The Absolute Galois Group of a Semi-Local Field (Springer Monographs in Mathematics) [1st ed. 2021] 3030891909, 9783030891909

Table of contents :
Introduction
Contents
Notation and Conventions
Chapter 1 Topologies
1.1 Profinite Spaces
1.2 Strict and Étale Topologies
1.3 Étale Hausdorff Sets
1.4 The Envelope of a Set of Closed Subgroups
1.5 Cantor Space
Chapter 2 Families of Subgroups
2.1 Continuous and Separated Families
2.2 Sheaves of Profinite Groups
2.3 Sheaf-Group Structures
Chapter 3 Free Products of Finitely Many Profinite Groups
3.1 Definition and First Properties
3.2 Finite Subgroups of Free Products of Finitely Many Profinite Groups
Chapter 4 Generalized Free Products
4.1 Inner Free Products
4.2 Outer Free Products
4.3 Equivalence of Inner and Outer Free Products
4.4 Free Profinite Products in the Sense of Binz–Neukirch–Wenzel
4.5 Properties of Inner Free Products
4.6 Further Properties of Inner Free Products
4.7 Free Products as Semi-direct Products
4.8 Closed Subgroups of Free Products
Chapter 5 Relative Embedding Problems
5.1 Embedding Problems and Projectivity
5.2 Strongly G-Projective Groups
5.3 Infinite Embedding Problems
5.4 The Closed Subgroup Theorem
5.5 The Set G_max
5.6 Prescribed Solutions
Chapter 6 Strong Proper Projectivity
6.1 Prosolvable Subgroups of G-Projective Groups
6.2 Groups of Local p-Type
6.3 Fields of Local p-Type
6.4 Groups of P-Type and G-Projectivity
Chapter 7 A Free Profinite Product over an Étale Compact Subset of Subgr(G)
7.1 A System of Representatives
7.2 The Generalized Iwasawa Theorem
Chapter 8 Fundamental Result
8.1 Classical Closures of a Field
8.2 Sections over Henselian Fields
8.3 G-Projectivity
8.4 Semi-Local Fields
Chapter 9 Main Result
9.1 The Groups Gal(K_tot,s[σ])
9.2 Finitely Generated Groups
References
Symbol Index
Subject Index

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Springer Monographs in Mathematics

Dan Haran Moshe Jarden

The Absolute Galois Group of a Semi-Local Field

Springer Monographs in Mathematics Editors-in-Chief Minhyong Kim, School of Mathematics, Korea Institute for Advanced Study, Seoul, South Korea; International Centre for Mathematical Sciences, Edinburgh, UK Katrin Wendland, Research group for Mathematical Physics, Albert Ludwigs University of Freiburg, Freiburg, Germany Series Editors Sheldon Axler, Department of Mathematics, San Francisco State University, San Francisco, CA, USA Mark Braverman, Department of Mathematics, Princeton University, Princeton, NY, USA Maria Chudnovsky, Department of Mathematics, Princeton University, Princeton, NY, USA Tadahisa Funaki, Department of Mathematics, University of Tokyo, Tokyo, Japan Isabelle Gallagher, Département de Mathématiques et Applications, Ecole Normale Supérieure, Paris, France Sinan Güntürk, Courant Institute of Mathematical Sciences, New York University, New York, NY, USA Claude Le Bris, CERMICS, Ecole des Ponts ParisTech, Marne la Vallée, France Pascal Massart, Département de Mathématiques, Université de Paris-Sud, Orsay, France Alberto A. Pinto, Department of Mathematics, University of Porto, Porto, Portugal Gabriella Pinzari, Department of Mathematics, University of Padova, Padova, Italy Ken Ribet, Department of Mathematics, University of California, Berkeley, CA, USA René Schilling, Institute for Mathematical Stochastics, Technical University Dresden, Dresden, Germany Panagiotis Souganidis, Department of Mathematics, University of Chicago, Chicago, IL, USA Endre Süli, Mathematical Institute, University of Oxford, Oxford, UK Shmuel Weinberger, Department of Mathematics, University of Chicago, Chicago, IL, USA Boris Zilber, Mathematical Institute, University of Oxford, Oxford, UK

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Dan Haran Moshe Jarden •

The Absolute Galois Group of a Semi-Local Field

123

Dan Haran School of Mathematics Tel Aviv University Tel Aviv, Israel

Moshe Jarden School of Mathematics Tel Aviv University Tel Aviv, Israel

ISSN 1439-7382 ISSN 2196-9922 (electronic) Springer Monographs in Mathematics ISBN 978-3-030-89190-9 ISBN 978-3-030-89191-6 (eBook) https://doi.org/10.1007/978-3-030-89191-6 Mathematics Subject Classification: 12E30, 12-02, 12F12, 20E06, 20E18 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

In Erinnerung an Wulf-Dieter Geyer (1939–2019)

Introduction

The main result of [HJP12], which is Theorem A below, describes the absolute Galois group of distinguished semi-local algebraic extensions of Q (among others) as free products of 𝐹ˆ𝜔 and local Galois groups. The proof of Theorem A depends on two results of Florian Pop from [Pop96] and on the main result of [Pop95]. The aim of this monograph is to work out proofs of the above mentioned result of [HJP12] along with the supporting results of Pop. In addition we follow Melnikov’s construction in [Mel90] of free products of profinite groups. Finally, we generalize the theory of free products of profinite groups and their subgroups developed in [Har87], and present results appearing in [HJP05] needed in the proofs.

Absolute Galois groups Our result is an instance of a positive answer to the generalized inverse problem of Galois theory. Originally, this problem asked whether every finite group occurs as a Galois group of a Galois extension of Q. For many groups this is the case [MaM99], but the general case is still wide open. One way to realize a finite group over Q is to do it in pieces. That is, one has to properly solve finite embedding problems over Q. Again, there are many examples of such problems which are properly solvable [MaM99]. But we do not have a characterization of all finite embedding problems over Q that are properly solvable. In particular, the structure of the absolute Galois group Gal(Q) of Q is unknown. Still, there are several families of fields with known absolute Galois groups. The most renowned example of a field with this property is C(𝑡), with 𝑡 transcendental over C, or, more generally, finite extensions of C(𝑡). The Riemann Existence Theorem [Voe96, p. 37, Thm. 2.13] implies that Gal(C(𝑡)) is the free profinite group on 2ℵ0 generators ([Rib70, p. 70, Thm. 8.1]). An analogous result holds for an arbitrary algebraically closed field of characteristic 0. Various “patching methods” give similar results in the case where 𝐾 is algebraically closed of positive characteristic (see [Hrb95], [Pop96], or [Jar11]).

vii

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Introduction

By definition, the absolute Galois group of a field 𝐾 is trivial if 𝐾 is algebraically closed or, more generally, separably closed. More subtle is the case where 𝐾 is the field R of real numbers or, more generally, real closed. In this case, Gal(𝐾) is isomorphic to the group with two elements [Lan93, p. 452, Thm. 2.2]. Much more difficult is the case where 𝐾 is a finite extension of Q 𝑝 or of F 𝑝 ((𝑡)) for some prime number 𝑝. In both cases 𝐾 is complete with respect to a discrete valuation and Gal(𝐾) = 𝑇 n 𝑊 is the semi-direct product of its “maximal tame quotient” 𝑇 and its “wild part” 𝑊 (Lemma 8.2.2). By Iwasawa, the tame group 𝑇 is generated by two elements 𝜎, 𝜏 satisfying the relation 𝜎𝜏𝜎 −1 = 𝜏 𝑞 . The wild group 𝑊 is a free pro-𝑝 group of rank ℵ0 . For the exact structure of Gal(𝐾) by generators, relations, and “conditions” we refer the reader to [NSW20, p. 418, Thm. 7.5.13] in the case where char(𝐾) = 𝑝 due to Helmut Koch [Koc67] and to [NSW20, p. 419, Thm. 7.5.14] for char(𝐾) = 0 and 𝑝 ≠ 2 due to Uwe Jannsen and Kay Wingberg [JaW82]. The case where char(𝐾) = 0 and 𝑝 = 2 was treated by Volker Diekert under the condition that √ 𝐾 ( −1)/𝐾 is unramified. See [NSW20, p. 431] or [Die84].

The field 𝑲tot,𝑺 From our point of view, more important than the fields Q 𝑝 and F𝑞 ((𝑡)) are their algebraic parts. We consider a classically local prime 𝔭 of a field 𝐾. Thus, the “completion” 𝐾ˆ 𝔭 of 𝐾 with respect to 𝔭 is either a finite extension of Q 𝑝 or of F𝑞 ((𝑡)), where 𝑝 is a prime number and 𝑞 is a power of a prime number, or 𝐾ˆ 𝔭 = R. Then, the algebraic part 𝐾𝔭 = 𝐾sep ∩ 𝐾ˆ 𝔭 of 𝐾ˆ 𝔭 is the Henselian (respectively, real) closure of 𝐾 with respect to 𝔭. This closure is uniquely defined up to 𝐾-isomorphism. By a lemma of Krasner (in the Henselian case), Gal(𝐾𝔭 ) is isomorphic to Gal( 𝐾ˆ 𝔭 ), so whatever information we have on Gal( 𝐾ˆ 𝔭 ) applies also to Gal(𝐾𝔭 ). This allows us to consider a finite set 𝑆 of classically local primes of 𝐾 and set Ñ Ñ 𝜌 𝐾tot,𝑆 = 𝔭∈𝑆 𝜌∈Gal(𝐾 ) 𝐾𝔭 . By [Pop96, Thm. 3], Ö Ö Gal(𝐾tot,𝑆 )  ∗ ∗ Gal(𝐾𝔭 ) 𝜌 . 𝔭∈𝑆 𝜌∈𝑅𝔭

Î Here, ∗ 𝜌∈𝑅𝔭 Gal(𝐾𝔭 ) 𝜌 stands for the free product of the profinite groups Gal(𝐾𝔭 ) 𝜌 Î (Definition 4.1.1), while ∗ 𝔭∈𝑆 denotes the free product of finitely many profinite groups following the operator. We refer to the fields 𝐾tot,𝑆 as fields of “semi-local type”.

The fields 𝑲sep (𝝈) and 𝑲sep [𝝈] Next we consider fields of another type, “with no arithmetic”. Basic Galois theory shows that the absolute Galois group of a finite field 𝐾 is isomorphic to Zˆ := lim Z/𝑛Z ←−− [FrJ08, p. 15, Sec. 1.5]. It is not difficult to show that the latter property extends

Introduction

ix

to non-principal ultra products [FrJ08, p. 141, Sec. 7.7] of finite fields. If 𝐹 is a ˜ is procyclic. Thus, there exists field of this type and char(𝐹) = 0, then Gal(𝐹 ∩ Q) ˜ is the fixed field Q(𝜎) ˜ ˜ Conversely, for a 𝜎 ∈ Gal(Q) such that 𝐹 ∩ Q of 𝜎 in Q. each 𝜎 ∈ Gal(Q) there exists a non-principal ultraproduct 𝐹 of finite fields such that ˜ ˜ [Ax67, Thm. 5]. Q(𝜎) = 𝐹∩Q ˜ Note that for an arbitrary 𝜎 ∈ Gal(Q) it may happen that Gal( Q(𝜎)) is not ˆ For example, this is the case if 𝜎 = 1 or 𝜎 is an involution. isomorphic to Z. ˜ However, Gal( Q(𝜎))  Zˆ for almost all 𝜎 ∈ Gal(Q) in the sense of the Haar measure of Gal(Q) [Ax67, Prop. 3]. The proof of the latter result uses the theory of cyclotomic extensions of Q. An alternative proof of this theorem applies Hilbert’s irreducibility theorem for Q, hence it holds for every Hilbertian field 𝐾. Moreover, the following result holds for every positive integer 𝑒 and for almost all 𝝈 := (𝜎1 , . . . , 𝜎𝑒 ) ∈ Gal(𝐾) 𝑒 : the group Gal(𝐾sep (𝝈)) is isomorphic to the free profinite group 𝐹ˆ𝑒 on 𝑒 generators [FrJ08, p. 379, Thm. 18.5.6]. If 𝐾 is also countable, then for almost all 𝝈 ∈ Gal(𝐾) 𝑒 the field 𝐾sep (𝝈) is, in addition, PAC [FrJ08, p. 380, Thm. 18.6.1]. This means that every geometrically integral variety over 𝐾sep (𝝈) has a 𝐾sep (𝝈)-rational point. The latter property implies that the Henselian closures (and the real closures) of almost all fields 𝐾sep (𝝈) are separably closed (a result of Frey–Prestel [FrJ08, p. 205, Cor. 11.5.5]). In this sense, these fields “lack arithmetic”. Digging further down, we denote the maximal Galois extension of 𝐾 in 𝐾sep (𝝈) by 𝐾sep [𝝈]. Under the latter assumptions on 𝐾 and 𝑒, [FrJ08, p. 669, Thm. 27.4.8] asserts that for almost all 𝝈 ∈ Gal(𝐾) 𝑒 the field 𝐾sep [𝝈] is PAC, the group Gal(𝐾sep [𝝈]) is isomorphic to the free profinite group 𝐹ˆ𝜔 on countably many generators, and 𝐾sep [𝝈] is Hilbertian.

The fields 𝑲tot,𝑺 [𝝈] As above, we consider a countable Hilbertian field 𝐾, a finite set 𝑆 of classically local primes of 𝐾, and a positive integer 𝑒. Given 𝝈 := (𝜎1 , . . . , 𝜎𝑒 ) ∈ Gal(𝐾) 𝑒 , we consider the fields 𝐾tot,𝑆 (𝝈) = 𝐾tot,𝑆 ∩ 𝐾sep (𝝈) and 𝐾tot,𝑆 [𝝈] = 𝐾tot,𝑆 ∩ 𝐾sep [𝝈] of “mixed type”. Our goal is to reproduce the description of Gal(𝐾tot,𝑆 [𝝈]) as it appears in [HJP12, Thm. 3.11] along with all supporting results from [Pop96], [Pop95], [Mel90], and [Har87].

The main result The main result of this monograph strengthens the main result of [HJP12]. Theorem A (Theorem 9.1.6) Let 𝐾 be a countable Hilbertian field, 𝑆 a finite set of classically local primes of 𝐾, and 𝑒 a positive integer. Then, for almost all 𝝈 ∈ Gal(𝐾) 𝑒 the field 𝐾tot,𝑆 [𝝈] is Hilbertian, P𝑆C, and ample. Moreover, for

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Introduction

each 𝔭 ∈ 𝑆Îthere exists a closed subset 𝑅𝔭 of Gal(𝐾) such that Gal(𝐾tot,𝑆 [𝝈])  Î 𝐹ˆ𝜔 ∗ ∗ 𝔭∈𝑆 ∗ 𝜌∈𝑅𝔭 Gal(𝐾𝔭 ) 𝜌 . Here an extension 𝑀 of 𝐾 in 𝐾tot,𝑆 is said to be P𝑆C if every geometrically integral 𝜌 curve Γ over 𝑀 with a simple 𝐾𝔭 -point for each 𝔭 ∈ 𝑆 and every 𝜌 ∈ Gal(𝐾) has infinitely many 𝑀-rational points. Also, one says that 𝑀 is ample if every geometrically integral curve over 𝑀 with a simple 𝑀-rational point has infinitely many 𝑀-rational points [Jar11, p. 68, Def. 5.3.2]. The Hilbertianity of 𝐾tot,𝑆 [𝝈] for almost all 𝝈 ∈ Gal(𝐾) 𝑒 follows from [BSF13, Thm. 1.1]. By [GeJ02], 𝐾tot,𝑆 [𝝈] is P𝑆C for almost all 𝝈 ∈ Gal(𝐾) 𝑒 . This implies that 𝑀 is ample [Pop96, Prop. 3.1]. Î Î Remark B The free factor 𝐶 := ∗ 𝔭∈𝑆 ∗ 𝜌∈𝑅𝔭 Gal(𝐾𝔭 ) 𝜌 appearing in Theorem A depends (up to isomorphism) only on 𝐾 and 𝑆 but not on the choice of the fields 𝐾𝔭 nor on 𝝈. In particular, this factor is isomorphic to Gal(𝐾tot,𝑆 ) (Remark 9.2.4). We call 𝐶 a Cantor free product over 𝑆, because each of the spaces 𝑅𝔭 is homeomorphic to the Cantor middle-third set (Section 1.5). Using the group-theoretic Lemma 4.7.5, Theorem A yields the following corollary. Corollary C (Remark 9.2.4 and Lemma 9.2.1) Let 𝐾 be a countable Hilbertian field, 𝑆 a finite set of classically local primes of 𝐾, and 𝑒 a non-negative integer. Then, for almost all 𝝈 ∈ Gal(𝐾) 𝑒 and for each 𝔭 ∈ 𝑆 there exists a closed subset 𝑅𝔭 of Gal(𝐾) such that Ö Ö Gal(𝐾tot,𝑆 (𝝈))  𝐹ˆ𝑒 ∗ ∗ ∗ Gal(𝐾𝔭 ) 𝜌 . 𝔭∈𝑆 𝜌∈𝑅𝔭

Remark D If 𝑆 is an empty set, then 𝐶 = 1. Thus, in this case, Corollary C and Theorem A say that for almost all 𝝈 ∈ Gal(𝐾) 𝑒 we have Gal(𝐾sep (𝝈))  𝐹ˆ𝑒 and Gal(𝐾sep [𝝈])  𝐹ˆ𝜔 , as mentioned in the Subsection “The fields 𝐾sep (𝝈) and 𝐾sep [𝝈]”.

A result of Pop The proof of Theorem A depends on [Pop96, Thm. 2.8]: Proposition E (The fundamental result, Proposition 8.4.3) Let 𝑆 be a finite set of classically local primes of a countable Hilbertian field 𝐾. Consider an infinite extension 𝑀 of 𝐾 in 𝐾tot,𝑆 which is ample and Hilbertian. Suppose that Gal(𝑀) is G𝐾 ,𝑆 -projective. Then, G𝐾 ,𝑆 = G𝐾 ,𝑆,max and G𝐾 ,𝑆 has an étale profinite system R Î of representatives for its Gal(𝑀)-orbits such that Gal(𝑀)  𝐹ˆ𝜔 ∗ ∗ Γ∈R Γ. Here, G𝐾 ,𝑆 is the set of all groups Gal(𝐾𝔭 ) 𝜌 with 𝔭 ∈ 𝑆 and 𝜌 ∈ Gal(𝐾) and the symbol G𝐾 ,𝑆,max stands for the set of all maximal elements of G𝐾 ,𝑆 . We say that Gal(𝑀) is G𝐾 ,𝑆 -projective if every finite G𝐾 ,𝑆 -embedding problem for Gal(𝑀) is solvable:

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Let 𝐺 be a profinite group and let G be a subset of the set of all closed subgroups of 𝐺. A finite G-embedding problem for 𝐺 is a triple (𝜑 : 𝐺 → 𝐴, 𝛼 : 𝐵 → 𝐴, B), where 𝛼 : 𝐵 → 𝐴 is an epimorphism of finite groups, 𝜑 : 𝐺 → 𝐴 is a homomorphism of profinite groups, and B is a set of subgroups of 𝐵 closed under 𝐵-conjugation and taking subgroups, such that for each Γ ∈ G there exists a homomorphism 𝛾Γ : Γ → 𝐵 with 𝛼 ◦ 𝛾Γ = 𝜑|Γ and 𝛾Γ (Γ) ∈ B. We say that a homomorphism 𝛾 : 𝐺 → 𝐵 is a solution if 𝛼 ◦ 𝛾 = 𝜑. The solution is strong if 𝛾(Γ) ∈ B for each Γ ∈ G. Finally, the embedding problem is proper if 𝜑 is surjective. In this case, a solution 𝛾 to the embedding problem is proper if 𝛾 is surjective. We write Subgr(𝐺) for the set of all closed subgroups of 𝐺 and equip Subgr(𝐺) with the étale topology. A base for this topology is the family of all open subgroups 𝐻 of 𝐺. A subset R of Subgr(𝐺) is said to be étale profinite if R is a profinite space under the induced étale topology of Subgr(𝐺).

Another result of Pop The proof of Proposition E depends on the following consequence of a variant of [Pop95, Thm. 3]: Proposition F (Proposition 6.4.8) Let 𝐺 be a profinite group and let G be a subset of Subgr(𝐺) of P-type. Suppose that every finite G-embedding problem for 𝐺 has a proper solution. Then, every finite G-embedding problem for 𝐺 has a proper strong solution. We do not repeat the definition of P-type here and only mention that by Definition 6.4.5, G𝐾 ,𝑆 is of P-type for every field 𝐾 and every finite set 𝑆 of classically local primes of 𝐾.

Supporting results In order to prove Proposition E we also need, in addition to Proposition F, the following result. Lemma G (Lemma 8.4.2) Let 𝑀 be an ample Hilbertian field and let G be a strictly closed Gal(𝑀)-invariant subset of Subgr(𝑀) of P-type. Suppose that Gal(𝑀) is G-projective. Then, every finite proper G-embedding problem for Gal(𝑀) has a proper strong solution. The proof of Lemma G uses the main Galois-theoretic property of ample fields: Every finite split embedding problem over 𝑀 (𝑡) with 𝑀 as in Lemma G and 𝑡 transcendental over 𝑀 is properly solvable (see [Pop96, Main Theorem B] or [Jar11, p. 89, Thm. 5.10.2]). The group-theoretic assumption in Lemma G is satisfied if we assume a stronger field-theoretic assumption on 𝑀:

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Lemma H (Lemma 8.3.5) Let 𝑀 be an infinite field and X a family of separable algebraic extensions of 𝑀. Suppose that G := {Gal(𝑀 0)} 𝑀 0 ∈X is étale compact and 𝑀 is PXC (Definition 8.3.2). Then, Gal(𝑀) is G-projective.

Generalized Iwasawa isomorphism theorem In addition to results F and G, the proof of Proposition E uses the following generalization of Iwasawa isomorphism theorem [Pop96, Thm. 4.5]: Proposition I (Proposition 7.2.2) Let 𝐺 and 𝐺 0 be profinite groups. Let G (resp. G 0) be a subset of Subgr(𝐺) (resp Subgr(𝐺 0)) that satisfies the following conditions: (a) rank(𝐺) ≤ ℵ0 (resp. rank(𝐺 0) ≤ ℵ0 ). (b) G (resp. G 0) is an étale compact set of representatives of the distinct conjugacy classes in (G 𝐺 )max . 0 (c) 𝐺 (resp. 𝐺 0) is properly stronglyÐG-projective Ð (resp. 0G -projective). (d) There is a homeomorphism 𝜇 : Γ∈ G Γ → Γ0 ∈ G0 Γ that satisfies the following condition: for every Γ ∈ G there is a Γ0 ∈ G 0 such that 𝜇|Γ : Γ → Γ0 is an isomorphism of groups. 0

Then, there is an isomorphism 𝜃 : 𝐺 → 𝐺 0 such that 𝜃 (G 𝐺 ) = (G 0) 𝐺 . Condition (b) in Proposition I is achieved by the following result. Lemma J (Lemma 7.1.3) Let 𝐺 be a profinite group of rank ≤ ℵ0 and let G be a 𝐺-invariant étale compact subset of Subgr(𝐺) such that G = Gmax and 𝐺 is strongly G-projective. Then, G has an étale compact subset of representatives for its 𝐺-orbits. Tel Aviv University August 2021

Dan Haran Moshe Jarden

Contents

Notation and Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xv 1

Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Profinite Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Strict and Étale Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.3 Étale Hausdorff Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.4 The Envelope of a Set of Closed Subgroups . . . . . . . . . . . . . . . . . . . . . 13 1.5 Cantor Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2

Families of Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Continuous and Separated Families . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Sheaves of Profinite Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Sheaf-Group Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

Free Products of Finitely Many Profinite Groups . . . . . . . . . . . . . . . . . . 33 3.1 Definition and First Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 3.2 Finite Subgroups of Free Products of Finitely Many Profinite Groups 37

4

Generalized Free Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Inner Free Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Outer Free Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Equivalence of Inner and Outer Free Products . . . . . . . . . . . . . . . . . . . 4.4 Free Profinite Products in the Sense of Binz–Neukirch–Wenzel . . . . 4.5 Properties of Inner Free Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Further Properties of Inner Free Products . . . . . . . . . . . . . . . . . . . . . . . 4.7 Free Products as Semi-direct Products . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Closed Subgroups of Free Products . . . . . . . . . . . . . . . . . . . . . . . . . . .

41 41 43 46 49 52 54 59 62

5

Relative Embedding Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Embedding Problems and Projectivity . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Strongly G-Projective Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Infinite Embedding Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 The Closed Subgroup Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

65 65 69 71 73

21 21 26 29

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5.5 5.6

The Set Gmax . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 Prescribed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

6

Strong Proper Projectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Prosolvable Subgroups of G-Projective Groups . . . . . . . . . . . . . . . . . . 6.2 Groups of Local p-Type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Fields of Local p-Type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Groups of P-Type and G-Projectivity . . . . . . . . . . . . . . . . . . . . . . . . . .

