Field Theory: Classical Foundations and Multiplicative Groups [1 ed.] 0824780299, 9780824780296

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Field Theory: Classical Foundations and Multiplicative Groups [1 ed.]
 0824780299, 9780824780296

Table of contents :
Preface
Contents
Chapter 1. Preliminaries
1. Notation and terminology
2. Polynomial algebras
3. Integral extensions
4. Tensor products
5. Module-theoretic prerequisites
6. Topological prerequisites
Chapter 2. Classical topics in field theory
1. Algebraic extensions
2. Normal extensions
3. Separable, purely inseparable and simple extensions
4. Galois extensions
5. Finite fields, roots of unity and cyclotomic extensions
6. Norms, traces and their applications
7. Discriminants and integral bases
8. Units in quadratic fields
9. Units in pure cubic fields
10. Finite Galois theory
11. Profinite groups
12. Infinite Galois theory
13. Witt vectors
14. Cyclic extensions
15. Kummer theory
16. Radical extensions and related results
17. Degrees of sums in a separable field extension
18. Galois cohomology
19. The Brauer group of a field
20. An interpretation of H_0^3(G, E*)
21. A cogalois theory for radical extensions
22. Abelian p-extensions over fields of characteristic p
23. Formally real fields
24. Transcendental extensions
Chapter 3. Valuation theory
1. Valuations
2. Valuation rings and places
3. Dedekind domains
4. Completion of a field
5. Extensions of valuations
6. Valuations of algebraic number fields
7. Ramification index and residue degree
8. Structure of complete discrete valued fields
A. Notation and terminology
B. The equal characteristic case
C. The unequal characteristic case
D. The inertia field
E. Cyclotomic extensions of p-adic fields
Chapter 4. Multiplicative groups of fields
1. Some general observations
2. Infinite abelian groups
3. The Dirichlet-Chevalley-Hasse Unit Theorem
4. The torsion subgroup
5. Global fields
6. Algebraically closed, real closed and the rational p-adic fields
7. Local fields
A. Preparatory results
B. The equal characteristic case
C. The unequal characteristic case
8. Extensions of algebraic number fields
9. Braridis's theorem
10. Fields with free multiplicative groups modulo torsion
11. A nonsplitting example
12. Embedding groups
13. Multiplicative groups under field extensions
14. Notes
Bibliography
Notation
Index

Citation preview

Field Theory

PURE AND APPLIED MATHEMATICS A Program of Monographs, Textbooks, and Lecture Notes

EXECUTIVE EDITORS

Earl J. Taft

Zuhair Nashed

Rutgers University New Brunswick, New Jersey

University of Deklware Newark, Delaware

CHAIRMEN OF THE EDITORIAL BOARD

S. Kobayashi

Edwin Hewitt

University of California, Berkeley Berkeley, California

University of Washington Seattle, Washington

EDITORIAL BOARD M. S. Baouendi Purdue University

Donald Passman University of Wisconsin-Madison

Jack K. Hale Brown University

Fred S. Roberts Rutgers University

Marvin Marcus University of California, Santa Barbara W. S. Massey Yale University Leopoldo Nachbin Centro Brasileiro de Pesquisas Fisicas and University of Rochester Anil Nerode Cornell University

Gian-Carlo Rota Massachusetts Institute of Technology David Russell University of Wisconsin-Madison Jane Cronin Scanlon Rutgers University Walter Schempp Universitat Siegen

Mark Teply University of Wisconsin-Milwaukee

MONOGRAPHS AND TEXTBOOKS IN PURE AND APPLIED MATHEMATICS I. 2. 3. 4. 5. 6. 7.

8. 9. I 0. 11. 12. 13. 14. I 5. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29.

K. Yano, Integral Formulas in Riemannian Geometry (1970) (out of print)

S. Kobayashi, Hyperbolic Manifolds and Holomorphic Mappings (1970) (out of print) V. S. Vladimirov, Equations of Mathematical Physics (A. Jeffrey, editor; A. Littlewood, translator) (1970) (out of print) B. N. Pshenichnyi, Necessary Conditions for an Extremum (L. Neustadt, translation editor; K. Makowski, translator) (1971) L. Narici, E. Beckenstein, and G. Bachman, Functional Analysis and Valuation Theory (1971) D. S. Passman, Infinite Group Rings (1971) L. Domhoff, Group Representation Theory (in two parts). Part A: Ordinary Representation Theory. Part B: Modular Representation Theory (1971, 1972) W. Boothby and G. L. Weiss (eds.}, Symmetric Spaces: Short Courses Presented at Washington University (1972) Y. Matsushima, Differentiable Manifolds (E.T. Kobayashi, translator) (1972) L. E. Ward, Jr., Topology: An Outline for a First Course (I 972) (out of print) A. Babakhanian, Cohomological Methods in Group Theory (1972) R. Gilmer, Multiplicative Ideal Theory (1972) J. Yeh, Stochastic Processes and the Wiener Integral (1973) (out of print) J. Barros-Neto, Introduction to the Theory of Distributions (1973) (out of print) R. Larsen, Functional Analysis: An Introduction (1973) (out of print) K. Yano and S. Ishihara, Tangent and Cotangent Bundles: Differential Geometry (1973) (out of print) C. Procesi, Rings with Polynomial Identities ( 197 3) R. Hermann, Geometry, Physics, and Systems (1973) N. R. Wallach, Harmonic Analysis on Homogeneous Spaces (1973) (out of print) J. DieudonnJ, Introduction to the Theory of Formal Groups (1973) l Vaisman, Cohomology and Differential Forms (1973) B.. -Y. Chen, Geometry of Sub manifolds (1973) M. Marcus, Finite Dimensional Multilinear Algebra (in two parts) (1973, 1975) R. Larsen, Banach Algebras: An Introduction (1973) R. 0. Kujala and A. L. Vitter (eds.), Value Distribution Theory: Part A; Part B: Deficit and Bez out Estimates by Wilhelm Stoll (1973) K. B. Stolarsky, Algebraic Numbers and Diophantine Approximation ( 1974) A. R. Magid, The Separable Galois Theory of Commutative Rings (1974) B. R. McDonald, Finite Rings with Identity (1974) J. Satake, Linear Algebra (S. Koh, T. A. Akiba, and S. Ihara, translators) (1975)

30. 31. 32. 33. 34. 3 5. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 54. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66.

J. S. Golan, Localization of Noncommutative Rings (1975) G. Klambauer, Mathematical Analysis (1975) M. K. Agoston, Algebraic Topology: A First Course (1976) K. R. Goodearl, Ring Theory: Nonsingular Rings and Modules (1976) L. E. Mansfield, Linear Algebra with Geometric Applications: Selected Topics (1976) N. J. Pullman, Matrix Theory and Its Applications (197 6) B. R. McDonald, Geometric Algebra Over Local Rings (1976) C. W. Groetsch, Generalized Inverses of Linear Operators: Representation and Approximation (1977) J.E. Kuczkowski and J. L. Gersting, Abstract Algebra: A First Look (1977) C. 0. Christenson and W. L. Voxman, Aspects of Topology (1977) M. Nagata, Field Theory (1977) R. L. Long, Algebraic Number Theory (1977) W. F. Pfeffer, Integrals and Measures (1977) R. L. Wheeden and A. Zygmund, Measure and Integral: An Introduction to Real Analysis (1977) J. H. Curtiss, Introduction to Functions of a Complex Variable (1978) K. Hrbacek and T. Jech, Introduction to Set Theory (1978) W. S. Massey, Homology and Cohomology Theory (1978) M. Marcus, Introduction to Modern Algebra (1978) E. C. Young, Vector and Tensor Analysis (1978) S. B. Nadler, Jr., Hyperspaces of Sets (1978) S. K. Segal, Topics in Group Rings (1978) A. C. M. van Rooij, Non-Archimedean Functional Analysis (1978) L. Corwin and R. Szczarba, Calculus in Vector Spaces (1979) C. Sadosky, Interpolation of Operators and Singular Integrals: An Introduction to Harmonic Analysis (1979) J. Cronin, Differential Equations: Introduction and Quantitative Theory (1980) C. W. Groetsch, Elements of Applicable Functional Analysis (1980) I. Vaisman, Foundations of Three-Dimensional Euclidean Geometry (I 980) H. I. Freedman, Deterministic Mathematical Models in Population Ecology (1980) S. B. Chae, Lebesgue Integration (1980) C. S. Rees, S. M. Shah, and C. V. Stanojevic, Theory and Applications of Fourier Analysis (1981) L. Nachbin, Introduction to Functional Analysis: Banach Spaces and Differential Calculus (R. M. Aron, translator) (1981) G. Orzech and M. Orzech, Plane Algebraic Curves: An Introduction Via Valuations (1981) R. Johnsonbaugh and W. E. Pfaffenberger, Foundations of Mathematical Analysis ( 1981) W. L. Voxman and R.H. Goetschel, Advanced Calculus: An Introduction to Modern Analysis (1981) L. J. Corwin and R. H. Szcarba, Multivariable Calculus (1982) V. I. Istratescu, Introduction to Linear Operator Theory (1981) R. D. Jarvinen, Finite and Infinite Dimensional Linear Spaces: A Comparative Study in Algebraic and Analytic Settings ( 1981)

67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. 100.

J. K. Beem and P. E. Ehrlich, Global Lorentzian Geometry ( 1981) D. L. Armacost, The Structure of Locally Compact Abelian Groups (1981) J. W. Brewer and M. K. Smith, eds., Emmy Noether: A Tribute to Her Life and Work (1981) K. H. Kim, Boolean Matrix Theory and Applications (1982) T. W. Wieting, The Mathematical Theory of Chromatic Plane Ornaments (1982) D. B. Gauld, Differential Topology: An Introduction (1982) R. L. Faber, Foundations of Euclidean and Non-Euclidean Geometry (1983) M. Carmeli, Statistical Theory and Random Matrices (1983) J. H. Carruth, J. A. Hildebrant, and R. J. Koch, The Theory of Topological Semigroups (1983) R. L. Faber, Differential Geometry and Relativity Theory: An Introduction (1983) S. Barnett, Polynomials and Linear Control Systems (1983) G. Karpilovsky, Commutative Group Algebras (1983) F. Van Oystaeyen and A. Verschoren, Relative Invariants of Rings: The Commutative Theory (1983) I. Vaisman, A First Course in Differential Geometry (1984) G. W. Swan, Applications of Optimal Control Theory in Biomedicine (1984) T. Petrie and J. D. Randall, Transformation Groups on Manifolds (1984) K. Goebel and S. Reich, Uniform Convexity, Hyperbolic Geometry, and Nonexpansive Mappings (1984) T. Albu and C. Nastasescu, Relative Finiteness in Module Theory (1984) K. Hrbacek and T. Jech, Introduction to Set Theory, Second Edition, Revised and Expanded (1984) F. Van Oystaeyen and A. Verschoren, Relative Invariants of Rings: The Noncommutative Theory (1984) B. R. McDonald, Linear Algebra Over Commutative Rings (1984) M. Namba, Geometry of Projective Algebraic Curves (1984) G. F. Webb, Theory of Nonlinear Age-Dependent Population Dynamics (1985) M. R. Bremner, R. V. Moody, and J. Patera, Tables of Dominant Weight Multiplicities for Representations of Simple Lie Algebras (1985) A. E. Fekete, Real Linear Algebra (1985) S. B. Chae, Holomorphy and Calculus in Normed Spaces (1985) A. J. Jerri, Introduction to Integral Equations with Applications (1985) G. Karpilovsky, Projective Representations of Finite Groups (1985) L. Narici and E. Beckenstein, Topological Vector Spaces (1985) J. Weeks, The Shape of Space: How to Visualize Surfaces and ThreeDimensional Manifolds (1985) P. R. Gribik and K. 0. K ortanek, Extremal Methods of Operations Research (1985) J.-A. Chao and W. A. Woyczynski, eds., Probability Theory and Harmonic Analysis (1986) G. D. Crown, M. H. Fenrick, and R. J. Valenza, Abstract Algebra (1986) J. H. Carruth, J. A. Hildebrant, and R. J. Koch, The Theory of Topological Semigroups, Volume 2 (1986)

101. R. S. Doran and V. A. Belfi, Characterizations of C*-Algebras: The Gelfand-Naimark Theorems (1986) 102. M. W. Jeter, Mathematical Programming: An Introduction to Optimization (1986) 103. M. Altman, A Unified Theory of Nonlinear Operator and Evolution Equations with Applications: A New Approach to Nonlinear Partial Differential Equations (1986) 104. A. Verschoren, Relative Invariants of Sheaves (1987) 105. R. A. Usmani, Applied Linear Algebra (1987) 106. P. Blass and J. Lang, Zariski Surfaces and Differential Equations in Characteristic p > 0 (198 7) 107. J. A. Reneke, R. E. Fennell, and R. B. Minton. Structured Hereditary Systems (1987) 108. H. Busemann and B. B. Phadke, Spaces with Distinguished Geodesics (1987) 109. R. Harte, Invertibility and Singularity for Bounded Linear Operators (1988) 110. G. S. Ladde, V. Lakshmikantham, and B. G. Zhang, Oscillation Theory of Differential Equations with Deviating Arguments (1987) 111. L. Dud kin, I. Rabinovich, and I. Vakhutinsky, Iterative Aggregation Theory: Mathematical Methods of Coordinating Detailed and Aggregate Problems in Large Control Systems ( 1987) 112. T. Okubo, Differential Geometry (1987) 113. D. L. Stancl and M. L. Stancl, Real Analysis with Point-Set Topology (1987) 114. T. C. Gard, Introduction to Stochastic Differential Equations (1988) 115. S. S. Abhyankar, Enumerative Combinatorics of Young Tableaux (1988) 116. H. Strade and R. Farnsteiner, Modular Lie Algebras and Their Representations ( 1988) 117. J. A. Huckaba, Commutative Rings with Zero Divisors ( 1988) 118. W. D. Wallis, Combinatorial Designs ( 1988) 119. W. Wi~sraw, Topological Fields (1988) 120. G. Karpilovsky, Field Theory: Classical Foundations and Multiplicative Groups ( 1988) 121. S. Caenepeel and F. Van Oystaeyen, Brauer Groups and the Cohomo!ogy of Graded Rings (1988) 122. W. Kozlowski, Modular Function Spaces ( 1988)

Other Volumes in Preparation

Field Theory Classical Foundations and Multiplicative Groups Gregory Karpilovsky University of the Witwatersrand Johannesburg, South Africa

Marcel Dekker, Inc.

New York • Basel

Library of Congress Cataloging-in-Publication Data Karpilovsky, Gregory Field theory: classical foundations and multiplicative groups/ Gregory Karpilovsky. p. cm. -- (Monographs and textbooks in pure and applied mathematics ; vol. 120) Bibliography: p. Includes index. ISBN 0-8247-8029-9 1. Class field theory. 2. Groups, Theory of. I. Title I I. Series. QA247.K324 1988 512'.32--dcl9 88-18914

Copyright© 1988 by MARCEL DEKKER, INC.

All Rights Reserved

Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage and retrieval system, without permission in writing from the publisher. MARCEL DEKKER, INC. 270 Madison Avenue, New York, New York 10016 Current printing (last digit): 10 9 8 7 6 5 4 3

2

1

PRINTED IN THE UNITED STATES OF AMERICA

For Helen, Suzann~, Elliott

Preface

The twin aims of this book are to provide a young algebraist with a reasonably comprehensive summary of the material on which research in field theory is based and to show how this material is used in the current research pertaining to multiplicative groups of fields.

The treatment is by no means exhaustive - this would

have been a Sisyphean task, since the subject has become so extensive and is growing almost from day to day. The book can be roughly divided into two parts, which will not preclude, however, some strong interrelations between these.

The first part (Chapters 1-3)

introduces classical foundations of the theory of fields, with an emphasis on refinements and extensions achieved in scope of relatively recent developments. Among these, we exhibit canonical fundamental units of certain classes of pure cubic fields, prove Kneser's theorem on torsion groups of separable field extensions and establish a theorem of Schinzel which provides necessary and sufficient conditions for the Galois group of a binomial

to be abelian.

xn- a

A separate

section is devoted to a result of Isaacs concerning the degrees of sums in a separable field extension. E/F,

If G is the Galois group of a finite Galois extension

an interpretation of a distinguished subgroup of

using the methods of Eilenberg and Maclane. pertaining to radical extensions.

H 3 (G,E*)

is provided by

We also include some recent results

Among them, we examine a cogalois correspon-

dence discovered by Greither and Harrison in 1986. The second part of the book (Chapter 4) is devoted exclusively to the study of multiplicative groups of fields.

After proving some preliminary results, we in-

vestigate the isomorphism class of F*, local or global.

where

F

is a distinguished field such as

Concentrating on field extensions V

E/F,

we then prove Brandi.s's

Preface

vi theorem which asserts that if F is infinite and Et F, finitely generated.

then

E*/F*

is not

The rest of the chapter is based on works of May, who made

fundamental contributions to the subject.

Numerous examples are given to illus-

trate that the results obtained are the best possible.

Special attention is

drawn to the study of fields whose multiplicative groups are free modulo torsion. These results have a number of important applications in the study of unit groups of group algebras and the isomorphism problem.

For this reason a complete ac-

count of May's contributions to the topic is given. This monograph is written on the assumption that the reader has had the equivalent of a standard first-year graduate algebra course.

Thus we assume a famil-

iarity with basic ring-theoretic and group-theoretic concepts and an understanding of elementary properties of modules, tensor products and fields.

Apart from a

few specific results and the general knowledge we have presupposed, the book is entirely self-contained.

We have included, for the convenience of the reader, a

chapter on algebraic preliminaries. topics needed later in the book. too technical.

This chapter provides a brief survey of We have tried to avoid making the discussion

With this view in mind, maximum generality has not been achieved

in those places where this would entail a loss of clarity or a lot of technicalities. A word about notation.

As is customary, Theorem 2.1.2 denotes the second

result of Section l of Chapter 2; however, for simplicity, all references to this result within Chapter 2 are designated as Theorem 1.2.

A systematic description

of the material is supplied by the introductions to individual chapters and therefore will not be repeated here. I would like to express my gratitude to my wife for the encouragement she has given me in the preparation of this book.

For answering specific queries on

topics contained in the text I am indebted to

W. May.

Finally, my thanks go to

Lucy Rich for her excellent typing. Gregory Karpilovsky

Co~e~s

PREFACE CHAPTER l.

V

PRELIMINARIES l.

2. 3. 4. 5. 6. CHAPTER 2.

CLASSICAL TOPICS IN FIELD THEORY l. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24.

CHAPTER 3.

Notation and terminology Polynomial algebras Integral extensions Tensor products Module-theoretic prerequisites Topological prerequisites

Algebraic extensions Normal extensions Separable, purely inseparable and simple extensions Galois extensions Finite fields, roots of unity and cyclotomic extensions Norms, traces and their applications Discriminants and integral bases Units in quadratic fields Units in pure cubic fields Finite Galois theory Profinite groups Infinite Galois theory Witt vectors Cyclic extensions Kummer theory Radical extensions and related results Degrees of sums in a separable field extension Galois cohomology The Brauer group of a field An interpretation of H~(G,E*) A cogalois theory for radical extensions Abelian p-extensions over fields of characteristic p Formally real fields Transcendental extensions

VALUATION THEORY l. 2. 3. 4. 5. 6.

l

7 29 32 37 41 47 47 57 63 79 85 103 117 135 148 167 172 184 192 205 214 221 242 247 263 282 303 323 333 348 353

Valuations Valuation rings and places Dedekind domains Completion of a field Extensions of valuations Valuations of algebraic number fields vii

353 368 376 389 400 411

Contents

viii 7. · Ramification index and residue degree 8. Structure of complete discrete valued fields A. Notation and terminology B. The equal characteristic case C. The unequal characteristic case D. The inertia field E. Cyclotomic ex~ensions of p-adic fields CHAPTER 4.

MULTIPLICATIVE GROUPS OF FIELDS 1. 2. 3. 4. 5. 6. 7.

8. 9. 10. 11. 12. 13. 14.

Some general observations Infinite abelian groups The Dirichlet-Chevalley-Hasse Unit Theorem The torsion subgroup Global fields Algebraically closed, real closed and the rational p-adic fields Local fields A. Preparatory results B. The equal characteristic case C. The unequal characteristic case Extensions of algebraic number fields Braridi s's theorem Fields with free multiplicative groups modulo torsion A nonsplitting example Embedding groups Multiplicative groups under field extensions Notes

414 421 421 422 427 431 436 439 439 443 449 461 463 468 474 475 481 482 487 496 501 517 519 525 531

BIBLIOGRAPHY

535

NOTATION

541

INDEX

547

Field Theory

1 Preliminaries The aim of this chapter is twofold.

First, to establish various notational

conventions for the rest of the book.

Second, to provide some preliminary re-

sults pertaining to polynomial rings.

Because we presuppose a familiarity with

various elementary group-theoretic and ring-theoretic terms, only a brief description of them is presented.

Many readers may wish to glance briefly at the con-

tents of this chapter, referring back to the relevant sections when they are needed later. 1. NOTATION AND TERMINOLOGY The aim of this section is to establish various notational conventions that we shall use throughout the book. Let A and B be arbitrary sets.

A map

f:A-.B

is a function that associates with each element a e A a unique element be B, f : a

i--+

b.

This b,

the image of a under

f: A be maps.

B and

The composite map A-. B -

f,

g: B -

is denoted by b

Let

= f(a).

C

c is denoted by gof·

Recall that

gof acts according to the rule (gof)(a) = g[f(a)]

If f: A-. B is any map, then set A'

of A.

flA'

forall

denotes the restriction of f

aEA

to a sub-

The inclusion map i : A'-->- A is defined by i(a) = a

for all

a e A'

If A,B are sets, we write Ac B if A is a su~set of B and Ac B if A is a proper subset of B.

If A~ B, B-A denotes as usual the set of elements

2

of

CHAPTER 1

B

not contatned i'n

A,

The cardinality of the set

is denoted by

A

IAI

The di'agram

f A----!•~B

th C signifies that h : C __.. B

case

are maps,

f = hog,

c

A,B and

are sets and that

f; A -

B,g:A -

c

and

The diagram commutes, or is said to be commutative, in

An arbitrary diagram is commutative if we get the same composite

maps whenever we fo 11 ow directed arrows a 1ong different paths from one set to another set in the diagram, Unless explicitely stated otherwise, all groups are assumed to be multiplicative.

We use the same symbol 1 for both the identity element and the identity

subgroup of a group

by

x

If

F*. X.

G,

is a subset of

G,

A cyclic group of order

n

wi 11 denote the subgroup of is denoted by

G.

while Let

If every element of

If

'1l. •

n

least common multiple of the orders of the elements of of

F is denoted by

The multiplicative group of a field

G is of finite order,

G

G

G

generated

ts finite, the

is called the exponent

G is called a torsion group,

G is torsion-free tf all Hs elements, except for 1, are of infinite order,

H be a subgroup of

from each 1eft co set

G.

A subset of

G containing just one element

is ca 11 ed a left transversal for

xH

H

in

G,

and right

transversals are defi'ned correspondingly, Let

{Gi}iEI. be a family of groups and let

The subgroup (£)1-'EIGi gi

of

n1-'EIGi

niEIGi

consisting of all

be their direct product

(g)

with finitely many

distinct from 1, is called the direct sum (or restricted direct product) of

the groups

Gi,

A sequence of groups and homomorphisms fi-1 fi __.. Gi-1 ~ Gi Gi+1 '''

is said to be exact at

G.

1-

if

NOTATION AND TERMINOLOGY

the sequence is called exact if it is exact at every group.

; Kerf.'2, = Imf. '!,- 1

particular, G

L

1 .- G

L

His exact if and only if

is exact if and only if

H .- 1

3

is injective, while

f

is surjective,

f

In

If we are given a 3-

terms exact sequence 1---+G

g

f

1

--a 2 --a 3

.-1

also called a short exact sequence, we see that G ~ f(G·) 1

N be a normal subgroup of

Let

morphism.

and

1

G

ea!-

3

G and let

G /f(G ) 2

l

n: G--+ G/N

be the natural homo-

Then the short exact sequence 7f

1--+N-->-G---->-G/N-->-l

is also called the natural exact sequence,

A is an abe 1 ian group.

Suppose now that finite order in

i's a subgroup of

A

A,

an

n > 0,

the map

A--->- A

We shall say that

n-divisible if

An = A.

p-divisible.

(i}

sending every element

is divisible if

An

An= A

of A

a

and

A [nl ,

for all

n

E

IN

to

and

Note that a p-group is divisible if and only if it is

Typical examples of divisible groups are:

The multiplicative group

(ii)

A

A.

if no confusion can arise.

A,

is a homomorphism whose image and kernel are denoted by

respectively.

of e 1ements of

called the maximal torsion subgroup of

It will also be called the torsi'on subgroup of For any integer

t(A)

Then the set

The group

z

00

of the

F*

pnth

of an algebraically closed field complex roots of 1 with

n

F

running over all

p

natural numbers. (iii}

The additive group ([)/'11. which is isomorphic to the group of all complex

roots of 1. Given a prime elements of

7f

we denote by

A

p

the p-component of A

A whose order is a non-negative power of p.

torsion, i.e. if A=EBpE

p,

·t(A) = A,

then there is a set

n

consisting of all

Note that if

A is

of prime numbers such that

A. p

A11 rings in this book are associative with 1 f O and subri ngs of a ring

R

CHAPTER l

4

are assumed to have the same identtty element as

Each ring homomorphism will

R.

be assumed to respect identity elements. Let

be a ring.

R

The mapping 'lL

defined by n

-->- R

homomorphism whose image is ca 11 ed the prime subring of R; ideal

m'lL for a unique m;;;, 0,

1-+

is a ring

n•l

its kernel is an

called the characteristic of R and denoted by

charR Let

be a ring.

R

positive integer n. potent, while

r

=

0,

An element x ER An ideal

is nilpotent if xn

=

in R is nil if every element of

J

is nilpotent if there is a pas iti ve integer n

J

where Jn

is the product of J with itself n

of R is idempotent if e 2

=

O for some

Two idempotents

e.

J

is nil-

such that

times.

An element e

are orthogonal if

u,v

A nonzero idempotent is primitive if it cannot be written as a sum

uv =vu= 0.

of two nonzero orthogonal idempotents. An element u of a ring R is said to be a unit if uv VER.

The set of all units of R,

denoted by U(R),

=

vu

= 1

for some

constitutes a group

called the unit group of R. Given an arbitrary ring R,

we denote by J(R)

the Jacobson radical of R,

i.e. the intersection of all maximal left ideals of R. simple if J(R)

Let

niE?i·

{Ri}iEI

=

0. be a family of rings and let R be the direct product set

We can define addition and multiplication on (x.) + (y.) ~· ~

It is straightforward to verify that direct product of the family

onto R.~

We say that R is semi-

R

R

(x.+y.)

=

~

by the rules (x.)(y.) ~ ~

~

is a ring; we shall refer to For all

(R.LiEI. ~

i EI,

is a ring homomorphism, the inj ecti ans R~. -

=

(x.y.) ~ ~ R

as the

the projection of

R

R preserve addition and

multiplication but not land so are not ring homomorphisms. Let R be a commutative ring.

Then an R-algebra is a ring A which is at

the same time an R-module such that r(xy)

=

(rx)y

=

for all

x(ry)

It follows directly from the definition that rx

=

(r•l)x

rER and x,yEA

for r ER, x E A,

and

5

NOTATION AND TERMINOLOGY that the map sending r of

to r•l

is a homomorphism of R into the centre

Conversely, if f: R -

A.

A is a homomorphism of R into

Z(A)

then

Z(A),

can be regarded as an R-algebra by setting

A

for all

rx = f(r)x

r ER, XE

A

We close this section by presenting some information pertaining to commutative In what follows

rings.

We say that

denotes a commutative ring.

R

R

is

noetherian if every ascending chain of ideals breaks off, or equivalently, if

every ideal is finitely generated.

A ring R is called artinian if every descenThe set N(R)

ding chain of ideals breaks off.

of all nilpotent elements of R

constitutes an ideal called the nilradical of R.

N(R)

= 0.

Note that N(R) '.:: J(R)

is a nil ideal.

We say that R is reduced if

N(R)

and thus

J(R)

=

if and only if J(R)

We say that R is local (respectively, semilocal) if it has

precisely one maximal ideal (respectively, if it has only finitely many maximal ideals).

An element x

y ER; in case

of R is a zero divisor if

x -f- 0 and

xy =

0 for some nonzero

is a zero divisor, we say that x

x

is a proper zero

divisor.

An integral domain is a commutative ring without proper zero divisors. ideal

P of a commutative ring

R is prime if R/P

P is primary if every zero divisor in

R/P

R5

r/s

and if, for any x,y

associated with a multiplicative set with r ER and

s

Es.

is an integral domain, while

is nilpotent.

Let S be a subset of a commutative ring R. cative if it contains

An

in

We say that s s, xy

Es.

is muUipZi-

The quotient ring

s consists of all elements of the form

By definition,

r/s = r'/s'

if there exists

Es such that

s l

s (s'r-sr')

=

0

l

Observe that if OE s,

then 0/1

we shall assume that Off. s. (r/s)(r'/s')

The mapping 1/Js : R -

Rs

=

is the only element of Rs;

for this reason

Multiplication and addition in Rs are defined by rr'/ss' ,

defined by

r/s+r'/s' = (s'r+sr')/ss' ¢ 5 (r) = r/1

is a homomorphism, called

CHAPTER 1

6

canonical;

1/Js

is injective if and only if S contains no zero divisors.

Suppose now that

A

cative subset of A.

is a commutative R-algebra and that s is a multipliThen the ring As may be endowed with an R-algebra struc-

ture by setting r(a/s)

=

r ER, a EA, s ES

ra/s

in which case 1/Js becomes a homomorphism of R-algebras.

The R-algebra As en-

joys the following easily verified universal property. 1.1. PROPOSITION.

Let

homomorphism f: A -

B

be a commutative R-algebra.

Then, for any R-algebra

B such that every element of f(S)

exists one and only one R-algebra homomorphism h : As -

is a unit, there B which renders com-

mutative the following diagram:

As

h f

A

.,B

Especially important is the case where s is the complement of a prime ideal Here one often writes RP instead of Rs. associate the expanded ideal

Ie

With every ideal

I

P.

in R we

of Rs given by

Ie = {i/sli EI, s Es}

The following property is easily verified. 1.2. PROPOSITION.

For any prime ideal

P of

R the correspondence

Ir->- Ie

is a bijection between the prime ideals of R contained in P and the prime ideals of

RP"

It follows from the above that Pe is a local ring.

is a unique maximal ideal of RP; hence RP

This ring RP is also called the local ring of R at P,

the process of forming

RP

is called localization.

and

POLYNOMIAL ALGEBRAS Finally, assume that R is an integral domain.

7

Then S

{O}

= R -

is a

multiplicative subset of R and Rs is obviously a field containing an isomorphic copy of R.

We shall refer to R8 as the quotient field of R.

We close by recording the following elementary fact. 1.3. PROPOSITION. Proof.

Let R be an artinian integral domain.

Fix O ,J:

and observe that the descending chain

x E R

Rx .::> Rx 2

xn {1-rx)

= 0.

Si nee

R

::J •••

for some n ~ 1,

so x n

= rx

has no proper zero dtvi sors,

1 - rx

= O and thus r

Hence Rxn = Rxn+i

must terminate.

Then R is a field.

n+1

or

ts a unit. • 2. POLYNOMIAL ALGEBRAS A monoid is a set G,

with an assoctatiVe binary operation, and having an iden-

tity element 1. Let RG of

R be a commutative ring and let G be a monoid. G over

The monoid algebra

R is the free R-module on the elements of

cation induced by that in

with multipli-

G,

More explicitly, RG consists of all formal

G.

linear combinations (x

Ex •g

g

g

with finitely many

x

g

,J:

Ex •g = Ey •g if and only if g g (ii) Ex •g + Z:1,1 •g = E(.x +y ) •g g ·g ' g g

(tv}

(Exg•g)(Eyh•h) = Ezt•t r(Exg •g).

=

I:(rxg ) •g

R, g

E

G)

O subject to x

(i)

(iii)

E

where for all

=

g

z

t

r

y

= E

for all

g

E

X

gh=t g

g

E G

yh

R

It is straightforward to verify that these operations define RG as an associati've R-algebra with R

and

G,

1 = 1R-1G,

respectively.

where

1R

and 1G are identity elements of

With the aid of the injective homomorphisms

CHAPTER l

8

we sha 11 in the future identify

and

R

with their images in

G

RG:

With

these identifications, the formal sums and products become ordinary sums and products. Let x

For this reason, from now on we drop the dot in xg·g. Then the support of x,

= i::x g ERG. g

Suppx

= {g E

G!xg

written Suppx,

is defined by

f O}

It is plain that Suppx is a finite subset of G that is empty if and only if X

= 0.

Let A be an R-algebra, let G be a monoid and let

2. l. PROPOSITION.

be any map satisfying iJJ(l) Then the map ijJ*

RG ->- A

=

and ijJ(xy)

1

Proof.

In particular, if

R-modules.

as a basts, then

Because

RG

for all

iJJ(x)ijJ(y)

RG ~

1jJ

is injective and A is

A.

is R-free with

G as a basis,

ijJ*

ts a homomorphism of

Let and

x = i:x g g

be two elements of

RG.

ijJ*(xy}

x,y E G

defined by

is a homomorphism of R-algebras. R-free with iJJ(G)

=

y = i:y g g

Then ¢*( Z

x ybab) =

a,bEG a

): X

ijJ(a)

aEG a

i::x y01)!(a)ijJ(b) a

[ y01jJ(b)

bEG

ijJ*(x)ijJ*(y)' as asserted. • Assume that G is a free abelian group freely generated by the set {X

.Ji E

1,

I}

Then each element of G can be uniquely written in the form

9

POLYNOMIAL ALGEBRAS n.

(n.'2,

nx.'2,

iEI

with only finitely many n.'2, ! 0.

> 0 for any

(1) in which

n.'2,

generated by

{X .. '2,

li

,z,

::l)

E

( 1)

The submonoid of G consisting of all elements is called a free commutative monoid freely

i EI

The corresponding monoid algebra over a commutative

EI}.

ring R is denoted by R[(Xi)iEil: nomial ring in the indeterminates

We shall refer to R[(Xi\Eil with coefficients in R.

X.'2,

as the poly-

For In partic-

ular, the polynomial ring in the indeterminate x with coefficients in R will n. be denoted by R[X]. > 0 and The elements of R[(Xi)iEI] (n. . '2,

niE~/

n. = 0 '2,

for all but finitely many i

definition,

R[(Xi)iEI]

are called monomials.

EI)

Thus by

is R-free freely generated by all monomials.

The degree

of the monomial (1) is defined by n.

deg ( II X •,z,} iEI

If f

is a nonzero polynomial in R[(Xi}iEil,

written

degf,

of f.

Thus

f = 0,

Let

~ n. iEI ,z,

=

,z,

degf = 0

if and only if f -

is a nonzero element of R.

We shall say that f

If

00

be a nonnegative integer and let

R[{Xi)iEI].

f,

to be the maximum of the degrees of the monomials in the support

then we say that its degree is d

then we define the degree of

be a nonzero polynomial in

f

is homogeneous of degree d

(or is a form of

degree d) if a11 monomi a1s in Sup pf a re of degree d. Let f(X)

be a nonzero polynomial tn R[Xl f(X) = a

with a.ER and with an f 0. ,z,

o

of degree n.

+ a X + ... +a 1 n

We call

a

n

r

the leading coefficient of

f

is manic if a n = 1. Let A be an R-algebra and let B be any subset of A. Then the elements its constant term.

We also say that

Then

f

(rV ER, v(b) > O) with finitely many r

V

and v(b)

This subalgebra, denoted by R[B],

distinct from zero form a subalgebra of (or simply R[a] if B

the smallest subalgebra containing B.

= {a})

A.

is obviously

For this reason, we refer to R[B]

as

CHAPTER 1

10

the subalgebra of A generated by B.

If A= R[BJ,

then we say that A is

gener>ated by B.

Let A be an R-algebra and let B be any subset of A.

We say that B is

algebr>aicaUy independent over R (or that the elements of B are algebraically

independent over R} tf the elements

(v(b) > O) are R-linearly independent,

In the special case where

B =

we see that a

{a},

i's algebraically independent over R if and only if {1,a,a 2 , . . . ,an, ... } is an R-linearly independent set.

If a

is algebraically independent over

EA

then we also say that a ts tmnscendental over R.

R,

In the case where a is

not transcendental over R, we say that a ts algebr>aic over R.

Thus a is

algebraic over

R

f{X)

such that f(a}

=

if and only if there exists a nonzero polynomial

E

R[XJ

0.

We next provide some elementary observations pertaining to polynomial rings. 2.2. PROPOSITION.

Let

Any map {x.'Z, Ji

( i)

of R-algebras.

A

be any commutative R-algebra.

E I} -+ A,

x.'Z,

In particular, if

can be extended to a homomorphism

a.'Z,

I---'---+

is generated by fo.li 'Z,

A

EI},

then

is a

A

h.omomorphic image of R[(Xi)iEIJ.

1

(ii}

The above homomorphism is tnjecti've if and only if the elements a.'Z, are

algebraically independent over (iti) A

A ~ R[(X)iEil

R

if and only if there is a generating set {ai Ii

which is algebraically independent over

only if

A

Proof.

In particular,

R.

A.~

E I}

R[XJ if and

ts generated by a transcendental element. (i)

i Let G be the monoid generated by the x., 'Z, n.

,jJ{nx/) iEI

it follows that w(1)

=

1 and w(xy)

,z,

=

=

n.'Z,

na.

iEI

Setting

(n.'Z, > O)

,z,

w(x)w(y)

EI.

for a11

x,y E

G,

(2) Now apply

Proposition 2.1 (ii}

for

Direct consequence of (2} and the definition of algebraic independence.

POLYNOMIAL (iii)

ALGEBRAS

Direct consequence of (i} and (ii).

2.3. PROPOSITION.

Let I=JUK

11



be a disjoint union and let S

R[(X.)

=

'EJ].

J J

Then as R-algebras. and, in particular, R [X

Proof.

1

, ...

,X ]

n

2e

and each

xk, k EK,

elements

xi,i EI

n

, ...

n-

X.

n

x1,., i

>--->-

1,

for all

l ][X ]

of R-algebras.

E I

n;;,

2

extends to a homo-

Since its image contains

the given homomorphism is surjective.

Now apply Proposition 2,2(ti).

s

Moreover, the

Let

f(X)



and g(X)

be nonzero polynomials in R[XJ

and bm be the leading coefficients of

at least one of a ,b n

m

f(X)

and g(X),

is not a zero divisor in R, deg (fg)

and the leading coefficient of fg Proof.

,X

of s[(Xk)kEKl are obviously algebraically independent over

2.4. PROPOSITION. a

1

By Proposition 2.2(t), the map

morphism R[(Xi)iEI]--->- S[(Xk)kEK]

R.

R [X

=

and let

respectively.

If

then

degf + degg

is ab . nm

Write +b

x1

m

Then

and the result follows. 2.5. COROLLARY. R[(Xi)iEI]

Proof.



Let R be an integral domain.

Then, for any set I,

is also an integral domain. We may harmlessly assume that I

Proposition 2.3, it suffices to show that R[XJ

is finite.

Furthermore, by

is an integral domain.

latter being a consequence of Proposition 2.4, the result follows. 2.6. PROPOSITION.

~~t

f(X}

and g(X)

The



be two nonzero polynomials in R[XJ

of

CHAPTER l

12

degrees m and n,

respectively.

leading coefficient of g(X).

Put k

max(m-n+l,O)

=

and denote by a

the

g(X) and r(X)

Then there exist polynomials

such that

ak f(X) where either r(X) in

R,

Proof.

