜎 ). Identify each 𝜎 ∈ 𝐺 (resp. 𝑎 ∈ 𝐴) with the pair (𝜎, 1) (resp. (1, 𝑎)). This embeds 𝐺 and 𝐴 into 𝐺 ⋉ 𝐴 such that 𝐴 is normal, 𝐺 ∩ 𝐴 = 1, and 𝐺 𝐴 = 𝐺 ⋉ 𝐴.

(15.1)

Each element of 𝐺 ⋉ 𝐴 has a unique presentation as 𝜎𝑎 with 𝜎 ∈ 𝐺 and 𝑎 ∈ 𝐴. The product rule becomes (𝜎𝑎) (𝜏𝑏) = (𝜎𝜏) (𝑎 𝜏 𝑏) and the action of 𝐺 on 𝐴 coincides with conjugation: 𝑎 𝜎 = 𝜎 −1 𝑎𝜎. The projection 𝜎𝑎 ↦→ 𝜎 of 𝐺 ⋉ 𝐴 on 𝐺 is an epimorphism with kernel 𝐴. Conversely, if 𝐻 is a profinite group and 𝐴, 𝐺 are closed subgroups satisfying 𝐴 ⊳ 𝐻, 𝐺 ∩ 𝐴 = 1, and 𝐺 𝐴 = 𝐻, then 𝐻 is the semidirect product 𝐺 ⋉ 𝐴. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 267 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_15

268

15 The Diamond Theorem

Likewise, consider a short exact sequence 𝛼

1 −→ 𝐴 −→ 𝐻 −→ 𝐺 −→ 1. Suppose that the sequence splits. That is, there exists a homomorphism 𝛼 ′: 𝐺 → 𝐻 satisfying 𝛼(𝛼 ′ (𝑔)) = 𝑔 for each 𝑔 ∈ 𝐺. Then, 𝛼 ′ is an embedding which we call a group-theoretic section of 𝛼. Identifying 𝐺 with 𝛼 ′ (𝐺), we have 𝐻 = 𝐺 ⋉ 𝐴. Suppose that 𝜑1 : 𝐺 → 𝐸 and 𝜑2 : 𝐴 → 𝐸 are homomorphisms of profinite groups and 𝜑2 (𝑎 𝜎 ) = 𝜑2 (𝑎) 𝜑1 ( 𝜎) for all 𝑎 ∈ 𝐴 and 𝜎 ∈ 𝐺. Then, 𝜑(𝜎𝑎) := 𝜑1 (𝜎)𝜑2 (𝑎) is a homomorphism 𝜑: 𝐺 ⋉ 𝐴 → 𝐸. Here is a Galois-theoretic interpretation of semidirect products: Let 𝐾, 𝐿, 𝐸, 𝐹 be fields, with 𝐿/𝐾 Galois, 𝐹/𝐾 Galois, 𝐸 𝐿 = 𝐹, and 𝐸 ∩ 𝐿 = 𝐾. Then, Gal(𝐹/𝐾) = Gal(𝐹/𝐸) ⋉ Gal(𝐹/𝐿). Definition 15.1.2 (Twisted wreath products) Let 𝐴 and 𝐺 be finite groups, 𝐺 0 a subgroup of 𝐺, and Γ a system of representatives for the right cosets 𝐺 0 𝜎, 𝜎 ∈ 𝐺. Thus, Ø Ø 𝐺 = · 𝐺 0 𝛾 = · 𝛾 −1 𝐺 0 . 𝛾 ∈Γ

𝛾 ∈Γ

Suppose that 𝐺 0 acts on 𝐴 from the right. Let 𝜎0 Ind𝐺 for all 𝜎 ∈ 𝐺 and 𝜎0 ∈ 𝐺 0 }. 𝐺0 ( 𝐴) = { 𝑓 : 𝐺 → 𝐴 | 𝑓 (𝜎𝜎0 ) = 𝑓 (𝜎)

Thus, for each subset {𝑎 𝛾 | 𝛾 ∈ Γ} of 𝐴 there exists a unique 𝑓 ∈ Ind𝐺 𝐺0 ( 𝐴) with 𝑓 (𝛾 −1 ) = 𝑎 𝛾 for 𝛾 ∈ Γ. Indeed, 𝑓 (𝛾 −1 𝜎0 ) = 𝑎 𝛾𝜎0 for 𝛾 ∈ Γ and 𝜎0 ∈ 𝐺 0 . 𝑔 Make Ind𝐺 𝐺0 ( 𝐴) a group by the rule ( 𝑓 𝑔) (𝜎) = 𝑓 (𝜎)𝑔(𝜎). Then, 𝑓 (𝜎) = 𝐺 𝑔 ( 𝜎) 𝑔 𝑓 (𝜎) , where 𝑓 denotes conjugation in Ind𝐺0 ( 𝐴) and the right-hand side is conjugation in 𝐴. 𝜎 The group 𝐺 acts on Ind𝐺 𝐺0 ( 𝐴) by 𝑓 (𝜏) = 𝑓 (𝜎𝜏). This gives rise to the semidirect product 𝐺 ⋉ Ind𝐺 𝐺0 ( 𝐴), which we also denote by 𝐴wr𝐺0 𝐺. Each element of this group has a unique presentation as a product 𝜎 𝑓 with 𝜎 ∈ 𝐺 and 𝑓 ∈ Ind𝐺 𝐺0 ( 𝐴). By Definition 15.1.1, the product and the inverse operation −1

in 𝐴wr𝐺0 𝐺 are given by (𝜎 𝑓 ) (𝜏𝑔) = 𝜎𝜏 𝑓 𝜏 𝑔 and (𝜎 𝑓 ) −1 = 𝜎 −1 ( 𝑓 −1 ) 𝜎 . The identifications 𝜎 = 𝜎 · 1 and 𝑓 = 1 · 𝑓 identify 𝐺 and Ind𝐺 𝐺0 ( 𝐴) as subgroups of 𝐺 𝐴wr𝐺0 𝐺. Under these identifications, Ind𝐺0 ( 𝐴) is normal, 𝐺 ∩ Ind𝐺 𝐺0 ( 𝐴) = 1, and 𝐺 −1 𝜎 𝐺 · Ind𝐺0 ( 𝐴) = 𝐴wr𝐺0 𝐺. Since (𝜎 · 1) (1 · 𝑓 ) (𝜎 · 1) = 1 · 𝑓 , conjugation of 𝑓 by 𝜎 in 𝐴wr𝐺0 𝐺 coincides with the action of 𝜎 on 𝑓 . The map 𝜎 𝑓 ↦→ 𝜎 is an epimorphism 𝜋: 𝐴wr𝐺0 𝐺 → 𝐺 with Ker(𝜋) = Ind𝐺 𝐺0 ( 𝐴). We call 𝐴wr𝐺0 𝐺 the twisted wreath product of 𝐴 and 𝐺 with respect to 𝐺 0 . Twisted wreath products are usually non-Abelian: Lemma 15.1.3 Let 𝐴 be a nontrivial finite group, 𝐺 a finite group, and 𝐺 0 a proper subgroup of 𝐺 which acts on 𝐴 from the right. Then, 𝐴wr𝐺0 𝐺 is not commutative. 𝜎 Proof. Choose 𝑎 ∈ 𝐴, 𝑎 ≠ 1. Define a function 𝑓 ∈ Ind𝐺 𝐺0 ( 𝐴) by 𝑓 (𝜎) = 𝑎 for 𝜎 ∖ ∖ 𝜎 ∈ 𝐺 0 and 𝑓 (𝜎) = 1 for 𝜎 ∈ 𝐺 𝐺 0 . Choose 𝜎 ∈ 𝐺 𝐺 0 . Then, 𝑓 (1) = 𝑓 (𝜎) = 1 ≠ 𝑎 = 𝑓 (1). Hence, 𝑓 𝜎 ≠ 𝑓 . Consequently, 𝐴wr𝐺0 𝐺 is not commutative. □

15.1 Twisted Wreath Products

269

Lemma 15.1.4 If 𝐴 and 𝐵 are proper subgroups of a group 𝐺, then 𝐴 ∪ 𝐵 ≠ 𝐺. Proof. We may assume that 𝐴 ≮ 𝐵 and 𝐵 ≮ 𝐴. Thus, there exist 𝑏 ∈ 𝐵 ∖ 𝐴 and 𝑎 ∈ 𝐴 ∖ 𝐵, so 𝑎𝑏 ∈ 𝐺 ∖ ( 𝐴 ∪ 𝐵). □ The next result shows that in general not only are twisted wreath products nonAbelian but their centers are small: Lemma 15.1.5 Let 𝜋: 𝐴wr𝐺0 𝐺 → 𝐺 be a twisted wreath product of finite groups, 𝐻1 ⊳ 𝐴wr𝐺0 𝐺, and ℎ2 ∈ 𝐴wr𝐺0 𝐺. Put 𝐼 = Ind𝐺 𝐺0 ( 𝐴) = Ker(𝜋) and 𝐺 1 = 𝜋(𝐻1 ). Suppose that 𝐴 ≠ 1. (a) Suppose that 𝜋(ℎ2 ) ∉ 𝐺 0 and (𝐺 1 𝐺 0 : 𝐺 0 ) > 2. Then, there is an ℎ1 ∈ 𝐻1 ∩ 𝐼 with ℎ1 ℎ2 ≠ ℎ2 ℎ1 . (b) Suppose that 𝐺 1 ̸ ≤ 𝐺 0 and 𝜋(ℎ2 ) ∉ 𝐺 1 𝐺 0 . Then, there is an ℎ1 ∈ 𝐻1 ∩ 𝐼 with ′ ℎ1ℎ2 ∉ ⟨ℎ1 ⟩ ℎ for all ℎ ′ ∈ 𝜋 −1 (𝐺 1 𝐺 0 ). In particular, ℎ1 ℎ2 ≠ ℎ2 ℎ1 . Proof. Put 𝜎2 = 𝜋(ℎ2 ). Consider 𝜎1 ∈ 𝐺 1 and 𝑔 ∈ 𝐼. By definition, there are 𝑓1 , 𝑓2 ∈ 𝐼 with 𝜎1 𝑓1 ∈ 𝐻1 and ℎ2 = 𝜎2 𝑓2 . Put ℎ1 = 𝑔 𝜎1 𝑓1 𝑔 −1 . Then, ℎ1 = [𝜎1 𝑓1 , 𝑔 −1 ] ∈ [𝐻1 , 𝐼] ≤ 𝐻1 ∩ 𝐼. For each 𝜏 ∈ 𝐺
ℎ1 (𝜏) = (𝑔 𝜎1 ) 𝑓1 (𝜏)𝑔(𝜏) −1 = 𝑔(𝜎1 𝜏) 𝑓1 ( 𝜏) 𝑔(𝜏) −1 . Hence, for all 𝜏 ∈ 𝐺 and 𝑓 ′ ∈ 𝐼 we have: (15.2a) ℎ1ℎ2 (1) = ℎ1 2 2 (1) = ℎ1 (𝜎2 ) 𝑓2 (1) = 𝑔(𝜎1 𝜎2 ) 𝑓1 ( 𝜎2 ) 𝑓2 (1) (𝑔(𝜎2 ) −1 ) 𝑓2 (1) , ′ ′ ′ 𝜏 𝑓′ (15.2b) ℎ1 (1) = ℎ1 (𝜏) 𝑓 (1) = 𝑔(𝜎1 𝜏) 𝑓1 ( 𝜏) 𝑓 (1) (𝑔(𝜏) −1 ) 𝑓 (1) , (15.2c) ℎ1 (1) = 𝑔(𝜎1 ) 𝑓1 (1) 𝑔(1) −1 . We apply (15.2a), (15.2b), and (15.2c) in the proofs of (a) and (b) to special values 𝜎1 and 𝑔. Choose 𝑎 ∈ 𝐴 with 𝑎 ≠ 1. 𝜎 𝑓

Proof of (a). Since (𝐺 1 𝐺 0 : 𝐺 0 ) > 2, there is a 𝜎1 ∈ 𝐺 1 with distinct cosets 𝜎1−1 𝐺 0 , 𝜎2 𝐺 0 , 𝐺 0 . Thus, none of the cosets 𝜎1 𝐺 0 , 𝜎2 𝐺 0 , 𝜎1 𝜎2 𝐺 0 is 𝐺 0 . Therefore, by definition of 𝐼, there is a 𝑔 ∈ 𝐼 with 𝑔(𝜎1 ) = 𝑔(𝜎2 ) = 𝑔(𝜎1 𝜎2 ) = 1 and 𝑔(1) = 𝑎. By (15.2a), ℎ1ℎ2 (1) = 1. By (15.2c), ℎ1 (1) ≠ 1. Consequently, ℎ1ℎ2 ≠ ℎ1 , as desired. 𝜎 −1

Proof of (b). Since 𝐺 1𝜎2 = 𝐺 1 ̸ ≤ 𝐺 0 , we have 𝐺 1 ̸ ≤ 𝐺 0 2 . Hence, 𝐺 1 ∩ 𝐺 0 and 𝜎 −1

𝐺 1 ∩ 𝐺 0 2 are proper subgroups of 𝐺 1 . Thus, by Lemma 15.1.4, there is an element 𝜎 −1

𝜎1 ∈ 𝐺 1 ∖ (𝐺 0 ∪ 𝐺 0 2 ). It follows that 𝜎2 ∉ 𝜎1 𝜎2 𝐺 0 . By assumption, 𝜎2 ∉ 𝐺 1 𝐺 0 . Therefore, there is a 𝑔 ∈ 𝐼 with 𝑔(𝐺 1 𝐺 0 ) = 1, 𝑔(𝜎1 𝜎2 ) = 1, and 𝑔(𝜎2 ) = 𝑎 −1 . Consider 𝜏 ∈ 𝐺 1 𝐺 0 and 𝑓 ′ ∈ 𝐼. By (15.2a), ℎ1ℎ2 (1) = 𝑎 𝑓2 (1) ≠ 1. By (15.2b), ′ ′ 𝜏 𝑓′ ℎ1 (1) = 1. Hence, (ℎ1𝑘 ) 𝜏 𝑓 (1) = 1 for all integers 𝑘. It follows that ℎ1ℎ2 ∉ ⟨ℎ1 ⟩ ℎ for all ℎ ′ ∈ 𝜋 −1 (𝐺 1 𝐺 0 ). □ Remark 15.1.6 (Decomposition of Ind𝐺 𝐺0 ( 𝐴) into a direct product) To each 𝑎 ∈ 𝐴 associate a function 𝑓 𝑎 : 𝐺 → 𝐴: ( 𝑎 𝜎 if 𝜎 ∈ 𝐺 0 𝑓 𝑎 (𝜎) = 1 if 𝜎 ∉ 𝐺 0 .

270

15 The Diamond Theorem

These functions satisfy the following rules: 𝑓 𝑎 𝑓𝑏 = 𝑓 𝑎𝑏 and 𝑔 −1 𝑓 𝑎 𝑔 = 𝑓 𝑎𝑔 (1) for all 𝑎, 𝑏 ∈ 𝐴 and 𝑔 ∈ Ind𝐺 𝐺0 ( 𝐴). Thus, the map 𝑎 ↦→ 𝑓 𝑎 identifies 𝐴 with the normal 𝐺 subgroup { 𝑓 ∈ Ind𝐺0 ( 𝐴) | 𝑓 (𝐺 ∖ 𝐺 0 ) = 1} of Ind𝐺 𝐺0 ( 𝐴). Applying 𝜎 ∈ 𝐺 on 𝐴 gives the following normal subgroup of Ind𝐺 𝐺0 ( 𝐴): ∖ −1 𝐴 𝜎 = { 𝑓 ∈ Ind𝐺 𝐺0 ( 𝐴) | 𝑓 (𝐺 𝜎 𝐺 0 ) = 1}. Î 𝐺 ( 𝐴) has a unique presentation 𝑓 = 𝛾 ∈Γ 𝑓 𝛾 , with An arbitrary element 𝑓 ∈ Ind𝐺 0 𝑓 𝛾 ∈ 𝐴 𝛾 . Specifically, 𝑓 𝛾 (𝐺 ∖ 𝛾 −1 𝐺 0 ) = 1 and 𝑓 𝛾 (𝛾 −1 𝜎0 ) = 𝑓 (𝛾 −1 𝜎0 ) for all 𝛾 ∈ Γ and 𝜎0 ∈ 𝐺 0 . It follows that Ö Ind𝐺 (15.3) 𝐴𝛾 . 𝐺0 ( 𝐴) = 𝛾 ∈Γ

The latter relation allows us to present 𝐴 as a quotient of Ind𝐺 𝐺0 ( 𝐴) in various 𝐺 ways: Let 𝑁 = { 𝑓 ∈ Ind𝐺0 ( 𝐴) | 𝑓 (1) = 1}. For each 𝜎 ∈ 𝐺 let 𝑁 𝜎 = { 𝑓 𝜎 | 𝑓 ∈ −1 −1 𝑁 } = { 𝑓 ∈ Ind𝐺 𝐺0 ( 𝐴) | 𝑓 (𝜎 ) = 1}. Then, the map 𝑓 ↦→ 𝑓 (𝜎 ) gives rise to 𝐺 𝜎 a short exact sequence 1 → 𝑁 → Ind𝐺0 ( 𝐴) → 𝐴 → 1. Like in the preceding paragraph, we find that Ö −1 𝐴𝛾 . 𝑁 𝜎 = { 𝑓 ∈ Ind𝐺 (15.4) 𝐺0 ( 𝐴) | 𝑓 (𝜎 ) = 1} = 𝛾∈Γ 𝐺0 𝛾≠𝐺0 𝜎

Note that 𝐺 0 leaves 𝑁 = 𝑁 1 invariant, so 𝑁 ⊳ Ind𝐺 𝐺0 ( 𝐴)𝐺 0 . We summarize some of the groups mentioned above in the following diagram: Ind𝐺 𝐺0 ( 𝐴)

Ind𝐺 𝐺0 ( 𝐴)𝐺 0

𝑁

𝑁𝐺 0

1

𝐺0

𝐴wr𝐺0 𝐺

(15.5)

𝐺

Remark 15.1.7 (Interpretation of twisted wreath products in Galois theory) ˆ ˆ (a) Let 𝐹/𝐾 be a finite Galois extension with Gal( 𝐹/𝐾)  𝐴wr𝐺0 𝐺 and 𝐴 ≠ 1. ˆ View Ind𝐺 ( 𝐴) and 𝐺 under this isomorphism as subgroups of Gal( 𝐹/𝐾). Let 𝐹, 𝐺0 𝐺 ˆ ˆ 𝐿, 𝐿 0 , and 𝐾 be the fixed fields in 𝐹, respectively, of the subgroups 𝑁, Ind𝐺0 ( 𝐴), Ind𝐺 𝐺0 ( 𝐴)𝐺 0 , and 𝐺. Galois theory interprets the various relations among the subgroups of 𝐴wr𝐺0 𝐺 as relations among fields: ˆ (15.6a) 𝐾 ⊆ 𝐿 0 ⊆ 𝐿 ⊂ 𝐹 ⊆ 𝐹. ˆ ˆ ˆ (15.6b) 𝐿 ∩ 𝐾 = 𝐾 and 𝐿 𝐾 = 𝐹. ˆ (15.6c) 𝐿/𝐾, 𝐹/𝐿 0 , and 𝐹/𝐾 are finite Galois extensions. 𝛾 (15.6d) The fields Î 𝐹 𝛾with 𝛾 ∈ Γ are linearly disjoint over 𝐿 (by (15.3)). Moreover, ˆ 𝐹 = 𝛾 ∈Γ 𝐹 . (15.6e) There is a field 𝐹0 with 𝐿 ∩ 𝐹0 = 𝐿 0 and 𝐹 = 𝐿𝐹0 .

15.1 Twisted Wreath Products

271

𝐺0

𝐹ˆ

𝑁

𝐴

𝐿

𝐹 𝐺0

𝐿0

𝐺0

𝐿 0 𝐾ˆ

𝐹0

𝐾ˆ

𝐾

The assertion “𝐹/𝐿 0 is Galois” follows from “𝑁 ⊳ Ind𝐺 𝐺0 ( 𝐴)𝐺 0 ”. Thus, Condition ˆ (15.6e) follows from (15.6a)–(15.6d) by taking 𝐹0 = 𝐹 ∩ 𝐿 0 𝐾. ˆ 𝐾ˆ that satisfy Conditions (15.6a)– (b) Conversely, consider fields 𝐾, 𝐿 0 , 𝐿, 𝐹, 𝐹, (15.6d). Let ˆ ˆ 𝐾) ˆ  Gal(𝐿/𝐾), 𝐹0 = 𝐹 ∩ (𝐿 0 𝐾), 𝐺 = Gal( 𝐹/ ˆ 0 𝐾) ˆ  Gal(𝐹/𝐹0 )  Gal(𝐿/𝐿 0 ), 𝐺 0 = Gal( 𝐹/𝐿 𝐴 = Gal(𝐹/𝐿)  Gal(𝐹0 /𝐿 0 ). ˆ In particular, Γ is a subset of Gal( 𝐹/𝐾). Suppose also that (15.6d) holds. Then, 𝐴 ⊳ Gal(𝐹/𝐿 0 ). Thus, 𝐺 0 viewed as a subgroup of Gal(𝐹/𝐿 0 ) acts on 𝐴 by conjugation. ˆ We construct an isomorphism 𝜑: 𝐴wr𝐺0 𝐺 → Gal( 𝐹/𝐾) which is the identity on 𝐺 𝐺 ˆ and maps Ind𝐺0 ( 𝐴) onto Gal( 𝐹/𝐿). ˆ is a Galois extension of degree | 𝐴| |Γ | . Hence, Construction of 𝜑. By (15.6d), 𝐹/𝐿 ˆ ˆ 𝐾) ˆ ⋉ Gal( 𝐹/𝐿). ˆ by (15.6b), Gal( 𝐹/𝐾) = Gal( 𝐹/ For each 𝜎 ∈ 𝐺, the group Ind𝐺 ( 𝐴) acts on 𝐹 𝜎 by the rule 𝐺0   −1
𝑓 ( 𝜎 −1 ) 𝜎 𝑧 𝑓 = 𝑧𝜎 , 𝑓 ∈ Ind𝐺 𝑧 ∈ 𝐹 𝜎. (15.7) 𝐺0 ( 𝐴), ′

This action does not depend on 𝜎. Indeed, assume 𝐹 𝜎 = 𝐹 𝜎 with 𝜎, 𝜎 ′ ∈ 𝐺. Write 𝜎 = 𝜎0 𝛾 and 𝜎 ′ = 𝜎0′ 𝛾 ′ with 𝜎0 , 𝜎0′ ∈ 𝐺 0 and 𝛾, 𝛾 ′ ∈ Γ. Since 𝐹/𝐹0 is Galois, ′ 𝐹 𝛾 = 𝐹 𝛾 , so by (15.6a) and (15.6d), 𝛾 = 𝛾 ′. Thus, 𝜎 ′ = 𝜌𝜎 with 𝜌 = 𝜎0′ 𝜎0−1 ∈ 𝐺 0 . Moreover, 𝜌 −1 𝑓 (𝜎 −1 𝜌 −1 ) 𝜌 = 𝑓 (𝜎 −1 𝜌 −1 ) 𝜌 = 𝑓 (𝜎 −1 𝜌 −1 𝜌) = 𝑓 (𝜎 −1 ). Hence, 

𝑧 (𝜌 𝜎)

−1


𝑓 ( (𝜌 𝜎) −1 )  𝜌 𝜎

=



𝑧𝜎

−1


𝑓 ( 𝜎 −1 )  𝜎

.

If 𝑧 ∈ 𝐿, then 𝑧 𝑓 = 𝑧, because 𝑓 (𝜎 −1 ) as an element of 𝐴 fixes 𝐿. Thus, the action 𝜎 (15.7) defines a homomorphism 𝜑 𝜎 : Ind𝐺 𝐺0 ( 𝐴) → Gal(𝐹 /𝐿). If 𝜑 𝜎 ( 𝑓 ) = 1, then −1

−1

−1

−1

(𝑧 𝜎 ) 𝑓 ( 𝜎 ) = 𝑧 𝜎 for each 𝑧 ∈ 𝐹 𝜎 . Hence, 𝑧 𝑓 ( 𝜎 ) = 𝑧 for each 𝑧 ∈ 𝐹. Therefore, 𝑓 (𝜎 −1 ) = 1. It follows that Ker(𝜑 𝜎 ) = 𝑁 𝜎 , with 𝑁 𝜎 as in (15.4). Using (15.6d), the 𝜑 𝛾 ’s, with 𝛾 ranging on Γ, define a homomorphism Ö ˆ 𝜑0 : Ind𝐺 ( 𝐴) → Gal(𝐹 𝛾 /𝐿) = Gal( 𝐹/𝐿) 𝐺0 𝛾 ∈Γ

Î

Ñ by 𝜑0 ( 𝑓 ) = 𝛾 ∈Γ 𝜑 𝛾 ( 𝑓 ). The kernel of 𝜑0 is 𝛾 ∈Γ 𝑁 𝛾 = 1 and |Ind𝐺 𝐺0 ( 𝐴)| = |Γ | ˆ | 𝐴| = |Gal( 𝐹/𝐿)|. Hence, 𝜑0 is an isomorphism.

272

15 The Diamond Theorem

Next we show that 𝜑0 is compatible with the action of 𝐺. To this end let 𝑧 ∈ 𝐹 𝜎 −1 −1 and 𝜏 ∈ 𝐺. Then, 𝑧 𝜏 ∈ 𝐹 𝜎 𝜏 and  −1
 𝜏    ( 𝜎 𝜏 −1 )  𝜏 −1 𝑓 𝜏 𝜏 −1 ( 𝜏 𝜎 −1 )
𝑓 ( 𝜏 𝜎 ) 𝑧 (𝑧 ) =   −1 −1
𝑓 𝜏 ( 𝜎 −1 ) 𝜎 −1
𝜎 𝜏 = (𝑧 𝜎 ) 𝑓 ( 𝜏 𝜎 ) = 𝑧 𝜎 = 𝑧𝑓 . Hence, 𝜏 −1 𝜑0 ( 𝑓 )𝜏 = 𝜑0 ( 𝑓 𝜏 ), as desired. This allows us to combine 𝜑0 with the ˆ identity map of 𝐺 and define the isomorphism 𝜑: 𝐴wr𝐺0 𝐺 → Gal( 𝐹/𝐾). ˆ ˆ Having established 𝜑, we say that the fields 𝐾, 𝐾, 𝐿 0 , 𝐿, 𝐹, 𝐹 realize the twisted wreath product 𝐴wr𝐺0 𝐺. We say that the fields 𝐾, 𝐿 0 , 𝐿, 𝐹, 𝐹ˆ realize the twisted ˆ 𝐿 0 , 𝐿, 𝐹, 𝐹ˆ realize the wreath product if there exists a field 𝐾ˆ such that 𝐾, 𝐾, twisted wreath product. ˆ 𝐿 0 , 𝐿, 𝐹, 𝐹ˆ that realize 𝐴wr𝐺0 𝐺. Suppose that 𝐸 is a (c) Consider fields 𝐾, 𝐾, Î Galois extension of 𝐿 0 with 𝐿 ⊆ 𝐸 ⊆ 𝐹. Let 𝐴¯ = Gal(𝐸/𝐿) and 𝐸ˆ = 𝛾 ∈Γ 𝐸 𝛾 . ˆ Let 𝐽 = 𝐸ˆ ∩ 𝐾. ˆ Then, 𝐸ˆ 𝐾ˆ = 𝐹. ˆ Hence, Then, 𝐸ˆ is a Galois extension of 𝐾 in 𝐹. ˆ = 𝐺 and Gal( 𝐸/𝐿 ˆ 0 𝐾) ˆ ˆ 𝐾) ˆ = 𝐺 0 . Moreover, ˆ 0 𝐽)  Gal( 𝐹/𝐿 Gal( 𝐸/𝐽)  Gal( 𝐹/ ˆ 𝐿 ∩ 𝐽 = 𝐾 and 𝐿𝐽 = 𝐸. ⑥⑥ ⑥⑥ ⑥⑥ 𝐿

𝐸

⑦ ⑦⑦ ⑦⑦ 𝐿0

𝐹 𝐸ˆ

①① ①① ① ①①

𝐿 0 𝐾ˆ ① ① ①① ①①

𝐹0

𝐸0

𝐾

𝐹ˆ

𝐿0 𝐽

𝐽

✇✇ ✇✇ ✇ ✇✇

𝐾ˆ

¯ 𝐺0 𝐺. By (b), 𝐾, 𝐽, 𝐿 0 , 𝐿, 𝐸, 𝐸ˆ realize 𝐴wr Remark 15.1.8 (Wreath products) Suppose that 𝐺 0 is the trivial subgroup of 𝐺. Then, the twisted wreath product 𝐴wr𝐺0 𝐺 simplifies to the (usual) wreath product 𝐴wr𝐺. 𝐺 In this case Ind𝐺 𝐺0 ( 𝐴) is just the group 𝐴 of all functions 𝑓 : 𝐺 → 𝐴. Multiplication is carried out componentwise. Again, 𝐺 acts on 𝐴𝐺 by the formula 𝑓 𝜎 (𝜏) = 𝑓 (𝜎𝜏). Thus, 𝐴wr𝐺 is the semidirect product 𝐺 ⋉ 𝐴𝐺 . Each element of 𝐺 ⋉ 𝐴𝐺 has a unique representation as a product 𝜎 𝑓 with 𝜎 ∈ 𝐺 and 𝑓 ∈ 𝐴𝐺 . The multiplication rule is 𝜎 𝑓 · 𝜏𝑔 = (𝜎𝜏)( 𝑓 𝜏 𝑔). Identify each 𝑎 ∈ 𝐴 with the function 𝑓 𝑎 : 𝐺 → 𝐴 given by Î 𝜎 𝑓 𝑎 (1) = 𝑎 and 𝑓 𝑎 (𝜎) = 1 for 𝜎 ≠ 1. Then, Ind𝐺 𝜎 ∈𝐺 𝐴 . 𝐺0 ( 𝐴) = Now suppose that 𝐾 ⊆ 𝐿 ⊆ 𝐹 ⊆ 𝐹ˆ is a tower of fields and 𝐾ˆ is a field satisfying these conditions: ˆ (15.8a) 𝐿/𝐾, 𝐹/𝐿, and 𝐹/𝐾 are finite Galois extensions.

15.2 The Diamond Theorem

273

ˆ ˆ Write 𝐺 = Gal( 𝐹/ ˆ 𝐾). (15.8b) 𝐿 ∩ 𝐾ˆ = 𝐾 and 𝐿 𝐾ˆ = 𝐹. Î (15.8c) The fields 𝐹 𝛾 with 𝛾 ∈ 𝐺 are linearly disjoint over 𝐿 and 𝐹ˆ = 𝛾 ∈𝐺 𝐹 𝛾 . ˆ → Gal( 𝐹/𝐾) By Remark 15.1.7(b), there is an isomorphism 𝜑: 𝐴wr𝐺 which Î ˆ ˆ and 𝜎≠1 𝐴 𝜎 onto Gal( 𝐹/𝐹). is the identity on 𝐺, maps 𝐴𝐺 onto Gal( 𝐹/𝐿), Moreover, by (15.8b), restriction to 𝐿 maps 𝐺 isomorphically onto Gal(𝐿/𝐾). We ˆ 𝐿, 𝐹, 𝐹ˆ realize the wreath product 𝐴wr𝐺. say that the fields 𝐾, 𝐾,

15.2 The Diamond Theorem The diamond theorem proved in this section says that all fields ‘captured’ between two Galois extensions of a Hilbertian field are Hilbertian. In particular, this theorem implies that a non-Hilbertian Galois extension 𝑁 of a Hilbertian field 𝐾 is not the compositum of two Galois extensions of 𝐾 which are properly contained in 𝑁. For example, 𝐾sep is not the compositum of two proper subfields which are Galois over 𝐾. Lemma 15.2.1 Let 𝐾 ⊆ 𝐿 0 ⊆ 𝐿 be fields with 𝐿/𝐾 finite and Galois. Let 𝑐 1 , . . . , 𝑐 𝑛 be a basis of 𝐿 0 /𝐾 and 𝑡 1 , . . . , 𝑡 𝑛 algebraically independent elements over 𝐾. Let 𝑓 ∈ 𝐿 0 [𝑈, 𝑋] be an absolutely irreducible polynomial which is monic in 𝑋 and Galois over 𝐿(𝑈) with deg𝑋 𝑓 ≥ 2. Put 𝐺 = Gal(𝐿/𝐾) = Gal(𝐿 (t)/𝐾 (t)), 𝐺 0 = Gal(𝐿/𝐿 0 ) = Gal(𝐿(t)/𝐿 0 (t)), and 𝐴 = Gal( 𝑓 (𝑈, 𝑋), 𝐿(𝑈)). Then, 𝐺 0 acts on 𝐴 and there exist fields 𝐹 and 𝐹ˆ with the following properties: (a) 𝐹ˆ is a regular extension of 𝐿. Moreover, 𝐹 = 𝐿(t, 𝑧) with irr(𝑧, 𝐿(t)) = Í𝑛 𝑓 ( 𝑖=1 𝑐 𝑖 𝑡 𝑖 , 𝑍) ∈ 𝐿 0 [t, 𝑍]. (b) 𝐾 (t), 𝐿 0 (t), 𝐿(t), 𝐹, 𝐹ˆ realize 𝐴wr𝐺0 𝐺. Proof. Fix 𝑥 ∈ 𝐿(𝑈)sep with 𝑓 (𝑈, 𝑥) = 0. Since 𝑓 is absolutely irreducible, [𝐿(𝑈, 𝑥) : 𝐿(𝑈)] = [𝐿 0 (𝑈, 𝑥) : 𝐿 0 (𝑈)], so 𝐿 (𝑈) is linearly disjoint from 𝐿 0 (𝑈, 𝑥) over 𝐿 0 (𝑈) and we may identify both Gal(𝐿 (𝑈)/𝐿 0 (𝑈)) and Gal(𝐿(𝑈, 𝑥)/𝐿 0 (𝑈, 𝑥)) with 𝐺 0 via restriction. Since 𝑓 (𝑈, 𝑋) is Galois over 𝐿 (𝑈) with respect to 𝑋, the extension 𝐿 (𝑈, 𝑥)/𝐿(𝑈) is Galois with Galois group 𝐴. In addition, 𝐴 ≠ 1, because deg𝑋 𝑓 ≥ 2. The fixed field in 𝐿 (𝑈, 𝑥) of the subgroup of Aut(𝐿(𝑈, 𝑥)) generated by 𝐴 and 𝐺 0 is 𝐿 0 (𝑈). By Artin [Lan97, p. 264, Thm. 1.8], 𝐿 (𝑈, 𝑥)/𝐿 0 (𝑈) is Galois and, by Definition 15.1.1, Gal(𝐿 (𝑈, 𝑥)/𝐿 0 (𝑈))  Gal(𝐿(𝑈, 𝑥)/𝐿 0 (𝑈, 𝑥)) ⋉ Gal(𝐿(𝑈, 𝑥)/𝐿(𝑈))  𝐺 0 ⋉ 𝐴. This defines an action of 𝐺 0 on 𝐴. As in Definition 15.1.2, choose a system of representatives Γ for the right cosets 𝐺 0 𝛾 of 𝐺 0 in 𝐺. Let S = {𝐺 0 𝛾 | 𝛾 ∈ Γ}. Choose algebraically independent elements 𝑢 𝑆 , 𝑆 ∈ S, over 𝐾. Write 𝑄 = 𝐿 (𝑢 𝑆 | 𝑆 ∈ S). For each 𝑆 ∈ S write 𝑐 𝑖𝑆 = 𝑐 𝑖𝜎 and 𝑓 𝑆 = 𝑓 𝜎 for some 𝜎 ∈ 𝑆. Since 𝑐 𝑖 ∈ 𝐿 0 and 𝑓 ∈ 𝐿 0 (𝑈, 𝑋), both 𝑐 𝑖𝑆 and 𝑓 𝑆 are independent of 𝜎. Then, choose a root 𝑧 𝑆 of 𝑓 𝑆 (𝑢 𝑆 , 𝑍) in 𝑄 sep and write 𝐹 𝑆 = 𝑄(𝑧 𝑆 ). Since 𝑓 𝑆 is absolutely irreducible, 𝐿 0𝑆 (𝑢 𝑆 , 𝑧 𝑆 ) is a regular extension of 𝐿 0𝑆 (Example 3.4.11). Hence, 𝐿 (𝑢 𝑆 , 𝑧 𝑆 ) is a regular extension of 𝐿. Since 𝐿(𝑢 𝑆 , 𝑧 𝑆 ), 𝑆 ∈ S, are algebraically independent over 𝐿, they are linearly disjoint over 𝐿 (Lemma 3.4.7).

274

15 The Diamond Theorem

Moreover, their compositum 𝐹ˆ = 𝐿(𝑢 𝑆 , 𝑧 𝑆 | 𝑆 ∈ S) = 𝑄(𝑧 𝑆 | 𝑆 ∈ S) is a regular extension of 𝐿. By Lemma 3.1.14, in each rectangle of the following diagram, the fields lying in the left upper corner and the right lower corner are linearly disjoint over the field in the left lower corner and their compositum is the field in the right upper corner: ′

𝐿 (u)(𝑧 𝑆 | 𝑆 ′ ≠ 𝑆)

𝐿(𝑢 𝑆 | 𝑆 ′ ≠ 𝑆)

𝐿(t) = 𝐿 (u) = 𝑄

𝐴

𝐿 (u, 𝑧 𝑆 ) = 𝐹 𝑆

𝐿(𝑢 𝑆 )

𝐴

𝐿

𝐿(𝑢 𝑆 , 𝑧 𝑆 )

𝐿 0𝑆

𝐿 0𝑆 (𝑢 𝑆 )

𝐴

𝐿 0𝑆 (𝑢 𝑆 , 𝑧 𝑆 )

′

𝐿 (𝑢 𝑆 , 𝑧 𝑆 | 𝑆 ′ ≠ 𝑆) ′

′

′ ′ 𝐿 (u𝑆 , 𝑧 𝑆 | 𝑆 ′ ∈ S) = 𝐹ˆ

𝐴

It follows that 𝐹 𝑆 /𝑄 is a Galois extension with Galois group isomorphic to 𝐴, the ˆ 𝐹 𝑆 , 𝑆 ∈ S, are linearly disjoint over 𝑄, and 𝐹ˆ is their compositum. Thus, 𝐹/𝑄 is Galois. The matrix (𝑐 𝑖𝑆 ) ∈ 𝑀𝑛 (𝐿) is invertible [Lan97, p. 286, Cor. 5.4]. Hence, the system of equations 𝑛 ∑︁ 𝑐 𝑖𝑆 𝑇𝑖 = 𝑢 𝑆 , 𝑆∈S (15.9) 𝑖=1

has a unique solution (𝑡1′ , . . . , 𝑡 𝑛′ ). It satisfies 𝐿 (𝑡 1′ , . . . , 𝑡 𝑛′ ) = 𝐿 (𝑢 𝑆 | 𝑆 ∈ S) = 𝑄. Since trans.deg(𝑄/𝐾) = 𝑛, the elements 𝑡1′ , . . . , 𝑡 𝑛′ are algebraically independent over 𝐾. We may therefore assume that 𝑡𝑖′ = 𝑡 𝑖 , 𝑖 = 1, . . . , 𝑛, so 𝑄 = 𝐿 (t). Use the linear disjointness of the 𝐹 𝑆 ’s over 𝑄 to extend the action of 𝐺 on 𝐿 to ˆ an action on 𝐹: (𝑢 𝑆 ) 𝜏 = 𝑢 𝑆 𝜏 and (𝑧 𝑆 ) 𝜏 = 𝑧 𝑆 𝜏 . In particular, 𝜏 permutes the equations of (15.9). Hence, (𝑡1𝜏 , . . . , 𝑡 𝑛𝜏 ) is a solution of (15.9), so it coincides with (𝑡1 , . . . , 𝑡 𝑛 ). In other words, the action of 𝐺 on 𝑄 is the unique extension of the given action on 𝐿 that fixes 𝑡 1 , . . . , 𝑡 𝑛 . In particular, 𝐾 (t) is the fixed field of 𝐺 in 𝑄. Now write 𝑢 = 𝑢 𝐺0 ·1 , Î 𝑧 = 𝑧 𝐺0 ·1 , and 𝐹 = 𝐹 𝐺0 ·1 . Then, 𝐹 𝑆 =Í𝐹 𝜎 for all 𝑆 ∈ S and 𝑛 ˆ each 𝜎 ∈ 𝑆. Hence, 𝐹 = 𝛾 ∈Γ 𝐹 𝛾 . Moreover, by (15.9), 𝑢 = 𝑖=1 𝑐 𝑖 𝑡 𝑖 ∈ 𝐿 0 (t), 𝑢 is transcendental over 𝐾, 𝑓 (𝑢, 𝑧) = 0, and 𝐹 = 𝑄(𝑧). Hence, 𝑓 (𝑢, 𝑍) ∈ 𝐿 0 (𝑢) [𝑍], and 𝑓 (𝑢, 𝑍) is Galois over 𝐿(𝑢). Since 𝑄 = 𝐿(u) is a purely transcendental extension of 𝐿 (𝑢), 𝑓 (𝑢, 𝑍) is irreducible and Galois over 𝑄. Therefore, 𝑓 (𝑢, 𝑍) = irr(𝑧, 𝑄) (this settles (a)), and 𝐹 is a Galois extension of 𝐿 0 (t). ˆ Finally, since 𝐹/𝐿(t) and 𝐿(t)/𝐾 (t) are Galois and each 𝜏 ∈ 𝐺 = Gal(𝐿 (t)/𝐾 (t)) ˆ the extension 𝐹/𝐾 ˆ (t) is Galois. Let 𝐾ˆ be the fixed extends to an automorphism of 𝐹, ˆ ˆ ˆ field of 𝐺 in 𝐹. Then, 𝐿(t) ∩ 𝐾 = 𝐾 (t) is the fixed field of 𝐺 in 𝐿(t) and 𝐿 (t) 𝐾ˆ = 𝐹. ˆ ˆ By Remark 15.1.7(b), 𝐾 (t), 𝐾, 𝐿 0 (t), 𝐿(t), 𝐹, 𝐹 realize 𝐴wr𝐺0 𝐺. □

15.2 The Diamond Theorem

275

Proposition 15.2.2 (Realization of Twisted Wreath Products) Let 𝐾 ⊆ 𝐿 0 ⊆ 𝐿 be a tower of fields with 𝐾 Hilbertian and 𝐿/𝐾 finite Galois. Consider an absolutely irreducible polynomial 𝑓 (𝑇, 𝑋) ∈ 𝐿 0 [𝑇, 𝑋] which is Galois over 𝐿(𝑇). Let 𝐺 = Gal(𝐿/𝐾), 𝐺 0 = Gal(𝐿/𝐿 0 ), and 𝐴 = Gal( 𝑓 (𝑇, 𝑋), 𝐿(𝑇)). Then, 𝐺 0 acts on 𝐴 and there are fields 𝑀, 𝑁, such that 𝐾, 𝐿 0 , 𝐿, 𝑀, 𝑁 realize 𝐴wr𝐺0 𝐺. Proof. Let 𝑡 1 , . . . , 𝑡 𝑛 be algebraically independent elements over 𝐾 with ˆ ˆ 𝐹, and 𝐹ˆ such that 𝐾 (t), 𝐾, 𝑛 = [𝐿 0 : 𝐾]. Lemma 15.2.1 gives fields 𝐾, ˆ 𝐿 0 (t) , 𝐿(t), 𝐹, 𝐹 realize 𝐴wr𝐺0 𝐺. Thus, Conditions (15.6a)–(15.6d) hold for ˆ Since 𝐾 is Hilbertian, Lemma 𝐾 (t), 𝐿 0 (t), 𝐿(t), 𝐹, 𝐹ˆ rather than for 𝐾, 𝐿 0 , 𝐿, 𝐹, 𝐹. 𝑟 14.1.1 gives a ∈ 𝐾 such that those conditions hold for the residue fields under each 𝐿-place 𝜑 of 𝐹ˆ with 𝜑(𝐾 (t)) = 𝐾 ∪ {∞}. Lemma 3.4.6 gives such a place. Moreover, the residue fields of 𝐿 0 (t) and 𝐿(t) under 𝜑 are 𝐿 0 and 𝐿, respectively. ˆ 𝐹, and 𝐹, ˆ respectively, under 𝜑. By Let 𝐾ˆ ′, 𝐹 ′, and 𝐹ˆ ′ be the residue fields of 𝐾, □ Remark 15.1.7(b), they realize 𝐴wr𝐺0 𝐺. Theorem 15.2.3 (Diamond Theorem [Har99a], Thm. 4.1) Let 𝐾 be a Hilbertian field, 𝑀1 and 𝑀2 Galois extensions of 𝐾, and 𝑀 an intermediate field of 𝑀1 𝑀2 /𝐾. Suppose that 𝑀 ̸ ⊆ 𝑀1 and 𝑀 ̸ ⊆ 𝑀2 . Then, 𝑀 is Hilbertian. Proof. Corollary 13.2.3 allows us to assume that [𝑀 : 𝐾] = ∞. Part A of the proof strengthens this assumption: Part A: We may assume [𝑀 : (𝑀1 ∩ 𝑀)] = ∞. Otherwise, [𝑀 : (𝑀1 ∩ 𝑀)] < ∞. Then, 𝐾 has a finite Galois extension 𝑀2′ such that 𝑀 ⊆ (𝑀1 ∩ 𝑀) 𝑀2′ . Then, 𝑀 ⊆ 𝑀1 𝑀2′ and [𝑀 : 𝑀 ∩ 𝑀2′ ] = ∞. Replace 𝑀1 by 𝑀2′ and 𝑀2 by 𝑀1 , if necessary, to restore our assumption. Part B: Reduction to an absolutely irreducible Galois polynomial. Let 𝑀 ′ be a finite Galois extension of 𝑀 and let 𝑓 ∈ 𝑀 [𝑇, 𝑋] be an absolutely irreducible polynomial which is monic and separable in 𝑋 and Galois over 𝑀 ′ (𝑇). By Corollary 14.2.4, it suffices to find 𝑎 ∈ 𝑀 such that 𝑓 (𝑎, 𝑋) is irreducible over 𝑀 ′. We may assume that 𝑀 ′ ⊆ 𝑀1 𝑀2 . Indeed, 𝐾 has a finite Galois extension 𝐾 ′ such that 𝑀 ′ ⊆ 𝑀𝐾 ′. If 𝑀 ′ ̸ ⊆ 𝑀2 𝐾 ′, replace 𝑀2 by 𝑀2 𝐾 ′. If 𝑀 ′ ⊆ 𝑀2 𝐾 ′, replace 𝑀1 by 𝐾 ′. In the latter case we still have [𝑀 : (𝐾 ′ ∩ 𝑀)] = ∞, because [𝑀 : 𝐾] = ∞ and [𝐾 ′ : 𝐾] < ∞. Thus, we may assume that 𝑓 (𝑇, 𝑋) is Galois over 𝑀1 𝑀2 . It suffices to produce 𝑎 ∈ 𝑀 such that 𝑓 (𝑎, 𝑋) is irreducible over 𝑀1 𝑀2 . Part C: Finite Galois extensions. Write 𝑀0 = 𝑀 and 𝑁 = 𝑀1 𝑀2 . Then, 𝑁/𝐾 is Galois. For each finite Galois extension 𝐿 of 𝐾 in 𝑁 let 𝐿 𝑖 = 𝑀𝑖 ∩ 𝐿, 𝑖 = 0, 1, 2. Then, 𝐿 𝑖 /𝐾 is Galois, 𝑖 = 1, 2. Use the assumptions 𝑀0 ̸ ⊆ 𝑀𝑖 , 𝑖 = 1, 2 and [𝑀0 : 𝑀1 ∩ 𝑀0 ] = ∞ (Part A) to choose 𝐿 large with 𝐿 0 ̸ ⊆ 𝐿 𝑖 for 𝑖 = 1, 2, [𝐿 0 : 𝐿 1 ∩ 𝐿 0 ] > 2, 𝑓 ∈ 𝐿 0 [𝑇, 𝑋], and 𝑓 (𝑇, 𝑋) is Galois over 𝐿 (𝑇).

(15.10)

276

15 The Diamond Theorem

Set 𝐺 = Gal(𝐿/𝐾). The conditions on the fields 𝐿 𝑖 translate into conditions on the groups 𝐺 𝑖 = Gal(𝐿/𝐿 𝑖 ), 𝑖 = 0, 1, 2: 𝐺 1 , 𝐺 2 ̸≤ 𝐺 0 . (𝐺 0 𝐺 1 : 𝐺 0 ) > 2.

(15.11) (15.12)

Part D: Twisted wreath products. Let 𝐴 ′ = Gal( 𝑓 , 𝐿(𝑇)) = Gal( 𝑓 , 𝐾sep (𝑇)). Choose a basis 𝑏 1 , . . . , 𝑏 𝑛 for 𝐿 0 /𝐾 and algebraically independent elements 𝑡 1 , . . . , 𝑡 𝑛 over 𝐾. By 15.10 and Lemma 15.2.1, the group 𝐺 0 acts on 𝐴 ′. Moreover, there are fields 𝐹, 𝐹ˆ such that 𝐾 (t), 𝐿 0 (t), 𝐿(t), 𝐹, 𝐹ˆ realize 𝐴 ′wr𝐺0 𝐺 and 𝑛 ∑︁ 𝐹 = 𝐿(t, 𝑧) with irr(𝑧, 𝐿(t)) = 𝑓 ( 𝑏 𝑖 𝑡 𝑖 , 𝑋).

(15.13) (15.14)

𝑖=1

Since 𝐾 is Hilbertian, Lemma 14.1.1 gives 𝑐 1 , . . . , 𝑐 𝑛 ∈ 𝐾 such that the specialization t → c extends to an 𝐿-place of 𝐹ˆ onto a Galois extension 𝐹ˆ ′ of 𝐾 with Galois ˆ (t)). Thus, 𝐹ˆ ′ has a subfield 𝐹 ′ with these properties: group isomorphic to Gal( 𝐹/𝐾 𝐾, 𝐿 0 , 𝐿, 𝐹 ′, 𝐹ˆ ′ realize 𝐴 ′wr𝐺0 𝐺 and 𝑛 ∑︁ 𝐹 ′ = 𝐿(𝑧 ′) with irr(𝑧 ′, 𝐿) = 𝑓 ( 𝑏 𝑖 𝑐 𝑖 , 𝑋).

(15.15) (15.16)

𝑖=1

Í𝑛 𝑏 𝑖 𝑐 𝑖 . Then, 𝑎 is in 𝐿 0 , hence in 𝑀. If we prove that 𝑁 ∩ 𝐹 ′ = 𝐿, it Let 𝑎 = 𝑖=1 will follow from 15.16 that [𝑁 (𝑧 ′) : 𝑁] = [𝐹 ′ : 𝐿] = deg( 𝑓 (𝑎, 𝑋)). Thus, 𝑓 (𝑎, 𝑋) will be irreducible over 𝑁, as desired. Part E: Conclusion of the proof. Let 𝐸 = 𝑁 ∩ 𝐹 ′ and 𝐴 = Gal(𝐸/𝐿). By Remark 15.1.7(c), 𝐺 0 acts on 𝐴 and there is a Galois extension 𝐸ˆ of 𝐾 such that 𝐾, 𝐿 0 , 𝐿, 𝐸, 𝐸ˆ realize 𝐻 = 𝐴wr𝐺0 𝐺.

(15.17)

In particular, all conjugates of 𝐸 over 𝐾 are in 𝑁. Hence, 𝐸ˆ ⊂ 𝑁. ˆ ˆ such that res𝐸/𝐿 Identify 𝐻 with Gal( 𝐸/𝐾) → Gal(𝐿/𝐾) coincides ˆ : Gal( 𝐸/𝐾) with the projection 𝜋: 𝐻 → 𝐺. Then, 𝜋 ◦ res 𝑁 /𝐸ˆ = res 𝑁 /𝐿 . For 𝑖 = 1, 2 let 𝐻𝑖 = res 𝑁 /𝐸ˆ (Gal(𝑁/𝑀𝑖 )). Then, 𝐻𝑖 ⊳ 𝐻 and 𝜋(𝐻𝑖 ) = res 𝑁 /𝐿 (Gal(𝑁/𝑀𝑖 )) = 𝐺 𝑖 . Since Gal(𝑁/𝑀1 ) and Gal(𝑁/𝑀2 ) are normal subgroups of Gal(𝑁/𝐾) with a trivial intersection, they commute. Hence, 𝐻1 and 𝐻2 commute. By (15.11), there exists an ℎ2 ∈ 𝐻2 with 𝜋(ℎ2 ) ∉ 𝐺 0 . If 𝐴 were not trivial, then by (15.12) and Lemma 15.1.5 there would be an ℎ1 ∈ 𝐻1 which does not commute with ℎ2 . We conclude from this contradiction that 𝐴 = 1 and therefore 𝑁 ∩ 𝐹 ′ = 𝐿, □ as desired.

15.3 Weissauer’s Theorem

277

Remark 15.2.4 In the notation of Theorem 15.2.3, suppose that 𝐾 is the quotient field of a Hilbertian ring 𝑅. Then, in Part D of the proof of that theorem, one could choose the basis 𝑏 1 , . . . , 𝑏 𝑛 in the integral Í𝑛closure 𝑅 𝐿0 of 𝑅 in 𝐿 0 . In addition one could choose 𝑐 1 , . . . , 𝑐 𝑛 in 𝑅. Then, 𝑎 = 𝑖=1 𝑏 𝑖 𝑐 𝑖 would be in the integral closure 𝑅 𝑀 of 𝑅 in 𝑀. This would imply that 𝑅 𝑀 is a Hilbertian ring. See also [JaR20, Thm. B], whose proof follows the original proof of [Har99a]. We conclude this section with an application of the diamond theorem that solves Problems 12.18 and 12.19 of [FrJ86]: Corollary 15.2.5 ([HaJ91]) Let 𝐾 be a Hilbertian field and let 𝑁 be a Galois extension of 𝐾 which is not Hilbertian. Then, 𝑁 is not the compositum of two Galois extensions of 𝐾 neither of which is contained in the other. In particular, this conclusion holds for 𝐾sep .

15.3 Weissauer’s Theorem The most useful application of the diamond theorem is part (b) of the following result: Theorem 15.3.1 (Weissauer) Let 𝐾 be a Hilbertian field. (a) Let 𝑀 be a separable algebraic extension of 𝐾 and 𝑀 ′ a finite proper separable extension of 𝑀. Suppose that 𝑀 ′ is not contained in the Galois closure of 𝑀/𝐾. Then, 𝑀 ′ is Hilbertian. (b) Let 𝑁 be a Galois extension of 𝐾 and 𝑁 ′ a finite proper separable extension of 𝑁. Then, 𝑁 ′ is Hilbertian. (c) Let 𝑁 be a Galois extension of a Hilbertian field 𝐾 and 𝐿 a finite proper separable extension of 𝐾. Suppose that 𝑁 ∩ 𝐿 = 𝐾. Then, 𝑁 𝐿 is Hilbertian. Proof. Statement (c) is a special case of Statement (b). Statement (b) is a special case of Statement (a). Statement (a) follows from Corollary 13.2.3 if [𝑀 : 𝐾] < ∞. Suppose therefore that [𝑀 : 𝐾] = ∞. Then, 𝐾 has a finite Galois extension 𝐿 with 𝑀 ′ ⊆ 𝑀 𝐿. Let 𝑁 be the Galois closure of 𝑀/𝐾. Then, 𝑀 ′ is contained in 𝑁 𝐿 but neither in 𝑁 nor in 𝐿. By Theorem 15.2.3, 𝑀 ′ is Hilbertian. □ Proposition 15.3.3 below is a stronger version of Theorem 15.3.1(c) which gives information about the Hilbert sets of 𝑁 𝐿. That version implies Theorem 15.3.1(a), hence also Theorem 15.3.1(b). The proof of Proposition 15.3.3 depends on the following lemma rather than on the diamond theorem: Lemma 15.3.2 Let 𝐿 = 𝐾 (𝛼) be a finite proper separable extension of a field 𝐾 and let 𝐿ˆ be the Galois hull of 𝐿/𝐾. Consider an absolutely irreducible polynomial ℎ ∈ 𝐿 [𝑇, 𝑋] with deg𝑋 (ℎ) ≥ 2 which is separable with respect to 𝑋. For 𝑢, 𝑣 algebraically independent elements over 𝐾, let 𝐸 be a Galois extension of 𝐾 (𝑢, 𝑣) such that 𝐸 ∩ 𝐿(𝑢, 𝑣) = 𝐾 (𝑢, 𝑣). Then, the polynomial ℎ(𝑢 + 𝛼𝑣, 𝑋) has no root in ˆ the field 𝐸 𝐿.

278

15 The Diamond Theorem

ˆ By assumption, 𝐸 is linearly disjoint from Proof. Put 𝑡 = 𝑢 + 𝛼𝑣 and 𝐹 = 𝐸 𝐿. 𝐿 over 𝐾. Hence, we may choose 𝜎 ∈ Gal(𝐹/𝐸) with 𝛼 ′ = 𝜎𝛼 ≠ 𝛼. Then, put 𝑡 ′ = 𝜎𝑡 = 𝑢 + 𝛼 ′ 𝑣 and ℎ ′ = 𝜎ℎ. Assume that there exists an 𝑥 ∈ 𝐹 with ℎ(𝑡, 𝑥) = 0. ˆ ˆ 𝑡 ′) = 𝐿(𝑢, Then, 𝑥 ′ = 𝜎𝑥 ∈ 𝐹 and ℎ ′ (𝑡 ′, 𝑥 ′) = 0. Since 𝛼 ≠ 𝛼 ′, we have 𝐿(𝑡, 𝑣). Hence, 𝑡, 𝑡 ′ are algebraically independent over 𝐾. 𝐿ˆ (𝑡, 𝑥)

𝐿ˆ (𝑡, 𝑡 ′, 𝑥)

𝐹

𝐿ˆ (𝑡)

ˆ 𝑡 ′) 𝐿(𝑡,

ˆ 𝑡 ′, 𝑥 ′) 𝐿(𝑡,

𝐿ˆ

ˆ ′) 𝐿(𝑡

ˆ ′, 𝑥 ′) 𝐿(𝑡

ˆ ′, 𝑥 ′) are algebraically independent over 𝐿. ˆ Since they are Therefore, 𝐿ˆ (𝑡, 𝑥) and 𝐿(𝑡 ˆ regular extensions of 𝐿, they are linearly disjoint (Lemma 3.4.7). By Lemma 3.1.14, ˆ 𝑡 ′, 𝑥) and 𝐿(𝑡, ˆ 𝑡 ′, 𝑥 ′) are linearly disjoint over 𝐿(𝑡, ˆ 𝑡 ′). In particular, 𝐿(𝑡, ˆ 𝑡 ′, 𝑥) ∩ 𝐿(𝑡, ˆ 𝑡 ′, 𝑥 ′) = 𝐿(𝑡, 𝑡 ′). 𝐿(𝑡,

(15.18)

ˆ 𝑡 ′) = 𝐹, because 𝐸 𝐿ˆ = 𝐹. Hence, with 𝐸 0 = 𝐸 ∩ 𝐿(𝑡, ˆ 𝑡 ′), On the other hand, 𝐸 · 𝐿(𝑡, ′ ˆ 𝑡 )). Therefore, 𝜎𝜏 = 𝜏𝜎 for all we have Gal(𝐹/𝐸 0 ) = Gal(𝐹/𝐸) × Gal(𝐹/ 𝐿(𝑡, ˆ 𝑡 ′, 𝑥)). In particular, 𝜏𝑥 ′ = 𝜏𝜎𝑥 = 𝜎𝜏𝑥 = 𝜎𝑥 = 𝑥 ′. It follows that 𝜏 ∈ Gal(𝐹/ 𝐿(𝑡, ˆ 𝑡 ′, 𝑥). By (15.18), 𝐿ˆ (𝑡, 𝑡 ′, 𝑥 ′) = 𝐿(𝑡, ˆ 𝑡 ′), so 𝑥 ′ ∈ 𝐿(𝑡, ˆ 𝑡 ′). But, since 𝑡 ′ is 𝑥 ′ ∈ 𝐿(𝑡, transcendental over 𝐿ˆ and ℎ(𝑇, 𝑋) is absolutely irreducible, ℎ(𝑡 ′, 𝑋) is irreducible over 𝐿ˆ (𝑡 ′). This contradicts ℎ ′ (𝑡 ′, 𝑥 ′) = 0 and deg𝑋 (ℎ ′) ≥ 2. □ Proposition 15.3.3 Let L be a finite proper separable extension of a Hilbertian field 𝐾 and let 𝑀 be a Galois extension of 𝐾 such that 𝑀 ∩ 𝐿 = 𝐾. Then, the field 𝑁 = 𝑀 𝐿 is Hilbertian. Moreover, every separable Hilbert subset of 𝑁 contains elements of 𝐿. Proof. By Lemma 14.1.2 and Lemma 14.2.3, it suffices to consider absolutely irreducible polynomials ℎ1 , . . . , ℎ 𝑚 ∈ 𝑁 [𝑇, 𝑋], separable monic and of degree at least 2 in 𝑋 and to find 𝑐 ∈ 𝐿 such that ℎ𝑖 (𝑐, 𝑋) has no root in 𝑁, 𝑖 = 1, . . . , 𝑚. Let 𝑢 and 𝑣 be algebraically independent over 𝐾. Choose a primitive element 𝛼 for 𝐿/𝐾 and let 𝐿 ′ be a finite extension of 𝐿 which is contained in 𝑁 and contains all coefficients of ℎ1 , . . . , ℎ 𝑚 . Put 𝐾 ′ = 𝑀 ∩ 𝐿 ′ and let 𝐹 be a finite Galois extension of 𝐾 ′ (𝑢, 𝑣) that contains 𝐿 ′ (𝑢, 𝑣) and over which all polynomials ℎ1 (𝑢 + 𝛼𝑣, 𝑋), . . . , ℎ 𝑚 (𝑢 + 𝛼𝑣, 𝑋) decompose into linear factors. Let 𝑔 ∈ 𝐾 [𝑢, 𝑣] be the product of the discriminants of ℎ1 (𝑢 + 𝛼𝑣, 𝑋), . . . , ℎ 𝑚 (𝑢 + 𝛼𝑣, 𝑋) with respect to 𝑋. Let 𝐵 ′ be the set of all (𝑎, 𝑏) ∈ (𝐾 ′) 2 with 𝑔(𝑎, 𝑏) ≠ 0 satisfying the following condition: (15.19) The 𝐿 ′-specialization (𝑢, 𝑣) → (𝑎, 𝑏) extends to a place 𝜑 of 𝐹 which induces an isomorphism 𝜑 ′: Gal(𝐹/𝐾 ′ (𝑢, 𝑣)) → Gal(𝐹 ′/𝐾 ′) (with 𝐹 ′ being the residue field of 𝐹) such that (𝜑 ′ 𝜎) (𝜑𝑥) = 𝜑(𝜎𝑥) for all 𝜎 ∈ Gal(𝐹/𝐾 ′ (𝑢, 𝑣)) and 𝑥 ∈ 𝐹 with 𝜑𝑥 ≠ ∞. In particular, 𝜑 maps the set of all zeros of ℎ𝑖 (𝑢 + 𝛼𝑣, 𝑋) bijectively onto the set of zeros of ℎ𝑖 (𝑎 + 𝛼𝑏, 𝑋), 𝑖 = 1 . . . , 𝑚. Also, since 𝜑 is an 𝐿 ′-specialization, 𝐿 ′ ⊆ 𝐹 ′.

15.3 Weissauer’s Theorem

279

By Lemma 14.1.1 and Example 3.4.10, 𝐵 ′ contains a separable Hilbert subset 𝐴 ′ of (𝐾 ′) 2 . By Corollary 13.2.3, 𝐴 ′ contains a separable Hilbert subset of 𝐾 2 . Thus, there are 𝑎, 𝑏 ∈ 𝐾 satisfying (15.19). Let 𝑐 = 𝑎 + 𝛼𝑏. Assume there is an 𝑖 between 1 and 𝑚 such that the polynomial ℎ𝑖 (𝑐, 𝑋) has a root 𝑥¯ in 𝑁. Then, 𝑥¯ ∈ 𝐹 ′ ∩ 𝑁. Since res: Gal(𝑁/𝐿 ′) → Gal(𝑀/𝐾 ′) is an isomorphism, there exists a Galois extension 𝐸 ′ of 𝐾 ′ contained in 𝑀 such that 𝐸 ′ 𝐿 ′ = 𝐹 ′ ∩ 𝑁. Since 𝐸 ′ ∩ 𝐿 ′ = 𝐾 ′ = 𝑀 ∩ 𝐿 ′, Gal(𝐹 ′/𝐸 ′) · Gal(𝐹 ′/𝐿 ′) = Gal(𝐹 ′/𝐾 ′) and Gal(𝐹 ′/𝐸 ′) ∩ Gal(𝐹 ′/𝐿 ′) = Gal(𝐹 ′/𝐹 ′ ∩ 𝑁).

(15.20)

Therefore, 𝐾 ′ (𝑢, 𝑣) has a Galois extension 𝐸 in 𝐹 such that 𝜑 ′ (Gal(𝐹/𝐸)) = Gal(𝐹 ′/𝐸 ′). From 15.20, 𝐸 ∩ 𝐿 ′ (𝑢, 𝑣) = 𝐾 ′ (𝑢, 𝑣) and 𝜑 ′ (Gal(𝐹/𝐸 𝐿 ′)) = Gal(𝐹 ′/𝐹 ′ ∩ 𝑁). Moreover, the polynomial ℎ𝑖 (𝑢 + 𝛼𝑣, 𝑋) has a root 𝑥 such that 𝜑(𝑥) = 𝑥. ¯ For each 𝜎 ∈ Gal(𝐹/𝐸 𝐿 ′) we have 𝜑(𝜎𝑥) = 𝜑 ′ (𝜎) ( 𝑥) ¯ = 𝑥¯ = 𝜑(𝑥). Hence, 𝜎𝑥 = 𝑥. In particular, 𝑥 ∈ 𝐸 𝐿 ′, a contradiction to Lemma 15.3.2. □ We deduce Theorem 15.3.1(a) from Proposition 15.3.3: Proposition 15.3.4 Let 𝑀 be a separable algebraic extension of a Hilbertian field 𝐾 and 𝑀 ′ a proper finite separable extension of 𝑀. Suppose that 𝑀 ′ is not contained in the Galois closure of 𝑀/𝐾. Then, 𝑀 ′ is Hilbertian. ˆ With no loss replace 𝑀 by 𝑀ˆ ∩ 𝑀 ′. Proof. Denote the Galois closure of 𝑀/𝐾 by 𝑀. ′ Choose a primitive element 𝛼 for 𝑀 /𝑀. Put 𝐿 ′ = 𝐾 (𝛼) and 𝐾 ′ = 𝑀ˆ ∩ 𝐿 ′. Then, the conditions of Proposition 15.3.3 are satisfied for 𝐾 ′, 𝐿 ′, 𝑀ˆ replacing 𝐾, 𝐿, 𝑀, respectively. Let 𝑓 ∈ 𝑀 ′ [𝑇, 𝑋] be an irreducible polynomial, monic and separable in 𝑋 with deg𝑋 ( 𝑓 ) ≥ 2. By Lemma 14.1.2, there exist ℎ1 , . . . , ℎ 𝑚 ∈ 𝑀 ′ [𝑇, 𝑋] abso′ (ℎ , . . . , ℎ ) ⊆ 𝐻 ′ ( 𝑓 ). Since lutely irreducible and separable in 𝑋 such that 𝐻 𝑀 ′ 𝑚 1 𝑀 ′ ˆ ℎ1 , . . . , ℎ 𝑚 are irreducible in 𝑀 𝐿 [𝑇, 𝑋], Proposition 15.3.3 gives 𝑎 ∈ 𝐿 ′ with ℎ1 (𝑎, 𝑋), . . . , ℎ 𝑚 (𝑎, 𝑋) irreducible in 𝑀ˆ (𝛼) [𝑋] and therefore in 𝑀 ′ [𝑋]. By Lemma 14.2.2, 𝑀 ′ is Hilbertian. □

˜ whose finite proper extensions are Remark 15.3.5 (Non-Hilbertian subfield of Q Hilbertian) Denote the compositum of all finite solvable extensions of Q by Qsolv . It is not Hilbertian: there exists no 𝑎 ∈ Qsolv such that 𝑋 2 − 𝑎 is irreducible over Qsolv . However, Qsolv is a Galois extension of Q. Hence, by Corollary 15.3.1(b), every finite proper extension of Qsolv is Hilbertian. By Corollary 15.2.5, Qsolv is not the compositum of two Galois extensions of Q neither of which is contained in the other.

280

15 The Diamond Theorem

Exercises 1. Prove the following result: Let 𝐴 be a finite group and let 𝑘 be a positive integer. Let 𝐺 be a subgroup of 𝑆 𝑘 which we regard as acting on 𝐴 𝑘 by permutation of the coordinates. Consider the semidirect product 𝐺 ⋉ 𝐴 𝑘 and let 𝛼: 𝐺 ⋉ 𝐴 𝑘 → 𝐺 be the canonical projection. Suppose that 𝐿/𝐾 is a Galois extension with Gal(𝐿/𝐾)  𝐺 and that there exists a Galois extension 𝐾 (𝑢, 𝑣)/𝐾 (𝑢), with 𝑢 transcendental over 𝐾, 𝐾 (𝑢, 𝑣) regular over 𝐾, and Gal(𝐾 (𝑢, 𝑣)/𝐾 (𝑢))  𝐴. Prove that there exists a Galois ˆ such that 𝐸 is a finitely generated purely transcendental extension of extension 𝐹/𝐸 ˆ 𝐾, 𝐹ˆ is a regular extension of 𝐿 and there is an isomorphism 𝜃: Gal( 𝐹/𝐸) → 𝐺 ⋉ 𝐴𝑘 such that 𝛼 ◦ 𝜃 = res 𝐿 . ˆ be finite Galois extensions. Suppose that there exists a homomor2. Let 𝐹/𝐸 and 𝐹/𝐹 ˆ such that res𝐹 (ℎ(𝜎)) = 𝜎 for each 𝜎 ∈ Gal(𝐹/𝐸). phism ℎ: Gal(𝐹/𝐸) → Aut( 𝐹) ˆ ˆ Prove that 𝐹/𝐸 is a Galois extension and that Gal( 𝐹/𝐸) is isomorphic to the semidiˆ rect product Gal(𝐹/𝐸) ⋉ Gal( 𝐹/𝐹). Hint: Show that 𝑥 ∈ 𝐹ˆ belongs to 𝐸 if and only if 𝑥 is fixed by the elements of ˆ the group generated by Gal( 𝐹/𝐹) and ℎ(Gal(𝐹/𝐸)).

Notes Weissauer proves that every finite proper separable extension of a Galois extension of a Hilbertian field is Hilbertian [Wei82, Satz 9.7]. Section 15.3 replaces Weissauer’s “nonstandard” proof by a simpler “standard” one still using Weissauer’s auxiliary variable trick as in Lemma 15.3.2 [Fri85, Thm. 1.3]. Haran’s diamond theorem 15.2.3 appears in [Har99a, Thm. 4.2]. It generalizes Weissauer’s theorem (Theorem 15.3.1) and Theorem 2.4 of [HaJ91]. Its proof is an offspring of the proofs of the two earlier results.

Chapter 16

Nonstandard Structures

A. Robinson invented “nonstandard” methods in order to supplement the Weierstrass 𝜀, 𝛿 formalism of the calculus by a rigorous version of the classical calculus of infinitesimals in the spirit of Leibniz and other formalists. We will apply the nonstandard approach to algebra in the next chapter in order to find new Hilbertian fields. Its main virtue, from an algebraic point of view, is that it creates additional algebraic structures to which well-known theorems can be applied. Here we present the basics of the nonstandard method: the higher order structure on a set 𝑀 (Section 16.1); the concept of an enlargement 𝑀 ∗ of 𝑀 (Sections 16.2 and 16.3); and the existence of 𝑀 ∗ (Section 16.4) via ultraproducts.

16.1 Higher Order Predicate Calculus Sentences that speak of arbitrary subsets, functions, relations, etc., are common in mathematics, even though they are usually outside the scope of first order languages. Here we introduce a language which includes such sentences. First the notion of a type (of a higher order object) is inductively defined by the following rules: (16.1a) The number 0 is a type. (16.1b) If 𝑛 is a positive integer and 𝜏(1), . . . , 𝜏(𝑛) are types, then the 𝑛-tuple (𝜏(1), . . . , 𝜏(𝑛)) is a type. Denote the set of all types by 𝑇. Given a set 𝑀, attach a set 𝑀 𝜏 to each type 𝜏 as follows: (16.2a) 𝑀0 = 𝑀. (16.2b) If 𝜏(1), . . . , 𝜏(𝑛) are types and 𝝉 = (𝜏(1), . . . , 𝜏(𝑛)), then 𝑀𝝉 = P (𝑀 𝜏 (1) × · · · × 𝑀 𝜏 (𝑛) ) := {all subsets of 𝑀 𝜏 (1) × · · · × 𝑀 𝜏 (𝑛) }. Elements of 𝑀 𝜏 are objects of type 𝜏 over 𝑀. We call them sets (or relations) of type 𝜏 if 𝜏 ≠ 0. A higher order set is a set of type 𝜏 for some 𝜏 ≠ 0. Call the system M = ⟨𝑀 𝜏 | 𝜏 ∈ 𝑇⟩ the higher order structure over 𝑀. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 281 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_16

282

16 Nonstandard Structures

For each higher order set 𝐴 of 𝑀 introduce a sequence of variables 𝑋 𝐴1 , 𝑋 𝐴2 , . . . . Inductively define formulas of the higher order language L∞ (𝑀) as follows: (16.3a) 𝑋 𝐴𝑖 = 𝑋 𝐴 𝑗 , 𝑋 𝐴𝑖 = 𝑎, and 𝑎 = 𝑏 are formulas for each higher order set 𝐴, all 𝑎, 𝑏 ∈ 𝐴, and 𝑖, 𝑗 ∈ N. (16.3b) If 𝐴(1), . . . , 𝐴(𝑛) are higher order sets, 𝐴(0) is a subset of P ( 𝐴(1) × · · · × 𝐴(𝑛)), and for 𝜈 between 0 and 𝑛 either 𝑡 𝜈 is a variable 𝑋 𝐴(𝜈),𝑖 (𝜈) , with 𝑖(𝜈) ∈ N, or an element of 𝐴(𝜈), then (𝑡 1 , . . . , 𝑡 𝑛 ) ∈ 𝑡0 is a formula. (16.3c) Negations and disjunctions (hence also conjunctions and implications) of formulas are formulas. (16.3d) If 𝜑 is a formula, then (∃𝑋 𝐴𝑖 ∈ 𝐴) [𝜑] (hence also (∀𝑋 𝐴𝑖 ∈ 𝐴) [𝜑]) is a formula for all higher order sets 𝐴 and 𝑖 ∈ N. Define free variables of a formula as usual (Section 8.1). In particular, a sentence is a formula without free variables. Likewise, a substitution is a function 𝑓 that, for each higher order set 𝐴 of 𝑀, replaces each 𝑋 𝐴𝑖 by an element 𝑥 𝐴𝑖 of 𝐴 and fixes the elements of 𝐴. Interpret the truth of a formula 𝜑 under 𝑓 by reading “(𝑡1 , . . . , 𝑡 𝑛 ) ∈ 𝑡 0 ” as “( 𝑓 (𝑡 1 ), . . . , 𝑓 (𝑡 𝑛 )) belongs to 𝑓 (𝑡0 ).” Note that each object 𝑡 of type 𝝉 = (𝜏(1), . . . , 𝜏(𝑛)) ≠ 0 has two roles in the language L∞ (𝑀). First, it is a constant: 𝑡 is equal or not equal to another object of the same type. Second, it is a relation between objects of types 𝜏(1), . . . , 𝜏(𝑛). If the set 𝑀 has a first order structure, embed this in the higher order structure of 𝑀 in a natural way. For example, if 𝑀 is a field, then addition corresponds to a subset 𝐴 of 𝑀 × 𝑀 × 𝑀, and 𝑋1 + 𝑏 = 𝑐 in the first order language of fields becomes (𝑋 𝑀1 , 𝑏, 𝑐) ∈ 𝐴. Remark 16.1.1 Let N∗ be a proper elementary extension of N. Then, N (0) (resp. (N∗ ) (0) ) is the collection of all subsets of N (resp. N∗ ). The second order structure N∗ ∪ (N∗ ) (0) is not an elementary extension of the second order structure N ∪ N (0) . For example the induction axiom   (∀𝑋 ∈ N (0) ) [1 ∈ 𝑋 ∧ (∀𝑥 ∈ N) [𝑥 ∈ 𝑋 → 𝑥 + 1 ∈ 𝑋]] → (∀𝑥 ∈ N) [𝑥 ∈ 𝑋] holds in N ∪ N (0) but it fails in N∗ ∪ (N∗ ) (0) if we replace N and N (0) by N∗ and (N∗ ) (0) , respectively. Indeed, the induction axiom does not hold for 𝑋 = N. In order to restore the elementary extension property for higher order structures of the elementary extension N∗ of N, we must restrict quantification of subsets of N∗ to a proper subcollection of (N∗ ) (0) , to be denoted N∗(0) , the “inner” subset of N∗ . We define these in the next section.

16.2 Enlargements We consider a set 𝑀 together with its higher order structure and define an enlargement of 𝑀 as a special model of the higher order theory of 𝑀 satisfying Conditions I and II below and III of Section 16.3. The enlargement will be saturated with respect to all higher order relations.

16.2 Enlargements

283

The underlying set 𝑀 ∗ of this model has the property that to each object 𝑎 of type 𝜏 of 𝑀 there corresponds an object 𝑎 ∗ of 𝑀 ∗ of the same type (i.e. 𝑎 ∗ is an element of (𝑀 ∗ ) 𝜏 ). Call such an 𝑎 ∗ standard. In particular, 𝑀 𝜏 itself, an element of 𝑀 ( 𝜏) , corresponds to an element (𝑀 𝜏 ) ∗ of (𝑀 ∗ ) ( 𝜏) . This means that (𝑀 𝜏 ) ∗ is a subset of (𝑀 ∗ ) 𝜏 . Call the elements of (𝑀 𝜏 ) ∗ the internal objects of type 𝜏 of 𝑀 ∗ . All other elements of (𝑀 ∗ ) 𝜏 are external. Note: For 𝜏 = 0 we have (𝑀0 ) ∗ = (𝑀 ∗ )0 = 𝑀 ∗ . Thus, every element of 𝑀 ∗ is internal, although (𝑀 ∗ ) (0) itself may have external subsets. To simplify the notation write 𝑀 𝜏∗ for (𝑀 𝜏 ) ∗ . If 𝜏(1), . . . , 𝜏(𝑛) are types and if 𝑅 ⊆ 𝑀 𝜏 (1) × · · · × 𝑀 𝜏 (𝑛) , then 𝑅 ∈ 𝑀𝝉 for 𝝉 = (𝜏(1), . . . , 𝜏(𝑛)). Hence, 𝑅 ∗ ∈ 𝑀𝝉∗ , that is, 𝑅 ∗ ⊆ (𝑀 ∗ ) 𝜏 (1) × · · · × (𝑀 ∗ ) 𝜏 (𝑛) . We demand, however, that an enlargement satisfies a stronger condition: I. Internal condition on 𝑛-ary relations. If 𝜏(1), . . . , 𝜏(𝑛) are types and if 𝑅 ⊆ 𝑀 𝜏 (1) × · · · × 𝑀 𝜏 (𝑛) , then 𝑅 ∗ ⊆ 𝑀 𝜏∗ (1) × · · · × 𝑀 𝜏∗ (𝑛) . That is, the elements of a standard 𝑛-ary relation are 𝑛-tuples with internal coordinates. Call them internal 𝑛-tuples. Suppose that 𝜏 is a type and 𝐴 is a subset of 𝑀 𝜏 . By Condition I, 𝐴∗ ⊆ 𝑀 𝜏∗ . So, each element of 𝐴∗ is an element of 𝑀 𝜏∗ , hence internal. Consequently, each element of a standard set is internal. On the other hand, each subset of 𝐴∗ is an element of (𝑀 ∗ ) ( 𝜏) . It is internal exactly when it belongs to 𝑀 (∗𝜏) . To interpret a formula of L∞ (𝑀) in 𝑀 ∗ consider only internal substitutions 𝑓 . These satisfy the condition 𝑓 (𝑋 𝐴𝑖 ) ∈ 𝐴∗ for each higher order set 𝐴. Define the truth value of the formula 𝜑 under 𝑓 first by placing an asterisk to the right of each constant or relation symbol that appears in 𝜑 to obtain a formula 𝜑∗ . Then, interpret the formula 𝜑∗ (under 𝑓 ) as usual. Note: If (∃𝑋 ∈ 𝐴) is part of 𝜑, then (∃𝑋 ∈ 𝐴∗ ) is a part of 𝜑∗ . Since each element of 𝐴∗ is internal, this means that one quantifies only on internal objects. For example, consider a formula 𝜑(𝑋1 , . . . , 𝑋𝑛 ) where 𝑋𝜈 = 𝑋 𝑀𝜏 (𝜈) ,𝑖 (𝜈) , 𝜈 = 1, . . . , 𝑛. Let 𝑅 = {(𝑎 1 , . . . , 𝑎 𝑛 ) ∈ 𝑀 𝜏 (1) × · · · × 𝑀 𝜏 (𝑛) | 𝑀 |= 𝜑(𝑎 1 , . . . , 𝑎 𝑛 )}. Then, the sentence 𝜃 (∀𝑋1 ∈ 𝑀 𝜏 (1) ) · · · (∀𝑋𝑛 ∈ 𝑀 𝜏 (𝑛) ) [(𝑋1 , . . . , 𝑋𝑛 ) ∈ 𝑅 ↔ 𝜑(𝑋1 , . . . , 𝑋𝑛 )] is true in 𝑀. The sentence 𝜃 ∗ takes the form (∀𝑋1 ∈ 𝑀 𝜏∗ (1) ) · · · (∀𝑋𝑛 ∈ 𝑀 𝜏∗ (𝑛) ) [(𝑋1 , . . . , 𝑋𝑛 ) ∈ 𝑅 ∗ ↔ 𝜑∗ (𝑋1 , . . . , 𝑋𝑛 )]. It is reasonable to ask that 𝜃 ∗ will be true in 𝑀 ∗ . Since Condition I implies that 𝑅 ∗ contains only elements with internal coordinates, we may rephrase 𝜃 ∗ as 𝑅 ∗ = {(𝑎 1 , . . . , 𝑎 𝑛 ) ∈ 𝑀 𝜏∗ (1) × · · · × 𝑀 𝜏∗ (𝑛) | 𝑀 ∗ |= 𝜑∗ (𝑎 1 , . . . , 𝑎 𝑛 )}.

284

16 Nonstandard Structures

Here is the condition guaranteeing this indeed holds: II. Elementary extension condition Let 𝐴1 , . . . , 𝐴𝑛 be higher order sets, 𝜑(𝑋 𝐴1 ,𝑖 (1) , . . . , 𝑋 𝐴𝑛 ,𝑖 (𝑛) ) a formula of L∞ (𝑀), and 𝑎 𝜈 ∈ 𝐴 𝜈 , 𝜈 = 1, . . . , 𝑛. Then, 𝜑(𝑎 1 , . . . , 𝑎 𝑛 ) is true in 𝑀 if and only if 𝜑∗ (𝑎 1∗ , . . . , 𝑎 ∗𝑛 ) is true in 𝑀 ∗ . For example, if 𝑎 ∈ 𝑀 𝜏 , then 𝑎 ∗ ∈ 𝑀 𝜏∗ . Thus, standard objects are internal. Also, if 𝑎, 𝑏 ∈ 𝑀 𝜏 and 𝑎 ≠ 𝑏, then 𝑎 ∗ ≠ 𝑏 ∗ : the canonical map 𝑎 ↦→ 𝑎 ∗ of 𝑀 𝜏 into 𝑀 𝜏∗ is injective. Therefore, we occasionally regard 𝑀 𝜏 as a subset of 𝑀 𝜏∗ . In addition, for 𝑅 ⊆ 𝐴1 × · · · × 𝐴𝑛 , if a ∈ 𝐴1 × · · · × 𝐴𝑛 belongs to 𝑅 ∗ , then a ∈ 𝑅. Thus, 𝑅 ∗ ∩ ( 𝐴1 × · · · × 𝐴𝑛 ) = 𝑅 and the relation 𝑅 ∗ is an extension of 𝑅. In particular, if 𝑀 has a first order structure, then it is an elementary substructure of a natural extension to 𝑀 ∗ . Now consider the notion of an internal function. Let 𝐴 and 𝐵 be two higher order sets. We view a function 𝑓 : 𝐴 → 𝐵 as a subset of 𝐴 × 𝐵 that satisfies these two conditions: (16.4a) (∀𝑎 ∈ 𝐴) (∃𝑏 ∈ 𝐵) [(𝑎, 𝑏) ∈ 𝑓 ]; (16.4b) (∀𝑎 ∈ 𝐴) (∀𝑏 1 , 𝑏 2 ∈ 𝐵) [(𝑎, 𝑏 1 ) ∈ 𝑓 ∧ (𝑎, 𝑏 2 ) ∈ 𝑓 → 𝑏 1 = 𝑏 2 ]. The subset 𝑓 ∗ of 𝐴∗ × 𝐵∗ satisfies the corresponding conditions: it is a function from 𝐴∗ to 𝐵∗ . Call it a standard function. For 𝐹, the set of all functions from 𝐴 to 𝐵, the elements of 𝐹 ∗ are the internal functions from 𝐴∗ to 𝐵∗ . Lemma 16.2.1 The image of each internal subset of 𝐴∗ under an internal function 𝑓 : 𝐴∗ → 𝐵∗ is an internal subset of 𝐵∗ . Proof. For each 𝑓 ∈ 𝐹 the following statement holds in 𝑀:  (∀𝑋 ⊆ 𝐴) (∃𝑌 ⊆ 𝐵) [(∀𝑎 ∈ 𝑋) 𝑓 (𝑎) ∈ 𝑌 ]  ∧ [(∀𝑏 ∈ 𝑌 ) (∃𝑎 ∈ 𝑋) 𝑓 (𝑎) = 𝑏] .

(16.5)

The close “∀𝑋 ⊆ 𝐴” is not part of the language L∞ (𝑀). However, it can be reinterpreted as “(∀𝑋 ∈ 𝑀 ( 𝜏) ) (∀𝑥 ∈ 𝑀 𝜏 ) [𝑥 ∈ 𝑋 → 𝑥 ∈ 𝐴]”, assuming 𝐴 ⊆ 𝑀 𝜏 and with 𝑋 = 𝑋 𝑀( 𝜏) ,1 and 𝑥 = 𝑋 𝑀𝜏 ,1 . The interpretation of this in 𝑀 ∗ is “(∀𝑋 ∈ 𝑀 (∗𝜏) ) (∀𝑥 ∈ 𝑀 𝜏∗ ) [𝑥 ∈ 𝑋 → 𝑋 ∈ 𝐴∗ ]”. This means “for all internal subsets 𝑋 of 𝐴∗ ”. Similarly, “∃𝑌 ⊆ 𝐵” interprets in 𝑀 ∗ as “there is an internal subset of 𝐵∗ ”. Thus, □ our claim is the interpretation of (16.5) in 𝑀 ∗ .

16.3 Concurrent Relations Suppose that a higher order relation 𝑅 of 𝑀 is finite with elements 𝑎 1 , . . . , 𝑎 𝑛 . Then, 𝑀 |= (∀𝑋 ∈ 𝑅) [𝑋 = 𝑎 1 ∨ · · · ∨ 𝑋 = 𝑎 𝑛 ], so that 𝑀 ∗ |= (∀𝑋 ∈ 𝑅 ∗ ) [𝑋 = 𝑎 ∗1 ∨ · · · ∨ 𝑋 = 𝑎 ∗𝑛 ]. That is, each element of 𝑅 ∗ is standard. We impose a final condition on 𝑀 ∗ guaranteeing 𝑅 ∗ will be a true “enlargement” of 𝑅 if 𝑅 is an infinite relation. Definition 16.3.1 Let 𝐴 and 𝐵 be two higher order sets of 𝑀. Call a binary relation 𝑅 ⊆ 𝐴× 𝐵 concurrent if for all 𝑎 1 , . . . , 𝑎 𝑛 ∈ 𝐴 there exists a 𝑏 ∈ 𝐵 with (𝑎 𝑖 , 𝑏) ∈ 𝑅, 𝑖 = 1, . . . , 𝑛.

16.3 Concurrent Relations

III. Compactness (or saturation) condition.

285

If 𝐴, 𝐵 are two higher order sets of

𝑀 and if 𝑅 ⊆ 𝐴 × 𝐵 is a concurrent relation, then there exists a 𝑏 ∈ 𝐵∗ such that (𝑎, 𝑏) ∈ 𝑅 ∗ for each 𝑎 ∈ 𝐴. Lemma 16.3.2 Let 𝐴 be a higher order set of 𝑀. (a) If 𝐴 = {𝑎 1 , . . . , 𝑎 𝑛 } is a finite set, then 𝐴∗ = {𝑎 1∗ , . . . , 𝑎 ∗𝑛 } and 𝐴 is identified with 𝐴∗ as higher order sets. (b) If 𝐴 is infinite, then 𝐴 is properly contained in 𝐴∗ . Ô𝑛 Proof of (a). The sentence (∀𝑋 𝐴,1 ∈ 𝐴) [ 𝑖=1 by Ô𝑛𝑋 𝐴,1 = 𝑎 1 ] ∗holds in 𝑀. Hence, Condition II of Section 16.2, (∀𝑋 𝐴,1 ∈ 𝐴∗ ) [ 𝑖=1 𝑋 𝐴,1 = 𝑎 1 ] is true in 𝑀 ∗ . Thus, 𝐴∗ = {𝑎 1∗ , . . . , 𝑎 ∗𝑛 }. Proof of (b). For every 𝑎 1 , . . . , 𝑎 𝑛 in 𝐴 there exists a 𝑏 ∈ 𝐴 with 𝑎 𝑖 ≠ 𝑏, 𝑖 = 1, . . . , 𝑛. That is, the inequality relation on 𝐴 is concurrent. Hence, there exists a 𝑏 ∈ 𝐴∗ such □ that 𝑎 ≠ 𝑏 for each 𝑎 ∈ 𝐴 (although not for each 𝑎 ∈ 𝐴∗ ). Remark 16.3.3 (Warning) Let 𝐴 be a higher order set of 𝑀. Suppose that 𝐴 is an element of another higher order set 𝐵. In Section 16.2, we have identified 𝐵 with a subset of 𝐵∗ . Under this identification we have identified the element 𝐴 of 𝐵 with the element 𝐴∗ of 𝐵∗ . Lemma 16.3.2 shows that the latter identification identifies 𝐴 with 𝐴∗ as higher order sets if and only if 𝐴 is finite. Remark 16.3.4 The case 𝑀 = N, the natural numbers, provides the first example of an external object. Replacing 𝑀 with N ∪ 𝑀, if necessary, we tacitly assume from now on that N ⊆ 𝑀. Lemma 16.3.5 The set N is an external subset of N∗ . Proof. By Lemma 16.3.2, N∗ contains a nonstandard element 𝑐. Since there are no elements of N between 𝑛 and 𝑛 + 1, the same is true for N∗ . In the extension 𝐼0 > 𝐼1 > 𝐼2 > · · · is a strictly descending sequence. 15.(a) Prove that N∗ ∖ N is not an internal subset of N∗ . (b) Prove that for each 𝑎 ∈ N the set {𝑥 ∈ N∗ | 𝑥 > 𝑎} is internal.

Notes This chapter is in the spirit of Chapter 2 of [RoR75], which contains a nonstandard proof of the Siegel–Mahler theorem.

Chapter 17

The Nonstandard Approach to Hilbert’s Irreducibility Theorem

We use the nonstandard methods of Chapter 16 to give a new criterion for a field 𝐾 to be Hilbertian: There exists a nonstandard element 𝑡 of an enlargement 𝐾 ∗ of 𝐾 such that 𝑡 has only finitely many poles in 𝐾 (𝑡)sep ∩ 𝐾 ∗ . From this there results a second and uniform proof (Corollary 17.3.4) that the classical Hilbertian fields are indeed Hilbertian. In addition, a formal power series field, 𝐾0 ((𝑋1 , . . . , 𝑋𝑛 )) of 𝑛 ≥ 2 variables over an arbitrary field 𝐾0 , is also Hilbertian (Example 17.5.2).

17.1 Criteria for Hilbertianity We give two criteria for a field 𝐾 to be Hilbertian in terms of an enlargement 𝐾 ∗ of 𝐾. The first is a straightforward application of the compactness property of 𝐾 ∗ . Proposition 17.1.1 (Gilmore–Robinson) The field 𝐾 is Hilbertian if and only if there exists a 𝑡 ∈ 𝐾 ∗ ∖ 𝐾 such that 𝐾 (𝑡)sep ∩ 𝐾 ∗ = 𝐾 (𝑡). Proof. Suppose that 𝐾 is Hilbertian. If 𝑓1 , . . . , 𝑓𝑚 ∈ 𝐾 [𝑇, 𝑋] are irreducible polynomials which are separable in 𝑋 and 𝑔1 , . . . , 𝑔𝑚 ∈ 𝐾 [𝑇] are nonzero polynomials, then there exists an 𝑎 ∈ 𝐾 with 𝑓𝑖 (𝑎, 𝑋) irreducible in 𝐾 [𝑋] and 𝑔𝑖 (𝑎) ≠ 0, 𝑖 = 1, . . . , 𝑚. The compactness property (Condition III of Section 16.3) gives 𝑡 ∈ 𝐾 ∗ such that for each irreducible 𝑓 ∈ 𝐾 [𝑇, 𝑋] which is separable in 𝑋 and each 0 ≠ 𝑔 ∈ 𝐾 [𝑇], the polynomial 𝑓 (𝑡, 𝑋) is irreducible in 𝐾 ∗ [𝑋] and 𝑔(𝑡) ≠ 0. The second condition implies that 𝑡 ∉ 𝐾. Consider 𝑥 ∈ 𝐾 (𝑡)sep ∩ 𝐾 ∗ . Let 𝑓 ∈ 𝐾 [𝑇, 𝑋] be an irreducible polynomial which is separable in 𝑋 with 𝑓 (𝑡, 𝑥) = 0. Since 𝑓 (𝑡, 𝑋) is irreducible over 𝐾 ∗ , it is linear. Hence, 𝑥 ∈ 𝐾 (𝑡). Conversely, suppose that 𝑡 is an element of 𝐾 ∗ ∖ 𝐾 with 𝐾 (𝑡)sep ∩ 𝐾 ∗ = 𝐾 (𝑡). Let 𝑓 ∈ 𝐾 [𝑇, 𝑋] be an irreducible polynomial which is separable, monic, and of degree at least 2 in 𝑋. If 𝑓 (𝑡, 𝑋) is reducible over 𝐾 ∗ , the coefficients of its factors lie in 𝐾 (𝑡)sep ∩ 𝐾 ∗ = 𝐾 (𝑡). Thus, 𝑓 (𝑡, 𝑋) is reducible over 𝐾 (𝑡). But since 𝑡 is transcendental over 𝐾, this means that 𝑓 (𝑇, 𝑋) is reducible over 𝐾, contrary to our assumption. It follows from this contradiction that 𝑓 (𝑡, 𝑋) is irreducible over 𝐾 ∗ . 291 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_17

292

17 The Nonstandard Approach to Hilbert’s Irreducibility Theorem

Since 𝐾 ∗ is an elementary extension of 𝐾 (Example 16.5.3) and 𝑡 ∈ 𝐾 ∗ , there exists an 𝑎 ∈ 𝐾 such that 𝑓 (𝑎, 𝑋) is irreducible over 𝐾. By Lemma 14.2.2, 𝐾 is Hilbertian. □ One may reapproach the classical Hilbertian fields through Proposition 17.1.1 [Roq75b], but it is easier to apply the following weaker condition of Weissauer. Consider 𝑡 ∈ 𝐾 ∗ ∖ 𝐾 and let Ω𝑡 = 𝐾 (𝑡)sep ∩ 𝐾 ∗ . By Example 16.5.3, 𝐾 ∗ /𝐾 is a regular extension, hence so is Ω𝑡 /𝐾. By definition, Ω𝑡 is a separable algebraic (possibly infinite) extension of 𝐾 (𝑡). Hence, we may consider Ω𝑡 as a generalized function field of one variable over 𝐾. It is a union of function fields of one variable over 𝐾. Call an equivalence class of valuations of Ω𝑡 which are trivial on 𝐾 a prime divisor of Ω𝑡 /𝐾. Refer to a prime divisor 𝔭 as a pole of 𝑡 if 𝑡 has a negative value at each representative of 𝔭. Finally, call 𝑡 a polefinite element of 𝐾 ∗ if 𝑡 has a pole in Ω𝑡 and the number of those poles is finite. In other words, 𝑡 ∈ 𝐾 ∗ ∖ 𝐾 and there is an integer 𝑚 such that the number of poles of 𝑡 in any function field 𝐹 with 𝐾 (𝑡) ⊆ 𝐹 ⊆ Ω𝑡 is at most 𝑚. Folgerung 3.2 of [Wei82] says that if 𝐾 ∗ has a polefinite element, then 𝐾 is Hilbertian. In order to generalize this result to integral domains we use Lemma 14.2.2. Proposition 17.1.2 Let 𝑅 be an integral domain with quotient field 𝐾. If 𝑅 ∗ contains a polefinite element of 𝐾 ∗ , then 𝑅 is Hilbertian. Proof. The existence of a polefinite element of 𝐾 ∗ implies 𝐾 ≠ 𝐾 ∗ . Hence, by Lemma 16.3.2(a), 𝐾 is infinite, so 𝑅 is also infinite. We assume that 𝑅 is not Hilbertian and prove that there is no polefinite element of 𝐾 ∗ in 𝑅 ∗ . Indeed, Lemma 14.2.2 gives an irreducible polynomial 𝑓 ∈ 𝐾 [𝑇, 𝑋], separable, monic, and of degree at least 2 in 𝑋, such that 𝑓 (𝑎, 𝑋) is reducible over 𝐾 for each 𝑎 ∈ 𝑅. The same statement holds for 𝑅 ∗ . Bearing in mind that each 𝑡 ∈ 𝑅 ∗ ∖ 𝐾 is transcendental over 𝐾, it follows that (17.1) 𝑓 (𝑡, 𝑋) is irreducible over 𝐾 (𝑡), but reducible over 𝐾 ∗ . Consider now an element 𝑡 of 𝑅 ∗ ∖ 𝐾. The remaining parts of the proof consider properties of the prime divisors of 𝐾 (𝑡)/𝐾. Part A: Removing ramification over ∞. Let 𝐹 be the splitting field of 𝑓 (𝑡, 𝑋) over 𝐾 (𝑡). For each 𝑎 ∈ 𝐾˜ ∪ {∞} let 𝔭𝑎 be the prime divisor of 𝐾 (𝑡)/𝐾 defined by 𝑡 ↦→ 𝑎. In particular, 𝔭∞ is the unique pole of 𝑡 in 𝐾 (𝑡)/𝐾. Since 𝐾 is infinite, we ¯ ¯ In particular, 𝐹/𝐹 may choose 𝑎 ∈ 𝐾 with 𝔭𝑎 unramified in 𝐹 with residue field 𝐹. ¯ ¯ is separable. By Lemma 7.1.1(a), 𝐹/𝐹 is normal, hence 𝐹/𝐹 is Galois. Apply the 1 𝐾-automorphism of 𝐾 (𝑡) defined by 𝑡 ↦→ 𝑡−𝑎 to replace 𝑓 (𝑇, 𝑋) by 𝑔(𝑇, 𝑋) := 1 𝑓 ( 𝑇−𝑎 , 𝑋), the field 𝐹 by the splitting field of 𝑔(𝑡, 𝑋), and 𝔭𝑎 by 𝔭∞ . Thus assume, along with (17.1), that (17.2) 𝔭∞ is unramified in 𝐹. Each 𝑎 ∈ 𝐾 defines a 𝐾-automorphism 𝜎𝑎 of 𝐾 (𝑡) by 𝑡 ↦→ 𝑡 +𝑎. It fixes 𝔭∞ . Extend 𝜎𝑎 to an automorphism of 𝐾 (𝑡)sep . Denote 𝜎𝑎 (𝐹) by 𝐹𝑎 . Thus, 𝔭∞ is unramified in 𝐹𝑎 and 𝐹¯𝑎 is Galois over 𝐾. By Corollary 2.3.7(c), (17.3) 𝔭∞ is unramified in the compositum, 𝐹 ′, of 𝐹𝑎 for all 𝑎 ∈ 𝐾.

17.2 Arithmetical Primes Versus Functional Primes

293

Part B: The finiteness of 𝐹 ′ over ∞. Let 𝔓 be a prime divisor of 𝐹 ′ over 𝔭∞ . Denote reduction modulo 𝔓 by a bar. The residue fields 𝐹¯ and 𝐹¯𝑎 are conjugate over 𝐾 (third paragraph before the end of Section 2.1). Since both are Galois extensions of 𝐾 (Lemma 7.1.1), they coincide. The compositum of the residue class fields of unramified extensions of 𝐾 (𝑡) is the residue class field of their compositum (Lemma 2.5.8). Thus, 𝐹¯ ′ = 𝐹¯ is a finite extension of 𝐾 which is independent of 𝔓. Part C: The infinitude of [𝐹 ′ ∩ 𝐾 ∗ : 𝐾 (𝑡)]. By (17.1), 𝐹𝑎 is not linearly disjoint from 𝐾 ∗ over 𝐾 (𝑡). Hence, 𝐸 𝑎 := 𝐹𝑎 ∩ 𝐾 ∗ , a regular extension of 𝐾 contained in Ω𝑡 , is a proper extension of 𝐾 (𝑡). Let R (𝐸 𝑎 ) (resp. R (𝐹𝑎 )) be the set of prime divisors of 𝐾 (𝑡)/𝐾 that ramify in 𝐸 𝑎 (resp. in 𝐹𝑎 ). By Remark 4.6.2(b), R (𝐸 𝑎 ) is finite and nonempty. Note: for 𝑐, 𝑐 ′ ∈ 𝐾˜ we have 𝔭𝑐 = 𝔭𝑐′ if and only if 𝑐 ′ is conjugate to 𝑐 over 𝐾. If R (𝐹) = {𝔭𝑐1 , . . . , 𝔭𝑐𝑛 }, then R (𝐹𝑎 ) = {𝔭𝑐1 −𝑎 , . . . , 𝔭𝑐𝑛 −𝑎 }. For 𝐿 a finite separable ˜ Since 𝐾 is an extension of 𝐾 (𝑡) let R (𝐿) be {𝔭𝑑1 , . . . , 𝔭𝑑𝑟 } with 𝑑1 , . . . , 𝑑𝑟 ∈ 𝐾. infinite field, we may choose 𝑎 ∈ 𝐾 with 𝑐 𝑖 − 𝑎 ≠ 𝜏(𝑑 𝑗 ) for all 𝑖 and 𝑗 and for every 𝐾 (𝑡)-isomorphism 𝜏 of 𝐿. Hence, R (𝐸 𝑎 ) ∩ R (𝐿) ⊆ R (𝐹𝑎 ) ∩ R (𝐿) = ∅. Since R (𝐸 𝑎 ) ≠ ∅, the field 𝐸 𝑎 is not contained in 𝐿. Therefore, the compositum 𝐸 ′ of 𝐸 𝑎 for all 𝑎 ∈ 𝐾 is an infinite extension of 𝐾 (𝑡) contained in Ω𝑡 ∩ 𝐹 ′. Part D: Conclusion of the proof. Assume that 𝑡 has only 𝑚 poles in Ω𝑡 . Use Part C to choose a finite extension 𝑁 of 𝐾 (𝑡) in 𝐸 ′ with [𝑁 : 𝐾 (𝑡)] > 𝑚 [ 𝐹¯ : 𝐾]. Let 𝔮1 , . . . , 𝔮𝑘 be all extensions of 𝔭∞ to 𝑁. Each of them extends to a pole of 𝑡 in Ω𝑡 , ¯ Since by (17.3), 𝔭∞ is unramified in 𝐹 ′, hence in 𝐸 ′, so 𝑘 ≤ 𝑚. By Part B, 𝑁¯ 𝔮𝑖 ⊆ 𝐹. we have by Statement (2.10) that [𝑁 : 𝐾 (𝑡)] =

𝑘 ∑︁

[ 𝑁¯ 𝔮𝑖 : 𝐾] ≤ 𝑚 [ 𝐹¯ : 𝐾].