89 89 93 96 99

7

A Free Profinite Product over an Étale Compact Subset of Subgr(𝑮) 105 7.1 A System of Representatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 7.2 The Generalized Iwasawa Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

8

Fundamental Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 8.1 Classical Closures of a Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 8.2 Sections over Henselian Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 8.3 G-Projectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 8.4 Semi-Local Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

9

Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 9.1 The Groups Gal(𝐾tot,𝑆 [𝝈]) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 9.2 Finitely Generated Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 Symbol Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Subject Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

Notation and Conventions

– Given disjoint subsets 𝐴 and 𝐵 of a common set 𝐶, we write 𝐴 ∪· 𝐵 for the union of 𝐴 and 𝐵. In this case we say that 𝐴 ∪· 𝐵 is the disjoint union of 𝐴 and 𝐵. – We write card( 𝐴) for the cardinality of a set 𝐴. Ð – For each 𝑖 in a set 𝐼 let 𝐴𝑖 be a subset of a common set 𝐴. We write · 𝑖 ∈𝐼 𝐴𝑖 for the union of the 𝐴𝑖 ’s to indicate that 𝐴𝑖 ∩ 𝐴 𝑗 = ∅ if 𝑖 ≠ 𝑗. – If 𝑎 𝑖 , for 𝑖 ∈ 𝐼, are elements of a set 𝐴, then we denote the subset {𝑎 𝑖 | 𝑖 ∈ 𝐼} of 𝐴 that consists of all the 𝑎 𝑖 ’s also by {𝑎 𝑖 }𝑖 ∈𝐼 . – We mainly work with the category of profinite groups [FrJ08, p. 4–9, Sec. 1.2]. Occasionally we speak about abstract groups, i.e. groups with no additional structure. – If 𝐺 𝑖 , for 𝑖 ∈ 𝐼, are closed subgroups of a profinite group 𝐺, then h𝐺 𝑖 i𝑖 ∈𝐼 is the smallest closed subgroup of 𝐺 that contains each of the 𝐺 𝑖 ’s. We then say that h𝐺 𝑖 i𝑖 ∈𝐼 is the group generated by the 𝐺 𝑖 ’s. – We denote the trivial subgroup of a multiplicative group 𝐺 by 1𝐺 or simply by 1 if 𝐺 is clear from the context. – We write 𝐺 ≤ 𝐻 (resp. 𝐺 < 𝐻) for groups 𝐺 and 𝐻 to indicate that 𝐺 is a subgroup (resp. proper subgroup) of 𝐻. If in addition 𝐺 and 𝐻 are profinite groups, then 𝐺 ≤ 𝐻 (resp. 𝐺 < 𝐻) means that 𝐺 is a closed subgroup (resp. proper closed subgroup) of 𝐻. – If 𝐺 is a subgroup of 𝐻, then 𝑁 𝐻 (𝐺) = {ℎ ∈ 𝐻 | 𝐺 ℎ = 𝐺} denotes the normalizer of 𝐺 in 𝐻. – Whenever we speak about a homomorphism 𝛼 : 𝐺 → 𝐻 between profinite groups, we tacitly assume that 𝛼 is continuous. – Given a field 𝐾, we fix an algebraic closure 𝐾˜ of 𝐾, write 𝐾sep for the separable ˜ and let 𝐾ins be the maximal purely inseparable algebraic closure of 𝐾 in 𝐾, ˜ extension of 𝐾 in 𝐾. – If 𝑁/𝐾 is a Galois extension of fields, then Gal(𝑁/𝐾) denotes the Galois group of 𝑁/𝐾. Accordingly, Gal(𝐾) := Gal(𝐾sep /𝐾) is the absolute Galois group of 𝐾.

xv

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Notation and Conventions

– As usual, Q denotes the field of rational numbers, F𝑞 the field with 𝑞-elements where 𝑞 is a power of a prime number 𝑝, Q 𝑝 the field of 𝑝-adic numbers, R the field of real numbers, and C the field of complex numbers. Also, Z denotes the ring of integers and Zˆ := lim Z/𝑛Z its completion. ←−−

Chapter 1

Topologies

Given a profinite group 𝐺, we consider the set Subgr(𝐺) of all closed subgroups of 𝐺 and equip it with two topologies. The first one, called the strict topology, has a basis that consists of all sets {Γ ∈ Subgr(𝐺) | Γ𝑁 = 𝐻𝑁 }, where 𝐻 is a closed subgroup of 𝐺 and 𝑁 is an open normal subgroup of 𝐺. Under this topology Subgr(𝐺) is the inverse limit of all Subgr(𝐺/𝑁), where 𝑁 ranges over all open normal subgroups of 𝐺 and Subgr(𝐺/𝑁) is a finite space equipped with the discrete topology. Thus, Subgr(𝐺) is a “profinite space”. We introduce this notion in Section 1.1 and discuss some of its basic properties there. The second topology on Subgr(𝐺) is called the étale topology. A basis of this topology consists of all sets Subgr(𝐻), where 𝐻 ranges over all open subgroups of 𝐺. In particular, every étale open (resp. closed) subset of Subgr(𝐺) is strictly open (resp. closed). Hence, every strictly compact subset of Subgr(𝐺) is also étale compact. Thus, being étale compact is a weaker condition which we often impose on a subset G of Subgr(𝐺). We prove, among other things, that if G is an étale compact subset of Subgr(𝐺), Ð then Γ∈ G Γ is a closed subset of 𝐺 (Lemma 1.2.5). We also prove that every Γ ∈ G is contained in a maximal element of G (Lemma 1.3.5) and the set Gmax of all maximal elements of G is also étale compact (Lemma 1.3.6(a)).

1.1 Profinite Spaces We start by recalling some basic facts about compact spaces and Hausdorff spaces. We adopt the convention of [HeR63, p. 11, (3.2)] that a topological space 𝑋 is compact if every covering of 𝑋 by open subsets has a finite subcovering. Thus, in contrast to the terminology of [Bou89, p. 83, Def. 1], 𝑋 need not be Hausdorff. We often use compactness of a space 𝑋 in the following form:

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 D. Haran and M. Jarden, The Absolute Galois Group of a Semi-Local Field, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-030-89191-6_1

1

2

1 Topologies

Lemma 1.1.1 A topological space 𝑋 is compact if and only if it satisfies the following condition. (*) Let 𝑈 be an open subset of 𝑋 and let 𝐶 ⊆ 𝑈 be the intersection of a family {𝐶𝑖 }𝑖 ∈𝐼 of closed subsets of 𝑋, closed under finite intersections. Then, there is an 𝑖 ∈ 𝐼 such that 𝐶𝑖 ⊆ 𝑈. Proof We show that compactness of 𝑋 implies (*) and leave the converse to the reader: Ð The assumptions in (*) imply 𝑈 ∪ 𝑖 ∈𝐼 (𝑋 r 𝐶𝑖 ) =Ð𝑈 ∪ (𝑋 r 𝐶) = 𝑋. So, if 𝑋 is 𝑛 compact, Ñ𝑛 there are 𝑖 1 , . . . , 𝑖 𝑛 ∈ 𝐼 such that 𝑋 = 𝑈 ∪ 𝑘=1 (𝑋 r 𝐶𝑖𝑘 ). Let 𝑖 ∈ 𝐼 such that 𝑘=1 𝐶𝑖𝑘 = 𝐶𝑖 . Then, 𝑋 = 𝑈 ∪ (𝑋 r 𝐶𝑖 ), and hence 𝐶𝑖 ⊆ 𝑈.  Definition 1.1.2 A topological space 𝑋 is said to be Hausdorff if any two distinct points of 𝑋 have disjoint open neighborhoods. In particular, every finite subset of 𝑋 is closed [Bou89, p. 77, Prop. 4]. Fact 1.1.3 We list some well known statements about topological spaces. (a) Every closed subset of a compact space is compact [Bou89, p. 86, §9.3 Prop. 3]. (b) Every compact subset of a Hausdorff space is closed [Bou89, p. 86, §9.3, Prop. 4]. (c) A subset of a compact Hausdorff space is closed if and only if it is compact [Bou89, p. 86, §9.3, Cor.]. (d) A continuous map 𝑓 : 𝑋 → 𝑌 of topological spaces maps compact sets onto compact sets [Bou89, p. 87, §9.4, Thm. 2]. (e) A continuous map from a compact space 𝑋 into a Hausdorff space 𝑌 is closed, that is, it maps closed subsets of 𝑋 onto closed subsets of 𝑌 [Bou89, p. 87, §9.4, Cor. 2]. In particular, ifÎ this map is a bijection, then it is a homeomorphism. (f) The cartesian product 𝑖 ∈𝐼 𝑋𝑖 of compact sets isÎ compact with respect to the product topology whose base consists of all sets 𝑖 ∈𝐼 𝑈𝑖 , with 𝑈𝑖 open in 𝑋𝑖 for all 𝑖 ∈ 𝐼 and 𝑈𝑖 = 𝑋𝑖 for all but finitely many 𝑖 ∈ 𝐼 (Tychonoff’s theorem) [Bou89, p. 88, §9.5, Thm. 3]. (g) Let 𝑇 and 𝐺 be topological spaces with 𝐺 compact. Then, the coordinate projection 𝑇 × 𝐺 → 𝑇 is closed [Sin13, p. 30, Thm. 6.1.16]. Remark 1.1.4 Let (𝐼, ≤) be a directed partially ordered set [FrJ08, p. 1, Sec. 1] and let (𝑋𝑖 , 𝜋 𝑗𝑖 ) 𝑗 ≥𝑖 be an inverse system of topological spaces, with connecting maps 𝜋 𝑗𝑖 : 𝑋 𝑗 → 𝑋𝑖 for all 𝑖, 𝑗 ∈ 𝐼 such that 𝑗 ≥ 𝑖. Its inverse limit 𝑋 = lim 𝑋𝑖 is the ←−−𝑖 ∈𝐼 Î subspace of 𝑖 ∈𝐼 𝑋𝑖 (with the Tychonoff topology) consisting of all 𝑥 = (𝑥𝑖 )𝑖 ∈𝐼 ∈ Î 𝑖 ∈𝐼 𝑋𝑖 (we sometimes write 𝑥 = lim 𝑥 𝑖 ) such that 𝜋 𝑗𝑖 (𝑥 𝑗 ) = 𝑥 𝑖 whenever 𝑗 ≥ 𝑖. The ←−− Î restriction to 𝑋 of the projection on the 𝑖th component of 𝑖 ∈𝐼 𝑋𝑖 is a continuous map 𝜋𝑖 : 𝑋 → 𝑋𝑖 , called the projection of 𝑋 on 𝑋𝑖 . It satisfies 𝜋𝑖 = 𝜋 𝑗𝑖 ◦ 𝜋 𝑗 whenever 𝑗 ≥ 𝑖. Since an open set is a union of elements of a basis for a topology, Lemma 1.1.1 of [FrJ08] characterizes the topology on 𝑋 as follows: For every 𝑖 ∈ 𝐼 let B𝑖 be a basis for the topology on 𝑋𝑖 . Then, {𝜋𝑖−1 (𝑈) | 𝑈 ∈ B𝑖 , 𝑖 ∈ 𝐼} is a basis for the topology on 𝑋.

1.1 Profinite Spaces

3

Example 1.1.5 Let { 𝐴𝑖 }𝑖 ∈𝐼 be a family of subsets of a set 𝐴. Suppose that for all 𝑖, 𝑗 ∈ 𝐼 there exists a 𝑘 ∈ 𝐼 such that 𝐴 𝑘 ⊆ 𝐴𝑖 ∩ 𝐴 𝑗 . We write 𝑖 ≤ 𝑗 for elements 𝑖, 𝑗 ∈ 𝐼 if 𝐴 𝑗 ⊆ 𝐴𝑖 . Then, (𝐼, ≤) is a directed partially ordered set [FrJ08, p. 1, Sec. 1.1]. Given 𝑖 ≤ 𝑗 in 𝐼 we define 𝜋 𝑗𝑖 : 𝐴 𝑗 → 𝐴𝑖 to be the inclusion map. Then, ( 𝐴𝑖 , 𝜋 𝑗𝑖 )𝑖Ñ ≤ 𝑗 is an inverse system of sets (or also of discrete topological spaces) and lim 𝐴𝑖 = 𝑖 ∈𝐼 𝐴𝑖 . ←−− Definition 1.1.6 A topological space 𝑋 is profinite if (a) 𝑋 is an inverse limit 𝑋 = lim 𝑋𝑖 of finite discrete spaces 𝑋𝑖 . ←−−𝑖 ∈𝐼 Equivalently ([FrJ08, pp. 2–3, Lemmas 1.1.3 and 1.1.7]), (b) 𝑋 is compact, Hausdorff, and has a basis for its topology that consists of openclosed sets. Remark 1.1.7 Characterization (b) yields several useful consequences: (a) For a subset 𝑌 of a profinite space the following are equivalent: (i) 𝑌 is closed; (ii) 𝑌 is compact; (iii) 𝑌 is profinite. (b) Direct products of profinite spaces equipped with the Tychonoff topology are again profinite spaces. (c) Every continuous map of profinite spaces is closed (apply Fact 1.1.3(e)), in particular, every continuous bijection of profinite spaces is a homeomorphism. (d) Furthermore, if 𝑋 is the inverse limit of finite spaces 𝑋𝑖 with maps 𝜋 𝑗𝑖 and 𝜋𝑖 as in Remark 1.1.4, we may replace 𝑋𝑖 by 𝜋𝑖 (𝑋) to get that a profinite space is an inverse limit of finite spaces with surjective connecting maps and surjective projections. Remark 1.1.8 Here is a more explicit elaboration of Remark 1.1.7(d). Let 𝑇 be a profinite space and denote the collection of all partitions of 𝑇 into disjoint open-closed sets by U. Since 𝑇 is compact, every U ∈ U is finite. We define a partial relation on U by setting U ≤ U 0 if U 0 is finer than U. In other words, every 𝑈 0 ∈ U 0 is contained in some 𝑈 ∈ U. In particular, the map 𝜋 U 0 ,U : U 0 → U, that maps every 𝑈 0 ∈ U 0 onto the unique 𝑈 ∈ U that contains 𝑈 0, is surjective. Then, 𝑇 is homeomorphic to the inverse limit lim U. In this isomorphism, every 𝑡 ∈ 𝑇 ←−− corresponds to the unique system (𝑈𝑡 , U )U ∈U , where 𝑈𝑡 , U is the unique set in U that contains 𝑡. Lemma 1.1.9 A compact space 𝑋 is profinite if and only if for all distinct 𝑥1 , 𝑥2 ∈ 𝑋 there is an open-closed subset 𝑋 0 of 𝑋 such that 𝑥 1 ∈ 𝑋 0, 𝑥2 ∉ 𝑋 0. Proof Suppose that the condition holds. Since the complement of an open-closed subset 𝑋 0 of 𝑋 is also open-closed, 𝑋 is Hausdorff. By Definition 1.1.6(b) it remains to show that given 𝑥 ∈ 𝑋 and an open neighborhood 𝑈 of 𝑥, there is an open-closed neighborhood of 𝑥 contained in 𝑈. Put 𝐶 = 𝑋 r 𝑈. For every 𝑦 ∈ 𝐶 there is an open-closed subset 𝑋 𝑦 of 𝑋 such that 𝑦 ∈ 𝑋 𝑦 , 𝑥 ∉ 𝑋 𝑦 . Notice that 𝐶 is closed in 𝑋 and {𝑋 𝑦 } 𝑦 ∈𝐶 is an open-closed covering of 𝐶. Since 𝑋 is compact, there is a finite subset 𝐶 0 ⊆ 𝐶 such that {𝑋 𝑦 } 𝑦 ∈𝐶 0

4

1 Topologies

Ð is a covering of 𝐶. Then, 𝑌 = 𝑦 ∈𝐶 0 𝑋 𝑦 is open-closed, 𝑥 ∉ 𝑌 , and 𝐶 ⊆ 𝑌 . Hence, 𝑉 = 𝑋 r 𝑌 is open-closed, 𝑥 ∈ 𝑉, and 𝑉 ⊆ 𝑈. The converse implication follows from Definition 1.1.6(b).  Lemma 1.1.10 Let 𝑇 be a topological space and T = {𝑇𝑖 }𝑖 ∈𝐼 a finite cover of 𝑇 by open-closed subsets. Then, 𝑇 has a finite cover U = {𝑈 𝑗 } 𝑗 ∈𝐽 by open-closed disjoint subsets such that U is finer than T . Proof Choose 𝑖 0 ∈ 𝐼 and consider the decomposition of 𝑇 Ø Ø Ø Ø (1.1) 𝑇= 𝑇𝑖 = (𝑇𝑖0 r 𝑇𝑖 ) ∪· (𝑇𝑖 ∩ 𝑇𝑖0 ) ∪· (𝑇𝑖 r 𝑇𝑖0 ). 𝑖 ∈𝐼

𝑖≠𝑖0

𝑖≠𝑖0

𝑖≠𝑖0

The first set on the right-hand side of (1.1) is an open-closed subset of 𝑇. The second and third sets are unions of card(𝐼) − 1 open-closed subsets of 𝑇. Now apply induction on the order of 𝐼.  A profinite group is an inverse limit 𝐺 := lim 𝐺 𝑖 of finite discrete groups. It ←−− is assumed that the connecting maps 𝜋 𝑗𝑖 : 𝐺 𝑗 → 𝐺 𝑖 commute with the group operations. This implies that 𝐺 is a compact group whose projections 𝜋𝑖 : 𝐺 → 𝐺 𝑖 are continuous homomorphisms. Example 1.1.11 A continuous action (from the right) of a profinite group 𝐺 on a profinite space 𝑋 is a continuous map (𝑥, 𝜎) ↦→ 𝑥 𝜎 of 𝑋 × 𝐺 into 𝑋 such that (𝑥 𝜎 ) 𝜏 = 𝑥 𝜎 𝜏 and 𝑥 1 = 𝑥 for all 𝑥 ∈ 𝑋 and 𝜎, 𝜏 ∈ 𝐺. If 𝐶 is a closed subset of 𝑋, then, by Fact 1.1.3(e), 𝐶 𝐺 := {𝑐 𝜎 | 𝑐 ∈ 𝐶 and 𝜎 ∈ 𝐺} is a closed subset of 𝑋. The following basic information about profinite groups is quite useful. Lemma 1.1.12 ([FrJ08, p. 6, Lemma 1.2.2])ÑLet {𝐻𝑖 }𝑖 ∈𝐼 be a directed family of closed subsets of a profinite group 𝐺. Set 𝐻 = 𝑖 ∈𝐼 𝐻𝑖 . (a) For Ñ every open subgroup Ñ 𝑈 of 𝐺 containing 𝐻 there is an 𝑖 ∈ 𝐼 with 𝐻𝑖 ≤ 𝑈. (b) 𝑖 ∈𝐼 𝐾 𝐻𝑖 = 𝐾 𝐻 and 𝑖 ∈𝐼 𝐻𝑖 𝐾 = 𝐻𝐾 for every closed subset 𝐾 of 𝐺. Corollary 1.1.13 Let 𝐺 be a profinite group, 𝑉 an open-closed subset of 𝐺, and Γ ∈ Subgr(𝐺) with Γ ⊆ 𝑉. Then, there is an open 𝐾 ⊳ 𝐺 such that Γ𝐾 ⊆ 𝑉. Proof By [FrJ08, p. 6, Remark 1.2.1(a)], the set K of all open normal subgroups Ñ Ñ of 𝐺 satisfies 𝐾 ∈K 𝐾 = 1. Hence, by Lemma 1.1.12(b), 𝐾 ∈K Γ𝐾 = Γ ⊆ 𝑉. Therefore, by Lemma 1.1.12(a), there exists a 𝐾 ∈ K with Γ𝐾 ⊆ 𝑉, as desired.  Lemma 1.1.14 ([FrJ08, p. 7, Lemma 1.2.3]) Every closed subgroup 𝐻 of a profinite group 𝐺 is an intersection of open subgroups of 𝐺. Here is the analogue of Lemma 1.1.14 for profinite spaces. Lemma 1.1.15 A closed subset of a profinite space 𝑇 is an intersection of openclosed subsets of 𝑇.

1.1 Profinite Spaces

5

Proof Let 𝑆 be a closed subset of 𝑇. Then, 𝑇 r 𝑆 is open. Hence, by DefiniÐ tion 1.1.6(b), 𝑇 r 𝑆 = 𝑖 ∈𝐼 𝑈𝑖 , where 𝑈𝑖 is open-closed for each 𝑖 ∈ 𝐼. Then, 𝑇 r 𝑈𝑖 Ñ is also open-closed for each 𝑖 ∈ 𝐼 and 𝑆 = 𝑖 ∈𝐼 (𝑇 r 𝑈𝑖 ), as claimed.  Lemma 1.1.16 Suppose that 𝑋 is the inverse limit of an inverse system (𝑋𝑖 , 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 of profinite spaces. For every 𝑖 ∈ 𝐼 let 𝜋𝑖 be the projection 𝑋 → 𝑋𝑖 . Let 𝑓 : 𝑋 → 𝐴 be a continuous map into a finite discrete space 𝐴. Then: (a) Given open-closed subsets 𝑈1 , . . . , 𝑈𝑛 of 𝑋, there exists a 𝑗 ∈ 𝐼 and open-closed subsets 𝑉1 , . . . , 𝑉𝑛 of 𝑋 𝑗 such that 𝑈𝑟 = 𝜋 −1 𝑗 (𝑉𝑟 ), for 𝑟 = 1, . . . , 𝑛. (b) There exists a 𝑗 ∈ 𝐼 and a continuous map 𝑔 : 𝑋 𝑗 → 𝐴 such that 𝑔 ◦ 𝜋 𝑗 = 𝑓 . (c) If 𝑔 0 : 𝑋 𝑗 → 𝐴 is another continuous map such that 𝑔 0 ◦ 𝜋 𝑗 = 𝑓 , there exists a 𝑘 ≥ 𝑗 such that 𝑔 ◦ 𝜋 𝑘 𝑗 = 𝑔 0 ◦ 𝜋 𝑘 𝑗 . Proof Proof of (a). Let 1 ≤ 𝑟 ≤ 𝑛. Since 𝑈𝑟 is open in 𝑋, it can be written as a Ð union 𝑈𝑟 = 𝑘 ∈𝐾 (𝑟 ) 𝑊 𝑘0 of basic open subsets of 𝑋. Since 𝑈𝑟 is also closed and hence compact, we may assume that 𝐾 (𝑟) is finite. By Remark 1.1.4, we may assume that for every 𝑘 ∈ 𝐾 (𝑟) there is an 𝑖 𝑘 ∈ 𝐼 and a basic open subset 𝑊 𝑘 of 𝑋𝑖𝑘 such that 𝑊 𝑘0 = 𝜋𝑖−1 (𝑊 𝑘 ). By Definition 1.1.6(b), we may assume that 𝑊 𝑘 is open-closed 𝑘 Ð in 𝑋𝑖𝑘 . Thus, 𝑈𝑟 = 𝑘 ∈𝐾 (𝑟 ) 𝜋𝑖−1 (𝑊 𝑘 ). 𝑘 Ð Choose 𝑗 ∈ 𝐼 such that 𝑗 ≥ 𝑖 𝑘 for all 𝑘 ∈ 𝑟𝑛=1 𝐾 (𝑟). Then, Ø 𝑉𝑟 = 𝜋 −1 𝑗𝑖𝑘 (𝑊 𝑘 ) 𝑘 ∈𝐾 (𝑟 )

is open-closed in 𝑋 𝑗 and

𝜋 −1 𝑗 (𝑉𝑟 )

= 𝑈𝑟 , for every 𝑟.

Proof of (b). The sets 𝑓 −1 ({𝑎}), for 𝑎 ∈ 𝐴, are open-closed in 𝑋. Hence, by (a), there is a 𝑗 ∈ 𝐼 and open-closed subsets 𝑉𝑎 of 𝑋 𝑗 , for 𝑎 ∈ 𝐴, such that 𝑓 −1 ({𝑎}) = 𝜋 −1 𝑗 (𝑉𝑎 ). WeÐmay assume that the sets 𝑉𝑎 are disjoint, otherwise replace each 𝑉 𝑎 with Ð 𝑉𝑎 r 𝑏∈ 𝐴 𝑉𝑏 . The complement 𝑉 0 = 𝑋 𝑗 r ( · 𝑎 ∈ 𝐴 𝑉𝑎 ) is also open-closed in 𝑋 𝑗 . 𝑏≠𝑎 Ð Choose 𝑐 ∈ 𝐴 and replace 𝑉𝑐 with 𝑉𝑐 ∪· 𝑉 0 to assume that 𝑋 𝑗 = · 𝑎 ∈ 𝐴 𝑉𝑎 . Then, the map 𝑔 : 𝑋 𝑗 → 𝐴, defined by 𝑔(𝑦) = 𝑎 if 𝑦 ∈ 𝑉𝑎 , is continuous and 𝑔 ◦ 𝜋 𝑗 = 𝑓 . Proof of (c). For every 𝑘 ≥ 𝑗 put 𝑔 𝑘 = 𝑔 ◦ 𝜋 𝑘 𝑗 . We first claim that (1.2)

there is a 𝑘 ≥ 𝑗 such that 𝑔 𝑘 (𝑋 𝑘 ) ⊆ 𝑓 (𝑋).

Indeed, since 𝐴 is finite and 𝑔𝑙 (𝑋𝑙 ) ⊆ 𝑔 𝑘 (𝑋 𝑘 ) for 𝑙 ≥ 𝑘, (1.2) is equivalent to Ñ 𝑘 ≥ 𝑗 𝑔 𝑘 (𝑋 𝑘 ) ⊆ 𝑓 (𝑋).
Ñ So let 𝑎 ∈ 𝑘 ≥ 𝑗 𝑔 𝑘 (𝑋 𝑘 ). Then, 𝑔 −1 𝑘 (𝑎) 𝑘 ≥ 𝑗 is an inverse system of non-empty profinite spaces. Their inverse limit is non-empty [FrJ08, p. 2, Lemma 1.1.3], and every element of the inverse limit belongs to 𝑓 −1 (𝑎). Hence 𝑎 ∈ 𝑓 (𝑋). This proves (1.2). We have the following commutative diagram /𝐴

𝑓

𝑋 𝜋𝑗

 𝑋𝑗

Δ

 / 𝐴×𝐴,

(𝑔,𝑔0 )

6

1 Topologies

in which Δ is the diagonal map 𝑎 ↦→ (𝑎, 𝑎). By (1.2), applied to (𝑔, 𝑔 0), Δ ◦ 𝑓 instead of 𝑔, 𝑓 , there is a 𝑘 ≥ 𝑗 such that (𝑔, 𝑔 0) ◦ 𝜋 𝑘 𝑗 (𝑋 𝑘 ) ⊆ Δ ◦ 𝑓 (𝑋). But this means that 𝑔 ◦ 𝜋 𝑘 𝑗 (𝑥) = 𝑔 0 ◦ 𝜋 𝑘 𝑗 (𝑥) for every 𝑥 ∈ 𝑋 𝑘 .  Corollary 1.1.17 Let 𝑌 be a profinite space, 𝑋 a closed subset of 𝑌 , and 𝑓 a continuous map of 𝑋 into a finite set 𝐴 equipped with the discrete topology. Then: (a) Given an open-closed subset 𝑈 of 𝑋, there is an open-closed subset 𝑉 of 𝑌 such that 𝑉 ∩ 𝑋 = 𝑈. (b) 𝑓 extends to a continuous map 𝑔 : 𝑌 → 𝐴. (c) If 𝑔 0 : 𝑌 → 𝐴 is another continuous map that extends 𝑓 , then 𝑔, 𝑔 0 coincide on an open-closed subset of 𝑌 containing 𝑋. Proof By Lemma 1.1.15 and Example 1.1.5, 𝑋 is an inverse limit of open-closed subsets 𝑋 𝑗 of 𝑌 containing 𝑋. Hence (a), (b), (c) follow from the corresponding assertions of Lemma 1.1.16: Proof of (a). There is an 𝑋 𝑗 and an open-closed subset 𝑉 of 𝑋 𝑗 such that 𝑉 ∩ 𝑋 = 𝑈. But 𝑉 is open-closed in 𝑌 as well. Proof of (b). There is an 𝑋 𝑗 and a continuous map 𝑔 : 𝑋 𝑗 → 𝐴 that extends 𝑓 . But 𝑌 r 𝑋 𝑗 is also open-closed. Choose 𝑐 ∈ 𝐴 and extend 𝑔 to 𝑌 by 𝑔(𝑦) = 𝑐 for every 𝑐 ∈ 𝑌 r 𝑋𝑗. Proof of (c). There is an 𝑋 𝑗 on which 𝑔 and 𝑔 0 coincide.