0 or degr(X) < n.

=

then q(X)

q(X)g(X)

=

and r(X)

If m < n,

Moreover, if a is not a zero divisor

then

and we may take q(X)

k = 0

in which case k

first part, we argue by induction on m.

the leading coefficient of f(X). and r I (X).

=

I

m-n + 1.

=

has degree at most m- 1,

=

f(X).

We

To prove the

where b is

Applying induction hypothesis, there exist

such that

a(m-I)-n+i(_af(.X) -b";m-ng(X)) x where either r (x)

r(X)

= 0,

The case m = n - 1 being trivial,

Then af(X) - bI'-n g(X)

polynomials q I (X) .

r(X)

are uniquely determined.

may therefore assume that m;;. n-1

assume m ;;. n.

+

O or degr (X) < n. I

q 1 ( X)g () X + r

=

1

(X)

This proves the first assertion, by

taking

Assume that a is not a zero divisor and that

ak f(X) where either r'(X)

=

q 1 (X)g(X) + r'(X)

0 or degr'(X) < n. (q(X} - q '(X) )g(X)

If q(X} - q'(X) f 0,

and so r'(X)

=

2.7. COROLLARY.

r'(X) - r(X)

then the left side has degree at least n,

leading coefficient of g(X) since r'(X)-r(X)

Then

is not a zero divisor.

However, this is impossible

is either zero or has degree less than n.

r(X).

Hence q(X)

=

q'(X)



Let f(X)

and g(X)

let the leading coefficient of g(X) polynomials q(X), r(X)

since the

E

be two nonzero polynomials in R[Xl and be a unit of R.

R[Xl such that f(X)

=

g(X)q(X)

+

r(X)

Then there exist unique

13

POLYNOMIAL ALGEBRAS

where either r(Xl Proof.

If f(X)

0.

=

Let f(X)

be a nonzero polynomial in R[X]. x-r

R, f(r) = 0 if and only if

E

Proof. f(r)

< degg(X).

Direct consequence of Proposition 2.6. •

2.8. COROLLARY. given r

0 or degr(X)

=

=

is a divisior of f(X).

for some g(X)

(X-r)g(X)

Then, for any

E

R[X],

then obviously

Conversely, assume that f(r) = 0 and put g(X)

=

x- r.

Since g(X)

is a monic polynomial of degree l, it follows from Proposition 2.6 that f(X) = q(X)g(X) + r

for some r

1

as required.

0 = f{r) = r , 1

E R 1

and some q(X)

But then

E R[XJ.



If X is an indeterminate, an element r ER such that f(r)

0 will be

called a root of f(X) . 2.9. COROLLARY.

Let

If a , ... ,a

(i}

be an integral domain and let f(X)

are distinct roots of f(X)

m

1

R

(x-a )(x-a ) ... (X-a ) 1

(ii)

If f(X)

t

0,

m

2 ·

in R, divides

then the number of roots of f(X)

E

R[Xl.

then f(X)

in R does not exceed

degf(X).

Proof.

We first note that (ii} is a consequence of (i) from considerations of To prove (i), we argue by induction on m.

degree.

The case m =

1

consequence of Corollary 2.8, assume that the statement is true for m -

being a 1

roots.

Then

and therefore domain,

f(am) = (a -a ) ... (a -a 1 )q(a ) . m 1 m ~ m q(a) = 0 so that X-a divides q(X), m m

completes the proof.

Since R is an integral by Corollary 2.8.

This



Recall that a principal ideal domain

(PID)

is an integral domain in which all

ideals are principal. 2.10. COROLLARY. Proof. I

Let

F

be a field.

Then

F[X]

We know, from Corollary 2.5, that F[X]

be a nonzero ideal of F[X] and let g

is a principal ideal domain. is an integral domain.

be a nonzero element of I

Let

of smallest

CHAPTER 1

14

If f f O is an element of I,

possible degree. r

f = qg +

r· = f

-

for some q,r E F[X],

qg EI,

Let

R

where degr




degh(X) >

1,

1.

f(X)

=

g(X)h(X)

with

By hypothesis

Since R/P is an integral domain, by looking at the image of f(X)

in

(RIP)

[XI,

we deduce that g(X) for some g(x)

and

1,;;; k

h(X)

= J(mod

< n

and some

belong P,

P[XJ) a,S

and h(X) E

R-P.

hence r n

E p2 ,

2.23. COROLLARY. (Eisenstein's Criterion).

= Si1-k(mod

P[XJ)

This implies that the constant terms of a contradiction. • Let

R

be a

UFD,

let

F

be the

25

POLYNOMIAL ALGEBRAS quotient field of R and let

f(X)

=

r

r

o

+r

1

Assume that there is a prime p of

r-

1

R

such that

+ ... + r

E

n

R[X]

(n;;,,

and r.1, = O(mod p), Then f(X} Proof.

Apply Theorems 2.22 and 2.2l(it}. • Let

be a nonzero square-free integer I±

k

r-k

the polynomial

n;;,, 1,

prime divisor

p

2.25. EXAMPLE.

of Let

k p

with p 2

is irreducible over a).

Jk

is irreductble over

=

XP

-1

+

•.v-2 A"

(I).

divisible by p.

Indeed, take any •

Then the polynomial + .•. + X + 1

Indeed, it suffices to prove that f{X+l)

To this end, note that the binomial coefficients

Q).

Then, for any

1.

and apply Corollary 2.23.

be a prime number.

f(X}

over

i..; n

is irreducible in F[XJ.

2.24. EXAMPLE. integer

1-.;·

1)

(~),

is irreducible

1 ..;

t ..;

p-1

are

Since

(P) f( ·X+l)=(X+l)p-!.=xp·l+(P}xp-2 {x+ 1) - l 1. + • • . + p-1

the required assertion follows from Corollary 2.23. 2.26. EXAMPLE.

Let

F

taining F such that a fi'e 1d of

F [a I.



be a field and let a be an element of some field conis transcendental over F.

Then, for any integer n

r- a

Denote by K the quotient

the polynomi a1

~ 1,

is irreducible in K[Xl

This is a direct consequence of Coro 11 ary 2. 23 and the fact that the ring ts a

UFD

such that a is a prime element of

2.27. PROPOSITION. (Reduction Criterion}. domain

R,

let

F

Let

R. P

R[X]-+ R[Xl

[al



be a prime ideal of an integral

and F be the quotient fields of

tively, and let

R = F

R

and R =

R/P,

respec-

CHAPTER l

26

If O I f

be the natural map. irreducible in F[X l,

E

then f

is such that degf = deg]'

RIX]

and ]' is

cannot be decomposed in R [X l as a product of

two nonconstant factors. Proof.

S-uppose f(X} = g(X}h(X)

wi'th g(X} ,h(X) e R[X].

Since degg·..; degg and degh..; degh,

our hypothesis implies that we must have

equality in these degree relations.

F[Xl we conclude that Let

R::, s

Hence from the irreducibility of ]' in

or h is an element of R,

g

Then ]' = gh .

as required.



be commutative rings and let a 1 , ... ,an be algebraically inde-

pendent elements of

s over

Let

R.

be a variable over R[a 1 , ... ,anl.

X

Then the polynomial f(X} = (X-a} ... (X-a 1

n

l over R[a 1 , ..• ,an I

can be written in the form •.n

f(X)

X

. .n-1

- 8 X 1

+ .. , + (·1}

is a polynomial in

where each s.1, = s.{a , ... ,al 1,·1 n

Ci

1

n

S

n

, •••

,a. n

In fact

s=a+a+ ..• +a 1

9

s

The polynomials s ,, .. ,s 1

a, ... ,a. 1

n

2

n

n

1

n

2

= a 1 a 2 + a 1 a s + , .• + an-1an

= a a

1 2

a

n

are called the elementary symmetrie polynomials

It is clear thats. is homogeneous of degree i 1,

Let sn be the symmetric group of degree n.

If

cr

esn

in a, ... ,a. n

1

and

f(a , ••. ,a } e R[a , ... ,a ] , 1 n 1 n

define /5 by /5(a 1 , ... ,al= f(a cr ( 1 ), •.. ,acr (n )) n

We say that the polynomial f

is symmetrie if

/5 = f

for all

O E S •

n

It is

27

POLYNOMIAL ALGEBRAS obvious that the set of symmetric polynomials is a subring of Ria ,.,.,a J n

1

containing

and the elementary symmetric polynomials

R

The result

s , ... ,s. 1 n

below shows that it contains nothing else. Let x , ... ,x

n

1

be variables.

We define the -weight of a monomial t

ti

x to be

t

+ 2t

1

n

I

The -weight of an arbitrary polynomial

+ .•. + nt. n

2

xn g(x, ... ,x) I n

is defined as the maximum of the weights of the monomials occuring in 2.28. THEOREM.

Let

Rs s

be commutative rings and let a, ... ,a 1

cally independent elements of s over R. symmetric of degree d,

E

n

1

be algebrai-

n R[a , ... ,a] n

I

is

then there exists a polynomial g(X , ... ,X}

E

n

1

of weight .a;;

If f(a , ... ,a)

g,

R[X , ... ,X] n

1

such that

d

f(a , •.. ,al= g(s , ... ,s) 1 n I n

Furthermore, the elementary symmetric polynomials s , ... ,s l

of a , ... ,a

n

n

I

are

algebraically independent over R. Proof.

We argue by induction on n,

that the result is true for n -

The case n

variables.

1

being trivial, assume

= 1

Substituting

a

n

in

= 0

we find

I' -

(X-a) ... (X-a · n- 1 Jx =, 1

s

I'- 1 +

1,0

··· +

where s.. is the expression obtained by substituting ,z,,O

(-lt-\n- 1 , 0 X a

n

= 0

in

s .• 'Z,

that s , ... ,s , are all elementary symmetric polynomials in 1 ,O n- •,O The case To prove the first assertion, we now proceed by induction on d. being trivial, assume d degree




Note

1,

.a;; d

0)

of degree d,

such that

=g I

(s

l

,o

, ... ,s _ 10 } n '

we may find a poly-

28

CHAPTER 1

Observe that

g (s , •.. ,s 1

1

n- 1

has degree


l

and some

s '. ~. ,s 1

in

n

We have equations ss. = Z'.\. 1,,

(:\ .. E

,S.

1,,J J

1,,J

Let us bring a11 the terms in (2 l to the 1eft-hand side.

R, 1

< i,j < n)

(2)

The theory of 1i near

equations applies, and if L is the determinant s-:\

-:\ 12

11

-:\ 21

s-:\

-'.\ 22

-:\ n2

we find that

L =

integral over R. 3.2. COROLLARY.

0.

-:\ 1n I1

s-:\

2n

nn

Expanding L gives us an equation that shows that

s

is



If R ~ s,

then the elements of s integral over R form a

subri ng of s. Proof.

Assume that u,v ES are integral over R.

We must show that both

30

Cl:JAPTER l and

u + v

are also integral over

uv

for some finitely generated R-submodules consisting of all finite sums R-submodule of

s.

Since

virtue of Lemma 3.1.

Zmini

By Lemma 3,1,

R.

s.

of

M,N

, mi EM, ni EN

and

(u+v)MN::: MN

and

vN

Then the product

MN

uM:::: M

is a finitely generated

(uv)MN::: MN,

the result follows by

s.

Then the subring of S consisting of elements

integral over R is ca 11 ed the integral closure of R ; n S.

We say that R

is integraZly closed in S

if R is the integral closure of R in S.

3.3. PROPOSITION.

be a subring of

Let

R

is integral over

u- 1

Proof.

If u

N



Let R be a subrtng of

Then

c

-1

s and let

if and only if

R

ts integral over R,

u- 1

u

be a unit of

s.

E R[ul.

then

u-n + r u-(n-1) + •.. + r 1

0

n

(r. ER)

(3)

1,,

Multiply (3) through by un and rearrange to get n-1

u(r +r u+ ... +r u 1 2 n

whence

u

-1

The argument is reversible.

E R[u].

3.4. COROLLARY.

=

Proof.

U(R}

uE R,

For u

Proposition 3.3, 3.6. PROPOSITION.

In particular,

we have

R[u]

Now apply Proposition 3.7.

= R.

Let R be an integral domain contained in a field

is integral over R, Proof.

is integral over R.

if S ts integral over R.

Since

3.5. COROLLARY.



and let u ER be a unit of s.

Let R be a subring of S

Then u- 1 ER if and only if u- 1 Rn u(s}

) = -1

u

F.

(ii)

s s

IO -1

tn R we have that

E R[u]

Let

R

= R,

u

as required.

be a subring of

s.

- I

is integral over R.

Then the following conditions

is a finitely generated ring over R and is integral over R (i)"" (ii):

By



is a finitely generated R-module

Proof.

lf

then R ts a field.

are equivalent: (i)



This is a direct consequence of Lemma 3.1

F

INTEGRAL EXTENSIONS (ii)• (i]:

31

Assume that S ts generated as a rtng over R by u 1 , ••• ,uk and

suppose that the equation showi'ng ui to be integral has degree ni.

Then the

elements (O,.; 1'.,.; n.-1) 1,

1,

span s over R. • 3. 7. PROPOSITION.

s.

Let R c s be rings and u an element of a ring containing

Suppose that u is integral over s and that s is integral over R.

Then u is integral over R. Proof.

By hypothesis, un + a un~i + ... +a = 0 1 n

for some n R

1

Let R

~ 1,

1

=

R[a , ... ,a ,u]. 1 n

is a finitely generated R•module.

(a. E S) 1,

Then one immediately verifies that

Now apply Lemma 3,1. •

An integral domain is said to be integraUy closed tf its integrally closed in tts quottent field. 3.8. PROPOSITION,

Let

s

be a multiplicative subset of an integral domain

R.

If R is integrally closed, then so ts R8 . Proof.

Suppose that the element u in the quotient field of

gral over Rs; we have to show that u

Rs·

E

is inte-

By hypothesis,

. un + (a /s 1 )un-.1 +, .. +(an I s] = O · 1 n

Put s = s 1 s 2 ... s n and t.1, = s/s1,,.

R8

(a.1, E R,s.1, ES)

Then

n n-1 su + t 1 a 1 u + ... + tnan = O

Multi'plying the above equality by

gives an equation asserting that su

integral over R.

Hence su ER and u E Rs,

as required.

3.9. PROPOSITION.

Let R be a GCD domain.

Then R is integrally closed.

Proof.

Let F be the quotient field of R. An +

Write

\=alb

with a,b ER.

'I'

1

An-i + ... + r

n

Suppose \ = 0

is



E

F is such that (4)

Then there exist s,t ER such that a= (a,b)s,

32 b

CHAPTER l (a,b}t,

whence Hence

t

and

Substl'tuting

(.,,t)=l.

divides

Since

sn

(sn,t) =

:\ER and the result follows.

1,

A=s/t

in(4),weobtain

it follows that t

is a unit of R.



4. TENSOR PRODUCTS Let

R

be a commutative ring and u,v,w any R-modules.

A map

f:uxv-w is said to be bilinear if f f(r

u

11

+r

(l )

is linear in each argument, i.e.

u , v)

22

f(u,r 1v1 +r 2v) 2

=

r

= r

l

f(u 1 ,v)

+ r

l

f(u,v) 1

+ r

2

f( u2 ,v)

2

f(u,v) 2

Our aim is to construct an R-module M and a bilinear map :\: u xv--+ M which is universal for all bilinear mapping (1), in the sense that to any bilinear map f

as in (l) there exists a unique homomorphism

r : M->- w

which renders

commutative the following diagram:

U xV--A_ _,..._M f*

w An R-module

M

with these properties is called a tensor product of u and

and is denoted by

u ® v or simply u ® v.

v

If it exists it is clearly unique

R

up to isomorphism, and we shall speak of the tensor product. The following standard procedure illustrates the existence of M. a free R-module with

u xv as a basis and let

all elements of the form

N

Let F be

be the submodule generated by

33

TENSOR PRODUCTS (r u +r u ,v} - r (u ,v} - r (u ,v) 1122

11

22

(u,r 1v1 +r 2v) - r 1 (u,v) -r (u,v) 2 l 2 2 for u,u ,u Eu, v,v ,v Ev and r ,r ER. 2

1

factor module

2

1

Then u 0 v is defined as the R

under the natural homomorphism

Thus the R-module u 0

is denoted by u 0 v.

F--+ F/N

(u,v}

The image of

F/N.

2

consists of all finite

V

R

sums l:u.'1,

0

v.'1,

(u . '1,

V)

E U, V • E '1,

and the elements u 0 v satisfy the relations

The map

=

r 1(u1 0 v)- + r 2(u2 0 v)

u 0 (r 11 v +r 2v) 2

=

r (u 0 v) + r (u 0 v) 1

l

\ : u x V->- u ® V defined by A(u,v)

2

f : u x v-

---+

(2)

2

0 v is therefore bilinear.

= u

R

Furthermore if f' : F

(r u +r u) 0 v '1122

w is any map, then f

determines a homomorphism

w given by

(u,v)) f '(Zr . U,V If we assume that

l:rU,V f'(u,v)

=

is bilinear, then NS Kerf'

f

f*

and so the map

U®V->-W R

defined by

f'(u

0

vl

f(u,v)

is a required homomorphism. Let v and

4.1. PROPOSITION. respectively.

Then

every element

x

V® R

w

be free R-modules with bases

{v.}

and

{w.}, J

'1,

w ts a free R-module with basis

{v.

,z,

0 w.}. J

Moreover,

E v 0 w can be written uniquely as the finite sum R ( V • E V, X. E

x=i:v.®x.

i

'1,

'1,

'1,

'1,

W)

and as X =

l: j

Given

Proof.

suitable r.,r\ ER, '1,

J

v

Ev and w E w,

X 1•

J

0

(x \ E V, W. E W)

W,

J

J

we have

v

=

i:r.v. '1,

'1,

J

and w = l:r\w. for

where both sums are, of course, finite.

J J

Therefore, by (2),

CHAPTER l

34

we have v © w =

L

i ,j

r.r'.(v. 0 w.) 'I,

J

v © w is generated by

Thus the R-module

J

'I,

Suppose now that

{v. © w.}.

R

J

'I,

(r.,

Lr. , ( v . © w •) = 0 1,J .

'I,

J

and fix some subscripts, say i = k, j

= s.

there exists an R- linear map

Similarly, there exists

µ

and µ8 (w)

1.

Then the map

x

is bilinear.

Since

{v.} 'I,

Ak : V->- R such that

Ak(vk) = 1.

f

: V

S

:

Hence there exists an R-linear map

is a basis for

µ

8

(w.) J

= 0 for

defined by f(v,w)

f*:

V© w->- R

R)

v,

i f: k

0 for

:,\(vi)

with

W ->-,R

W-->- R

E

1,J

and

j ,f- s

\(v)µ 8 (w) with

Finally,

and

{v. ©w.}

is a basis for

J

'I,



w.

The remaining assertions follow easily by grouping terms. then

x

isuniquelygivenasthefinitesum x=

Thus if x

Zx .. (v.©w.) 'I, J

i ,j 1,J

EV©

w,

and hence by

(2)' X

= LV. ©

i

'I,

(fa ..w.) j 1,J J .

= L(fa ..v.) ©w,

j i 1,J

as required.



Let B and for all

b

E

B, c

C be corrmuting subalgebras 0f an algebra E

with b.'I, EB, c.'I, E A= BC

R

w

Then the set

C.

c,

A is generated by

v and

w

i.e.

A,

let be= cb

BC consisting of all finite sums

is the smallest subalgebra containing

then we say that

Now assume that V©

J

'I,.

B and

are R-algebras.

B

and

c.

Lb .c. 'I,

If

c.

Because they are R-modules,

exists, and we assert that this R-module is also an R-algebra.

'I,

TENSOR PRODUCTS

w be R-algebras.

Let v and

4.2. PROPOSITION.

35

Then V® w is also an RR

algebra with multiplication given by (v

®

1

w ) (v

Furthermore, the maps v of V and w,

w )

i---->-



2

1

2

and w

1

1 and



(w

1l)

1

2

l

(v

®w

--+ 1

V E 1' 2

V w '

1l)

1'

2

E

W)

are R-a l gebra homomorphisms

which are injective if V and W are

w are commuting s uba l gebras that generate

1 ®

w. Let F be a free R-module with V x

Proof. plication in

F

,w )(v 2,w) 2

11

Define multi-

V ®

R

,w w)

= (v v

1212

Furthermore, the submodule

is obviously an R-algebra.

definition of

w as a basis.

distributively using

F

(v

Then

= (v V ) ®

2

respectively, into V ® W,

R-free, and the images v@

®

1

w is an ideal of

Therefore

F.

(v,w)

In particular, since

algebra structure of F.

N

V ® W = F/N

F

used in the

inherits the

maps to v ®w this says

that (v

®

1

w ) (v ® w 1

2

2

l

= (v v ) ® (w 12

w )

12

Finally, by the foregoing and (2) it is clear that the maps v w ,-;- 1 ®

1l)

are R-algebra homomorphisms of

Moreover, if V and are injective.

V

and w,

i-;-

v ® 1

and

respectively, into

V ®

w.

w are R-free then by Proposition 4.1, these homomorphisms

The remaining assertion being obvious, the result follows.

We shall refer to the homomorphisms in Proposition 4. 2 as canonicaZ.

• The next

result provides a universal characterization of tensor product of algebras. 4.3. PROPOSITION.

Let A ,A 1

R-algebra homomorphism, subalgebras of B.

i

=

2

and B be R-algebras and let ~-:A._,. B be an ~

1,2,

such that ~(A} 1

l

~

and ~(A}, are commuting 2

2

Then there exists one and only one R-algebra homomorphism h: A ®A ---,.3 1 R 2

which renders commutative the following diagram:

CHAPTER l

36

B i

= 1, 2

(f.1, is the canontcal homomorphism) Proof. the map h

The map

A 1

x

A

1

R

ism of R-modules.

2

iJJ

12

defined by h(a

A ® A -- B

~

(a ,a )

----->- B, 2

Furthermore, if

x = a

1

1

(a )iJJ (a )

® a ) = iJJ 2

® a

and

2

is bilinear and so

1122 1

(a )iJJ (a ) 1

2

is a homomorph-

2

y =a'® a', I

2

then

h(xy) = h(a a' @a a')= ijJ (a }iJJ (a'liJJ (a )1/J (a') 11

= [ 1/J =

22

11112222

(a )iJJ (a ) l [iJJ (a 1 ) iJJ ( a 1 )]

1122

1122

h(x)h(y)

Since h(f.(a.)) = 1/J.(a.), a.EA., 1, - 1, 1, 1, 1, 1,

so that h is a homomorphism of R-algebrgs. the above diagram ts commutattve,

Finally, the uniqueness of

quence of the fact any R-algebra homomorphism mined by its restriction to

{a

1

A

® A --. B

1 R

2

®a Ja. EA., i = 1,2}. 2

1,

h

ts a conse-

is uniquely deter-



1,

There ts a simple criterion for an algebra over a field to be a tensor product which is often useful.

Let A be an algebra over a field

subspaces of A,

u and

Then

V are said to be linearly disjoint over

for any linearly independent elements u. in 1,

u.v. 1,

J

in A are linearly independent over F.

natural mapping ~U®V--+A ( U

is an injective F-homomorphism.

F and let

0

V

1-----4 UV

u,v be F

if

u and v. in v, the elements J

Clearly this just means that the

MODULE-THEORETIC PREREQUISITES 4.4. PROPOSITION. B,C of A,

Let

tf (i)

be an algebra over a field

A

B and

37

F.

Given subalgebras

c are linearly disjoint over F,

(ii)

A= BC

and (iii} B and c commute elementwise, then A~ B ® c · and B n c F

By Proposition 4.3, the map B

Proof.

morphism of F-algebras. A~ B ® F

c.

Ae

c

A , b ® c ..- be

-->-

F.

is a homo-

Thts is i"njective by (i) and surjective by (ii}, hence

Assume that A e B n c.

independent, hence by (i'), Thus

®

=

1,A,A,A 2

If Ai F,

then 1,A are F-linearly

are F-linearly independent, which is absurd.

and the result follows. •

F

4.5. PROPOSITION.

Let A be a finite-dimensional algebra over a field F and

let B,C be subalgebras which commute elementwise. ly disjoint over

Then B and C are linear-

if and only if

F

dtmBC = (dimB)(dimC} F

Proof.

F

F

The image BC of the natural map B

® C-->- A, b ® c ~ be

has the

F

same dimension as

B ® C

if and only if the map is injective, as required.



F

5. MODULE-THEORETIC PREREQUISITES Let R be an arbi'trary ring and let M be a (left} R-module.

We say that M

is noetherian (arttnian} tf every ascendtng (descending) chain of submodules of M stops. ~

module

The ring R is said to be (left)_ noetherian (artinian} if the regular is noethertan (arttnian}; in other words,

R

is a noethertan

(arti'ntan} ri'ng if and only if every ascending (descending} chain of left ideals of

stops.

R

5. l . PROPOSITION.

Let

M

be an R-modul e.

Then the fo 11 owing con di ti ons are

equivalents (i} (ii) (iii"}

M

ts noetherian

Every submodule of

M

is finitely generated

Every nonempty collection of submodules of

M

contains a maximal element,

that is, a submodule which is not properly contained in any other submodule in the collection.

CHAPTER l

38

M

Assume that (it] is fa1se, and let

(i] • (ii):

Proof.

which ts not finitely generated.

Then, by induction, we can construct an

inftnite sequence of elements a 1 ,a: 2 , . . . then for every

so (i] •

in

such that if

N

is properly contained in

k, Mk

be a submodule of

N

Mk= Ra/ ••• +Rak'

This contradicts (i) and

Mk+l.

(it] ts true.

Assume that (tttl does not hold.

(ii} • (Hi}:

Then there exists a nonempty

collection X of submodules which has no maximal element. find submodules tained tn

tn x such that for every

M ,M , • . • 1

2

The union

Mk+i·

not necessarily a member of

w=

is properly con-

k, Mk.

00

u Mk is clearly a submodule of M although k=l

We show that w

X.

By induction, we can

is not finitely generated.

Assume the contrary, say

w for some s. the modules

Then each m. ~

= Rm

E Mk

i

we deduce that

Mk.'

l

+ ... +Rm s

for some

w = Mk.

~

~

and, chaos i ng the largest of

for some index

and hence that

ki

~

But this contradicts the way the submodules

Mk .+1 =Mk.· ~

k •,

were constructed,

Mk

~

so the implication is true. (tit]• (il:

Deny the statement.

Then there exists an infinite properly ascen-

ding chain of submodules, and ft ts clear that this collection of submodules can have no maximal element.

The proof i's therefore complete. •

Let u be a submodule of an R-module v.

5.2. PROPOSITION.

noetherian ff and only if both u and

V/U

Then v is

are noetherian.

Furthermore, if

R

is noetherian and if v is finitely generated, then v is a noetherian R-module. Proof.

If

V

i·s noetherian, then, clearly, so are u and

Convers-

V/U.

ely, assume that u and V/V are noetherian, and let

be an ascending chain .of R-submodules of v. {(An+U)/u}

Then both sequences

eventually stabilize, and hence for some integer A

n

u=

A

An

=3.

At

we infer from the modular law that

n

Accordingly, since

t

n

u and

A

n

+

u=

t

At

+

u

{A

n

nu}

and

we have for all n

~

t

MODULE-THEORETIC PREREQUISITES

39

An = An n (An +U} = An n (At+ U} = At =

Hence the sequence

At

stabilizes, and v is noetherian.

{A}

n

Finally, assume that a homomorphic image of

+ (An n U} = At + (At n U)

R

is noetl'lerian.

Then every one-generator R-module is

and hence is noethertan.

~

It therefore follows from

the preceding, by induction on the number of generators, that every finitely generated R-module ts also noetherian. • A submodule W of V ts said to be maximai, if W 1' v and there is no proper submodule of

V

strictly containing

w.

A ftnitely generated module over

any ring has an important maximaltty property, which is a consequence of Zorn's lemma. 5.3. PROPOSITION.

Let

be any ring and v a finitely generated R-module.

R

Then every proper submodule of Proof.

Write

Jf = Rv

1

ts contai'ned in a maximal submodule.

V

+ .•. + Rv . n

Given a proper submodule w of v,

w.

consider the set X of proper submodules of V which contain tnductive: module.

it contains w and if Assume uwA

=

v.

This is

cw,) ts a chain in x~ then uwA is a sub-

Then each vi EWA, for some Ai.



Let

be the

'1,

largest of the modules

WA , ••• 1

n

which contradicts the definition of X to be inductive.

Then w11 contains each vi and so v

,P.'). .

x.

=

Therefore uwA is proper and this shows

We can therefore apply Zorn's lemma and deduce that x has

a maximal element; thi's ts a proper submodule of v containing

w,

as we had to

construct. • 5.4. COROLLARY.

Let

R

be any ring.

contained in a maximal left ideal of

Then any proper left ideal of R.

In particular,

R

R

is

has maximal left

ideals. Proof.

The module

~

is generated by the single element 1.

Now apply

Proposition 5.3. • The most important source of noetherian rings is the following classical result:

wif

CHAPTER 1

40

5.5. THEOREM. (Htlbert basts theorem). then so is R[X , ... ,x J Proof.

By induction on n,

Let I

R[X].

for any integer

n

I

If

is a commutattve noethertan rtng,

R

n;;. 1.

it suffices to consider the polynomial ring

be an ideal in R[X] and denote by An

coefficients of polynomials of degree
->- X

X

of G x G into G is continuous

>->- xy - l

of G into

G

A group structure and a topology on a set

is continuous

G are said to be compatible if they

satisfy (i) and (ii) above. Let

R

be a ring.

We say that

is a topological ring if

R

R

is also a

topological space such that: The mapping

(i)

(ii}

(x,y) ....-

The mapping

(Hi}

Let

x ,--,. -x

The mapping

of Ax

y

of A

are homeomorphisms. of sets gX, x

g

E

into

A

is continuous

A

into

A

then the mappings

G,

is

continuous

n be the neighbourhood filter of the iden-

G be a topological group and let

If

A

into A is continuous of A x

(x,y} ,--,. xy

tity element l of G.

gn

x +

x

.-+

gx

and x

It follows that the neighbourhood filter of g

En and also the family

ng

of sets

Xg,

x En.

-+-

xg

is the family We close

this section by quoting the following standard properties (see Bourbaki, 1966}). 6.1. PROPOSITION.

Let G be a topological group and let n be the neighbour-

hood filter of 1.

Then

(i)

Given any x

En, there exists

(_ii l Given any x (iii).

For all

E

n, we have x- 1

g E G

and all

x

y

En such that

E

n

En, we have

gXg

-1

Y•Y

c x

En

Conversely, if G is a group and n a filter on G satisfying (i) - (iii), then there is a unique topology on

G compatible with the group structure of

G,

for which n is the neighbourhood filter of 1. 6.2. PROPOSITION. are equivalent: (i)

G is Hausdorff

Let G be a topological group.

Then the following conditions

46

CHAPTER l

(ii}

The set

(iii}

is closed

{1}

The intersection of the neighbourhoods of 1 consists only of the point 1.

6.3. PROPOSITION.

A subgroup of a topological group is open if and only if it

has an interior point. Let

X ., 1,

Every open subgroup is closed.

be an arbitrary collection of topological spaces.

i EI,

definition, the product topology on

n

the form

U.

with

U.

~I1,

1[ X.

iEI

has a basis consisting of the sets of

1,

an open subset of

1,

for all but finitely many i 6.4. PROPOSITION.

X.

for every i

and u.1,

EI,

1,

E

Then, by

=

x.1,

I.

A subspace of a Hausdorff space is Hausdorff.

of Hausdorff spaces is Hausdorff.

Every product

Conversely, if a product of nonempty spaces is

Hausdorff, then each factor ts a Hausdorff space. A collection

of subsets of a topological space x is said to have the

Y

finite intersection property if every finite subcollection

n s

~

0.

Y

1

of

Y

satisfies

For the proof of the following results we refer to Pontryagin (1977).

SEY 1

A topological space x ts compact if and only if every col-

6.5. PROPOSITION.

lection of closed subsets of

X

possessing the finite intersect ton property has

nonempty intersection. 6.6. PROPOSITION. (ii) (iii) (iv)

(i}

A closed subset of a compact topological space is compact.

A compact subset of a Hausdorff topological space is closed. A continuous image of a compact topological space is compact. Any continuous btjection x _,. Y of a compact topological space x onto

a Hausdorff space 6.7. PROPOSITION.

Y

is a homeomorphism, Let G be a compact totally disconnected topological group

and 1et u be an arbitrary neighbourhood of the identity.

Then there is an open

normal subgroup N of G such that N ~ u. 6.8. PROPOSITION. (Tychonoff's Theorem). of compact topological spaces is compact.

The product of an arbitrary collection

2 Classical topics in field theory

Our aim here is to introduce classical foundations of the theory of fields, with an emphasis on refinements and extensions achieved in scope of relatively recent developments.

Among these, we exhibit canonical fundamental units of certain

classes of pure cubic fields, prove Kneser 1 s theorem on torsion groups of separable field extensions and establish a theorem of Schinzel which provides necessary and sufficient conditions for the Galois group of a binomial abelian.

to be

A separate section is devoted to a result of Isaacs concerning the

degrees of sums in a separable field extension. finite Galois extension E/F, H3 (G,E*l

r -a

If G is the Galois group of a

an interpretation of a distinguished subgroup of

is provided by using the methods of Eilenberg and MacLane.

We also

include some recent results pertaining to radical extensions, especially a cogalois correspondence discovered by Greither and Harrison(l986}. 1. ALGEBRAIC EXTENSIONS Let E be a field and F a subfield; we write E/F for the field E considered as an extension of F. over F;

We can view E as an F-algebra and hence a vector space

its dimension is either a positive integer or an infinite cardinal and

is cal led the degree of

E

degree of E over F.

We say that E/F is a finite or an infinite extens1,"on

accordingly as

(E:F}

over

F.

In what follows we write

for the

is finite or infinite.

If S is a subset of E, we write F(S}

(respectively,

smallest subfield (respectively, subring) containing s and S =

(E:F}

F[Sl}

for the

In case

F.

{s ,s , ••• ,s }, we also write F(s , ••• ,s ) and [F s , ••• ,s 1 instead of 1.2

F(Sl and F[SJ,

n

1

respectively.

n

1

It is clear that F(s 1 , . . .,sn} 47

n

is the quotient

CHAPTER 2

48

field of F[e 1 , ... ,en I and that F[e 1 , ... ,sn J consists of all polynomials in s , ... ,s 1

F by 8

over F.

n

s , ... ,s 1

, , , , ,B

1

n

We shall refer to

F(s , ... ,s)

as the field generated over

n

1

or the field obtained by adjoining to F the elements

n

,

The field extension E/F is said to be finitely generated if E

=

F(s , ... ,s) n

1

for suitable ftnitely many elements a , ... ,s

of E. The extension E/F is 1 n called sirrrple if there is an element a in E, called a primitive element over

F,

such that

F(a).

E =

When we have a chain, also called a tower of fields: FCECK

we can regard K as an extension of E or F. E/F, K/F and K/E;

their degrees are related by the important product formula:

1.1. PROPOSITION.

Let F c E c K be a chain of fields.

bases for E/F and K/E, for K/F.

Thus we have three extensions

In particular,

respectively, then AB= {abla

(K:F}

If A and B are E

A,b

is a basis

EB}

(E:F}(K:E}.

=

m

Proof. some

is an element of

If " J

~

1.

dent over F. j=l J J = l:l,

have e. J

µ.= Z:a ..a, J i=l ~J ~

Hence A= E

i=l

n

Let

m

E e.b.

for some m ~ 1 and

(a .. EF,a.EA) ~

~J

m

E a ..a.b. which shows that

j=l ~J ~ J

K

is spanned by

It remains to show that the elements of AB are linearly indepen-

AB over F.

Then

E 'J.i .b. j=1 J J

n n

for some n

=

Furthermore, we have

µ . E E ,b . E B. J

then "

K,

E

i=l =

O, e. EE, J

1..: j ..:m.

E A..a.b.

j=l ~J ~ J

=

0 with

A •• E F. ~J

We set e . J

=

E A. .a ..

i= 1 ~J

~

and since the b. are E-1 i nearly independent, we must J

The latter in turn implies that the A •• are zero, since

the a.~ are F·linearly independent. 1.2. COROLLARY.

n

m

~J

This completes the proof.

Let F c E c K be a chain of fields.



Then K/F is finite if

and only if both E/F and K/E are finite. Proof.

Direct consequence of Proposition 1.1.

Let E/F be a field extension.



An element a of E is said to be algebraic

over F if a is a root of a nonzero polynomial over F.

If f(X)

is a nonzero

49

ALGEBRAIC EXTENSIONS

polynomial over then

F w1'th

g(X) = A-lf(X}

braic over

:,._

is the leading coefficient of

is a monic polynomial with

if and only if

F

and tf

f(ci,1 = 0

ci,

g(ci,}

=

is integral over F.

0.

Thus

ci,

f(X),

is alge-

We say that the extension

E/F is algebraic if all elements of E are algebraic over F. 1.3. PROPOSITION. F

E/F be a field extensi'on, let a

Let

and let

f(X} e F[Xl

f(X)

is irreducible

(il ( i il (iii)

F(a) = F[al

F(a).

g(X)

Proof.

(i)

F [X] with

E

and if n = degf(X)

In particular,

f(a} = 0.

be of least degree such that

di Vides each

f(X)

be algebraic over

EE

Then

g(a) = 0

then

l ,a, ... ,an-I

is an F-basis for

(F(a} :F} = n.

f(X) = f (X)f (X), f. (X) e F [X ].

Suppose that

1

2

Then

1,,

f (.a}f (.a)= 0, 1

and since

2

f (a)= 0 or f (a}= 0.

is a field, we have

F

f (al= 0.

iteness that

Since

1

1

Assume for defin-

2

degf (X}..; degf(X),

we must have

1

degf (X) = degf(X) 1

degf 2 (X} .

which forces (ii)

Let

= 0.

f(X)

This shows that

g(X) e F[Xl be such that g(a) = 0.

is irreducible.

Owing to Corollary 1.2.7, we

have g(X} = q(X)f(X) + r(X} where either r(X) = 0 r(X} = 0,

degr(X) < degf(X).

or

artinian.

Since

F[a]

of Proposition 1.1.3. linear span of with

A.1,, e F

over

F

r(a) = 0,

F

F[X]}

we must have

and not all

A.1,,

}.

Furthermore, if equal to

0,

which is impossible.

then

and hence

F(a) = F[al

by virtue

belongs to the F-

It ts clear that each element of F[a] n-·1

- E be

f(X)

has a root a

l

in F,

f(X)

so there exists g(X) e F[Xl = (X-a

1

)g(X)

If degg > l,

we can repeat this argument inductively, and express

desired form.

Note that A is the leading coefficient of f(X).

coefficients of f(X)

such that

lie in a subfield K of F,

then

f(X)

in the

Hence if the

\EK.

The following result, whose proof is due to Artin, guarantees the existence of algebraically closed fields. 1.13. THEOREM. field

E containing

Proof. F[X]

Let

F

be a field.

Then there exists an algebraically closed

F as a subfield.

We first construct an extension

of degree

~

1 has a root.

We form the polynomial ring the polynomials

f(Xf)

in

F[S] F[S].

in which every polynomial in

E /F 1

To this end, put s = {Xflf and denote by

I

We claim that I

there is a finite combination of elements in I

E

F[XJ,degf

~

l}.

the ideal generated by all ,j F[S].