𝑖=1

This contradiction to the choice of 𝑁 proves that 𝑡 is not a polefinite element of 𝐾 ∗ , as claimed. □

17.2 Arithmetical Primes Versus Functional Primes To apply Proposition 17.1.2 we start with a field 𝐾 that carries arithmetic structure and extend this structure to an enlargement 𝐾 ∗ of 𝐾. For each 𝑡 ∈ 𝐾 ∗ ∖ 𝐾 we consider finite extensions 𝐹 of 𝐾 (𝑡) in 𝐾 ∗ . Then, we compare the function field structure of 𝐹/𝐾 with the arithmetic structure induced on 𝐹 from 𝐾 ∗ . The goal is to find conditions on 𝑡 to be a polefinite element. Let 𝑆 be a set of primes of 𝐾. Thus, 𝑆 is a set of equivalent classes of absolute values of 𝐾 (Section 14.3). For each 𝔭 ∈ 𝑆 choose a representative | |𝔭 . Define a map 𝑣𝔭 : 𝐾 → R ∪ {∞} by 𝑣𝔭 (𝑎) = − log(|𝑎|𝔭 ). Conditions (14.5a)–(14.5d) on the absolute values imply the following properties of 𝑣𝔭 : (17.4a) 𝑣𝔭 (𝑎) = ∞ if and only if 𝑎 = 0. (17.4b) 𝑣𝔭 (𝑎𝑏) = 𝑣𝔭 (𝑎) + 𝑣𝔭 (𝑏). (17.4c) 𝑣𝔭 (𝑎 + 𝑏) ≥ min(𝑣𝔭 (𝑎), 𝑣𝔭 (𝑏)) − log 2.

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17 The Nonstandard Approach to Hilbert’s Irreducibility Theorem

(17.4d) There is an 𝑎 ∈ 𝐾 × with 𝑣𝔭 (𝑎) ≠ 0. We call 𝑣𝔭 an additive absolute value of 𝐾. If 𝔭 is ultrametric, then |𝑎 + 𝑏|𝔭 ≤ max(|𝑎|𝔭 , |𝑏|𝔭 ), hence 𝑣𝔭 (𝑎 + 𝑏) ≥ min(𝑣𝔭 (𝑎), 𝑣𝔭 (𝑏)) for all 𝑎, 𝑏 ∈ 𝐾. Thus, 𝑣𝔭 is a valuation of 𝐾. The following lemma gives a simple criterion for 𝔭 to be metric: Proposition 17.2.1 A prime 𝔭 of a field 𝐾 is metric if and only if 𝑣𝔭 (2) < 0. Proof. Suppose first that 𝔭 is metric. By Remark 14.3.4, char(𝐾) = 0 and the restriction of | |𝔭 to Q is a metric absolute value. By a theorem of Ostrowski, | |𝔭 is equivalent to the ordinary absolute value | | of Q [CaF67, p. 45, Thm.]. Thus, there is a positive real number 𝑐 such that |𝑥|𝔭 = |𝑥| 𝑐 for each 𝑥 ∈ Q. In particular, |2|𝔭 = |2| 𝑐 > 1. Consequently, 𝑣𝔭 (2) < 0. Conversely, if 𝑣𝔭 (2) < 0, then |2|𝔭 > 1. Hence, 𝔭 is metric (Remark 14.3.4). □ We assume now that 𝑆 satisfies the following finiteness condition: (17.5) If 𝑎 ∈ 𝐾 × , then {𝔭 ∈ 𝑆 | 𝑣𝔭 (𝑎) ≠ 0} is a finite set. By Proposition 17.2.1, there are only finitely many archimedean primes in 𝑆. Consider again an enlargement 𝐾 ∗ of 𝐾. Then, 𝑆 extends to a set 𝑆 ∗ of arithmetical primes of 𝐾 ∗ . To each 𝔭 ∈ 𝑆 ∗ there corresponds a star-additive absolute value 𝑣𝔭 : (𝐾 ∗ ) × → R∗ . In particular, each prime in 𝑆 extends to an element 𝔭 of 𝑆 ∗ , a standard prime. The elements of 𝑆 ∗ ∖ 𝑆 are the nonstandard primes. Condition (17.5) becomes: (17.6) If 0 ≠ 𝑎 ∈ 𝐾 ∗ , then the set {𝔭 ∈ 𝑆 ∗ | 𝑣𝔭 (𝑎) ≠ 0} is starfinite. (Example 16.5.1). In particular, if 0 ≠ 𝑎 ∈ 𝐾, then the finite set {𝔭 ∈ 𝑆 | 𝑣𝔭 (𝑎) ≠ 0} does not grow in the enlargement. Thus, 𝑣𝔭 (𝑎) = 0 for all 𝔭 ∈ 𝑆 ∗ ∖ 𝑆 and 𝑎 ∈ 𝐾 × . Therefore, by Proposition 17.2.1 and (17.5), there are only finitely many archimedean primes in 𝑆 ∗ ; they are all standard. For an arbitrary prime 𝔭 ∈ 𝑆 ∗ consider the convex hull of 𝑣𝔭 (𝐾 × ): Γ𝔭 = {𝑟 ∈ R∗ | ∃𝑥 ∈ 𝐾 × with |𝑟 | ≤ 𝑣𝔭 (𝑥)}. In the notation of Example 16.5.2, (17.7) Γ𝔭 = Rfin if 𝔭 ∈ 𝑆 and Γ𝔭 = 0 if 𝔭 ∈ 𝑆 ∗ ∖ 𝑆. Unlike nonstandard primes, standard primes form valuations that are nontrivial on 𝐾. We modify them so that they will “behave” like the nonstandard primes. However, these “modified” primes will not be internal objects. Combining internal and external objects is the key to the nonstandard machinery. For 𝔭 ∈ 𝑆 ∗ , consider the ordered group R¤ 𝔭 = R∗ /Γ𝔭 . By (17.7), R¤ 𝔭 = R¤ (Example 16.5.2) if 𝔭 is standard and R¤ 𝔭 = R∗ otherwise. Define the modified valuation 𝑣¤𝔭 : (𝐾 ∗ ) × → R¤ 𝔭 by 𝑣¤𝔭 (𝑥) = 𝑣𝔭 (𝑥) + Γ𝔭 . Suppose that 𝔭 is standard. Then, 𝔭 may be metric or ultrametric. In the second case 𝑣𝔭 is a valuation, whereas in the first case it is not. In both cases, however, 𝑣𝔭 satisfies (17.7). Since log 2 ∈ R ⊆ Γ𝔭 , 𝑣¤𝔭 (𝑎 + 𝑏) ≥ min{ 𝑣¤𝔭 (𝑎), 𝑣¤𝔭 (𝑏)}. Hence, 𝑣¤𝔭 is a valuation. In addition, if 0 ≠ 𝑥 ∈ 𝐾, then 𝑣𝔭 (𝑥) ∈ R. Hence, 𝑣¤𝔭 (𝑥) = 0. In summary:

17.3 Fields with the Product Formula

295

Lemma 17.2.2 For each 𝔭 ∈ 𝑆 ∗ the function 𝑣¤𝔭 : (𝐾 ∗ ) × → R¤ 𝔭 defined by 𝑣¤𝔭 (𝑥) = 𝑣𝔭 (𝑥) + Γ𝔭 is a valuation which is trivial on 𝐾. Moreover, if 𝔭 ∈ 𝑆 ∗ ∖ 𝑆, then 𝑣¤𝔭 = 𝑣𝔭 , so 𝑣𝔭 is trivial on 𝐾. Consider an element 𝑡 ∈ 𝐾 ∗ ∖ 𝐾. Let 𝐹 be a finite extension of 𝐾 (𝑡) in 𝐾 ∗ . By Example 16.5.3, 𝐹 is a regular extension of 𝐾, hence a function field over 𝐾. Consider 𝔭 ∈ 𝑆 ∗ with 𝑣¤𝔭 not vanishing on 𝐹 × . Then, by Lemma 14.2.2, the restriction of 𝑣¤𝔭 to 𝐹 is a valuation which is trivial on 𝐾. Thus, it defines a prime 𝑃 of 𝐹/𝐾. We say that 𝑃 is induced from 𝔭. Conversely, we may ask if a prime divisor 𝑃 of 𝐹/𝐾 (i.e. a functional prime) is induced by an arithmetical prime. This question is inspired by the following observation. Proposition 17.2.3 Let 𝑆 be a set of primes of a field 𝐾 with {𝔭 ∈ 𝑆 | 𝑣𝔭 (𝑎) ≠ 0} being a finite set for each 𝑎 ∈ 𝐾 × . Let 𝑡 be a nonstandard element of an enlargement 𝐾 ∗ of 𝐾. Suppose that 𝑆(𝑡) = {𝔭 ∈ 𝑆 ∗ | 𝑣¤𝔭 (𝑡) < 0} is a finite set and for each finite separable extension 𝐹 of 𝐾 (𝑡) in 𝐾 ∗ all poles of 𝑡 in 𝐹 are induced by arithmetical primes (i.e. elements of 𝑆 ∗ ). Then, 𝑡 is a polefinite element. Proof. Let 𝑃 be a pole of 𝑡 in Ω𝑡 = 𝐾 (𝑡)sep ∩ 𝐾 ∗ . For each finite extension 𝐹 of 𝐾 in Ω𝑡 the restriction of 𝑃 to 𝐹 is also a pole of 𝑡. Hence, there is a 𝔭𝐹 ∈ 𝑆 ∗ which induces 𝑃| 𝐹 . In particular, 𝑣¤𝔭𝐹 (𝑡) < 0, so 𝔭𝐹 ∈ 𝑆(𝑡). Since 𝑆(𝑡) is finite, there is a 𝔭 ∈ 𝑆(𝑡) which induces 𝑃| 𝐹 for each 𝐹 as above (Use compactness of 𝑆(𝑡).) Therefore, 𝔭 determines 𝑃. Thus, the number of poles of 𝑡 in Ω𝑡 is at most |𝑆(𝑡)|. Consequently, 𝑡 is a polefinite □ element.

17.3 Fields with the Product Formula As in Section 17.2, suppose that 𝑆 is a nonempty set of primes of 𝐾. For each 𝔭 ∈ 𝑆 choose an absolute value | |𝔭 representing 𝔭 and let 𝑣𝔭 be the corresponding additive absolute value. We say that 𝑆 satisfies a product formula if the following statement holds: for each 𝔭 ∈ 𝑆 there exists a positive real number 𝜆𝔭 with the following property: (17.8) For each 𝑎 ∈ 𝐾 × the set {𝔭 ∈ 𝑆 | |𝑎|𝔭 ≠ 1} is finite and Î 𝜆𝔭 𝔭∈𝑆 |𝑎|𝔭 = 1. In this case call 𝐾 a field with a product formula. Example 17.3.1 (Basic examples of fields with product formulas) (a) For 𝐾 = Q and 𝑝 a prime number, define | | 𝑝 by |𝑎| 𝑝 = 𝑝 −𝑟 for 𝑎 = 𝑦𝑥 · 𝑝 𝑟 with 𝑟 ∈ Z and 𝑥, 𝑦 ∈ Z relatively prime to 𝑝. The infinite absolute value | | ∞ is the usual absolute value. This set 𝑆 of primes satisfies the product formula with 𝜆𝔭 = 1 for each 𝔭 ∈ 𝑆.

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17 The Nonstandard Approach to Hilbert’s Irreducibility Theorem

(b) For 𝐾0 (t), with 𝐾0 an arbitrary field and t = (𝑡1 , . . . , 𝑡 𝑛 ), 𝑛 ≥ 1, choose a real number 0 < 𝑐 < 1. For each irreducible polynomial 𝑝 ∈ 𝐾0 [t] define the absolute value | | 𝑝 by |𝑢| 𝑝 = 𝑐𝑟 for 𝑢 = 𝑔𝑓 · 𝑝 𝑟 with 𝑓 , 𝑔 polynomials relatively prime to 𝑝. Let 𝜆 𝑝 be deg( 𝑝). For a quotient 𝑓 /𝑔 of polynomials, define the infinite absolute value as | 𝑓 /𝑔| ∞ = 𝑐deg(𝑔)−deg( 𝑓 ) and let 𝜆∞ = 1. Now recall that two polynomials 𝑓 , 𝑔 ∈ 𝐾0 [t] are associate if there exists a 𝑢 ∈ 𝐾0× such that 𝑔 = 𝑢 𝑓 . Let 𝑆 be the set of representatives of the associate classes of all irreducible Î polynomials in 𝐾 [t]. Then, every 𝑓 ∈ 𝐾 (t) has a unique factorization 𝑓 = 𝑢 𝑝 ∈𝑆 𝑝 𝑖 𝑝 with 𝑢 ∈ 𝐾0× , 𝑖 𝑝 ∈ Z and 𝑖 𝑝 = 0 for all but finitely many 𝑝’s (Example 2.2.1(b)). Hence, Ö 𝜆 Ö 𝑖 Ö 𝑖 deg( 𝑝) Ö Ö | 𝑓 | 𝜆∞∞ · | 𝑓 | 𝑝𝑝 = | 𝑝| ∞𝑝 · | 𝑝| 𝑝𝑝 = 𝑐−𝑖 𝑝 deg( 𝑝) · 𝑐𝑖 𝑝 deg( 𝑝) = 1 𝑝 ∈𝑆

𝑝 ∈𝑆

𝑝 ∈𝑆

𝑝 ∈𝑆

𝑝 ∈𝑆

for each 𝑓 ∈ 𝐾0 (t). Thus, 𝐾0 (t) is a field with a product formula. (c) A product formula of a field, if it exists, need not be unique. Indeed, in the situation of (b) with 𝑛 ≥ 2, we may consider 𝐾0 (t) as the field 𝐾𝑛−1 (𝑡 𝑛 ) of rational functions in 𝑡 𝑛 over the field 𝐾𝑛−1 = 𝐾0 (𝑡 1 , . . . , 𝑡 𝑛−1 ). Then, every prime element of the ring 𝐾𝑛−1 [𝑡 𝑛 ] is also a prime element of 𝐾0 [t] but not conversely. Therefore, the product formulas that (b) gives with respect to the rings 𝐾0 [t] and 𝐾𝑛−1 [𝑡 𝑛 ] are not the same. (d) If a set 𝑆 of primes of a field 𝐾 satisfies the product formula and if 𝑆 ′ is the set of primes of a finite extension 𝐾 ′/𝐾 lying over 𝑆, then 𝑆 ′ satisfies the product formula [Lan62, p. 20]. We conclude from (a) and (b) that every number field and every function field of several variables over an arbitrary field has a product formula. Again, let 𝐾 be any field with the product formula with respect to a nonempty set of primes 𝑆. Rewrite condition (17.8) additively: (17.9) Í For each 𝑎 ∈ 𝐾 × , {𝔭 ∈ 𝑆 | 𝑣𝔭 (𝑎) ≠ 0} is finite and 𝔭∈𝑆 𝜆𝔭 𝑣𝔭 (𝑎) = 0. For an enlargement 𝐾 ∗ of 𝐾, Condition (17.9) has a similar form: (17.10) Í For each 𝑎 ∈ (𝐾 ∗ ) × , the set {𝔭 ∈ 𝑆 ∗ | 𝑣𝔭 (𝑎) ≠ 0} is starfinite and ∗ 𝔭∈𝑆 Í∗ 𝜆𝔭 𝑣𝔭 (𝑎) = 0, where 𝜆𝔭 ∈ R and 𝜆𝔭 > 0 are independent of 𝔭 and here is the starfinite summation (Example 16.5.1). Proposition 17.3.2 Let 𝑡 be a nonstandard element of 𝐾 ∗ and 𝐹 a finite separable extension of 𝐾 (𝑡) in 𝐾 ∗ . If there exists a 𝔮 ∈ 𝑆 ∗ with 𝑣¤ 𝔮 (𝑡) < 0, then each functional prime of 𝐹 is induced by an arithmetical prime of 𝐾 ∗ . Proof. Denote the set of all functional primes of 𝐹 which are induced by elements of 𝑆 ∗ by P. Let 𝐷 = {𝑥 ∈ 𝐹 | 𝑣 𝑃 (𝑥) ≥ 0 for all 𝑃 ∈ P} = {𝑥 ∈ 𝐹 | 𝑣¤𝔭 (𝑥) ≥ 0 for all 𝔭 ∈ 𝑆 ∗ } be the holomorphy ring of P. The assumption 𝑣¤ 𝔮 (𝑡) < 0 implies that 𝑡 ∉ 𝐷 and P ≠ ∅, so 𝐷 ≠ 𝐹. Assume that P does not contain all the functional primes of 𝐹. Then, 𝐹 = Quot(𝐷) (Proposition 4.3.2(a)). We now show that 𝐷 is a field. This will give a contradiction to 𝐷 ≠ 𝐹 and establish the theorem.

17.3 Fields with the Product Formula

297

We need to prove that if 𝑥 ∈ 𝐷 and 𝑥 ≠ 0, then 𝑥 −1 ∈ 𝐷. Indeed, by (17.10), the set 𝑆(𝑥) = {𝔭 ∈ 𝑆 ∗ | 𝑣𝔭 (𝑥) < 0} is starfinite. For 𝔭 ∈ 𝑆 ∗ ∖ 𝑆, Lemma 17.2.2 shows that 𝑣𝔭 (𝑥) = 𝑣¤𝔭 (𝑥) ≥ 0. This means that 𝑆(𝑥) contains only standard primes. Hence, by Corollary 16.3.6(a), 𝑆(𝑥) is a finite set. For 𝔭 ∈ 𝑆(𝑥) we have 𝑣𝔭 (𝑥) < 0 and 𝑣¤𝔭 (𝑥) ≥ 0, so 𝑣¤𝔭 (𝑥) = 0. Now use the summation formula (17.10) and the additivity of starfinite summation (Example 16.5.1): ∑︁ ∑︁ 𝜆𝔭 𝑣𝔭 (𝑥) =¤ (17.11) 𝜆𝔭 𝑣𝔭 (𝑥) =¤ 0. 𝔭∈𝑆 ∗ ∖ 𝑆 ( 𝑥)

𝔭∈𝑆 ∗

If 𝔭 ′ ∈ 𝑆 ∗ ∖ 𝑆(𝑥) and 𝑣¤𝔭′ (𝑥) ≠ 0, then 𝑣¤𝔭′ (𝑥) > 0. This means that ∑︁ 𝔭∈𝑆 ∗

𝜆𝔭 𝑣𝔭 (𝑥) ≥ 𝜆𝔭′ 𝑣𝔭′ (𝑥) >¤ 0,

∖ 𝑆 ( 𝑥)

a contradiction to (17.11). Thus, 𝑣¤𝔭 (𝑥) = 0 for each 𝔭 ∈ 𝑆 ∗ ∖ 𝑆(𝑥) and therefore for each 𝔭 ∈ 𝑆 ∗ . It follows that 𝑣¤𝔭 (𝑥 −1 ) = 0 for each 𝔭 ∈ 𝑆 ∗ . Consequently, 𝑥 −1 ∈ 𝐷, as desired. □

Theorem 17.3.3 (Weissauer) Let 𝑅 be an integral domain with quotient field 𝐾 with a product formula. Then, 𝑅 is Hilbertian, hence so is 𝐾. Proof. Let 𝑆 be a nonempty set of primes of 𝐾 satisfying the product formula, as in (17.8). Choose 𝔭 ∈ 𝑆 and an element 𝑥 ∈ 𝐾 × such that 𝑣𝔭 (𝑥) ≠ 0 (by (17.4d)). Then, write 𝑥 = 𝑏𝑎 with nonzero 𝑎, 𝑏 ∈ 𝑅. Without loss assume that 𝑣𝔭 (𝑎) ≠ 0. By (17.9), Í 𝔮 ∈𝑆 𝜆𝔮 𝑣𝔮 (𝑎) = 0. Since 𝜆𝔮 > 0 for each 𝔮 ∈ 𝑆, there exists a 𝔮 ∈ 𝑆 with 𝑣𝔮 (𝑎) < 0. In an enlargement 𝐾 ∗ of 𝐾, we use Lemma 16.3.2(b) to pick a nonstandard positive integer 𝜔 ∈ N∗ ∖ N. We prove that the nonstandard element 𝑡 = 𝑎 𝜔 of 𝑅 ∗ is polefinite. Indeed, 𝑆(𝑡) = {𝔭 ∈ 𝑆 ∗ | 𝑣𝔭 (𝑡) < 0} = {𝔭 ∈ 𝑆 ∗ | 𝑣𝔭 (𝑎) < 0} = {𝔭 ∈ 𝑆 ∗ | 𝑣¤𝔭 (𝑡) < 0}. Therefore, 𝑆(𝑡) is a starfinite set that contains 𝔮. In particular, 𝑣¤ 𝔮 (𝑡) < 0. By Lemma 17.2.2 and by the definition of 𝑆, if 𝔭 ′ ∈ 𝑆 ∗ ∖ 𝑆, then 𝑣¤𝔭′ (𝑎) = 𝑣𝔭′ (𝑎) = 0. Thus, 𝑆(𝑡) consists only of standard primes, so, by Lemma 16.3.6(b), 𝑆(𝑡) is a finite set. Suppose that 𝐹 is a finite separable extension of 𝐾 (𝑡) in 𝐾 ∗ . By Proposition 17.3.2, each functional prime 𝑃 of 𝐹/𝐾 (𝑡) is induced by some 𝔭 ∈ 𝑆 ∗ . If 𝑃 is a pole of 𝑡, then 𝔭 ∈ 𝑆(𝑡). By Proposition 17.2.3, 𝑡 is polefinite. Consequently, by Proposition 17.1.2, 𝑅 is Hilbertian. □ An application of Example 17.3.1 leads to the following special case of Theorem 17.3.3. Corollary 17.3.4 Let 𝑅 be an integral domain with quotient field 𝐾. Suppose that 𝐾 is either a number field or a function field of several variables over an arbitrary field. Then, 𝑅 (hence, also 𝐾) is Hilbertian.

298

17 The Nonstandard Approach to Hilbert’s Irreducibility Theorem

17.4 Generalized Krull Domains Non-standard methods produce new Hilbertian fields: The quotient field of each ‘generalized Krull domain of dimension at least 2’ is Hilbertian (Theorem 17.4.6). In particular, fields of formal power series of at least two variables over arbitrary fields are Hilbertian (Example 17.5.2). Let 𝑅 be an integral domain with quotient field 𝐾. We call 𝑅 a generalized Krull domain if 𝐾 has a nonempty set 𝑆 of primes satisfying Conditions (17.12a)–(17.12d) below: (17.12a) For each 𝔭 ∈ 𝑆, 𝑣𝔭 is a real valuation. (17.12b) The valuation ring 𝑂𝔭 of 𝑣𝔭 is the local ring of 𝑅 relative to 𝔪𝔭 := {𝑎 ∈ 𝑅 | 𝑣𝔭 (𝑎) > 0}. Ñ (17.12c) 𝑅 = 𝔭∈𝑆 𝑂𝔭 . (17.12d) For each 𝑎 ∈ 𝐾 × the set {𝔭 ∈ 𝑆 | 𝑣𝔭 (𝑎) ≠ 0} is finite. The dimension of a ring 𝑅 is the maximal integer 𝑛 for which there is a sequence 𝔭0 ⊂ 𝔭1 ⊂ · · · ⊂ 𝔭𝑛 of 𝑛 + 1 distinct prime ideals. Thus, dim(𝑅) ≥ 2 if and only if (17.12e) 𝑅 has a maximal ideal 𝑀 which properly contains a nonzero prime ideal. Hence, 𝑅 is a generalized Krull ring of dimension exceeding 1 if and only if it satisfies Conditions (17.12a)–(17.12e). Lemma 17.4.1 Let 𝑅 be a generalized Krull domain of dimension exceeding 1. Then: (a) For each nonunit 𝑏 of 𝑅, 𝑏 ≠ 0, there exists a 𝔭 ∈ 𝑆 with 𝑣𝔭 (𝑏) > 0. (b) For each 𝔭 ∈ 𝑆, 𝔪𝔭 is minimal among nonzero prime ideals of 𝑅 and 𝔪𝔭 ≠ 𝑀. (c) If 𝔭, 𝔮 are distinct primes in 𝑆, then 𝔪𝔭 ̸ ⊆ 𝔪𝔮 . Ñ Proof of (a). Otherwise 𝑏 −1 ∈ 𝔭∈𝑆 𝑂𝔭 = 𝑅. Proof of (b). Let 𝔫 ⊆ 𝔪𝔭 be a nonzero prime ideal. Choose 0 ≠ 𝑎 ∈ 𝔫. Since 𝑣𝔭 is real, for each 𝑏 ∈ 𝔪𝔭 there is a positive integer 𝑛 with 𝑣𝔭 (𝑎) ≤ 𝑛𝑣𝔭 (𝑏) = 𝑣𝔭 (𝑏 𝑛 ), so 𝑏𝑛 𝑐 𝑏𝑛 𝑛 𝑎 ∈ 𝑂 𝔭 . By (17.12b), there are 𝑐, 𝑑 ∈ 𝑅 with 𝑑 ∉ 𝔪𝔭 and 𝑎 = 𝑑 , so 𝑑𝑏 = 𝑎𝑐 ∈ 𝔫. Since 𝑑 ∉ 𝔫, we have 𝑏 ∈ 𝔫. Consequently, 𝔫 = 𝔪𝔭 . By (17.12e), 𝑀 is not minimal, hence 𝔪𝔭 ≠ 𝑀. Proof of (c). By (17.12b), 𝔪𝔭 ≠ 𝔪𝔮 . By (b), 𝔪𝔭 ⊄ 𝔪𝔮 Hence, 𝔪𝔭 ̸ ⊆ 𝔪𝔮 .

□

Lemma 17.4.2 The local ring 𝑅 𝑀 of 𝑅 at 𝑀 is a generalized Krull domain of dimension exceeding 1 with respect to 𝑆 ′ = {𝔭 ∈ 𝑆 | 𝔪𝔭 ⊂ 𝑀 }. Proof. Conditions (17.12a), (17.12b), (17.12d), and (17.12e) followÑfrom the basic definitions of the local ring 𝑅 𝑀 . It remains to prove that 𝑅 𝑀 = 𝔭∈𝑆′ 𝑂𝔭 . This follows if we show, for 𝑥 ∈ 𝐾 with 𝑣𝔭 (𝑥) ≥ 0 for each 𝔭 ∈ 𝑆 ′, that 𝑥 ∈ 𝑅 𝑀 . By (17.12d), the set 𝑇 = {𝔮 ∈ 𝑆 | 𝑣𝔮 (𝑥) < 0} is finite. If 𝔮 ∈ 𝑇, then 𝑣𝔮 (𝑥) < 0, hence 𝔮 ∉ 𝑆 ′ (by the condition imposed on 𝑥), so 𝔪𝔮 ̸ ⊆ 𝑀. Choose 𝑎𝔮 ∈ 𝔪𝔮 ∖ 𝑀. By (17.12a), there is a positive integer 𝑛(𝔮) with 𝑣𝔮 (𝑎𝔮𝑛(𝔮) ) > −𝑣𝔮 (𝑥). Let 𝑎 = Î 𝑛(𝔮) and 𝑦 = 𝑎𝑥. Then, 𝑎 ∈ 𝑅 ∖ 𝑀 and 𝑣𝔭 (𝑦) ≥ 0 for each 𝔭 ∈ 𝑆. Therefore, 𝔮 ∈𝑇 𝑎𝔮 by (17.12c), 𝑦 ∈ 𝑅. Consequently, 𝑥 ∈ 𝑅 𝑀 . □

17.4 Generalized Krull Domains

299

For the goal of this section – a proof that 𝐾 is Hilbertian – we may replace 𝑅 by 𝑅 𝑀 to assume that (17.13) 𝑅 is a local ring and 𝑀 is its maximal ideal. Consider an enlargement 𝐾 ∗ of 𝐾. Then, 𝑅 ∗ is a local ring with maximal ideal Also, 𝑆 ∗ has these properties: (17.14a) For 𝔭 ∈ 𝑆 ∗ , 𝑣𝔭 is a valuation of 𝐾 ∗ with values in R∗ . (17.14b) The valuation ring 𝑂𝔭∗ of 𝑣𝔭 is the local ring of 𝑅 ∗ relative to the prime ideal 𝔪𝔭∗ = {𝑎 ∈ 𝑅 ∗ | 𝑣𝔭 (𝑎) > 0}. Ñ (17.14c) 𝑅 ∗ = 𝔭∈𝑆 ∗ 𝑂𝔭∗ . (17.14d) For each 𝑎 ∈ 𝐾 ∗ ∖{0} the set {𝔭 ∈ 𝑆 ∗ | 𝑣𝔭 (𝑎) ≠ 0} is starfinite. (17.14e) 𝑀 ∗ properly contains the prime ideals 𝔪𝔭∗ .

𝑀 ∗.

As with fields with a product formula, consider the modified valuations 𝑣¤𝔭 and ∗ ¤ their valuation rings ring of these Ñ 𝑂𝔭¤ = {𝑥 ∈ 𝐾 | 𝑣¤𝔭 (𝑥) ≥ 0}. The holomorphy ¤ valuations, 𝑅 = 𝔭∈𝑆 ∗ 𝑂𝔭 , contains both 𝐾 (Lemma 17.2.2) and 𝑅 ∗ . Moreover: Lemma 17.4.3 The ring 𝑅¤ equals 𝐾 · 𝑅 ∗ . Proof. We have to show only that each 𝑥 ∈ 𝑅¤ lies in the composite 𝐾 · 𝑅 ∗ . Indeed, 𝑣¤𝔭 (𝑥) ≥ 0 for each 𝔭 ∈ 𝑆 ∗ . From (17.14b) and lemma 17.2.2, the starfinite set 𝑆(𝑥) = {𝔭 ∈ 𝑆 ∗ | 𝑣𝔭 (𝑥) < 0} contains only standard primes. Hence, (Lemma 16.3.6(b)), it is actually finite. For each 𝔭 ∈ 𝑆(𝑥) there exists an 𝑟𝔭 ∈ R with 𝑣𝔭 (𝑥) ≥ 𝑟𝔭 (because 𝑣¤𝔭 (𝑥) ≥ 0). By (17.14b), we may find 0 ≠ 𝑎𝔭 ∈ 𝑅 with Î 𝑣𝔭 (𝑎𝔭 ) ≥ −𝑣𝔭 (𝑥). Let 𝑎 = 𝔭∈𝑆 ( 𝑥) 𝑎𝔭 and 𝑦 = 𝑎𝑥. Then, 𝑣𝔭 (𝑦) ≥ 0 for each 𝔭 ∈ 𝑆 ∗ . □ By (17.14c), 𝑦 ∈ 𝑅 ∗ . Consequently, 𝑥 = 𝑎 −1 𝑦 ∈ 𝐾 · 𝑅 ∗ . ¤ Suppose that for each 𝔭 ∈ 𝑆 we have Lemma 17.4.4 Let 𝑥 and 𝑦 be nonunits of 𝑅. ¤ 𝑣¤𝔭 (𝑥) > 0 or 𝑣¤𝔭 (𝑦) > 0. Then, 𝛼𝑥 + 𝛽𝑦 ≠ 1 for all 𝛼, 𝛽 ∈ 𝑅. Proof. Assume that there exist 𝛼, 𝛽 ∈ 𝑅¤ with 𝛼𝑥 + 𝛽𝑦 = 1. Apply Lemma 17.4.3 to write 𝑎𝑥 𝑏𝑦 + = 1 with 𝑎, 𝑏 ∈ 𝑅 ∗ and 𝑐 ∈ 𝐾 × . (17.15) 𝑐 𝑐 We show that both summands on the left-hand side of (17.15) belong to 𝑀 ∗ . This contradiction to (17.15) will conclude the proof of the lemma. Indeed, let 𝔭 ∈ 𝑆 ∗ with 𝑣𝔭 (𝑐) ≠ 0. Then, 𝑣¤𝔭 (𝑐) = 0 and 𝔭 is standard (Lemma 17.2.2). Hence, by the assumption of the lemma, 𝑣¤𝔭 (𝑥) > 0 or 𝑣¤𝔭 (𝑦) > 0. If 𝑏𝑦 𝑎𝑥 𝑣¤𝔭 (𝑥) > 0, then 𝑣¤𝔭 ( 𝑎𝑥 𝑐 ) > 0. Hence, 𝑣𝔭 ( 𝑐 ) > 0. So, by (17.15), 𝑣𝔭 ( 𝑐 ) = 0. 𝑏𝑦 Similarly, if 𝑣¤𝔭 (𝑦) > 0, then 𝑣𝔭 ( 𝑐 ) > 0 and 𝑣𝔭 ( 𝑎𝑥 𝑐 ) = 0. 𝑏𝑦 ∗ Next consider 𝔭 ∈ 𝑆 with 𝑣𝔭 (𝑐) = 0. Then, 𝑣𝔭 ( 𝑎𝑥 𝑐 ) ≥ 0 and 𝑣𝔭 ( 𝑐 ) ≥ 0. 𝑏𝑦 ∗ It follows from (17.14c) that both 𝑎𝑥 𝑐 and 𝑐 belong to 𝑅 . ∗ ¤ Since 𝑥 is a nonunit in 𝑅, there exists a 𝔭 ∈ 𝑆 with 𝑣¤𝔭 (𝑥) > 0. Hence 𝑣¤𝔭 ( 𝑎𝑥 𝑐 ) > 0 ∗ . Similarly 𝑏𝑦 ∈ 𝑀 ∗ . and therefore 𝑎𝑥 ∈ 𝑀 □ 𝑐 𝑐

300

17 The Nonstandard Approach to Hilbert’s Irreducibility Theorem

Our next lemma is “standard”: Lemma 17.4.5 Let {𝔭1 , . . . , 𝔭𝑚 } and {𝔮1 , . . . , 𝔮𝑛 } be two disjoint finite subsets of 𝑆. Then, there exists an element 𝑎 ∈ 𝑅 such that 𝑣𝔭𝑖 (𝑎) = 0 for 𝑖 = 1, . . . , 𝑚 and 𝑣𝔮 𝑗 (𝑎) > 0 for 𝑗 = 1, . . . , 𝑛.

(17.16)

Proof. Proceed by induction on 𝑚. Suppose that 𝑚 = 1. By Lemma 17.4.1(c), 𝔪𝔮 𝑗 ̸ ⊆ 𝔪𝔭1 for 1 ≤ 𝑗 ≤ 𝑛. Let 𝑎 𝑗 ∈ 𝔪𝔮 𝑗 ∖ 𝔪𝔭1 . Then, 𝑎 = 𝑎 1 · · · 𝑎 𝑛 satisfies (17.16). Suppose that 𝑚 > 1. The induction hypothesis gives 𝑎 1 ∈ 𝑅 with 𝑣𝔭1 (𝑎 1 ) = · · · = 𝑣𝔭𝑚−1 (𝑎 1 ) = 0 and 𝑣𝔭𝑚 (𝑎 1 ), 𝑣𝔮1 (𝑎 1 ), . . . , 𝑣𝔮𝑛 (𝑎 1 ) > 0. By the case 𝑚 = 1, there exists an 𝑎 2 ∈ 𝑅 with 𝑣𝔭𝑚 (𝑎 2 ) = 0 and 𝑣𝔭1 (𝑎 2 ), . . . , 𝑣𝔭𝑚−1 (𝑎 2 ), 𝑣𝔮1 (𝑎 2 ), . . . , 𝑣𝔮𝑛 (𝑎 2 ) > 0. The element 𝑎 = 𝑎 1 + 𝑎 2 satisfies 17.16.

□

We now prove the main theorem of this section. Theorem 17.4.6 (Weissauer) The quotient field of a generalized Krull domain of dimension exceeding 1 is Hilbertian. Proof. As previously, let 𝐾, 𝑅, 𝑆, 𝑀 satisfy (17.12a)–(17.12e) and (17.13). Let 𝔮 ∈ 𝑆 and choose 𝑏 ∈ 𝑀 ∖ 𝔪𝔮 (by (Lemma 17.4.1(b)). By (17.12d) and Lemma 17.4.1(a), the set 𝑇 (𝑏) = {𝔭 ∈ 𝑆 | 𝑣𝔭 (𝑏) > 0} is finite and nonempty.

(17.17)

For each finite set 𝑇 with 𝑇 (𝑏) ⊆ 𝑇 ⊆ 𝑆 there exists an 𝑎𝑇 ∈ 𝑅 such that 𝑣𝔭 (𝑎𝑇 ) = 0 for 𝔭 ∈ 𝑇 (𝑏) and 𝑣𝔭 (𝑎𝑇 ) > 0 for each 𝔭 ∈ 𝑇 ∖ 𝑇 (𝑏) (Lemma 17.4.5). Proposition 16.3.7 gives 𝑎 ∈ 𝑅 ∗ such that 𝑣𝔭 (𝑎) = 0 for each 𝔭 ∈ 𝑇 (𝑏) and 𝑣𝔭 (𝑎) > 0 for each 𝔭 ∈ 𝑆 ∖ 𝑇 (𝑏).

(17.18)

By Lemma 17.2.2, 𝑣𝔭 (𝑏) = 0 for each 𝔭 ∈ 𝑆 ∗ ∖ 𝑆. Thus, 𝑇 (𝑎) : = {𝔭 ∈ 𝑆 ∗ | 𝑣𝔭 (𝑎) > 0} is disjoint from 𝑇 (𝑏) = {𝔭 ∈ 𝑆 ∗ | 𝑣𝔭 (𝑏) > 0}.

(17.19)

Choose 𝜔 ∈ N∗ ∖ N. Put 𝑥 = 𝑎 𝜔 and 𝑦 = 𝑏 𝜔 . We conclude the proof in two parts, from Lemma 17.1.2, by showing that 𝑡 := 𝑦𝑥 is polefinite. Part A: 𝑆(𝑡) = {𝔭 ∈ 𝑆 ∗ | 𝑣𝔭 (𝑡) < 0} is a finite set. Indeed, since 𝔮 ∈ 𝑆 ∖ 𝑇 (𝑏), we have by (17.18) that 𝑣𝔮 (𝑎) > 0. Hence, 𝑣𝔮 (𝑥) > 0 and even 𝑣¤ 𝔮 (𝑥) > 0.

(17.20)

Next choose 𝔮 ′ ∈ 𝑇 (𝑏). By definition, 𝑣𝔮′ (𝑏) > 0, so 𝑣¤ 𝔮′ (𝑦) > 0. Thus, 𝑥 and 𝑦 ¤ Moreover, if 𝔭 ∈ 𝑇 (𝑏), then 𝑣¤𝔭 (𝑦) > 0 and if 𝔭 ∈ 𝑆 ∖ 𝑇 (𝑏), then are nonunits of 𝑅. 𝑣¤𝔭 (𝑥) > 0 (by 17.18). By Lemma 17.4.4, ¤ 𝛼𝑥 + 𝛽𝑦 ≠ 1 for all 𝛼, 𝛽 ∈ 𝑅. In particular, 𝛼𝑡 + 𝛽 ≠ 1 for all 𝛼, 𝛽 ∈ 𝐾, so 𝑡 ∉ 𝐾.

(17.21)

17.5 Examples

301

For each 𝔭 ∈ 𝑆 ∗ , (17.19) gives (17.22a) 𝑣𝔭 (𝑥) > 0 =⇒ 𝑣𝔭 (𝑎) > 0 =⇒ 𝑣𝔭 (𝑦) = 0 and 𝑣𝔭 (𝑡) = 𝑣𝔭 (𝑥); and (17.22b) 𝑣𝔭 (𝑦) > 0 =⇒ 𝑣𝔭 (𝑏) > 0 =⇒ 𝑣𝔭 (𝑥) = 0 and 𝑣𝔭 (𝑡) = −𝑣𝔭 (𝑦). Therefore, by (17.17), 𝑆(𝑡) = 𝑇 (𝑏) is a nonempty finite set. Part B: Application of Proposition 17.2.3. To complete the proof it suffices to show that if 𝐹/𝐾 (𝑡) is a finite separable extension in 𝐾 ∗ , then the functional primes of 𝐹 are induced by arithmetical primes. Let P be the set Ñ of functional primes of 𝐹 which are induced by arithmetical primes. Let 𝐷 = 𝑃 ∈ P 𝑂 𝑃 be the corresponding holomorphy ring in 𝐹. Since, by (17.20) and (17.22a), 𝑣𝔮 (𝑡) = 𝑣𝔮 (𝑥) > 0, the set P is nonempty. Assume that P excludes a functional prime of 𝐹. Then, the holomorphy ring theorem (Proposition 4.3.2) says that 𝐷 is a Dedekind domain. Note that 𝑎, 𝑏 ∈ 𝑅 ∗ . Hence, 𝑣¤𝔭 (𝑥), 𝑣¤𝔭 (𝑦) ≥ 0 for each 𝔭 ∈ 𝑆 ∗ . Therefore, by (17.22a)–(17.22b), 𝐴:= = 𝐵:= =

{𝑧 {𝑧 {𝑧 {𝑧

∈ ∈ ∈ ∈

𝐷| 𝐷| 𝐷| 𝐷|

𝑣¤𝔭 (𝑧) ≥ 𝑣¤𝔭 (𝑥) for all 𝔭 ∈ 𝑆 ∗ } 𝑣 𝑃 (𝑧) ≥ max(𝑣 𝑃 (𝑡), 0) for all 𝑃 ∈ P}; and 𝑣¤𝔭 (𝑧) ≥ 𝑣¤𝔭 (𝑦) for all 𝔭 ∈ 𝑆 ∗ } 𝑣 𝑃 (𝑧) ≥ max(−𝑣 𝑃 (𝑡), 0) for all 𝑃 ∈ P}.

Every maximal ideal of 𝐷 is the center of a prime 𝑃 ∈ P (Proposition 4.3.2(d)). Thus, 𝐴 is the ideal of zeros of 𝑡 and 𝐵 is the ideal of poles of 𝑡. That is, 𝐴 = 𝑃1𝑘1 · · · 𝑃𝑟𝑘𝑟 and 𝐵 = 𝑄 𝑙11 · · · 𝑄 𝑙𝑠𝑠 , where 𝑃1 , . . . , 𝑃𝑟 , 𝑄 1 , . . . , 𝑄 𝑠 are distinct maximal ideals of 𝐷 and 𝑘 1 , . . . , 𝑘 𝑟 , 𝑙1 , . . . , 𝑙 𝑠 are positive integers with 𝑣 𝑃𝑖 (𝑡) = 𝑘 𝑖 , 𝑖 = 1, . . . , 𝑟, and 𝑣 𝑄 𝑗 (𝑡) = −𝑙 𝑗 , 𝑗 = 1, . . . , 𝑠. In particular, 𝐴 and 𝐵 are relatively prime ideals of 𝐷. Hence, 𝐴 + 𝐵 = 𝐷. Thus, there exist 𝜆 ∈ 𝐴 and 𝜇 ∈ 𝐵 such that 𝜆 + 𝜇 = 1. ¤ a ¤ Thus, 𝜆 = 𝛼𝑥 and 𝜇 = 𝛽𝑦 with 𝛼, 𝛽 ∈ 𝑅, By definition, 𝐴 ⊆ 𝑥 𝑅¤ and 𝐵 ⊆ 𝑦 𝑅. □ contradiction to 17.21.

17.5 Examples Let 𝑅 be an integral domain with quotient field 𝐾. Suppose that 𝑆 is a nonempty set of primes of 𝐾 which satisfies Conditions (17.12b)–(17.12d). Suppose in addition that 𝑣𝔭 is a discrete valuation, 𝔭 ∈ 𝑆. Then, 𝑅 is a Krull domain, hence a generalized Krull domain. Every Dedekind domain 𝑅 is a Krull domain with 𝑆 being the set of primes of 𝐾 associated with the maximal ideals of 𝑅. Since each nonzero prime ideal of 𝑅 is maximal, dim(𝑅) = 1. Thus, 𝑅 does not satisfy Condition (17.12e). Example 17.5.1 (Polynomial rings over fields) Every unique factorization domain 𝑅 is a Krull domain. Here 𝑆 corresponds to the set of nonzero prime ideals of 𝑅. For example, the polynomial ring 𝑅 = 𝐾0 [𝑋1 , . . . , 𝑋𝑛 ] over an arbitrary field 𝐾0 is a unique factorization domain [ZaS58, p. 38, Thm. 13]. When 𝑛 ≥ 2, 𝑅/(𝑅𝑋1 + 𝑅𝑋2 )  𝐾0 [𝑋3 , . . . , 𝑋𝑛 ] is an integral domain. Hence, 𝑅𝑋1 + 𝑅𝑋2 is a prime ideal of 𝑅 which properly contains each of the prime ideals 𝑅𝑋1 and 𝑅𝑋2 . Thus,

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dim(𝑅) ≥ 2. (Indeed, dim(𝑅) = 𝑛 [Mat94, p. 117, Thm. 15.4].) By Theorem 17.4.6, 𝐾0 (𝑋1 , . . . , 𝑋𝑛 ) is Hilbertian. This is a weaker result than Theorem 14.4.2 which says that 𝐾0 (𝑋1 , . . . , 𝑋𝑛 ) is Hilbertian for each 𝑛 ≥ 1. Example 17.5.2 ([Wei82], p. 203, Formal power series over a field) Let 𝐾 be a field and 𝑛 ≥ 2. The ring of formal power series 𝑅 = 𝐾 [[𝑋1 , . . . , 𝑋𝑛 ]] is a local integral domain with the maximal ideal 𝑀=

∞  ∑︁

𝑓𝑖 | 𝑓𝑖 ∈ 𝐾 [𝑋1 , . . . , 𝑋𝑛 ] is a form of degree 𝑖 .

𝑖=1

Denote its quotient field by 𝐾 ((𝑋1 , . . . , 𝑋𝑛 )): the field of formal power series over 𝐾 in 𝑋1 , . . . , 𝑋𝑛 . The Weierstrass preparation theorem implies that 𝑅 is a unique factorization domain [ZaS60, p. 148]. For each prime element 𝑝 of 𝑅 one of the elements 𝑋1 or 𝑋2 does not belong to 𝑅 𝑝. Thus, 𝑅 𝑝 is properly contained in 𝑀, so dim(𝑅) ≥ 2. (Again, dim(𝑅) = 𝑛. See [ZaS60, p. 218] or [Mat94, p. 117, Thm. 15.4].) By Theorem 17.4.6, 𝐾 ((𝑋1 , . . . , 𝑋𝑛 )) is Hilbertian. This settles a problem of [Lan62, p. 142, end of third paragraph]. Example 17.5.3 (Formal power series over a ring) Let 𝐴 be a Krull domain and 𝑛 ≥ 1. Then, the ring of polynomials 𝐴[𝑋1 , . . . , 𝑋𝑛 ] and the ring of formal power series 𝐴[[𝑋1 , . . . , 𝑋𝑛 ]] in the variables 𝑋1 , . . . , 𝑋𝑛 over 𝐴 are Krull domains (see [Mat94, p. 89] for the case 𝑛 = 1; the general case follows by induction). Suppose in addition that 𝐴 is not a field. Let 𝔭0 be a nonzero prime ideal of 𝐴. Then, 𝑃=

∞  ∑︁

𝑓𝑖 ∈ 𝐴[𝑋1 , . . . , 𝑋𝑛 ] is a form of degree 𝑖 and

𝑖=1

𝑃0 =

∞  ∑︁

𝑓𝑖 ∈ 𝐴[𝑋1 , . . . , 𝑋𝑛 ] is a form of degree 𝑖 and 𝑓0 ∈ 𝔭0



𝑖=0

are nonzero prime ideals of 𝑅 = 𝐴[[𝑋1 , . . . , 𝑋𝑛 ]] and 𝑃0 ⊂ 𝑃. Indeed, 𝑅/𝑃  𝐴 and 𝑅/𝑃0  𝐴/𝔭0 are integral domains. Thus, dim( 𝐴[[𝑋1 , . . . , 𝑋𝑛 ]]) ≥ 2 (Actually, dim( 𝐴) = dim( 𝐴) + 𝑛 [Mat94, p. 117]). Consequently, by Theorem 17.4.6, the quotient field of 𝐴[[𝑋1 , . . . , 𝑋𝑛 ]] is a Hilbertian field. For example, the quotient fields of Z[[𝑋1 , . . . , 𝑋𝑛 ]] and 𝑂 [[𝑋1 , . . . , 𝑋𝑛 ]], where 𝑂 is a discrete valuation ring, are Hilbertian. Lemma 17.5.4 (Geyer) No Henselian field is Hilbertian. Proof. Let 𝐾 be a Henselian field with valuation ring 𝑅 and maximal ideal 𝔪. Choose 𝑚 ∈ 𝔪, 𝑚 ≠ 0, and a prime number 𝑝 ≠ char(𝐾). Consider the irreducible polynomials 𝑓 (𝑇, 𝑋) = 𝑋 𝑝 + 𝑚𝑇 − 1 and 𝑔(𝑇, 𝑋) = 𝑋 𝑝 + 𝑇 −1 − 1 of 𝐾 (𝑇) [𝑋]. Assume that 𝐾 is Hilbertian. Then, there exists an 𝑎 ∈ 𝐾 × with both 𝑓 (𝑎, 𝑋) and 𝑔(𝑎, 𝑋) irreducible in 𝐾 [𝑋]. In particular, none of them has a zero in 𝐾.

17.5 Examples

303

But either 𝑎 ∈ 𝑅 or 𝑎 −1 ∈ 𝔪. Suppose first that 𝑎 ∈ 𝑅. Then, 𝑓 (𝑎, 1) ≡ 0 mod 𝔪 𝜕𝑓 and 𝜕𝑋 (𝑎, 1) ̸≡ 0 mod 𝔪. Since 𝐾 is Henselian, 𝑓 (𝑎, 𝑋) has a zero in 𝐾. Similarly, if 𝑎 −1 ∈ 𝔪, then 𝑔(𝑎, 𝑋) has a zero in 𝐾. This contradiction to the preceding paragraph proves that 𝐾 is not Hilbertian. □ Example 17.5.5 Q 𝑝 and the formal power series 𝐾0 ((𝑋)) are complete discrete valuation fields. Hence, they are Henselian (Proposition 4.5.2). By Lemma 17.5.4, they are not Hilbertian. Thus, the assumption “𝑛 ≥ 2” in Example 17.5.2 is necessary.

Example 17.5.6 (Geyer) In the notation of Example 17.5.2, the ring 𝑅 = 𝐾0 [[𝑋1 , . . . , 𝑋𝑛 ]] of formal power series is not Hilbertian. Indeed, let 𝐾 = Quot(𝑅) and check Í for char(𝐾0𝑗) ≠ 2 (resp. char(𝐾0 ) ≠ 3) that every power series 𝑓 (𝑋1 , 𝑋2 ) = 1 + 𝑖+ 𝑗 ≥1 𝑎 𝑖 𝑗 𝑋1𝑖 𝑋2 is a square (resp. a cube) in 𝑅. Thus, the polynomial 𝑍 2 − (1 + 𝑋1𝑇) (resp. 𝑍 3 − (1 + 𝑋1𝑇)) is irreducible in 𝐾 [𝑇, 𝑍] but 𝑍 2 − (1 + 𝑋1 𝑡) (resp. 𝑍 3 − (1 + 𝑋1 𝑡)) is reducible for each 𝑡 ∈ 𝑅. Lemma 17.5.7 Let 𝑅0 be a unique factorization domain with quotient field 𝐾0 . Consider a set 𝑃 of unequivalent prime elements of 𝑅0 . For each 𝑝 ∈ 𝑃 let 𝑣 𝑝 be the corresponding discrete valuation of 𝐾0 . Let 𝐾 be an algebraic extension of 𝐾0 . Suppose that each 𝑣 𝑝 with 𝑝 ∈ 𝑃 is unramified in 𝐾. For each 𝑝 ∈ 𝑃 choose an extension 𝑤 𝑝 of 𝑣 𝑝 to 𝐾. Then, the holomorphy ring 𝑅 = {𝑥 ∈ 𝐾 | 𝑤 𝑝 (𝑥) ≥ 0 for each 𝑝 ∈ 𝑃} is a unique factorization domain with quotient field 𝐾. Proof. The assumptions imply that 𝑤 𝑝 ( 𝑝) = 1 and 𝑤 𝑝′ ( 𝑝) = 0 for all distinct 𝑝, 𝑝 ′ ∈ 𝑃. Consider 𝑥 ∈ 𝑅. Then, 𝑤 𝑝 (𝑥) ≥ 0 for each 𝑝 ∈ 𝑃. Moreover, there are only finitely many 𝑝 ∈ 𝑃 with 𝑤 𝑝 (𝑥)Î> 0. Indeed, if 𝑓 = irr(𝑥, 𝐾0 ) and 𝑣 𝑝 ( 𝑓 (0)) = 0, then 𝑤 𝑝 (𝑥) = 0. Thus, 𝑢 = 𝑥 𝑝 ∈𝑃 𝑝 −𝑤 𝑝 ( 𝑥) is an element of 𝐾 and Î 𝑤 𝑝 (𝑢) = 0 for each 𝑝 ∈ 𝑃. Hence, 𝑢 is a unit of 𝑅 and 𝑥 = 𝑢 𝑝 ∈𝑃 𝑝 𝑤 𝑝 ( 𝑥) is the desired decomposition of 𝑥. Finally, observe that 𝑅 contains the integral closure of 𝑅0 in 𝐾. Therefore, 𝐾 = Quot(𝑅). □ The following example generalizes [CoZ98, Theorem 1 (i) and (ii)]: Example 17.5.8 (Unique factorization domain with a non-Hilbertian quotient field) Let 𝑅0 be either Z or 𝐹 [𝑡] for some finite field 𝐹. It is a unique factorization domain. Put 𝐾0 = Quot(𝑅0 ). The proof of Theorem 14.6.2 (with 𝐾0 replacing 𝐾) gives prime numbers 𝑙 and 𝑞, an element 𝑢 ∈ 𝐾0× , an infinite set 𝑃 of nonequivalent prime elements of 𝑅0 , and a field extension 𝐾 of 𝐾0 (in the notation of Theorem 14.6.2 equals to 𝑀 (𝐾0 )) with these properties: (17.23a) For every 𝑥 ∈ 𝐾 there is a 𝑦 ∈ 𝐾 with 𝑦 𝑙 − 𝑥 𝑙𝑞 + 𝑢 = 0. (17.23b) For each 𝑝 ∈ 𝑃, the discrete valuation 𝑣 𝑝 is unramified in 𝐾. Since 𝑌 𝑙 − 𝑋 𝑙𝑞 + 𝑢 is absolutely irreducible, Condition (17.23a) implies that 𝐾 is not Hilbertian. Let 𝑤 𝑝 , 𝑝 ∈ 𝑃, and 𝑅 be as in Lemma 17.5.7. Then, 𝐾 = Quot(𝑅), {𝑤 𝑝 | 𝑝 ∈ 𝑃} is an infinite set of discrete valuations, and 𝑅 is a unique factorization domain.

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17 The Nonstandard Approach to Hilbert’s Irreducibility Theorem

This answers negatively the questions posed in Problem 14.20 of [FrJ86] and Problem 14.21 of [FrJ86] (which is a problem posed in [Lan62, p. 142, end of the second paragraph]). More generally, every Noetherian integrally closed domain 𝑅 is a Krull domain [Bou89a, p. 482, Cor.], so 𝑅 is also a generalized Krull domain. Thus, by Theorem 17.4.6, Quot(𝑅) is a Hilbertian field if in addition dim(𝑅) ≥ 2. The following example asks for a generalization of Example 17.5.3: Example 17.5.9 Problem 15.5.9 of [FrJ08] asks the following question: Let 𝐴 be a generalized Krull domain which is not a field. (17.24a) Is 𝐴[[𝑋]] a generalized Krull domain? (17.24b) Is Quot( 𝐴[[𝑋]]) Hilbertian? It is classically known that if 𝐴 is a Krull domain, then so is 𝐴[[𝑋]] [Mat94, Thm. 12.4(iii)]. However, if 𝐴 is a generalized Krull domain which is not a Krull domain (e.g. 𝐴 is a rank-1 non discrete valuation ring), then surprisingly 𝐴[[𝑋]] is not a generalized Krull domain [PaT10, Thm. 1.1]. This gives a precise negative answer to Question (17.24a). Nevertheless, [Par12, Thm. 2.4] asserts that in each case Quot( 𝐴[[𝑋]]) is Hilbertian, answering Question (17.24b) affirmatively. Moreover, [FeP13, Prop. 4.1] strengthens this result by proving that whenever 𝐴 is an integral domain which is not a field and 𝐴[[𝑋]] is integrally closed, then Quot( 𝐴[[𝑋]]) is Hilbertian. In particular, this is the case whenever 𝐴 is a generalized Krull domain [FeP13, Remark 4.2].

Exercises 1. Let 𝐾 be a countable Hilbertian field. List the irreducible polynomials in 𝐾 [𝑇, 𝑋] as 𝑓1 (𝑇, 𝑋), 𝑓2 (𝑇, 𝑋), . . .. For each 𝑖 choose 𝑡𝑖 ∈ 𝐾 for which 𝑓1 (𝑡 𝑖 , 𝑋), . . . , 𝑓𝑖 (𝑡𝑖 , 𝑋) are irreducible in 𝐾 [𝑋]. (a) Observe that the infinite set 𝐻 = {𝑡 1 , 𝑡2 , . . .} has the universal Hilbert subset property: 𝐻 ∖ 𝐻𝐾 (𝑔1 , . . . , 𝑔𝑚 ) is finite for every collection {𝑔1 , . . . , 𝑔𝑚 } of irreducible polynomials in 𝐾 [𝑇, 𝑋]. (b) Let 𝐾 ∗ be an enlargement of 𝐾. Prove that if 𝑡 ∈ 𝐻 ∗ ∖ 𝐻, then 𝐾 (𝑡) is algebraically closed in 𝐾 ∗ . 2. Let 𝐴 = {1, 22! , 33! , . . .} andÐconsider an enlargement Q∗ of Q. Prove for each 1/𝑛 ) ⊆ Q∗ . nonstandard element 𝑡 ∈ 𝐴∗ that ∞ 𝑛=1 Q(𝑡 3. Let 𝑓1 , 𝑓2 , 𝑓3 , . . . be a sequence in Q(𝑌 ). Put 𝑔𝑛 (𝑌 ) = 𝑓1 (. . . ( 𝑓𝑛−1 ( 𝑓𝑛 (𝑌 )))). Suppose that if 𝑦 is transcendental over Q and 𝑥 = 𝑔𝑛 (𝑦), then 𝑥 has at least 𝑛 distinct poles in Q(𝑦). Let 𝐴 = {𝑔1 (1), 𝑔2 (2), 𝑔3 (3), . . .}. Consider a nonstandard element 𝑡 of 𝐴∗ . Prove that the equation 𝑡 = 𝑔𝑛 (𝑌 ) is solvable in Q∗ for each 𝑛. Conclude that 𝑡 is not polefinite (Section 17.1).

17.5 Examples

305

4. [Wei82, Satz 2.3] Generalize the Gilmore–Robinson criterion as follows. Let 𝐾 ∗ be an enlargement of a field 𝐾. Then, 𝐾 is Hilbertian if and only if there is a Hilbertian field 𝐹 with 𝐾 ⊆ 𝐹 ⊆ 𝐾 ∗ and 𝐹 is separably closed in 𝐾 ∗ . 5. Consider the set 𝐴 = {2, 3!, 4!, 5!, . . .} and let 𝑡 be a nonstandard element of 𝐴∗ . Show that the metric absolute value of Q induces via Q∗ (Section 17.2) the infinite prime 𝑡 → ∞ of Q(𝑡), while every ultrametric absolute value of Q induces the prime 𝑡 → 0 of Q(𝑡). Conclude from Proposition 17.3.2 that all other primes of Q(𝑡) are induced by nonstandard arithmetical primes of Q∗ . 6. Consider the field 𝐾 = 𝐾0 ((𝑋1 , . . . , 𝑋𝑛 )) of formal power series in 𝑛 variables over a basic field 𝐾0 . Construct a 𝐾0 -place 𝜑: 𝐾 → 𝐾0 ∪ {∞} with 𝜑(𝑋𝑖 ) = 0, 𝑖 = 1, . . . , 𝑛. Conclude that 𝐾/𝐾0 is a regular extension. Hint: The case 𝑛 = 1 is easy. For arbitrary 𝑛 embed 𝐾 in the field of iterated power series 𝐿 = 𝐾0 ((𝑋1 )) ((𝑋2 )) · · · ((𝑋𝑛 )).

Notes The nonstandard characterization of Hilbertian fields appears in [GiR55]. [Roq75b] has exploited the Gilmore–Robinson criterion through a nonstandard interpretation of the Siegel–Mahler theorem (compare with the remarks at the beginning of Section 14.3). Most of this Chapter is from Weissauer’s thesis [Wei82]. The proof of Proposition 17.3.2 for number fields that appears in [Rob73] is based on a nonstandard interpretation of the “distributions” that appear in Weil’s thesis [Wei28]. The influence of [RoR75] in Section 17.2 and parts of Section 17.3 should be obvious. The polefinite property of Section 17.1 and its relation to Hilbert’s irreducibility theorem appears in a standard form based on 𝑝-adic analysis in the case 𝐾 = Q in [Spr80]. Standard simplified proofs of both Sprindz¸uk’s result and Weissauer’s approach to Hilbertianity of fields with a product formula (Section 17.3), featuring their common concepts, appear in [Fri85], which also gives the concept of a universal Hilbert subset (e.g. in Exercise 1) whose existence is a consequence of the Gilmore– Robinson observation (Proposition 17.1.1). As far as we know [Spr81] is the first to give an explicit universal Hilbert subset (over Q): √︁ 2 𝐻 = {[exp( log(log(𝑚))] + 𝑚!2𝑚 }∞ 𝑚=1 . It has the property that 𝐻 ∖ 𝐻Q (𝑔1 , . . . , 𝑔𝑡 ) is finite for every collection {𝑔1 , . . . , 𝑔𝑡 } of irreducible polynomials in Q[𝑇, 𝑋] [Fri85, Thm. 4.9]. Our proof of Theorem 17.3.3 for the integral domain 𝑅 follows the proof of [Wei82, Satz 6.2] for the field 𝐾 := Quot(𝑅) with the product formula. Similarly, Theorem 17.4.6 is [Wei82, Satz 7.2]. Finally, [Kle82] gives “standard” proofs of Weissauer’s results, Theorems 17.3.3 (again for 𝐾 rather than for 𝑅) and 17.4.6.

Chapter 18

Galois Groups over Hilbertian Fields

Given a field 𝐾, one may ask which finite groups occur as Galois groups over 𝐾. If 𝐾 is Hilbertian, then every finite group that occurs over 𝐾 (𝑡), with 𝑡 transcendental over 𝐾, also occurs over 𝐾. Moreover, suppose that 𝐹/𝐾 (𝑡) is Galois with a finite Galois group 𝐺 and 𝐹/𝐾 is regular. Then, 𝐾 has a linearly disjoint sequence of Galois extensions, 𝐿 1 , 𝐿 2 , 𝐿 3 , . . ., with Gal(𝐿 𝑖 /𝐾)  𝐺, 𝑖 = 1, 2, 3, . . . (Lemma 18.2.6). We prove that this set up occurs for symmetric groups (Corollary 18.2.7), Abelian groups (Proposition 18.3.5), and when char(𝐾) = 0 for alternating groups (Proposition 18.7.6). If 𝐾 is PAC (but not necessarily Hilbertian), every finite group is regular over 𝐾 (Proposition 18.10.2). If 𝐾 is Hilbertian, then Z 𝑝 occurs over 𝐾 (Corollary 18.6.7) but is not necessarily regular over 𝐾. For example, Z 𝑝 is not regular over Q (Corollary 18.6.11). Realization of a finite nonsimple group over 𝐾 is usually done in steps. First one realizes a quotient of the group and then embeds the solution field in a larger Galois extension with the given Galois group. The latter step is always possible when 𝐾 is Hilbertian and the kernel is a product of non-Abelian simple groups each of which has a GAR realization over 𝐾 (Sections 18.8 and 18.9). For example, 𝐴𝑛 with 𝑛 = 5 or 𝑛 ≥ 7 is GAR over 𝐾 when char(𝐾) ∤ (𝑛 − 1)𝑛 (Corollary 18.9.2). Chapter 21 will exploit the result about the regularity of 𝑆 𝑛 .

18.1 Galois Groups of Polynomials We prove two theorems about preservation of Galois groups of polynomials under specializations of parameters. One of them (Proposition 18.1.5) is a polynomial analog of Lemma 14.1.1(b). It assumes the ground field to be Hilbertian. The other one (Proposition 18.1.4) is an application of Bertini–Noether. Here the ground field is arbitrary but the polynomial in question is absolutely irreducible and Galois. We start with an analog of Lemma 14.1.1(a) for Galois groups of polynomials:

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 307 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_18

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Let 𝑓 ∈ 𝐾 [𝑋] be a separable polynomial of degree 𝑛. By definition, 𝑓 has 𝑛 distinct roots 𝑥1 , . . . , 𝑥 𝑛 in 𝐾sep . Thus, 𝐿 = 𝐾 (𝑥1 , . . . , 𝑥 𝑛 ) is a finite Galois extension. Restriction of elements of Gal(𝐿/𝐾) to {𝑥1 , . . . , 𝑥 𝑛 } gives an embedding of Gal(𝐿/𝐾) into the group of all permutations of {𝑥1 , . . . , 𝑥 𝑛 }, called a permutation representation. The image of Gal(𝐿/𝐾) under this representation is Gal( 𝑓 , 𝐾). Next consider another field 𝐾¯ and a separable polynomial 𝑓¯ ∈ 𝐾¯ [𝑋] of degree ¯ to indicate that the two groups in question are 𝑛. We write Gal( 𝑓 , 𝐾)  Gal( 𝑓¯, 𝐾) ¯ of isomorphic as permutation groups. Thus, there exists an isomorphism 𝜎 ↦→ 𝜎 ¯ as abstract groups and there is a listing 𝑥¯1 , . . . , 𝑥¯ 𝑛 of the Gal( 𝑓 , 𝐾) and Gal( 𝑓¯, 𝐾) roots of 𝑓¯ with 𝜎𝑥𝑖 = 𝜎 ¯ 𝑥¯𝑖 , 𝜎 ∈ Gal( 𝑓 , 𝐾), 𝑖 = 1, . . . , 𝑛. Similarly, we speak about ¯ into Gal( 𝑓 , 𝐾) as permutation groups. Finally, we write an embedding of Gal( 𝑓¯, 𝐾) Gal( 𝑓 , 𝐾)  𝐺 for a finite group 𝐺 when Gal( 𝑓 , 𝐾) and 𝐺 are isomorphic as abstract groups. Lemma 18.1.1 Let 𝐿/𝐾 be a finite Galois extension, 𝑓 ∈ 𝐾 [𝑋] a separable polynomial, and 𝜑 a place of 𝐿. Denote the residue field of 𝐾 (resp. 𝐿) under 𝜑 by ¯ Suppose that 𝐿 is the splitting field of 𝑓 over 𝐾 and 𝑓¯ = 𝜑( 𝑓 ) is a 𝐾¯ (resp. 𝐿). separable polynomial in 𝐾¯ [𝑋] with deg( 𝑓¯) = deg( 𝑓 ). ¯ 𝐾) ¯  Gal( 𝑓¯, 𝐾) ¯ → Gal( 𝑓 , 𝐾). (a) Then, there is an embedding 𝜑∗ : Aut( 𝐿/ (b) Suppose in addition that 𝑓 is Galois and 𝑓¯ is irreducible. Then, 𝑓¯ is Galois, 𝜑∗ ¯ is an isomorphism, and 𝐿¯ is the splitting field of 𝑓¯ over 𝐾. ¯ 𝐾¯ is separable. Then, 𝐿¯ is the splitting field of 𝑓¯ (c) Alternatively, suppose that 𝐿/ ¯ over 𝐾. Î𝑛 Proof of (a). Let 𝑎 be the leading coefficient of 𝑓 . Then, 𝑓 (𝑋) = 𝑎 𝑖=1 (𝑋 − 𝑥𝑖 ) with distinct 𝑥 1 , . . . , 𝑥 𝑛 ∈ 𝐾sep and 𝐿 = 𝐾 (𝑥 1 , . . . , 𝑥 𝑛 ) is the splitting field of 𝑓 over 𝐾. Since deg( 𝑓¯) = deg( 𝑓 ), we Î have 𝑎¯ = 𝜑(𝑎) ∈ 𝐾¯ × . Hence, 𝑥¯𝑖 := 𝜑(𝑥 𝑖 ), 𝑛 × 𝑖 = 1, . . . , 𝑛, are in 𝐾¯ sep and 𝑓¯(𝑋) = 𝑎¯ 𝑖=1 (𝑋 − 𝑥¯𝑖 ). By assumption, 𝑥¯1 , . . . , 𝑥¯ 𝑛 are distinct. Therefore, the map 𝑥 𝑖 ↦→ 𝑥¯𝑖 , 𝑖 = 1, . . . , 𝑛, is bijective.

(18.1)

Lemma 7.1.1 gives an epimorphism 𝜎 ↦→ 𝜎 ¯ of the decomposition group 𝐷 𝜑 onto ¯ Suppose that ¯ 𝐾). ¯ It satisfies 𝜎 Aut( 𝐿/ ¯ 𝑦¯ = 𝜎𝑦 for each 𝑦 ∈ 𝐿 with 𝑦¯ = 𝜑(𝑦) ∈ 𝐿. 𝜎| ¯ 𝐾¯ ( 𝑥¯1 ,..., 𝑥¯ 𝑛 ) = 1. Then, 𝜎𝑥 𝑖 = 𝜎 ¯ 𝑥¯𝑖 = 𝑥¯𝑖 . Since 𝜎 permutes 𝑥1 , . . . , 𝑥 𝑛 , we have, by (18.1), that 𝜎𝑥 𝑖 = 𝑥 𝑖 , 𝑖 = 1, . . . , 𝑛. Hence, 𝜎 = 1. ¯ 𝐾¯ ( 𝑥¯1 , . . . , 𝑥¯ 𝑛 )). Then, there is a 𝜎 ∈ 𝐷 𝜑 with 𝜎 Moreover, let 𝜏 ∈ Aut( 𝐿/ ¯ = 𝜏. By ¯ 𝐾) ¯ → the preceding paragraph, 𝜎 = 1, hence 𝜏 = 𝜎 ¯ = 1. It follows that res: Aut( 𝐿/ ¯ ¯ Gal( 𝐾 ( 𝑥¯1 , . . . , 𝑥¯ 𝑛 )/𝐾) is an isomorphism. Finally, consider each 𝜎 ∈ Gal(𝐿/𝐾) as a permutation of {𝑥1 , . . . , 𝑥 𝑛 }. Likewise, consider 𝜎 ¯ as a permutation of {𝑥¯1 , . . . , 𝑥¯ 𝑛 }. Then, 𝜎 ¯ ↦→ 𝜎 is an embedding ¯ → Gal( 𝑓 , 𝐾). 𝜑∗ : Gal( 𝑓¯, 𝐾) Proof of (b). Since 𝑓 is Galois, in particular irreducible, and since 𝐿 is the splitting field of 𝑓 over 𝐾, we have deg( 𝑓 ) = |Gal( 𝑓 , 𝐾)| = |Gal(𝐿/𝐾)|. Since 𝑓¯ is irre¯ Since, in addition, 𝑓¯ is separable, 𝐾¯ ( 𝑥¯1 , . . . , 𝑥¯ 𝑛 ) is ducible, deg( 𝑓¯) ≤ |Gal( 𝑓¯, 𝐾)|. ¯ ¯ so Gal( 𝑓¯, 𝐾) ¯  Gal( 𝐾¯ ( 𝑥¯1 , . . . , 𝑥¯ 𝑛 )/𝐾). ¯ Therefore, the splitting field of 𝑓 over 𝐾, ¯ = [ 𝐾¯ ( 𝑥¯1 , . . . , 𝑥¯ 𝑛 ) : 𝐾] ¯ deg( 𝑓 ) = deg( 𝑓¯) ≤ |Gal( 𝑓¯, 𝐾)| ¯ ¯ ≤ [ 𝐿 : 𝐾] ≤ [𝐿 : 𝐾] = |Gal( 𝑓 , 𝐾)| = deg( 𝑓 ).

18.1 Galois Groups of Polynomials

309

¯ = |Gal( 𝑓 , 𝐾)| and deg( 𝑓¯) = [ 𝐾¯ ( 𝑥¯1 , . . . , 𝑥¯ 𝑛 ) : 𝐾]. ¯ Thus, 𝑓¯ is Hence, |Gal( 𝑓¯, 𝐾)| ¯ Galois and 𝜑∗ is an isomorphism. In addition, 𝐿¯ = 𝐾¯ ( 𝑥¯1 , . . . , 𝑥¯ 𝑛 ) is Galois over 𝐾. ¯ 𝐾¯ is normal and separable, it is Galois. By the proof of (a), Proof of (c). Since 𝐿/ ¯ 𝐾¯ ( 𝑥¯1 , . . . , 𝑥¯ 𝑛 )) = 1. Therefore, 𝐿¯ = 𝐾¯ ( 𝑥¯1 , . . . , 𝑥¯ 𝑛 ). Gal( 𝐿/ □ An essential assumption in Lemma 18.1.1 is the irreducibility of 𝑓¯. Proposition 18.1.4 below uses Bertini–Noether to satisfy this assumption. Proposition 18.1.5 applies Hilbert irreducibility theorem to achieve the same goal. Lemma 18.1.2 Let 𝑉 be a variety defined in A𝑛 over a field 𝐾 and x a generic point of 𝑉 over 𝐾. Then, 𝑉 has a nonempty Zariski 𝐾-open subset 𝑉0 with the following ˜ there is a 𝐾-place 𝜑 of 𝐾 (x) with 𝜑(x) = a and with property: For each a ∈ 𝑉0 ( 𝐾) residue field 𝐾 (a). Proof. By Corollary 11.2.2(a), 𝐾 (x)/𝐾 is a regular extension. Assume without loss that 𝑥 1 , . . . , 𝑥𝑟 form a separating transcendence basis for 𝐾 (x)/𝐾. Example 3.4.10 produces a 𝐾-place 𝜑0 of 𝐾 (𝑥 1 , . . . , 𝑥𝑟 ) with 𝜑(𝑥 𝑖 ) = 𝑎 𝑖 , 𝑖 = 1, . . . , 𝑟, and with residue field 𝐾 (𝑎 1 , . . . , 𝑎𝑟 ). Use Remark 7.1.5 to find a nonzero polynomial 𝑔 ∈ 𝐾 [𝑋1 , . . . , 𝑋𝑟 ] such that 𝐾 [𝑥1 , . . . , 𝑥 𝑖+1 , 𝑔(𝑥1 , . . . , 𝑥𝑟 ) −1 ]/𝐾 [𝑥 1 , . . . , 𝑥 𝑖 , 𝑔(𝑥1 , . . . , 𝑥𝑟 ) −1 ] is a ring cover, 𝑖 = 𝑟, . . . , 𝑛 − 1. Let 𝑉0 = {a ∈ 𝑉 | 𝑔(𝑎 1 , . . . , 𝑎𝑟 ) ≠ 0}. ˜ Let 𝜑 be a place of 𝐾 (x) which extends 𝜑0 by 𝜑(x) = a. Suppose that a ∈ 𝑉0 ( 𝐾). □ By Remark 7.1.7, the residue field of 𝐾 (x) at 𝜑 is 𝐾 (a). Remark 18.1.3 (Simple points) The conclusion of Lemma 18.1.2 is actually true for each simple point a of 𝑉 [JaR80, Cor. A2]. Proposition 18.1.4 Let 𝑉 be a variety defined over a field 𝐾0 in A𝑚 , u a generic point of 𝑉 over 𝐾0 , 𝐾 = 𝐾0 (u), and ℎ1 , . . . , ℎ 𝑘 ∈ 𝐾0 [𝑈1 , . . . , 𝑈𝑚 , 𝑇1 , . . . , 𝑇𝑟 , 𝑋] polynomials. Suppose that ℎ 𝑗 (u, T, 𝑋) are absolutely irreducible as polynomials in (T, 𝑋) over 𝐾 and Galois as polynomial in 𝑋 over 𝐾 (T). Then, 𝑉 has a nonempty Zariski 𝐾0 -open subset 𝑉0 with the following property: For each a ∈ 𝑉0 (𝐾0 ) and for 𝑗 = 1, . . . , 𝑘 the polynomial ℎ 𝑗 (a, T, 𝑋) is absolutely irreducible, Galois over 𝐾0 (T), and Gal(ℎ 𝑗 (a, T, 𝑋), 𝐾0 (T))  Gal(ℎ 𝑗 (u, T, 𝑋), 𝐾 (T)).

(18.2)

Proof. Bertini–Noether (Proposition 10.4.3) gives a nonempty Zariski 𝐾0 -open subset 𝑉1 of 𝑉 such that ℎ 𝑗 (a, T, 𝑋) are absolutely irreducible for each a ∈ 𝑉1 ( 𝐾˜ 0 ), 𝑗 = 1, . . . , 𝑘. Lemma 18.1.2, applied to 𝑉 over 𝐾0 (T), gives a nonempty Zariski  𝐾0 (T)-open subset 𝑉2′ of 𝑉 satisfying this: For each a ∈ 𝑉2′ ( 𝐾 0 (T)) there is a 𝐾0 (T)-place 𝜑 of 𝐾0 (u, T) with residue field 𝐾0 (a, 𝑇). Choose a nonempty Zariski 𝐾0 -open subset 𝑉2 of 𝑉 such that 𝑉2 ( 𝐾˜ 0 ) ⊆ 𝑉2′ ( 𝐾˜ 0 ). Finally, there is a nonempty Zariski 𝐾0 -open subset 𝑉3 of 𝑉 with ℎ 𝑗 (a, T, 𝑋) separable and deg𝑋 (ℎ 𝑗 (a, T, 𝑋)) = deg𝑋 (ℎ 𝑗 (u, T, 𝑋)) for each a ∈ 𝑉3 ( 𝐾˜ 0 ), 𝑗 = 1, . . . , 𝑘.

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𝑉0 = 𝑉1 ∩𝑉2 ∩𝑉3 is a nonempty Zariski 𝐾0 -open subset of 𝑉. For each a ∈ 𝑉0 (𝐾0 ) let 𝜑 be a 𝐾0 -place of 𝐾 (T) with residue field 𝐾0 (T). By Lemma 18.1.1, ℎ 𝑗 (a, T, 𝑋) is Galois over 𝐾0 (T) and (18.2) holds. □ Proposition 18.1.5 Let 𝐾 be a Hilbertian field and ℎ 𝑗 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑟 , 𝑋] a separable polynomial in 𝑋, 𝑗 = 1, . . . , 𝑘. Then, 𝐾 𝑟 has a separable Hilbert subset 𝐻 with Gal(ℎ 𝑗 (a, 𝑋), 𝐾)  Gal(ℎ 𝑗 (T, 𝑋), 𝐾 (T)) for each a ∈ 𝐻, 𝑗 = 1, . . . , 𝑘. Proof. Denote the splitting field of ℎ 𝑗 (T, 𝑋) over 𝐾 (T) by 𝐹 𝑗 . Lemma 14.1.1(a) gives a nonempty Zariski 𝐾-open subset 𝑈1 of A𝑟 satisfying the following assertion: For each a ∈ 𝑈1 (𝐾) there is a 𝐾-place 𝜑 𝑗 of 𝐹 𝑗 extending T ↦→ a such that the residue field 𝐹¯ 𝑗 of 𝐹 𝑗 under 𝜑 𝑗 is Galois over 𝐾, 𝑗 = 1, . . . , 𝑘. Lemma 14.1.1(b) gives a separable Hilbert subset 𝐻 ′ of 𝐾 𝑟 such that for each a ∈ 𝐻 ′ ∩𝑈1 (𝐾) and every 𝐾-place 𝜑 of 𝐹 𝑗 with residue field 𝐹¯ 𝑗 we have Gal( 𝐹¯ 𝑗 /𝐾)  Gal(𝐹 𝑗 /𝐾 (T)). Finally, there is a nonempty Zariski open subset 𝑈2 of A𝑟 satisfying this: ℎ 𝑗 (a, 𝑋) is separable and deg𝑋 (ℎ 𝑗 (a, 𝑋)) = deg𝑋 (ℎ 𝑗 (T, 𝑋)) for each a ∈ 𝑈2 (𝐾), 𝑗 = 1, . . . , 𝑘. Put 𝐻 = 𝐻 ′ ∩ 𝑈1 (𝐾) ∩ 𝑈2 (𝐾). Consider a ∈ 𝐻 and let 𝜑 𝑗 be as above. Then, Lemma 18.1.1 gives an isomorphism 𝜑∗𝑗 : Gal(ℎ 𝑗 (a, 𝑋), 𝐾) → Gal(ℎ 𝑗 (T, 𝑋), 𝐾 (T)), 𝑗 = 1, . . . , 𝑘, as claimed.

□

18.2 Stable Polynomials Let 𝐾 be a field and 𝐺 a profinite group. Suppose that 𝐾 has a Galois extension 𝐿 with Gal(𝐿/𝐾)  𝐺. Then, 𝐺 occurs (or is realizable) over 𝐾 and 𝐿 is a 𝐺-extension of 𝐾. If 𝐺 belongs to a family G of profinite groups, we also say that 𝐿 is a G-extension of 𝐾. For example, when 𝐺 is Abelian, we say 𝐿 is an Abelian extension of 𝐾. If 𝐺 is an inverse limit of finite solvable groups, we say 𝐿 is a prosolvable extension of 𝐾. The main problem of Galois theory is whether every finite group occurs over Q. Even if this holds, it is not clear whether the same holds for every Hilbertian field. To approach the latter problem, consider algebraically independent elements 𝑡 1 , . . . , 𝑡𝑟 over 𝐾. If 𝐾 is Hilbertian, 𝐺 is finite, and 𝐺 occurs over 𝐾 (t), then 𝐺 occurs over 𝐾 (Lemma 14.1.1(b)). If we want 𝐺 to occur more than once over 𝐾, we have to assume more, as we now explain. We say that 𝐺 is regular over 𝐾 if there exist algebraically independent elements 𝑡 1 , . . . , 𝑡𝑟 over 𝐾 such that 𝐾 (t) has a Galois extension 𝐹 which is regular over 𝐾 with Gal(𝐹/𝐾 (t))  𝐺. This stronger property is inherited by all extensions of 𝐾: Lemma 18.2.1 Let 𝐾 be a field and 𝐺 a profinite group. Suppose that 𝐺 is regular over 𝐾. Then, (a) every quotient 𝐺¯ of 𝐺 is regular over 𝐾 and (b) 𝐺 is regular over every extension 𝐿 of 𝐾.

18.2 Stable Polynomials

311

Proof. Let t = (𝑡1 , . . . , 𝑡𝑟 ) and 𝐹 be as above. Let 𝐹¯ be the fixed field of a kernel of ¯ Then, Gal( 𝐹/𝐾 ¯ ¯ (t))  𝐺¯ and 𝐹/𝐾 an epimorphism 𝐺 → 𝐺. is regular (Corollary 3.4.2(b)). This proves (a). For (b) consider a field extension 𝐿 of 𝐾. Assume without loss that 𝑡 1 , . . . , 𝑡𝑟 are algebraically independent over 𝐿. Then, 𝐹 is algebraically independent from 𝐿 over 𝐾 (Lemma 3.4.4), so 𝐹 is linearly disjoint from 𝐿 over 𝐾 (Lemma 3.4.7). Hence, 𝐹 𝐿 is a regular extension of 𝐿 (Corollary 3.4.8(a)), 𝐹 is linearly disjoint from 𝐿(t) over 𝐾 (t) (Lemma 3.1.3), and Gal(𝐹 𝐿/𝐿(t))  𝐺. Consequently, 𝐺 is regular over 𝐿. □ The regular inverse Galois problem asks whether every finite group is regular over every field. By Lemma 18.2.1, it suffices to check the problem over Q and over each of the fields F 𝑝 . In this generality the problem is still wide open. Nevertheless, there are many special cases where groups 𝐺 are proved to be regular over specific fields. This chapter discusses cases when this happens. Remark 18.2.2 (Reinterpretation of ‘regularity’ by polynomials) Consider an irreducible polynomial 𝑓 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑟 , 𝑋], separable with respect to 𝑋. Call 𝑓 𝑋-stable over 𝐾 if Gal( 𝑓 (T, 𝑋), 𝐾 (T))  Gal( 𝑓 (T, 𝑋), 𝐿(T))

(18.3)

for every extension 𝐿 of 𝐾. In this case denote the splitting field of 𝑓 (T, 𝑋) over 𝐾 (T) by 𝐹. Then, Gal(𝐹/𝐾 (T))  Gal(𝐹 𝐿/𝐿(T)). Hence, 𝐹 is linearly disjoint from 𝐿 over 𝐾 (Lemma 3.1.3). Thus, 𝐹 is a regular extension of 𝐾 and 𝐺 = Gal( 𝑓 (T, 𝑋), 𝐾 (T)) is regular over 𝐾. In addition, 𝑓 is absolutely irreducible (Corollary 11.2.2(b)). Conversely, suppose that 𝐾 (𝑇1 , . . . , 𝑇𝑟 ) has a Galois extension 𝐹 which is regular over 𝐾 and with Galois group 𝐺. Choose a primitive element 𝑥 for 𝐹/𝐾 (T) and let 𝑓 ∈ 𝐾 [T, 𝑋] be an irreducible polynomial with 𝑓 (T, 𝑥) = 0. Then, 𝑓 (T, 𝑋) is 𝑋-stable over 𝐾, Galois with respect to 𝑋 over 𝐾 (T), and Gal( 𝑓 (T, 𝑋), 𝐾 (T))  𝐺.

Lemma 18.2.3 Let 𝐾 be a field and 𝑓 a polynomial in 𝐾 [𝑇1 , . . . , 𝑇𝑟 , 𝑋], separable in 𝑋. Suppose that Gal( 𝑓 (T, 𝑋), 𝐾 (T))  Gal( 𝑓 (T, 𝑋), 𝐾sep (T)). Then, 𝑓 is 𝑋-stable over 𝐾. Proof. Condition (18.3) holds when 𝐿 is a purely inseparable or a regular extension of 𝐾. Hence, we have to consider only the case where 𝐿 is a separable algebraic extension of 𝐾. In this case both maps res: Gal( 𝑓 (T, 𝑋), 𝐾sep (T)) → Gal( 𝑓 (T, 𝑋), 𝐿(T)) and res: Gal( 𝑓 (T, 𝑋), 𝐿(T)) → Gal( 𝑓 (T, 𝑋), 𝐾 (T)) are injective. By assumption, their compositum is bijective. Hence, each of them is bijective. □

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18 Galois Groups over Hilbertian Fields

Lemma 18.2.4 Let 𝐾 be a field and 𝑁 a Galois extension of 𝐾. (a) Suppose that 𝑓 is a polynomial in 𝑁 [𝑇1 , . . . , 𝑇𝑟 , 𝑋] separable in 𝑋. Then, 𝐾 has a finite Galois extension 𝐿 in 𝑁 with 𝑓 ∈ 𝐿 [T, 𝑋] and Gal( 𝑓 (T, 𝑋), 𝐿(T))  Gal( 𝑓 (T, 𝑋), 𝑁 (T)). (b) If 𝑓 is Galois over 𝑁 (T), then 𝑓 is Galois over 𝐿(T). (c) If 𝑓 is 𝑋-stable over 𝑁, then 𝑓 is 𝑋-stable over 𝐿. Proof of (a). Choose a finite extension 𝐿 0 of 𝐾 in 𝑁 which contains the coefficients of 𝑓 . Denote the splitting field of 𝑓 (T, 𝑋) over 𝐿 0 (T) by 𝐹. Then, 𝐿 1 = 𝐹 ∩ 𝑁 is a finitely generated extension of 𝐾 (Lemma 11.5.1). In addition, 𝐿 1 /𝐾 is a separable algebraic extension. Hence, 𝐿 1 /𝐾 is finite. Also, Gal(𝐹/𝐿 1 (T))  Gal(𝐹 𝑁/𝑁 (T)). Finally, let 𝐿 be the Galois closure of 𝐿 1 /𝐾. Then, 𝐿 satisfies the requirements of (a). Proof of (b). Suppose that 𝑓 and 𝐿 satisfy the conclusion of (a) and 𝑓 is Galois over 𝑁 (T). Let 𝑥1 , . . . , 𝑥 𝑛 be the roots of 𝑓 (T, 𝑋) in 𝐾 (T)sep . Consider 𝜎 ∈ Gal( 𝑓 (T, 𝑋), 𝐿(T)). Suppose that 𝑥 1𝜎 = 𝑥 1 . By (a), 𝜎 extends to an element 𝜎 in Gal( 𝑓 (T, 𝑋), 𝑁 (T)). Since 𝑓 is Galois over 𝑁 (t), we have 𝑥𝑖𝜎 = 𝑥𝑖 , 𝑖 = 1, . . . , 𝑛. Therefore, 𝑓 is Galois over 𝐿(T). Proof of (c). By assumption, Gal( 𝑓 (T, 𝑋), 𝑁 (T))  Gal( 𝑓 (T, 𝑋), 𝐾sep (T)). Hence, by (a), Gal( 𝑓 (T, 𝑋)), 𝐿(T))  Gal( 𝑓 (T, 𝑋), 𝐾sep (T)). By Lemma 18.2.3, 𝑓 is 𝑋-stable over 𝐿. □ Example 18.2.5 (Stable polynomials) (a) The general polynomial of degree 𝑛 is 𝑓 (𝑇1 , . . . , 𝑇𝑛 , 𝑋) = 𝑋 𝑛 + 𝑇1 𝑋 𝑛−1 + · · · + 𝑇𝑛 . It satisfies Gal( 𝑓 (T, 𝑋), 𝐾 (T))  𝑆 𝑛 for every positive integer 𝑛 and for every field 𝐾 [Lan97, p. 272, Example 4]. Thus, 𝑓 is 𝑋-stable over every field. (b) The polynomial 𝑋 𝑛 − 𝑇 satisfies Gal(𝑋 𝑛 − 𝑇, 𝐾 (𝑇))  Z/𝑛Z for every field 𝐾 with char(𝐾) ∤ 𝑛 that contains 𝜁 𝑛 . Thus, 𝑋 𝑛 − 𝑇 is 𝑋-stable over 𝐾. If however 𝜁 𝑛 ∉ 𝐾, then Gal(𝑋 𝑛 − 𝑇, 𝐾 (𝑇))  Gal(𝐾 (𝜁 𝑛 )/𝐾) ⋉ Z/𝑛Z. Therefore, 𝑋 𝑛 − 𝑇 is not 𝑋-stable over 𝐾. (c) The polynomial 𝑋 𝑝 − 𝑋 − 𝑇 satisfies Gal(𝑋 𝑝 − 𝑋 − 𝑇, 𝐾 (𝑇))  Z/𝑝Z for every field 𝐾 of characteristic 𝑝 [Lan97, p. 290, Thm. 6.4]. Therefore, 𝑓 is 𝑋-stable over 𝐾. (d) Suppose that Gal( 𝑓 (T, 𝑋), 𝐾 (T)) is simple. Let 𝐹 be the splitting field of 𝑓 (T, 𝑋) over 𝐾 (T). Suppose that 𝐹 ̸ ⊆ 𝐾sep (T). Then, 𝐹 ∩ 𝐾sep (T) = 𝐾 (T). Therefore, 𝑓 is 𝑋-stable over 𝐾. (e) Suppose that 𝑓 ∈ 𝐾 [T, 𝑋] is absolutely irreducible and Galois over 𝐾 (T). Then, 𝑓 is 𝑋-stable over 𝐾. Indeed, let 𝐿 be an extension of 𝐾. Then, Gal( 𝑓 (T, 𝑋), 𝐿(T)) is a subgroup of Gal( 𝑓 (T, 𝑋), 𝐾 (T)). On the other hand,

18.2 Stable Polynomials

313

𝑓 (T, 𝑋) is also Galois over 𝐿(T). Hence, |Gal( 𝑓 (T, 𝑋), 𝐿(T))| = deg𝑋 ( 𝑓 (T, 𝑋)) = |Gal( 𝑓 (T, 𝑋), 𝐾 (T))|. Thus, Gal( 𝑓 (T, 𝑋), 𝐿(T))  Gal( 𝑓 (T, 𝑋), 𝐾 (T)), as claimed. We use stable polynomial to construct linearly disjoint sequences of Galois extensions over Hilbertian fields with given Galois groups: Lemma 18.2.6 Let 𝑓𝑖 (𝑇1 , . . . , 𝑇𝑟 , 𝑋) be an 𝑋-stable polynomial over a Hilbertian field 𝐾 and 𝐺 𝑖 = Gal( 𝑓𝑖 (T, 𝑋), 𝐾 (T)), 𝑖 = 1, 2, 3, . . . . Given a finite separable extension 𝐿 0 of 𝐾, there is a sequence, 𝐿 1 , 𝐿 2 , 𝐿 3 , . . . of Galois extensions of 𝐾 with the following properties: (a) Gal(𝐿 𝑖 /𝐾)  𝐺 𝑖 and 𝑓𝑖 has an 𝐿 𝑖 -rational zero c𝑖 , 𝑖 = 1, 2, 3, . . . . (b) c1 , c2 , c3 , . . . are distinct. (c) The sequence 𝐿 0 , 𝐿 1 , 𝐿 2 , . . . is linearly disjoint over 𝐾. Proof. Suppose by induction that there are Galois extensions 𝐿 1 , . . . , 𝐿 𝑛 of 𝐾 such that Gal(𝐿 𝑖 /𝐾)  𝐺 𝑖 , 𝑖 = 1, . . . , 𝑛, and 𝐿 0 , 𝐿 1 , . . . , 𝐿 𝑛 are linearly disjoint over 𝐾. In addition suppose that there are a𝑖 ∈ 𝐾 𝑟 and 𝑏 𝑖 ∈ 𝐿 𝑖 such that 𝑓𝑖 (a𝑖 , 𝑏 𝑖 ) = 0 and a1 , . . . , a𝑛 are distinct. Then, 𝐿 = 𝐿 0 𝐿 1 · · · 𝐿 𝑛 is a finite separable extension of 𝐾. By assumption, Gal( 𝑓𝑛+1 (T, 𝑋), 𝐿(T))  Gal( 𝑓𝑛+1 (T, 𝑋), 𝐾 (T)) = 𝐺 𝑛+1 . Apply Proposition 18.1.5 and Corollary 13.2.3 to find a𝑛+1 ∈ 𝐾 𝑟 with Gal( 𝑓𝑛+1 (a𝑛+1 , 𝑋), 𝐿)  Gal( 𝑓𝑛+1 (a𝑛+1 , 𝑋), 𝐾)  𝐺 𝑛+1 and a𝑛+1 ≠ a1 , . . . , a𝑛 . Let 𝐿 𝑛+1 be the splitting field of 𝑓𝑛+1 (a, 𝑋) over 𝐾. Then, Gal(𝐿 𝑛+1 /𝐾)  𝐺 𝑛+1 and 𝐿 𝑛+1 is linearly disjoint from 𝐿 over 𝐾. In particular, 𝐿 0 , . . . , 𝐿 𝑛+1 are linearly disjoint over 𝐾. Finally, choose a zero 𝑏 𝑛+1 of 𝑓𝑛+1 (a𝑛+1 , 𝑋). Then, 𝑏 𝑛+1 ∈ 𝐿 𝑛+1 . □ Our next result is an immediate application of this lemma to the polynomials of Example 18.2.5(a),(b),(c): Corollary 18.2.7 Let 𝐾 be a Hilbertian field and 𝐺 a finite group. Then, 𝐾 has a linearly disjoint sequence of Galois extensions with 𝐺 as Galois group in each of the following cases: (a) 𝐺 = 𝑆 𝑛 , 𝑛 ∈ N. (b) 𝜁 𝑛 ∈ 𝐾 and 𝐺 = Z/𝑛Z, 𝑛 ∈ N, char(𝐾) ∤ 𝑛. (c) char(𝐾) = 𝑝 > 0 and 𝐺 = Z/𝑝Z. We prove in Section 18.4 that every finite Abelian group 𝐴 satisfies the conclusion of Corollary 18.2.7. But the latter simple corollary already has interesting implications to closed normal subgroups of Gal(𝐾) (Section 18.10).

314

18 Galois Groups over Hilbertian Fields

The following result shows that in many cases it suffices to take 𝑟 = 1 in the definition of “𝐺 is regular over 𝐾”: Proposition 18.2.8 Let 𝐾 be an infinite field and 𝐺 a finite group. Suppose that 𝐺 is 𝐾-regular. Then: (a) There is an 𝑋-stable polynomial ℎ ∈ 𝐾 [𝑇, 𝑋] which is Galois with respect to 𝑋 and with Gal(ℎ(𝑇, 𝑋), 𝐾 (𝑇))  𝐺. Thus, the splitting field 𝐹 of ℎ(𝑇, 𝑋) over 𝐾 (𝑇) is regular over 𝐾 and Gal(𝐹/𝐾 (𝑇))  𝐺. (b) If in addition 𝐾 is Hilbertian, then 𝐾 has a linearly disjoint sequence 𝐿 1 , 𝐿 2 , 𝐿 3 , . . . of Galois extensions such that Gal(𝐿 𝑖 /𝐾)  𝐺 for all 𝑖. Proof. Statement (b) follows from (a) by Lemma 18.2.6. We prove (a): Remark 18.2.2 gives a polynomial 𝑓 ∈ 𝐾 [𝑇1 , . . . , 𝑇𝑟 , 𝑋] which is 𝑋-stable over 𝐾 with 𝑟 ≥ 1, Galois with respect to 𝑋 over 𝐾 (T) such that Gal( 𝑓 (T, 𝑋), 𝐾 (T))  𝐺, and absolutely irreducible. Assume without loss that 𝑓 is monic with respect to 𝑋. If 𝑟 = 1, take ℎ = 𝑓 . So, assume 𝑟 ≥ 2. Let 𝑢 1 , 𝑢 2 be algebraically independent elements over 𝐾 and put 𝐿 = 𝐾 (𝑢 1 , 𝑢 2 ), T′ = (𝑇1 , . . . , 𝑇𝑟−1 ), and 𝑔(T′, 𝑋) = 𝑓 (𝑇1 , . . . , 𝑇𝑟−1 , 𝑢 1 + 𝑢 2𝑇𝑟−1 , 𝑋). By Proposition 11.5.4, 𝑔(T′, 𝑋) is absolutely irreducible. Extend the map T → (T′, 𝑢 1 + 𝑢 2𝑇𝑟−1 ) to an 𝐿 (T′)-place of 𝐿 (T) with residue field 𝐿(T′) (Lemma 2.2.7). By Lemma 18.1.1, 𝐺  Gal( 𝑓 (T, 𝑋), 𝐿(T))  Gal(𝑔(T′, 𝑋), 𝐿(T′)). By assumption, 𝐾 is infinite. Hence, Proposition 18.1.4 gives 𝑏 1 , 𝑏 2 ∈ 𝐾 with 𝑔(T ¯ ′, 𝑋) = 𝑓 (𝑇1 , . . . , 𝑇𝑟−1 , 𝑏 1 + 𝑏 2𝑇𝑟−1 , 𝑋) absolutely irreducible, Galois over ¯ ′, 𝑋), 𝐾 (T′))  Gal(𝑔(T′, 𝑋), 𝐿(T′))  𝐺. By the last para𝐾 (T′), and Gal( 𝑔(T ¯ ′, 𝑋) is also 𝑋-stable over 𝐾. graph of Remark 18.2.2, 𝑔(T Finally, use induction on 𝑟 to find an absolutely irreducible polynomial ℎ ∈ 𝐾 [𝑇1 , 𝑋] which is 𝑋-stable, Galois with respect to 𝑋, and with Gal(ℎ(𝑇1 , 𝑋), 𝐾 (𝑇1 ))  𝐺.

□

Problem 18.2.9 Does Proposition 18.2.8 hold if 𝐾 is finite?

18.3 Regular Realization of Finite Abelian Groups The inverse problem of Galois theory has an affirmative solution for every finite Abelian group 𝐴 and every Hilbertian field 𝐾 (Corollary 18.3.6). Moreover, every finite Abelian group is even regular over 𝐾 (Proposition 18.3.5). The proof of the latter result moves from the case where 𝐴 is cyclic to the general case. In the cyclic case we have to distinguish between the case where char(𝐾) does not divide the order of the group and where the order of the group is a power of char(𝐾). In the general case, we distinguish between the cases where 𝐾 is finite and 𝐾 is infinite.

18.3 Regular Realization of Finite Abelian Groups

315

Lemma 18.3.1 Let 𝐾 be a field, 𝑛 a positive integer with char(𝐾) ∤ 𝑛, and 𝑡 an indeterminate. Then, 𝐾 (𝑡) has a cyclic extension 𝐸 of degree 𝑛 which is contained in 𝐾 ((𝑡)). Proof. Choose a root of unity 𝜁 𝑛 of order 𝑛 in 𝐾sep . Let 𝐿 = 𝐾 (𝜁 𝑛 ) and 𝐺 = 𝜒 ( 𝜎) Gal(𝐿/𝐾). Then, there is a map 𝜒: 𝐺 → {1, . . . , 𝑛 − 1} such that 𝜎(𝜁 𝑛 ) = 𝜁 𝑛 . Hence, gcd( 𝜒(𝜎), 𝑛) = 1 and 𝜒(𝜎𝜏) ≡ 𝜒(𝜎) 𝜒(𝜏) mod 𝑛

(18.4)

for all 𝜎, 𝜏 ∈ 𝐺. By Example 4.5.1, 𝐾 ((𝑡)) is a regular extension of 𝐾 and 𝐿((𝑡)) = 𝐾 ((𝑡)) (𝜁 𝑛 ). Thus, we may identify 𝐺 with Gal(𝐿 ((𝑡))/𝐾 ((𝑡))). Choose a primitive element 𝑐 of 𝐿/𝐾. Consider the element Ö

𝑔(𝑡) =

1 + 𝜎(𝑐)𝑡


𝜒 ( 𝜎 −1 )

𝜎 ∈𝐺

of 𝐿 [𝑡]. Since char(𝐾) ∤ 𝑛, Hensel’s lemma (Proposition 4.5.2) gives an 𝑥 ∈ 𝐿 [[𝑡]] with 𝑥 𝑛 = 1 + 𝑐𝑡. Then, Ö −1 (18.5) 𝑦 := 𝜎(𝑥) 𝜒 ( 𝜎 ) ∈ 𝐿 [[𝑡]] 𝜎 ∈𝐺

and 𝑦𝑛 =

Ö

𝜎(𝑥 𝑛 ) 𝜒 ( 𝜎

−1 )

Ö

=

𝜎 ∈𝐺

(1 + 𝜎(𝑐)𝑡) 𝜒 ( 𝜎

−1 )

= 𝑔(𝑡).

𝜎 ∈𝐺

Since 𝜁 𝑛 ∈ 𝐿, 𝐹 = 𝐿 (𝑡, 𝑦) is a cyclic extension of degree 𝑑 of 𝐿(𝑡), where 𝑑|𝑛 and 𝑦 𝑑 ∈ 𝐿 (𝑡) [Lan97, p. 289, Thm. 6.2(ii)]. The distinct linear factors of 𝑔(𝑡) are not associated in 𝐿 [𝑡]. In particular, (1 + 𝑐𝑡)|𝑔(𝑡) and (1 + 𝑐𝑡) 2 ∤ 𝑔(𝑡). Hence, by Eisenstein’s criterion (Lemma 2.3.10), 𝑌 𝑛 − 𝑔(𝑡) is irreducible over 𝐿(𝑡). Therefore, 𝑑 = 𝑛 and the Galois group Gal(𝐹/𝐿 (𝑡)) is generated by an element 𝜔 satisfying 𝜔(𝑦) = 𝜁 𝑛 𝑦. By (18.4) there exists for each 𝜏, 𝜌 ∈ 𝐺 an integer 𝑘 (𝜏, 𝜌) and a rational function 𝑓 𝜏 (𝑡) ∈ 𝐿(𝑡) such that 𝜏(𝑦) =

Ö

𝜏𝜎(𝑥) 𝜒 ( 𝜎

−1 )

𝜎 ∈𝐺

=

Ö

=

Ö

𝜌(𝑥) 𝜒 (𝜌

−1 𝜏)

𝜌∈𝐺

𝜌(𝑥)

𝜒 (𝜌−1 ) 𝜒 ( 𝜏)+𝑘 ( 𝜏,𝜌)𝑛

𝜌∈𝐺

=

Ö

𝜌(𝑥) 𝜒 (𝜌

= 𝑦 𝜒 ( 𝜏)


𝜒 ( 𝜏) Ö

(𝜌(𝑥 𝑛 )) 𝑘 ( 𝜏,𝜌)

𝜌∈𝐺

𝜌∈𝐺 (18.5)

−1 )

Ö

(1 + 𝜌(𝑐)𝑡) 𝑘 ( 𝜏,𝜌) = 𝑦 𝜒 ( 𝜏) 𝑓 𝜏 (𝑡).