Lemma 1.1.18 Let 𝑓 : 𝑋 → 𝑌 be a map of sets and let 𝑋0 be a subset of 𝑋. We set 𝑌0 = {𝑦 ∈ 𝑌 | 𝑓 −1 (𝑦) ⊆ 𝑋0 }. Then, 𝑓 (𝑋 r 𝑋0 ) = 𝑌 r 𝑌0 . In particular, if 𝑋 and 𝑌 are topological spaces, 𝑓 is a closed map, and 𝑋0 is an open subset of 𝑋, then 𝑌0 is an open subset of 𝑌 . Proof If 𝑦 ∈ 𝑓 (𝑋 r 𝑋0 ), then there exists an 𝑥 ∈ 𝑋 r 𝑋0 with 𝑓 (𝑥) = 𝑦. In particular, 𝑥 ∈ 𝑓 −1 (𝑦) r 𝑋0 , so 𝑓 −1 (𝑦) 6 ⊆ 𝑋0 , hence 𝑦 ∈ 𝑌 r 𝑌0 . Conversely, let 𝑦 ∈ 𝑌 r 𝑌0 . Then, 𝑓 −1 (𝑦) 6 ⊆ 𝑋0 , hence there exists an 𝑥 ∈ 𝑋 such that 𝑓 (𝑥) = 𝑦 and 𝑥 ∉ 𝑋0 . In particular, 𝑦 ∈ 𝑓 (𝑋 r 𝑋0 ). Now, if 𝑓 is a closed map of topological spaces and 𝑋0 is open in 𝑋, then 𝑋 r 𝑋0 is closed, hence, by the preceding paragraph, 𝑌 r 𝑌0 = 𝑓 (𝑋 r 𝑋0 ) is closed in 𝑌 , whence 𝑌0 is open, as claimed.  Given a topological space 𝑋 and a topological group 𝐺 that acts continuously on 𝑋 (from the right), one considers the quotient space 𝑋/𝐺. Its elements are the 𝐺-orbits 𝑥 𝐺 := {𝑥 𝜎 } 𝜎 ∈𝐺 with 𝑥 ∈ 𝑋. The quotient map 𝜋 : 𝑋 → 𝑋/𝐺 maps each 𝑥 ∈ 𝑋 onto its orbit 𝑥 𝐺 . The topology on 𝑋/𝐺 is the quotient topology, that is, a ¯ is open (resp. closed) in 𝑋. subset 𝑈¯ of 𝑋/𝐺 is open (resp. closed) if 𝜋 −1 (𝑈) Lemma 1.1.19 Let 𝑋 be a compact Hausdorff space and 𝐺 a compact group continuously acting on 𝑋 (from the right). Then, the quotient map 𝜋 : 𝑋 → 𝑋/𝐺 is continuous, open, and Ð Ñ closed. In particular, if 𝑈 ⊆ 𝑋 is open (resp. closed), then so are 𝜎 ∈𝐺 𝑈 𝜎 and 𝜎 ∈𝐺 𝑈 𝜎 .

1.1 Profinite Spaces

7

Proof Continuity is a part of the definition of a quotient map. If 𝑈 ⊆ 𝑋 is open in Ð 𝑋, then 𝜋 −1 (𝜋(𝑈)) = 𝜎 ∈𝐺 𝑈 𝜎 is a union of open sets, hence open in 𝑋, so 𝜋(𝑈) is open in 𝐺/𝑋. Thus, 𝜋 is open. The action 𝜓 : 𝑋 × 𝐺 → 𝑋 of 𝐺 on 𝑋 is a continuous map from a compact space into a Hausdorff space. Thus, if 𝑈 is closed in 𝑋, then, by Fact 1.1.3(e), Ð 𝜋 −1 (𝜋(𝑈)) = 𝜎 ∈𝐺 𝑈 𝜎 = 𝜓(𝑈 × 𝐺) is closed in 𝑋, whence 𝜋(𝑈) is closed in 𝐺/𝑋. Therefore, 𝜋 is closed. Ñ Ð For every 𝑈 ⊆ 𝑋 we have 𝜎 ∈𝐺 𝑈 𝜎 = 𝑋 r 𝜎 ∈𝐺 (𝑋 r𝑈) 𝜎 , so, by the preceding paragraph, this set is open (resp. closed) if 𝑈 is open (resp. closed).  Lemma 1.1.20 (a) Let 𝜋 : 𝑋 → 𝑌 be a continuous open and closed map of a profinite space 𝑋 onto a topological space 𝑌 . Then, 𝑌 is also a profinite space. (b) If 𝑋 is a profinite space and 𝐺 is a profinite group acting on 𝑋, then the quotient space 𝑋/𝐺 is also profinite. Proof By definition, the quotient map 𝑋 → 𝑋/𝐺 is surjective. By Lemma 1.1.19, the quotient map is continuous, open, and closed. Hence, (b) follows from (a). For (a), we first observe that since 𝑋 is compact and 𝜋 is continuous and surjective, 𝑌 is also compact (Fact 1.1.3(d)). Next, let 𝑦, 𝑦 0 be distinct points of 𝑌 and choose 𝑥, 𝑥 0 ∈ 𝑋 that 𝜋 maps, respectively, onto 𝑦, 𝑦 0. Since {𝑥 0 } is closed, so is {𝑦 0 }, hence 𝑉 = 𝑌 r{𝑦 0 } is an open neighborhood of 𝑦. Then, 𝑈 = 𝜋 −1 (𝑉) is an open neighborhood of 𝑥. Choose an open-closed neighborhood 𝑈0 of 𝑥 in 𝑈. Then, 𝜋(𝑈0 ) is an open-closed neighborhood of 𝑦 in 𝑉, so 𝜋(𝑈0 ) does not contain 𝑦 0. By Lemma 1.1.9, 𝑌 is a profinite space.  Remark 1.1.21 Let 𝑋 be a topological space and 𝐻 a topological group that acts on 𝑋 from the left. Then, we denote the corresponding quotient space by 𝐻\𝑋 and introduce a topology on 𝐻\𝑋 in an analogous manner to the case of right action. Similarly, the obvious analogs of Lemma 1.1.19 and Lemma 1.1.20 hold for 𝐻\𝑋 if 𝐻 is a profinite group. If, in addition, 𝐺 is a profinite group that acts on 𝑋 from the right and (𝜂𝑥)𝜎 = 𝜂(𝑥𝜎) for all 𝜂 ∈ 𝐻, 𝑥 ∈ 𝑋, and 𝜎 ∈ 𝐺, then the double quotient space 𝐻\𝑋/𝐺 is profinite. In the sequel, we freely use Lemmas 1.1.19 and 1.1.20, and the present remark, and only occasionally explicitly refer to them. Remark 1.1.22 Recall that a topological space 𝑋 is said to be 𝑇1 if for every pair (𝑥1 , 𝑥2 ) of distinct points of 𝑋 there exists an open subset 𝑈 of 𝑋 such that 𝑥1 ∈ 𝑈 and 𝑥2 ∉ 𝑈. Equivalently, every point of 𝑋 is closed in 𝑋 [Sin13, p. 109, Prop. 4.4.7]. Thus, if 𝑋0 is a closed subset of 𝑋 and 𝑋 is 𝑇1 , then 𝑋0 is 𝑇1 . Lemma 1.1.23 Let 𝑋, 𝑇, 𝐺 be compact topological 𝑇1 -spaces and let 𝜔 : 𝑋 → 𝐺 and 𝜏 : 𝑋 → 𝑇 be continuous maps. Assume that 𝐺 is profinite and 𝜏 is a closed map. Then, the map 𝜃 : 𝑋 → 𝐺 × 𝑇, defined by 𝜃 (𝑥) = (𝜔(𝑥), 𝜏(𝑥)), is closed. Proof Let 𝐶 be a closed subset of 𝑋 and let (𝑔, 𝑡) ∈ 𝐺 × 𝑇 such that (𝑔, 𝑡) ∉ 𝜃 (𝐶). Then, 𝜔−1 (𝑔) ∩ 𝜏 −1 (𝑡) ∩ 𝐶 = ∅. We have to find an open neighborhood of (𝑔, 𝑡) in 𝐺 × 𝑇 disjoint from 𝜃 (𝐶).

8

1 Topologies

Indeed, since 𝐺 is a 𝑇1 -space, {𝑔} is closed in 𝐺. By Lemma 1.1.15, there is a Ñ family {𝑈𝑖 }𝑖 ∈𝐼 of open-closed neighborhoods of 𝑔 in 𝐺 such that 𝑖 ∈𝐼 𝑈𝑖 = {𝑔}. Without loss, we may assume that the family {𝑈𝑖 }𝑖 ∈𝐼 is closed under finite intersections. Then, by the preceding paragraph, Ù
Ù 𝑈𝑖 ∩ 𝜏 −1 (𝑡) ∩ 𝐶 = 𝜔−1 (𝑔) ∩ 𝜏 −1 (𝑡) ∩ 𝐶 = ∅. 𝜔−1 (𝑈𝑖 ) ∩ 𝜏 −1 (𝑡) ∩ 𝐶 = 𝜔−1 𝑖 ∈𝐼

𝑖 ∈𝐼

𝜏 −1 (𝑡)

Since 𝑇 is a 𝑇1 -space, is a closed subset of 𝑋. Thus, the left-hand side is an intersection of closed sets in 𝑋. Since 𝑋 is compact, there is an 𝑖 ∈ 𝐼 such that 𝜔−1 (𝑈𝑖 ) ∩ 𝜏 −1 (𝑡) ∩ 𝐶 = ∅ (Lemma 1.1.1). Set 𝑋0 = 𝑋 r (𝜔−1 (𝑈𝑖 ) ∩ 𝐶). Then, 𝑋0 is open in 𝑋 and 𝜏 −1 (𝑡) ⊆ 𝑋0 . By Lemma 1.1.18, 𝑉 = {𝑠 ∈ 𝑇 | 𝜏 −1 (𝑠) ⊆ 𝑋0 } is open in 𝑇. Clearly 𝑡 ∈ 𝑉 and 𝜏 −1 (𝑉) ⊆ 𝑋0 = 𝑋 r (𝜔−1 (𝑈𝑖 ) ∩ 𝐶). Thus, 𝑈𝑖 × 𝑉 is an open neighborhood of (𝑔, 𝑡) in 𝐺 × 𝑇, and 𝜃 −1 (𝑈𝑖 × 𝑉) ∩ 𝐶 = 𝜔−1 (𝑈𝑖 ) ∩ 𝜏 −1 (𝑉) ∩ 𝐶 = ∅. It follows that (𝑈𝑖 × 𝑉) ∩ 𝜃 (𝐶) = ∅, as desired.



1.2 Strict and Étale Topologies Let 𝐺 be a profinite group. We denote the set of all closed (resp. open, open normal) subgroups of 𝐺 by Subgr(𝐺) (resp. Open(𝐺), OpenNormal(𝐺)). We introduce two topologies on Subgr(𝐺) called the “strict topology” and the “étale topology” and relate them to each other. A basis for the strict topology of Subgr(𝐺) is the collection of all sets (1.3)

𝜈(Δ, 𝑁) = {Γ ∈ Subgr(𝐺) | Γ𝑁 = Δ𝑁 },

with Δ ∈ Subgr(𝐺) and 𝑁 ∈ OpenNormal(𝐺). When 𝐺 is finite, Subgr(𝐺) is finite and the strict topology on Subgr(𝐺) is the discrete topology. In general, Subgr(𝐺)  lim Subgr(𝐺/𝑁) with 𝑁 ranging over all ←−− open normal subgroups of 𝐺. Thus, Subgr(𝐺) is a profinite space under the strict topology. In particular, Subgr(𝐺) is a compact Hausdorff space (Definition 1.1.6). We use the adverb “strictly” as a substitute for the clause “in the strict topology”. For example, given a subset G of Subgr(𝐺), we say that G is strictly open (resp. closed, compact, Hausdorff, etc.) if it is open (resp. closed, compact, Hausdorff, etc.) in the strict topology. Likewise, we say that a function 𝑓 from a topological space 𝑋 into Subgr(𝐺) is strictly continuous if 𝑓 is continuous with respect to the strict topology on Subgr(𝐺). In particular, if 𝑋 = Subgr(𝐻) for a profinite group 𝐻, then we assume in this context that 𝑓 is continuous with respect to the strict topologies of both Subgr(𝐻) and Subgr(𝐺). A basis for the étale topology on Subgr(𝐺) is {Subgr(𝑈)}𝑈 ∈Open(𝐺) . Note that an open subgroup 𝑈 of 𝐺 is also closed [FrJ08, p. 6, Remark 1.2.1(a)], hence Subgr(𝑈) is a subset of Subgr(𝐺). As above, for a subset G of Subgr(𝐺) we say that G is étale open (closed, compact, Hausdorff, etc.) if G is open (closed, compact, Hausdorff, etc.) in the étale topology. Likewise we say that a function from a topological space

1.2 Strict and Étale Topologies

9

𝑋 into Subgr(𝐺) is étale continuous if 𝑓 is continuous when Subgr(𝐺) is equipped with the étale topology. In particular, if 𝑋 = Subgr(𝐻) for a profinite group 𝐻, then we assume in this context that 𝑓 is continuous with respect to the étale topologies of both Subgr(𝐻) and Subgr(𝐺). Remark 1.2.1 (Maps in the strict and the étale topologies) (a) Homomorphism: Let 𝜑 : 𝐻 → 𝐺 be a homomorphism of profinite groups. In particular, 𝜑 maps closed subgroups of 𝐻 onto closed subgroups of 𝐺 [FrJ08, p. 6, Remark 1.2.1(e)]. Hence, 𝜑 induces a map 𝜑∗ : Subgr(𝐻) → Subgr(𝐺). If 𝑈 is an open subgroup of 𝐺, then 𝑉 = 𝜑−1 (𝑈) is an open subgroup of 𝐻 and 𝜑−1 ∗ (Subgr(𝑈)) = Subgr(𝑉). It follows that 𝜑∗ is étale continuous. If 𝜑 is an epimorphism, Δ ∈ Subgr(𝐺), and 𝑁 an open normal subgroup of 𝐺, then 𝜑−1 (𝑁) is an open normal subgroup of 𝐻 and 𝜑−1 (𝜈(Δ, 𝑁)) = 𝜈(𝜑−1 (Δ), 𝜑−1 (𝑁)). It follows that 𝜑∗ is strictly continuous. (b) Inclusion: Let 𝐻 be a closed subgroup of 𝐺. By (a), the inclusion Subgr(𝐻) → Subgr(𝐺) is étale continuous. Moreover, a subgroup 𝑉 of 𝐻 is open in 𝐻 if and only if 𝑉 = 𝐻 ∩ 𝑈 for some 𝑈 ∈ Open(𝐺) [FrJ08, p. 8, Lemma 1.2.5(b)]. Clearly, Subgr(𝑉) = Subgr(𝐻) ∩ Subgr(𝑈). Thus, the étale topology of Subgr(𝐻) is the one induced from the étale topology of Subgr(𝐺). (c) Restriction: Let 𝐻 be a closed subgroup of 𝐺. Then, the restriction 𝑟 : Subgr(𝐺) → Subgr(𝐻), given by Γ ↦→ Γ ∩ 𝐻, is étale continuous. Indeed, let Γ ∈ Subgr(𝐺) and let 𝑉 be an open subgroup of 𝐻 such that 𝑟 (Γ) = Γ ∩ 𝐻 ∈ Subgr(𝑉). By Lemma 1.1.14 and Lemma 1.1.1, there is an open subgroup 𝑈 of 𝐺 containing Γ such that 𝑈 ∩ 𝐻 ≤ 𝑉. Thus, 𝑟 maps the neighborhood Subgr(𝑈) of Γ into the neighborhood Subgr(𝑉) of Γ ∩ 𝐻. In particular, by Fact 1.1.3(d), if G is an étale compact subset of Subgr(𝐺), then 𝑟 (G) = {Γ ∩ 𝐻 | Γ ∈ G} is étale compact in Subgr(𝐻). Lemma 1.2.2 Let 𝐻 be an open (closed) subgroup of a profinite group 𝐺. Then, Subgr(𝐻) is strictly open (closed) in Subgr(𝐺). Thus, the strict topology of Subgr(𝐺) is finer than the étale topology of Subgr(𝐺). Proof First let 𝐻 be open in 𝐺. We choose an open normal subgroup 𝑁 of 𝐺 contained in 𝐻 and let 𝐻1 , . . . , 𝐻𝑚 , . . . , 𝐻𝑛 be the subgroups between 𝑁 and 𝐺, Ð𝑚 where 𝐻1 , . . . , 𝐻𝑚 are those that are contained in 𝐻. Then, Subgr(𝐻) = 𝑖=1 {Γ ∈ Subgr(𝐺) Ð𝑛 | Γ𝑁 = 𝐻𝑖 𝑁 = 𝐻𝑖 }. Hence, Subgr(𝐻) is strictly open. But V = 𝑖=𝑚+1 {Γ ∈ Subgr(𝐺) | Γ𝑁 = 𝐻𝑖 } is also strictly open. Since Subgr(𝐺) = Subgr(𝐻) ∪· V, it follows that Subgr(𝐻) is strictly closed as well. Now let 𝐻 be closed in 𝐺. By Lemma 1.1.14, Ù Ù 𝐻= 𝑈, hence Subgr(𝐻) = Subgr(𝑈). 𝑈 ≥𝐻 𝑈 ∈Open(𝐺)

𝑈 ≥𝐻 𝑈 ∈Open 𝐺

Hence, by the preceding paragraph, applied to each of the Subgr(𝑈)’s, the set Subgr(𝐻) is strictly closed in Subgr(𝐺).  Lemma 1.2.3 Let 𝐺 be a profinite group, let 𝐶 be a subset of 𝐺, and let G be an étale compact subset of Subgr(𝐺).

10

1 Topologies

(a) If 𝐶 6 ⊆ Γ for every Γ ∈ G, then there is an 𝑁 ∈ OpenNormal(𝐺) such that 𝐶 6 ⊆ Γ𝑁 for every Γ ∈ G. Equivalently, (b) if for every 𝑁 ∈ OpenNormal(𝐺) there is an Γ ∈ G such that 𝐶 ⊆ Γ𝑁, then there is an Γ ∈ G such that 𝐶 ⊆ Γ. Proof Assume that 𝐶 * Γ for every Γ ∈ G. For every Γ ∈ G there is an 𝑁Γ ∈ OpenNormal(𝐺) such that 𝐶 6⊆ Γ𝑁Γ (Lemma 1.1.14). Thus, Ð G ⊆ Γ∈ G Subgr(Γ𝑁Γ ). By compactness, there are Γ1 , . . . , Γ𝑛 ∈ G such that Ð𝑛 Ñ𝑛 G ⊆ 𝑖=1 Subgr(Γ𝑖 𝑁Γ𝑖 ). Then, 𝑁 = 𝑖=1 𝑁Γ𝑖 has the required property. Indeed, for every Γ ∈ G there is an 𝑖 such that Γ ≤ Γ𝑖 𝑁Γ𝑖 . Since 𝐶 6 ⊆ Γ𝑖 𝑁Γ𝑖 , also 𝐶 6 ⊆ Γ𝑁.  Notation 1.2.4 For a profinite group 𝐺 and a subset G of Subgr(𝐺) we write Ð 𝑈 (G) = Γ∈ G Γ. Lemma 1.2.5 ([Gim95, Lemma 1.4]) Let 𝐺 be a profinite group and let G be an étale compact subset of Subgr(𝐺). Then, 𝑈 (G) is closed in 𝐺. Proof Let 𝑥 ∈ 𝐺 be in the closure of 𝑈 (G). Then, for every open normal subgroup 𝑁 of 𝐺 there is a Γ ∈ G such that 𝑥𝑁 ∩ Γ ≠ ∅, that is, 𝑥 ∈ Γ𝑁. Hence, by Lemma 1.2.3(b), there is a Γ ∈ G such that 𝑥 ∈ Γ ⊆ 𝑈 (G). Thus, 𝑈 (G) is closed.  Remark 1.2.6 Since 1 is contained in every étale open subset of Subgr(𝐺), a subset G of Subgr(𝐺) is étale compact if and only if G r {1} is. Remark 1.2.7 (The étale topology versus the strict topology) Since Subgr(𝐺) is strictly profinite, Lemma 1.2.2 and Lemma 1.2.5 give the following chain of implications for a subset G of Subgr(𝐺): G is étale closed =⇒ G is strictly closed ⇐⇒ G is strictly compact =⇒ G is étale compact =⇒ 𝑈 (G) is a closed subset of 𝐺. Denote the strict closure of G in Subgr(𝐺) by StrictClosure(G) and the étale ´ closure of G by EtaleClosure(G). Lemma 1.2.8 Let 𝐻 be a closed subgroup of 𝐺. Then, ´ EtaleClosure({𝐻}) = {𝐵 ∈ Subgr(𝐺) | 𝐻 ≤ 𝐵}. ´ Proof First suppose that 𝐵 ∈ EtaleClosure({𝐻}). Then, 𝐻 belongs to every étale open neighborhood of 𝐵 in Subgr(𝐺). In particular, if 𝐵 ≤ 𝑈 ∈ Open(𝐺), then 𝐻 ≤ 𝑈. By Lemma 1.1.14, 𝐻 ≤ 𝐵. Conversely, suppose that 𝐻 ≤ 𝐵. Then, 𝐻 belongs to each of the sets Subgr(𝑈) with 𝐵 ≤ 𝑈 ∈ Open(𝐺). Thus, 𝐻 belongs to every basic étale open neighborhood ´ of 𝐵. Therefore, 𝐵 ∈ EtaleClosure({𝐻}).  Lemma 1.2.9 Let 𝜑 : 𝐺 → 𝐻 be an epimorphism of profinite groups and let 𝐺 0 be a closed subgroup of 𝐺. Then, the set B = {𝐵 ∈ Subgr(𝐺) | 𝜑(𝐺 0 ) ≤ 𝜑(𝐵)} is étale closed.

1.3 Étale Hausdorff Sets

11

Proof By Lemma 1.2.8, the subset C = {𝐶 ∈ Subgr(𝐻) | 𝜑(𝐺 0 ) ≤ 𝐶} of Subgr(𝐻) is étale closed. By Remark 1.2.1(a), the map 𝜑∗ : Subgr(𝐺) → Subgr(𝐻) induced by 𝜑 : 𝐺 → 𝐻 is étale continuous. Hence, B = 𝜑−1 ∗ (C) is étale closed, as claimed. Lemma 1.2.10 Let 𝐺 be a profinite group and G a subset of Subgr(𝐺). Let {𝑁𝑖 }𝑖 ∈𝐼 be a collection of open subgroups of 𝐺, closed under finite intersections, such that Ñ 𝑖 ∈𝐼 𝑁𝑖 = 1. Then, G is étale compact if and only if G r Subgr(𝑁𝑖 ) is étale compact for every 𝑖 ∈ 𝐼. Proof First assume that G is étale compact. Since Subgr(𝑁𝑖 ) is étale open in Subgr(𝐺), the set G r Subgr(𝑁𝑖 ) is étale closed in G. Hence, by Fact 1.1.3(a), G r Subgr(𝑁𝑖 ) is étale compact. Conversely, assume that G r Subgr(𝑁𝑖 ) is étale compact for every 𝑖 ∈ 𝐼. Consider an étale open covering {Subgr(𝐻 𝑗 )} 𝑗 ∈𝐽 of G. Let 𝑗 0 ∈ 𝐽. Then, there is an 𝑖 such that 𝑁𝑖 ≤ 𝐻 𝑗0 (Lemma 1.1.12(a)). Since G r Subgr(𝑁 Ð 𝑖 ) is étale compact, there is a finite subset 𝐽 0 of 𝐽 such that G r Subgr(𝑁𝑖 ) ⊆ 𝑗 ∈𝐽 0 Subgr(𝐻 𝑗 ). Without loss of generality, 𝑗0 ∈ 𝐽 0. If Γ ∈ G ∩ Subgr(𝑁𝑖 ), then Γ ∈ Subgr(𝐻 𝑗0 ). We conclude that Ð G ⊆ 𝑗 ∈𝐽 0 Subgr(𝐻 𝑗 ). Therefore, G is étale compact, as claimed. 

1.3 Étale Hausdorff Sets The intersection of two étale open basic sets contains the set {1}. Hence, if 𝐺 ≠ 1, the étale topology of Subgr(𝐺) is not Hausdorff. However, a subset G of Subgr(𝐺) can be étale Hausdorff. Indeed, we will be looking for such G’s which are even étale profinite. The latter means that G is étale compact, étale Hausdorff, and has a base for its étale topology that consists of étale open-closed sets. Remark 1.3.1 Let Δ ∈ Subgr(𝐺) and let 𝑁 ∈ OpenNormal(𝐺). Note that if 1 ∈ 𝜈(Δ, 𝑁), then 𝜈(Δ, 𝑁) = Subgr(𝑁) (notation (1.3)). Therefore, the following conditions are equivalent: (a) 1 ∉ StrictClosure(G), (b) 𝐺 has an open normal subgroup 𝑁 that contains no Γ ∈ G, (c) 𝐺 has an open subgroup 𝐻 that contains no Γ ∈ G, ´ (d) 1 ∉ EtaleClosure(G). Lemma 1.3.2 Let G be an étale Hausdorff subset of Subgr(𝐺) with card(G) ≥ 2. Then, G satisfies the equivalent conditions of Remark 1.3.1. Proof Let Γ1 , Γ2 ∈ G be distinct. Then,
there are 𝑈1 , 𝑈2 ∈
Open(𝐺) such that Γ1 ≤ 𝑈1 , Γ2 ≤ 𝑈2 , and G ∩ Subgr(𝑈1 ) ∩ G ∩ Subgr(𝑈2 ) = ∅. In particular, 𝐻 = 𝑈1 ∩ 𝑈2 satisfies G ∩ Subgr(𝐻) = ∅, which is Condition (c) of Remark 1.3.1. Definition 1.3.3 Let 𝐺 be a profinite group. We say that a subset G of Subgr(𝐺) is stellate if Γ1 ∩ Γ2 = 1 for all distinct Γ1 , Γ2 ∈ G. Lemma 1.3.4 Let G be a stellate subset of Subgr(𝐺) that satisfies the equivalent conditions of Remark 1.3.1. Then, G is étale Hausdorff.

12

1 Topologies

Proof Let Γ1 , Γ2 ∈ G be distinct. Put U𝑖 = {𝑈 ∈ Open(𝐺) | Γ𝑖 ≤ 𝑈}, for 𝑖 = 1, 2. Ñ Let 𝐻 be as in Condition (c) of Remark 1.3.1. Since 𝑈 ∈U𝑖 𝑈 = Γ𝑖 for 𝑖 = 1, 2, we Ñ have 𝑈1 ∈U1 , 𝑈2 ∈U2 𝑈1 ∩𝑈2 = Γ1 ∩ Γ2 = 1 ≤ 𝐻. By the compactness of 𝐺, there are 𝑈1 ∈ U1 and 𝑈2 ∈ U2 such that 𝑈1 ∩ 𝑈2 ≤ 𝐻 (Lemma 1.1.1). Thus, G ∩ Subgr(𝑈1 ) and G ∩ Subgr(𝑈2 ) are étale open neighborhoods of Γ1 and Γ2 in G, respectively. Their intersection is empty, because G ∩ Subgr(𝑈1 ∩ 𝑈2 ) ⊆ G ∩ Subgr(𝐻) = ∅.  Lemma 1.3.5 Let G be an étale compact subset of Subgr(𝐺). Then, every Γ ∈ G is contained in a maximal element of G. Proof By Zorn’s lemma, it suffices to prove that every ascending chain G0 in G is bounded by an element of G. To this end consider Γ1 , . . . , Γ𝑛 ∈ G0 . Any two Γ𝑖 ’s are comparable. Hence, we may assume that Γ1 Ñ ≤ Γ2 ≤ · · · ≤ Γ𝑛 . By Lemma 1.2.8, Ñ𝑛 ´ 𝑛 ´ Γ𝑛 ∈ 𝑖=1 EtaleClosure({Γ𝑖 }). Therefore, G ∩ 𝑖=1 EtaleClosure({Γ𝑖 }) ≠ ∅. Ñ ´ Since G is étale compact, G ∩ Γ∈ G0 EtaleClosure({Γ}) is non-empty. Again, by Lemma 1.2.8, every element of the latter set is an element of G that bounds G0 .  If a subset G of Subgr(𝐺) contains groups Γ and Δ with Γ < Δ, then G is not étale Hausdorff. Thus, removing all non-maximal elements from G is the only way to make G étale Hausdorff while preserving the essential information stored in G. We denote the set of all maximal elements of G by Gmax . Lemma 1.3.6 Let G be an étale compact subset of Subgr(𝐺). Then: (a) Gmax is also étale compact. (b) If G is 𝐺-invariant, then Gmax is 𝐺-invariant. Proof Proof of (a). Let 𝔘 be an open étale covering of Gmax . We may assume without loss that 𝔘 consists of basic étale open subsets of Subgr(𝐺). Thus, Ð 𝔘 = (Subgr(𝑈𝑖 ))𝑖 ∈𝐼 with 𝑈𝑖 ∈ Open(𝐺) for 𝑖 ∈ 𝐼 such that Gmax ⊆ 𝑖 ∈𝐼 Subgr(𝑈𝑖 ). Every Γ ∈ G is contained in a group Γ0 that belongs to Gmax . Hence, Ð G ⊆ Ð𝑖 ∈𝐼 Subgr(𝑈𝑖 ). Since G is étale compact, 𝐼Ðhas a finite subset 𝐼0 such that G ⊆ 𝑖 ∈𝐼0 Subgr(𝑈𝑖 ). Therefore, Gmax ⊆ G ⊆ 𝑖 ∈𝐼0 Subgr(𝑈𝑖 ). It follows that Gmax is étale compact. Proof of (b). Let Γ ∈ Gmax and let 𝜎 ∈ 𝐺. By assumption, Γ 𝜎 ∈ G, hence by −1 Lemma 1.3.5 there is a Γ0 ∈ Gmax such that Γ 𝜎 ≤ Γ0. Then, Γ ≤ (Γ0) 𝜎 ∈ G. Since −1 0 𝜎 𝜎 0 𝜎 Γ is maximal, we have Γ = (Γ ) , and hence Γ = Γ . Thus, Γ ∈ Gmax .  Here is a case in which the strict topology of a subset G of Subgr(𝐺) coincides with the étale topology of G. Lemma 1.3.7 ([HJP09, Lemma 2.1]) Let 𝐺 be a profinite group and G a strictly closed subset of Subgr(𝐺) that satisfies G = Gmax . Then, the étale topology of G coincides with its strict topology. Proof Since the strict topology of Subgr(𝐺) is finer than its étale topology, it suffices to prove for every Γ0 ∈ G that every basic strictly open neighborhood N of Γ0 in G contains an étale open neighborhood of Γ0 .