Indeed, otherwise

which is equal to

l:

54

CHAPTER 2

x1 . and observe that the po lynomta 1s g.~ involve ~ actually only a finite number of variables, say X ,x , ... ,x , with m ~ n.

with

g. E F (SJ. ~

Put

X• = ~

1

m

2

Hence we may rewrite our relation as: n E g.(X , ••• ,X )f.(X.) = 1.

i=l

~

m

1

~

~

Owing to Corollary 1.11, we may choose a finite extension K/F in which each

f., 1 ..; i ~

polynomial

~

has a root, say a.. ~

n,

Substituting a. for x., 1..; i..; m, ~

Put a.= 0 for n < i..; m. ~

in our relation, we get O = 1, a contra-

~

diction. Let

M

be a maximal ideal of

F[Sl

r.

containing

field containing an isomorphic copy (F+M}/M of F, a field extension of F. ;;,,, l , xf + M E

E1

Then E

l

Hence we may regard

Furthermore, for every polynomial

is a

= F(S]/M

f E F[Xl

E

l

as

of degree

is a root of f.

Inductively, we can form a chain of fields ECEC . .,CEC

-n-

1-2-

such that each polynomial in

E [Xl

n

E be the union of all fields

and every polynomial f

1.14. COROLLARY. E/F such that

~

Let

E,

~

E

is obviously a field But then

Then there exists an algebraic extension

Let E be the algebraic closure of F in K.

1,

then f(X)

Moreover, if f(X}

has a root, say a,

in

E E [X l

F

be a field.

E containing

Hence a is algebraic

K.

F.

Thus



By an algebraic closure of

F such that

Then

is a po lynomi a1 of

and therefore, by Proposition l .5(ii}, a is algebraic over

a EE and the result follows. Let

E

E is algebraically closed.

is an a1gebra i c extension.

degree

Then

Let

as asserted. • be a field.

F

is algebraically closed.

over

E, n = 1,2, ... n

1 has a root in En+i·

By Theorem 1.13, there exists a field extension K/F such that K

Proof.

E/F

~

E[Xl has its coefficients in some field E. n

f E

has a root tn En+J.

of degree

E/F

F

we understand any field

is algebraic and E is algebraically closed.

Owing to Corollary 1.14, such E always exists.

It will be shown below that E

ALGEBRAIC EXTENSIONS is determi'ned uniquely, up to an F-isomorphism.

55 The following preliminary result

wtll clear our path. 1.15. LEMMA.

Let

be a field and let cr :

F

tnto an algebraically closed field L. field extension of F,

and let f(X) e F[X]

then there exist exactly t Proof.

coefficients of f(X}.

form w(a)

F

be the minimal polynomial for a.

extensions of cr

(in the algebraic closure of F), to a homomorphism F(a)

-->-

L.

t 0(x} e L[Xl be obtained from f(X) by applying cr to the

Let

t 0(x} in

be a homomorphism of

Let a be an algebraic element in a

is the number of distinct roots of f(X)

If t

of

F-->- L

Since

is algebraically closed, there exists a root

L

Given an element of F(a)

L.

with some polynomial

=

Fial,

w(x} e F[X].

B

we can write it in the

We define an extension of cr by

mapping

This is well defined, i.e. independent of the choice of the polynomial 1

w(a) (w-w )(al = 0,

w0 (x)

-

l

w01 (x), .

morphism F(a) f(Xl

whence

and thus -->- L

wO'(B). 1

=

inducing cr on

in the algebraic closure of

t 0(x) e

L[X].

a root of

w1 (a},

=

f(X) divides

l'CB) .

w(X) - w (x).

1.16. PROPOSITION.

F.

Now the number of distinct roots of

(ii)

--+ L

There exists an extension cr'

of cr is



be an algebraic extension and let cr

E/F

divides

is equal to the number of distinct roots of

F

the result follows. Let

f 0(x)

It is now clear that our map is a homo-

a homomorphism into an algebraically closed field (i)

Hence

l

Since the image of a under any extension F(a)

t 0(X),

used

Indeed, if w (X) e F(X] is such that

to express our element in F[al.

then

~(X)

F---+ L

be

L.

of O' to a homomorphism

E--+ L

If E is algebraically closed and L is algebraic over cr(F),

then cr'

is an isomorphism. Proof.

Let us show first that (ii}

is algebraically closed, and

L

is a consequence of (i).

is algebraic over cr(F),

Indeed, if

then cr'(E)

is alge-

E

56

CHAPTER 2

braically closed and L is algebratc over a'(E),

hence L

To prove (i), denote by s the set of all pairs field of E containing F, If

K -->-L.

and

Kc K'

and

(K,A)

Note that

A'JK = A,

ely ordered:

If

to a homomorphism

0

we write

and s

so SI 0,

(F,0} Es ,

if

(K,A),,,;; (K',A')

is a totally ordered subset, we let

{(K.,A.)} 1,,

9efine

are in s,

(K 1 ,A')

where K is a sub-

(K,A)

and A is an extension of

a'(El,

=

is inductivu

K =

K~

1,,

and

"

Then

A on K to be equal to A.1,, on each K1,,..

(K ,A)

is an upper bound

for the totally ordered subset.

Owi'ng to Zorn's 1emma, we may therefore choose a

maximal element, say

s.

claim that T

=

in

(T,µ),

Then

Otherwise, there exists

E.

has an extension to

Let

a

E

be a field.

F

0

and, by Lemma 1.15,

E-T

to E,

and we

0,

thereby contradicting the maximality of

T(a),

proves that there exists an extension of 1.17. COROLLARY.

is an extension of

µ

as required.

µ

This

(T,µ).



Then any two algebraic closures of

F

are F-isomorphic. Proof.

Let E and E 1 be algebraic closures of F and let

be the inclusion map. phism

E -

Then, by Propositton l.16(ii},

as required.

E',

0

0

:

F -

E'

extends to an isomor-



Let E/F be an algebraic field extension and let F be the

1 . 18. COROLLARY.

a1gebra i c closure of

F

con ta i ni ng

E.

Then any F-homomorph ism

E -

F

is

extendible to an automorphism of ffe. Proof.

Since F/E is an algebraic extension and

F

is algebraically close~

the result follows by virtue of Propositions 1. 16(i) and 1.12. • We close this section by proving the following observation. 1.19

PROPOSITION.

finite.

Then

Proof.

JEJ

Let E/F be an algebraic field extension and let F be in=

JFJ.

Let s be the subset of F[Xl

positive degree and, for any n

~

0,

of all polynomials of degree n +

1.

r+l + A r

so

1

+ ... + A ' A' E n

1,,

F,

consisting of monic polynomials of

let S(n)

be the subset of s consisting

The elements of S(n)

have the form

NORMAL EXTENSIONS

ls(n) I = IF x ... x Fl n factors

+-

57

IFI

=

+

Also

= I u S(n) I = IFI

Isl We now map each in

f(X)

Es

wO

into the finite set

Because every element of

E.

(possibly empty) of its roots

Kf

is algebraic,

E

UK

E =

fES f

Since each IEI.

is finite the cardinality of the collection

Kf

Therefore

IEI

l{Rf}I ,;;;; Isl

=

=

IFI

{Kf}

is the same as

which clearly implies that

IEI

=

IFI .•

2. NORMAL EXTENSIONS Let

degree ;;,, 1. of

F

{f.li 1,

be a field and let

F

By a

EI}

splitting field

such that every f.1,

be a family of polynomials in

splits into linear factors in

F[Xl

of

for this family we understand an extension

generated by all the roots of all the polynomials family of polynomials in

p[xl

f.,i 1,

and

E[X],

EI.

E

E

is

We note that any

always has a splitting field, namely the field

generated by the roots of the given family in any algebraic closure of F. turns out that the splitting field of

{f.li 1,

EI}

It

is determined uniquely up to an

F-isomorphism, as we shall see below. 2.1. LEMMA. (i)

E

(ii)

If

Let

and F

K

c

E

f(X)

E

and let

F[X]

E,K

be two splitting fields of

are F-isomorphic c F where

F is an algebraic closure of

morphism K-+ F is an isomorphism of K onto Proof.

(i}

f(X).

Let

E

exists an F-homomorphism

0

:

K--+

E.

F.

Then

E

is algebraic

Owing to Proposition 1.16, there

We are therefore left to verify that

o(K) = E

We have a factorization

then any F-homo-

E.

be an algebraic closure of E.

over F, hence is an algebraic closure of

F,

(l )

CHAPTER 2

58

f(X) with

Bi EK, 1


-

F is extendible to an

provides, by virtue of Proposition 2.5,

E



Kc E

c

be a chain of fields, where

L

If the number of K-homomorphisms

then every F-homomorphtsm K __.. L

has exactly

E ~ L

is a

is equal to

n extensions to homomorph-

E__,_ L

Proof.

Let

(respectively,

G be the group of all F-automorphisms of L and let G(K) G(E))

be the subgroup of G consisting of those automorphisms

of L which leave fixed every element of K (respectively, of E). that

L/F

G(E)

is a subgroup of G(K).

Let

n

G(K) = U ~.G(E) and i=l ~

m G =

u ~.G(K)

j=l J

It is clear

63

SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS be cosets decompositions of G(K]

and G. G =

U

i,j

Then

ijJ ,c/J .G(E) J 1,

is a coset decomposition of G with respect to G(E}. automorphisms any element

have distinct restrictions to

ip. J

of

ip

G

to

K

K

It is clear that the m

and that the restriction of

cotnci'des with the restriction of one of the

iJJ .. J

Since by Propositi'on 2.9, every F-homomorphism of K into L is the restriction of some automorphism of L,

it follows that K has exactly m F-homomorphisms

into L and that these are given by the restrictions of

to K.

ip , ••• ,ip 1

m

A similar argument shows that E has exactly mn F-homomorphisms into L and that these are given by the restrictions of the each c/Ji restricts to the identity on trictions to

K

if j t-

products

Thus each F-homomorphism of

k.

which are F-homomorphisms of

i'nto

E ijJ

to E.

i)!.¢. J 1,

Now

and ipj and ipk have distinct res-

K

one represented by the restriction of ipj'

to E.

nm

K

into

L,

say the

has exactly n extensions to

E

namely the restrictions of

L,

.c/l ,ip .¢ , ... ,ijJ .c/J

J 1

J n

J 2



3. SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS Let

F

be a field and f

a polynomial in

If

F[X].

:X.

is a root of f

in

F,

then

for some g(X) and say that

E :X.

F[Xl with

g(::\} t- 0.

is a multipZe root if m >

aim is to be able to test whether F,

We call

f

without having to go outside F.

m

the nruUiplicity of

:X.

and a simpZe root if m = 1.

in F, Our

has multiple roots in some field containing To this end, we first define derivatives

formally. Given any polynomial we define its derivative

f(X) = a +ax+ ... + a x1 o 1 n Df or f' by the rule f' (X) = a

1

+ 2a X + 2

over a commutative ring R,

CHAPTER 2

64

The mappi'ng ~R[XJ->- R[X]

~

ff-+['

enjoys the fo 11 owing properties (f +g J I = f

(i )

(ii}

+ gI

(f,g E R[Xl)

(r

(rf) ' = rf'

(ii i ) (fg) (tv)

I

I

=

E

R)

f Ig + fg I

X' = l.

Conversely, f'

is completely determined by (i) - (iv) and this provides an alter-

native definition. 3.1. PROPOSITION. f

Let F be a field, let f

in some field extension of F.

Proof.

Dividing

Then \

by (X-A) 2

f

f(X)

E

is a multiple root if and only if

we obtain

(X-A) 2 g(X) + h(X)

=

where h(X) = 0 or degh(X) < 1.

Putting

we see that f(\) = h(\) = 0.

X = A,

Differentiating and putting X

= \,

which shows that

(X-\) 2 \J(X}

H and only if f'(A)

3.2, COROLLARY.

Let f

splitting field of

f'.

E

we find that h'(\)

degf > 1.

F[X] with

are simple if and only if

f)

In particular, if

and let A be any root of

F[Xl

=

=

f 1 (A).

Hence

O. •

Then all the roots of f

(in

is prime to its derivative

f

f is irreducible, then all the roots of f are simple

if and only if f' 1 0. Proof.

By Proposition 3.1,

have a common factor. 3.3. PROPOSITION. (i) (ii)

If If

charF

=

f

has a multiple root if and only if f



Let F be a field and let f(X) 0,

and f'

E

F[X]

be of degree > 1.

then f' 1 0

charF=p>O,

then f'=O

if and only if f

has the form g(xP).

65

SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS

(t l

Proof.

f' = 0,

then

(ii)

If

that

f'

f

f

f( ·X) = a0 + a 1 X + . • • + a n x" 0, which is impossible.

Write

=

na

n has the form ~

f'

then obvtously

1 ~

ia. = 0,

Then

= O.

g(xP),

i

~

n,

with

a:

0.

=

n

f. 0, n ~ 1 .

Conversely, assume

t

a. = 0 if p

and hence

If

~

i.

Thus

has the form

f(X) = a in other words,

0

+ a

p

xP

+ a

2p

is now a polynomial in

f

+ arp rt;

x2 P +

xP. •

In what follows by the roots of a polynomial we mean those lying in a splitting field of that polynomial. 3.4. COROLLARY. (i}

If

O;

charF =

(ii).

If

form

g(xt'}.

charF = p

Proof. (ii}

Let

(il

then r>

F.

be an irreducible polynomial over a field

f

has only simple roots

f

0,

f

then

has multiple roots if and only if

Apply Corollary 3.2

f

has .the

and Proposition 3.3(i}.

Apply Corollary 3.2 and Proposition 3.3(ii}.



The next observation is a useful companion to Corollary 3.4(ii}. 3.5. PROPOSITION.

f(Xl

F[Xl

E

F

Let

be a field of characteristic p > 0

be an i'rreducible polynomial.

and let

Choose the nonnegative integer

e

such that e

f(X}

E

F[xP 1

e+l but f(X} ff F[xP . 1

e

(i) (ti}

If

f(X} =

w(xP ) ,

All roots of

then

f(X)

1/J

is 1'rreducible and has only simple roots.

have the same multiplicity equal to

f(X).

equal to the number of distinct roots of

pe

and

degi/J

is

In particular,

degf = ( degi/!)pe Propf. otherwise

( i)

1/J(y)

the assumption. (ii}

It is clear that 1/J is of the form

and hence

Hence, by Corollary 3.2,

By (i), we may split

extension of F):

g(yP}

1/J(y}

. Moreover, 1/J' f. 0, since e+l f(X) E F[xP ] , contrary to

is irreducible.

1/J

has only simple roots.

into distinct linear factors (in a suitable

66

CHAPTER 2 m

rl (y-S .)

1/J(y}

i=l

1,

Then m

i=l

e

be a root of i? - Si.

Let ai

p

e

1,

Then e

s.

(J;i

;f -

e

n (Jf -S .)

f(X} =

s. 1,

(a , ... ,a are distinct ) 1 m

1,

e

- c:l1,.

= ;f

e

=

Hence

f(X} as required.



The degree of the polynomial

(in the notation of Proposition 3.5) is

1)J

called the separable degree of f(X), rability of f(X}.

ial

f(Xl,

If

o:,

while e is called the exponent of insepa·

is an algebraic element over

then the separable degree of

a over F)

of inseparability of

a

over

F

=

0,

with minimal polynom-

(respectively, the exponent

is deftned to be the separable degree of f(X)

(respectively, the exponent of inseparability of f(X)). charF

F

the separable degree of a

In the case where

is defined to be the degree of a.

Thus,

in all cases, the separable degree of a ts equal to the number of distinct roots of the minimal polynomial of a. Let E/F be a finite extensi'on and let K be an algebraic closure of E. The separable degree

morphisms of E tnto K. 3.6. LEMMA.

of E/F is defined to be the number of F-homo-

(E:F) 8

This ts obviously independent of the choice of K.

Let F(a)/F be a finite extension.

the separable degree of a over morphtsms Proof.

F(a)

--+

E

F

Then

(F(a):F} s is equal to

and ts also equal to the number of F-homo-

where E is any normal extension of F containing a.

The first assertion is a particular case of Lemma 1.15 in which cr

is equal to the inclusion map. algebraic closure of E, Since E/F is normal,

To prove the second assertion, let

K

be an

Then K is obviously an algebraic closure of F(a). the image of any F-homomorphism F(a}--+ K is contained

SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS in

as required.

E,

67



We now introduce a property of field extensions which will play an important role in our subsequent investigations. Let F be a field and let f

E

if all irreducible factors of f field of f).

over

is separable over F

have only simple roots (in a splitting

F

In particular, an irreducible polynomial is separable if and only

if its roots are simple.

An algebraic element a of a field extension

F is called separable over

If

We say that f

F[X].

over F, we call

a separable extension.

E/F

the definition that a is separable over separable degree of a over 3.7, PROPOSITION.

Let

of

F if its minimal polynomial over F is separable.

is an algebraic extension such that all elements of

E/F

E

F

F

E

It is an immediate consequence of

if and only if the degree and the

cotncide.

be a field of characteristic 0.

F

ials over F are separable.

are separable

Then all polynom-

In particular, any algebraic extension of F is

separable. Proof.

Apply Corollary 3.4(;}. •

3,8. PROPOSITION.

Let

be a field of characteristic p,

F

braic element in a field extenston of rability of a, (F(a) :F)

(i}

= pe(F(a) :F)

(iii)

a is separable over

Proof.

(il Let

f(X)

the separable degree of a,

F

F

E

if and only if F(a)

(F(a}:F)

F(ar).

Then =

degf(X)

by Proposition 3.5(ii).

= pem,

=

F[Xl be the minimal polynomial of a and let m be

(F(a):F} and so

be the exponent of insepa-

e

s

is separable over

aP

and let

Then

e

Ci'il

F

let a be an alge-

Now apply Lemma 3.6. e

(ii}

In the notation of the proof of Proposition 3.5, aP

irreducible polynomial

iJ;

which has only simple roots.

is a root of the Hence

aP

e

is separable

over F. (iii} Assume that a is not separable.

Then, by Proposition 3.5(ii},

68

CHAPTER 2

aF is a root of g(X).

and

f(Xl = g(xPl

Hence

(F(cl) :F) ,;;;;· degg(X) < degf(X)

=

(F(a) :F)

which shows that F(aP}; F(a). Now suppose a is separable so that f(X) be the minimal polynomial of a over F(aP). distinct roots.

Then g(X)if(X),

Also a is a root of the polynomial

divides xP-aP

g(X)

has distinct roots.

that g(_X) = x-a.

=

(X-a)P.

Hence a

E

Since g(X) F(aP)

xP-aP

Let g(X)

so g(X)

has

over F(aP),

so

has distinct roots, this implies

and therefore F(a} = F(aP).



The next observation will enable us to take full advantage of the results so far obtained. 3.9. LEMMA.

Let F .'.:. K .'.:. E be a chain of fields, where E is an algebraic

closure of F and let A: K--+ E be an F-homomorphism. there exist exactly Proof.

(K(a}:K} . s F-homomorphisms K(a1-+

Let f(X)

(K(a):K} s

Then, for any a EE, E

extending A,

be the minimal polynomial of a over K.

is equal to the separable degree of a over K,

of distinct roots of f(X)

in E.

By Lemma 3.6,

hence to the number

Now apply Lemma 1. 15. •

As an application of the above result, we now prove 3.10. PROPOSITION.

Let K/F be a finite field extenston, say K

r

(i}

(K:F) =

(ii)

(K:F) 8

S

(iii}

n i=

(F(a , ... ,a.) : F(a ,, .. ,a, 1 )) 1 1 1,, 1 1,,8 = (K:F) if and only if each ai is separable over

If charF =

F(a , ... ,a.1,- -1 ) ,

=

F(a , ... ,a ). 1

r

F(a 1 , . . . ,ai_ 1 )

and e.1,, is the exponent of inseparability of a.1,, over

p > 0

then

1

e +_., .+e

(K:F). = (K:F) 8 p

Proof.

(i}

Put n. 1,, m. = 1,

Then, by Lemma . 3. 6, a.1,,-1 }.

=

1

r

(F(a 1 , ... ,a1,,..) : F(a l , .. , ,a.1,,- 1 )} 8

(F(a 1 , ... ,a.} : F(a , ... ,a.1,,- 1 ) l, .· 1,, . 1 n.1,,

1,;;;;

and

i ,;;;; r.

is equa 1 to the separsib 1e d_ egree of a.1,, over .. F(a , ... , . 1

This, together with Proposition l_. l, ensures that r (K:F) = 1]m. . i=l 1,,

and n.1,, ,;;;; m.1,,

(1 ,;;;; i ,;;;; r)

(1)

SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS The case r

betng obvious, we argue by i'nductton on r.

= :i.

If

ar-1 ) and E is an algebraic closure of K(hence of K1 ), there exist exactly n 1n 2 .. . n r-1 F-homomorphisms

K

it follows from Lemma 3.9 that each F-homomorphism extensions to F-homomorphisms K-+ E. morphism (K:Fl -

(_ii)

S

=

to

K->- E

1

-

Since

F(a , ... , l

K = K

(ar ) ,

1

has exactly n

E

1

1

then by induction

E.

K -

=

K

r

Because the restriction of an F-homo-

ts an F-homomorphism

K

1

69

K

-

it follows that

E,

l

n l .:.n, as required. !'

Di'rect consequence of Lemma 3. 6, ( i) and ( 1).

(ii i ) By Proposition 3.8(t}, m.'1, (_K:F}

as required.

=

e. '1,

e 1 +, . • +er

'_1:'_

1'__

r Jm. = Cl

bl'I,

hence by (1) and (i),

= p 1,n.,

= (K:F)

Jn.)p

bl'I,

e 1 + •.• +e p

r

S



The result above has a number of important applicattons.

Before putting it

to use, we make the followtng observation: If F c E c K ts a chain of fields and a EK ts separable over F, separ~ble over

then a is (2)

E

Indeed, if f(Xl

E

with irreducible

F[Xl

ts the mtnimal polynomial of a and f(X) = f 1 (X) .. • fn(X) then a is

fiX) E E[Xl,

must have only simple roots stnce so does 3. 11 . COROLLARY.

Let

a root

of some Thus

f(X)..

be a fie 1d extension.

E/F

fi(x)

and

f .(X)

a is separable over E.

Then the set of a 11 e1ements

of E whtch are separable over F is a subfield of E containing F. parttcular, i"f E/F

is generated over

E

'1,

In

by a family of separable elements, then

F

is separable. Proof.

Assume that a , a 1

are separab 1e over

2

{a

\ E

1

Then F(a ,a} 1

2

=

F(A,a ,a) l

2

by Proposition 3. lO(ti),

2

2

2

l

1

ts separable over F(a) l

E

by (2).

Since all elements of

F.

the result follows.

Let E/F be a field extension. understand the subfield of

1

is separable over

obviously separable over F,

and 1et

a , a +a , a-1 } ,a f O.

and a

;>,.

F

Hence, F

are



By the separable closure of F in E we

consisting of all elements which are separable over

CHAPTER 2

70

F.

If

J

F is called the closure of F Let

F,

ts an algebraic c1osure of separable aZosu'I'e of

F.

then the separable closure of F tn Due to Corollary 1.17, the separable

ts uniquely determined up to an F·isomorphism.

KIF be a finite extenston.

We defi'ne the inseparable degree

(K:F).

'Z,

of K over F by (K:F).'Z, = (K:F}/(K:F} 8 If

charF = 0,

charF = p > 0,

then by Propositions 3.lO(it) and 3.7,

(K:F}. . 'Z,

then by Proposi'tion 3. lO(iii),

ts a power of

3. 12. COROLLARY.

If F c E c ]{

Proof.

(K:F) 8

while if p.

= (K:F}

if

is algebraic over

F.

is a finite extension, then

KIF is separable

and only if (ii}

If KIF

(i}

(K:F}.'Z,

= 1,

(i).

is a chain of fi'elds and

Write

K = F(a. l , ••• ,a.} 'I'

KIF

is finite, then

where each

ll•. 'Z,

Then, by Propos i'tion 3, 1O( ii}, the condition that each F(a. 1 , ••• ,a.'!,• 1 }

a.'Z,

is separab 1e over

is equivalent to the condi'tton that each ai

(stmply rewrite

is separable over F

F(a. , ..• ,a.'I' } as F(a'Z-.,a. 1 , •.. ,a.'!,•1 ,0\'Z,·+1,, .•. ,a'I' } . . 1

condition, in view of Corollary 3.11, is equtvalent to

The latter

KIF being separable.

Now apply Proposi'tton 3. lO(tt). (itl

Apply Proposition 3.10(1),(iii}.

3.13. COROLLARY.

If

field extension of Proof. forces

F ts a field of characteristic p

F with p) (E:F),

(E:F}

8

=

(E:F).

If EIF

(i}

lf F ~E

is separable and

tained in some field, then

separable.

and

(i)

If KIF

Assume that

KE!/K

~

E a finite

EIF is separable.

Now apply Corollary 3.12.

rable if and only if both EIF

Proof.

then

> 0

Owing to Proposition 3. lO(iii}, the assumption that

3.14. COROLLARY.

(ii}



p ) (E:F}



K is a chain of fields, then KIF is sepa-

and KIE

are separable.

KIF is any extension with both E and K conis separable.

is seoarable, then obviously both

EIF and KIE are separable.

EIF and KIE are

If KIF

is finite, then

SEPARABLE, PURELY INSEPARABLE ANO SIMPLE EXTENSIONS

71

the result is true by vty,tue of the second equaltty in Corollary 3.12(ii). K/F is infinite, let a.EK.

Then a. is a root of a separable polynomial

Let a 0 ,a 1 , •.• ,an be the coefficients of f(X}

f(X) E E[Xl,

If

and let

E9 "' F(a 0 , ... ,an).

Then we have F -c E0 -c E0 (a.) where E0 (a.)/F ts finite and both E0/F and E0 (a.}/E0 are separable.

rable over F,

Hence E0 (a.)/F is separable and therefore a. is sepa-

proving that KIF is separable.

(ii} Every element of E ts separable over F, hence separable over K. KE is generated over K by the elements of E,

Since

it follows that KE/K is sepa-

rable by Corollary 3. 11. • If

F

c Kc E is a chain of fields with

but KIF need not be normal.

EIF normal, then EIK is normal

The following result provides a condition under

which KIF is normal. Let EIF be a normal fteld extension and let F8

3.15. PROPOSITION.

separable closure of F in E. Proof.

Let

F-homomorphism. E -

E.

F

8

IF is norma 1.

be an algebraic closure of E and let cr; F8

E.

--+

E

be an

By Proposttton 1,16, we may extend cr tq an F-homomorphism

Since EIF ts normal, we see that cr(E}

morphi sm of in F8 •

E

Then

be the

= E,

Furthermore, cr(F8 ) is separable over

Thus cr(F8 } = F8

so cr ts an f-autohence is contained

F,

and, by Proposition 2.5, F8 IF is normal. •

We next provide a class of fields over which all extensions are separable. Let

F

be a field of characteristic

p >

0 and let iP

:e

{a.Pix e F},

Then

the mapping

is obviously an injective homomorphism whose image is the subfield iP of Ifdt is also surjective (i.e .. if F is said to be perfeat,

p-th power.

F =

F,

#), and hence an automorphism, the field

Thus F is perfect if and only if every element is a

In addition, every field of characteristic O is perfect by defin-

CHAPTER 2

72

ition. Let

F

be a field of characteristic

closure of F.

For each

Then pP

-n

be an algebraic·

F

define

n;,,, 1,

pl?

0 and let

p >

-n

FlxP

= {x E

n E

F}

is obviously a subfield of F and we have a chain F

of subfields of F.

-1

-2

pP . c ·Fp

C

C

Fp

"' C

-n C

Put -oo

00

up?

-n

n=1

3. 16. LEMMA.

With the notation above,

ts the smallest perfect subfield of

pi? -

containing

F

F.

Furthermore, if

Fi'. pi?

..-00

(lC)

,

then #

is an infinite

/F

extension. Proof.

Let K be a perfect subfield of F containing F.

n

then xP E F ~ K,

so xP

n

=

E

#

-n

n =

for some A EK since K is perfect.

AP

-yt

x

If x

Hence

..-00

A EK and therefore pl?

~K,

proving that

F

-00

The fact that pl?

c K.

is perfect is a consequence of the equality (#

-n p

1

-oo·

n;,,, 1,

hence

.

(n ;,,, 1)

..-co

-oo

If pl? /F is finite, then #

Finally, assume that some

-(n-1)

= pi?

F =

(#

-n

n )P

=

We sha 11 refer to the fte l d pP

#

for

-n

a contradiction.



as the perfect c7,osure of

F.

By Coro 11 ary

-cp

1. 17,

pl?

is determined uniquely, up to an F-tsomorphi sm.

3.17. PROPOSITION. Proof.

Every finite field is perfect,

Direct consequence of the fact that an injection of a finite set into

itself is a bijection. • The simplest example of an imperfect field is the field functions in x over a field

F

cannot be a p-th power in F(X).

of charatteristic

p >

0.

of rational

This is so sihce x

We now provide some criteria for a field to be

perfect. 3.18. PROPOSITION.

F(X)

The following conditions are equivalent:

SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS The field

(i)

73

is perfect

F

Any finite extension of F is separable

(ii)

(ii i ) Any algebraic extension of F is separable Any po 1ynomi a1 in F[X]

(iv)

is separable.

For characteristic O there is nothing to prove:

Proof.

in that case every

field is perfect and, by Proposition 3.7, (ii), (iii) and (iv) always hold. Assume that charF

0.

= p >

that (iii} implies (ii).

It is obvious that (iii) and (iv) are equivalent and Furthermore, (ii) implies (iii) by Corollary 3.11.

We are therefore left to veri'fy that (i} is equivalent to (ii}. Let F be a perfect field and let f If

f

is not separable, then by Coro 11 ary 3. 2,

3.3(ii},

f(X)

=

g(iP),

is perfect,

F

f' = O.

Hence, by Propos i'ti on

say f(X)

Since

be any irreducible polynomial over F.

a iPr + a jP(r-l) + o 1

=

for some b.

a.= b~ ~

~

=

~

(b

r

·o

+b

1

E

+ a

r

F and hence

.f'-L + ... + br )P

But this contradicts the irreducibility of f.

Thus every finite extension of

is separable.

AE

Conversely, assume that every finite extension of

F

and consider the splitting field of iP - :\.

If

F

is separable. µ

and so the minimal polynomial of µ over F is of degree 1. A = µP,

as required.

is a root, then

Hence µ

E

F and



Let F be a field of characteristic extension.

Fix

p >

0 and let E/F be an algebraic

We say that E/F is purely inseparable if the only elements of E

that are separable over

F

are the elements of F.

Expressed otherwise,

E/F

is purely inseparable if and only if the separable closure of F in E is F. It is clear that E/F can be inseparable without being purely inseparable. An algebraic element a of a field containing F is said to be purely

F

74

CHAPTER 2 n

JJ e

insepa.PabZe over F tf

P for some n

;i,,-

If E/F is an algebraic

0.

fi'eld extension, then the set of purely inseparable elements of E over F is obviously a subfield of

containing

F.

We shall refer to this subfield as

the pm>e inseparabZe aZosm>e of F in E.

The next result shows that it is a

E

purely tnseparable fteld extension of F containing any other such extension K/F With

KC E.

3.19. PROPOSITION.

Let

an algebraic extension.

be a field of characteristic p » 0

F

and let E/F be

Then the following conditions are equivalent:

E/F is purely inseparable

(i)

(.ii} All elements of (.iii)

E

are purely tnseparable over

is generated over

E

F

by a family of purely inseparab 1e elements

P

(iv} There exists exactly one F-homomorphism of

E

into an algebraic closure

of E. Proof.

Since the elements of E which are purely inseparable over F form

a subfield of E containing F, (i) * (ii}:

If a.EE,

then

ti

(ii) and (iii} are equivalent. e

is separable over

F

for some e;.. 0 (Propo-

e

sitton 3.8(ii)), hence a.P E (ii} * (ivl:

Let

F.

Thus a. is purely inseparable over

be a set of purely inseparable elements of

$

and let E be an algeoraic closure of E.

E = F(S}

n

root of the polynomial

x? -

"E F[Xl

E

E -

E is a root of

(iv} * (i): rable.

for some n ~ O and some

"E F.

hence a. -

Then (F(a.) :F) = (F(a.) :F) 8

E -

E.

f(X)

Since e F[Xl

Now the image of a. under any F-homomorphism a.

proving (iv).

Assume by way of contradiction that a. E

morphisms F(a.) phism

f(X),

such that

Each element a.Es is a

all the roots of this polynomial are the same, the minimal polynomial of a. has a. as the only root.

F.

>

E,

a. If:. F

and a. is sepa-

1 and hence there are more than one F-homo-

But each such homomorphism is extendible to a homomor-

E by Proposition 2.9.

This provides the desired contradiction and

completes the proof. • 3.20. COROLLARY.

Let F be a field of characteristic p

be a chain of fields.

>

0 and let F ~ E ~ K

Then K/F is purely inseparable if and only if both E/F

SEPARABLE, PURELY INSEPARABLE AND SIMP4E EXTENSIONS

and

are purely inseparable.

K/E

Proof.

Direct consequence of Proposition 3.19(ii). •

3.21. PROPOSITION.

Let

be a field of characteristic

F

an algebraic field extension. then

If Fs

is separable and

F /F

s

Proof. Let

That Fs /F

E/F

Corollary 3.14(i).

braic extension

Let

E/F

be

is the separable closure of F in E,

is separable is a consequence of the definition of F. s

Hence

3.22. COROLLARY.

O and let

p >

is purely inseparable.

s

be the separable closure of

K

75

K = F

s

F

in

s

Then

E.

and therefore

E/F

K/F

'

is purely inseparable. •

s

be a field of characteristic

F

is separable by

p >

0.

Then every alge-

may be obtained by taking a separable extension followed by

E/F

a purely inseparable extension. Proof.

Direct consequence of Proposition 3.21.



The following result provides a criterion under which the order of the extensions in Corollary 3.22 can be reversed. 3.23. PROPOSITION.

Let

tic

K

p >

0 and let

rable closure of Proof.

F

be an algebraic extension of fields of characteris-

E/F

and

in

F

Then

E.

Assume that

be respectively the pure inseparable and sepa-

s

is separable if and only if

E/K

is separable.

E/K

and, by Corollary 3.20 and Proposition 3.21, Hence

Conversley, if

E = KF •

s

E =

KF

s

E = KF . s

Then E is separable over KFs E

then

is purely inseparable over

KF. s

is separable by Corollary

E/K

3.14(ii). • 3.24. PROPOSITION.

Let

be a field of characteristic

F

p >

0 and let

E/F

be

a finite extension. (i)

E/F

is purely inseparable if and only if

Proposition 3. lO(iii}, if

(E:F)

s

=

is purely inseparable, then

E/F

In particular, by

1.

(E:F)

is a power of

P· (ii}

(E:F) Proof.

(ii)

s

= (F

(i)

s

:F), where

F

s

is the separable closure of

F

in

E.

Direct consequence of Proposition 3.19(iv).

By Proposition 3.21,

E/F8

is purely inseparable, hence by (i), (E:F8 \= 1.

CHAPTER 2

76 Now (E:F)

by Corollary 3.12(ii), and

=

s

(F :F) (E:F } s

Hence

(F :F)

=

S

Let F be a field of characteristic

finite field extension of F. Proof.

Now E/F mi

by Proposition 3.19,

a~1,

we have a~1, E

F

E F

for all

s

EPF.

=

Then E/EPF is both separable and =

Let

EPF.

F

s

Since E/F is finite, we may write

is purely inseparable (Proposition 3.21) and hence,

s

forsome

s

m =

rn

0 and let E be a

Conversely, assume that E

EPF.

=

be the separable closure of F in E. F(a , ... ,a ) . 1 r

p >

Then E/F is separable if and only if E

Assume that E/F is separable.

purely i'nseparable, hence E

=

by Corollary 3.12(i).



3.25. PROPOSITION.

E

s s

(F :F) = (F :F) s s s

(E:F) 3

as required.

s

m.;;,,0,1.;;;i.;;;r. 1,

Setting

max{m.}, 1, m

Hence EP c

i E {1,2, ... ,r}.

-

F

But E

s

=

EPF

obviously implies

m E = EP F,

3.26. COROLLARY.

Let E/F be a finite separable extension of fields of charac-

teristic

If u , ... ,u

p

!>

0.

hence E

is separable over F.

F

s

is a basis for E over F,

n

1

=

m



then so is

m

up , ... ,uP n

J

for any m ~

By Proposition 3.25,

Proof. over F,

1.

and so is a basis.

E

=

EPF.

Therefore uP, ... ,uP again spans E 1

n

Now the result follows by induction on m. •

As a preliminary to the next result, we record the following observation: 3.27. LEMMA.

Let A be a finite-dimensional commutative algebra over a field F.

If A has no nonzero nilpotent elements and no nontrivial idempotents, then A is a field. Proof. J(A)

Since

of A is

0.

A

has no nonzero nilpotent elements, the Jacobson radical Hence

A

is a finite direct product of fields.

has no nontrivial idempotents, hence

A

is a field.



But

A

SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS 3.28. PROPOSITION.

Let F be a fteld of charactertstic

77

O and let EIF

p >

and KIF be respectively a separable and purely inseparable field extensions. Then E 0 K is a field. F

Proof.

O ix EE@ K,

Given

we may, by Proposition 1.4.1, write

F

n

Icc0S.

x=

i=l

for some F-linearly independent a.1,, E

E

1,,

1,,

S,EK,l 1. 1, over F.

Stnce E/F ts separable, each a.1,

Now apply Proposition 3.30.

is separable



The result above can also be derived as a consequence of the following general assertion due to Artin. 3.33. PROPOSITION.

Let E/F be a finite-dimensional field extension, where F

is an infinite field.

Then E/F is a simple extension if and only if there are

only a finite number of intermediate fields between Proof.

Assume that E/F is simple, say E

mediate field.

Let

f(X)

= F(0)

(E:K') =

K'

coefficients

K and let

Then

K'

and

K' c K

Therefore

= degf(X) = (E:K)

is generated by the coefficients of f(X).

divides the minimal polynomial Because g(X)

e over

and the coefficients of f(X).

F

is also the minimal polynomial of e over K'.

and thus K

F.

and let K be an inter-

be the minimal polynomial of

f(X)

be the field generated by

and

E

Note also that f(X)

of e over F and both

g(X)

f(X) ,g(X) E E[Xl.

has only a finite number of distinct factors in E[Xl with leading 1,

the number of intermediate fields

is finite.

K

Conversely, assume that there are only a finite number of intermediate fields between

E

s imp 1e.

and

F.

Given a,S

To this end, let

y

have an infinite number of y exist y,o

in F,y 'I

E

F

E F

and consider the subfield Ey = F(a+yS). and a finite number of EY.

o, such that

s=

we rieed only verify that F(a,S)/F is

E E,

(y-o)

and therefore a= a+ yS-yS EE. y

EY -1

=

E0 .

We

Therefore, there

Then

(a+yS-a-cSS) E Ey

Thus F(a,S) = F(a+yS), as required.

11

4. GALOIS EXTENSIONS Let F be a field and let G be a group of automorphisms of F. FG

the subfield of F consisting of all elements x

F

such that for al 1 a

a(x) = x

We shall refer to FG as the fixed field of

E

We denote by

G.

E G

CHAPTER 2

80

Let

ca 11 ed the is a

E/F

be a field extension.

E/F

of the extension and is written Ga 1(E/F).

Galois group

is a1gebra i c over

if E

Galois extension

of the group

The group of all F-automorphfsms of

and

F

is

E

We say that

is the fixed field

F

Gal(E/F).

4.1. PROPOSITION.

Let

be an algebraic extension.

E/F

Then the following

conditions are equivalent: (i}

E/F

(ii)

E/F

(iii)

E

is Galois is normal and separable is the splitting field of a family of separable polynomials over F

Proof.

(i) • (ii):

over f.