𝜌∈𝐺

It follows that 𝐺 leaves 𝐹 invariant. Let 𝐸 be the fixed field of 𝐺 in 𝐹.

(18.6)

316

18 Galois Groups over Hilbertian Fields

𝐾 ((𝑡))

𝐺

𝐿 ((𝑡))

𝐺

𝐸

𝐹= 𝐿(𝑡, 𝑦) 𝑛

𝐾 (𝑡)

𝐺

𝐺

𝐾

𝐿 (𝑡) 𝐿= 𝐾 (𝜁 𝑛 )

Denote the subgroup of Aut(𝐹/𝐾 (𝑡)) generated by 𝐺 and Gal(𝐹/𝐿(𝑡)) by 𝐻. Then, the fixed field of 𝐻 is 𝐾 (𝑡), so 𝐹/𝐾 (𝑡) is a Galois extension with Gal(𝐹/𝐾 (𝑡)) = 𝐺 · Gal(𝐹/𝐿 (𝑡)). Moreover, given 𝜏 ∈ 𝐺, put 𝑚 = 𝜒(𝜏). Then, (18.6)

𝜏𝜔(𝑦) = 𝜏(𝜁 𝑛 𝑦) = 𝜁 𝑛𝑚 𝑦 𝑚 𝑓 𝜏 (𝑡) = 𝜔(𝑦) 𝑚 𝑓 𝜏 (𝑡) = 𝜔(𝑦 𝑚 𝑓 𝜏 (𝑡)) = 𝜔𝜏(𝑦). Thus, 𝜏𝜔 = 𝜔𝜏, so 𝐺 commutes with Gal(𝐹/𝐿(𝑡)). Therefore, 𝐸/𝐾 (𝑡) is a Galois extension with Gal(𝐸/𝐾 (𝑡))  Gal(𝐹/𝐿 (𝑡))  Z/𝑛Z. □ Lemma 18.3.2 Suppose that 𝑝 = char(𝐾). Let 𝐿 be a cyclic extension of degree 𝑝 𝑛 , 𝑛 ≥ 1, of 𝐾. Then, 𝐾 has a Z/𝑝 𝑛+1 Z-extension 𝐿 ′ which contains 𝐿. Proof. Define 𝐿 ′ to be 𝐿 (𝑥) where 𝑥 is a zero of 𝑋 𝑝 − 𝑋 − 𝑎 with 𝑎 ∈ 𝐿. The three parts of the proof produce 𝑎, and then show that 𝐿 ′ has the desired properties. Part A: Construction of 𝑎. Since 𝐿/𝐾 is separable, there is a 𝑏 1 ∈ 𝐿 with 𝑐 := trace 𝐿/𝐾 (𝑏 1 ) ≠ 0 [Lan97, p. 286, Thm. 5.2]. Put 𝑏 = 𝑏𝑐1 . Then, trace 𝐿/𝐾 (𝑏) = 1 and trace 𝐿/𝐾 (𝑏 𝑝 − 𝑏) = (trace 𝐿/𝐾 (𝑏)) 𝑝 − trace 𝐿/𝐾 (𝑏) = 0. With 𝜎 a generator of Gal(𝐿/𝐾), the additive form of Hilbert’s Theorem 90 [Lan97, p. 290, Thm. 6.3] gives 𝑎 ∈ 𝐿 with 𝜎𝑎 − 𝑎 = 𝑏 𝑝 − 𝑏. (18.7) Part B: Irreducibility of 𝑋 𝑝 − 𝑋 − 𝑎. Assume that 𝑋 𝑝 − 𝑋 − 𝑎 is reducible over 𝐿. Then, the root 𝑥 of the latter polynomial is in 𝐿 [Lan97, p. 290, Thm. 6.4(b)]. Thus, (𝜎𝑥 − 𝑥) 𝑝 − (𝜎𝑥 − 𝑥) − (𝑏 𝑝 − 𝑏) = (𝜎𝑥 − 𝑥) 𝑝 − (𝜎𝑥 − 𝑥) − (𝜎𝑎 − 𝑎) = (𝜎𝑥 𝑝 − 𝜎𝑥 − 𝜎𝑎) − (𝑥 𝑝 − 𝑥 − 𝑎) = 0.

(18.8)

Since 𝑏 is a root of 𝑋 𝑝 − 𝑋 − (𝑏 𝑝 − 𝑏), there is an integer 𝑖 between 0 and 𝑝 − 1 with 𝜎𝑥 − 𝑥 = 𝑏 + 𝑖 [Lan97, p. 290, Thm. 6.4(b)]. Apply trace 𝐿/𝐾 to both sides to get 0 on the left and 1 on the right. This contradiction proves that 𝑋 𝑝 − 𝑋 − 𝑎 is irreducible. Part C: Extension of 𝜎 to 𝜎 ′ that maps 𝑥 to 𝑥 + 𝑏. Equality (18.7) implies that 𝑥 + 𝑏 is a zero of 𝑋 𝑝 − 𝑋 − 𝜎𝑎. Thus, by Part B, 𝜎 extends to an automorphism 𝜎 ′ of 𝐿 ′ with 𝜎 ′ (𝑥) = 𝑥 + 𝑏. We need only prove that 𝜎 ′ has order 𝑝 𝑛+1 . Induction shows that (𝜎 ′) 𝑗 (𝑥) = 𝑥 + 𝑏 + 𝜎𝑏 + · · · + 𝜎 𝑗−1 𝑏. In particular, (𝜎 ′) 𝑝 (𝑥) = 𝑥 + trace 𝐿/𝐾 (𝑏) = 𝑥 + 1. 𝑛

𝑛 (𝜎 ′) 𝑖 𝑝 (𝑥)

Hence, contended.

= 𝑥 + 𝑖, 𝑖 = 1, . . . , 𝑝. Therefore, the order of

(18.9) 𝜎′

is

𝑝 𝑛+1 ,

as □

18.3 Regular Realization of Finite Abelian Groups

317

Remark 18.3.3 (Lemma 18.3.2 is a special case of a theorem of Witt.) Suppose that char(𝐾) = 𝑝. Then, AS(𝐾) := {𝑥 𝑝 − 𝑥 | 𝑥 ∈ 𝐾 } is a subgroup of the additive group of 𝐾 and 𝐾/𝐴𝑆(𝐾) is a vector space over F 𝑝 of dimension, say, 𝑟. Consider an embedding problem 𝐺 → Gal(𝐿/𝐾) over 𝐾 with 𝐺 a finite 𝑝-group which is generated by 𝑟 elements. Witt’s theorem says that this problem is solvable (If 𝑟 = ∞, there is no restriction on the number of generators of 𝐺.) The technique of Galois cohomology [Rbs70, p. 257, Cor. 3.4] simplifies Witt’s original proof [Wit36]. Lemma 18.3.4 Let 𝐾 be a field, 𝑡 an indeterminate, and 𝐴 a finite cyclic group. Then, 𝐾 (𝑡) has a Galois extension 𝐹 such that Gal(𝐹/𝐾 (𝑡))  𝐴 and 𝐹/𝐾 is regular. Proof. We put 𝑝 = char(𝐾) and divide the proof into three parts: Part A: 𝐴  Z/𝑚Z and 𝑝 ∤ 𝑚. By Lemma 18.3.1, 𝐾 (𝑡) has a cyclic extension 𝐸 𝑚 of degree 𝑚 which is contained in 𝐾 ((𝑡)). By Example 4.5.1, 𝐾 ((𝑡)) is a regular extension of 𝐾. Hence, so is 𝐸 𝑚 (Corollary 3.4.2(b)). Part B: 𝐴  Z/𝑝 𝑘 Z. Assume without loss that 𝑘 ≥ 1. By Example 2.3.11, the polynomial 𝑋 𝑝 − 𝑋 − 𝑡 is irreducible over 𝐾˜ (𝑡). Let 𝑥 be a root of 𝑋 𝑝 − 𝑋 − 𝑡 in 𝐾 (𝑡)sep . In particular, 𝐾 (𝑥) = 𝐾 (𝑡, 𝑥). Then, by Artin–Schreier, [Lan97, p. 290, Thm. 6.4(b)], 𝐾 (𝑥) is a cyclic extension of degree 𝑝 of 𝐾 (𝑡). Lemma 18.3.2 gives a cyclic extension 𝐸 𝑝 𝑘 of 𝐾 (𝑡) of degree 𝑝 𝑘 which contains 𝐾 (𝑥). By the preceding paragraph, 𝐾 (𝑥) ∩ 𝐾˜ (𝑡) = 𝐾 (𝑡). Since Gal(𝐸 𝑝 𝑘 /𝐾 (𝑡)) is a cyclic group of order 𝑝 𝑘 , each subextension of 𝐸 𝑝 𝑘 which properly contains 𝐾 (𝑡) must contain 𝐾 (𝑥). Hence, 𝐸 𝑝 𝑘 ∩ 𝐾˜ (𝑡) = 𝐾 (𝑡). Thus, 𝐸 𝑝 𝑘 is linearly disjoint from 𝐾˜ (𝑡) over 𝐾 (𝑡). By the tower property (Lemma 3.1.3), 𝐸 𝑝 𝑘 is linearly disjoint from 𝐾˜ over 𝐾; that is, 𝐸 𝑝 𝑘 /𝐾 is regular. Part C: 𝐴  Z/𝑛Z, 𝑛 = 𝑚 𝑝 𝑘 , 𝑝 ∤ 𝑚. The compositum 𝐸 𝑛 = 𝐸 𝑚 𝐸 𝑝 𝑘 is a cyclic extension of 𝐾 (𝑡) of degree 𝑛 and [𝐸 𝑛 : 𝐸 𝑝 𝑘 ] = [𝐸 𝑚 : 𝐾 (𝑡)] = 𝑚. Moreover, since 𝐸 𝑝 𝑘 /𝐾 is regular, 𝐸 𝑝 𝑘 is linearly disjoint from 𝐾˜ (𝑡) over 𝐾 (𝑡) (by the tower property), so [𝐸 𝑝 𝑘 𝐾˜ : 𝐾˜ (𝑡)] = 𝑝 𝑘 . Similarly, [𝐸 𝑚 𝐾˜ : 𝐾˜ (𝑡)] = 𝑚. Since gcd(𝑚, 𝑝 𝑘 ) = 1, [𝐸 𝑛 𝐾˜ : 𝐾˜ (𝑡)] = 𝑛, hence 𝐸 𝑛 is linearly disjoint from 𝐾˜ (𝑡) over 𝐾 (𝑡). Hence, by the tower property, 𝐸 𝑛 is linearly disjoint from 𝐾˜ over 𝐾. Thus, 𝐸 𝑛 is a regular extension of 𝐾, as desired. □ We generalize Lemma 18.3.4 from cyclic groups to arbitrary Abelian groups: Proposition 18.3.5 Let 𝐾 be a field, 𝑡 an indeterminate, and 𝐴 a finite Abelian group. Then, 𝐾 (𝑡) has a Galois extension 𝐹 with Gal(𝐹/𝐾 (𝑡))  𝐴 and 𝐹/𝐾 regular. Proof. The first two parts of the proof prove the proposition in the case where 𝐴 = (Z/𝑞Z) 𝑛 , 𝑞 = 𝑝 𝑟 , 𝑝 is a prime number, and 𝑛, 𝑟 are positive integers. Part A: 𝐴 = (Z/𝑞Z) 𝑛 is as above and 𝐾 is infinite. Choose algebraically independent elements 𝑡1 , . . . , 𝑡 𝑛 over 𝐾. For each 𝑖 between 1 and 𝑛, Lemma 18.3.4 gives a finite cyclic extension 𝐸 𝑖 of 𝐾 (𝑡𝑖 ) of degree 𝑞 which is regular over 𝐾. Then, 𝐸 1 , . . . , 𝐸 𝑛 are algebraically independent over 𝐾. Hence, by Corollary 3.4.8, 𝐸 = 𝐸 1 · · · 𝐸 𝑛 is a regular extension of 𝐾. In addition, by Lemma 3.4.7, 𝐸 1 , . . . , 𝐸 𝑛

318

18 Galois Groups over Hilbertian Fields

are linearly disjoint over 𝐾. Hence, by Lemma 3.1.14, 𝐸 1 (t), . . . , 𝐸 𝑛 (t) are linearly disjoint over 𝐾 (t) and each 𝐸 𝑖 (t) is a cyclic extension of 𝐾 (t) of degree 𝑞. It follows that Gal(𝐸/𝐾 (t))  (Z/𝑞Z) 𝑛 . Consequently, by Proposition 18.2.8, 𝐾 (𝑡) has a Galois extension 𝐹 with Galois group (Z/𝑞Z) 𝑛 which is regular over 𝐾. Part B: 𝐴 = (Z/𝑞Z) 𝑛 is as above and 𝐾 is finite. Choose a transcendental element 𝑢 over 𝐾 (𝑡). Lemma 18.3.4 gives a cyclic extension 𝐸 of 𝐾 (𝑡, 𝑢) of degree 𝑞 which is regular over 𝐾 (𝑡). Choose a polynomial ℎ ∈ 𝐾 (𝑡) [𝑢, 𝑋] which is 𝑋stable with Gal(ℎ(𝑢, 𝑋), 𝐾 (𝑡, 𝑢))  Z/𝑞Z (e.g. use Proposition 18.2.8). Denote the unique extension of 𝐾 of degree 𝑞 by 𝐾𝑞 . By Theorem 14.4.2, 𝐾 (𝑡) is Hilbertian. Hence, by Lemma 18.2.6, 𝐾 (𝑡) has a sequence 𝐹1 , 𝐹2 , 𝐹3 , . . . of cyclic extensions of degree 𝑞 such that 𝐾𝑞 (𝑡), 𝐹1 , 𝐹2 , . . . , 𝐹𝑛 are linearly disjoint over 𝐾 (𝑡). Then, 𝐹 = 𝐹1 𝐹2 · · · 𝐹𝑛 is a Galois extension of 𝐾 (𝑡) with Gal(𝐹/𝐾 (𝑡))  (Z/𝑞Z) 𝑛 . Moreover, 𝐹 ∩ 𝐾𝑞 (𝑡) = 𝐾 (𝑡). ˜ The group 𝐵 = Gal(𝐹 ∩ 𝐾/𝐾) is a cyclic quotient of Gal(𝐹/𝐾 (𝑡)), hence the exponent of 𝐵 divides 𝑞. Therefore, 𝐹 ∩ 𝐾˜ ⊆ 𝐾𝑞 . It follows from the preceding paragraph that 𝐹 ∩ 𝐾˜ = 𝐾. Consequently, 𝐹/𝐾 is regular. Î𝑚 (Z/𝑞 𝑖 Z) 𝑛𝑖 , where 𝑞 𝑖 = 𝑝 𝑟𝑖 𝑖 , 𝑚, 𝑛𝑖 , 𝑟 𝑖 , 𝑟 𝑖 𝑗 are positive integers, and Part C: 𝐴  𝑖=1 𝑝 1 , . . . , 𝑝 𝑟 are distinct prime numbers. Parts A and B give for each 𝑖 a Galois extension 𝐹𝑖 of 𝐾 (𝑡) which is regular over 𝐾 and with Gal(𝐹𝑖 /𝐾 (𝑡))  (Z/𝑞 𝑖 Z) 𝑛𝑖 . Since 𝑞 1 , . . . , 𝑞 𝑚 are pairwise relatively prime, 𝐹1 , . . . , 𝐹𝑚 are linearly disjoint over 𝐾 (𝑡). Hence, 𝐹 = 𝐹1 · · · 𝐹𝑚 is a Galois extension with Gal(𝐹/𝐾 (𝑡))  𝐴. Let 𝐸 = 𝐹 ∩ 𝐾˜ (𝑡). Since 𝐹𝑖 is a regular extension of 𝐾, we have 𝐹𝑖 ∩ 𝐸 = 𝐾 (𝑡), so 𝑞 𝑖𝑛𝑖 | [𝐹 : 𝐸]Îfor 𝑖 = 1, . . . , 𝑚. Since 𝑞 1 , . . . , 𝑞 𝑚 are pairwise relatively prime, 𝑚 𝑞 𝑖𝑛𝑖 divides [𝐹 : 𝐸]. Hence, 𝐸 = 𝐾 (𝑡). Consequently, 𝐹 is a [𝐹 : 𝐾 (𝑡)] = 𝑖=1 regular extension of 𝐾. Part D: 𝐴 is an arbitrary finite Abelian group. Then, 𝐴=

𝑚𝑖 𝑚 Ö Ö

𝑟

(Z/𝑝 𝑖 𝑖 𝑗 Z) 𝑛𝑖 𝑗

𝑖=1 𝑗=1

where 𝑝 1 , . . . , 𝑝 𝑚 are distinct prime numbers, 𝑚 ≥ 0, and 𝑚 𝑖 , 𝑛𝑖 𝑗 , 𝑟 𝑖 𝑗 are positive integers. For each 𝑖 let 𝑟 𝑖 = max(𝑟 𝑖1 , . . . , 𝑟 𝑖,𝑚𝑖 ) and 𝑛𝑖 = 𝑛𝑖1 + · · · + 𝑛𝑖,𝑚𝑖 Part C ˆ (𝑡))  gives extension 𝐹ˆ of 𝐾 (𝑡) which is regular over 𝐾 and with Gal( 𝐹/𝐾 Î𝑚 a Galois 𝑟𝑖 𝑛𝑖 . By construction, 𝐴 is a quotient of Gal( 𝐹/𝐾 ˆ (𝑡)). Hence, 𝐾 (𝑡) has (Z/𝑝 Z) 𝑖=1 𝑖 ˆ a Galois extension 𝐹 in 𝐹ˆ with Gal(𝐹/𝐾 (𝑡))  𝐴. Since 𝐹/𝐾 is regular, so is 𝐹/𝐾. □ Corollary 18.3.6 Let 𝐾 be a Hilbertian field and 𝐴 a finite Abelian group. Then, 𝐴 is realizable over 𝐾. Moreover, 𝐾 has a linearly disjoint sequence 𝐿 1 , 𝐿 2 , 𝐿 3 , . . . of Galois extensions with Gal(𝐿 𝑖 /𝐾)  𝐴 for each 𝑖. Proof. By Proposition 18.3.5, 𝐴 is regular over 𝐾. Hence, by Proposition 18.2.8, 𝐾 has a linearly disjoint sequence 𝐿 1 , 𝐿 2 , 𝐿 3 , . . . of Galois extensions with Gal(𝐿 𝑖 /𝐾)  𝐴 for each 𝑖. □

18.4 Split Embedding Problems with Abelian Kernels

319

18.4 Split Embedding Problems with Abelian Kernels Attempts to realize a finite Abelian group 𝐴 over a Hilbertian field 𝐾 usually lead to an extension of 𝐾 with roots of unity. This gives a “split embedding problem with Abelian kernel”. The main result of this section is that each such problem is solvable. Consequently, 𝐴 occurs over 𝐾. Indeed, for 𝐴 to be regular over 𝐾 it is not necessary that 𝐾 is Hilbertian. We prove that it is true for an arbitrary field. Definition 18.4.1 (Embedding problems) Let 𝐿/𝐾 be a Galois extension, 𝐺 a profinite group, and 𝛼: 𝐺 → Gal(𝐿/𝐾) an epimorphism. The embedding problem associated with 𝛼 consists of embedding 𝐿 in a Galois extension 𝑁 of 𝐾 with an isomorphism 𝛽: Gal(𝑁/𝐾) → 𝐺 satisfying 𝛼 ◦ 𝛽 = res 𝑁 /𝐿 . Refer to 𝛼 as an embedding problem over 𝐾, 𝛽 its solution, and 𝑁 the solution field. Call the problem finite if 𝐺 is finite. The problem splits if 𝛼 has a section, that is, an embedding 𝛼 ′: Gal(𝐿/𝐾) → 𝐺 with 𝛼 ◦ 𝛼 ′ = idGal(𝐿/𝐾) . The latter case occurs when 𝐺 = Gal(𝐿/𝐾) ⋉ Ker(𝛼) and 𝛼 is the projection of 𝐺 onto Gal(𝐿/𝐾). Let 𝑡 1 , . . . , 𝑡 𝑛 be algebraically independent elements over 𝐾. Then, res: Gal(𝐿 (t)/𝐾 (t)) → Gal(𝐿/𝐾) is an isomorphism. Hence, 𝛼: 𝐺 → Gal(𝐿/𝐾) gives rise to an embedding problem 𝛼t : 𝐺 → Gal(𝐿 (t)/𝐾 (t)) over 𝐾 (t) with res 𝐿 (t)/𝐿 ◦ 𝛼t = 𝛼. Refer to a solution of 𝛼t as a solution of 𝛼 over 𝐾 (t). Refer to a solution field 𝐹 of 𝛼t as a regular solution of 𝛼 if 𝐹/𝐿 is regular. We say that 𝛼 is regularly solvable if there are 𝑡 1 , . . . , 𝑡 𝑛 as above and 𝛼t has a solution field 𝐹 which is regular over 𝐿.

Lemma 18.4.2 Let 𝐾 be a Hilbertian field, 𝛼: 𝐻 → Gal(𝐿/𝐾) a finite embedding problem, 𝑡 1 , . . . , 𝑡 𝑛 algebraically independent elements over 𝐾, and 𝑀 a finite separable extension of 𝐿. If 𝛼 is solvable over 𝐾 (t), then 𝛼 is also solvable over 𝐾. If 𝛼 is regularly solvable, then 𝛼 has a solution field 𝑁 over 𝐾 which is linearly disjoint from 𝑀 over 𝐿. Proof. Let 𝐹 be a solution field of 𝛼 over 𝐾 (t). Thus, 𝐹 is a Galois extension of 𝐾 (t) which contains 𝐿 and there is an isomorphism 𝜃: Gal(𝐹/𝐾 (t)) → 𝐻 with 𝛼 ◦ 𝜃 = res𝐹/𝐿 . Lemma 14.1.1 gives a separable Hilbert subset 𝐴 of 𝐾 𝑛 having the following property: For each a ∈ 𝐴 there is a 𝐾-place 𝜑: 𝐹 → 𝐾˜ ∪ {∞} satisfying these conditions: (18.10a) 𝜑(t) = a, 𝐾 (t) 𝜑 = 𝐾, and 𝐿(t) 𝜑 = 𝐿. (18.10b) There is an isomorphism 𝜑∗ : Gal( 𝐹¯𝜑 /𝐾) → Gal(𝐹/𝐾 (t)) with res𝐹/𝐿 ◦ 𝜑∗ = res𝐹¯𝜑 /𝐿 . The map 𝛽 = 𝜃 ◦ 𝜑∗ solves embedding problem 𝛼. Suppose now that 𝐹/𝐿 is regular. Then, Gal(𝐹 𝑀/𝑀 (t))  Gal(𝐹/𝐿(t)). Lemma 14.1.1 gives a separable Hilbert subset 𝐴 ′ of 𝑀 𝑛 satisfying this: For each a ∈ 𝐴 ∩ 𝐴 ′ and each 𝑀-place 𝜑 of 𝐹 𝑀 satisfying (18.10a) and (18.10b), we have Gal(𝐹 𝑀/𝑀 (t))  Gal( 𝐹¯𝜑 𝑀/𝑀). By Corollary 13.2.3, there is an a ∈ 𝐴 ∩ 𝐴 ′. The preceding paragraph gives 𝜑 satisfying (18.10a) and (18.10b). The corresponding field 𝐹¯𝜑 is linearly disjoint from 𝑀 over 𝐿. □

320

18 Galois Groups over Hilbertian Fields

Let 𝐴 and 𝐺 be finite groups. Recall that 𝐴𝐺 is the group of all functions 𝑓 : 𝐺 → 𝐴 and 𝐺 acts on 𝐴𝐺 by the rule 𝑓 𝜏 (𝜎) = 𝑓 (𝜏𝜎). The semidirect product 𝐺 ⋉ 𝐴𝐺 is the wreath product 𝐴wr𝐺 (Remark 15.1.8). Lemma 18.4.3 Let 𝐺 and 𝐴 be finite groups with 𝐴 Abelian and 𝐺 acting on 𝐴. Then, there is an epimorphism 𝐴wr𝐺 → 𝐺 ⋉ 𝐴 with kernel in 𝐴𝐺 . Î −1 Proof. Define a map 𝛼: 𝐴𝐺 → 𝐴 by 𝛼( 𝑓 ) = 𝜎 ∈𝐺 𝑓 (𝜎) 𝜎 . Since 𝐴 is Abelian, the right-hand side is a well-defined homomorphism. For each 𝑎 ∈ 𝐴 define a function 𝑓 𝑎 : 𝐺 → 𝐴 by 𝑓 𝑎 (1) = 𝑎 and 𝑓 𝑎 (𝜎) = 1 for 𝜎 ∈ 𝐺, 𝜎 ≠ 1. Then, 𝛼( 𝑓 𝑎 ) = 𝑎, so 𝛼 is surjective. Next consider 𝜏 ∈ 𝐺. Then, Ö Ö Ö −1 −1 −1 𝑓 (𝜌) 𝜌 𝜏 = 𝛼( 𝑓 ) 𝜏 . 𝑓 (𝜏𝜎) 𝜎 = 𝛼( 𝑓 𝜏 ) = 𝑓 𝜏 (𝜎) 𝜎 = 𝜎 ∈𝐺

𝜎 ∈𝐺

𝜌∈𝐺

Thus, 𝛼 respects the action of 𝐺. Therefore, 𝛼 extends to an epimorphism 𝛼: 𝐺 ⋉ 𝐴𝐺 → 𝐺 ⋉ 𝐴 satisfying 𝛼(𝜎 𝑓 ) = 𝜎𝛼( 𝑓 ) for all 𝜎 ∈ 𝐺 and 𝑓 ∈ 𝐴𝐺 . This gives rise to the following commutative diagram: 1

/ 𝐴𝐺 𝛼

1

 /𝐴

/ 𝐺 ⋉ 𝐴𝐺

/𝐺

/1

/𝐺

/1

𝛼

 /𝐺⋉𝐴

Every element of 𝐺 ⋉ 𝐴𝐺 has the form 𝜎 𝑓 with 𝜎 ∈ 𝐺 and 𝑓 ∈ 𝐴𝐺 . By definition, 𝛼(𝜎 𝑓 ) = 𝜎𝛼( 𝑓 ) ∈ 𝐺 ⋉ 𝐴. Thus, if 𝛼(𝜎 𝑓 ) = 1, then 𝜎 = 1, so 𝜎 𝑓 = 𝑓 ∈ 𝐴𝐺 . Thus, Ker(𝛼) ≤ 𝐴𝐺 . □

Proposition 18.4.4 Let 𝐿/𝐾 be a finite Galois extension of degree 𝑛 with Galois group 𝐺. Suppose that 𝐺 acts on a finite Abelian group 𝐴. Let 𝜋: 𝐺 ⋉ 𝐴 → 𝐺 be the projection map. Then, 𝜋 is regularly solvable. Proof. By Proposition 18.3.5, 𝐾 (𝑇) has a Galois extension 𝑁 such that Gal(𝑁/𝐾 (𝑇))  𝐴 and 𝑁/𝐾 is regular. By Proposition 18.2.8, there exists an 𝑋-stable polynomial ℎ ∈ 𝐾 [𝑇, 𝑋] with Gal(ℎ(𝑇, 𝑋), 𝐾 (𝑇))  𝐴. In particular, ℎ(𝑇, 𝑋) is absolutely irreducible and Gal(ℎ(𝑇, 𝑋), 𝐿(𝑇))  𝐴. Let 𝜋: ˆ 𝐺 ⋉ 𝐴𝐺 → 𝐺 be the projection on 𝐺. Let 𝑡1 , . . . , 𝑡 𝑛 be algebraically independent elements over 𝐾 with 𝑛 = [𝐿 : 𝐾]. Lemma 15.2.1, with 𝐺 0 trivial, gives a Galois extension 𝐹ˆ of 𝐾 (t) ˆ (t)) → 𝐺 ⋉ 𝐴𝐺 with 𝜋ˆ ◦ 𝛾 = res. which contains 𝐿 and an isomorphism 𝛾: Gal( 𝐹/𝐾 ˆ Moreover, 𝐹 is a regular extension of 𝐿. Lemma 18.4.3 gives an epimorphism 𝛼: 𝐺 ⋉ 𝐴𝐺 → 𝐺 ⋉ 𝐴 which is the identity map on 𝐺. Thus, 𝜋ˆ = 𝜋 ◦ 𝛼. Let 𝐹 be the fixed field in 𝐹ˆ of Ker(𝛼 ◦ 𝛾). Then, 𝐿 (t) ⊆ 𝐹 ⊆ 𝐹ˆ and there is an isomorphism 𝛽: Gal(𝐹/𝐾 (t)) → 𝐺 ⋉ 𝐴 with 𝛼 ◦ 𝛾 = 𝛽 ◦ res𝐹/𝐹 and 𝜋 ◦ 𝛽 = res𝐹/𝐿 . ˆ

18.4 Split Embedding Problems with Abelian Kernels

𝐹ˆ 𝐹 𝐾 (t)

𝐺

𝐿 (t)

𝐺

𝐾

𝐿

321

ˆ (t)) Gal( 𝐹/𝐾 ⑤ ⑤ ⑤⑤  res 𝛾 ⑤⑤ ⑤⑤ Gal(𝐹/𝐾 (t)) ⑤⑤ ⑤ ⑤ 𝛽 ④④④ res }⑤⑤  ④④ 𝜋ˆ ④ 𝐺 ⋉ 𝐴𝐺 ④④/④Gal(𝐿 (t)/𝐾 (t)) ④④ 𝛼 ④  }④④ 𝜋 /𝐺 𝐺⋉𝐴

Thus, 𝛽 is a solution of 𝜋 over 𝐾 (t) with 𝐹 being the solution field. By construction, ˆ Since 𝐹/𝐿 ˆ is regular, so is 𝐹/𝐿 (Corollary 3.4.2(b)). 𝐿 ⊆ 𝐹 ⊆ 𝐹. □ Proposition 18.4.5 ([Ike60, p. 126]) Let 𝐾 be a Hilbertian field. Then, every finite split embedding problem over 𝐾 with Abelian kernel is solvable. In particular, every finite Abelian group occurs as a Galois group over 𝐾. Proof. Combine Lemma 18.4.2 and Proposition 18.4.4.

□

Remark 18.4.6 Proposition 18.4.5 does not hold for an arbitrary profinite Abelian group. For example, Z 𝑝 is regular over Q for no 𝑝 (Proposition 18.6.10), but it is not known if every finite 𝑝-group is regular over Q. Nevertheless, every finite 𝑝-group occurs over Q. More generally, every finite solvable group occurs over every global field. This is a theorem of Shafarevich. Its proof does not use Hilbert irreducibility theorem but rather class field theory and complicated combinatorial arguments. See [NSW00, Section 9.5] for a proof that uses cohomological arguments and for a reference to the original articles. Class field theory is not available over an arbitrary Hilbertian field 𝐾. Thus, Shafarevich’s proof does not apply to 𝐾. However, when 𝑝 = char(𝐾) > 0, every finite 𝑝-group occurs over 𝐾. This follows from a theorem of Witt (Remark 18.3.3), but it is not known if each finite 𝑝-group occurs over 𝐾 when 𝐾 is a Hilbertian field of characteristic 0. Amazingly enough, both realization problems raised in Remark 18.4.6 are easy consequences of Shafarevich’s theorem when char(𝐾) > 0. This is the content of the following result: Theorem 18.4.7 Let 𝐺 be a finite 𝑝-group and 𝐾 a field of positive characteristic. Then, 𝐺 is regular over 𝐾. If in addition 𝐾 is Hilbertian, then 𝐺 is realizable over 𝐾. Proof. The second statement of the theorem follows from the first one by Hilbert (Lemma 14.1.1). In order to prove that 𝐺 is regular over every field of positive characteristic, it suffices to prove that 𝐺 is regular over every finite field 𝐾 (Lemma 18.2.1). Assume without loss that 𝐺 is nontrivial. Let 𝑡 be an indeterminate. By Shafarevich, 𝐺 × 𝐺 occurs over 𝐾 (𝑡). Thus, 𝐾 (𝑡) has linearly disjoint Galois extensions 𝐹1 and 𝐹2 with Gal(𝐹𝑖 /𝐾 (𝑡))  𝐺, 𝑖 = 1, 2. Assume none of them is regular over 𝐾.

322

18 Galois Groups over Hilbertian Fields

Then, for 𝑖 = 1, 2, the field 𝐾˜ ∩ 𝐹𝑖 is a cyclic extension of 𝐾 of degree 𝑝 𝑘𝑖 with 𝑘 𝑖 ≥ 1. Hence, 𝐹𝑖 contains the unique extension 𝐾 𝑝 of 𝐾 of degree 𝑝. Therefore, both 𝐹1 and 𝐹2 contain 𝐾 𝑝 (𝑡), which is a proper extension of 𝐾 (𝑡). This contradiction to the linear disjointness of 𝐹1 and 𝐹2 over 𝐾 (𝑡) proves that one of them is regular over 𝐾. □ Corollary 18.4.8 Let 𝐾 be a field, 𝐺 a finite group, and 𝐴 a finite Abelian group. Suppose that 𝐺 is regular over 𝐾 and 𝐺 acts on 𝐴. Then, 𝐺 ⋉ 𝐴 is regular over 𝐾. Proof. There exist algebraically independent elements 𝑡1 , . . . , 𝑡 𝑛 over 𝐾 and a finite Galois extension 𝐸 of 𝐾 (t) such that Gal(𝐸/𝐾 (t))  𝐺 and 𝐸/𝐾 is regular. Choose an indeterminate 𝑢. Proposition 18.4.4 gives a Galois extension 𝐹 of 𝐾 (t, 𝑢) such that Gal(𝐹/𝐾 (t, 𝑢))  𝐺 ⋉ 𝐴 and 𝐹 is a regular extension of 𝐸. By Corollary 3.4.2(a), 𝐹 is a regular extension of 𝐾. Thus, 𝐺 ⋉ 𝐴 is regular over 𝐾. □ Remark 18.4.9 (Realization of 𝑝-groups of low order) The results of this section imply that each 𝑝-group of order at most 𝑝 4 is regular over every field 𝐾. To this end let A be the smallest family of all finite groups satisfying this: (18.11a) Every Abelian group belongs to A. (18.11b) Suppose that 𝐻 = 𝐺 · 𝐴 with 𝐺 ∈ A and 𝐴 Abelian and normal. Then, 𝐻 ∈ A. Note that the group 𝐻 in (18.11b) is a quotient of 𝐺 ⋉ 𝐴. Hence, if 𝐺 is regular over 𝐾, so is 𝐻 (Proposition 18.4.8 and Lemma 18.2.1(a)). It follows by induction on the order of the group that each 𝐺 ∈ A is regular over 𝐾. Thus, if 𝐾 is Hilbertian, every 𝐺 ∈ A occurs over 𝐾 (Lemma 14.1.1(b)). The family A contains each finite group 𝐺 which satisfies one of the following conditions: (18.12a) 𝐺 has nilpotence class at most 2, i.e. [𝐺, 𝐺] ≤ 𝑍 (𝐺) (Thompson [MaM99, p. 277, Prop. 2.9(a)]). (18.12b) 𝐺 is solvable and every Sylow subgroup of 𝐺 is Abelian (Thompson [MaM99, p. 277, Prop. 2.9(b)]). (18.12c) 𝐺 is a 𝑝-group of order at most 𝑝 4 (Dentzer [MaM99, p. 278, Cor. 2.10]). (18.12d) 𝐺 is a 2-group of order 25 (Dentzer [MaM99, p. 278, Cor. 2.10]). However, there are groups of order 𝑝 5 (for 𝑝 ≠ 2) and 26 which do not belong to A [MaM99, p. 278].

18.5 Embedding Quadratic Extensions in Z/2 𝒏 Z-Extensions Nonsplit embedding problems with Abelian√kernel over Hilbertian fields need not be solvable. For example Z/4Z → Gal(Q( √ −1)/Q) is not solvable. Otherwise, Q has a Galois extension 𝑁√ containing Q( −1) with Gal(𝑁/Q)  Z/4Z. The only subfields of 𝑁 are Q, Q( −1), and 𝑁. Hence, 𝑁 ∩ R = Q, so [𝑁R : R] = 4, which is a contradiction.

18.5 Embedding Quadratic Extensions in Z/2𝑛 Z-Extensions

323

Here is a general criterion for a quadratic extension to be embeddable in a Z/4Zextension. Proposition 18.5.1 Let 𝐾 be a field with char(𝐾) √ ≠ 2 and let 𝑎 be a nonsquare in 𝐾. Then, the embedding problem Z/4Z → Gal(𝐾 ( 𝑎)/𝐾) is solvable if and only if there are 𝑥, 𝑦 ∈ 𝐾 with 𝑎 = 𝑥 2 + 𝑦 2 . √ Proof. By assumption, 𝐿 := 𝐾 ( 𝑎) is a quadratic extension of 𝐾. Suppose that 𝐾 has a Galois extension 𝑁 containing 𝐿 with Gal(𝑁/𝐾)  Z/4Z. Then, there is a 𝑢 ∈ 𝐿 √ with 𝑁 = 𝐿( 𝑢). Let 𝜎 be a generator of Gal(𝑁/𝐾). Then, the √𝐾-automorphism √ of 𝐿 defined by 𝑢 ↦→ 𝜎𝑢 = 𝐿( 𝑢) √→ 𝐿 ( 𝜎𝑢). √ extends√to a 𝐾-isomorphism 𝑁 √ Since 𝑁/𝐾 is Galois, 𝐿 ( 𝑢)√= 𝐿 ( √ 𝜎𝑢) and we may choose 𝜎𝑢 as 𝜎 𝑢. It follows from Kummer theory that 𝜎 𝑢 = 𝑣 𝑢 with 𝑣 ∈ 𝐿 [Lan97, p. 295, Thm. 8.2]. Then, √ √ √ 𝜎 2 𝑢 = 𝑣 · 𝜎𝑣 𝑢 = 𝑏 𝑢 with 𝑏 = 𝑣 · 𝜎𝑣. Thus, 𝑏 is an element of 𝐾 which is a 2 generates Gal(𝐿 ( √𝑢)/𝐿), so 𝜎 2 √𝑢 = −√𝑢. norm from 𝐿. On the other hand, 𝜎 √ √ Therefore, 𝑏 = −1, 𝑣 · 𝜎𝑣 𝑢√= − 𝑢, and 𝑣 · 𝜎𝑣 = −1. Finally, write √ 𝑣 = √𝑐 + 𝑑 𝑎 with 𝑐, 𝑑 ∈ 𝐾. √First suppose that 𝑑 = 0. Then, 𝑣 ∈ 𝐾 and 𝑣 2 𝑢 = − 𝑢. Hence, 𝑣 2 = −1 and −1 ∈ 𝐾. Therefore, the identity √
2
2 𝑎 = −1(1 − 𝑎4 ) + 1 + 𝑎4 yields a representation of 𝑎 as a sum of two squares in
2
2 𝐾. Now suppose that 𝑑 ≠ 0. Then, −1 = 𝑣 · 𝜎𝑣 = 𝑐2 − 𝑎𝑑 2 . Hence, 𝑎 = 𝑑𝑐 + 𝑑1 , as desired. Conversely, suppose that 𝑎 = 𝑥 2 + 𝑦 2 with√ 𝑥, 𝑦 ∈ 𝐾. Then, 𝑥, 𝑦 ≠ 0 and −1 = 2 𝑐 −𝑎𝑑 2 , where 𝑐 = 𝑦𝑥 and 𝑑 = 1𝑦 . Put 𝑣 = 𝑐+𝑑 𝑎. Then, norm 𝐿/𝐾 𝑣 = 𝑐2 −𝑎𝑑 2 = −1. Hence, norm 𝐿/𝐾 𝑣 2 = 1. Let 𝜎 be an element of Gal(𝐾) whose restriction to 𝐿 generates Gal(𝐿/𝐾). Hilbert’s Theorem 90 gives 𝑢 ∈ 𝐿 with 𝑣 2 = 𝜎𝑢 𝑢 [Lan97, √ √ 𝜎 𝑢
2 𝜎 𝑢 2 p. 288, Thm. 6.1]. Thus, 𝑣 = √𝑢 and √𝑢 = ±𝑣. Replacing 𝑣 by −𝑣, if √ √ √ √ √ √ √ necessary, we√may assume 𝜎 √ 𝑢=𝑣 √ 𝑢, 𝜎 2 𝑢√ = 𝑣 · 𝜎𝑣 𝑢 = − 𝑢, 𝜎 3 𝑢 =√−𝑣 𝑢, √ √ 2 3 and 𝜎 4 𝑢 = 𝑢. Thus, 𝑢, 𝜎 𝑢, √ 𝜎 𝑢, 𝜎 𝑢 are distinct conjugates of 𝑢 over 𝐾. All of them belong to 𝑁 = 𝐿( 𝑢). Therefore, 𝑁 is a cyclic extension of degree 4 √ of 𝐾 which contains 𝐿. Consequently, Z/4Z → Gal(𝐾 ( 𝑎)/𝐾) is solvable. □ Remark 18.5.2 (Embedding in cyclic extensions of higher order.) It is possible to slightly generalize √ Proposition 18.5.1: Suppose that 𝐾 is a field and 𝑎 is a nonsquare in 𝐾. Then, 𝐾 ( 𝑎) can be embedded in a Z/8Z-extension of 𝐾 if and only if 𝑎 is a sum of two squares in 𝐾 and there are 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 in 𝐾, not all zero, with 𝑥 12 + 2𝑥 22 + 𝑎𝑥32 − 2𝑎𝑥42 = 0. This result is proved in [Kim90, Thm. 3] in a slightly different form. The formulation we give appears in [GJe07, 21◦ ]. Let 𝐾 be a number field and 𝑎 ∈ 𝐾. Suppose that −1 is a √nonsquare in 𝐾 and √ 𝐾 ( −1) is embeddable in a Z/16Z-extension of 𝐾. Then, 𝐾 ( −1) is embeddable in a Z/2𝑛 Z-extension of 𝐾 for each positive integer 𝑛 [GJe07, Thm. 4]. The proof requires class field theory. √ √ As an example, let 𝐾 = Q( −14). Then, 𝐾 ( −1) is embeddable in a√Z/8Zextension of 𝐾 ([AFS89, p. 846, Cor. 4] or [GJe07, Remark 19◦ ]). But 𝐾 ( −1) is not embeddable in a Z/16Z-extension of 𝐾 [GJe07, Example to Proposition 3]. Let 𝑂 𝐾 be the ring of integers of 𝐾. Consider a nonzero prime ideal 𝔭 of 𝑂 𝐾 such that

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18 Galois Groups over Hilbertian Fields

(18.13) −1 is a nonsquare but −1 is the sum of two squares in the completion 𝐾ˆ 𝔭 . Since 𝑋 2 + 𝑌 2 + 1 is absolutely irreducible, −1 is a sum of two squares in F 𝑝 for all large 𝑝. By Hensel’s lemma, −1 is a sum of two squares in 𝐾ˆ 𝔭 for all but finitely √ many 1 ˆ 𝔭. Hence, by Chebotarev, Condition (18.13) holds with density 2 . Then, 𝐾𝔭 ( −1) is embeddable in a Z/2𝑛 Z-extension of 𝐾ˆ 𝔭 for all 𝑛 [GJe07, Thm. 3]. Thus, there could be no criterion for embedding a quadratic extension of a field in a Z/16Z-extension similar to the one we gave above for a Z/8Z-extension. √ embedding in √ Consider now the field 𝐾 = Q( −17). Then, 𝐾 ( −1) is embeddable in a Z/2𝑛 Zextension for each positive integer 𝑛 but is not embeddable in a Z2 -extension [GJe96, p. 371].

18.6 Z 𝒑 -Extensions of Hilbertian Fields Lemma 18.3.2 implies that if 𝐾 is a field of characteristic 𝑝, then every Z/𝑝Zextension of 𝐾 can be embedded in a Z 𝑝 -extension. This result does not generalize to the general case. Remark 18.5.2 gives an example of a quadratic extension 𝐿/𝐾 (with char(𝐾) ≠ 2) which cannot be embedded in a Z2 -extension. So, we have to settle for less. We fix a field 𝐾, a prime number 𝑝, and let 𝑞 = 𝑝 if 𝑝 ≠ 2 and 𝑞 = 4 if 𝑝 = 2. We ask when Z 𝑝 occurs over 𝐾. We prove this is the case when Z/𝑞Z occurs over 𝐾. This condition is satisfied when 𝐾 is Hilbertian. Lemma 18.6.1 Suppose that 𝑝 ≠ char(𝐾). Let 𝐿 be a cyclic extension of 𝐾 of degree 𝑝 𝑛 with 𝑛 ≥ 1. (a) Suppose that 𝜁 𝑝 𝑛+1 ∈ 𝐾. Then, 𝐾 has a Z/𝑝 𝑛+1 Z-extension which contains 𝐿. (b) Suppose that 𝜁 𝑝 𝑘 ∈ 𝐾 for all 𝑘. Then, 𝐾 has a Z 𝑝 -extension that contains 𝐿. Proof. Statement (b) follows from (a) by induction and taking inverse limit. It remains to prove Statement (a). 𝑛 The theory of cyclic extensions gives 𝑎 ∈ 𝐾 with 𝐿 = 𝐾 (𝑎 1/ 𝑝 ) [Lan97, p. 289, 𝑛+1 Thm. 6.2(i)]. In particular, 𝑎 1/ 𝑝 ∉ 𝐾. The field 𝐿 ′ = 𝐾 (𝑎 1/ 𝑝 ) contains 𝐿. By [Lan97, p. 289, Thm. 6.2(ii)], 𝐿 ′ is a cyclic extension of degree 𝑝 𝑘 for some 𝑘 ≤ 𝑛+1 𝑛+1−𝑘 𝑛+1−𝑘 𝑝 𝑛−𝑘 and 𝑎 1/ 𝑝 ∈ 𝐾. If 𝑘 ≤ 𝑛, then 𝑎 1/ 𝑝 = (𝑎 1/ 𝑝 ) ∈ 𝐾. We conclude from this contradiction that 𝑘 = 𝑛 + 1, as desired. □ Remark 18.6.2 (Abelian pro-𝑝 groups as Z 𝑝 -modules) Let 𝐴 be a finite Abelian additive 𝑝-group of exponent 𝑝 𝑚 . Then, Z/𝑝 𝑚 Z acts on 𝐴 through the rule (𝑘 + 𝑝 𝑚 Z)𝑎 = 𝑘𝑎, 𝑎 ∈ 𝐴. Since Z/𝑝 𝑚 Z is a quotient of Z 𝑝 , this defines a continuous action of Z 𝑝 on 𝐴 with the discrete topology. This action commutes with homomorphisms of finite Abelian 𝑝-groups. Hence, it defines a continuous action of Z 𝑝 on projective limits of Abelian 𝑝-groups (also called Abelian pro-𝑝 groups) which commutes with homomorphisms. Consider elements 𝑎 1 , . . . , 𝑎𝑟 of 𝐴. Then, the map Z𝑟𝑝 → 𝐴 given by (𝑧 1 , . . . , 𝑧𝑟 ) ↦→

𝑟 ∑︁ 𝑖=1

𝑧𝑖 𝑎𝑖

18.6 Z 𝑝 -Extensions of Hilbertian Fields

325

Í On the other hand, one is continuous. Its image 𝑟𝑖=1ÍZ 𝑝 𝑎 𝑖 is compact, hence closed. Í can approximate each sum 𝑟𝑖=1 𝑧𝑖 𝑎 𝑖 with 𝑧 𝑖 ∈ Z 𝑝 by sums 𝑟𝑖=1 𝑘 𝑖 𝑎 𝑖 with 𝑘 𝑖 ∈ Z. Therefore, the closed subgroup of 𝐴 generated by 𝑎 1 , . . . , 𝑎𝑟 coincides with the Z 𝑝 -submodule of 𝐴 generated by 𝑎 1 , . . . , 𝑎𝑟 . Lemma 18.6.3 Let 0 → 𝐵 → 𝐴 → Z/𝑝Z → 0 be an exact sequence of Abelian pro-𝑝 groups with 𝐵  Z 𝑝 . Then, either 𝐴  Z 𝑝 or 𝐴 = 𝐵 ⊕ 𝐴0 with 𝐴0  Z/𝑝Z. Proof. Consider 𝐴 as a Z 𝑝 -module (Remark 18.6.2). The exact sequence yields 𝑝 𝐴 ≤ Z 𝑝 . By Lemma 1.4.2(e), 𝑝 𝐴 is generated by one element. On the other hand, 𝐴 is generated by two elements, one which generates 𝐵 and the other with image in Z/𝑝Z generating that module. Assume 𝐴  Z 𝑝 and 𝐴  Z 𝑝 ⊕ Z/𝑝Z. Then, by Proposition 2.2.3, 𝐴  Z 𝑝 ⊕ Z/𝑝 𝑘 Z with 𝑘 ≥ 2 or 𝐴  Z 𝑝 ⊕ Z 𝑝 . Then, again by Lemma 1.4.2(e), 𝑝 𝐴  𝑝Z 𝑝 ⊕ Z/𝑝 𝑘−1 Z  Z 𝑝 ⊕ Z/𝑝 𝑘−1 Z or 𝐴  𝑝Z 𝑝 ⊕ 𝑝Z 𝑝  Z 𝑝 ⊕ Z 𝑝 . In each case 𝑝 𝐴 is not generated by one element, contradicting the preceding paragraph. Therefore, 𝐴  Z 𝑝 or 𝐴  Z 𝑝 ⊕ Z/𝑝Z. In the second case 𝐴 has a subgroup 𝐴0 isomorphic to Z/𝑝Z. Since Z 𝑝 contains no nontrivial closed subgroups of finite order (Lemma 1.4.2(c)), 𝐴0 is not contained in 𝐵. Consequently, 𝐴 = 𝐵 ⊕ 𝐴0 . □ Lemma 18.6.4 Suppose that 𝑝 ≠ char(𝐾). Let 𝐿 = 𝐾 (𝜁 𝑝 𝑛 | 𝑛 = 1, 2, 3, . . .). Then, 𝐿/𝐾 is an Abelian extension and there is a field 𝐾∞ satisfying the following conditions: (a) 𝐾∞ (𝜁𝑞 ) = 𝐿 and 𝐾∞ ∩ 𝐾 (𝜁𝑞 ) = 𝐾. (b) Gal(𝐿/𝐾) = Gal(𝐿/𝐾 (𝜁𝑞 )) × Gal(𝐿/𝐾∞ ). (c) Gal(𝐿/𝐾∞ )  Gal(𝐾 (𝜁𝑞 )/𝐾) and Gal(𝐾 (𝜁𝑞 )/𝐾) is isomorphic to a subgroup of Z/( 𝑝 − 1)Z (resp. Z/2Z) if 𝑝 ≠ 2 (resp. 𝑝 = 2). (d) If 𝐿 ≠ 𝐾 (𝜁𝑞 ), then Gal(𝐾∞ /𝐾)  Gal(𝐿/𝐾 (𝜁𝑞 ))  Z 𝑝 . Proof. Let 𝑛 be a positive integer. Then, 𝐾 (𝜁 𝑝 𝑛 ) is the splitting field of the separable 𝑛 polynomial 𝑋 𝑝 − 1 over 𝐾. Hence, 𝐾 (𝜁 𝑝 𝑛 )/𝐾 is Galois. Embed Gal(𝐾 (𝜁 𝑝 𝑛 )/𝐾) 𝑛 × into (Z/𝑝 Z) by mapping each 𝜎 ∈ Gal(𝐾 (𝜁 𝑝 𝑛 )/𝐾) onto the element 𝑠(𝜎) of (Z/𝑝 𝑛 Z) × satisfying 𝜁 𝑝𝜎𝑛 = 𝜁 𝑝𝑠 (𝑛𝜎) . Thus, 𝐾 (𝜁 𝑝 𝑛 )/𝐾 is Abelian. Hence, 𝐿/𝐾 is also Abelian. Recall that 𝑞 = 𝑝 if 𝑝 ≠ 2 and 𝑞 = 4 if 𝑝 = 2. Thus, Gal(𝐾 (𝜁𝑞 )/𝐾) is isomorphic to a subgroup of Z/( 𝑝 − 1)Z if 𝑝 ≠ 2 and of Z/2Z if 𝑝 = 2. If 𝐾 (𝜁𝑞 ) = 𝐿, then 𝐾∞ = 𝐾 satisfies Conditions (a)–(c). Assume from now on 𝐾 (𝜁𝑞 ) ≠ 𝐿. Let 𝑚 be the maximal positive integer with 𝜁 𝑝 𝑚 ∈ 𝐾 (𝜁𝑞 ). Then, 𝜁 𝑝 𝑚 ≠ 𝑥 𝑝 for all 𝑥 ∈ 𝐾 (𝜁𝑞 ). If 𝑝 = 2, then 𝑚 ≥ 2 and −1 = 𝜁42 is a square in 𝐾 (𝜁𝑞 ), so 𝜁 𝑝 𝑚 ≠ −4𝑦 4 for all 𝑦 ∈ 𝐾 (𝜁4 ). In both cases, consider 𝑛−𝑚 𝑛−𝑚 𝑛 ≥ 𝑚. Then, 𝐾 (𝜁 𝑝 𝑛 ) = 𝐾 (𝜁𝑞 ) (𝜁 𝑝1/𝑚𝑝 ) and 𝑋 𝑝 − 𝜁 𝑝 𝑚 is irreducible over 𝐾 (𝜁𝑞 ) [Lan97, p. 297, Thm. 9.1]. So, [𝐾 (𝜁 𝑝 𝑛 ) : 𝐾 (𝜁𝑞 )] = 𝑝 𝑛−𝑚 . Divide the rest of the proof into two parts.

(18.14)

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18 Galois Groups over Hilbertian Fields

Part A: Suppose that 𝑝 ≠ 2 and 𝑞 = 𝑝. Then, (Z/𝑝 𝑛 Z) ×  Z/( 𝑝 − 1)Z × Z/𝑝 𝑛−1 Z.

(18.15)

Since the factors on the right-hand side of 18.15 are cyclic of relatively prime orders, (Z/𝑝 𝑛 Z) × is a cyclic group. Hence, by the first paragraph of the proof, Gal(𝐾 (𝜁 𝑝 𝑛 )/𝐾) is also cyclic. Let 𝜎 ∈ Gal(𝐾 (𝜁 𝑝 𝑛 )/𝐾 (𝜁 𝑝 )). Then, 𝜁 𝑝𝑠 ( 𝜎) = 𝜁 𝑝𝜎 = 𝜁 𝑝 , so 𝑠(𝜎) ≡ 1 mod 𝑝. 𝑛−1 Hence, 𝑠(𝜎) 𝑝 ≡ 1 mod 𝑝 𝑛 [LeV58, p. 50, Thm. 4-5], so Gal(𝐾 (𝜁 𝑝 𝑛 )/𝐾 (𝜁 𝑝 )) is isomorphic to a subgroup of Z/𝑝 𝑛−1 Z. It follows from (18.14) that Gal(𝐾 (𝜁 𝑝 𝑛 )/𝐾 (𝜁 𝑝 ))  Z/𝑝 𝑛−𝑚 Z. Therefore, Gal(𝐾 (𝜁 𝑝 𝑛 )/𝐾) is cyclic of order 𝑑𝑝 𝑛−𝑚 with 𝑑 = [𝐾 (𝜁 𝑝 ) : 𝐾] dividing 𝑛−𝑚 𝑝 −1. Let 𝜏 be a generator of this group. Then, ord(𝜏 𝑝 ) = 𝑑. Denote the fixed field 𝑛−𝑚 of 𝜏 𝑝 in 𝐾 (𝜁 𝑝 𝑛 ) by 𝐾𝑛 . Then, 𝐾𝑛 ∩ 𝐾 (𝜁 𝑝 ) = 𝐾 and 𝐾𝑛 (𝜁 𝑝 ) = 𝐾 (𝜁 𝑝 𝑛 ). Moreover, 𝐾𝑛 is the unique field with these properties. Hence, 𝐾𝑛 ⊆ 𝐾𝑛′ for 𝑛 ′ ≥ 𝑛 ≥ 𝑚. Now Ð let 𝐾∞ = ∞ 𝑛=𝑚 𝐾 𝑛 . Then, 𝐾∞ ∩ 𝐾 (𝜁 𝑝 ) = 𝐾 and 𝐾∞ (𝜁 𝑝 ) = 𝐿. It follows that Gal(𝐿/𝐾∞ )  Gal(𝐾 (𝜁 𝑝 )/𝐾) and Gal(𝐾∞ /𝐾)  Gal(𝐿/𝐾 (𝜁 𝑝 ))  Z 𝑝 . 𝑛−3

Part B: Suppose that 𝑝 = 2 and 𝑞 = 4. Let 𝑛 ≥ max(3, 𝑚). Then, 52 ≡ 1 + 2𝑛−1 mod 2𝑛 and the order of 5 modulo 2𝑛 is 2𝑛−2 [LeV58, p. 54]. In addition, −1 + 2𝑛 Z does not belong to the subgroup of (Z/2𝑛 Z) × generated by 5 + 2𝑛 Z. Otherwise, 5 𝑘 ≡ −1 mod 2𝑛 for some 1 ≤ 𝑘 < 2𝑛−2 . Raising both sides to an 𝑙+1 odd power, we may assume 𝑘 = 2𝑙 with 1 ≤ 𝑙 ≤ 𝑛 − 3. Then, 52 ≡ 1 mod 2𝑛 . 𝑛−3 ≡ −1 mod 2𝑛 . It follows that Hence, 𝑛 − 2 ≤ 𝑙 + 1, so 𝑙 = 𝑛 − 3 and 52 𝑛 𝑛−1 −1 ≡ 1 + 2 mod 2 , which is a contradiction, because 𝑛 ≥ 3. Consequently, (Z/2𝑛 Z) ×  Z/2Z × Z/2𝑛−2 Z with −1 generating the first factor and 5 generating the second factor. Consider 𝜎 ∈ Gal(𝐾 (𝜁2𝑛 )/𝐾 (𝜁4 )). Then, 𝜁4𝑠 ( 𝜎) = 𝜁4𝜎 = 𝜁4 . Hence, 𝑠(𝜎) ≡ 1 mod 4. By the preceding paragraph, 𝑠(𝜎) ≡ (−1) 𝑖 5 𝑗 mod 2𝑛 with 0 ≤ 𝑖 ≤ 1 and 0 ≤ 𝑗 ≤ 𝑛 − 1. Then, (−1) 𝑖 ≡ 𝑠(𝜎) ≡ 1 mod 4, so 𝑖 = 0 and 𝑠(𝜎) ≡ 5 𝑗 mod 2𝑛 . Thus, Gal(𝐾 (𝜁2𝑛 )/𝐾 (𝜁4 )) is isomorphic to a subgroup of Z/2𝑛−2 Z. By (18.14), Gal(𝐾 (𝜁2𝑛 /𝐾 (𝜁4 )))  Z/2𝑛−𝑚 Z. Taking inverse limit, we get that Gal(𝐿/𝐾 (𝜁4 ))  Z2 . This gives a short exact sequence of Abelian pro-2 groups 0 → Z2 → Gal(𝐿/𝐾) → Gal(𝐾 (𝜁4 )/𝐾) → 1 with Gal(𝐾 (𝜁4 )/𝐾) trivial or isomorphic to Z/2Z. By Lemma 18.6.3, Gal(𝐿/𝐾) = Gal(𝐿/𝐾 (𝜁4 )) × 𝐴0 with 𝐴0 trivial or of order 2. Denote the fixed field of 𝐴0 in 𝐿 □ by 𝐾∞ . It satisfies Conditions (a)–(d). The case where 𝑝 ≠ char(𝐾) and 1 < [𝐿 : 𝐾] < ∞ is the most complicated. We need some concepts and facts from group theory. Let 𝐺 be a profinite group and 𝑥, 𝑦, 𝑧 ∈ 𝐺. Define the commutator of 𝑥, 𝑦 by [𝑥, 𝑦] = 𝑥 −1 𝑦 −1 𝑥𝑦. It satisfies the following identities: [𝑥, 𝑦] −1 = [𝑦, 𝑥], [𝑥, 𝑦] 𝑧 = [𝑥 𝑧 , 𝑦 𝑧 ], [𝑥, 𝑦𝑧] = [𝑥, 𝑧] [𝑥, 𝑦] 𝑧 , [𝑥𝑦, 𝑧] = [𝑥, 𝑧] 𝑦 [𝑦, 𝑧].

(18.16)

18.6 Z 𝑝 -Extensions of Hilbertian Fields

327

The commutator subgroup of 𝐺 is [𝐺, 𝐺] = ⟨[𝑥, 𝑦] | 𝑥, 𝑦 ∈ 𝐺⟩. Suppose that 𝑁 is a closed subgroup of 𝐺 which contains [𝐺, 𝐺]. For each 𝑛 ∈ 𝑁 and 𝑔 ∈ 𝐺 we have 𝑛𝑔 = 𝑛[𝑛, 𝑔]. Hence, 𝑁 ⊳ 𝐺 and 𝐺/𝑁 is Abelian. Conversely, if 𝑁 is a closed normal subgroup of 𝐺 and 𝐺/𝑁 is Abelian, then [𝐺, 𝐺] ≤ 𝑁. Consider now a profinite group 𝐶 and an epimorphism 𝑔: 𝐺 → 𝐶. Suppose that 𝐴 = Ker(𝑔) is Abelian. Define an action of 𝐶 on 𝐴 in the following way: For each 𝛾 ∈ 𝐶 choose 𝛾˜ ∈ 𝐺 with 𝑔( 𝛾) ˜ = 𝛾. Then, let 𝑎 𝛾 = 𝛾˜ −1 𝑎 𝛾˜ for 𝑎 ∈ 𝐴. Since 𝐴 is Abelian, this action is independent of 𝛾. ˜ finite. The is 𝐶 ring Z[𝐶] consists of all formal sums group that Suppose Í 𝛾 with Z. 𝑘 𝑘 is defined componentwise. Multiplication in Z[𝐶] Addition ∈ 𝛾 𝛾 ∈𝐶 𝛾 is a linear extension of multiplication in 𝐶. Thus, ∑︁ ∑︁ ∑︁ ∑︁
𝑘𝛾𝛾 · 𝑙𝛿𝛿 = 𝑘 𝛾 𝑙 𝛿 𝜀. 𝛾 ∈𝐶

𝛿 ∈𝐶

𝜀 ∈𝐶

𝛾 𝛿=𝜀

The action of 𝐶 on 𝐴 naturally extends to an action of Z[𝐶] on 𝐴: Ö Í 𝑎 𝛾∈𝐶 𝑘𝛾 𝛾 = 𝑎 𝑘𝛾 ) 𝛾 . 𝛾 ∈𝐶

Let 𝛾 be an element of 𝐶 with 𝛾 𝑚 = 1. Put 𝑐 = 1 − 𝛾 𝑚 = 0. Hence, (1 − 𝛾 𝑘 )𝑐 =

𝑘−1 ∑︁

𝛾 𝑖 (1 − 𝛾)𝑐 = 0,

Í𝑚−1 𝑖=0

𝛾 𝑖 . Then, (1 − 𝛾)𝑐 =

𝑘 = 0, . . . , 𝑚 − 1.

(18.17)

𝑖=0

Lemma 18.6.5 Let 𝑝 be a prime number, 𝐶 a finite cyclic group, 𝐺 a profinite group, and 𝑔: 𝐺 → 𝐶 an epimorphism. Suppose that 𝐴 := Ker(𝑔) is an Abelian pro-𝑝 group and 𝑓0 : 𝐴 → Z 𝑝 is an epimorphism. Let 𝑞 ′ be a power of 𝑝 and let 𝜋: Z 𝑝 → Z/𝑞 ′Z an epimorphism. Put 𝛼 = 𝜋 ◦ 𝑓0 . Suppose that one of the following conditions holds: (a) 𝑞 ′ ≠ 1, 𝑝 ∤ |𝐶 |, and 𝛼 extends to an epimorphism 𝛽: 𝐺 → Z/𝑞 ′Z × 𝐶. (b) 𝑝 = |𝐶 | = 2, 𝑞 ′ ≥ 4, and 𝛼 extends to an epimorphism 𝛽: 𝐺 → Z/𝑞 ′Z × Z/2Z. (c) 𝑝 = 𝑞 ′ = |𝐶 | = 2 and 𝛼 extends to an epimorphism 𝛽: 𝐺 → Z/4Z. Then, there exists an epimorphism 𝑓 : 𝐴 → Z 𝑝 with Ker( 𝑓 ) ⊳ 𝐺 such that 𝐺/Ker( 𝑓 ) is Abelian. Proof. The first two parts of the proof are common to all cases. The rest of the proof handles each case separately. Part A: The commutator of 𝐺. Since 𝐴 is Abelian, 𝐶 acts on 𝐴 by lifting and conjugating. Extend this to an action of Z[𝐶] on 𝐴. Since 𝐶 is Abelian, [𝐺, 𝐺] ≤ 𝐴. Moreover, [𝐺, 𝐺] = ⟨𝑎 1−𝛾 | 𝑎 ∈ 𝐴, 𝛾 ∈ 𝐶⟩. (18.18) Indeed, choose a generator 𝛾0 of 𝐶 and an element 𝛾˜ 0 in 𝐺 with 𝑔( 𝛾˜ 0 ) = 𝛾0 . For each 𝑖 let 𝛾e0𝑖 = 𝛾˜ 0𝑖 . Then,

328

18 Galois Groups over Hilbertian Fields

˜ = 1 for all 𝛾, 𝛿 ∈ 𝐶 [ 𝛾, ˜ 𝛿]

(18.19)

and 𝑎 1−𝛾 = [ 𝛾, ˜ 𝑎]

for each 𝑎 ∈ 𝐴.

(18.20)

Thus, the right-hand side of (18.18) (which we denote by 𝐺 1 ) is contained in [𝐺, 𝐺]. To prove the other inclusion, it suffices to prove that [𝑢, 𝑣] ∈ 𝐺 1 for all 𝑢, 𝑣 ∈ 𝐺. To this end, write 𝑢 = 𝑎 𝛾˜ and 𝑣 = 𝑏 𝛿˜ with 𝑎, 𝑏 ∈ 𝐴 and 𝛾, 𝛿 ∈ 𝐶. Now use the hypothesis that 𝐴 and 𝐶 are Abelian: (18.16)

˜ = [𝑎 𝛾, ˜ [𝑎 𝛾, [𝑢, 𝑣] = [𝑎 𝛾, ˜ 𝛿] ˜ 𝑏 𝛿] ˜ 𝑏] 𝛿 (18.18),(18.20)

(18.19) ˜ [𝑎, 𝑏] 𝛾 𝛿 [ 𝛾, ˜ [ 𝛾, ˜ 𝛾 [ 𝛾, = [𝑎, 𝛿] ∈ 𝐺1. ˜ 𝑏] 𝛿 = [𝑎 𝛾 , 𝛿] ˜ 𝑏 𝛿] ˜ 𝛿] Í𝑚−1 𝑖 Part B: Twist of 𝐴. Let 𝑚 = |𝐶 | and put 𝑐 = 𝑖=0 𝛾0 . Let 𝜇: 𝐴 → 𝐴 be the homomorphism given by 𝜇(𝑎) = 𝑎 𝑐 . Since 𝛾0𝑚 = 1, (18.17) implies that (1 − 𝛾)𝑐 = 0 for each 𝛾 ∈ 𝐶. Hence, (𝑎 1−𝛾 ) 𝑐 = 𝑎 (1−𝛾)𝑐 = 1 for each 𝑎 ∈ 𝐴. Therefore, by (18.18), 𝜇(𝑤) = 𝑤 𝑐 = 1 for each 𝑤 ∈ [𝐺, 𝐺]. By (18.20) and since 𝛽 is an epimorphism of 𝐺 onto an Abelian group, 𝛼(𝑎 1−𝛾 ) = 𝛽( [ 𝛾, ˜ 𝑎]) = 0. Hence, 𝛼(𝑎 𝛾 ) = 𝛼(𝑎) for all 𝑎 ∈ 𝐴 and 𝛾 ∈ 𝐶. Consequently, Í𝑚−1 𝑖 𝛼(𝑎 𝛾0 ) = 𝑚𝛼(𝑎). 𝛼(𝜇(𝑎)) = 𝛼(𝑎 𝑐 ) = 𝑖=0 (18.16)

Part C: Suppose that 𝑞 ′ > 1 and 𝑝 ∤ |𝐶 |. Then, 𝑚 is an invertible element of Z 𝑝 and Z/𝑞 ′Z. Let 𝑚 −1 : Z 𝑝 → Z 𝑝 and 𝑚 −1 : Z/𝑞 ′Z → Z/𝑞 ′Z be multiplications by 𝑚 −1 . Define a homomorphism 𝑓 : 𝐴 → Z 𝑝 by 𝑓 = 𝑚 −1 ◦ 𝑓0 ◦ 𝜇. By Part B, [𝐺, 𝐺] ≤ Ker(𝜇) ≤ Ker( 𝑓 ). Thus, Ker( 𝑓 ) ⊳ 𝐺 and 𝐺/Ker( 𝑓 ) is Abelian. By Part B, this establishes the following commutative diagram: 𝑓

𝜇

𝐴 𝛼

 Z/𝑞 ′Z

𝑓0 /𝐴 / Z𝑝 ❉❉ ❉❉ ❉❉𝛼 ❉❉ 𝜋 ❉!  𝑚 / Z/𝑞 ′Z

'/

𝑚−1

Z𝑝 𝜋

𝑚−1

 / Z/𝑞 ′Z

In particular, 𝛼 = 𝜋 ◦ 𝑓 . Thus, 𝜋( 𝑓 ( 𝐴)) = 𝛼( 𝐴) = Z/𝑞 ′Z. It follows from Lemma 1.4.3 that 𝑓 is surjective. Part D: Suppose that 𝑝 = |𝐶 | = 2 and 𝑞 ′ ≥ 4. In this case Part B gives the following commutative diagram: 𝜇

𝐴 𝛼

 Z/𝑞 ′Z

𝑓0 /𝐴 / Z2 ❉❉ ❉❉ ❉❉𝛼 ❉❉ 𝜋 ❉"  2 / Z/𝑞 ′Z

Let 𝐻 = 𝑓0 (𝜇( 𝐴)). By Remark 1.2.1(g), 𝐻 is a closed subgroup of Z2 . It satisfies, 𝜋(𝐻) = 2(Z/𝑞 ′Z) = 𝜋(2Z2 ). Hence, 𝐻 + 𝑞 ′Z2 = 2Z2 + 𝑞 ′Z2 = 2Z2 . By Lemma 1.4.2(b),(c), 𝐻 is trivial or 𝐻 = 2𝑛 Z2 for some positive integer 𝑛. It follows from 𝑞 ′ ≥ 4 that 𝐻 = 2Z2 .

18.6 Z 𝑝 -Extensions of Hilbertian Fields

329

Part E: Suppose that 𝑝 = 𝑞 ′ = 2 = |𝐶 | and 𝛼 extends to an epimorphism 𝛽: 𝐺 → Z/4Z. Then, (𝐺 : 𝐴) = |𝐶 | = 2. Hence, the generator 𝛾0 of 𝐶 satisfies 𝛾02 = 1, so its lifting 𝛾˜ 0 to 𝐺 satisfies 𝑎 := 𝛾˜ 02 ∈ 𝐴, So, 𝑎 𝛾0 = 𝑎. Hence, in the notation of Part B, 𝜇(𝑎) = 𝑎 1+𝛾0 = 𝑎𝑎 𝛾0 = 𝑎 2 . The existence of 𝛽 yields a commutative diagram 1

0

/𝐴

/𝐺

𝛼

𝛽

 / Z/2Z

 / Z/4Z

𝑔

/𝐶

/1

𝛽¯

𝜋¯

 / Z/2Z

/0

¯ 𝛾˜ 0 )) = 1 + 2Z. Hence, 𝛽( 𝛾˜ 0 ) = with both rows exact. In particular, 𝜋(𝛽( ¯ 𝛾˜ 0 )) = 𝛽(𝑔( ±1 + 4Z, so 𝜋( 𝑓0 (𝑎)) = 𝛼(𝑎) = 𝛽( 𝛾˜ 02 ) = 1 + 2Z. Hence, by Lemma 1.4.3, ⟨ 𝑓0 (𝑎)⟩ = Z2 . Therefore, ⟨ 𝑓0 (𝜇(𝑎))⟩ = ⟨ 𝑓0 (𝑎 2 )⟩ = 2Z2 . It follows that 𝑓0 (𝜇( 𝐴)) = Z2 or 𝑓0 (𝜇( 𝐴)) = 2Z2 . Suppose first that 𝑓0 (𝜇( 𝐴)) = Z2 . Then, 𝑓 := 𝑓0 ◦ 𝜇 maps 𝐴 onto Z2 . For each 𝑤 ∈ [𝐺, 𝐺], Part B implies that 𝑓 (𝑤) = 𝑓0 (𝜇(𝑤)) = 0. Therefore, [𝐺, 𝐺] ≤ Ker( 𝑓 ), Ker( 𝑓 ) ⊳ 𝐺, and 𝐺/Ker( 𝑓 ) is Abelian. The case 𝑓0 (𝜇( 𝐴)) = 2Z2 (which we henceforth assume), will be handled in Part F. Part F: Conclusion of the proof in cases (b) and (c). Multiplication by 2 gives an isomorphism of Z2 onto 2Z2 . Hence, by Parts D and E, there is an epimorphism 𝑓 : 𝐴 → Z2 with 2 𝑓 = 𝑓0 ◦ 𝜇. For each 𝑤 ∈ [𝐺, 𝐺], Part B implies 2 𝑓 (𝑤) = 𝑓0 (𝜇(𝑤)) = 0. Hence, 𝑓 (𝑤) = 0. Therefore, [𝐺, 𝐺] ≤ Ker( 𝑓 ), Ker( 𝑓 ) ⊳ 𝐺, and 𝐺/Ker( 𝑓 ) is Abelian. □ Theorem 18.6.6 ([Wha57], Thm. 2) Let 𝐾 be a field and 𝑝 a prime number. Put 𝑞 = 𝑝 if 𝑝 ≠ 2 and 𝑞 = 4 if 𝑝 = 2. Suppose that Z/𝑞Z occurs over 𝐾. Then, Z 𝑝 occurs over 𝐾. Proof. Let 𝐾 ′ be a Z/𝑞Z-extension of 𝐾. Suppose first that 𝑝 = char(𝐾). Lemma 18.3.2 embeds 𝐾 ′ in a Z 𝑝 -extension of 𝐾. Assume from now on, 𝑝 ≠ char(𝐾). Let 𝐿 = 𝐾 (𝜁 𝑝 𝑛 | 𝑛 = 1, 2, 3, . . . ). When 𝐿 = 𝐾, Lemma 18.6.1(b) gives a Z 𝑝 extension of 𝐾. When 𝐿 ≠ 𝐾 (𝜁𝑞 ), Lemma 18.6.4(d) gives a Z 𝑝 -extension of 𝐾. Assume from now on, 𝐿 = 𝐾 (𝜁𝑞 ) and 𝐿 ≠ 𝐾. Let 𝐶 = Gal(𝐿/𝐾). Then, 𝐶 is a cyclic group of order which divides 𝑝 − 1 if 𝑝 ≠ 2 and of order 2 if 𝑝 = 2. Put 𝐿 ′ = 𝐿𝐾 ′. Then, as we shall see in the three cases below, 𝐿 ′ is a finite Abelian extension of 𝐾. Moreover, 𝐿 ′/𝐿 is a cyclic extension of degree 𝑞 ′ which is a power of 𝑝. We distinguish between three cases: Case A: 𝑝 ≠ 2. Then, 𝐾 ′/𝐾 is a cyclic extension of degree 𝑝 and [𝐿 : 𝐾] | 𝑝 − 1. Hence, 𝑞 ′ := [𝐿 ′ : 𝐿] = 𝑝 and 𝑝 ∤ |𝐶 |. Case B: 𝑝 = 2 and 𝐿 ∩ 𝐾 ′ = 𝐾. Then, 𝐾 ′/𝐾 is a cyclic extension of degree 4, |𝐶 | = 2, and 𝑞 ′ := [𝐿 ′ : 𝐿] = 4.

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18 Galois Groups over Hilbertian Fields

Case C: 𝑝 = 2 and 𝐿 ∩ 𝐾 ′ ̸ ⊆ 𝐾. Then, 𝐿 ⊆ 𝐾 ′, 𝐾 ′ = 𝐿 ′, and 𝑞 ′ := |𝐶 | = 2. In each case Lemma 18.6.1(b) gives a Z 𝑝 -extension 𝐹 of 𝐿 which contains 𝐿 ′. Denote the compositum of all finite Abelian extensions of 𝐿 of a 𝑝-power order by 𝑁. In particular, 𝐹 ⊆ 𝑁. Since 𝐿/𝐾 is Galois, so is 𝑁/𝐾. Put 𝐺 = Gal(𝑁/𝐾) and 𝐴 = Gal(𝑁/𝐿). Let 𝑓0 be res 𝑁 /𝐹 : Gal(𝑁/𝐿) → Gal(𝐹/𝐿) and let 𝜋 be res𝐹/𝐿′ : Gal(𝐹/𝐿) → Gal(𝐿 ′/𝐿). Then, 𝛼 := 𝜋 ◦ 𝑓0 = res 𝑁 /𝐿′ : Gal(𝑁/𝐿) → Gal(𝐿 ′/𝐿). In Case A, Gal(𝐿 ′/𝐾)  Z/𝑝Z × 𝐶. In Case B, Gal(𝐿 ′/𝐾)  Z/4Z × Z/2Z. In Case C, Gal(𝐿 ′/𝐾)  Z/4Z. In all cases, 𝛽 := res 𝑁 /𝐿′ : Gal(𝑁/𝐾) → Gal(𝐿 ′/𝐾) extends 𝛼. In each case, Lemma 18.6.5 gives an extension 𝐹 ′ of 𝐿 in 𝑁 with 𝐹 ′/𝐾 Abelian and Gal(𝐹 ′/𝐿)  Z 𝑝 . Suppose first that 𝑝 ≠ 2. Then, [𝐿 : 𝐾] | 𝑝 − 1. So, Gal(𝐹 ′/𝐾) has a unique subgroup of order [𝐿 : 𝐾]. This is a special case of a profinite version of the Schur–Zassenhaus theorem ([Hup67, p. 126, Hauptsatz 16.1] or Lemma 25.11.1). Its fixed field 𝐸 in 𝐹 ′ satisfies Gal(𝐸/𝐾)  Z 𝑝 . Now suppose that 𝑝 = 2. Lemma 18.6.3 gives a Z2 -extension 𝐸 of 𝐾 in 𝐹 ′. Consequently, in each case 𝐾 has a Z 𝑝 -extension. □ Corollary 18.6.7 Let 𝐾 be a Hilbertian field and 𝑝 a prime number. Then, Z 𝑝 occurs over 𝐾. Proof. By Corollary 18.3.6, Z/𝑝Z and Z/𝑝 2 Z occur over 𝐾. Hence, by Theorem 18.6.6, Z 𝑝 occurs over 𝐾. □ Remark 18.6.8 The assumption in Theorem 18.6.6 that Z/4Z rather than Z/2Z occurs over 𝐾 is necessary for the theorem to hold. Indeed, Gal(C/R)  Z/2Z but Z2 does not occur over R. Proposition 18.6.10 below shows it is impossible to conclude that Z 𝑝 is regular over 𝐾 in Corollary 18.6.7. Lemma 18.6.9 Let 𝐸 be a field, 𝑝 a prime number, 𝑣 a discrete valuation of 𝐸, and 𝐹 a Z 𝑝 -extension of 𝐸. Suppose that 𝑝 ∤ char( 𝐸¯ 𝑣 ) and 𝐸¯ 𝑣 (𝜁 𝑝 , 𝜁 𝑝2 , 𝜁 𝑝3 , . . .) is an infinite extension of 𝐸¯ 𝑣 . Then, 𝑣 is unramified in 𝐹. Proof. Let 𝑤 be a valuation of 𝐹 extending 𝑣. Denote reduction at 𝑤 by a bar. Assume 𝑤 is ramified over 𝐸. Then, its inertia group 𝐼 𝑤/𝑣 is nontrivial. By Lemma 1.4.2, 𝐼 𝑤/𝑣 is an open subgroup of Gal(𝐹/𝐸). Replace 𝐸 by the fixed field of 𝐼 𝑤/𝑣 in 𝐹, if necessary, to assume 𝐼 𝑤/𝑣 = Gal(𝐹/𝐸). Let 𝑛 be a positive integer. Denote the unique extension of 𝐸 in 𝐹 of degree 𝑝 𝑛 by 𝐸 𝑛 . Let 𝑣 𝑛 be a normalized valuation of 𝐸 𝑛 over 𝑣. It is totally ramified over 𝐸. Consider the completion 𝐸ˆ of 𝐸 under 𝑣. Then, 𝐸ˆ 𝑛 = 𝐸 𝑛 𝐸ˆ is the completion of 𝐸 𝑛 ˆ  Gal(𝐸 𝑛 /𝐸)  Z/𝑝 𝑛 Z [CaF67, p. 41, under 𝑣 𝑛 (Proposition 4.5.3) and Gal( 𝐸ˆ 𝑛 /𝐸) Prop. 3]. Choose 𝑦 ∈ 𝐸ˆ 𝑛 and 𝑥 ∈ 𝐸ˆ with 𝑣 𝑛 (𝑦) = 1 and 𝑣(𝑥) = 1. Then, 𝑣 𝑛 (𝑥) = 𝑝 𝑛 = 𝑛 𝑛 ¯ there is an 𝑣 𝑛 (𝑦 𝑝 ). Hence, 𝑦 𝑝 = 𝑢𝑥 with 𝑢 ∈ 𝐸ˆ 𝑛 and 𝑣 𝑛 (𝑢) = 0. Since 𝐸¯ 𝑛 = 𝐸, ′ −1 ′ ˆ ˆ Since 𝑎 ∈ 𝐸 with 𝑎¯ = 𝑢. ¯ Let 𝑢 = 𝑢𝑎 and 𝑥1 = 𝑎𝑥. Then, 𝑣 𝑛 (𝑢 −1) > 0 and 𝑥 1 ∈ 𝐸. 𝑝𝑛 ¯ ˆ 𝑝 ∤ char( 𝐸 𝑣 ), Hensel’s lemma (Proposition 4.5.2(a)) gives 𝑢 1 ∈ 𝐸 𝑛 with 𝑢 1 = 𝑢 ′.

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𝑛

Put 𝑦 1 = 𝑢 1−1 𝑦. Then, 𝑦 1𝑝 = 𝑥1 . Since 𝐸ˆ 𝑛 /𝐸ˆ is cyclic of degree 𝑝 𝑛 , this implies that 𝜁 𝑝 𝑛 ∈ 𝐸ˆ 𝑛 . Taking residues, we find 𝜁 𝑝 𝑛 ∈ 𝐸¯ 𝑣 . Thus, 𝐸¯ 𝑣 (𝜁 𝑝 , 𝜁 𝑝2 , 𝜁 𝑝3 , . . .) = 𝐸¯ 𝑣 , contrary to the assumption of the lemma. Consequently, 𝑣 is unramified in 𝐹. □ Proposition 18.6.10 Let 𝐾 be a field and 𝑝 a prime number. Suppose that 𝑝 ≠ char(𝐾) and 𝐾 (𝜁 𝑝 , 𝜁 𝑝2 , 𝜁 𝑝3 , . . .) is an infinite extension of 𝐾. Then, Z 𝑝 is not regular over 𝐾. Proof. Assume Z 𝑝 is regular over 𝐾. Then, there are algebraically independent elements 𝑡1 , . . . , 𝑡𝑟 over 𝐾 and there is a Z 𝑝 -extension 𝐹 of 𝐸 := 𝐾 (𝑡1 , . . . , 𝑡𝑟 ). Induction on 𝑟 proves it suffices to consider the case where 𝑟 = 1. Put 𝑡 = 𝑡 1 . Let 𝐸 1 be the unique extension of 𝐸 in 𝐹 of degree 𝑝. Then, 𝐸 1 /𝐾 is regular. Remark 4.6.2(b) gives a prime divisor 𝔭 of 𝐸/𝐾 which is ramified in 𝐸 1 . The valuation 𝑣𝔭 of 𝐸 associated with 𝔭 is discrete (Section 4.1). Its residue field 𝐸¯𝔭 is a finite extension of 𝐾. Hence, 𝐸¯𝔭 (𝜁 𝑝 , 𝜁 𝑝2 , 𝜁 𝑝3 , . . .) is an infinite extension of 𝐾. In addition, char( 𝐸¯𝔭 ) = char(𝐾) ∤ 𝑝. Hence, by Lemma 18.6.9, 𝔭 is unramified in 𝐹. In particular, 𝔭 is unramified in 𝐸 1 . It follows from this contradiction that Z 𝑝 is not □ regular over 𝐾. Corollary 18.6.11 Let 𝐾 be a field and 𝑝 a prime number. Suppose that 𝑝 ≠ char(𝐾) and 𝐾 is finitely generated over its prime field. Then, Z 𝑝 is not regular over 𝐾.

18.7 Symmetric and Alternating Groups over Hilbertian Fields The Galois group of the general polynomial of degree 𝑛 is 𝑆 𝑛 (Example 18.2.5(a)). There is a standard strategy to construct special polynomials with Galois groups 𝑆 𝑛 . Lemma 18.7.1 Let 𝐾 be a field, 𝑡1 , . . . , 𝑡𝑟 algebraically independent elements over 𝐾, and 𝐹 a separable extension of 𝐾 (t) of degree 𝑛. Denote the Galois closure of ˜ 𝐾˜ (t))  𝑆 𝑛 . Then, Gal( 𝐹/𝐾 ˆ Suppose that Gal( 𝐹ˆ 𝐾/ ˆ (t))  𝑆 𝑛 . 𝐹/𝐾 (t) by 𝐹. ˆ (t)) is isomorphic to a Proof. Since [𝐹 : 𝐾 (t)] = 𝑛, the group 𝐺 = Gal( 𝐹/𝐾 subgroup of 𝑆 𝑛 , On the other hand, |𝑆 𝑛 | = [ 𝐹ˆ 𝐾˜ : 𝐾˜ (t)] ≤ [ 𝐹ˆ : 𝐾 (t)] = |𝐺 |. Consequently, 𝐺  𝑆 𝑛 . □ Consider for example the case when 𝑟 = 1. Put 𝑡 = 𝑡1 . Let 𝑥 be a primitive element of 𝐹/𝐾 (𝑡) and 𝑓 ∈ 𝐾 [𝑇, 𝑋] an absolutely irreducible polynomial which is separable and of degree 𝑛 in 𝑋 with 𝑓 (𝑡, 𝑥) = 0. Then, 𝐺˜ = Gal( 𝑓 (𝑡, 𝑋), 𝐾˜ (𝑡)) acts transitively on the 𝑛 distinct roots of 𝑓 (𝑡, 𝑋) in 𝐾g (𝑡). Inspecting inertia groups of prime divisors ˜ 𝐾˜ (𝑡) in 𝐹ˆ 𝐾, ˜ one may prove that 𝐺˜ contains cycles which generate 𝑆 𝑛 . Then, of 𝐹 𝐾/ ˆ (𝑡))  𝑆 𝑛 . by Lemma 18.7.1, Gal( 𝐹/𝐾 ˆ It is a quadratic In the latter case one may consider the fixed field 𝐸 of 𝐴𝑛 in 𝐹. extension of 𝐾 (𝑡). If we prove that 𝐸 is rational over 𝐾, then 𝐴𝑛 becomes regular over 𝐾. We are able to do it, for example, if char(𝐾) ≠ 2 and 𝐾 (𝑡)/𝐾 has at most three prime divisors which ramify in 𝐹, each of degree 1 (Lemma 18.7.5). We start with a valuation-theoretic condition that yields 𝑒-cycles for the Galois group of a polynomial. By an 𝑒-cycle of 𝑆 𝑛 we mean a cycle of length 𝑒.

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Lemma 18.7.2 Let (𝐸, 𝑣) be a discrete valued field, 𝑓 ∈ 𝑂 𝑣 [𝑋] a monic separable irreducible polynomial over 𝐸, and 𝑥 a root of 𝑓 in 𝐸 sep . Denote reduction at 𝑣 by ¯ 𝑒 𝜂(𝑋) with 𝑎 ∈ 𝑂 𝑣 , a bar. Suppose that 𝐸¯ is separably closed, 𝑓¯(𝑋) = (𝑋 − 𝑎) ¯ ∤ 𝑒, and 𝜂 ∈ 𝐸¯ [𝑋] is a monic separable polynomial with 𝜂( 𝑎) 𝑒 ≥ 1, char( 𝐸) ¯ ≠ 0. Suppose that 𝑣 extends to a valuation 𝑤 of 𝐸 (𝑥) with 𝑤(𝑥 − 𝑎) > 0 and 𝑒 𝑤/𝑣 = 𝑒. Then, Gal( 𝑓 , 𝐸) contains an 𝑒-cycle. ˆ 𝑣ˆ ). Embed 𝐸 sep into 𝐸ˆ sep and extend Proof. Denote the completion of (𝐸, 𝑣) by ( 𝐸, ˆ By assumption, 𝜂(𝑋) is 𝑣ˆ to a valuation 𝑣ˆ of 𝐸ˆ sep . Let 𝑂ˆ be the valuation ring of 𝐸. ¯ 𝑒 . Hence, Hensel’s lemma (Proposition 4.5.2(b)) gives relatively prime to (𝑋 − 𝑎) = (𝑋 − 𝑎) ¯ 𝑒 , and a factorization 𝑓 (𝑋) = 𝑔(𝑋)ℎ(𝑋) with 𝑔, ℎ ∈ 𝑂ˆ [𝑋] monic, 𝑔(𝑋) ¯ ¯ = 𝜂(𝑋). ℎ(𝑋) Let 𝑔(𝑋) = 𝑔1 (𝑋) · · · 𝑔 𝑘 (𝑋) with 𝑔1 , . . . , 𝑔 𝑘 distinct monic irreducible polynomials in 𝐸ˆ [𝑋]. For each 𝑖 choose a root 𝑦 𝑖 of 𝑔𝑖 in 𝐸 sep . Then, there is a 𝜎𝑖 ∈ Gal(𝐸) with 𝜎𝑖 𝑥 = 𝑦 𝑖 . The formula 𝑤 𝑖 (𝑧) := 𝑣ˆ (𝜎𝑖 𝑧) for 𝑧 ∈ 𝐸 (𝑥) defines a valuation 𝑤 𝑖 of 𝐸 (𝑥) which extends 𝑣 and with 𝐸ˆ (𝑦 𝑖 ) as completion (Proposition 4.5.3(a),(b)). Similarly, for each root 𝑦 of ℎ(𝑋) there is a 𝜎 ∈ Gal(𝐸) with 𝜎𝑥 = 𝑦 and ¯ 𝑦¯ ) = 0 𝑣 ′ (𝑧) := 𝑣ˆ (𝜎𝑧) for 𝑧 ∈ 𝐸 (𝑥) defines an extension of 𝑣. By assumption, ℎ( ¯ 𝑎) and ℎ( ¯ Hence, 𝑣 ′ (𝑥 − 𝑎) = 𝑣ˆ (𝑦 − 𝑎) = 0. Since 𝑤(𝑥 − 𝑎) > 0, this ¯ ≠ 0, so 𝑦¯ ≠ 𝑎. implies that 𝑤 ≠ 𝑣 ′. Thus, 𝑤 is one of the 𝑤 𝑖 ’s, say 𝑤 = 𝑤 1 (Proposition 4.5.3(c)). Therefore, 𝑒 = 𝑒 𝑤/𝑣 ≤

𝑘 ∑︁ 𝑖=1

𝑒 𝑤𝑖 /𝑣 ≤

𝑘 ∑︁ 𝑖=1

ˆ = [ 𝐸ˆ (𝑦 𝑖 ) : 𝐸]

𝑘 ∑︁

deg(𝑔𝑖 ) = deg(𝑔) = 𝑒.

𝑖=1

ˆ List the roots of 𝑔 in 𝐸 sep as It follows that 𝑘 = 1 and 𝑔 is irreducible over 𝐸. 𝑥1 , . . . , 𝑥𝑒 . ¯ ∤ 𝑒, ˆ = deg(𝑔) = 𝑒. Since char( 𝐸) By the preceding paragraph, [ 𝐸ˆ (𝑥) : 𝐸] 1/𝑒 ˆ ˆ ˆ ˆ ˆ 𝐸 (𝑥)/𝐸 is tamely ramified. Hence, 𝐸 (𝑥) = 𝐸 (𝜋 ) for some 𝜋 ∈ 𝐸 [Lan70, p. 52, ˆ Prop. 12]. Since 𝐸¯ is separably closed, it contains 𝜁𝑒 . By Hensel’s lemma, 𝜁𝑒 ∈ 𝐸. ˆ is a cyclic Hence, 𝐸ˆ (𝑥)/𝐸ˆ is a cyclic group of order 𝑒. In other words, Gal(𝑔, 𝐸) group of order 𝑒. It acts transitively on 𝑥1 , . . . , 𝑥 𝑒 . Therefore, it is generated by an 𝑒-cycle, say (𝑥1 𝑥 2 · · · 𝑥 𝑒 ). ¯ Finally, by assumption, ℎ(𝑋) is separable. Hence, it is a product of distinct monic lemma, ℎ(𝑋) decomposes into distinct monic linear factors in 𝐸¯ [𝑋]. By Hensel’s Î 𝑛 ˆ acts (𝑋 − 𝑥𝑖 ). Therefore, Gal( 𝐸ˆ (𝑥)/𝐸) linear factors in 𝐸ˆ [𝑋], say, ℎ(𝑋) = 𝑖=𝑒+1 ˆ is generated by the cycle (𝑥1 𝑥2 · · · 𝑥 𝑒 ). trivially on 𝑥 𝑒+1 , . . . , 𝑥 𝑛 . Thus, Gal( 𝑓 , 𝐸) ˆ ≤ Gal( 𝑓 , 𝐸), the group Gal( 𝑓 , 𝐸) contains an 𝑒-cycle □ Since Gal( 𝑓 , 𝐸) Lemma 18.7.3 Let 𝐾 be a field, 𝑞 ∈ 𝐾 [𝑋] with 𝑛 := deg(𝑞), and 𝑥 a transcendental element over 𝐾. Suppose that char(𝐾) ∤ 𝑛. Put 𝑡 = 𝑞(𝑥) and 𝑓 (𝑇, 𝑋) = 𝑞(𝑋) − 𝑇. Then: (a) 𝑓 (𝑡, 𝑋) is irreducible over 𝐾˜ (𝑡), separable in 𝑋, and [𝐾 (𝑥) : 𝐾 (𝑡)] = 𝑛. (b) Denote the derivative of 𝑞 by 𝑞 ′. Let 𝜑 be a 𝐾-place of 𝐾 (𝑡). Denote the corresponding prime divisor of 𝐾 (𝑡)/𝐾 by 𝔭. Suppose that 𝜑(𝑡) ≠ ∞ and 𝜑(𝑡) ≠ 𝑞(𝑏) for each root 𝑏 of 𝑞 ′ (𝑋). Then, 𝔭 is unramified in 𝐾 (𝑥). (c) Gal( 𝑓 (𝑡, 𝑋), 𝐾˜ (𝑡)) contains an 𝑛-cycle.

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(d) Suppose that there is 𝑏 ∈ 𝐾˜ such that 𝑞(𝑋) − 𝑞(𝑏) = (𝑋 − 𝑏) 𝑒 𝜂(𝑋), char(𝐾) ∤ 𝑒, 𝜂(𝑋) ∈ 𝐾˜ [𝑋], and 𝜂(𝑏) ≠ 0. Then, Gal( 𝑓 (𝑡, 𝑋), 𝐾˜ (𝑡)) contains an 𝑒-cycle. ˜ 𝜆 ∈ 𝐾˜ [𝑋], and 𝜆(𝑏) ≠ 0. Then, (e) Suppose that 𝑞 ′ (𝑋) = (𝑋 − 𝑏)𝜆(𝑋) with 𝑏 ∈ 𝐾, Gal( 𝑓 (𝑡, 𝑋), 𝐾˜ (𝑡)) contains a 2-cycle. Proof. Let 𝑐 be the coefficient of 𝑋 𝑛 in 𝑞. Replace 𝑥 by 𝑐𝑥 and 𝑡 by 𝑐 𝑛−1 𝑡 to assume 𝑞 is monic. Proof of (a). Since 𝑓 (𝑇, 𝑋) is linear and monic in 𝑇, it is absolutely irreducible. Hence, 𝑓 (𝑡, 𝑋) is irreducible over 𝐾 (𝑡) and over 𝐾˜ (𝑡). Therefore, [𝐾 (𝑥) : 𝐾 (𝑡)] = 𝑛. The separability of 𝑓 (𝑡, 𝑋) follows from the assumption char(𝐾) ∤ 𝑛. Proof of (b). Extend 𝜑 to a place 𝜓 of 𝐾 (𝑥). Then, 𝜑(𝑡) = 𝜓(𝑞(𝑥)) = 𝑞(𝜓(𝑥)). Hence, by assumption, 𝜓(𝑞 ′ (𝑥)) = 𝑞 ′ (𝜓(𝑥)) ≠ 0. Therefore, the value of 𝜑 at the product of all 𝐾 (𝑡)-conjugates of 𝑞 ′ (𝑥) is not 0. In other words, 𝜑(𝑁 𝐾 ( 𝑥)/𝐾 (𝑡) 𝑞 ′ (𝑥)) ≠ 0. By (a), 𝑓 (𝑡, 𝑋) = irr(𝑥, 𝐾 (𝑡)). In addition, 𝑞 ′ (𝑥) = 𝜕𝑓 𝜕𝑋 (𝑡, 𝑥). It follows from Lemma 7.1.8(b) that 𝔭 is unramified in 𝐾 (𝑥). Proof of (c). Let 𝑞(𝑋) = 𝑋 𝑛 + 𝑐 𝑛−1 𝑋 𝑛−1 + · · · + 𝑐 0 with 𝑐 0 , . . . , 𝑐 𝑛−1 ∈ 𝐾. Then, 𝑥 𝑛 + 𝑐 𝑛−1 𝑥 𝑛−1 + · · · + 𝑐 0 = 𝑡. Let 𝑣 ∞ be the valuation of 𝐾˜ (𝑡)/𝐾˜ with 𝑣 ∞ (𝑡) = −1 and let 𝑤 be an extension of 𝑣 ∞ to 𝐾˜ (𝑥). Put 𝑒 = 𝑒 𝑤/𝑣∞ . By Example 2.3.11, 𝑤(𝑥) = −1 and 𝑒 = 𝑛. Now make a change of variables: 𝑢 = 𝑡 −1 and 𝑦 = 𝑡 −1 𝑥. Put 𝑔(𝑈, 𝑌 ) = 𝑌 𝑛 + 𝑐 𝑛−1𝑈𝑌 𝑛−1 + · · · + 𝑐 0𝑈 𝑛 − 𝑈 𝑛−1 . Then, 𝐾˜ (𝑢) = 𝐾˜ (𝑡), 𝐾˜ (𝑦) = 𝐾˜ (𝑥), 𝑤(𝑢) = 𝑛, 𝑤(𝑦) = 𝑛 − 1, 𝑔(𝑢, 𝑌 ) is irreducible over 𝐾˜ (𝑢), separable in 𝑌 , 𝑔(𝑢, 𝑦) = 0, and 𝑔(0, 𝑌 ) = 𝑌 𝑛 . Assume without loss that 𝑛 ≥ 2, so that 𝑤(𝑦) > 0. By Lemma 18.7.2, with 𝑔 replacing 𝑓 of that lemma, Gal(𝑔(𝑢, 𝑌 ), 𝐾˜ (𝑢)) contains an 𝑛-cycle. Therefore, Gal( 𝑓 (𝑡, 𝑋), 𝐾˜ (𝑡)) which is isomorphic to Gal(𝑔(𝑢, 𝑌 ), 𝐾˜ (𝑢)) has an 𝑛-cycle. Proof of (d). Put 𝑎 = 𝑞(𝑏). Let 𝑣 𝑎 be the valuation of 𝐾˜ (𝑡)/𝐾˜ with 𝑣 𝑎 (𝑡 − 𝑎) = 1. By assumption, 𝑓 (𝑎, 𝑋) = (𝑋 − 𝑏) 𝑒 𝜂(𝑋) with 𝜂 ∈ 𝐾˜ [𝑋], 𝜂(𝑏) ≠ 0, and char(𝐾) ∤ 𝑒. Let 𝐸ˆ := 𝐾˜ ((𝑡 − 𝑎)) be the completion of ( 𝐾˜ (𝑡), 𝑣 𝑎 ). Let 𝑂ˆ be the valuation ring ˆ By Hensel’s lemma, 𝑓 (𝑡, 𝑋) = 𝑔(𝑋)ℎ(𝑋) with 𝑔, ℎ monic polynomials of 𝐸. in 𝑂ˆ [𝑋] whose residues at 𝑣 𝑎 are (𝑋 − 𝑏) 𝑒 , 𝜂(𝑋), respectively. Embed 𝐾 (𝑥) in 𝐸ˆ sep such that 𝑥¯ = 𝑏. Let 𝑤 be the corresponding extension of 𝑣 𝑎 to 𝐾 (𝑥). Then, ˆ ≤ deg(𝑔) = 𝑒. 𝑒 𝑤/𝑣𝑎 ≤ [ 𝐸ˆ (𝑥) : 𝐸] By the preceding paragraph, (𝑥 − 𝑏) 𝑒 𝜂(𝑥) = 𝑞(𝑥) − 𝑎 = 𝑡 − 𝑎. Since 𝜂(𝑏) ≠ 0, we have 𝑤(𝜂(𝑥)) = 0. Hence, 𝑒𝑤(𝑥 − 𝑏) = 𝑤(𝑡 − 𝑎) = 𝑒 𝑤/𝑣𝑎 ≤ 𝑒. Thus, 𝑤(𝑥 − 𝑏) = 1 and 𝑒 𝑤/𝑣𝑎 = 𝑒. We conclude from Lemma 18.7.2 that Gal( 𝑓 (𝑡, 𝑋), 𝐾˜ (𝑡)) has an 𝑒-cycle. Proof of (e). Write 𝑞(𝑋) − 𝑞(𝑏) = (𝑋 − 𝑏) 𝑒 𝜂(𝑋) with 𝑒 ≥ 1, 𝜂 ∈ 𝐾˜ [𝑋], and 𝜂(𝑏) ≠ 0. Then,
𝑞 ′ (𝑋) = (𝑋 − 𝑏) 𝑒−1 𝑒𝜂(𝑋) + (𝑋 − 𝑏)𝜂 ′ (𝑋) . (18.21) If 𝑒 = char(𝐾), then 𝑞 ′ (𝑋) = (𝑋 − 𝑏) 𝑒 𝜂 ′ (𝑋). Hence, 𝑒 = 1, which is impossible. Therefore, 𝑒 ≠ char(𝐾), so the second factor on the right-hand side of (18.21) does not vanish on 𝑏. In addition, 𝑞 ′ (𝑋) = (𝑋 − 𝑏)𝜆(𝑋) with 𝜆(𝑏) ≠ 0. Hence, 𝑒 = 2. Consequently, by (d), Gal( 𝑓 (𝑡, 𝑋), 𝐾˜ (𝑡)) has a 2-cycle. □

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Lemma 18.7.4 Let 𝐾 be a field, 𝑛 ≥ 2 an integer, 𝑥 an indeterminate, and 𝑡 = 𝑥 𝑛 − 𝑥 𝑛−1 . Suppose that char(𝐾) ∤ (𝑛 − 1)𝑛. Then, 𝑋 𝑛 − 𝑋 𝑛−1 − 𝑡 is separable, irreducible over both 𝐾 (𝑡) and 𝐾˜ (𝑡), and Gal(𝑋 𝑛 − 𝑋 𝑛−1 − 𝑡, 𝐾 (𝑡)) = Gal(𝑋 𝑛 − 𝑋 𝑛−1 − 𝑡, 𝐾˜ (𝑡)) = 𝑆 𝑛 . Moreover, 𝐾 (𝑡) has at most three prime divisors which ramify in 𝐾 (𝑥). Each of them has degree 1. Proof. The polynomial 𝑞(𝑋) = 𝑋 𝑛 − 𝑋 𝑛−1 factors as 𝑞(𝑋) = 𝑋 𝑛−1 (𝑋 − 1). By Lemma 18.7.3(d) (with 𝑏 = 0), Gal(𝑋 𝑛 − 𝑋 𝑛−1 − 𝑡, 𝐾˜ (𝑡)) contains an (𝑛 − 1)-cycle. The derivative of 𝑞(𝑋) is 𝑞 ′ (𝑋) = 𝑛𝑋 𝑛−2 (𝑋 − 𝑛−1 𝑛 ). By Lemma 18.7.3(e) (with 𝑛 − 𝑋 𝑛−1 − 𝑡, 𝐾 ˜ (𝑡)) contains a 2-cycle. Hence, by [Wae37, p. 201], 𝑏 = 𝑛−1 ), Gal(𝑋 𝑛 Gal(𝑋 𝑛 − 𝑋 𝑛−1 − 𝑡, 𝐾˜ (𝑡)) = 𝑆 𝑛 . It follows from Lemma 18.7.1 that Gal(𝑋 𝑛 − 𝑋 𝑛−1 − 𝑡, 𝐾 (𝑡)) = 𝑆 𝑛 .
𝑛−1 The derivative 𝑞 ′ (𝑋) has exactly two roots, 0 and 𝑛−1 and 𝑛 . Put 𝑎 = 𝑞 𝑛 observe that 𝑞(0) = 0. Let 𝜑 be a 𝐾-place of 𝐾 (𝑡) and 𝔭 the prime divisor of 𝐾 (𝑡)/𝐾 corresponding to 𝜑. Suppose first that 𝜑(𝑡) ≠ 𝑎, 0, ∞. By Lemma 18.7.3(b), 𝔭 is unramified in 𝐾 (𝑥). Thus, 𝔭 ramifies in 𝐾 (𝑥) in at most three cases, when 𝜑(𝑡) is 𝑎, or 0, or ∞. In each of these cases deg(𝔭) = 1. □ Lemma 18.7.5 Let 𝐾 be a field, 𝑡 an indeterminate, and 𝐸 a quadratic extension of 𝐾 (𝑡). Suppose that char(𝐾) ≠ 2, 𝐸/𝐾 is regular, and 𝐾 (𝑡)/𝐾 has at most three prime divisors which ramify in 𝐸, each of degree 1. Then, 𝐸 = 𝐾 (𝑢) with 𝑢 transcendental over 𝐾. Moreover, 𝐾 (𝑡)/𝐾 has exactly two prime divisors which are ramified in 𝐸 and each of them is of degree 1. Proof. Denote the genus of 𝐸 by 𝑔. Let 𝔭1 , . . . , 𝔭𝑘 be the prime divisors of 𝐾 (𝑡)/𝐾 which ramify in 𝐸. By assumption, 𝑘 ≤ 3 and deg(𝔭𝑖 ) = 1, 𝑖 = 1, . . . , 𝑘. Since [𝐸 : 𝐾 (𝑡)] = 2, each 𝔭𝑖 extends uniquely to a prime divisor 𝔮𝑖 of 𝐸/𝐾 of degree 1. Since char(𝐾) ≠ 2, the ramification of 𝔭𝑖 in 𝐸 is tame. By Riemann–Hurwitz (Remark 4.6.2(c)), 2𝑔 − 2 = −4 + 𝑘 ≤ −1. Hence, 𝑔 ≤ 12 . This implies that 𝑔 = 0, so 𝑘 = 2. It follows from Example 4.2.4, third paragraph, that 𝐸 = 𝐾 (𝑢) with 𝑢 transcendental over 𝐾. □ Proposition 18.7.6 Let 𝐾 be a field and 𝑛 ≥ 2 an integer with char(𝐾) ∤ (𝑛 − 1)𝑛. Then, 𝐴𝑛 is regular over 𝐾. Specifically, there is a tower of fields 𝐾 (𝑡) ⊆ 𝐾 (𝑢) ⊆ 𝐹 satisfying: 𝐹/𝐾 is regular, 𝐹/𝐾 (𝑡) is Galois, Gal(𝐹/𝐾 (𝑡))  𝑆 𝑛 , Gal(𝐹/𝐾 (𝑢))  𝐴𝑛 , and 𝐾 (𝑡)/𝐾 has at most three prime divisors which ramify in 𝐹, each of degree 1. Proof. Let 𝑡 and 𝑥 be transcendental elements over 𝐾 with 𝑥 𝑛 −𝑥 𝑛−1 = 𝑡. By Lemma 18.7.4, 𝐾 (𝑥)/𝐾 (𝑡) is separable. Denote the Galois closure of 𝐾 (𝑥)/𝐾 (𝑡) by 𝐹. By ˜ 𝐾˜ (𝑡))  𝑆 𝑛 . In particular, 𝐹/𝐾 is regular. Lemma 18.7.4, Gal(𝐹/𝐾 (𝑡))  Gal(𝐹 𝐾/ In addition, 𝐾 (𝑡)/𝐾 has at most three prime divisors 𝔭1 , 𝔭2 , 𝔭3 which ramify in 𝐾 (𝑥), each of degree 1. Denote the fixed field of 𝐴𝑛 in 𝐹 by 𝐸. Then, 𝐸 is a quadratic extension of 𝐾 (𝑡). By Corollary 2.3.7(c), 𝔭1 , 𝔭2 , 𝔭3 are the only prime divisors of 𝐾 (𝑡)/𝐾 which may ramify in 𝐹. Hence, they are the only prime divisors of 𝐾 (𝑡)/𝐾 which may ramify in 𝐸. By Lemma 18.7.5, 𝐸 = 𝐾 (𝑢) with 𝑢 transcendental over 𝐾 and Gal(𝐹/𝐾 (𝑢))  𝐴𝑛 , as desired. □

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Next we use the √ polynomial 𝑋 𝑛 − 𝑋 𝑛−1 − 𝑡 to solve embedding problems of the form 𝑆 𝑛 → Gal(𝐾 ( 𝑎)/𝐾) when 𝐾 is Hilbertian. Remark 18.7.7 (On the discriminant of a polynomial) LetÎ𝐾 be a field and 𝑓 ∈ 𝐾 [𝑋] a 𝑛 monic separable polynomial of degree 𝑛. Write 𝑓 (𝑋) = 𝑖=1 (𝑋 −𝑥𝑖 ) with 𝑥 𝑖 ∈ 𝐾sep . Put 𝑁 = 𝐾 (𝑥1 , . . . , 𝑥 𝑛 ) and embed Gal(𝑁/𝐾) into 𝑆 𝑛 by 𝜎(𝑖) = 𝑗 if 𝜎(𝑥𝑖 ) = 𝑥 𝑗 . Formula (7.1) for the discriminant disc( 𝑓 ) of 𝑓 is disc( 𝑓 ) = (−1)

𝑛(𝑛−1) 2

Ö

(𝑥 𝑖 − 𝑥 𝑗 ) = (−1)

𝑛(𝑛−1) 2

𝑖≠ 𝑗

𝑛 Ö

𝑓 ′ (𝑥 𝑗 ),

(18.22)

𝑗=1

with 𝑓 ′ the derivative of 𝑓 . It can be rewritten as disc( 𝑓 ) = Ö √︁ disc( 𝑓 ) = (𝑥𝑖 − 𝑥 𝑗 ).