1.4 The Envelope of a Set of Closed Subgroups

13

We may assume that N = {Γ ∈ G | Γ𝑁 = Γ0 𝑁 } for some open normal subgroup 𝑁 of 𝐺. Let H be the set of all open subgroups 𝐻 of 𝐺 with Γ0 ≤ 𝐻. It suffices to find 𝐻 ∈ H such that Subgr(𝐻) ∩ G ⊆ N . Since N 0 = {Γ ∈ G | Γ𝑁 ≠ Γ0 𝑁 } is strictly closed in G, it is strictly closed in Subgr(𝐺). Every Γ ∈ N 0 satisfies Γ ≠ Γ0 , and hence Γ 6 ≤ Γ0 , because G = Gmax . Thus,Ñthere is an 𝐻Γ ∈ H such that Γ 6 ≤ 𝐻Γ . Otherwise, by 1.1.15, we would have Γ ≤ 𝐻 ∈H 𝐻 = Γ0 , which is a contradiction. Ñ Thus, Γ∈N0 Subgr(𝐻Γ ) ∩ N 0 = ∅. The sets Subgr(𝐻Γ ) ∩ N 0 are strictly closed in Subgr(𝐺). Hence, by the compactness, there are 𝐻1 , . . . , 𝐻𝑛 ∈ H such that Ñ𝑛 Ñ𝑛 0 0 𝑖=1 Subgr(𝐻𝑖 ) ∩ N = ∅. Let 𝐻 = 𝑖=1 𝐻𝑖 , then Subgr(𝐻) ∩ N = ∅. Since 0 N = G r N , we have Subgr(𝐻) ∩ G ⊆ N , as desired.  Example 1.3.8 Let 𝐺 be a profinite group and 𝐺 0 a closed subgroup of 𝐺. The 𝑔 map 𝐺 → Subgr(𝐺) given by 𝑔 ↦→ 𝐺 0 is strictly continuous, hence its image 𝑔 G = {𝐺 0 | 𝑔 ∈ 𝐺} is strictly closed, by Fact 1.1.3(e). Since Gmax = G, the étale topology on G coincides by Lemma 1.3.7 with its strict topology. In particular, since a strictly closed subset of Subgr(𝐺) is a profinite space, G is étale profinite.

1.4 The Envelope of a Set of Closed Subgroups Let 𝐺 be a profinite group. The envelope, Env(G), of a subset G of Subgr(𝐺) is the set of all Γ0 ∈ Subgr(𝐺) which are contained in some Γ ∈ G. We use Env(G) to relate the strict topology and the étale topology of Subgr(𝐺) to each other. Lemma 1.4.1 A subset G of Subgr(𝐺) is étale compact if and only if Env(G) is strictly closed. Proof First suppose that G is étale compact. Let Δ ∈ Subgr(𝐺) r Env(G). Then, Δ 6 ≤ Γ for every Γ ∈ G. Hence, by Lemma 1.2.3, there is an 𝑁 ∈ OpenNormal(𝐺) with Δ ∉ Subgr(Γ𝑁) for every Γ ∈ G. Put Ø U= Subgr(Γ𝑁). Γ∈ G

Then, Env(G) ⊆ U, but Δ ∉ U. Moreover, if (1.3)

Δ0 ∈ 𝜈(Δ, 𝑁) = {Δ0 ∈ Subgr(𝐺) | Δ0 𝑁 = Δ𝑁 }, then Δ0 ∉ U as well. Thus, the strictly open neighborhood 𝜈(Δ, 𝑁) of Δ is disjoint from U and hence also from Env(G). It follows that Env(G) is strictly closed. Conversely, suppose that Env(G) is strictly closed, hence strictly Ð compact (Remark 1.2.7). Let 𝑈𝑖 , 𝑖 ∈ 𝐼, be open subgroups of 𝐺 with G ⊆ 𝑖 ∈𝐼 Subgr(𝑈𝑖 ). Ð Then, by definition, Env(G) ⊆ 𝑖 ∈𝐼 Subgr(𝑈𝑖 ). Since eachÐof the sets Subgr(𝑈𝑖 ) is strictly Ð open, 𝐼 has a finite subset 𝐼0 with Env(G) ⊆ 𝑖 ∈𝐼0 Subgr(𝑈𝑖 ). Thus, G ⊆ 𝑖 ∈𝐼0 Subgr(𝑈𝑖 ). Therefore, G is étale compact.  Corollary 1.4.2 Let G be an étale compact subset of Subgr(𝐺). Then, StrictClosure(G) is contained in Env(G).

14

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Let G be a subset of Subgr(𝐺) and 𝐻 a closed subgroup of 𝐺. We set G 𝐻 = {Γℎ | Γ ∈ G, ℎ ∈ 𝐻} and write Con(G) for the set of all Γ0 ∈ Subgr(𝐺) that are contained in Γ𝑔 for some Γ ∈ G and 𝑔 ∈ 𝐺. Thus, Con(G) = Env(G 𝐺 ) = Env(G) 𝐺 . Corollary 1.4.3 Let G be an étale compact subset of Subgr(𝐺). Then, the sets G 𝐺 , Env(G), and Con(G) are étale compact. Moreover, both Env(G) and Con(G) are also strictly compact. Finally, if G is strictly compact, then so is G 𝐺 . Proof The set G 𝐺 is the image in Subgr(𝐺) of the compact space G × 𝐺 under the continuous map (Γ, 𝑔) ↦→ Γ𝑔 . Hence, by Fact 1.1.3(d), G 𝐺 is étale compact. Likewise, if G is strictly compact, then so is G 𝐺 . By Lemma 1.4.1, Env(G) is strictly closed, hence both strictly compact and étale compact, by Remark 1.2.7. Therefore, by the preceding paragraph, Con(G) = Env(G) 𝐺 is both strictly compact and étale compact.  Corollary 1.4.4 Let 𝐻1 , . . . , 𝐻𝑛 be closed subgroups of a profinite group 𝐺. Then, Con({𝐻1 , . . . , 𝐻𝑛 }) is strictly closed in Subgr(𝐺). Proof By definition, Con({𝐻1 , . . . , 𝐻𝑛 }) =

𝑛 Ø

Subgr(𝐻𝑖 )


𝐺

.

𝑖=1

Ð𝑛 By Lemma 1.2.2, 𝑖=1 Subgr(𝐻𝑖 ) is strictly closed. Hence, by Remark 1.2.7 and Corollary 1.4.3, Con({𝐻1 , . . . , 𝐻𝑛 }) is strictly closed.  Lemma 1.4.5 Let 𝐺 be a profinite group and G a stellate étale compact subset of Subgr(𝐺). Let G 0 ⊆ G. Then, G 0 is étale compact if and only if 𝑈 (G 0) is closed. Proof We may assume without loss that G 0 is nonempty. Assume that 𝑈 (G 0) is closed. By Lemma 1.4.1 we have to show that Env(G 0) is strictly closed, that is, that every Δ ∈ Subgr(𝐺) r Env(G 0) has a strictly open neighborhood disjoint from Env(G 0). Since G 0 ≠ ∅, we have 1 ∈ Env(G 0), hence we have Δ ≠ 1. By definition, Env(G 0) ⊆ Env(G). Hence, if Δ ∉ Env(G), then, as Env(G) is strictly closed (again by Lemma 1.4.1), Δ has a strictly open neighborhood disjoint from Env(G) and hence also from Env(G 0). Otherwise, Δ ∈ Env(G), so there is a Γ ∈ G such that Δ ≤ Γ. Since Δ ∉ Env(G 0), this implies that Γ ∉ G 0. Also, the assumption that G is stellate implies that (1.4)

Γ0 ∩ Δ = 1 for every Γ0 ∈ G 0 .

Let 1 ≠ 𝑔 ∈ Δ. By (1.4), 𝑔 ∉ Γ0 for all Γ0 ∈ G 0. Hence, 𝑔 ∉ 𝑈 (G 0). Since 𝑈 (G 0) is closed, there is an open 𝑁 ⊳ 𝐺 such that 𝑔𝑁 ∩ 𝑈 (G 0) = ∅. Now, if Δ0 ≤ 𝐺 satisfies Δ0 𝑁 = Δ𝑁, then there is a 𝑔 0 ∈ Δ0 such that 𝑔 0 ∈ 𝑔𝑁. Hence 𝑔 0 ∉ 𝑈 (G 0), whence Δ0 ∉ Env(G 0). We conclude that the strictly open neighborhood 𝜈(Δ, 𝑁) (notation (1.3)) of Δ is disjoint from Env(G 0). The converse is Lemma 1.2.5. 

1.5 Cantor Space

15

1.5 Cantor Space Recall that the Cantor middle-third set is the set ℭ of all real numbers 𝑥 in the unit interval [0, 1] that, when expressed in base 3, have no 1’s in their ternary expansion. Specifically, let       1 2 1 2 7 8 𝐽1 = , and 𝐽2 = , ∪ , . 3 3 9 9 9 9 For an arbitrary integer 𝑛 ≥ 2 let 𝐽𝑛 be the Ð union of the 2𝑛−1 intervals of the form 𝑛−1 1+3𝑘 2+3𝑘 in [0, 1] r 𝑖=1 𝐽𝑖 . The Cantor middle-third set is 3𝑛 , 3𝑛 ) which are contained Ð∞ defined as ℭ = [0, 1] r 𝑛=1 𝐽𝑛 . This is the set of all points in [0, 1] possessing at Í 𝑎𝑛 least one triadic expansion ∞ 𝑛=1 3𝑛 with 𝑎 𝑛 ∈ {0, 2} for all 𝑛 [HoY61, p. 97, first paragraph of Sec. 2-15]. By definition, ℭ is closed in [0, 1], hence compact under the topology induced on it by the usual Euclidean topology of [0, 1]. Lemma 1.5.1 The Cantor middle-third set ℭ is homeomorphic to the profinite space {0, 2}N . In particular, ℭ is a profinite space. Proof We follow [Sin13, p. 73, Example 3.2.2].
Í 𝑥𝑛 Consider the function 𝑓 : {0, 2}N → ℭ defined by 𝑓 (𝑥 𝑛 )𝑛∈N = ∞ 𝑛=1 3𝑛 . In addition to being surjective, 𝑓 is also injective. Indeed, assume by contradiction that ∞ ∞ Õ Õ 𝑎𝑛 𝑏𝑛 = , 𝑛 3 3𝑛 𝑛=𝑚 𝑛=𝑚+1 Í 𝑏𝑛 −𝑎𝑛 with 𝑚 ≥ 1 and 𝑎 𝑛 , 𝑏 𝑛 ∈ {0, 2} for all 𝑛 and 𝑎 𝑚 = 2. Then, 32𝑚 = ∞ 𝑛=𝑚+1 3𝑛 . Hence, ∞ ∞ Õ Õ |𝑎 𝑛 − 𝑏 𝑛 | 2 1 2 ≤ ≤ = 𝑚, 𝑚 𝑛 𝑛 3 3 3 3 𝑛=𝑚+1 𝑛=𝑚+1 which is a contradiction. Next we prove that 𝑓 is continuous. To this end we consider x ∈ {0, 2}N and 𝜀 > 0. We choose a positive integer 𝑚 with 3−𝑚 < 𝜀. Then, 𝑈 = {(𝑦 𝑛 )𝑛∈N ∈ {0, 2}N | 𝑦 𝑖 = 𝑥𝑖 for 𝑖 = 1, . . . , 𝑚} is an open neighborhood of x in {0, 2}N and | 𝑓 (y) − 𝑓 (x)| < 𝜀 for all y ∈ 𝑈. As a subset of the Hausdorff space [0, 1], ℭ is also Hausdorff. By Remark 1.1.7(b), {0, 2}N is a profinite space, in particular Hausdorff compact. Hence, by Fact 1.1.3(e), 𝑓 is a homeomorphism. Therefore, ℭ is a profinite space, as claimed. Having proved that ℭ is a profinite space, we call ℭ from now on the Cantor space. Every topological space which is homeomorphic to ℭ, alternatively to {0, 2}N , is said to be a Cantor space. Recall that a point 𝑥 of a topological space 𝑋 is isolated if {𝑥} is an open subset of 𝑋. Í 𝑎𝑛 We claim that ℭ has no isolated points. Indeed, let 𝑥 = ∞ 𝑛=1 3𝑛 with 𝑎 𝑛 ∈ {0, 2} be an element of ℭ. Given a positive integer 𝑚, we consider the element 𝑥 0 ∈ ℭ obtained from 𝑥 by replacing 𝑎 𝑚 by 2 − 𝑎 𝑚 . Then, 𝑥 0 ∈ ℭ and |𝑥 − 𝑥 0 | = 32𝑚 . Since this holds for each 𝑚, the point 𝑥 is not isolated in ℭ.

16

1 Topologies

Note that the family 𝐵 of all open intervals (𝑎, 𝑏), with 𝑎, 𝑏 ∈ Q such that 0 ≤ 𝑎, 𝑏 ≤ 1, is an infinite countable base for the topology of [0, 1]. Hence, the family {(𝑎, 𝑏) ∩ ℭ | (𝑎, 𝑏) ∈ 𝐵} is an infinite countable base for the topology of ℭ. It turns out that these two properties of the profinite space ℭ determine it up to homeomorphism. This depends on the following observation: Lemma 1.5.2 Let 𝑋 be a profinite space with an infinite countable base U for the topology T of 𝑋. Then, the family OC of all open-closed subsets of 𝑋 is an infinite countable base for T . Proof By Definition 1.1.6, OC is a base for T . Every 𝑉 ∈ OC is open in 𝑋, so Ð 𝑉 = 𝑖 ∈𝐼 𝑈𝑖 with 𝑈𝑖 ∈ U for some subset 𝐼 of N. Since 𝑉 is also closed in 𝑋 and 𝑋 is compact, we may take 𝐼 to be finite. Since the collections of finite subsets of N is infinite countable, so is OC, as claimed.  Lemma 1.5.3 Let 𝑋 be a profinite space without isolated points that has an infinite countable base B = {𝑉1 , 𝑉2 , 𝑉3 , . . .} for its topology, Then, 𝑋 is homeomorphic to {0, 2}N . Thus, 𝑋 is a Cantor space. Proof By Lemma 1.5.2, we may assume that the elements of B are open-closed in 𝑋. For every 𝑛 ≥ 0 let 𝑌𝑛 = {0, 2} {1,...,𝑛} . We construct a partition P𝑛 = {𝑈 𝑓 | 𝑓 ∈ 𝑌𝑛 } of 𝑋 into 2𝑛 disjoint non-empty open-closed sets 𝑈 𝑓 such that: (1.5a) P𝑛 is finer than P𝑛−1 ; more precisely, if 𝑔, ℎ ∈ 𝑌𝑛 are the distinct extensions of 𝑓 ∈ 𝑌𝑛−1 , then 𝑈 𝑓 = 𝑈𝑔 ∪· 𝑈ℎ . (1.5b) For every positive integer 𝑛 and every 𝑓 ∈ 𝑌𝑛 either 𝑈 𝑓 ⊆ 𝑉𝑛 or 𝑈 𝑓 ⊆ 𝑋 r 𝑉𝑛 . The construction is by induction on 𝑛. For 𝑛 = 0 the set 𝑌0 = {0, 2} ∅ contains only one element, and we let P0 = {𝑋 }. The induction step: Suppose 𝑛 ≥ 1 and we have constructed P𝑛−1 . Let 𝑓 , 𝑔, ℎ be as in (1.5a). If either 𝑈 𝑓 ⊆ 𝑉𝑛 or 𝑈 𝑓 ⊆ 𝑋 r𝑉𝑛 , choose 𝑥 ∈ 𝑈 𝑓 . It is not isolated, hence there is a 𝑦 ∈ 𝑈 𝑓 such that 𝑥 ≠ 𝑦. Choose 𝑉 𝑗 such that 𝑥 ∈ 𝑉 𝑗 but 𝑦 ∉ 𝑉 𝑗 . Then, 𝑈𝑔 := 𝑈 𝑓 ∩ 𝑉 𝑗 and 𝑈ℎ := 𝑈 𝑓 r 𝑉 𝑗 are non-empty open-closed subsets of 𝑈 𝑓 and 𝑈 𝑓 = 𝑈𝑔 ∪· 𝑈ℎ . Since 𝑈𝑔 , 𝑈ℎ are contained in 𝑈 𝑓 , they are contained either in 𝑉𝑛 or in 𝑋 r 𝑉𝑛 . If 𝑈 𝑓 6 ⊆ 𝑉𝑛 and 𝑈 𝑓 6 ⊆ 𝑋 r 𝑉𝑛 , then 𝑈𝑔 := 𝑈 𝑓 ∩ 𝑉𝑛 and 𝑈ℎ := 𝑈 𝑓 r 𝑉𝑛 are non-empty open-closed subsets of 𝑈 𝑓 and 𝑈 𝑓 = 𝑈𝑔 ∪· 𝑈ℎ . By construction, 𝑈𝑔 ⊆ 𝑉𝑛 and 𝑈ℎ ⊆ 𝑋 r 𝑉𝑛 . This completes the induction. Now we set 𝑌 = {0, 2}N and for every 𝑓 ∈ 𝑌 and every 𝑛 ≥ 0 we denote the restriction of 𝑓 to 𝑌𝑛 by 𝑓𝑛 . Claim A: 𝑈 𝑓𝑛+1 ⊆ 𝑈 𝑓𝑛 for every 𝑛 ≥ 0. Indeed, by definition, 𝑓𝑛+1 extends 𝑓𝑛 . Hence, by (1.5a), 𝑈 𝑓𝑛+1 ⊆ 𝑈 𝑓𝑛 . Ñ Claim B: The set 𝐶 𝑓 := ∞ 𝑛=0 𝑈 𝑓𝑛 consists of one element. Indeed, 𝑈 𝑓0 ⊇ 𝑈 𝑓1 ⊇ 𝑈 𝑓2 ⊇ · · · are non-empty, hence this sequence has the finite intersection property. Since 𝑋 is compact and the 𝑈 𝑓𝑛 are closed, 𝐶 𝑓 ≠ ∅.

1.5 Cantor Space

17

Assume there are distinct 𝑥, 𝑦 ∈ 𝐶 𝑓 . For every 𝑛 we have 𝑥, 𝑦 ∈ 𝐶 𝑓 ⊆ 𝑈 𝑓𝑛 , hence, by (1.5b), either 𝑥, 𝑦 ∈ 𝑉𝑛 or 𝑥, 𝑦 ∈ 𝑋 r 𝑉𝑛 . But this is impossible, because B is a basis for a Hausdorff topology. Having proved Claim B, we define a map 𝜑 : 𝑌 → 𝑋 as follows: For each 𝑓 ∈ 𝑌 , 𝜑( 𝑓 ) is the unique point of 𝐶 𝑓 . Claim C: The map 𝜑 : 𝑌 → 𝑋 is injective. Indeed, if 𝑓 , 𝑔 ∈ 𝑌 are distinct, there is an 𝑛 such that the restrictions 𝑓𝑛 , 𝑔𝑛 ∈ 𝑌𝑛 are distinct. Thus, by (1.5a), 𝑈 𝑓𝑛 , 𝑈𝑔𝑛 ∈ P𝑛 are disjoint. Since 𝜑( 𝑓 ) ∈ 𝑈 𝑓𝑛 and 𝜑(𝑔) ∈ 𝑈𝑔𝑛 , we have 𝜑( 𝑓 ) ≠ 𝜑(𝑔). Claim D: 𝜑 is surjective. Let 𝑥 ∈ 𝑋 and let 𝑛 ≥ 1 be an integer. Since P𝑛 is a partition of 𝑋, there exists a unique 𝑓 (𝑛) ∈ 𝑌𝑛 such that 𝑥 ∈ 𝑈 𝑓 (𝑛) . Let 𝑔 = 𝑓 (𝑛) |𝑌𝑛−1 . By (1.5a), 𝑥 ∈ 𝑈 𝑓 (𝑛) ⊆ 𝑈𝑔 . Since 𝑥 ∈ 𝑈 𝑓 (𝑛−1) , it follows from the uniqueness of 𝑓 (𝑛−1) that 𝑓 (𝑛−1) = 𝑔. Thus, Ñ there is an 𝑓 ∈ 𝑌 such that 𝑓𝑛 := 𝑓 |𝑌𝑛 = 𝑓 (𝑛) for every 𝑛. Since 𝑥 ∈ ∞ 𝑛=0 𝑈 𝑓𝑛 = 𝐶 𝑓 , we have 𝑥 = 𝜑( 𝑓 ), as claimed. Claim E: 𝜑 is continuous. First notice that {𝑈 𝑓 } 𝑓 ∈𝑌𝑛 , 𝑛≥0 is a basis for the topology on 𝑋. Indeed, let 𝑥 ∈ 𝑋 and let 𝑈 be an open neighborhood of 𝑥 in 𝑋. There is a 𝑉𝑛 such that 𝑥 ∈ 𝑉𝑛 ⊆ 𝑈. Since P𝑛 is a partition of 𝑋, there is 𝑓 ∈ 𝑌𝑛 such that 𝑥 ∈ 𝑈 𝑓 . In particular, 𝑉𝑛 ∩ 𝑈 𝑓 ≠ ∅. By (1.5b), 𝑈 𝑓 ⊆ 𝑉𝑛 . Now, let 𝑔 ∈ 𝑌𝑛 . The definition of 𝜑 and Claim A imply that ∞ Ù 𝜑−1 (𝑈𝑔 ) = { 𝑓 ∈ 𝑌 | 𝜑( 𝑓 ) ∈ 𝑈𝑔 } = { 𝑓 ∈ 𝑌 | 𝑈 𝑓𝑚 ⊆ 𝑈𝑔 } 𝑚=0

= {𝑓 ∈𝑌 |

∞ Ù

𝑈 𝑓𝑚 ⊆ 𝑈𝑔 } =

𝑚=𝑛

∞ Ø

{ 𝑓 ∈ 𝑌 | 𝑈 𝑓𝑚 ⊆ 𝑈𝑔 }

𝑚=𝑛

= { 𝑓 ∈ 𝑌 | 𝑈 𝑓𝑛 ⊆ 𝑈𝑔 } = { 𝑓 ∈ 𝑌 | 𝑈 𝑓𝑛 = 𝑈𝑔 } (1.5a)

= { 𝑓 ∈ 𝑌 | 𝑓𝑛 = 𝑔}.

Thus,

𝜑−1 (𝑈

𝑔)

is an open subset of 𝑌 , so 𝜑 is continuous, as claimed.

End of proof: By Claims C, D, and 𝐸, 𝜑 : 𝑌 → 𝑋 is a continuous bijection. Since both 𝑋 and {0, 2}N are profinite spaces, Fact 1.1.3(e) implies that 𝜑 is a homeomorphism. This concludes the proof of the lemma.  Corollary 1.5.4 Let 𝑋 be a profinite space with a countable base and let 𝑌 be a Cantor space. Then, 𝑋 × 𝑌 is a Cantor space. Proof Let {𝑈𝑖 }𝑖 ∈𝐼 be a countable base for the topology of 𝑋. By the discussion preceding Lemma 1.5.2, the topology of 𝑌 has an infinite countable base {𝑉 𝑗 } 𝑗 ∈𝐽 . Then, {𝑈𝑖 × 𝑉 𝑗 } (𝑖, 𝑗) ∈𝐼 ×𝐽 is an infinite countable base for the topology of 𝑋 × 𝑌 . Without loss we assume that 𝑉 𝑗 ≠ ∅ for all 𝑗 ∈ 𝐽. If (𝑥, 𝑦) ∈ 𝑋 × 𝑌 and {(𝑥, 𝑦)} is open in 𝑋 × 𝑌 , then there exist 𝑖 ∈ 𝐼 and 𝑗 ∈ 𝐽 such that 𝑈𝑖 × 𝑉 𝑗 ⊆ {(𝑥, 𝑦)}, so 𝑉 𝑗 ⊆ {𝑦}. Since 𝑉 𝑗 ≠ ∅, the set {𝑦} = 𝑉 𝑗 is open. This contradiction to the

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assumption that 𝑌 is a Cantor space (see the discussion preceding Lemma 1.5.2) proves that no point of 𝑋 × 𝑌 is isolated. It follows from Lemma 1.5.3 that 𝑋 × 𝑌 is a Cantor space, as claimed.  The following special case of [Rib70, p. 31, Prop. 3.5] turns out to be quite useful. Lemma 1.5.5 Let 𝐻 be a closed subgroup of a profinite groups 𝐺. Then, the quotient map 𝜋 : 𝐺 → 𝐺/𝐻 of profinite spaces (Lemma 1.1.20(b)) defined by 𝜋(𝑔) = 𝑔𝐻 has a continuous section 𝜋 0 : 𝐺/𝐻 → 𝐺. Thus, 𝜋(𝜋 0 (𝑔𝐻)) = 𝑔𝐻 for each 𝑔 ∈ 𝐺. Recall that the rank of a profinite group 𝐺 (denoted rank(𝐺)) is the smallest cardinality of a subset of 𝐺 that generates 𝐺 and converges to 1 [FrJ08, p. 339, Rem. 17.1.3]. Lemma 1.5.6 Let 𝐺 be a profinite group with rank(𝐺) ≤ ℵ0 , 𝐻 a closed subgroup of infinite index, and 𝑅 a closed subset of representatives for the left cosets of 𝐺 modulo 𝐻. Then, 𝐺/𝐻 and 𝑅 are Cantor spaces. Proof By assumption, 𝐺 has an infinite countable base U for its open sets. Then, {𝑈𝐻/𝐻 | 𝑈 ∈ U} is an infinite countable base for the open sets of 𝐺/𝐻. If 𝑔𝐻 is isolated in 𝐺/𝐻, for some 𝑔 ∈ 𝐺, then 𝑔𝐻 is open in 𝐺. Hence, 𝐻 is an open subgroup of 𝐺, which means (𝐺 : 𝐻) < ∞. This contradiction to our assumption proves that 𝐺/𝐻 is a Cantor space (Lemma 1.5.3). Finally, as a closed subspace of 𝐺, the set 𝑅 is a profinite space under the topology induced from that of 𝐺. Moreover, the map 𝑟 ↦→ 𝑟 𝐻 is a continuous bijection 𝑅 → 𝐺/𝐻 of profinite spaces. Hence, by Fact 1.1.3(e), this map is a homeomorphism. It follows that 𝑅 is also a Cantor space.  For a profinite group 𝐺 and closed subgroups 𝐻1 , 𝐻2 we consider the space of the double cosets 𝐻1 \𝐺/𝐻2 with its quotient topology (Remark 1.1.21). Lemma 1.5.7 Let 𝐺 be a profinite group, 𝑔 ∈ 𝐺, and 𝐻1 , 𝐻2 closed subgroups of 𝐺. Then, 𝐻1 𝑔𝐻2 is an isolated point in the quotient space 𝐻1 \𝐺/𝐻2 if and only if 𝑔 𝐻1 𝐻2 is open in 𝐺. Proof By definition of the quotient topology, the point 𝐻1 𝑔𝐻2 ∈ 𝐻1 \𝐺/𝐻2 is isolated if and only if its preimage 𝐻1 𝑔𝐻2 in 𝐺 is open. Since multiplication from 𝑔 the left by 𝑔 −1 is a homeomorphism of 𝐺, this is equivalent to 𝐻1 𝐻2 being open in 𝐺.  Lemma 1.5.8 Let 𝐺 be a profinite group with rank(𝐺) ≤ ℵ0 , 𝐻 a closed subgroup of 𝐺 of infinite index, and 𝑁 a closed normal subgroup of 𝐺 contained in 𝐻. Let 𝐽1 , . . . , 𝐽𝑛 be closed subgroups of 𝑁, let G = {𝐽𝑖𝜎 }𝑖=1,...,𝑛, 𝜎 ∈𝐺 , and let R be a strictly closed set of representatives for the 𝐻-orbits of G such that: (a1) 𝐽𝑖 ≠ 1 and 𝐽𝑖𝜎 ∩𝐽𝑖 = 1 if 𝜎 ∈ 𝐺 r𝐽𝑖 ; in particular, 𝑁𝐺 (𝐽𝑖 ) = 𝐽𝑖 for 𝑖 = 1, . . . , 𝑛. (a2) 𝐽𝑖𝜎 ∩ 𝐽 𝑗 = 1 for all 𝑖 ≠ 𝑗 and 𝜎 ∈ 𝐺. Then, Ð𝑛 (b1) R = · 𝑖=1 R 𝑖 , where each R 𝑖 is a strictly closed set of representatives for the 𝐻-orbits of G𝑖 := {𝐽𝑖𝜎 } 𝜎 ∈𝐺 .