Let a e

in E.

f

fixes g

g = (X-a

E F[Xl.

Since

In this way each rr 1

}(x-a) ... (X-a ).

Hence

2

Y'

Gal(E/F)

E

Since

E/F

Let

be all distince roots

is extendible to an auto-

permutes the

and hence

a.1,

is Galois, we deduce that gJf

proving that f =

g.

permutes the a1,. transitively,

as required.

Since

is separable,

E/F

{aJ

of separable elements.

over

F.

norma 1 ,

Y'

2

and, by the definition of g,

fig

a.EE, 1,

(ii)• (iii):

a,a , ... ,a

= 1

is irreducible, the group Gal(E/F)

g

hence each

1,

and let a

are distinct and lie in E.

By Corollary 1.18, each a e Gal(E/F)

E.

morphism of

E

be the minimal polynomial of a

f

We must show that all roots of f

E be an algebraic closure of of

and let

E

Then each

E

Ne denote by

is generated over

hence

E,

E

by a family

f.1, the minimal polynomial of a.1,

fi is a separable polynomial over

f.1, splits over

F

F.

Since

E/F

is

is the splitting field of the family

{f1,.}

of separable polynomials over F. (tit)• (i):

Assume that

separable polynomials over a EE -

F.

Since

a

homomorphism 0 : F(a) sition 2.9, Galois.

is the splitting field of a family

E

F.

Then

E/F

{f.Ji 1,

is normal and separable..

EI} of Fix

is separable, it follows from Lemma 3.6 that there is an --+ E

for which 0(a) f. a.

F-

On the other hand, by Propo-

0 is extendible to an automorphism of E,

proving that E/F is



Owing to Corollary 3.22, any algebraic extension may be obtained by taking a

GALOIS EXTENSIONS

81

separable extenston followed by a purely l'nseparable extension. cannot reverse the order of the tower.

Usually, one

However, there ts an important case when

it can be done. 4.2. PROPOSITION. (i)

Let

be a normal extension and let

E/F

is purely inseparable and

EG/F

(i i l lf

F

are ltnearly dtsjotnt over F. E

~

i1

®

F 8

p

n

Proof.

If charF

=

0,

Then

in

F

then

E,

= F s EG and

E

F

s'

EG

In particular,

then AP e

(iii)

= Gal (E/F).

ts separable

E/EG

ts the separable closure of

B

G

B

n

Er; = F

and some n

p

~

1.

= F B and hence, by Proposition 4.1,

E

which clearly yields the result.

F

for some prime

F

then

and

Assume that charF

= p ~

0.

EG

=F

Owing to Propo-

sitions 3.29 and 3.19, we need only verify (i). Since Gal (E/F) = Gal (E/EG), for all

a

E:

Gal (E/EG},

then 1i

it follows that if A E E is such that cr(A) = A E

i 1•

a : Ea -- E be an F-homomorphism.

i.

is normal.

ts Galois and therefore, by

E/EG

Let E be an algebratc closure of

Proposition 4.1, is separable.

automorphism of

Hence

E

and let

By Corollary 1.18, cr is extendible to an

Its restriction to E is an F-automorphism of E since

Hence cr(x)

= x

for al1

r'.

x E EJ

E/F

and the result follows by virtue of

Proposition 3.19. • We next provide a useful characterization of finite Galois extensions. 4.3. PROPOSITION.

Let E/F be a finite field extension. IGal(E/F)I

~

Then

(E:F}

and IGal(E/F)I if and only if Proof. Assume that separable. whi'le since

E/F

(E:F)

is Galois.

The first inequality follows from E/F

is Galois.

Then, by Proposition 4.1,

Since E/F is separable, we have E/F

Gal(E/F) ~· (E:F)s

is normal we have

(E:F)

s

=

E/F

(E:F)

~

(E:F}.

is both normal and (Corollary 3.12),

(E:F} 5 = IGal (E/F) I - Hence IGal (E/F} I= (E:F).

CHAPTER 2

82

JGal(E/F} I = (E:F).

Conversely, assume that Since

(E:F) 8

= (E:F),

IGa 1 (E/F} I = (E:F) B = (E:F).

is separable by Corollary 3.12.

E/F

must also be normal,

E/F

Then

Since

JGal(E/F)J

=

Thus

is Galois, by Proposition 4,1, •

E/F

(E:F} 8 ,

As a preliminary to the next result, we prove 4.4. LEMMA.

Let

ger

n;;;. 1

E/F

is finite and Proof.

be a separable extension.

E/F

Assume that there is an inte~

such that every element a of E is of degree .;;; (E:F}

over F.

Then

.;;;n.

Let a EE be such that the degree

We claim that E = F(a). such that Bi F(~),

n

Deny the claim.

(.F(a} :F)

is maximal, say

m.;;; n.

Then there exists an element BEE

and by Corollary 3.32, F(a,B)

=

F(y)

for some y E F(a,B).

!>

m over

But from the chain F ~

we see that tion.

(F(a,B) :F}

> m

F(a)

~

F(a,B)

whence y has degree

a contradic-

F,



We are now ready to prove the followtng classical result. 4.5. PROPOSITION. (Artin). automorphisms of F.

Let F be a field and let G be a finite group of

Then F/~ ts a finite Galois extension,

G = Gal(F/FG}

and

Proof.

In view of Proposition 4.3, it suffices to verify that

finite Galois extension and maximal set of elements of crE

G

then

(F:FG}o;;;JGJ. G

is a

F/FG

Let ~EF and let cr, ... ,cr 1

such that cr 1 (a}, ... ,cr. !' (a)

(crcr 1 (a), ... ,crcrr(a)) differs from

are distinct.

(cr 1 (a), ... ,cr:r>(a)) 1

If

by a permu-

tation, s i nee cr is injective and each crcr .(a} E fo (a), ... ,cr (a)}. ~

bea

!'

!'

Therefore

· a is a root of the polynomial

n (x-cr .(a)) i=l !'

f(X) =

and for any cr E G,

f

=



~

Thus the coefficients of f

G lie in F.

Moreover,

GALOIS EXTENSIONS

f

ts separable.

It follows that every element

rable polynomtal over Fa of degree into linear factors in F.

~Jal.

83

a.

of

is a root of a sepa·

F

Furthermore, this polynomial splits

Thus F/Fa is both separable and normal, hence F/Fa

is Galois, by Proposition 4.1.

By Lemma 4.4, we have

(F:fi)..;;

Jal

and the

result follows.• Let E/F be a finite Galois extension and let a be its Galois

4.6. COROLLARY. group.

Then, for any subgroup H of

a, there exists an intermediate field

K

such that E/K is Galois and H = Gal(E/K). Proof.

Put

~ and apply Proposition 4.5. •

K =

We close by exami'ntng some properties of automorphisms of field extensions.

s be a set and

Let

a field.

F

to be linearly independent over

The maps F

:\f 1

then all

O.

:>... = 'Z,

1

f.'Z, : s----+-

F, 1..;; i

~

are said

n,

if whenever we have a relation +, •• +:X.f =O n n

(LE F) 'Z,

Recall that a monoid ts a set G, with an associative binary

operation and having an identity element.

x1 ,x 2 , ... ,Y''n be dtsttnct homomorphisms from a to the multiplicative group F* of a field F. Then x1 , ••• ,x,_. are

4.7. PROPOSITION. (.Artin}.

monoid

G

Let

1inearly independent over F.

x1 ,x 2 , ... ,y''n x = nr A ·X. for If

Proof. assume that

i=2

1

independent.

are linearly dependent, then we may harmlessly some '>..i e

F

and that

i, -i,

x2 , ••• ,x,_.

are

1i nearly

Then e a

for al 1

g

for all

g,x

( 1)

Replace g by gx in (1):

Multiply (1) by x (x} 1

Ea

and subtract the result from (2}: for all

Hence

n

r

i=2

:>..

.(x .(xl -i,

since x2 , •••

-i,

(2)

x1 (x) )xi

=

O and so \Cxi(x} •

,x,_. are linearly independent.

x1 (x))

Since xi f: \

0, 2

~

for i f:

i

g,x ~

1,

n,

it

E

G

84

CHAPTER 2

follows that ;..,

-z.,

O for if:

=

and (1) reduces to the form x (g}

1

1

0. which is

impossible. • Let f ,f , ... ,f

4. 8. COROLLARY.

1

into another field F. Proof.

Put G =

Then

f

n

Since

E J...f.

i=l

=

, •••

I

,f

G

E

are linearly independent over F.

n

and denote by X·-z., the restriction of f.-z.,

E*

x.-z., is a homomorphism of

each

be di sti net homomorphisms of a field

n

2

to G.

Then

into F* and the X·-z., are obviously distinct.

0 obviously implies

-z., -z.,

n

E ;.. .x.

i=l

=

0,

the result follows by virtue

-z., -z.,

of Proposition 4.7. • 4.9. COROLLARY. (E:F)

= n
1,

and if bf- O,(b-l)q=

(a-b)q = aq - bq = a - b, (ab}q = clbq = ab,

Let

be a field of q

F

=

pm

elements.

Then, for any

there exists a field extension K of F of dimension

namely the

n,

splitting field of n

I' Moreover, any two field extensions of Proof.

of dimension n are F-isomorphic.

F

Assume that E is a field extension of F of dimension n.

has qn elements and, by Proposition 5.1 (ii), the elements of

E

ly the roots of f1

of

F

n

-

r

splitting field of

r

over F

X

of dimension n

n - X

X

i'n an algebraic closure of

n - X

over

F.

r

e1ements and therefore has degree n The field of

q

are preciseis a

E

This shows that any two field extensions

are F-isomorphic.

are roots of

F

Therefore

IF • p

Conversely, the splitting field of

over F is the same as the splitting field of I'

all elements of

E

Then

n

n

- x. over

elements is denoted by

-

X

over

IF,

p

since

Hence it consists of precisely q F.

n



Jfj' •

q

For

q = p

this agrees with

FINITE FIELDS

the notatton IF

.87

i'ntroduced earlter.

p

5.3. PROPOSITION.

If q = pn,

where p

is a prime and n;;;,

a Galois extensi'on with cyclic Galois group of order n, beni us automorphism x Proof.

:-+

xP

The extension

is

generated by the Fro-

of IF • q

IF/ffi'p

being a finite extension of the perfect field

must be separable (Proposition 3.18}.

JFP

then IF/IFP

1,

On the other hand, IF/IFP

by Propositton 5.l(ii} and hence is Galois by Proposition 4.1.

is normal

Therefore, by

Gal (1E' /IF ) is of order n. Since xP = x for all X EIF , p q p the Frobenius automorphism a. of IF is an IF -automorphism of JF • Finally, q p q since Proposition 4.3,

!'

a.!'(x) = and thi's is the identity for that

a.

has order n.

r = n

xP

and not the identity for

Gal (1E' /IF ) = q p

Thus

for all 1 ..;

as required.

x e IF

q

r < n,

we see



All subfields of IF m are of the form IF n where nlm and, p p n there is exactly one subfie1 d of IF of order p • Furthermore,

5,4. PROPOSITION.

n Im,

for any if nlm,

min,

pm

then IF

,IIF' n

generated b~ Proof.

a.~,

is a Galois extension with cyclic Galois group of order where

Assume that nlm.

m

xJ? -

ts the Frobenius automorphism of IF m' n p Then any root of xP - X is also a root of

a.

n

Hence, by Proposition 5.l(ii}, an the roots of xJ? - X form a n Conversely, if F is a subfield of IF m unique subfi e 1d of IF m of order p . p n . p and IFI = p, then X.

m=

(JF m:F) (F:IFP) = p

provi'ng that nlm. (1E' m:IF n } = min and, by Proposition 5.3, IF rrf/Tri' n is Galois p p . p p n hence its Galois group is of order min. Since a.n is of order min and a.

By the above

fixes IF

elementwise, the result follows. • n We now turn our attention to roots of unity. p

ment

Let

F

be a field.

A E F is said to be an n-th root of 1 (or of unity) if

charF = p > 0

and

n = pms

with

(s,p) = 1,

then

"An = 1

An= 1. implies

An eleIf m (A 8 )P =

88

CHAPTER 2

or As

1.

=

Hence, if '.\

is an n-th root of unity then '.\

is an s-th root of

We shall therefore assume in what follows that the characteristic of the

unity.

field does not divide n. 5.5. PROPOSITION. (i)

Let

be a field and let

F

Then-th roots of unity in

(ii)

r -1

If

unity in

F

Proof.

F

n

> 1 be such that charF l

form a cyclic group whose order divides n.

sp 1its into li'near factors over

F,

then the n-th roots of

form a cyclic group of order n. It is clear that (i} is a consequence of (ii).

To prove (ii), note

that then-th roots of unity in F certainly form a group, say G. (r-1)

1

=

Hence

contains exactly n n-th roots of unity, i.e.

F

a.

Let m be the exponent of

r - 1,

of 1i C,

Since

nx"- 1 r 0, it follows from Proposition 3.1, that all roots of

are simple.

n.

hence

JaJ < m.

r

- 1

Jal

= n.

Then m < n and all elements of G are roots This shows that m = n and therefore

G

is cyc-



Let F be a field.

We say that :\

if the order of A in F*

F is a primitive n-th root of unity

E

is precisely n.

Given n

> 1, by a

primitive n-th

root of unity over F we mean a primitive n-th root of unity in the splitting field of i7' - 1 over

By Proposition 5.5, if charF

F.

n-th root of unity over

always exists.

F

n-th root of unity over F,

i!7'

over F.

then F(s} n

The field F(s) n

l

n,

then a primitive

Furthermore, if sn i's a primitive is obviously the splitting field of

is ca1led then-th cycZotomic extension of E

By the maximal cycZotomic extension of F we understand the field obtained from F

by adjoining n-th roots of unity for all

5.6. COROLLARY.

Let

F

n

> 1.

be a finite field of q elements, let m > 1 be an

integer coprime to q and 1et

s be a primitive m-th root of unity over

F.

Then (i) (ii)

(F(s):F)

is equal to the order of q modulo m

Gal (F(s)/F) Proof.

is cyclic generated by s '--'" sq

It is clear that (i) is a consequence of (ii).

Let a be the n

Frobenius automorphism of F.

If q = pn, p prime, then an(s) = sP = sq.

FINITE FIELDS

89

Therefore (it} is a consequence of Proposition 5.4. 5.7. PROPOSITION.

E and

let

sions of

Then both Since

noti'ce that

n ~ 1

be a field, let

F

be such that charF

l

and

n

K be respectively then-th cyclotomic and maximal cyclotomic exten-

F.

Proof.

Let



charF

E/F

and

K/F

l

i1 -

1

n,

are Ga 1 ois exte.ns ions. has no multiple roots.

I' -

is the splitting field of

E

1

x" -

ting field of the family of all such polynomials Proposition 4.1).

over

and

F

1

with

It remains to K

is the split-

charF

l

(see

n



The following observation is often useful. 5.8. LEMMA.

Let

be a Galois extension and let

E/F

µ ,µ , ... ,µ 1

of

E

such that each

CT

E

Gal(E/F)

permutes the

2

µ , ... ,µ. n

l

n Then

be elements

(X-µ } (X-µ ) ... (X-µn) 1

is a monic polynomial over

2

If, furthermore,

F.

an integrally closed integral domain (X-µ ) ... (X-µ ) 1

and each

R

is a monic polynomial over

is the quotient field of

F

µ.

is integral over

1,,

then

R,

R.

8

Proof.

f(X) = (X-µ 1 )

Put

(X-µn} .

•••

Then

f ( X,1 = x.n - s 1 x.n-1 + s 2 x.n-2 + •.• + (-l)ks· kx.n~k +

, .. + ( - l)nsn

where s = µ + µ + .•. +µ 1

s

s

Hence

CT(s .) .

1,,

=

s.

1,,

for all .

2

n

CT

is Galois, we deduce that each al hypotheses, each integrally closed,

s.

1,,



1

=µµ 1

2

2

+µµ 1

=µµ ..• µ 1

E

2

Gal(E/F) s.

1,,

E

F,

is integral over

3

n

+ ... +µ

n-1



n

n

i

E

{1,2, ... ,n} .

f(X)

E

F[XJ.

and all hence R,

hence belongs to

Since

E/F

Under the addition-

R since

R is

90

CHAPTER 2

('lL/n'll,)*

In what follows we write

order of this group is denoted by

for the unit group of the ring

¢(n),

where

¢

('ll,/n'll,}* = {µ+n'll,Jl,;;;; µ < n, (µ,n) = l}

Note that

and that

n.

number of primitive n-th roots of unity is equal to

¢(n).

n ~ 1

be a field, let

F

r = ¢(n},

¢(n)

be all primitive n-th roots of unity over

coincides with

In particular, the

charF X n

be such that

The

is known as the Eiler function.

the number of generators of a cyclic group of order

Let

'll,/n'll..

F.

and lets , ... ,s, 1 r Then, by Proposition

5.7 and Lemma 5.8, r

f -J (X-s ,)

i=l

1,,

F and hence over F.

is a manic polynomial over the prime subfield of charF = 0,

more, if

Further-

then by Lemma 5.8, 1> (X} E 'll,[X]

n

We shall refer to

cJi/Xl

5.9. PROPOSITION.

Let

let

n

to a subgroup of Ga 1 (Fn/Fl cJi n (X)

(ii} (iii}

2e

(F :F) n

=

Let

a E Gal(F /F), n

a

be such that

n;,,, 1

Then

F.

F. charF X n

Gal (F /F) n

and

is isomorphic

Furthermore, the foll owing conditions are equ iva 1ent:

('11./n'll.)*. ('11./n'll,)*

is irreducible over

Proof.

more,

be a field, let

F

be the n-th cyclotomic extension of

F

( i}

as the n-th cycfotomic polynomial over

F

¢(n) s then

be a primitive n-th root of unity so that

0(2)

=



for some

is uniquely determined by

µ

1,;;;; µ < n

modulo

n

with

F

n

If

= F(s)-.

(µ,n) = 1.

and may be denoted by

FurtherCT • ).l

Then the map

l

Gal (Fn/F)->- ('11./n'll.)* CT

).l

t-->- ).l

is obviously an injective homomorphism. above and Propositions 4.3 and 5.7. Let

E/F

cyclicl if

be a field extension.

E/F

is Galois and

+ n'li. The second assertion follows from the

• We say that

Gal(E/F)

E/F

is abelain (respectively,

is abelian (respectively, cyclic).

91

FINITE FIELDS

5.10. PROPOSITION. extension of

Let Then

F.

Proof.

a ,a

Let

1

be a fteld and let

F

E

Ga 1 (E/F).

But if

E.

µ. ai(s) = s ~

required. Let

n

(µi,n)

with

Si nee

is generated over

E

(a a }(s) = (a a )Cs)

unity, it suffices to show that unity in

be the maximal cyclotomic

ts abelian.

E/F

2

E

1

2

2

is the order of Therefore

= 1.

where

1

s, ai(s)

s

by roots of

F

is a root of

is also of order

a 1 a 2 (s)

a a (s) 2

hence

n,

as

1

• n1 n2

n = p

p

1

2

nk ••• pk be the canonical decomposition of

n.

·Then

and therefore

For this reason, to determine the isomorphism class of ly assume that

n

(Z:'./nZ:'.)*,

we may harmless-

ts a power of a prime.

5.11. PROPOSITION.

(i)

If

ts an odd prime, then

p

(Z:'./pnZ:'.)*

is cyclic of

order pn- 1 (p-1} (ii)

Both

(Z:'./2Z:'.)*

Proof. (i} Since pn-l(p-1).

are cyclic, and if

(Z:'./4Z:'.}*

n;;. 3,

then

cp(pn) = pn-l(p-1) (for any prime p), (Z:'./pnZ:'.)* is of order

ts a direct product of its subgroup H of order n-1 pn-l consisting of the elements which satisfy d? = 1 and the subgroup K order

K

Hence

and

p - l

(Z:'./pnZ:'.)*

of the elements satisfying

d?- 1

are coprime, it suffices to verify that both If

n = 1,

then

(Z:'./pZ:'.}* = K

= 1. H

Since the orders of and

K

and this is cyclic by Proposition 5.l(i). 2

m + pn'll, E (Z:'./pnZ:'.)*. kp-l

which implies

k + pnZ:'. EK.

Also

(ef

and

are cyclic.

v-1

Hence we can choose an integer m such that m + pa'., m + pa'., •.. ,M n-1 distinct in Z:'./pZ:'.. Set k = rrf Because and

H

of

n-1 p-i

}

Since

( n)

= mcp P

- 1 (mod pn)

+ pa'. are

CHAPTER 2

92

n-1

= m (mod

k = rrl

p)

the elements k + p'll,, k 2 + p'll,, •.• ,kp-l + p'll, are distinct.

are distinct.

This implies that the order of

But the order of

K

is

hence

p - 1,

n

Hence also

is precisely p - 1.

k + p '1l,

is cyclic generated by

K

+ pn'll,.

k

To prove that H is cyclic, we may assume that n;;.;. 2 since otherwise H = 1 n.

Then H is a direct product of t ~ l

cyclic groups of order

Hence the number of solutions of the equation :J? fore suffices to show that the number of integers

sP = 1 (mod

pn}

then, since

sP

does not exceed p.

=s

(mod p},

p

is

= 1, x EH

ni;;.;. l.

1,,,

rt there-

pt.

satisfying

s, O < s < pn,

Now if s satisfies these conditions,

we have

s

=1

(mod p}.

Then if

s

f

1,

we may

write s

= 1

+ ypf + zpf+l

(1 ~·

f ~n-1,0 < y < p, z;;.;. 0)

in which case + (il(y+zp}pf + {1i) (y+zp) 2pd

sP = 1

+ ... + (y+zp)pppf - 1 + ypf+\mod Pft2)

If

sP

= 1 (mod

pn}

and

yp

so y = 0 (mod p}

this gives

f < n-l., f +l

= 0 (mod

contrary to O < y < s = 1 + yp

n-1

(ii)

The order of

('ll/2n'71,}*

is

)

Thus, if

p.

H

satisfies ,

1 < s < pn

This gives altogether at most

, 0 < y < p.

solutions including 1 and proves that

p

,~2

p

is cyclic. If n

¢(2n) = 2n-l.

or

=

2,

orders are 1. and 2, respectively, and there ts nothing to prove. n;;.;.

3.

We first claim that there are four distinct elements

satisfying x 2

l; tf sustained, it will follow that

=

product of at least two distinct cyclic groups n-1

a = l + 2 3

n-1

, a = -1 + 2 4

, x .. 1,,

=

n

a • + 2 '71,, 1,,

t

1.

('ll/2n'71,)*

Put a l

Then the

then these So assume that

x E

('ll/2n'71,)*

is a direct = 1,

a = -l, 2

.

xi are distinct and

FINITE FIELDS

xi

satisfy

93

which substantiates our claim.

= 1,

r

('ll./2nZ)* is a direct product of at least two cyclic groups

Since

the order of

is

('1l,/2nZ)*

2n-i, we see that, if x

E

(Z/2nZ)*,

or, what is the same thing, if m is an odd integer, then

m2

n-2

then

1 and

x2

n-2

=

= 1 (mod 2n).

2n-3 To complete the proof, we are therefore left to exhibit an x such that X f 1. n7/. 52n-3 = 5 =/- 1(mod 2n) Put x = 5 + 2 £ Observe first that, if n = 3, then n-3

but 52 k

- 1 (mod 2n- 1 ).

Now let

f;;.

be the largest integer

such that ?f-3

5Then we have k(3} y

3 and let k(f)

ts odd.

2.

=

k

= 1 (mod 2 )

Also for any

2f-3 3 we have 5

f ~

=

k(f)

1 + y2 ·

where

This implies that 52(f+l}-3 = (52!- 3/ = l + y2k(f)+l + y22k(f)

whence k(f+1)

~

so k(f)

k(f),

~

2

if f

52(!+1)-3 where

z = y

+ 2kCfl-\ 2 ts odd.

we deduce that k(f)

3.

Then the relation shows that

1 + z2k(f)+l

=

Thus

1 for al 1

= f -

~

f;;.

k(f+l) 3.

= k(f)

+ 1.

Since

k(3) = 2,

Hence

n-3

52

11 (mod 2n}

for n ;;. 3

and the result follows. • It will be shown below (Proposition 5,131 that Gal(a)/(f)}

~

('lL/nZ)*.

With

this tn mind, we now derive 5.12. COROLLARY.

Let n

k = p , p

a pri'me, let

F

be a field with

and let Fn be then-th cyclotomtc extension of F. unless p

=

2 and

k

~

3,

charF

rp

Then F/F is cyclic n

in which case Fn /F is either cyclic or Gal(Fn /F)

is a direct product of a cyclic group of order 2 and one of order 2t- 2 where t

< k.

Furthermore if p is odd if p =

2

94

CHAPTER 2 Proof.

Direct consequence of Propositions 5.9 and 5.11.

We now turn our attention to the case where 5.13. PROPOSITION.

((l)n :(Q) =

(ii }

¢

n

F = (Q.

be then-th cyclotomic extension of

(Q

n

be then-th cyclotomic polynomial over

n(X)

( i)

Let



and let

aJ'

variables, has a non-

r.

5.22. PROPOSITION. (Chevalley). Proof.

xF

Er

f(- End(S), sf--->-£

be the regular representation of

s

R

By the trace (norm,characteristic polynomial) of s ES we understand the

trace (determinant, characteri sttc po 1ynomi a1 of

The trace and norm of

£ )• s·

s

will be denoted by

respectively. Now assume that

E/F

is a finite field extension.

dimensional F-algebra and so for any ;\EE,

Then

E

is a finite-

we may define the trace (norm, char-

acteristic polynomial) as above. 6.1. LEMMA.

Let

E/F

be a finite field extenston, let :\EE and let f

be

the characteristic polynomial of :\. (i)

is a root of f

;\

:\ over (ii)

and, if

E =

F(:\),

then f

is the minimal polynomial of

F.

If K/F is any field extension in which

f

splits into linear factors,

n

f =

il (x-:\.), then

i=l

1,

n Z :\. i=l

1,

NORMS, TRACES AND THEIR APPLICATIONS

Proof.

Let

(i)

be the matrtx for multi'plicatton by ;\ tn

M

respect to a given F-basis of E. This implies that If E

=

F(>..),

f(">.)

and

0

when

f(X) = g(X)m,

(ii) NE/F(">.)

=

(iii} TrE/F( ">.) Proof.

Let g(X)

over F,

">.

hence the assertion.

when

and let f(X)

Then

[Nx;/1.) ]m = m [TrK/p{:\)

l

Properties (ii) and (iii) follow from (i) and Lemma 6.l(ii).

is a basis of E over F.

Let M = (aih)

tn K with respect to the basis 'A(yizj)

To

be a basis of K over F and let {zjl1 < j


Ea.,n-1 )p -1 1 n-

(b. E Z)

o

Let j

1,,

be the minimal index for

Then the number

J

s = (bl

= {b

are divisible by p.

b~s

r

i=o r

and not all the

is not divisible by

(-l)na 0

exists r ER such that

which

dividing DF;,)

p

n-1 -1

+ ... + bn-i'>--

)p

=

r- ((b/p) + (b/p);, + ... + (bj_/p):\

j-1

)

ts an algebraic integer and so ts also _.

s1 -

n-1-1 _

bi

p

- SA

n-j-1

.n A (b j+ l + b j+ zil +

-

+b

n-1

"n-j- 2) P

1

rt follows that n

(

p NF/(!') s 1

Since

p2

,!' NF/,),

)

= NF/cr:)(p1\ )

(.

n-1 ) = bl n n-1 Fft;(;,)

= NF/IJ) b}'-

the latter impli'es that plbj'

a contradicti'on. •

We close this section by providing a number of examples of the calculation of a discriminant. 7.18. PROPOSITION. n

> 2,

Let E/F be a finite separable field extension of degree

let A EE be such that E = F(X)

of :\ over

.F

and assume that the minimal polynomial

is

f(X)

=

i1

+ax+

b

rhen DE;p(X) = (-1 t(n-1)/2(nnbn-1 + (-1 t-l(n- t-1an)

Proof. µ = f' (:\) ,

By Proposition 7.8, we have

DE/F(A) =

(-lt(n-l)/ 2NE/F(f'(J,,)).

Putting

DISCRIMINANTS AND INTEGRAL BASES

0 implies n11.n-1

since 11.n +a\+ b

11. = -nb(µ

The minimal polynomial

of µ

g(X)

-1

b

1

= -na - nb\- .

129

It follows from this that

+ {n-l}a) -l

over F is therefore the numerator of .

-1

f(-nb(X+(n-l)a))

Hence we must have

The norm of

is

µ

ttmes the constant term of this po lynomi a 1 , i . e.

(-1) n

NE/F ( µ )

= n

which clearly yields the result.

nbn-1 + (-l)n-l(n-l)n-l~n ~



We now turn our attentton to quadratic fieZds, i.e. algebraic number fields of degree two.

An integer d

is said to be square-free if d

E 'll.

not divisible by any square except 1. d = -1

or

Id I

Thus

and d is

is square-free tf and only if

d

is a product of di sti net primes.

a square can be uniquely written in the form n We refer to s

t

Every integer n which is not

= m2 s

where s

is square-free.

as the square-free part of n.

For convenience, we assume that all quadrati'c fields are embedded in ..) = 2A-l = la,

it follows from Proposition 7.8 that

-:"'-.-.··,.·

DISCRIMlNANTS AND INTEGRAL BASES

Since

131

is square-free, i't follows from Proposttton 7.12 that

d

and

d(F} = d

l,A is an integral basis of F.

that

Assume that

d

= 2(mod

g(X) = X2 + 2X + l - d. f'(./a} = 2tl'a,

4}

or

Since

= 3(mod 4).

d

Let

f(X) = X2

-

Id and

i's the minimal polynomial of

f(X)

and

d

we have DF;tf)la} = -NF;fI}21J} = 4d

Let p

d.

be any prime dividing

Proposition 7. 17 wtth respect to and

are the same.

d(F)

by Proposi tton 7; 12.

If d

Proposition 7 .17 for

p,

satisfies the hypotheses of

hence the highest powers of p

= 2(mod

d

= 3(mod

p = 2.

Since

If

f(X)

Then

4},

4),

then

4d

d(F) = 4d

is even, hence

d

g(X}

then

dividing

sattsftes the hypotheses of

Id - 1 is a root of

and

g(X}

DF/a 1

I

2

and

x2

-

2

+bla=a

2



+ b Id and

1

1

dy 2 = ± m,

then

a

a

1

2

1

>a ""'b 2

l + 15

are positi've solutions with

>b

l

2

+ b ./J

3 + /5

1

x2 !>

-

a + b la need not imply

!>

2

d

=2

or

3(mod 4)

2

5y 2 = ± 4,

l + /5

Our next result provides a useful charactertza tton of possible to treat the cases

(_2)

2

+ b la are, positive solutions of

Indeed, consider the double equation

> b •

are

la and a 2 + b 2 /if are two solutions of (1}, then

We warn the reader that tf a

b

b

and

The positive solutions will be ordered by the size of + b

1

will be called positive if a

and

but b Ed'

Then l

= b

= 1.

A1though it is

=l(mod 4)

d

2

3 + /5

simultan-

eously by applying Remark 8.2, we consider them separably for practical purposes. 8.5. PROPOSITION,

where (ti)

(i}

If

d

=2

or

3(mod 4), then

a + bid ts the minimal positive solution of x 2 If

d

= 1 (mod

dy 2 = ± 1.

-

4}, then Ed = (1/2}-(a+b./J)

where

a + bid is the minimal positive soluti'on of x 2

Proof. satisfying

(i)

Owing to Lemma 8.l(i), we may write

-

dy 2 = ± 4.

Ed= a+ bid with

a,b E 'll

UNITS IN QUADRATIC FIELDS a: 2 ~ dii2 =

Since Ed > 1,

solution of x 2

± 1

Lemma 8.4(ii) implies that a > 0

is a positive solution of x 2 -

-

139

dy 2 = ± 1.

Hence a + bid

and b > 0.

Let a:

l

+ b Id be any positive l

Then a + b Id> 1 and, by Lemma 8.l(i),

dy 2 = ± 1.

l

1

+ b/d E U(R}.

a:

Therefore, by Lemma 8.4(i), a

Because (ii}

Ed > 1,

+

l

this shows that

a

b

1

Id=

+ b Id ;;,,

1

for some n;;,, 1

End

l

+ bid,

a

as required.

Apply arguments of (i} with Lemma 8.l(t} replaced by Lemma 8.l(ii). a Observe that if N(Ed) = 1, then

if d = 2 or 3(mod 4}

and

= 1 for all units

N(u)

u

of

Hence,

R.

then by Proposition 8.5(i), the follow-

Ed= a+ bid,

ing holds: (a)

a+ bid is the minimal positive solutl'on of

according to whether N(Ed)

1 or N(Ed)

=

=

x2

-1.

-

dy 2

= 1 or x 2

In particular,

-

N(Ed}

dy 2 = 1

= -1 if

and only if the equation x2

-

-1

dy 2

has no solution in Z. Similarly, if d = 1 (mod 4)

and

Ed= (1/2) (a+b/d),

then by Propositton

8.5(til, we have (b)

a+ bid is the mini'ma1 positive solution of x 2

-

dy 2 = 4

or x 2

-

dy 2 = -4

according to whether N(Ed} = 1 or N(Ed) = -1. In particular,

N(Ed) = 1

ff and only tf the equation x2

has no solution in

Assume that d = 2 or

(i}

with a ,b EN.

Then the sequence

l

dy 2 = -4

Z.

8.6. PROPOSITION. l

-

3(mod 4}

a + b Id = (a +b Id) n

n

gives all positive solutions of x 2

1

-

1

dy 2 = ± 1.

n

and let

Ed= a 1 + b/d

(n ;;:> 1)

CHAPTER 2

140

Assume that d= l(mod 4)

(it)

and let Ea= (1/21Ca/b 1 v'd)

with a ,b EN. 1

1

Then the sequence (n;.. 1)

gives all positive solutions of x 2 Proof.

of x 2

n

;;i,

and

U(R}

E

a

n

+ bn IJ, n;.. l,

Moreover, by Lemma 8.l(i) and 8.4(ii), they are positive solutions If z = a + bid ts any positive solution of x 2

dy 2 = ± 1.

-

then z

Because Ed is of infinite order, all the

(i}

are distinct.

dy 2 = ± 4.

-

z

1 so that by Lemma 8.4(i},

l>

z.

= E~ = an +

-

= ± 1,

dy 2

bnld for some

1.

(til Apply arguments of(.,'). with Lemma 8.l(i'). replaced by Lemma 8.l(ii). • We now apply the result above to provide a rather crude method for calculating the canonical fundamental unit Ed of R. cases d

=2

or 3(mod 4)

8.7. PROPOSITION.

and d

= l(mod

It will be convenient to treat the 4)

separately.

Assume that d = 2 or 3(.mod 4}.

In order to calculate

Ed = a 1 + b 1 Id it suffices to write down the sequence

first number db 21 Proof.

1

db 2 , b = 1,2, •.. ,

of thi's sequence which differs from

Let an+bnld

(a ,b

±

1

E

lN)

and to stop at the l

by a square a 21 •

be the sequence defined in Proposition 8.6(t}.

It

suffices to verify that b

n+l

l>

b

for all

n

n

;;i,

Indeed, in this case Ed can be characterized as the unique positive solution a + bid

of x2

-

dy 2 =

±

1 with smallest possible b.

an+l + bn+lid = (a 1 +b 1 la)(an +bn Id)·

we have bn+l-=ba +ab >bn 1 n 1 n

as required.



Because

l

UNITS IN QUADRATIC FIELDS

To tllustrate the case d = 2 or 3(mod 41, the sequence db 2 , b a

8,

= 1

1,2, ...

=

we see that

8 +

3./T

stmilar argument shows that

Ed=

1 +

12,

7,28, 7.3 2

=

let d 82

1,

-

{7,6,2}.

E

If d

so taking b

1

=

7,

=

3 and

is the canonical fundamental unit of (Q(/7).

5 + 2/6

8 + 317

Note that both units and hence

ts

141

ts the canonical fundamental unit of (Q(/6).

5 + 2/6

and

A

1.

have norm

By taking d

2,

=

we obtain an example of a canonical fundamental unit of

norm -1. Turning to the case d = l(mod 4], observe that tt is not true in general that b

n+l

> b

for all

n

n > 1.

Indeed, it will be shown below that

the canonical fundamental unit of (Q(/51. + bn 15

a

n

then b

= b

l

= 2

21 -n(l+/S'n l

=

d

= l(mod

the

= l(mod 4}.

method of Proposition 8.7 tn order to treat the case d Assume that

In order to calculate

4}.

(a ,b 1

it suffices to write down the sequence dcl ,b term db 2

is

Hence, if

This remark explains why we have to modtfy slightly

1.

8.8. PROPOSITION.

(l/2}(1+/51

1 , 2, ... ,

=

of this sequence which dtffers from

JN)

and to stop at the first

by a square a 2

± 4

E 1

1

(if there

1

are two such squares, then as Proof.

bn +l = (l/21(a · · 1bn +b 1 an }

and so b

=ab

2

1 l

8,9. REMARK.

~

as required.

b , l

u.

a

1

posstble

+b

1

d f 5.

and that

If

(1/2}(a +b lcf}

E~ =

then



d = l(mod 41

Assume that

1

1

Id is the unique postttve solution of

= ± 4 with smallest

x 2 -dy 2

:xl -

Indeed, if a + bid is a postttve solution of

b . 1

with sma 11 est poss tb 1e b,

2

+b

Accordingly, 1 - db 2 = -4

2

l

+b

1

Id

or a + bid= a

Id, then b l

= b

2

and so db 2 = 5.

l

dy 2 = ± 4

then by the allove

a + bid = a

Hence, i'f a+ bid= a

11

=ab

11

+ b Id

2

2

tn which case

Because d;.,, 2,

1

a

l

=

1.

we infer that

d= . 5,

as claimed. • The remark above together with Proposition 8.7 shows that for any square-free integer d

;i;,,

2 distinct from

positive solution of x 2

-

d=l(mod4),

first term of the sequence 1

= 1 and

a = min{l,3} 1

Ed= (l/2}(a+b/J}

where

a+ bid is a unique

dy 2 = ± 4 with smallest possible b.

To illustrate the case

b

5,

db 2 ,b =

1.

let

dE{5,13,17,69}.

= 1,2, ... , is 5 = l2 + 4 Therefore

(1/2}(1+/5)

= 32

-

If

d=5,

4.

Hence

the

is the canonical funda-

UNITS IN QUADRATIC FIELDS

143

mental unit of (J?(/5), If d 13

=

then the first term of the sequence dh 2 , b

13,

=

32 + 4.

Thus

b

=

l

1 and a

fundamental unit of

1 be coprime integers.

m if x 2 = a(mod m)

for some x

Let a

We say that a is a quadratia residue moduZo E

Z.

Otherwise, we say that a is a quadratia

nonres~aue moduZo m.

For each n e group of Z/mZ.

1et ii be its image in 'll,/mZ and let U(m}

'll,,

be the unit

a

Then a is a quadratic residue modulo m if and only if

a square in U(m}. Let p be an odd prime and let a be an integer coprime to p. the Legendre symbol

(alp}

1, (alrl

of a,

relative to p,

We define

as follows:

if a is a quadratic residue modulo

p

==

-1,

ff a ts a quadratic nonresidue modulo

p

A straightforward verification shows that

(i] (ii}

If a= b(mod p),

then

(ablp} = (alrl(blp)

8.10. LEMMA,

(alrl = (blp)

(in particular,

(a 2 lp)

=

1)

Let p be an odd prime and let a be an integer coprime to p.

is

144

CHAPTER 2

Then

and, in particular, (-lip)= (-a2Jp) = (-l)(p-1)/2

Proof. 0



It is now easy to derive the following properties: 8.16. PROPOSITION.