Î

𝑖< 𝑗 (𝑥 𝑖

− 𝑥 𝑗 ) 2 . Thus,

𝑖< 𝑗

For each 𝜎 ∈ Gal(𝑁/𝐾) we have Ö √︁ √︁ 𝜎( disc( 𝑓 )) = (𝑥 𝜎 (𝑖) − 𝑥 𝜎 ( 𝑗) ) = (−1) sgn( 𝜎) disc( 𝑓 ), 𝑖< 𝑗

where sgn(𝜎) is the number of pairs (𝑖, 𝑗) with 𝑖 < 𝑗 and 𝜎(𝑖) > 𝜎( 𝑗). If char(𝐾) ≠ 2, then (−1) sgn( 𝜎) ≠ 1 for odd 𝜎 and (−1) sgn( 𝜎) = 1 for even 𝜎. In this case, √︁
Gal 𝑁/𝐾 ( disc( 𝑓 ) ) = 𝐴𝑛 ∩ Gal(𝑁/𝐾). √︁ In particular, if Gal(𝑁/𝐾) = 𝑆 𝑛 , then 𝐾 ( disc( 𝑓 )) is the unique quadratic extension √︁ of 𝐾 in 𝑁 and Gal(𝑁/𝐾 ( disc( 𝑓 ))) = 𝐴𝑛 . Î ˜ Suppose now that char(𝐾) ∤ 𝑛. Then, 𝑓 ′ (𝑋) = 𝑛 𝑛−1 𝑗=1 (𝑋 − 𝑦 𝑗 ) with 𝑦 𝑗 ∈ 𝐾. Substituting in (18.22) and changing the order of multiplication gives an alternative formula for the discriminant: disc( 𝑓 ) = (−1)

𝑛(𝑛+1) 2

𝑛𝑛

𝑛+1 Ö

𝑓 (𝑦 𝑘 ).

(18.23)

𝑘=1

The proof of part (a) of the following result gives an explicit polynomial over 𝐾 (𝑡) with Galois group 𝐴𝑛 if char(𝐾) ∤ (𝑛 − 1)𝑛. Proposition 18.7.8 (Brink) Let 𝐾 be a Hilbertian field, let 𝑎 be a nonsquare in 𝐾, and let 𝑛 ≥ 2 be an integer with char(𝐾) ∤ (𝑛 − 1)𝑛. Then, (a) 𝐴𝑛 is regular over 𝐾 and √ (b) the embedding problem 𝑆 𝑛 → Gal(𝐾 ( 𝑎)/𝐾) is solvable. Proof. Let 𝑡 be an intermediate, 𝑓 (𝑡, 𝑋) = 𝑋 𝑛 − 𝑋 𝑛−1 −𝑡, and 𝐹 be the splitting field of 𝑓 (𝑡, 𝑋) over 𝐾 (𝑡). 18.7.4, Gal(𝐹/𝐾 (𝑡))  𝑆 𝑛 . Since char(𝐾) ∤ 𝑛, we  By Lemma  𝑛−2 . Hence, by (18.23), have 𝑓 ′ (𝑡, 𝑋) = 𝑛 𝑋 − 𝑛−1 𝑋 𝑛

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𝑛 − 1 𝑛𝑛 𝑓 𝑓 (0) 𝑛−2 𝑛 h  𝑛 − 1  𝑛  𝑛 − 1  𝑛−1 i 𝑛(𝑛+1) = (−1) 2 𝑛𝑛 − − 𝑡 (−𝑡) 𝑛−2 𝑛 𝑛 h  𝑛 − 1  𝑛−1 1 i 𝑛(𝑛+1) = (−1) 2 𝑛𝑛 − − 𝑡 (−𝑡) 𝑛−2 𝑛 𝑛

disc( 𝑓 ) = (−1)

𝑛(𝑛+1) 2

(𝑛+1) (𝑛+2)

2 = (−1) [(𝑛 − 1) 𝑛−1 𝑡 𝑛−2 + 𝑛𝑛 𝑡 𝑛−1 ]. √︁ By Remark 18.7.7, 𝐾 ( disc( 𝑓 )) is a quadratic extension of 𝐾 (𝑡) in 𝐹 and √︁
Gal 𝐹/𝐾 ( disc( 𝑓 ) )  𝐴𝑛 . Let (𝑛+1) (𝑛+2)   2   disc( 𝑓 )𝑡 1−𝑛 = (−1) [(𝑛 − 1) 𝑛−1 𝑡 −1 + 𝑛𝑛 ] if 𝑛 is odd 𝑢= (𝑛+1) (𝑛+2)  2  disc( 𝑓 )𝑡 2−𝑛 = (−1) [(𝑛 − 1) 𝑛−1 + 𝑛𝑛 𝑡] if 𝑛 is even  √ and 𝑢 ′ = 𝑢. Then, 𝐾 (𝑢) = 𝐾 (𝑡), so both 𝑢 and 𝑢 ′ are transcendental √︁ over 𝐾. Moreover, 𝑢 is a product of disc( 𝑓 ) with a square of 𝐾 (𝑡), so 𝐾 (𝑢 ′) = 𝐾 ( disc( 𝑓 ) ). Thus, Gal(𝐹/𝐾 (𝑢 ′))  𝐴𝑛 . This proves (a). Next, express 𝑡 in terms of 𝑢 and substitute into 𝑓 (𝑡, 𝑋) to get the irreducible polynomial for the roots of 𝑓 (𝑡, 𝑋) in terms of 𝑢:  𝑛 (𝑛−1) 𝑛−1  if 𝑛 is odd   𝑋 − 𝑋 𝑛−1 − (−1) (𝑛+1) (𝑛+2) /2 𝑢−𝑛𝑛 𝑔(𝑢, 𝑋) = (𝑛+1) (𝑛+2) /2 𝑛−1 𝑢−(𝑛−1)   𝑋 𝑛 − 𝑋 𝑛−1 − (−1) if 𝑛 is even. 𝑛𝑛 

Then, Gal(𝑔(𝑢, 𝑋), 𝐾 (𝑢)) =√𝑆 𝑛 and Gal(𝑔((𝑢 ′) 2 , 𝑋), 𝐾 (𝑢 ′)) = 𝐴𝑛 . Let 𝑣 = 𝑢𝑎 and 𝑣 ′ = 𝑢 ′ 𝑎. The map 𝑢 ↦→ 𝑣 extends to an automorphism of 𝐾 (𝑢) over 𝐾. This automorphism further extends to an isomorphism of 𝐾 (𝑢 ′) onto 𝐾 (𝑣 ′) and further to an isomorphism of 𝐹 onto a field 𝐹 ′. Thus, Gal(𝐹 ′/𝐾 (𝑣))  𝑆 𝑛 and Gal(𝐹 ′/𝐾 (𝑣 ′))  𝐴𝑛 . Since 𝑎 is not a square in 𝐾, there is no 𝑐 ∈ 𝐾 with 𝑣 = 𝑐2 𝑢. Hence, 𝐾 (𝑢 ′) and 𝐾 (𝑣 ′) are distinct quadratic extensions of 𝐾 (𝑢). Therefore, 𝐾 (𝑢 ′) ̸ ⊆ 𝐹 ′. By definition, 𝑣 = 𝑢𝑎 = (𝑢 ′) 2 𝑎, so 𝐾 (𝑣) ⊆ 𝐾 (𝑢 ′) and 𝐹 ′ ∩ 𝐾 (𝑢 ′) = 𝐾 (𝑣). Put 𝐹 ′′ = 𝐾 (𝑢 ′)𝐹 ′. Then, Gal(𝐹 ′′/𝐾 (𝑢 ′))  Gal(𝐹 ′/𝐾 (𝑣)) √ 𝑆 𝑛 ′ ′ ′ and Gal(𝐹 ′′/𝐾 (𝑢 ′, 𝑣 ′))  Gal(𝐹 ′/𝐾 (𝑣 ′))  𝐴𝑛 . In addition, √ 𝐾 (𝑢 , 𝑣 ) = 𝐾 (𝑢 , 𝑎). By Lemma 18.4.2, the embedding problem 𝑆 𝑛 → Gal(𝐾 ( 𝑎)/𝐾) is solvable. □

18.8 GAR-Realizations Let 𝐾 be a field, 𝐿 a finite Galois extension of 𝐾, and 𝛼

1 −→ 𝐶 −→ 𝐺 −→ Gal(𝐿/𝐾) −→ 1

(18.24)

an embedding problem over 𝐾 (Definition 18.4.1) with kernel 𝐶 having a trivial center. We give a sufficient condition for the embedding problem to have a regular solution. As usual, Aut(𝐶) denotes the group of all automorphisms of 𝐶. For each 𝑐 ∈ 𝐶 let 𝜄(𝑐) be the inner automorphism of 𝐶 induced by conjugation with 𝑐. The map 𝑐 ↦→ 𝜄(𝑐) identifies 𝐶 with the group Inn(𝐶) of all inner automorphisms of 𝐶.

18.8 GAR-Realizations

337

Let 𝐹/𝐾 be an extension of fields. We say that 𝐹 is rational over 𝐾 (or 𝐹 is 𝐾-rational) if 𝐹 = 𝐾 (𝑇) with 𝑇 being a set of algebraically independent elements over 𝐾. Definition 18.8.1 (GAR-Realizations) Let 𝐶 be a finite group with a trivial center. We say that 𝐶 is GA over 𝐾 if there are algebraically independent elements 𝑡 1 , . . . , 𝑡𝑟 over 𝐾 satisfying: (18.25) 𝐾 (t) has a finite extension 𝐹 which is regular over 𝐾 such that Aut(𝐶) acts on 𝐹 with 𝐾 (t) being the fixed field of 𝐶. Denote the fixed field of Aut(𝐶) in 𝐹 by 𝐸. We say that 𝐶 is GAR over 𝐾 if in addition to (18.25): (18.26) Every extension 𝐸 ′ of 𝐸 satisfying 𝐸 ′ 𝐾sep = 𝐾sep (t) is a purely transcendental extension of 𝐸 ′ ∩ 𝐾sep . 𝐹 𝐶 Aut(𝐶)

𝐾sep (t)= 𝐸 ′ 𝐾sep ♠ ♠ ♠ ♠♠♠ ♠♠♠ ♠ ♠ ♠♠♠ 𝐸 ′= (𝐸 ′ ∩ 𝐾sep ) (u)

𝐾 (t)

𝐸

𝐾

𝐸 ′ ∩ 𝐾sep

𝐾sep .

Remark 18.8.2 (a) The “𝐺” in GAR abbreviates “Galois”, “𝐴” abbreviates “Automorphisms”, and “𝑅” abbreviates “Rational”. (b) The GAR-condition implies that the fixed field 𝐸 of Aut(𝐶) is regular over 𝐾. But, it does not require 𝐸 to be rational over 𝐾. This is however the case when 𝑟 = 1 (Lüroth’s theorem, Remark 4.6.2(a)). Lemma 18.8.3 Let 𝐶1 , . . . , 𝐶𝑟 be finite non-Abelian simple groups. Put 𝐶 = Î𝑟 𝑖=1 𝐶𝑖 . Then: (a) Each normal simple subgroup 𝑁 of 𝐶 coincides with 𝐶𝑖 for some 𝑖 between 1 and 𝑟. (b) Suppose Î that 𝛼𝑖 : 𝐶1 → 𝐶𝑖 is an isomorphism, 𝑖 = 1, . . . , 𝑟. Then, Aut(𝐶)  𝑆𝑟 ⋉ 𝑟𝑖=1 Aut(𝐶𝑖 ). Proof of (a). Choose 𝑛 ∈ 𝑁, 𝑛 ≠ 1. Write 𝑛 = 𝑐 1 · · · 𝑐𝑟 with 𝑐 𝑖 ∈ 𝐶𝑖 . Assume, without loss, 𝑐 1 ≠ 1. Since 𝐶1 is simple and non-Abelian, its center is trivial. Hence, there is a 𝑐 1′ ∈ 𝐶1 with 𝑐 1 𝑐 1′ ≠ 𝑐 1′ 𝑐 1 . For each 𝑖 ≥ 2 we have 𝑐 1′ 𝑐 𝑖 = 𝑐 𝑖 𝑐 1′ . Therefore, 𝑛𝑐 1′ ≠ 𝑐 1′ 𝑛, so (𝑐 1′ ) −1 𝑛−1 𝑐 1′ 𝑛1 ≠ 1. It follows that 𝑁 ∩ 𝐶1 ≠ 1. Consequently, 𝑁 = 𝐶1 . Proof of (b). To simplify notation identify 𝐶𝑖 with 𝐶1 via 𝛼𝑖 . Thus, each element of 𝐶 is an 𝑟-tuple c = (𝑐 1 , . . . , 𝑐𝑟 ) with 𝑐 𝑖 ∈ 𝐶1 . Each 𝐶𝑖 is then the group of all c with 𝑐 𝑗 = 1 for 𝑗 ≠ 𝑖. We embed Aut(𝐶1 ) 𝑟 and 𝑆𝑟 in Aut(𝐶) by the rules: c𝜸 = (𝑐 1 1 , . . . , 𝑐𝑟 𝑟 ) 𝛾

𝛾

and

(𝑐 1 , . . . , 𝑐𝑟 ) 𝜎 = (𝑐 1 𝜎−1 , . . . , 𝑐𝑟 𝜎−1 ).

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This gives: 𝜎 −1 𝜸𝜎 = (𝛾1 𝜎−1 , . . . , 𝛾𝑟 𝜎−1 ).

(18.27)

Thus, 𝑆𝑟 normalizes Aut(𝐶1 ) 𝑟 . In addition observe that Aut(𝐶1 ) 𝑟 ∩ 𝑆𝑟 = 1. 𝛾 Finally, consider 𝛾 ∈ Aut(𝐶). By (a), there is a unique 𝜎 ∈ 𝑆𝑟 with 𝐶𝑖 = 𝐶𝑖 𝜎 , −1 𝑖 = 1, . . . , 𝑟. So, there are 𝛾1 , . . . , 𝛾𝑟 in Aut(𝐶1 ) with 𝛾 = (𝛾1 , . . . , 𝛾𝑟 )𝜎 . Thus, Aut(𝐶) is the semidirect product of Aut(𝐶1 ) 𝑟 with 𝑆𝑟 , where the action of 𝑆𝑟 on Aut(𝐶1 ) 𝑟 is given by (18.27). □

Remark 18.8.4 (Minimal normal subgroups) Let 𝐺 be a finite group and 𝐶 a normal subgroup. Suppose that 𝐶 is a minimal normal subgroup of 𝐺. Thus, 𝐶 ≠ 1, 𝐶 ⊳ 𝐺, and 𝐺 has no normal subgroup in 𝐶 other than 1 and 𝐶. Let 𝐶1 be a simple normal subgroup of 𝐶. The conjugates 𝐶1 , 𝐶2 , . . . , 𝐶𝑟 of 𝐶1 in 𝐺 generate a nontrivial normal subgroup of 𝐺 which is contained in 𝐶. Hence, they generate 𝐶. 𝑘 be an integer between 1 and Suppose in addition that 𝐶1 is non-Abelian. LetÎ 𝑘 𝐶𝑖 . By Lemma 18.8.3(a), 𝑟 − 1. Suppose by induction that ⟨𝐶1 , . . . , 𝐶 𝑘 ⟩ = 𝑖=1 Î𝑘 Î 𝑘+1 Î 𝐶 𝑘+1 ̸ ≤ 𝑖=1 𝐶𝑖 . Hence, ⟨𝐶1 , . . . , 𝐶 𝑘+1 ⟩ = 𝑖=1 𝐶𝑖 . In particular, 𝐶 = 𝑟𝑖=1 𝐶𝑖 . By construction, 𝐺 acts transitively on the set {𝐶1 , . . . , 𝐶𝑟 } by conjugation.

Lemma 18.8.5 (Semilinear Rationality Criterion) Let 𝐾, 𝐿, 𝐸, 𝐹 be fields with 𝐿/𝐾 Galois, 𝐾 ⊆ 𝐸, 𝐸 ∩ 𝐿 = 𝐾, and 𝐸 𝐿 = 𝐹. Let 𝑉 be an 𝐿-subspace of 𝐹. Consider the 𝐾-subspace 𝑈 := {𝑣 ∈ 𝑉 | 𝑣 𝜎 = 𝑣 for all 𝜎 ∈ Gal(𝐹/𝐸)} of 𝐸. (a) Suppose that 𝑉 is Gal(𝐹/𝐸)-invariant. Then, each 𝐾-basis of 𝑈 is an 𝐿-basis of 𝑉. Thus, 𝑉  𝑈 ⊗𝐾 𝐿. (b) Suppose, in addition to the assumption of (a), that 𝐹 = 𝐿 (𝑉). Then, 𝐸 = 𝐾 (𝑈). (c) Suppose, in addition to the assumptions of (a) and (b), that dim(𝑉) is finite and equal trans.deg(𝐹/𝐿). Then, 𝐹 is 𝐿-rational and 𝐸 is 𝐾-rational. Proof of (a). By assumption, 𝐸 is linearly disjoint from 𝐿 over 𝐾. Thus, each 𝐾-basis of 𝑈 is linearly independent over 𝐿. Hence, it suffices to prove that 𝑈 spans 𝑉 over 𝐿. To this end consider 𝑣 ∈ 𝑉. Choose a finite Galois extension 𝐹0 of 𝐸 in 𝐹 containing 𝑣. Let 𝐿 0 = 𝐹0 ∩ 𝐿. By assumption, 𝐹0 𝐿 = 𝐹 and res: Gal(𝐹/𝐸) → Gal(𝐿/𝐾) is an isomorphism. Hence, res(Gal(𝐹/𝐹0 )) = Gal(𝐿/𝐿 0 ) and res: Gal(𝐹0 /𝐸) → Gal(𝐿 0 /𝐾) is an isomorphism. Put 𝑚 = [𝐹0 : 𝐸] = [𝐿 0 : 𝐾] and Í 𝐺 = Gal(𝐹0 /𝐸). Choose a basis 𝑐 1 , . . . , 𝑐 𝑚 for 𝐿 0 /𝐾. For each 𝑖 let 𝑢 𝑖 = 𝜎 ∈𝐺 𝑐 𝑖𝜎 𝑣 𝜎 . Then, 𝑢 𝑖 ∈ 𝑈. Since det(𝑐 𝑖𝜎 ) ≠ 0 [Lan97, p. 286, Cor. 5.4] and |𝐺 | = 𝑚, each 𝑣 𝜎 is a linear combination of the 𝑢 𝑖 with coefficients which are rational functions in 𝑐 𝑖𝜏 , 𝜏 ∈ 𝐺, 𝑖 = 1, . . . , 𝑚. In particular, 𝑣 is in the 𝐿-vector space spanned by 𝑈, as desired. Proof of (b). By the tower property (Lemma 3.1.3), 𝐸 is linearly disjoint from 𝐿 (𝑈) over 𝐾 (𝑈). By (a), 𝐹 = 𝐿 (𝑉) = 𝐿(𝑈). Therefore, by Corollary 3.1.4, 𝐸 = 𝐾 (𝑈). Proof of (c). Let 𝑢 1 , . . . , 𝑢 𝑛 be a 𝐾-basis of 𝑈. By (a), 𝑢 1 , . . . , 𝑢 𝑛 is an 𝐿-basis of 𝑉. By (b), 𝐹 = 𝐿 (𝑢 1 , . . . , 𝑢 𝑛 ). Since 𝑛 = trans.deg(𝐹/𝐿), the elements 𝑢 1 , . . . , 𝑢 𝑛 are algebraically independent over 𝐿. Therefore, 𝐹 is 𝐿-rational.

18.8 GAR-Realizations

339

By (b), 𝐸 = 𝐾 (𝑢 1 , . . . , 𝑢 𝑛 ). Since 𝐹/𝐸 and 𝐿/𝐾 are algebraic, trans.deg(𝐸/𝐾) = trans.deg(𝐹/𝐿) = 𝑛. Consequently, 𝐸 is 𝐾-rational.

□

Proposition 18.8.6 (Matzat) Consider theÎ embedding problem (18.24), where 𝐶 is a minimal normal subgroup of 𝐺 and 𝐶 = 𝑟𝑖=1 𝐶𝑖 with 𝐶𝑖 simple non-Abelian and conjugate to 𝐶1 in 𝐺, 𝑖 = 1, . . . , 𝑟. Suppose that 𝐶1 is GAR over 𝐾. Then, embedding problem (18.24) is regularly solvable over 𝐾. Proof. We break the proof into several parts. Part A: Group theory. For each 𝑔 ∈ 𝐺, let 𝜄(𝑔) be the automorphism of 𝐶 induced by conjugation with 𝑔. The map 𝑔 ↦→ (𝜄(𝑔), 𝛼(𝑔)) embeds 𝐺 into Aut(𝐶)×Gal(𝐿/𝐾) with 𝐺 ∩ Aut(𝐶) = 𝐶, 𝛼 = pr2 , pr2 (𝐺) = Gal(𝐿/𝐾). (18.28) Indeed, (𝜄(𝑔), 𝛼(𝑔)) ∈ 𝐺 ∩Aut(𝐶) means that 𝛼(𝑔) = 1, so 𝑔 ∈ 𝐶, hence 𝜄(𝑔) ∈ 𝐶 by the assumption made on 𝐶 in the first paragraph of this section. Thus, (𝜄(𝑔), 𝛼(𝑔)) ∈ 𝐶. Ñ For each 𝑖 let 𝐷 𝑖 = 𝑁𝐺 (𝐶𝑖 ). Put 𝐷 = 𝑟𝑖=1 𝐷 𝑖 . Then, 𝐶 ≤ 𝐷 ≤ 𝐷 𝑖 ≤ 𝐺. By (18.28), 𝐷 𝑖 ∩ Aut(𝐶𝑖 ) = 𝐶𝑖 . (18.29) Indeed, if 𝑗 ≠ 𝑖, then 𝐶 𝑗 acts trivially on 𝐶𝑖 by conjugation. Hence, considering 𝐺 as a subgroup of Aut(𝐶) × Gal(𝐿/𝐾), we have 𝐷 𝑖 ∩ Aut(𝐶𝑖 ) = 𝐷 𝑖 ∩ Aut(𝐶𝑖 ) ∩ 𝐺 ∩ Aut(𝐶) (18.28)

= 𝐷 𝑖 ∩ Aut(𝐶𝑖 ) ∩ 𝐶 = 𝐷 𝑖 ∩ 𝐶𝑖 = 𝐶𝑖 .

For each 𝜎 ∈ 𝐺, Lemma 18.8.3(a) gives a unique 𝜋 ∈ 𝑆𝑟 with 𝐶𝑖𝜎 = 𝐶𝑖 𝜋 , 𝑖 = 1, . . . , 𝑟. Thus, 𝐷 𝑖𝜎 = 𝐷 𝑖 𝜋 , 𝑖 = 1, . . . , 𝑟. Therefore, 𝐷 is normal in 𝐺. Î Finally, as in Lemma 18.8.3(b), we embed 𝑟𝑖=1 Aut(𝐶𝑖 ) in Aut(𝐶) by the rule: (𝑐 1 , . . . , 𝑐𝑟 ) (𝛾1 ,...,𝛾𝑟 ) = (𝑐 1 1 , . . . , 𝑐𝑟 𝑟 ). 𝛾

𝛾

Part B: GA-realization of 𝐶. The assumption that 𝐶1 is GAR gives algebraically independent elements 𝑡 1 , . . . , 𝑡 𝑚 over 𝐾 and fields 𝐸 1 , 𝐹1 satisfying this (see Diagram in (18.26)): (18.30a) 𝐾 ⊆ 𝐸 1 ⊆ 𝐾 (𝑡 1 , . . . , 𝑡 𝑚 ) ⊆ 𝐹1 , 𝐹1 /𝐾 regular, 𝐹1 /𝐸 1 Galois, (18.30b) Gal(𝐹1 /𝐸 1 ) = Aut(𝐶1 ), and Gal(𝐹1 /𝐾 (𝑡 1 , . . . , 𝑡 𝑚 )) = 𝐶1 . (18.30c) Let 𝑀 be an extension of 𝐸 1 with 𝑀𝐾sep = 𝐾sep (𝑡 1 , . . . , 𝑡 𝑚 ). Then, 𝑀 is a field of rational functions over 𝑀 ∩ 𝐾sep . Choose algebraically independent elements 𝑡 𝑖 𝑗 , 𝑖 = 1, . . . , 𝑟, 𝑗 = 1, . . . , 𝑚, over 𝐾 with 𝑡 1 𝑗 = 𝑡 𝑗 , 𝑗 = 1, . . . , 𝑚. For each 𝑖 let t𝑖 = (𝑡 𝑖1 , . . . , 𝑡 𝑖𝑚 ). Also, let 𝛼𝑖 : 𝐾 (t1 ) → 𝐾 (t𝑖 ) be the 𝐾-isomorphism with 𝑡 𝛼𝑗 𝑖 = 𝑡 𝑖 𝑗 , 𝑗 = 1, . . . , 𝑚. Put 𝐸 𝑖 = 𝐸 1𝛼𝑖 and extend 𝛼𝑖 to an isomorphism of 𝐹1 on a field 𝐹𝑖 . The latter induces an isomorphism of Gal(𝐹1 /𝐾 (t1 )) and Gal(𝐹1 /𝐸 1 ) onto Gal(𝐹𝑖 /𝐾 (t𝑖 )) and Gal(𝐹𝑖 /𝐸 𝑖 ), respectively. They are given by 𝛾 ↦→ 𝛼𝑖−1 𝛾𝛼𝑖 . Identify Gal(𝐹𝑖 /𝐾 (t𝑖 )) and Gal(𝐹𝑖 /𝐸 𝑖 ) through this isomorphism with 𝐶𝑖 and Aut(𝐶𝑖 ), respectively. Hence, 𝐸 𝑖 , 𝐾 (t𝑖 ), 𝐹𝑖 satisfy

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18 Galois Groups over Hilbertian Fields

(18.30a)–(18.30c) with 𝑖 replacing 1. Also, 𝐹1 , . . . , 𝐹𝑟 are algebraically independent over 𝐾 and regular over 𝐾 (by (18.30a)). Therefore, they are linearly disjoint over 𝐾 (Lemma 3.4.7). Next let 𝐸 = 𝐸 1 · · · 𝐸𝑟 , t = (t1 , . . . , t𝑟 ), and 𝐹 = 𝐹1 · · · 𝐹𝑟 . Then, (18.31) 𝐾 ⊆ 𝐸 ⊆ 𝐾 (t) ⊆ 𝐹, 𝐹/𝐾 is regular, 𝐹/𝐸 is Galois, Î Î (18.32) Gal(𝐹/𝐸) = 𝑟𝑖=1 Aut(𝐶𝑖 ), and Gal(𝐹/𝐾 (t)) = 𝑟𝑖=1 𝐶𝑖 . The group 𝑆𝑟 acts on 𝐹 by permuting the triples (𝐸 𝑖 , t𝑖 , 𝐹𝑖 ), 𝑖 = 1, . . . , 𝑟. Namely, −1 𝜋 𝑥 = 𝑥 𝛼𝑖 𝛼 𝑗 for 𝜋 ∈ 𝑆𝑟 with 𝑖 𝜋 = 𝑗 and for 𝑥 ∈ 𝐹𝑖 . This induces an action on Gal(𝐹/𝐸) which coincides with the action given by (18.27). (Recall: to obtain (18.27) we identify 𝐶𝑖 with 𝐶1 via 𝛼𝑖 and then identify 𝛼𝑖 with the identity map.) Î So, by Lemma 18.8.3(b), the action of 𝑟𝑖=1 Aut(𝐶𝑖 ) on 𝐹 extends to an action of Aut(𝐶) on 𝐹 over 𝐾. Denote the fixed field of Aut(𝐶) in 𝐹 by 𝐸 0 . Part C: The field crossing argument. By (18.31), 𝐹 is linearly disjoint from 𝐿 over 𝐾. Put 𝑄 = 𝐹 𝐿. Then, Gal(𝑄/𝐸 0 )  Aut(𝐶) × Gal(𝐿/𝐾) with pr2 = res𝑄/𝐿 . Also, 𝑄/𝐿 is regular. Identify Gal(𝑄/𝐸 0 𝐿) with Aut(𝐶) and Gal(𝑄/𝐿(t)) with 𝐶. Part A identifies 𝐺 with a subgroup of Gal(𝑄/𝐸 0 ). Denote the fixed field of 𝐺 in 𝑄 by 𝑃, so Gal(𝑄/𝑃) = 𝐺. Also, by (18.28), Gal(𝑄/𝑃) ∩ Gal(𝑄/𝐸 0 𝐿) = 𝐺 ∩ Aut(𝐶) = 𝐶 and res𝑄/𝐸0 𝐿 (Gal(𝑄/𝑃)) = Gal(𝐸 0 𝐿/𝐸 0 ). Therefore, 𝑃𝐿 = 𝐿(t)

𝑃∩𝐿 =𝐾

and

(18.33)

and we have the following partial diagram of fields and Galois groups: 𝐹

𝑄

𝐶 Î𝑟 𝑖=1

Aut(𝐶𝑖 )

𝐶

𝐾 (t)

𝐿(t)= 𝑃𝐿 ⑤ ⑤ ⑤⑤ ⑤⑤ ⑤⑤

Aut(𝐶)

𝐸

④④ ④④ ④ ④ ④④

𝑃

𝐸0

𝐸0 𝐿

𝐾

𝐿

Thus, all that remains to be proved is the rationality of 𝑃 over 𝐾. Part D: New transcendence basis for 𝐿 (t)/𝐿. The group 𝐷 1 = 𝑁𝐺 (𝐶1 ) acts on 𝐿 because 𝐿/𝐾 is Galois. Denote the fixed field by 𝑀0 . Each 𝜎 ∈ 𝐷 1 satisfies 𝐶1𝜎 = 𝐶1 . Hence, the permutation of {1, . . . , 𝑟 } corresponding to 𝜎 fixes 1, so 𝐹1𝜎 = 𝐹1 . Thus, 𝐷 1 acts on 𝐹1 𝐿. Denote the fixed field by 𝑀. By (18.29), 𝐷 1 ∩ Aut(𝐶1 ) = 𝐶1 . The fixed field of 𝐶1 in 𝐹1 𝐿 is 𝐿 (t1 ) and that of Aut(𝐶1 ) in 𝐹1 𝐿 is 𝐸 1 𝐿. Therefore, 𝑀 𝐸 1 𝐿 = 𝐿 (t1 ). Next observe that the restriction of 𝐷 1 to 𝐹1 maps 𝐷 1 into Aut(𝐶1 ), hence into Gal(𝐹1 /𝐸 1 ). Hence, 𝐸 1 ⊆ 𝑀. Thus, 𝑀 𝐿 = 𝐿(t1 ), so 𝑀𝐾sep = 𝐾sep (t1 ). By construction, 𝑀 ∩ 𝐿 = 𝑀0 . It follows from (18.30c) that there are algebraically independent elements 𝑣 1 , . . . , 𝑣 𝑚 over 𝐾 with 𝑀 = 𝑀0 (𝑣 1 , . . . , 𝑣 𝑚 ). They satisfy

18.8 GAR-Realizations

341

𝐿(𝑣 1 , . . . , 𝑣 𝑚 ) = 𝐿(t1 ).

(18.34)

Here is a partial diagram of the fields and Galois groups involved:

𝐹1

𝐹1 𝐿 ☛ ☛ ☛ 𝐶1 ☛ 𝐷1 ☛☛ ☛ ☛☛ ②𝐿 (t1 ) ☛ ☛ ②② ☛☛②②② ☛ ☛ ②②

𝐶1 Aut(𝐶1 )

 𝐾 (t1 )

𝐸1

𝑀

𝐾

𝑀0

𝐿

For each 𝑖 between 1 and 𝑟 choose 𝜎𝑖 ∈ 𝐺 with 𝐶1𝜎𝑖 = 𝐶𝑖 (end of Remark 18.8.4). 𝑖 Then, 𝐷 1𝜎𝑖 = 𝐷 𝑖 . Put 𝑣 𝑖 𝑗 = 𝑣 𝜎 𝑗 , 𝑗 = 1, . . . , 𝑚, and v𝑖 = (𝑣 𝑖1 , . . . , 𝑣 𝑖𝑚 ). Then, let v = (v1 , . . . , v𝑟 ). By (18.34), 𝐿 (v𝑖 ) = 𝐿 (t𝑖 ), 𝑖 = 1, . . . , 𝑟, so 𝐿(v) = 𝐿 (t). Hence, the 𝑣 𝑖 𝑗 are algebraically independent over 𝐾. Since 𝑣 𝑗 ∈ 𝑀, we have 𝑣 𝛿𝑗 = 𝑣 𝑗 for all 𝛿 ∈ 𝐷 1 and 𝑗 = 1, . . . , 𝑚. If 1 ≤ 𝑖 ≤ 𝑟 ( 𝜎𝑖 𝛿 𝜎𝑖−1 ) 𝜎𝑖

and 𝛿 ∈ 𝐷 𝑖 , then 𝜎𝑖 𝛿𝜎𝑖−1 ∈ 𝐷 1 , hence 𝑣 𝑖𝛿𝑗 = 𝑣 𝑗 Ñ 𝛿 ∈ 𝐷 = 𝑟𝑖=1 𝐷 𝑖 that 𝑣 𝑖𝛿𝑗 = 𝑣 𝑖 𝑗 for all 𝑖 and 𝑗.

𝑖 = 𝑣𝜎 𝑗 = 𝑣 𝑖 𝑗 . It follows for

Part E: Rationality of 𝑃. Denote the fixed field of 𝐷 in 𝑄 by 𝑁. Since Gal(𝑄/𝑃) = 𝐺 (Part C) and 𝐷 is normal in 𝐺 (Part A), 𝑁/𝑃 is Galois. Put 𝑁0 = 𝑁 ∩ 𝐿. By Part D, 𝑣 𝑖𝛿𝑗 = 𝑣 𝑖 𝑗 for each 𝛿 ∈ 𝐷 and all 𝑖, 𝑗. Thus, 𝑁0 (v) ⊆ 𝑁. By Part D, 𝐿 (t) = 𝐿(v). Also, by (18.33), 𝑃𝐿 = 𝐿(t) = 𝐿 (v). Hence, [𝐿(v) : 𝑁0 (v)] = [𝐿 : 𝑁0 ] = [𝐿(v) : 𝑁], so 𝑁 = 𝑁0 (v). By (18.33), 𝑃𝑁0 = 𝑁.

𝑃

𝑁

𝐿 (t)

𝐾

𝑁0

𝐿

𝑄

Let 𝑉 be the vector space spanned by the 𝑣 𝑖 𝑗 ’s over 𝑁0 . For each 𝜎 ∈ 𝐺 and 𝜎𝑖−1 𝜎𝑖′

each 𝑖 there exists an 𝑖 ′ with 𝐶𝑖𝜎 = 𝐶𝑖′ = 𝐶𝑖 −1 ( 𝜎 𝜎𝑖−1 ′ 𝜎𝑖 ) 𝜎𝑖 𝜎𝑖′

𝜎𝑖−1 𝜎𝑖′

. Hence, 𝜎𝜎𝑖−1 ′ 𝜎𝑖 ∈ 𝐷 𝑖 , so

𝑣 𝑖𝜎𝑗 = 𝑣 𝑖 𝑗 = 𝑣𝑖 𝑗 = 𝑣 𝑖′ , 𝑗 . Thus, Gal(𝑁/𝑃) leaves 𝑉 invariant. By Part D, dim(𝑉) = 𝑟𝑚 = trans.deg(𝑃/𝐾). It follows from Lemma 18.8.5 applied to 𝐾, 𝑁0 , 𝑃, 𝑁 rather than to 𝐾, 𝐿, 𝐸, 𝐹 that 𝑃 is 𝐾-rational. □

342

18 Galois Groups over Hilbertian Fields

18.9 Embedding Problems over Hilbertian Fields An affirmative solution to the inverse problem of Galois theory, the realization of all finite groups over Q, seems at present to be out of reach. Matzat’s method of GAR-realization gives an effective tool for a partial solution of the problem, namely for the realization of finite groups with non-Abelian composition factors. Every finite group 𝐺 has a sequence 𝑁0 , . . . , 𝑁 𝑛 of subgroups with 𝑁0 = 𝐺, 𝑁 𝑛 = 1, 𝑁𝑖 ⊳ 𝑁𝑖−1 , and 𝑁𝑖−1 /𝑁𝑖 simple, 𝑖 = 1, . . . , 𝑛. For each finite simple group 𝐶, the number of 𝑖 with 𝐶  𝑁𝑖−1 /𝑁𝑖 depends only on 𝐺 and not on the sequence [Hup67, p. 64, Satz 11.7]. If this number is positive, 𝐶 is a composition factor of 𝐺. Proposition 18.9.1 (Matzat) Let 𝛼: 𝐺 → Gal(𝐿/𝐾) be a finite embedding problem over a field 𝐾. Suppose that every finite embedding problem for each quotient of 𝐺 over 𝐾 with an abelian kernel is solvable in Case (a) (respectively, is regularly solvable in Case (b) below). (a) If 𝐾 is Hilbertian and each composition factor of Ker(𝛼) is GAR over 𝐾, then 𝛼 is solvable. (b) If each composition factor of Ker(𝛼) is GAR over every extension of 𝐾, then 𝛼 is regularly solvable. Proof. Assume without loss that Ker(𝛼) is nontrivial. Choose a minimal normal subgroup 𝐶 of 𝐺 in Ker(𝛼). Then, 𝛼 induces an epimorphism 𝛼 ′: 𝐺/𝐶 → Gal(𝐿/𝐾). Proof of (a). An induction hypothesis gives a Galois extension 𝐿 ′ of 𝐾 containing 𝐿 and an isomorphism 𝛽 ′: Gal(𝐿 ′/𝐾) → 𝐺/𝐶 with 𝛼 ′ ◦ 𝛽 ′ = res 𝐿′ /𝐿 . Let 𝜋: 𝐺 → 𝐺/𝐶 be the quotient map. The kernel of (𝛽 ′) −1 ◦ 𝜋: 𝐺 → Gal(𝐿 ′/𝐾) is 𝐶. By Remark 18.8.4, 𝐶  ⟨𝐶1 , . . . , 𝐶𝑟 ⟩ with 𝐶𝑖 simple normal subgroup of 𝐶 and 𝐶𝑖 conjugate to 𝐶1 in 𝐺, 𝑖 = 1, . . . , 𝑟. In particular, 𝐶1 is a composition factor of 𝐺. By assumption, (𝛽 ′) −1 ◦ 𝜋 is solvable if 𝐶 is Abelian. Otherwise, Î 𝐶1 non-Abelian and, by assumption, 𝐶1 is GAR over 𝐾. By Remark 18.8.4, 𝐶  𝑟𝑖=1 𝐶𝑖 . Hence, by Proposition 18.8.6, (𝛽 ′) −1 ◦ 𝜋 is regularly solvable. Since 𝐾 is Hilbertian, (𝛽 ′) −1 ◦ 𝜋 is solvable (Lemma 18.4.2). In other words, 𝐾 has a Galois extension 𝑁 containing 𝐿 ′ and there is an isomorphism 𝛽: Gal(𝑁/𝐾) → 𝐺 with (𝛽 ′) −1 ◦ 𝜋 ◦ 𝛽 = res 𝑁 /𝐿′ . Then, 𝛼 ◦ 𝛽 = res 𝑁 /𝐿 and 𝛽 solves embedding problem 𝛼. Gal(𝑁/𝐾) ✝✝ ✝✝ res ✝ ✝  𝛽 ✝✝ ✝ Gal(𝐿 ′/𝐾) ✝ ✝ ✝ ✝✝ 𝛽′ ✝✝ res ✝ ✝ ✝ ✝  ✝✝ 𝛼 ✝✝ ✝ / ✝ 𝐺 Gal(𝐿/𝐾) ✝✝ ✝✝ ✝ 𝜋 ✝  ✝✝ 𝛼′ / Gal(𝐿/𝐾) 𝐺/𝐶

18.9 Embedding Problems over Hilbertian Fields

343

Proof of (b). An induction hypothesis gives algebraically independent elements 𝑡 1 , . . . , 𝑡 𝑚 over 𝐾, a Galois extension 𝐸 of 𝐾 (t), and an isomorphism 𝛽 ′: Gal(𝐸/𝐾 (t)) → 𝐺/𝐶 satisfying: 𝐸 is a regular extension of 𝐿 and 𝛼 ′ ◦ 𝛽 ′ = res𝐸/𝐿 . As above, let 𝜋: 𝐺 → 𝐺/𝐶 be the quotient map. By assumption, (𝛽 ′) −1 ◦𝜋 is regularly solvable over 𝐾 (t) if 𝐶 is Abelian. Otherwise 𝐶 is non-Abelian, so each composition factor of 𝐶 is non-Abelian. By assumption, each composition factor of 𝐶 is GAR over 𝐾 (t). By Proposition 18.8.6, (𝛽 ′) −1 ◦ 𝜋 is regularly solvable over 𝐾 (t). In other words, there exist algebraically independent elements 𝑢 1 , . . . , 𝑢 𝑛 over 𝐾 (t), a Galois extension 𝐹 of 𝐾 (t, u), and an isomorphism 𝛽: Gal(𝐹/𝐾 (t, u)) → 𝐺 satisfying: 𝐹 is a regular extension of 𝐸 and (𝛽 ′) −1 ◦ 𝜋 ◦ 𝛽 = res𝐹/𝐸 . Then, 𝐹 is a regular extension of 𝐿 and 𝛽 is a regular solution of embedding problem 𝛼. □ Proposition 18.9.2 Let 𝐾 be a field and 𝑛 a positive char(𝐾) ∤ (𝑛 − 1)𝑛 and 𝑛 ≠ 2, 3, 6. Then, 𝐴𝑛 is GAR over 𝐾.

integer

with

Proof. Use the notation of Proposition 18.7.6. If 𝑛 ≥ 5, then 𝐴𝑛 is simple [Suz82, p. 295, Thm. 2.11], in particular the center of 𝐴𝑛 is trivial. The group 𝐴4 is solvable and nonsimple. However, its center is nevertheless trivial. Indeed, the Klein group 𝑉4 = {(1), (1 2) (3 4), (1 3) (2 4), (1 4) (2 3)} is the only non-trivial proper normal subgroup of 𝐴4 and the action of 𝐴3 = {(1), (1 2 3), (1 3 2)} on 𝑉4 by conjugation fixes no element but 1. The assumption 𝑛 ≠ 2, 3, 6 implies that 𝑆 𝑛 = Aut( 𝐴𝑛 ) [Suz82, p. 299, Statement 2.17]. Hence, by Proposition 18.7.6, 𝐹/𝐾 (𝑢), with 𝑢 transcendental over 𝐾, is a GA-realization of 𝐴𝑛 , that is, Condition (18.25) holds for 𝐴𝑛 . We must still prove Condition (18.26). Consider an extension 𝐸 ′ of 𝐾 (𝑡) with 𝐸 ′ 𝐾sep = 𝐾sep (𝑢) (notation as in Proposition 18.7.6). Let 𝐿 = 𝐸 ′ ∩ 𝐾sep . We have to prove that 𝐸 ′ is 𝐿-rational. To begin, note by the tower property that 𝐸 ′ ∩ 𝐾sep (𝑡) = 𝐿 (𝑡) and [𝐸 ′ : 𝐿 (𝑡)] = [𝐾sep (𝑢) : 𝐾sep (𝑡)] = [𝐾 (𝑢) : 𝐾 (𝑡)] = 2. In particular, 𝐸 ′ is a function field of one variable over 𝐿. Since the genus of 𝐾sep (𝑢) is 0, so is the genus of 𝐸 ′ (Proposition 4.4.3(b)). Suppose that 𝔭 ′ is a prime divisor of 𝐿(𝑡)/𝐿 which ramifies in 𝐸 ′. Then, deg(𝔭 ′) = 1. Hence, there is exactly one prime divisor 𝔭sep of 𝐾sep (𝑡)/𝐾sep over 𝔭 ′ (Proposition 4.4.3(e)) and 𝔭sep ramifies in 𝐾sep (𝑢) (Proposition 4.4.3(c)). Let 𝔭 be the common restriction of 𝔭sep and of 𝔭 ′ to 𝐾 (𝑡). Then, 𝔭 ramifies in 𝐾 (𝑢) (Proposition 4.4.3(c)). But, by Proposition 18.7.6, there are at most three (actually two) such 𝔭’s and each of them is of degree 1. Hence, there are at most three possibilities for 𝔭 ′ (Proposition 4.4.3(e)). It follows from Lemma 18.7.5 that 𝐸 ′ is 𝐿-rational. □ The combination of Propositions 18.9.1 and 18.9.2 gives concrete solvable embedding problems with non-Abelian kernels. Proposition 18.9.3 Let 𝐾 be a field and 𝛼: 𝐺 → Gal(𝐿/𝐾) a finite embedding problem over 𝐾. Suppose that each composition factor of Ker(𝛼) is 𝐴𝑛 with char(𝐾) ∤ (𝑛 − 1)𝑛, and 𝑛 ≠ 2, 3, 6. Then, 𝛼 is regularly solvable over 𝐾. If, in addition, 𝐾 is Hilbertian, then 𝛼 is solvable over 𝐾.

344

18 Galois Groups over Hilbertian Fields

Take 𝐿 = 𝐾 in Proposition 18.9.3: Corollary 18.9.4 Let 𝐾 be a field and 𝐺 a finite group. Suppose that each composition factor of 𝐺 is 𝐴𝑛 with char(𝐾) ∤ (𝑛 − 1)𝑛 and 𝑛 ≠ 2, 3, 6. Then, 𝐺 is regular over 𝐾. If, in addition, 𝐾 is Hilbertian, then 𝐺 is realizable over 𝐾. Remark 18.9.5 (More GAR-realizations) There is a long list of finite non-Abelian simple groups which are known to be GAR over Q. Beside 𝐴𝑛 (with 𝑛 ≠ 2, 3, 6), this list includes PSL2 (F 𝑝 ) with 𝑝 odd and 𝑝 ̸≡ ±1 mod 24 and all sporadic groups with the possible exception of 𝑀23 . See [MaM99, Thm. IV.4.3]. The list becomes longer over the maximal Abelian extension Qab of Q. In addition to the groups that are GAR over Q it contains 𝐴6 , PSL2 (F 𝑝 ) with 𝑝 odd, and 𝑀23 . See [MaM99, Thm. IV.4.6]. It is still unknown whether every finite non-Abelian simple group is GAR over Qab . If this is the case, each finite embedding problem over Qab would be solvable, as we will see in Example 27.8.5(a). It is even unknown if every finite non-Abelian simple group is GAR over each ˜ An affirmative answer to this question would enable us to field 𝐾 containing Q. ˜ 1 , 𝑡2 )) which is Hilbertian by Example solve embedding problems over the field Q((𝑡 17.5.2. The lists over Q and Qab have been established in large part by using the Riemann Existence Theorem. This partially explains the lack of knowledge of GAR realizations in characteristic 𝑝. An exception to our lack of knowledge of GAR realization is the family 𝐴𝑛 of alternative groups. Theorem 15 of [Bri04] improves Proposition 18.9.2 and proves that 𝐴𝑛 is GAR over every field 𝐾 if 𝑛 ≠ 2, 6 and char(𝐾) ≠ 2. The case char(𝐾) = 2 is left open.

18.10 Regularity of Finite Groups over Complete Discrete-Valued Fields The results of this section depends on a theorem of Harbater whose proof is unfortunately outside the scope of this book. Proposition 18.10.1 Suppose that 𝐾 is a complete field under a discrete valuation and 𝐺 is a finite group. Then, 𝐺 is regular over 𝐾. On the proof. Harbater’s proof of the proposition uses the language of formal schemes [Hrb87, Thm. 2.3]. Serre [Ser92, Thm. 8.4.6] and Liu [Liu95] translate the proof into the language of rigid analytic geometry. Both approaches rely on general GAGA theorems relating formal (resp. rigid analytic) geometry to algebraic geometry. A short cut proof which is more algebraic than the former ones appears in [HaV96] and in [Vol96, p. 239]. Each of these proofs actually provides a Galois extension 𝐹 of 𝐾 (𝑡) (with 𝑡 transcendental over 𝐾), regular over 𝐾, with Gal(𝐹/𝐾 (𝑡))  𝐺. □

18.10 Regularity of Finite Groups over Complete Discrete-Valued Fields

345

Proposition 18.10.2 Let 𝐾 be a PAC field and 𝐺 a finite group. Then, 𝐺 is regular over 𝐾. Proof. The field 𝐾ˆ = 𝐾 ((𝑧)) of formal power series in 𝑧 is complete under a ˆ By discrete valuation (Example 4.5.1). By Proposition 18.10.1, 𝐺 is regular over 𝐾. ˆ Proposition 18.2.8, there exists an absolutely irreducible polynomial 𝑓 ∈ 𝐾 [𝑇, 𝑋], monic and Galois in 𝑋, with Gal( 𝑓 (𝑇, 𝑋), 𝐾ˆ (𝑇))  𝐺. Choose 𝑢 1 , . . . , 𝑢 𝑛 ∈ 𝐾ˆ such that 𝐾 [u] contains the coefficients of 𝑓 and 𝑓 is Galois over 𝐾 (u, 𝑇) with Gal( 𝑓 (𝑇, 𝑋), 𝐾 (u))  𝐺. Write 𝑓 (𝑇, 𝑋) = 𝑔(u, 𝑇, 𝑋) with 𝑔 a polynomial with coefficients in 𝐾. ˆ Denote the 𝐾-variety that u generates in A𝑛 by 𝑉. Since 𝐾/𝐾 is regular (Example 4.5.1), so is 𝐾 (u)/𝐾 (Corollary 3.4.2(b)). Hence, 𝑉 is absolutely irreducible (Corollary 11.2.2(a)). Proposition 18.1.4 gives a nonempty Zariski 𝐾-open subset 𝑉0 of 𝑉 satisfying: For each a ∈ 𝑉0 (𝐾), the polynomial 𝑔(a, 𝑇, 𝑋) is absolutely irreducible and Galois over 𝐾 (𝑇). Moreover, Gal(𝑔(a, 𝑇, 𝑋), 𝐾 (𝑇))  Gal(𝑔(u, 𝑇, 𝑋), 𝐾 (u, 𝑇))  𝐺. Since 𝐾 is PAC, there exists a ∈ 𝑉0 (𝐾). By the preceding paragraph, 𝑔(a, 𝑇, 𝑋) is absolutely irreducible, monic and Galois in 𝑋, and Gal(𝑔(a, 𝑇, 𝑋), 𝐾 (𝑇))  𝐺. Consequently, 𝐺 is regular over 𝐾.

□

Remark 18.10.3 (Ample fields) The assumption “𝐾 is PAC” in the proof of Proposition 18.10.2 is used only to find a 𝐾-rational point of 𝑉0 . In many instances the condition “𝐾 is PAC” can be replaced by “𝐾 is ample” (Remark 12.7.9). Indeed, the former condition is equivalent to “𝐾 is existentially closed in each regular extension” (Proposition 12.3.5), while the latter condition is equivalent to “K is existentially closed in 𝐾 ((𝑡))” [Jar11, p. 67, Lemma 5.3.1]. Thus, PAC fields are ample. In addition, Henselian fields are ample. For more details see [Jar11, Chap. 5]. Proposition 18.10.4 Let 𝐾 be a field and 𝐺 a finite group. (a) Then, 𝐺 is regular over a finite Galois extension 𝐿 of 𝐾. (b) Moreover, suppose that 𝐾 is Hilbertian. Then, 𝐿 has a finite Galois extension 𝑁 with Gal(𝑁/𝐿)  𝐺. Proof of (a). 𝐾sep is PAC (Section 12.1). By Proposition 18.10.2, 𝐺 is regular over 𝐾sep . Hence, there is an irreducible Galois polynomial 𝑓 ∈ 𝐾sep [T, 𝑋] in 𝑋 with T = (𝑇1 , . . . , 𝑇𝑟 ) and Gal( 𝑓 (T, 𝑋), 𝐾sep (T))  𝐺. By Lemma 18.2.4(a),(b), there is a finite Galois extension 𝐿 of 𝐾 such that 𝑓 (T, 𝑋) is Galois over 𝐿(T) and Gal( 𝑓 (T, 𝑋), 𝐿(T))  𝐺. Consequently, by Lemma 18.2.4(c), 𝐺 is regular over 𝐿. Proof of (b). By Corollary 13.2.3, 𝐿 is Hilbertian. Proposition 18.1.5 gives a ∈ 𝐿 𝑟 with 𝑓 (a, 𝑋) Galois over 𝐿 and Gal( 𝑓 (a, 𝑋), 𝐿)  𝐺. Let 𝑁 be the splitting field of 𝑓 (a, 𝑋) over 𝐿. Then, Gal(𝑁/𝐿)  𝐺. □ Remark 18.10.5 (An evaluation of Proposition 18.10.4(b)) The proof of Proposition 18.10.4(b) relies on the deep result, Proposition 18.10.1. If we ask 𝐿/𝐾 only to be finite and separable, the proof becomes quite elementary. To this end embed 𝐺 in

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𝑆 𝑛 , say, for 𝑛 = |𝐺 |. Then, use the general polynomial of degree 𝑛 to find a Galois extension 𝑁 of 𝐾 with Gal(𝑁/𝐾)  𝑆 𝑛 . Denote the fixed field of 𝐺 in 𝑁 by 𝐿. Then, Gal(𝑁/𝐿)  𝐺. However, unless 𝐺 is trivial, 𝐺  Z/2Z, or 𝐺 = 𝑆 𝑛 itself, 𝐿 will not be Galois over 𝐾.

Exercises 1. Let 𝐾 be a Z 𝑝 extension of Q and 𝑙 ≠ 𝑝 a prime number. Use Lemma 18.6.9 to prove that 𝑙 is unramified in 𝐾. 2. Prove that Z 𝑝 × Z 𝑝 does not occur over Q. Hint: Use Exercise 1 and Kronecker-Weber: Every finite Abelian extension of Q is contained in Q(𝜁 𝑛 ) for some positive integer 𝑛. Alternatively, use Lemma 18.6.4 ˆ where 𝐺 is a product of finite groups. Then, use to prove that Gal(Qab /Q)  𝐺 × Z, Kronecker-Weber. 3. Suppose that each of the direct powers 𝐺 𝑛 of a finite group 𝐺 can be realized over the field 𝐾. Prove that there exists a linearly disjoint sequence 𝐿 1 , 𝐿 2 , 𝐿 3 , . . . of Galois extensions of 𝐾 with Gal(𝐿 𝑖 /𝐾)  𝐺, 𝑖 = 1, 2, 3, . . . . 4. Use Corollary 18.2.7 to prove that if 𝐾 is a Hilbertian field, then the group Gal(𝐾ab /𝐾) is not finitely generated.

Notes Let 𝛼: 𝐺 → Gal(𝐿/𝐾) be an embedding problem over a field 𝐾. Put 𝐺˜ = Gal(𝐾), 𝐴 = Gal(𝐿/𝐾), and 𝜑 = res𝐾sep /𝐿 . We call a homomorphism 𝛾: 𝐺˜ → 𝐺 satisfying 𝛼 ◦ 𝛾 = 𝜑 a solution of the embedding problem if 𝛾 is surjective, because we are exclusively looking for solutions of embedding problems over fields. However, in the framework of profinite groups, it is of great interest to consider also homomorphisms 𝛾: 𝐺˜ → 𝐴 satisfying 𝛼 ◦ 𝛾 = 𝜑 which are not necessarily surjective. In Chapter 25 we call such 𝛾’s weak solutions of the embedding problem (𝜑, 𝛼) and define 𝐺˜ to be projective if every finite embedding problem for 𝐺˜ has a weak solution. Other authors (e.g. [MaM99, p. 296]) prefer the terminology “solution” and “proper solution” rather than “weak solution” and “solution”. Likewise, what we call a “regular solution” of the embedding problem 𝛼: 𝐺 → Gal(𝐿 (t)/𝐾 (t)) in Definition 18.4.1, is called a “proper parametric solution” in [MaM99, p. 296]. The use of the group epimorphism 𝐴wr𝐺 → 𝐺 ⋉ 𝐴 (Lemma 18.4.3) replaces Uchida’s proof of Proposition 18.4.4 as presented in [FrJ86, Lemma 24.46]. We have borrowed this approach from [MaM99, Section IV.2.2] [Ser83, Thm. 1.2.1] gives a cohomological proof of Proposition 18.5.1. Our proof of Whaple’s theorem (Theorem 18.6.6) is an elaboration of [KuL75].

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The special case of Proposition 18.6.10 when 𝐾 = Q appears in [Ser83, p. 36, Exer. 1]. The article [GJe96] studies the number of linearly disjoint Z 𝑝 -extensions of a given field 𝐾. Theorem 2.7 of that article says this number does not grow in a finitely generated regular extension of a finitely generated extension of Q. In characteristic 0 this is essentially Proposition 18.6.10. The realization of symmetric and alternating groups over Q using Hilbert irreducibility theorem goes back to [Hil92]. Our treatment of this subject in Section 18.7 is a workout of [Ser92, pp. 42-43]. The concept of GAR-realization and its application to solve embedding problems with kernels having trivial centers go back to [Mat85]. See also [MaM99, Chap. IV.3]. The proof of Proposition 18.10.2 essentially depends on the following result: (*) For a field 𝐾 and a finite group 𝐺 there is an absolutely irreducible variety 𝑉 over 𝐾 satisfying this: If 𝑉 (𝐾) ≠ ∅, then 𝐺 is regular over 𝐾. [FrV91, Thm. 2] proves (*) in characteristic 0 as an application of the parametrization of Galois extensions of C(𝑡) by moduli spaces. The proof of Proposition 18.10.2 assumes that 𝐺 is regular over a regular extension of 𝐾 (e.g. over 𝐾 ((𝑡)) ) and construct 𝑉 satisfying the conclusion of (*). Pop observed that the condition “𝐾 is existentially closed in 𝐾 ((𝑡))” is common to PAC fields and to Henselian fields as well as to other families of fields. He then observed that “𝐺 is regular over 𝐾” is an elementary property of 𝐾 and deduced Proposition 18.10.2 for ample fields. This is included in a letter from P. Roquette to W.-D. Geyer in February, 1991. Pop’s observation replaces the use of Bertini– Noether in the proof of Proposition 18.10.2. Pop actually uses the terminology “large fields” instead of “ample fields” [Pop96, p. 4]. We prefer the latter terminology because the adjective “large” in the naive sense has been attached to algebraic extensions of Hilbertian fields in several works (e.g. in [Jar69], [FyJ74], [Jar75], [GeJ78], [Jar79b], [LuD81]).

Chapter 19

Small Profinite Groups

A finitely generated profinite group 𝐺 has for each positive integer 𝑛 only finitely many open subgroups of index at most 𝑛 (Lemma 19.1.2). Groups that satisfy the latter condition are “small”. Small groups have in the category of profinite groups a property similar to the one that finite sets have in the category of all sets: Every epimorphism of a small profinite group onto itself is an isomorphism (Proposition 19.1.6(a)). Galois extensions of Hilbertian fields with small Galois groups share a property with finite Galois extensions: If 𝑁 is a Galois extension of a Hilbertian field and Gal(𝑁/𝐾) is small, then every separable Hilbert subset of 𝑁 contains an element of 𝐾 (Proposition 19.2.1). That property leads to a quick proof of a theorem of Kuyk: If 𝑁 is an abelian extension of a Hilbertian field 𝐾, then 𝑁 is also Hilbertian (Theorem 19.2.3). A Z 𝑝 -extension 𝑁 of 𝐾 is an example of a Galois extension with finitely generated Galois group. If 𝐾 is Hilbertian, then so is 𝑁 (Proposition 19.2.1). This is one ingredient of the proof that each Abelian extension of 𝐾 is Hilbertian (Theorem 19.2.3). Finally, the regularity of Z/𝑝Z over a Hilbertian field 𝐾 has some implications for the structure of Gal(𝐾). For example, Gal(𝐾) has no closed prosolvable normal subgroup. In particular, the center of Gal(𝐾) is trivial.

19.1 Finitely Generated Profinite Groups Let 𝑆 be a subset of a profinite group 𝐺. Denote the closed subgroup generated by 𝑆 by ⟨𝑆⟩. We say that 𝑆 generates 𝐺 if ⟨𝑆⟩ = 𝐺. In this case each map 𝜑0 of 𝑆 into a profinite group 𝐻 has at most one extension to a (continuous) homomorphism 𝜑: 𝐺 → 𝐻. A profinite group 𝐺 is finitely generated if it has a finite set of generators. In this case, the minimal number of generators of 𝐺 is the rank of 𝐺. Note that the rank of a quotient of 𝐺 does not exceed the rank of 𝐺.

349 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_19

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Example 19.1.1 Consider the group (Z/𝑝Z) 𝑛 as a vector space over F 𝑝 of dimension 𝑛. Each group-theoretic set of generators of (Z/𝑝Z) 𝑛 generates (Z/𝑝Z) 𝑛 as a vector space over F 𝑝 . Hence, rank(Z/𝑝Z) 𝑛 = 𝑛. Since (Z/𝑝Z) 𝑛 is a quotient of Z𝑛𝑝 , the rank of the latter group is at least 𝑛. For each 𝑖 between 1 and 𝑛 consider the element 𝑒 𝑖 = (0, . . . , 1, . . . , 0) with 1 in the 𝑖th coordinate and 0 elsewhere. Then, 𝑒 1 , . . . , 𝑒 𝑛 generates Z𝑛𝑝 . It follows that rank(Z𝑛𝑝 ) = 𝑛. Lemma 19.1.2 A finitely generated profinite group 𝐺 has, for each positive integer 𝑛, only finitely many open subgroups of index at most 𝑛. Proof. Each open subgroup 𝑀 of 𝐺 of index ≤ 𝑛 contains an open normal subgroup 𝑁 with 𝐺/𝑁 isomorphic to a subgroup of 𝑆 𝑛 . Indeed, suppose 𝑚 = (𝐺 : 𝑀) ≤ 𝑛. Then, 𝐵 = {𝑔𝑀 | 𝑔 ∈ 𝐺} is a set of order 𝑚. Multiplication from the left with an element 𝑥 of 𝐺 induces a permutation 𝜋(𝑥) of 𝐵. Specifically, 𝜋(𝑥) (𝑔𝑀) = 𝑥𝑔𝑀. Thus, 𝜋 is a homomorphism of 𝐺 into 𝑆 𝑛 with Ñ Ker(𝜋) = 𝑥 ∈𝐺 𝑀 𝑥 . Therefore, Ker(𝜋) is an open normal subgroup of 𝐺 which is contained in 𝑀. Moreover, 𝐺/Ker(𝜋) is isomorphic to a subgroup of 𝑆 𝑚 , hence to a subgroup of 𝑆 𝑛 . Thus, it suffices to prove that 𝐺 has only finitely many open normal subgroups 𝑁 with 𝐺/𝑁 isomorphic to a subgroup of 𝑆 𝑛 . The map 𝛼 ↦→ Ker(𝛼) maps the set of all homomorphisms 𝛼: 𝐺 → 𝑆 𝑛 onto the set of all open normal subgroups 𝑁 of 𝐺 such that 𝐺/𝑁 is isomorphic to a subgroup of 𝑆 𝑛 . Hence, the number 𝜈 of those 𝑁’s does not exceed the number of the 𝛼’s. Let 𝑆 be a finite set of generators of 𝐺. Then, every homomorphism 𝛼: 𝐺 → 𝑆 𝑛 is determined by its values on 𝑆. Therefore, 𝜈 ≤ (𝑛!) |𝑆 | . □ We call a profinite group 𝐺 small if for each positive integer 𝑛 the group 𝐺 has only finitely many open subgroups of index 𝑛. By Lemma 19.1.2, every finitely generated profinite group is small. Thus, each of the results we prove in this section for small profinite groups holds for finitely generated profinite groups (Remark 19.1.9). Remark 19.1.3 (Small profinite groups and open subgroups) Let 𝐺 be a profinite group. (a) Denote the intersection of all open subgroups of 𝐺 of index at most 𝑛 by 𝐺 𝑛 . Then, 𝐺 𝑛 is a closed normal subgroup of 𝐺. Moreover, 𝐺 is small if and only if 𝐺 𝑛 is open in 𝐺 for all 𝑛. (b) Let 𝐻 be an open subgroup of index 𝑚 of 𝐺. Consider open subgroups 𝐺 ′ and 𝐻 ′ of 𝐺 and 𝐻, respectively. Then, (𝐺 : 𝐻 ′) = (𝐺 : 𝐻) (𝐻 : 𝐻 ′) and (𝐻 : 𝐻 ∩ 𝐺 ′) ≤ (𝐺 : 𝐺 ′). Thus, 𝐺 𝑚𝑛 ≤ 𝐻𝑛 ≤ 𝐺 𝑛 . By (a), 𝐺 is small if and only if 𝐻 is small. (c) Let 𝐺 and 𝐻 be as in (b). Suppose that 𝐻 is finitely generated, say by ℎ1 , . . . , ℎ 𝑑 . Let 𝑔1 , . . . , 𝑔𝑚 be representatives of 𝐺/𝐻. Then, 𝑔1 , . . . , 𝑔𝑚 , ℎ1 , . . . , ℎ 𝑑 generate 𝐺. Conversely, if 𝐺 is finitely generated, then 𝐻 is finitely generated (Exercise 3). Corollary 20.6.3 gives a qualitative version of this result.

19.1 Finitely Generated Profinite Groups

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(d) Let 𝛼: 𝐺 → 𝐻 be an epimorphism of profinite groups. If 𝑔1 , . . . , 𝑔𝑛 are generators of 𝐺, then 𝛼(𝑔1 ), . . . , 𝛼(𝑔𝑛 ) are generators of 𝐻. If 𝐺 is small, so is 𝐻. Indeed, let 𝑚 be a positive integer and 𝐻0 an open subgroup of 𝐻 of index 𝑚. Then, 𝐺 0 = 𝛼−1 (𝐻0 ) is an open subgroup of 𝐺 of index 𝑚. The map 𝐻0 ↦→ 𝛼−1 (𝐻0 ) is injective. Since there are only finitely many 𝐺 0 ’s, there are only finitely many 𝐻0 ’s. Example 19.1.4 (A small profinite group which is not finitely generated) Let 𝐴 = Î 𝑝 Z 𝑝 with 𝑝 ranging 𝑁 of 𝐴 of Î over all prime numbers. Consider an open subgroup Î index 𝑛 with 𝑛 = 𝑝 ≤𝑚 𝑝 𝑘 𝑝 . Then, 𝑁 contains the open subgroup 𝑝 ≤𝑚 𝑝 𝑘 𝑝 Z 𝑝𝑝 × Î 𝑝 𝑝>𝑚 Z 𝑝 . Hence, there are only finitely many possibilities for 𝑁. Thus, 𝐴 is small. On the other hand, rank( 𝐴) ≥ rank(Z 𝑝𝑝 ) = 𝑝 for each 𝑝 (Example 19.1.1). Therefore, 𝐴 is not finitely generated. Remark 19.1.5 (Characteristic subgroups) A closed subgroup 𝑁 of a profinite group 𝐺 is characteristic if it is invariant under every automorphism of 𝐺. In particular, 𝑁 is normal in 𝐺. Suppose that 𝐺 is small. Then, 𝐺 𝑛 is a characteristic open subgroup. The decreasing sequence 𝐺 ≥ 𝐺 2 ≥ 𝐺 3 ≥ · · · intersects in 1. Thus, it consists of a basis of open neighborhoods of 1 in 𝐺. Let 𝑁 be an open characteristic subgroup of 𝐺. Then, Aut(𝐺/𝑁) is a finite group. For each 𝛼 ∈ Aut(𝐺) define 𝛼 𝑁 ∈ Aut(𝐺/𝑁) by 𝛼 𝑁 (𝑔𝑁) = 𝛼(𝑔)𝑁. The map 𝛼 ↦→ 𝛼 𝑁 gives a homomorphism Φ 𝑁 : Aut(𝐺) → Aut(𝐺/𝑁). By the preceding paragraph, the intersection of all these 𝑁 is the trivial group. Hence, the Φ 𝑁 combine to an embedding Aut(𝐺) → lim Aut(𝐺/𝑁) which is actually surjective. Thus, ←− Aut(𝐺) is a profinite group. The most useful properties of small profinite groups are embodied in the following result: Proposition 19.1.6 Let 𝐺 be a small profinite group. Then: (a) Every epimorphism of 𝐺 onto itself is an automorphism. (b) Let 𝛼: 𝐺 → 𝐻 and 𝛽: 𝐻 → 𝐺 be epimorphisms. Then, both 𝛼 and 𝛽 are isomorphisms. Proof of (a). Let 𝜃: 𝐺 → 𝐺 be an epimorphism. Let G𝑛 be the finite set of all open subgroups of 𝐺 of index at most 𝑛. The map 𝐻 ↦→ 𝜃 −1 (𝐻) maps G𝑛 injectively into itself. Hence, it maps G𝑛 onto itself. Therefore, in the notation of Remark 19.1.3 Ù Ù Ù 𝐺𝑛 = 𝐻= 𝜃 −1 (𝐻) = 𝜃 −1 ( 𝐻) = 𝜃 −1 (𝐺 𝑛 ). 𝐻 ∈ G𝑛

If 𝜃 (𝑔) = 1, then 𝑔 ∈

Ñ∞ 𝑛=1

𝐻 ∈ G𝑛

𝜃 −1 (𝐺

𝑛)

𝐻 ∈ G𝑛

=

Ñ∞

𝑛=1 𝐺 𝑛

= 1. Therefore, 𝜃 is injective.

Proof of (b). By (a), 𝛽 ◦ 𝛼 is an isomorphism. Hence, both 𝛼 and 𝛽 are injective. □ For a profinite group 𝐺, we denote the set of all finite quotients (up to an isomorphism) of 𝐺 by Im(𝐺).

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19 Small Profinite Groups

Proposition 19.1.7 Let 𝐺 and 𝐻 be profinite groups with 𝐺 small. (a) If Im(𝐻) ⊆ Im(𝐺), then 𝐻 is a quotient of 𝐺. (b) If Im(𝐻) = Im(𝐺), then 𝐻 is isomorphic to 𝐺. Proof of (a). First we prove that 𝐻 is small. Indeed, let 𝑛 be a positive number and 𝐵1 , . . . , 𝐵𝑟 distinct open subgroups of 𝐻 ofÑindex at most 𝑛. Choose an open normal subgroup 𝑁 of 𝐻 which is contained in 𝑟𝑖=1 𝐵𝑖 . Then, 𝐻/𝑁 ∈ Im(𝐻). By assumption, 𝐺 has an open normal subgroup 𝑀 with 𝐺/𝑀  𝐻/𝑁. Hence, 𝐺 has 𝑟 open subgroups of index at most 𝑛. Consequently, 𝑟 is bounded. By Remark 19.1.3(a), the finite group 𝐻/𝐻𝑛 belongs to Im(𝐻), and therefore to Im(𝐺). Thus, 𝐺 has an open normal subgroup 𝐾 with 𝐺/𝐾  𝐻/𝐻𝑛 . In particular, 𝐾 is an intersection of open subgroups of index ≤ 𝑛. Hence, 𝐺 𝑛 ≤ 𝐾. Therefore, there is an epimorphism from 𝐺/𝐺 𝑛 to 𝐻/𝐻𝑛 . Denote the finite nonempty set of all epimorphisms of 𝐺/𝐺 𝑛 → 𝐻/𝐻𝑛 by Φ𝑛 . Let 𝜑: 𝐺/𝐺 𝑛+1 → 𝐻/𝐻𝑛+1 be an epimorphism. It maps the set of all subgroups of 𝐺/𝐺 𝑛 of index at most 𝑛 onto the set of all subgroups of 𝐻/𝐻𝑛 of index at most 𝑛. Hence, 𝜑(𝐺 𝑛 /𝐺 𝑛+1 ) ≤ 𝐻𝑛 /𝐻𝑛+1 . Therefore, 𝜑 induces an epimorphism 𝜑: ¯ 𝐺/𝐺 𝑛 → 𝐻/𝐻𝑛 . This defines a map Φ𝑛+1 → Φ𝑛 . By Corollary 1.1.4, lim Φ𝑛 is nonempty. Each element in lim Φ𝑛 gives a com←− ←− patible system of epimorphisms 𝛽𝑛 : 𝐺/𝐺 𝑛 → 𝐻/𝐻𝑛 . It defines an epimorphism 𝛽: 𝐺 → 𝐻, as desired. Proof of (b). Statement (a) gives epimorphisms 𝜑: 𝐺 → 𝐻 and 𝜓: 𝐻 → 𝐺. Hence, by Proposition 19.1.6(b), both 𝜑 and 𝜓 are isomorphisms. □

Corollary 19.1.8 Let 𝛼: 𝐺 → 𝐻 be an epimorphism of profinite groups. Suppose that 𝐺 is small and Im(𝐺) ⊆ Im(𝐻). Then, 𝛼 is an isomorphism. In particular, every epimorphism 𝛼: 𝐺 → 𝐺 is an isomorphism. Proof. Since 𝐺 is small, so is 𝐻 (Remark 19.1.3(d)). By Proposition 19.1.7(a), 𝐺 □ is a quotient of 𝐻. Therefore, by Proposition 19.1.6(b), 𝛼 is an isomorphism. By Lemma 19.1.2, every finitely generated profinite group is small. Hence, each of the above results about small profinite groups holds for finitely generated profinite groups: Remark 19.1.9 Let 𝐺 and 𝐻 be profinite groups with 𝐺 finitely generated. Then: (a) Aut(𝐺) is a profinite group (by Remark 19.1.5). (b) Every epimorphism of 𝐺 onto itself is an isomorphism (by Proposition 19.1.6(a)). (c) Let 𝛼: 𝐺 → 𝐻 and 𝛽: 𝐻 → 𝐺 be epimorphisms. Then, both 𝛼 and 𝛽 are isomorphisms (by Proposition 19.1.6(b)). (d) If Im(𝐻) ⊆ Im(𝐺), then 𝐻 is a quotient of 𝐺 (by Proposition 19.1.7(a)). (e) If Im(𝐻) = Im(𝐺), then 𝐻 is isomorphic to 𝐺 (by Proposition 19.1.7(b)).

19.1 Finitely Generated Profinite Groups

353

Example 19.1.10 (Small Galois groups) (a) For each finite field 𝐾, Gal(𝐾)  Zˆ (Section 1.5). Thus, Gal(𝐾) is generated by one element. (b) Let 𝑝 be a prime number. The local compactness of Q 𝑝 and Krasner’s lemma imply that Gal(Q 𝑝 ) is small [Lan70, p. 54, Prop. 14]. Deeper arguments show that Gal(Q 𝑝 ) is generated by 4 elements [Jan82, Satz 3.6]. (c) Let 𝐾 be a number field, 𝑂 𝐾 its ring of integers, and 𝑆 a finite number of prime ideals of 𝑂 𝐾 . Denote the maximal algebraic extension of 𝐾 unramified outside 𝑆 by 𝐾𝑆 . It is a Galois extension of 𝐾 (Corollary 2.3.7(c)). We prove that Gal(𝐾𝑆 /𝐾) is a small profinite group. Let 𝑇 be the set of all prime numbers which ramify in 𝐾 or lie under a prime ideal belonging to 𝑆. Then, 𝑇 is finite and 𝐾𝑆 ⊆ Q𝑇 . Suppose we already know that Gal(Q𝑇 /Q) is small. Then, by Remark 19.1.3(b), Gal(Q𝑇 /𝐾) is small. Hence, by Remark 19.1.3(d), Gal(𝐾𝑆 /𝐾) is also small. We may therefore assume 𝐾 = Q and 𝑆 is a finite set of prime numbers. Suppose that 𝐿 is a finite extension of Q in Q𝑆 of degree at most 𝑛. By [Ser72, p. 130, Proposition 6],

log discriminant(𝐿/Q) ≤ (𝑛 − 1)

∑︁

log 𝑝 + 𝑛|𝑆| log 𝑛.

𝑝 ∈𝑆

Thus, discriminant(𝐿/Q) is bounded. By Hermite–Minkowski [Lan70, p. 121, Thm. 5], there are only finitely many extensions of Q with a given discriminant. Consequently, there are only finitely many possibilities for 𝐿. Alternatively, one may follow [Ser90, p. 107] and first observe that 𝑑 := discriminant(𝐿/Q) is divisible only by 𝑝 ∈ 𝑆. For each such 𝑝, [𝐿Q 𝑝 : Q 𝑝 ] ≤ 𝑛. By (b), there are only finitely many possibilities for 𝐿Q 𝑝 , hence for the 𝑝th part of the different of 𝐿/Q. Therefore, there are only finitely many possibilities for 𝑑. It is, however, not clear whether Gal(𝐾𝑆 /𝐾) is finitely generated [Sha62, §3]. (d) There Îare small absolute Galois groups which are not finitely generated. The group 𝐴 = Z 𝑝𝑝 of Example 19.1.4 is one example. To construct a field with absolute Galois group 𝐴, we start from a field 𝐾 of characteristic 0 that contains all roots of unity. Then, Gal(𝐾 ((𝑡)))  Gal(𝐾) × Zˆ [GeJ88, Cor. 4.2]. Thus, for each 𝑝 there is an algebraic extension of 𝐾 ((𝑡)) with absolute Galois group Gal(𝐾) ×Z 𝑝 . In particular, taking 𝐾 to be algebraically closed, we find a field 𝐾 ′ with Gal(𝐾 ′)  Z 𝑝 . Induction gives fields 𝐾 𝑝,𝑖 with 𝑝 ranging over all prime numbers and 𝑖 = 1, . . . , 𝑝 Î satisfying these conditions: Gal(𝐾 𝑝,𝑖 ) = 𝑙< 𝑝 Z𝑙𝑙 × Z𝑖𝑝 and 𝐾𝑙,𝑖 ⊆ 𝐾 𝑝, 𝑗 if 𝑙 < 𝑝 or Ð Ð𝑝 𝑙 = 𝑝 and 𝑖 ≤ 𝑗. Let 𝐿 = 𝑝 𝑖=1 𝐾 𝑝,𝑖 . Then, Gal(𝐿) = lim Gal(𝐾 𝑝,𝑖 )  𝐴. ←−

354

19 Small Profinite Groups

19.2 Abelian Extensions of Hilbertian Fields We give here some results about Galois extensions of Hilbertian fields which involve small groups: Proposition 19.2.1 Let 𝑁 be a Galois extension of a Hilbertian field 𝐾. Suppose that Gal(𝑁/𝐾) is small. Then, each separable Hilbert subset 𝐻 of 𝑁 𝑟 contains a separable Hilbert subset of 𝐾 𝑟 . In particular, 𝑁 is Hilbertian. Proof. By definition 𝐻 = 𝐻 𝑁 ( 𝑓1 , . . . , 𝑓 𝑘 ; 𝑔), where 𝑓𝑖 ∈ 𝑁 (𝑇1 , . . . , 𝑇𝑟 ) [𝑋] is irreducible and separable, 𝑖 = 1, . . . , 𝑘, and 𝑔 ∈ 𝑁 [T], 𝑔 ≠ 0. Let 𝑛 = max(deg𝑋 ( 𝑓1 ), . . . , deg𝑋 ( 𝑓 𝑘 )). Choose a finite extension 𝐿 of 𝐾 in 𝑁 that contains all coefficients of 𝑓1 , . . . , 𝑓 𝑘 , 𝑔. Put 𝑚 = [𝐿 : 𝐾]. Denote the compositum of all extensions of 𝐾 in 𝑁 of degree at most 𝑚𝑛 by 𝑀. By assumption, [𝑀 : 𝐾] < ∞. By Corollary 13.2.3, 𝐻 𝑀 ( 𝑓1 , . . . , 𝑓 𝑘 ; 𝑔) contains a separable Hilbert subset 𝐻𝐾 of 𝐾𝑟 . Let a ∈ 𝐻𝐾 . Consider 𝑖 between 1 and 𝑘. Then, 𝑔(a) ≠ 0 and 𝑓𝑖 (a, 𝑋) is irreducible over 𝑀. Let 𝑏 be a zero of 𝑓𝑖 (a, 𝑋) in 𝐾sep . Then, 𝐿(𝑏) is linearly disjoint from 𝑀 over 𝐿. But, [𝑁 ∩ 𝐿 (𝑏) : 𝐾] ≤ 𝑚𝑛. Hence, 𝑁 ∩ 𝐿(𝑏) ⊆ 𝑀 ∩ 𝐿 (𝑏) = 𝐿. Thus, 𝑓𝑖 (a, 𝑋) is irreducible over 𝑁. Consequently, a ∈ 𝐻. □ Theorem 19.2.2 Let 𝐾 be a Hilbertian field and 𝐴 a finite Abelian group. Then, Zˆ × 𝐴 occurs over 𝐾. Proof. For each prime number 𝑝, Corollary 18.6.7 gives a Z 𝑝 -extension 𝐿 𝑝 of 𝐾. The sequence of all these extensions is linearly disjoint over 𝐾. Hence, the ˆ compositum 𝐿 of all 𝐿 𝑝 ’s is a Z-extension of 𝐾 (Lemma 1.4.5). By Proposition 18.3.5 and Remark 18.2.2, there is an 𝑋-stable polynomial 𝑓 ∈ 𝐾 [𝑇, 𝑋] which is Galois in 𝑋 with Gal( 𝑓 (𝑇, 𝑋), 𝐾 (𝑇))  𝐴. In particular, Gal( 𝑓 (𝑇, 𝑋), 𝐿(𝑇))  𝐴. Let 𝐻1 (resp. 𝐻2 ) be the set of all 𝑎 ∈ 𝐾 with Gal( 𝑓 (𝑎, 𝑋), 𝐾)  𝐴 (resp. Gal( 𝑓 (𝑎, 𝑋), 𝐿)  𝐴). Since Zˆ is small, 𝐿 is Hilbertian (Proposition 19.2.1). Moreover, each Hilbert subset of 𝐿 contains a Hilbert subset of 𝐾. By Lemma 14.1.1, 𝐻1 (resp. 𝐻2 ) contains a Hilbert subset of 𝐾 (resp. 𝐿). Hence, 𝐻1 ∩ 𝐻2 contains a Hilbert subset 𝐻 of 𝐾. Choose 𝑎 ∈ 𝐻. Let 𝑀 be the splitting field of 𝑓 (𝑎, 𝑋) over 𝐾. Put 𝑁 = 𝐿 𝑀. □ Then, Gal(𝑀/𝐾)  Gal(𝑁/𝐿)  𝐴. Therefore, Gal(𝑁/𝐾)  Zˆ × 𝐴. Theorem 19.2.3 ([Kuy70], p. 113) Every Abelian extension 𝑁 of a Hilbertian field 𝐾 is Hilbertian. Proof. [Wei82, Satz 9.8] We may assume that 𝑁 ≠ 𝐾 and choose 𝜎 ∈ Gal(𝑁/𝐾), 𝜎 ≠ 1. The fixed field 𝐿 of 𝜎 in 𝑁 is a proper subfield of 𝑁. Hence, 𝐿 has a finite proper extension 𝑀 in 𝑁. Let 𝑚 = [𝑀 : 𝐿]. Since Gal(𝑁/𝐿) is a closed subgroup of an Abelian group, 𝐿/𝐾 is Galois. By Theorem 15.3.1(b), 𝑀 is a Hilbertian field. Now observe that 𝜎 𝑚 generates Gal(𝑁/𝑀). Therefore, by Proposition 19.2.1, 𝑁 is Hilbertian. □

19.2 Abelian Extensions of Hilbertian Fields

355

Remark 19.2.4 We cannot draw the conclusion, in Theorem 19.2.3, that every separable Hilbert subset of 𝑁 contains a separable Hilbert subset of 𝐾. Indeed, consider 𝑁 = 𝐾ab , the maximal Abelian extension of 𝐾. Suppose that char(𝐾) ≠ 2. Then, there exists no 𝑎 ∈ 𝐾 with 𝑋 2 − 𝑎 irreducible over 𝐾ab , even though 𝑋 2 − 𝑇 is absolutely irreducible. When char(𝐾) = 2, there is no 𝑎 ∈ 𝐾 with 𝑋 2 − 𝑋 − 𝑎 irreducible over 𝐾ab although 𝑋 2 − 𝑋 − 𝑇 is absolutely irreducible and separable in 𝑋. Lemma 19.2.5 Let 𝐾 be a Hilbertian field. Then, Gal(𝐾) is neither prosolvable nor small. Proof. By Corollary 18.2.7(a), 𝐾 has a Galois extension 𝐿 with Galois group 𝑆5 . Hence, Gal(𝐾) is not prosolvable. Again, by Corollary 18.2.7(b),(c), 𝐾 has infinitely many quadratic extensions. Therefore, Gal(𝐾) is not small. □ Proposition 19.2.6 Let 𝑁 be a Galois extension of a Hilbertian field 𝐾. Suppose that 𝑁 ≠ 𝐾sep . Then, Gal(𝑁) is neither prosolvable ([Sch33], [Kuy70, Thm. 2]) (in particular, the center of Gal(𝐾) is trivial) nor it is contained in a closed small subgroup of Gal(𝐾). Proof. By assumption, 𝑁 has a proper finite separable extension 𝑁 ′. By Theorem 15.3.1(b), 𝑁 ′ is Hilbertian. Hence, by Lemma 19.2.5, Gal(𝑁 ′) is not prosolvable. Consequently, Gal(𝑁) is not prosolvable. Now consider an extension 𝑀 of 𝐾 in 𝑁. Let 𝑀 ′ be a finite extension of 𝑀 with 𝑁 𝑀 ′ = 𝑁 ′. In particular, 𝑀 ′ ̸ ⊆ 𝑁. Hence, by Theorem 15.3.1(a), 𝑀 ′ is Hilbertian. Hence, by Lemma 19.2.5, Gal(𝑀 ′) is not small. It follows from Remark 19.1.3(b) that Gal(𝑀) is not small. □ Example 19.2.7 (A Hilbertian field which is not a proper finite extension of any field) For each 𝑝, Q has a Galois extension 𝑁 with Gal(𝑁/Q)  Z 𝑝 (Lemma 18.6.4(d) or Corollary 18.6.7). By Proposition 19.2.1, 𝑁 is Hilbertian. Since Z 𝑝 has no nontrivial finite subgroups (Lemma 1.4.2(c)), 𝑁 is not a proper finite extension of any field.

Exercises 1. Strengthen Theorem 19.2.6 to prove that if 𝐿 is a separable algebraic extension of a Hilbertian field 𝐾 and Gal(𝐿) is a prosolvable group, then the Galois closure of 𝐿/𝐾 is 𝐾sep . 2. Let 𝐺 be a profinite group, 𝑁 a closed characteristic subgroup of 𝐺, and 𝛼 an automorphism of 𝐺. Suppose that the centralizer of 𝑁 in 𝐺 is trivial. Prove that the map 𝛼 ↦→ 𝛼| 𝑁 is an embedding of Aut(𝐺) into Aut(𝑁). Hint: Use the identity 𝛼(𝑛−1 𝑔𝑛) = 𝛼(𝑛) −1 𝛼(𝑔)𝛼(𝑛) for 𝑔 ∈ 𝐺 and 𝑛 ∈ 𝑁. 3. Let 𝐺 be a finitely generated group and 𝐻 an open subgroup. Prove that 𝐻 is finitely generated. Ð𝑛 Hint: Let 𝑥 1 , . . . , 𝑥 𝑒 be generators of 𝐺 and write 𝐺 = · 𝑖=1 𝐻𝑔𝑖 with 𝑔1 = 1. −1 Prove that 𝐻 is generated by all elements 𝑔𝑖 𝑥 ±1 𝑗 𝑔 𝑘 which belong to 𝐻.

356

19 Small Profinite Groups

Notes The concept of small profinite groups appears in [Kli74]. Theorem 25.10.7 partially generalizes Theorem 19.2.3 to pronilpotent extensions of 𝐾. Geyer (private communication) uses wreath products to prove that if 𝐾 is a Hilbertian field, then Gal(𝐾) has no nontrivial finitely generated normal closed subgroups. Proposition 19.2.6 generalizes this result. Example 19.2.7 settles a question of Sonn.

Chapter 20

Free Profinite Groups

We continue the discussion on profinite groups of Chapters 1, 18, and 19. Central to this chapter is a discussion on free profinite groups. In particular, we prove that an open subgroup of a free profinite group is free (Proposition 20.6.2).

20.1 The Rank of a Profinite Group The rank of a finitely generated profinite group is defined to be the minimal number of generators of the group (Section 19.1). The definition of the rank of arbitrary profinite group 𝐺 puts a topological condition on the minimal set of generators of 𝐺. Once this condition is satisfied, the cardinality 𝑚 of that set depends only on the group. Indeed, if 𝐺 is not finitely generated, then 𝑚 is the cardinality of the set of all open subgroups of 𝐺 (Proposition 20.1.2) and rank(𝐺) is defined to be 𝑚. The above mentioned topological condition generalizes convergence of a sequence. A subset 𝑋 of a profinite group 𝐺 is said to converge to 1, if 𝑋 ∖ 𝑁 is a finite set for every open normal subgroup 𝑁 of 𝐺. Proposition 20.1.1 (Douady) Every profinite group 𝐺 has a system of generators that converges to 1. Proof. Without loss assume 𝐺 is infinite. Well order the set N of open normal subgroups of 𝐺: ÑN = {𝑁 𝛼 | 𝛼 < 𝑚}, where 𝑚 is an infinite cardinal number. For 𝛽 ≤ 𝑚 let 𝑀𝛽 = 𝛼 0 with the same imperfect degree. Then, 𝐸 and 𝐹 are elementarily equivalent. Proof. Set 𝐿 = 𝑀 = F˜ 𝑝 in Theorem 23.3.3.

□

23.4 On 𝒆-Free PAC Fields In Theorem 23.3.3 we may assume without loss that 𝐸/𝐿 and 𝐹/𝑀 are regular extensions (e.g. as in the proof of Lemma 23.2.2). It is the relation between the groups Gal(𝐸) (resp. Gal(𝐹)) and Gal(𝐿) (resp. Gal(𝑀)) via restriction that complicates applications. When Gal(𝐸) and Gal(𝐹) are isomorphic to the free profinite group, 𝐹ˆ𝑒 , on 𝑒 generators, Gaschütz’s lemma (Lemma 20.7.3) comes to our aid: Proposition 23.4.1 Let 𝐸 and 𝐹 be 𝑒-free PAC fields with the same imperfect degree and with a common subfield 𝐾. Suppose that 𝐸/𝐾˜ ∩ 𝐸 and 𝐹/𝐾˜ ∩ 𝐹 are separable extensions and 𝐾˜ ∩ 𝐸 𝐾 𝐾˜ ∩ 𝐹. Then, 𝐸 ≡𝐾 𝐹. Proof. Put 𝐿 = 𝐾˜ ∩ 𝐸. By Proposition 20.7.4, there exists an isomorphism 𝜑: Gal(𝐹) → Gal(𝐸) satisfying res𝐸sep /𝐿sep ◦ 𝜑 = res𝐹sep /𝐿sep . Hence, by Theorem □ 23.3.3, 𝐸 ≡𝐾 𝐹. If 𝐸 is perfect, then 𝐾˜ ∩ 𝐸 is also a perfect field and therefore 𝐸/𝐾˜ ∩ 𝐸 is a separable extension. This simplifies Proposition 23.4.1: ˜ Corollary 23.4.2 Let 𝐾 be a subfield of perfect 𝑒-free PAC fields 𝐸 and 𝐹. If 𝐾∩𝐸 𝐾 𝐾˜ ∩ 𝐹, then 𝐸 ≡𝐾 𝐹. Corollary 23.4.3 Let 𝐸 and 𝐹 be 𝑒-free PAC fields with the same imperfect degree. If 𝐹 is a regular extension of 𝐸, then 𝐹 is an elementary extension of 𝐸.

23.4 On 𝑒-Free PAC Fields

463

We axiomatize the concept of 𝑒-free PAC fields for model-theoretic applications: Proposition 23.4.4 Let 𝐾 be a field and 𝑒 a positive integer. Then, there exists a set Ax(𝐾, 𝑒) of axioms in the language L (ring, 𝐾) such that a field extension 𝐹 of 𝐾 satisfies the axioms if and only if it is perfect, PAC and 𝑒-free. The axioms are sentences that interpret the field axioms, perfectness axioms, [ 𝑝 ≠ 0] ∨ (∀𝑋) (∃𝑌 ) [𝑌 𝑝 = 𝑋], as 𝑝 ranges over primes, the positive diagram of 𝐾 (Example 8.3.1), and the following axioms: (a) PAC axioms: Every absolutely irreducible polynomial 𝑓 (𝑋, 𝑌 ) of degree 𝑑 has a zero, 𝑑 = 1, 2, . . . . (b) 𝑒-free axioms: The finite groups which appear as Galois groups over 𝐹 are exactly the groups of rank bounded by 𝑒. Proof. Section 12.3 translates the PAC axioms into elementary statements. Thus, it suffices to translate the 𝑒-free axioms into elementary statements. For this, consider a polynomial 𝑓 (𝑋) = 𝑋 𝑛 + 𝑢 1 𝑋 𝑛−1 + · · · + 𝑢 𝑛 with indeterminate coefficients 𝑢 1 , . . . , 𝑢 𝑛 and a subgroup 𝐺 of 𝑆 𝑛 which is given by its action on {1, 2, . . . , 𝑛}. Suppose that the following assertion is an elementary statement on 𝑢 1 , . . . , 𝑢 𝑛 : “The polynomial 𝑓 is irreducible, separable, normal, and has 𝐺 as a Galois group.” Then, consider all subgroups 𝐺 1 , . . . , 𝐺 𝑟 of 𝑆 𝑛 which are generated by 𝑒 elements. Restate axiom (b): “For each 𝑛 and for every irreducible, separable, and normal polynomial 𝑓 of degree ≤ 𝑛, the Galois group of 𝑓 is isomorphic to one of the groups 𝐺 1 , . . . , 𝐺 𝑟 ; and for each subgroup 𝐺 of 𝑆 𝑛 of rank ≤ 𝑒, there exists an irreducible, separable, normal polynomial of degree at most 𝑛 with Galois group isomorphic to 𝐺.” The normality condition on 𝑓 means that a root 𝑧 of 𝑓 (𝑋) (in the algebraic closure) gives all other roots as polynomials in 𝑧 of degree at most 𝑛 − 1 with coefficients in ˜ use congruences modulo 𝑓 (𝑍) as follows: 𝐹. To eliminate the reference to 𝐹, There exist polynomials 𝑝 1 (𝑍) = 𝑍, 𝑝 2 (𝑍), . . . , 𝑝 𝑛 (𝑍) of degree at most 𝑛 − 1 with 𝑛 Ö (𝑋 − 𝑝 𝑖 (𝑍)) mod 𝑓 (𝑍). 𝑓 (𝑋) ≡ (23.13) 𝑖=1

Of course, (23.13) is actually 𝑛 congruence conditions on the coefficients of the powers of 𝑋 on both sides. For example, equating the free coefficients on both sides gives the condition 𝑢 𝑛 ≡ (−1) 𝑛 𝑝 1 (𝑍) · · · 𝑝 𝑛 (𝑍) mod 𝑓 (𝑍), which is equivalent to the existence of a polynomial 𝑔 of degree at most 𝑛(𝑛 − 1) with 𝑢 𝑛 = (−1) 𝑛 𝑝 1 (𝑍) · · · 𝑝 𝑛 (𝑍) + 𝑔(𝑍) 𝑓 (𝑍). Thus, the normality condition on 𝑓 is elementary. The condition “Gal( 𝑓 , 𝐹) is isomorphic to 𝐺 as a permutation group” may be shown to be elementary by considering the action on the roots of 𝑓 . Then, eliminate the reference to 𝐹˜ as before. Indeed, suppose 𝑓 is monic, irreducible, separable, normal, and 𝑝 1 (𝑧) = 𝑧 = 𝑧1 , 𝑝 2 (𝑧) = 𝑧2 , . . ., 𝑝 𝑛 (𝑧) = 𝑧 𝑛 are the roots of 𝑓 with 𝑝 𝑖 ∈ 𝐹 [𝑍] a polynomial of degree at most |Gal( 𝑓 , 𝐹)| − 1. Suppose that 𝜎 ∈ Gal( 𝑓 , 𝐹).