1.5 Cantor Space

19

(b2) R is a Cantor space. 𝜌 (b3) For 𝑖 = 1, . . . , 𝑛 there exists a closed subset 𝑅𝑖 of 𝐺 such that R 𝑖 = {𝐽𝑖 | 𝜌 ∈ 𝑅𝑖 } 𝜌 and the map 𝜌 ↦→ 𝐽𝑖 is a homeomorphism of 𝑅𝑖 onto R 𝑖 . Proof Proof of (b1). let G𝑖 := {𝐽𝑖𝜎 } 𝜎 ∈𝐺 and R 𝑖 = G𝑖 ∩ R. By Ð𝑛 For 𝑖 = 1, . . . , 𝑛 Ð 𝑛 definition, G = 𝑖=1 G𝑖 . Hence, R = 𝑖=1 R 𝑖 . By Ð𝑛(a1) and (a2), G1 , . . . , G𝑛 are disjoint. Hence, so are R 1 , . . . , R 𝑛 . Therefore, R = · 𝑖=1 R𝑖 . By Example 1.3.8, each of the G𝑖 ’s is strictly closed in Subgr(𝐺). Since R is strictly closed in Subgr(𝐺), so are R 1 , . . . , R 𝑛 . Since distinct groups in R are non-𝐻-conjugate, so are distinct groups in each R 𝑖 . Given 𝑖 ∈ {1, . . . , 𝑛} and 𝜎 ∈ 𝐺, there exist 𝑗 ∈ {1, . . . , 𝑛}, 𝜏 ∈ 𝐺, and 𝜂 ∈ 𝐻 such 𝜏𝜂 that 𝐽 𝑗𝜏 ∈ R and 𝐽𝑖𝜎 = 𝐽 𝑗 . By (a2), 𝑖 = 𝑗, so 𝐽 𝑗𝜏 ∈ R ∩ G𝑖 = R 𝑖 , hence 𝐽𝑖𝜎 ∈ R 𝑖𝐻 . It follows that R 𝑖 is a set of representatives for the 𝐻-orbits of G𝑖 , as claimed. Proof of (b2). Since rank(𝐺) ≤ ℵ0 and 𝐺 is infinite (because (𝐺 : 𝐻) = ∞), the strict topology of Subgr(𝐺) has an infinite countable base. Hence, the topology of the strictly closed subset R of Subgr(𝐺) also has an infinite countable base. Thus, by Lemma 1.5.3, we have Ð𝑛 only to prove that R has no isolated points. Since, by (b1), R = 𝑖=1 R 𝑖 , it suffices to prove that R 𝑖 has no isolated points for each 𝑖 ∈ {1, . . . , 𝑛}. Since, by (b1), R 𝑖 is a set of representatives of the 𝐻-orbits of G𝑖 , (1.6) the map R 𝑖 → G𝑖 /𝐻 := {Γ 𝐻 }Γ∈R𝑖 defined by Γ ↦→ Γ 𝐻 is a strictly continuous bijection of profinite spaces (Lemma 1.1.20(b)). By Remark 1.1.7(c), this map is a homeomorphism. Next note that the map 𝑔 : 𝐺 ↦→ Subgr(𝐺) defined by 𝑔(𝜎) = 𝐽𝑖𝜎 is a continuous map of profinite spaces. Since, 𝑁𝐺 (𝐽𝑖 ) = 𝐽𝑖 (by (a1)), we have 𝐽𝑖𝜎 = 𝐽𝑖𝜏 if and only if 𝜏 ∈ 𝐽𝑖 𝜎. Therefore, by Remark 1.1.7(c), 𝑔 induces a homeomorphism 𝑔¯ : 𝐽𝑖 \𝐺 := {𝐽𝑖 𝜎} 𝜎 ∈𝐺 → G𝑖 mapping 𝐽𝑖 𝜎 onto 𝐽𝑖𝜎 . The map 𝑔¯ is compatible with the action of 𝐻 on both spaces (from the right). Hence, 𝑔¯ induces a homeomorphism 𝐽𝑖 \𝐺/𝐻 → G𝑖 /𝐻. Since 𝑁 ⊳ 𝐺 and (𝐺 : 𝑁) = ∞, the relations 𝐽𝑖𝜎 𝐻 ≤ 𝑁 𝜎 𝐻 = 𝑁 𝐻 = 𝐻 and (𝐺 : 𝐻) = ∞ imply that the set 𝐽𝑖𝜎 𝐻 is not open in 𝐺. Hence, by Lemma 1.5.7, the point 𝐽𝑖 𝜎𝐻 of 𝐽𝑖 \𝐺/𝐻 is not isolated. Therefore, by the three preceding paragraphs, none of the points of R 𝑖 is isolated, as needed. Proof of (b3). Given 𝑖 ∈ {1, . . . , 𝑛}, the map 𝜎 ↦→ 𝐽𝑖𝜎 is a continuous surjection 𝑓 : 𝐺 → G𝑖 . By the paragraph following (1.6), 𝑓 decomposes into the quotient map 𝐺 → 𝐽𝑖 \𝐺 and a homeomorphism 𝐽𝑖 \𝐺 → G𝑖 . By Lemma 1.5.5, the quotient map has a continuous section. Hence, also 𝑓 has a continuous section 𝑓 0 : G𝑖 → 𝐺. Since R 𝑖 is closed in G𝑖 , its image 𝑅𝑖 under 𝑓 0 is a closed subset of 𝐺 (Fact 1.1.3(e)). It satisfies the requirements of (b3). 

Chapter 2

Families of Subgroups

This chapter deals with different notions of continuity of families of closed subgroups of a profinite group. We introduce an “étale continuous family” which is indexed by a topological (usually profinite) space. A “separated family” is closely related to this. Then, we consider “sheaves of subgroups” and “morphisms from sheaves into profinite groups”.

2.1 Continuous and Separated Families As a preparation for the construction of free products we introduce two types of families of Subgr(𝐺) for a given profinite group 𝐺: “continuous families” and “separated families”. Under certain favorable conditions, these concepts are closely related (Propositions 2.1.7 and 2.1.8). Definition 2.1.1 Let 𝑇 be a topological space and let 𝐺 be a profinite group. For every 𝑡 ∈ 𝑇 let Γ𝑡 be a closed subgroup of 𝐺. We say that the family {Γ𝑡 }𝑡 ∈𝑇 is étale continuous if the following conditions are satisfied: (2.1a) Γ𝑠 ∩ Γ𝑡 = 1 for all distinct 𝑠, 𝑡 ∈ 𝑇; in particular, Γ𝑠 ≠ Γ𝑡 , unless Γ𝑠 = Γ𝑡 = 1. (2.1b) The map 𝑡 ↦→ Γ𝑡 from 𝑇 into Subgr(𝐺) is étale continuous. In other words, for every open subgroup 𝐻 of 𝐺 the subset {𝑡 ∈ 𝑇 }Γ𝑡 ≤𝐻 of 𝑇 is open. Lemma 2.1.2 Let 𝑇 be a profinite space and G = {Γ𝑡 }𝑡 ∈𝑇 an étale continuous family of subgroups of a profinite group 𝐺. Then, G is étale compact and 𝑈 (G) (Notation 1.2.4) is closed in 𝐺. Proof Since 𝑇 is compact and the map 𝑡 ↦→ Γ𝑡 is étale continuous, G is étale compact (Fact 1.1.3(d)). By Lemma 1.2.5, 𝑈 (G) is closed in 𝐺. See also [Rib17, p. 145, Lemma 5.2.1]. 

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 D. Haran and M. Jarden, The Absolute Galois Group of a Semi-Local Field, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-030-89191-6_2

21

22

2 Families of Subgroups

Lemma 2.1.3 Let 𝑇 be a profinite space and 𝐺 a finite group. For every 𝑡 ∈ 𝑇 let Γ𝑡 be a subgroup of 𝐺 and put G = {Γ𝑡 }𝑡 ∈𝑇 . Assume that the map 𝑡 ↦→ Γ𝑡 is étale Ð𝑛 continuous. Then, there is a partition 𝑇 = · 𝑖=1 𝑇𝑖 of 𝑇 into disjoint open-closed subsets and for every 1 ≤ 𝑖 ≤ 𝑛 there is a 𝑡 𝑖 ∈ 𝑇𝑖 such that Γ𝑡 ≤ Γ𝑡𝑖 for every 𝑡 ∈ 𝑇𝑖 . Proof By induction on card(G). If card(G) = 0 (i.e., 𝑇 = ∅) or card(G) = 1, let 𝑛 = 1 and 𝑇1 = 𝑇. Thus, we assume that card(G) > 1. We use that 𝐺 is finite to choose 𝑡 1 ∈ 𝑇 such that Γ𝑡1 is maximal among the groups that belong to G. By Lemma 1.2.8, {Γ𝑡1 } is étale closed in G. Since the map 𝑡 ↦→ Γ𝑡 is étale continuous, 𝐶 = {𝑡 ∈ 𝑇 | Γ𝑡 = Γ𝑡1 } is closed in 𝑇, 𝑈 = {𝑡 ∈ 𝑇 | Γ𝑡 ≤ Γ𝑡1 } is open in 𝑇 (because Subgr(Γ𝑡1 ) is open in Subgr(𝐺)), and 𝐶 ⊆ 𝑈. By Lemma 1.1.15, 𝐶 is an intersection of open-closed subsets of 𝑇. Hence, since 𝐶 is compact, it follows from Lemma 1.1.1 that there is an open-closed 𝑇1 ⊆ 𝑇 such that 𝐶 ⊆ 𝑇1 ⊆ 𝑈. Thus, Γ𝑡 ≤ Γ𝑡1 for every 𝑡 ∈ 𝑇1 . Now, 𝑇 0 = 𝑇 r 𝑇1 is open-closed in 𝑇 and Γ𝑡1 ∉ {Γ𝑡 }𝑡 ∈𝑇 0 . Otherwise, there exists a 𝑡 ∈ 𝑇 0 with Γ𝑡 = Γ𝑡1 . Thus, 𝑡 ∈ 𝐶, so by the preceding paragraph 𝑡 ∈ 𝑇1 , which is a contradiction. Therefore, card({Γ𝑡 }𝑡 ∈𝑇 0 ) < card(G). By the induction hypothesis, there is a Ð𝑛 partition 𝑇 0 = · 𝑖=2 𝑇𝑖 of 𝑇 0 into disjoint open-closed subsets and Ð𝑛for every 2 ≤ 𝑖 ≤ 𝑛 there is a 𝑡𝑖 ∈ 𝑇𝑖 such that Γ𝑡 ≤ Γ𝑡𝑖 for every 𝑡 ∈ 𝑇𝑖 . Then, 𝑇 = · 𝑖=1 𝑇𝑖 is the required partition.  Lemma 2.1.4 Let 𝑇 be a profinite space, let G = {Γ𝑡 }𝑡 ∈𝑇 beÐan étale continuous 𝑛 family of closed subgroups of a profinite group 𝐺, and let 𝑇 = · 𝑖=1 𝑇𝑖 be a partition of 𝑇 into disjoint open-closed subsets. Let 𝐾 ⊳ 𝐺 be open. Then, there is an open 𝑁 ⊳ 𝐺 such that Γ𝑠 𝑁 ∩ Γ𝑡 𝑁 ≤ 𝐾 for all 𝑠, 𝑡 ∈ 𝑇 which are not in the same 𝑇𝑖 . Proof Let 1 ≤ 𝑖, 𝑗 ≤ 𝑛 be distinct. It suffices to find an open 𝑁 (𝑖,Ñ 𝑗) ⊳ 𝐺 such that Γ𝑠 𝑁 (𝑖, 𝑗) ∩ Γ𝑡 𝑁 (𝑖, 𝑗) ≤ 𝐾 for all 𝑠 ∈ 𝑇𝑖 and 𝑡 ∈ 𝑇 𝑗 , since then 𝑁 := (𝑖, 𝑗) 𝑁 (𝑖, 𝑗) has the required property. Indeed, for every (𝑠, 𝑡) ∈ 𝑇𝑖 ×𝑇 𝑗 we have Γ𝑠 ∩Γ𝑡 = 1. Hence, by Lemma 1.1.12(b), we have Ù Ù Ù (Γ𝑠 𝐿 ∩ Γ𝑡 𝐿) = Γ𝑠 𝐿 ∩ Γ𝑡 𝐿 = Γ𝑠 ∩ Γ𝑡 = 1, 𝐿

𝐿

𝐿

where, in all cases, 𝐿 ranges over OpenNormal(𝐺). Therefore, by Lemma 1.1.12(a), there exists an 𝐿 (𝑠,𝑡) ∈ OpenNormal(𝐺) such that Γ𝑠 𝐿 (𝑠,𝑡) ∩ Γ𝑡 𝐿 (𝑠,𝑡) ≤ 𝐾. If (𝑠 0, 𝑡 0) ∈ 𝑇𝑖 × 𝑇 𝑗 is sufficiently close to (𝑠, 𝑡), then Γ𝑠0 ≤ Γ𝑠 𝐿 (𝑠,𝑡) and Γ𝑡 0 ≤ Γ𝑡 𝐿 (𝑠,𝑡) , so we can take 𝐿 (𝑠0 ,𝑡 0 ) = 𝐿 (𝑠,𝑡) to get Γ𝑠0 𝐿 (𝑠0 ,𝑡 0 ) ∩ Γ𝑡 0 𝐿 (𝑠0 ,𝑡 0 ) ≤ 𝐾. Since 𝑇𝑖 × 𝑇 𝑗 is compact, there are open 𝐿 1 , . . . , 𝐿 𝑚 ⊳ 𝐺 suchÑ that for every (𝑠, 𝑡) ∈ 𝑇𝑖 × 𝑇 𝑗 there is a 𝑘 with Γ𝑠 𝐿 𝑘 ∩ Γ𝑡 𝐿 𝑘 ≤ 𝐾. Then, 𝑁 (𝑖, 𝑗) := 𝑚 𝑘=1 𝐿 𝑘 has the required property.  Definition 2.1.5 Let 𝐺 be a profinite group and G a subset of Subgr(𝐺). We say that G is separated if (2.2a) G is stellate, i.e. Γ1 ∩ Γ2 = 1 for all distinct Γ1 , Γ2 ∈ G, and

2.1 Continuous and Separated Families

23

(2.2b) for all distinct groups Γ1 , Γ2 ∈ G there exist disjoint subsets G1 , G2 of G such that G = G1 ∪ G2 , Γ1 ∈ G1 , Γ2 ∈ G2 , and 𝑈 (G1 ), 𝑈 (G2 ) are closed subsets of 𝐺; in particular, 𝑈 (G) is closed in 𝐺. Remark 2.1.6 (a) An equivalent formulation of Definition 2.1.5 is obtained if (2.2b) is required to hold only for Γ1 , Γ2 ∈ G r {1}. Indeed, if Γ1 = 1, then (2.2b) holds with G1 = {1} and G2 = G r {1}. (b) Likewise, we may replace the requirement G1 ∩ G2 = ∅ by G1 ∩ G2 ⊆ {1}. Indeed, if G1 ∩ G2 = {1}, and, say, Γ1 ≠ 1, replace G1 by G10 = G1 r {1}, noticing that 𝑈 (G10 ) = 𝑈 (G1 ) and 1 ∈ G2 . (c) A subset G of Subgr(𝐺) is separated if and only if G 0 = G r {1} is separated. Indeed, assume that 1 ∈ G, otherwise G 0 = G. Following (a), let Γ1 , Γ2 ∈ G with Γ1 , Γ2 ≠ 1, that is, Γ1 , Γ2 ∈ G 0. Assume (2.2a). If (2.3) G1 , G2 ⊆ G, Γ1 ∈ G1 , Γ2 ∈ G2 , 𝑈 (G1 ), 𝑈 (G2 ) are closed, G = G1 ∪ G2 , and G1 ∩ G2 = ∅, put G10 = G1 r {1} and G20 = G2 r {1}. Then, 𝑈 (G10 ) = 𝑈 (G1 ) and 𝑈 (G20 ) = 𝑈 (G2 ), hence (2.4) G10 , G20 ⊆ G 0, Γ1 ∈ G10 , Γ2 ∈ G20 , 𝑈 (G10 ), 𝑈 (G20 ) are closed, and G 0 = G10 ∪ G20 , G10 ∩ G20 = ∅. Conversely, if (2.4) holds, then (2.3) holds with G1 = G10 ∪ {1}, G2 = G20 . Proposition 2.1.7 Let G = {Γ𝑡 | 𝑡 ∈ 𝑇 } be an étale continuous family of closed subgroups of a profinite group 𝐺 (Definition 2.1.1). If 𝑇 is a profinite space, then G is separated. Proof Let 𝑡 1 , 𝑡2 be elements of 𝑇 such that Γ𝑡1 , Γ𝑡2 ∈ G are distinct. Then, 𝑡1 ≠ 𝑡 2 , hence (2.2a) follows from (2.1a). Let 𝐻 be an open subgroup of 𝐺 containing Γ𝑡1 . By (2.1b), 𝑈 0 = {𝑡 ∈ 𝑇 | Γ𝑡 ≤ 𝐻} is open in 𝑇. Since 𝑇 is a profinite space, 𝑡 1 has an open-closed neighborhood 𝑈1 in 𝑇 such that 𝑈1 ⊆ 𝑈 0 and 𝑡 2 ∉ 𝑈1 . Then, 𝑈2 = 𝑇 r𝑈1 is an open-closed neighborhood of 𝑡 2 such that 𝑇 = 𝑈1 ∪· 𝑈2 . For 𝑖 = 1, 2, the set G𝑖 = {Γ𝑡 | 𝑡 ∈ 𝑈𝑖 } is a subset of G that contains Γ𝑡𝑖 . It is étale compact, since it is the image of the compact set 𝑈𝑖 (use Condition (2.1b) and Fact 1.1.3(d)). By Lemma 1.2.5, 𝑈 (G𝑖 ) is closed in 𝐺. Finally, G1 ∪ G2 = {Γ𝑡 | 𝑡 ∈ 𝑈1 ∪ 𝑈2 } = G, and, by (2.1a), G1 ∩ G2 ⊆ {1}. Thus, by Remark 2.1.6(b), G is separated.  Proposition 2.1.8 Let G be a separated family of closed subgroups of a profinite group 𝐺 (Definition 2.1.5). Then, there exists a profinite space 𝑇 and an étale continuous family {Γ𝑡 }𝑡 ∈𝑇 of closed subgroups of 𝐺 such that {Γ𝑡 | 𝑡 ∈ 𝑇 } is either G or G ∪ {1}.

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2 Families of Subgroups

Proof Let Σ = {G 0 ⊆ G | 𝑈 (G 0) and 𝑈 (G r G 0) are closed in 𝐺}. For subfamilies G10 , . . . , G𝑛0 of G the identities 𝑈

𝑛 Ù

𝑛
Ù G𝑖0 = 𝑈 (G𝑖0)

𝑖=1

𝑖=1

and 𝑈 Gr

𝑛 Ù

𝑛 𝑛 Ø Ø
G𝑖0 = 𝑈 (G r G𝑖0)) = 𝑈 (G r G𝑖0)

𝑖=1

𝑖=1

𝑖=1

show that Σ is closed under finite intersections. A Σ-partition of G is a finite collection 𝑇𝑖 of disjoint non-empty elements of Σ whose union is G. A Σ-partition 𝑇 𝑗 of G is finer than 𝑇𝑖 if every G 𝑗 ∈ 𝑇 𝑗 is contained in a (necessarily unique) subset G𝑖 ∈ 𝑇𝑖 . Under this partial ordering, the Σ-partitions form an inverse system of finite sets (𝑇𝑖 )𝑖 ∈𝐼 . Therefore, 𝑇 = lim 𝑇𝑖 is a profinite ←−− space. Explicitly, Ö 𝑇 = {(G𝑖 )𝑖 ∈𝐼 ∈ 𝑇𝑖 | G 𝑗 ⊆ G𝑖 if 𝑇 𝑗 is finer than 𝑇𝑖 }. 𝑖 ∈𝐼

Ñ Notice that if 𝑡 = (G𝑖 )𝑖 ∈𝐼 ∈ 𝑇, then card( 𝑖 ∈𝐼 G𝑖 ) ≤ 1. Indeed, by (2.2b), for distinct Γ1 , Γ2 ∈ G there is a partition {H1 , H2 }, such that Γ1 ∈ H1 and Γ2 ∈ H2 . Since {𝑇𝑖 }𝑖 ∈𝐼 is the set of all Σ-partitions of ÑG, there exists a 𝑗 ∈ 𝐼 such thatÑ{H1 , H2 } = 𝑇 𝑗 . If G 𝑗 = H1 , then Γ2 ∉ G 𝑗 , so Γ2 ∉ 𝑖 ∈𝐼 G𝑖 ; if G 𝑗 = H2 , then Γ1 ∉ 𝑖 ∈𝐼 G𝑖 . We define a map 𝜓 : G → 𝑇 as follows: 𝜓(Γ) = (G𝑖 )𝑖 ∈𝐼 , where, for every 𝑖 ∈ 𝐼, G𝑖 is the unique element of 𝑇𝑖 such that Γ ∈ G𝑖 . By the preceding paragraph, 𝜓 is injective. It follows that if 𝑡 = (G𝑖 )𝑖 ∈𝐼 ∈ 𝑇, then either Ñ Ñ (a) 𝑖 ∈𝐼 G𝑖 contains a single element Γ ∈ G; then 𝑡 = 𝜓(Γ) and 𝑖 ∈𝐼 𝑈 (G𝑖 ) = Γ; or (b)

Ñ

𝑖 ∈𝐼

G𝑖 = ∅; then 𝑡 ∉ 𝜓(G) and

Ñ

𝑖 ∈𝐼

𝑈 (G𝑖 ) = 1.

We define a map 𝜑 : 𝑇 → Subgr(𝐺) by 𝜑(𝑡) = Γ in case (a) and 𝜑(𝑡) = 1 in case (b). Then, 𝜑 is étale continuous. Indeed, let 𝐻 be an open subgroup of 𝐺 and consider 𝑡 = (G𝑖 )𝑖 ∈𝐼 ∈ 𝑇 with Ñ 𝜑(𝑡) ≤ 𝐻. Then, in both cases 𝑖 ∈𝐼 𝑈 (G𝑖 ) ⊆ 𝐻. Since the sets 𝑈 (G𝑖 ) are closed and 𝐺 is compact, Lemma 1.1.1 yields an 𝑖 ∈ 𝐼 such that 𝑈 (G𝑖 ) ⊆ 𝐻, and hence G𝑖 ⊆ Subgr(𝐻). Thus, if 𝑡 0 ∈ 𝑇 is close to 𝑡, namely, has the same 𝑖-th coordinate G𝑖 , then either (a) 𝜑(𝑡 0) = Γ0 ∈ G𝑖 ; or (b) 𝜑(𝑡 0) = 1. In both cases 𝜑(𝑡 0) ≤ 𝐻. It follows that 𝜑(𝑇) = G or 𝜑(𝑇) = G ∪ {1}.  We give an alternative proof of [Har87, Cor. 3.8]. Lemma 2.1.9 Let 𝐺 be a profinite group and G a separated subset of Subgr(𝐺) (Definition 2.1.5). Then, G is étale compact. Proof By Remark 1.2.6 and Remark 2.1.6(c), we may assume that 1 ∈ G. By Proposition 2.1.8 there exists a profinite space 𝑇 and an étale continuous surjective map 𝜌 : 𝑇 → G. Since 𝑇 is compact, G is étale compact (Fact 1.1.3(d)). 