If d = n 2 + 4 for some n

(t)

Ed= (n +

(ii}

If d

=

2 f m E lN,

m2 + 1 with

=m +

Ed

Proof.

( i}

/n 2 +4)/2

Suppose that

and n 2 + 4 are the same.

mE

/m

2

E

and N(Ed}

lN, =

then -1

then +l

and N(Ed} = -1

lN is such that the square-free parts of m2 + 4

Because d ts square-free, if m In,

then

n 2 + 4 < m2 + 4

and so n (ii)


2,

determine

Ed

exclusively in

terms of d. In this vein we quote the fo 11 owi·ng result due to Ri chaud (1866} and Degert (1958)_. 8.18. PROPOSITION,

Let d > 2 be a square .. free integer and write d

-n < r..; n.

divides

If r

n+IJ with Ed =

(n + /al/2

4n,

n 2 + r,

then

N(Ed) = .. sgnr

wtth N(Ed1

=

for

= .. sgnr

[(2n 2 +r) + 2n/a]/r wtth

lrl

=

for

lz,I = 4 for lrl

N(Ed} = 1

1

(except for

d=5,n=2,r=l}

t- 1,4

9. UNITS IN PURE CUBIC FIELDS A rational integer d > 1 ts said to be aube-free i'f d cube of a prtme.

The field

F =

integer and ¥1, is rea 1, ts ca 11 ed a pure aubia field. the ring of integers of the pure cubic field

F =

5,

then

divisible by

p

p,

3 % (3k+1) (3k+2)

x = ± 1

and stnce tt is clear from (4) that

x = -y.

we have

Now assume that p p

(3k+l )(3k+2}

n = 2 (mod3),

and when

xy < 0,

(x,y) = 1.

1.

y = ±

When

153

ak

d with paid but pa+l} d.

is a prime di'visor of

k > 5 :;;. 3k + 2

for

k ;;, 1 ,

which is impossible since

If

so each term in the sum ( 5) is (x,d) = 1.

we reach the same contrad1'ction.

If

Therefore

p = 3,

r°'

=

then since 2 or

5, and

d E {2,5,10}.

The informatton obtained so far shows that (n-2) 2d 2 ( n-2) 2d 3 45 + · 6 7-8 If

0

(6)

d = 10, this becomes

This equation is true for mentioned in the lemma. and obtain

n = 5,

and leads to the first of the exceptions

For other values of

n,

we may divide through by (n-5)/6

154

CHAPTER 2

n2

4n

-

+

= _

6

k+l z: (-l)k(n-6) (n-2)(n-3l(n-4)•12•10 3k-l 3k(3k+ll(3k+2)(3k+3)(3k+4)(3k+5) 71

The highest power of 5 which divides the denominator of a term in the sum is obviously at most 5(3k+5), n2

-

and because 5k+l

>

5(3k+5)

for k;;;, 2, we have

4n + 6 = (.n-2} 2 + 2

= (n 26 l (n-2 l(n-3 l(n-4). • 12 • l 0 3•4•5•6•7•8

= O(mod5) which is impossible since -2 When d

=

2 or 5,

is a quadratic nonresidue of 5.

equation (6) implies the congruence

which is again impossible, by Lemma 9,3, We are therefore left to examine the possibility Y

=

0.

The proof that this

happens only in the case of the second exception mentioned in the lemma is completely similar to what has just been done for the case z

=

0 (the only variation

lies in the fact that d may now have the sole pri'me divisor 2, so that d or 4l.

=

2

The 1emma i's therefore es tab li'shed. •

Let

d >

1 be a cube-free integer and

the rl'ng of integers of 1.

In the second case we have

But then

x,y,z

are not d1'visible by

This i's impossible, since i'f The case in which

p

=

ab[6

3,

then

so that . is fixed by all elements If

K

C

1 -

then every

K , 2

K ' l

then H

l

=E

and

is also the identity map on

K

H c H ,

K

H

cp

If

This proves (a).

K •

>, proving (b}.

H

as asserted.

2

which is the identtty map on

H c H •

elementwise if and only if

K K

then A. is fixed by all elements of

hence by

2

2-

1

fixes

E

Conversely, if :>..EE

2,

a E Gal(EIF}

i.e.

K

of

H

H

nE

cp

E

2

= K2



Let EIF be a Galois extension.

We shall sometimes call the group Gal(EIK)

of an intermediate fteld K the group associated with K.

We also say that a

subgroup H of Gal (EIF) · beZongEJ to an intermediate field K if H = Gal (EIK). If K ,K 1

2

are two intermediate fields, then we say that K

l

gate under an automorphism aE Gal(EIF) if

10.2. LEMMA.

cr(K) l

=

and K

2

are conju-

K, 2

Let EIF oe a normal extension and let K E>e an intermediate

field such that K/F ts normal, ~

Then the restriction map

Gal (EIF) -

l

a

Gal (KIF)

1-+

alK

induces an isomorphts~ Gal (EIF)/Gal (E/K) Proof.

Gal (KIF)

Si nee K/F i's norma 1 , the restriction of any cr E Gal (EIF)

is· an automorphism of

K,

homomorphism wtth kernel

hence alK Gal(EIK).

tion 2,9, the result follows. 10.3. LEMMA.

~

E

Gal (KIF).

Since cr

1-+

to

K

The given map is obviously a crlx is surjective by Proposi-



Let EIF be a Galois extension, let K ,K l

2

be two intermediate

.""':. , .

-

FINITE GALOIS THEORY

fields and 1et

H. = Gal(E/K,l,i = 1,2. 1,

1,

(i1 Then, for any given (ii)

a E Gal(E/Fl, cr(K l = K 1

is a normal subgroup of

(.i)

Proof. = 1,2.

Bydeftnttton,

Assume that

w- 1 (.y)=o- 1 (y}

K

1, 1

= aH

1

a- 1 .

and when this is so,

Then, for any

y

e K , a-l (y} e K , 2

CITcr- 1 (.y}=y,

i'.e,

,EH,

2

tfandonlyif ,(:x:)=:x: for all

1,

= a{K

H

Gal(E/F)/Gal(E/K)

,Elf.,

2

forall

:!!

if and only if

ts normal if and only if

K/F

Gal (E/F}

Gal(K/Fl

i

2

If K ts any intermediate fi'eld, then

Gal (E/Kl

169

:x:EK., 1,

hence

1

whichmeansthat

l

OTO

-1

E lf , 2

aH a-l c H

Thus

1

-

H = aH a- 1 • 2

K = a- 1 (K l

2

1,

tt follows that

o· 1H o c H, 2

-

1

i.e.

Retracing our steps, we obtain the converse,

1

(i'l'l

and since

2

Apply Ci 1 and Lemma 10.2.



We are now ready to record our matn result. 10.4, THEOREM, (Main theorem of fi'ntte Galots theory}.

The map

Kf---+

subgroups of H 1--r

E8.

(iil

If

G

Gal(E/Kl

be a finite

from the set of intermediate fields into the set of

is an i'ncluston.,.reversing bijection with the inverse map given by

B is a subgroup of G,

IHI= (ii il

E/F

G.

Galois extenston wtth Galois group (i}

Let

The extension

K/F

then

(E:rl

and

(G:Bl

= cE8:F)

ts norma 1 if and only if

Gal (E/K) - gi (i ,e, the restri'ction of the i .. th coordinate projection

In fact of

1[ Gi

to

satisfles this condttion,

G}

These

fi

are called canonical

homomorphisms.

(bl

If every

fij

f,. 1,,J

Assume that each Given

j EI,

is injective, then so are all the

let

is injective, and

k EI satisfy

fikh(Yl,

whence by our hypothesi's,

fjkfk(y) =

f/Yl

(cl

If

j EJ

J

for every

i: ,;;; j],

such that

and so

if we define

Now

fikfk(.x)

x

=

x,y E G.

fi(x)

Therefore

= fi(_y)

fj(x)=fjkh(x)

=

= y,

(_i .e, if for every

I

i EI there exists

then

Hm

G, 1,

lim

2cc

G.

+-jEJ J

na.,

G. C there exists a unique g Elim G. such +-j6J J - jEJ' J +-iEI 1, j E J', the j~th coordinates of g 1 and g are equal, In fact,

g1

that, for every

= fAYl for some v

= fk(y)_,

fk(xl

--i'EI

Indeed, given

v

i,j,;;; k,

j EI,

i's a cofinal subset of

fAxl

fi,

E Hm

g = (gi)

with

gi = fij(gJ) (i ,;;; j),

then

g' .-- g is a required

i'somorphism. The following result provides a universal characterization of projective 1imi ts.

11,4, PROPOSITION.

The projective limit

G = 1 im G. of the projective sys tern .,,_1-

CHAPTER 2

176

{G.,f,.\i,j EI} ~

~

of groups (rtngsl sattsftes: .

there are homomorphisms

if lI ts a group (rtngl and if

o. : H-t- G •. with commutative diagrams: ~

~

H ( 1)

(i ,;;;; j)

then there exists a unique homomorphtsm

cr : H __.. G for which all the datgrams

H - -0 ----i,..~G (2) (i

are commutati've (where characteri'zes Proof,

ft ts the canoni'ca 1 homomorphtsml.

Gi'ven

cri(_h} = fi(.a(.h}l,

h EH,

o(h1

E G,

set

cr(hl = (o,tLhl}

Thus

Th i's property

E

nai,

Because of the commu-

r:r : H--+ G i's a homomorphism satisfying

whence the commutativity of (.21 results.

also makes (21 commutative, then er=

I)

G up to isomorphism,

tat1'vity of (11,

Thus

E

fi [o(_h)l = fi[cr'(h}l

If

for all

o' : H

-+ G

h EH, i EI,

0 1•

In order to establish the second asserti'on, assume

and maps ' i : H0 - + Gi

H0

have the property formulated in the first statement of the theorem, extst untque homomorphisms

O' :

H0 -

a and o-0

: G-

H0

Then there

satisfying

fia = 'i

PROFINITE GROUPS

and T/r0 = f,c 00 0 CJ'

177

We tnfer fi = f/rcro and Ti= Tp0o for all

and o0 o are the identity maps of

G

and H0 ,

i EI,

Thus

respectively, provtng that

fs an isomorphism. • We next intrdduce the notion of homomorphism for projective systems of topolo-

gical groups (rings).

Let

be two projective systems of topological groups (rings) indexed by the same directed set I.

A homomoPphiem

{¢. : G.1,· ---+ H1,., i E.I} 1,

¢: X -

Y is a set of continuous homomorphisms

subject to the condition that the diagrams

G·l - - - - - - -fiji ) l l l a . - Gi (i..;;

be commutative. 11.5. PROPOSITION, i; ..;;·j}

... ff.

H·J

l

i ~j} and Y. = {H.,o.,l{,j EI, w 'Z, ~ be two projective systems of topo1ogtca1 groups (rings} and 1et ¢:X --

be a homomorphi'sm.

Let X

j)

= {G.,f .. li:,j EI, 1,

Then, there is a untque homomorphism (which is necessarily

conttnuousl

¢*: lima.- limn. + - 1,

such that, for every

i; E I,

¾-·-1,

the diagram

(3)

Y

178

CHAPTER 2 (f.1, ,P.1,

is commutative

denote the canoni ca 1 homomorphisms).

furthermore, if every ¢.

is injective, then

1,

Proof.

The homomorphisms ¢1,.,i EI,

¢*

In fact

is injective.

induce a homomorphism

~: na.- na. 1,

given by ¢({gi)}

1,

The commutativity of (3} implies that if

(¢i(gi)).

g = (g .}

E

1,

i:hen ¢(g l

E

1imH. ; hence define -

¢*

Wi'th thi's

qi*,

then

If \)J : limG.

p .¢* = p .\)J 1,

1,

¢*(x}=¢*(y),

injective, then

Let

=

{K.,?-. .. \ i·,j 1,

E

1,J

= (¢i)

ltmG. ,

makes (3) commutative for

+-- 1,

thus

¢* = \)J.

=qi.f.(y}]. 1, 1,

i

for all

E."

Therefore,ifevery

and thus

I

x = y.

¢.1,



¢

=

(i)

Y

to

be a homomorphism from x to z,

Assume that for all

i

E

Y

J,

and the

sequence

Then the induced sequence 1-+ limG. -

L

lima.

1,

L

limx. -

-1,

1

.,__1,

is also exact, Proof.

By Proposition 11.5,

cation shows that surjecttve. J

To this end, fix

Th.en all the

1,

A. J

¢*

Ker¢*= Imcp*.

A.= {(h.)\(rz.) 1,

E

is

be three projective systems of compact topolog-

I,i..; j}

a homomorphism from

is exact.

¢*

X = {G.,f .• \i,j EJ,i,;;; J'}, y = {H.,a .. \i,j EJ,i,;;; j}, 1, ~ ' ' 1, ~

ical groups (rings} and let \)!

cp.[f.(xl 1, 1,

f/Y1

limH.

-->-

1,

for every i,

then

f/x) =

11.6. PROPOSITION,

z

to

the diagram (3) becomes commutative and

-

and

-qi

to be the restriction of

-1,

is obviously continuous,

If

1,

1,

i·.e. by formula (4).

every i',

limG. , -

is injective.

A straightforward verifi-

We are therefore left to verify that \)J* (k.} 1,

in

-

limK. 1,

and consider the sets

f-\H., \)J,(h.) = k., a .. (h.) = h., i EI 1, J J J 1,J J . 1,

are closed and nonempty.

for all

i,;;; j}

Furthermore, the collection

is

179

PROFINITE GROUPS

{A .},j EI, J

possesses the finite intersection property,

Since

n

H.

iEI 1, Tychonoff' s theorem, it fol lows from Proposition 1.6.5, that each element of

n A.

result follows.

jEI J •

belongs to

is compact, by

and is mapped by

~Hi

nA. f 0.

jf::;I J l/;* to

For the rest of th.e section, we are interested in the case

Since the

(ki),

{Gi,fi)i,j EI}

is a projective system of finite groups, each endowed with the discrete topology. We call the projective li'mit

-" compact group to

gives a criterion for a Let

11.7. PROPOSITION.

G = limG.

G

a profinite group.

The following result

be profinite.

be a compact topo 1ogi ca 1 group and 1et

{N,

1,

Ii

E I}

be a family of closed normal subgroups of finite index such that Cil

for every finite subset

of

J'

there exists

I,

i

such that

EI

c: n N. jEJ J

1,-

ilN.=1

(ii}

iEI

1,

G ~ HmG/N.

Then

-

Proof. means

{13/Ni,fij li,j E I}

Conditi'on (_i} guarantees that and

N. ::i N.

" -

topologically and algebraically.

1,·

J

f •• : G/N, 1,,J J

@/N. 1,'

tive system of topological groups.

l

(where

is the natural projection) is a projec.

.

Condition (iil ensures that the map

G

.JL. ~ l. i·m

g

i-r

GIN. 1,

(gNi}

is an injective homomorphism (which is obviously continuous).

(g1,·.N1,,• } then the closed sets compact, there exists

e

N.

g .N. 1,,

1,,

g

E

E

1i m G/N1,. ,

+--

have the finite intersection property.

n g,N,.

iEI is a continuous bijection.

1,

This element satisfies

Since

6(g) = (giNi)'

G

is so

1,

Since

G

is compact and

lim G/N.

+--

(Proposition 1.6.4), it follows from Proposition 1.6,6 that as required.

If

e

is Hausdorff

1,,

is a homeomorphism,



We now provide some useful characterizations of profinite groups.

11.8. PROPOSITION. tions are equivalent:

Let

G be a topological group.

Then the following condi-

180

CHAPTER 2

(il

G is a proftnite group

(ii}

G

is a Hausdorff, compact group which has a bas is of open neighbourhoods of

l consisting of normal subgroups (iii)

is a Hausdorff, compact, totally disconnected group.

G

Assume that G is a profinite group.

Proof.

is compact and Hausdorff.

The normal subgroups U8 =

S

a finite subset of

I, N, 1,,

neighbourhoods of l. (ii)~ (i)_

N

iES

1,,

where

1,,

a normal subgroup of G., 1,,

form a basis of open

This proves that (i) implies (ii).

If

be the connected component of 1.

The implication Assume that (ii)

{u} Cl,

let Na= ua n N for all as

are open normal subgroups of N; a.

G

nc. X nN.,

iES

of all open normal subgroups of G,

for all

u8 n G,

ts a consequence of Propositions 11.7 and 11.3.

holds and let

NCl,

By Proposition ll.2(ii),

denotes the family a.

The groups

is connected, we must have

N

N

Cl,

= N

But then

N = nN = n(N nu)= N n(n u l = N n 1 act

a

a

a

the last equality bei'ng true by hypothesis (Hl,

a

Thus

N

=

and so (iii} holds.

The implication (iii)~ (iil being a consequence of Proposition 1.6.7, the result fol lows.

11

11.9. PROPOSITION.

Let G be a proftntte group.

(closed normal subgroup} of

Then any closed subgroup

i's the i'ntersection of some open subgroups (open

G

normal subgroups}. Proof. pose that tains H.

Let H be a closed subgroup (closed normal subgroup) of G and supg E

G belongs to every open subgroup (open normal subgroup) that con-

Then, for any open normal subgroup N of G,

1

S

of open normal subgroups of G,

s

s

n (gN. n H) = g( n N.) n H

i=l

Stnce the sets

gN

i=l

1,,

n H are closed and

and so

0.

gN n Hf

Hence for any finite family N , ••• ,N

g EHN

G

we have

t- 0

1,,

is compact, it follows from Proposition

PROFINITE GROUPS 1.6.5 that there exists h

such that h

EH

181 for all open normal subgroups

E gN

But then g-lh belongs to all open normal subgroups of G.

N.

Since G is

Hausdorff and some open normal subgroups of G form a basis of open neighbourhoods of 1 (Proposition 11.81, it follows from Proposition 1.6.2 that g-lh

1.

=

Thus g EH and the result follows. • 11.10. PROPOSITION. then

(il

If

is a closed subgroup of a profinite group G,

H

is proftntte.

H

(iil If N is a closed normal subgroup of a profinite group G,

then G/N is

profinite

(nn

The direct product

G

=

n

G.

iEI

of an arbitrary family

{G.

i

Ii

of pro-

E r}

i

fini'te groups ts profi'ntte (i'vl Every projective Hmit

(il If

Proof.

N. i

open normal subgroup of

limG. of profinite groups is profinite.

G =

_i

ts an open norma 1 subgroup of The family of all groups

ll.

l\ypotheses of Proposi'tton 11,7 and thus A closed normal subgroup

(ii')

by Propostti on 11 • 9,

G,

N

normal subgroup of

G., J

of G of finite index. ly, G ~ ltm

~-

H~

lim

then

H .

is an

n N.~

satisfies the

ll n Ni

n Ni} is a profinite group.

H/(H

ts an intersection of open normal subgroups of

Now apply Proposition 11 • 7.

Let J be a fi'nite subset of I.

(iii}

G,

Then

NJ=

Given j

r\N. x iEJ,,-J n Gi jEJ' J

E J,

let N, denote an open J

is an open normal subgroup

But the intersection of all such

ts 1,

NJ

According-

i's a proftntte group, by Proposition 11,7,

G/NJ

(ivl We know, from Proposi'tton 11.2(;}, that G ts a closed subgroup of i!Gi. By (iii-1,

1\G.

iEI

ts profi'nite,

Now apply (il,



i'EI i

An important result for the cohomology theory of profinite groups ts: 11,11, PROPOSITION, group of o{Hl

=

Let G be a profinite group and let H be a closed sub-

Then there exists a continuous cross-section a :

G.

1, l',e.

there exists a continuous map a:

spaces such that the composite map Proof.

G/H-+ G

G/H ~ G --. G/H

G/H--+

with

G,

of topological

is the identity map,

Let K ~ s be subgroups of G such that S/K is finite.

show that there ts a continuous cross..-sectton o : G/H __. G/K with o(H}

We first = K,

182

CHAPTER 2

Indeed, let uG be the family of all normal subgroups of G of finite index. {UKIK\U E u0} is a basis of open neighbourhoods of

Then

is fi'nite, there exists u E uG such that

(UKIK) n (SIK}

l

E

=

1.

Because SIK

GIK.

Then the natura 1

map : UK/K __.. UHIH

is a homeomorphism.

Let

be a transversal for u in

I\ = l,. . .,gk

G

and put

(n = 1 , 2,. , • , k, u

a/H to GIK with

The map cr is obviously a continuous cross-secti'on from cr(H}

U}

E

= K.

Let x be the set of all pairs {S,cr}, where s is a subgroup of is a continuous cross-section from GIB to GIS. {S,cr} < {S',cr'}

if S'

cs and

We may order x by setting

cr is induced by cr',

Indeed, let {S.,cr.},i EI be a chain in 1, 1,

X

the set {S.,cr.},i EI. 1, 1, In this case s index.

=

1.

The set X is inductive

and let s

The cross-

n Si,

=

1,"EI

sections cr.1, : GIB-->- GISi induce a continuous map a continuous cross-section cr : GIB -

and cr

B

Q" I

: GIB -

lim GIS ., -

i.e.

1,

Then {S, cr} is an upper bound for

G/S.

By Zorn's lemma, there exists a maximal element

{s,o}.

Indeed, if sf 1, then s has a proper subgroup of finite

Then, by the foregoing,

{S,cr}

is not maximal.

This implies the re-

quired assertion. • We close this section by providi-ng a number of examples, (al

If p

is a pri'me number, then the rings '11/pn'll.,n

E IN

form a projective

system with respect to the canonical projections 7/l

n

4p 'll, -

m

(n;;.. m)

'11./p '11.

The corresponding projective limit '11. = lim '11.lpn'll. p

i's the :nng of p--adic 1,ntegers.

-

Every element of '11.

p

is a sequence

x = (x.+pi'll.),i EN, 1,

where x.1,

E '11.

and x.J = x 1,.(mod pi'll.l for .

J' ;,i,

i.

A basic neighbourhood of

x

PROFINITE GROUPS

i's gi'ven by an tnteger m ~ O and

183

tt conststs of all elements



=

(y .+piZ} ,i E lN 'l,

with

The map

i's an embeddt-ng and we identify Z wl'th its image in

(xil,i i'n

E

x = (xi+pill,i

lN converges to

Z. p

lN in the p-adic topology,

E

-

Z is dense

Z/piz

?(x~+piZ}

1----+

'l,

has kernel

(b}

Because the series

The canonical projection

iZ

index

zp •

i pl.

i. The ri'ngs

tions

p

Thus

Z/nZ, n

piZP

~

Z/nZ-. Z/mZ for

ibtlity

min.

lN,

x .. + 'l,

iz

i's an open subgroup of (the additive group)

ZP

of

form a projective system with respect to the projec-

min,

where the order in lN is now given by the divis-

The projecti've limit

Z=

ltm Z/nZ A

is called the Prufer ring,

The groups

nZ,n E lN,

are precisely the open sub-

~

groups of the profl'nite group

Z,

and it is easily verified that ~

(;'I

Z/nZ

£;

Z/nZ

np p of p \) np Z/p Pz \)

Given a canonical decomposftton

n =

Z/nZ

£;

by virtue of the chinese remainder theorem.

n E lN,

we have the decomposition

Passing to projective limits we

obtain a canonical decomposition A

Z=

(_c}

Let JFP

be the field of

rable closure of JFP,

p

For each

nz p

p

elements, n E JN,

p

prime,

and let iff'P

be the sepa-

I/le have a canonical isomorphism

184

CHAPTER 2 Ga 1(lF' /JF 1 9! '11./n'll. pn p

wlii'ch maps the Frobenius automorphism

of IF

¢

n

p

to

n

l

mod n'll.,

Passing to

projective limits we obta tn a canoi1i ca 1 isomorphism

which sends the Frobentus automorphism the group

of ffi

¢

to

p

,,., 1 E '71.

and therefore maps

A

to the dense subgroup '71. of 'll..

12. INFINITE GALOIS THEORY The usual Galois correspondence between subgroups of Galois groups of finite Galois extensions and tntermedi'ate fields is not valid for infinite Galois extensions (_see Example 10,5}.

The Krull topology introduced below restores this

correspondence for closed subgroups (Theorem 12.2}, Throughout, group

G can be endowed with the Krull topology defined as follows.

{E./Fli EI} 'l,

rr

G denotes the Galois group of the Galois extension E/F. Let

be the set of all finite Galoi,; subextensions of E/F,

For each

we take the cosets

E G,

crGa 1(E/E , )

(i

'l,

as a basts of neighbourhoods of rr.

t---+ CJ'

1

2

o1 o 2 Gal(E/E.) 'l,

contains the open neighbourhood 0

of

I)

o2

is cont1'nuous, since the pre-tmage of the basic open neighbourhood 1

E

The multiplicatton map

G, (rr l ,cr} 2

GX G-

of rr o

The

(o ,rr2 ),. . 1

1

Gal (E/E j . 'l,

S1'milarly, the map

becomes a topologi'cal group.

X CJ'

2

G-->- G,

Gal (E/E'l,_} o

1-+

o-l

is continuous, so

Of course, if E/F is a finite extension, then the Note also that, by Lemma 10,2,

Krull topology is discrete. toe map

i

G

G/Gal (E/Ei) - - Gal (E/F) Ga 1(. E/E·z.. ) t---- o IE'(.,

Gal(E/E.) . 'l,

- W(R}

W(R), set . - -

, r

n

: R ->- W

n

(R}

E

Wn(R)

WITT VECTORS

199

00

r

a • = (.a 0 ,a0 +a , • o • ,a0 +a + •• , + a 1 , •• , · 1 1 n-

i=o i. 13,4, THEOREM (.i).

The maps

and vn

s,sn

For x = (x 0 ,,,,,xn-·l) E Wn(R)., n-1 . .

(Ji}

preserve addition

y = (.x 0 ,x , ... ,x

n-·

l

f

~ si.(r (x.)),y = si.(r(x.}} i=o n n i. i~o i.

(a).

x =

(b).

1'n (alx = (ax 0 ,aPx1

(.c)_

n-1 r(a).y = (ax 0 ,aP\,.··•aP xn-l''"

1 , ... ) E W(R), a ER,

n-1

(tii)

(iv).

x,y

E

(vl

xn_ 1 )

for all

are such that xi = 0 or yi

W(R)_

+y

X

,aP

and rn(.ab). = rn(a)rn(b).

r(ab) = r(a)r(b} lf

, , ..

If x,yEWn(R)

= (x +y ,x +y , , , , ,x

o o

1

n-

1

are such that xi=O

l

= 0

a,b ER, n E lN

for all

i ;;;., 1,

then

+ y n- l'.,, )

or yi=O

for all

1.;;;·i,;;;n-l,

then

Ci"l

Pro0f.

We first note that s (x ,x , ••• ,x 1 ). = a s(x0 ,x , • , • ,x 1 ,.,, n o 1 nn n1

and

Stnce both

crn

vertfy that

s

and

an+l

preserve additton (Theorem l3,3(t)), i't suffi'ces to

preserves addi'tion,

To prove the latter, we may harmlessly

assume that R = A in which case s(x)_ -_ [0 ,px Col , • , • ,px (n•2) , • • • ] , s

and hence

preserves addition.

(.i·i'}

Since crn(y). = x

for x.

and

crn(r(_a).y) = rn(.a).x,

Again, we may assume that R = A, rn(a)_ =

it suffices to prove the formulas

in which case

[a,aP,. .. ,aP

n-1

and hence . . n-1-i l [0, ... ,0,p '1,a, ••. ,p '1,a'P

Therefore

(O ,;;; i ,;;; n-1)

CHAPTER 2

200

[x(o), •.• ,x(n-l)l = x

and the latter sum is

by (2).

Also

p pn-l (o) (l) (n-1) [a,a , •.• ,a ] [x ,x · , ••• ,x ]

(by (2)

and

(3))

as required. (iii}

x = rn(b)

Apply (_b) and (c) for

and

y = r(b1,

respecttvely,

It suffi'ces to prove (vl, since (iv) follows from (v) by applying

(i'vl and (v)_.

By (i)_,

rrn'

n-1 •

x +y

= =

stnce

~- = i

z s"(r

n-1 .

n-1 .

=

(x.Jl + Z s~(rn(yi)j

i=o n n "

i=o

.z

i=o

s~(rn(xi) + rn(yi))

n-1 . Z s"(r (x .+y ,1)

i~o n n ·

0 or y.i

i

i

0 for O < i < n-1. •

=

Next we introduce the followi'ng two maps 1T

~· W(R)

- 'IT_,. W(R)

I '(xi)

f-----;.

(~)

13,5, LEMMA.

(i')

If

W (R) __!!_.. W (R) n n (xo,'' • ,xn-1)

x,y e W(R),

then for all

t---T ( { , • • ' ' ~ - l)

i;;;, 0,

(1r(x+t1) )i = (1r(x) + 1r(y) )i (mod pR) (1r(xy))i = (1r(x)1r(y))i (mod pR) (i'i')_

If

x,yEWn(R),

then for all

(1rn(x+y) )i

=

iE{O,l,. •• ,n-1},

(1r{x} + 1r(y) )i (mod pR)

(1Tn(xy))i= (1r}x)1rn(y))i (mod pR) Proof,

This ts a dtrect consequence of (8),



From now on, we concentrate on the very important special case where charR = p.

The prime subring of R is the field IF

13.6. THEOREM.

.

Let

R

p

of

p

elements.

be an arbitrary commutative ring of characteristic

p > 0

WITT VECTORS

201

and let n;.,, 1, (tl

The maps

(ttl

IT

and IT

p"°'x=s~(IT!(x)}

(ttt)

wn (R1

and iy=si(ITi(y))

(_v)

If

R

Proof, (i'i'}

xEWn(R),yEW(R),i;;;.l

forall

W (r;? 1 = '11./pn'll, is the prime subring of Wn(R), n p is pn and the characteristic of W(R) is O, For.any xE Wn(R)

(i'v}

are rtng homomorphisms

n

Let

and yE Wn+/Rl, vn(x)y = vn(xfn,n+lrrn+l(y})

i's a field of characteristic p,

then

Direct consequence of Lemma 13, 5,

( i} X =

the characteristic of

(x0 ,x 1 , •• ,,xn~i'"' }

W(A).

E

We have

(pX). (i) = pxCi} = p(TI(x) (i-l} t- ,ix,r)

= p11{x}(i-l}

( by ( 1) )

= (_sIT(x}l(i) (mod pi+l'll,[X]}

Thus

(pXli = (1rn(x}li (mod p'll.[X] by Lemma 13.1, whtch clearly impHes have

p\

\i(y})

= s1

for all

now obtatned by applyi'ng

i ~ 1.

The pri'me subri'ng of

Wn(R)

Stnce

pnl = s~(l l = 0 and

pml

Thts proves that

Iterating the latter, we

The correspondtng assertion for

x

is

rr. n

(i'i'i}

pn.

py = srr(y),

1

is addittvely generated by

=

(1,0,.,.,0).

= s~(l l 1 0 for m 0

and

Since

µm E F.

n = m/,

Setting

= 1.

by Theorem 14,4(1).

m (hence n},

Proposition 3.8, hence the result.

= charp,

and let

(i)

E/F

E/F

t:/F

Our point of departure is the case where

14.7. THEOREM. (Artin-Schreir).

F(µ}/F

F(A)

But

=

is cyclic t F(AP ), by



We now turn to the case of cyclic extensions p

we have

µ=AP,

contains a primitive m-th root of unity,

F

of degree dividing

(m,p)

where

Let

F

of degree n

pn

where

= 1,

be a field of characteristic

p ~

0

be a field extension.

and

H and only if

p

is cyclic of degree

that

AP -· A E F

(ii}

The minimal polynomial over

for some

E = F(?J

A E E such

A ¢= F.

F

of any

\

E

E

with

AP - \ E F

and

/\ ff. F

is

(iii)_ over

For any given

F or splits into Proof,

then

a

a+ i

E F,

p

the polynomial

xP - x -

a

distinct linear factors over

a

We first note that if is also a root for

i

=

E F

and

a

0,1, •.• ,p-1.

is either irreducible

F.

is a root of

xP - x ·- a,

This is so since

CYCLIC EXTENSIONS

Thus

xP ..

fore

F(.al/F

X - a

splits tnto

p

dtst1'nct linear factors over

F(al

and there-

is Galois.

Now assume that E = F(A) Si'nce

209

xP -

A i's a root of

for some

A EE

X - ()l-A), E/F

i (J E {0,1, •.• ,p-1},

there exi'sts a unique

such that

AP - A E F

is Galois.

and

Ai F,

a E Gal (EJF),

Given

a{AP,..A). = AP - A + i (J .

such that

Consider the mapping

!

Gal (.E/F} (J

'll,/p'll,

i (J +

I--->-

This is obviously an injecti've homomorphism, and

we conclude that

EI F,

cyclic of degree

F,

Stnce

'll,/p'll,

is cyclic of order

ts cyclic of order

pl-A}

Thus

p.

is

E/F

is the minimal polynomial of

p

A over

This proves (ii) and the ~if" part of (t}. Assume that

Gal(E/Fl, some

xP - x -

and

p

Gal(.E/Fl

p'll,

so that

a has order p.

so that

A E F,

is cyclic of degree

F,/F

Then

"JP-\ E F,

a(tl} But

= a(.11t

cr be a generator of crp,)

Owing to Lemma 14,3(it),

= ,._P + 1

and

=" + 1

a(AP-A) = (\P+1) - {Ml)

cr(\) "' A + 1 I

since

:>,, '/; F

Let

p.

Hence

:>,,,

E =

for

= AP-A,

F(A)

and

(i) is established, To prove (itt), we need only verify that if no root of xP - X - a

F,

then

f(X} =

xP - x - a

ts trreductble over

F.

lies in

Assume by way of contra-

di'cti'on that f(X) = g(X}.h(X}. with

g,h E F[X]

1..;; degg ~ p.

and

Since

~ f(X}. = I I (X-a-i)

i=o

(_a is a root of integers

i

E

sum of terms to

-da+j

f(X}l,

we see that

{0,1,. •. ,p-1}. -(a+i)

If

g(X)

for some integer

cause the coefficients of

j.

g(Xl

then the coefficient of

d = degg,

taken over precisely But lie in

d

d

integers

I O tn F,

(X-a-i}

is a product of

F,

i.

is a

~-l

Hence it is equal

hence

a

a contradiction.

Our next aim is to characterize. the cyclic extensions

over certain

E/F

lies t n F,

be-

• of degree

p

n

210

CHAPTER 2

where p = charF,

This will be achieved by analyztng conditions for the exis~

tence of an injective homomorphism Gal (E/F)

--+ 'll./pn,lL

Precisely this was done in Theorem 14,7 in the case n =

Since 'll./pn'll.

1,

isomorphi'c to the prime subring of any ring R of characteristic pn, morphisms from

Gal(E/F}

Gal(E/Fl---+- Pn' able ring

where Pn

is the additive group of the prime subring of a suitOur ring

shall use a simplified notation pertaining to (instead of

1r

n

will be the n-th Wi'tt ring

R

of E defined tn the previous section.

wn(El

all homo-

to 'll./pn'll. are identifiable with the homomorphisms

of characteristic pn.

R

is

Since our n will be fixed, we wn(E}.,

Namely, we shall write

1r

l for the homomorphism

Wn (E}---+- Wn (El, (x 0- ,x 1 ,,,,,x 1 ~ (xP ,J?,, .. ,fn-1 } · n-1 · 0 1 and

s

(instead of sn}

wn (E)

for the shift homomorphism

Wn (El, (x ,x , , • , ,xn-1 ) -0 1

---+-

For any homomorphism a:

E---+- E,

.--+

(O ,x0 ,,, , ,xn-- 2 }

we let

cY(x) = ( cY(x 0 } ,a(x } ,. .. ,cY(x 1 ) 1 n-

1 e wn (E},

for x = (x 0 ,x ,,,.,x 1

n-1

by polynomials wi'th coeffici'ents in

er on

IF

p

and si'nce a(al .

wn (E)- to wn-(E).

is a homomorphism from duced

Since additton and multiplication are defined

Ftnally, if

for all

=

a

O'

e Gal (E/F},

a EJFp , er

the in-

wn.(E)- ts an automorphism leaving fixed the elements of the subring

wn (F).14,8, LEMMA.

Let

E

fi'ni'te order d,

be a field of characteristic p and let ere Aut(.E)

Then for some

= ~1

Let A e

Proof,

0

n

Setting y

=

have

E

be such that y = 0

x e Wn(E)

d i r a (A) f O and let

i=l

0

.

1,

Ea (A}, we see that the zeroth coordinate of y is y0 ,

Hence,

i=l

.

- -~~;:.·:.:,

CYCLIC EXTENSIONS by Corollary 13,10,

is a unit of W~(El,

y

211

Because

o(y) = y,

we therefore

have d

'

E o-i (\y- 1

1

)

i=l

and we may take

x = -\y

Given m,;;; n,

-1





note that the addi'tive order of sn-m(1)

co, ... ,o,1,0, •.• ,o)

is pm and that m n-m

p s

14.9, LEMMA.

Let

fi'nite order

Proof,

Applying

pm

be a field of characteristic

E

dividing

pn.

xE

wn (E)

Aut(E)

have

for all

(x} = 0

p

and let

CJ E

Then

Owing to Lemma 14.8, we may choose

\

E

wn (E) such that

we have

sn-m,

m

CJ(x)

p ·+1 E 1:CJ-i (sn-m(\))

i=l Pm



i j~ (sn-m(\11 j=1 = x + sn-m(l)

since pm sn-m(\l

=

0. •

We are now ready to describe cyclic extensions of degree dividing 14.10. THEOREM.

Let

F

be a field of characteristic

integer and let E/F be an algebraic field extension, degree

pm

dividing

l

if and only if there exist

p >

0,

let n

pn.

be an

~ 1

Then E/F is cyclic of \ 0 , \ 1 , ••• , \n-1

in

E

such

CHAPTER 2

212

E

that

F(\ o ,\ 1 ,.,.,An-1 ] and the element \=(A,\ , ... ,\n~1 ] - o 1

=

wn (El

of

sati'sfies

Furthermore, for each di·vtdi-ng

and an element

pn

Proof,

Step 1,

there exi'sts a cyclic extension

a E Wn(F),

A = (1.0 ,'.\

1

,•, •

·

,A

)

n-1

of degree

K/F

wn (K) such that

in

For the sake of clartty, we divide the proof into three steps.

Here we prove the first assertion under the assumption that

separable.

Assume that

is such that

n(\) - '.\ e -

E = F(A 0 ,A 1 , • , , ,),n~1 l,

wn (F1,

Denote by

E

where

E/F is

A= (\ 0 ,A· 1 , , •• ,An~1 ) E Wn (E)

E,

an algebraic closure of

Note

(E1, (n(x) • x) - (TI(\} - \) = 0 if and only if TI(x•\} = x - ,\ n that is, if and only i'f x - A E w (IF } , Hence {\ + i Ii e 'll./pn'll.} is the set of n p all roots of n(x) - x - (TI(A) - \) in w (E}, si nee w (JF ) = 'll.Jpn'll. by Theorem n · n p that, for

x E

13,6(iii}. under

w-_

Because

Gal (1;/p),

generated over

TI(\) - ;\ E

Hence F

E

wn (F),

i's mapped i'nto i-tself under

E/F

i's Galois. ~ Ga 1(EJF)

? cr(\) = \ + ,r;cr'

fore cyclic of order

Since

Gal(E/F},

rJ

'll./pn'll.

pm dividing

Conversely, assume that be a generator for

A + i,if!, 'll./pn'll.,

by the coordinates of

rable, we conclude that

where

these roots are mapped into themselves Gal (E/F), Si_nce

since

E/F

E

is

i-s sepa-

We have an injective homomorphism

---+ 'll.Jpn'll. I->-

i rJ

ts cyclic of order

pn, Gal(E/F)

is there-

pn,

E/F is cyclic of degree so that

o

has order

pm dtviding pm,

pn.