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23 The Elementary Theory of 𝑒-Free PAC Fields

Then, for each 𝑖, 𝜎𝑧𝑖 = 𝑝 𝑖 (𝜎𝑧). Hence, 𝑧 𝜎 (𝑖) = 𝑝 𝑖 (𝑧 𝜎 (1) ). Hence, 𝑝 𝜎 (𝑖) (𝑧) = 𝑝 𝑖 (𝑧 𝜎 (1) ). Conversely, suppose that 𝜎 ∈ 𝑆 𝑛 satisfies 𝑝 𝜎 (𝑖) (𝑧) = 𝑝 𝑖 (𝑧 𝜎 (1) ) for 𝑖 = 1, . . . , 𝑛. Let 𝜏 be the unique element of Gal( 𝑓 , 𝐹) with 𝜏(1) = 𝜎(1). Then, 𝑝 𝜎 (𝑖) (𝑧) = 𝑝 𝑖 (𝑧 𝜎 (1) ) = 𝑝 𝑖 (𝑧 𝜏 (1) ) = 𝑝 𝜏 (𝑖) (𝑧). Hence, 𝜎(𝑖) = 𝜏(𝑖) for each 𝑖. Therefore, 𝜎 = 𝜏 ∈ Gal( 𝑓 , 𝐹). Consequently, “Gal( 𝑓 , 𝐹)  𝐺” is equivalent to 𝑛 Û Û 𝜎 ∈𝐺 𝑖=1

[ 𝑝 𝜎 (𝑖) (𝑧) = 𝑝 𝑖 (𝑧 𝜎 (1) )] ∧

Û

𝑛 Ü

[ 𝑝 𝜎 (𝑖) (𝑧) ≠ 𝑝 𝑖 (𝑧 𝜎 (1) )].□

𝜎 ∈𝑆𝑛 ∖𝐺 𝑖=1

Remark 23.4.5 (a) If, in an application, 𝐾 is the quotient field of a distinguished subring 𝑅, we may replace the positive diagram of 𝐾 by the positive diagram of 𝑅. (b) When desired, axioms indicating that the imperfect exponent of 𝐹 is 𝑚 (0 ≤ 𝑚 ≤ ∞) may replace the perfect axioms. (c) If 𝐾 is presented with elimination theory, then Ax(𝐾, 𝑒) can be effectively presented. (d) Let 𝐾 be a field and 𝐺 a subgroup of 𝑆 𝑛 . Then, “𝐺 occurs as a Galois group over 𝐾” is an elementary statement on 𝐾. Indeed, the proof of Proposition 23.4.4 presents the equivalent statement “there is an irreducible monic Galois polynomial 𝑓 ∈ 𝐾 [𝑋] with Gal( 𝑓 , 𝐾)  𝐺” as an elementary statement. Proposition 23.4.6 ([Kli74]) Let 𝐾 and 𝐿 be elementarily equivalent fields. Suppose that Gal(𝐾) is a small profinite group. Then, Gal(𝐾)  Gal(𝐿). Proof. By Remark 23.4.5, a finite group 𝐺 occurs as a Galois group over 𝐾 if and only if 𝐺 occurs as a Galois group over 𝐿. Thus, Gal(𝐾) and Gal(𝐿) have the same finite quotients. It follows from Proposition 19.1.7 that Gal(𝐾)  Gal(𝐿). □

23.5 The Elementary Theory of Perfect 𝒆-Free PAC Fields We interpret the elementary theory of perfect 𝑒-free PAC fields that contain a field 𝐾 in the following cases: (23.14a) 𝐾 is finite and 𝑒 = 1. (23.14b) 𝐾 is countable and Hilbertian, and 𝑒 ≥ 1. In each of these cases (𝐾, 𝑒) is called a Hilbertian pair. For 𝝈 = (𝜎1 , . . . , 𝜎𝑒 ) in Gal(𝐾) 𝑒 , let 𝐾˜ (𝝈) = 𝐾sep (𝝈)ins be the maximal purely inseparable extension of 𝐾sep (𝝈). It is the fixed field in 𝐾˜ of the unique extension ˜ The fields 𝐾sep (𝝈) and 𝐾˜ (𝝈) have the of 𝝈 to an 𝑒-tuple of automorphisms of 𝐾. same absolute Galois group. If 𝐾sep (𝝈) is PAC, so is 𝐾˜ (𝝈). But 𝐾˜ (𝝈) is a perfect field. Apply Corollary 21.5.9 and Proposition 21.6.4 to case (23.14a) and Theorems 21.5.6 and 21.6.1 to case (23.14b): Theorem 23.5.1 Suppose that (𝐾, 𝑒) is a Hilbertian pair. Then, 𝐾˜ (𝝈) is a perfect 𝑒-free PAC field for almost all 𝝈 ∈ Gal(𝐾) 𝑒 .

23.6 The Probable Truth of a Sentence

465

Recall the regular ultrafilters of Section 8.7, when the index set 𝑆 is Gal(𝐾) 𝑒 and “small sets” are the subsets of Gal(𝐾) 𝑒 of measure zero. In particular, a regular ultrafilter on Gal(𝐾) 𝑒 contains all subsets of Gal(𝐾) 𝑒 of measure 1. We will compare an arbitrary 𝑒-free PAC field with a regular ultraproduct of the fields 𝐾˜ (𝝈). Denote the theory of all sentences of L (ring, 𝐾) which are true in 𝐾˜ (𝝈) for almost all 𝝈 ∈ Gal(𝐾) 𝑒 by Almost(𝐾, 𝑒). Lemma 23.5.2 Let (𝐾, 𝑒) be a Hilbertian pair. (a) A field 𝐹 is a model of Almost(𝐾, 𝑒) if and only if it is 𝐾-elementarily equivalent to a regular ultraproduct of the fields 𝐾˜ (𝝈). (b) Every regular ultraproduct of the fields 𝐾˜ (𝝈) is a perfect 𝑒-free PAC field. Proof. Statement (a) is a special case of Proposition 8.8.1(b). Statement (b) follows □ from Proposition 23.4.4 and Theorem 23.5.1. We define the corank of a field 𝐾 as the rank of Gal(𝐾). Lemma 23.5.3 Let 𝐾 be a field and 𝑒 a positive integer. Then, for every perfect field 𝐹 of corank at most 𝑒 that contains 𝐾, there exists a regular ultraproduct 𝐸 of the 𝐾˜ (𝝈)’s with 𝐾˜ ∩ 𝐸 𝐾 𝐾˜ ∩ 𝐹. Proof. Let 𝜏1 , . . . , 𝜏𝑒 be generators of Gal(𝐹). For each finite Galois extension 𝐿 of 𝐾 the set 𝑆(𝐿) = {𝝈 ∈ Gal(𝐾) 𝑒 | res 𝐿 (𝝈) = res 𝐿 (𝝉)} has a positive Haar measure, so 𝑆(𝐿) is not small. If 𝐿 is contained in a larger finite Galois extension 𝐿 ′ of 𝐾, then 𝑆(𝐿 ′) ⊆ 𝑆(𝐿). By Lemma 8.7.1, there exists a regular ultrafilter D on Gal(𝐾) 𝑒 which contains Î the sets 𝑆(𝐿) as 𝐿 runs over all finite Galois extensions of 𝐾. Let 𝐸 = 𝐾˜ (𝝈)/D. Then, 𝐿 ∩ 𝐸 = 𝐿 ∩ 𝐹 for every finite Galois extension 𝐿 of 𝐾. Therefore, 𝐾sep ∩ 𝐸 = 𝐾sep ∩ 𝐹. Since 𝐾˜ ∩ 𝐸 and 𝐾˜ ∩ 𝐹 are perfect, they are equal. □ Theorem 23.5.4 If (𝐾, 𝑒) is a Hilbertian pair, then Ax(𝐾, 𝑒) (Proposition 23.4.4) is a set of axioms for Almost(𝐾, 𝑒). Specifically, a field 𝐹 is perfect, 𝑒-free, PAC, and contains 𝐾 if and only if it satisfies each sentence 𝜃 of Ax(𝐾, 𝑒) which is true in 𝐾˜ (𝝈) for almost all 𝝈 ∈ Gal(𝐾) 𝑒 . Proof. Suppose that 𝐹 |= Ax(𝐾, 𝑒). By Lemma 23.5.3, there exists a regular ultraproduct 𝐸 of the 𝐾˜ (𝝈)’s such that 𝐾˜ ∩ 𝐸 𝐾 𝐾˜ ∩ 𝐹. By Lemma 23.5.2(b), 𝐸 |= Ax(𝐾, 𝑒). Corollary 23.4.2 now gives 𝐸 ≡𝐾 𝐹. From Lemma 23.5.2(a), 𝐹 |= Almost(𝐾, 𝑒). Theorem 23.5.1 gives the converse. □

23.6 The Probable Truth of a Sentence Let 𝐾 be a field and let 𝑒 be a positive integer. For a sentence 𝜃 of L (ring, 𝐾), consider the truth set of 𝜃: 𝑆(𝐾, 𝑒, 𝜃) = {𝝈 ∈ Gal(𝐾) 𝑒 | 𝐾˜ (𝝈) |= 𝜃}.

(23.15)

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23 The Elementary Theory of 𝑒-Free PAC Fields

Refer to the case where 𝐾 has elimination theory (Definition 22.2.8) and 𝑒 and 𝜃 are explicitly given, as the explicit case. Regard the measure of 𝑆(𝐾, 𝑒, 𝜃) (if it exists) as the probability that 𝜃 is true among the 𝐾˜ (𝝈)’s. Call a sentence 𝜆 of the form 𝑃((∃𝑋) [ 𝑓1 (𝑋) = 0], . . . , (∃𝑋) [ 𝑓𝑚 (𝑋) = 0])

(23.16)

with 𝑓1 , . . . , 𝑓𝑚 ∈ 𝐾 [𝑋] separable polynomials and 𝑃 a Boolean polynomial (Section 8.7), a test sentence. In this case it is fairly easy to describe the set 𝑆(𝐾, 𝑒, 𝜆). Indeed, the splitting field 𝐿 of the polynomial 𝑓1 · · · 𝑓𝑚 is a finite Galois extension of 𝐾. Denote the set of all 𝝉 ∈ Gal(𝐿/𝐾) 𝑒 with 𝐿(𝝉) |= 𝜆 by 𝑆0 . Then, 𝑆(𝐾, 𝑒, 𝜆) = {𝝈 ∈ Gal(𝐾) 𝑒 | res 𝐿 (𝝈) ∈ 𝑆0 }.

(23.17)

Indeed, if 𝜆 has the form (∃𝑋) [ 𝑓1 (𝑋) = 0], then 𝑆0 consists exactly of the 𝝉 ∈ Gal(𝐿/𝐾) 𝑒 for which 𝐿 (𝝉) contains at least one root of 𝑓1 (𝑋). Since 𝐿 contains all roots of 𝑓1 (𝑋), this gives (23.17). An induction on the structure of 𝑃 gives (23.17) in general. Let 𝜇 be the Haar measure of Gal(𝐾) 𝑒 . From (23.17), 𝜇(𝑆(𝐾, 𝑒, 𝜆)) =

|𝑆0 | . [𝐿 : 𝐾] 𝑒

(23.18)

In the explicit case the right-hand side of (23.18) can be computed effectively from Lemma 22.3.2. Lemma 23.6.1 Let 𝐾 be a field, 𝑒 a positive integer, and 𝜆 a test sentence. Then, 𝜇(𝑆(𝐾, 𝑒, 𝜆)) is a rational number which, in the explicit case, can be effectively computed. The reduction of arbitrary sentences to test sentences depends on a general result of field theory: Lemma 23.6.2 Let 𝐾 ⊆ 𝐿 ⊆ 𝐿 ′ be a tower of fields. Suppose that 𝐿 ′/𝐾 is algebraic and 𝐿 𝐾 𝐿 ′. Then, 𝐿 = 𝐿 ′. Proof. Consider 𝑥 ∈ 𝐿 ′ and let 𝑓 = irr(𝑥, 𝐾). Denote the set of all zeros of 𝑓 in 𝐿 (resp. 𝐿 ′) by 𝑍 (resp. 𝑍 ′). Then, 𝑍 ⊆ 𝑍 ′ and |𝑍 | = |𝑍 ′ |, so 𝑍 = 𝑍 ′. Therefore, 𝑥 ∈ 𝐿. □ The proof of Lemma 23.6.4 uses the following well-known result from linear algebra: Lemma 23.6.3 Let 𝑉 be a vector space over a field 𝐾. (a) If 𝐾 has at least 𝑚 elements, then 𝑉 is not the union of 𝑚 proper subspaces. (b) If 𝐾 is infinite, then every finite union of proper subspaces of 𝑉 is a proper subset of 𝑉. (c) If 𝐾 is infinite and 𝑊, 𝑉1 , . . . , 𝑉𝑛 are subspaces of 𝑉 such that 𝑊 ⊆ 𝑉1 ∪ · · · ∪𝑉𝑛 , then there exists a 𝑗 with 𝑊 ⊆ 𝑉 𝑗 . Proof of (a). Statement (a) is obvious for 𝑚 = 1. Let 𝑚 ≥ 2 and assume that Statement (a) holds for 𝑚 − 1. Ð𝑚 Assume that 𝑉1 , . . . , 𝑉𝑚 are proper subspaces of 𝑉 with 𝑉 = 𝑖=1 𝑉𝑖 . The induction hypothesis implies that 𝑉2 ∪ · · · ∪ 𝑉𝑚 < 𝑉. Hence, there exists

23.6 The Probable Truth of a Sentence

467

𝑣 1 ∈ 𝑉1 ∖ 𝑉2 ∪ 𝑉3 ∪ · · · ∪ 𝑉𝑚 .

(23.19)

𝑣 2 ∈ 𝑉2 ∖ 𝑉1 ∪ 𝑉3 ∪ · · · ∪ 𝑉𝑚 .

(23.20)

Similarly, there exists By (23.19) and (23.20), 𝛼𝑣 1 + 𝑣 2 ∉ 𝑉1 for each 𝛼 ∈ 𝐾. Since 𝐾 has at least 𝑚 elements, there exist, by the pigeonhole principle, distinct elements 𝛼, 𝛽 ∈ 𝐾 and 2 ≤ 𝑗 ≤ 𝑚 such that 𝛼𝑣 1 + 𝑣 2 ∈ 𝑉 𝑗 and 𝛽𝑣 1 + 𝑣 2 ∈ 𝑉 𝑗 . Thus, (𝛼 − 𝛽)𝑣 1 ∈ 𝑉 𝑗 , so 𝑣 1 ∈ 𝑉 𝑗 , contradicting (23.19). Proof of (b). Statement (b) is a special case of Statement (a). Proof of (c). Assume toward contradiction that 𝑊 ̸ ⊆ 𝑉 𝑗 for 𝑗 = 1, . . . , 𝑛 Then, 𝑊 ∩ 𝑉1 , . . . , 𝑊 ∩ 𝑉𝑛 are proper subspaces of 𝑊 and 𝑊 = (𝑊 ∩ 𝑉1 ) ∪ · · · ∪ (𝑊 ∩ 𝑉𝑛 ). □ This contradicts (b). Lemma 23.6.4 Let 𝐸 and 𝐹 be fields having a common subfield 𝐾. (a) Suppose that each irreducible polynomial 𝑓 ∈ 𝐾 [𝑋] which has a root in 𝐸 has a root in 𝐹. Then, there exists a 𝐾-embedding of 𝐾˜ ∩ 𝐸 into 𝐾˜ ∩ 𝐹. (b) Suppose that an irreducible polynomial 𝑓 ∈ 𝐾 [𝑋] has a root in 𝐸 if and only if it has a root in 𝐹. Then, 𝐾˜ ∩ 𝐸 𝐾 𝐾˜ ∩ 𝐹. Proof. We may assume in both parts of the lemma that 𝐸 and 𝐹 are algebraic over 𝐾. Proof of (a). If 𝐾 is finite and 𝑥 ∈ 𝐸, then 𝐾 (𝑥) is a Galois extension of 𝐾. By assumption, irr(𝑥, 𝐾) has a root in 𝐹. Hence, 𝐾 (𝑥) ⊆ 𝐹. Therefore, 𝐸 ⊆ 𝐹. Now assume that 𝐾 is infinite. Let 𝐿 be a finite extension of 𝐾 in 𝐸. Choose a finite normal extension 𝐿ˆ of 𝐾 which contains 𝐿. Put 𝐿 ′ = 𝐿ˆ ∩ 𝐹. List the 𝐾-isomorphisms of 𝐿 ′ into 𝐾˜ as 𝜎1 , . . . , 𝜎𝑛 . Put 𝐿 𝑖 = 𝜎𝑖 (𝐿 ′), 𝑖 = 1, . . . , 𝑛. By assumption, for each 𝑥 ∈ 𝐿 the polynomial 𝑓 = irr(𝑥, 𝐾) has a root 𝑥 ′ ∈ 𝐿 ′. Extend the map 𝑥 ↦→ 𝑥 ′ to a ˆ Then, 𝜏 −1 | 𝐿′ = 𝜎𝑖 for some 𝑖, so 𝑥 = 𝜏 −1 (𝑥 ′) = 𝜎𝑖 (𝑥 ′) ∈ 𝐿 𝑖 . 𝐾-automorphism 𝜏 of 𝐿. It follows that 𝐿 ⊆ 𝐿 1 ∪ · · · ∪ 𝐿 𝑛 . Now consider each of the fields 𝐿, 𝐿 1 , . . . , 𝐿 𝑛 as a subspace of the finite dimensional vector space 𝐿ˆ over 𝐾. Since 𝐾 is infinite, there ′ exists a 𝑗 with 𝐿 ⊆ 𝐿 𝑗 (Lemma 23.6.3(c)). Therefore, 𝜎 −1 𝑗 (𝐿) ⊆ 𝐿 ⊆ 𝐹. Denote the finite nonempty set of all 𝐾-embeddings of 𝐿 into 𝐹 by 𝐼 (𝐿). If 𝐿 is contained in another finite extension 𝐿 1 of 𝐾, contained in 𝐸, then restriction defines a canonical map of 𝐼 (𝐿 1 ) into 𝐼 (𝐿). Take the inverse limit of the 𝐼 (𝐿)’s to establish the existence of a 𝐾-embedding of 𝐸 into 𝐹 (Corollary 1.1.4), as desired. Proof of (b). By (a), the assumption of (b) gives a 𝐾-embedding 𝛼: 𝐸 → 𝐹 and a 𝐾-embedding 𝛽: 𝐹 → 𝐸. Then, 𝛼(𝛽(𝐹)) 𝐾 𝐹. Hence, by Lemma 23.6.2, 𝛼(𝛽(𝐹)) = 𝐹. Thus, 𝛼 is surjective, hence 𝛼 is a 𝐾-isomorphism, as claimed. □ Corollary 23.6.5 Let 𝐸 and 𝐹 be perfect fields with a common subfield 𝐾. (a) Suppose that each separable irreducible polynomial 𝑓 ∈ 𝐾 [𝑋] which has a root in 𝐸 has a root in 𝐹. Then, 𝐾˜ ∩ 𝐸 can be 𝐾-embedded into 𝐾˜ ∩ 𝐹. (b) Suppose that each separable irreducible polynomial 𝑓 ∈ 𝐾 [𝑋] has a root in 𝐸 if and only if it has a root in 𝐹. Then, 𝐾˜ ∩ 𝐸 𝐾 𝐾˜ ∩ 𝐹.

468

23 The Elementary Theory of 𝑒-Free PAC Fields

Proof. Lemma 23.6.4 covers the case where char(𝐾) = 0, so we assume char(𝐾) = 𝑝 > 0. Let 𝑓 ∈ 𝐾 [𝑋] be an irreducible polynomial. Then, there exists a separable irreducible polynomial 𝑔 ∈ 𝐾 [𝑋] and a power 𝑞 of 𝑝 such that 𝑓 (𝑋) = 𝑔(𝑋 𝑞 ). Since 𝐸 (resp. 𝐹) is perfect, 𝑓 has a root in 𝐸 (resp. 𝐹) if and only if 𝑔 has one, as well. Therefore, we may apply Lemma 23.6.4 to conclude both (a) and (b). □ We combine Corollary 23.4.2 with Corollary 23.6.5: Lemma 23.6.6 Let (𝐾, 𝑒) be a Hilbertian pair and 𝐸 and 𝐹 models of Almost(𝐾, 𝑒). Then, 𝐸 ≡𝐾 𝐹 is and only if 𝐸 and 𝐹 satisfy exactly the same test sentences. Proof. Suppose that 𝐸 and 𝐹 satisfy the same test sentences. Then, a separable polynomial 𝑓 ∈ 𝐾 [𝑋] has a root in 𝐸 if and only if 𝑓 has a root in 𝐹. Since 𝐸 and 𝐹 are models of Almost(𝐾, 𝑒), they are perfect (Theorem 23.5.4). Hence, by Corollary 23.6.5, 𝐾˜ ∩ 𝐸 𝐾 𝐾˜ ∩ 𝐹. Thus, by Corollary 23.4.2, 𝐸 ≡𝐾 𝐹. □ Proposition 23.6.7 Let (𝐾, 𝑒) be a Hilbertian pair. For each sentence 𝜃 of L (ring, 𝐾) there exists a test sentence 𝜆 satisfying: (a) The sets 𝑆(𝐾, 𝑒, 𝜃) and 𝑆(𝐾, 𝑒, 𝜆) differ only by a zero set; the sentence 𝜃 ↔ 𝜆 belongs to Almost(𝐾, 𝑒). (b) There is a formal proof (𝛿1 , . . . , 𝛿 𝑛 ) of 𝜃 ↔ 𝜆 from the set of axioms Ax(𝐾, 𝑒); both 𝜆 and (𝛿1 , . . . , 𝛿 𝑛 ) can be found in the explicit case in a recursive way by checking all proofs from Ax(𝐾, 𝑒). Proof. Proposition 8.8.2 gives a test sentence 𝜆 satisfying (a). By Theorem 23.5.4, Ax(𝐾, 𝑒) is a set of axioms for Almost(𝐾, 𝑒). Therefore, by Corollary 9.2.6, Ax(𝐾, 𝑒) ⊢ 𝜃 ↔ 𝜆, which is the first part of (b). The second part of (b) follows □ from Proposition 9.7.2. We use Proposition 23.6.7 to generalize Lemma 23.6.1 to arbitrary sentences: Theorem 23.6.8 ([JaK75]) Let (𝐾, 𝑒) be a Hilbertian pair and 𝜃 a sentence of L (ring, 𝐾). Then, 𝜇(𝑆(𝐾, 𝑒, 𝜃)) is a rational number which, in the explicit case, can be recursively computed. In particular, the theory Almost(𝐾, 𝑒) is recursively decidable. Proof. Indeed, 𝜇(𝑆(𝐾, 𝑒, 𝜃)) = 1 if and only if the sentence 𝜃 belongs to Almost(𝐾, 𝑒). □ Remark 23.6.9 Chapter 34 uses algebraic geometry Almost(𝐾, 𝑒) is primitive recursive (Theorem 34.7.2).

to

prove

that

23.7 Change of Base Field Suppose that 𝜃 is a sentence of L (ring, 𝐾) and 𝐾 ′ is a field containing 𝐾. Then, 𝜃 is also a sentence of L (ring, 𝐾 ′). It is therefore natural, to relate 𝑆(𝐾, 𝑒, 𝜃) and 𝑆(𝐾 ′, 𝑒, 𝜃). More generally, we relate 𝑆(𝐾, 𝑒, 𝜃) and 𝑆(𝐾 ′, 𝑒, 𝜃) when 𝜃 is an “infinite

23.7 Change of Base Field

469

Ô∞ to the language sentence”. To define the latter concept, we adjoin the symbol 𝑖=1 and define the set of infinite sentences to be the smallest set of strings that satisfy the following rules: (23.21a) Every sentence of L (ring, 𝐾) is an infinite sentence. (23.21b) If 𝜃 is an infinite sentence, then so is ¬𝜃. Ô∞ (23.21c) For 𝜃 1 , 𝜃 2 , 𝜃 3 , . . . a sequence of infinite sentences, 𝑖=1 𝜃 𝑖 is an infinite sentence. The interpretation of infinite Ô∞ sentences is given by the following rule: For 𝐹 a field extension of 𝐾, 𝐹 |= 𝑖=1 𝜃 𝑖 if 𝐹 |= 𝜃 𝑖 for some integer 𝑖. If 𝜃 is an infinite sentence, use (23.15) to define its truth set, 𝑆(𝐾, 𝑒, 𝜃). Observe that the operator 𝑆(𝐾, 𝑒, ∗) commutes with infinite disjunctions. Also, if fields 𝐸 and 𝐹 containing 𝐾 are 𝐾-elementarily equivalent, then the same infinite sentences are true in both of them. Consider a regular extension 𝐾 ′ of 𝐾. Let 𝜌: Gal(𝐾 ′) 𝑒 → Gal(𝐾) 𝑒 be the restriction map. Then, 𝐾˜ ∩ 𝐾˜ ′ (𝝈) = 𝐾˜ (𝜌(𝝈)) = 𝐾˜ (𝝈) for each 𝝈 ∈ Gal(𝐾 ′) 𝑒 . Denote the measure of Gal(𝐾 ′) 𝑒 by 𝜇 ′ and use the rule 𝜇 ′ (𝜌 −1 ( 𝐴)) = 𝜇( 𝐴) for each measurable subset 𝐴 of Gal(𝐾) 𝑒 (Proposition 21.2.2). Theorem 23.7.1 ([JaJ84], Thm. 1.1) Let (𝐾, 𝑒) and (𝐾 ′, 𝑒) be Hilbertian pairs such that 𝐾 ′ is a regular extension of 𝐾. Then: (a) For almost all 𝝈 ∈ Gal(𝐾 ′) 𝑒 , 𝐾˜ (𝝈) ≺ 𝐾˜ ′ (𝝈). (b) For each infinite sentence 𝜃 of L (ring, 𝐾), 𝑆(𝐾 ′, 𝑒, 𝜃) and 𝜌 −1 (𝑆(𝐾, 𝑒, 𝜃)) differ only by a zero set. (c) 𝜇 ′ (𝑆(𝐾 ′, 𝑒, 𝜃)) = 𝜇(𝑆(𝐾, 𝑒, 𝜃)). Proof. Denote the set of all 𝝈 in Gal(𝐾) 𝑒 (resp. in Gal(𝐾 ′) 𝑒 ) with 𝐾˜ (𝝈) 𝑒-free and PAC by 𝑆 (resp. 𝑆 ′). By Theorem 23.5.1, 𝜇(𝑆) = 1 and 𝜇 ′ (𝑆 ′) = 1. Hence, 𝜇 ′ (𝜌 −1 (𝑆) ∩ 𝑆 ′) = 1. By Corollary 23.4.3, 𝐾˜ (𝝈) is an elementary subfield of 𝐾˜ ′ (𝝈) for every 𝝈 ∈ 𝜌 −1 (𝑆) ∩ 𝑆 ′. This completes the proof of (a). Statement (b) follows □ from (a); and (c) follows from (b). Consider the special case where 𝐾 ′ = 𝐾 (𝑡 1 , . . . , 𝑡𝑟 ) is the field of rational functions over 𝐾 in the variables 𝑡 1 , . . . , 𝑡𝑟 . Let 𝑅 = 𝐾 [𝑡 1 , . . . , 𝑡𝑟 ] be the corresponding ring of polynomials. Regard a sentence of L (ring, 𝑅) as a formula 𝜃 (𝑡 1 , . . . , 𝑡𝑟 ) of L (ring, 𝐾) involving the variables 𝑡1 , . . . , 𝑡𝑟 . If 𝑎 1 , . . . , 𝑎𝑟 ∈ 𝐾, then 𝜃 (a) is a sentence of L (ring, 𝐾). The next theorem generalizes Hilbert’s irreducibility theorem (Exercise 4): Theorem 23.7.2 Let 𝐾 be a countable Hilbertian field, 𝑒 a positive integer, 𝜃 (𝑡 1 , . . . , 𝑡𝑟 ) a sentence of L (ring, 𝑅). Then, there exists a separable Hilbert subset 𝐻 of 𝐾 𝑟 such that 𝜇 ′ (𝑆(𝐾 ′, 𝑒, 𝜃 (t))) = 𝜇(𝑆(𝐾, 𝑒, 𝜃 (a))) (23.22) for each a ∈ 𝐻. Proof. Using test sentences divides the proof into two parts. Part A: Reduction to test sentences. Proposition 23.6.7 gives a test
sentence 𝜆(t) for 𝜃 (t) of the form 𝑃 (∃𝑋) [ 𝑓1 (t, 𝑋) = 0], . . . , (∃𝑋) [ 𝑓𝑚 (t, 𝑋) = 0] with 𝑓1 , . . . , 𝑓𝑚 ∈ 𝐾 (t) [𝑋] separable polynomials and 𝑃 a boolean polynomial. Moreover, there exists

470

23 The Elementary Theory of 𝑒-Free PAC Fields

a formal proof (𝛿1 (t), . . . , 𝛿 𝑛 (t)) of 𝜃 (t) ↔ 𝜆(t) from Ax(𝐾 ′, 𝑒). The axioms of the positive diagram of 𝐾 ′ (Example 8.3.1) involved in this proof have the form 𝑟 1 (t) + 𝑟 2 (t) = 𝑟 3 (t) or 𝑟 1 (t) · 𝑟 2 (t) = 𝑟 3 (t) where 𝑟 1 , 𝑟 2 , 𝑟 3 ∈ 𝐾 (t). Define 𝑈 to be the 𝐾-Zariski open set of all a ∈ A𝑟 at which none of the denominators of the 𝑓𝑖 ’s and the 𝑟 𝑗 ’s vanishes. Thus, 𝐾˜ (𝝈) |= 𝛿𝑖 (a) for each a ∈ 𝑈 (𝐾), 𝑖 = 1, . . . , 𝑛. Therefore, for each a ∈ 𝑈 (𝐾) and for each 𝝈 ∈ Gal(𝐾) 𝑒 with 𝐾˜ (𝝈) 𝑒-free and PAC, 𝐾˜ (𝝈) |= 𝜃 (a) ↔ 𝜆(a). Now apply Theorem 23.5.1 to both 𝐾 and 𝐾 ′ to obtain (23.23a) 𝜇 ′ (𝑆(𝐾 ′, 𝑒, 𝜃 (t))) = 𝜇 ′ (𝑆(𝐾 ′, 𝑒, 𝜆(t))) and (23.23b) 𝜇(𝑆(𝐾, 𝑒, 𝜃 (a))) = 𝜇(𝑆(𝐾, 𝑒, 𝜆(a)))

for each a ∈ 𝑈 (𝐾).

Part B: Test sentences. Let 𝐿 ′ be the splitting field of 𝑓1 (t, 𝑋) · · · 𝑓𝑚 (t, 𝑋) over 𝐾 ′. Take a primitive element 𝑧 for 𝐿 ′/𝐾 ′ which is integral over 𝑅 and let 𝑔(t, 𝑋) = irr(𝑧, 𝐾 ′). Put ℎ = 𝑔 𝑓1 · · · 𝑓𝑚 . Denote the set of all a ∈ 𝑈 (𝐾) such that the discriminants (and therefore also the leading coefficients) of 𝑓1 , . . . , 𝑓𝑚 , 𝑔 remain nonzero under the specialization t → a by 𝐻 ′. Make 𝐻 ′ smaller, if necessary, to assume that the specialization t → a induces an isomorphism of Gal(𝐿 ′/𝐾 ′) onto Gal(𝐿/𝐾), where 𝐿 is the splitting field of ℎ(a, 𝑋) over 𝐾, that preserves the operation on the roots of 𝑓1 , . . . , 𝑓𝑛 (Lemma 14.1.1(a)). For a ∈ 𝐻 ′, the number of 𝝈 ′ ∈ Gal(𝐿 ′/𝐾 ′) 𝑒 with 𝐿 ′ (𝝈 ′) |= 𝜆(t) is equal to the number of 𝝈 ∈ Gal(𝐿/𝐾) 𝑒 such that 𝐿(𝝈) |= 𝜆(a). Hence 𝜇 ′ (𝑆(𝐾 ′, 𝑒, 𝜆(t))) = 𝜇(𝑆(𝐾, 𝑒, 𝜆(a))). Therefore, (23.22) follows from (23.23a)–(23.23b). By Lemma 14.1.1, 𝐻 ′ contains a Hilbert subset of 𝐾 𝑟 . The theorem follows. □

Remark 23.7.3 With 𝐾 fixed, Section 34.7 analyzes the effect of a change in 𝑒 on 𝜇(𝑆(𝐾, 𝑒, 𝜃)).

23.8 The Fields 𝑲sep (𝝈1 , . . . , 𝝈𝒆 ) The free generators theorem (Theorem 21.5.6) and the PAC Nullstellensatz (Theorem 21.6.1) establish properties satisfied by almost all fields 𝐾sep (𝝈). As applications, however, the last sections developed a theory of properties shared by almost all fields 𝐾˜ (𝝈). The next section explains this shift. Comparison of the theory of these fields to the theory of finite (and therefore perfect) fields forces us to consider the perfect fields 𝐾˜ (𝝈) rather than the imperfect fields 𝐾sep (𝝈). Nevertheless, if we replace 𝐾˜ (𝝈) by 𝐾sep (𝝈) and make some obvious changes, the results in Sections 23.4, 23.5, and 23.6 as well as their proofs remain valid. First, the analog of Corollary 23.4.2: Proposition 23.8.1 Let 𝐸 and 𝐹 be separable extensions of a field 𝐾 with the same imperfect degree. Suppose that 𝐸 and 𝐹 are 𝑒-free, PAC, 𝐾sep ∩ 𝐸 𝐾 𝐾sep ∩ 𝐹. Then, 𝐸 ≡𝐾 𝐹.

23.8 The Fields 𝐾sep ( 𝜎1 , . . . , 𝜎𝑒 )

471

Now replace Ax(𝐾, 𝑒) by a set of axioms Ax′ (𝐾, 𝑒). A field extension 𝐹 of 𝐾 satisfies Ax′ (𝐾, 𝑒) if and only if 𝐹 is PAC, 𝑒-free, [𝐹 : 𝐹 𝑝 ] = [𝐾 : 𝐾 𝑝 ], and 𝐹/𝐾 is separable. To express the separability of 𝐹/𝐾 by sentences of L (ring, 𝐾), choose a 𝑝-basis 𝐵 of 𝐾 over 𝐾 𝑝 . Then, “𝐹/𝐾 is separable” if and only if “𝐵0 is 𝑝-independent over 𝐹 𝑝 ” for all finite subsets 𝐵0 of 𝐵. Having done this, we write the analog of Theorem 23.5.1: Theorem 23.8.2 Suppose that (𝐾, 𝑒) is a Hilbertian pair. Then, for almost all 𝝈 ∈ Gal(𝐾) 𝑒 , the field 𝐾sep (𝝈) is 𝑒-free, PAC, separable over 𝐾, and [𝐾sep (𝝈) : 𝐾sep (𝝈) 𝑝 ] = [𝐾 : 𝐾 𝑝 ]. We denote the theory of all sentences of L (ring, 𝐾) which are true in 𝐾sep (𝝈) for almost all 𝝈 ∈ Gal(𝐾) 𝑒 by Almost ′ (𝐾, 𝑒). Then, the analog of Lemma 23.5.2 holds: Lemma 23.8.3 Let (𝐾, 𝑒) be a Hilbertian pair. (a) A field 𝐹 is a model of Almost ′ (𝐾, 𝐸) if and only if it is 𝐾-elementarily equivalent to a regular ultraproduct of the fields 𝐾sep (𝝈). (b) Every regular ultraproduct of the fields 𝐾sep (𝝈) is 𝑒-free and PAC, has the same imperfect degree as 𝐾, and is separable over 𝐾. Now we present the analog of Lemma 23.5.3: Lemma 23.8.4 Let 𝐾 be a field, 𝑒 a positive integer, and 𝐹 a field of corank at most 𝑒 which is separable over 𝐾. Then, there exists a regular ultraproduct 𝐸 of the 𝐾sep (𝝈)’s with 𝐾sep ∩ 𝐸 𝐾 𝐾sep ∩ 𝐹. This gives the analog of Theorem 23.5.4: Theorem 23.8.5 Let (𝐾, 𝑒) be a Hilbertian pair. Then, Ax′ (𝐾, 𝑒) is a set of axioms for Almost ′ (𝐾, 𝑒). Finally, for a sentence 𝜃 of L (ring, 𝐾) let 𝑆 ′ (𝐾, 𝑒, 𝜃) = {𝝈 ∈ Gal(𝐾) 𝑒 | 𝐾sep (𝝈) |= 𝜃}. Then, the following analog of Theorem 23.6.8 holds: Theorem 20.8.6: Let (𝐾, 𝑒) be a Hilbertian pair and 𝜃 a sentence of L (ring, 𝐾). Then, 𝜇(𝑆 ′ (𝐾, 𝑒, 𝜃)) is a rational number which, in the explicit case, can be recursively computed. In particular, the theory Almost ′ (𝐾, 𝑒) is recursively decidable. The results of Section 23.7 are in general false for the fields 𝐾sep (𝝈). Suppose for example that 𝐾 = F 𝑝 (𝑡) and 𝐾 ′ = F 𝑝 (𝑡, 𝑢) with 𝑡, 𝑢 algebraically independent elements over F 𝑝 . Then, the separable extensions of 𝐾 have imperfect exponent 1 while the separable extensions of 𝐾 ′ have imperfect exponent 2. Therefore, no ′ (𝝈) is elementarily equivalent to 𝐾 (𝝈). 𝐾sep sep

472

23 The Elementary Theory of 𝑒-Free PAC Fields

23.9 The Transfer Theorem This section connects the elementary theory of finite fields with the elementary theory of the fields 𝐾˜ (𝜎). Let 𝐾 be a global field and 𝑂 𝐾 the ring of integers of 𝐾. Denote the set of nonzero prime ideals of 𝑂 𝐾 by 𝑃(𝐾). It is equipped with the Dirichlet density 𝛿 (Section 7.3). We consider models of the language L (ring, 𝑂 𝐾 ) that are field extensions either of 𝐾 or of one of the residue fields 𝐾¯ 𝔭 , for 𝔭 ∈ 𝑃(𝐾). By Proposition 8.8.1, a sentence 𝜃 of L (ring, 𝑂 𝐾 ) is true in 𝐾¯ 𝔭 for almost all if and only if 𝜃 is true in every nonprincipal ultraproduct of the 𝐾¯ 𝔭 ’s. If 𝔭 ∈ 𝑃(𝐾) Î 𝐹 = 𝐾¯ 𝔭 /D is one of these ultraproducts, then, by Proposition 8.9.1 and Corollary 12.3.4, 𝐹 is a perfect, 1-free, PAC field that contains 𝐾. Since 𝐾 is a global field, it is Hilbertian (Theorem 14.4.2). Hence, by Theorem 23.5.4, 𝐹 is a model of Almost(𝐾, 1). We have therefore proved: Lemma 23.9.1 If a sentence 𝜃 of L (ring, 𝑂 𝐾 ) is true in 𝐾˜ (𝜎) for almost all 𝜎 ∈ Gal(𝐾), then 𝜃 is true in 𝐾¯ 𝔭 for almost all 𝔭 ∈ 𝑃(𝐾). For a given sentence 𝜃 of L (ring, 𝑂 𝐾 ) we compare the sets 𝑆(𝜃) = 𝑆(𝐾, 1, 𝜃) = {𝜎 ∈ Gal(𝐾) | 𝐾˜ (𝜎) |= 𝜃} 𝐴(𝜃) = 𝐴(𝐾, 𝜃) = {𝔭 ∈ 𝑃(𝐾) | 𝐾¯ 𝔭 |= 𝜃},

and

using the Dirichlet density 𝛿 of 𝑃(𝐾) and the Haar measure 𝜇 of Gal(𝐾). Lemma 23.9.2 Let 𝜆 be the test sentence
𝑝 (∃𝑋) [ 𝑓1 (𝑋) = 0], . . . , (∃𝑋) [ 𝑓𝑚 (𝑋) = 0] ,

(23.24)

where 𝑓1 , . . . , 𝑓𝑚 ∈ 𝐾 [𝑋] are separable polynomials and 𝑝 is a Boolean polynomial. Let 𝐵 be the set of all 𝔭 ∈ 𝑃(𝐾) such that all coefficients of 𝑓𝑖 are 𝔭-integral and the leading coefficient and the discriminant of the 𝑓𝑖 ’s are 𝔭-units, 𝑖 = 1, . . . , 𝑚. Denote the splitting field of 𝑓1 · · · 𝑓𝑚 over 𝐾 by 𝐿. Then: (a) For each 𝔭 ∈ 𝐵, every 𝔓 ∈ 𝑃(𝐿) over 𝔭, every 𝜎 ∈ 𝐷 𝔓 , and every field extension ¯ is the image of 𝜎 under the map ¯ (where 𝜎 𝐹 of 𝐾¯ 𝔭 satisfying 𝐿¯ 𝔓 ∩ 𝐹 = 𝐿¯ 𝔓 ( 𝜎) 𝐷 𝔓 → Gal( 𝐿¯ 𝔓 /𝐾¯ 𝔭 ) induced by 𝔓) we have 𝐿(𝜎) |= 𝜆 ⇐⇒ 𝐹 |= 𝜆. | 𝑆¯ (𝜆) | ¯ = {𝜎 ∈ Gal(𝐿/𝐾) | 𝐿 (𝜎) |= 𝜆}. Then, 𝛿( 𝐴(𝜆)) = [𝐿:𝐾 (b) Let 𝑆(𝜆) ]. Proof of (a). First note that 𝐵 is a cofinite set, because each 𝑓𝑖 is separable. Suppose that 𝔭 ∈ 𝐵. Then, 𝔭 is unramified in 𝐿 (Section 7.2). If 𝜆 is (∃𝑋) [ 𝑓𝑖 (𝑋) = 0], statement (a) is a reinterpretation of Lemma 7.1.8(a). The general case follows by induction on the structure of 𝜆. Proof of (b). Use (a) and the Chebotarev density theorem (Theorem 7.3.1).

□

Theorem 23.9.3 (The Transfer Theorem) Let 𝑂 𝐾 be the ring of integers of a global field 𝐾 and 𝜃 a sentence of L (ring, 𝑂 𝐾 ). Then, 𝑆(𝜃) is measurable, 𝐴(𝜃) has a Dirichlet density, and 𝜇(𝑆(𝜃)) = 𝛿( 𝐴(𝜃)).

23.9 The Transfer Theorem

473

Proof. Proposition 23.6.7 provides a test sentence 𝜆 of the form (23.24) with 𝜃 ↔ 𝜆 true in 𝐾˜ (𝜎) for almost all 𝜎 ∈ Gal(𝐾). Without loss assume the coefficients of the polynomials 𝑓1 , . . . , 𝑓𝑚 belong to 𝑂 𝐾 . Thus (Lemma 23.9.1), 𝜃 ↔ 𝜆 is true in 𝐾¯ 𝔭 , for almost all 𝔭 ∈ 𝑃(𝐾). Hence, 𝑆(𝜃) ≈ 𝑆(𝜆) (i.e. 𝑆(𝜃) and 𝑆(𝜆) differ by a set of measure zero) and 𝐴(𝜃) ≈ 𝐴(𝜆) (i.e. 𝐴(𝜃) and 𝐴(𝜆) differ by a finite set). Therefore, it suffices to prove the theorem for 𝜆, rather than for 𝜃. Let 𝐿 be the splitting field of the polynomial 𝑓1 · · · 𝑓𝑚 . Then, 𝐿 is a finite Galois ¯ := {𝜏 ∈ Gal(𝐿/𝐾) | 𝐿 (𝜏) |= 𝜆} is a union of conjugacy classes extension of 𝐾, 𝑆(𝜆) | 𝑆¯ (𝜆) | ¯ Hence, 𝜇(𝑆(𝜆)) = [𝐿:𝐾 of Gal(𝐿/𝐾), and 𝑆(𝜆) = {𝜎 ∈ Gal(𝐾) | res 𝐿 𝜎 ∈ 𝑆(𝜆)}. ]. By Lemma 23.9.2,   𝐿/𝐾 ¯ 𝐴(𝜆) ≈ {𝔭 ∈ 𝑃(𝐾) | ⊆ 𝑆(𝜆)} (23.25) 𝔭 and 𝛿( 𝐴(𝜆)) =

| 𝑆¯ (𝜆) | [𝐿:𝐾 ] .

Consequently, 𝜇(𝑆(𝜆)) = 𝛿( 𝐴(𝜆)).

□

¯ If 𝛿( 𝐴(𝜆)) = 0, then 𝜇(𝑆(𝜆)) = 0 and 𝑆(𝜆) is empty. Therefore, by (23.25), 𝐴(𝜆), hence 𝐴(𝜃), are finite sets. Therefore, Theorem 23.6.8 gives the following supplement to Theorem 23.9.3: Theorem 23.9.4 ([Ax67], p. 161, Cor.) Consider a sentence 𝜃 of L (ring, 𝑂 𝐾 ). Then, 𝛿( 𝐴(𝜃)) is a rational number which is positive if 𝐴(𝜃) is infinite. In the explicit case 𝛿( 𝐴(𝜃)) can be recursively computed. The special case 𝐾 = Q gives the decidability of the theory of sentences which are true in F 𝑝 for almost all 𝑝. The next result represents this theory in two more ways. Consider the set Q of all powers of prime numbers. Call a subset 𝐵 of Q small if only finitely many prime numbers divide the elements of 𝐵. Proposition 23.9.5 The following three statements about a sentence 𝜃 of L (ring) are equivalent: ˜ (a) Q(𝜎) |= 𝜃 for almost all 𝜎 ∈ Gal(Q). (b) F 𝑝 |= 𝜃 for almost all 𝑝 ∈ 𝑃(Q). (c) F𝑞 |= 𝜃 for almost all 𝑞 ∈ Q (i.e. for all but a small set). Proof. The equivalence “(a) ⇐⇒ (b)” is a special case of the transfer theorem. The implication “(c) =⇒ (b)” follows from the definitions. Finally, the implication “(a) Î =⇒ (c)” follows from Lemma 23.5.2(a), because every regular ultraproduct F𝑞 /D, where D contains all complements of small sets of Q, is a 1-free PAC field of characteristic zero (Proposition 8.9.1 and Corollary 12.3.4). □ Corollary 23.9.6 ([Ax68], p. 265) Let 𝐾 be a given global field. Then, the theory of all sentences in L (ring, 𝑂 𝐾 ) which are true in 𝐾¯ 𝔭 for almost all 𝔭 ∈ 𝑃(𝐾) is recursively decidable. Proof. By Theorem 23.9.4, the set of all sentences 𝜃 of L (ring, 𝐾) with 𝛿( 𝐴(𝜃)) = 1 is recursive. □ If 𝛿( 𝐴(𝜃)) = 1, then there are only finitely many 𝔭 ∈ 𝑃(𝐾) for which 𝜃 may be false in 𝐾¯ 𝔭 . The proof of the next theorem gives a recursive procedure for displaying these primes:

474

23 The Elementary Theory of 𝑒-Free PAC Fields

Theorem 23.9.7 ([Ax68], p. 264) Let 𝐾 be a given global field. Then, the theory of all sentences true in 𝐾¯ 𝔭 for all 𝔭 ∈ 𝑃(𝐾) is recursively decidable. Proof. We follow the pattern of proof of Theorem 23.9.3. Part A: Finding a test sentence. Let 𝜃 ∈ L (ring, 𝑂 𝐾 ). Proposition 23.6.7 recursively gives a test sentence 𝜆 of the form (23.16) and a formal proof (𝛿1 , . . . , 𝛿 𝑛 ) from Ax(𝐾, 1) (in the language L (ring, 𝐾)) of 𝜃 ↔ 𝜆. Let 𝐴0 be the set of all 𝔭 ∈ 𝑃(𝐾) that divide one of the denominators of the elements of 𝐾 involved in 𝛿1 , . . . , 𝛿 𝑛 . If a field 𝐹 of characteristic not in 𝐴0 contains a homomorphic image of 𝑂 𝐾 and if the axioms among the 𝛿𝑖 are true in 𝐹, then (𝛿1 , . . . , 𝛿 𝑛 ) is a valid proof in 𝐹. In particular, 𝜃 ↔ 𝜆 is true in 𝐹. Part B: Reduction modulo 𝔭 of the PAC axioms. Let 1 ≤ 𝑖 ≤ 𝑛. If 𝛿𝑖 is the axiom “each absolutely irreducible polynomial 𝑓 (𝑋, 𝑌 ) of degree 𝑑 has a zero”, define 𝐴𝑖 to be the set of all 𝔭 ∈ 𝑃(𝐾) with | 𝐾¯ 𝔭 | ≤ (𝑑 −1) 4 . By Corollary 6.4.2, if 𝔭 ∈ 𝑃(𝐾) ∖ 𝐴𝑖 , then 𝐾¯ 𝔭 |= 𝛿𝑖 . Otherwise, let 𝐴𝑖 = 𝐴0 . Let 𝐵1 = 𝐴0 ∪ 𝐴1 ∪ · · · ∪ 𝐴𝑛 . Since 𝐾¯ 𝔭 is perfect and 1-free, 𝐾¯ 𝔭 |= 𝜃 ↔ 𝜆 for each 𝔭 ∈ 𝑃(𝐾) ∖ 𝐵1 . Part C: Exceptional primes for 𝜆. Construct the splitting field 𝐿 of the product of the polynomials 𝑓1 , . . . , 𝑓𝑚 appearing in 𝜆 and check if there exists a 𝜏 ∈ Gal(𝐿/𝐾) with 𝐿(𝜏) ̸ |= 𝜆. In this case, 𝜆 (and therefore 𝜃) is false in 𝐾¯ 𝔭 for almost all 𝔭 ∈ 𝑃(𝐾)
with 𝜏 ∈ 𝐿/𝐾 (Lemma 23.9.2). By Chebotarev, infinitely many primes 𝔭 satisfy the 𝔭 latter condition. Assume therefore that 𝐿 (𝜏) |= 𝜆 for each 𝜏 ∈ Gal(𝐿/𝐾). Compute a finite subset 𝐵2 of 𝑃(𝐾) including all 𝔭 which are either ramified in 𝐿 or divide one of the discriminants (or the leading coefficients) of 𝑓1 , . . . , 𝑓𝑚 . Let 𝐵 = 𝐵1 ∪ 𝐵2 . Then, 𝐾¯ 𝔭 |= 𝜆 (Lemma 23.9.2), so 𝐾¯ 𝔭 |= 𝜃 for each 𝔭 ∈ 𝑃(𝐾) ∖ 𝐵. Complete the procedure by checking whether 𝐾¯ 𝔭 |= 𝜃 for each of the finitely many exceptional primes 𝔭 ∈ 𝐵. □

23.10 The Elementary Theory of Finite Fields We conclude with a discussion of the theory of finite fields and its decidability. Call a field 𝐹 pseudo finite if 𝐹 is perfect, Gal(𝐹)  Zˆ (i.e. 1-free), and PAC. Lemma 23.10.1 Every nonprincipal ultraproduct of distinct finite fields is pseudo finite. Proof. Let 𝐹1 , 𝐹2 , 𝐹3 , . . . be distinct finite fields. Consider a nonprincipal ultrafilter Î D on N and let 𝐹 = 𝐹𝑖 /D. Then, 𝐹 is perfect, 1-free, and PAC (Proposition 8.9.1 and Corollary 12.3.4). Thus, 𝐹 is pseudo finite. □ The following result is a special case of Corollary 23.4.2 in the case 𝑒 = 1: Proposition 23.10.2 Let 𝐸 and 𝐹 be pseudo finite fields. ˜ Then, (a) Suppose that 𝐸 and 𝐹 contain a common field 𝐾 and 𝐸 ∩ 𝐾˜ 𝐾 𝐹 ∩ 𝐾. 𝐸 ≡𝐾 𝐹. (b) Suppose that 𝐹 is a regular extension of 𝐸. Then, 𝐸 ≺ 𝐹.

23.10 The Elementary Theory of Finite Fields

475

Lemma 23.10.3 Let 𝐾 be a finite field and 𝐿 an algebraic extension of 𝐾. For each positive integer 𝑛 denote the unique extension of 𝐾 of degree 𝑛 by 𝐾𝑛 . Then, there Î exists a nonprincipal ultraproduct 𝐷 = 𝐾𝑛 /D such that 𝐷 ∩ 𝐾˜ = 𝐿. Proof. For each positive integer 𝑑 the set 𝐴𝑑 = {𝑛 ∈ N | 𝐾𝑛 ∩ 𝐾 𝑑 = 𝐿 ∩ 𝐾 𝑑 } is infinite. If 𝑑|𝑑 ′, then 𝐴𝑑′ ⊆ 𝐴𝑑 . Thus, given 𝑑1 , . . . , 𝑑𝑟 , we put 𝑑 = lcm(𝑑1 , . . . , 𝑑𝑟 ) and conclude from the relation 𝐴𝑑 ⊆ 𝐴𝑑1 ∩ · · · ∩ 𝐴𝑑𝑟 that the latter intersection is an infinite set. By Lemma 8.5.4, Î there exists a nonprincipal ultrafilter D on N which contains each 𝐴𝑑 . Let 𝐷 = 𝐾𝑛 /D be the corresponding ultraproduct. We prove that 𝐷 ∩ 𝐾˜ = 𝐿: Consider a positive integer 𝑑, let 𝑥 be a primitive element for 𝐾 𝑑 over 𝐾, and put 𝑓 = irr(𝑥, 𝐾). Then, 𝑓 is Galois over 𝐾. If 𝐾 𝑑 ⊆ 𝐿, then 𝐾 𝑑 ⊆ 𝐾𝑛 for each 𝑛 ∈ 𝐴𝑑 , so 𝑓 has a root in 𝐾𝑛 for each 𝑛 ∈ 𝐴𝑑 . By Łoš (Proposition 8.6.1), 𝑓 has a root in 𝐷, so 𝐷 ∩ 𝐾 𝑑 = 𝐾 𝑑 ⊆ 𝐿. If 𝐾 𝑑 ̸ ⊆ 𝐿, then 𝐿 ∩ 𝐾 𝑑 ⊂ 𝐾 𝑑 , so 𝑓 has no root in 𝐾𝑛 for each 𝑛 ∈ 𝐴𝑑 . By Łoš, 𝑓 has no root in 𝐷, so 𝐷 ∩ 𝐾 𝑑 ̸ ⊆ 𝐷. It follows that 𝐷 ∩ 𝐾˜ = 𝐿. □ Proposition 23.10.4 Let 𝜃 be a sentence of L (ring). Then, 𝜃 is true in almost all (i.e. all but finitely many) finite fields if and only if 𝜃 is true in every pseudo finite field. Proof. First suppose that 𝜃 is true in almost all finite fields. Consider a pseudo finite Î field 𝐹. By Lemma 23.10.3, there exists a nonprincipal ultraproduct 𝐷 = F 𝑝 𝑛 /D such that 𝐹 ∩ F˜ 𝑝 = 𝐷 ∩ F˜ 𝑝 . By Lemma 23.10.1, 𝐷 is pseudo finite. Hence, by Proposition 23.10.2, 𝐹 ≡𝐾 𝐷. By Łoš (Proposition 8.6.1), 𝜃 is true in 𝐷. Therefore, 𝜃 is true in 𝐹. Conversely, suppose that 𝜃 is false in a sequence 𝐹1 , 𝐹2 , 𝐹3 , . . . of distinct finite Î fields. Choose a nonprincipal ultrafilter D on N. Put 𝐹 = 𝐹𝑖 /D. By Łoš, 𝜃 is false in 𝐹. By Lemma 23.10.1, 𝐹 is pseudo finite. □ Corollary 23.10.5 ([Ax68], p. 240) A field 𝐹 is pseudo finite if and only if 𝐹 is an infinite model of the theory of finite fields. Proof. First suppose that 𝐹 is pseudo finite. Then, by Proposition 23.10.4, each sentence 𝜃 which holds in every finite field holds in 𝐹. Conversely, suppose that 𝐹 is an infinite model of the theory of finite fields. Then, 𝐹 is perfect and Gal(𝐹)  Zˆ (Proposition 23.4.4). For each positive integer 𝑑, let 𝜃 𝑑 be the sentence “there are at most (𝑑 − 1) 4 distinct elements or every absolutely irreducible polynomial in the variables 𝑋, 𝑌 of degree at most 𝑑 has a zero”. By Corollary 6.4.2(b), 𝜃 𝑑 holds in each finite field. Hence, 𝜃 𝑑 holds in 𝐹. But 𝐹 is infinite. Hence, every absolutely irreducible polynomial of degree 𝑑 has a zero in 𝐹. □ Consequently, 𝐹 is PAC and therefore pseudo finite. Like the theory of all residue fields of a given global field, the theory of finite fields is decidable. This was a problem raised by Tarski and solved by Ax: Theorem 23.10.6 ([Ax68], p. 264) The theory of all sentences of L (ring) which are true in every finite field is recursively decidable.

476

23 The Elementary Theory of 𝑒-Free PAC Fields

Proof. Consider 𝜃 ∈ L (ring). Follow the proof of Theorem 23.9.7 in the case 𝐾 = Q to check if 𝜃 is true in F 𝑝 for all 𝑝 ∈ 𝑃(Q). In the affirmative case, choose an integer 𝑛 greater than (𝑑 − 1) 4 for all 𝑑’s that appear in Part B and greater than the primes belonging to Part C of that proof. Then, conclude from Lemma 23.9.2, as at the end of Part C of the proof of Theorem 23.9.7, that F 𝑝𝑖 |= 𝜃 for each 𝑝 ≥ 𝑛 and each 𝑖. We therefore only need to check if, for a given prime 𝑝, the sentence 𝜃 is true in F 𝑝𝑖 for all 𝑖 ∈ N. This is equivalent to checking if 𝜃 is true in F 𝑝 (𝑡)𝔭 for each 𝔭 ∈ 𝑃(F 𝑝 (𝑡)). This, again, is a special case of Theorem 23.9.7. Now, the proof is complete. □

Example 23.10.7 (Pseudo finite fields) (a) Let 𝐾 be a global field and 𝜃 a sentence of L (ring, 𝑂 𝐾 ). Suppose that 𝜃 holds in 𝐾¯ 𝔭 with 𝔭 ranging over an infinite set 𝐴 of prime divisors of 𝐾. Then, 𝜃 holds in each nonprincipal ultraproduct 𝐹 of the 𝐾¯ 𝔭 ’s. Reduction modulo 𝔭 embeds 𝑂 𝐾 in 𝐹. Thus, 𝐹 extends 𝐾. By Lemma 23.10.1, 𝐹 is pseudo finite. (b) Let 𝐾 be a global field. Then, 𝐾˜ (𝜎) is pseudo finite for almost all 𝜎 ∈ Gal(𝐾) (Theorem 23.5.1 for 𝑒 = 1). Consider a sentence 𝜃 of L (ring, 𝑂 𝐾 ) which holds in 𝐾¯ 𝔭 for infinitely many prime 𝔭 of 𝐾. By Theorem 23.9.4, 𝛿( 𝐴(𝜃)) > 0. Hence, by the transfer theorem, 𝜇(𝑆(𝜃)) > 0. In particular, there is a 𝜎 ∈ Gal(𝐾) such that 𝐾˜ (𝜎) is pseudo finite and 𝜃 holds in 𝐾˜ (𝜎). (c) Let 𝐹 be an infinite algebraic extension of F 𝑝 . Then, 𝐹 is perfect and PAC Î (Corollary 12.2.4). Moreover, Gal(𝐹)  𝑙 ∈𝑆 Z𝑙 for some set 𝑆 of prime numbers (Exercise 8 of Chapter 1)). Suppose that 𝐹 has an extension of degree 𝑙 for every ˆ Hence, 𝐹 is a pseudo finite field. prime number 𝑙. Then, Gal(𝐹)  Z.

TheoremÎ23.10.8 Let 𝐾 be a global field. Denote the set of all nonprincipal ultraproducts 𝐾¯ 𝔭 /D, where 𝔭 ranges over 𝑃(𝐾), by F . Let 𝑇 be the set of all sentences 𝜃 ∈ L (ring, 𝑂 𝐾 ) which hold in almost all residue fields 𝐾¯ 𝔭 . Then: (a) Each 𝐹 ∈ F is pseudo finite. (b) A sentence 𝜃 of L (ring, 𝑂 𝐾 ) belongs to 𝑇 if and only if 𝜃 holds in every 𝐹 ∈ F . (c) Every model of 𝑇 is elementarily equivalent in L (ring, 𝑂 𝐾 ) to some 𝐹 ∈ F . (d) For every 𝜎 ∈ Gal(𝐾) there is an 𝐹 ∈ F with 𝐹 ∩ 𝐾˜ 𝐾 𝐾˜ (𝜎). Proof. Statement (a) is a special case of Lemma 23.10.1. Statement (b) is a special case of Proposition 8.8.1(a). Statement (c) is a special case of Proposition 8.8.1(b). Statement (d) can be proved directly from the Chebotarev density theorem, but we deduce it here from Lemma 23.5.3. By that lemma, there is a regular ultraproduct 𝐸 of the 𝐾˜ (𝜏)’s with 𝐸 ∩ 𝐾˜ 𝐾 𝐾˜ (𝜎). By Lemma 23.5.2, 𝐸 is a pseudo finite field. By Proposition 23.10.4, 𝐸 is a model of 𝑇. By (c), there is an 𝐹 ∈ F which is elementarily ˜ equivalent in L (ring, 𝑂 𝐾 ) to 𝐸. It follows from 23.6.4, that 𝐸 ∩ 𝐾˜ 𝐾 𝐹 ∩ 𝐾. ˜ ˜ Consequently, 𝐹 ∩ 𝐾 𝐾 𝐾 (𝜎). □

23.10 The Elementary Theory of Finite Fields

477

Exercises 1. Give an example showing that the hypothesis of Corollary 23.3.4 can hold nontrivially, by taking 𝐾 to be a PAC field for which Gal(𝐾) is finitely generated. Then, let 𝐹 be a nonprincipal ultraproduct of countably many copies of 𝐾. 2. Let 𝜆 be the test sentence (Section 23.6) (∃𝑋) [ 𝑓1 (𝑋) = 0∨ 𝑓2 (𝑋) = 0∨ 𝑓3 (𝑋) = 0] with 𝑓𝑖 (𝑋) = 𝑋 2 − 𝑎 𝑖 and nonzero integers 𝑎 𝑖 ∈ Z for 𝑖 = 1, 2, 3. What is the exact condition on 𝑎 1 , 𝑎 2 , 𝑎 3 such that 𝜇(𝑆(𝐾, 1, 𝜆)) = 1? Now answer the same question for 𝜇(𝑆(𝐾, 2, 𝜆)) = 1. 3. [Jar80a, p. 149] It is a consequence of Proposition 23.6.7(a) that if (𝐾, 𝑒) is a Hilbertian pair and 𝜃 is a sentence of L (ring, 𝐾), then 𝑆(𝐾, 𝑒, 𝜃) is a measurable set. We outline an alternative proof, valid for every countable field 𝐾 and every positive integer 𝑒: Let 𝜑(𝑋1 , . . . , 𝑋𝑛 ) be a formula of L (ring, 𝐾). For 𝑥1 , . . . , 𝑥 𝑛 ∈ 𝐾˜ define 𝑆(𝐾, 𝑒, 𝜑(x)) = {𝝈 ∈ Gal(𝐾) 𝑒 | 𝑥1 , . . . , 𝑥 𝑛 ∈ 𝐾˜ (𝝈) and 𝐾˜ (𝝈) |= 𝜑(x)}. (a) Use induction on structure to show that 𝑆(𝐾, 𝑒, 𝜑(x)) is a Borel set. Ð Hint: Use the identity 𝑆(𝐾, 𝑒, (∃𝑌 )𝜑(x, 𝑌 )) = 𝑦 ∈ 𝐾˜ 𝑆(𝐾, 𝑒, 𝜑(x, 𝑦)). (b) Deduce that 𝑆(𝐾, 𝑒, 𝜃) is a Borel set for each infinite sentence of L (ring, 𝐾). 4. Prove the converse of Theorem 23.7.2: Let 𝐾 be a field and 𝑡 an indeterminate. Consider the ring 𝑅 = 𝐾 [𝑡] and the field 𝐾 ′ = 𝐾 (𝑡). Suppose that for each positive integer 𝑒 and each sentence 𝜃 (𝑡) of L (ring, 𝑅) there exists a nonempty subset 𝐻 of 𝐾 with 𝜇 ′ (𝑆(𝐾 ′, 𝑒, 𝜃 (𝑡))) = 𝜇(𝑆(𝐾, 𝑒, 𝜃 (𝑎))) for each 𝑎 ∈ 𝐻. Prove that 𝐾 is Hilbertian. Hint: Let 𝑓 ∈ 𝐾 [𝑋] be separable. Then, 𝑓 is irreducible over 𝐾 if and only if for all large 𝑒 there exists an 𝑆 ⊆ Gal(𝐾) 𝑒 of positive measure such that 𝑓 is irreducible over 𝐾˜ (𝝈) for all 𝝈 ∈ 𝑆. 5. ([Ax68, p. 260]) Let 𝐾 be a global field and 𝜏 ∈ Gal(𝐾). Prove that there exists a Î nonprincipal ultraproduct 𝐹 = 𝐾¯ 𝔭 /D with 𝐾˜ ∩ 𝐹  𝐾˜ (𝜏). Hint: Either reproduce Ax’s direct application of the Chebotarev density theorem, or combine the transfer theorem, Lemma 23.5.3, and Proposition 8.8.1. 6. Give an example of an infinite sentence of L (ring) which is true in each field ˜ Q(𝜎) but is false in each field F 𝑝 . Thus, the transfer theorem does not generalize to infinite sentences. Indeed, the Haar measure is 𝜎-additive while the Dirichlet density is only finitely-additive. 7. Verify the transfer theorem over Q for the test sentence (∃𝑋) [𝑋 2 = 𝑎], where 𝑎 is an integer, by using quadratic reciprocity and the Dirichlet density theorem (Corollary 7.3.2) for primes in arithmetic progressions. 8. Let 𝐾 be a global field and 𝑓 ∈ 𝐾 [𝑋] be separable and irreducible with deg( 𝑓 ) = 𝑛 > 1. Denote the set of prime numbers 𝑝 such that 𝑓 has no zero mod 𝑝 by 𝐴. Put 𝐺 = Gal( 𝑓 , 𝐾) and let 𝑆 be the set of all 𝜎 ∈ 𝐺 which fix no zero of 𝑓 . |𝑆 | (a) Apply the transfer theorem to prove: 𝛿( 𝐴) = |𝐺 | . Alternatively, use the Chebotarev density theorem to prove the equality directly. 1 (b) Suppose that Gal( 𝑓 , 𝐾)  𝑆 𝑛 . Prove: 𝛿( 𝐴) = 2!1 − 3!1 + · · · + (−1) 𝑛 𝑛! .

478

23 The Elementary Theory of 𝑒-Free PAC Fields

9. Let 𝑓 ∈ Z[𝑋, 𝑌 ] be an absolutely irreducible polynomial. Prove that for almost all primes 𝑝 there exists an 𝑎 ∈ F 𝑝 such that 𝑓 (𝑎, 𝑌 ) decomposes into linear factors in F 𝑝 [𝑌 ]. Hint: Let 𝐹 = Q(𝑥, 𝑦) be the function field of 𝑉 ( 𝑓 ). Choose a stabilizing element 𝑡 for the extension 𝐹/Q (Theorem 21.11.3). Let 𝐹ˆ be the Galois hull of 𝐹/Q(𝑡). One way to complete the proof is to specialize 𝑡 to elements of Q infinitely often so that 𝑥, 𝑦 are integral over the corresponding local ring and such that the residue fields of 𝐹ˆ form a linearly disjoint sequence of Galois extensions of Q (as in the proof of Theorem 21.12.3). Use Borel–Cantelli (Lemma 21.3.5) to prove that for almost ˜ all 𝜎 ∈ Gal(𝐾) there exists an 𝑎 ∈ Q(𝜎) such that 𝑓 (𝑎, 𝑌 ) decomposes into linear factors. Now use the transfer principle. ˆ Alternatively, let 𝑧 be a primitive element for the extension 𝐹/Q(𝑡). Let 𝑔 ∈ Z[𝑇, 𝑍] be an absolutely irreducible polynomial with 𝑔(𝑡, 𝑧) = 0. Now find an ℎ ∈ Z[𝑋], ℎ ≠ 0, such that for almost all 𝑝 and for each 𝑏 ∈ F 𝑝 with ℎ(𝑏) ≠ 0 and 𝑔(𝑏, 𝑇) has zeros in F 𝑝 the specialization 𝑡 → 𝑏 extends to a specialization (𝑡, 𝑥) → (𝑏, 𝑎) and the polynomial 𝑓 (𝑎, 𝑌 ) decomposes into linear factors. Thus, the result follows directly from Corollary 6.4.2. 10. Show that the small sets of Q in Section 23.9 cannot be replaced by finite sets, by giving a sentence 𝜃 ∈ L (ring) for which 𝜃 is true for all 𝑝 ∈ 𝑃(Q) but the 𝑞 ∈ Q for which F𝑞 ̸ |= 𝜃 is infinite. 11. Let 𝐾 be a number field. Denote the set of all prime ideals of 𝑂 𝐾 by 𝑃(𝐾). For each finite extension 𝐿 of 𝐾 let Splt(𝐿/𝐾) be the set of all 𝔭 ∈ 𝑃(𝐾) which split completely in 𝐿. (a) Prove that there exists an ultrafilter L on 𝑃(𝐾) which contains Splt(𝐿/𝐾) for each finite extension 𝐿 of 𝐾Î and each subset of 𝑃(𝐾) of Dirichlet density 1. ˜ ⊆ 𝐹. (b) Let L be as in (a). Put 𝐹 = 𝔭∈𝑃 (𝐾) 𝐾¯ 𝔭 /L. Prove that Q ′ (c) Let L be another ultrafilter on 𝑃(𝐾) satisfying the conditions of (a). Put 𝐹 ′ = Î ¯ 𝐾𝔭 /L ′. Prove that 𝐹 ′ ≡𝐾 𝐹.

Notes Ax initiated the investigation of free PAC fields in [Ax67] and [Ax68] by connecting the theory of finite fields with the theory of pseudo finite fields. This connection is based on the special case of the embedding lemma (Lemma 23.2.2), where 𝐸 and 𝐹 are 1-free and Gal(𝐿) is procyclic (Ax’s proof [Ax68, p. 248] is very complicated), and on Exercise 5. The transfer theorem [Jar72] strengthens this connection. The treatment of perfect 𝑒-free PAC fields can be found in [JaK75]. [CDM82] introduces the treatment of imperfect fields that are 𝑒-free and PAC. Finally, we follow [JaJ84] and include finite fields, with 𝑒 = 1, as base fields, in addition to the countable Hilbertian fields. In [Ax67, Lemma 5], Ax proves Lemma 23.6.4(b) essentially for the case where both 𝐾˜ ∩𝐸/𝐾 and 𝐾˜ ∩𝐹/𝐾 are separable extensions. His reduction of the general case to the separable case is vague. The argument which appears in our proof and which uses vector spaces is due independently to B. Poizat [Poi80] and to H. W. Lenstra.

Chapter 24

Problems of Arithmetical Geometry

We apply the model theory – measure theory technique of Chapter 23 to concrete field arithmetic problems. The transfer principle between properties of finite fields and properties of the fields 𝐾˜ (𝜎) can often be accomplished through direct application of the Chebotarev density theorem. Usually, however, application of the transfer theorem avoids a repetition of arguments. This chapter includes the theory of 𝐶𝑖 -fields, Kronecker conjugacy of global field extensions, Davenport’s problem on value sets of polynomials over finite fields, a solution of Schur’s conjecture on permutation polynomials, and a solution of the generalized Carlitz conjecture on the degree of a permutation polynomial in characteristic 𝑝. Each of these concrete problems focuses our attention on rich historically motivated concepts that could be overlooked in an abstract model-theoretic viewpoint.

24.1 The Decomposition-Intersection Procedure The classical diophantine concern is the description of the Q-rational (resp. Zrational) points of a Zariski Q-closed set 𝐴. The decomposition-intersection procedure reduces this concern to the study of a union 𝐴∗ of subvarieties (Section 11.2) of 𝐴 defined over Q. We start with an arbitrary base field 𝐾. To each nonempty Zariski 𝐾-closed set 𝐴 in the affine space, A𝑛 , or in the projective space, P𝑛 , there corresponds a canonical ′ 𝐴 into its 𝐾-components, 𝐾-closed Ð subset 𝐴 defined as follows: First, decompose Ð ˜ 𝑉𝑖 = 𝑗 𝑊𝑖 𝑗 . For a 𝐴 = 𝑖 𝑉𝑖 . Then, decompose each 𝑉𝑖 into its 𝐾-components, fixed 𝑖, {𝑊𝑖 𝑗 } 𝑗 is a complete Ñ set of conjugate varieties, defined over 𝐾˜ (Section 11.2). 𝑈𝑖 = 𝑗 𝑊𝑖 𝑗 is invariant under the action of Gal(𝐾): 𝑈𝑖 is a Thus, the intersectionÐ 𝐾-closed set. Denote 𝑖 𝑈𝑖 by 𝐴 ′. Continue the procedure to obtain a descending sequence 𝐴 ⊇ 𝐴 (1) ⊇ 𝐴 (2) ⊇ · · · ⊇ 𝐴 (𝑚) ⊇ 𝐴 (𝑚+1) ⊇ · · · of 𝐾-closed sets, with 𝐴 (𝑚+1) = ( 𝐴 (𝑚) ) ′. Hilbert’s basis theorem (Lemma 11.1.1) gives an integer 𝑚 with 𝐴 (𝑟) = 𝐴 (𝑚) for all 𝑟 ≥ 𝑚. Denote 𝐴 (𝑚) by 𝐴∗ . 479 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_24

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˜ Let 𝑊1 , . . . , 𝑊𝑠 be the list of all 𝐾-components of 𝐴, 𝐴 (1) , . . . , 𝐴 (𝑚) . The compositum, 𝐿 ′, of the fields of definition of 𝑊1 , . . . , 𝑊𝑠 is a finite normal extension of 𝐾. The maximal separable extension 𝐿 of 𝐾 in 𝐿 ′ is the smallest Galois extension of 𝐾 such that each of 𝑊1 , . . . , 𝑊𝑠 is 𝐿-closed. Call 𝐿 the Galois splitting field of 𝐴 over 𝐾. Lemma 24.1.1 Let 𝑀 be a field extension of 𝐾 with 𝐿 ∩ 𝑀 = 𝐾. Then: (a) 𝐴∗ contains all 𝑀-closed subvarieties of 𝐴. In particular, 𝐴∗ (𝑀) = 𝐴(𝑀). (b) 𝐴∗ is nonempty if and only if 𝐴 contains an 𝑀-closed nonempty subvariety. Proof of (a). Let 𝑊 be a nonempty 𝑀-closed subvariety of 𝐴. Choose a generic point x of 𝑊 over 𝑀. Then, x is a generic point of 𝑊 over 𝐿 𝑀. Also, there exist 𝑖 and 𝑗 with x ∈ 𝑊𝑖 𝑗 . Hence, 𝑊 ⊆ 𝑊𝑖 𝑗 . Each 𝜎 ∈ Gal(𝐿/𝐾) extends to an element of Gal(𝐿 𝑀/𝑀). Since, for a fixed 𝑖, the 𝑊𝑖 𝑗 ’s are conjugate under the action of Ñ Gal(𝐿/𝐾), we have 𝑊 ⊆ 𝑗 𝑊𝑖 𝑗 = 𝑈𝑖 ⊆ 𝐴 ′. Proceed by induction to show that ∗ 𝑊⊆𝐴 . In particular, each 𝑀-rational point of 𝐴 belongs to 𝐴∗ . Therefore, 𝐴(𝑀) = ∗ 𝐴 (𝑀). Proof of (b). Suppose first that 𝐴∗ is nonempty. Assume without loss that 𝐴 = 𝐴∗ . Then, 𝐴 = 𝐴 ′ and 𝐴 is nonempty. For each 𝑖 we have 𝑈𝑖 ⊆ 𝑉𝑖 . Conversely, let x be a generic point of 𝑉𝑖 over 𝐾. Then, there exists an 𝑖 ′ such that x ∈ 𝑈𝑖′ . Hence, x ∈ 𝑉𝑖′ , so 𝑉𝑖 = 𝑉𝑖′ . Thus, 𝑖 = 𝑖 ′ and 𝑉𝑖 ⊆ 𝑈𝑖 . It follows that 𝑈𝑖 = 𝑉𝑖 . This implies that 𝑉𝑖 is absolutely irreducible. Otherwise there would be at least two 𝑊𝑖 𝑗 ’s and hence we Ñ would have the contradiction dim(𝑈𝑖 ) = dim( 𝑗 𝑊𝑖 𝑗 ) < dim(𝑉𝑖 ) (Lemma 11.1.2). Thus, 𝑉𝑖 is a variety which is 𝐾-closed, hence also 𝑀-closed. The converse follows from (a). □

24.2 𝑪𝒊 -Fields and Weakly 𝑪𝒊 -Fields Recall that a form 𝑓 (𝑋0 , . . . , 𝑋𝑛 ) with coefficients in a field 𝐾 defines a projective Zariski 𝐾-closed set in P𝑛 (Section 11.7). If x = (𝑥0 , . . . , 𝑥 𝑛 ) ≠ (0, . . . , 0) is a nontrivial zero of 𝑓 , then the (𝑛 + 1)-tuple (𝑎𝑥0 , . . . , 𝑎𝑥 𝑛 ) defines the same point of P𝑛 for each 𝑎 ∈ 𝐾 × . Since projective hypersurfaces are the simplest projective sets, they occupy a special place in diophantine investigations. We follow the lead of a problem due to Artin: Does each form 𝑓 ∈ F𝑞 [𝑋0 , . . . , 𝑋𝑑 ] of degree 𝑑 have a nontrivial zero? Chevalley’s affirmative solution [Che36] motivated Lang [Lan52] to explore the concept of a 𝐶𝑖 -field: Definition 24.2.1 The field 𝐾 is called 𝐶𝑖,𝑑 if each form 𝑓 ∈ 𝐾 [𝑋0 , . . . , 𝑋𝑛 ] of degree 𝑑 with 𝑑 𝑖 ≤ 𝑛 has a nontrivial zero in 𝐾 𝑛+1 . Call it 𝐶𝑖 if it is 𝐶𝑖,𝑑 for each 𝑑 ∈ N. In this case, 𝐾 is 𝐶 𝑗 for each 𝑗 ≥ 𝑖. Example 24.2.2 Every algebraically closed field 𝐾 is 𝐶0 . Indeed, every form 𝑓 ∈ 𝐾 [𝑋0 , . . . , 𝑋𝑛 ] of positive degree with 𝑛 ≥ 1 has a nontrivial zero in 𝐾 𝑛+1 .

24.2 𝐶𝑖 -Fields and Weakly 𝐶𝑖 -Fields

481

Lemma 24.2.3 (Chevalley–Warning) Let 𝑞 be a power of a prime number 𝑝 and let 𝑓1 , . . . , 𝑓𝑚 ∈ F𝑞 [𝑋1 , . . . , 𝑋𝑛 ] be polynomials with deg( 𝑓1 · · · 𝑓𝑚 ) < 𝑛. Put 𝐴 = 𝑉 ( 𝑓1 , . . . , 𝑓𝑚 ). Then, | 𝐴(F𝑞 )| ≡ 0 mod 𝑝.
Î 𝑞−1 . If x ∈ 𝐴(F ), then 𝑓 (x) = 0, Proof. Consider 𝑔(X) = 𝑚 𝑗 𝑞 𝑗=1 1 − 𝑓 𝑗 (X) 𝑗 = 1, . . . , 𝑚. Hence, 𝑔(x) = 1. If x ∈ F𝑞𝑛 ∖ 𝐴(F𝑞 ), then 𝑓 𝑗 (x) ≠ 0 for some 𝑗. Therefore, 𝑓 𝑗 (x) 𝑞−1 = 1 and 𝑔(x) = 0. This gives ∑︁ (24.1) 𝑔(x). | 𝐴(F𝑞 )| + 𝑝Z = x∈F𝑞𝑛

Í Rewrite 𝑔(X) as i∈𝐼 𝑐 i 𝑋1𝑖1 · · · 𝑋𝑛𝑖𝑛 where 𝐼 is a subset of {0, . . . , 𝑞 − 1} 𝑛 and 𝑐 i ∈ F×𝑞 for each i ∈ 𝐼. Then,    ∑︁ ∑︁ ∑︁  ∑︁ 𝑔(x) = 𝑥 𝑛𝑖𝑛 . 𝑥1𝑖1 · · · 𝑐i (24.2) x∈F𝑞𝑛

𝑥𝑛 ∈F𝑞

𝑥1 ∈F𝑞

i∈𝐼

Í 𝑖 Let i ∈ 𝐼. If there exists a 𝑗 with 𝑖 𝑗 = 0, then 𝑥 𝑗 ∈F𝑞 𝑥 𝑗𝑗 = 𝑞 = 0. Otherwise, Í𝑚 𝑛 ≤ 𝑖 1 + · · · + 𝑖 𝑛 ≤ deg(𝑔) ≤ 𝑗=1 deg( 𝑓 𝑗 ) (𝑞 − 1) < 𝑛(𝑞 − 1), so there is a 𝑗 with 1 ≤ 𝑖 𝑗 < 𝑞 − 1. Put 𝑘 = 𝑖 𝑗 . Since the polynomial 𝑋 𝑘 − 1 has at most 𝑘 roots in F×𝑞 , there is an 𝑎 ∈ F×𝑞 with 𝑎 𝑘 ≠ 1. Hence, ∑︁ ∑︁ ∑︁ (𝑎𝑥) 𝑘 = 𝑎 𝑘 𝑥𝑘 = 𝑥𝑘, 𝑥 ∈F𝑞

𝑥 ∈F𝑞

𝑥 ∈F𝑞

Í

so 𝑥 ∈F𝑞 𝑥 𝑘 = 0. Thus, the right-hand side of (24.2) is 0. By (24.1), | 𝐴(F𝑞 )| ≡ 0 mod 𝑝, as claimed. □ If 𝑓1 , . . . , 𝑓𝑚 are forms, (0, . . . , 0) ∈ 𝐴(F𝑞 ). Hence, by Lemma 24.2.3, 𝑉 (F𝑞 ) has at least 𝑝 points. An application of this argument to the case when 𝑚 = 1 gives Chevalley’s result: Proposition 24.2.4 Every finite field is 𝐶1 . Ax applied ultraproducts to deduce from Chevalley’s result that each perfect PAC field with Abelian absolute Galois group is 𝐶1 (Theorem 25.10.6). He left unsolved this question: Problem 24.2.5 (Ax) Is every perfect PAC field 𝐶1 ? If Ax’s problem has an affirmative solution, then each form of degree 𝑑 over Q in 𝑑 + 1 variables has a nonempty Q-closed subvariety (Proposition 24.3.2). This motivates the introduction of the weak 𝐶𝑖 condition. We use it to prove some results about PAC fields which are 𝐶𝑖 . Definition 24.2.6 A field 𝐾 is called weakly 𝐶𝑖,𝑑 , if for each form 𝑓 ∈ 𝐾 [𝑋0 , . . . , 𝑋𝑛 ] of degree 𝑑, with 𝑑 𝑖 ≤ 𝑛, the Zariski 𝐾-closed set 𝑉 ( 𝑓 ) of P𝑛 contains a subvariety 𝑊 which is Zariski 𝐾-closed. If 𝐾 is weakly 𝐶𝑖,𝑑 for each 𝑑 ∈ N, we say that 𝐾 is weakly 𝐶𝑖 .

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Remark 24.2.7 By definition, every 𝐶𝑖,𝑑 field is also weakly 𝐶𝑖,𝑑 . In addition, every perfect PAC field 𝐾 which is weakly 𝐶𝑖,𝑑 is also 𝐶𝑖,𝑑 . Perfectness here guarantees that a Zariski 𝐾-closed variety is defined over 𝐾 (Lemma 11.2.3). Example 24.2.8 (Separably closed fields) Let 𝐾 be a field of positive characteristic 𝑝. Suppose that 𝐾 has an infinite sequence 𝑎 0 , 𝑎 1 , 𝑎 2 , . . . of 𝑝-independent elements 𝑝 over 𝐾 𝑝 (Section 3.6). Then, 𝑎 0 , 𝑎 2 , 𝑎 2 , . . . are 𝑝-independent over 𝐾sep (Lemma Í 𝑝𝑖 3.6.3). Hence, for each 𝑖 and for all 𝑥0 , . . . , 𝑥 𝑝𝑖 ∈ 𝐾sep the relation 𝑗=0 𝑎 𝑗 𝑥 𝑝𝑗 = 0 implies 𝑥 0 , . . . , 𝑥 𝑝𝑖 = 0. Therefore, 𝐾sep is not 𝐶𝑖 although 𝐾sep is PAC (Section 12.1). On the other hand, each form 𝑓 ∈ 𝐾 [𝑋0 , 𝑋1 ] has a nontrivial zero (𝑥 0 , 𝑥1 ) ∈ 𝐾˜ 2 . Hence, 𝐾sep is weakly 𝐶0 (Lemma 24.2.9). For example, take 𝐾 = F 𝑝 (𝑡1 , 𝑡2 , 𝑡3 , . . .) with 𝑡1 , 𝑡2 , 𝑡3 , . . . algebraically independent over F 𝑝 . Then, 𝑡1 , 𝑡2 , 𝑡3 , . . . are 𝑝-independent over 𝐾 𝑝 (proof of Lemma 3.6.2). Thus, 𝐾sep is 𝐶𝑖 for no positive integer 𝑖. We explore the behavior of the weakly 𝐶𝑖 property under field extensions. Recall: An extension 𝐹 of 𝐾 is primary if 𝐾sep ∩ 𝐹 = 𝐾 (Lemma 3.5.1). Lemma 24.2.9 A field 𝐾 is weakly 𝐶𝑖,𝑑 if and only if every form 𝑓 ∈ 𝐾 [𝑋0 , . . . , 𝑋𝑛 ] of degree 𝑑 with 𝑑 𝑖 ≤ 𝑛 has a nontrivial zero x such that 𝐾 (x) is a primary extension of 𝐾. Proof. Suppose first that 𝐾 is weakly 𝐶𝑖,𝑑 . Let 𝑓 be as stated in the lemma. Then, 𝑉 ( 𝑓 ) contains a 𝐾-closed variety 𝑊. It is defined over a purely inseparable extension 𝐾 ′ of 𝐾. Let p be a generic point of 𝑊 over 𝐾 ′. Then, p is represented by a zero (𝑥 0 , . . . , 𝑥 𝑛 ) of 𝑓 where 𝑥0 , say, is transcendental over 𝐾 ′ (p) = 𝐾 ′ ( 𝑥𝑥10 , . . . , 𝑥𝑥𝑛0 ) and 𝐾 ′ (p)/𝐾 ′ is regular. Therefore, 𝐾sep ∩ 𝐾 (𝑥 0 , . . . , 𝑥 𝑛 ) = 𝐾. Conversely, suppose that 𝑓 (x) = 0 and that 𝐾 (x)/𝐾 is a primary extension. Denote the 𝐾-closed subset of P𝑛 generated by x by 𝑊. Since 𝐾 (x) is linearly disjoint from 𝐾sep over 𝐾, the set 𝑊 is 𝐾sep -irreducible (lemma 11.2.1). It follows that 𝑊 is a variety (Lemma 11.2.4). □ The form that appears in the next lemma might aptly be called a weakly normic form. Lemma 24.2.10 Assume that 𝐾 is not separably closed. Let 𝑒 0 be an integer. Then, there exist an integer 𝑒 > 𝑒 0 and a form ℎ ∈ 𝐾 [𝑌1 , . . . , 𝑌𝑒 ] of degree 𝑒 satisfying this: Suppose that 𝐾 (𝑦 1 , . . . , 𝑦 𝑒 ) is a primary extension of 𝐾. Then, ℎ(𝑦 1 , . . . , 𝑦 𝑒 ) = 0 =⇒ 𝑦 1 = · · · = 𝑦 𝑒 = 0.

(24.3)

Proof. By assumption, there exists a nontrivial Galois extension 𝐿/𝐾. Denote its degree by 𝑙 and its Galois group by 𝐺. Let 𝑤 1 , . . . , 𝑤 𝑙 be a basis for 𝐿/𝐾 and consider the form Ö 𝑔(X) = (𝑤 1𝜎 𝑋1 + · · · + 𝑤 𝑙𝜎 𝑋𝑙 ) 𝜎 ∈𝐺

of degree 𝑙 and with coefficients in 𝐾. Suppose that 𝐾 (y)/𝐾 is a primary extension and 𝑔(y) = 0. Then, there exists a 𝜎 ∈ 𝐺 with 𝑤 1𝜎 𝑦 1 + · · · + 𝑤 𝑙𝜎 𝑦 𝑙 = 0. Moreover, 𝐾 (y) is linearly disjoint from 𝐿 over 𝐾. Hence, 𝑦 1 = · · · = 𝑦 𝑙 = 0.

24.2 𝐶𝑖 -Fields and Weakly 𝐶𝑖 -Fields

483

Next consider the form 𝑔2 = 𝑔(𝑔(𝑋1 , . . . , 𝑋𝑙 ), 𝑔(𝑋𝑙+1 , . . . , 𝑋2𝑙 ), . . . , 𝑔(𝑋 (𝑙−1)𝑙+1 , . . . , 𝑋𝑙2 )). It is of degree 𝑙 2 and it has property (24.3) for 𝑒 = 𝑙 2 . Iterate this procedure to obtain, □ finally, a form ℎ of degree exceeding 𝑒 0 . The next results treat 𝐶𝑖 -fields and weakly 𝐶𝑖 -fields simultaneously: Lemma 24.2.11 Let 𝐾 be a 𝐶𝑖 -field (resp. weakly 𝐶𝑖 -field) and 𝑓1 , 𝑓2 , . . . , 𝑓𝑟 forms over 𝐾 of degree 𝑑 in 𝑛 variables with 𝑛 > 𝑟 𝑑 𝑖 . In the first case, suppose that either 𝐾 is algebraically closed or 𝐾 is not separably closed. Then, 𝑓1 , 𝑓2 , . . . , 𝑓𝑟 have a common nontrivial zero in 𝐾 (resp. some primary extension of 𝐾). Proof. First suppose that 𝐾 is separably closed, so by assumption either 𝐾 is an algebraically closed 𝐶𝑖 -field or 𝐾 is a separably closed weakly 𝐶𝑖 -field. By the projective dimension theorem (Section 11.7), 𝑉 ( 𝑓1 , . . . , 𝑓𝑟 ) is a nonempty Zariski ˜ which is a primary closed subset of P𝑛−1 . Thus, 𝑓1 , . . . , 𝑓𝑟 have a nontrivial zero in 𝐾, extension of 𝐾. Now assume that 𝐾 is not separably closed. Then, Lemma 24.2.10 gives a weakly normic form ℎ ∈ 𝐾 [𝑌1 , . . . , 𝑌𝑒 ] of degree 𝑒 > 𝑟 satisfying 𝑛 h𝑒i 𝑛 −1 ≥ 𝑛 + 1. (24.4) 𝑟 𝑟 For each positive integer 𝑗, let X 𝑗 = (𝑋 𝑗1 , . . . , 𝑋 𝑗𝑛 ) be a vector of variables. Let ℎ1 (X1 , . . . , X𝑞 ) = ℎ( 𝑓1 (X1 ), . . . , 𝑓𝑟 (X1 ), . . . , 𝑓1 (X𝑞 ), . . . , 𝑓𝑟 (X𝑞 ), 0, . . . , 0)     with 𝑞 = 𝑟𝑒 . Then, ℎ1 is a form of degree 𝑑1 = 𝑑𝑒 in 𝑛1 = 𝑛 𝑟𝑒 variables. If (x1 , . . . , x𝑞 ) is a nontrivial zero of ℎ1 with coordinates in 𝐾 (resp. a primary extension of 𝐾), then there exists a 𝑗 between 1 and 𝑞 with x 𝑗 ≠ 0 and 𝑓1 (x 𝑗 ) = · · · = 𝑓𝑟 (x 𝑗 ) = 0. 2 Replace  𝑛ℎ1 by ℎ1 in the above definition and define a form ℎ2 of degree 𝑑2 = 𝑑 𝑒 in 𝑛2 = 𝑛 𝑟 variables. Continue by induction to define, for every positive integer   𝑘, a form ℎ 𝑘 of degree 𝑑 𝑘 = 𝑑 𝑘 𝑒 in 𝑛 𝑘 = 𝑛 𝑛𝑘−1 variables. Every nontrivial zero of 𝑟 ℎ 𝑘 in 𝐾 (resp. in a primary extension 𝐸 of 𝐾) defines a common nontrivial zero of 𝑓1 , . . . , 𝑓𝑟 in 𝐾 (resp. in 𝐸). By assumption, a zero of that type exists if 𝑘 satisfies 𝑛 𝑘 > 𝑑 𝑖𝑘 . The lemma is done, then, if we choose 𝑘 such that 𝑛 𝑘 > 𝑑 𝑖𝑘 . Indeed, h𝑛 i 𝑛 𝑘 𝑛 𝑘+1 = 𝑛 > 𝑛 𝑘 − 𝑛. (24.5) 𝑟 𝑟 Use (24.5) for 𝑛 𝑘 rather than for 𝑛 𝑘+1 and substitute into the right-hand side of (24.5) to obtain 𝑛 𝑘+1 > ( 𝑛𝑟 ) 2 𝑛 𝑘−1 − 𝑛( 𝑛𝑟 + 1). Continue in this manner, inductively, to obtain the inequality 𝑛 𝑘   𝑛  𝑘−1  𝑛  𝑘−2  𝑛 𝑘+1 > 𝑛1 − 𝑛 + +···+1 𝑟 𝑟 𝑟 (24.6)  𝑛 𝑘  (24.4) 1   𝑛  𝑘  1  h 𝑒 i = 𝑛𝑏 −𝑛 +𝑛 ≥ +𝑛 , 𝑏 𝑟 𝑟 𝑏 𝑟

484

with 𝑏 =

24 Problems of Arithmetical Geometry 𝑛 𝑟

− 1. Therefore,

1 (( 𝑟 𝑛𝑑 𝑖 ) 𝑘 + 𝑑𝑛𝑖𝑘 ). Since 𝑛 > 𝑟 𝑑 𝑖 , the right-hand 𝑑 𝑖 𝑒𝑖 𝑏 Consequently, 𝑛 𝑘 > 𝑑 𝑖𝑘 for all large 𝑘, as claimed. □

𝑛𝑘+1 𝑖 𝑑𝑘+1

side tends to infinity with 𝑘.

>

Proposition 24.2.12 Let 𝐾 be a 𝐶𝑖 -field (resp. weakly 𝐶𝑖 -field) and 𝐸 an extension of 𝐾 of transcendence degree 𝑗. In the first case, assume that either 𝐾 is algebraically closed or 𝐾 is not separably closed. Then, 𝐸 is 𝐶𝑖+ 𝑗 (resp. weakly 𝐶𝑖+ 𝑗 ). Proof. It suffices to prove the proposition for simple transcendental extensions and then for finite separable extensions. Part A: 𝐸 = 𝐾 (𝑡), 𝑡 is transcendental over 𝐾. Let 𝑓 ∈ 𝐾 (𝑡) [𝑋0 , . . . , 𝑋𝑛 ] be a form of degree 𝑑 with 𝑛 ≥ 𝑑 𝑖+1 . Assume without loss that the coefficients of 𝑓 lie in 𝐾 [𝑡]. Denote the maximum of the degrees of the coefficients of 𝑓 by 𝑟. Let 𝑠 be a positive integer with (𝑛 − 𝑑 𝑖+1 + 1)𝑠 > 𝑑 𝑖 (𝑟 + 1) − (𝑛 + 1). Thus, (𝑛 + 1) (𝑠 + 1) > 𝑑 𝑖 (𝑑𝑠 + 𝑟 + 1).

(24.7)

Consider the (𝑛 + 1) (𝑠 + 1) variables 𝑌 𝑗 𝑘 , 𝑗 = 0, . . . , 𝑛 and 𝑘 = 0, . . . , 𝑠. There exist forms 𝑓0 (Y), . . . , 𝑓 𝑑𝑠+𝑟 (Y) over 𝐾 of degree 𝑑 with 𝑓(

𝑠 ∑︁ 𝑘=0

𝑌0𝑘 𝑡 𝑘 , . . . ,

𝑠 ∑︁ 𝑘=0

𝑌𝑛𝑘 𝑡 𝑘 ) =

𝑑𝑠+𝑟 ∑︁

𝑓𝑙 (Y)𝑡 𝑙 .

𝑙=0

By (24.7) and Lemma 24.2.11, there exists a nonzero y in 𝐾 (𝑛+1) (𝑠+1) (resp. Í 𝐾 (y)/𝐾 is a primary extension) with 𝑓0 (y) = · · · = 𝑓 𝑑𝑠+𝑟 (y) = 0. Put 𝑥 𝑗 = 𝑠𝑘=0 𝑦 𝑗 𝑘 𝑡 𝑘 , 𝑗 = 0, . . . , 𝑛. These elements are not all zero and 𝑓 (x) = 0. When 𝐾 is 𝐶𝑖 , x is 𝐾 (𝑡)-rational. Therefore, 𝐾 (𝑡) is 𝐶𝑖+1 . When 𝐾 is weakly 𝐶𝑖 , we may choose y with 𝐾 (y) algebraically independent from 𝐾 (𝑡) over 𝐾. By Lemma 3.5.4(a), 𝐾 (y, 𝑡) is a primary extension of 𝐾 (𝑡). Hence, 𝐾 (x, 𝑡)/𝐾 (𝑡) is primary. It follows from Lemma 24.2.9 that 𝐾 (𝑡) is weakly 𝐶𝑖+1 . Part B: 𝐸/𝐾 is finite. Let 𝑓 ∈ 𝐸 [𝑋0 , . . . , 𝑋𝑛 ] be a form of degree 𝑑, with 𝑛 ≥ 𝑑 𝑖 . With 𝑤 1 , . . . , 𝑤 𝑒 a basis for 𝐸/𝐾, introduce new variables 𝑍 𝑗 𝑘 , 𝑗 = 0, . . . , 𝑛 and 𝑘 = 1, . . . , 𝑒. Then, there exist forms 𝑓1 (Z), . . . , 𝑓𝑒 (Z) over 𝐾 of degree 𝑑 with 𝑒 𝑒 𝑒 ∑︁ ∑︁ ∑︁ 𝑓( 𝑍0𝑘 𝑤 𝑘 , . . . , 𝑍 𝑛𝑘 𝑤 𝑘 ) = 𝑓 𝑘 (Z)𝑤 𝑘 . 𝑘=1

𝑘=1

𝑘=1

Since (𝑛 + 1)𝑒 > 𝑒𝑑 𝑖 , Lemma 24.2.11 again gives a nonzero z such that 𝑓1 (z) = · · · = 𝑓𝑒 (z) = 0 and 𝐾 (z) = 𝐾 (resp. 𝐾 (z)/𝐾) is primary). The elements 𝑥 𝑗 = Í 𝑒 𝑘=1 𝑧 𝑗 𝑘 𝑤 𝑘 , 𝑗 = 0, . . . , 𝑛, satisfy 𝑓 (x) = 0 and not all of them are zero. If 𝐾 is 𝐶𝑖 , then 𝐸 (x) = 𝐸. Therefore, 𝐸 is 𝐶𝑖 . If 𝐾 is weakly 𝐶𝑖 , then 𝐸 (z)/𝐸 is primary (Lemma 3.5.4(a)). Hence, 𝐸 (x)/𝐸 is primary. Consequently, 𝐸 is weakly 𝐶𝑖 . □ Weakly 𝐶𝑖 -fields have a large class of subfields that are also weakly 𝐶𝑖 . Note, however, that this result does not hold for 𝐶𝑖 fields (Exercise 5). Proposition 24.2.13 If 𝐿 is a primary extension of a field 𝐾 and 𝐿 is weakly 𝐶𝑖 , then 𝐾 is also weakly 𝐶𝑖 . Proof. Since a primary extension 𝐿 (x) of 𝐿 is, under the hypotheses, a primary extension of 𝐾, this is immediate from Lemma 24.2.9. □

24.3 Perfect PAC Fields which are 𝐶𝑖

485

24.3 Perfect PAC Fields which are 𝑪𝒊 First we note a relation between weakly 𝐶𝑖 -fields, 𝐶𝑖 -fields, and the fields 𝐾˜ (𝜎): Lemma 24.3.1 Let 𝐾 be a countable Hilbertian field. Then, the following are equivalent: (a) 𝐾 is weakly 𝐶𝑖 . (b) For each 𝑒 ∈ N and for almost all 𝝈 ∈ Gal(𝐾) 𝑒 , 𝐾˜ (𝝈) is 𝐶𝑖 . Proof. If (a) holds, then every algebraic extension of 𝐾 is weakly 𝐶𝑖 (Proposition 24.2.12). By Theorem 21.6.1, 𝐾˜ (𝝈) is a perfect PAC field for almost all 𝝈 ∈ Gal(𝐾) 𝑒 . Since a perfect weakly 𝐶𝑖 PAC field is 𝐶𝑖 (Remark 24.2.7), this gives (b). Assume now that (b) holds. Let 𝑓 ∈ 𝐾 [𝑋0 , . . . , 𝑋𝑛 ] be a form of degree 𝑑 with 𝑑 𝑖 ≤ 𝑛. Apply the decomposition-intersection procedure to the subset 𝐴 = 𝑉 ( 𝑓 ) of P𝑛 (Section 24.1). Let 𝐿 be the Galois splitting field of 𝐴 over 𝐾. Choose generators 𝜎1′ , . . . , 𝜎𝑒′ for Gal(𝐿/𝐾). Then, there exist 𝜎1 , . . . , 𝜎𝑒 ∈ Gal(𝐾) that respectively extend 𝜎1′ , . . . , 𝜎𝑒′ , and for which 𝐾˜ (𝝈) is 𝐶𝑖 . In particular, 𝐿 ∩ 𝐾˜ (𝝈) = 𝐾 and 𝐴 has a 𝐾˜ (𝝈)-rational point. By Lemma 24.1.1(a), 𝐴∗ is nonempty. With 𝐾 replacing 𝑀 in Lemma 24.1.1(b), conclude that 𝐴 contains a nonempty subvariety which is 𝐾-closed. Hence, 𝐾 is weakly 𝐶𝑖 . □ We do not expect a general field extension of a 𝐶𝑖 -field to be 𝐶𝑖 (Exercise 6). But, under simple conditions this is true for weakly 𝐶𝑖 fields: Proposition 24.3.2 The following conditions on a countable Hilbertian field 𝐾 are equivalent: (a) 𝐾 is weakly 𝐶𝑖 . (b) Every field extension 𝐹 of 𝐾 is weakly 𝐶𝑖 . (c) Every perfect PAC field extension 𝐹 of 𝐾 is 𝐶𝑖 . Proof of “(a) =⇒ (b)”. Assume without loss 𝐹 is also countable. If 𝐹/𝐾 is algebraic, then 𝐹 is weakly 𝐶𝑖 (Proposition 24.2.12). Suppose that 𝐹/𝐾 is transcendental. Let 𝐾 ′ be a purely transcendental extension of 𝐾 with 𝐹 algebraic over 𝐾 ′. By Proposition 14.2.1, 𝐾 ′ is Hilbertian. By Lemma 24.3.1, the field 𝐾˜ (𝝈) is 𝐶𝑖 for each 𝑒 ∈ N and almost all 𝝈 ∈ Gal(𝐾) 𝑒 . Hence, by Theorem 23.7.1, 𝐾˜ ′ (𝝈) is 𝐶𝑖 for each 𝑒 ∈ N and almost all 𝝈 ∈ Gal(𝐾 ′) 𝑒 . It follows from Lemma 24.3.1 that 𝐾 ′ is weakly 𝐶𝑖 . Consequently, 𝐹 is weakly 𝐶𝑖 (Proposition 24.2.12). Proof of “(b) =⇒ (c)”. Use Remark 24.2.7. Proof of “(c) =⇒ (a)”. By Theorem 21.6.1 and assumption (c), 𝐾˜ (𝝈) is 𝐶𝑖 for each 𝑒 ∈ N and for almost all 𝝈 ∈ Gal(𝐾) 𝑒 . Hence, by Lemma 24.3.1, 𝐾 is weakly 𝐶𝑖 . □ Proposition 24.2.4 says that every finite field is 𝐶1 . Hence, by Proposition 24.2.12, every algebraic extension of a finite field is 𝐶1 . Every other field contains either Q or F 𝑝 (𝑡) for some 𝑝 and transcendental element 𝑡. Each of the latter fields is countable and Hilbertian (Theorem 14.4.2). An application of Proposition 24.3.2 to 𝑖 = 1 therefore gives this reformulation of Ax’s problem 24.2.5:

486

24 Problems of Arithmetical Geometry

Corollary 24.3.3 The following conditions are equivalent: (a) Every field is weakly 𝐶1 . (b) Each of the fields Q and F 𝑝 (𝑡) is weakly 𝐶1 . (c) Every perfect PAC field is 𝐶1 . Lemma 24.3.4 Let 𝐾 be a weakly 𝐶𝑖 -field and 𝐹 an extension of 𝐾. Then, 𝐹 is weakly 𝐶𝑖+1 . If, in addition, 𝐹 is perfect and PAC, then 𝐹 is 𝐶𝑖+1 . Proof. The second statement follows from the first by remark 24.2.7. To prove the first statement, suppose first that 𝐹/𝐾 is algebraic. By Proposition 24.2.12, 𝐹 is weakly 𝐶𝑖 . Hence, 𝐹 is weakly 𝐶𝑖+1 . Now suppose that 𝐹/𝐾 is transcendental. Use Proposition 24.2.13 and replace 𝐾 by a smaller field to assume that 𝐾 is countable. Choose 𝑡 ∈ 𝐹 transcendental over 𝐾. Then, 𝐾 (𝑡) is Hilbertian (Theorem 14.4.2) and weakly 𝐶𝑖+1 (Proposition 24.2.12). It follows from Proposition 24.3.2 that 𝐹 is weakly 𝐶𝑖+1 . □ Lemma 24.3.5 Let 𝐹 be a field. (a) Suppose that 𝐹 contains an algebraically closed field. Then, 𝐹 is weakly 𝐶1 . (b) Suppose that 𝐹 has a positive characteristic. Then, 𝐹 is weakly 𝐶2 . (c) Suppose that Gal(𝐹) is procyclic. Then, 𝐹 is weakly 𝐶1 . ˜ (d) Suppose that 𝐹 contains Q(𝜎) with 𝜎 ∈ Gal(Q). Then, 𝐹 is weakly 𝐶2 . Proof of (a). By Example 24.2.2, each algebraically closed field is 𝐶0 . Hence, by Lemma 24.3.4, 𝐹 is weakly 𝐶1 . Proof of (b). By assumption, 𝐹 contains a finite field. The latter is 𝐶1 (Proposition 24.2.4). Hence, by Lemma 24.3.4, 𝐹 is weakly 𝐶2 . Proof of (c). Let 𝐾0 be the prime field of 𝐹. Consider a form 𝑓 ∈ 𝐹 [𝑋0 , . . . , 𝑋𝑑 ] of degree 𝑑. The coefficients of 𝑓 are algebraic over a finitely generated extension 𝐾 of 𝐾0 . Also, Gal( 𝐾˜ ∩ 𝐹) is procyclic. Suppose that 𝑉 ( 𝑓 ) has a subvariety 𝑊 which is 𝐾˜ ∩ 𝐹-closed. Then, 𝑊 is also 𝐹-closed. So, assume without loss, 𝐹 is algebraic over 𝐾. Finally, replace 𝐹 by 𝐹ins , if necessary, to assume 𝐹 is perfect (Proposition 24.2.13). Now 𝐾 is either a finite field, or algebraic over Q, or transcendental over 𝐾0 . In all cases (𝐾, 1) is a Hilbertian pair in the sense of Section 23.5 (Theorem 14.4.2). By Lemma 23.5.3, there is a regular ultraproduct 𝐸 of the fields 𝐾˜ (𝜎) with 𝐾˜ ∩ 𝐸 𝐾 𝐹. By Lemma 23.5.2(b), 𝐸 is pseudo finite (Section 23.10). For each positive integer 𝑑 the 𝐶1,𝑑 property of a field is elementary. Since every finite field has this property (Proposition 24.2.4), so does every pseudo finite field (Proposition 23.10.2). Therefore, 𝐸 is 𝐶1 . Consequently, by Proposition 24.2.13, 𝐹 is weakly 𝐶1 . Proof of (d). Use (c) and Lemma 24.3.4.