2.1 Continuous and Separated Families

25

Corollary 2.1.10 Let 𝐺 be a profinite group and G a stellate (Definition 1.3.3) étale compact Hausdorff subset of Subgr(𝐺). Let G 0 ⊆ G. Then, the following three conditions are equivalent: (a) G 0 is étale compact; (b) G 0 is étale closed in G; (c) 𝑈 (G 0) is closed. Moreover, (d) G is separated if and only if G is étale profinite. Proof The equivalence (a)⇔(b) follows from Fact 1.1.3(c), while (a)⇔(c) is Lemma 1.4.5. To prove Condition (d) we may assume that 1 ∉ G and card(G) ≥ 2. Indeed, if card(G) = 1, then G trivially satisfies Conditions (2.2a) and (2.2b), so G is separated. Also, G is trivially étale profinite. Thus, we may assume that card(G) ≥ 2. By Lemma 1.3.2, G satisfies the equivalent conditions of Remark 1.3.1. In particular, 1 ∉ StrictClosure(G), so 1 ∉ G, as claimed. To proceed we assume that G is separated, and let Γ1 , Γ2 ∈ G be distinct. Then, there exist G1 , G2 ⊆ G such that G = G1 ∪· G2 , Γ1 ∈ G1 , Γ2 ∈ G2 , and 𝑈 (G1 ), 𝑈 (G2 ) are closed. By (b)⇔(c), G1 , G2 are étale closed in G, and since their respective complements are étale closed, they are also étale open. By Lemma 1.1.9, G is étale profinite. Assume that G is étale profinite and let Γ1 , Γ2 ∈ G be distinct. By Lemma 1.1.9 there are étale open-closed G1 , G2 ⊆ G such that G = G1 ∪· G2 and Γ1 ∈ G1 , Γ2 ∈ G2 . By Lemma 1.2.5, 𝑈 (G1 ), 𝑈 (G2 ) are closed. Thus, G is separated.  Problem 2.1.11 Is every stellate étale compact subset of Subgr(𝐺) separated? We give a detailed proof of the first statement in the proof of [Har87, Thm. 5.1]. Lemma 2.1.12 Let 𝐺 be a profinite group, 𝐻 a closed subgroup, and G a separated subset of Subgr(𝐺). Then, H := {Γ ∩ 𝐻 | Γ ∈ G} is a separated subset of Subgr(𝐻). Proof Let Δ1 , Δ2 be distinct groups in H . Then, there exist Γ1 , Γ2 ∈ G such that Δ1 = Γ1 ∩ 𝐻 and Δ2 = Γ2 ∩ 𝐻. It follows that Γ1 ≠ Γ2 , so, by (2.2a), Γ1 ∩ Γ2 = 1, hence Δ1 ∩ Δ2 = 1. By (2.2b), there exist disjoint subsets G1 , G2 of G such that G1 ∪ G2 = G, Γ𝑖 ∈ G𝑖 , and 𝑈 (G𝑖 ) is a closed subset of 𝐺 for 𝑖 = 1, 2. For 𝑖 = 1, 2 we set H𝑖 = {Γ ∩ 𝐻 | Γ ∈ G𝑖 }; then Δ𝑖 ∈ H𝑖 . Let Δ ∈ H1 ∩ H2 . Then, there exist Γ10 ∈ G1 and Γ20 ∈ G2 such that Γ10 ∩ 𝐻 = Δ = Γ20 ∩ 𝐻. Since G1 ∩ G2 = ∅, we have Γ10 ≠ Γ20 . Since G is separated, Γ10 ∩ Γ20 = 1. Hence, Δ = 1. It follows that H1 ∩ H2 ⊆ {1}. Finally, 𝑈 (G𝑖 ) is a closed subset of 𝐺, hence 𝑈 (H𝑖 ) = 𝑈 (G𝑖 ) ∩ 𝐻 is a closed subset of 𝐻. By Remark 2.1.6(b), H is a separated subset of Subgr(𝐻).

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2.2 Sheaves of Profinite Groups We introduce “sheaves of profinite groups” and prove some of their properties. Definition 2.2.1 A sheaf (of profinite groups) is a triple X = (𝑋, 𝜏, 𝑇) in which 𝑋 and 𝑇 are compact 𝑇1 -topological spaces, 𝜏 : 𝑋 → 𝑇 is a closed continuous surjective map, and 𝑋, 𝜏, 𝑇 satisfy the following conditions: (2.5a) For every 𝑡 ∈ 𝑇 the fiber 𝑋𝑡 = 𝜏 −1 (𝑡) is a profinite group under the Ð topology induced by that of 𝑋 (in particular, 𝑋 = · 𝑡 ∈𝑇 𝑋𝑡 ). (2.5b) The group operations of 𝑋𝑡 are uniformly continuous. In other words, writing 𝑋 (2) = {(𝑥1 , 𝑥2 ) ∈ 𝑋 × 𝑋 | 𝜏(𝑥1 ) = 𝜏(𝑥2 )}, the structure map 𝜇 𝑋 : 𝑋 (2) → 𝑋, defined by 𝜇 𝑋 (𝑥1 , 𝑥2 ) = 𝑥1−1 𝑥 2 , is continuous. Note that the 𝑇1 -property of 𝑋 follows from the 𝑇1 -property of 𝑇 and from the fact that 𝜏 : 𝑋 → 𝑇 is continuous and each of the spaces 𝑋𝑡 is profinite, hence 𝑇1 . Similarly, the compactness of 𝑇 follows from the compactness of 𝑋 and the continuity and the surjectivity of 𝜏 (Fact 1.1.3(d)). Since the sheaves of profinite groups are the only type of sheaves we consider in this monograph, we henceforth speak about “sheaves” rather than about “sheaves of profinite groups”. We say that a sheaf X is profinite if 𝑋 and 𝑇 are profinite spaces. In this case the closedness of 𝜏 follows from its continuity by Fact 1.1.3(e). In particular, X is finite if 𝑋 and 𝑇 are finite discrete spaces. Example 2.2.2 Let 𝑋 = 𝐺 × 𝑇, where 𝐺 is a profinite group and 𝑇 a compact 𝑇1 -space. Then, 𝑋 is a compact 𝑇1 -space. We write pr𝑇 : 𝑋 → 𝑇 for the projection on the second factor. It is a closed map by Fact 1.1.3(g). Thus, for every 𝑡 ∈ 𝑇 we have pr𝑇−1 (𝑡) = 𝐺 × {𝑡}. We define a group structure on the latter fiber by (𝑔, 𝑡) (ℎ, 𝑡) = (𝑔ℎ, 𝑡). Then, 𝐺 × {𝑡} is a profinite group and the projection (𝑔, 𝑡) ↦→ 𝑔 on the first factor is an isomorphism 𝐺 × {𝑡}  𝐺. If 𝑇0 is an open neighborhood of 𝑡 in 𝑇 and 𝑁 is an open normal subgroup of 𝐺, then 𝜇 𝑋 maps the open neighborhood ((𝑔𝑁 ×𝑇0 ) × (ℎ𝑁 ×𝑇0 )) ∩ 𝑋 (2) of ((𝑔, 𝑡), (ℎ, 𝑡)) onto 𝑔 −1 ℎ𝑁 × 𝑇0 . Thus, the group operations in the group 𝐺 × {𝑡} are uniformly continuous. It follows that the triple (𝑋, pr𝑇 , 𝑇) is a sheaf called the constant sheaf. Example 2.2.3 Let X = (𝑋, 𝜏, 𝑇) be a sheaf and let 𝑇 0 be a closed subspace of 𝑇. Then, 𝑇 0 is compact and 𝑇1 . Hence, 𝑋 0 = 𝜏 −1 (𝑇 0) is a closed 𝑇1 -subset of 𝑋, so 𝑋 0 is also compact. We set 𝜏 0 = 𝜏|𝑋 0 . Then, X0 := (𝑋 0, 𝜏 0, 𝑇 0) is a sheaf and we say that X0 is a closed subsheaf of a sheaf X or that X contains X0. Note that in this case, if X is a profinite sheaf, then so is X0. We say that X0 is open-closed in X if X0 is a closed subsheaf of X and 𝑇 0 is open-closed in 𝑇. We make the class of sheaves a category by defining morphisms between sheaves.

2.2 Sheaves of Profinite Groups

27

Definition 2.2.4 A morphism from a sheaf X = (𝑋, 𝜏, 𝑇) into a sheaf Y = (𝑌 , 𝜎, 𝑆) is a pair 𝛼 = (𝛼1 , 𝛼2 ) of continuous maps 𝛼1 : 𝑋 → 𝑌 and 𝛼2 : 𝑇 → 𝑆 that satisfy the following conditions: (2.6a) 𝜎 ◦ 𝛼1 = 𝛼2 ◦ 𝜏; thus, 𝛼1 induces a continuous map 𝛼1(2) : 𝑋 (2) → 𝑌 (2) by (𝑥1 , 𝑥2 ) ↦→ (𝛼1 (𝑥1 ), 𝛼1 (𝑥2 )). (2.6b) 𝜇𝑌 ◦ 𝛼1(2) = 𝛼1 ◦ 𝜇 𝑋 ; here 𝜇 𝑋 , 𝜇𝑌 are the structure maps of X, Y, respectively. The latter condition can be expressed as follows: (2.6c) If 𝛼2 (𝑡) = 𝑠, then the restriction of 𝛼1 to 𝑋𝑡 is a group homomorphism 𝑋𝑡 → 𝑌𝑠 . By abuse of notation we usually write 𝛼 for both 𝛼1 and 𝛼2 . The rest of this section is devoted to properties of profinite sheaves. Remark 2.2.5 We consider a directed partially ordered set (𝐼, ≤) and an inverse system (X𝑖 , 𝜋 𝑗𝑖 )𝑖 ≤ 𝑗 of profinite sheaves with X𝑖 = (𝑋𝑖 , 𝜏𝑖 , 𝑇𝑖 ) for every 𝑖 ∈ 𝐼. Thus, for all 𝑗 ≥ 𝑖 in 𝐼 we have the commutative diagrams: 𝜏𝑗

𝑋𝑗

/ 𝑇𝑗

𝑋 𝑗(2) 𝜋 𝑗𝑖

𝜋 𝑗𝑖

 𝑋𝑖

𝜏𝑖

𝜇𝑗

/ 𝑋𝑗

𝜇𝑖

 / 𝑋𝑖 ,

(2)

𝜋 𝑗𝑖

𝜋 𝑗𝑖

 / 𝑇𝑖



𝑋𝑖(2)

in which 𝜇𝑖 : 𝑋𝑖(2) → 𝑋𝑖 is the structure map of X𝑖 , 𝑋𝑡𝑖 = 𝜏𝑖−1 (𝑡𝑖 ), and 𝜋 (2) 𝑗𝑖 is the map induced by 𝜋 𝑗𝑖 . The inverse limit of this system of profinite sheaves is a triple X = (𝑋, 𝜏, 𝑇) in which 𝑋 = lim 𝑋𝑖 and 𝑇 = lim 𝑇𝑖 are profinite spaces, and 𝜏 = lim 𝜏𝑖 is a continuous ←−− ←−− ←−− map from 𝑋 to 𝑇. Moreover, 𝑋 (2) = lim 𝑋𝑖(2) and the 𝜇𝑖 define a continuous map ←−− 𝜇 𝑋 = lim 𝜇𝑖 : 𝑋 (2) → 𝑋. If we denote the projections 𝑋 → 𝑋𝑖 and 𝑇 → 𝑇𝑖 by 𝜋𝑖 , ←−− we get the commutative diagrams: 𝜏

𝑋

/𝑇 𝜋𝑖

𝜋𝑖

 𝑋𝑖

𝑋 (2)

𝜏𝑖

 / 𝑇𝑖

𝜇𝑋

/𝑋

𝜇𝑖

 / 𝑋𝑖 ,

(2)

𝜋𝑖



𝑋𝑖(2)

𝜋𝑖

in which 𝜋𝑖(2) is the map induced by 𝜋𝑖 . If 𝑡 ∈ 𝑇, 𝑡 𝑖 = 𝜋𝑖 (𝑡), 𝑋𝑡𝑖 = 𝜏𝑖−1 (𝑡𝑖 ), and 𝑋𝑡 = 𝜏 −1 (𝑡), then 𝑋𝑡 = lim 𝑋𝑡𝑖 is a profinite group. It follows that X is a profinite ←−− sheaf. We write X = lim X𝑖 . ←−− Lemma 2.2.6 Let X be the inverse limit of an inverse system of profinite sheaves, (X𝑖 , 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 , 𝑖 ≤ 𝑗 , with projections 𝜋𝑖 : X → X𝑖 . Let 𝛼 : X → Y be a morphism into a finite sheaf Y. Then, (a) there are 𝑗 ∈ 𝐼 and a morphism 𝛼 𝑗 : X 𝑗 → Y such that 𝛼 = 𝛼 𝑗 ◦ 𝜋 𝑗 . (b) If 𝛼 0𝑗 : X 𝑗 → Y is another morphism such that 𝛼 = 𝛼 0𝑗 ◦ 𝜋 𝑗 , there exists a 𝑘 ≥ 𝑗 such that 𝛼 𝑗 ◦ 𝜋 𝑘 𝑗 = 𝛼 0𝑗 ◦ 𝜋 𝑘 𝑗 .

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Proof Proof of (a). We set X = (𝑋, 𝜏, 𝑇), X𝑖 = (𝑋𝑖 , 𝜏𝑖 , 𝑇𝑖 ), for every 𝑖 ∈ 𝐼, Y = (𝑌 , 𝜎, 𝑆), and 𝛼 = (𝛼1 , 𝛼2 ) (assuming 1, 2 ∉ 𝐼). Let 𝜇 𝑋 : 𝑋 (2) → 𝑋, 𝜇𝑌 : 𝑌 (2) → 𝑌 , and 𝜇𝑖 : 𝑋𝑖(2) → 𝑋𝑖 be the structure maps of the respective sheaves. By Lemma 1.1.16(b), there is a 𝑗 ∈ 𝐼 and continuous maps 𝛼1 𝑗 : 𝑋 𝑗 → 𝑌 and 𝛼2 𝑗 : 𝑇 𝑗 → 𝑆 such that 𝛼1 = 𝛼1 𝑗 ◦ 𝜋 𝑗 and 𝛼2 = 𝛼2 𝑗 ◦ 𝜋 𝑗 . For 𝑘 ≥ 𝑗 let 𝛼1𝑘 = 𝛼1 𝑗 ◦ 𝜋 𝑘 𝑗 and 𝛼2𝑘 = 𝛼2 𝑗 ◦ 𝜋 𝑘 𝑗 . Then, 𝛼1 = 𝛼1𝑘 ◦ 𝜋 𝑘 and 𝛼2 = 𝛼2𝑘 ◦ 𝜋 𝑘 as well. Thus, by Lemma 1.1.16(c), we may replace 𝑗 by a suitable 𝑘 ≥ 𝑗 to assume that the following two diagrams 𝜏𝑗 𝜇𝑗 / 𝑇𝑗 / 𝑋𝑗 𝑋 (2) 𝑋𝑗 𝑗

𝛼1 𝑗

𝛼2 𝑗

 𝑌

𝜎

 /𝑆

(2) 𝛼1 𝑗



𝑌 (2)

𝛼1 𝑗 𝜇𝑌

 /𝑌

commute. Then, 𝛼 𝑗 = (𝛼1 𝑗 , 𝛼2 𝑗 ) is a morphism of sheaves. Proof of (b). Again, this follows from Lemma 1.1.16(c).



Example 2.2.7 Let X := (𝑋, 𝜏, 𝑇) be a closed subsheaf of a profinite sheaf Y := (𝑌 , 𝜎, 𝑆) (Example 2.2.3). Let {𝑇𝑖 }𝑖 ∈𝐼 be the family of all open subsets of 𝑆 that contain 𝑇. For each 𝑖 ∈ 𝐼 let 𝑋𝑖 = 𝜎 −1 (𝑇𝑖 ), and set 𝜏𝑖 : 𝑋𝑖 → 𝑇𝑖 as the restriction of 𝜎 to 𝑋𝑖 . Then, X𝑖 = (𝑋𝑖 , 𝜏𝑖 , 𝑇𝑖 ) is an open-closed subsheaf of Y that Ñ contains X. As in Example 1.1.5, X = 𝑖 ∈𝐼 X𝑖 = lim X𝑖 . ←−− An application of Lemma 2.2.6 to this situation gives: Corollary 2.2.8 Let X be a profinite sheaf, X0 a closed subsheaf (Example 2.2.3), and let 𝛼 : X0 → Y be a morphism into a finite sheaf Y. Then: (a) There is an open-closed subsheaf U of X containing X0 such that 𝛼 extends to a morphism 𝛼𝑈 : U → Y. (b) If V is another open-closed subsheaf of X containing X0 and 𝛼𝑉 : V → Y is a morphism extending 𝛼, then there exists an open-closed sheaf W such that X0 ⊆ W ⊆ U, V and 𝛼𝑈 and 𝛼𝑉 coincide on W. Lemma 2.2.9 Let X be a profinite sheaf and let X0 be a closed subsheaf. Let 𝛼 : X0 → Y be a morphism into a finite sheaf Y. Then, 𝛼 extends to a morphism 𝛼 : X → Y. Proof We set X = (𝑋, 𝜏, 𝑇), X0 = (𝑋 0, 𝜏 0, 𝑇 0), Y = (𝑌 , 𝜎, 𝑆), and 𝛼 = (𝛼1 , 𝛼2 ), By Corollary 2.2.8(a), 𝛼 extends to an open-closed subsheaf of X containing X0. Thus, we may assume that X0 is open-closed, that is, 𝑇 0 is open-closed in 𝑇 and 𝑋 0 = 𝜏 −1 (𝑇 0) is open-closed in 𝑋. We choose 𝑠 ∈ 𝑆, denote the unit element of 𝑌𝑠 = 𝜎 −1 (𝑠) by 1𝑠 , and extend 𝛼1 , 𝛼2 to maps 𝛼1 : 𝑋 → 𝑌 and 𝛼2 : 𝑇 → 𝑆 by 𝛼1 (𝑥) = 1𝑠 , if 𝑥 ∈ 𝑋 r 𝑋 0, and 𝛼2 (𝑡) = 𝑠, if 𝑡 ∈ 𝑇 r 𝑇 0. Since 𝑋 0 and 𝑇 0 are open-closed in 𝑋 and 𝑇 respectively, the extended maps 𝛼1 and 𝛼2 are continuous. By assumption, 𝜎 ◦ 𝛼1 coincides with 𝛼2 ◦ 𝜏 on 𝑋 0. By our definition, 𝜎 ◦ 𝛼1 coincides with 𝛼2 ◦ 𝜏 on 𝑋 r 𝑋 0. Thus, 𝜎 ◦ 𝛼1 = 𝛼2 ◦ 𝜏 on 𝑋.

2.3 Sheaf-Group Structures

29

Finally, if 𝑡 ∈ 𝑇 r 𝑇 0, 𝑥1 ∈ 𝑋 r 𝑋 0, and 𝑥 2 ∈ 𝑋 satisfy 𝜏(𝑥1 ) = 𝑡 = 𝜏(𝑥2 ), then 𝑥 1 , 𝑥2 ∈ 𝑋𝑡 , 𝛼1 (𝑥1 𝑥2 ) = 1𝑠 = 1𝑠 · 1𝑠 = 𝛼1 (𝑥1 )𝛼1 (𝑥2 ). This verifies Condition (2.6c). Therefore, the extended map (𝛼1 , 𝛼2 ) : X → Y is indeed a morphism of sheaves, as desired.  The following result will not be needed. Loosely speaking, it assures that every “finite embedding problem” for a profinite sheaf is “solvable” if it is “locally solvable”. Proposition 2.2.10 Let X = (𝑋, 𝜏, 𝑇) be a profinite sheaf and let Y = (𝑌 , 𝜎, 𝑆) and Z be finite sheaves. Let 𝛼 : Y → Z and 𝜑 : X → Z be morphisms of sheaves. Assume that for every 𝑡 ∈ 𝑇 there exist 𝑠 = 𝑠(𝑡) ∈ 𝑆 such that 𝛼(𝑠) = 𝜑(𝑡) and a group homomorphism 𝛾𝑡 : 𝑋𝑡 → 𝑌𝑠 such that 𝛼 ◦ 𝛾𝑡 = 𝜑|𝑋𝑡 . Then, there exists a morphism 𝛾 : X → Y such that 𝛼 ◦ 𝛾 = 𝜑. Moreover, given a finite subset 𝑇0 of 𝑇, we may construct 𝛾 so that it extends 𝛾𝑡 for every 𝑡 ∈ 𝑇0 . Ð Proof Let 𝑋0 = 𝜏 −1 (𝑇0 ) = · 𝑡 ∈𝑇0 𝑋𝑡 . Then, X0 = (𝑋0 , 𝜏|𝑋0 , 𝑇0 ) is a closed subsheaf of X and {𝛾𝑡 }𝑡 ∈𝑇0 , together with the map 𝑇0 → 𝑆 given by 𝑡 ↦→ 𝑠(𝑡) ∈ 𝑆, define a morphism 𝛾0 : X0 → Y such that 𝛼 ◦ 𝛾0 = 𝜑|X0 . By Corollary 2.2.8(a), 𝛾0 extends to a morphism 𝛾1 : X1 → Y, where X1 = (𝑋1 , 𝜏|𝑋1 , 𝑇1 ) is an open-closed subsheaf of X containing X0 . Since 𝜑|X0 = 𝛼 ◦ 𝛾0 = 𝛼 ◦ 𝛾1 |X0 , it follows from Corollary 2.2.8(b) that X has an open-closed subsheaf X10 such that X0 ⊆ X10 ⊆ X1 and 𝛼 ◦ 𝛾1 |X01 = 𝜑|X01 . Replacing X1 by X10 , if necessary, we may assume that 𝛼 ◦ 𝛾1 = 𝜑|X1 . Let 𝑋2 = 𝑋 r 𝑋1 and 𝑇2 = 𝑇 r 𝑇1 . Then, X2 = (𝑋2 , 𝜏|𝑋2 , 𝑇2 ) is an open-closed subsheaf of X, and it suffices to construct a morphism 𝛾2 : X2 → Y such that 𝛼 ◦ 𝛾2 = 𝜑|X2 . Thus, replacing X2 by X, we may assume that 𝑇0 = ∅. Let 𝑡 ∈ 𝑇. As in the first paragraph of this proof (with {𝑡} instead of 𝑇0 ) there is an open-closed subsheaf X0 = (𝑋 0, 𝜏|𝑋 0 , 𝑇 0) of X such that 𝑋𝑡 ⊆ 𝑋 0 and there is a morphism 𝛾 0 : X0 → Y such that 𝛼 ◦ 𝛾 0 = 𝜑|X0 . The collection of all 𝑇 0, as 𝑡 runs through the points of 𝑇, is an open-closed covering of 𝑇. Since 𝑇 is compact, there are finitelyÐ many open-closed Ð subsheaves X𝑖 = (𝑋𝑖 , 𝜏𝑖 , 𝑇𝑖 ), for 𝑖 = 1, . . . , 𝑛, of X such that 𝑇 = 𝑖 𝑇𝑖 and 𝑋 = 𝑖 𝑋𝑖 , and there are morphisms 𝛾𝑖 : X𝑖 → Y such that 𝛼 ◦ 𝛾𝑖 = 𝜑|X𝑖 for each 𝑖. Applying Lemma 1.1.10, we may replace the 𝑇𝑖 ’s by their intersections to assume that the 𝑇𝑖 , and hence also the 𝑋𝑖 , are disjoint. Then, there is a unique morphism 𝛾 : X → Y such that 𝛾|X𝑖 = 𝛾𝑖 . In particular, 𝛼 ◦ 𝛾 = 𝜑. 

2.3 Sheaf-Group Structures The objects introduced in this section will be instrumental in the construction of outer free products in Chapter 4. Remark 2.3.1 Every profinite group 𝐺 can be regarded as a sheaf (𝐺, 𝜏𝐺 , 𝑇𝐺 ), where 𝑇𝐺 is the one-point space and 𝜏𝐺 : 𝐺 → 𝑇𝐺 is the unique map onto 𝑇𝐺 . We have 𝐺 𝑡 = 𝐺, for the unique 𝑡 in 𝑇𝐺 .

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Definition 2.3.2 Let X = (𝑋, 𝜏, 𝑇) be a sheaf and let 𝐺 be a profinite group. A morphism 𝜔 : X → 𝐺 is a morphism of sheaves, where 𝐺 is regarded as a sheaf (Remark 2.3.1). In other words, if 𝜔 = (𝜔1 , 𝜔2 ), then 𝜔1 : 𝑋 → 𝐺, usually written as 𝜔, is a continuous map whose restriction to every fiber 𝑋𝑡 is a homomorphism (in particular 𝜔(𝑋𝑡 ) is a subgroup of 𝐺) and 𝜔2 is the unique map from 𝑇 onto the one-point space 𝑇𝐺 . Let us apply Lemma 2.2.9 to this setup: Corollary 2.3.3 Let X = (𝑋, 𝜏, 𝑇) be a profinite sheaf. Consider distinct elements 𝑠, 𝑡 ∈ 𝑇. Then, every homomorphism 𝛼𝑡 of 𝑋𝑡 into a finite group 𝐴 extends to a morphism 𝛼 : X → 𝐴 such that 𝛼(𝑋𝑠 ) = 1. Proof Let 𝑋 0 = 𝑋𝑠 ∪· 𝑋𝑡 , let 𝑇 0 = {𝑠, 𝑡}, and let 𝜏 0 : 𝑋 0 → 𝑇 0 be the restriction of 𝜏 to 𝑋 0. Then, 𝑋 0 = 𝜏 −1 ({𝑠, 𝑡}) and X0 = (𝑋 0, 𝜏 0, 𝑇 0) is a closed subsheaf of X. Let 𝛼 0 : 𝑋 0 → 𝐴 be the map that maps 𝑋𝑠 onto the unit element of 𝐴 and coincides with 𝛼𝑡 on 𝑋𝑡 . We regard 𝛼 0 as a morphism of sheaves 𝛼 0 : X0 → 𝐴, as in Definition 2.3.2. By Lemma 2.2.9, 𝛼 0 extends to a morphism 𝛼 : X → 𝐴. In particular, 𝛼|𝑋𝑠 is the trivial homomorphism and 𝛼|𝑋𝑡 = 𝛼𝑡 .  Definition 2.3.4 Let X = (𝑋, 𝜏, 𝑇) be a sheaf, 𝐺 a profinite group, and 𝜔 : X → 𝐺 a morphism. We call (X, 𝜔, 𝐺), sometimes written more explicitly as (𝑋, 𝜏, 𝑇, 𝜔, 𝐺), a sheaf-group structure. A morphism of sheaf-group structures 𝜃 : (X, 𝜔, 𝐺) → (X0, 𝜔 0, 𝐺 0) is a pair (𝜃 𝑠 , 𝜃 𝑔 ), where 𝜃 𝑠 : X → X0 is a morphism of sheaves and 𝜃 𝑔 : 𝐺 → 𝐺 0 is a homomorphism of profinite groups (regarded as a morphism of sheaves) such that 𝜃 𝑔 ◦𝜔 = 𝜔 0 ◦𝜃 𝑠 in the category of sheaves. An isomorphism of sheaf-group structures is a morphism that has an inverse. Lemma 2.3.5 Every sheaf-group structure (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) satisfies the following conditions. (a) The map 𝜌 : 𝑇 → Subgr(𝐺) defined by 𝜌(𝑡) = 𝜔(𝑋𝑡 ) is étale continuous. (b) The map 𝜉 : 𝑋 → 𝑇 × 𝐺 defined by 𝜉 (𝑥) = (𝜏(𝑥), 𝜔(𝑥)) is closed. (c) The map 𝜔 is closed. (d) If 𝜏 is an open map, then 𝜌 is strictly continuous. (e) If 𝜌 is strictly continuous and 𝜉 is injective, then 𝜏 is an open map. Proof By definition 𝑋 is compact. Proof of (a). Let 𝐻 be an open subgroup of 𝐺. Then, 𝜔−1 (𝐻) is an open subset of 𝑋. For each 𝑡 ∈ 𝑇 we have 𝑋𝑡 = 𝜏 −1 (𝑡), so 𝜔(𝑋𝑡 ) ≤ 𝐻 if and only if 𝜏 −1 (𝑡) ≤ 𝜔−1 (𝐻). Thus, 𝑆 := {𝑡 ∈ 𝑇 | 𝜔(𝑋𝑡 ) ≤ 𝐻} = {𝑡 ∈ 𝑇 | 𝜏 −1 (𝑡) ≤ 𝜔−1 (𝐻)}. Since 𝜏 is a closed map, Lemma 1.1.18 with 𝑇, 𝜏, 𝜔−1 (𝐻), 𝑆 replacing 𝑌 , 𝑓 , 𝑋0 , 𝑌0 , respectively, implies that 𝑆 is open in 𝑇. Therefore, 𝜌 is étale continuous. Proof of (b). This is Lemma 1.1.23.