Let

o

Owing to Lemma 14,9,

we have for some Then

and therefore

\

E

wn (E)

CYCLIC EXTENSIONS

213

TI(A} - A E w (F). Thus F(A 0 ,A •···•" n 1 n-1 )/F is cyclic of degree pk dividing pn, by the first part of the proof, Since sn-m(l) has additive

whence

pm,

order

pk= pm

E = F(\ ,A ,, •• ,A

and

o

1

K/F of degree dividing TI(A) -

A= a.

pn

and

cm

(TI-1)\ = a

= 1,

n

and

we take

(Theorem 14,7(i)}.

A=

there exists a cyclic extension (1,. 0,1,.

,.,.,A J in 1 n-1

'.A. E

o

"a

E

~

such the

that there exists

O ..;. i ..; n-1.

K =

The requtred assertion

F(\,"i···•\_ 1 },

s to be a root of

Next, let n

n

,A 1 , ••• ,An-1 }

for

S

1,

W CK)

(TI-llA = TI(\) - A and denote by S

We prove by i'nductfon on n

wi'll then follow by Step 1 bv taking If n

E W (F),

element

~ = (1,.

such that

a

We use the notation

separab 1e closure of F.

Thus

}.

n-1 Here ~e show that, for each

Step 2,

that

a to F(A 0 ,\ 1 ,, •• ,An-1 } has order pm.

the restrtctton of

xP -

and again choose

1

and

X - a0 :>,

0

Es

K

= F(\)

such that

Then

for suitable sure of

bk Es.

If

K'

=

F(b . 1 , ..• ,bn-1 ),

then

K' and we can find, by i.nduction 1\,,,, ,µn-i

s E

ts the separable cloS

such that

(b1'"''bn_) = (TI-lH11i,,..,µn-1) = (P;,.,,,µ~-1) - (µ1'"''µn-1) But then (0,b , .. ,,b } · 1 n-1

(by Theorem 13,4(i))

(Q,µP, .. ,,µP } - (O,µ ,. .. ,µ 1 } 1 n-1 1 n-

(rr-1)(0,µ ,,,,,µ 1

n- 1

)

He now have

(a0· ,a ,.,.,a ) 1 n-1

for suitable

( TI-1} (("a, 0, •. , , 0) + ( 0 , µ 1 ,

"i E S, 1 ..; i..; n-1 ,

Thus

(TI-1)'.A =

a

.. , ,

for

µ n-i) )

A = (\,\,..

""n-1'

CHAPTER 2

214

as requi'red, Step 3,

In view of Steps l and 2, we are left to

Completion of the proof.

verifythatif E=F(A 0 ,A, .. ,,A

n- 1

1

n(A) . . - AE w n.(F),

such that

Step 2, to find

1

n

of F,

Thus

p

and therefore

n-1

)

)\=()\ 0 ,A, .. ,,A 1

is separable. such that

is separable over

that each µ., O < i < n-1, 1., µ - A E w /;IF ) •

E/F

then

µ = (µ 0 ,µ , ••. ,µ

where

),

E/F is separable,

)EW(E)

is

n

To this end, we apply

n(µ) - µ = n(A) - A and such But then

F.

A = µ - (µ-A) E wn(s),

n- 1

where

n(µ-A)

µ - A and

=

s is the separable closure



15, KUMMER THEORY

F be a field and n

Let

to be of exponent n for a 11

0

E Gal (E/F),

abel ian extension

E/F

n-th root of unity. F

a positive integer.

if the exponent of

A Galois extension

Gal (E/F)

divi'des n,

E/F

i.e.

if

By a Kwnmer extension of exponent n one understands an of finite exponent n,

where

F

contains a primitive

Throughout the discussi'on, the ground field

n

then

Moreover,

con ta i'ns a primitive n~th root of unity, which wi 11 be denoted

Our aim is to survey all Kummer extensions

E/F.

G endowed with the Krull topology is a compact group.

Hom c (G, < sn >) = {f : Then

will be

F

fixed, and all extensions 1ie i'n a 9iven algebraic closure P of F.

by s.

an = 1

It follows that for any such extension the characteristic of

does not divide n,

we assume that F

is said

Hom/G,)

G-

< sn > I f

G = Gal(E/F),

If

We write

; s a continuous homomorphism}

is a group under multiplication of values: Cas)(g} = a(g)B(g)

We denote by morphisms from

Hom(G, < sn >) G to

the group of a 11 (not necessarily continuous) homo-

. n

Then

Hom (G, n

C

l is a subgroup of

Hom(G, < s >) n

which can be characterized by Homc(G,) ={XE Hom(G,)I Kerx

is an open subgroup of

G}

(l)

The latter l's true, since a homomorphism from a topologi'cal group into a discrete

KUMMER THEORY

215

group ts conttnuous tf and only tf tts kernel ts an open subgroup, is finite (hence discrete), we have

Homc(G,)

In case

Hom(G,).

=

G

Some further

i'nformation is provided below, G be a finite abelian group.

Let

homomorphisms from

to ..) 2 •

>.. =



We are now ready to tackle the prime power case (the prime case will be excluded because of Lemma 16,3), 16.5. LEMMA.

Let

and let

p

be a prime.

(i)

p

i's odd, or

F

If

only i'f

and

charF f 2,

a (/:. F 2

and

a (/:. -4F 4 ,

(i}

Proof. n-1

xP

by

and

p = 2

If

an arbitrary element in F. n charF = 2, then xP -a is irreducible over

a

Denote by p = 2

If

a= 1l

2n

then

for some

X

µ E

- a

F,

i's irreducible over

Conversely assume that a ef:. pP. n-1 µ = ;\P Then µ ts a root of xP

xP

then

- µ.

and let

be a positive integer

2

a(/:. FP.

if and only if

(i'i)

n:;;,,

be an arbi'trary field, let

F

Let

- a,

n

~a= /\

xP

F

if and

n -µP

is dividible n

be a root of

xP

-

a

hence by Lemma 16,3,

(F(µ) :F) = p We clatm that

'.\

has degree

p

n-1

F(µ); if sustained, it wil 1 fol low that

over n

>..

has degree

degree

p n-1

over

pn

over

F(µ)

F

and

xP - a

is irreducible over

will be true by induction on

n,

F.

That

provided

A has

µ '!, F ( µ )p .

The desired conclusion is therefore a consequence of Lemma l6.4(i). (ii')

If

a E -4F 4

a

E F2

,

then obviously

and write

a= -4a 4 , a E

x

2n

- a is reducible.

F, Y

=

x2

n-2



Then

Suppose next that

225

RADICAL EXTENSIONS AND RELATED RESULTS Conversely, assume that a ff. that

and a ff. -4F 4 •

F2

-4a is not a fourth power in F. Since a ff.

by Lemma 16.3.

We must show that

X2 = 2

and for n

In the latter case, -µ

over F.

-+

1-

(F(µ) :F)



For n

2

and

= 2

this will be

this will be true by induc-

and -4µ

is not a fourth So it suffices

is a square in



-a

we have

is a square tn F(µ).

these two statements are equivalent since µ

2n

a;

> 2

to rule out the possibility that either µ or -µ

F(µ}

-

n-1

(F(:\):F(µ))

provided 11 is not a square in F(µ}

power in F(µ).

it follows

Again, let :\ be a root of x

and µ is a root of

F2

true if µ is not a square in F(µ}, tion on n,

Because a ff. -4F 4 ,

F(µ).

Now

induces an automorphism of

Hence, by applying Lemma 16.4(ii), the result follows.



Applying the foregoing, we now deduce the following result. Let F be an arbitrary field, let n P

16.6. THEOREM.

is irreducible over

;(I - a if and only if

Proof. a

E pP,

a If:.

If p8 -a

xP

If

Conversely, assume that a ff; Let p8

4\n

and

F.

di vi ding n

p

a=

-4:\ 4 ,

p

Hence

and if

r-a

s xP - a

is

then

:\ E F,

dividing n and a ff. -4F 4

be the highest power of a prime p dividing n.

Lemma 16.2, it suffices to show that

Then

dividing n and a ff. -4F 4 whenever 4\n.

for all primes

FP

E

F

is reduci'ble by Lemmas 16.3 and 16,5.

reducible, by Lemma 16.2.

whenever 4\n,

p

is the hi gbest power of a prime

s

then

for all primes

pP

and let a

1

is irreducible.

By

The desired con-

clusion is therefore a consequence of Lemmas 16.5 and 16.3. • Our next aim is to prove that if intermediate field

K,

Gal(K/F)

E/F

is a radical extension, then for any

is solvable.

The following preliminary obser-

vations will clear our path. Let

16.7. LEMMA. subextensions. Proof. E2 =

E/F

Then

be a field extension and let (E E ••• E 1

2

n

n

1

be radical

)/F is a radical extension.

It suffices to treat the case n

F(µ 1 , ••• ,µk)

E /F, ••• ,E /F

exhibit the assumption that

= 2. E 1 /F

Let and

E

1

=

E 2 /F

F(:\ 1 , .•• ,:\) m are radical

and

226

CHAPTER 2

extenstons.

Then

EE 1

EE /F

shows that

1

(_i)

Let

If E/F

(ii}

If

F c Kc E

1



be a chain of ftelds.

is radical, then so is is radical and

KIF

m

1

is a radical extenston.

2

16.8. LEMMA.

= F(\ , •• ,,\ ,µ , ••• ,µk)

2

E/K,

K over

ts the normal closure of

E

F,

then

E/F

ts radical.

(_i)

Proof.

n. E = F(A •• ,.,A) with A.1., E F(A _., ..• ,L 1l, 1,;;; i,;;; m, 1 n. m 1., 1 1.,l with A! EK{\, .. ,,\. 1 ), as required.

If

then

E = K(\- 1 , ... ,\m

Cti'l

Si'nce

is fintte, we may write

K/F

fi(X)

of an irreducible polynomial

f.(X), 1.,

then

radical.

F(A , ••• ,A) - l s

Stnce

Then

Gal(8/F)

(til

If

Let

be a prime and

p

i1' ~ 1

and

F

g

If

s.

\.1.,

is a root

is any root of

F(\ 1 , . , . , \8 )/F

a splttttng field of

is



x'P-1

over

a

an arbi-

splits into linear factors,

the splitting fteld of

8

charF = p,

If

i1'-a

over

F,

F.

then

then

E =

F(s),

E = F and there is nothing to prove.

then

where

s

is a prtmttive p-th root of untty.

Gal {_E/Fl

is cyclic, by Corollary 5. 12.

(iil

u

r-1!1,,

is one root of

and so

morpntsm of

a(u)

y'""x+y>x'+y'

(b)

x > O,y > 0 ""xy > 0

In discussing the ordering on R, we shall write x > and use

x = y,

for the opposite ordering.

to mean x

y

or

> y

An element x ER is said to 0.




Indeed, otherwise by Lemma 16.3,

If charF

is a power of

charF

If

Thus we may harm-

Furthermore, by passing to a normal clo-

we may assume that E/F

is Galois.

Let P be a Sylow p-

subgroup of G = Gal(E/F},

and let K be the corresponding subfield.

(K:F}

p.

(E:F}

=

is prime to

(G:Pl

is a power of

16.21. THEOREM. a square root. a root in Proof.

be an ordered field in which every positive element has

F

Assume further that every polynomtal of odd degree over Then F(i}, i 2

F.

Thus

K = F.

as required. •

p,

Let

By our hypothesis, this implies

Then

By hypothesis,

=

F

-1,

has

F

is algebraically closedo

has no finite nontrivial extensions of odd degree.

Invoki'ng Lemma 16.20, we infer that the degree of any finite extension of

F

a power of 2.

is

F(il,

We claim that the only nontrivial finite extension of

which will clearly yield the result.

let G = Gal(E/F}.

By Lemma 16.17,

passi'ng to the normal closure of

E,

charF

Let E/F be a finite extension and =

0 and so E/F is separable.

we may assume that

know that the order of G is a power of 2,

If

(E:F}

is Galois.

E/F ~

2,

F

c

F 1

c

F 2

of fields with

is the only quadratic extension of F, easily tmplies that every element in

(F :F} l

=

(F :F ) 2

we must have F

1

F

1

2.

=

1

=

We

F

2

This Since F(i}

The latter

F(i).

is a square, hence

By

then G has a sub-

group of index 4 which in turn is contained in a subgroup of index 2. yields a chain

F

is

cannot exist, a

contradiction. • Let E/F be a field extension and let M be a subgroup of E* F*M/F*

is finite.

When is it the case that

IF*M/F* I

= (F(M)

:F)?

such that An answer

to this question is given by the following result. 16.22. THEOREM. (Kneser(_l975)}.

Let E/F be a separable field extension and let

CHAPTER 2

234

M be a subgroup of

primes p,

such that FM/F*

F7

ts finite,

each p-th root of 1 which lies in F*M also lies in

i = ,J.:i_ E F

if

i E

1 ±

Proof.

(F(M} :F}

We first observe that F(Ml

consists of all F-linear combinations of

Therefore we may choose finitely many

basis for

F(M}.

Then

and that

F

Then

F* •

IF'

N

the norm of

C-11P-1aP, we have

((-l}P-1J'lq = N(alpd-p

For odd p, (61,

aP

In case

is then a p-th power of an element in F(N8 _ 11, which contradicts p = 2,

-a 2 = ;.. 2 with ;.. E F(N8 _ 11 which shows that

we have

= ad = ±i1,d.

i E F(N8 }, i ¢ F(N8 _ 11 and a 2 a== g +

Then a 2 == (g 2 -h 2 } + 2ghi

Let us write

ih

= ±i'Ad

which impli'es g 2

-=

h 2 and hence a == (l±i}g.

We conclude that

which ensures, by applying twice the induction hypothesis, that then 1

± i

E F*M and so, by hypothesis,

i E F ::_ F(N8 _ 1 ).

g E Ns- 1 •

But

This contradiction

completes the proof of the theorem. • Returning to radical extensions, we next investigate conditions under which the Galois group of a binomial

r -a

16.23. THEOREM. (Schinzel(l977}1.

Let

is abel ian. F

be a field,

n

a positive integer not

divisible by charF and let m be the number of n-th roots of unity contained in

236

CHAPTER 2 Then, for any gi'ven a

F.

only if am= \n for some \ Proof.(Wojcik(l982}}.

E

F.

For any integer t

not divisible by charF,

denote a primitive t-th root of unity over F.

Since

Let rb

If am= An for some A E F,

extension of F.

ts afielian if and

the Galots group of Jf-a

E J!,

be the maximal abelian

then Z7'a

O ~ j ~· n-1.

= sjI:!}'5::,

n

contains a primitive m-th root of unity, ~Erb by virtue of Lemma

F

15.9(.iiil.

Thus '{;fa E



and F(Z7'a,s n }/F is abelian, Conversely, assume that the Galol's group of i"- a is abelian,

the m.aximal divisor of n for which am verify that k = n, that k

let st

and m\k.

of p dividing m,

be

We need only

F.

Assume by way of contradiction

Fix a prime p dividing n/k and denote by p6

n.


0.

with root A and define the

F

E

F(e )} m

I' ~.11!

(E:F} Proof.

(m,p}

f>y s

lf

Let F be an arbitrary

=

over F, n(F(e }:F} sm

then

Our first step is to reduce the genera 1 case to the case where m = n,

charF X n,

So assume that charF = p E

= F(\,e} m = F(Ap

>

O.

Then

k

,A~E) m k

wh.tch in turn is the composite of the separaf>le extension F(\P ,e )/F and a m k purely inseparable extension F(Am}/F. By Lemma 16,2, both Jfl _ a: and xP - a: k

are irreducible over

F,

Stnce Am

is a root of

iP - a, we have

CHAPTER 2

238

(F(\ml:Fl

=

l.

Hence, by Propo$ttton 3.28, k

k

(E:F} = p (F(\P ,sm}:F} k

Thus, if the result holds for i" - a (whose splitttng fteld is

We may therefore assume that m = n and hence that charF Put L

= F(E

n

1n

and s,

F(\}

Since

= (L:FL

r- a

F(\P

,sm1, then

%n, is irreducible,

(F('A} :F} = n

On the other hand, by Proposition 12,5,

It therefore follows that

and we must show that s

= s',

Now by the definition of

in Lemma 16,24,

1'

F(s1' 1 -c F(sn 1 n F(Al and hence, oy Lemma 16,24,

L

= F(\ql for

q

= L

c F(\1

-

= (F(\l:L}.

(F(\l:Fl = n,

Since

we have (L:F} = n/q =

Thts snows that \q e F(snl and q

=

n/s',

On the other hand, since \n/s e F(s 1 n F(\l n

s' Hence, by the definition of s,

= L,

we have

(F(\}:L} = q..;; n/s

whence

s

1

~

s, as requtred, •

Let E/F be a simple radical extension, say E = F(\) with power of \

in I!*.

Then we obviously have

( < ;>., > t(E*ll!*l/1!*

~ t(.E* /J!*l

;>.,

e E* and a

239

RADICAL EXTENSIONS AND RELATED RESULTS It is therefore natural to tnvesttgate ctrcumstances under whtch the equality holds, 16.26. THEOREM. (May(l979}}. sion with charFt-p. further assume that

Let p oe a prime and let E/F be a fteld exten-

Assume that

t- F(i'} {i 2 = -1}.

E

t(E* /F*}

Proof. (E:F} = p.

Let

µ

= (

xP -

Owtng to Lemma 16. 3,

where )..PEF-iP.

E=F(\l

If p=2,

Then

t(E*}F*}/F* 1..P ts trreducib 1e over

F and hence

e E* oe an element of prtme-power order modulo F*.

First

1:'

assume that the order is qr for some prtme q :j p, and 1et µq l'

is the norm map from

N

y e F*. l

E

It follows that

then N(µlq

y e F*.

=

If

1'

·,l, hence

to

F,

µ =

y E for some root of unity Ee F* which shows

=

for some

y = yq l

1

that µF* E t(E*}F*. 1'

Now assume that µ has order p:i> modulo F*, claim that y ¢ if?.

and let

Assume the contrary and write y

Let E be a prfmi ti ve p-th root of unity.

=

i/

=ye

r.

We

yP1 for some y l e F*.

Then for some m

We must have p ) m, than pr.

for otherwtse the order of

modulo

µ

For the same reason, we must have e fJ F*,

wi'th (F(e:} :F}

But

> 1.

(E:F}

=

r

would be 1ess

It follows that

p and (F(E} :Fl

divides p - 1.

This contra-

and i rt

we have (F(µ) :F}

dtction substantiates our claim. Except possibly when p = 2 ,r

!>

1

virtue of Lemmas 16.5 and 16.3, and thus r i.e. assume that p

=

2 ,r > 1,i

t

and

F

F,

= 1.

(F(µ}:F}

=

pr by

Consider the exceptional case, < 2r.

stances we know, from Lemma 16.5, that y =-4d 4 for some

Under these circum8

e

F.

From

1'

µ2

=

we see that i e E and hence E = F(i).

-4d4,

Since this case is excluded in the

hypothesis, we may therefore return to the situation where

CHAPTER 2

240

µP = y

and

(F(µ} :F} = p

Then we have

= E = F(:\)

F(µ} Hence, by Corollary 15.7, µF*

F*

E

µ = )...ka

for some

and the result follows.

E

Z and some

a

E

Thus

F.



We return to the study of the field ci'ble polynomial

k

F(al

where

a

is a root of an irredu-

As an easy application of Corollary 15.7, we

xP-a E F[X].

first show that the adjunction of two ~genuinely different~ p-th roots result in an extension of degree

16.27. PROPOSITION. If

a,S

p2 • Let

be a prime and let

p

arerootsoftrreduciblepolynomials

charF t- p.

be a field with

F

over

xP-a,xP-b

then

F,

(F(a,S} :F} = p 2 unless

b = c?ak

Proof.

for some

k E Z

If the polynomial

(F(a,Sl:F} = p 2 , (Lemma 16. 3 l. for some

k e

b = cP ak.



Assume that

i.P-b

yP =

c

e P,

a,S

b E F,

c E F,

remains irreducible over

If it is reducible over Since

z,

and some

F(a},

F(a},

then

then it has a root

in

y

it follows from Corollary 15.7 that

y = ca

are roots of irreducible polynomials a+ S,

provided

over

xP-a, xP-b

(F(a,S}:F1

=

p2?

To

answer this question, we first maRe the following preliminary observation.

elements of

Let E

be a fintte Galois extension and let

E/F

of degree

m and

n,

respectively, over

F

a

and

S be

such that

(F(a,S} :F) = mn. (_t l

For any conjugate

a,

~

of

a

and

e,.J

of

S,

there exists

0 E

Gal(E/F)

such that o(a)=a. ~

(ii)

k

Raising this equation to the p-th power, we obtain

What can be said about the degree of

16.28. LEMMA.

F(a~

and

If no difference of two conjugates of

o(s)=s. J

a

equals a difference of two

F,

241

RADICAL EXTENSIONS AND RELATED RESULTS

conjugates of

B,

then F(a,S) = F(a+S)

(i}

Proof.

Since

f E Gal (E/F).

Corollary 6.5(i), choose for some conjugate

f 1 (a)

that taktng

0

=

a

is a finite Galois extension, we may, by virtue of

E/F

Bk

of

and

f

1

B.

f(ai) = a.

such that

f(B)

Write

f 1 E Gal(E/F)

We claim that there exists

=

Bk

such

if sustained, the assertion will follow by

(Bk} = B;

(J f}- 1 •

=

I

Let

B ,.,,,B I n that the degree of polynomial over

be all conjugates of B over

F(o;l

F(al

are still

F.

B over

is still

n,

B1 , .. .,Bn •

Our hypothesis implies

so that the roots of its minimal Since

E/F(o;}

is Galois, the

required automorphtsm exists,

(_ift Let a 1 , ..• ,am be all conjugates of o; over F.

Then, by (il,

o;. + B • are all conjugates of two conjugates of

o;

by hypothesis, all Hence

mn.

a

over p2

and

over

If a coincidence occurs, then the difference of

1,

+ B.

are distinct, provtng that the degree of

J

Let

F

as asserted,

then

Thus,

a + B is



be a field and let

(F(a,,S}:F} = p 2 ,

Since

p

F(a+Bl

# charF, a, and

a fi'nite Galois extension

p

be a prime di.sttnct from

charF.

xP-a, Jf.J-b

= F(ct,Bl. . i,e. a,+ B has degree

with

E/F

B are separable over a,S EE.

to verify that no difference of two conjugates of conjugates of

(si - sj)a,,

B,

where

s

a,/B E F(s}.

But

a

a

has the form

Therefore, if a differ-

equals a difference of two conjugates of

(F(s}:F} m

Then deg(a+\S} Proof. Step 1.

= mn

for all

Of

AE F

For the sake of clarity, we divide the proof into a number of steps, Here we fix notation and make some preliminary observations.

243

DEGREES OF SUMS IN A SEPARABLE FIELD EXTENSION Let the fi'e l d [3 , ••• 1

be obta tned from

E

,Bn of a and

B,

Then

ts a fi'nite Galois extension

E/F

The group G acts transitively on the sets {a , •.. ,a}

and

m

1

B = {[3 ,, •• 1

Let H and K be the stabilizers in G of a and (G:Hl

and stnce

(m,nl = l,

=

m,

,B} n

B,

respectivelyo

Then

(G:K} = n

(l )

a standard argument yields (G: (.H n Kl} = mn

Hence the size of the orbit of A

x

is mn.

B

and

m

1

respectively.

and we put G = Gal(E/F). A=

by adjointng a11 conjugates a , ••• ,a

P

(a,B1 EA x B under the induced action of G on

It follows that

(.1.;; i ,;;;;· m, 1.;; j,;;;; n}.

x

A

B

and hence

ts a conjugate of a+ B

J

Since

ts transitive on

G

each a.+ B, 1.,

(.2)

[HK[ = [H[ [Kl/[H n

K\,

(3)

it follows from (l} and

(.21 that

[HK\ Hence

(.K: (H n K))

=

fPr

\~I • I~\

\G\

=

and therefore

= m

K acts transitively on A

Let W and W 1

be the F-ltnear spans of A and B,

2

V=W

and let

u = W1 n

and w ,w ,u .1

2

w. 2·

1

+W

are G-tnvariant subspaces,

2

deg(a+Bl

=

mn.

Replacing

there is nothing to prove. m0

B by

We use induction on

F-dtmenston of the normal closure of F(a,B).

are of coprtme degrees

respectively, let

Then G acts as a group of linear transformations of V

it suffices to show that

Li F,

(4)

If

[G[

=

[3 0

over L.

derive a contradiction,

if necessary,

[a\, i.e. on the then

a,B

E F

and

By induction, we may therefore assume that if a,B and n0 ,

respectively, over an intermediate field

and (a), (b} or (cl hold with respect to m0 ,n, 0

degree of a 0 +

1,

\[3,

From now on, we assume that

then m0 n 0

is the

deg(a+BJ < mn

and

CHAPTER 2

244

Owtng to (3}, not a11

s.J

a.+ 1,,

are dtsttnct and so for some k e {1,.,.,m},

s E {1, •.. ,n}

(5)

and so u t- O. Let M be the kernel of the action of G on v and let L be the fixed Then F.:: F(a,Sl::, L and L/F is Galois, so E

field of M.

L and hence

=

Thus

M = 1.

G acts fatthfully on Step 2,

(6)

Here -we prove that ther>e is no pr>oper subgr>oup

that the orbits

where m0

=

A0

IA 0

and B0 of a

and n 0

J

=

Assume the contrary.

L

@0

Stnce

L.

S under G0

(m 0 ,n0 1 = 1

Then

is the ftxed field of

respectively, over

and

G0

of

G which ac-ts so

satisfy

IB0 l,

the number of elements in the where

v

and since, by (5}, c1, + f3

-orbit of a+ f3

Now G0

+ f3s'

Gal(E/L}

=

and m0 ,n 0 are the degrees of a and 13,

@0 ,

1@0 1 < Jal,

(c} holds with respect to m0 ,n0 ,

is< m0 n 0 ,

= ak

it follows that none of (a)., (b} or

But then none of (a}, (bl or (c) holds with

respect to m ,n, a contradiction. Step 3,

Let

N neZ of the action of G on nf3 = u

fixes

N

(u

+y

2

tatn equal number of elements of A.

H ~ G0

A0 ,

Let

and hence

G0

Indeed, if a.1,, e

w1

and ts transitive on the

All of these, therefore, con-

Now a,ak ea+ u and if

A

0

be the stabilizer of the coset a+ u in

G0

is transitive on A, 0

(7)

(8)

nS

set of those cosets which contatn elements of A.

JA 0 J Im.

We now show that

ts fixed by G}

e U, ye W

Note that G permutes the sets of cosets of u in

then

A.

then

ag =

B.

We claim that

a.1,, for some

g

e G,

G0

=An (a+U), G.

Clearly,

is transitive on

Hence

245

DEGREES OF SUMS IN A SEPARABLE FIELD EXTENSION

and so g

E G0 •

G0 = G.

Thus

B

c

This establtshes transitivity and, by Step 2, we must have stabilizes

G

B + u and so B. J

=

a +

u

u.

and so Ac a+

B + u. for some

uJ,

J

By a similar argument,

Summing over

Eu,

s.J

we

EB,

derive

rs.J y,

Hence nB = u +

where

=

Eu. + nB J

J

proves (71 and (81, since

fixed by G,

2

-

tf x EL~ K,

r

1 = -

a.

a~ = 1,

rrm

Because y and

sm,

o

so

B and hence JcJ Jm!

N = 1,

of

y

It!,

charF)

(9)

we have

!>

1

are ,,;;; min(m,n).

B is

B # B. s·

G on A.

Thus

If

G

Therefore all prime divisors of

Since n

we conclude that

Be!'e oJe dePiVe a final contradiction.

charF

8

f

nm,

charF

%n,

(in particular, those of

this certainly excludes (c).

by hypothesis.

Jal Jn!

then by (8),

ts isomorphic to a subgroup of Jal

B# B ),

(recall that

and B in the above argument we obtain

IcJ

n

; A-z,.o + .1tl 'Y n

by (61.

Since (c} is excluded, we have A

1

-z,

i=l

This is a contradiction since

Completion of the pPoof.

nl are ,,;;; m.

i,

10i m

are fixed fiy G and

N be the kernel of the action of

fixes

X

;\.a,+ -

it therefore follows that

L,

IL IB = ill

N

r

L ts transitive on A,

and note that since

Summing (91 over

Let

"

1,

r

Step 5,

m

n i= 1

xEL

fixed by G.

0•

1, 1,

we have

m

i=1

IL I,

i=l

1

B = BX =

This

By (71, we have nB = u + y for some u Eu and some m As u cw, we have u = r ;\.a. for some A. E F. Now

Assume the contrary.

yE w

G.

2

then there ir, no subgPoup L c K

chaX'P ,I'

r.,ihi'Ca is t!'ansitive on A and tJuch that

w ts fixed by

w1 -~ u.

fixes all elements of

N

E

J

BePe we shaw that if charP ,I' rrm,

Step 4,

Put er

y = rs.

u = .. zu. Eu and

By interchanging

and hence all prime divisors

Hence, if charF = p > 0 and (a) holds or if

246

CHAPTER 2

charF

=

th.en cflarF) \GI.

0,

must have charF = p

>

But thts contradtcts (41 and Step 4.

We may assume that m = qe, q prtme, and let

0 and (b}.

Q be a Sylow q-subgroup of K.

it fol lows from (4) that

(K:(K n H}} = qe so K =(Kn H}Q and

Then

ts transittve on

Q

Hence we

Since charF ) q,

A.

by hypo-

thesis, we derive a ftnal contradiction by applying Step 4. • The following simple observation illustrates that to establish the best possible improvement of the above theorem, with conditions given in terms of m,n and charF, 17.2. LEMMA,

it suffices to consider only group representations Let v be a finite-dimensional vector space over a field

let a finite group Assume that u,v

G

act faithfully on

Ev are permuted oy

V

as linear transformations of

G into orbits of sizes m and n

tively and that u + v lies in an orbit of size k.

K

and

v. respec-

Then there exists a finite

Galois extension E/F (with K~ Fl and elements o.,S EE such that dego. = m, degS = n and deg(o.+S) = k Let t

Proof.

R = K[X1 , , •• ,xtJ

=

dtm v and let x ,x ,,,,,xt be indeterminates. K

1

and let E be the quoHent fteld of R. · We may identify v

wi'th the K-1 inear span of

X

l

,x , , . , ,xt in 2

R,

group of K-automorpftisms of R and hence of E. G in

E

and v.

Put

2

(so K::, F}

and let a and

Then

G

acts faithfully as a

Let F be the fixed field of

S be elements of E corresponding to u

By Proposition 4,5, E/F is a finite Galois extension and G ts the

Galois group of E/F. y under G,

Since for any ye E, degy is the size of the orbit of

the result follows. •

We close by giving some limitations on possiole improvements of Theorem 17.1. The fo 11 owing two ex amp 1es in conjunction with Lemma 17. 2 i 11 ustra te that the conclusion of Theorem 17.1 fails in each of the cases below: (il

(ti}

charF = 2, m = 3 and n charF

= 3,

m=3

and n

17.3. EXAMPLE.(Isaacs(l970)}. Let V*

= 4

=4 Let G be the alternating group of degree 4.

be a four dimensional vector space over Z/2Z and let G permute a

247

GALOIS COHOMOLOGY basis,

tn the natural manner.

{w,x,y,z},

v0 = {O, w +

x +

Put and V = V*/v0

y + z}

Then G acts faithfully on v and we put u

=

w+

v0 and

x +

v = w+

v0 •

Then u and v are permuted by a into orbits of sizes 3 and 4, respectively, However, the size of the orbit of u 17,4. EXAMPLE.(Isaacs(l970)).

with basis

'll./3'1!.,

a ,a ,a l

2

GL(V)

E 3

0 a

{w ,x ,y ,z}.

is

+ v

4 f 12.

Let V be a four dimensional vector space over Let G oe the group generated by the elements

whose matrices are

0

0

0

0

0

0

0

0

-1

0

0

0 0

a

(J'

l

0

0

0

0

2

0

0

0

0

{w,w

Put u = w and v = y.

+ x,w - x},

3

0

0

Then G is the direct product of the subgroups of order 2.



l

2

0

0

0

0

0

0

0

0

0

0

of order 6 and

3

Then the orbit of u under a is

and the orbit of v under a is

However,

{y,y + x,z,z + x}.

the oroit of u + v is {w + y, w + y + x, w + y - x, w + z, w + z + x, w + z -

x}

which has six elements. • 18, GALOIS COHOMOLOGY

Let

G

be a group.

A a-module is an aoelian group

A

on which

G

operates

such that (i}

la = a

for all

a EA

(it)

g(a+b) = ga + gb

for all

g E G, a,b EA

(iii)

(gg )a= g(g a)

for all

g,g

l

l

E G, a EA 1

Note that if A is a a-module, then A is a 'll.G-module via (fa g)a g

= fa

g

(ga)

(x g

E 'll., g E G,a EA)

Conversely, any given 'll.G-module A can be regarded in the obvious way as a a-

248

CHAPTER 2

The fixed module of a G-modul e A

module.

AG= {a E Alga= a

It is clear that AG If

g E G}

for all

is the largest submodule of A on which

G

acts trivially.

are a-modules, the group of all abelian group homomorphisms

A,B

is denoted by as follows:

i's defined to be the subgroup

Hom(A,B), if

The group

E Hom(A,B), g¢

¢

Hom(A,B)

A->- B

has a a-module structure defined

is the mapping

ai----g¢(g- 1a), a EA, g E G.

ln particular, if x is any abeHan group, we can form the G-module

Hom(JZ'.G,X).

A a-module of this type is said to be co-induced. Let A copies of

be any a-module. G to

A

G,

f

from the direct product of n

is cal led an n-cocha1,,"n of G in A.

said to be normalized if ti ty e 1ement of

A function

f(g , •• , ,g ) = 0 1 n

whenever any of the g.1,,

The set of a 11 n-cochatns, written

abeltan group under the multiplication of values, d1(G,AL

to the case

n = 0,

An n-cochain

c12 (G,A),

f

is

is the i denbecomes an

Extending the definition of

we put

c0 (G,A)_

= A

The formula

( 1)

determines a homomorphism

It is a standard fact that d d

n n-1

Zn(G,A)

and refer to the elements of aries, respectively.

A

for all

= 0

= Kerdn and zn(G,A)

and

n ~ 1.

Bn(G,A)

= Imdn-1

(n;;;,, 1)

as n-cocycles and n-cobound-

Bn(G,A)

Then-th cohomo'logy group

We set

~(G,A)

with coefficients in

is defined by ~(G,A)

= Zn(G,A}/Bn(G,A)

H°(G,A) = AG

(n ;;;,, 1)

GALOIS COHOMOLOGY

249

rt can be easi'ly shown that the conomology group Hn (G,A} restrict ourselves to normalized n-cochains.

is unaffected if we

Let us recall the standard fact

which says that gn(G,A) = 0

n ~ 1

for all

if

A

is co-induced

(2)

Formula (1) shows that a 1-cocycle is a crossed homomorphism, i.e. a map G'--+A

f

satisfyi'ng f(gg

I

l = gf (g 1 + f

f (g l

Similarly a 1-coboundary is a principaZ crossed homomorphism,

i.e. a map

f:G-A

for which there exists a EA such that f(gl = ga - a

In particular, if G acts trivially on A, H1 (G,A} =

then

Hom(G,A)

From (11 we see also that a 2-cocycle is a function f: G x G -

A such that (3)

while a 2-coboundary is a function map t :

G --'--->- A

f

: G

x G' -

A

for which there exists a

such that

Let E/F be a finite Galois extension with G = Gal(E/F). and multi'plicattve groups

E+

Then the additive

and EJi' of E are a-modules and we may consider

the groups for all Our first aim is to show that the group gn(G,E+l

is always trivial.

n

~ 1

This will

be derived as a consequence of a general result of independent interest, namely the normal basis theorem.

CHAPTER 2

250

Gl'ven a fi'ni'te Galois extension

E/F,

fo(a1 Ier E a EE,

for some

any F-basts gEG g

where a e

E

F)

is such that

fo(a) la

E

G}

l:: ;\ g(a} gEG g

is a normal basis.

Invoking (2)_, we

CHAPTER 2

252 are therefore left to verify tnat

Hom(U,F) For each A E Hom(a'.G,F)

and each

put Ag

g E G,

l

as a-modules

~ FG

Hom(U,F) -

Then the map

= A(g). .

FG

A i-- EA/

is a required isomorphism. • We now turn our attention to the study of ff"(G,E*), more complicated:

the ftrst cohomology is always trivial, but in contrast to

Corollary 18.2, the groups /l(G,E*} for n 18,3, THEOREM.

Then H1 (G,E*}

are not generally trivial.

> 1

Let E/F be a finite Galots extension with Galois group Go = 1

Let f: a-

Proof.

Here the situation is

E*

be a crossed homomorphism. a =

For \EE* we put

I: f(cr}crA CJEG

Because of the linear independence of the automorphisms o(Corollary 4.81, we may choose A E E* such that a ; 0, w

=

We then obtafo for ,

i ,f(crl(wAl = i f(,1- 1f(wl(,CJ"),} crEG

E (}

= f(,)- 1 a

crEG

whi'ch implies that f(, l

-1

,a_l

=

0\

Therefore f

E B 1 (.G,E*)

and so 8

1 (@,E*l

= 1,



In what follows, we shall f1'x the following notation: E/F G =

of

a finite Galois extension Ga 1 (EJF)., N . a norma 1 subgroup of G E B2 (G,E*l

ts a coboundary corresponding to f :

G

-+

(elf}( cr ,cr ) = er (f(cr )lf(cr cr )- 1f(cr ) 1

2

1

2

12

·

1

E*,

i.e. (cr ,cr E G) 1

2

K the fixed field of N.

The above assumptions guarantee that K/F ts a Galois extension whose Galois group is identifiable with G/N.

Each a

E

Z2 (G/N,K*)

determines a unique

253

GALOIS COHOMOLOGY

a*

z 2 (G,E*)

E

given by a* ( 0 , 0 } = a ( CT , CT ) 1

where

The map

ai = aiN' i = 1,2,

ca 11 ed inffotion map.

2

1

2

induces a homomorphism

at-->- a*

By restricting the co chains to

N x N,

we al so have a

homomorphism

ca 11 ed the restriction map.

These two homomorphisms a re connected by the fo 11 ow-

ing theorem. 18,4, THEOREM. (Hochschild(l950), Eilenberg and MacLane(l948}). l

-->-

The sequence

~ 8 2 (N,E*)

H2 (G/N,K*) J..!!f.,_ H2 (G,E*)_

is exact, Proof. Step 1.

For the sake of clarity, we divide the proof into three steps. Here we show that rnf is injective,

Assume that a

is such that a*

E 2 2 (GIN ,K*)

a*(0 ,0 1

Then

a(1,cr}

= t(1}

so that

2

1=

=

ot where t

0 (t(0 ))t(0 0 )- 1t(cr 1

2

12

l

(0 ,CT 1

t (0} = 0(t(1)),

Now set

t(d E K*.

1

i.e.

G-->- E*,

E 2

G)

Then

l

t

(0) E 1

x*

for all

implies that t

a

= 1

aos- 1 , We have

1

0

= 13*

E

c,

and t (an}

t (0) · for all

=

1

for some

1

S: GIN- K*,

it suffices to verify that a

where

0

G, n

E

E

This

N.