□

Suppose that 𝐹 in Lemma 24.3.5 is PAC and perfect. Then, we may replace “weakly 𝐶𝑖 ” by “𝐶𝑖 ” in each of the statement of that lemma. This gives the main result of this section:

24.4 The Existential Theory of PAC Fields

487

Theorem 24.3.6 Let 𝐹 be a perfect PAC field. ˜ Then, 𝐹 is 𝐶1 . (a) Suppose that 𝐹 contains an algebraically closed field 𝐾. (b) Suppose that 𝐹 has a positive characteristic. Then, 𝐹 is 𝐶2 . (c) Suppose that Gal(𝐹) is procyclic. Then, 𝐹 is 𝐶1 . ˜ (d) Suppose that 𝐹 contains a field Q(𝜎), where 𝜎 ∈ Gal(Q). Then, 𝐹 is 𝐶2 . Remark 24.3.7 Theorem 25.10.6 strengthens part (c) of Theorem 24.3.6 relaxing the condition “Gal(𝐹) is procyclic” to “Gal(𝐹) is Abelian.” János Kollár proves in [Kol07b, Thm. 1] that every PAC field of characteristic 0 is 𝐶1 . This settles Problem 24.2.5 in characteristic 0.

24.4 The Existential Theory of PAC Fields In Section 32.10 it is shown that the theory of perfect PAC fields is undecidable. It is therefore of interest to observe that the decomposition-intersection procedure gives a decision procedure for the existential theory of PAC fields: Definition 24.4.1 Call a sentence of L (ring) existential if it has the form hÜÛ i (∃𝑋1 ) · · · (∃𝑋𝑛 ) [ 𝑓𝑖 𝑗 (X) = 0 ∧ 𝑔𝑖 (X) ≠ 0] 𝑖

(24.8)

𝑗

with 𝑓𝑖 𝑗 , 𝑔𝑖 ∈ Z[X], and 𝑖, 𝑗 range over finite sets. Replace each inequality 𝑔𝑖 (X) ≠ 0 in (24.8) by the equivalent formula (∃𝑌𝑖 ) [𝑌𝑖 𝑔𝑖 (X) − 1 = 0] to assume that the sentence has the form hÜÛ i (∃𝑋1 ) · · · (∃𝑋𝑛 ) 𝑓𝑖 𝑗 (X) = 0 . (24.9) 𝑖

𝑗

The bracketed expression in (24.9) defines a Zariski Q-closed subset 𝐴 of A𝑛 . Rewrite (24.9) as (∃𝑋1 ) · · · (∃𝑋𝑛 ) [X ∈ 𝐴]. (24.10) To test if (24.10) is true in every PAC field of characteristic 0, apply the decomposition-intersection procedure (Section 24.1, effective by Proposition 22.5.6). Let 𝐿 be the Galois splitting field of 𝐴 over Q and 𝜏1 , . . . , 𝜏𝑒 generators for Gal(𝐿/Q). The set of 𝝈 ∈ Gal(Q) 𝑒 whose restriction to 𝐿 is 𝝉 has positive measure (= ˜ 1/[𝐿 : Q] 𝑒 ). Hence, by Theorem 21.6.1, there is a 𝝈 such that 𝑀 = Q(𝝈) is PAC and 𝐿 ∩ 𝑀 = Q. If 𝐴∗ is empty, then 𝐴 has no 𝑀-rational point (Lemma 24.1.1(a)). Thus, there is a PAC field of characteristic 0 for which 24.10 is false. If 𝐴∗ is nonempty, then (Lemma 24.1.1(b)) each Q-component 𝑊 of 𝐴∗ is a subvariety of 𝐴 defined over Q (effectively computable). That variety has an 𝑀rational point in each PAC field 𝑀 of characteristic 0. In this case, (24.10) is true in every PAC field of characteristic 0.

488

24 Problems of Arithmetical Geometry

Now we check the PAC fields of positive characteristic. Proposition 11.4.2 gives us a finite set of primes 𝑆 such that for each 𝑝 ∉ 𝑆 the variety 𝑊 (as above) is well defined and remains a variety when considered as a Zariski F 𝑝 -closed set. For each 𝑝 ∈ 𝑆, repeat the decomposition-intersection procedure for 𝐴 over F 𝑝 . If 𝐴∗ is nonempty in each of these cases, then (24.10) is true in every PAC field. If, however, there exists a 𝑝 ∈ 𝑆 for which 𝐴∗ is empty over F 𝑝 , then, as above, there exists a PAC field 𝑀 such that 𝐴(𝑀) is empty. In this case (24.10) fails in some PAC field. We summarize: Theorem 24.4.2 Both the existential theory of PAC fields of a given characteristic and the existential theory of all PAC fields are primitive recursively decidable.

24.5 Kronecker Classes of Number Fields Our next example discusses the “Kronecker conjugacy” of polynomials. We place it in the framework of classical algebraic number theory and mention some results and open problems, with partial reformulation in terms of group theory. It is customary to credit Kronecker (in 1880) for the impetus to investigate the decomposition of primes in number field extensions [Jeh77]. Throughout this section fix a global field 𝐾. Definition 24.5.1 Let 𝐿 be a finite separable extension of 𝐾. Denote the set of all ¯ 𝔭 ∈ 𝑃(𝐾) that have a prime divisor 𝔓 ∈ 𝑃(𝐿) of relative degree 1 (i.e. 𝐿¯ 𝔓 = 𝐾𝔭) by 𝑉 (𝐿/𝐾). Call a finite separable extension 𝑀 of 𝐾 Kronecker conjugate to 𝐿 (over 𝐾) if 𝑉 (𝐿/𝐾) and 𝑉 (𝑀/𝐾) differ by only finitely many elements. The set of all finite separable extensions 𝑀 of 𝐾 which are Kronecker conjugate to 𝐿 is called the Kronecker class of 𝐿/𝐾. We denote it by K (𝐿/𝐾). Lemma 24.5.2 (Dedekind, Kummer) Let 𝑅 be a Dedekind domain with quotient field 𝐾. Consider a finite separable extension 𝐿 = 𝐾 (𝑥) of 𝐾 and let 𝑆 be the integral closure of 𝑅 in 𝐿. Suppose that 𝑥 is integral over 𝑅 and let 𝑓 = irr(𝑥, 𝐾). Let 𝔭 be a nonzero prime ideal of 𝑅. If 𝑅𝔭 [𝑥] is the integral closure of 𝑅𝔭 in 𝐿 (by Lemma 7.1.2, this holds for almost all 𝔭) and 𝑓 (𝑋) ≡ 𝑓1 (𝑋) 𝑒1 · · · 𝑓𝑟 (𝑋) 𝑒𝑟 mod 𝔭 is the factorization of 𝑓 (𝑋) as a product of powers of distinct monic irreducible polynomials modulo 𝔭, then 𝔭𝑆 = 𝔓1𝑒1 · · · 𝔓𝑟𝑒𝑟 , where 𝔓1 , . . . , 𝔓𝑟 are distinct prime ideals of 𝑆 with 𝑓 (𝔓𝑖 /𝔭) = deg( 𝑓𝑖 ), 𝑖 = 1, . . . , 𝑟. Proof. See [Lan70, p. 27] or [Jan73, p. 32].

□

The arithmetic condition for Kronecker conjugacy is equivalent to a Galoistheoretic condition:

24.5 Kronecker Classes of Number Fields

489

Lemma 24.5.3 Let 𝐿 1 and 𝐿 2 be finite separable extensions of 𝐾 and 𝐿 a finite Galois extension of 𝐾 which contains both 𝐿 1 and 𝐿 2 . Then, 𝐿 1 and 𝐿 2 are Kronecker conjugate over 𝐾 if and only if, with 𝐺 = Gal(𝐿/𝐾), Ø Ø Gal(𝐿/𝐿 1 ) 𝜎 = (24.11) Gal(𝐿/𝐿 2 ) 𝜎 . 𝜎 ∈𝐺

𝜎 ∈𝐺

Proof. For 𝑖 = 1, 2 let 𝑥𝑖 ∈ 𝑂 𝐿𝑖 be a primitive element of the extension 𝐿 𝑖 /𝐾 and let 𝑓𝑖 = irr(𝑥𝑖 , 𝐾). By Lemma 24.5.2, 𝑉 (𝐿 𝑖 /𝐾) differs from the set of primes 𝔭 for which (∃𝑋) [ 𝑓𝑖 (𝑋) = 0] is true in 𝐾¯ 𝔭 by only finitely many primes. Let 𝜃 be the sentence (∃𝑋) [ 𝑓1 (𝑋) = 0] ↔ (∃𝑋) [ 𝑓2 (𝑋) = 0]. Then, Kronecker conjugacy of 𝐿 1 and 𝐿 2 over 𝐾 is equivalent to the truth of 𝜃 in almost all 𝐾¯ 𝔭 . By the transfer theorem applied to test sentences (Lemma 23.9.2), this is equivalent to the truth of 𝜃 in 𝐿 (𝜎), for all 𝜎 ∈ Gal(𝐿/𝐾). But this is just a □ restatement of (24.11). If 𝐿 𝑖 /𝐾 is Galois, 𝑖 = 1, 2, and 𝐿 1 and 𝐿 2 are Kronecker conjugate over 𝐾, Lemma 24.5.3 implies that 𝐿 1 = 𝐿 2 . This is Bauer’s theorem (see also Exercise 5 of Chapter 7). But, in general, Kronecker conjugacy does not imply conjugacy of the fields 𝐿 1 and 𝐿 2 over 𝐾. This is demonstrated in the following example: degrees [Sch68], Example 24.5.4 (Kronecker conjugate extensions √︃ of Q of different 

Lemma 4) Let 𝐿 = Q(cos 27𝜋 ) and 𝑀 = Q 2 cos 27𝜋 . Then, 𝐿/Q is a cyclic extension of degree 3, 𝑀/𝐿 is a quadratic extension, 𝑀 is not Galois over Q, but 𝑀 is Kronecker conjugate to 𝐿 over Q. We give proofs of these statements.

Part A: irr(2 cos 27𝜋 , Q) = 𝑋 3 + 𝑋 2 − 2𝑋 − 1. Indeed, let 𝜁 = 𝑒 2 𝜋𝑖/7 and 𝜂 = 2 cos 27𝜋 = 𝜁 + 𝜁 −1 . Then, Q(𝜁)/Q is a cyclic extension of degree 6 and 𝜂 is fixed by complex conjugation. Hence, Q(𝜂)/Q is a cyclic extension of degree 3. The conjugates of 𝜂 over Q are 𝜂𝑖 = 𝜁 𝑖 + 𝜁 −𝑖 , 𝑖 = 1, 2, 3, and they satisfy 𝜂1 𝜂2 𝜂3 = 1,

𝜂1 𝜂2 + 𝜂1 𝜂3 + 𝜂2 𝜂3 = −2,

𝜂1 + 𝜂2 + 𝜂3 = −1.

(24.12)

Hence, irr(𝜂, Q) = 𝑋 3 + 𝑋 2 − 2𝑋 − 1. √ Part B: 𝜂 ∉ 𝐿. Assume that 𝜂 = 𝜃 2 with 𝜃 in 𝐿. Then, irr(𝜃, Q) is a polynomial of degree 3 with integral coefficients. The element 𝜃 is a zero of 𝑓 (𝑋) := 𝑋 6 + 𝑋 4 − 2𝑋 2 − 1. Therefore, irr(𝜃, Q) divides 𝑓 (𝑋): 𝑋 6 + 𝑋 4 − 2𝑋 2 − 1 = (𝑋 3 + 𝑎𝑋 2 + 𝑏𝑋 + 1) (𝑋 3 + 𝑐𝑋 2 + 𝑑𝑋 − 1) with 𝑎, 𝑏, 𝑐, 𝑑 ∈ Z. Thus, 𝑎 + 𝑐 = 0, 𝑑 − 𝑏 = 0, 𝑎𝑑 + 𝑏𝑐 = 0, 𝑎𝑐 + 𝑏 + 𝑑 = 1, and 𝑐 − 𝑎 + 𝑏𝑑 = −2. Eliminate 𝑏 and 𝑐 to come down to the equations 2𝑑 = 𝑎 2 + 1 and 2𝑎 = 𝑑 2 + 2. Then, 𝑑 ≥ 𝑎 from the first and 𝑎 > 𝑑 from the second, a contradiction. ˆ Part C: The structure of the Galois group 𝐺 := Gal( 𝑀/Q). By Part B, 𝑀 = √ √ Q( 𝜂1 ) is a quadratic extension of 𝐿. Similarly, Q( 𝜂2 ) is a quadratic extension of √ 𝐿 which we claim to be different from Q( 𝜂1 ). Otherwise, there exists an 𝑥 ∈ 𝐿 such that 𝜂2 = 𝑥 2 𝜂1 . By (24.12), 𝜂3−1 = (𝑥𝜂1 ) 2 , a contradiction to Part B. Also √︃ √ √ √ 𝜂3 = 𝜂1−1 𝜂2−1 implies that 𝑀ˆ := 𝐿 ( 𝜂1 , 𝜂2 ) is a Galois extension of 𝐿 of degree

490

24 Problems of Arithmetical Geometry

ˆ  Z/2Z × Z/2Z. The generator of the cyclic group Gal(𝐿/Q) acts 4 with Gal( 𝑀/𝐿) ˆ on Gal( 𝑀/𝐿) by permuting the three subgroups of order 2 in a cyclic way. So, 𝐺 is the semidirect product of Z/3Z with Z/2Z × Z/2Z. Part D: The Kronecker conjugacy of 𝐿 and 𝑀. Ø

𝜎 ˆ = Gal( 𝑀/𝑀)

𝜎 ∈𝐺

3 Ø

By part C we have

√ ˆ ˆ Gal( 𝑀/Q( 𝜂𝑖 )) = Gal( 𝑀/𝐿) =

Ø

𝜎 ˆ Gal( 𝑀/𝐿) .

𝜎 ∈𝐺

𝑖=1

Hence, by Lemma 24.5.3, 𝑀 and 𝐿 are Kronecker conjugate. Lemma 24.5.3 immediately gives 𝐿 1 ⊆ 𝐿 2 if 𝐿 1 /Q is Galois and 𝐿 1 and 𝐿 2 are Kronecker conjugate — just as we saw in Schinzel’s example. Call a field 𝑀 of a Kronecker class K (𝐿/𝐾) minimal if it contains no proper subfields of the same class. Lemma 24.5.5 Let 𝐿 1 and 𝐿 2 be minimal fields of a Kronecker class over 𝐾. Then, their Galois closures 𝐿ˆ 1 and 𝐿ˆ 2 over 𝐾 are equal. Proof. Let 𝑁 be a finite Galois extension of 𝐾 that contains 𝐿ˆ 1 and 𝐿ˆ 2 . By Lemma 24.5.3 Ø Ø Gal(𝑁/𝐿 1 ) 𝜎 = Gal(𝑁/𝐿 2 ) 𝜎 , 𝜎 ∈𝐺

𝜎 ∈𝐺

with 𝐺 := Gal(𝑁/𝐾). Restrict both sides to 𝐿ˆ 2 to conclude that Ø Ø Gal( 𝐿ˆ 2 /𝐿 1 ∩ 𝐿ˆ 2 ) 𝜎 = Gal( 𝐿ˆ 2 /𝐿 2 ) 𝜎 , 𝜎 ∈𝐺2

𝜎 ∈𝐺2

with 𝐺 2 := Gal( 𝐿ˆ 2 /𝐾). By Lemma 24.5.3, 𝐿 1 ∩ 𝐿ˆ 2 is Kronecker conjugate to 𝐿 2 and therefore to 𝐿 1 . Minimality of 𝐿 1 implies that 𝐿 1 = 𝐿 1 ∩ 𝐿ˆ 2 ⊆ 𝐿ˆ 2 . Therefore, 𝐿ˆ 1 ⊆ 𝐿ˆ 2 ; and, by symmetry, 𝐿ˆ 1 = 𝐿ˆ 2 . □ Corollary 24.5.6 Each Kronecker class contains only finitely many minimal fields. Remark 24.5.7 (On the size of Kronecker classes) Fix a number field 𝐾. (a) Call the number of nonconjugate minimal fields of K (𝐿/𝐾) the width and denote it by 𝜔(𝐿/𝐾). [Jeh77, §4] gives 𝐿 with 𝜔(𝐿/𝐾) arbitrarily large. The proof incorporates Exercise 7 and the theorem of Scholz–Shafarevich that each 𝑝-group is a Galois group over 𝐾. (b) On the other hand, [Jeh77, Theorem 3] gives examples of number fields 𝐾, where K (𝐿/𝐾) is infinite. Indeed, this is the case if Aut(𝐿/𝐾) contains either a nontrivial automorphism of odd order, or a cyclic group of order 8, or a quaternion group of order 8. (c) For a quadratic extension 𝐿/𝐾 [Jeh77, §6] conjectures that K (𝐿/𝐾) consists only of 𝐿. This conjecture has been proved by Saxl [Sax88, Prop. 2]. To prove the conjecture, assume 𝑀 is another field in K (𝐿/𝐾). By Lemma 24.5.3, 𝐿 ⊂ 𝑀. Replace 𝑀 by a smaller field to assume 𝑀 is a minimal extension ˆ ˆ of 𝐿. Let 𝑀ˆ be the Galois closure of 𝑀/𝐾, 𝐺 = Gal( 𝑀/𝐾), 𝐻 = Gal( 𝑀/𝐿), ˆ and 𝑈 = Gal( 𝑀/𝑀). Then, (𝐺 : 𝐻) = 2, 𝑈 is a maximal subgroup of 𝐻, and

24.6 Davenport’s Problem

491

Ð 𝐻 = 𝑔 ∈𝐺 𝑈 𝑔 . By [Jeh77, Thm. 5], 𝐻 has a unique minimal normal subgroup 𝑁 which is non-Abelian and simple. Moreover, 𝑈𝑁 = 𝐻. So, 𝑉 = 𝑈 ∩ 𝑁 is a proper subgroup of 𝑁. The minimality of 𝑁 implies 𝑁 ⊳ 𝐺. Choose 𝜎 ∈ 𝐺 ∖ 𝐻. View 𝜎 Ð Ð as an automorphism of 𝑁 acting as conjugation. Then, 𝑁 = ℎ∈𝐻 𝑉 ℎ ∩ ℎ∈𝐻 𝑉 𝜎ℎ . But this contradicts the main result of [Sax88]: Let 𝑁 be a finite simple group and 𝑉 a proper subgroup. Then, there exists an 𝑎 ∈ 𝑁 lying in no Aut(𝑁)-orbit of 𝑉. Saxl’s proof uses the classification of finite simple groups, proving the result case by case. The case where 𝑁 = PSL(2, 𝑝 𝑟 ) with 𝑝 ≠ 2 appears already in [Jeh77, Thm. 5]. The case 𝑁 = 𝐴𝑛 with 𝑛 ≥ 5 is due to [Kli79]. (d) Saxl’s solution of Jehne’s conjecture has an interesting arithmetic consequence: Let 𝑑 be a √square free integer and 𝑓 ∈ Z[𝑋] an irreducible polynomial without a root in Q( 𝑑). Then, there are infinitely many prime numbers 𝑝 such that
𝑑 = 1 but 𝑓 (𝑋) ≡ 0 mod 𝑝 has no solution. 𝑝 √ ˜ 𝑀 = Q( 𝑑, 𝑥), 𝑀ˆ the Galois closure Otherwise, let 𝑥 be a root of 𝑓 (𝑋) in Q,
ˆ of 𝑀/Q, and 𝐺 = Gal( 𝑀/Q). Then, for almost 𝑝 with 𝑑𝑝 = 1 the equation 𝑓 (𝑋) ≡ 0 mod 𝑝 has a solution. By the transfer principle, Ø √ 𝜎 ˆ ˆ Gal( 𝑀/Q( 𝑑)) = Gal( 𝑀/𝑀) . 𝜎 ∈𝐺

√

Thus, 𝑀 ∈ K (Q( 𝑑)/Q), in contrast to Jehne–Saxl. Another problem of Jehne is still open (see the notes at the end of the chapter for a profinite version): Problem 24.5.8 ([Jeh77], §7) Are there fields 𝐾 ⊂ 𝐿 ⊂ 𝑀, with 𝐾 global, 𝐿/𝐾 finite separable, 𝑀/𝐾 infinite separable, such that 𝑀0 ∈ K (𝐿/𝐾) for every intermediate field 𝐿 ⊂ 𝑀0 ⊂ 𝑀 of finite degree over 𝐾?

24.6 Davenport’s Problem Let 𝐾 be a global field of characteristic 𝑝 and let 𝑓 , 𝑔 ∈ 𝐾 [𝑌 ] be nonconstant polynomials which are not 𝑝-powers. For each prime 𝔭 ∈ 𝑃(𝐾) for which 𝑓 is defined modulo 𝔭, consider the value set of 𝑓 : 𝑉𝔭 ( 𝑓 ) = { 𝑓 (𝑦) | 𝑦 ∈ 𝐾¯ 𝔭 }. If 𝑔(𝑌 ) = 𝑓 (𝑎𝑌 + 𝑏) with 𝑎 ∈ 𝐾 × and 𝑏 ∈ 𝐾, then 𝑔 and 𝑓 are said to be strictly linearly related. In this case, 𝑉𝔭 ( 𝑓 ) = 𝑉𝔭 (𝑔) for almost all 𝔭 ∈ 𝑃(𝐾).

(24.13)

Call the pair ( 𝑓 , 𝑔) exceptional if 𝑓 and 𝑔 satisfy (24.13) but they are not strictly linearly related.

492

24 Problems of Arithmetical Geometry

Remark 24.6.1 (𝑉 𝑝 (𝑋 8 ) = 𝑉 𝑝 (16𝑋 8 ) for every prime number 𝑝) To prove this statement, verify the identity

𝑋 8 − 16 = (𝑋 2 − 2) (𝑋 2 + 2) (𝑋 − 1) 2 + 1 (𝑋 + 1) 2 + 1 . (24.14)



2 −2 −1 4 By the multiplicativity of the Legendre symbol, 𝑝 𝑝 𝑝 = 𝑝 = 1 for each odd prime number 𝑝. Hence, at least one of the first three factors on the right hand side of (24.14) is divisible by 𝑝 for some value of 𝑋. It follows that 16 is an 8th power modulo every odd 𝑝. Thus, for each 𝑥 ∈ F 𝑝 there is a 𝑦 ∈ F 𝑝 with 𝑥 8 = 16𝑦 8 , as claimed. Consider now a polynomial ℎ(𝑋) = 𝑐 𝑛 𝑋 𝑛 + 𝑐 𝑛−1 𝑋 𝑛−1 + · · · + 𝑐 0 in Q[𝑋] with 𝑛 ≥ 1 and 𝑐 𝑛 ≠ 0. Then, 𝑉 𝑝 (ℎ(𝑋 8 )) = 𝑉 𝑝 (ℎ(16𝑋 8 )) for each 𝑝 which divides no denominator of 𝑐 0 , . . . , 𝑐 𝑛−1 , 𝑐 𝑛 . Assume there were 𝑎, 𝑏 ∈ Q with 𝑎 ≠ 0 and ℎ(16𝑋 8 ) = ℎ((𝑎𝑋 + 𝑏) 8 ). Comparing the coefficients of 𝑋 8𝑛 on both sides this would give 16 = 𝑎 8 , which is a contradiction. It follows that (ℎ(𝑋 8 ), ℎ(16𝑋 8 )) is an exceptional pair. Problem 24.6.2 (a) Are all exceptional pairs in Q[𝑋] of the form (ℎ(𝑋 8 ), ℎ(16𝑋 8 )) with ℎ ∈ Q[𝑋] (Müller)? (b) For a given global field 𝐾 find all exceptional pairs ( 𝑓 , 𝑔) with 𝑓 , 𝑔 ∈ 𝐾 [𝑋] (Davenport). A complete solution of this problem is still unavailable. Transfer principles, however, allow us to reduce (24.13) to a condition on a finite Galois extension of 𝐾 (𝑡), in complete analogy with the Galois-theoretic version of Kronecker conjugacy of global field extensions. From this point we derive certain relations between 𝑓 and 𝑔 (e.g. deg( 𝑓 ) = deg(𝑔) if the degrees are relatively prime to char(𝐾)). In the special case where 𝐾 = Q and deg( 𝑓 ) is a prime, we prove that there is no 𝑔 ∈ Q[𝑋] such that ( 𝑓 , 𝑔) is exceptional. But we mention the existence of exceptional pairs ( 𝑓 , 𝑔) with deg( 𝑓 ) prime over other number fields. Clearly (24.13) is equivalent to the truth of the sentence   (∀𝑇) (∃𝑋) [ 𝑓 (𝑋) = 𝑇] ↔ (∃𝑌 ) [𝑔(𝑌 ) = 𝑇] (24.15) in almost all fields 𝐾¯ 𝔭 . By Theorem 23.9.3, this is equivalent to the truth of (24.15) in 𝐾˜ (𝜎), for almost all 𝜎 ∈ Gal(𝐾). First we eliminate (∀𝑇) to reduce the problem to a test sentence over 𝐾 ′ = 𝐾 (𝑡). To this end we choose 𝑥, 𝑦 ∈ 𝐾g (𝑡) with 𝑓 (𝑥) = 𝑡 and 𝑔(𝑦) = 𝑡 and put 𝐸 = 𝐾 (𝑡, 𝑥) and 𝐹 = 𝐾 (𝑡, 𝑦). Assume that 𝐸/𝐾 (𝑡) and 𝐹/𝐾 (𝑡) are separable extensions. When 𝑝 = char(𝐾) > 0 this means that 𝑓 (𝑋) and 𝑔(𝑋) are not 𝑝th powers in 𝐾˜ [𝑋]. Lemma 24.6.3 Let 𝑁 a finite Galois extension of 𝐾 (𝑡) containing 𝐸 and 𝐹. Put 𝐺 = Gal(𝑁/𝐾 (𝑡)). Then, (24.13) is equivalent to Ø Ø Gal(𝑁/𝐸) 𝜈 = Gal(𝑁/𝐹) 𝜈 . (24.16) 𝜈 ∈𝐺

𝜈 ∈𝐺

Proof. Put = 𝐾 (𝑡). If (24.15) holds in 𝐾˜ (𝜎) for almost all 𝜎 ∈ Gal(𝐾), then (24.15) holds in 𝐾˜ ′ (𝜎) for almost all 𝜎 ∈ Gal(𝐾 ′) (Theorem 23.7.2). In particular, the formula in the outer brackets of (24.15) holds for 𝑡. That is, the sentence 𝐾′

24.6 Davenport’s Problem

493

(∃𝑋) [ 𝑓 (𝑋) = 𝑡] ↔ (∃𝑌 ) [𝑔(𝑌 ) = 𝑡]

(24.17)

is true in 𝐾˜ ′ (𝜎) for almost all 𝜎 ∈ Gal(𝐾 ′). Therefore, 24.17 holds in 𝑁 (𝜎) for all 𝜎 ∈ 𝐺. Consequently, (24.16) is true. Conversely, suppose that the elements 𝜏 ∈ 𝐺 that fix a root of 𝑓 (𝑋) − 𝑡 are exactly the elements that fix a root of 𝑔(𝑌 ) − 𝑡. We prove that (24.15) holds in 𝐾˜ (𝜎) for all 𝜎 ∈ Gal(𝐾). Indeed, let 𝜎 ∈ Gal(𝐾) and 𝑎 ∈ 𝐾˜ (𝜎). Extend the 𝐾-specialization 𝑡 → 𝑎 to a homomorphism 𝜑 of the ˜ The residue field 𝑁¯ = 𝜑(𝑅) is a normal integral closure 𝑅 of 𝐾 [𝑡] in 𝑁 into 𝐾. extension of 𝐾 (𝑎). Denote the decomposition group {𝜏 ∈ 𝐺 | 𝜏(Ker(𝜑)) = Ker(𝜑)} ¯ (𝑎)) through of 𝜑 by 𝐻. Then, 𝜑 induces an epimorphism 𝜑∗ of 𝐻 onto Aut( 𝑁/𝐾 the formula (Section 7.1) 𝜑∗ (𝜏) (𝜑(𝑧)) = 𝜑(𝜏(𝑧)) for each 𝜏 ∈ 𝐻 and 𝑧 ∈ 𝑅.

(24.18)

Observe that the roots of 𝑓 (𝑋) − 𝑡 and 𝑔(𝑌 ) − 𝑡 are in 𝑅. Hence, 𝜑 maps the roots of 𝑓 (𝑋) − 𝑡 (resp. 𝑔(𝑌 ) − 𝑡) onto the roots of 𝑓 (𝑋) − 𝑎 (resp. 𝑔(𝑌 ) − 𝑎). Therefore, if 𝑏 ∈ 𝐾˜ (𝜎) satisfies 𝑓 (𝑏) = 𝑎, then there exists a zero 𝑥 of 𝑓 (𝑋) − 𝑡 with 𝜑(𝑥) = 𝑏. Note that 𝐻 ∩ Gal(𝑁/𝐾 (𝑥)) is the decomposition group of 𝜑 over 𝐾 (𝑥). Hence, by Lemma 7.1.1(a), the restriction of 𝜑∗ to 𝐻 ∩ Gal(𝑁/𝐾 (𝑥)) maps that subgroup onto ¯ (𝑏)). In particular, there exists a 𝜏 ∈ 𝐻 ∩ Gal(𝑁/𝐾 (𝑥)) with Aut( 𝑁/𝐾 𝜑∗ (𝜏) = res 𝑁¯ (𝜎).

(24.19)

Since 𝜏(𝑥) = 𝑥, the assumptions give a zero 𝑦 of 𝑔(𝑌 ) − 𝑡 with 𝜏(𝑦) = 𝑦. Thus, □ 𝑐 = 𝜑(𝑦) is a zero of 𝑔(𝑌 ) − 𝑎 for which (24.18) and (24.19) give 𝜎(𝑐) = 𝑐. Remark 24.6.4 The switch from the surjectivity of 𝜑∗ on 𝐻 to its surjectivity on 𝐻 ∩ Gal(𝑁/𝐾 (𝑥)) is a subtlety in the proof of Lemma 24.6.3. From the surjectivity of 𝜑∗ on 𝐻 alone we could choose 𝜏 ∈ 𝐻 such that (24.19) holds. It follows that there exist zeros 𝑥 and 𝑥 ′ of 𝑓 (𝑋) − 𝑡 with 𝜏(𝑥) = 𝑥 ′ and 𝜑(𝑥) = 𝜑(𝑥 ′) = 𝑏. If 𝑓 (𝑋) − 𝑎 is separable, this implies that 𝑥 = 𝑥 ′ and conclude the proof easily. Nothing, however, in the assumptions guarantees separability of 𝑓 (𝑋) − 𝑎. This same subtlety occurs in the stratification procedure of Chapter 34. Following Section 24.5 we say that fields 𝐸 and 𝐹 satisfying (24.16) (and the polynomials 𝑓 and 𝑔 satisfying (24.15)) are Kronecker conjugate over 𝐾 (𝑡). This is independent of 𝑁. If 𝐸 and 𝐹 are minimal extensions of 𝐾 (𝑡) with this property, then as in Lemma 24.5.5, they have the same Galois closure over 𝐾 (𝑡). The next results, however, allow us the same conclusion under a less restrictive assumption than minimality. Throughout let deg( 𝑓 ) = 𝑚, deg(𝑔) = 𝑛. Corollary 24.6.5 Suppose that 𝐸 and 𝐹 are Kronecker conjugate over 𝐾 (𝑡) and ˆ be the [𝐸 : 𝐾 (𝑡)] and [𝐹 : 𝐾 (𝑡)] are not divisible by char(𝐾). Let 𝐸ˆ (resp. 𝐹) Galois closure of 𝐸/𝐾 (𝑡) (resp. 𝐹/𝐾 (𝑡)). Then, [𝐸 : 𝐾 (𝑡)] = [𝐹 : 𝐾 (𝑡)] and ˆ 𝐸ˆ = 𝐹. Proof. By assumption, 𝔭∞ is tamely ramified in all fields conjugate to 𝐸 or to 𝐹 over 𝐾 (𝑡). Hence, 𝔭∞ is tamely ramified in the compositum 𝑁 of all these fields [CaF67, p. 31, Cor. 2]. Let 𝔓∞ be a prime divisor of 𝑁 that lies over 𝔭∞ . Then, the inertia

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group of 𝔓∞ is cyclic [CaF67, p. 29, Thm. 1]. Let 𝜏 be a generator of it. Then, 𝔭∞ is unramified in 𝑁 (𝜏). Since 𝔭∞ is totally ramified in 𝐸 (Example 2.3.11), the two fields are linearly disjoint over 𝐾 (𝑡) (Lemma 3.1.11). In particular, the polynomial 𝑓 (𝑋) −𝑡 (a root of which generates 𝐸 over 𝐾 (𝑡)) remains irreducible over 𝑁 (𝜏). Thus, the group ⟨𝜏⟩ operates transitively on the roots 𝑥1 , . . . , 𝑥 𝑚 of 𝑓 (𝑋) − 𝑡. Therefore, 𝜏 is an 𝑚-cycle on 𝑥1 , . . . , 𝑥 𝑚 . By symmetry, 𝜏 is an 𝑛-cycle on 𝑦 1 , . . . , 𝑦 𝑛 , the roots of 𝑔(𝑌 ) − 𝑡. If 𝑚 < 𝑛, then 𝜏 𝑚 would fix each of 𝑥1 , . . . , 𝑥 𝑚 but would move each of 𝑦 1 , . . . , 𝑦 𝑛 . This contradicts (24.16). Thus, 𝑚 ≥ 𝑛; and by symmetry, 𝑚 = 𝑛. Next restrict relation (24.16) to 𝐹ˆ to conclude that 𝐸 ∩ 𝐹ˆ and 𝐹 are Kronecker conjugate over 𝐾 (𝑡). By the first part of the corollary, [𝐸 ∩ 𝐹ˆ : 𝐾 (𝑡)] = [𝐹 : 𝐾 (𝑡)] = ˆ By symmetry 𝐸ˆ = 𝐹. ˆ ˆ Hence, 𝐸ˆ ⊆ 𝐹. [𝐸 : 𝐾 (𝑡)]. Therefore, 𝐸 = 𝐸 ∩ 𝐹ˆ ⊆ 𝐹. □ Lemma 24.6.6 Assume the hypotheses of Corollary 24.6.5. Then, the polynomial 𝑓 (𝑋) − 𝑔(𝑌 ) is reducible in 𝐾 [𝑋, 𝑌 ]; and there are constants 𝑎, 𝑏 ∈ 𝐾 and distinct roots 𝑥1 , . . . , 𝑥 𝑘 of 𝑓 (𝑋) − 𝑡 with 1 ≤ 𝑘 < deg( 𝑓 ) and 𝑎𝑦 + 𝑏 = 𝑥1 + · · · + 𝑥 𝑘 .

(24.20)

Proof. Assume 𝑓 (𝑋) − 𝑔(𝑌 ) is irreducible in 𝐾 [𝑋, 𝑌 ]. Then, since 𝑦 is transcendental over 𝐾, 𝑓 (𝑋) − 𝑡 = 𝑓 (𝑋) − 𝑔(𝑦) is irreducible in 𝐹 [𝑋]. Lemma 14.3.2 gives 𝜎 ∈ Gal(𝑁/𝐹) that moves each root of 𝑓 (𝑋) − 𝑡. By (24.16), 𝜎 moves each root of 𝑔(𝑌 ) − 𝑡. In particular, 𝜎𝑦 ≠ 𝑦, a contradiction. Thus, 𝑓 (𝑋) − 𝑔(𝑌 ) is reducible. Now consider a nontrivial factorization 𝑓 (𝑋) − 𝑔(𝑌 ) = 𝑝(𝑋, 𝑌 )𝑞(𝑋, 𝑌 ) in 𝐾 [𝑋, 𝑌 ]. By Corollary 24.6.5, deg( 𝑓 (𝑋)) = deg(𝑔(𝑌 )), so deg𝑋 ( 𝑝) + deg𝑋 (𝑞) = deg𝑋 ( 𝑓 (𝑋) − 𝑔(𝑌 )) = deg( 𝑓 (𝑋) − 𝑔(𝑌 )) = deg( 𝑝) + deg(𝑞). Since deg𝑋 ( 𝑝) ≤ deg( 𝑝) and deg𝑋 (𝑞) ≤ deg(𝑞), it follows that deg𝑋 ( 𝑝) = deg( 𝑝) = 𝑘, with 1 ≤ 𝑘 < deg( 𝑓 ). Write 𝑝(𝑋, 𝑌 ) as 𝑐 00 𝑋 𝑘 + (𝑐 10𝑌 + 𝑐 11 ) 𝑋 𝑘 + 𝑝 2 (𝑌 ) 𝑋 𝑘−2 + · · · + 𝑝 𝑘 (𝑌 ),

(24.21)

with 𝑐 00 , 𝑐 10 , 𝑐 11 ∈ 𝐾, 𝑐 00 ≠ 0, and 𝑝 2 , . . . , 𝑝 𝑘 ∈ 𝐾 [𝑌 ]. Since, 𝑓 (𝑋) − 𝑡 = 𝑝(𝑋, 𝑦)𝑞(𝑋, 𝑦), the 𝑘 roots of 𝑝(𝑋, 𝑦), say 𝑥 1 , . . . , 𝑥 𝑘 , are roots of 𝑓 (𝑋) − 𝑡. Since 𝑓 (𝑋) − 𝑡 is irreducible and separable over 𝐾 (𝑡), 𝑥1 , . . . , 𝑥 𝑘 are distinct. By (24.21), −1 □ 𝑎𝑦 + 𝑏 = 𝑥 1 + · · · + 𝑥 𝑘 , where 𝑎 = −𝑐−1 00 𝑐 10 and 𝑏 = −𝑐 00 𝑐 11 , as claimed. We summarize: Proposition 24.6.7 Let 𝐾 be a global field and 𝑓 , 𝑔 ∈ 𝐾 [𝑋] nonconstant polynomials of degree not divisible by char(𝐾). Suppose that 𝑉𝔭 ( 𝑓 ) = 𝑉𝔭 (𝑔) for almost all 𝔭 ∈ 𝑃(𝐾). Then: (a) deg( 𝑓 ) = deg(𝑔). (b) 𝑓 (𝑋) − 𝑡 and 𝑔(𝑋) − 𝑡 are Kronecker conjugate over the rational function field 𝐾 (𝑡). (c) The splitting fields of 𝑓 (𝑋) − 𝑡 and 𝑔(𝑋) − 𝑡 over 𝐾 (𝑡) coincide. (d) 𝑓 (𝑋) − 𝑔(𝑌 ) is reducible in 𝐾 [𝑋, 𝑌 ]. (e) There exist constants 𝑎, 𝑏 ∈ 𝐾 and distinct roots 𝑥 1 , . . . , 𝑥 𝑘 of 𝑓 (𝑋) − 𝑡 with 1 ≤ 𝑘 < deg( 𝑓 ) and 𝑎𝑦 + 𝑏 = 𝑥1 + · · · + 𝑥 𝑘 .

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495

Note that conditions (a)–(e) are all necessary for 𝑓 and 𝑔 to be strictly linearly related. Here is a case when they are also sufficient. Let 𝐾 be a field. Call a polynomial 𝑓 ∈ 𝐾 [𝑋] decomposable over 𝐾 if 𝑓 (𝑋) = 𝑓1 ( 𝑓2 (𝑋)) with 𝑓1 , 𝑓2 ∈ 𝐾 [𝑋], deg( 𝑓1 ) > 1, and deg( 𝑓2 ) > 1. (Then, deg( 𝑓 ) = deg( 𝑓1 ) deg( 𝑓2 ).) Otherwise, 𝑓 is indecomposable over 𝐾 (e.g. if 𝑓 is of prime degree). Theorem 24.6.8 ([Fri73b]) Let 𝑓 , 𝑔 ∈ Q[𝑋]. Suppose that 𝑉 𝑝 ( 𝑓 ) = 𝑉 𝑝 (𝑔) for almost all prime numbers 𝑝 and 𝑓 is indecomposable. Then, 𝑔 and 𝑓 are strictly linearly related. Proof. We give an elementary proof in the case that deg( 𝑓 ) = 𝑙 is a prime. The proof in the general case depends on deeper group-theoretic assertions, specifically the theory of doubly transitive permutation groups. By Proposition 24.6.7(a), deg(𝑔) = deg( 𝑓 ) = 𝑙. Suppose without loss that 𝑓 is monic and let 𝑐 be the leading coefficient of 𝑔. By Proposition 24.6.7(d), 𝑓 (𝑋) − 𝑔(𝑌 ) = 𝑟 (𝑋, 𝑌 )𝑠(𝑋, 𝑌 ) withÍ𝑟, 𝑠 ∈ Q[𝑋, 𝑌 ] and 𝑚 = deg(𝑟) ≥ 1, 𝑛 = deg(𝑠) ≥ 1. Write 𝑟 = 𝑛 𝑠 = 𝑖=0 𝑠𝑖 with 𝑟 𝑖 , 𝑠𝑖 ∈ Q[𝑋, 𝑌 ] homogeneous of degree 𝑖. By (24.22), 𝑋 𝑙 − 𝑐𝑌 𝑙 = 𝑟 𝑚 (𝑋, 𝑌 )𝑠 𝑛 (𝑋, 𝑌 ).

(24.22) Í𝑚

𝑖=0 𝑟 𝑖

and

(24.23)

𝑋 −1 Let 𝑍 = 𝑐1/𝑙 . Since 𝑍𝑍−1 is irreducible in Q[𝑍] [Lan97, p. 184], one of the factors 𝑌 𝑟 𝑚 or 𝑠 𝑚 must be linear (and 𝑐 an 𝑙-th power in Q). Thus, either 𝑚 = 1 or 𝑛 = 1. Suppose for example that 𝑚 = 1. Then, 𝑟 (𝑋, 𝑌 ) = 𝑎 1 𝑋 + 𝑎 2𝑌 + 𝑎 3 with 𝑎 1 , 𝑎 2 , 𝑎 3 ∈ Q and, say, 𝑎 1 ≠ 0. Let 𝑎 = − 𝑎𝑎21 and 𝑏 = − 𝑎𝑎31 . Substituting 𝑎𝑌 + 𝑏 for 𝑋 in (24.22) gives 𝑓 (𝑎𝑌 + 𝑏) = 𝑔(𝑌 ), as claimed. □ 𝑙

Let 𝑛 be one of the integers 7, 11, 13, 15, 21, or 31. [Fri80] and [Fri99] give indecomposable polynomials 𝑓 and 𝑔 of degree 𝑛 and a number field 𝐾 such that ( 𝑓 , 𝑔) is an exceptional pair over 𝐾. For indecomposable polynomials, these numbers are the only exceptional degrees: a consequence of the classification of finite simple groups.

24.7 On Permutation Groups We gather in this section various definitions and results about permutation groups which enter the proof of the Schur conjecture in the next section. Remark 24.7.1 (The affine linear group) (a) Let 𝐺 be a group. Suppose that 𝐺 acts on a set 𝑋 from the left. For each 𝑥 ∈ 𝑋 put 𝐺 𝑥 = {𝜎 ∈ 𝐺 | 𝜎𝑥 = 𝑥}. Call 𝐺 transitive if for all 𝑥, 𝑦 ∈ 𝑋 there is a 𝜎 ∈ 𝐺 with 𝜎𝑥 = 𝑦. Call 𝐺 regular if 𝐺 𝑥 = 1 for each 𝑥 ∈ 𝑋. Note that if 𝐺 is both Ñ transitive and regular, then |𝐺 | = |𝑋 |. Finally, call 𝐺 faithful if 𝑥 ∈𝑋 𝐺 𝑥 = 1. (b) Let 𝐺 be a permutation group of a set 𝑋 and 𝐻 a subgroup of 𝐺. Suppose that 𝐻 is regular and transitive. Then, 𝐻 ∩ 𝐺 𝑥 = 1 and 𝐺 = 𝐻𝐺 𝑥 = 𝐺 𝑥 𝐻 for each 𝑥 ∈ 𝑋.

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Indeed, if 𝜂 ∈ 𝐻 ∩ 𝐺 𝑥 , then 𝜂𝑥 = 𝜂, so 𝜂 = 1. For each 𝜎 ∈ 𝐺 there is an 𝜂 ∈ 𝐻 with 𝜂𝑥 = 𝜎𝑥. Hence, 𝜎 = 𝜂 · 𝜂−1 𝜎 ∈ 𝐻𝐺 𝑥 . Thus, 𝐺 = 𝐻𝐺 𝑥 . Applying the same argument to 𝜎 −1 , we find that 𝐺 = 𝐺 𝑥 𝐻. If in addition 𝐻 is normal in 𝐺, then 𝐺 = 𝐺 𝑥 ⋉ 𝐻. (c) Denote the cyclic group of order 𝑛 by 𝐶𝑛 and the dihedral group of order 2𝑛 by 𝐷 𝑛 . The latter group is generated by elements 𝜎, 𝜏 with defining relations 𝜏 2 = 1, 𝜎 𝑛 = 1, and 𝜎 𝜏 = 𝜎 −1 . (d) Let 𝑙 be a prime number. Denote the group of all affine maps of F𝑙 by AGL(1, F𝑙 ). It consists of all maps 𝜎𝑎,𝑏 : F𝑙 → F𝑙 with 𝑎 ∈ F𝑙× and 𝑏 ∈ F𝑙 defined by 𝜎𝑎,𝑏 (𝑥) = 𝑎𝑥 + 𝑏. Thus, |AGL(1, F𝑙 )| = (𝑙 − 1)𝑙. The subgroup 𝐿 = {𝜎1,𝑏 | 𝑏 ∈ F𝑙 } is the unique 𝑙-Sylow subgroup of AGL(1, F𝑙 ). Thus, 𝐿  𝐶𝑙 and 𝐿 is normal. Its action on F𝑙 is transitive and regular. Hence, by (b), AGL(1, F𝑙 ) = AGL(1, F𝑙 ) 𝑥 ⋉ 𝐿. Moreover, AGL(1, F𝑙 ) 𝑥  F𝑙× . Thus, AGL(1, F𝑙 ) is a metacyclic group. It particular, AGL(1, F𝑙 ) is solvable. Claim: Each 𝜎 ∈ AGL(1, F𝑙 ) ∖ 𝐿 fixes exactly one 𝑥 ∈ F𝑙 . Indeed, 𝜎 = 𝜎𝑎,𝑏 with 𝑏 𝑎 ≠ 1. Hence, 𝜎𝑥 = 𝑥 if and only if 𝑥 = 1−𝑎 . The following lemma establishes a converse to Remark 24.7.1(d): Lemma 24.7.2 Let 𝑙 be an odd prime number and 𝐺 a nontrivial finite solvable group acting faithfully and transitively on a set 𝑋 of 𝑙 elements. Let 𝐿 be a minimal normal subgroup of 𝐺. Then: (a) 𝐿 is transitive and regular. (b) 𝐿  𝐶𝑙 and 𝐺 = 𝐺 𝑥 ⋉ 𝐿 for each 𝑥 ∈ 𝑋. (c) 𝐿 is its own centralizer in 𝐺. (d) 𝐺/𝐿 is a cyclic group of order dividing 𝑙 − 1 and 𝐺 𝑥  𝐺/𝐿 for each 𝑥 ∈ 𝑋. (e) 𝐿 is the unique 𝑙-Sylow subgroup of 𝐺. (f) 𝐺 is isomorphic as a permutation group to a subgroup of AGL(1, F𝑙 ). (g) Each 𝜌 ∈ 𝐺 ∖ 𝐿 belongs to exactly one 𝐺 𝑥 . (h) Let 𝐼 be a subgroup of 𝐺 which contains 𝐿 and 𝑃 a 𝑝-Sylow subgroup of 𝐼 with 𝑝 ≠ 𝑙. Suppose that 𝑃 ⊳ 𝐼. Then, 𝑃 = 1. (i) Suppose that (𝐺 : 𝐿) = 2 and let 𝑥 ∈ 𝑋. Then, 𝐺 is the dihedral group 𝐷 𝑙 generated by the involution 𝜏 of 𝐺 𝑥 and a generator 𝜆 of 𝐿 with the relation 𝜆 𝜏 = 𝜆−1 . Proof of (a) and (b). Since 𝐺 is solvable, 𝐿 is Abelian. Ð Choose of the 𝐿-orbits of 𝑋. Then, 𝑋 = · 𝑦 ∈𝑌 𝐿𝑦 Í a system 𝑌 of representatives ′ ′ and 𝑙 = 𝑦 ∈𝑌 |𝐿𝑦|. Given 𝑦, 𝑦 ∈ 𝑌 , choose 𝜎 ∈ 𝐺 with 𝜎𝑦 = 𝑦 . Then, 𝜎(𝐿𝑦) = 𝜎𝐿𝜎 −1 𝜎𝑦 = 𝐿𝑦 ′. Thus, all 𝐿-orbits have the same length 𝑛. Therefore, 𝑙 = |𝑌 |𝑛. By definition, 𝐿 ≠ 1. Choose 𝜆 ∈ 𝐿, 𝜆 ≠ 1. Then, there is an 𝑥 ∈ 𝑋 with 𝜆𝑥 ≠ 𝑥, because 𝐺 is faithful. Hence, 𝐿 𝑥 < 𝐿. Thus, 𝑛 = |𝐿𝑥| = (𝐿 : 𝐿 𝑥 ) > 1. Since 𝑙 is prime, 𝑛 = 𝑙 and |𝑌 | = 1. Therefore, 𝐿 is transitive. Next consider 𝜆 ∈ 𝐿 and 𝑥 ∈ 𝑋 with 𝜆𝑥 = 𝑥. For each 𝑥 ′ ∈ 𝑋 choose 𝜆 ′ ∈ 𝐿 with ′ 𝜆 𝑥 = 𝑥 ′. Then, 𝜆𝑥 ′ = 𝜆𝜆 ′𝑥 = 𝜆 ′𝜆𝑥 = 𝜆 ′𝑥 = 𝑥 ′, so 𝜆 = 1. Thus, 𝐿 𝑥 = 1 and 𝐿 is regular. Hence, by Remark 24.7.1(b), 𝐺 = 𝐺 𝑥 ⋉ 𝐿. In addition, |𝐿| = (𝐿 : 𝐿 𝑥 ) = 𝑙. Therefore, 𝐿  𝐶𝑙 .

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497

Proof of (c). Assume 𝜎 ∈ 𝐶𝐺 (𝐿). Fix 𝑥 ∈ 𝑋 and use (b) to write 𝜎 = 𝜆𝜌 with 𝜆 ∈ 𝐿 and 𝜌 ∈ 𝐺 𝑥 . Since 𝐿 is Abelian, 𝜌 ∈ 𝐶𝐺 (𝐿). For each 𝑥 ′ ∈ 𝑋 choose, by (a), 𝜆 ′ ∈ 𝐿 with 𝜆 ′𝑥 = 𝑥 ′. Then, 𝜌𝑥 ′ = 𝜌𝜆 ′𝑥 = 𝜆 ′ 𝜌𝑥 = 𝜆 ′𝑥 = 𝑥 ′. Since 𝐺 is faithful, this implies that 𝜌 = 1 and 𝜎 ∈ 𝐿, as asserted. Proof of (d). Conjugation of 𝐿 by elements of 𝐺 embeds 𝐺/𝐿 into Aut(𝐿) (by (c)). The latter is isomorphic to F𝑙× . Hence, 𝐺/𝐿 is a cyclic group of order dividing 𝑙 − 1. By (b), 𝐺 𝑥  𝐺/𝐿. Proof of (e). By (d), 𝑙 2 ∤ |𝐺 |, so 𝐿 is an 𝑙-Sylow subgroup of 𝐺. Since 𝐿 is normal, it is the unique 𝑙-Sylow subgroup of 𝐺. Proof of (f) and (g). By (b), 𝐿  𝐶𝑙 . Let 𝜆 be a generator of 𝐿. Choose 𝑥 0 ∈ 𝑋. By (a) and (b), 𝑋 = {𝑥0 , 𝜆𝑥 0 , . . . , 𝜆𝑙−1 𝑥 0 }. The bijection 𝑋 → F𝑙 mapping 𝜆 𝑏 𝑥0 onto 𝑏 for 𝑏 = 0, . . . , 𝑙 − 1 identifies 𝐿 as a permutation group of F𝑙 such that 𝜆 𝑏 0 = 𝑏 for all 𝑏 ∈ F𝑙 . For an arbitrary 𝑥 ∈ F𝑙 we have 𝜆 𝑏 𝑥 = 𝜆 𝑏 𝜆 𝑥 0 = 𝜆 𝑏+𝑥 0 = 𝑥 + 𝑏. Let 𝜎 ∈ 𝐺 and 𝑎 ∈ F𝑙 with 𝜎𝜆𝜎 −1 = 𝜆 𝑎 . Suppose first that 𝜎0 = 0. Then, 𝜎𝑥 = 𝜎𝜆 𝑥 0 = 𝜎𝜆 𝑥 𝜎 −1 0 = 𝜆 𝑎𝑥 0 = 𝑎𝑥. In the general case there is a 𝑏 ∈ F𝑙 with 𝜎0 = 𝜆 𝑏 0. Then, (𝜆−𝑏 𝜎)𝜆(𝜆−𝑏 𝜎) −1 = 𝜆−𝑏 (𝜎𝜆𝜎 −1 )𝜆 𝑏 = 𝜆−𝑏 𝜆 𝑎 𝜆 𝑏 = 𝜆 𝑎 , so by the preceding case, 𝜆−𝑏 𝜎𝑥 = 𝑎𝑥. Therefore, 𝜎𝑥 = 𝑎𝑥 + 𝑏. This gives an embedding of 𝐺 into AGL(1, F𝑙 ) as permutation groups. In particular, each 𝜎 ∈ 𝐺 ∖ 𝐿 belongs to exactly one 𝐺 𝑥 (Remark 24.7.1(d)). Proof of (h). Under the assumptions of (h), 𝑃 ∩ 𝐿 = 1 and 𝑃, 𝐿 ⊳ 𝐼. Hence, by (c), 𝑃 ≤ 𝐶𝐺 (𝐿) = 𝐿. Therefore, 𝑃 = 1. Proof of (i). Suppose that (𝐺 : 𝐿) = 2. Choose a generator 𝜆 of 𝐿 and a generator 𝜏 of 𝐺 𝑥 (use (d)). By (b), |𝐺 𝑥 | = (𝐺 : 𝐿) = 2. Hence, 𝜏 2 = 1, 𝜆𝑙 = 1, and 𝜆 𝜏 = 𝜆𝑡 for some 𝑡 ∈ F𝑙 satisfying 𝑡 2 = 1. By (c), 𝜏 ∉ 𝐿 = 𝐶𝐺 (𝐿). Hence, 𝑡 = −1. Consequently, □ 𝐺  𝐷𝑙 . Definition 24.7.3 (Primitive permutation groups) Let 𝐺 be a transitive permutation group of a nonempty set 𝑋. A block of 𝐺 is a subset 𝑌 of 𝑋 with the following property: For each 𝜎 ∈ 𝐺 either 𝜎𝑌 = 𝑌 or 𝑌 ∩𝜎𝑌 = ∅. Let 𝐺𝑌 = {𝜎 ∈ 𝐺 | 𝜎𝑌 = 𝑌 }. Choose Ð a system of representatives 𝑅 for the left cosets of 𝐺𝑌 in 𝐺. Then, 𝑋 = · 𝜌∈𝑅 𝜌𝑌 and 𝐺 acts transitively on the set {𝜌𝑌 | 𝜌 ∈ 𝑅} = {𝜎𝑌 | 𝜎 ∈ 𝐺}. The set 𝑋 itself, the empty set, and each {𝑥} with 𝑥 ∈ 𝑋 are trivial blocks. Call 𝐺 imprimitive if it has a nontrivial block, otherwise 𝐺 is primitive. Lemma 24.7.4 Let 𝐺 be a transitive permutation group of a set 𝑋 and let 𝑥 ∈ 𝑋. Then, 𝐺 is primitive if and only if 𝐺 𝑥 is maximal in 𝐺. Proof. Suppose first that 𝐺 is imprimitive. Then, 𝐺 has a nontrivial block 𝑌 . Since 𝐺 is transitive, we may assume 𝑥 ∈ 𝑌 . Then, 𝐺 𝑥 ≤ 𝐺𝑌 . Since 𝑌 ⊂ 𝑋 and 𝐺 is transitive, 𝐺𝑌 < 𝐺. Now choose 𝑦 ∈ 𝑌 , 𝑦 ≠ 𝑥, and 𝜎 ∈ 𝐺 with 𝜎𝑥 = 𝑦. Then, 𝜎𝑌 = 𝑌 but 𝜎𝑥 ≠ 𝑥, so 𝜎 ∈ 𝐺𝑌 ∖ 𝐺 𝑥 . Therefore, 𝐺 𝑥 is not maximal. Conversely, suppose that 𝐺 has a subgroup 𝐻 with 𝐺 𝑥 < 𝐻 < 𝐺. Put 𝑌 = 𝐻𝑥. Then, 𝐻 = 𝐺𝑌 . Indeed, if 𝜂 ∈ 𝐺𝑌 , then 𝜂𝐻𝑥 = 𝜂𝑌 = 𝑌 = 𝐻𝑥. Hence, there exists an 𝜂 ′ ∈ 𝐻 with (𝜂 ′) −1 𝜂𝑥 = 𝑥, so (𝜂 ′) −1 𝜂 ∈ 𝐺 𝑥 < 𝐻. Therefore, 𝜂 ∈ 𝐻. It follows that 𝐺𝑌 ≤ 𝐻. Conversely, if 𝜂 ∈ 𝐻, then 𝜂𝑌 = 𝜂𝐻𝑥 = 𝐻𝑥 = 𝑌 , so 𝜂 ∈ 𝐺𝑌 .

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Moreover, 𝑌 is a 𝐺-block. Indeed, let 𝜎 ∈ 𝐺 and 𝑦, 𝑦 ′ ∈ 𝑌 with 𝜎𝑦 = 𝑦 ′. Then, 𝜎𝜂𝑥 = 𝜂 ′𝑥 for some 𝜂, 𝜂 ′ ∈ 𝐻. Hence, (𝜂 ′) −1 𝜎𝜂 ∈ 𝐺 𝑥 < 𝐻, so 𝜎 ∈ 𝐻. Therefore, 𝜎𝑌 = 𝑌 . Further, 𝜎𝑌 ≠ 𝑌 for each 𝜎 ∈ 𝐺 ∖ 𝐻. Hence, 𝑌 ⊂ 𝑋. By definition, 𝑥 ∈ 𝑌 . In addition, 𝜎𝑥 ≠ 𝑥 for each 𝜎 ∈ 𝐻 ∖ 𝐺 𝑥 . Hence, 𝑌 contains more than one element. Thus, 𝑌 is a nontrivial 𝐺-block. Consequently, 𝐺 is imprimitive. □ Lemma 24.7.5 Let 𝐺 be a permutation group of a set 𝐴. We say that 𝐺 is doubly transitive if one of the following equivalent conditions hold: (a) 𝐺 is transitive on 𝐴 and there exists an 𝑎 ∈ 𝐴 such that 𝐺 𝑎 is transitive on 𝐴 ∖{𝑎}. (b) 𝐺 is transitive on 𝐴 and for each 𝑎 ∈ 𝐴, 𝐺 𝑎 is transitive on 𝐴 ∖{𝑎}. (c) Let 𝑎 1 , 𝑎 2 , 𝑏 1 , 𝑏 2 ∈ 𝐴 with 𝑎 1 ≠ 𝑎 2 and 𝑏 1 ≠ 𝑏 2 . Then, there is a 𝜏 ∈ 𝐺 with 𝜏𝑎 1 = 𝑏 1 and 𝜏𝑎 2 = 𝑏 2 . Proof. Statements (a) and (b) are equivalent because 𝐺 is transitive. Thus, it suffices to prove that (b) and (c) are equivalent: Suppose that (b) holds. Let 𝑎 1 , 𝑎 2 , 𝑏 1 , 𝑏 2 be as in (c). Choose 𝑎 ∈ 𝐴 and 𝛼, 𝛽 ∈ 𝐺 with 𝛼𝑎 1 = 𝑎 and 𝛽𝑏 1 = 𝑎. Then, 𝛼𝑎 2 ≠ 𝑎 and 𝛽𝑏 2 ≠ 𝑎. Hence, there is a 𝜎 ∈ 𝐺 𝑎 with 𝜎𝛼𝑎 2 = 𝛽𝑏 2 . Put 𝜏 = 𝛽−1 𝜎𝛼. Then, 𝜏𝑎 1 = 𝑏 1 and 𝜏𝑎 2 = 𝑏 2 . Conversely, suppose that (c) holds. Let 𝑎 ∈ 𝐴 and 𝑏, 𝑏 ′ ∈ 𝐴 ∖{𝑎}. Then, there is □ a 𝜎 ∈ 𝐺 with 𝜎𝑎 = 𝑎 and 𝜎𝑏 = 𝑏 ′. Remark 24.7.6 (Doubly transitive subgroups of AGL(1, F 𝑝 )) If 𝑥 1 , 𝑥2 , 𝑦 1 , 𝑦 2 are elements in F 𝑝 with 𝑥 1 ≠ 𝑥 2 and 𝑦 1 ≠ 𝑦 2 , then there are unique 𝑎 ∈ F×𝑝 and 𝑏 ∈ F 𝑝 such that 𝑎𝑥1 + 𝑏 = 𝑦 1 and 𝑎𝑥 2 + 𝑏 = 𝑦 2 . Thus, AGL(1, F 𝑝 ) is doubly transitive. Conversely, let 𝐺 be a doubly transitive subgroup of AGL(1, F 𝑝 ). Then, for each (𝑎, 𝑏) ∈ F×𝑝 × F 𝑝 there is a 𝜎 ∈ 𝐺 with 𝜎(1) = 𝑎 + 𝑏 and 𝜎(0) = 𝑏. Then, 𝜎(𝑥) = 𝑎𝑥 + 𝑏 for all 𝑥 ∈ F 𝑝 . It follows that 𝐺 = AGL(1, F 𝑝 ). One of the main tools in the proof of Schur’s proposition below is the group ring Z[𝐻] of a finite Abelian group 𝐻 (Section 18.6). Each 𝜌 ∈ Z[𝐻] has a unique Í presentation as a sum 𝜂 ∈𝐻 𝑐 𝜂 𝜂 with coefficients 𝑐 𝜂 ∈ Z. Define the support of 𝜌 to be the setÍ Supp(𝜌) = {𝜂 ∈ 𝐻 | 𝑐 𝜂 ≠ 0}. Then, ⟨Supp(𝜌)⟩ = ⟨𝜂 ∈ 𝐻 | 𝑐 𝜂 ≠ 0⟩. Let 𝜌 ′ = 𝜂 ∈𝐻 𝑐 ′𝜂 𝜂 be another element of Z[𝐻]. Suppose that 𝑐 𝜂 , 𝑐 ′𝜂 ≥ 0 for all
Í Í ′ 𝜂. Then, 𝜌𝜌 ′ = 𝜏 ∈𝐻 𝜂 𝜂 ′ =𝜏 𝑐 𝜂 𝑐 𝜂 ′ 𝜏 and there are no cancellations among the coefficients of 𝜌𝜌 ′. Hence, Supp(𝜌𝜌 ′) = Supp(𝜌)Supp(𝜌 ′). Consider a prime number 𝑝. We write 𝜌 ≡ 𝜌 ′ mod 𝑝 if 𝑐 𝜂 ≡ 𝑐 ′𝜂 mod 𝑝 for all Í 𝜂 ∈ 𝐻. Since Z[𝐻] is a commutative ring, 𝜌 𝑝 ≡ 𝜂 ∈𝐻 𝑐 𝜂 𝜂 𝑝 mod 𝑝. Proposition 24.7.7 (Schur) Let 𝐺 be a primitive permutation group of 𝐴 = {1, 2, . . . , 𝑛}. Suppose that 𝐺 contains an 𝑛-cycle 𝜈. Then, 𝐺 is doubly transitive or 𝑛 is a prime number. Proof. We assume that 𝐺 is not doubly transitive, 𝑛 is a composite number, and draw a contradiction. Denote the identity map of 𝐴 by 𝜀. Let 𝐻 = ⟨𝜈⟩.

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499

Claim A: 𝐺 = 𝐻𝐺 1 = 𝐺 1 𝐻 and 𝐻 ∩ 𝐺 1 = {𝜀}. Indeed, 𝐻 is transitive. Hence, for each 𝜎 ∈ 𝐺 there is an 𝜂 ∈ 𝐻 with 𝜂(1) = 𝜎(1). Therefore, 𝜎 = 𝜂 · 𝜂−1 𝜎 ∈ 𝐻𝐺 1 . Consequently, 𝐺 = 𝐻𝐺 1 . Likewise, 𝐺 = 𝐺 1 𝐻. In order to prove that 𝐻 ∩ 𝐺 1 = {𝜀} we may assume that 𝜈 is the cycle (1 2 · · · 𝑛). Then, 𝜈 𝑖 (1) = 𝑖 + 1 for each 𝑖 ∈ {1, 2, . . . , 𝑛 − 1}. Thus, if 𝜈 𝑖 (1) = 1, then 𝑖 = 0, so 𝜈 𝑖 = 𝜀, as asserted. It follows that each 𝜎 ∈ 𝐺 can be uniquely written as 𝜎 = 𝜂𝜎1 with 𝜂 ∈ 𝐻 and 𝜎1 ∈ 𝐺 1 . Likewise, there are unique 𝜂 ′ ∈ 𝐻 and 𝜎1′ ∈ 𝐺 1 with 𝜎 = 𝜎1′ 𝜂 ′. Í Claim B: Consider Z[𝐻] as a subring of Z[𝐺]. Ð Let 𝛾 = 𝜎 ∈𝐺1 𝜎 and 𝑅 = {𝜌 ∈ 𝐻 = · 𝑟𝑗=1 𝐻 𝑗 with 𝐻1 = {𝜀} and 𝑟 ≥ 3 Z[𝐻] | 𝛾𝜌 = 𝜌𝛾}. Í Í𝑟 Then, there is a partition such that 𝑅 = 𝑗=1 Z𝛼 𝑗 where 𝛼 𝑗 = 𝜂 ∈𝐻 𝑗 𝜂. Let 𝐴1 , . . . , 𝐴𝑟 with 𝐴1 = {1} be the 𝐺 1 -orbits of 𝐴. Since 𝐺 is not doubly transitive, 𝐺 1 is not transitive on {2, . . . , 𝑛} (Lemma 24.7.5). Hence, 𝑟 ≥ 3. Next define a map 𝑓 : 𝐻 → 𝐴 by 𝑓 (𝜂) = 𝜂(1) for each 𝜂 ∈ 𝐻. Check that 𝑓 is bijective and 𝐺 1 𝜂𝐺 1 ∩𝐻 = 𝑓 −1 (𝐺 1 𝜂(1)). For each 𝑗 choose 𝜂 𝑗 ∈ 𝐻 with 𝜂 𝑗 (1) ∈ 𝐴 𝑗 . Then,Ð𝐺 1 𝜂 𝑗 (1) = 𝐴 𝑗 . Hence, 𝐻 𝑗 := 𝐺 1 𝜂 𝑗 𝐺 1 ∩𝐻 = 𝑓 −1 (𝐺 1 𝜂 𝑗 (1)) = 𝑓 −1 ( 𝐴 𝑗 ). Then, 𝐻 = · 𝑟𝑗=1 𝐻 𝑗 and 𝐻1 = {𝜀}. Í Now consider an element 𝜌 = 𝜂 ∈𝐻 𝑐 𝜂 𝜂 in Z[𝐻]. Then, 𝜌 ∈ 𝑅 if and only if ∑︁ ∑︁ ∑︁ ∑︁ 𝑐 𝜂 𝜎𝜂 = 𝑐 𝜂 𝜂𝜎. (24.24) 𝜎 ∈𝐺1 𝜂 ∈𝐻

𝜎 ∈𝐺1 𝜂 ∈𝐻

By Claim A, the 𝜎𝜂 (resp. 𝜂𝜎) on the left- (resp. right-)hand side of (24.24) are distinct. Hence, (24.24) holds if and only if for all 𝜎, 𝜎 ′ ∈ 𝐺 1 and 𝜂, 𝜂 ′ ∈ 𝐻 the equality 𝜎𝜂 = 𝜂 ′ 𝜎 ′ implies 𝑐 𝜂 = 𝑐 𝜂′ . In other words, 𝑐 𝜂 is constant as 𝜂 ranges on 𝐺 1 𝜂 ′𝐺 1 ∩ 𝐻. In particular, 𝑐 𝜂 is constant on each 𝐻 𝑗 . Thus, there are 𝑐 1 , . . . , 𝑐𝑟 ∈ Z Í Í Í with 𝜌 = 𝑟𝑗=1 𝑐 𝑗 𝜂 ∈𝐻 𝑗 𝜂 = 𝑟𝑗=1 𝑐 𝑗 𝛼 𝑗 . This concludes the proof of Claim B. Claim C: Let 𝜌Í∈ 𝑅. Put 𝐾 = ⟨Supp(𝜌)⟩. Then, 𝐾 = {𝜀} or 𝐾 = 𝐻. Write Í 𝜌 = 𝜂 ∈𝐻 𝑐 𝜂 𝜂. By Claim B, 𝑐 𝜂 are constant on 𝐻 𝑗 , 𝑗 = 1, . . . , 𝑟. Put 𝜅 = 𝜂 ∈Supp(𝜌) 𝜂. Then, Supp(𝜅) = Supp(𝜌) and the coefficients of 𝜅 (which are 0 or 1) are also constant on each 𝐻 𝑗 . Hence, by Claim B, 𝜅 ∈ 𝑅. This means that 𝜅𝛾 = 𝛾𝜅. Since the coefficients of both 𝜅 and 𝛾 are nonnegative, Supp(𝜅)Supp(𝛾) = Supp(𝛾)Supp(𝜅). Therefore, 𝐾𝐺 1 = 𝐺 1 𝐾. This implies that 𝐾𝐺 1 is a subgroup of 𝐺 which contains 𝐺 1 . Since 𝐺 is primitive, 𝐺 1 is a maximal subgroup of 𝐺 (Lemma 24.7.4). Hence, 𝐾𝐺 1 = 𝐺 1 or 𝐾𝐺 1 = 𝐺. Since 𝐾 is a subgroup of 𝐻 and 𝐻 ∩ 𝐺 1 = {𝜀}, we have |𝐾𝐺 1 | = |𝐾 | · |𝐺 1 |. Thus, |𝐾 | · |𝐺 1 | = |𝐺 1 | or |𝐾 | · |𝐺 1 | = |𝐺 | = |𝐻| · |𝐺 1 | (by Claim A). Therefore, 𝐾 = {𝜀} or 𝐾 = 𝐻, as claimed. Now choose a prime divisor 𝑝 of 𝑛. By definition, 𝐻 is a cyclic group of order 𝑛. Hence, 𝐻 has a unique subgroup 𝑃 of order 𝑝. Since 𝑛 is composite, 𝑃 is a proper subgroup of 𝐻. Claim D: For each 𝑗, 𝐻 𝑗 ∖ 𝑃 is a union of cosets of 𝑃. Indeed, 𝑃 = {𝜂 ∈ 𝐻 | 𝜂 𝑝 = 1}. Hence, for each 𝜂0 ∈ 𝐻 the set {𝜂 ∈ 𝐻 | 𝜂 𝑝 = 𝜂0 } is either empty or a coset of 𝑃. In the latter case, it consists of 𝑝 elements. If we prove for each 𝜂0 ∈ 𝐻 with 𝜂0 ≠ 𝜀 that

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𝑝|#{𝜂 ∈ 𝐻 𝑗 | 𝜂 𝑝 = 𝜂0 },

(24.25)

then {𝜂 ∈ 𝐻 𝑗 | 𝜂 𝑝 = 𝜂0 } is either empty or coincides with {𝜂 ∈ 𝐻 | 𝜂 𝑝 = 𝜂0 }. In the latter case it is a coset of 𝑃. This will proveÍClaim D. To prove (24.25), consider an element 𝜌 = 𝜂 ∈𝐻 𝑐 𝜂 𝜂 in Z[𝐻]. For each 𝜂 ∈ 𝐻 let 𝑐 ′𝜂 be the unique integer between 0 and 𝑝 − 1 satisfying 𝑐 ′𝜂 ≡ 𝑐 𝜂 mod 𝑝. Put Í 𝜌 ′ = 𝜂 ∈𝐻 𝑐 ′𝜂 𝜂. If 𝑐 𝜂 is constant on each 𝐻 𝑗 , then so is 𝑐 ′𝜂 . So, by Claim B, 𝜌 ∈ 𝑅 implies 𝜌 ′ ∈ 𝑅. By definition, Supp(𝜌 ′) ⊆ Supp(𝜌). Also, 𝜌1 ≡ 𝜌2 mod 𝑝, with ′ ′ 𝜌1 , 𝜌2 ∈ Z[𝐻], is equivalent Í to 𝜌1𝑝 = 𝜌2 . Specifically, let 𝜌 = 𝜂 ∈𝐻 𝑗 𝜂 . Then, Supp(𝜌 ′) ⊆ Supp(𝜌) ⊆ {𝜏 𝑝 | 𝜏 ∈ 𝐻}. The right-hand side is a proper subgroup of 𝐻 (because 𝑝|𝑛). Hence, ⟨Supp(𝜌 ′)⟩ < 𝐻. In addition, 𝜌 ≡

Í

𝑝 𝜂 ∈𝐻 𝑗

𝜂

(24.26)


′ ≡ 𝛼 𝑝𝑗 mod 𝑝. Hence, 𝜌 ′ = 𝛼 𝑝𝑗 . By definition, 𝑅

is a subring of Z[𝐻]. Since 𝛼 𝑗 is in 𝑅 (Claim B), so is 𝛼 𝑝𝑗 . Therefore, 𝜌 ′ ∈ 𝑅. It follows from (24.26) and Claim C that Supp(𝜌 ′) ⊆ ⟨Supp(𝜌 ′)⟩ ≤ {𝜀}. In particular, the coefficient of 𝜂0 in 𝜌 ′ is 0. In other words, the coefficient of 𝜂0 in 𝜌 is divisible by 𝑝. But the latter coefficient is exactly the right-hand side of (24.25). Consequently, (24.25) is true. Claim E: Suppose that 1 < 𝑗 ≤ 𝑟 and |𝐻 𝑗 ∩ 𝑃| ≤ 𝑝−1 𝜂 ∈ 𝑃. 2 . Then, 𝛼 𝑗 𝜂 = 𝛼 𝑗 for all Í Í Ð𝑚 Put 𝛼 = 𝛼 𝑗 , 𝜋 = 𝜂 ∈𝑃 𝜂, and 𝜇 = 𝜂 ∈𝐻 𝑗 ∩𝑃 𝜂. By Claim D, 𝐻 𝑗 ∖ 𝑃 = · 𝑖=1 𝜆𝑖 𝑃 Í𝑚 with 𝜆1 , . . . , 𝜆 𝑚 ∈ 𝐻. Put 𝜆 = 𝑖=1 𝜆𝑖 . Then, ∑︁ ∑︁ ∑︁ 𝜆 ∈ Z[𝐻] and 𝛼 = 𝜂= 𝜂+ 𝜂 = 𝜆𝜋 + 𝜇. (24.27) 𝜂 ∈𝐻 𝑗

𝜂 ∈𝐻 𝑗 ∖ 𝑃

𝜂 ∈𝐻 𝑗 ∩𝑃

For each 𝜂 ∈ 𝑃, multiplication from the right by 𝜂 yields a bijection of 𝑃 onto itself. Hence, 𝜋𝜂 = 𝜋. Thus, in order to prove that 𝛼𝜂 = 𝛼, it suffices (by (24.27)) to prove that 𝜇 = 0. Í Í To this end observe that 𝜋 2 = 𝜂 ∈𝑃 𝜋𝜂 = 𝜂 ∈𝑃 𝜋 = 𝑝𝜋. Similarly, 𝜋𝜇 = 𝑘 𝜋 with 𝑘 = |𝐻 𝑗 ∩ 𝑃|. Since Z[𝐻] is a commutative ring, this implies that 𝛼2 − 2𝑘𝛼 = 𝑝𝜆2 𝜋 + 𝜇2 − 2𝑘 𝜇. Hence, (𝜇2 − 2𝑘 𝜇) ′ = (𝛼2 − 2𝑘𝛼) ′. By Claim B, 𝛼 is in 𝑅. Therefore, by the proof of Claim D, (𝛼2 − 2𝑘𝛼) ′ ∈ 𝑅. Thus, (𝜇2 − 2𝑘 𝜇) ′ ∈ 𝑅. In addition, Supp(𝜇2 − 2𝑘 𝜇) ′ ⊆ Supp(𝜇2 − 2𝑘 𝜇) ⊆ 𝑃. Since
𝑝 ≠ 𝑛, 𝑃 is a proper subgroup of 𝐻. Therefore, by Claim C, Supp (𝜇2 − 2𝑘 𝜇) ′ ⊆ {𝜀}. It follows that 𝜇2 − 2𝑘 𝜇 = 𝑝𝜌 + 𝑙𝜀

(24.28) Í𝑘

with 𝜌 ∈ Z[𝐻], 𝜀 ∉ Supp(𝜌), and 𝑙 ∈ Z. By definition, 𝜇 = 𝑖=1 𝜂𝑖 , where 𝜂1 , . . . , 𝜂 𝑘 are distinct elements of 𝐻. Each row of the matrix 𝐵 = (𝜂𝑖 𝜂 𝑗 )1≤𝑖, 𝑗 ≤𝑘 consists of distinct elements of 𝐻. Hence, each element Í of 𝐻 appears at most 𝑘 times as an entry of 𝐵 and therefore as a summand in 𝑖,𝑘 𝑗=1 𝜂𝑖 𝜂 𝑗 = 𝜇2 . Thus, the coefficients of 𝜇2 − 2𝑘 𝜇 lie between −2𝑘 and 𝑘. Since 2𝑘 < 𝑝, this implies that 𝜌 in (24.28) is 0. It follows that (24.28) may be rewritten as 𝜇2 = 2𝑘 𝜇 + 𝑙𝜀. Thus, 𝜂𝑖 𝜂 𝑗 ∈ {𝜂1 , . . . , 𝜂 𝑘 , 𝜀} for all 𝑖, 𝑗. This means that Supp(𝜇) ∪ {𝜀} = {𝜂1 , . . . , 𝜂 𝑘 , 𝜀} is a subgroup of 𝑃. But 𝑘 + 1 < 𝑝. Hence, Supp(𝜇) ∪ {𝜀} = {𝜀}. As 𝑗 > 1, 𝜀 ∉ Supp(𝜇) (Claim B). Thus, Supp(𝜇) = ∅. Consequently, 𝜇 = 0, as claimed.

24.7 On Permutation Groups

End of proof:

501

Í Ð By Claim B, · 𝑟𝑗=2 𝐻 𝑗 ∩ 𝑃 = 𝑃 ∖{𝜀}. Hence, 𝑟𝑗=2 |𝐻 𝑗 ∩ 𝑃| = 𝑝 − 1.

Since 𝑟 ≥ 3, there is a 𝑗 ≥ 2 with |𝐻 𝑗 ∩ 𝑃| ≤ 𝑝−1 2 . Hence, by Claim E, 𝑃 is contained in the subgroup 𝐿 := {𝜎 ∈ 𝐺 | 𝛾𝛼 𝑗 𝜎 = 𝛾𝛼 𝑗 } of 𝐺. By Claim B, 𝛾𝛼 𝑗 = 𝛼 𝑗 𝛾. By definition of 𝛾, 𝛾𝜎 = 𝛾 for each 𝜎 ∈ 𝐺 1 . Therefore, 𝐺 1 ≤ 𝐿 and 𝑃𝐺 1 ≤ 𝐿. But, 𝑃 ∩ 𝐺 1 = 1. Hence, 𝐺 1 < 𝐿. By the primitivity of 𝐺, the group 𝐺 1 is maximal in 𝐺 (Lemma 24.7.4), hence 𝐿 = 𝐺. Ð On the other hand, since 𝐻 = · 𝑟𝑗=1 𝐻 𝑗 , 𝑟 ≥ 3, |𝐻1 | = 1 (Claim B), and 𝑗 ≥ 2, we have by Claims A and B that |Supp(𝛾𝛼 𝑗 )| = |Supp(𝛾)Supp(𝛼 𝑗 )| ≤ |Supp(𝛾)| · |Supp(𝛼 𝑗 )| = |𝐺 1 | · |𝐻 𝑗 | < |𝐺 1 | · |𝐻| = |𝐺 |. Choose 𝜁 ∈ 𝐺 ∖ Supp(𝛾𝛼 𝑗 ) and 𝜏 ∈ Supp(𝛾𝛼 𝑗 ). Then, 𝜏 · 𝜏 −1 𝜁 = 𝜁. In particular, 𝛾𝛼 𝑗 ·𝜏 −1 𝜁 ≠ 𝛾𝛼 𝑗 , so 𝐿 ≠ 𝐺. This contradiction to the preceding paragraph concludes the proof of the proposition. □ The following result completes Schur’s Proposition: Proposition 24.7.8 (Burnside) Let 𝐺 be a transitive permutation group of a set 𝐴 of 𝑝 elements for some prime number 𝑝. Then, 𝐺 is doubly transitive or isomorphic (as a permutation group) to a subgroup of AGL(1, F 𝑝 ). Proof. For each 𝑎 ∈ 𝐴 we have (𝐺 : 𝐺 𝑎 ) = | 𝐴| = 𝑝. Hence, 𝑝 divides |𝐺 |. By Cauchy’s theorem, 𝐺 has an element 𝜋 of order 𝑝. Decompose 𝜋 as a product of disjoint cycles. The order of 𝜋, namely 𝑝, is the least common multiple of the lengths of these cycles. Thus, 𝜋 must be a cycle of length 𝑝. We may therefore identify 𝐴 with F 𝑝 such that 𝜋(𝑎) = 𝑎 + 1 for each 𝑎 ∈ F 𝑝 . Denote the vector space of all functions from F 𝑝 to F 𝑝 by 𝑉. The operations in 𝑉 are defined as follows: ( 𝑓 +𝑔) (𝑥) = 𝑓 (𝑥)+𝑔(𝑥) (𝑎 𝑓 ) (𝑥) = 𝑎 𝑓 (𝑥). For each 𝑓 ∈ 𝑉 Í 𝑝−1 and Í 𝑝−1 there is a unique polynomial ℎ(𝑋) = 𝑖=0 𝑎 𝑖 𝑋 𝑖 in F 𝑝 [𝑋] with 𝑖=0 𝑎 𝑖 𝑥 𝑖 = 𝑓 (𝑥) for all 𝑥 ∈ F 𝑝 . Indeed, the latter condition gives a Vandermonde system of linear equations for 𝑎 0 , . . . , 𝑎 𝑝−1 which determines them uniquely. Define the degree of 𝑓 to be deg(ℎ). Then, let 𝑉𝑘 be the subspace of all 𝑓 ∈ 𝑉 with deg( 𝑓 ) ≤ 𝑘. The action of 𝐺 on F 𝑝 naturally gives an action of 𝐺 on 𝑉: (𝜎 𝑓 ) (𝑥) = 𝑓 (𝜎 −1 𝑥). The rest of the proof is divided into four parts: Part A: A basis of 𝑉𝑘 . Let 𝑘 be an integer between 0 and 𝑝 − 1. Then, 1, 𝑋, . . . , 𝑋 𝑘 give a basis of 𝑉𝑘 . Thus, dim(𝑉𝑘 ) = 𝑘 + 1. Conversely, suppose that 𝑓0 , . . . , 𝑓 𝑘 are elements of 𝑉 with deg( 𝑓 𝑗 ) = 𝑗, then they are linearly independent, so they form a basis of 𝑉𝑘 . Suppose now that 1 ≤ 𝑘 ≤ 𝑝 − 1. Consider 𝑓 ∈ 𝑉 of degree 𝑘 given by 𝑓 (𝑥) = 𝑎 𝑘 𝑥 𝑘 + 𝑎 𝑘−1 𝑥 𝑘−1 + · · · + 𝑎 0 with 𝑎 𝑖 ∈ F 𝑝 and 𝑎 𝑘 ≠ 0. Then, (𝜋 𝑓 ) (𝑥) = 𝑎 𝑘 (𝜋 −1 𝑥) 𝑘 + 𝑎 𝑘−1 (𝜋 −1 𝑥) 𝑘−1 + · · · + 𝑎 0 = 𝑎 𝑘 (𝑥 − 1) 𝑘 + 𝑎 𝑘−1 (𝑥 − 1) 𝑘−1 + · · · + 𝑎 0 = 𝑎 𝑘 𝑥 𝑘 + (−𝑘𝑎 𝑘 + 𝑎 𝑘−1 )𝑥 𝑘−1 + lower terms. Hence, 𝑓 (𝑥) − (𝜋 𝑓 ) (𝑥) = 𝑘𝑎 𝑘 𝑥 𝑘−1 + lower terms. Thus, deg(𝜋 𝑓 ) = 𝑘 and deg( 𝑓 − 𝜋 𝑓 ) = 𝑘 − 1.

502

24 Problems of Arithmetical Geometry

Inductively define 𝑓 𝑘 = 𝑓 and 𝑓 𝑗−1 = 𝑓 𝑗 − 𝜋 𝑓 𝑗 for 𝑗 = 𝑘, 𝑘 − 1, . . . , 1. Then, deg( 𝑓 𝑗 ) = 𝑗, so 𝑓0 , . . . , 𝑓 𝑘 form a basis of 𝑉. Since 𝑓0 , . . . , 𝑓 𝑘 are linear combinations of 𝑓 , 𝜋 𝑓 , . . . , 𝜋 𝑘 𝑓 , the latter 𝑘 + 1 functions also form a basis of 𝑉𝑘 all of its elements have degree 𝑘. Part B: The fixed space of 𝐺 0 . Assume from now on that 𝐺 is not doubly transitive. Then, by Lemma 24.7.5, 𝐺 0 is not transitive on F 𝑝 ∖{0}. Hence, with 𝐴0 = {0}, F 𝑝 has at least three 𝐺 0 -orbits, 𝐴0 , 𝐴1 , 𝐴2 . Let 𝑔𝑖 : F 𝑝 → F 𝑝 be the characteristic function of 𝐴𝑖 , 𝑖 = 0, 1, 2. Then, 𝜎𝑔𝑖 = 𝑔𝑖 for each 𝜎 ∈ 𝐺 0 and 𝑔0 , 𝑔1 , 𝑔2 are linearly independent. Thus, at most one of the functions 𝑔0 , 𝑔1 , 𝑔2 is of degree 0 and each of them belongs to the subspace 𝑈 = { 𝑓 ∈ 𝑉 | 𝜎 𝑓 = 𝑓 for each 𝜎 ∈ 𝐺 0 } of 𝑉. Hence, at least two of the functions 𝑔0 , 𝑔1 , 𝑔2 are of degree ≥ 1. We choose one of them, denote it by 𝑓 , and note that 𝑓 ∈ 𝑈 and 𝑘 := deg( 𝑓 ) ≥ 1. Part C: The subspaces 𝑉𝑘 and 𝑉𝑘−1 are 𝐺-invariant. By part A, 𝑓 , 𝜋 𝑓 , . . . , 𝜋 𝑘 𝑓 form a basis of 𝑉𝑘 and ⟨𝜋⟩ leaves 𝑉𝑘 invariant. In addition, ⟨𝜋⟩ is transitive on F 𝑝 . Hence, 𝐺 = ⟨𝜋⟩𝐺 0 = 𝐺 0 ⟨𝜋⟩. Moreover, 𝜎 𝑓 = 𝑓 for each 𝜎 ∈ 𝐺 0 . Therefore, 𝐺 0𝑉𝑘 = 𝑉𝑘 . Consequently, 𝐺𝑉𝑘 = 𝑉𝑘 . Next consider the 𝐺-invariant subspace o n ∑︁ ∑︁ 𝑊= 𝑐𝜎 = 0 𝑐𝜎 · 𝜎 𝑓 | 𝜎 ∈𝐺

𝜎 ∈𝐺

of 𝑉. We claim that 𝑊 ⊆ 𝑉𝑘−1 . Indeed, write each 𝜎 ∈ 𝐺 as 𝜎 = 𝜋 𝑖 𝜎0 with 𝜎0 ∈ 𝐺 0 . Then, in the notation of Part A, (𝜎 𝑓 ) (𝑥)Í = (𝜋 𝑖 𝜎0 𝑓 ) (𝑥) = (𝜋 𝑖 𝑓 )Í(𝑥) = 𝑓 (𝜋 −𝑖 𝑥) = 𝑓 (𝑥 − 𝑖) = 𝑎 𝑘 𝑥 𝑘 + lower terms. Hence, if 𝜎 ∈𝐺 𝑐 𝜎 = 0, then deg( 𝜎 ∈𝐺 𝑐 𝜎 · 𝜎 𝑓 ) ≤ 𝑘 − 1, as claimed. In particular, dim(𝑊) ≤ dim(𝑉𝑘−1 ) = 𝑘. Finally, note that 𝑓 − 𝜋 𝑓 , 𝜋 𝑓 − 𝜋 2 𝑓 , . . . , 𝜋 𝑘−1 𝑓 − 𝜋 𝑘 𝑓 are in 𝑊 and they are linearly independent. Hence, dim(𝑊) = 𝑘. Therefore, 𝑊 = 𝑉𝑘−1 . It follows that 𝑉𝑘−1 is 𝐺-invariant. Part D: End of proof. Define the product 𝑔ℎ of two functions 𝑔, ℎ ∈ 𝑉 by (𝑔ℎ) (𝑥) = 𝑔(𝑥)ℎ(𝑥). Then, 𝜎(𝑔ℎ) = 𝜎(𝑔)𝜎(ℎ) for each 𝜎 ∈ 𝐺 and 𝑉1 = {𝑔 ∈ 𝑉 | 𝑔ℎ ∈ 𝑉𝑘 for all ℎ ∈ 𝑉𝑘−1 }. Since both 𝑉𝑘−1 and 𝑉𝑘 are 𝐺-invariant, so is 𝑉1 . In particular, applying 𝜎 −1 to the identity map of F 𝑝 , we find 𝑎 ∈ F×𝑝 and 𝑏 ∈ F 𝑝 with 𝜎(𝑥) = 𝑎𝑥+𝑏 for all 𝑥 ∈ F 𝑝 . This gives the desired embedding of 𝐺 into AGL(1, F 𝑝 ). □ The combination of Schur’s result and Burnside’s result gives the following characterization of transitive proper subgroups of AGL(1, F 𝑝 ): Corollary 24.7.9 Let 𝐺 be a primitive permutation group of a set 𝐴 of 𝑛 elements. Suppose that 𝐺 contains a cycle of length 𝑛 but 𝐺 is not doubly transitive. Then, 𝑛 is a prime number 𝑝 and 𝐺 is isomorphic (as a permutation group) to a subgroup of AGL(1, F 𝑝 ). In particular, 𝐺 is solvable. Proof. By Schur (Proposition 24.7.7), 𝑛 = 𝑝 is a prime number. By Burnside (Proposition 24.7.8), 𝐺 is isomorphic to a subgroup of AGL(1, F 𝑝 ). It follows from □ Remark 24.7.1(d) that 𝐺 is solvable.

24.8 Schur’s Conjecture

503

24.8 Schur’s Conjecture Let 𝐾 be a field and 𝑓 ∈ 𝐾 [𝑋]. We say that 𝑓 permutes 𝐾, 𝑓 is a permutation polynomial on 𝐾, or 𝑓 is bijective on 𝐾, if the map 𝑥 ↦→ 𝑓 (𝑥) is a bijection of 𝐾 onto itself. Let 𝑅 be a ring, 𝑓 ∈ 𝑅[𝑋], and 𝑃 a maximal ideal of 𝑅. We say that 𝑓 is a permutation polynomial modulo 𝑃, if the reduction of 𝑓 modulo 𝑃 is a permutation polynomial on 𝑅/𝑃. In this section we consider a global field 𝐾 and a polynomial 𝑓 ∈ 𝑂 𝐾 [𝑋] with char(𝐾) ∤ deg( 𝑓 ) which is a permutation polynomial modulo infinitely many prime ideals of 𝑂 𝐾 . We prove a conjecture of Schur: 𝑓 is composed of linear polynomials and Dickson polynomials 𝐷 𝑛 (𝑎, 𝑋). The latter are defined in Z[𝑎, 𝑋] by induction on 𝑛: 𝐷 0 (𝑎, 𝑋) = 2, 𝐷 1 (𝑎, 𝑋) = 𝑋, and 𝐷 𝑛 (𝑎, 𝑋) = 𝑋 𝐷 𝑛−1 (𝑎, 𝑋) − 𝑎𝐷 𝑛−2 (𝑎, 𝑋) for 𝑛 ≥ 2.

(24.29)

Thus, 𝐷 2 (𝑎, 𝑋) = 𝑋 2 − 2𝑎, 𝐷 3 (𝑎, 𝑋) = 𝑋 3 − 3𝑎𝑋, and 𝐷 4 (𝑎, 𝑋) = 𝑋 4 − 4𝑎𝑋 2 + Call 𝐷 𝑛 (𝑎, 𝑋) the Dickson polynomial of degree 𝑛 with parameter 𝑎.

2𝑎 2 .

Lemma 24.8.1 Let 𝐾 be a field, 𝑎, 𝑏 ∈ 𝐾, and 𝑛 a nonnegative integer. Use 𝑋 and 𝑍 for variables. Then: (a) 𝐷 𝑛 (𝑎, 𝑍 + 𝑎𝑍 −1 ) = 𝑍 𝑛 + 𝑎 𝑛 𝑍 −𝑛 . (b) Suppose that 𝑓 ∈ 𝐾 (𝑍) satisfies 𝑓 (𝑍 + 𝑎𝑍 −1 ) = 𝑍 𝑛 + 𝑎 𝑛 𝑍 −𝑛 . Then, 𝑓 (𝑋) = 𝐷 𝑛 (𝑎, 𝑋). (c) 𝑏 𝑛 𝐷 𝑛 (𝑎, 𝑋) = 𝐷 𝑛 (𝑏 2 𝑎, 𝑏𝑋). (d) For 𝑛 ≥ 3 we have 𝐷 𝑛 (𝑎, 𝑋) = 𝑋 𝑛 − 𝑛𝑎𝑋 𝑛−2 + 𝑔(𝑋) with 𝑔 ∈ Z[𝑎, 𝑋] of degree at most 𝑛 − 3. Proof of (a). Check (a) for 𝑛 = 0 and 𝑛 = 1. Next assume (a) by induction for 𝑛 − 2 and 𝑛 − 1. Then, 𝐷 𝑛 (𝑎, 𝑍 + 𝑎𝑍 −1 ) = (𝑍 + 𝑎𝑍 −1 )𝐷 𝑛−1 (𝑎, 𝑍 + 𝑎𝑍 −1 )−𝑎𝐷 𝑛−2 (𝑎, 𝑍 + 𝑎𝑍 −1 ) = (𝑍 + 𝑎𝑍 −1 ) (𝑍 𝑛−1 + 𝑎 𝑛−1 𝑍 1−𝑛 )−𝑎(𝑍 𝑛−2 + 𝑎 𝑛−2 𝑍 2−𝑛 ) = 𝑍 𝑛 + 𝑎 𝑛 𝑍 −𝑛 . Proof of (b). Put 𝑋 ′ = 𝑍 + 𝑎𝑍 −1 . Then, 𝑓 (𝑋 ′) = 𝐷 𝑛 (𝑎, 𝑋 ′) (by (a)) and [𝐾 (𝑍) : 𝐾 (𝑋 ′)] ≤ 2. Hence, 𝑋 ′ is a variable. Therefore, 𝑓 (𝑋) = 𝐷 𝑛 (𝑎, 𝑋). Proof of (c) and (d). Carry out induction on 𝑛.

□

Definition 24.8.2 (Linear relation of polynomials) Let 𝐾 be a field and 𝑓 , 𝑔 ∈ 𝐾 [𝑋] polynomials. We say that 𝑓 and 𝑔 are linearly related over 𝐾 if there exist 𝑎, 𝑏 ∈ 𝐾 × and 𝑐, 𝑑 ∈ 𝐾 with 𝑓 (𝑋) = 𝑎𝑔(𝑏𝑋 + 𝑐) + 𝑑. Lemma 24.8.3 Let 𝐾 be a field, 𝑓 ∈ 𝐾 [𝑋], 𝐾 ′ an extension of 𝐾, and 𝑛 a positive integer. Suppose that char(𝐾) ∤ 𝑛 and 𝑓 (𝑋) is linearly related over 𝐾 ′ to 𝐷 𝑛 (𝑎 ′, 𝑋) for some 𝑎 ′ ∈ 𝐾 ′. Then, 𝑓 (𝑋) is linearly related over 𝐾 to 𝐷 𝑛 (𝑎, 𝑋) for some 𝑎 ∈ 𝐾. Proof. We prove the lemma only for 𝑛 ≥ 3. The case 𝑛 ≤ 2 may be checked directly.

504

24 Problems of Arithmetical Geometry

By assumption, there are 𝛼 ′, 𝛽 ′ ∈ (𝐾 ′) × and 𝑎 ′, 𝛾 ′, 𝛿 ′ ∈ 𝐾 ′ with 𝑓 (𝑋) = 𝛼 ′ 𝐷 𝑛 (𝑎 ′, 𝛽 ′ 𝑋 + 𝛾 ′) + 𝛿 ′. Put 𝑎 = (𝛽 ′) −2 𝑎 ′, 𝛼 = 𝛼 ′ (𝛽 ′) 𝑛 , 𝛾 = (𝛽 ′) −1 𝛾 ′, and 𝛿 = 𝛿 ′. By Lemma 24.8.1(c), 𝑓 (𝑋) = 𝛼𝐷 𝑛 (𝑎, 𝑋 + 𝛾) + 𝛿. By Lemma 24.8.1(d), 𝑓 (𝑋) = 𝛼(𝑋 + 𝛾) 𝑛 − 𝑛𝛼𝑎(𝑋 + 𝛾) 𝑛−2 + lower terms

= 𝛼𝑋 𝑛 + 𝑛𝛼𝛾 𝑋 𝑛−1 + 𝛼 𝑛2 𝛾 2 − 𝑛𝑎 𝑋 𝑛−2 + lower terms.