2.3 Sheaf-Group Structures

31

Proof of (c). Since the coordinate projection 𝜋 0 : 𝑇 × 𝐺 → 𝐺 is closed by Fact 1.1.3(g), for 𝐺, 𝑇 replacing 𝑇, 𝐺, and 𝜉 is closed by (b), the map 𝜔 = 𝜋 0 ◦ 𝜉 is closed. Proof of (d). Assume that 𝜏 is an open map and let 𝑡 ∈ 𝑇 and 𝑁 ∈ OpenNormal(𝐺). We have to show that there is an open neighborhood 𝑉 of 𝑡 in 𝑇 such that 𝜔(𝑋𝑠 )𝑁 = 𝜔(𝑋𝑡 )𝑁 for every 𝑠 ∈ 𝑉. Let 𝜋 : 𝐺 → 𝐺/𝑁 be the quotient map. Replacing 𝜔 by 𝜋 ◦ 𝜔 : 𝑋 → 𝐺/𝑁 we may assume that 𝐺 is finite and 𝑁 = 1. In particular, 𝜔(𝑋𝑡 ) is finite, say, 𝜔(𝑋𝑡 ) = {𝜔(𝑥1 ), . . . , 𝜔(𝑥 𝑛 )}. As 𝜌 is étale continuous by (a), there is an open neighborhood 𝑉0 of 𝑡 such that 𝜔(𝑋𝑠 ) ≤ 𝜔(𝑋𝑡 ) for every 𝑠 ∈ 𝑉0 . Let 1 ≤ 𝑖 ≤ 𝑛. Since 𝜔 is continuous, 𝑈𝑖 = 𝜔−1 (𝜔({𝑥 𝑖 })) is open in 𝑋. By assumption, 𝑉𝑖 := 𝜏(𝑈𝑖 ) is open in 𝑇. If 𝑠 ∈ 𝑉𝑖 , there is an 𝑥 0 ∈ 𝑈𝑖 such that 𝜏(𝑥 0) = 𝑠, that is, 𝑥 0 ∈ 𝑈𝑖 ∩ 𝑋𝑠 . Since 𝜔(𝑥 0) ∈ 𝜔(𝑈𝑖 ) = 𝜔({𝑥𝑖 }), we have 𝜔(𝑥 0) = 𝜔(𝑥 𝑖 ). Thus, 𝜔(𝑥𝑖 ) ∈ 𝜔(𝑋 Ñ𝑠𝑛). Let 𝑉 = 𝑖=0 𝑉𝑖 . Then, 𝑉 has the required property: If 𝑠 ∈ 𝑉, then 𝜔(𝑋𝑠 ) ≤ 𝜔(𝑋𝑡 ) and 𝜔(𝑥 1 ), . . . , 𝜔(𝑥 𝑛 ) ∈ 𝜔(𝑋𝑠 ), so 𝜔(𝑋𝑠 ) = 𝜔(𝑋𝑡 ). Proof of (e). Let 𝑥 ∈ 𝑋 and let 𝑈 0 be an open neighborhood of 𝑥 in 𝑋. We have to find an open neighborhood 𝑈 ⊆ 𝑈 0 of 𝑥 such that 𝜏(𝑈) is open in 𝑇. Let 𝑡 = 𝜏(𝑥) and 𝑔 = 𝜔(𝑥). Then, 𝑔 ∈ 𝜔(𝑋𝑡 ). Put 𝑉 0 = (𝑇 × 𝐺) r 𝜉 (𝑋 r 𝑈 0). Since 𝑥 ∈ 𝑈 0 and 𝜉 is injective, (𝑡, 𝑔) = 𝜉 (𝑥) ∉ 𝜉 (𝑋 r 𝑈 0) , so (𝑡, 𝑔) ∈ 𝑉 0. Since 𝑋 r 𝑈 0 is closed in 𝑋 and 𝜉 is a closed map by (b), 𝑉 0 is open in 𝑇 × 𝐺. Thus, there is an open neighborhood 𝑉 of 𝑡 in 𝑇 and open normal subgroup 𝑁 of 𝐺 such that (2.7)

(𝑡, 𝑔) ∈ 𝑉 × 𝑔𝑁 ⊆ 𝑉 0 .

Since 𝜌 : 𝑇 → Subgr(𝐺) is strictly continuous, the set 𝑇0 := {𝑠 ∈ 𝑇 | 𝜔(𝑋𝑠 )𝑁 = 𝜔(𝑋𝑡 )𝑁 } = {𝑠 ∈ 𝑇 | 𝜌(𝑠)𝑁 = 𝜌(𝑡)𝑁 } is open in 𝑇. Replacing 𝑉 by 𝑉 ∩ 𝑇0 , we may assume that (2.8)

𝜔(𝑋𝑠 )𝑁 = 𝜔(𝑋𝑡 )𝑁 for every 𝑠 ∈ 𝑉 .

Since 𝜉 is continuous and 𝜉 (𝑥) = (𝑡, 𝑔) ∈ 𝑉 × 𝑔𝑁, the set 𝑈 := 𝜉 −1 (𝑉 × 𝑔𝑁) is an open neighborhood of 𝑥. Since 𝜉 is injective, we have that 𝜉 −1 (𝑉 0) = 𝑈 0. Hence, (2.7)

𝑈 = 𝜉 −1 (𝑉 × 𝑔𝑁) ⊆ 𝜉 −1 (𝑉 0) = 𝑈 0 . It now suffices to show that 𝜏(𝑈) = 𝑉. On one hand, 𝜉 (𝑈) ⊆ 𝑉 × 𝑔𝑁, so 𝜏(𝑈) ⊆ 𝑉. Conversely, if 𝑠 ∈ 𝑉, then (2.8)

𝑔𝑁 ∈ 𝜔(𝑋𝑡 )𝑁 = 𝜔(𝑋𝑠 )𝑁, so there is an 𝑥 0 ∈ 𝑋𝑠 such that 𝑔𝑁 = 𝜔(𝑥 0)𝑁. Then, 𝜉 (𝑥 0) = (𝑠, 𝜔(𝑥 0)) ∈ 𝑉 × 𝑔𝑁, so 𝑥 0 ∈ 𝑈 and 𝜏(𝑥 0) = 𝑠. Hence, 𝑠 ∈ 𝜏(𝑈). Thus, 𝑉 ⊆ 𝜏(𝑈).  Proposition 2.3.6 Let {𝐺 𝑡 }𝑡 ∈𝑇 be a family of closed subgroups of a profinite group 𝐺. Assume that 𝑇 is a compact 𝑇1 -space and the map 𝜌 : 𝑇 → Subgr(𝐺), given by 𝑡 ↦→ 𝐺 𝑡 , is étale continuous.

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Then, there exists a sheaf-group structure (𝑋, 𝜏, 𝑇, 𝜔, 𝐺), unique up to isomorphism, such that 𝜔 maps 𝑋𝑡 isomorphically onto 𝐺 𝑡 , for every 𝑡 ∈ 𝑇. We call X = (𝑋, 𝜏, 𝑇) the associated sheaf and (X, 𝜔, 𝐺) the associated sheafgroup structure of {𝐺 𝑡 }𝑡 ∈𝑇 . If 𝑇 is a profinite space, then X is a profinite sheaf. Proof Set 𝑋 = {(𝑡, 𝑔) ∈ 𝑇 × 𝐺 | 𝑔 ∈ 𝐺 𝑡 } and let 𝜏 : 𝑋 → 𝑇 be the restriction to 𝑋 of the projection of 𝑇 × 𝐺 on the first factor. In other words, 𝜏(𝑡, 𝑔) = 𝑡 for every (𝑡, 𝑔) ∈ 𝑋. For every 𝑡 ∈ 𝑇 we have 𝑋𝑡 = 𝜏 −1 (𝑡) = {𝑡} × 𝐺 𝑡 . Then, the multiplication rule (𝑡, 𝑔) (𝑡, ℎ) = (𝑡, 𝑔ℎ) makes 𝑋𝑡 into a group which is isomorphic to 𝐺 𝑡 . Claim A: The group operations in 𝑋𝑡 are uniformly continuous, i.e. the structure map 𝜇 : 𝑋 (2) → 𝑋 is continuous. Indeed, an element of 𝑋 (2) has the form ((𝑡, 𝑔), (𝑡, ℎ)) with 𝑔, ℎ ∈ 𝐺 𝑡 , and 𝜇((𝑡, 𝑔), (𝑡, ℎ)) = (𝑡, 𝑔 −1 ℎ). Every open neighborhood of (𝑡, 𝑔 −1 ℎ) in 𝑋 contains a set of the form (𝑇 0 × 𝑔 −1 ℎ𝑁) ∩ 𝑋, where 𝑁 ∈ OpenNormal(𝐺) and 𝑇 0 is an open neighborhood of 𝑡 in 𝑇. Then, ((𝑇 0 × 𝑔𝑁) × (𝑇 0 × ℎ𝑁)) ∩ 𝑋 (2) is an open neighborhood of ((𝑡, 𝑔), (𝑡, ℎ)) in 𝑋 (2) which 𝜇 maps into (𝑇 0 × 𝑔 −1 ℎ𝑁) ∩ 𝑋. Claim B: 𝑋 is closed in 𝑇 × 𝐺. Indeed, let (𝑡, 𝑔) ∈ (𝑇 × 𝐺) r 𝑋. Then, 𝑔 ∉ 𝐺 𝑡 . Since 𝐺 𝑡 is closed in 𝐺, there exists an open normal subgroup 𝑁 of 𝐺 such that 𝑔𝑁 ∩ 𝐺 𝑡 = ∅, that is, 𝑔𝑁 ∩ 𝐺 𝑡 𝑁 = ∅.

(2.9) 𝑇0

Since 𝜌 is étale continuous, = {𝑡 0 ∈ 𝑇 | 𝐺 𝑡 0 ≤ 𝐺 𝑡 𝑁 } is an open neighborhood 0 of 𝑡 in 𝑇. Hence, the set 𝑇 × 𝑔𝑁 is an open neighborhood of (𝑡, 𝑔) in 𝑇 × 𝐺. If (𝑡 0, 𝑔 0) ∈ 𝑇 0 × 𝑔𝑁, then 𝑔 0 ∈ 𝑔𝑁 and 𝐺 𝑡 0 ≤ 𝐺 𝑡 𝑁. If 𝑔 0 ∈ 𝐺 𝑡 0 , then 𝑔 0 ∈ 𝑔𝑁 ∩ 𝐺 𝑡 0 ⊆ 𝑔𝑁 ∩𝐺 𝑡 𝑁. This contradiction to (2.9) implies that 𝑔 0 ∉ 𝐺 𝑡 0 , so (𝑡 0, 𝑔 0) ∉ 𝑋. It follows that (𝑇 × 𝐺) r 𝑋 is open, so 𝑋 is closed. By Fact 1.1.3(a), 𝑋 is compact. Since 𝑇 ×𝐺 is a 𝑇1 -space, so is 𝑋. By Fact 1.1.3(g), the projection 𝑇 × 𝐺 → 𝑇 is a closed map. Hence, by Claim B, its restriction 𝜏 : 𝑋 → 𝑇 to 𝑋 is also closed. It follows from Claim A that X = (𝑋, 𝜏, 𝑇) is a sheaf. The projection 𝑇 × 𝐺 → 𝐺, given by (𝑡, 𝑔) ↦→ 𝑔, is a continuous map. It maps every 𝑋𝑡 isomorphically and homeomorphically onto 𝐺 𝑡 . In particular, its restriction to 𝑋 is a morphism 𝜔 : X → 𝐺. Thus, (X, 𝜔, 𝐺) is a sheaf-group structure. If 𝑇 is profinite, then so is 𝑇 × 𝐺 and hence also 𝑋. Claim C: (X, 𝜔, 𝐺) is unique up to isomorphism. Let X0 = (𝑋 0, 𝜏 0, 𝑇) be a sheaf and let 𝜔 0 : X0 → 𝐺 be a morphism such that 𝜔 0 maps every 𝑋𝑡0 := (𝜏 0) −1 (𝑡) isomorphically onto 𝐺 𝑡 . Let 𝜉 0 : 𝑋 0 → 𝑇 × 𝐺 be the continuous map 𝑥 0 ↦→ (𝜏 0 (𝑥 0Ð ), 𝜔 0 (𝑥 0)). 0 0 Then,Ð𝜉 maps every 𝑋𝑡 isomorphically onto {𝑡} × 𝐺 𝑡 = 𝑋𝑡 . Since 𝑋 = · 𝑡 𝑋𝑡 and 𝑋 0 = · 𝑡 𝑋𝑡0, we see that 𝜉 0 : 𝑋 0 → 𝑋 is a bijection. Since 𝜉 0 is continuous and hence, by Lemma 2.3.5(b), closed, (𝜉 0, id𝑇 ) : X0 → X is an isomorphism of sheaves. More

over, for every 𝑡 ∈ 𝑇 and every 𝑥 0 ∈ 𝑋𝑡0 we have 𝜔 𝜉 0 (𝑥 0) = 𝜔 (𝑡, 𝜔 0 (𝑥 0)) = 𝜔 0 (𝑥 0), hence 𝜔 ◦ 𝜉 0 = 𝜔 0. Thus, (𝜉 0, id𝐺 ) : (X0, 𝜔 0, 𝐺) → (X, 𝜔, 𝐺) is an isomorphism of sheaf-group structures. 

Chapter 3

Free Products of Finitely Many Profinite Groups

Given Î a finite system {𝐺 𝑖 }𝑖 ∈𝐼 of profinite groups, we construct the free product 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 𝑖 in the category of profinite groups. It is the unique (up to isomorphism) profinite group which contains each of the 𝐺 𝑖 ’s and has the property that every system of homomorphisms {𝜑𝑖 : 𝐺 𝑖 → 𝐴}𝑖 ∈𝐼 into a profinite group 𝐴 uniquely extends to a homomorphism 𝜑 : 𝐺 → 𝐴. This implies that 𝐺 𝑖𝑥 ∩ 𝐺 𝑗 = 1 if 𝑖 ≠ 𝑗 and 𝑥 ∈ 𝐺 and 𝑁𝐺 (𝐺 𝑖 ) = 𝐺 𝑖 for each 𝑖 (Lemma 3.1.10). Then, we prove a theorem of Wolfgang Herfort and Luis Ribes saying that every finite subgroup of 𝐺 is already contained in a conjugate of one of the free factors 𝐺 𝑖 (Proposition 3.2.7). The chapter partly overlaps with Section 9.1 of [RiZ10].

3.1 Definition and First Properties We prove the existence of free products of finitely many profinite groups and establish a few of their properties. Proposition 3.1.1 Let 𝐼 be a finite set and for each 𝑖 ∈ 𝐼 let 𝐺 𝑖 be aÎ profinite group. Then, there exists a unique (up to isomorphism) profinite group 𝐺ˆ = ∗ 𝑖 ∈𝐼 𝐺 𝑖 , called the free product (in the category of profinite groups) of the groups 𝐺 𝑖 , 𝑖 ∈ 𝐼, with the following properties: ˆ (a) Every 𝐺 𝑖 is a closed subgroup of 𝐺. (b) Let 𝐴 be a profinite group and for each 𝑖 ∈ 𝐼 let 𝜑𝑖 : 𝐺 𝑖 → 𝐴 be a homomorphism. Then, there exists a unique homomorphism 𝜑ˆ : 𝐺ˆ → 𝐴 such that 𝜑| ˆ 𝐺𝑖 = 𝜑𝑖 for each 𝑖 ∈ 𝐼. Proof Let 𝐺 be the free product of the 𝐺 𝑖 ’s in the category of abstract groups, that is, the unique (up to isomorphism) abstract group which is generated by all of the 𝐺 𝑖 ’s and has the property that for every abstract group 𝐴 and every system of abstract homomorphisms 𝛼𝑖 : 𝐺 𝑖 → 𝐴, 𝑖 ∈ 𝐼, there exists a unique abstract homomorphism 𝛼 : 𝐺 → 𝐴 such that 𝛼|𝐺𝑖 = 𝛼𝑖 for all 𝑖 ∈ 𝐼. Denote the family of normal subgroups 𝑁 of 𝐺 of finite index such that 𝑁 ∩ 𝐺 𝑖 is open in 𝐺 𝑖 for each 𝑖 ∈ 𝐼 by N . Then, 𝑁 ∩ 𝑁 0 ∈ N for all 𝑁, 𝑁 0 ∈ N . Hence, N is, © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 D. Haran and M. Jarden, The Absolute Galois Group of a Semi-Local Field, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-030-89191-6_3

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3 Free Products of Finitely Many Profinite Groups

in the terminology of [FrJ08, Sec. 17.2], a directed family. Let 𝐺ˆ = lim 𝐺/𝑁, with ←−− 𝑁 ranging over N , be the corresponding inverse limit. We define an abstract homomorphism 𝜃 : 𝐺 → 𝐺ˆ by 𝜃 (𝑔) = (𝑔𝑁)𝑁 ∈N for each 𝑔 ∈ 𝐺. Thus, if 𝜋 𝑁 : 𝐺 → 𝐺/𝑁 and 𝜋ˆ 𝑁 : 𝐺ˆ → 𝐺/𝑁 are the corresponding projections and 𝑁ˆ = Ker( 𝜋ˆ 𝑁 ), then (3.1) 𝜋ˆ 𝑁 ◦ 𝜃 = 𝜋 𝑁 and 𝜃 induces the canonical isomorphism ˆ 𝑁. ˆ 𝜃 𝑁 : 𝐺/𝑁 → 𝐺/

Given an 𝑖 ∈ 𝐼 and 𝑁𝑖 ∈ OpenNormal(𝐺 𝑖 ), let 𝜈 : 𝐺 → 𝐺/𝑁𝑖 be the homomorphism whose restriction to 𝐺 𝑖 is the quotient map onto 𝐺 𝑖 /𝑁𝑖 and 𝜈|𝐺 𝑗 is the trivial homomorphism for each 𝑗 ∈ 𝐼 r {𝑖}. Then, Ker(𝜈) Ñ ∩ 𝐺 𝑖 = 𝑁𝑖 and Ker(𝜈) ∩ 𝐺 𝑗 = 𝐺 𝑗 for each 𝑗 ≠ 𝑖. Hence, Ker(𝜈) ∈ N . It follows that 𝑁 ∈N 𝑁 ∩ 𝐺 𝑖 = 1, so 𝜃 maps 𝐺 𝑖 ˆ We identify each 𝐺 𝑖 with its image 𝜃 (𝐺 𝑖 ) in 𝐺. ˆ injectively into 𝐺. ˆ By [FrJ08, p. 341, Lemma 17.2.1(a4)], 𝜃 (𝐺) is dense in 𝐺. Hence, ˆ (3.2) the closed subgroup h𝐺 𝑖 i𝑖 ∈𝐼 of 𝐺ˆ generated by the 𝐺 𝑖 ’s is 𝐺. The N -topology: If 𝑁1 , 𝑁2 ∈ N and 𝑔, 𝑔1 , 𝑔2 are elements of 𝐺 with 𝑔 ∈ 𝑔1 𝑁1 ∩𝑔2 𝑁2 , then 𝑁1 ∩ 𝑁2 ∈ N and 𝑔 ∈ 𝑔(𝑁1 ∩ 𝑁2 ) ⊆ 𝑔1 𝑁1 ∩ 𝑔2 𝑁2 . Thus, {𝑔𝑁 | 𝑔 ∈ 𝐺, 𝑁 ∈ N } forms a base for a group topology on 𝐺 that we call the N -topology. By [FrJ08, Sec. 1.2], the family { 𝜋ˆ −1 𝑁 (𝑔𝑁) | 𝑔 ∈ 𝐺, 𝑁 ∈ N } is a base for the ˆ For every 𝑔 and 𝑁 as above, 𝜃 −1 ( 𝜋ˆ −1 (𝑔𝑁)) = 𝑔𝑁 is an N -open topology of 𝐺. 𝑁 subset of 𝐺. Hence, 𝜃 is N -continuous. Proof of (a): Given an 𝑖 ∈ 𝐼, the paragraph below (3.1) shows that the restriction of the N -topology of 𝐺 to 𝐺 𝑖 coincides with the profinite topology of 𝐺 𝑖 . Since 𝜃 is continuous, so is 𝜃|𝐺𝑖 . But 𝐺 𝑖 is profinite, hence 𝜃 (𝐺 𝑖 ) is closed in 𝐺ˆ (Remark 1.1.7(c)). Proof of (b): Let 𝐴 be a finite group and for each 𝑖 ∈ 𝐼 let 𝜑𝑖 : 𝐺 𝑖 → 𝐴 be a homomorphism. Then, let 𝜑 : 𝐺 → 𝐴 be the homomorphism whose restriction to 𝐺 𝑖 is 𝜑𝑖 for each 𝑖 ∈ 𝐼. Then, 𝑁 = Ker(𝜑) ∈ N , so there is a homomorphism 𝜑¯ : 𝐺/𝑁 → 𝐴 such that 𝜑 = 𝜑¯ ◦ 𝜋 𝑁 . It follows that 𝜑ˆ = 𝜑¯ ◦ 𝜋ˆ 𝑁 : 𝐺ˆ → 𝐴 satisfies 𝜑ˆ ◦ 𝜃 = 𝜑. Hence, 𝜑| ˆ 𝐺𝑖 = 𝜑𝑖 . By (3.2), 𝜑ˆ is unique. Using the uniqueness of 𝜑, ˆ an inverse limit argument proves the consequence of the preceding paragraph for each profinite group 𝐴.  Remark 3.1.2 By [RiZ10, p. 358, Prop. 9.1.8], the natural map 𝜃 : 𝐺 → 𝐺ˆ mentioned in the proof of Proposition 3.1.1 is injective. Remark 3.1.3 The uniqueness in the proof of Proposition 3.1.1 follows from Statement (3.2). We establish here a generalization of the converse of this implication. (3.3) Let 𝐺 be a profinite group and 𝐻 a closed subgroup of 𝐺. Suppose that every homomorphism of 𝐻 into a profinite group 𝐴 uniquely extends to a homomorphism 𝐺 → 𝐴. Then, 𝐺 = 𝐻.

3.1 Definition and First Properties

35

Indeed, by assumption, the identity map id 𝐻 : 𝐻 → 𝐻 extends to a homomorphism 𝜂 : 𝐺 → 𝐻. Let 𝑖 : 𝐻 → 𝐺 be the inclusion map. Then, both id𝐺 and 𝑖 ◦ 𝜂 extend 𝑖. Hence, 𝑖 ◦ 𝜂 = id𝐺 . Thus, 𝑔 = 𝑖(𝜂(𝑔)) ∈ 𝐻 for each 𝑔 ∈ 𝐺. Therefore, 𝐺 = 𝐻. In particular, if 𝐺 1 , . . . , 𝐺 𝑛 are closed subgroups of a profinite group 𝐺 such that for each system (𝛼𝑖 : 𝐺 𝑖 → 𝐴)𝑖=1,...,𝑛 of homomorphisms into a profinite group 𝐴 there exists a unique homomorphism 𝛼 : 𝐺 → 𝐴 such that 𝛼|𝐺𝑖 = 𝛼𝑖 for 𝑖 = 1, . . . , 𝑛, then, by (3.3), 𝐺 = h𝐺 1 , . . . , 𝐺 𝑛 i. Lemma 3.1.4 In the notation of Proposition Î 3.1.1, let 𝐽 be a subset of 𝐼. Then, 𝐺 𝐽 = h𝐺 𝑗 i 𝑗 ∈𝐽 is the free profinite product ∗ 𝑗 ∈𝐽 𝐺 𝑗 . Proof Let 𝐴 be a profinite group and for each 𝑗 ∈ 𝐽 let 𝛼 𝑗 : 𝐺 𝑗 → 𝐴 be a homomorphism. For each 𝑖 ∈ 𝐼 r 𝐽 let 𝛼𝑖 : 𝐺 𝑖 → 𝐴 be the trivial homomorphism. By Proposition 3.1.1, there exists a homomorphism 𝛼 : 𝐺 → 𝐴 with 𝛼|𝐺𝑖 = 𝛼𝑖 for each 𝑖 ∈ 𝐼. Hence, 𝛼 𝐽 = 𝛼|𝐺 𝐽 is a homomorphism of 𝐺 𝐽 into 𝐴 that extends each of the 𝛼 𝑗 ’s with 𝑗 ∈ 𝐽. Since the groups 𝐺 𝑗 with 𝑗 ∈ 𝐽 generate 𝐺 𝐽 , the homomorphism 𝛼 𝐽 is unique with the above property. It follows from the definition of the free product Î (Proposition 3.1.1) that 𝐺 𝐽 = ∗ 𝑗 ∈𝐽 𝐺 𝑗 .  Lemma 3.1.5 Let 𝐴, 𝐵, 𝐶 be three profinite groups considered as closed subgroups of 𝐴 ∗ 𝐵 ∗ 𝐶. Then, ( 𝐴 ∗ 𝐵) ∩ (𝐵 ∗ 𝐶) = 𝐵. Proof It suffices to prove that ( 𝐴 ∗ 𝐵) ∩ (𝐵 ∗ 𝐶) ≤ 𝐵. To this end we consider the homomorphism 𝜑 : 𝐴 ∗ 𝐵 ∗ 𝐶 → 𝐴 ∗ 𝐵 ∗ 𝐶 whose restrictions to 𝐴 and 𝐵 are the identity maps and its restriction to 𝐶 is the trivial map. By Lemma 3.1.4, 𝜑 maps 𝐴 ∗ 𝐵 identically onto itself. In particular, if 𝑔 ∈ ( 𝐴 ∗ 𝐵) ∩ (𝐵 ∗ 𝐶), then 𝑔 ∈ 𝐴 ∗ 𝐵, so 𝜑(𝑔) = 𝑔. On the other hand 𝑔 ∈ 𝐵 ∗ 𝐶 and 𝜑(𝐵 ∗ 𝐶) = 𝐵, so 𝜑(𝑔) ∈ 𝐵. Hence, 𝑔 = 𝜑(𝑔) ∈ 𝐵. It follows that ( 𝐴 ∗ 𝐵) ∩ (𝐵 ∗ 𝐶) ⊆ 𝐵.  Remark 3.1.6 In the special case of Proposition 3.1.1, where each 𝐺 𝑖 is isomorphic Î to Zˆ and 𝑔𝑖 is a generator of 𝐺 𝑖 , the free product 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 𝑖 is the free profinite group with a system of free generators h𝑔𝑖 i𝑖 ∈𝐼 . The following result, due to Binz, Neukirch, and Wenzel, is a special case of the main result of [BNW71]. It is based on the Kurosh subgroup theorem for free products of abstract groups. Î Proposition 3.1.7 ([RiZ10, p. 359, Thm. 9.1.9]) Let 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 𝑖 be a free product of finitely many profinite groups and let 𝐻 be an open subgroup of 𝐺. Then, there exists a finitely generated free profinite group 𝐹 and for each 𝑖 ∈ 𝐼 there exists a Ð finite subset 𝑆𝑖 of 𝐺 that contains 1 such that 𝐺 = · 𝜎 ∈𝑆𝑖 𝐺 𝑖 𝜎𝐻 and Ö Ö (3.4) 𝐻=𝐹∗ ∗ ∗ (𝐺 𝑖𝜎 ∩ 𝐻). 𝑖 ∈𝐼

Moreover, rank(𝐹) = 1 − (𝐺 : 𝐻) +

Í

𝜎 ∈𝑆𝑖

𝑖 ∈𝐼


(𝐺 : 𝐻) − card(𝑆𝑖 ) .