Setting

S*(cr) = S(CT).

is a coboundary. 1

a*= ds,

where

s(cr} = t(0}cr(t(1))- 1 •

If n ,n EN then

1

1

2

(ds)(n ,n ) = 1 1

By Theorem 18,3, we can find Let us set µ(nl = 1

µ(0)

'.\EE*

= s (0) '.\cr{'.\)- 1 •

for all

n EN,

Then

2

such that



=

os,

s(n} = n(>.)r 1

so that

a* 1

for all

= oµ.

Furthermore,

This gives (er

t.e.

µ(0n) = µ(0)

for all

n E N.

0 E G, n EN,

Also

E G,

n

E

N)

CHAPTER 2

254

Therefore n(µ(cr}} Thus we have µ

= µ(cr}

for all

for some u: G/N -

= u*

which means that µ(cr)

cr E G, n E N

E

K*,

which shows that

K*,

ct = ct oS = . er }t.. q-1 •

l

2

l

l

Owing to Theorem 18.3, we can find Because sa depends only on

2

1 . l

,cr}

Acr EE*

& and s 1

= 1,

such that we may take

Acr

GALOIS COHOMOLOGY

-

to depend only on

and ii. = 1.

(J

= a (a ,a la (ii.. )- 1\

a(a ,o} then a

If we deftne

1

1

2

1

1

2

a2

1

a a ii.~l l

2

ts obviously tn the same cohomology class as a

2

the same element of B2 (G,E*) only on a

and the coset cr

1

a (a n,a 2

1

2

l

as do a of a

2

=

mod N.

2

=

2

2

1

Cl early,

a (a ,no ]a (ii.

=

a 2 (a 1 ,al 2

1 · 1

l-1 ii.

2

1 · 0

2

1

2

cr

1-l Aa

l

a

ii.-

-

2

l

a

2

and o .

1

-1

1

2

1 2

l

l

Finally, by our construe-

lf

a2 (a l ,a 2 1

=

for some t E Z2 (G/N,K*).

= t*

2

means that the cohomology class of a belongs to

H2 (G,E*),

2

ii.-l

a2 (no 1 ,a2 la 2 (n,o 1a2 )_ a2 (.n,a 1 l

2

2

18.5. COROLLARY.

a (a ,a) depends

Moreover,

1

=

Thus a (a ,a l EK* and therefore a 2

and hence represents

1 which implies

n(a (a ,a J) = 2

l l

a ao a 2 1 2_11 l a (a n,a }a (ii. a (n,a )l A ii.: 1·1 2 1 CT 1· 2 O'CT u 1

provi'ng that a 2 (a- 1 ,a). depends only on 2 •

a (n,a}

and a.

1

a (a n,a ]a n(ii. 1

=

tion,

255

This

Im Inf, as required. •

ts of order m and if c is an arbitrary element of

N

then cm e lm(rnf)_ where

f:nf : Ii2(G/N,K*)--+ H2 (G,E*)

is the in-

flation map. Proof.

Let a E Z2 (G,E*}.

Define t : G -

t(a)

=

E*

by

naca,nl

nEN

Then we have (dt) (a

,0 ) = 1

2

= =

11 (a1 (a(a 2 ,n) la(a 1 er2 ,nl- 1 a(cr 1 ,n))

nEN

11 Ca(a1,a2n1- 1 a(cr1,a2 )a(cr1 ,n))

nEN

a(o

)m

,CJ 1

2

n (a(cr

nEN

,a n)- 1 a(a ,n)) 1

2

1

In particular, if a EN this yields 2

(8t)(a ,er}= a(a ,a )m l

2

l

2

whence cm is in Ker Res , where Res : H2 (G,E*l->- H2 (N,E*) tion map,

is the restric-

The desired conclusion now follows by virtue of Theorem 18,4. •

CHAPTER 2

256 Two Galots extensions

are sai'd to be isomorpMc H

E /F and E /F l

2

E

1

is

F-isomorphtc to E. 2

18.6. THEOREM.

let

Let

and

E /F 1

2

= Gal(E./F), i = 1,2. 1, .

G. 1,

be isomorphic finite Galois extensions and

E /F

Then there is an isomorphism

f

such that for any F-isomorphism g: E _,. E 1

p

2

a (CT ,CT g

l

Proof.

2

1=

a

then

Ag

1

l

,E*) 1

onto

g

a

1-r

a

g

is a homomorphism which maps

H 2 (G

,E*).

2

A is another F-isomorphism

If

2

E---+ E 1

t(CT}

=

-1

2

SE z (G ,E*1, SS,., 2

2

,-.

is a coboundary.

a, g

In fact,

S(CT,µ)_S(11,µ- 1CTµ).- 1 , then we have

Since

as can easily be verified. as required.

The unique isomorphism

a A = (ag ) µ ,

-1

we conclude that a,/\

is cohomo-

• of Theorem 18.6 i's called the natural isomorphism.

f

Next we introduce the following important homomorphism. 18.7. LEMMA.

for any a E cEH 2 (G,E*),

Let E/F be a finite Galois extension with Galois group Z2

(G,E*},

let

say c=a,

a

denote the cohomology class of a,

defi'ne

~

c' G -

( CT Then

c'

2

= µ, say, is an automorphism of E/F, and we have aA = (o;g)µ,

(ml (CT1,CT2 1 = S(CT ,CT }S (CT ,CT ) . ··12)112 logous to

f.

Because it has an inverse, it induces an isomor-

We now claim that, for any if we put

a

,e:)

induces the isomorphism

2

It is clear that the mapping

H 2 (G -1

I--->-

g( a(g -lCT g ,g-lCT g) 1,

coboundaries to coboundartes. phism of

the mapping

Cc 1 ,E7l- z2 (G 2

? where

2

i---+

F*/NE/F(E*) (1la(T,CT))NE/F(E*) ,EG

is a we 11-defi ned homomorphism and the map ~H 2 (G,E*)

?

--,. Hom(G,F*/NE/p{E*))

c1---rc 1

G and,

Given

,

GALOIS COHOMOLOGY

257

ts a homomorphtsm. Proof.

Given

z 2{G, E"'},

a. E

put '&(cr}

It is easy to verify that, for all

na.('r, a) TEG cr , cr E G, we have =

1

2

a (a( a } )_ = a( cr } ·1

which means that say

-1

as = olj'

a,

maps

into

G

2

2

If

F*.

SE Z 2 (G,E*)

is such that

B=

then

=

a(cr)S(cr)- 1

Thi's shows that the map

= NE;ig(a))

n,(g(cr)}g(wl- 1g(,)_

,EG

is well-deftned,

c'

By the multiplicative version of (3}, we have rn( a , a )_ = a(Tcr , a ) a( T, a a ). - 1 a( ,, cr } l

2

l

2

12

l

whence NE/F( a(cr 1 , a)) =

n

,EG

T( a( 0 , 0 l

}1

= a,( 0

2

a ) - la,( (J'

) a( - En~ (V)

?r

fr

,---+

We say that R acts dense Zy on V if for each family v ,v , .•. ,v 1

E End (V)

s

and each finite

v there exists r ER such that

in

n

2

8

8(1\1 = fl'(vi)

Let V be an R-module, let s

19.1. LEMMA.

generated as s-modu 1e.

Then

R

(1.;;; i.;;; n)

End (V)_

=

R V

acts densely on

and let V be finitely

if and only if the homomor-

phism ~ R-,,

?r

En~ (V)_

.- f r

is surjective. Proof. densely on

If the given homomorphi'sm i's surjective, then obvi'ously

Conversely, assume that

V.

v.

be a generating set for the s-module

acts densely on

R

If

8

E

End (V),

V

R

acts

and let V 1 ,.o.,Vn

then by hypothesis

s

there exists r ER with

proving that f r

=

8(v.) = f r-(v.). Thus, for any 1,· 'i n n 8( I: \ .v .) = f ( L \.v .) , i=l 1, 1, r i=l 1, 1,

in

s,

8. •

Assume that an R-module v is a finite direct sum of irreducible

19.2. LEMMA. submodules.

Then R acts densely on Put s

Proof.

=

End (v)

v.

and fix v E v,

R

E;

R.

Because

f

E End (V).

We claim that

S r1J =

for some r

:\ 1 ,\ 2 , ••. ,\n

f( V)

v is completely reducible, v =

Rv EB

w for some

R-

265

THE BRAUER GROUP OF A FIELD

w,

submodule

Let

V -

1r

!'.le tfte projecti'on.

R1!J

Then

71' E

and so

S

f( V) = f (TrV) = 'ITf ( V)

This shows that f(v}

E

as claimed.

Rv,

e E End (V} and let

Let

v , ••• ,v

and put s 1

=

End (fl}. R

Then

v,

and the assumption on

n

1

s

Define en

Ev.

fl -

i's a full matrix ring over s.

S'

one immedl'ately verifies that e

n r ER

Hence, by the first paragraph, there exists an element

by

fl

Using this

lies in End (~}. S

such that

which is what we wanted to prove. • For any F-algebra A we can form the so-called enveloping algebra A ®A 0 , F

where Ao denotes the F-algebra opposite to A. left multiplication by a on A,

and ar

i

A -,. At ::_

la

1-->-

Given a EA,

right multiplicationi En~

let ai denote Then the maps

(A)

a1

and iAo _.,Ar~ En~ (A)

?a

I-·--+ a

r

where A ={ala EA} 1

t

and Ar ={ala EA} r

are injective homomorphisms of p.-algebras.

By the associativity of for all

ab = b a Q; r r 1

Hence

A

A,

we have a,b EA

becomes a left (A® A0 )-modu1e upon defining F

(a,b,x EA}

(a ® b )x = (airl (x) = axb

19.3. LEMMA.

Let A be an F-algebra and let R

= A ®A 0 F

(t}

End (A)_ R

=

Z(Ali

:!!!

Z(Al

266

CHAPTER 2

(i't}

ts sim13le if and only if

A

(iii)

If A is simple, then

Proof.

i's an i'rreduci'ble R,-module is a field

Z(A)

The inclusion one way is clear.

(i)

f E End (A).

A

f

Then, by an easy computation,

Conversely, assume that and one then checks that

= f{l) ,Q,,

R

since

f(l) E Z(A)

End (A).

f E

R

The two-sided ideals in A are precisely the R-submodules of A

(ii)

(iii)

If

is simple, then for any Of a

A

a

as required.

A 0 A0

Z(A), Aa

= Aa = A.

Hence

U(A) n Z(A) = U(Z(A)),



19.4. LEMMA. (i)

E

E

Let A

be a central simple F-algebra

acts densely on A

F

A 0 A0 ~ Mn(F)

(ii)

where n = dim A.

F

Proof.

F

Put R

( i}

= A

0 A

and note that A is an irreduci b1e R-modul e by

0

F

Lemma 19.3(ii). (ii)

By Lemma 19.2, it follows that

By Lemma l9.3(i), Ind (A}~

F

acts densely on

R

A.

and so, by (i} and Lemma 19.l, the map

R

R -+ End (A)., r

f-->-

is a surjecttve homomorphism of F-algebras.

fr•

Since

F

End A

~

Mn (F),

we have

F

di'm R

dtm (End A)

= n2 =

F

and the result follows. 19.5. LEMMA. of u and

Let V,

F



u,v

be vector spaces over F and let

respectively,

V',V'

be subspaces

Then

(vi 0 V) n (V 0 V') F

Proof.

F

U1 0 V 1

F

F

We can write

u = u' EB for some subspaces

U",V" of

u

and

U"

v,

and

v = v' EB

V"

respectively.

Accordingly,

U © V = (U' 0 V') EB (U 1 0 V") EB (U" © V') EB (U" © V") F

F

F

Now the left-hand side of the required equality is

F

F

267

THE BRAUER GROUP OF A FIELD [(V' ® V'} EB (U" ® F

and ts clearly

r'll

n [(U' ®

F

V"l

EB (U 1' ®

F

V'll

F

U' ® V', • F

The next observati'on descrtbes a useful lattice isomorphism. 19.6. LEMMA.

Let

be a central stmple F-algebra and let

A

Then the map I -

B

be il,ny F-algebra.

is a lattice tsomorphism between the ideals of B and

A® I F

those of A~ B.

where

B

The inverse of this isomorphism is given by

ts 1'denttfied with 1 ®

Proof.

B

We claim that the given maps are inverse to each other.

This will

prove the result, for they establish a bijection which is clearly order-preserving and hence a lattice isomorphism. Let I

be an ideal in B,

Then A® I

F (A ®I) n B

ts obviously an ideal in A® B and F

= I

F

by virtue of Lemma 19,5, Conversely, let

so we need only verify that F nonzero a E J' can be written as a = a

F Jc A®

A® I c J',

and wtth

{a ,.,.,a}

acts densely on A,

A® A

0

-

1

F

® b

1

-:1-

n B,

= J

Then clearly

To this end, note that each

I, , ..

a ® b

-:1-

r

being a subset of an F~basts of A.

r

1

r

be an tdeal fo A® B and put

J

r

with O ,; b • E i

By Lemma l9,4(t},

Hence there extst xj,YJ EA for j E {1,2, •• ,,r}

F

such that Exja ,.y. = j

J

'Z,

& • , 1 .;;; i .;;;

Ex.ay.=

Therefore b 1 e JC A® I, -

I

r,

But then

'Z,

J

J

ExJa,.y.®b~=l®b EJ i; ,j 'Z, J 'Z, l

and simtlarly b '2,.. e

I

for i e

{2, ... ,p},

proving that



F

19,7. LEMMA. A,,i=l,2. '2,

Let

A

and

l

be F~algebras and let

2

B. '2,

® B ) = CA (B ) ® CA (B 2 }

® A (B 1

Proof.

A

Then CA

F

B

2

It is obvious that

1

F

2

1

1

2

be a subalgebra of

CHAPTER 2

268

To prove the opposite inclusion, fix an F-basis element of A 0 A F

1

of

{v.} 1,

can be uniquely written in the form

Then every

A 2

0

Ea.

1,

2

v.

1,

with

a.EA 1,

Given any b EB, we have 1

Therefore, if Hence a.1,

0

Eai

(B 1 0 B),

A

l

(B" 1 1 and so

E CA

0

vi E CA

then

for all

aib = bai

b

E B , 1

2

1

Ea • 0 v • E CA (B ) 0 A 1, 1, 1 1 F z

Applying Lemma 19.5, we i'nfer that CA

0

® A (B 1 F

F

1

2

B ) 2

C

(A

-

1

0

F

(B ) ) n

CA

(CA

2

2

1

(B ) 0 1 F

A

) 2

CA (B ) 0 CA (B ) , 2 1 1 F 2

as asserted, • 19,8, COROLLARY.

Let A

and A

l

Z(A

Proof,

be F-algebras,

2

0 A ) = Z(A

1 F

2

1

l

0 Z(A ) F

Apply Lemma 19,7 for the case B

1

19.9, COROLLARY.

Then

=

2

A

1

and B =A. • 2

2

Let A be a central simple F-algebra and let B be any F-

algebra.

(i'l A 0 B i's simple i'f and only i'f B i's simple F

( i i )_

Z(A 0 B) == Z(B) F

In particular, the tensor product of central simple F-algebras is again central simple. Proof.

Apply Lemma 19.6 and Corollary 19,8, •

We are now ready to prove 19.10. PROPOSITION.

The Brauer classes of central simple F-algebras form an

abelian group with respect to the multiplication induced by the tensor product, Proof.

Let

A

and

B

be central simple F-a 1gebras.

Then

A

0 F

is again a central simple F-algebra by virtue of Corollary 19.9,

B

== B 0 A F

Suppose that

1

THE BRAUER GROUP OF A FIELD

A "" D ® M (Fl

B ~ D

and

n

1

2

269

® M (F) F

m

Then A ® B ~ (D ® M (_F)) ® (D ® M (F)) ~ (D ® D ) ® M (F) F 1Fn 2Fm 1F2 nm

and so

® D.

A® B - D

F

1

F

Hence, if

A - A', B - B 1 ,

then

2

A'®B'-D

F

®D 1

provtng that the multiplication

F

-A®B, 2

[Al [Bl = [A® Bl

F is well defined.

F

The multiplication of Brauer classes is associative by the corresponding law for tensor products.

Taking i'nto account that

A® F"" A

and that, by Lemma 19,4

F

(iil,

A® A0 ="' M (Fl,

F

n = dim A,

F

n

we also have

[Al [F] = [A] ,

This shows that

[Fl

[A] [A 0

[F]

]

is the identity element and that

[A 0 ]

is the inverse of

[Al, •

The group of Brauer classes of central simple F-algebras is called the Brauer group of

Let

F and is denoted by

be an F-algebra.

A

field for

A

Br(FL A field extension

E)F

is said to be a splitting

if A® E =" M (El F r·

for some posi'tive integer

r.

Our next aim ts to show that

E which is a ftntte extension of

splitting field

F,

A

always has a

The following basic

result ts known as the Skolem-Noether theorem

19.11. THEOREM. suba 1gebra of

Let A.

A

be a central simple F-algebra and let

B

be a simple

Then every isomorphism of F-a 1gebras ¢ : B-,. B' ~. A

can be extended to an inner automorphism of in

A

A,

that is, there exists a unit

a

such that ¢(b ) = aba

-1

In particular, isomorphic simple subalgebras of

for all A

b E B

are conjugate and hence have

CHAPTER 2

270

tsomorpntc centraltzers, Let V be an irreductble (left} A-module, and let D = End (V).

Proof.

A

Then D ts a division algebra with centre F~lv and B 0 D is a simple F-algebra F

by Corollary l9.9(i).

We may view

V

as a

D-module via

B0 F

(b E B, d E D, v E V)

(b 0 d)v = b(dv)

Using the automorphism ¢,

we can deftne a second left B 0 D•module V 1 ,

whose

F

underlying vector space over F i's v, and witn tne acti'on of B

0 D

given by

F

(b

Then v and V'

are of tne same F·dimension.

Because B

B, d E D,

E

ts simple artin-

® D

F

e v-->- V'.

tan, we infer that there exists an tsomorphtsm of B ® D-modules

e E End

Since

F

(_v),

plication at for some a EA,

Furthermore, a E U{.Al

pnism.

is a

B ® F

D•tsomorphism, we have

a(o(dvU = ¢(o)d(avl

that ab

= 1,

e is a left multisince e is an isomor-

it follows from Lemmas 19,l and 19.2 tnat

D

Taking d

for all

b EB, d ED, VE V

and applytng the fact that A acts fattnfully on so ¢(bl

= ¢(_bla,

19, 12. COROLLARY,

VI E V )

Let

A

subalgebra with centre E.

for all

= aba- 1

b EB. •

be a central si'mp1e F..-algebra and let If dim B = k, p

v, tt follows

B

be a simple

tnen

A ~ B0 ~ CA(Bl ~ Mk(Pl ~ Mk(CiBlJ.

In particular, tf E ts a subfteld of A containing F, A® E ~ CA(El 0 Mk(F)_ , F F

Proof.

We may identify Mk(F)

ations of B, and

B

r

= (E;F)

witn the algebra of all F-ltnear transfrom-

As such it contains the subalgebras Bi of left multiplications

of right multiplications.

Clearly Bi~ B and Br~ B 0

simple,

where k

tnen

and, by Corollary 19.9, A

j

Mk(F)

is central

Now B ® F and F 0 B1 are isomorphic simple subalgebras of A® Mk(F). F

F

Hence, by Theorem 19,11 and LenJTia 19.7, CA(Bl 0 M,-.(F) ~ A ® B ~ A 0 8° F K. F r F

F

THE BRAUER GROUP OF A FIELD

19,13. COROLLARY,

271

Let A be a central simple F-algebta and let B be a stmple

subalgebra wtth centre E, (t}

CA(Bl

(ti'l

is simple with centre E

CA(C/BD = B

(.tiil dim A = {.dtm Bl(dim CA(Bl} F

Proof.

F

F

Comparing dimensions tn Corollary 19.12, we have

Bl= (dtm

(dim A}(dtm F

proving (tiil,

F

Since A® Tf

B)2 dim CA(.B},

F

F

is stmple, so ts

by Corollaries 9.12 and

CA(B}

F

9,9,

Replacing B by CiB1

tn (tit},

we therefore find

dim A = {.dim CA(BU(dim CA(C/Bll F

F

F

·

whence dim CA(C/Bll = dim B F

Since we have clearly cA(C/BU

F ~ B,

tt follows that CA(cA(B11 = B,

proving

(i i 1.

Finally, i'f K ts the centre of cA(Bl,

we also have

E ~ K,

19, 14, PROPOSITION.

t.e.

E = K.

Proof. CD(El = E,

and stnce



Let A be a central simple F-algebra, say A = M (D} n

D ts a divtsion rtng wtth centre

Then A.® E ~ Mnk(El F perfect square.

then K ~ E,

where k ·

=

F,

(p;:F},

and let E be a maximal subfield of D. In particular,

Applytng Lemma 19,12 for the case

A = D

dtm A= n 2 k 2 F

Hence

F

2!!

M (D ® E} ~ M (Mk(E}l n F n

£!!

is a

and taking into account that

we deduce that

A© E

where

M k(E) n

272

CHAPTER 2

as requ i'red. • Let

A

be a central simple F-algebra.

It is a consequence of Proposition

19.13 that

dim

A=

r2

F

for some integer r

and that there is a finite extension E of F such that A 0 E = Mr(E) F

Thus A always has a splitting field E which is a finite extension of F. say that

[A]

is sp tit by E,

We

or that E is a split ting field for the class

[Al,

if E is a splitting field for A.

for

[Al

That the notion of a splitting field

is well defined is a consequence of the following simple observation.

19.15. LEMMA.

Let E be a field extension of F.

Then the map

} Br(F)_--,. Br(.E) ( [Al ,..__,.

[A ©E] F

is a homomorphism whose kernel consists of those Proof,

[Al

which are split by E.

Let A be a central simple F~algebra and write A~ D ©F Mn (F). ..

Then A 0 E F

and hence

D 0 M (F) 0 E F n F

2e

D 0 M (F) F n

Thus the given map is well defined.

[A 0 El = [D 0 El,

F

2e

It is also

F

a homomorphism, since (A 0 B)_ 0 E F

2e

(.A 0 E) 0 (B 0 E)

F

F

F

The final assertion is a consequence of the definition of the splitting field for [Al•

• We denote by Br(E/F)

the kernel of the homomorphism } Br(F) (

Hence, by Lemma 19.15, [Al

which are split by

Br(E/F) E.

Br(E)

[A] f->- [A 0 E] F

is the subgroup of Br(F)

consisting of those

THE BRAUER GROUP OF A FIELD

[Al E Br(F)

We next present a criteri'on for 19.16. LEMMA.

Let

field extension.

Proof.

Let

Of c

be the division algebra in the class

D

E::. M/D).

invertible. 19.13(i),

such

[Al

n, v

C/E),

n

be

is an irreducible E-module.

is a dtvision algebra.

is the centre of

and let

[Al ,

determines a nonzero endomorphism of

c3 (E)

Thus E

BE

B = Mn(D}, v = Dn and observe that

Put

By the minimality of

c3 (E}

E

be a finite

E/F

E as a self-centralizing subfield.

is an E-module.

Now any

Br(E/F).

if and only if there exists

[Al E Br(E/F)

the least integer such that V

to be an e1ement of

be a central simple F-algebra and let

A

Then

B contains

that

273

v,

hence

c

is

Furthermore, by Corollary

l'nvoking Corollary 19.12, we also have

c3 (E) "'B 0 E - D 0 E F F E

By Lemma 19.15, this shows that

splits

D

E ts a self-centralizing subfield of

and on1y if

c8 (E)

if and only if

= E,

i.e. if

B. •

Our final observation on central simple F-algebras ts given by 19.17, LEMMA. idempotent of

Let

be a central simple F-algebra and let

A

A.

=

vector space over

D

D on which

an inner automorphism of

where

I

r

eAe

=a Mr(D).

A

A

=a Mn (Dl

Br(Fl for some

and

By changing the basis of an n-dtmensional

1.

acts irreductbly and faithfully, i.e. applying

A, we may assume that

is the identity

eAe

and so

n:;,,,

and some

in

[eAe]

By Wedderburn 1 s theorem,

divisfon algebra

be a nonzero

Then [A]

Proof.

e

=a[:r Thus

r x r-ma trix.

:J [eAel

Then

Mn(D)

[:r

[DJ

[A]

:J

[:r(D)

as required,

:J JI

CHAPTER 2

274

Let group

be a finite Ga 1ois extension and let

E/F

with respect to the natural action of

Z2 (G,E*)

remarked earlier, 2-cocycles.

G = Ga 1 (E/Fl. G on

Consider the As has been

E*.

is unaffected if we restrict ourselves to normalized

H2 (G,E*)

For this reason, we shall assume that all 2-cocycles are normalized.

For any given

ct E Z 2 (G,E*},

we shall construct a central simple F-algebra

called the crossed product of G over

This crossed product

E,

FfiG

lxG,

was intro-

duced by Noether and played a si'gni'ftcant role i'n the classical theory of central simple algebras.

r@

That the crossed products

occur in an unavoidable way in

the study of centra 1 simp 1e a 1gebras will be tll ustra ted by the fact that any central simple F-algebra is isomorphic to

such that E

A

for some

lJ,G

the cohomology class of

a

with basis

Given

G.

{g Ig

e: G}.

E Z 2 (G,E*L

In what follows, we write

& for

E Z2 (G,E*l,

ct

For the rest of this section, Galois group

is a self-centralizing subfteld of A

E/F denotes a finite Galois extension 1,Jith

ct E z 2 (G,E* l,

we denote by EctG

Thus each e 1ement of

EctG

a free 1eft E-modul e

can be uniquely written in

the form

Zxj We define the multiplication on

EaG

(xg E E, g E G}.

distributtve1y by using

(r ,r EE, x,y E G) (l} 1

2

where for all Recall that for all

A EE, g E G

x,y,z E G (2)

ct(x,ylct(xy,z). = xa(y,z)a(x,yz)

(3)

ct(x,11 = ct(l,x) = 1

Note also that, by (1),

gttg 19, 18. LEMMA.

l.

For any

ct

E

-1

=git'

xy=

Z2 (G,E*}, EaG

ct(x,y)xy

(g,x,y E G, A EE)

(4)

is an F-algebra with identity element

275

THE BRAUER GROUP OF A FIELD

It suffices to veri'fy that the multiplication given by (1} is asso-

Proof. ciative.

Indeed, in this case

is obviously a ring with identity 1 (the

EaG

latter can be seen by applying (l} and (3)_) and, by (4), morphic copy of

F

r ,r ,r

Fix

1

2

contained in

EE

and

Thus

Z(EaG).

{\•ll\

is an iso-

E F}

is an F-algebra.

EaG

We first show that

x,y,z E G.

3

for a 11

\ E E

( 5)

Indeed, applying (41 we have

x(y\jj

x(y\) =

proving (51.

-1

-1

)x

= a(x.,y)xy

\xy

-1

a(x,y)- 1

= a(x,y)xy \ a(x,y)- 1

We now have

(by (5}} = rxr a(x,y)xyr a(x,y)- 1a(x,y)a(xy,z)xyz 1

2

(by (2))

3

= [(r x}Cr 1

2

ifll Cr 3 zL

as required.• In what follows, we identify E wtth its copy 19.19. LEMMA.

For any a E Z2 (G,E*1,

Lemma 19.16,

(i}

Proof.

In particular, by

CEaaCE} = E.

[EaG] E Br(E/F).

EaG ~ Mn(F)

(ii'}

ectc.

in

the following properties hold:

is a central simple F-algebra and

(i)__ EaG

{\•ll\ EE}

if and only tf

for some n;;;. 1 Given

x = ix

g

gE

and

EaG

a is a coboundary.

\EE,

it follows from (l} and (4)

that \x

Hence

\x = x\

and

xg

t-

0

i.e.

g

=

1.

if and only tf

for some

This shows that

the other hand, if for all

g E G,

g E G,

µ E Z(EaG)

hence

µ E F,

= t\x - g g\x

g

and

= g\x

then

Thus

= Ixgg\g-

for all

x E CEa0 (E)

CEa 0 (E)

then

g

x\

= E.

g E G.

implies

Therefore, if \ = g\

It is clear that

µEE= CEa 0 (E).

Z(eGtGl = F.

for all F

:=

Z(EaG).

Furthermore, by (4)

x

t-

0

\ E E, On gµ=µ

276

CHAPTER 2 Gtven

that

x =

w:i El\;,

let !1,(xl denote the number of nonzero

Ffa.

ts a nonzero ideal of

I

choose such that has the smallest

Among the nonzero elements x Multiplying

i(x).

if necessary, we may assume that x 1 # 0.

Si'nce

y

g

Thus, by hypothesis,

1

i, e,

Ax - xA = 0

x

r.xj

of

I

Now for any A EE

f' 0 implies x . 1' 0 and since y = Of, x ,

g

=

by an appropriate g:... 1 ,

x

= Ey_q rr EI, where yg = Axg

Ax - xA

Assume

xg.

1

- x gA g

we have

E cg1iE} = E.

!L(Ax-xA) < i(x)

Hence I = Ea,G and so

ts simple,

EaG

ts a coboundary, say a = at for some t :

Assume that a

(it)

t(g}

=

1.

For each

g

E

t(g}- 1g

g=

put

G,

.

Then

{gig

E G}

G -

with

E*

is an E-basi's

of Ea,G and, for x,y E G,

xy

t(x)- 1 xt(y)- 1 y = t(xl-lxt(y)- 1 a(_x,y)xy = t(xyl- 1 Xb' =

Given

AE

E,

let

xy

k.

such that g (\) f g (\), 1

(A}s- ,,, SA = 0 1

since each

one easily verifies that

2

g

1

gE

u(A).

278

CHAPTER 2

gtves a shorter nontrtvtal relatton connecttng

Thts ts a contradtc.

tion and thus (91 ts establtshed. -1

Suppose now that mutes with each

Then the equation (81 shows that

x,y E G.

:>..EE.

Since

is its own centralizer in

E

xy

cocycle and thus we have shown that

with

implies that

A

A~ Ea.G

for some

com-

we obtain

A,

= a.(x,y}.iy

The associativity of multtplicatton in

xyxy

a.(x,y)

a. : G x G - - E* a. E Z2 (G,E*).

E*

E

ts a



The next property ts crucial for the proof of the main result. 19.21. LEMMA.

a.,S e z 2 (G,E*),

For all

there is a stmtlarity of F•algebras

Ea.G ® ESG - Ea.SG F

Proof.

Let

c = A ®B, where F

EB Eg and B = ia = EB Eg g€G gE-G is a central simple F-algebra by Lemma 19.20 and Corollary 19,9. A = Ea.G =

c

Then

a,'m ts to find an idempotent

e e E ®Es F

c

such that

eCe

~ & 6a,

Our

This will

yield the destred result by appealing to Lemma 19.17, We ftrst observe that the subfields

E®1

and

1 ®

E of

E ®E

commute

F

elementwtse, E = F(),1.

where

E ts separable over F,

Since Let

n = (_E: Fl.

f(X} E F[X]

there exists

be the minimal polynomial of

g

e

G - {l},

inator ts not zero stnce. A f. g(A.}

for each

also distinct from zero in

since the elements

1 inearly independent over

Then

degf(X) = n,

E ® E, F 1 ® E,

Hence

e

Observe that the denom-

g E G - {l}.

f. 0 in

E ® E,

F

f(X} = {.X-:>.)

ranges over

G - {l}.

The numerator is

{;\i ® 110

Next we note that

g

A•

such that

Now deftne

where these products are taken over all

where

A EE

n (X-g(:>..)}

Therefore, we must have

< i,;;;

n-1}

are

THE BRAUER GROUP OF A FIELD which shows that

(1 0 >.)e

=

(>. 0 l)e

i'n

E

0

279

Thus, by induction,

E,

F

>.he

(>.i 0 lle = (1 0 n-1

Because E

=

for all

i;;. 0

.

EB F>.1, i=O

and multiplication in E 0 E is commutative, we derive F

= e(l 0

(µ 0 lle = e(µ 0 1)

µ}

= (1 0 µ)e for allµ e

E

(10)

Hence e2

=

e1\(>. 01 - 10 g(;\l)/n(>.-g(>.)) 01

=

en ((>.-g(>.}) 0 1)/n (>.-g(>.)) 0 1

= e,

proving that e

is an idempotent in E 0 E. F

We are left to veri'fy that (11)

To this end, note that e(E 0 E}(; 0 y)e

?

eCe =

x,yE.G

t e(E 0 1)e•e(10 E)e(; 0 y)e

=

x,yE@

But e(l 0 E)e

=

e(E 0 l)e

by (10), and

We now compute e(; 0 y)e,

to E.

e(E 0 l)e = E'

is a field F-isomorphic

by using the formulas (µ

E E, x,y E G)

We have

where

If

x

f

From (10), we obtain

ranges over all elements of c - {1}.

z

y,

then

x(>.) - yz(>,) =

0

for

e(; On the other hand, when

x = y

z = y

0

y)e

=

x

0

we obtain e(; 0 y)e =

A stmilar argument proves that

-1

c; 0

y)e

E G-

{l},

and so

CHAPTER 2

280

e(x © yle = e(x ©

y)

The foregoing shows that

=

eCe

EBE'

g

gE-G

where and g = e(g

E'= e(E © l}e

©

g)e

(g

G}

E

and also shows that

g Clearly,

Gal (.E'/F).

e,,

g e(11 Hence, conjugation by

~fl

Thus the

=

(g © g).e

G and, for all © l).e =

e(g ©

g acts a~

µ

e(g © g).

E;

E, g E G,

g) (J.1

© l).e =

on E 1 •

g

:i:y }e

=

=

e(xy ©

=

/\ e(a(x,y}B(x,y). © l}e xy

result follows.

e(g(J.1) © l).e

E

G)

g

Moreover, for all

x,y

E

G,

= e(a(x,y)xy © B(x,y}xy)e

multiply according to the cocycle

g's

(g

aB,

This proves (.11) and the



We have now come to the demonstratton for which this section has been devel~ oped, 19.22. THEOREM.

Let E/F be a finite Galois extension, let G = Gal (E/F)

for any a E Z2 (G,E*).,

let

l"a

and

be the crossed product of G over E with

respect to the natural action of G on E,

Then the map

~H 2 (G,E*).->- B:r{E/F)

?

a

I-+ [~Gl

is an isomorphism, Proof.

Owing to Lemma 19. 19 (;), [EaG]

Then, by Lemma 19,21,

f

~ z2 (G,E*)

L

?

~

a

Br(E/F).

E

Consider the map

Br(E/F) [EaG]

is a homomorphism.

Furthermore, by Lemma 19, 19{tt},

THE BRAUER GROUP OF A FIELD

Kerf and thus

f(al

by

(G,E*)_

induces an injective homomorphism

f =

= B2

Now assume that

[EaG] •

exists BE [Al

f

Then, by Lemma 19. 16, there

E as a self-centralizing subfield.

B ~

voking Lemma 19.20, we deduce that

given

H2 (G,E*)-->- Br(E/F)

[Al E Br(E/F}.

such that B contains

281

EaG

f(a} = [EaG]

=

for some a

z2 (G,E*).

E

In-

Thus

[B] = [A]

and the result follows. • Our next aim is to show that every Brauer class contains a crossed product. This will be achieved with the aid of the following result. 19.23, PROPOSITION.

Every Brauer class of a field

has a splitting field which

F

is a finite Galois extension of F, Proof.

Let D be a division algebra in a given Brauer class of F,

We

claim that D contains a maximal subfield E such that E/F is a separable extension; if sustained it will follow, by Lemma 19.16, that E

is contained in some finite Galois extension

K/F, K

splits

E

Since

D,

will also split

D,

as

required, To substantiate our claim, we may harmlessly assume that charF

O.

= p >

Our first task is to show that

field extension of F, over F.

Hence, if there exists

are no separable extensions over

£

arable, i.e. that

AP

and

;\ if'. F

Z(D)

Choose

x ED,

but /(bl

=

f(c)

i.e.

and

A~ F such that

=

0,

O.

F,

A ts separable over F,

Assume by way of contradiction, that there Then, each element x e

D

is purely insep-

r

for some r = r(x).

E F

Let

E F,

contains a proper separable

D

To this end, note that each element of D is algebraic

F(Al is the required extenston.

then

DI F and that

f

be the mapping

We may therefore choose

x >--+ x\ - ;\x •

On the other band, we have fP(x) = xAP

b ED

such that

Then c

cA

=

Ac,

=/-

1

(bl has the form

Setting v

=

A- 1 c,

i

F

such

Then f f O since

= F,

f(bl f O and let

A !/'.

- Apx =

be such that

c = uA - AU

0 for all / -1 (b) # 0

and satisfies

we see that v commutes with

;\

CriAPTl:.R 2

282

u'\ .,.

C = /\ V =

Hence, putting w

=

uv

-1

we derive

,

\ -1 /\ = ( u'\-'\u,v = UV -1 /\

Thus w

+ '\w'\- 1 ,

= 1

but wq wq

=1 +

l

0,

=

-

/\UV

for some q

E F

(:\w).-l)q

a contradiction.

=

-1

= WA

pn.

-

/\W

But then

= 1 + :\r,F'\-l

(since wq

= 1 + wq

and so

/\U

E F)

Consequently, D contains a proper separable

field extension of F. Let E/F be a separable field extension in D of maximal degree. is a maximal separable extension of simple with centre E. Therefore, if cvtE} t- E, first part.

in

F

D,

By Corollary l9.13(i},

But D is a dtvis1'on algebra, hence so is

This would imply that

El'

is separable over

Thus

19.24, COROLLARY.

Let A be a central simple F-algebra.

CD(E)_ = E,

i'.e.

F,

is similar to the crossed product E°'G.

is

CD(E).

by the

contrary to the

i's a maximal subfield of D. •

E

Then there exists a

finite Galois extension E/F and a E Z2 (G',E*), where G = Gal(E/F),

Br(F}

CnfE)

then E has a proper separable ex tens ion E ',

maximality of E.

A

Then E

such that

In particular, every Brauer class in

contains a crossed product.

Proof. such that

By Proposition 19.23, there exists a finite Galois extension E/F [Al E

Br(E/F}.

Now apply Theorem 19.22, •

20. AN INTERPRETATION OF H~(G',E*} Throughout this section,

E/F

is a fi'nite Galois extension with Galois group G.

As usual, we define then-th cohomology group ~(G,E*) n-cochains.

This does not affect the isomorphism class of ~(G',E*)

useful in computations.

K/F

and is very

To motivate our discussion, we first introduce a dis-

tinguished subgroup ~(G,E*) Let

with respect tonormalized

of ~(G,E*).

be a finite Galois extension with H = Ga l(KJF) '

s=

K ::i E

Gal( K/E)_

and put

AN INTERPRETATION OF Hi(G,E*) Given

crE

H,

we put a'= aiE

283

so tn.at the map ( 1)

s.

is a surjective homomorphism with kernel aK

c!1' (H ,K*)

E

Each a

cfl(G,E*)

E

determines

given by ( 2)

aK( er , ••• ,a ) = a (er', ••• ,a') 1 n 1 n It is clear tnat if

a E

is split by K if aK

zn(G,E*), then

aK

is a coboundary.

E

We say that

zn(H,K*).

a E

zn(G,E*)

The map

izn(G,E*)--+ Zn(H,K*) (

a

aK

1--+

is obvi'ously a homomorphism (which carries coboundaries to coboundaries); hence it induces a homomorpnism

whose kernel consists of all

a

E

zn(G,E*l

which are split by K,

We put

zn(G E*) = u Ker¢ o ' - K K

where

K ranges over all finite Galois extensions of F containing E.

ed otherwise,

z~(G,E*)

consists of those

is a subgroup of

It is clear that

E.

I:f (G,E*).

containi'ng

zn(G,E*)

which are split by a

a E zn(G,E*)

suitable finite Galois extension of F containing

Express-

z~(G,E*)

We put

~(_G,E*) = Z~(G,E*)/ffl(G,E*)

Our aim is to provi'de an interpretation of of E (the group H~ (G,E*)

H~(G,E*)

in terms of the Brauer group

turns out to be trivia 1).