Hence, 𝛼 ∈ 𝐾, 𝑛𝛼𝛾 ∈ 𝐾, and 𝛼 𝑛2 𝛾 2 − 𝑛𝑎 ∈ 𝐾. By assumption, 𝑛 ≠ 0 in 𝐾. Therefore, 𝛾 ∈ 𝐾 and 𝑎 ∈ 𝐾. Finally, 𝛿 = 𝑓 (0) − 𝛼𝐷 𝑛 (𝑎, 𝛾) ∈ 𝐾. Consequently, □ 𝑓 (𝑋) is linearly related to 𝐷 𝑛 (𝑎, 𝑋) over 𝐾. Special cases of Dickson polynomials give rise to well known families of polynomials. For 𝑎 = 0 and 𝑛 ≥ 1, induction or Lemma 24.8.1(a) give 𝐷 𝑛 (0, 𝑋) = 𝑋 𝑛 . The special case 𝑎 = 1 gives the Chebyshev polynomials: 𝑇𝑛 (𝑋) = 𝐷 𝑛 (1, 𝑋). We extract properties of Chebyshev polynomials from Lemma 24.8.1: Lemma 24.8.4 Let 𝐾 be a field and 𝑛 a nonnegative integer. Then: (a) 𝑇0 (𝑋) = 2, 𝑇1 (𝑋) = 𝑋, and 𝑇𝑛 (𝑋) = 𝑋𝑇𝑛−1 (𝑋) − 𝑇𝑛−2 (𝑋) for 𝑛 ≥ 2. (b) 𝑇𝑛 (𝑍 + 𝑍 −1 ) = 𝑍 𝑛 + 𝑍 −𝑛 . (c) Suppose that 𝑓 ∈ 𝐾 (𝑍) satisfies 𝑓 (𝑍 + 𝑍 −1 ) = 𝑍 𝑛 + 𝑍 −𝑛 . Then, 𝑓 = 𝑇𝑛 . (d) For each 𝑛 ≥ 3 there is a 𝑔 ∈ Z[𝑋] of degree at most 𝑛 − 3 satisfying 𝑇𝑛 (𝑋) = 𝑋 𝑛 − 𝑛𝑋 𝑛−2 + 𝑔(𝑋). (e) Suppose that char(𝐾) ∤ 𝑛 and 𝑛 ≥ 3. Then, for each 𝑎 ∈ 𝐾˜ there are distinct 𝑥 1 , 𝑥2 ∈ 𝐾˜ with 𝑇𝑛 (𝑥1 ) = 𝑇𝑛 (𝑥2 ) = 𝑎. Proof of (e). Choose 𝑦 ∈ 𝐾˜ satisfying 𝑦 2 − 𝑎𝑦 + 1 = 0. Then, 𝑦 + 𝑦 −1 = 𝑎. Since char(𝐾) ∤ 𝑛, there are distinct 𝑧 1 , . . . , 𝑧 𝑛 ∈ 𝐾˜ with 𝑧 𝑖𝑛 = 𝑦, 𝑖 = 1, . . . , 𝑛. Since 𝑛 ≥ 3, there are 𝑖, 𝑗 such that 𝑥𝑖 := 𝑧𝑖 + 𝑧𝑖−1 and 𝑥 𝑗 := 𝑧 𝑗 + 𝑧−1 𝑗 are distinct. They satisfy □ 𝑇𝑛 (𝑥𝑖 ) = 𝑇𝑛 (𝑥 𝑗 ) = 𝑎. Remark 24.8.5 (Automorphisms of 𝐾 (𝑧)) Let 𝐾 be a field and 𝑧 an indeterminate. Recall that PGL(2, 𝐾) is the quotient
of GL(2, 𝐾) by the group of scalar  matrices.  Denote the image of a matrix 𝑎𝑐 𝑑𝑏 ∈ GL(2, 𝐾) in PGL(2, 𝐾) by 𝑎𝑐 𝑑𝑏 . Thus,  𝑎′ 𝑏′  𝑎 𝑏

𝑎′ 𝑏′ 𝑒𝑎 𝑒𝑏 × 𝑐 𝑑 = 𝑐′ 𝑑′ means that there is an 𝑒 ∈ 𝐾 with 𝑐′ 𝑑′ = 𝑒𝑐 𝑒𝑑 . Remark 12.7.4 identifies PGL(2, 𝐾) with Aut(𝐾 (𝑧)/𝐾), where the action of 𝑎𝑐 𝑑𝑏 on 𝐾 (𝑧)   is defined by the formula: 𝑎𝑐 𝑑𝑏 𝑧 = 𝑎𝑧+𝑏 𝑐𝑧+𝑑 . Lemma 24.8.6 Let 𝐾 be an algebraically closed  field and 𝑛 ≥  3 an integer with char(𝐾) ∤ 𝑛. Consider the elements 𝜎 = 10 𝜁0𝑛 and 𝜏 = 01 10 of PGL(2, 𝐾). Put Δ = ⟨𝜎, 𝜏⟩. Then: (a) 𝜎 𝑛 = 𝜏 2 = 1, 𝜎 𝜏 = 𝜎 −1 , and Δ  𝐷 𝑛 . (b) Let Δ1 = ⟨𝜎1 , 𝜏1 ⟩ be a subgroup of PGL(2, 𝐾) with 𝜎1𝑛 = 𝜏12 = 1 and 𝜎1𝜏1 = 𝜎1−1 (so, Δ1  𝐷 𝑛 ). Then, there is a 𝜆 ∈ PGL(2, 𝐾) with ⟨𝜎1 ⟩ 𝜆 = ⟨𝜎⟩, 𝜏1𝜆 = 𝜏, and Δ𝜆1 = Δ. (c) Let 𝑧 be an indeterminate. Then, the fixed field of 𝜏 (resp. Δ) in 𝐾 (𝑧) is 𝐾 (𝑧 + 𝑧−1 ) (resp. 𝐾 (𝑧 𝑛 + 𝑧−𝑛 )).

24.8 Schur’s Conjecture

505

Proof of (a). Put 𝜁 = 𝜁 𝑛 . All we need to do is to verify the relation 𝜎 𝜏 = 𝜎 −1 :         1 0 01 10 01 𝜁0 = = . 10 0𝜁 10 01 0 𝜁 −1 Proof of (b). Denote the image of 𝜌 ∈ GL(2, 𝐾) in PGL(2, 𝐾) by [𝜌]. Since 𝐾 ˜ 1 , 𝜏˜1 ∈ GL(2, 𝐾) with [ 𝜎 ˜ 1 ] = 𝜎1 , [ 𝜏˜1 ] = 𝜏1 , and is algebraically closed, there are 𝜎 𝜎 ˜ 1 divides 𝑋 𝑛 − 1. Since char(𝐾) ∤ 𝑛, ˜ 1𝑛 = 𝜏˜12 = 1. The minimal polynomial of 𝜎
𝑋 𝑛 − 1 has 𝑛 distinct roots in 𝐾. Hence, 𝜎 ˜ 1 is conjugate to a diagonal matrix 𝑎0 𝑑0   with 𝑎 𝑛 = 𝑑 𝑛 = 1. Therefore, [ 𝜎 ˜ 1 ] is conjugate to 10 𝑎−10 𝑑 with 𝑎 −1 𝑑 = 𝜁 𝑘 and 𝑘 gcd(𝑘, 𝑛) = 1. Replace ˜ 1 ] = 𝜎. 𝜁 by 𝜁
, if necessary, to assume [ 𝜎  Let 𝜏1 = 𝑤𝑦 𝑥𝑧 with 𝑤𝑦 𝑥𝑧 ∈ GL(2, 𝐾). Since 𝜏1 𝜎 = 𝜎 −1 𝜏1 , there is an 𝑎 ∈ 𝐾 × with       𝑎 0 𝑤𝑥 10 𝑤𝑥 . = 𝑦 𝑧 0𝜁 0 𝑎𝜁 −1 𝑦 𝑧 Therefore, 𝑤 = 𝑎𝑤, 𝑥𝜁 = 𝑎𝑥, 𝑦 = 𝑎𝜁 −1 𝑦, and 𝑧𝜁 = 𝑎𝜁 −1 𝑧. Assume 𝑤 ≠ 0. Then, 𝑎 = 1, 𝑥 = 0, 𝑦 = 0, and 𝑧 = 0 (use 𝑛 ≠ 2). It follows that 𝜏12 ≠ 1. This contradiction proves  that 𝑤 = 0. Hence, 𝑥 ≠ 0 and 𝑦 ≠ 0. Therefore, 𝑎 = 𝜁 and 𝑧 = 0. Consequently, 𝜏1 = 0𝑦 0𝑥 . √  Put 𝜌 = 0𝑥 √0𝑦 . Then, 𝜌 −1 𝜏1 𝜌 = 𝜏 and 𝜌 −1 𝜎𝜌 = 𝜎. So, there exists a 𝜆 ∈ PGL(2, 𝐾) as claimed. Proof of (c). Let 𝑥 = 𝑧 + 𝑧 −1 and 𝑡 = 𝑧 𝑛 + 𝑧−𝑛 . Denote the fixed field of 𝜏 (resp. Δ) in 𝐾 (𝑧) by 𝐸 1 (resp. 𝐸 2 ). Then, [𝐾 (𝑧) : 𝐸 1 ] = 2 and [𝐾 (𝑧) : 𝐸 2 ] = 2𝑛. Deduce from 𝜏𝑧 = 𝑧−1 and 𝜏𝑧−1 = 𝑧 that 𝜏𝑥 = 𝑥. Hence, 𝐾 (𝑥) is contained in 𝐸 1 . On the other hand, 𝑧 is a root of 𝑋 2 − 𝑥 𝑋 + 1. Therefore, [𝐾 (𝑧) : 𝐾 (𝑥)] ≤ 2. Combined with the preceding paragraph, this implies that 𝐾 (𝑥) = 𝐸 1 . Similarly, 𝐾 (𝑡) ⊆ 𝐸 2 and 𝑧 is a root of the equation 𝑋 2𝑛 − 𝑡 𝑋 𝑛 + 1 = 0. Hence, 𝐾 (𝑡) = 𝐸 2 . □ Proposition 24.8.7 (Müller) Let 𝐾 be an algebraically closed field of characteristic 𝑝, 𝑓 ∈ 𝐾 [𝑋] a polynomial of prime degree 𝑙 ≠ 𝑝, and 𝑡 an indeterminate. Suppose that 𝐺 := Gal( 𝑓 (𝑋) − 𝑡, 𝐾 (𝑡)) is solvable. Then, either 𝐺  𝐶𝑙 and 𝑓 is linearly related to 𝑋 𝑙 or 𝐺  𝐷 𝑙 and 𝑓 is linearly related to the Chebyshev polynomial 𝑇𝑙 . Proof. By assumption, 𝑓 (𝑋) − 𝑡 is a separable polynomial of degree 𝑙 which is irreducible. List its zeros in 𝐾 (𝑡)sep as 𝑥1 , . . . , 𝑥𝑙 . Then, 𝐹 = 𝐾 (𝑥1 , . . . , 𝑥𝑙 ) is the Galois closure of 𝐾 (𝑥 𝑗 )/𝐾 (𝑡), 𝐺 := Gal(𝐹/𝐾 (𝑡)) acts faithfully and transitively on {𝑥1 , . . . , 𝑥𝑙 }. When 𝑙 = 2, 𝐺  𝐶2 and 𝑓 (𝑋) is linearly related to 𝑋 2 (use that 2 ≠ 𝑝). So, assume from now on 𝑙 > 2. Then, we may apply Lemma 24.7.2 with {𝑥1 , . . . , 𝑥𝑙 } replacing 𝑋 and 𝐻 𝑗 := Gal(𝐹/𝐾 (𝑥 𝑗 )) replacing 𝐺 𝑥 𝑗 , 𝑗 = 1, . . . , 𝑙. In particular, let 𝐿 be a minimal normal subgroup of 𝐺. Then, by Lemma 24.7.2(b,e), 𝐿  𝐶𝑙 and 𝐿 is the unique 𝑙-Sylow subgroup of 𝐺. Let 𝑚 = [𝐹 : 𝐾 (𝑥 𝑗 )] and 𝑔 = genus(𝐹/𝐾). In the remaining parts of the proof we apply Riemann–Hurwitz to prove that 𝑔 = 0 and 𝑚 ≤ 2. We then show: If 𝑚 = 1, then 𝐺  𝐶𝑙 and 𝑓 is linearly related to 𝑋 𝑙 . If 𝑚 = 2, then 𝐺  𝐷 𝑙 and 𝑓 is linearly related to 𝑇𝑙 .

506

24 Problems of Arithmetical Geometry

Part A: The Riemann–Hurwitz formula. Denote the set of all prime divisors of 𝐹/𝐾 which ramify over 𝐾 (𝑡) by 𝑅. It includes all prime divisors of 𝐹/𝐾 which ramÍ Í ify over 𝐾 (𝑥 𝑗 ). Hence, Diff(𝐹/𝐾 (𝑡)) = 𝔮 ∈𝑅 𝑑𝔮 𝔮 and Diff(𝐹/𝐾 (𝑥 𝑗 )) = 𝔮 ∈𝑅 𝑑𝔮, 𝑗 𝔮 (Section 4.6). Here 𝑑𝔮 (resp. 𝑑𝔮, 𝑗 ) are the different exponents of 𝔮 over 𝐾 (𝑡) (resp. 𝐾 (𝑥 𝑗 )). Since 𝐾 is algebraically closed, deg(𝔮) = 1 for each 𝔮 ∈ 𝑅. Therefore, the Riemann–Hurwitz formula for 𝐹/𝐾 (𝑡) and 𝐹/𝐾 (𝑥 𝑗 ) becomes: ∑︁ (24.30) 2𝑔 − 2 = −2𝑚𝑙 + 𝑑𝔮 𝔮 ∈𝑅

2𝑔 − 2 = −2𝑚 +

∑︁

𝑑𝔮, 𝑗 ,

𝑗 = 1, . . . , 𝑙

(24.31)

𝔮 ∈𝑅

(Formula (4.18)). Subtract the sum of the 𝑙 equations (24.31) from (24.30): 2(𝑔 − 1) (1 − 𝑙) =

∑︁ 𝔮 ∈𝑅

𝑑𝔮 −

𝑙 ∑︁

𝑑𝔮, 𝑗 ).

(24.32)

𝑗=1

Now consider 𝔮 ∈ 𝑅. Denote the inertia group of 𝔮 over 𝐾 (𝑡) by 𝐼𝔮 . Then, the inertia group of 𝔮 over 𝐾 (𝑥 𝑗 ) is 𝐻 𝑗 ∩ 𝐼𝔮 . To compute the 𝔮th term in (24.32) we distinguish between two cases. Case A1: 𝐼𝔮 acts intransitively on {𝑥 1 , . . . , 𝑥𝑙 }. By Lemma 24.7.2(a), 𝐿 ̸ ≤ 𝐼𝔮 . Since 𝐿  𝐶𝑙 , we have 𝐿 ∩ 𝐼𝔮 = 1. Thus, 𝐼𝔮  𝐿𝐼𝔮 /𝐿 ≤ 𝐺/𝐿. Since 𝐺/𝐿 is cyclic (Lemma 24.7.2(d), 𝐼𝔮 is cyclic. Choose a generator 𝜏 of 𝐼𝔮 . Then, 𝜏 ∉ 𝐿. Hence, by Lemma 24.7.2(g), there is a 𝑘 between 1 and 𝑙 with 𝜏 ∈ 𝐻 𝑘 and therefore 𝐼𝔮 ≤ 𝐻 𝑘 . This implies that 𝔮| 𝐾 ( 𝑥𝑘 ) is unramified over 𝐾 (𝑡). Therefore, 𝑑𝔮 = 𝑑𝔮,𝑘 (Lemma 4.5.8). For 𝑗 ≠ 𝑘 we have 𝐼𝔮 ∩ 𝐻 𝑗 = 1 (Lemma 24.7.2(g)). Hence, 𝔮 is unramified over 𝐾 (𝑥 𝑗 ) and 𝑑𝔮, 𝑗 = 0 (Remark 4.5.7(b)). Consequently, the 𝔮th term in the right-hand side of (24.32) is 0. Case A2: 𝐼𝔮 acts transitively on {𝑥1 , . . . , 𝑥𝑙 }. Then, for each 𝑗 we have (𝐼𝔮 : 𝐼𝔮 ∩ 𝐻 𝑗 ) = 𝑙. Hence, 𝑙 divides |𝐼𝔮 |. Therefore, by Lemma 24.7.2(e), 𝐿 ≤ 𝐼𝔮 . Let 𝑃 be the trivial group if 𝑝 = 0 and a 𝑝-Sylow subgroup of 𝐼𝔮 if 𝑝 ≠ 0. Then, 𝑃 is a normal subgroup of 𝐼𝔮 [CaF67, p. 29, Thm. 1(ii)]. Hence, by Lemma 24.7.2(h), 𝑃 is trivial. Thus, 𝐼𝔮 is cyclic [CaF67, p. 31, Cor. 1]. In particular, each Sylow subgroup of 𝐼𝔮 other than 𝐿 is normal, hence trivial (Lemma 24.7.2(h)). Therefore, 𝐼𝔮 = 𝐿. Denote the fixed field of 𝐿, hence of 𝐼𝔮 , in 𝐹 by 𝐸. By Lemma 24.7.2(b), 𝐸/𝐾 (𝑡) is Galois of degree 𝑚, 𝐾 (𝑥 𝑗 )𝐸 = 𝐹, and 𝐾 (𝑥 𝑗 ) ∩ 𝐸 = 𝐾 (𝑡). By the preceding paragraph, 𝔮 is totally ramified over 𝐸 and 𝔮| 𝐸 is unramified over 𝐾 (𝑡). Hence, 𝔮 is unramified over 𝐾 (𝑥 𝑗 ) and 𝔮| 𝐾 ( 𝑥 𝑗 ) is totally and tamely ramified over 𝐾 (𝑡). Therefore, 𝑑𝔮 = 𝑙 − 1 and 𝑑𝔮, 𝑗 = 0 (Remark 4.5.7). There are exactly 𝑚 prime divisors of 𝐹/𝐾 lying over 𝔮| 𝐾 (𝑡) . Hence, they contribute 𝑚(𝑙 − 1) to the right-hand side of (24.32). Part B: Computation of 𝑔. Consider a prime divisor 𝔭 of 𝐾 (𝑡)/𝐾. Let 𝔮 be a prime divisor of 𝐹/𝐾 over 𝔭. Suppose that 𝐼𝔮 acts transitively on {𝑥1 , . . . , 𝑥𝑙 }. By Case A2, 𝔭 totally ramifies in 𝐾 (𝑥1 ). Hence, there are only finitely many such 𝔭. Denote their number by 𝑟.

24.8 Schur’s Conjecture

507

Conversely, suppose that 𝐼𝔮 acts intransitively on {𝑥 1 , . . . , 𝑥𝑙 }. Case A1 gives 𝑘 with 𝔮| 𝐾 ( 𝑥𝑘 ) unramified over 𝐾 (𝑡). Choose 𝜌 ∈ 𝐺 with 𝜌𝑥 𝑘 = 𝑥1 . Then, 𝜌𝔮| 𝐾 ( 𝑥1 ) is unramified over 𝐾 (𝑡). Hence, 𝔭 is not totally ramified in 𝐾 (𝑥1 ). Thus, by Case A1, only the 𝑟 prime divisors 𝔭 with 𝐼𝔮 transitive contribute to the right-hand side of (24.32). The contribution of each of them is, by Case A2, 𝑚(𝑙 −1). Thus, (24.32) simplifies to 2(𝑙 − 1) (1 − 𝑔) = 𝑟𝑚(𝑙 − 1) and furthermore to 2(1 − 𝑔) = 𝑟𝑚.

(24.33)

By Example 2.3.11, the infinite prime divisor 𝔭∞ of 𝐾 (𝑡)/𝐾 totally ramifies in 𝐾 (𝑥1 ). Hence, 𝑟 ≥ 1. Therefore, by (24.33), 𝑔 = 0 and 𝑟𝑚 = 2. It follows that either 𝑚 = 1 and 𝑟 = 2 or 𝑚 = 2 and 𝑟 = 1. Part C: Suppose that 𝑚 = 1 and 𝑟 = 2. Then, 𝐾 (𝑥 1 ) = 𝐹 is a Galois extension of degree 𝑙 of 𝐾 (𝑡). Hence, 𝐺  𝐶𝑙 . Since 𝑟 = 2, 𝐾 (𝑡) has a finite place 𝔭𝑐 , mapping 𝑡 to 𝑐 ∈ 𝐾, which totally ramifies in 𝐹. In particular, 𝔭𝑐 has only one prime divisor in 𝐹. Hence, 𝑓 (𝑋) − 𝑐 has only one root in 𝐾. It follows that there are 𝑎 ∈ 𝐾 × and 𝑏 ∈ 𝐾 with 𝑓 (𝑋) − 𝑐 = (𝑎𝑋 + 𝑏) 𝑙 . Thus, 𝑓 (𝑋) is linearly related to 𝑋 𝑙 . Part D: Suppose that 𝑚 = 2 and 𝑟 = 1. By Lemma 24.7.2(i), 𝐺 is the dihedral group 𝐷 𝑙 generated by the involution 𝜏 of 𝐻1 and a generator 𝜎 of 𝐿 with the relation 𝜎 𝜏 = 𝜎 −1 . By part B, 𝑔 = 0. Since 𝐾 is algebraically closed, each prime divisor of 𝐹/𝐾 has degree 1. Hence, 𝐹 = 𝐾 (𝑧) for some 𝑧 (Example 4.2.4).  Lemma 24.8.6(b)   gives an automorphism 𝜆 of Aut(𝐹/𝐾) with 𝜆𝜎𝜆−1 = ⟨ 10 𝜁0 ⟩ (and 𝜁 = 𝜁𝑙 ) and   𝜆𝜏𝜆−1 = 01 10 . Put 𝑧 ′ = 𝜆𝑧, 𝑥 ′ = 𝜆𝑥 1 , and 𝑡 ′ = 𝜆𝑡. Then, 𝐾 (𝑥 ′) is the fixed field in D    E  𝐾 (𝑧 ′) of 10 10 and 𝐾 (𝑡 ′) is the fixed field in 𝐾 (𝑧 ′) of 10 𝜁0 , 10 10 . By Lemma 24.8.6(c), the former field is 𝐾 (𝑧 ′ + (𝑧 ′) −1 ) and the latter field is 𝐾 ((𝑧 ′) 𝑙 + (𝑧 ′) −𝑙 ). Therefore, there are 𝜅1 ∈ Aut(𝐾 (𝑥 ′)/𝐾) and 𝜅2 ∈ Aut(𝐾 (𝑡 ′)/𝐾) with 𝑥 ′ = 𝜅1 (𝑧 ′ + (𝑧 ′) −1 ) and 𝑡 ′ = 𝜅 2 ((𝑧 ′) 𝑙 + (𝑧 ′) −𝑙 ). Identify 𝜆, 𝜅 1 , 𝜅2 with Möbius transformations having coefficients in 𝐾. Put 𝜇 = 𝜆−1 𝜅1 and 𝜈 = 𝜆−1 𝜅2 . Then, 𝑥 1 = 𝜇(𝑧 ′ + (𝑧 ′) −1 ) and 𝑡 = 𝜈((𝑧 ′) 𝑙 + (𝑧 ′) −𝑙 ). Hence, 𝜈 −1 ◦ 𝑓 ◦ 𝜇(𝑧 ′ + (𝑧 ′) −1 ) = (𝑧 ′) 𝑙 + (𝑧 ′) −𝑙 . Since 𝑧 ′ is transcendental over 𝐾, Lemma 24.8.4(c) implies that 𝜈 −1 ◦ 𝑓 ◦ 𝜇 = 𝑇𝑙 . Part E: Conclusion of the proof. It remains to prove that 𝜇 and 𝜈 −1 are linear polynomials. Assume first that 𝜈 −1 is not a polynomial. Thus, 𝜈 −1 (𝑋) = 𝑎𝑋+𝑏 𝑐𝑋+𝑑 with 𝑎, 𝑏, 𝑐, 𝑑 ∈ 𝐾 and 𝑐 ≠ 0. It follows that 𝜈 −1 (∞) = 𝑎𝑐 . Use Lemma 24.8.4(e) to find distinct 𝑒 1 , 𝑒 2 ∈ 𝐾 with 𝑇𝑙 (𝑒 𝑖 ) = 𝑎𝑐 for 𝑖 = 1, 2. Then, 𝑓 (𝜇(𝑒 𝑖 )) = 𝜈(𝑇𝑙 (𝑒 𝑖 )) = 𝜈( 𝑎𝑐 ) = ∞. Since 𝑓 is a polynomial, 𝜇(𝑒 𝑖 ) = ∞. This contradiction to the injectivity of 𝜇 on 𝐾 ∪ {∞} proves that 𝜈 −1 is a linear polynomial. Finally, assume that 𝜇 is not a polynomial. Then, 𝑒 := 𝜇(∞) ∈ 𝐾. Hence, by the preceding paragraph, 𝑇𝑙 (∞) = 𝜈 −1 ( 𝑓 (𝜇(∞))) = 𝜈 −1 ( 𝑓 (𝑒)) ∈ 𝐾, which is a contradiction. Consequently, 𝜇 is a linear polynomial. □ Lemma 24.8.8 Let 𝐾 be an algebraically closed field, 𝑓 ∈ 𝐾 [𝑋] a polynomial of degree 𝑛, 𝑥 a transcendental element over 𝐾, and 𝑡 = 𝑓 (𝑥). Suppose that char(𝐾) ∤ 𝑛. Then, Gal( 𝑓 (𝑋) − 𝑡, 𝐾 (𝑡)) contains an 𝑛-cycle.

508

24 Problems of Arithmetical Geometry

Proof. By assumption, 𝑓 (𝑋) − 𝑡 has 𝑛 distinct roots 𝑥1 , . . . , 𝑥 𝑛 in 𝐾 (𝑡)sep . Assume without loss that 𝑥 = 𝑥1 and put 𝐹 = 𝐾 (𝑥1 , . . . , 𝑥 𝑛 ). Let 𝑣 ∞ be the valuation of 𝐾 (𝑡)/𝐾 with 𝑣 ∞ (𝑡) = −1. By Example 2.3.11, 𝑣 ∞ is totally ramified in 𝐾 (𝑥). Thus, 𝑣 ∞ has a unique extension 𝑤 to 𝐾 (𝑥). Denote the completion of 𝐸 = 𝐾 (𝑡) at 𝑣 ∞ by 𝐸ˆ (By Example 4.5.1, 𝐸ˆ  𝐾 ((𝑡 −1 )).) By Proposition 4.5.3, 𝑓 (𝑋) − 𝑡 is irreducible over 𝐸ˆ and 𝐸ˆ (𝑥)/𝐸ˆ is a totally ramified ˆ extension of degree 𝑛. Since char(𝐾) ∤ 𝑛, 𝐸ˆ (𝑥)/𝐸ˆ is tamely ramified. Put 𝐹ˆ = 𝐹 𝐸. ˆ Then, 𝐹ˆ = 𝐸ˆ (𝑥 1 ) · · · 𝐸ˆ (𝑥 𝑛 ) is a compositum of tamely ramified extensions of 𝐸. ˆ 𝐸ˆ is tamely ramified [CaF67, p. 31, Cor. 2]. Since 𝐾 is algebraically Hence, 𝐹/ ˆ 𝐸ˆ is 1. Therefore, 𝐹/ ˆ 𝐸ˆ is totally and tamely ramified. closed, the residue degree of 𝐹/ ˆ is cyclic [CaF67, p. 31, Cor. 1]. It follows that 𝐸ˆ (𝑥)/𝐸ˆ is a ˆ 𝐸) Therefore, Gal( 𝐹/ ˆ 𝐸) ˆ  Gal( 𝐹/ ˆ is cyclic of Galois extension. Hence, 𝐹ˆ = 𝐸ˆ (𝑥) and Gal( 𝑓 (𝑋) − 𝑡, 𝐸) ˆ is transitive. order 𝑛. It follows that Gal( 𝑓 (𝑋) − 𝑡, 𝐸) ˆ It decomposes into disjoint cycles Choose a generator 𝜎 of Gal( 𝑓 (𝑋) − 𝑡, 𝐸). whose lengths sum up to 𝑛. Since ⟨𝜎⟩ is transitive, there is only one cycle. Thus, 𝜎 □ is an 𝑛-cycle. Finally, note that 𝜎 ∈ Gal( 𝑓 (𝑋) − 𝑡, 𝐾 (𝑡)). Lemma 24.8.9 Let 𝐾 be a field, 𝑡 an indeterminate, and 𝑓 ∈ 𝐾 [𝑋] an indecomposable polynomial over 𝐾 of degree 𝑛 with char(𝐾) ∤ 𝑛. Then, Gal( 𝑓 (𝑋) − 𝑡, 𝐾 (𝑡)) is a primitive group. Proof. Let 𝑥 be a root of 𝑓 (𝑋) −𝑡 in 𝐾 (𝑡)sep and 𝐹 the Galois closure of 𝐾 (𝑥)/𝐾 (𝑡). By Lemma 24.7.4, it suffices to prove that Gal(𝐹/𝐾 (𝑥)) is a maximal subgroup of Gal(𝐹/𝐾 (𝑡)). In other words, we have to prove that 𝐾 (𝑥)/𝐾 (𝑡) is a minimal extension. Assume that there is a field 𝐸 which lies strictly between 𝐾 (𝑡) and 𝐾 (𝑥). Let 𝑑 = [𝐾 (𝑥) : 𝐸] and 𝑚 = [𝐸 : 𝐾 (𝑡)]. Then, 𝑑 > 1 and 𝑚 > 1. Let 𝑣 ∞ be the valuation of 𝐾 (𝑡)/𝐾 with 𝑣 ∞ (𝑡) = −1. Then, 𝑣 ∞ is totally ramified in 𝐾 (𝑥). Thus, 𝑣 ∞ has a unique extension 𝑤 ′ to 𝐾 (𝑥), and 𝑤 ′ satisfies 𝑤 ′ (𝑥) = −1 and has the same residue field as 𝑣 ∞ , namely 𝐾 (Example 2.3.11). Let 𝑤 be the restriction of 𝑤 ′ to 𝐸. Then, the residue field of 𝐸 at 𝑤 is 𝐾. By Lüroth (Remark 4.6.2(a)), 𝐸 is a rational function field over 𝐾. Hence, there is a 𝑦 ∈ 𝐸 with 𝑤(𝑦) = −1 and 𝑣(𝑦) ≥ 0 for all other valuations 𝑣 of 𝐸/𝐾. Therefore, 𝑤 ′ (𝑦) = −𝑑 and 𝑣(𝑦) ≥ 0 for all other valuations 𝑣 of 𝐾 (𝑥)/𝐾. It follows that 𝑦 = ℎ(𝑥) for some ℎ ∈ 𝐾 [𝑋]. The degree of the pole divisor of 𝑦 as an element of 𝐾 (𝑥) is 𝑑. Therefore, [𝐾 (𝑥) : 𝐾 (𝑦)] = 𝑑. Since 𝐾 (𝑦) ⊆ 𝐸, this implies that 𝐸 = 𝐾 (𝑦). Therefore, deg(ℎ) = 𝑑. Repeat the above arguments for 𝑡, 𝑦 replacing 𝑦, 𝑥 to find a polynomial 𝑔 ∈ 𝐾 [𝑋] of degree 𝑚 with 𝑡 = 𝑔(𝑦). This gives 𝑓 (𝑥) = 𝑡 = 𝑔(𝑦) = 𝑔(ℎ(𝑥)). Hence, 𝑓 (𝑋) = 𝑔(ℎ(𝑋)). This contradiction to the indecomposability of 𝑓 proves that 𝐾 (𝑥)/𝐾 (𝑡) is □ minimal. Let 𝐾 be a field and 𝑓 ∈ 𝐾 [𝑋] a polynomial of positive degree. Put 𝑓 ∗ (𝑋, 𝑌 ) =

𝑓 (𝑋) − 𝑓 (𝑌 ) . 𝑋 −𝑌

24.8 Schur’s Conjecture

509

Proposition 24.8.10 Let 𝐾 be an algebraically closed field and 𝑓 ∈ 𝐾 [𝑋] a polynomial of degree 𝑙. Suppose that char(𝐾) ∤ 𝑙, 𝑓 is indecomposable over 𝐾, and 𝑓 ∗ (𝑋, 𝑌 ) is reducible. Then, 𝑙 is prime and 𝑓 (𝑋) is linearly related to 𝑋 𝑙 or to 𝑇𝑙 (𝑋). Proof. Let 𝑡 be an indeterminate. Since char(𝐾) ∤ 𝑙, 𝑓 (𝑋) − 𝑡 has 𝑙 distinct roots 𝑥 1 , . . . , 𝑥𝑙 in 𝐾 (𝑡)sep . Hence, 𝐹 := 𝐾 (𝑥1 , . . . , 𝑥𝑙 ) is a Galois extension of 𝐾 (𝑡). Put 𝐺 = Gal(𝐹/𝐾 (𝑡) and 𝐻 𝑗 = Gal(𝐹/𝐾 (𝑥 𝑗 )), 𝑗 = 1, . . . , 𝑙. View 𝐺 as a permutation group of {𝑥1 , . . . , 𝑥𝑙 }. Since 𝑓 (𝑋) − 𝑡 is irreducible in 𝐾 (𝑡) [𝑋], 𝐺 is transitive. Since 𝑓 is indecomposable, 𝐺 is primitive (Lemma 24.8.9). Further, by Lemma 24.8.8, 𝐺 contains an 𝑙-cycle. Since 𝑓 ∗ (𝑋, 𝑌 ) is reducible, 𝑙 − 1 ≥ 2. By Gauss, 𝑓 ∗ (𝑥1 , 𝑌 ) is reducible in 𝐾 (𝑥 1 ) [𝑌 ] of degree 𝑙 − 1. Therefore, 𝐻1 is intransitive on the roots 𝑥 2 , . . . , 𝑥𝑙 of 𝑓 ∗ (𝑥1 , 𝑌 ). This means that 𝐺 is not doubly transitive (Lemma 24.7.5). It follows from Corollary 24.7.9 that 𝑙 is a prime number and 𝐺 is solvable. Consequently, by Proposition 24.8.7, 𝑓 is linearly related to 𝑋 𝑙 or to 𝑇𝑙 (𝑋). □ Lemma 24.8.11 ([FrM69], Thm. 3.5) Let 𝐾 be a field, 𝑓 ∈ 𝐾 [𝑋] a polynomial, and 𝐿 a field extension of 𝐾. Suppose that char(𝐾) ∤ deg( 𝑓 ) and 𝑓 is decomposable over 𝐿. Then, 𝑓 is decomposable over 𝐾. Í𝑚 Í 𝑎 𝑖 𝑋 𝑖 and ℎ(𝑋) = 𝑛𝑗=0 𝑏 𝑗 𝑋 𝑗 in 𝐿 [𝑋] with 𝑓 (𝑋) = Proof. There are 𝑔(𝑋) = 𝑖=0 𝑔(ℎ(𝑋)) and deg(𝑔), deg(ℎ) < deg( 𝑓 ). Assume without loss that 𝐿 is algebraically closed. Divide 𝑔 and 𝑓 by 𝑎 𝑚 , if necessary, to assume 𝑎 𝑚 = 1. Then, replace 𝑋 by 𝑋, if necessary, to assume 𝑏 𝑛 = 1. It follows that 𝑓 , 𝑔, and ℎ are monic. Finally, 𝑏 −1/𝑛 𝑛 choose 𝛽 ∈ 𝐿 with ℎ(𝛽) = 0. Then, replace 𝑋 by 𝑋 + 𝛽 to assume 𝑏 0 = 0. We prove that 𝑔, ℎ ∈ 𝐾 [𝑋]. Indeed, 𝑓 (𝑋) = (𝑋 𝑛 + 𝑏 𝑛−1 𝑋 𝑛−1 + · · · + 𝑏 1 𝑋) 𝑚 +

𝑚−1 ∑︁

𝑎 𝑖 (𝑋 𝑛 + 𝑏 𝑛−1 𝑋 𝑛−1 + · · · + 𝑏 1 𝑋) 𝑖 .

(24.34)

𝑖=0

The highest power of 𝑋 involved in the second term on the right-hand side of (24.34) is 𝑋 (𝑚−1)𝑛 . Hence, for each 𝑘 between 1 and 𝑛 − 1, only the first term on the right-hand side of (24.34) contributes to the coefficient of 𝑋 𝑚𝑛−𝑘 . Therefore, this coefficient is 𝑚𝑏 𝑛−𝑘 + 𝑝 𝑘 (𝑏 𝑛−𝑘+1 , . . . , 𝑏 𝑛−1 , 1) with 𝑝 𝑘 ∈ Z[𝑋𝑛−𝑘+1 , . . . , 𝑋𝑛 ]. Observe that deg( 𝑓 ) = 𝑚𝑛, so char(𝐾) ∤ 𝑚. Induction on 𝑘 proves that 𝑏 𝑛−𝑘 ∈ 𝐾. Thus, ℎ ∈ 𝐾 [𝑋]. Finally, suppose that 𝑎 𝑚−1 , . . . , 𝑎 𝑚−𝑟+1 have been proved to be in 𝐾. Then, 𝑎 𝑚−𝑟 ℎ(𝑋) 𝑚−𝑟 + 𝑎 𝑚−𝑟−1 ℎ(𝑋) 𝑚−𝑟−1 + · · · + 𝑎 0 is a polynomial in 𝐾 [𝑋]. Its lead□ ing coefficient is 𝑎 𝑚−𝑟 . Hence, 𝑎 𝑚−𝑟 ∈ 𝐾. Consequently, 𝑔 ∈ 𝐾 [𝑋]. Let 𝐾 be a field and 𝑓 ∈ 𝐾 [𝑋]. We say that 𝑓 is injective (resp. surjective) on 𝐾 if the map 𝑥 ↦→ 𝑓 (𝑥) of 𝐾 into itself is injective (resp. surjective). Let 𝐾 be a field and 𝑓 ∈ 𝐾 [𝑋] a polynomial with deg( 𝑓 ) > 1. Then, 𝑓 = 𝑓1 ◦ 𝑓2 ◦ · · · ◦ 𝑓𝑟 with 𝑓𝑖 ∈ 𝐾 [𝑋] indecomposable and deg( 𝑓𝑖 ) > 1. Call each 𝑓𝑖 a composition factor of 𝑓 over 𝐾.

510

24 Problems of Arithmetical Geometry

Proposition 24.8.12 Let 𝐾 be a field and 𝑓 ∈ 𝐾 [𝑋] a polynomial of degree 𝑙 with char(𝐾) ∤ 𝑙. (a) Suppose that 𝐾 is PAC and 𝑓 is injective on 𝐾 and indecomposable over 𝐾. Then, 𝑙 is a prime number and 𝑓 is linearly related over 𝐾 to a Dickson polynomial of degree 𝑙. (b) Suppose that 𝐾 is pseudo finite and 𝑓 is injective or surjective on 𝐾. Then, each composition factor of 𝑓 is linearly related over 𝐾 to a Dickson polynomial of a prime degree. Proof of (a). By assumption, there are no 𝑥, 𝑦 ∈ 𝐾 with 𝑓 (𝑥) = 𝑓 (𝑦) and 𝑥 ≠ 𝑦. Hence, there are no 𝑥, 𝑦 ∈ 𝐾 with 𝑓 ∗ (𝑥, 𝑦) = 0 and 𝑥 ≠ 𝑦. By Proposition 12.1.1, ˜ ˜ By Lemma 24.8.11, 𝑓 is indecomposable over 𝐾. 𝑓 ∗ (𝑋, 𝑌 ) is reducible over 𝐾. ˜ Therefore, by Proposition 24.8.10, 𝑙 is prime and 𝑓 is linearly related over 𝐾 to 𝑋 𝑙 or to 𝑇𝑙 (𝑋). Both 𝑋 𝑙 and 𝑇𝑙 (𝑋) are Dickson polynomials. It follows from Lemma 24.8.3 that 𝑓 is linearly related over 𝐾 to a Dickson polynomial of degree 𝑙. Proof of (b). Every injective (resp. surjective) polynomial on a finite field 𝐹 of order 𝑛 is bijective. Since this is an elementary statement, it holds for every pseudo finite field (Proposition 23.10.4). In particular, 𝑓 is bijective on 𝐾. Suppose that 𝑓 (𝑋) = 𝑔(ℎ(𝑋)) with 𝑔, ℎ ∈ 𝐾 [𝑋]. Since 𝑓 is bijective on 𝐾, the polynomial 𝑔 is surjective on 𝐾 and ℎ is injective. Hence, by the preceding paragraph, both 𝑔 and ℎ are bijective on 𝐾. It follows that each composition factor 𝑞 of 𝑓 over 𝐾 is bijective on 𝐾. Since every pseudo finite field is PAC (Section 23.10), it follows from (a) that each composition factor of 𝑓 is linearly related to a Dickson polynomial of a prime degree. □ Theorem 24.8.13 (Schur’s Conjecture) Let 𝐾 be a global field and 𝑓 ∈ 𝐾 [𝑋] a polynomial with char(𝐾) ∤ deg( 𝑓 ). Suppose that 𝑓 is a permutation polynomial modulo 𝔭 for infinitely many prime divisors 𝔭 of 𝐾. Then, each composition factor of 𝑓 is linearly related over 𝐾 to a Dickson polynomial of a prime degree. Proof. The statement “ 𝑓 is a permutation polynomial” is elementary. By Example 23.10.7(a), there is a pseudo finite field 𝐹 containing 𝐾 such that 𝑓 is a permutation polynomial on 𝐹. By Proposition 24.8.12(b), each composition factor of 𝑓 is linearly related over 𝐹, hence over 𝐾 (Lemma 24.8.3), to a Dickson polynomial of a prime □ degree. Theorem 24.8.14 Let 𝐾 be a finite field and 𝑓 ∈ 𝐾 [𝑋] a polynomial with char(𝐾) ∤ deg( 𝑓 ). Suppose that 𝑓 permutes infinitely many finite extensions 𝐿 of 𝐾. Then, each composition factor of 𝑓 is linearly related over 𝐾 to a Dickson polynomial of a prime degree. Proof. Take a nonprincipal ultraproduct 𝐹 of those finite extensions of 𝐾 on which 𝑓 is a permutation polynomial. Then, 𝑓 is a permutation polynomial on 𝐹. By Example 23.10.7(a), 𝐹 is pseudo finite and 𝐾 ⊆ 𝐹. By Proposition 24.8.12(b), each composition factor of 𝑓 is linearly related over 𝐹, hence over 𝐾 (Lemma 24.8.3) to □ a Dickson polynomial of a prime degree. We end this section with examples which give converses to Proposition 24.8.12 and Theorem 24.8.14.

24.8 Schur’s Conjecture

511

Example 24.8.15 (Cyclic polynomials) Let 𝑛 be a positive integer and 𝑞 a prime number. Then, F×𝑞 is a cyclic group of order 𝑞 − 1. Hence, the map 𝑥 ↦→ 𝑥 𝑛 is bijective on F𝑞 if and only if gcd(𝑛, 𝑞 − 1) = 1. Suppose that 𝑛 ≥ 3 is odd. Let 𝑞 1 , . . . , 𝑞 𝑟 be the prime divisors of 𝑛. Then, both 1 and 2 are relatively prime to 𝑞 𝑖 , 𝑖 = 1, . . . , 𝑟. Choose an integer 𝑎 with 𝑎 ≡ 1 mod 𝑞 𝑖 , 𝑖 = 1, . . . , 𝑟. Then, both 𝑎 and 𝑎 + 1 are relatively prime to 𝑛. Dirichlet’s theorem (Corollary 7.3.2) gives infinitely many prime numbers 𝑝 ≡ 𝑎 + 1 mod 𝑛. By the preceding paragraph, 𝑋 𝑛 is a permutation polynomial modulo 𝑝 for each of these 𝑝.

The next lemma will be used to give more examples of Dickson polynomials which are permutation polynomials modulo infinitely many primes: Lemma 24.8.16 Let 𝑛 be a positive integer, 𝑞 a power of a prime number, and 𝑎 ∈ F𝑞 . Suppose that gcd(𝑛, 𝑞 2 − 1) = 1. Then, 𝐷 𝑛 (𝑎, 𝑋) permutes F𝑞 . Proof. Put 𝐾 = F𝑞 and 𝐿 = F𝑞2 . Then, 𝐿 × is a cyclic group of order 𝑞 2 − 1. Hence, the map 𝑧 ↦→ 𝑧 𝑛 is bijective on 𝐿. Now consider 𝑥 1 , 𝑥2 ∈ 𝐾 with 𝐷 𝑛 (𝑎, 𝑥1 ) = 𝐷 𝑛 (𝑎, 𝑥2 ). Choose 𝑧1 , 𝑧2 ∈ 𝐿 with −1 −1 𝑧 𝑖 + 𝑎𝑧−1 𝑖 = 𝑥 𝑖 , 𝑖 = 1, 2. Then, 𝐷 𝑛 (𝑎, 𝑧 1 + 𝑎𝑧 1 ) = 𝐷 𝑛 (𝑎, 𝑧 2 + 𝑎𝑧 2 ). By Lemma 𝑛 𝑛 −𝑛 𝑛 𝑛 −𝑛 24.8.1(a), 𝑧1 + 𝑎 𝑧1 = 𝑧2 + 𝑎 𝑧2 . Denote the common value of the two sides of the latter equality by 𝑏. The product of the two solutions of the equation 𝑋 + 𝑎 𝑛 𝑋 −1 = 𝑏 is 𝑎 𝑛 . Both 𝑧 1𝑛 and 𝑧2𝑛 are solutions. Hence, 𝑧 1𝑛 = 𝑧 2𝑛 or 𝑧1𝑛 = 𝑎 𝑛 𝑧−𝑛 2 . By the preceding −1 paragraph, 𝑧1 = 𝑧2 or 𝑧 1 = 𝑎𝑧 2 . In both cases, 𝑥 1 = 𝑥2 . Consequently, 𝐷 𝑛 (𝑎, 𝑋) is a permutation polynomial on 𝐾. □ Example 24.8.17 (Permutation polynomials modulo infinitely many primes) (a) Let 𝑛 be a positive integer, 𝐾 a number field, and 𝑎 ∈ 𝑂 𝐾 . Suppose that gcd(6, 𝑛) = 1 and 𝐾 ∩ Q(𝜁 𝑛 ) = Q. Then, 𝐷 𝑛 (𝑎, 𝑋) is a permutation polynomial modulo infinitely many prime ideals of 𝑂 𝐾 . Indeed, list the prime divisors of 𝑛 as 𝑞 1 , . . . , 𝑞 𝑟 . Choose an integer 𝑏 with 𝑏 ≡ 2 mod 𝑞 𝑖 , 𝑖 = 1, . . . , 𝑟. Then, gcd(𝑏, 𝑛) = 1 and gcd(𝑏 2 − 1, 𝑛) = 1. Denote the Galois closure of 𝐾 (𝜁 𝑛 )/Q by 𝐿. Then, choose 𝜎 ∈ Gal(𝐿/𝐾) with 𝜎𝜁 𝑛 = 𝜁 𝑛𝑏 . Chebotarev density theorem (Theorem 7.3.1) gives infinitely many prime ideals 𝔓   of 𝑂 𝐿 with 𝐿/Q = 𝜎. Let 𝔓 be one of them. Then, the decomposition group of 𝔓 𝔓 over Q is in Gal(𝐿/𝐾). Put 𝔭 = 𝔓| 𝐾 . Then, 𝑝 := 𝑁𝔭 is a prime number. By the proof of Corollary 7.3.2, 𝑝 ≡ 𝑏 mod 𝑛. Hence, by the choice of 𝑏, gcd(| 𝐾¯ 𝔭 | 2 − 1, 𝑛) = 1. Consequently, by Lemma 24.8.16, 𝐷 𝑛 (𝑎, 𝑋) is a permutation polynomial modulo 𝔭. (b) Let 𝑛 be a positive integer, 𝑝 a prime number, 𝑞 a power of 𝑝, 𝐾 a finite separable extension of F𝑞 (𝑡) which is regular over F𝑞 , and 𝑎 ∈ 𝑂 𝐾 . Suppose that gcd( 𝑝(𝑞 2 −1), 𝑛) = 1. Then, 𝐷 𝑛 (𝑎, 𝑋) is a permutation polynomial modulo infinitely many prime divisors of 𝐾/F𝑞 . Indeed, let 𝐿 be the Galois closure of 𝐾/F𝑞 (𝑡), F𝑞 𝑚 the algebraic closure of F𝑞 in 𝐿, and 𝜑(𝑛) the Euler totient function. Then, 𝑞 𝜑 (𝑛) ≡ 1 mod 𝑛. Consider a large positive integer 𝑘. Choose 𝜏 in Gal(𝐿/𝐾) whose restriction to F𝑞 𝑚 coincides with 𝑘 𝜑 (𝑛)+1 the restriction of Frob𝑞 . Denote the conjugacy class of 𝜏 in Gal(𝐿/F𝑞 (𝑡)) by Con(𝜏). Proposition 7.4.8 gives a prime divisor 𝔭 of F𝑞 (𝑡) with deg(𝔭) = 𝑘 𝜑(𝑛) + 1

512

24 Problems of Arithmetical Geometry

 𝐿/F𝑞 (𝑡)  𝐿/F𝑞 (𝑡)
and = Con(𝜏). Hence, there is a prime divisor 𝔔 of 𝐿 with = 𝜏. 𝔭 𝔔 Denote the restriction of 𝔔 to 𝐾 by 𝔮. Then, deg(𝔮) = deg(𝔭) = 𝑘 𝜑(𝑛) + 1. Hence, | 𝐾¯ 𝔮 | 2 − 1 = 𝑞 2(𝑘 𝜑 (𝑛)+1) − 1 ≡ 𝑞 2 − 1 mod 𝑛. Consequently, by Lemma 24.8.16, 𝐷 𝑛 (𝑎, 𝑋) is a permutation polynomial modulo 𝔮.

24.9 The Generalized Carlitz Conjecture Let 𝑞 be a power of a prime number 𝑝 and 𝑓 ∈ F𝑞 [𝑋] a polynomial of degree 𝑛. Suppose that 𝑓 is a permutation polynomial on F𝑞 𝑘 for infinitely many 𝑘 but 𝑓 is not a 𝑝-power in F𝑞 [𝑋]. Theorem 24.8.14 describes the composition factors of 𝑓 when 𝑝 ∤ 𝑛. In the general case the generalized Carlitz conjecture states that gcd(𝑛, 𝑞 − 1) = 1. In particular, for 𝑝 ≠ 2, the conjecture predicts that 𝑛 is odd (Carlitz’s conjecture). The paper [FGS93] gives more precise information about permutation polynomials than Carlitz’s conjecture does. Not only does it prove that conjecture for 𝑝 > 3, but it gives information about Gal( 𝑓 (𝑋) − 𝑡, F˜ 𝑞 (𝑡)). Suppose without loss that 𝑓 is indecomposable over F𝑞 . Then, there are three cases: (a) 𝑝 ∤ 𝑛 and 𝐺 is cyclic or isomorphic to the dihedral group 𝐷 𝑛 . This is essentially contained in Section 24.8. (b) 𝑛 = 𝑝 𝑚 and 𝐺  𝐻 ⋉ F𝑚 𝑝 , where 𝐻 is a subgroup of GL(𝑚, F 𝑝 ) acting linearly on F𝑚 . 𝑝 𝑎 𝑎 (c) 𝑝 ∈ {2, 3}, 𝑛 = 𝑝 ( 𝑝2 −1) with 𝑎 ≥ 3 odd, and 𝐺 normalizes the simple group PSL(2, F 𝑝 𝑎 ). The proof that no other cases arise uses the classification of finite simple groups and is beyond the scope of this book. One can find an elementary proof of Carlitz’s conjecture in [Len85]. That proof argues with decomposition and inertia groups over F𝑞 (𝑡). Here we follow [CoF95] which, following a suggestion of [Len85, Remark 1], takes place over F𝑞 ((𝑡)). The first step in the proof translates the assumption “ 𝑓 permutes infinitely many finite extensions of F𝑞 ” into a statement of a general nature: Lemma 24.9.1 Let 𝑓 ∈ F𝑞 [𝑋] be a polynomial which permutes infinitely many finite 𝑓 (𝑌 ) extensions of F𝑞 . Then, 𝑓 ∗ (𝑋, 𝑌 ) = 𝑓 (𝑋)− has no absolutely irreducible factor 𝑋−𝑌 which belongs to F𝑞 [𝑋, 𝑌 ]. Proof. Assume ℎ ∈ F𝑞 [𝑋, 𝑌 ] is an absolutely irreducible factor of 𝑓 ∗ (𝑋, 𝑌 ). By Corollary 6.4.2 there is an 𝑛0 such that for every integer 𝑛 > 𝑛0 there are distinct 𝑥, 𝑦 ∈ F𝑞 𝑛 with ℎ(𝑥, 𝑦) = 0. Then, 𝑓 (𝑥) = 𝑓 (𝑦). Hence, 𝑓 does not permute F𝑞 𝑛 , in contradiction to our assumption. □ The converse of Lemma 24.9.1 is also true (see Exercise 14). However, it is not used in the proof of the generalized Carlitz conjecture.

24.9 The Generalized Carlitz Conjecture

513

Lemma 24.9.2 Let 𝑁/𝑀 be a finite cyclic extension, 𝜎 an element of Gal(𝑀) whose restriction to 𝑁 generates Gal(𝑁/𝑀), and ℎ ∈ 𝑀 [𝑋] a separable irreducible polynomial which becomes reducible over 𝑁. Then, 𝜎𝑥 ≠ 𝑥 for every root 𝑥 of ℎ. Proof. Assume ℎ has a root 𝑥 with 𝜎𝑥 = 𝑥. By assumption, 𝑁 ∩ 𝑀sep (𝜎) = 𝑀. Hence, 𝑁 ∩ 𝑀 (𝑥) = 𝑀, so 𝑁 and 𝑀 (𝑥) are linearly disjoint over 𝑀. Therefore, [𝑁 (𝑥) : 𝑁] = [𝑀 (𝑥) : 𝑀]. It follows that ℎ = irr(𝑥, 𝑀) is irreducible over 𝑁, a contradiction. □

Lemma 24.9.3 Let 𝐿/𝐾 be a finite cyclic extension, 𝑓 ∈ 𝐾 [𝑋] a polynomial of degree at least 2 with a nonzero derivative 𝑓 ′, and 𝑧 an indeterminate. Suppose 𝑓 (𝑌 ) that each irreducible factor of 𝑓 ∗ (𝑋, 𝑌 ) = 𝑓 (𝑋)− in 𝐾 [𝑋, 𝑌 ] is reducible over 𝑋−𝑌 𝐿. Then, 𝑓 (𝑋) − 𝑧 is separable and each 𝜌 ∈ Gal(𝐾 (𝑧)) whose restriction to 𝐿 generates Gal(𝐿/𝐾) fixes exactly one root of 𝑓 (𝑋) − 𝑧. Proof. By assumption, ( 𝑓 (𝑋) − 𝑧) ′ = 𝑓 ′ (𝑋) ≠ 0. All roots of 𝑓 (𝑋) − 𝑧 are transcendental over 𝐾 while all roots of 𝑓 ′ (𝑋) are algebraic over 𝐾. Hence, 𝑓 (𝑋) − 𝑧 is relatively prime in 𝐾 (𝑧) [𝑋] to its derivative. Therefore, 𝑓 (𝑋) − 𝑧 is separable. Let 𝐹ˆ be the finite Galois extension of 𝐸 = 𝐾 (𝑧) generated by 𝐿(𝑧) and all roots of 𝑓 (𝑋) − 𝑧. ˆ Choose a generator 𝜌0 of Gal(𝐿/𝐾) and let 𝐺 ∗ = {𝜌 ∈ Gal( 𝐹/𝐸) | 𝜌| 𝐿 = 𝜌0 }. For each root 𝑥 of 𝑓 (𝑋) − 𝑧 let 𝐺 ∗𝑥 = {𝜌 ∈ 𝐺 ∗ | 𝜌𝑥 = 𝑥}. We have to prove that Ð 𝐺 ∗ = · 𝑥 𝐺 ∗𝑥 , where 𝑥 ranges over all roots of 𝑓 (𝑋) − 𝑧. We divide the rest of the proof into two parts. ˆ Then, 𝑓 (𝑥) = Part A: Disjointness. Let 𝑥 and 𝑦 be distinct roots of 𝑓 (𝑋) − 𝑧 in 𝐹. 𝑧 and 𝑓 (𝑦) = 𝑧, so 𝑥, 𝑦 are transcendental over 𝐾. By the above, 𝑓 ∗ (𝑋, 𝑦) = 𝑓 (𝑋)− 𝑓 ( 𝑦) ∗ = 𝑓 (𝑋)−𝑧 𝑋−𝑦 𝑋−𝑦 ∈ 𝐾 (𝑦) [𝑋] is a separable polynomial and 𝑥 is a root of 𝑓 (𝑋, 𝑦). ∗ Let ℎ ∈ 𝐾 [𝑦, 𝑋] be an irreducible factor of 𝑓 (𝑋, 𝑦) with ℎ(𝑥) = 0. In particular, ℎ is primitive. By assumption, ℎ(𝑋) is reducible in 𝐿 [𝑦, 𝑋]. Hence, by Gauss’ lemma, ℎ(𝑋) is reducible in 𝐿 (𝑦) [𝑋]. ˆ (𝑦)) and 𝜌| 𝐿 ( 𝑦) generates the cyclic Now consider 𝜌 ∈ 𝐺 ∗𝑦 . Then, 𝜌 ∈ Gal( 𝐹/𝐾 group Gal(𝐿 (𝑦)/𝐾 (𝑦)). By Lemma 24.9.2, 𝜌 fixes no root of ℎ. In particular, 𝜌𝑥 ≠ 𝑥. Thus, 𝐺 ∗𝑥 ∩ 𝐺 ∗𝑦 = ∅. Part B: Every 𝜌 ∈ 𝐺 ∗ fixes a root of 𝑓 (𝑋)−𝑧. Indeed, extend 𝜌0 in the unique possiˆ ble way to a generator 𝜌0∗ of Gal(𝐿 (𝑧)/𝐾 (𝑧)). Then, 𝐺 ∗ = {𝜌 ∈ Gal( 𝐹/𝐸) | 𝜌| 𝐿 (𝑧) = ∗ ∗ 𝜌0 }. Hence, |𝐺 | = [ 𝐹ˆ : 𝐿 (𝑧)]. Next let 𝑥 be a root of 𝑓 (𝑋) − 𝑧 in 𝐹ˆ and extend 𝜌0∗ in the unique possible way to a generator 𝜌1∗ of Gal(𝐿 (𝑥)/𝐾 (𝑥)). Then, ˆ (𝑥)) | 𝜌| 𝐿 ( 𝑥) = 𝜌 ∗ }. 𝐺 ∗𝑥 = {𝜌 ∈ Gal( 𝐹/𝐾 1 Hence, |𝐺 ∗𝑥 | = [ 𝐹ˆ : 𝐿(𝑥)].

514

24 Problems of Arithmetical Geometry

𝐾 (𝑥)

𝐿(𝑥)

𝑛

𝐹ˆ

𝑛

𝐾 (𝑧)

𝐿 (𝑧)

𝐾

𝐿

Finally, let 𝑥1 , . . . , 𝑥 𝑛 be the distinct roots of 𝑓 (𝑋) − 𝑧. Then, 𝑛 = deg( 𝑓 ) = [𝐾 (𝑥 𝑖 ) : 𝐾 (𝑧)] = [𝐿(𝑥 𝑖 ) : 𝐿(𝑧)]. By Part A, |

𝑛 Ø

𝐺 ∗𝑥𝑖 | =

𝑖=1

𝑛 ∑︁

|𝐺 ∗𝑥𝑖 | = 𝑛[ 𝐹ˆ : 𝐿 (𝑥𝑖 )]

𝑖=1

= [𝐿(𝑥𝑖 ) : 𝐿(𝑧)] [ 𝐹ˆ : 𝐿(𝑥 𝑖 )] = [ 𝐹ˆ : 𝐿(𝑧)] = |𝐺 ∗ |. Hence, 𝐺 ∗ =

∗ 𝑖=1 𝐺 𝑥𝑖 .

Ð𝑛

Thus, each 𝜌 ∈ 𝐺 ∗ fixes some 𝑥𝑖 .

□

Lemma 24.9.4 Let 𝐾 be a field, 𝑓 ∈ 𝐾 [𝑋] a monic polynomial of degree 𝑛 > 1, 𝑟 a divisor of 𝑛 with char(𝐾) ∤ 𝑟, and 𝑧 an indeterminate. Put 𝐸 = 𝐾 ((𝑧−1 )) and let 𝑥 be ˜ Then, 𝐸 (𝑥) contains an 𝑟th root √𝑟 𝑧 and [𝐸 ( √𝑟 𝑧) : 𝐸] = 𝑟. a root of 𝑓 (𝑋) − 𝑧 in 𝐸. Proof. Let 𝑣 be the unique discrete complete valuation of 𝐸/𝐾 such that 𝑣(𝑧−1 ) = 1 (Example 4.5.1). Extend 𝑣 to 𝐸 (𝑥) in the unique possible way. By Proposition 4.5.2, both 𝐸 and 𝐸 (𝑥) satisfy Hensel’s lemma. Now write 𝑓 (𝑋) = 𝑋 𝑛 + 𝑎 𝑛−1 𝑋 𝑛−1 + · · · + 𝑎 0 . Put 𝑔(𝑌 ) = 1 + 𝑎 𝑛−1𝑌 + · · · + 𝑎 1𝑌 𝑛−1 + 𝑎 0𝑌 𝑛 and 𝑦 = 𝑥 −1 . Then, 𝑥 𝑛 𝑔(𝑦) = 𝑓 (𝑥). Consider the polynomial ℎ(𝑇) = 𝑇 𝑟 − 𝑔(𝑦) ∈ 𝐸 (𝑥) [𝑇]. Then, ℎ(1) = −𝑎 𝑛−1 𝑦 − · · · − 𝑎 1 𝑦 𝑛−1 − 𝑎 0 𝑦 𝑛 and ℎ ′ (1) = 𝑟 ≠ 0. By assumption, 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 + · · · + 𝑎 0 = 𝑧. Hence, 𝑣(𝑥) < 0, so 𝑣(𝑦) > 0. It follows that 𝑣(ℎ(1)) = 𝑣(1 − 𝑔(𝑦)) > 0 and 𝑣(ℎ ′ (1)) = 0. Hensel’s lemma gives 𝑡 ∈ 𝐸 (𝑥) with 𝑡 𝑟 = 𝑔(𝑦). Finally, let 𝑤 = 𝑥 𝑛/𝑟 𝑡. Then, 𝑤 ∈ 𝐸 (𝑥) and 𝑤 𝑟 = 𝑥 𝑛 𝑡 𝑟 = 𝑥 𝑛 𝑔(𝑦) = 𝑓 (𝑥). Moreover, 𝐸 (𝑤)/𝐸 is a totally ramified extension of degree 𝑟 (Example 2.3.8). □ Theorem 24.9.5 (Lenstra) Let 𝑝 be a prime number, 𝑞 a power of 𝑝, and 𝑓 ∈ F𝑞 [𝑋] a polynomial of degree 𝑛 > 1. Suppose that 𝑓 ≠ 𝑓1𝑝 for all 𝑓1 ∈ F𝑞 [𝑋] and 𝑓 permutes infinitely many finite extensions of F𝑞 . Then, gcd(𝑛, 𝑞 − 1) = 1. Proof. Put 𝐾 = F𝑞 . Since 𝑓 is not a 𝑝th power of a polynomial in 𝐾 [𝑋] and 𝐾 𝑓 (𝑌 ) is perfect, the derivative of 𝑓 is nonzero. Put 𝑓 ∗ (𝑋, 𝑌 ) = 𝑓 (𝑋)− . By Lemma 𝑋−𝑌 24.9.1, no absolutely irreducible factor of 𝑓 ∗ (𝑋, 𝑌 ) belongs to 𝐾 [𝑋, 𝑌 ]. Choose a finite, necessarily cyclic, extension 𝐿 of 𝐾 such that all absolutely irreducible factors of 𝑓 ∗ (𝑋, 𝑌 ) belong to 𝐿 [𝑋, 𝑌 ]. Then, every irreducible factor of 𝑓 ∗ (𝑋, 𝑌 ) over 𝐾 is reducible over 𝐿. Let 𝑧 be an indeterminate. Put 𝐸 = 𝐾 ((𝑧−1 )) and 𝐹 = 𝐿((𝑧 −1 )). Then, 𝐸/𝐾 is a regular extension and 𝐸 𝐿 = 𝐹 (Example 4.5.1). Hence, res: Gal(𝐹/𝐸) → Gal(𝐿/𝐾) is an isomorphism. By Lemma 24.9.3, 𝑓 (𝑋) −𝑧 is separable. Choose a finite Galois extension 𝐹ˆ of 𝐸 which contains 𝐹 and all roots of 𝑓 (𝑋) − 𝑧 in 𝐸 sep .

24.9 The Generalized Carlitz Conjecture

515

Now assume gcd(𝑛, 𝑞 − 1) > 1. Choose a common prime divisor 𝑟 of 𝑛 and √ 𝑞 − 1. Then, 𝑝 ≠ 𝑟 and 𝐾 contains all roots of 1 of order 𝑟. Put 𝐸 ′ = 𝐸 ( 𝑟 𝑧) and √ 𝐹 ′ = 𝐹 ( 𝑟 𝑧). By Lemma 24.9.4, 𝐸 ′/𝐸 and 𝐹 ′/𝐹 are cyclic extensions of degree 𝑟. Therefore, both maps res: Gal(𝐹 ′/𝐹) → Gal(𝐸 ′/𝐸) and res: Gal(𝐹 ′/𝐸 ′) → ˆ Gal(𝐹/𝐸) are isomorphisms. In addition, Lemma 24.9.4 implies that 𝐸 ′, 𝐹 ′ ⊆ 𝐹. ˆ We may therefore choose 𝜎, 𝜏 ∈ Gal( 𝐹/𝐸) such that ⟨res𝐹 ′ 𝜎⟩ = Gal(𝐹 ′/𝐸 ′) and ⟨res𝐹 ′ 𝜏⟩ = Gal(𝐹 ′/𝐹). Put 𝜌 = 𝜎𝜏. Then, res 𝐿 𝜌 generates Gal(𝐿/𝐾). 𝐸 (𝑥)

𝐹 (𝑥)

𝐸′

𝐹′

𝐸

𝐹

𝐾

𝐿

𝐹ˆ

ˆ By Lemma 24.9.4, 𝐸 (𝑥) By Lemma 24.9.3, 𝜌 fixes a root 𝑥 of 𝑓 (𝑋) − 𝑧 in 𝐹. ′ contains a root of 𝑧 of order 𝑟, so 𝐸 ⊆ 𝐸 (𝑥). Hence, res𝐸 ′ 𝜌 = id. On the other hand, res𝐸 ′ 𝜌 = res𝐸 ′ 𝜏 ≠ id. This contradiction proves that gcd(𝑛, 𝑞 − 1) = 1. □

Exercises 1. Let 𝐾 be a perfect field and 𝑉 a Zariski 𝐾-closed set with 𝑉 (𝐾) infinite. Use the decomposition-intersection procedure and Corollary 11.5.3 to prove that 𝑉 contains a curve defined over 𝐾. 2. For each positive integer 𝑖, give an example of a PAC field 𝐾 which is 𝐶𝑖+1 but not 𝐶𝑖 . Hint: Take 𝐾 to be imperfect, follow Example 24.2.8 and use Propositions 24.2.4 and 24.2.12. 3. Let 𝑓 ∈ Q[𝑋0 , . . . , 𝑋𝑙 ] be an irreducible form of prime degree 𝑙. Prove that 𝑉 ( 𝑓 ) contains a variety defined over Q. Hint: The intersection of 𝑙 hyperplanes in P𝑙 is nonempty. ˜ 4. Prove that almost all 𝜎 ∈ Gal(Q) satisfy this: Let 𝑓 ∈ Q(𝜎) [𝑋0 , . . . , 𝑋𝑛 ] be a ˜ polynomial of degree 𝑑 ≤ 𝑛 with a Q(𝜎)-zero. Then, 𝑉 ( 𝑓 ) contains a curve defined ˜ over Q(𝜎). Hint: Use Chevalley–Warning, the transfer theorem and Exercise 1. 5. Give an example of fields 𝐾0 and 𝐾 such that 𝐾0 is algebraically closed in 𝐾, 𝐾 is 𝐶1 but 𝐾0 is not. ˜ Suggestion: Take 𝐾0 to be Q(𝜏) where 𝜏 is the conjugation on C. Then, use Lemmas 23.5.3 and 24.3.5. 6. Show that F 𝑝 (𝑡), with 𝑡 a transcendental, is not 𝐶1 . Suggestion: Choose a nonsquare 𝑎 ∈ F 𝑝 and prove that 𝑋 2 + 𝑎𝑌 2 + 𝑡𝑍 2 has no nontrivial zero in F 𝑝 (𝑡).

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7. Let 𝑀 and 𝐿 be finite separable extensions of a global field 𝐾. Suppose that for almost all 𝔭 ∈ 𝑃(𝐾) the number of primes 𝔓 ∈ 𝑃(𝐿) lying over 𝔭 and having a relative degree 1 is equal to the number of primes 𝔓 ∈ 𝑃(𝑀) lying over 𝔭 and having relative degree 1. Prove that [𝐿 : 𝐾] = [𝑀 : 𝐾] and the Galois closure 𝐿ˆ of 𝐿/𝐾 is equal to the Galois closure 𝑀ˆ of 𝑀/𝐾 [Kro80]. Hint: Follow these steps. (a) Choose primitive elements 𝑥 and 𝑦 for 𝐿/𝐾 and 𝑀/𝐾, respectively, integral over 𝑂 𝐾 . Put 𝑓 = irr(𝑥, 𝐾) and 𝑔 = irr(𝑦, 𝐾). Use Lemma 24.5.2 to prove that for almost all 𝔭 ∈ 𝑃(𝐾) the numbers of roots of 𝑓 and 𝑔 in 𝐾¯ 𝔭 are equal. (b) Let 𝑁 be a finite Galois extension of 𝐾 that contains both 𝐿 and 𝑀. Use the transfer theorem (or alternatively the Chebotarev density theorem) to prove that every 𝜎 ∈ Gal(𝑁/𝐾) fixes the same numbers of roots of 𝑓 and of 𝑔. ˆ ˆ and the elements 𝜏 ∈ Gal(𝑁/ 𝑀) (c) Apply (b) to the elements 𝜎 ∈ Gal(𝑁/ 𝐿) and prove that deg( 𝑓 ) = deg(𝑔). ˆ ˆ all roots of 𝑔 belong to 𝑁 (𝜎), so 𝑀ˆ ⊆ 𝐿. (d) Observe: For 𝜎 ∈ Gal(𝑁/ 𝐿), 8. Let 𝐺 beÐa finite group and 𝐻, 𝐼, 𝑁 subgroups. Suppose that 𝐺 = 𝐻 ⋉ 𝑁 and Ð 𝑥 𝑥 𝑥 ∈𝐺 𝐼 = 𝑥 ∈𝐺 𝐻 . Prove that 𝐺 = 𝐼 ⋉ 𝑁. Deduce that 𝐻 is not a proper subgroup of 𝐼. Ð Hint: Observe that 𝑥 ∈𝐺 (𝐼 𝑁) 𝑥 = 𝐺. Ð 9. (a) Let 𝐴 ⊳ 𝐵 ≤ 𝐺 be a tower of finite groups. Prove that | 𝑥 ∈𝐺 𝐴 𝑥 | ≤ (𝐺 : 𝐵)| 𝐴|. (b) Let 𝐾 ⊆ 𝐿 be a finite separable extension of a global field. Observe: The Dirichlet density 𝛿 of the set 𝑉 (𝐿/𝐾) (Definition 24.5.1) is equal to the Haar Ð measure of 𝜎 ∈Gal(𝐾) Gal(𝐿) 𝜎 . (c) Use (a), (b) and Lemma 24.5.3 to prove that if 𝑀 is a finite Galois extension of 𝐿 which is Kronecker conjugate to 𝐿 over 𝐾, then [𝑀 : 𝐿] ≤ 𝛿−1 [Kli78]. 10. Exercise 1(c) of Chapter 14 gives a pair of polynomials 𝑓 (𝑋) = 𝑋 4 + 2𝑋 2 , 𝑔(𝑋) = −4𝑋 4 − 4𝑋 2 − 1 for which 𝑓 (𝑋) − 𝑔(𝑌 ) is reducible. Show however that 𝑓 (𝑋) − 𝑡 and 𝑔(𝑌 ) − 𝑡 are not Kronecker conjugate over Q(𝑡). Hint: Use Corollary 24.6.5. 11. Prove for nonzero integers 𝑎, 𝑏 1 , . . . , 𝑏 𝑛 that the following two statements are equivalent. (a) There exist 𝜀1 , . . . , 𝜀 𝑛 ∈ {0, 1} and 𝑐 ∈ Z satisfying 𝑎 = 𝑏 1𝜀1 · · · 𝑏 𝑛𝜀𝑛 𝑐2 . (b) For almost all primes 𝑝, if 𝑏 1 , . . . , 𝑏 𝑛 are quadratic residues modulo 𝑝, then so is 𝑎. Hint: Combine the transfer theorem (or directly the Chebotarev density theorem) with Kummer theory for quadratic extensions over Q. 12. Let 𝑛 be a positive integer, 𝑞 a power of a prime number 𝑝, and 𝑎 ∈ F𝑞 . Suppose that gcd(𝑛, 𝑞 − 1) > 1. Find distinct 𝑥, 𝑦 ∈ F𝑞 with 𝐷 𝑛 (𝑎, 𝑥) = 𝐷 𝑛 (𝑎, 𝑦). Combine this with Theorem 24.8.13 to supply a proof of the generalized Carlitz conjecture (Theorem 24.9.5) in the case where 𝑝 ∤ deg( 𝑓 ). Hint: Choose a common prime divisor 𝑙 of 𝑛 and 𝑞 − 1. Then, choose 𝑥 = 𝑧 + 𝑎𝑧 −1 and 𝑦 = 𝜁 𝑧 + 𝑎 −1 𝜁 𝑧 with 𝜁, 𝑧 ∈ F𝑙 , 𝜁 𝑙 = 1, and 𝜁 ≠ 1. 13. Let 𝑝, 𝑙 be distinct prime numbers, 𝑞 a power of 𝑝, and 𝑎 ∈ F×𝑞 . Suppose that 𝑙 | 𝑝 2 − 1. Then, 𝐷 𝑙 (𝑎, 𝑋) permutes only finitely many fields F𝑞 𝑘 .

24.9 The Generalized Carlitz Conjecture

517

Hint: Use the hint of Exercise 12 to reduce to the case 𝑙 |𝑞 + 1 and 𝑙 ≠ 2. Then, (𝑙−1)/2 Ö 𝐷 𝑙 (𝑎, 𝑋) − 𝐷 𝑙 (𝑎, 𝑌 ) = (𝑋 2 − 𝛼𝑖 𝑋𝑌 + 𝑌 2 + 𝛽𝑖2 𝑎), 𝑋 −𝑌 𝑖=1

(24.35)

where 𝛼𝑖 = 𝜁𝑙𝑖 + 𝜁𝑙−𝑖 and 𝛽𝑖 = 𝜁𝑙𝑖 − 𝜁𝑙−𝑖 [Sch00, p. 52]. Observe that 𝜁𝑙𝑞 = 𝜁𝑙−1 , so 𝛼𝑖 , 𝛽𝑖2 ∈ F𝑞 . Next note that each of the factors on the right-hand side of (24.35) is absolutely irreducible. Therefore, it has F𝑞 𝑘 -zeros off the diagonal if 𝑘 is large. 14. Prove the following generalization of a theorem of MacCluer (see also [Fri74b, Thm. 1]): Let 𝐾 be a pseudo finite field, 𝑓 ∈ 𝐾 [𝑋] a separable polynomial of degree 𝑛 > 0, 𝑥 an indeterminate. Put 𝑡 = 𝑓 (𝑥). Suppose that no 𝐾-irreducible factor of 𝑓 (𝑌 ) 𝑓 ∗ (𝑋, 𝑌 ) = 𝑓 (𝑋)− is absolutely irreducible. Then, 𝑓 permutes 𝐾. 𝑋−𝑌 Hint: Let 𝑥 1 , . . . , 𝑥 𝑛 be the 𝑛 distinct roots of 𝑓 (𝑋) = 𝑡. Put 𝐹 = 𝐾 (𝑥1 , . . . , 𝑥 𝑛 ) ˆ and consider 𝑎 ∈ 𝐾. Then, find a 𝐾-place 𝜓: 𝐹 → 𝐾˜ ∪ {∞} with 𝜓(𝑡) = 𝑎. Further, choose 𝜎 in the decomposition group of 𝜓 which belongs to 𝑇. Let 𝐾ˆ be the algebraic closure of 𝐾 in 𝐹, 𝑇 = {𝜏 ∈ Gal(𝐹/𝐾 (𝑡)) | res𝐾ˆ 𝜏 = res𝐾ˆ 𝜎}, and 𝑇𝑖 = 𝑇 ∩ Gal(𝐹/𝐾 Í𝑛(𝑥𝑖 )), 𝑖 = 1, . . . , 𝑛. Ð𝑛 Prove: |𝑇 | = 𝑖=1 |𝑇𝑖 | and 𝑇𝑖 ∩ 𝑇 𝑗 = ∅ for 𝑖 ≠ 𝑗. Conclude that 𝑇 = · 𝑖=1 𝑇𝑖 . There is an 𝑖 with 𝜎𝑥 𝑖 = 𝑥 𝑖 . Hence, 𝑏 = 𝜓(𝑥𝑖 ) is in 𝐾 and satisfies 𝑓 (𝑏) = 𝑎. Consequently, 𝑓 is surjective on 𝐾. Since 𝐾 is pseudo finite, 𝑓 is also injective.

Notes The decomposition-intersection procedure has been used by several authors. For example, [Gre65] uses it to prove that for each 𝑑 > 0 almost all fields Q 𝑝 are 𝐶2,𝑑 . Also, it appears in the first version, [FrS76], of the stratification procedure. [Kli98] gives a comprehensive survey on Kronecker classes of number fields. Problem 24.5.8 has a negative solution if the following statement holds for each infinite profinite Ð group 𝐺 of rank ≤ ℵ0 : For each closed subgroup 𝐻 of 𝐺 of infinite index the set 𝑔 ∈𝐺 𝐻 𝑔 does not contain an open neighborhood of 1. There is an attempt in [Kli78, Thm. 6] to give a negative answer to Problem 24.5.8. Unfortunately there is an error in the proof: A closed subgroup 𝐻 of a profinite group 𝐺 of countable rank may have uncountably many, rather than countably many (as claimed in [Kli78]), conjugates in 𝐺. For example, there are uncountably many involutions in 𝐺 (Q), all conjugate. Thus, Problem 24.5.8 is still open. Davenport’s original problem [FrJ86, Problem 19.26] asked if 𝑉 𝑝 ( 𝑓 ) = 𝑉 𝑝 (𝑔) for nonconstants 𝑓 , 𝑔 ∈ Q[𝑋] and almost all 𝑝 implies that 𝑓 and 𝑔 are linearly related. Remark 24.6.1 supplies counter examples to that problem. Davenport’s present problem 24.6.2 is a modification of the older one. Lemma 24.7.2 overlaps with a result attributed to Galois [Hup67, p. 163, Satz 3.6]. More on Dickson’s polynomials can be found in [Sch00, §1.4-1.5] and [LMT93]. The original proof of Proposition 24.7.7 appears in [Scr33]. Our proof is an elaboration of [LMT93, p. 126]. Likewise, the original proof of Proposition 24.7.8 appears in [Bur06]. We have elaborated [LMT93, p. 127].

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[Scr23] proves that every polynomial 𝑓 ∈ Z[𝑋] of prime degree which is a permutation polynomial modulo infinitely many prime numbers 𝑝 is a composition ˜ to cyclic polynomials or Chebyshev of polynomials which are linearly related over Q polynomials. Schur conjectured that his result holds for an arbitrary number field 𝐾. [Fri70] uses the theory of Riemann surfaces to prove Schur’s conjecture. [Tur95, Thm. 2] uses algebraic methods to generalize Fried’s result to arbitrary global field (see also [LMT93]). Our proof of Schur’s conjecture (Theorem 24.8.13) is based on a result of Müller (Proposition 24.8.7). Our version of Cohen and Fried’s proof of the generalized Carlitz conjecture (Theorem 24.9.5) uses improvements of Bensimhoun and Haran. Partially building on ideas of [Len85], [GuM97, §8] further generalizes the generalized Carlitz conjecture to nonconstant separable morphisms between smooth projective curves over F𝑞 .

Chapter 25

Projective Groups and Frattini Covers

Every profinite group is a Galois group of some Galois extension (Leptin, Proposition 3.4.12). However, not every profinite group is an absolute Galois group. For example, the only finite groups that appear as absolute Galois groups are the trivial group and Z/2Z (Artin [Lan97, p. 299, Cor. 9.3]). This raises the question of characterizing the absolute Galois groups among all profinite groups. This question is still wide open. A more restrictive question finds a complete solution in projective groups: By Theorem 12.6.2, the absolute Galois group of a PAC field is projective. Conversely, for each projective group 𝐺 there exists a PAC field 𝐾 such that Gal(𝐾)  𝐺 (Corollary 26.1.2). Thus, a profinite group 𝐺 is projective if and only if it is isomorphic to the absolute Galois group of a PAC field. In this chapter we define a projective group as a profinite group 𝐺 for which every embedding problem is weakly solvable. By Gruenberg’s theorem (Lemma 25.3.2), it suffices to weakly solve only finite embedding problems. This leads to the second characterization of projective groups as those profinite groups which are isomorphic to closed subgroups of free profinite groups (Corollary 25.4.6). Projective groups also appear as the universal Frattini covers of profinite groups (Proposition 25.7.1). Thus, as a preparation for decidability and undecidability results about families of PAC fields, we introduce in this chapter the basis properties of Frattini covers.

25.1 The Frattini Group of a Profinite Group Consider a profinite group 𝐺. Denote the intersection of all maximal open subgroups of 𝐺 by Φ(𝐺). Here we call a subgroup 𝑀 maximal if there is no subgroup 𝑀 ′ of 𝐺 such that 𝑀 < 𝑀 ′ < 𝐺. The characteristic (and therefore normal) closed subgroup Φ(𝐺) is called the Frattini group of 𝐺. We characterize the elements of Φ(𝐺) as “dispensable generators” of 𝐺: Lemma 25.1.1 Let 𝐺 be a profinite group. An element 𝑔 ∈ 𝐺 belongs to Φ(𝐺) if and only if there is no proper closed subgroup 𝐻 of 𝐺 for which ⟨𝐻, 𝑔⟩ = 𝐺. Also, if 𝐻 is a closed subgroup for which 𝐻 · Φ(𝐺) = 𝐺, then 𝐻 = 𝐺. 519 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. D. Fried, M. Jarden, Field Arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 11, https://doi.org/10.1007/978-3-031-28020-7_25

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Proof. Let 𝑔 ∈ Φ(𝐺) and let 𝐻 be a closed subgroup of 𝐺 for which ⟨𝐻, 𝑔⟩ = 𝐺. If 𝐻 ≠ 𝐺, then 𝐻 is contained in an open subgroup (Lemma 1.2.3), hence in a maximal open subgroup 𝑀 of 𝐺. Therefore, 𝐺 = ⟨𝐻, 𝑔⟩ ≤ 𝑀 < 𝐺, a contradiction. Conversely, suppose that an element 𝑔 ∈ 𝐺 satisfies the above condition. Let 𝑀 be a maximal open subgroup of 𝐺. If 𝑔 ∉ 𝑀, then 𝐺 = ⟨𝑀, 𝑔⟩. Hence, 𝑀 = 𝐺, a contradiction. Therefore, 𝑔 ∈ 𝑀, and since 𝑀 is arbitrary, 𝑔 ∈ Φ(𝐺). Finally, let 𝐻 be a closed subgroup of 𝐺 such that 𝐻 · Φ(𝐺) = 𝐺. If 𝐻 < 𝐺, then 𝐻 is contained in a maximal open subgroup 𝑀 of 𝐺 and 𝑀 = 𝑀 · Φ(𝐺) = 𝐺, a contradiction. Thus, 𝐻 = 𝐺. □ Lemma 25.1.2 The Frattini group of a profinite group 𝐺 is pronilpotent. Proof. Assume first that 𝐺 is finite. Let 𝑃 be a 𝑝-Sylow group of Φ(𝐺). For 𝑔 ∈ 𝐺 the group 𝑃 𝑔 is also a 𝑝-Sylow subgroup of Φ(𝐺). Hence, there exists an 𝑎 ∈ Φ(𝐺) such that 𝑃 𝑔 = 𝑃 𝑎 . Therefore, 𝑔𝑎 −1 is in the normalizer, 𝑁𝐺 (𝑃), of 𝑃 in 𝐺. Thus, 𝐺 = 𝑁𝐺 (𝑃)Φ(𝐺). By Lemma 25.1.1, 𝐺 = 𝑁𝐺 (𝑃). Hence, 𝑃 is normal in 𝐺, and therefore in Φ(𝐺). It follows that Φ(𝐺) is the direct product of its 𝑝-Sylow groups. That is, 𝐺 is nilpotent [Hup67, p. 260]. In the general case observe that for any open normal subgroup 𝑁 of 𝐺, Φ(𝐺)/(𝑁 ∩ Φ(𝐺))  𝑁Φ(𝐺)/𝑁 ≤ Φ(𝐺/𝑁). Since Φ(𝐺/𝑁) is a nilpotent finite group, so is Φ(𝐺)/(𝑁 ∩ Φ(𝐺)). Therefore, Φ(𝐺) = lim Φ(𝐺)/(𝑁 ∩ Φ(𝐺)) is a pronilpotent ←− group. □ Lemma 25.1.3 Let 𝐴 be a minimal normal subgroup of a finite group 𝐺. If 𝐴 ≤ Φ(𝐺), then 𝐴 is 𝑝-elementary Abelian for some prime number 𝑝. Proof. By Lemma 25.1.2, 𝐴 is a nilpotent group. Hence, the center 𝑍 ( 𝐴) of 𝐴 is nontrivial [Hup67, p. 260] and normal in 𝐺. Thus, 𝑍 ( 𝐴) = 𝐴 and 𝐴 is Abelian. As such, 𝐴 is a direct product of its Sylow subgroups and each of them is normal in 𝐺. Therefore, 𝐴 is an Abelian 𝑝-group for some prime number 𝑝. The subgroup of 𝐴 consisting of all elements 𝑥 with 𝑥 𝑝 = 1 is normal in 𝐺. Consequently, 𝐴 = (Z/𝑝Z) 𝑚 for some 𝑚 ≥ 0. □ As an operation, taking the Frattini group of a group has functorial properties: Lemma 25.1.4 (a) Let 𝜃: 𝐺 → 𝐻 be an epimorphism of profinite groups. Then, 𝜃 (Φ(𝐺)) ≤ Φ(𝐻). If Ker(𝜃) ≤ Φ(𝐺), then 𝜃 (Φ(𝐺)) = Φ(𝐻). (b) Suppose that 𝑈 is a closed subgroup of 𝐺 and 𝑁 is a closed subgroup of Φ(𝑈), normal in 𝐺. Then, 𝑁 ≤ Φ(𝐺). (c) Let 𝑁 be a Î closed normal subgroup of 𝐺. Then, Φ(𝑁) ≤ Φ(𝐺). = (d) Let 𝐺 𝑖 ∈𝐼 𝐺 𝑖 be a direct product of profinite groups. Then, Φ(𝐺) = Î Φ(𝐺 ). 𝑖 𝑖 ∈𝐼 Proof of (a). The inverse image of each maximal open subgroup of 𝐻 is a maximal open subgroup of 𝐺. Hence, 𝜃 (Φ(𝐺)) ≤ Φ(𝐻). Proof of (b). Assume 𝑁 ̸ ≤ Φ(𝐺). Then, 𝐺 has a maximal open subgroup 𝑀 with 𝑁 ̸ ≤ 𝑀. Thus, 𝑀 𝑁 = 𝐺. Hence, 𝑈 = 𝑈 ∩ 𝑀 𝑁 = (𝑈 ∩ 𝑀)𝑁. Since 𝑁 ≤ Φ(𝑈), Lemma 25.1.1 implies that 𝑈 = 𝑈 ∩ 𝑀. Therefore, 𝑁 ≤ 𝑀, contrary to the assumption.

25.2 Cartesian Squares

521

Proof of (c). Apply (b) to 𝑁 and Φ(𝑁) instead of to 𝑈 and 𝑁. Proof of (d). EachÎ𝐺 𝑖 is a closed normal subgroup of 𝐺. Therefore, (c) gives Φ(𝐺 𝑖 ) ≤ Φ(𝐺) and 𝑖 ∈𝐼 Φ(𝐺 𝑖 ) ≤ Φ(𝐺). Î Conversely, let 𝑀 be a maximal open subgroup of 𝐺 𝑗 . Then, 𝑀 × 𝑖≠ 𝑗 𝐺 𝑖 is a Î maximal open subgroup Î of 𝐺. Thus Φ(𝐺) ≤ 𝑀 × 𝑖≠ 𝑗 𝐺 𝑖 . Running over all 𝑀, this Φ(𝐺 gives Φ(𝐺) ) × ≤ 𝑗 𝑖≠ 𝑗 𝐺 𝑖 . Since this is true for each 𝑗 ∈ 𝐼, we conclude that Î Φ(𝐺) ≤ 𝑖 ∈𝐼 Φ(𝐺 𝑖 ). □

Remark 25.1.5 (The map 𝜃: Φ(𝐺) → Φ(𝐻) in Lemma 25.1.4(a) need not be surjective) Let 𝐺 = ⟨𝑏⟩ ⋉ ⟨𝑎⟩ be the semidirect product of a cyclic group ⟨𝑎⟩ of order 5 with a cyclic group 𝐻 = ⟨𝑏⟩ of order 4 with the relation 𝑎 𝑏 = 𝑎 2 . Let 𝜃: 𝐺 → 𝐻 be the quotient map. Both ⟨𝑎⟩ and ⟨𝑏⟩ are maximal subgroups of 𝐺, so Φ(𝐺) = 1. On the other hand, 𝐻 has order 4, so Φ(𝐻) is of order 2. Thus, 𝜃 (Φ(𝐺)) < Φ(𝐻).

25.2 Cartesian Squares The usual direct product of group theory has a useful generalization: Let 𝛼: 𝐵 → 𝐴 and 𝛾: 𝐶 → 𝐴 be homomorphisms of profinite groups. Let 𝐵 × 𝐴 𝐶 = {(𝑏, 𝑐) ∈ 𝐵 × 𝐶 | 𝛼(𝑏) = 𝛾(𝑐)}. Since 𝐴 is Hausdorff, 𝐵 × 𝐴 𝐶 is a closed subgroup of 𝐵 × 𝐶. It is therefore a profinite group called the fiber product of 𝐵 and 𝐶 over 𝐴. There are natural projection maps pr 𝐵 : 𝐵 × 𝐴 𝐶 → 𝐵 and pr𝐶 : 𝐵 × 𝐴 𝐶 → 𝐶 defined by pr 𝐵 (𝑏, 𝑐) = 𝑏 and pr𝐶 (𝑏, 𝑐) = 𝑐, respectively. It is possible to change 𝐴, 𝐵, and 𝐶 such that the projection maps will be surjective: Put 𝐴0 = {𝑎 ∈ 𝐴 | ∃(𝑏, 𝑐) ∈ 𝐵 × 𝐶: 𝛼(𝑏) = 𝛾(𝑐) = 𝑎}, 𝐵0 = 𝛼−1 ( 𝐴0 ), and 𝐶0 = 𝛾 −1 ( 𝐴0 ). Then, 𝐴0 (resp. 𝐵0 , 𝐶0 ) is a closed subgroup of 𝐴 (resp. 𝐵, 𝐶), 𝐵0 × 𝐴0 𝐶0 = 𝐵 × 𝐴 𝐶, and the projections maps of 𝐵0 × 𝐴0 𝐶0 are surjective. Proposition 25.2.1 Consider a commutative diagram of profinite groups 𝛿

𝐷

/𝐶

(25.1) 𝛾

𝛽

 𝐵 𝛼

 /𝐴

where 𝛽, 𝛼, 𝛾, and 𝛿 are homomorphisms. The following statements are equivalent:

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25 Projective Groups and Frattini Covers

(a) There exists an isomorphism 𝜃: 𝐷 → 𝐵× 𝐴𝐶 with 𝛽◦𝜃 −1 = pr 𝐵 and 𝛿◦𝜃 −1 = pr𝐶 . (b) Let 𝐺 be a profinite group and 𝜑: 𝐺 → 𝐵 and 𝜓: 𝐺 → 𝐶 homomorphisms. Suppose that 𝛼◦𝜑 = 𝛾◦𝜓. Then, there exists a unique homomorphism 𝜋: 𝐺 → 𝐷 which makes the following diagram commutative: 𝐺✵ ❅❖❖ ✵✵❅❅❖❖❖❖ 𝜓 ✵✵ ❅❅❅ 𝜋❖❖❖❖ ❖❖❖ ✵✵ ❅ 𝛿 ❖/' ✵ 𝐷 𝐶 𝜑 ✵ ✵✵ ✵✵ 𝛽 𝛾 ✵   𝐵 𝛼 /𝐴

(25.2)

Proof of “(a)=⇒(b)”. Without loss assume 𝐷 = 𝐵 × 𝐴 𝐶, 𝛽 = pr 𝐵 , and 𝛿 = pr𝐶 . Define a map 𝜋: 𝐺 → 𝐷 by 𝜋(𝑔) = (𝜑(𝑔), 𝜓(𝑔)). Then, 𝜋 is the unique homomorphism which makes (25.2) commutative. Proof of “(b)=⇒(a)”. Let 𝐺 = 𝐵 × 𝐴 𝐶, 𝜑 = pr 𝐵 , and 𝜓 = pr𝐶 . Then, (b) gives a unique 𝜋 that makes (25.2) commutative. Define 𝜃: 𝐷 → 𝐵 × 𝐴 𝐶 by 𝜃 (𝑑) = (𝛽(𝑑), 𝛿(𝑑)) for 𝑑 ∈ 𝐷. Apply the uniqueness property to maps from 𝐷 to 𝐷 and from 𝐺 to 𝐺. It implies that the map 𝜋 ◦ 𝜃 (resp. 𝜃 ◦ 𝜋) is the identity, and 𝜃 satisfies (a). □ Definition 25.2.2 A commutative diagram (25.1) satisfying the conditions of Proposition 25.2.1 is said to be a cartesian square. Lemma 25.2.3 Let (25.1) be a cartesian square of homomorphisms of profinite groups. (a) If 𝑏 ∈ 𝐵 and 𝑐 ∈ 𝐶 satisfy 𝛼(𝑏) = 𝛾(𝑐), then there exists a unique 𝑑 ∈ 𝐷 with 𝛽(𝑑) = 𝑏 and 𝛿(𝑑) = 𝑐. In particular, if 𝑑 ∈ 𝐷 satisfies 𝛽(𝑑) = 1 and 𝛿(𝑑) = 1, then 𝑑 = 1. (b) If 𝛼 (resp. 𝛾) is surjective, then 𝛿 (resp. 𝛽) is surjective. Proof of (a). Assume without loss that 𝐷 = 𝐵 × 𝐴 𝐶. If 𝑏 ∈ 𝐵 and 𝑐 ∈ 𝐶 satisfy 𝛼(𝑏) = 𝛾(𝑐), then 𝑑 = (𝑏, 𝑐) is the unique element of 𝐷 satisfying 𝛽(𝑑) = 𝑏 and 𝛿(𝑑) = 𝑐. Proof of (b). Consider 𝑐 ∈ 𝐶. By assumption, there is a 𝑏 ∈ 𝐵 with 𝛼(𝑏) = 𝛾(𝑐). By (a), there is a 𝑑 ∈ 𝐷 with 𝛿(𝑑) = 𝑐. Thus, 𝛿 is surjective. □ In view of Lemma 25.2.3, we may say that the fiber product 𝐵 × 𝐴 𝐶 defined by homomorphisms 𝛼: 𝐵 → 𝐴 and 𝛾: 𝐶 → 𝐴 is a fiber product with surjective homomorphisms if both 𝛼 and 𝛾 are surjective. Lemma 25.2.4 Let (25.1) be a commutative square of epimorphisms of profinite groups. Then, (25.1) is cartesian if and only if Ker(𝛼 ◦ 𝛽) = Ker(𝛽) × Ker(𝛿). Proof. First suppose that (25.1) is a cartesian square. Clearly Ker(𝛽) · Ker(𝛿) ≤ Ker(𝛼 ◦ 𝛽).

25.2 Cartesian Squares

523

Assume without loss that 𝐷 = 𝐵 × 𝐴 𝐶, 𝛽 = pr 𝐵 , and 𝛿 = pr𝐶 . Let (𝑏, 𝑐) ∈ Ker(𝛼 ◦ 𝛽). Then, 𝛼(𝑏) = 1 = 𝛾(𝑐). Hence, (1, 𝑐) and (𝑏, 1) belong to 𝐷. Moreover, (𝑏, 𝑐) = (1, 𝑐) · (𝑏, 1) ∈ Ker(𝛽) ·Ker(𝛿). Finally, Ker(𝛽) ∩Ker(𝛿) = 1. Consequently, Ker(𝛼 ◦ 𝛽) = Ker(𝛽) × Ker(𝛿). Now suppose that Ker(𝛼◦𝛽) = Ker(𝛽)×Ker(𝛿). Define a homomorphism 𝜃: 𝐷 → 𝐵 × 𝐴 𝐶 by 𝜃 (𝑑) = (𝛽(𝑑), 𝛿(𝑑)). If 𝜃 (𝑑) = (1, 1), then 𝑑 ∈ Ker(𝛽) ∩ Ker(𝛿) = 1. Thus, 𝜃 is injective. To prove that 𝜃 is surjective, consider (𝑏, 𝑐) ∈ 𝐵 × 𝐴 𝐶. Then, 𝛼(𝑏) = 𝛾(𝑐). There exist 𝑑1 , 𝑑2 ∈ 𝐷 with 𝛽(𝑑1 ) = 𝑏 and 𝛿(𝑑2 ) = 𝑐. Then, 𝛼(𝛽(𝑑1 𝑑2−1 )) = 𝛼(𝛽(𝑑1 )) · 𝛾(𝛿(𝑑2 )) −1 = 𝛼(𝑏)𝛾(𝑐) −1 = 1. By assumption, 𝑑1 𝑑2−1 = 𝑑3−1 𝑑4 for some 𝑑3 ∈ Ker(𝛽) and 𝑑4 ∈ Ker(𝛿). Let 𝑑 = 𝑑3 𝑑1 = 𝑑4 𝑑2 . Then, 𝛽(𝑑) = 𝑏 and 𝛿(𝑑) = 𝑐. □ Therefore, 𝜃 (𝑑) = (𝑏, 𝑐). Lemma 25.2.5 Let (25.1) be a cartesian square of profinite groups. Then, 𝛽 maps Ker(𝛿) isomorphically onto Ker(𝛼). In particular, if 𝛼 is injective, then so is 𝛿. Proof. We may assume by Proposition 25.2.1 that 𝐷 = 𝐵 × 𝐴 𝐶. Then, Ker(𝛿) = {(𝑏, 𝑐) ∈ 𝐵 × 𝐶 | 𝛼(𝑏) = 𝛾(𝑐), 𝑐 = 1} = {(𝑏, 1) ∈ 𝐵 × 𝐶 | 𝛼(𝑏) = 1} = Ker(𝛼) × 1. Hence, 𝛽(Ker(𝛿)) = Ker(𝛼). In addition, Ker(𝛿) ∩ Ker(𝛽) = 1. Hence, 𝛽: Ker(𝛿) → Ker(𝛼) is an isomorphism. □ Lemma 25.2.6 Let (25.1) and (25.2) be commutative diagrams of homomorphisms of profinite groups. Suppose that (25.1) is cartesian and all maps in (25.2) except possibly 𝜋 are surjective. Then: (a) If 𝜋 is surjective, then Ker(𝛼 ◦ 𝜑) = Ker(𝜑)Ker(𝜓). (b) If Ker(𝛼 ◦ 𝜑) ≤ Ker(𝜑)Ker(𝜓), then 𝜋 is surjective. Proof of (a). The condition 𝛼 ◦ 𝜑 = 𝛾 ◦ 𝜓 implies Ker(𝜑)Ker(𝜓) ≤ Ker(𝛼 ◦ 𝜑). Now suppose that 𝜋 is surjective. Let 𝑔 ∈ Ker(𝛼 ◦ 𝜑). Put 𝑏 = 𝜑(𝑔). Then, 𝛼(𝑏) = 1. Hence, by Lemma 25.2.5, there is a 𝑑 ∈ 𝐷 with 𝛽(𝑑) = 𝑏 and 𝛿(𝑑) = 1. Choose an ℎ ∈ 𝐺 with 𝜋(ℎ) = 𝑑. Then, 𝜓(ℎ) = 𝛿(𝜋(ℎ)) = 𝛿(𝑑) = 1 and 𝜑(𝑔ℎ−1 ) = 𝑏𝛽(𝜋(ℎ)) −1 = 𝑏𝛽(𝑑) −1 = 1. Therefore, 𝑔 = 𝑔ℎ−1 · ℎ ∈ Ker(𝜑)Ker(𝜓). Proof of (b). Suppose that Ker(𝛼 ◦ 𝜑) ≤ Ker(𝜑)Ker(𝜓). Consider 𝑑 ∈ 𝐷. Put 𝑏 = 𝛽(𝑑) and 𝑐 = 𝛿(𝑑). Choose 𝑔, ℎ ∈ 𝐺 with 𝜑(𝑔) = 𝑏 and 𝜓(ℎ) = 𝑐. Then, 𝛼(𝜑(𝑔ℎ−1 )) = 𝛼(𝜑(𝑔))𝛾(𝜓(ℎ)) −1 = 𝛼(𝑏)𝛾(𝑐) −1 = 𝛼(𝛽(𝑑))𝛾(𝛿(𝑑)) −1 = 1. Thus, there are 𝑔1 ∈ Ker(𝜑) and ℎ1 ∈ Ker(𝜓) with 𝑔ℎ−1 = 𝑔1 ℎ1 . The element 𝑔1−1 𝑔 = ℎ1 ℎ of 𝐺 satisfies 𝜑(𝑔1−1 𝑔) = 𝑏 and 𝜓(ℎ1 ℎ) = 𝑐. Therefore, by Lemma 25.2.3(a), 𝜋(𝑔1−1 𝑔) = 𝑑. □ Example 25.2.7 (Fiber products) Here are four examples where fiber products naturally appear. The verification that the occurring commutative squares are cartesian follows from Lemma 25.2.4: (a) Let 𝑀 and 𝑀 ′ be Galois extensions of a field 𝐾. Put 𝐿 = 𝑀 ∩ 𝑀 ′ and 𝑁 = 𝑀 𝑀 ′. Then, Gal(𝑁/𝐾) is the fiber product of Gal(𝑀/𝐾) and Gal(𝑀 ′/𝐾) over Gal(𝐿/𝐾) with respect to the restriction maps.

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25 Projective Groups and Frattini Covers

(b) Similarly, let 𝐺 be a profinite group and 𝐾, 𝐿, 𝑀, 𝑁 closed normal subgroups. Suppose that 𝐾 ∩ 𝐿 = 𝑁 and 𝐾 𝐿 = 𝑀. Then, 𝐺/𝑁 = 𝐺/𝐾 ×𝐺/𝑀 𝐺/𝐿 with respect to the quotient maps. (c) Let 𝜑: 𝐵 → 𝐴 be an epimorphism of profinite groups, 𝐶 = Ker(𝜑), and 𝐵0 a closed normal subgroup of 𝐵. Suppose that 𝐵0 ∩ 𝐶 = 1. Put 𝐵¯ = 𝐵/𝐵0 , 𝐴0 = 𝜑(𝐵0 ), ¯ 𝐵¯ → 𝐴¯ the map 𝐴¯ = 𝐴/𝐴0 , 𝛼: 𝐴 → 𝐴¯ and 𝛽: 𝐵 → 𝐵¯ the canonical maps, and 𝜑: induced from 𝜑. Then, 𝐵  𝐵¯ × 𝐴¯ 𝐴. Indeed, if 𝑏 ∈ 𝐵 satisfies 𝛼(𝜑(𝑏)) = 1, then 𝜑(𝑏) ∈ 𝐴0 . Hence, there exists a 𝑏 0 ∈ 𝐵0 with 𝜑(𝑏 0 ) = 𝜑(𝑏). Therefore, 𝑏 = 𝑏 0 · 𝑏 −1 0 𝑏 ∈ Ker(𝛽)Ker(𝜑). Since Ker(𝛽) ∩ Ker(𝜑) = 𝐵0 ∩ 𝐶 = 1, we have Ker(𝛼 ◦ 𝜑) = Ker(𝛽) × Ker(𝜑). It follows from Lemma 25.2.4 that 𝐵  𝐵¯ × 𝐴¯ 𝐴. Suppose now that 𝐵0 ∩ 𝐶 is not necessarily trivial. Then, by the preceding paragraph, 𝐵/𝐵0 ∩ 𝐶  𝐵¯ × 𝐴¯ 𝐴. Hence, for each 𝑎 ∈ 𝐴 and 𝑏 ∈ 𝐵 satisfying = 𝛼(𝑎) there is a 𝑏 ′ ∈ 𝐵 (not necessarily unique) with 𝛽(𝑏 ′) = 𝛽(𝑏) and 𝜑(𝛽(𝑏)) ¯ ′ 𝜑(𝑏 ) = 𝑎. (d) Condition (20.3b) on a family C of finite groups is equivalent to “C is closed under fiber products with surjective homomorphisms”. Thus, a formation is a family of finite groups which is closed under taking quotients and fiber products with surjective homomorphisms.