A precise description of

the main result is given below. For convenience, we first recall that the elements cisely those maps

G x G x G--+ E*

1

1

234

,a

of

Z3 (G,E*)

are pre-

satisfying

a(a ,a ,er )

a a(_a ,a ,o )a(a ,a

a

2

if any

= 1

3

,a ,a )

CT' )a(a 1234 123

=

a.1, =

a(a a ,a ,er )a(a ,a ,a a ) 1234

1234

1

(3)

(4)

CHAPTER 2

284

Also, for any normalized 2-cochain S: G x G ary oS: G x G x G---+ E*

cosH01,02,03 i

=

the corresponding cobound-

E*,

is defined by

01 sc02,03 1sc0102,03 r 1 sc0 1,CT2 CT3 is(0 1,CT2 i-1

(5)

is defined to be any (fi'ni'te-di'mensional) central simple E-al-

AG-normal algebra

gebra A for which every F-automorphism of E can be extended to a (rtng} automorphism of A.

It wi 11 be shown that the Brauer cl asses of G-norma 1 a1gebras

form a subgroup BrG(E)

of Br(El containing the image BrF{E)

of the homo-

morphism ~ Br(Fl. --.- Br(El

?

[A] t--->- [A ® E] F

The main result asserts that

20.1. LEMMA.

Let A and B be E-algebras such that the given automorphism CT

of E can be extended to rtng automorphisms tively,

and

f

Then there i·s a unique ring automorphi'sm

g

f ® g

of A and B, of

A® B

respec-

such that

E

Cf® glCa ® bl = f(i!I)_ ® g(bl

Proof.

Si'nce any ring automorphi'sm of

A® B

i's uni'quely determined by its

E

values on a® b, a EA, b EB, morphi'sm of

A ® B.

i't suffices to show that

Consider the map

A

x

B

--L. A

E

f ®g

® B, (a ,b)

i----+

is a ring autof(a) ® g(b).

E

Then iJ; is obviously biaddittve and, for any >.EE, iJ;(a,)J,1

= f(a)

® g(:\bl

= f(al

® 0(:\)g(b}

= CT(;\)f(a)

® g(b)

= f(>.a) ® g(b) = iJ;(:\a,b}

Hence

f ®g

A® B --+A® B

E

Since

f ®g

E

is a z-module homomorphtsm with (·.f ® g 1-l = f-1 ® g -1

obviously preserves multiplication, the result follows.•

20.2. COROLLARY.

If A and B are G-normal algebras, then so is A ®B. E

Proof,

Let CT be any P-automorphism of

E.

Then there are ring automor-

AN INTERPRETATION OF H~(G,E*) phisms

and g of

f

20,1, we see that

A

f ® g

and

285

respecttvely, whtch extend cr,

B,

is a ring automorphi'sm of

A® B

Applying Lenuna

such that

E

(f

as required.

for al 1 A E

1} = cr(J,.) ® 1 = cr(J,.)(1 ® 1)

Let A and B be E-algebras and let

f,,g. 1., 1.,

be ring automorphisms

respectively, extending the automorphism Ai of E for i = 1,2.

Proof,

E



20.3. LEMMA. of A,B

® g}(\ ®

By Lemma 20. l, both sides are rl'ng automorphisms of

A® B. E

(f1 ® g 1 ) (! ® g 1(a ® b 1 . 2 2

Since

(a EA, b EB) = (f

the result follows.

Then

!

1 2

® g g

1 2

1(a

® b) ,



As a preltmtnary to the next result, let us recall the following standard fact, Assume that {e .. } and

(i}

r =

~

1'

2

and D

D

1

2

n

1

are dtvtsion rings and let

1 and

respectively.

M (D }, r 2

and there extsts a untt u of R such that u

= ~-lD

D

1

be the matrl'x units of M (D

1.,

2

Ci i l

n

{v.J}

1.,J

Then

= M (D) = M (D 1 where

R

1

vij -- u -l eiju

for a 11

i,j E {1,2,.,.,n}

(see Jacobson (1968,p.26}1. Now assume that R = M (D) n

morphism of R.

cp(Dl

= u- 1Du

Then R

where

D

is a dt-vtsion ri'ng and let

where D' = 'f'A-(D).

= Mn.(D'l

for some unit u of R,

be an auto-

cp

Hence, by the above,

It follows that 1/J

= i'

u

cp,

where

i (x) = uxu- 1 , u

is an automorphism of

D

such that

1jJ(D} 20.4. LEMMA. (ii)

(il

= D

and 1/JIZ(D)

=

cp!Z(D)

If an algebra A ts a-normal, then so are all similar algebras

The Brauer classes of a-normal algebras form a subgroup BrG(E)

the i'mage Br>iE)

(6)

of the homomorphism

containing

CHAPTER 2

286

~ Br(F) ..- Br{E) (

[A]

[A ® E]

f---->

F

Proof.

(i)

Let D be a division algebra over E with centre E.

suffices to show that Mn(D)

is a-normal if and only D is a-normal.

G-normal, then obviously so is M)D). ma 1 and 1et

;\

of

Replacing

M

n

(iil

(D).

By hypothesis,

E G.

by

¢

It is clear that if

20.2,

;\

is a-normal, then so is A

0

¢,

Hence, by Corollary



Let A be a central simple F-algebra Then, the map

i)! : A ® E _,. A ® E F F

is a ring automorphism which restricts to '.\

= a® '.\(e)

as required.

is G-nor-

as i'n (52, the assertton follows.

i)!

A

If D is

extends to a ring automorphism, say

be an F-automorpntsm of E.

given by Ha® el 1 ® E,

;\

is a subgroup of Br(El,

BraCEl

and 1et

Conversely, assume that Mn(D)

It

on



The following observation will allow us to tie together G-normal algebras and certain E*-valued 3-cocycles, 20.5. LEMMA.

Let A be a G-normal algebra and, for each

a (rtngl automorphism of A,

define

i(ul f

E

AutA

A

extending

with

x,

by i(u)Ca) = ucm- 1 , a and

: G X G _,. U(A).

t : G

X

f(1,y)

= f(x,1) = 1

let

cr(x)

Tnen there exist maps

EA,

such that

E*

for all

x,y

E G

(7)

for all

x,y

E

G

(8} (9)

0(xl [f(y,z),Jf(x,yzl = t(x,y,zlf(x,y).f(xy,z)

ff

t(x,y,z) = 1

Proof.

The product

on the centre

0(xl0(y)0(xy)- 1

E the identity

0(1). = 1,

we may and do choose

x,y,

or

( 10)

z = 1

is an automorphism of A which induces

1 = xy(xy)- 1 ;

phism (by Theorem 19. 11 ) , say i [f(x,y }]

be

For any unit u bf

0(1} = 1.

GX G -

0(x)0{y1 = i[f(x,y)J0(xyl

x E G,

it is therefore an inner automor-

for some

f(1,y) = f(x,1),

f(x,y)

E

U(A).

Si nee

which proves (7) and (8).

Using the equation (7). and the identities i(uv). = i(u)i(v), ;\i(u) = i( ;\(u)) ;\

we may calculate the triple product

cr(xlcr(y).0(21

(;\ E

in two ways as

G,

u,v E

U(A))

287

AN INTERPRETATION OF H~(G,E*) [cr(x)cr(y)Icr(zl = i[f(x,y}f(xy,zl] cr(xyz} cr(x) [0(y)0(z)J = ifo(x) [f(y,z)J f(x,yz)}a(xyz)

These results must be tdentical; hence if we write b ,b 1

of i

on the right, we have i(b ) 1

b b- 1 = t 2

1

= i(b ) 2

2

or i'(.b b- 1 ) 1

is an element in the centre E of A,

2

for the two arguments which means that

= 1,

This proves (9).

Finally,

the special choices made above also prove (10]. • The following result will be derived at the end of the section as a consequence of some general facts of cohomology theory. 20.6. LEMMA.

Let t : G x c; x G

-->-

be as in Lemma 20.5.

E*

Then

t E Z3 (G,E*)

(i).

(ii)

For fi'xed choices cr{xl a different choice of f(x,y)

U(A)

E

replaces t

by a cohomologous cocycle, and a suitable different choice of f(x,y) t

replaces

by any specifted cohomologous cocycle

(iii)_

If the choice of the extensions cr(x)

of x

altered, and if suitable new values of f(x,y)

to automorphisms of A is

are chosen, the cocycle t

is

unaltered. In what follows we call

t

a TeichmuUer cocycle of the a-normal algebra A.

Our next aim is to provide a complete descriptton of the normality of a crossed product and the form of its Tel'chmuller cocycle,

This will be achieved with the

aid of the followtng three prelimtnary results, Let K/F Be a finite Galois extension wi'th k

s=

Gal (K/E),

For any a

E

E and put

li =

Gal(K/F),

z2 (s,K*l, we denote by Ka,S the corresponding

crossed product of s over K.

Recall, from Lemma 19, 19, that Jf's i's a

central simple E-algebra with c

rs

20.7. LEMMA.

~

(_K) = K,

With the notation above, assume that K /E is a field extension l

and that there is an isomorphtsm f: K morphism of E. Proof.

Then K and K

Put A= flE

1

such that f\E

is an F-auto-

are E-isomorphic.

and note that, since K/F is normal,

extended to an p-automorphi sm, say ly an E-isomorphism, •

K

1jJ,

of

K.

Then

fi}! : K

--->-

-1 A

K

can be is obvious-

288

CHAPTER 2

a.. norma1 tf and only tf every F-automorphi·sm can be extended to a ring automorphtsm of Ifs.

20,8, LEMMA. of

K

Proof. K,

The algebra ifs ts

Since each g E G = Gal(E/F}

can be extended to an F-automorphism of Suppose conversely that Ifs

the given conditi'on ts obviously suffi'cient.

ts a-normal. fi'eld

For each :>.EH,

K of ifs

the extension

cr(:>.'}

onto an tsomcirphtc subfield Kl

by Lemma 20,7, there exists an E-tsomorphism K

containing :>.'(E) = E, . .

The product ¢cr(A '1

morphism of Ifs whose restrtctton to K is an element, say :>. that :>. µ

l

IE

= A',

Thus

?>.A

-1



l

ts an E-automorphism of

can be extended to an E-automorphi'sm

of Ka,S.

1/J

Hence

which by Theorem 19, 11,

K,

->l

¢ of Ifs,

can be extended to an automorphism

of :>.'=:>.IE maps the sub-

ts an auto-

of H,

l

such

By Theorem 19,11,

K.

Finally,

iJ;qia(?,, 1 )

is a

ring automorphism of ifs carrying K tnto K by the automorphism ?>.?>.- 1?>.

~" = I

l

= :>.;

l

it is the required extenston of A to ifs.



In vtew c,f (11, we have an exact sequence 1 ;~here 11 -

--+

S

Ii

--+

G i's the restriction map,

--+

l

--+

Hence 11 may be described as a group

extension of s by G by chciostng for each v(.1)_ = 1,

G

an extension v(g} E H,

g E (}

and defi'ni'ng the automorphism 8 i---- g*s of the normal subgroup

with S

by g*3 = v (g )_s v(g)_

Then there ts a "factor set''

-1

of elements of

e(x,y)_

s

such that (x,y

v(x)v(y} = e(x,y)v(.xy)

(g

v(g)s = (g*s)v(g)

Any elements

E

G)

( 11)

G, s E S)

(12)

E

h ,h EH can be expressed uniquely as l

2

h

l

= s 1 v(g) 1

,

h

2

= s 2 v(.g) 2

G)

(13)

1e(g l ,g 2 ) )

( 14)

(s ,s ES, g ,g E l

2

l

2

Their product, in view of (11} and (12), is given by [s

1

v(g }J l

[s

2

v(.g

2

11

= s 'v(g g l

12

l

(8 , = s (g *$ l

1

l

2

AN INTERPRETATION OF H3 (G,E*)

289

0

The factor set 0 satisfies the associ'ativity condition (x ,y ,z

[x*G (y ,z )J 0 (x ,yz) = 0 (x ,y )~ (xy ,z)

G)

{15)

E S)

(16)

E

and is related to the * operation by (x ,y

20.9. LEMMA. s Es

The crossed product

Ko;S

s

E G,

is G-normal if and only if for any g

G,

E

there exist m(_g,s) EK* such that m(l,s)

=

m(g,1)

1

and ( 17)

m(g,fJS ) [v(g)o;(s,s )_] = m(g,s) [(g*s)m(g,s 1 )]a(g*S,g*s ) 1

Proof.

1

Suppose f1'rst that

1

i's G-normal,

Ko;S

Then, by Lemma 20.8, v(g)

can be extended to a ring automorphism, say a(_g),

of KaS,

with 0(1)

=

We

1,

know that multipli'cation in Ko;S l's determined by

; s a(s,s e/\ = s(';\)_$

)ss

=

l

In (191 set ;\

=

1

ES)

( 18)

(;\EK)

(19)

(s,s

l

1

v(_g)- 1µ for any µEK and apply the automorphism a(g)

to

obtain [a(g)_s] µ = [(g*s)µ] [a(g)e)

On the other hand, (19) implies that g*B-

µ = [(g*s1ril

g*s

-1

These two equations assert that l tes in

K,

[o-(g)s] g*s·

commutes with every µ

E

K,

hence

so we may write (20)

a(g)s = m(g,s)g*S

for some m(g,s) E

K*

with m(g,1)

Applying

= m(1,s) = 1.

to both sides

a(g)

of (18), we obtain m(g,s)

gn

m(g,s } g*'s 1

= 1

[v(g)a(s,s )]m(g,ss ) g*sfJ

The left side, by applying (18} and (19}, is

1

1

1

CHAPTER 2

290

Compartson with the rtght side gtves the required equation (17}. Conversely, given such a functton

s,

on

a(g)

and to defi'ne a(g} :

we may use (20) to define the effect of

m,

additively by using

Ka,S--.- KaS

a(g}(),s) = [v(g)\] [o(g)s]

(\

By reversing the above computation, one immedi'ately verifies that automorphism of

extending the given F-automorphism

Ka,S

of

g

K)

is a ring

a(g) E.

E



We are now ready to provtde a complete description of the normality of a crossed product and the form of its Teichmuller cocycle. 20.10.THEOREM. (Eilenberg and MacLane(19481)_, Galois extensions with and for a given

s

over K.

a

E

Gi'ven

is G-normal with

KaS

exists a cochain

let

K:: E,

(:;

z3 (H,K*)

tK E

SE C2 (H,K*)

1

of

KaS,

Assume that

Ka,S

2

(for all

2

t

i's G-normal with

f(x,y}

of

KaS

s ,s ES) 1

2

(22)

as one of its Teichmuller to an automorphism o(g}

v(g), g E G,

t

Owing to Lemmas 20.5 and 20.6,

0(1) = 1.

from a suitable unit

Then

(21)

K 1

We may extend, by Lemma 20.8, with

be defined by (2).

such that

S(s ,s )_ = a,(s ,s )

cocycles.

Gal(K/E),

=

as one of its Teichmuller cocycles if and only if there

t

8S = t

Proof.

be finite

KIF

be the corresponding crossed product Of

let

t E Z3 (G,E*},

and

Gal(E/Fl, H = Gal(K/F}, S

=

let ifs

Z2 (S,K*),

Let E/F

satisfying (71,

can be obtained

Consider the map

given by (s

n(sv(g)J = i'(s)a(g)

i(sl

where

S.

If

rs

denotes the inner automorphism of

h ,h E H, 12

say h

n(h

1

1

= s v(g ) , h 1

)n(h)

1

=

2

=

i(; i[i

1

= s 2

)a(g

2

1

then

2

2

)a(g) 2

(a(g ); Da(g )a(g) 1

l

2

S, g E G)

obtained by conjugation with

v(g ) ,

)i(;

E

l

2

AN INTERPRETATION OF H3 (G,E*)

291

0

Define S(h

1

{by (7})

i[; (cr(g }; 1f(g ,g }] a(g g )

=

1

1

2

1 '2

1

2

using the notation of (14) for the product h h ' I 2

,h ) E v(Kas), 2

=;

by

(a(g ); )f(g ,g)

{23)

~(h )~(h ) = '[S(Tz ,h )J ~(h h )

(24)

B{h ,h )s•

121

1

12

12

Then the previous equation becomes I

2

This shows that conjugation by S = S(.h ~(h )_~(h )~(h h I

SE

2

K0,S

SC1,h

2

12

1-1 which is on

I

1,,

,h 1

2

2

12

1 induces on

the identity h h (_h h )-1; in other words,

K

12

commutes with all elements of

hence

K,

S(h

g 1 = g2

= 1,

ted to

s

x

and

s

S(s ,s 1

2

12

,h ) l

l = S(h 1 ,11 = 1, so that B is a 2-cochain of

2

E K*.

}ss l 2

I

ss .

=

coincides with a,

2

then s'

2

s 1s 2 ,

=

1

It follows from (18) that

2

l

1

Clearly

over K*.

H

B choose h = s, h = s;

In the definition (23} of

an automorphism

KaS

B,

restric-

as asserted in (22},

as,

Now we calculate the coboundary

for arguments

h ,h • l

2

and

h

= 3

s v(g ), 3

3

and with the notation 03), (141, and h h 2

= 3

s'v(g g l, 2

Because multiplication in

2

2

2

* s 3 ) e(g 2 , g 3 1

= Tz (h h) = EV(g g g)

123

EE

2

ts associative,

H

{h h )Tz

where

s ' = s (g

3

123

123

S is determined from both products as E

= s'[(g g )*s ]0(g g ,g) = s (_g *s'}e(g ,g g). 1 12 3 123 112 123

Using the definition (23) repeatedly, we calculate that S(h ,h )B(h h ,h )s 12

= B(h ,h

123

12

=

B(h ,h

)S(h h ,h )s' [{g 1231

ls'"

121

(cr(g g

g

12

)*S ] 0(g 3

)s )f(g g ,g}

123

123

=

s1 (cr(g 1)s2 )f(g 1,g2 )[a(_g 1g 2)s3l f{g 1g 2,g3 )

=

xf(g 1 ,g 2 lf(g 1 g2 ,g 3 l

(25)

where X

=

S (a(g ) S )f(_g ,g ){a(g g ) S /f(g ,g 1

=

1

g ,g )

123

2

l

2

12

sI (a(g 1 )s 2 )(_a(g l )a(g 2 )s 3 )

3

I

2

rl

CHAPTER 2

292

s,

On the otfier fiand, using tfie second express i'on for

we have

[h S(h ,h )l S(h ,h h )s = [h S(h ,h )ls [o(g )s'l f(g ,g g ) 1

2

3

1

23

1

2

3

1

1

= ;; [s- 1 h S(h ,h 111

23

2

1

23

)l [o(g )s'l f(g ,g g ) 12

123

=

s o(g

=

sl o(g l } [s 2 (o(g 2 )s 3 )f(g 2 ,g 3 )l f(g l ,g 23 g )

=

X

1

1

) [S(h ,h )s'l f(g ,g g ) 2

3

2

1

23

[O(g lf(g ,g }] f(g ,g g ) 1

2

3

1

2

3

By the relation (9), this becomes xt(g1

,g2

Applying (25} and the fact that

,g )f(_g1 ,g 2 lf(g 1g 2 ,g)

lies i'n the centre

t(g ,g ,g ) 1

2

3

E

of

KCY,S,

we

deduce that t(g ,g ,g }B(h ,h )B(h h ,h ) = [h B(h ,h )l B(h ,h h ) 123

h'

But

=

1

oB

g, h' 1

= tK'

=

2

12

g, h' 2

3

=

g,

123

1

23

123

so by (51 and (21 this equation proves that

3

which is (21) in the theorem,

Conversely, assume that there exists a cochain and(22),foragiven

tEZ 3 (G,E*).

Define

BE C2 (H,K*) for

m(g,s)EK*

satisfying (21) gEG,sES

(26)

m(g,s)S(g*s,v(gl)_ = S(v(g) ,s)

rs

To show that

by definition of

by

is c~normal, we must demonstrate that

satisfies (17), which

m

m reduces to the identi'ty

B(v(g),ss ) [v(g)S(s,s }] S(g*s,v(gl} [(g*s)B(g*s ,v(g)}] 1

1

=

Since

&S = tK'

1

B(v (g l,s )[ {g*s 1B(v (g ),s

H fol lows that

lies in the subgroup

8S(h ,h ,h 1

s.

2

3

1

1

ll

B(g*s ,g*s ) S(g* (ss ) , v (g)} 1

1

vanishes when any one argument

Hence

1 = ..) = 0'(\ 1 } ® '.\

Lemma 20.l, there is an automorphism on the respective factors with

295

cr('.\ 1 )

is extendible to an automorphism of

and

'.\

of

and which has proving that

A® K,

A® K F

which agrees

cr*(l) = 1.

A® K

E

Thus

>..

is H-normal.

E

Applying Lemma 20.3 twice, we have O'*('.\)cr*(µ)

= [cr('.\ 1 } =

® '.\]

[O'(µ '} ®

= cr(\ ')cr(ii '}

µ]

®

'.\µ (by (7))

i[f('.\',µ'}l cr(\'µ')@ \µ

= (i [f(\, ,µI)] ® l )CO'((\µ) I l ® AP)

(i[f(Y,11 1 )1 ® lHcr*(>..µ).}

The automorphism i[f(\',µ')l ® l f*('.\,µ)

=

f(). ',µ ')

is conjugation in

A® K

by the element

E ®

1

and thus cr* (\)O'* (µ) = i [f* (\ ,µ)J cr* (\µ}

f*

i.e.

satisfi'es condition (7) with respect to

muller cocycle

t*

of

A ® K

H.

The corresponding Teich-

can therefore be derived from

f*

by formula (9)

E

as

t*(\,µ,v) Thus

t* = tK

t(>..',µ',v') ® 1

=

and the result follows.



We have now come to the demonstration for which this section has been developed. 20.12. THEOREM. (Eilenberg and Maclane(l948). TeichmUller(l940)}. fi'nite Galois extension with Gale>1's group Br(El

let

G,

BrG(E)

t(A)

iBrG(B 1___!_,. H (G,E*) 3

?

[A]

I---->-

t (A)

is a homomorphism such that KerT In particular

= BriE)

and

ImT

t(A)

of a a-normal algebra

the map

= H~(G,E*)

be a

E/F

be the subgroup of

consisting of Brauer classes of a-normal algebras, and let

cohomology class of a Teichmuller cocycle

Let

be the A.

Then

CHAPTER 2

296

Here

BrF(E)

form

[B ® E],

Br(E)

is the subgroup of where

[Bl

consisti'ng of all Brauer classes of the

Br(F).

E

F

Proof.

Let

respectively.

1

2

We claim that

A ® A I

Indeed, choose extensions satisfy (7) and (9) for

t ,t,

be two G-normal algebras with Teichmuller cocycles

A ,A

CT .(x) ~

.

E

2

l

in

~

A ®A.

to

1 E

2

2

f .(x,y)

Then, by Lemma 20, l, x

1

as its TeichmUller cocycle.

t t

and elements

i = 1,2.

an extension of the automorphism

has

to

A . , x ,y E G, ~

G(x)

G (x) ® CT (x)

=

1

is

2

Moreover,

2

f(x ,y l = f (x ,y) ® f (x ,y) l

A ®A

ts a unit of

I E

2

and

2

i[f(x,yll = i"[f (x,ylJ ® i[f (x,y)l 1

Applying Lemma 20.3, we deduce that lating (91 one deduces that parti~ular, for any

n > 1,

CT

t = t t 1

2

2

and

f

are related by (7), while calcu-

is a Tei chrnu'll er cocycl e of

I

Mn (E) - - has t

E

2

as a Teichmuller cocycle.

= 1

In

® A •

A

There-

fore, any two similar c-normal algebras have the same Teichmuller cocycles,

Thus

is indeed a homomorphism.

T

To prove that

KerT = BrF(El,

we must show that

(a)_

For any centra 1 simple F-a 1gebra

(b)

If

for

k = (E:F)

A

B, B ® E F

has a Tei chmull er cocyc 1e l B ® E ~ Mk(A)

is a c-normal algebra with Teichmuller cocycle 1, then and for some central simple F-algebra

F

.

In particular,

B.

[Al = [B ® E] F

Property (a) is the special case consider

A with

t(A) = 1,

K = E, E = F

E

To prove (b),

Using (9), we can construct an algebra

elements uniquely represented as sums g

of Lemma 20.10,

Z:a

a

~

with coefficients

a

g

B with

EA

for each

G with multiplication table determined by the distributive law and the rules

ga= xy

[CT(g )al

g

= f(x,y)xy

(g

E

(x,y

G, a E

E

A)

(29)

G)

This construction is exactly analogous to the construction of a crossed product

AN INTERPRETATION OF H~(G,E*)

algebra over a fl'eld. (_9). that

Si'nce

t(x,y,z) =

for all

1

i

is an associaUve algebra with

B

particular,

of the argument in Lemma 19.19 shows that

C/El

forward computation shows that is a central simple F-algebra.

E

{a•i\a

is simple.

B

= A;

x,y,z

G,

tt follows from

as the identity element.

is identifiable with the subalgebra

A

297

EA}.

In

A repetition

. Furthermore, a straight-

hence by (29),

= F.

Z(B)

Applying Corollary 19.12, we have

Thus

B ® E

,:!!

B

Mk(A}

F

k = (E:F),

where

which clearly implies (bl.

ItnT = H~(G,E*)

We now claim that the assertion

(K

following two statements (c)

ts a finite Galois extension of

A is split by

If the a-normal algebra

cocycle

t

(dl

t

If

A, tK

of

K ~ E,

tK

is such that

Indeed, if (c) is true, then choosing as si'tion 19.23}, we see that

A ® K

!:!!

t

ts the Teich-

K,

K a splitting field for A

sJ(G,E*). 2 I'mT,

(see Propo-

On the other hand, (dl ensures that

as claimed.

Property (c) ts a consequence of Lemma 20.11. then

then for any Teichmuller

ts a coboundary, then

muller cocycle of some G-normal algebra split by

ImT 2 HJ(G,E*).,

F}

ts a coboundary.

Z3 (G,E*)

E

is a consequence of the

for some

M (E)

E n (with arguments i'n

n ;,i, 1

Indeed, if

A

ts split by

and every Tei chmull er cocycl e for

g) is a coooundary, and Lemma 20.11 shows that

tK

K,

A ® K

E ts a co-

boundary.

To prove (.d}., suppose that Then

tK =

oS

t E z 3 (G,E*l

for some 2-cochatn

s··

of

is such that

H ,~n

K*,

is a coboundary.

tK

In parttcul ar,

dS(x,y ,z 1 = 1 for all

x,y,z

Z2 (.s,K*)..

E

so that the definition

Si

We may therefore construct the crossed product

Applying Theorem 20.9, we conclude that Teichmuller cocycles. 20.13. COROLLARY. 20.12.

a(x,yl = s(x,y)

If

K/F

Let

~S of S over K.

~s is a-normal with t

as one of its

This proves (d) and hence the result. • T: BrG(E) -

H3 (G,E*)

be the homomorphism of Theorem

is a finite Galois extension of

carries the elements of

gives an element of

Br/81

~b.ich are split by

F

with

K

2

E,

then

T

K onto those cohomology

298

CHAPTER 2

classes

t

for whi'ch t e

e H3 (G,E*).

Proof.

Z 3 {6,E*l

ts spltt by K,

Thts ts a direct consequence of properties (c) and (d} established fn

the course of the proof of Theorem 20.12. • 20.14. COROLLARY.

Further to the assumptions and notations of Theorem 20.1_2,

assume that G ts cyclic. central stmple F-algebra

Then, for any G-normal algebra A, B

there exists a

such that where k

Proof.

The general reduction theory for cohomology groups (see Etlenberg

and MacLane(l 947a,§16} 1 proves that H3 (G,E*l =

= (E:F}

8 3 (G,E*l ="' H1 (G,E*),

Hence, by Theorem 18.3,

Thus the Tetchmull er cocycl e of A ts 1 ,

1.

The destred con cl u-

s ion now follows by virtue of property (bl established tn the course of the proof of Theorem 20.12. • Turning our attention te o~(.G,E*l, we now prove 20.15. THEOREM. (Eilenberg and Maclane(1948).l.

~et E/F be a fintte Galois

extensf'on wtth Galots group a.

=

Let

Proof. K/F

aE

Then s:(G,E*l

z 02 (G,E*l, .

t.e,

aK

= &(3

1,

for some f3

E

i's a suitable ftntte Galots extenston with Galets group

s.

c1 (H,K*l,

and with

ll

Define f3 0 (sl = f3(s)

for all

and

Hence, by Theorem 18.3, there ts a constant

(3 0

e Z1 (S,K*l,

that f3 0 (slb h EH.

subgroup

=

Then

s.

s(bl (3 1

for all

s

s-

e

where K.:: E.

Then, stnce a ts normalized, of30 = l

f\

Deftne

es,

is cohomologous to

(3,

and

as

b EK*

B/h) = B(h)b/h(b)

~(3 1 = CJ.K'

such for all

whtle f3 1 ts 1 on the

In parttcular, because a ts normalized, for all

s E S

and so f3 1 (hs} = [hf3 (slJ f3 (h)., B (sh) = [sf3 (h).J S (s} . 1 1 1 1 l

Because B1 (s)

= 1, B (hs) = B (h) = B (sh} 1

Hence each value 13 1 (h} . tension all

v(.g) E H

x,y E G,

and

1

1

lies tn E*. v(.x)v(y)

ts an element invariant under s.

Now each automorphism

= nv(xy} for some

n

= n(x,y)

g E G E

s.

has an exThen, for

AN INTERPRETATION OF

H 3 (G,E*)

299

0

= aK(v(x) ,v(y)1 = &(\ (v(xl ,v(y)}

a(x,y}

= v(x)B 1

(v(yl)

ll - 1 B1 (v(x))

[S (nv(xy) 1

= xB (_v(yl} [S (v(xy) 1] - 1 B (v(x)) 1

and therefore

1

1

= oB 2 , where B2 (x) = B1 (_v(x}l is a cochain of

a

G

over E*.

So the theorem is true. • We close by showing how Lemma 20.6 can be derived as a consequence of a general cohomology theory. 20.16. LEMMA.

The following simple observation will clear our path.

Let A be a central simple algebra over a field E. Z(U(A})

E*

Proof.

It suffices to show that z(u(All c E*.

represented as the algebra of all a division algebra of

A

n x n

with centre

D

Then

The algebra A can be

matrices, with matrix units e 1,,J .. , over . Then e

E.

is the identity element

= 'Ze..

1,,1,,

and, for any r # s, (e+e

) (e-e ) rs rs -

=e

¼

e

rs

- e

rs

=e

(e-e rs l(e+e - · rs l = e - e rs + e rs = e

Proving that

Z(U(A)l,

e + e

rs

If a= 'Zd .je ,., d •.

E U(Al.

·

then a(e-te rs }

1,,

= (e+e

rs-

la or

ae

1,,J

rs

= e

ED,

1,,J

rs

a.

is any element of

Multiplication gives

E d • e • = ae = e a = 'Z d ,e , 1,,r 1,,s rs rs j SJ rJ

i

Comparing the coefficients, it follows that drr for i-! r. have O-! d

Hence, i'f d EE,

=

is defined to be a group 8 :

where

In(N)

e0 (x)

G -->-

N

E,

we

together with a homomorphism

Out(Nl

=

Aut(N}/In(N)

Each inner automorphism of

fixed, so that each coset e(x}

of Z(N).

=

0

A G-kernel

is the normal subgroup of the automorphism group Aut(N)

elements of Z(N)

=

H'

Because Z(D)

G denotes an arbitrary group.

consisting of inner automorphisms.

phism

and d.



For the rest of th i's section, (N,8}

d8$ for r-! s

then a = 'Z.de ii = de.

drr'

as required.

=

N

of

N

leaves the

determines a unique automor-

Thus we have a homomorphism

CHAPTER 2

300

eo : a -

Z(N)

and therefore The pair

g

Let

E/F

for all

U(A)

U(A).

Setting

e(x)

N

= U(A).

is the coset of

(U(A) .e) (N.e)

0(xl

of

N

A.

0(x).0(y) =0(xy)

g Ea.

0(g)

Then

0(xl

let

(z(N), with

0(g)

be an

is also an automorph-

Z(N)

= E*.

Thus.

U(A).

modulo _the inner automorphisms of

6(x)

G.

modulo the inner automorphsims of

then

(E* .e 0 )

is a a-kernel with centre

in

a.

(N.e).

it follows from Lemma 20.16 that

be a a-kernel with centre

automorphism

n E Z(N). g E

be a finite Galois extension with Galois group

to a rtng automorphism of

and. by (7} •

ism of

Let

e0 (g)n

be a a-normal algebra and. for each automorphism

A

extension of

if

g•n =

is called the centre of the a-kernel

(Z(N).e 0 }

20.17. EXAMPLE. let

is a G-module via

Aut(Z(N11

e 0 ).

For each

0(1) = 1.

Since

e(x).

e:

choose an

G ___,. Out(N),

is a homomorphism. 0(x)0(y)0(xy)-l E ln(N) Hence, we may choose elements

f(l,y) = f(x,1) = 1

f(x,yl E N with

0Cxl0(yl = i'[f(x,y)] a(xy)

x,y E G

for all

for all

such that

x,y E a

(30)

where

i[f(x,y)] (n) = f(x,y)n f(x,yf 1 Since

0(x) [cr(y)0(2)] = [0{x}0{y)] 0{2),

i[f(x,y)f(xy,2)] Thus there exist elements

=

t(x,y,2)

n

for a 11

E N

an application of (30) yields

i'[(0(x)f(y,2}}f(x,y2)] in

Z(N)

such that (31)

[ a(x)f(y,2)] f(x,yz) = t(x,y,2)f(x,y)f(xy,2) Thus

t

is normalized 3-cochain of

G with values in

Z(N).

Hence, in the

situation of Example 20.17, we have 20.18, LEMMA.

Each Teichmuller 3-cochain

cochain of the corresponding G-kernel Proof.

t

of a G-normal algebra

(U(Al,e)

Compare (30, (31) with (71, (9).

with centre •

(E*,8 0 ).

A

is a 3-

301

AN INTERPRETATION OF H~(G,E*)

.we can now deri've Lemma 20,6 as a consequence of Lemma 20.18 and the following general result. 20.19, THEOREM. (Eflenberg and MacLane(l947b}). wfth values tn Z(N)

correspondfng to the G-kernel

(il The cochain t (ti)

Let t

be a 3-cochain of

(N,9).

ts a cocycle

If the chotce of

f

tn (30} is changed, then t

is changed to a cohomo-

By suftable changi'ng the choice of f,

logous cocycle.

G

the cocycle t

may be

changed to any cohomologous cocycle. (tiil

If the chotce of the automorphisms a is changed, then a suitable new

selection of Proof.

leaves the cocycle t

f

unaltered,

Let us calculate the expression

(i}

A= a(x).[fo(y}f(z,u}}f(y,zu)Jf(x,yzu}

in two ways.

First using (_31} three times, and recalling that t

li'es tn Z(N),

we have A = a(_x} [t(y,z,ulf(_y,z}f(yz,ulif(x,yzu}

= a(x l [t(y ,z ,u}f(y,z 11 t(x,yz ,u}f(x,yz )f{xyz ,u} =

where

B

(t:t(y,z,u}It(x,yz,u)_t(x,y,z}B

i's given by B = f(x,y}f(_xy,z}f(::cyz,u)_

Alternatively, by first applying (30)_, we have A = f(x,y) [a(xy }f(z ,u}If(x,y 1-1 [cr{_x}f{_y,zu}J f(x,yzu} = f(x,y 1[a{xy )f(z ,u}J t(x,y ,zu}f(xy ,zu) = t(x,y,zu}t(xy,z,u)B

Comparison gives [xt(_y,z,u)J t(x,yz;u}t(x,y,z) = t(x,y,zu}t(xy,z,u)

which, by (4), shows that t

E

z3 (G,Z(N1).

(ii} Any other choice of f(x,y l in (301 has the form f 1 (x,y} = g(x,y)f(x,y}

(g(x,y}

E

Z(N})

CHAPTER 2

302

where

t'

g

from

ts any normali'zed 2-cochai"n of

f',

G tn

Z(N}.

Using (311 to calculate

we have t 1 (x,y ,z) = t(x,y ,z 1{ [xg(_y ,z 11 g(xy,z }- 1 g(x,yz )g(x,y )-1 }-l = t(x,y,z}

Cl and, by Corollary 3.13,

Lemma 21.l(i}, (E:F) = !Cog(_E/F}I, p,

i,e.

Then E/F is coseparable by

A¢ F*

and

APE F,

p!(E:F}.

there is an element 'AF

E

Since, by

Cog(E/F)

Then for allµ E F. µ + /\ ¢ F by

Then

of order

CHAPTER 2

308

(l1+11)F* f (d+"-}F*

One immediately verifies that Cog(E/F)

contains infinitely many elements

for

µ ,j, d

(µ+:\}F*, µ E F,

E

F.

Therefore

which is a contra-

diction. To pr0ve purity, let We must show that then

s = l

and

s

t-

s

sp = 1

with

s E E,

Assume first that

E F.

and there ts nothing to prove.

p

where p

p = 4.

is a prime or

ts a prime.

If

charF = p, charF t- p

Hence we may assume that

Since

1.

+ s + ••• + s p-1 -- 0 1•F* , s F* ,. • .,sp-lF*

it is not the case that (seeLemma21.l(il), Now assume that s2

t-

1.

si/sjEF(-Ct-j),

Hence some p = 4.

F*.

E

1 + s - (l+s}

distinct (see Lemma 21.l(i)). s

E

F,

choose a finite subgroup G = Cog(E/F}L

Since

sEF. charF f 2,

But

O, are

Since any of the three possible equality relati'ons purity is established.

E/F

Conversely, assume that

Cog(E/F)

l•F;., sF*, (1+s}F* of Cog(E/F}

so it is not the case that the elements

between them implies

so

Again we may harmlessly assume that (1+s )_ 4 = ~4

One checks that

are d.1s t.inc t elements in

G of

is coseparable, separable and pure. Cog(E/F}

such that

We may

EG = E (e.g, take

G is abelian, there i's a chain l = H C H c ... cH=G 0-1-· -n

of subgroups of Hence

EH /EH. 1, 1,-1

coseparable. EH. i--1

G

such that

HJ H, 1, 1,-1

is pure (since

Because

H ./H ., 1, i--1

E/F

1,

E/F.

and thus conormal.

p

0 .,,

v

1 .;;· i .;; n

is pure} and, by construction, it is also

is cyclic of order

by adjoining a p ..-th root.

since so is

is cyclic of prime order

p.,, EH 1,

• 1,

is obtained from

is separable, EH.JEE. 1, i,-1 is cogalois, conclude that EH_JEH. 1, i,-1

Note also that

Invoking Lemma 21.3, we

Hence, by inductive application of Lemma 21.4, we deduce that

E/F is also conormal.

Thus

E/F

is cogalois and the result follows.



With the aid of the above result, we now provide the following examples of

309

A COGALOIS THEORY FOR RADICAL EXTENSIONS cogalots extenstons, 21 . 6. EXAMPLE.

Let

be positive raUonal numbers, let n.1,

lN and

E

let n1

ns

E=(f;)(va, •• ,,f;;.}clR 1

Since E

5: JR,

E;(Q

21 .?.EXAMPLE. p

-

It is also clear that E;(Q is coseparable

E;tf) is cogalois, by virtue of Theorem 21.5,

Thus

E

is obviously pure.

8

Let p

be an odd prime, let n

be a positive integer and let

be a primitive pn ~th root of uni"ty over