The homomorphism 𝜋 of diagram (25.2) need not be surjective even if all other maps are surjective. Here is a condition for this to happen: Lemma 25.2.8 Let 𝜑: 𝐺 → 𝐵 and 𝜓: 𝐺 → 𝐶 be epimorphisms of profinite groups. Then, there is a commutative diagram (25.2), unique up to an isomorphism, such that (25.1) is a cartesian square with 𝛽, 𝛼, 𝛾, 𝛿, 𝜋 epimorphisms. Proof. Let 𝑀 = Ker(𝜑)Ker(𝜓), 𝑁 = Ker(𝜑) ∩ Ker(𝜓), 𝐴 = 𝐺/𝑀, 𝐷 = 𝐺/𝑁, and 𝜋: 𝐺 → 𝐺/𝑁 the quotient map. Now find epimorphisms 𝛽, 𝛼, 𝛾, 𝛿 that make (25.1) cartesian and (25.2) commutative. □

The following lemma gives a useful criterion for 𝛿 in the cartesian diagram (25.1) to have a group-theoretic section: Lemma 25.2.9 Let (25.1) be a cartesian diagram of homomorphisms of profinite groups. Suppose that there exists a homomorphism 𝜑: 𝐶 → 𝐵 with 𝛼 ◦ 𝜑 = 𝛾. Then, there exists a monomorphism 𝛿 ′: 𝐶 → 𝐷 such that 𝛿 ◦ 𝛿 ′ = id𝐶 . Proof. Let 𝜓 = id𝐶 . Then, 𝛼 ◦ 𝜑 = 𝛾 ◦ 𝜓. Hence, by Proposition 25.2.1(b), there exists a homomorphism 𝛿 ′: 𝐶 → 𝐷 such that 𝛿 ◦ 𝛿 ′ = id𝐶 (and 𝛽 ◦ 𝛿 ′ = 𝜑), as claimed. Note that the existence of 𝜑 implies that 𝛿 is surjective. □

25.3 On C-Projective Groups

525

25.3 On C-Projective Groups Embedding problems and projective groups have already appeared, for example, in Sections 18.5 and 20.7. Now we consider the subject in detail: Definition 25.3.1 (Embedding problems) An embedding problem for a profinite group 𝐺 is a pair (𝜑: 𝐺 → 𝐴, 𝛼: 𝐵 → 𝐴) (25.3) in which 𝜑 and 𝛼 are epimorphisms of profinite groups. We call Ker(𝛼) the kernel of the problem. We call the problem finite if 𝐵 is finite. We say that (25.3) splits if 𝛼 has a group-theoretic section. That is, there is a homomorphism 𝛼 ′: 𝐴 → 𝐵 with 𝛼 ◦ 𝛼 ′ = id 𝐴. Embedding problem (25.3) is said to be solvable (resp. weakly solvable) if there exists an epimorphism (resp. homomorphism) 𝛾: 𝐺 → 𝐵 with 𝛼 ◦ 𝛾 = 𝜑. The map 𝛾 is a solution (resp. weak solution) to (25.3). Suppose that C is a Melnikov formation of finite groups (Section 20.3). Let 𝐺 be a pro-C group. Then, call (25.3) a C-embedding problem (resp. pro-C embedding problem), if 𝐵 is a C-group (resp. pro-C). Call 𝐺 C-projective if every pro-C embedding problem (25.3) for 𝐺 is weakly solvable. Lemma 25.3.2 (Gruenberg) Let C be a Melnikov formation of finite groups and 𝐺 be a pro-C group. Suppose that every C-embedding problem for 𝐺 is weakly solvable. Then, 𝐺 is C-projective. Proof. Consider embedding problem (25.3) for 𝐺 with 𝐵 a pro-C group. Let 𝐶 = Ker(𝛼). The transition from the finite case to the general case in Part C of the proof forces us to consider pairs (25.3) in which 𝜑 is a homomorphism which is not necessarily surjective. Part A: Suppose that 𝜑: 𝐺 → 𝐴 is a homomorphism, 𝐵 ∈ C, and 𝛼: 𝐵 → 𝐴 is an epimorphism. Then, there is a homomorphism 𝛽: 𝐺 → 𝐵 with 𝛼 ◦ 𝛽 = 𝜑. Indeed, 𝐵 and 𝐴0 = 𝜑(𝐺) are in C. Hence, so is 𝐶 and therefore also 𝐵0 = 𝛼−1 ( 𝐴0 ). By assumption, there is a homomorphism 𝛽: 𝐺 → 𝐵0 with 𝛼 ◦ 𝛽 = 𝜑. Part B: 𝐶 is finite. Choose an open normal subgroup 𝑈 of 𝐵 with 𝐶 ∩𝑈 = 1. Since 𝐵/𝑈 ∈ C, Part A gives a homomorphism 𝛽: 𝐺 → 𝐵/𝑈 for which

1

/𝐶

1

/𝐶

𝐺 ✞ ✞ ✞ 𝜑 ✞✞  ✞ 𝛼 ✞✞ /𝐵 /𝐴 ✞✞ ✞ 𝛽 ✞ ✞✞  ✞✞ 𝛼¯  / 𝐵/𝑈 / 𝐴/𝛼(𝑈)

/1 /1

is a commutative diagram. The right square is cartesian (Example 25.2.7(c)). Therefore (Proposition 25.2.1(b)), there is a homomorphism 𝛾: 𝐺 → 𝐵 with 𝛼 ◦ 𝛾 = 𝜑.

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25 Projective Groups and Frattini Covers

Part C: The general case. Apply Zorn’s lemma to Part B. Let Λ be the set of pairs (𝐿, 𝜆), where 𝐿 is a closed normal subgroup of 𝐵 contained in 𝐶 and 𝜆: 𝐺 → 𝐵/𝐿 is a homomorphism which makes the following diagram commutative: 𝐺 ④④ ④ ④ 𝜑 ④④ ④ }④  / 𝐵/𝐿 /𝐴

(25.4)

𝜆

1

/ 𝐶/𝐿

𝛼𝐿

/ 1.

−1 ◦ 𝜑) belongs to Λ, so Here 𝛼 𝐿 is the epimorphism induced by 𝛼. The pair (𝐶, 𝛼𝐶 ′ ′ Λ is nonempty. Partially order Λ by (𝐿 , 𝜆 ) ≤ (𝐿, 𝜆) if 𝐿 ′ ≤ 𝐿 and the following triangle is commutative: 𝐺 ✇✇ 𝜆′ ✇✇✇ 𝜆 ✇✇ {✇✇  / 𝐵/𝐿. 𝐵/𝐿 ′

Suppose that Λ0 = {(𝐿 𝑖 , 𝜆𝑖 ) | 𝑖 ∈ 𝐼} is a descending chain in Λ. Then, lim 𝐵/𝐿 𝑖 = ←− Ñ 𝐵/𝐿 with 𝐿 = 𝑖 ∈𝐼 𝐿 𝑖 . The 𝜆𝑖 ’s define a homomorphism 𝜆: 𝐺 → 𝐵/𝐿 with (25.4) commutative. Therefore, (𝐿, 𝜆) is a lower bound for Λ0 . By Zorn’s lemma, Λ has a minimal element (𝐿, 𝜆). It suffices to prove that 𝐿 = 1. Assume 𝐿 ≠ 1. Then, 𝐵 has an open normal subgroup 𝑁 with 𝐿 ̸ ≤ 𝑁. Therefore, 𝐿 ′ := 𝑁 ∩ 𝐿 is a proper open subgroup of 𝐿 which is normal in 𝐵. Part B gives a homomorphism 𝜆 ′: 𝐺 → 𝐵/𝐿 ′ which makes the following diagram commutative: ①①

𝜆′ ①①①

1

/ 𝐿/𝐿 ′

①① {①①

/ 𝐵/𝐿 ′

(25.5)

𝐺 𝜆

 / 𝐵/𝐿

/ 1.

Therefore, (𝐿 ′, 𝜆 ′) is an element of Λ less than (𝐿, 𝜆), a contradiction.

□

Remark 25.3.3 (A variant of Lemma 25.3.2) Consider an arbitrary embedding problem (25.3) for a profinite group 𝐺. Put 𝐶 = Ker(𝛼). Suppose that for all 𝐿, 𝐿 ′ ≤ 𝐶 such that 𝐿, 𝐿 ′ ⊳ 𝐵 and 𝐿 ′ open in 𝐿 and for each homomorphism 𝜆: 𝐺 → 𝐵/𝐿 there is a homomorphism 𝜆 ′: 𝐺 → 𝐵/𝐿 ′ making (25.5) commutative. The proof of Lemma 25.3.2 proves that (25.3) is weakly solvable. Lemma 25.3.4 Let C be a Melnikov formation of finite groups and 𝐺 a pro-C group. Suppose that every C-embedding problem (25.3) in which Ker(𝛼) is a minimal normal subgroup of 𝐵 is weakly solvable. Then, 𝐺 is C-projective. Proof. By Lemma 25.3.2, it suffices to weakly solve each C-embedding problem (25.3). Put 𝐶 = Ker(𝛼). If 𝐶 is minimal normal in 𝐵, then (25.3) is solvable by assumption. Suppose that 𝐶 is not minimal normal in 𝐵 and proceed by induction on |𝐶 |. By assumption, there is a nontrivial normal subgroup 𝐿 of 𝐵 which is properly contained in 𝐶. Consider the diagram

25.3 On C-Projective Groups

527

𝐿

𝐺 𝜑

1

/𝐶

1

 / 𝐶/𝐿

 /𝐵

𝛼

 /𝐴

/1

𝛼¯

/𝐴

/1

𝜋

 / 𝐵/𝐿

in which 𝛼¯ is induced by 𝛼. Since |𝐶/𝐿| < |𝐶 |, the induction hypothesis gives a homomorphism 𝛽: 𝐺 → 𝐵/𝐿 with 𝛼¯ ◦ 𝛽 = 𝜑. Also, |𝐿| < |𝐶 |. Using the induction hypothesis again, there is a homomorphism 𝛾: 𝐺 → 𝐵 with 𝜋 ◦ 𝛾 = 𝛽. Therefore, 𝛼 ◦ 𝛾 = 𝜑. □ Proposition 25.3.5 Let C be a full formation of finite groups and 𝐺 a pro-C group. Suppose that each C-embedding problem (25.3) for 𝐺, where Ker(𝛼) is a minimal Abelian 𝑝-elementary normal subgroup of 𝐵, is weakly solvable. Then, 𝐺 is Cprojective. Proof. Let 𝐶 = Ker(𝛼). By Lemma 25.3.4, it suffices to prove that every embedding problem (25.3) for 𝐺, where 𝐵 ∈ C and 𝐶 is a minimal normal subgroup of 𝐵, is weakly solvable. We distinguish between two cases: Case A: Suppose that 𝐶 ̸ ≤ Φ(𝐵) Then, 𝐵 has a maximal subgroup 𝐵1 with 𝐶 ̸ ≤ 𝐵1 . Let 𝛼1 be the restriction of 𝛼 to 𝐵1 . Then, 𝐵1 𝐶 = 𝐵 and 𝛼1 (𝐵1 ) = 𝛼(𝐶 𝐵1 ) = 𝛼(𝐵) = 𝐴. Consider the commutative diagram 𝐺 𝜑

1

/𝐶 O

/𝐵 O

𝛼

 /𝐴

1

/ 𝐶1

/ 𝐵1

𝛼1

/𝐴

/1 /1

with 𝐶1 = 𝐶 ∩ 𝐵1 . Since C is full, 𝐵1 ∈ C. An induction hypothesis on |𝐵| gives a homomorphism 𝛾: 𝐺 → 𝐵1 with 𝛼1 ◦ 𝛾 = 𝜑, hence 𝛼 ◦ 𝛾 = 𝜑. Case B: 𝐶 ≤ Φ(𝐵) By Lemma 25.1.3, 𝐶 is a minimal Abelian 𝑝-elementary normal subgroup of 𝐵. Hence, by assumption, (25.3) is solvable. □ The following lemma gives the basic example for projective C-groups: Lemma 25.3.6 Let C be a Melnikov formation of finite groups. Then, each free pro-C group 𝐹 is C-projective. Proof. Choose a basis 𝑋 for 𝐹. Consider a C-embedding problem (𝜑: 𝐹 → 𝐴, 𝛼: 𝐵 → 𝐴). Choose 𝑥1 , . . . , 𝑥 𝑛 ∈ 𝑋 such that 𝜑(𝑥1 ), . . . , 𝜑(𝑥 𝑛 ) generate 𝐴. Then, choose 𝑏 1 , . . . , 𝑏 𝑛 ∈ 𝐵 with 𝛼(𝑏 𝑖 ) = 𝜑(𝑥𝑖 ), 𝑖 = 1, . . . , 𝑛. Map 𝑋 into 𝐵 by 𝑥𝑖 ↦→ 𝑏 𝑖 , 𝑖 = 1, . . . , 𝑛 and 𝑥 ↦→ 1 for 𝑥 ∈ 𝑋 ∖{𝑥 1 , . . . , 𝑥 𝑛 }. This map extends to a homomorphism 𝛾: 𝐹 → 𝐵 with 𝛼 ◦ 𝛾 = 𝜑. Consequently, 𝐹 is C-projective. □

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25 Projective Groups and Frattini Covers

25.4 Projective Groups A profinite group 𝐺 is projective if every embedding problem for 𝐺 is weakly solvable (i.e. 𝐺 is projective with respect to the formation of all finite groups). Lemma 25.4.1 If C is a full formation of finite groups and 𝐺 is C-projective, then 𝐺 is projective. Proof. By Proposition 25.3.5, it suffices to prove that every finite embedding problem (𝜑: 𝐺 → 𝐴, 𝛼: 𝐵 → 𝐴) (25.6) with Ker(𝛼) = (Z/𝑝Z) 𝑚 and 𝑝 a prime number is weakly solvable. If 𝑝 ∤ | 𝐴|, then Schur–Zassenhaus [Hup67, p. 126, Hauptsatz 18.1] gives a homomorphism 𝛼 ′: 𝐴 → 𝐵 with 𝛼 ◦ 𝛼 ′ = id 𝐴. Thus, 𝛼 ′ ◦ 𝜑 is a weak solution of (25.6). Now suppose that 𝑝 divides | 𝐴|. Then, 𝐴 contains an isomorphic copy of Z/𝑝Z (Cauchy). Since 𝐴 is in C, so is Z/𝑝Z. Hence, (Z/𝑝Z) 𝑚 ∈ C. Therefore, 𝐵 ∈ C. By □ assumption, (25.6) has a weak solution. Remark 25.4.2 Suppose that 𝐺 is projective and 𝜋: 𝐹 → 𝐺 is an epimorphism of profinite groups. Then, there exists an embedding 𝜋 ′: 𝐺 → 𝐹 such that 𝜋 ◦ 𝜋 ′ is the identity map (Take 𝐴 = 𝐺, 𝐵 = 𝐹, 𝛼 = 𝜋, and 𝜑 = id𝐺 in (25.6).) Therefore, 𝜋 ′ is injective, so 𝐺 is isomorphic to a closed subgroup of 𝐹. Corollary 25.4.3 below uses this observation to give a criterion for projectivity. Corollary 25.4.3 A profinite group 𝐺 is projective if and only if every short exact 𝛼 sequence 1 → 𝐶 → 𝐵 −→ 𝐺 → 1 in which 𝐶 is a finite Abelian 𝑝-elementary group splits. Proof. By Remark 25.4.2, we have only to prove sufficiency. By Proposition 25.3.5, it suffices to give a weak solution to each finite embedding problem (𝜑: 𝐺 → 𝐴, 𝛼: 𝐵 → 𝐴) with Ker(𝛼) = (Z/𝑝Z) 𝑚 for some prime number 𝑝 and a positive integer 𝑚. Complete the embedding problem to a commutative diagram 1

/ (Z/𝑝Z) 𝑚

/ 𝐵 ×𝐴 𝐺

/ (Z/𝑝Z) 𝑚

 /𝐵

/𝐺

/1 𝜑

1

𝛼

 /𝐴

/1 .

By assumption, the upper row splits. Hence, there is a homomorphism 𝛾: 𝐺 → 𝐵 with 𝛼 ◦ 𝛾 = 𝜑. □ Remark 25.4.4 (Cohomological interpretation of projectivity) In cohomological terms, Corollary 25.4.3 signifies that a profinite group 𝐺 is projective if and only if its cohomological dimension is bounded by 1 [Rbs70, p. 211].

25.4 Projective Groups

529

Corollary 25.4.5 Let C be a full formation of finite groups and 𝐹 a free pro-C-group. Then, 𝐹 is projective. Proof. By Lemma 25.3.6, 𝐹 is C-projective. Hence, by Lemma 25.4.1, 𝐹 is pro□ jective. Corollary 25.4.6 A profinite group 𝐺 is projective if and only if it is isomorphic to a closed subgroup of a free profinite group. Proof. By Proposition 20.4.8, 𝐺 is a quotient of a free profinite group 𝐹. Hence, by Remark 25.4.2, 𝐺 is isomorphic to a closed subgroup of 𝐹. Conversely, let 𝐻 be a closed subgroup of a free profinite group 𝐹. Let 𝜑: 𝐻 → 𝐴 and 𝛼: 𝐵 → 𝐴 be epimorphisms, with 𝐵 finite. Lemma 1.2.5(c) gives an open subgroup 𝐻 ′ of 𝐹 containing 𝐻 and an epimorphism 𝜑 ′: 𝐻 ′ → 𝐴 extending 𝜑. By Proposition 20.6.2, 𝐻 ′ is free. Hence, by Corollary 25.4.5, 𝐻 ′ is projective. Therefore, there exists a homomorphism 𝛾 ′: 𝐻 ′ → 𝐵 with 𝛼 ◦ 𝛾 ′ = 𝜑 ′. Denote the restriction of 𝛾 ′ to 𝐻 by 𝛾. Then, 𝛼 ◦ 𝛾 = 𝜑. It follows that 𝐻 is projective. □ Proposition 25.4.7 Let 𝐺 be a projective group and 𝐻 a closed subgroup. Then, 𝐻 is projective. Moreover, 𝐻 is either trivial or infinite. In particular, 𝐺 is torsion free. Proof. By Corollary 25.4.6, 𝐺 is isomorphic to a closed subgroup of a free profinite group 𝐹. Hence, 𝐻 is also isomorphic to a closed subgroup of 𝐹. A second application of Corollary 25.4.6 now proves that 𝐻 is projective. Assume 𝐺 has a nontrivial finite subgroup 𝐻. By Cauchy’s theorem, 𝐻 contains an isomorphic copy of Z/𝑝Z for some 𝑝. By the preceding paragraph, Z/𝑝Z is projective. It follows from Corollary 25.4.3 that the natural map Z/𝑝 2 Z → Z/𝑝Z has a group-theoretic section. Hence, Z/𝑝 2 Z  Z/𝑝Z×Z/𝑝Z. This is a contradiction. □ A combination of Corollary 25.4.5 and Proposition 25.4.7 yields the following result: Corollary 25.4.8 Let C be a full formation of finite groups and 𝐹 a free pro-C group. Then, every closed subgroup 𝐻 of 𝐹 is projective. In particular, 𝐻 is either trivial or infinite. Proposition 25.4.9 Let 𝐺 be a projective group and C a full formation of finite groups. Then, the maximal pro-C quotient of 𝐺 is projective. In particular, for each prime number 𝑝 the maximal pro-𝑝 quotient of 𝐺 is projective. Proof. Let 𝜋: 𝐺 → 𝐺¯ be the quotient map of 𝐺 onto its maximal pro-C quotient (Definition 20.3.2). Consider a C-embedding problem (𝜑: 𝐺¯ → 𝐴, 𝛼: 𝐵 → 𝐴) for ¯ Since 𝐺 is projective, there is a homomorphism 𝛾: 𝐺 → 𝐵 such that 𝛼 ◦ 𝛾 = 𝜑 ◦ 𝜋. 𝐺. Since 𝐵 ∈ C and C is full, 𝛾(𝐺) ∈ C. Hence, 𝛾 factors through a homomorphism 𝛾: ¯ 𝐺¯ → 𝐵; that is 𝛾¯ ◦ 𝜋 = 𝛾. It follows that 𝛼 ◦ 𝛾¯ = 𝜑. By Lemma 25.4.1, 𝐺¯ is projective. □

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25 Projective Groups and Frattini Covers

25.5 Free Products of Finitely many Profinite Groups We introduce “free products of finitely many profinite groups” and prove that if each of those groups is projective, then so is the free product. Proposition 25.5.1 Let 𝐼 be a finite set and for each 𝑖 ∈ 𝐼 let 𝐺 𝑖 be a Î profinite group. Then, there exists a unique (up to isomorphism) profinite group 𝐺ˆ = ∗ 𝑖 ∈𝐼 𝐺 𝑖 , called the free product (in the category of profinite groups) of the groups 𝐺 𝑖 , 𝑖 ∈ 𝐼, with the following properties: ˆ (a) Every 𝐺 𝑖 is a closed subgroup of 𝐺. (b) Let 𝐴 be a profinite group and for each 𝑖 ∈ 𝐼 let 𝜑𝑖 : 𝐺 𝑖 → 𝐴 be a homomorphism. Then, there exists a unique homomorphism 𝜑: ˆ 𝐺ˆ → 𝐴 such that 𝜑| ˆ 𝐺𝑖 = 𝜑𝑖 for each 𝑖 ∈ 𝐼. ˆ (c) If each of the 𝐺 𝑖 ’s is projective, then so is 𝐺. Proof. Let 𝐺 be the free product of the 𝐺 𝑖 ’s in the category of abstract groups. It is the unique (up to isomorphism) abstract group which is generated by all of the 𝐺 𝑖 ’s and for every abstract group 𝐴 and every system of abstract homomorphisms 𝛼𝑖 : 𝐺 𝑖 → 𝐴, 𝑖 ∈ 𝐼, there exists a unique abstract homomorphism 𝛼: 𝐺 → 𝐴 such that 𝛼| 𝐺𝑖 = 𝛼𝑖 for all 𝑖 ∈ 𝐼. Denote the family of normal subgroups 𝑁 of 𝐺 of finite index such that 𝑁 ∩ 𝐺 𝑖 is open in 𝐺 𝑖 for each 𝑖 ∈ 𝐼 by N . Then, 𝑁 ∩ 𝑁 ′ ∈ N for all 𝑁, 𝑁 ′ ∈ N . Hence, N is, in the terminology of Section 20.2, a directed family Let N -topology be the associated topology introduced in Section 20.2. Let 𝐺ˆ = lim 𝐺/𝑁, with 𝑁 ranging over N , be the corresponding inverse limit. ←− We define an abstract homomorphism 𝜃: 𝐺 → 𝐺ˆ by 𝜃 (𝑔) = (𝑔𝑁) 𝑁 ∈N for each 𝑔 ∈ 𝐺. Thus, if 𝜋 𝑁 : 𝐺 → 𝐺/𝑁 and 𝜋ˆ 𝑁 : 𝐺ˆ → 𝐺/𝑁 are the corresponding projections and 𝑁ˆ = Ker( 𝜋ˆ 𝑁 ), then (25.7) 𝜋ˆ 𝑁 ◦ 𝜃 = 𝜋 𝑁 and 𝜃 induces the canonical isomorphism ˆ 𝑁. ˆ 𝜃 𝑁 : 𝐺/𝑁 → 𝐺/

□

Given an 𝑖 ∈ 𝐼 and 𝑁𝑖 an open normal in 𝐺 𝑖 , let 𝜈: 𝐺 → 𝐺/𝑁𝑖 be the homomorphism whose restriction to 𝐺 𝑖 is the quotient map onto 𝐺 𝑖 /𝑁𝑖 and 𝜈| 𝐺 𝑗 is the trivial homomorphism for each 𝑗 ∈ 𝐼 ∖{𝑖}. Then, Ker(𝜈) ∩ Ñ𝐺 𝑖 = 𝑁𝑖 and Ker(𝜈) ∩ 𝐺 𝑗 = 𝐺 𝑗 for each 𝑗 ≠ 𝑖. Hence, Ker(𝜈) ∈ N . It follows that 𝑁 ∈N 𝑁 ∩ 𝐺 𝑖 = 1, so 𝜃 maps 𝐺 𝑖 ˆ ˆ We identify each 𝐺 𝑖 with its image 𝜃 (𝐺 𝑖 ) in 𝐺. injectively into 𝐺. ˆ Hence, By Lemma 20.2.1(a4), 𝜃 (𝐺) is dense in 𝐺. ˆ (25.8) the closed subgroup ⟨𝐺 𝑖 ⟩𝑖 ∈𝐼 of 𝐺ˆ generated by the 𝐺 𝑖 ’s is 𝐺. We endow 𝐺 with the N -topology (Section 20.2). By Lemma 20.2.1(a5), the map 𝜃 is continuous. Proof of (a). Given an 𝑖 ∈ 𝐼, the paragraph below (25.7) shows that the restriction of the N -topology of 𝐺 to 𝐺 𝑖 coincides with the profinite topology of 𝐺 𝑖 . Since 𝜃 is continuous, so is 𝜃| 𝐺𝑖 . But 𝐺 𝑖 and 𝐺ˆ are profinite groups, hence 𝜃 (𝐺 𝑖 ) is closed in 𝐺ˆ (use Remark 1.2.1(i)).

25.6 Frattini Covers

531

Proof of (b). Let 𝐴 be a finite group and for each 𝑖 ∈ 𝐼 let 𝜑𝑖 : 𝐺 𝑖 → 𝐴 be a homomorphism. Then, let 𝜑: 𝐺 → 𝐴 be the abstract homomorphism whose restriction to 𝐺 𝑖 is 𝜑𝑖 for each 𝑖 ∈ 𝐼. Then, 𝑁 = Ker(𝜑) ∈ N , so there is a ¯ 𝐺/𝑁 → 𝐴 such that 𝜑 = 𝜑¯ ◦ 𝜋 𝑁 . It follows that 𝜑ˆ = 𝜑¯ ◦ 𝜋ˆ 𝑁 : 𝐺ˆ → homomorphism 𝜑: 𝐴 satisfies 𝜑ˆ ◦ 𝜃 = 𝜑. Hence, 𝜑| ˆ 𝐺𝑖 = 𝜑𝑖 . By (25.8), 𝜑ˆ is unique. Using the uniqueness of 𝜑, ˆ an inverse limit argument proves the consequence of the preceding paragraph for each profinite group 𝐴. ˆ Then, Proof of (c). Let (𝜑: 𝐺ˆ → 𝐴, 𝛼: 𝐵 → 𝐴) be an embedding problem for 𝐺. for each 𝑖 ∈ 𝐼 there exists a homomorphism 𝛾𝑖 : 𝐺 𝑖 → 𝐵 with 𝛼 ◦ 𝛾𝑖 = 𝜑| 𝐺𝑖 . The homomorphism 𝛾: 𝐺ˆ → 𝐵 with 𝛾| 𝐺𝑖 = 𝛾𝑖 for each 𝑖 ∈ 𝐼, which exists by (b), is a weak solution of the given embedding problem. Hence, 𝐺 is projective. □

25.6 Frattini Covers Frattini covers allow us to form profinite groups that inherit properties from one of their finite quotients: Definition 25.6.1 (Frattini covers) A homomorphism 𝜑: 𝐻 → 𝐺 of profinite groups is called a Frattini cover if it satisfies one, hence all, of the following equivalent conditions (Lemma 25.1.1 and Lemma 25.1.4(a)): (25.9a) 𝜑 is surjective and Ker(𝜑) ≤ Φ(𝐻). (25.9b) A closed subgroup 𝐻0 of 𝐻 is equal to 𝐻 if and only if 𝜑(𝐻0 ) = 𝐺. (25.9c) A subset 𝑆 of 𝐻 generates 𝐻 if and only if 𝜑(𝑆) generates 𝐺. In particular, by Lemma 25.1.4(a), 𝜑(Φ(𝐻)) = Φ(𝐺). Lemma 25.6.2 Let 𝐺 be a profinite group and 𝑋 a subset of 𝐺 which contains 1 and converges to 1. Then, the following holds: (a) 𝑋 is closed. (b) Every closed subset 𝑋0 of 𝑋 which does not contain 1 is finite. (c) Let 𝜑: 𝐻 → 𝐺 be an epimorphism of profinite groups. Then, 𝐻 has a closed subset 𝑋 ′ which contains 1 and converges to 1 such that 𝜑 maps 𝑋 ′ homeomorphically onto 𝑋. Proof of (a). Let 𝑔 ∈ 𝐺 ∖ 𝑋. Then, 𝑔 ≠ 1. Hence, 𝐺 has an open normal subgroup 𝑁0 with 𝑔 ∉ 𝑁0 , so 𝑥𝑁0 ≠ 𝑔𝑁0 for each 𝑥 ∈ 𝑋 ∩ 𝑁0 . In addition, since 𝑋 converges to 1, 𝑋 ∖ 𝑁0 is finite and 𝑔 ∉ 𝑋 ∖ 𝑁0 . Therefore, there exists an open normal subgroup 𝑁1 of 𝐺 with 𝑁1 ≤ 𝑁0 and 𝑥𝑁1 ≠ 𝑔𝑁1 for each 𝑥 ∈ 𝑋 ∖ 𝑁1 . By the above, 𝑥𝑁1 ≠ 𝑔𝑁1 for all 𝑥 ∈ 𝑋 ∩ 𝑁1 . Thus, the set 𝑔𝑁1 is an open neighborhood of 𝑔 which is disjoint from 𝑋. Consequently, 𝑋 is closed. Proof of (b). By (a), 𝐺 ∖ 𝑋0 is an open neighborhood of 1. Hence, 𝐺 ∖ 𝑋0 contains an open normal subgroup 𝑁, so 𝑋0 ⊆ 𝑋 ∖ 𝑁. By assumption, 𝑋 ∖ 𝑁 is finite. Therefore, 𝑋0 is finite.

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25 Projective Groups and Frattini Covers

Proof of (c). By Lemma 1.2.7, there exists a continuous map 𝜑 ′: 𝐺 → 𝐻 such that 𝜑 ◦ 𝜑 ′ = id𝐺 . Put 𝐺 ′ = 𝜑 ′ (𝐺) and 𝑋 ′ = 𝜑 ′ (𝑋). Then, 𝐺 ′ and 𝑋 ′ are closed subsets of 𝐻 (Remark 1.2.1(h)) and 𝜑 maps 𝐺 ′ (resp. 𝑋 ′) homeomorphically onto 𝐺 (resp. 𝑋). Now consider an open normal subgroup 𝑁 of 𝐻. Then, 𝑋 ′ ∖ 𝑁 is a closed subset of 𝑋 ′ which does not contains 1. Hence, 𝜑(𝑋 ′ ∖ 𝑁) is a closed subset of 𝑋 which does not contain 1. By (b), 𝜑(𝑋 ′ ∖ 𝑁) is finite. Hence, 𝑋 ′ ∖ 𝑁 is finite. Consequently, 𝑋 ′ converges to 1. Alternatively, we could use Exercise 1 of Chapter 20. □ Corollary 25.6.3 (a) If 𝜑: 𝐻 → 𝐺 is a Frattini cover, then rank(𝐻) = rank(𝐺). (b) In particular, if a closed normal subgroup 𝑁 of 𝐻 is contained in Φ(𝐻), then rank(𝐻) = rank(𝐻/𝑁). Proof of (a). Choose a system of generators 𝑋 for 𝐺 in the following way: If rank(𝐺) < ∞, then |𝑋 | = rank(𝐺). If rank(𝐺) = ∞, then 𝑋 converges to 1 (Proposition 20.1.1) and 1 ∈ 𝑋. In the former case choose a subset 𝑋 ′ of 𝐻 which 𝜑 maps bijectively onto 𝑋. In the latter case, rank(𝐺) = |𝑋 | (by definition). Also, Lemma 25.6.2(c) gives a subset 𝑋 ′ of 𝐻 which converges to 1 and which 𝜑 maps homeomorphically onto 𝑋. By Definition (25.9c), 𝑋 ′ generates 𝐻. Therefore, in both cases, rank(𝐻) = |𝑋 ′ | = |𝑋 | = rank(𝐺). Proof of (b). By Definition 25.6.1, 𝐻 → 𝐻/𝑁 is a Frattini cover. Hence, (b) is a special case of (a). □ The following rules follow directly from Definition 25.6.1: 𝜓

𝜑

Lemma 25.6.4 Let 𝐻 −→ 𝐺 −→ 𝐴 be homomorphisms of profinite groups. (a) If 𝜑 and 𝜓 are Frattini covers, then 𝜑 ◦ 𝜓 is a Frattini cover. (b) If 𝜑 is a Frattini cover and 𝜑 ◦ 𝜓 is surjective, then 𝜓 is surjective. (c) If 𝜑 ◦ 𝜓 is a Frattini cover and 𝜓 is surjective, then both 𝜑 and 𝜓 are Frattini covers. (d) If both 𝜑 and 𝜑 ◦ 𝜓 are Frattini covers, then so is 𝜓. Lemma 25.6.5 Consider a cartesian square of epimorphisms (Definition 25.2.2): 𝛿

𝐷

/𝐶 𝛾

𝛽

 𝐵 𝛼

 /𝐴

If 𝛿 is a Frattini cover, then so is 𝛼. Proof. By Lemma 25.2.5, 𝛽(Ker(𝛿)) = Ker(𝛼). Since 𝛿 is a Frattini cover, Ker(𝛿) ≤ Φ(𝐷). By Lemma 25.1.4(a), 𝛽(Φ(𝐷)) ≤ Φ(𝐵). Hence, Ker(𝛼) ≤ Φ(𝐵) and 𝛼 is a Frattini cover. □

25.6 Frattini Covers

533

For each profinite group epimorphism, restriction of the domain gives a Frattini cover: Lemma 25.6.6 Let 𝜑: 𝐻 → 𝐺 be an epimorphism of profinite groups. Then, 𝐻 has a closed subgroup 𝐻 ′ such that 𝜑| 𝐻 ′ : 𝐻 ′ → 𝐺 is a Frattini cover. Proof. Let {𝐻𝑖 | 𝑖 ∈ 𝐼} be a decreasing chain of Ñ closed subgroups of 𝐻 with 𝜑(𝐻𝑖 ) = 𝐺 for each 𝑖 ∈ 𝐼. By Lemma 1.2.2(c), 𝜑( 𝑖 ∈𝐼 𝐻𝑖 ) = 𝐺. Applying Zorn’s lemma, we conclude that 𝐻 has a minimal closed subgroup 𝐻0 with 𝜑(𝐻0 ) = 𝐺. By Definition (25.9b), the restriction of 𝜑 to 𝐻0 is a Frattini cover of 𝐺. □ We combine the fiber product construction with Lemma 25.6.6: Lemma 25.6.7 Let 𝛼: 𝐵 → 𝐴 and 𝛾: 𝐶 → 𝐴 be Frattini covers. Then, there is a commutative diagram 𝛿 / 𝐷❅ 𝐶 ❅❅ ❅❅𝜑 𝛾 𝛽 ❅❅ ❅   𝐵 𝛼 /𝐴 in which 𝛽, 𝛿, and 𝜑 are Frattini covers. Proof. Let 𝐷 1 = 𝐵× 𝐴𝐶 (Section 25.2), 𝛽1 = pr 𝐵 , 𝛿1 = pr𝐶 , and 𝜑1 = 𝛾◦𝛿1 = 𝛼◦𝛽1 . This gives a cartesian square, but the maps in it may not be Frattini covers. Apply Lemma 25.6.6 to find a closed subgroup 𝐷 of 𝐷 1 such that 𝜑 = 𝜑1 | 𝐷 is a Frattini cover. Lemma 25.6.4(d) implies that 𝛽 = 𝛽1 | 𝐷 and 𝛿 = 𝛿1 | 𝐷 are Frattini covers. □ Let (𝜑: 𝐺 → 𝐴, 𝛼: 𝐵 → 𝐴)

(25.10)

be an embedding problem for a profinite group 𝐺. Call (25.10) a Frattini embedding problem if 𝛼 is a Frattini cover. Let C be a Melnikov formation of finite groups and 𝐺 a pro-C group. Then, call (25.10) a C-Frattini embedding problem (resp. pro-C-Frattini embedding problem) if in addition 𝐵 is a C-group (resp. pro-C-group). The following proposition shows that in order to solve an arbitrary embedding problem for 𝐺, it suffices to solve Frattini embedding problem followed by a split embedding problem: Proposition 25.6.8 Let C be a full formation of finite groups and let (25.10) be a C-embedding problem (resp. pro-C-embedding problem) for a pro-C group 𝐺. Suppose that the following two conditions are satisfied: (a) Every Frattini C-embedding problem (resp. Frattini pro-C-embedding problem) (𝜑: 𝐺 → 𝐴, 𝛼0 : 𝐵0 → 𝐴) where 𝐵0 is a subgroup of 𝐵 is solvable. (b) Every split C-embedding problem (resp. split pro-C-embedding problem) (𝛾0 : 𝐺 → 𝐵0 , 𝛼 ′: 𝐵 ′ → 𝐵0 ) for 𝐺 with Ker(𝛼 ′)  Ker(𝛼) is solvable. Then, (25.10) is solvable. Proof. Choose a closed subgroup 𝐵0 of 𝐵 such that 𝛼0 = 𝛼| 𝐵0 : 𝐵0 → 𝐴 is a Frattini cover (Lemma 25.6.6). Since C is full, 𝐵0 ∈ C (resp. by Lemma 20.3.1, 𝐵0 is pro-C). Thus,

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25 Projective Groups and Frattini Covers

(𝜑: 𝐺 → 𝐴, 𝛼0 : 𝐵0 → 𝐴)

(25.11)

is a Frattini C-embedding problem (resp. Frattini pro-C-embedding problem). By (a), there is an epimorphism 𝛾0 : 𝐺 → 𝐵0 with 𝛼0 ◦ 𝛾0 = 𝜑. Put 𝐶 = Ker(𝛼) and let 𝐵 ′ = 𝐵0 ⋉ 𝐶 with 𝐵0 acting on 𝐶 by conjugation. Since C is full, 𝐵 ′ ∈ C (resp., by Lemma 20.3.1, 𝐵 ′ is pro-C). Then, 𝐵0 𝐶 = 𝐵, so 𝜋: 𝐵 ′ → 𝐵 given by 𝜋(𝑏 0 , 𝑐) = 𝑏 0 𝑐 with 𝑏 0 ∈ 𝐵0 and 𝑐 ∈ 𝐶 is an epimorphism. Let 𝛼 ′: 𝐵 ′ → 𝐵0 be the projection on 𝐵0 . Then, 𝛼 ◦ 𝜋 = 𝛼0 ◦ 𝛼 ′ and (𝛾0 : 𝐺 → 𝐵0 , 𝛼 ′: 𝐵 ′ → 𝐵0 )

(25.12)

is a split C-embedding problem (resp. split pro-C-embedding problem) for 𝐺 with Ker(𝛼 ′) = 𝐶. Condition (b) gives an epimorphism 𝛾 ′: 𝐺 → 𝐵 ′ with 𝛼 ′ ◦ 𝛾 ′ = 𝛾0 : ′

𝛾 𝐵 ′ ❅o 𝐺 ❅❅ ♦♦♦⑧ 𝛾0 ♦♦♦ ⑧⑧ ❅ 𝜋 ❅❅ ♦♦♦ ⑧⑧ 𝜑 𝛼′ ❅♦ ⑧ 𝛾   w♦♦♦♦ ❅ ⑧⑧ 𝛼 /𝐵 3/ 𝐴 𝐵0 𝛼0

Then, 𝛾 := 𝜋 ◦ 𝛾 ′ solves embedding problem (25.10).

□

Proposition 25.6.8 is useful when 𝐺 is C-projective: Proposition 25.6.9 Let C be a Melnikov formation of finite groups and 𝐺 a Cprojective group. Then: (a) Every pro-C-embedding problem (25.10) in which 𝛼 is a Frattini cover is solvable. (b) Suppose that C is a full formation of finite groups. If every split C-embedding problem (resp. split pro-C-embedding problem) for 𝐺 is solvable, then every C-embedding problem (resp. pro-C-embedding problem) for 𝐺 is solvable. (c) If 𝜓: 𝐺 → 𝐺 is a Frattini cover, then 𝜓 is an isomorphism. Proof of (a). Since 𝐺 is C-projective, there exists a homomorphism 𝛾: 𝐺 → 𝐵 with 𝛼 ◦ 𝛾 = 𝜑. Then, 𝛼(𝛾(𝐺)) = 𝜑(𝐺) = 𝐴. By (25.9b), 𝛾(𝐺) = 𝐵. Thus, 𝛾 is a solution of (25.10). Proof of (b). Use (a) and Proposition 25.6.8. Proof of (c). Take 𝐵 = 𝐴 = 𝐺, 𝜑 = id𝐺 , and 𝛼 = 𝜓 to see that there exists an epimorphism 𝛾: 𝐺 → 𝐺 with 𝜓 ◦ 𝛾 = id𝐺 . Thus, 𝜓 is an isomorphism. □ Many properties of C-projective groups are determined by their quotients modulo their Frattini groups: Corollary 25.6.10 Let 𝐺 and 𝐻 be profinite groups with 𝐻 projective. (a) Each epimorphism 𝜃 0 : 𝐻/Φ(𝐻) → 𝐺/Φ(𝐺) has a lift to an epimorphism 𝜃: 𝐻 → 𝐺. (b) If in addition 𝐺 is projective, then each isomorphism 𝜃 0 : 𝐻/Φ(𝐻) → 𝐺/Φ(𝐺) has a lift to an isomorphism 𝜃: 𝐻 → 𝐺.

25.6 Frattini Covers

535

Proof of (a). Let 𝜋 𝐻 : 𝐻 → 𝐻/Φ(𝐻) and 𝜋𝐺 : 𝐺 → 𝐺/Φ(𝐺) be the quotient maps. Then, apply Proposition 25.6.9(a) to find an epimorphism 𝜃: 𝐻 → 𝐺 with 𝜋𝐺 ◦ 𝜃 = 𝜃 0 ◦ 𝜋 𝐻 . Proof of (b). Suppose that 𝜃 0 is an isomorphism. Note that both 𝜋𝐺 and 𝜋 𝐻 are Frattini covers. Hence, by Lemma 25.6.4, the map 𝜃: 𝐻 → 𝐺 given by (a) is a Frattini cover. Since 𝐺 is projective, a symmetrical argument gives a Frattini cover 𝜃 ′: 𝐺 → 𝐻. Since 𝜃 ′ ◦ 𝜃: 𝐻 → 𝐻 is also a Frattini cover, Proposition 25.6.9(c) implies that 𝜃 ′ ◦ 𝜃 is an isomorphism. Consequently, 𝜃 is an isomorphism. □ Let C be an arbitrary formation of finite groups. The proof of Proposition 25.6.9(b) does not extend to C-projective groups, because the proof of Proposition 25.6.8 involves a subgroup 𝐵0 of a C-group 𝐵 which need not be a C-group. Nevertheless, the result itself is true: Proposition 25.6.11 Let C be a formation of finite groups and 𝐺 a C-projective group. Suppose that every split C-embedding problem (25.10) for 𝐺 such that Ker(𝛼) is a minimal normal subgroup of 𝐵 is solvable. Then, every C-embedding problem for 𝐺 is solvable. Proof. Let (25.10) be a C-embedding problem for 𝐺. First suppose that 𝐶 = Ker(𝛼) is a minimal normal subgroup of 𝐵 but (25.10) does not necessarily split. Since 𝐺 is C-projective, there is a homomorphism 𝛾: 𝐺 → 𝐵 with 𝛼 ◦ 𝛾 = 𝜑. Put ¯ 𝐺¯ → 𝐵 and 𝜑: 𝐺¯ = 𝐺/Ker(𝛾). Let 𝜋: 𝐺 → 𝐺¯ be the quotient map and 𝛾: ¯ 𝐺¯ → 𝐴 the homomorphisms induced by 𝛾 and 𝜑, respectively. Then, 𝛼 ◦ 𝛾¯ = 𝜑. ¯ This leads to a diagram 𝐺 𝜋

 𝜓 / 𝐺¯ 𝐵 × 𝐴 𝐺¯ ① 𝛾¯ ①①① ① 𝜑¯ 𝛽 ① ①   {①①① 𝛼 /𝐴 𝐵 in which 𝜓 and 𝛽 are the projections on the coordinates. By Lemma 25.2.5, 𝛽 maps Ker(𝜓) isomorphically onto 𝐶. Hence, Ker(𝜓) is a minimal normal subgroup of ¯ Lemma 25.2.9 gives a group-theoretic section to 𝜓. Therefore, by assumption, 𝐵× 𝐴𝐺. there is an epimorphism 𝜁: 𝐺 → 𝐵 × 𝐴 𝐺¯ such that 𝜓 ◦ 𝜁 = 𝜋. The epimorphism 𝛽 ◦ 𝜁: 𝐺 → 𝐵 solves (25.10). Finally, suppose that 𝐶 is not a minimal normal subgroup of 𝐵. Then, 𝐶 has a ¯ 𝐵 → 𝐵¯ proper nontrivial subgroup 𝐶0 which is normal in 𝐵. Put 𝐵¯ = 𝐵/𝐶0 , 𝛽: ¯ the quotient map, and 𝛼: ¯ 𝐵 → 𝐴 the epimorphism induced by 𝛼. This gives a Cembedding problem (𝜑: 𝐺 → 𝐴, 𝛼: ¯ 𝐵¯ → 𝐴) whose kernel 𝐶/𝐶0 has a smaller order than |𝐶 |. An induction hypothesis gives an epimorphism 𝜑: ¯ 𝐺 → 𝐵¯ with 𝛼¯ ◦ 𝜑¯ = 𝜑. ¯ ¯ ¯ whose kernel 𝐶0 also This gives an embedding problem ( 𝜑: ¯ 𝐺 → 𝐵, 𝛽: 𝐵 → 𝐵) has a smaller order than |𝐶 |. Again, an induction hypothesis gives an epimorphism 𝛾: 𝐺 → 𝐵 with 𝛽¯ ◦ 𝛾 = 𝜑. ¯ The epimorphism 𝛾 solves (25.10). □

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25 Projective Groups and Frattini Covers

25.7 The Universal Frattini Cover The study of projective groups and the study of Frattini covers have a common subject: “universal Frattini covers”: Starting from a profinite group 𝐺, we partially order the epimorphisms of profinite groups onto 𝐺 (also called covers of 𝐺). Let 𝜃 𝑖 : 𝐻𝑖 → 𝐺, 𝑖 = 1, 2, be covers. We write 𝜃 2 ≥ 𝜃 1 and say that 𝜃 2 is larger than 𝜃 1 if there is an epimorphism 𝜃: 𝐻2 → 𝐻1 with 𝜃 1 ◦ 𝜃 = 𝜃 2 . If 𝜃 is an isomorphism, then 𝜃 1 is said to be isomorphic to 𝜃 2 . An epimorphism 𝜑: 𝐺˜ → 𝐺 is called a projective cover if 𝐺˜ is a projective group. In this case, 𝐺˜ is also called a projective cover of 𝐺. ˜ 𝐺˜ → 𝐺, unique up to Proposition 25.7.1 Each profinite group 𝐺 has a cover 𝜑: an isomorphism, called the universal Frattini cover and satisfying the following equivalent conditions: (a) 𝜑˜ is a projective Frattini cover of 𝐺. (b) 𝜑˜ is the largest Frattini cover of 𝐺. (c) 𝜑˜ is the smallest projective cover of 𝐺. In particular, each projective group is its own universal Frattini cover. Proof. By Proposition 20.4.8, there exists an epimorphism 𝜑: 𝐹 → 𝐺 with 𝐹 a free profinite group. Every closed subgroup of 𝐹 is projective (Corollary 25.4.6). Apply Lemma 25.6.6 to produce a closed subgroup 𝐺˜ of 𝐹 such that 𝜑˜ = 𝜑| 𝐺˜ is a Frattini (and projective) cover of 𝐺. The cover 𝜑: ˜ 𝐺˜ → 𝐺 appears throughout this proof. Proof of (a) =⇒ (b). Let 𝜃: 𝐺 1 → 𝐺 be a Frattini cover. Proposition 25.6.9 gives ˜ ˜ Thus, 𝜃 ≤ 𝜑. an epimorphism 𝛾: 𝐺˜ → 𝐺 1 with 𝜃 ◦ 𝛾 = 𝜑. Proof of (b) =⇒ (a). Let 𝜑 ′: 𝐺 ′ → 𝐺 be a Frattini cover that is larger than any ˜ 𝐺 ′ → 𝐺˜ such that 𝜑˜ ◦ 𝜃˜ = 𝜑 ′. Frattini cover. Thus, there exists an epimorphism 𝜃: By Lemma 25.6.4(c), 𝜃˜ is a Frattini cover. By “(a) =⇒ (b)”, 𝜑˜ is also a maximal ˜ Frattini cover of 𝐺. Thus, there exists a Frattini cover 𝜃 ′: 𝐺˜ → 𝐺 ′ with 𝜑 ′ ◦ 𝜃 ′ = 𝜑. It follows that 𝜃˜ ◦ 𝜃 ′: 𝐺˜ → 𝐺˜ is a projective Frattini cover. By Proposition 25.6.9(c), 𝜃˜ ◦ 𝜃 ′ is an isomorphism. Hence, 𝜃 ′: 𝐺˜ → 𝐺 ′ is an isomorphism. Consequently, 𝐺 ′ is projective and (a) holds. Proof of (a) =⇒ (c). Let 𝜑: 𝑃 → 𝐺 be a projective cover. Proposition 25.6.9 gives an epimorphism 𝛾: 𝑃 → 𝐺˜ with 𝜑˜ ◦ 𝛾 = 𝜑. Thus, 𝜑˜ ≤ 𝜑. Proof of (c) =⇒ (a). Let 𝜑 ′: 𝐺 ′ → 𝐺 be a projective cover which is smaller than any other projective cover of 𝐺. Since 𝐺˜ is projective, there is an epimorphism ˜ 𝐺˜ → 𝐺 ′ with 𝜑 ′ ◦ 𝜃˜ = 𝜑. ˜ By Lemma 25.6.4(c), 𝜑 ′ is a Frattini cover. 𝜃: Proof of the uniqueness of 𝜑. ˜ Suppose that 𝜑 ′: 𝐺˜ ′ → 𝐺 is another universal Frattini cover. In particular, by (b), there exist Frattini covers 𝜓: 𝐺˜ ′ → 𝐺˜ and ˜ By Proposition 25.6.9(c), 𝜓 ◦ 𝜌 is an 𝜌: 𝐺˜ → 𝐺˜ ′ with 𝜑˜ ◦ 𝜓 = 𝜑 ′ and 𝜑 ′ ◦ 𝜌 = 𝜑. □ isomorphism. Therefore, 𝜌 is an isomorphism, as claimed.

25.8 Projective Pro- 𝑝-Groups

537

Lemma 25.7.2 Let C be a full formation of finite groups and 𝐺 a pro-C-group. Then, ˜ = rank(𝐺). the smallest projective cover 𝐺˜ of 𝐺 is a pro-C-group and rank( 𝐺) Proof. Let 𝑚 = rank(𝐺). By Corollary 25.4.5, 𝐹ˆ𝑚 (C) is a projective cover of 𝐺. Hence, 𝐺˜ is a quotient of 𝐹ˆ𝑚 (C). In particular, 𝐺˜ is a pro-C group. By Proposition 25.7.1, 𝐺˜ is a Frattini cover of 𝐺. The equality of the ranks follows from Corollary □ 25.6.3(a). The next lemma characterizes the quotients of the universal Frattini cover of a profinite group: Lemma 25.7.3 Let 𝜑: ˜ 𝐺˜ → 𝐺 be the universal Frattini cover of a profinite group 𝐺. Then, a profinite group 𝐻 is a quotient of 𝐺˜ if and only if 𝐻 is a Frattini cover of a quotient of 𝐺. Proof. Suppose that 𝜓: 𝐺˜ → 𝐻 is an epimorphism. Lemma 25.2.8 gives a commutative diagram of epimorphisms 𝐺˜ ✴ ❄❖❖ ✴✴❄❄❖❖❖❖ 𝜑˜ ✴✴ ❄❄❄ ❖❖❖ ❖❖❖ ✴✴ ❄❄𝜋 ❖ 𝛿 ❖'/ ✴✴ 𝐺 𝜓 ✴ 𝐷 ✴✴ ✴✴ 𝛽 𝛾 ✴   𝐻 𝛼 / 𝐴, where the square is cartesian. By Lemma 25.6.4(c), 𝛿 is a Frattini cover. Thus (Lemma 25.6.5), so is 𝛼. Conversely, suppose 𝛼: 𝐻 → 𝐴 is a Frattini cover and 𝛾: 𝐺 → 𝐴 is an epimorphism. Apply the projectivity of 𝐺˜ and Proposition 25.6.9(a) to find an epimorphism 𝜓: 𝐺˜ → 𝐻. □

25.8 Projective Pro- 𝒑-Groups Fix a prime number 𝑝 for the whole section. By Example 20.3.3(b), the family of finite 𝑝-groups is full. Thus (Corollary 25.4.5), each free pro-𝑝-group is projective. We prove here a converse to this statement: Lemma 25.8.1 Let 𝑚 be a cardinal number and 𝑉 = F𝑚 𝑝 be the direct product of 𝑚 copies of the additive group of F 𝑝 . Then, rank(𝑉) = 𝑚 and |𝑉 | = 𝑝 𝑚 . Proof. Suppose first that 𝑚 is finite. Consider 𝑉 as a vector space over F 𝑝 . Then, rank(𝑉) = dim(𝑉) = 𝑚, so |𝑉 | = 𝑝 𝑚 . Now suppose that 𝑚 is infinite. By Lemma 1.2.6(b), 𝑉 is a profinite group with respect to the Tychonov topology. Choose a set 𝐼 of cardinality Î 𝑚 and for each Î 𝑖∈𝐼 an isomorphic copy 𝐹𝑖 of F 𝑝 . Each open subgroup 𝐻 of 𝑖 ∈𝐼 𝐹𝑖 contains 𝑖 ∈𝐽 𝐹𝑖

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25 Projective Groups and Frattini Covers

for some cofinite subset 𝐽 of 𝐼. The cardinality of the set of cofinite subsets of 𝐼 is Î Î 𝑚. For each such 𝐽 only finitely many open subgroups of 𝑖 ∈𝐼 𝐹𝑖 contain 𝑖 ∈𝐽 𝐹𝑖 . Î Hence, the cardinality of all open subsets of 𝑖 ∈𝐼 𝐹𝑖 is 𝑚, as claimed. □ Corollary 25.8.2 Let 𝑚 ≥ 2 be a cardinal number. (a) If 𝑚 is finite, then | 𝐹ˆ𝑚 | = 2ℵ0 . (b) If 𝑚 ≥ ℵ0 , then | 𝐹ˆ𝑚 | = 2𝑚 . Proof. Consider a set 𝑋 of cardinality 𝑚. Let 𝐹𝑚 be the free abstract group on 𝑋 and let N be the collection of all normal subgroups 𝑁 of 𝐹𝑚 of finite index such that 𝑋 ∖ 𝑁 is finite. By Proposition 20.4.2, 𝐹ˆ𝑚 = lim 𝐹/𝑁, where 𝑁 ranges over all ←− Î elements of N . In particular, by the definition of the inverse limit, 𝐹ˆ𝑚 ≤ 𝑁 ∈N 𝐹/𝑁, where the right-hand side is equipped with the Tychonov topology (Section 1.1). Î If 𝑚 is finite, then |N | = ℵ0 , hence | 𝐹ˆ𝑚 | ≤ 𝑁 ∈N |𝐹/𝑁 | = 2ℵ0 . On the other hand, by Proposition 20.4.8, Z2 is a quotient of 𝐹ˆ𝑚 . Hence, | 𝐹ˆ𝑚 | ≥ |Z2 | = 2ℵ0 , where the right-hand side follows from the presentation of Z2 as the set of all formal power Í∞ 𝑎 𝑖 2𝑖 with 𝑎 𝑖 ∈ {0, 1} for all 𝑖 (remark preceding Lemma 1.4.3). Using series 𝑖=0 the preceding estimate, we conclude that | 𝐹ˆ𝑚 | = 2ℵ0 . If 𝑚 is infinite, then rank( 𝐹ˆ𝑚 ) = 𝑚 (Lemma 20.4.6), hence Î |N | = 𝑚 (Section 20.1). Therefore, by the first paragraph of this proof, | 𝐹ˆ𝑚 | ≤ 𝑁 ∈N |𝐹/𝑁 | = 2𝑚 . Conversely, by Lemma 25.8.1, rank(F2𝑚 ) = 𝑚, hence again by Proposition 20.4.8, | 𝐹ˆ𝑚 | ≥ |F2𝑚 | = 2𝑚 . Using the preceding inequality, we conclude that | 𝐹ˆ𝑚 | = 2𝑚 , as claimed. □ Lemma 25.8.3 Let 𝐼 be a set, 𝐺 = (F 𝑝 ) 𝐼 , and 𝑁 a closed subgroup of 𝐺. Then, 𝐺 = 𝑁 × 𝑁 ′ for some closed subgroup 𝑁 ′ of 𝐺. Proof. Let 𝜑: 𝐺 → 𝐺/𝑁 be the quotient map. Lemma 25.6.6 gives a closed subgroup 𝑁 ′ of 𝐺 such that 𝜑 ′ = 𝜑| 𝑁 ′ : 𝑁 ′ → 𝐺/𝑁 is a Frattini cover. Thus, Ker(𝜑 ′) ≤ Φ(𝑁 ′) ≤ Φ(𝐺) = Φ(F 𝑝 ) 𝐼 = 1 (Lemma 25.1.4(c,d)). Consequently, 𝐺 = 𝑁 × 𝑁 ′. □ Lemma 25.8.4 Let 𝑚 be a cardinal number, 𝐺 = F𝑚 𝑝 , and 𝐻 a closed subgroup of 𝐺. Then, 𝐻  F 𝑘𝑝 for some cardinal number 𝑘 ≤ 𝑚. Proof. Choose a set 𝐽 of cardinality |𝐺 |. Let U be the set of all triples (𝑈, 𝐼, 𝜑) with 𝑈 ≤ 𝐻, 𝐼 ⊆ 𝐽, and 𝜑: 𝐻/𝑈 → F𝐼𝑝 is an isomorphism. Define a partial ordering on U by the following rule: (𝑈 ′, 𝐼 ′, 𝜑 ′) ≤ (𝑈, 𝐼, 𝜑) if 𝑈 ′ ≤ 𝑈, 𝐼 ⊆ 𝐼 ′, and the following diagram is commutative: 𝐻/𝑈 ′ 𝜑′

𝜋𝑈′ ,𝑈

/ 𝐻/𝑈

(25.13)

𝜑



′

F𝐼𝑝

𝜌𝐼 ′ ,𝐼



/ F𝐼 𝑝

.

Here 𝜋𝑈′ ,𝑈 is the quotient map and 𝜌 𝐼 ′ ,𝐼 is the projection.

25.8 Projective Pro- 𝑝-Groups

539

The triple (𝐻, ∅, id) belongs to U. Suppose that {(𝑈 𝛼 , 𝐼 𝛼 , 𝜑 𝛼 ) | 𝛼 ∈ 𝐴} is a Ñ Ð descending chain in U. Let 𝑈0 = 𝛼∈ 𝐴 𝑈 𝛼 , 𝐼0 = 𝛼∈ 𝐴 𝐼 𝛼 , and 𝜑0 = lim 𝜑 𝛼 (we ←− assume here 0 ∉ 𝐴). Then, (𝑈0 , 𝐼0 , 𝜑0 ) is an element of U which is smaller or equal to each (𝑈 𝛼 , 𝐼 𝛼 , 𝜑 𝛼 ). Zorn’s lemma gives a minimal element (𝑈, 𝐼, 𝜑) in U. Assume that 𝑈 is not trivial. Lemma 25.8.3 gives a closed subgroup 𝑉 of 𝐺 with 𝑉 × 𝑈 = 𝐺. Then, 𝑉 is a proper subgroup of 𝐺. Hence, 𝑉 is contained in an open subgroup 𝐺 ′ of 𝐺 of index 𝑝. Put 𝐻 ′ = 𝐺 ′ ∩ 𝐻 and 𝑈 ′ = 𝐺 ′ ∩ 𝑈. Then, 𝐻/𝐻 ′  𝐺/𝐺 ′  F 𝑝 and 𝐻/𝑈 ′  𝐻/𝑈 × 𝐻/𝐻 ′. Since |𝐺 | ≥ |𝐻/𝑈| = 𝑝 |𝐼 | > |𝐼 |, there is a 𝑗 ∈ 𝐽 ∖ 𝐼. Put ′ 𝐼 ′ = 𝐼 ∪ { 𝑗 }. Then, it is possible to lift 𝜑 to an isomorphism 𝜑 ′: 𝐻/𝑈 ′ → F𝐼𝑝 such that (25.13) is commutative. This contradicts the minimality of (𝑈, 𝐼, 𝜑) and proves that 𝑈 is trivial. It follows that 𝐻  (F 𝑝 ) 𝐼 . By Lemma 25.8.1 and Corollary 20.1.5, |𝐼 | = □ rank(𝐻) ≤ rank(𝐺) = 𝑚, as claimed. For a profinite group 𝐺 let 𝐺 𝑝 = ⟨𝑔 𝑝 | 𝑔 ∈ 𝐺⟩

and

 [𝐺, 𝐺] = [𝑔1 , 𝑔2 ] | 𝑔1 , 𝑔2 ∈ 𝐺 .

Both closed subgroups of 𝐺 are characteristic. Lemma 25.8.5 Let 𝐺 be a pro-𝑝-group of rank 𝑚. Then, Φ(𝐺) = 𝐺 𝑝 [𝐺, 𝐺] and 𝐺/Φ(𝐺) is isomorphic to the vector space F𝑚 𝑝. Proof. Maximal subgroups of finite 𝑝-groups are normal subgroups of index 𝑝. Therefore, maximal open subgroups of 𝐺 are open Î normal subgroups of index 𝑝. This gives a canonical embedding 𝐺/Φ(𝐺) → 𝐺/𝑁, where 𝑁 ranges over all open normal subgroups of 𝐺 of index 𝑝. By Lemma 25.8.4, 𝐺/Φ(𝐺)  F𝑚 𝑝 with 𝑚 = rank(𝐺/Φ(𝐺)) = rank(𝐺) (Corollary 25.6.3(b)). Now let 𝐺 0 = 𝐺 𝑝 [𝐺, 𝐺]. Since 𝐺/Φ(𝐺)  F𝑚 𝑝 , the canonical map 𝐺 → 𝐺/Φ(𝐺) maps each 𝑔 𝑝 [𝑎, 𝑏] to 1. Hence, 𝐺 0 ≤ Φ(𝐺). On the other hand let 𝑈 be an open normal subgroup of 𝐺 that contains 𝐺 0 . Then, 𝐺/𝑈 is an Abelian elementary 𝑝-group. In particular, 𝑈 is the intersection of all open normal subgroups of index 𝑝 that contain 𝑈. Hence, Φ(𝐺) ≤ 𝑈. Since 𝐺 0 ⊳ 𝐺, the intersection of all □ such 𝑈 is 𝐺 0 . Thus, Φ(𝐺) ≤ 𝐺 0 . Consequently, Φ(𝐺) = 𝐺 0 . Remark 25.8.6 (Subgroups of finite index) Let 𝑚 be an infinite cardinal number. By Lemma 25.8.1, the cardinality of the set of open subgroups of F𝑚 𝑝 of index 𝑝 is 𝑚. 𝑚 𝑚 𝑚 On the other hand, |F𝑚 𝑝 | = 2 . Hence, any basis 𝐵 of F 𝑝 , where F 𝑝 is considered as 𝑚 a vector space over F 𝑝 , has cardinality 2 . For each 𝑣 ∈ 𝐵 the subgroup generated by 𝐵 ∖{𝑣} has index 𝑝. All these subgroups are distinct. Therefore, F𝑚 𝑝 has subgroups of index 𝑝 which are not open. In contrast, Serre proved that every subgroup of finite index of a finitely generated pro-𝑝 group 𝐺 is open [Ser97, p. 32, Exercises 5 and 6] (see also [RiZ00, Thm. 4.2.8]) and wrote he did not know if that statement holds for arbitrary finitely generated profinite groups. Nikolov–Segal gave an affirmative answer to the question of Serre [NiS07]: (25.14) Every subgroup 𝑁 of finite index of a finitely generated profinite group 𝐺 is open.

540

25 Projective Groups and Frattini Covers

The proof of (25.14) depends on properties of certain “verbal subgroups”: Consider a group-theoretic word 𝑤(𝑋1 , . . . , 𝑋𝑚 ) in the variables 𝑋1 , . . . , 𝑋𝑚 . For each group 𝐺 let 𝑤(𝐺) be the subgroup generated (in the sense of abstract groups) by 𝑤(x) with x ∈ 𝐺 𝑚 . Given a positive integer 𝑑, we say that 𝑤 is 𝑑-locally finite if every group 𝐻 which is generated (in the abstract sense) by 𝑑 elements and satisfies 𝑤(𝐻) = 1 is finite. The existence of 𝑑-locally finite words is proved in the introduction of [NiS07]: (25.15) For each finite group 𝐴 which is generated by 𝑑 elements there exists a 𝑑-locally finite word 𝑤 with 𝑤( 𝐴) = 1. Ñ To prove (25.15) consider the free group 𝐹 on 𝑥1 , . . . , 𝑥 𝑑 . Let 𝑁 = Ker(𝜃), where 𝜃 ranges over all homomorphisms from 𝐹 to 𝐴. Then, 𝑁 is a normal subgroup of a finite index. By Corollary 20.5.8, 𝑁 is finitely generated. Let 𝑦 1 , . . . , 𝑦 𝑘 be generators of 𝑁. For each 𝑖 write 𝑦 𝑖 = 𝑤 𝑖 (x), where 𝑤 𝑖 is a word in 𝑥 1 , . . . , 𝑥 𝑑 . Then, consider the word 𝑤(X) = 𝑤 1 (X1 ) · · · 𝑤 𝑘 (X 𝑘 ), where X𝑖 = (𝑋𝑖1 , . . . , 𝑋𝑖𝑑 ) and X = (X1 , . . . , X 𝑘 ). Each of the generators 𝑦 𝑖 = 𝑤 𝑖 (x) of 𝑁 belongs to 𝑤(𝐹), so 𝑁 ≤ 𝑤(𝐹). Conversely, let 𝑥1′ , . . . , 𝑥 𝑑′ ∈ 𝐹. Then, the map 𝑥 𝑖 ↦→ 𝑥 𝑖′, 𝑖 = 1, . . . , 𝑑, extends to a homomorphism 𝜅: 𝐹 → 𝐹. Let 𝜃: 𝐹 → 𝐴 be an arbitrary homomorphism. Then, 𝜃 ◦ 𝜅 is also a homomorphism from 𝐹 to 𝐴. Hence, 𝜃 (𝑤 𝑖 (x′)) = 𝜃 (𝑤 𝑖 (𝜅(x))) = 𝜃 ◦ 𝜅(𝑤 𝑖 (x)) = 𝜃 ◦ 𝜅(𝑦 𝑖 ) = 1 for 𝑖 = 1, . . . , 𝑘. It follows that 𝑤(𝐹) ≤ 𝑁. Consequently, 𝑤(𝐹) = 𝑁. Now let 𝐻 be a group with 𝑑 generators and 𝑤(𝐻) = 1. Let 𝜋: 𝐹 → 𝐻 be an epimorphism. Then, 𝜋(𝑤(𝐹)) = 𝑤(𝐻) = 1. Hence, |𝐻| ≤ (𝐹 : 𝑤(𝐹)) = (𝐹 : 𝑁) < ∞. Consequently, 𝑤 is 𝑑-locally finite, as desired. The key result in the proof of (25.14) bounds the number of factors in the elements of 𝑤(𝐻): (25.16) Let 𝑑 be a positive integer and 𝑤(𝑋1 , . . . , 𝑋𝑚 ) a 𝑑-locally finite word. Then, there exists a positive integer 𝑟 such that for each finite group 𝐴 generated by 𝑑 elements and for each 𝑎 ∈ 𝑤( 𝐴) there are b1 , . . . , b𝑟 ∈ 𝐴𝑚 and 𝛽1 , . . . , 𝛽𝑟 ∈ {±1} such that 𝑎 = 𝑤(b1 ) 𝛽1 · · · 𝑤(b𝑟 ) 𝛽𝑟 [NiS07, Thm. 2.1]. The proof of (25.16) uses the classification of finite simple groups. Compactness arguments generalize (25.16) to profinite groups: (25.17) Let 𝑑 be a positive integer and 𝑤(𝑋1 , . . . , 𝑋𝑚 ) a 𝑑-locally finite word. Then, there exists a positive integer 𝑟 such that for each profinite group 𝐺 generated (in the profinite sense) by 𝑑 elements and for each 𝑔 ∈ 𝑤(𝐺) there are b1 , . . . , b𝑟 ∈ 𝐺 𝑚 and 𝛽1 , . . . , 𝛽𝑟 ∈ {±1} such that 𝑔 = 𝑤(b1 ) 𝛽1 · · · 𝑤(b𝑟 ) 𝛽𝑟 . In particular, 𝑤(𝐺) is closed. Now consider a profinite group 𝐺 generated (in the profinite sense) by elements 𝑥1 , . . . , 𝑥 𝑑 and let 𝑁 be a subgroup of finite index. Without loss assume that 𝑁 is normal in 𝐺. Then, 𝐺/𝑁 is finite. By (25.15), there exists a 𝑑-locally finite word 𝑤(𝑋1 , . . . , 𝑋𝑚 ) such that 𝑤(𝐺/𝑁) = 1. It follows that 𝑤(𝐺) ≤ 𝑁. Let 𝐺 0 be the abstract subgroup of 𝐺 generated by 𝑥1 , . . . , 𝑥 𝑑 . Then, 𝑤(𝐺 0 ) ⊳ 𝐺 0 , 𝑤(𝐺 0 /𝑤(𝐺 0 )) = 1, and 𝐺 0 /𝑤(𝐺 0 ) is generated (in the sense of abstract groups) by 𝑑 elements. By definition, 𝐺 0 /𝑤(𝐺 0 ) is finite, so (𝐺 0 : 𝑤(𝐺 0 )) < ∞. Therefore, (𝐺 0 𝑤(𝐺) : 𝑤(𝐺)) = (𝐺 0 : 𝐺 0 ∩ 𝑤(𝐺)) ≤ (𝐺 0 : 𝑤(𝐺 0 )) < ∞.

25.8 Projective Pro- 𝑝-Groups

541

By (25.17), 𝑤(𝐺) is closed, so 𝐺 0 𝑤(𝐺) is closed. Since 𝐺 0 is dense in 𝐺, we have 𝐺 0 𝑤(𝐺) = 𝐺. It follows that 𝑤(𝐺) has a finite index in 𝐺. Consequently, 𝑤(𝐺) is open. Since 𝑤(𝐺) ≤ 𝑁, also 𝑁 is open, as claimed. Proposition 25.8.7 (Tate) A pro-𝑝-group 𝐺 is projective if and only if it is pro-𝑝 free. Proof. By Corollary 25.4.5, it suffices to show that 𝐺 projective implies 𝐺 is pro-𝑝 free. Let 𝑚 = rank(𝐺). By Lemma 25.8.5, ˆ ˆ 𝐺/Φ(𝐺)  F𝑚 𝑝  𝐹𝑚 ( 𝑝)/Φ( 𝐹𝑚 ( 𝑝)). Therefore, by Corollary 25.6.10(b), 𝐺  𝐹ˆ𝑚 ( 𝑝).

□

Proposition 25.8.7 immediately gives an analog of Schreier’s theorem (Proposition 20.5.6) for discrete free groups: Corollary 25.8.8 (Tate) A closed subgroup of a free pro-𝑝-group is pro-𝑝 free. Proof. Let 𝐺 be a closed subgroup of a free pro-𝑝 group 𝐹. By Proposition 25.8.7, 𝐹 is projective. By Proposition 25.4.7, 𝐺 is projective. Therefore, by Proposition 25.8.7, 𝐺 is free pro-𝑝. □ Corollary 25.8.9 Let 𝐺 be a pro-𝑝 group of rank 𝑚. Then, the universal Frattini cover of 𝐺 is 𝐹ˆ𝑚 ( 𝑝). Proof. Let 𝐺˜ be the universal Frattini cover of 𝐺. By Lemma 25.7.2, 𝐺˜ is a pro-𝑝 group of rank 𝑚. By Proposition 25.7.1, 𝐺˜ is projective. Hence, by Proposition 25.8.7, 𝐺  𝐹ˆ𝑚 ( 𝑝). □ Remark 25.8.10 Let 𝑃 be a pro-𝑝 group of finite rank 𝑒 and let 𝜑: 𝐹ˆ𝑒 ( 𝑝) → 𝑃 be the universal Frattini cover of 𝑃. Given an arbitrary epimorphism 𝜑 ′: 𝐹ˆ𝑒 ( 𝑝) → 𝑃, there is an epimorphism 𝛾: 𝐹ˆ𝑒 ( 𝑝) → 𝐹ˆ𝑒 ( 𝑝) with 𝜑 ′ ◦ 𝛾 = 𝜑 (Proposition 20.7.4). By Proposition 19.1.6, 𝛾 is an automorphism. Hence, 𝜑 ′: 𝐹ˆ𝑒 ( 𝑝) → 𝑃 is also the universal Frattini cover of 𝑃. Here is a generalization of Lemma 25.1.4(c) for pro-𝑝 groups: Lemma 25.8.11 Let 𝐺 be a pro-𝑝 group and 𝐻 a closed subgroup. Then, Φ(𝐻) ≤ Φ(𝐺). Proof. By Lemma 25.8.5, Φ(𝐻) = 𝐻 𝑝 [𝐻, 𝐻] ≤ 𝐺 𝑝 [𝐺, 𝐺] = Φ(𝐺).

□

Proposition 25.8.12 Every pro-𝑝 group 𝐺 which is small is finitely generated. Proof. By Lemma 25.8.5, Φ(𝐺) is the intersection of all open subgroups of 𝐺 of index 𝑝. By assumption, there are only finitely many of them. Hence, 𝐺/Φ(𝐺) is a finite group. In particular, 𝐺/Φ(𝐺) is finitely generated. It follows from Lemma 25.1.1 that 𝐺 is also finitely generated. □

542

25 Projective Groups and Frattini Covers

Corollary 25.8.13 Let 𝐴 be an Abelian projective pro-𝑝 group. Then, either 𝐴 is trivial or 𝐴  Z 𝑝 . Proof. By Proposition 25.8.7, 𝐴 is pro-𝑝 free. Since 𝐴 is Abelian, rank( 𝐴) ≤ 1. Otherwise, 𝐴 would have a non-Abelian finite quotient of rank 2. So, either 𝐴 = 1 or 𝐴  Z 𝑝 . □ The following result surveys the structure of an arbitrary Abelian pro-𝑝 group: Proposition 25.8.14 Let 𝐴 be an additive Abelian pro-𝑝 group. Then: (a) 𝐴 is a Z 𝑝 -module. (b) Suppose that 𝐴 is finitely generated. Then, 𝐴  Z/𝑝 𝑘1 Z⊕· · ·⊕Z/𝑝 𝑘𝑚 Z⊕Z𝑟𝑝 , with 𝑘 1 ≥ · · · ≥ 𝑘 𝑚 ≥ 1 and 𝑟 is a nonnegative integer. The sequence 𝑘 1 , . . . , 𝑘 𝑚 , 𝑟 are uniquely determined by 𝐴. (c) Suppose that 𝐴 is finitely generated and torsion-free. Let 𝐵 be a subgroup. Then, 𝐴 and 𝐵 are free Z 𝑝 -modules, 𝐴 has a Z 𝑝 -basis 𝑎 1 , . . . , 𝑎 𝑛 , and there are positive integers 𝛼1 , . . . , 𝛼𝑚 with 0 ≤ 𝑚 ≤ 𝑛 satisfying this: 𝛼1 𝑎 1 , . . . , 𝛼𝑚 𝑎 𝑚 form a Z 𝑝 -basis of 𝐵 and 𝛼𝑖 |𝛼𝑖+1 , 𝑖 = 1, . . . , 𝑚 − 1. ′ ′ (d) In general, 𝐴  Z𝑚 𝑝 × 𝐴 , where 𝑚 is a cardinal number and 𝐴 is the intersection of all kernels of the epimorphisms of 𝐴 onto Z 𝑝 . (e) If 𝐴 is torsion-free, then 𝐴  Z𝑚 𝑝 for some cardinal number 𝑚. of 𝐴. Then, ( 𝐴 : 𝐵) = 𝑝 𝑛 with 𝑛 a Proof of (a). Let 𝐵 be an open subgroup Ñ 𝑛 𝑛 nonnegative integer. Hence, 𝑝 𝐴 ≤ 𝐵. Thus, ∞ 𝑛=1 𝑝 𝐴 = 0. 𝑛 𝑛 𝑛 The rule (𝑧 + 𝑝 Z) (𝑎 + 𝑝 𝐴) = 𝑧𝑎 + 𝑝 𝐴 defines a continuous action of Z/𝑝 𝑛 Z on 𝐴/𝑝 𝑛 𝐴. Going to the limit, the first paragraph gives an action of Z 𝑝 on 𝐴. Proof of (b). Statement (b) is a special case of the main theorem on finitely generated modules over principal ideal domains [Lan97, §III, Thms. 7.3 and 7.5]. Proof of (c). See [Lan97, §III, Thm. 7.8]. Proof of (d). See [GJe96, p. 337, (4)]. Proof of (e). See [RiZ00, p. 133, Thm. 4.3.3].

□

Example 25.8.15 (The rank of local Galois groups) Let 𝐾 be a local field, that is, 𝐾 is a finite extension of Q 𝑝 or of F 𝑝 ((𝑡)). Denote the maximal tamely ramified extension of 𝐾 by 𝐾tr . Let 𝐺 = Gal(𝐾) and 𝑃 = Gal(𝐾tr ). Then, 𝑃 is a closed normal pro-𝑝 group of 𝐺. By Lemma 25.1.4(c), Φ(𝑃) ≤ Φ(𝐺). By Lemma 25.8.5, Φ(𝑃) = 𝑃 𝑝 [𝑃, 𝑃]. Hence, rank(𝐺) = rank(𝐺/[𝑃, 𝑃]) (Corollary 25.6.3). This observation is implicitly used in [Jan82, Section 3] to bound the number of generators of 𝐺. Indeed, if 𝐾 is a finite extension of Q 𝑝 , then by results of Iwasawa and local class field theory, rank(𝐺/[𝑃, 𝑃]) ≤ [𝐾 : Q 𝑝 ] + 3. Hence, rank(𝐺) ≤ [𝐾 : Q 𝑝 ] + 3. Moreover, denote the fixed field of Φ(𝐺) in 𝐾˜ by 𝐹. It is proved in [JaR91] that 𝐹 is the compositum of all Galois extensions 𝑁 of 𝐾 containing 𝐾tr such that Gal(𝑁/𝐾tr )  Z/𝑝Z.

25.9 Supernatural Numbers

543

It turns out that the inequality rank(𝐺) ≤ [𝐾 : Q 𝑝 ] + 3 can be improved. Indeed, Theorem 7.4.1 of [NSW20] says that rank(Gal(𝐾)) is even bounded from above by [𝐾 : Q 𝑝 ] + 2. Moreover, if 𝜁 𝑝 ∈ 𝐾 and 𝐾 ( 𝑝) is the maximal 𝑝-extension of 𝐾, then Gal(𝐾 ( 𝑝) /𝐾) is a Demushkin group of rank [𝐾 : Q 𝑝 ] + 2 [Koc70, p. 96, Satz 10.4]. It follows that in this case rank(Gal(𝐾)) = [𝐾 : Q 𝑝 ] + 2. In the general case, we consider the field 𝐿 = 𝐾 (𝜁 𝑝 ) and conclude from the special case that rank(Gal(𝐿)) = [𝐿 : Q 𝑝 ] + 2. Then, the proof of [JaS21, Thm. 1.1] uses the Nielsen–Schreier inequality (Corollary 20.6.3) to conclude that in general, rank(Gal(𝐾)) = [𝐾 : Q 𝑝 ] + 2. In particular, rank(Gal(Q 𝑝 )) = 3. Finally, we note that if char(𝐾) = 𝑝, then the rank of 𝐺/[𝑃, 𝑃] and therefore also of Gal(𝐾) is infinite.

25.9 Supernatural Numbers The next definition generalizes the index and the order of finite groups to profinite groups: Definition 25.9.1 (Supernatural numbers) A supernatural number is a formal Î 𝑛𝑝 product 𝑛 = 𝑝 , where 𝑛 𝑝 is either a nonnegative integer or ∞, and 𝑝 ranges over all primes. Let 𝑚 and 𝑛 be supernatural numbers. We say that 𝑚 divides 𝑛 if 𝑚 𝑝 ≤ 𝑛 𝑝 for every prime 𝑝. Define the product of supernatural numbers 𝑛𝑖 , for 𝑖 ∈ 𝐼, by the formula ÖÖ Ö Í 𝑝 𝑛𝑖, 𝑝 = 𝑝 𝑖∈𝐼 𝑛𝑖, 𝑝 . 𝑖 ∈𝐼

𝑝

𝑝

Í

Here 𝑖 ∈𝐼 𝑛𝑖, 𝑝 = ∞ if 𝑛𝑖, 𝑝 > 0 for infinitely many 𝑖 ∈ 𝐼 or 𝑛𝑖, 𝑝 = ∞ for at least one 𝑖 ∈ 𝐼. Similarly, there is a greatest common divisor and a least common multiple of supernatural numbers: Ö Ö gcd(𝑚, 𝑛) = 𝑝 min(𝑚 𝑝 ,𝑛 𝑝 ) and lcm(𝑚, 𝑛) = 𝑝 max(𝑚 𝑝 ,𝑛 𝑝 ) . 𝑝

𝑝

Define the index of a closed subgroup 𝐻 in a profinite group 𝐺 to be
(𝐺 : 𝐻) = lcm (𝐺 : 𝑈) | 𝑈 is an open subgroup of 𝐺 that contains 𝐻 . Then, define the order of 𝐺 to be
#𝐺 = (𝐺 : 1) = lcm (𝐺 : 𝑈) | 𝑈 is an open subgroup of 𝐺 . Note that the index generalizes the usual index, (𝐺 : 𝐻), if 𝐻 is an open subgroup of 𝐺. Example 25.9.2 (Order of three classical Abelian groups) Ö Ö Ö #Z 𝑝 = 𝑝 ∞ ; #Zˆ = 𝑝 ∞ ; and # Z/𝑝Z = 𝑝. 𝑝

𝑝

𝑝

544

25 Projective Groups and Frattini Covers

The usual rules for indices of finite groups hold for profinite groups: Lemma 25.9.3 (a) Let 𝐾 ≤ 𝐻 ≤ 𝐺 be profinite groups. Then, (𝐺 : 𝐾) = (𝐺 : 𝐻) (𝐻 : 𝐾). (b) Let {𝐻𝑖 | 𝑖 ∈ 𝐼} be a directed family of closed subgroups of a profinite group 𝐺. Ñ Put 𝐻 = 𝑖 ∈𝐼 𝐻𝑖 . Then, (𝐺 : 𝐻) = lcm𝑖 ∈𝐼 (𝐺 : 𝐻𝑖 ). (c) Let ⟨𝐺 𝑖 , 𝜋 𝑗𝑖 ⟩ 𝑗,𝑖 ∈𝐼 be an inverse limit of profinite groups such that 𝜋 𝑗𝑖 is an epimorphism for 𝑗 ≥ 𝑖. Then, the inverse limit 𝐺 = lim 𝐺 𝑖 satisfies #𝐺 = ←− lcm Î𝑖 ∈𝐼 {#𝐺 𝑖 }.Î (d) # 𝑖 ∈𝐼 𝐺 𝑖 = 𝑖 ∈𝐼 #𝐺 𝑖 . Proof. Let 𝐻 be a closed subgroup of 𝐺. We use the following observation: Let H be a set of open subgroups of 𝐺 that contains 𝐻 such that each open subgroup 𝐻 ′ of 𝐺 that contains 𝐻 already contains some 𝐻0′ ∈ H . Then, (𝐺 : 𝐻) = lcm((𝐻 : 𝐻0′ ) | 𝐻0′ ∈ H ). Proof of (a). Let N be the
set of all open normal subgroups of 𝐺. Then, (𝐺 :
𝐻) = lcm (𝐺 : 𝐻𝑁) | 𝑁 ∈ N and (𝐻 : 𝐾) = lcm (𝐻 : 𝐾 (𝐻 ∩ 𝑁)) | 𝑁 ∈ N . The identities 𝐾 (𝐻 ∩ 𝑁) = 𝐻 ∩ 𝐾 𝑁 and (𝐻𝑁 : 𝐾 𝑁) = (𝐻 : 𝐻 ∩ 𝐾 𝑁) imply (𝐻 : 𝐾) = lcm{(𝐻𝑁 : 𝐾 𝑁) | 𝑁 ∈ N }.

(25.18)

(𝐺 : 𝐾 𝑁) = (𝐺 : 𝐻𝑁) (𝐻𝑁 : 𝐾 𝑁).

(25.19)

In addition, 𝑝 ∞ |(𝐺

: 𝐾), then (25.18) and (25.19) imply 𝑝 ∞ |(𝐺 : Let 𝑝 be a prime number. If ∞ 𝐻) or 𝑝 |(𝐻 : 𝐾). Suppose therefore that 𝑝 ∞ ∤ (𝐺 : 𝐾). Let 𝑛1 , 𝑛2 , and 𝑛3 be the maximal exponents of 𝑝-powers that divide (𝐺 : 𝐾), (𝐺 : 𝐻), and (𝐻 : 𝐾), respectively. Then, 𝑛1 , 𝑛2 , 𝑛3 < ∞. Also, there exists a group 𝑁 ∈ N such that 𝑝 𝑛1 (resp. 𝑝 𝑛2 , 𝑝 𝑛3 ) is the greatest power of 𝑝 dividing (𝐺 : 𝐾 𝑁) (resp. (𝐺 : 𝐻𝑁), (𝐻𝑁 : 𝐾 𝑁)). By (25.19), 𝑛1 = 𝑛2 + 𝑛3 , as desired. Proof of (b). By (a), (𝐺 : 𝐻𝑖 )|(𝐺 : 𝐻) for each 𝑖 ∈ 𝐼. Hence, lcm𝑖 ∈𝐼 (𝐺 : 𝐻𝑖 )|(𝐺 : 𝐻). Conversely, let 𝑈 be an open subgroup of 𝐺 that contains 𝐻. Then, there is an 𝑖 ∈ 𝐼 with 𝐻𝑖 ≤ 𝑈 (Lemma 1.2.2(a)). Therefore, (𝐺 : 𝑈)|(𝐺 : 𝐻𝑖 ), so (𝐺 : 𝑈)|lcm𝑖 ∈𝐼 (𝐺 : 𝐻𝑖 ). Thus, lcm𝑈 (𝐺 : 𝑈)|lcm𝑖 ∈𝐼 (𝐺 : 𝐻𝑖 ) with 𝑈 ranging over all open subgroups containing 𝐻. This gives (b). Proof of (c). Let 𝜋𝑖 : 𝐺 → Ñ 𝐺 𝑖 be the canonical epimorphism and 𝑁𝑖 = Ker(𝜋𝑖 ). Then, (𝐺 : 𝑁𝑖 ) = #𝐺 𝑖 and 𝑁𝑖 = 1. Apply (b): #𝐺 = lcm#𝐺 𝑖 . Proof of (d). Denote the family Î Î of finite subsets of 𝐼 by F . Partially order F by inclusion. Then, 𝐺 𝑖 = lim 𝑖 ∈𝐹 𝐺 𝑖 , where 𝐹 ranges over F . Hence, by (a) and ←− (c), Ö Ö Ö Ö 𝐺 𝑖 = lcm𝐹 ∈ F # 𝐺 𝑖 = lcm𝐹 ∈ F # #𝐺 𝑖 .□ #𝐺 𝑖 = 𝑖 ∈𝐼

𝑖 ∈𝐹

𝑖 ∈𝐹

𝑖 ∈𝐼

Lemma 25.9.4 Let 𝐺 be a profinite group and 𝐻 a closed subgroup. Suppose that 𝑝 ∤ (𝐺 : 𝐻) for all but finitely many 𝑝 and 𝑝 ∞ ∤ (𝐺 : 𝐻) for all 𝑝. Then, 𝐻 is open. Proof. For each 𝑝 let 𝑝 𝑛 𝑝 be the maximal power of 𝑝 which divides (𝐺 : 𝐻). By Î assumption, (𝐺 : 𝐻) = 𝑝 𝑛 𝑝 < ∞, so 𝐻 is open. □

25.10 The Sylow Theorems

545

Lemma 25.9.5 Let 𝐴 be a finitely generated Abelian profinite group and 𝐵 a closed subgroup of infinite index. Then, 𝐵 is contained in an open subgroup of index 𝑝 for infinitely many 𝑝 or 𝐵 is contained in a closed subgroup 𝐶 with 𝐴/𝐶  Z 𝑝 for some 𝑝. Proof. Divide out by 𝐵, if necessary, to assume 𝐵 = 0. We have to prove: Z/𝑝Z is a factor of 𝐴 for infinitely many 𝑝 or Z 𝑝 is a factor of 𝐴 for some 𝑝. Proposition 25.8.14(b) presents 𝐴 as Ê 𝑘 𝐴= (25.20) Z/𝑝 𝑘 𝑝,1 Z ⊕ · · · ⊕ Z/𝑝 𝑘 𝑝,𝑚( 𝑝) Z ⊕ Z 𝑝𝑝 , 𝑝

with nonnegative integers 𝑘 𝑝 , 𝑚( 𝑝) and positive integers 𝑘 𝑝,𝑖 . Each of the direct summands in (25.20) appears also as a factor group of 𝐴. In particular, if 𝑚( 𝑝) > 0, then Z/𝑝 𝑘 𝑝,1 Z is a factor of 𝐴. Hence, so is Z/𝑝Z. By Lemma 25.9.4, 𝑚( 𝑝) > 0 for infinitely many 𝑝 or there is a 𝑝 with 𝑘 𝑝 > 0. This together with the preceding arguments concludes the proof of the lemma. □ Remark 25.9.6 (Degree of algebraic extensions) Let 𝐿/𝐾 be an algebraic extension. Define [𝐿 : 𝐾] to be the least common multiple of all degrees [𝐸 : 𝐾] with 𝐸 ranging over all finite subextensions of 𝐿/𝐾. Call [𝐿 : 𝐾] the degree of 𝐿/𝐾. When 𝐿/𝐾 is separable, Galois correspondence implies that [𝐿 : 𝐾] = (Gal(𝐾) : Gal(𝐿)). Hence, we may rephrase Lemma 25.9.3 in field-theoretic terms: (25.21a) Let 𝐾 ⊆ 𝐿 ⊆ 𝑀 be fields with 𝑀/𝐾 separable algebraic. Then, [𝑀 : 𝐾] = [𝑀 : 𝐿] [𝐿 : 𝐾]. (25.21b) Let {𝐿 𝑖 | 𝑖 ∈ 𝐼} be a directed family of separable algebraic extensions of 𝐾. Thus, there is a partial ordering ≤ on 𝐼 that makes (𝐼, ≤) a directed partial ordered set in the sense of Section 1.1 such that 𝐿 𝑖 ⊆ 𝐿 𝑗 for 𝑖, 𝑗 ∈ 𝐼 Î with 𝑖 ≤ 𝑗. Put 𝐿 = 𝑖 ∈𝐼 𝐿 𝑖 . Then, [𝐿 : 𝐾] = lcm𝑖 ∈𝐼 [𝐿 𝑖 : 𝐾]. (25.21c) Suppose that {𝐿 𝑖 | 𝑖 ∈ 𝐼} is a linearly disjoint separable algebraic extension Î of 𝐾. Then, [𝐿 : 𝐾] = 𝑖 ∈𝐼 [𝐿 𝑖 : 𝐾]. Now suppose that 𝐿/𝐾 is an arbitrary algebraic extension. Let 𝐸 be a finite subextension. Denote the maximal separable extension of 𝐾 in 𝐸 (resp. L) by 𝐸 0 (resp. 𝐿 0 ). Then, 𝐸 is linearly disjoint from 𝐿 0 over 𝐸 0 , so [𝐸 : 𝐸 0 ] = [𝐿 0 𝐸 : 𝐿 0 ]. Thus, [𝐿 : 𝐾] = [𝐿 : 𝐿 0 ] [𝐿 0 : 𝐾]. This and similar considerations show that the separability assumption in (25.21a) is redundant.

25.10 The Sylow Theorems The concept of Sylow groups carries over to profinite groups: Definition 25.10.1 Let 𝑝 be a prime number. A closed subgroup 𝑃 of a profinite group 𝐺 is said to be a 𝑝-Sylow group of 𝐺 if 𝑃 is a pro-𝑝-group and 𝑝 ∤ (𝐺 : 𝑃).

546

25 Projective Groups and Frattini Covers

Proposition 25.10.2 Let 𝐺 be a profinite group and 𝑝 a prime number. Then: (a) 𝐺 has a 𝑝-Sylow group and every pro-𝑝 subgroup of 𝐺 is contained in a 𝑝-Sylow group of 𝐺. (b) The 𝑝-Sylow groups of 𝐺 are conjugate. (c) If 𝜃: 𝐺 → 𝐻 is an epimorphism of profinite groups and 𝑃 is a 𝑝-Sylow group of 𝐺, then 𝜃 (𝑃) is a 𝑝-Sylow group of 𝐻. (d) If 𝑁 is a closed normal subgroup of 𝐺 and 𝑃 a 𝑝-Sylow group of 𝐺, then 𝑁 ∩ 𝑃 is a Sylow group of 𝑁. Î (e) #𝐺 = 𝑝 #𝐺 𝑝 where 𝐺 𝑝 is a 𝑝-Sylow group of 𝐺 and 𝑝 ranges over all prime numbers. Proof of (a). Let N be the set of all open normal subgroups of 𝐺 and 𝐻 a pro-𝑝subgroup of 𝐺 (𝐻 may be trivial). For each 𝑁 ∈ N denote the set of all 𝑝-Sylow groups of 𝐺/𝑁 which contain 𝐻𝑁/𝑁 by P (𝑁). By Sylow’s theorem for finite groups, P (𝑁) is finite and nonempty. If 𝑀 ∈ N and 𝑀 ≤ 𝑁, then the quotient map 𝐺/𝑀 → 𝐺/𝑁 maps 𝑝-Sylow groups of 𝐺/𝑀 onto 𝑝-Sylow groups of 𝐺/𝑁. It therefore defines a canonical map of P (𝑀) into P (𝑁). By Corollary 1.1.4, the inverse limit of the sets P (𝑁) is nonempty. Thus, there is a (𝑃 𝑁 ) 𝑁 ∈N such that 𝑃 𝑁 ∈ P (𝑁) and 𝑃 𝑀 is mapped onto 𝑃 𝑁 for all 𝑀, 𝑁 ∈ N with 𝑀 ≤ 𝑁. The inverse limit 𝑃 = lim 𝑃 𝑁 is a pro-𝑝-subgroup of 𝐺 = lim 𝐺/𝑁 that contains 𝐻. If 𝑁 ∈ N , ←− ←− then 𝑃𝑁/𝑁 = 𝑃 𝑁 . Hence, 𝑝 ∤ (𝐺 : 𝑃𝑁). Therefore, 𝑝 ∤ (𝐺 : 𝑃). It follows that 𝑃 is a 𝑝-Sylow group of 𝐺 that contains 𝐻. Proof of (b). Let 𝑃 and 𝑃 ′ be 𝑝-Sylow groups of 𝐺. If 𝑁 ∈ N , then 𝑃𝑁/𝑁 and 𝑃 ′ 𝑁/𝑁 are 𝑝-Sylow groups of the finite group 𝐺/𝑁. Hence, they are conjugate. It follows from Lemma 1.2.2(e) that 𝑃 and 𝑃 ′ are conjugate. Proof of (c). The group 𝑄 = 𝜃 (𝑃) is a pro-𝑝-group and (𝐻 : 𝑄) = (𝐺 : 𝜃 −1 (𝑄)) divides (𝐺 : 𝑃), which is relatively prime to 𝑝. Hence, 𝑄 is a 𝑝-Sylow group of 𝐻. Proof of (d). Use (a) to choose a 𝑝-Sylow group 𝑄 of 𝑁 and a 𝑝-Sylow group 𝑃 ′ of 𝐺 which contains 𝑄. Then, (𝑁 ∩ 𝑃 ′ : 𝑄) divides (𝑃 ′ : 𝑄) (Lemma 25.9.3(a)), hence #𝑃 ′, so (𝑁 ∩ 𝑃 ′ : 𝑄) is a power of 𝑝. On the other hand, (𝑁 ∩ 𝑃 ′ : 𝑄) divides (𝑁 : 𝑄), so (𝑁 ∩ 𝑃 ′ : 𝑄) is relatively prime to 𝑝. It follows that (𝑁 ∩ 𝑃 ′ : 𝑄) = 1 and 𝑁 ∩ 𝑃 ′ = 𝑄. By (b), there exists a 𝑔 ∈ 𝐺 with 𝑃 𝑔 = 𝑃 ′. Since 𝑁 ⊳ 𝐺, we have (𝑁 ∩ 𝑃) 𝑔 = −1 𝑁 ∩ 𝑃 ′ = 𝑄, so 𝑁 ∩ 𝑃 = 𝑄 𝑔 is a 𝑝-Sylow group of 𝑁. Î Proof of (e). The formula #𝐺 = 𝑙 #𝐺 𝑙 follows from the formula #𝐺 = (𝐺 : 𝐺 𝑙 ) · #𝐺 𝑙 for each prime 𝑙 (Lemma 25.9.3(a)). □ Finite nilpotent groups are direct products of their unique 𝑝-Sylow groups as 𝑝 runs over all prime numbers. The next proposition is an immediate consequence of this and Lemma 25.1.2: Proposition 25.10.3 Each pronilpotent group 𝑁 is the direct product of its 𝑝-Sylow groups. In particular, if 𝐺 is a profinite group, Φ(𝐺) is a direct product of its 𝑝-Sylow groups and each of them is normal in 𝐺.

25.11 On Complements of Normal Subgroups

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Proof. For each prime number 𝑝 the group 𝑁 has a unique 𝑝-Sylow group, 𝑁 𝑝 , and it is characteristic in 𝑁, since this is the case for the finite quotients of 𝑁. The groups 𝑁 𝑝 generate 𝑁 and the intersection of 𝑁 𝑝 with Îthe group generated by all other Sylow groups of 𝑁 is trivial. It follows that 𝑁 = 𝑁 𝑝 . Since each 𝑝-Sylow group of Φ(𝐺) is characteristic, Lemma 25.1.2 gives the second part of the proposition. □ Proposition 25.10.4 Let 𝐴 Îbe an Abelian projective group. Then, there is a set 𝑆 of prime numbers with 𝐴  𝑝 ∈𝑆 Z 𝑝 . Î Proof. By Proposition 25.10.3, 𝐴  𝑝 𝐴 𝑝 , where 𝐴 𝑝 is the 𝑝-Sylow group of 𝐴. By Proposition 25.4.7, 𝐴 𝑝 is projective. Hence, by Corollary 25.8.13, 𝐴 𝑝 is either trivial or is isomorphic to Z 𝑝 . □ We conclude this section with some applications to fields: Corollary 25.10.5 Let 𝐾 be an algebraic Î extension of a finite field. Then, there exists a set 𝑆 of prime numbers with Gal(𝐾)  𝑝 ∈𝑆 Z 𝑝 . ˆ Hence, Gal(𝐾) is projective Proof. By Section 1.5, Gal(𝐾) is a subgroup of Z. Î (Corollary 25.4.6). It follows from Proposition 25.10.4 that Gal(𝐾)  𝑝 ∈𝑆 Z 𝑝 for □ some set 𝑆 of prime numbers. Theorem 25.10.6 ([Ax68], Thm. D) Let 𝐾 be a perfect PAC field with Gal(𝐾) Abelian. Then, 𝐾 is 𝐶1 . Proof. By Theorem 12.6.2, Gal(𝐾) is projective. Hence, by Proposition 25.10.4, □ Gal(𝐾) is procyclic. It follows from Theorem 24.3.6(c) that 𝐾 is 𝐶1 . And here is an application to Hilbertian fields: Theorem 25.10.7 ([Uch80], Thm. 3(ii)) Let 𝐾 ⊆ 𝐿 ⊆ 𝑁 be fields. Suppose that 𝐾 is Hilbertian, 𝑁 is a pronilpotent extension of 𝐾, and [𝐿 : 𝐾] is divisible by two distinct prime numbers. Then, 𝐿 is Hilbertian. Proof. By Proposition 25.10.3, Gal(𝑁/𝐾) is the direct product of its Sylow subgroups 𝐺 𝑝 . Choose a prime divisor 𝑞 of [𝐿 : 𝐾], let 𝑁𝑞 be the fixed field in 𝑁 of 𝐺 𝑞 , Î and 𝑁𝑞′ the fixed field in 𝑁 of 𝑝≠𝑞 𝐺 𝑝 . Then, 𝑁𝑞 𝑁𝑞′ = 𝑁, 𝐿 ̸ ⊆ 𝑁𝑞 , and 𝐿 ̸ ⊆ 𝑁𝑞′ . □ By the diamond theorem (Theorem 15.2.3), 𝐿 is Hilbertian.

25.11 On Complements of Normal Subgroups Let 𝑁 be a closed normal subgroup of a profinite group 𝐺. A closed subgroup 𝐻 of 𝐺 is called a complement to 𝑁 in 𝐺 if 𝑁 ∩ 𝐻 = 1 and 𝑁 𝐻 = 𝐺. The Schur– Zassenhaus theorem, whose finite version appears in the proof of Lemma 25.4.1, gives the existence of a complement, unique up to conjugation in 𝐺, under a simple index condition.

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Now suppose that 𝑁, 𝑀 ⊳ 𝐺, 𝐻 ≤ 𝐺, and 𝑀 ≤ 𝑁 ∩ 𝐻. We say that 𝐻 is a complement to 𝑁 in 𝐺/𝑀 if 𝐻/𝑀 is a complement to 𝑁/𝑀 in 𝐺/𝑀. In other words, 𝑁 ∩ 𝐻 = 𝑀 and 𝑁 𝐻 = 𝐺. Lemma 25.11.1 (Schur–Zassenhaus) Let 𝑁 be a closed normal subgroup of a profinite group 𝐺. Suppose that gcd(#𝑁, (𝐺 : 𝑁)) = 1. Then, 𝑁 has a complement in 𝐺. Furthermore, all complements to 𝑁 in 𝐺 are conjugate. Proof. First we prove the uniqueness part of the lemma: Let 𝐻 and 𝐻 ′ be two complements to 𝑁 in 𝐺. Let 𝑀 be an open normal subgroup of 𝐺. Then, (𝐻 𝑀 : 𝑀) = (𝐻 : 𝐻 ∩ 𝑀) = (𝐺 : (𝐻 ∩ 𝑀)𝑁)|(𝐺 : 𝑁) and (𝑁 𝑀 : 𝑀) = (𝑁 : 𝑀 ∩ 𝑁)|#𝑁. Hence, (𝐻 𝑀 : 𝑀) and (𝑁 𝑀 : 𝑀) are relatively prime. Therefore, 𝐻 𝑀 ∩ 𝑁 𝑀 = 𝑀. It follows that 𝐻 𝑀/𝑀 is a complement to 𝑁 𝑀/𝑀 in 𝐺/𝑀. Similarly, 𝐻 ′ 𝑀/𝑀 is a complement to 𝑁 𝑀/𝑀 in 𝐺/𝑀. By the uniqueness part of the finite group version of Schur–Zassenhaus [Hup67, p. 128, Hauptsatz 18.3], 𝐻 𝑀/𝑀 and 𝐻 ′ 𝑀/𝑀 are conjugate. Consequently, by Lemma 1.2.2(e), 𝐻 and 𝐻 ′ are conjugate. The proof of the existence of a complement is divided into two parts: Part A: 𝑁 is finite. Then, 𝐺 has an open normal subgroup 𝑀 with 𝑁 ∩ 𝑀 = 1. As in the proof of Lemma 25.4.1, apply Schur–Zassenhaus [Hup67, p. 126, Hauptsatz 18.1] to the finite group 𝐺/𝑀 to find a complement 𝐻 to 𝑁 𝑀 is 𝐺/𝑀. Then, 𝐻 is a complement to 𝑁 in 𝐺. Part B: 𝑁 is infinite. With 𝑈0 = 𝐺 let {𝑈 𝛼 | 𝛼 < 𝜆} beÑa well ordering of the open normal subgroups of 𝐺. For each 𝛽 ≤ 𝜆 let 𝑀𝛽 = 𝑁 ∩ 𝛼