The presentation (3.4) of 𝐻 is called the Kurosh decomposition of 𝐻.

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3 Free Products of Finitely Many Profinite Groups

Addendum 3.1.8 In the notation of Proposition 3.1.7, for every 𝜏 ∈ 𝐺 and each 𝑖 ∈ 𝐼 there is a 𝜎 ∈ 𝑆𝑖 such that the group 𝐺 𝑖𝜏 ∩ 𝐻 is 𝐻-conjugate to 𝐺 𝑖𝜎 ∩ 𝐻. Proof By the definition of 𝑆𝑖 , there exist 𝜁 ∈ 𝐺 𝑖 , 𝜎 ∈ 𝑆𝑖 , and 𝜂 ∈ 𝐻 such that 𝜁 𝜎𝜂 𝜏 = 𝜁 𝜎𝜂. Hence, 𝐺 𝑖𝜏 ∩ 𝐻 = 𝐺 𝑖 ∩ 𝐻 𝜂 = (𝐺 𝑖𝜎 ∩ 𝐻) 𝜂 , as claimed.  Î Corollary 3.1.9 Let 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 𝑖 be a free product of finitely many profinite groups. Let 𝐻 be an open subgroup of 𝐺 and for Î each 𝑖 ∈ 𝐼 set 𝐻𝑖 = 𝐺 𝑖 ∩ 𝐻. Then, 𝐺 has a closed subgroup 𝑊 such that 𝐻 = 𝑊∗ ∗ 𝑖 ∈𝐼 𝐻𝑖 . Proof We may write the Kurosh decomposition (3.4) of 𝐻 as  Ö   Ö Ö ∗ (𝐺 𝑖𝜎 ∩ 𝐻) ∗ ∗ (𝐺 𝑖 ∩ 𝐻) . (3.5) 𝐻= 𝐹∗ ∗ 𝑖 ∈𝐼

1≠𝜎 ∈𝑆𝑖

𝑖 ∈𝐼

Let 𝑊 be the free product Î of the factors in the first parentheses. Then, (3.5) obtains the desired form 𝐻 = 𝑊∗ ∗ 𝑖 ∈𝐼 𝐻𝑖 .  The following result is due to Herfort and Ribes [HeR85]: Î Lemma 3.1.10 Let (𝐺 𝑖 )𝑖 ∈𝐼 be a finite family of profinite groups, and let 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 𝑖 be its free profinite product. Then: (a) 𝐺 𝑖 ∩ 𝐺 𝑖𝜏0 = 1 for all distinct 𝑖, 𝑖 0 ∈ 𝐼 and all 𝜏 ∈ 𝐺. (b) 𝐺 𝑖 ∩ 𝐺 𝑖𝜏 = 1 for all 𝑖 ∈ 𝐼 and 𝜏 ∈ 𝐺 r 𝐺 𝑖 . If 𝐺 𝑖 ≠ 1, then 𝑁𝐺 (𝐺 𝑖 ) = 𝐺 𝑖 . Proof Proof of (a). Let 𝜑 : 𝐺 → 𝐺 𝑖 be the homomorphism that satisfies 𝜑|𝐺𝑖 = id𝐺𝑖 and 𝜑(𝐺 𝑗 ) = 1 for all 𝑗 ∈ 𝐼 r {𝑖}. If 𝑖, 𝑖 0 and 𝜏 are as in (a) and 𝑥 ∈ 𝐺 𝑖 ∩ 𝐺 𝑖𝜏0 , then −1 𝜑(𝑥) = 𝑥 because 𝑥 ∈ 𝐺 𝑖 , and 𝜑(𝑥) = 1 because 𝑥 𝜏 ∈ 𝐺 𝑖0 . It follows that 𝑥 = 1, as claimed. Proof of (b). Let 𝑖 ∈ 𝐼 and 𝜏 ∈ 𝐺 r 𝐺 𝑖 . By Lemma 1.1.14, 𝐺 has an open subgroup 𝐻 that contains 𝐺 𝑖 such that 𝜏 ∈ 𝐺 r 𝐻. Then, 𝜏 ∉ 𝐻 = 𝐺 𝑖 · 1 · 𝐻. Consider the Kurosh decomposition (3.4) of 𝐻 and let 𝜎 ∈ 𝑆𝑖 with 𝜏 ∈ 𝐺 𝑖 𝜎𝐻. By the preceding paragraph, 𝐺 𝑖 𝜎𝐻 and 𝐺 𝑖 · 1 · 𝐻 are distinct double cosets of 𝐺. Taking into account that 1 ∈ 𝑆𝑖 , we have that 𝐺 𝑖 = 𝐺 1𝑖 ∩ 𝐻 and 𝐺 𝑖𝑠 ∩ 𝐻 are distinct free factors in the right-hand side of (3.4) for each 𝑠 ∈ 𝑆𝑖 r {1}. The preceding paragraph yields 𝑔 ∈ 𝐺 𝑖 and ℎ ∈ 𝐻 with 𝜏 = 𝑔𝜎ℎ. It follows from (a), applied to the free factorization (3.4) of 𝐻, that 𝐺 𝑖 ∩ 𝐺 𝑖𝜏 = (𝐺 𝑖 ∩ 𝐻) ∩ 𝐺 𝑖𝜏 = 𝐺 𝑖 ∩ (𝐺 𝑖𝜏 ∩ 𝐻) = 𝐺 𝑖 ∩ (𝐺 𝑖𝜎ℎ ∩ 𝐻 ℎ ) = 𝐺 𝑖 ∩ (𝐺 𝑖𝜎 ∩ 𝐻) ℎ = 1, as claimed. Now assume that 𝐺 𝑖 ≠ 1. If 𝑥 ∈ 𝑁𝐺 (𝐺 𝑖 ), then 𝐺 𝑖𝑥 = 𝐺 𝑖 , so 𝐺 𝑖 ∩ 𝐺 𝑖𝑥 = 𝐺 𝑖 ≠ 1. It follows from the first assertion of this part that 𝑥 ∈ 𝐺 𝑖 .  Î Lemma 3.1.11 Let (𝐺 𝑖 )𝑖 ∈𝐼 be a finite family of profinite groups and let 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 𝑖 be their free profinite product. Let 𝐽 be a nonempty subset of 𝐼 such that 𝐺 𝑗 ≠ 1 for each 𝑗 ∈ 𝐽. Put G = {𝐺 𝜏𝑗 | 𝑗 ∈ 𝐽, 𝜏 ∈ 𝐺}. Then: (a) Gmax = G, the set G is strictly closed, any two distinct groups in G intersect in 1, and the étale and strict topologies coincide on G. (b) Let 𝑁 ⊳ 𝐺 be open, and put F = Env(G) r Subgr(𝑁). Then, the map 𝛾 : F → G, mapping every Δ ∈ F to the unique Γ ∈ G with Δ ≤ Γ, is étale continuous.

3.2 Finite Subgroups of Free Products of Finitely Many Profinite Groups

37

Proof Proof of (a). By Lemma 3.1.10, Γ1 ∩ Γ2 = 1 for any two distinct Γ1 , Γ2 ∈ G. In particular, Γ1 6 ⊆ Γ2 , because Γ1 ≠ 1. Thus, Gmax = G. It also follows from Lemma 3.1.10(a) that the sets G 𝑗 = {𝐺 𝜏𝑗 | 𝜏 ∈ 𝐺}, Ð for 𝑗 ∈ 𝐽, are disjoint. By Example 1.3.8, the G 𝑗 ’s are strictly closed. Hence, G = · 𝑗 ∈𝐽 G 𝑗 is strictly closed. By Lemma 1.3.7, the étale topology on G coincides with the strict topology. Proof of (b). The map is well defined: By definition, every Δ ∈ Env(G) is contained in some Γ ∈ G. If Δ ≠ 1, then this Γ is unique, by (a). So, let Δ ∈ F and let Γ := 𝛾(Δ) ∈ G be its image. Thus, Γ is the unique group in G with Δ ≤ Γ. Since, by (a), G is strictly closed, and Subgr(𝐺) is strictly profinite (third paragraph of Section 1.2), G is strictly profinite. Hence, by (a), G is also étale profinite. Therefore, it suffices to consider an étale open-closed neighborhood G1 of Γ in G and prove that there exists an étale open neighborhood of Δ in F that 𝛾 maps into G1 . To this end we set G2 = G r G1 , so that G = G1 ∪· G2 . Moreover, (3.6)

𝑈 (G1 ) ∩ 𝑈 (G2 ) = 1.

Indeed, if 𝑥 ∈ 𝑈 (G1 ) ∩ 𝑈 (G2 ), then 𝑥 ∈ Γ1 ∩ Γ2 for distinct groups Γ1 ∈ G1 and Γ2 ∈ G2 . By (a), 𝑥 = 1, as claimed. It follows that 𝑋 = 𝑋1 ∪· 𝑋2 , where (3.7) 𝑋 = 𝑈 (G) r 𝑁, 𝑋1 = 𝑈 (G1 ) r 𝑁, 𝑋2 = 𝑈 (G2 ) r 𝑁. By Lemma 1.2.5, 𝑋, 𝑋1 , 𝑋2 are closed in 𝐺. Hence, 𝑋 is a profinite space, and 𝑋1 is an open subset of 𝑋. We have Δ ≤ Γ ∈ G1 , hence Δ r 𝑁 is a closed subset of 𝑋1 , so also of 𝑋. Let H be Ñ the family of open subgroups 𝐻 of 𝐺 containing Δ. By Lemma 1.1.14, Δ = 𝐻 ∈H 𝐻, Ñ hence Δ r 𝑁 = 𝐻 ∈H 𝑋 ∩ 𝐻. This is an intersection of a family of closed subsets of 𝑋, and this family is closed under finite intersections. Since 𝑋 is compact and 𝑋1 is open in 𝑋, there is, by Lemma 1.1.1, an 𝐻 ∈ H such that 𝑋 ∩ 𝐻 ⊆ 𝑋1 . If Δ0 ∈ F ∩ Subgr(𝐻), then Δ0 6 ≤ 𝑁. Since Δ0 r 𝑁 ⊆ 𝑋, we have ∅ ≠ Δ0 r 𝑁 ⊆ 𝑋 ∩ 𝐻 ⊆ 𝑋1 . In particular, by (3.6) and (3.7), Δ0 is not contained in an element of G2 . Therefore, since G = G1 ∪· G2 , Δ0 is contained in an element Γ0 of G1 . Thus, 𝛾(Δ0) = Γ0 ∈ G1 . With this, we have proved that 𝛾 maps the étale open neighborhood F ∩Subgr(𝐻) of Δ into the étale open neighborhood G1 of Γ, as desired. 

3.2 Finite Subgroups of Free Products of Finitely Many Profinite Groups We present a result of Herfort and Ribes that a finite subgroup of a free product of finitely many profinite groups is contained in a conjugate of one of the factors of this product. Lemma 3.2.1 Every finite subgroup of a free profinite group is trivial.

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3 Free Products of Finitely Many Profinite Groups

Proof By [FrJ08, p. 507, Cor. 22.4.6], every free profinite group 𝐹 is projective. Hence, by [FrJ08, p. 507, Prop. 22.4.7], 𝐹 is torsion free. In particular, every finite subgroup of 𝐹 is trivial, as asserted.  Lemma 3.2.2 Let 𝐻 be a closed subgroup of a profinite group 𝐺 and let 𝐴 be a finite subgroup of 𝐺. Suppose that 𝐴 ∩ 𝐻 𝑥 = 1 for each 𝑥 ∈ 𝐺. Then, 𝐺 has an open normal subgroup 𝑁 such that 𝐴𝑁 ∩ 𝐻 𝑥 𝑁 = 𝑁 for each 𝑥 ∈ 𝐺. Ð Proof Let 𝑎 ≠ 1 be an element of 𝐴. By assumption, 𝑎 ∉ 𝑥 ∈𝐺 𝐻 𝑥 = 𝐻 𝐺 . Since 𝐻 𝐺 is a closed subset of 𝐺 (Lemma 1.1.19), Ñ 𝐺 has an open normal subgroup 𝑁 𝑎 such that 𝑎𝑁 𝑎 ∩ 𝐻 𝐺 = ∅. Since 𝐴 is finite, 𝑁 = 𝑎 ∈ 𝐴r1 𝑁 𝑎 is an open normal subgroup of 𝐺. Moreover, 𝑎𝑁 ∩ 𝐻 𝐺 = ∅ for each 𝑎 ∈ 𝐴 r 1. Therefore, 𝐴𝑁 ∩ 𝐻 𝑥 𝑁 = 𝑁 for each 𝑥 ∈ 𝐺, as claimed.  Î We consider a finite set 𝐼 and a free Î product 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 𝑖 of profinite groups 𝐺 𝑖 . We also consider the direct product 𝑖 ∈𝐼 𝐺 𝑖 of the groups 𝐺 𝑖 and the unique epimorphism Ö (3.8) 𝜅: 𝐺 → 𝐺𝑖 𝑖 ∈𝐼

whose restriction to each 𝐺 𝑖 is the identity map. We call 𝜅 the cartesian map of 𝐺 and Ker(𝜅) the cartesian kernel of 𝐺. Lemma 3.2.3 If each of the groups 𝐺 𝑖 is finite, then the cartesian kernel of 𝐺 is a free profinite group that is open and normal in 𝐺. Î Proof Since the 𝐺 𝑖 ’s are finite, so is 𝑖 ∈𝐼 𝐺 𝑖 . Hence, 𝐾 = Ker(𝜅) is an open normal Î Î subgroup of 𝐺. Let 𝐾  𝐹∗ ∗ 𝑖 ∈𝐼 ∗ 𝜎 ∈𝑆𝑖 (𝐺 𝑖𝜎 ∩ 𝐾) be the Kurosh decomposition of 𝐾 (Proposition 3.1.7). Since 𝜅 is injective on each 𝐺 𝑖 , it is injective on 𝐺 𝑖𝜎 , so 𝐺 𝑖𝜎 ∩ 𝐾 = 1 for each 𝑖 ∈ 𝐼 and 𝜎 ∈ 𝑆𝑖 . It follows that 𝐾 is isomorphic to the free profinite group 𝐹.  Lemma 3.2.4 Suppose that each of the free factors 𝐺 𝑖 of 𝐺 is finite and let 𝐴 be a non-trivial finite subgroup of 𝐺. Then, there exist 𝑖 ∈ 𝐼 and 𝑥 ∈ 𝐺 such that 𝐺 𝑖𝑥 ∩ 𝐴 ≠ 1. Proof Assume toward contradiction that (3.9)

𝐺 𝑖𝑥 ∩ 𝐴 = 1 for every 𝑖 ∈ 𝐼 and 𝑥 ∈ 𝐺.

By Lemma 3.2.2, 𝐺 has an open normal subgroup 𝑁 such that (3.10)

𝐺 𝑖𝑥 𝑁 ∩ 𝐴𝑁 = 𝑁 for every 𝑖 ∈ 𝐼 and 𝑥 ∈ 𝐺.

By Lemma 3.2.3, the cartesian kernel 𝐾 of 𝐺 is a free profinite group. Moreover, 𝐾 is an open normal subgroup of 𝐺. Hence, 𝐾 ∩ 𝑁 is a free profinite group [FrJ08, p. 358, Prop. 17.6.2], which is open and normal in 𝐺. We replace 𝑁 by 𝐾 ∩ 𝑁, if necessary, to assume that 𝑁 is free. Let Ö Ö (3.11) 𝐴𝑁 = 𝐹 ∗ ∗ ∗ (𝐺 𝑖𝜎 ∩ 𝐴𝑁) 𝑖 ∈𝐼

𝜎 ∈𝑆𝑖

3.2 Finite Subgroups of Free Products of Finitely Many Profinite Groups

39

be the Kurosh decomposition of 𝐴𝑁 (Proposition 3.1.7). By (3.10), for every 𝑖 ∈ 𝐼 and 𝜎 ∈ 𝑆𝑖 the group 𝐺 𝑖𝜎 ∩ 𝐴𝑁 is contained in the free profinite group 𝑁. On the other hand, 𝐺 𝑖𝜎 ∩ 𝐴𝑁 is contained in the finite group 𝐺 𝑖𝜎 . Hence, by Lemma 3.2.1, 𝐺 𝑖𝜎 ∩ 𝐴𝑁 = 1. Therefore, by (3.11), 𝐴𝑁  𝐹 is a free profinite group. But then, again, 𝐴 as a finite subgroup of the free profinite group 𝐹 is trivial. This contradicts the assumption on 𝐴.  Lemma 3.2.5 Suppose that each of the free factors 𝐺 𝑖 of 𝐺 is finite and let 𝐴 be a non-trivial finite subgroup of 𝐺. Then, there exists a unique 𝑖 ∈ 𝐼 with 𝐴 ⊆ 𝐺 𝑖𝐺 . Proof The uniqueness of 𝑖 follows from Lemma 3.1.10. We prove its existence. By Lemma 3.2.4, there exist 𝑖 ∈ 𝐼, 𝑥 ∈ 𝐺, and 𝑎 ∈ 𝐺 𝑖𝑥 ∩ 𝐴 r 1. We consider another element 𝑏 ∈ 𝐴 r 1 and apply Lemma 3.2.4 to the group h𝑏i in order to find 𝑗 ∈ 𝐼 and 𝑐 ∈ 𝐺 𝐺𝑗 ∩ h𝑏i r 1. Assume toward contradiction that 𝑐 ∉ 𝐺 𝑖𝐺 . In particular, Î Î 𝑖 ≠ 𝑗. We consider the cartesian map 𝜅 : 𝐺 → 𝐺 . Then, 𝐺 ∩ 𝐺 = 1 in 𝑘 𝑖 𝑗 𝑘 ∈𝐼 𝑘 ∈𝐼 𝐺 𝑘 . Since each 𝐺 𝑘 is Î normal in 𝑘 ∈𝐼 𝐺 𝑘 , we have 𝜅( [𝑎, 𝑐]) = [𝑎, 𝑐] ∈ 𝐺 𝑖 ∩𝐺 𝑗 = 1. Hence, [𝑎, 𝑐] belongs to the free profinite group 𝐾 = Ker(𝜅) (Lemma 3.2.3). On the other hand, [𝑎, 𝑐] belongs to the finite group 𝐴, so [𝑎, 𝑐] is of finite order. Therefore, by Lemma 3.2.1, [𝑎, 𝑐] = 1. It follows that 𝑎 = 𝑎 𝑐 ∈ 𝐺 𝑖𝑥𝑐 . On the other hand, 𝑎 ∈ 𝐺 𝑖𝑥 . Hence, 𝐺 𝑖𝑥 ∩ 𝐺 𝑖𝑥𝑐 ≠ 1. Therefore, by Lemma 3.1.10, 𝑥𝑐𝑥 −1 ∈ 𝐺 𝑖 , so 𝑐 ∈ 𝐺 𝑖𝑥 . The latter contradiction to our assumption implies that there exists a 𝑦 ∈ 𝐺 with 𝑦 𝑦𝑏 𝑦 𝑦𝑏 𝑐 ∈ 𝐺 𝑖 . Since 𝑐 is a power of 𝑏, we have 𝑐 = 𝑐 𝑏 ∈ 𝐺 𝑖 . Hence, 𝐺 𝑖 ∩ 𝐺 𝑖 ≠ 1. 𝑦 Again, by Lemma 3.1.10, 𝑦𝑏𝑦 −1 ∈ 𝐺 𝑖 . Therefore, 𝑏 ∈ 𝐺 𝑖 ⊆ 𝐺 𝑖𝐺 , as claimed.  The next result strengthens Lemma 3.2.5 to the case where the 𝐺 𝑖 ’s are arbitrary profinite groups. Lemma 3.2.6 Let 𝐴 be a non-trivial finite subgroup of 𝐺. Then, there exists a unique 𝑖 ∈ 𝐼 with 𝐴 ⊆ 𝐺 𝑖𝐺 . Proof Again, the uniqueness of 𝑖 follows from Lemma 3.1.10. We prove its existence. To this end we consider the following diagram of profinite groups: Î 𝛼 /Î 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 𝑖 ∗ 𝑖 ∈𝐼 𝐺 𝑖 /(𝐺 𝑖 ∩ 𝑁)= 𝐺¯ 𝜋

$ w

𝛽

𝐺/𝑁 in which 𝑁 is an open normal subgroup of 𝐺, the restriction of 𝛼 to each 𝐺 𝑖 is the quotient map 𝐺 𝑖 → 𝐺 𝑖 /(𝐺 𝑖 ∩ 𝑁), the restriction of 𝛽 to each of the groups 𝐺 𝑖 /(𝐺 𝑖 ∩ 𝑁) is the corresponding inclusion, and 𝜋 : 𝐺 → 𝐺/𝑁 is the quotient map. Since for each 𝑔 ∈ 𝐺 𝑖 we have 𝛽(𝛼(𝑔)) = 𝜋(𝑔), the diagram is commutative. We choose 𝑁 such that 𝜋( 𝐴) ≠ 1. Then, 𝛼( 𝐴) is a non-trivial finite subgroup ¯ Since all of the free factors 𝐺 𝑖 /(𝐺 𝑖 ∩ 𝑁) of 𝐺¯ are finite, Lemma 3.2.5 of 𝐺. ¯ gives a unique 𝑖 ∈ 𝐼 with 𝛼( 𝐴) ⊆ (𝐺 𝑖 /𝐺 𝑖 ∩ 𝑁) 𝐺 . Applying 𝛽 to both sides gives 𝐺 𝐺/𝑁 𝜋( 𝐴) ⊆ (𝐺 𝑖 𝑁/𝑁) , hence 𝐴 ≤ 𝐺 𝑖 𝑁.

40

3 Free Products of Finitely Many Profinite Groups

The element 𝑖 may depend on 𝑁. But, since 𝐼 is a finite set, there exist 𝑖 ∈ 𝐼 and a set N of open normal subgroups of 𝐺 whose intersection is 1 such that 𝐴 ≤ 𝐺 𝑖𝐺 𝑁 for each 𝑁 ∈ N . Since 𝐺 𝑖𝐺 is closed in 𝐺 (Lemma 1.1.19), we have 𝐴 ⊆ 𝐺 𝑖𝐺 (Lemma 1.1.12(b)), as claimed.  The following result due to Herfort and Ribes is proved in [Rib17, p. 223, Thm. 7.1.2] using the machinery of fundamental groups of graphs of groups. Proposition 3.2.7 For each 𝑖 in a finite set 𝐼 let 𝐺 𝑖 be a profinite group and set Î 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 𝑖 . Then, for each non-trivial finite subgroup 𝐴 of 𝐺 there exist 𝑖 ∈ 𝐼 and 𝑥 ∈ 𝐺 such that 𝐴 𝑥 ≤ 𝐺 𝑖 . Proof By Lemma 3.2.6, there exists a unique 𝑖 ∈ 𝐼 with 𝐴 ⊆ 𝐺 𝑖𝐺 . Thus, given −1 −1 𝑎 ∈ 𝐴 r 1 there exists an 𝑥 ∈ 𝐺 with 𝑎 ∈ 𝐺 𝑖𝑥 . Replacing 𝑎 by 𝑎 𝑥 and 𝐴 by 𝐴 𝑥 , we may assume that 𝐴0 = 𝐺 𝑖 ∩ 𝐴 ≠ 1 and 𝑎 ∈ 𝐴0 . Now consider another non-trivial element 𝑏 of 𝐴. Then, (3.12)

𝑏 = 𝑔 𝑦 for some 𝑔 ∈ 𝐺 𝑖 and 𝑦 ∈ 𝐺.

Claim: 𝑦 ∈ 𝐺 𝑖 𝐴. Otherwise, 𝐺 has an open normal subgroup 𝑁 with 𝑦𝑁 ∩ 𝐺 𝑖 𝐴 = ∅. In particular, 𝑦 ∉ 𝐺 𝑖 𝐴𝑁. We set 𝐻 = 𝐴𝑁 and let Ö Ö ∗ (𝐺 𝜎 (3.13) 𝐻𝐹∗ ∗ 𝑗 ∩ 𝐻) 𝑗 ∈𝐼

𝜎 ∈𝑆 𝑗

be the Kurosh decomposition of 𝐻 Ð (Proposition 3.1.7). In particular, 𝑆𝑖 is a finite subset of 𝐺 that contains 1 and 𝐺 = · 𝜎 ∈𝑆𝑖 𝐺 𝑖 𝜎𝐻. Thus, there are 𝛾 ∈ 𝐺 𝑖 , 𝜎 ∈ 𝑆𝑖 , and ℎ ∈ 𝐻 such that 𝑦 = 𝛾𝜎ℎ and 𝜎 ≠ 1. Then, 𝑏 = 𝑔 𝑦 ∈ 𝐺 𝑖𝜎ℎ and 𝑏 ∈ 𝐴 ≤ 𝐻 = 𝐻 ℎ , hence 𝑏 ∈ (𝐺 𝑖𝜎 ∩ 𝐻) 𝐻 . But 𝑎 ∈ 𝐺 𝑖 ∩ 𝐴 ≤ 𝐺 1𝑖 ∩ 𝐻. This contradicts Lemma 3.2.6 applied to the decomposition (3.13) of 𝐻. 0 It follows from the Claim that 𝑦 = 𝑔𝑖 𝑎 0 with 𝑔𝑖 ∈ 𝐺 𝑖 and 𝑎 0 ∈ 𝐴. Hence, 𝑏 = 𝑔 𝑔𝑖 𝑎 and 𝑔 𝑔𝑖 = 𝑎 0 𝑏(𝑎 0) −1 ∈ 𝐴. In addition 𝑔 𝑔𝑖 ∈ 𝐺 𝑖 , so 𝑔 𝑔𝑖 ∈ 𝐺 𝑖 ∩ 𝐴 = 𝐴0 . Therefore, 0 𝑏 = 𝑔 𝑔𝑖 𝑎 ∈ 𝐴0𝐴. Thus, 𝐴 = 𝐴0𝐴. Since 𝐴 is finite, the latter equality implies that 𝐴0 = 𝐴 [FrJ08, p. 238, Lemma 13.3.2]. Therefore, 𝐴 ≤ 𝐺 𝑖 , as claimed.  Î Corollary 3.2.8 Let 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 𝑖 be a free product of finitely many profinite groups 𝐺 𝑖 . Consider 𝑖 ∈ 𝐼 and let 𝐴 be a finite subgroup of 𝐺. Then, the subset 𝑆 := {𝜎 ∈ 𝐺 | 𝐴 ≤ 𝐺 𝑖𝜎 } of 𝐺 is closed. Proof If 𝐴 = 1, then 𝑆 = 𝐺, so 𝑆 is closed. Otherwise, if 𝜎, 𝜏 ∈ 𝑆, then 𝐺 𝑖𝜎 ∩𝐺 𝑖𝜏 ≠ 1, −1 so 𝐺 𝑖 ∩ 𝐺 𝑖𝜏 𝜎 ≠ 1. Hence, by Lemma 3.1.10, 𝜏𝜎 −1 ∈ 𝐺 𝑖 , so 𝜏 ∈ 𝐺 𝑖 𝜎. It follows that either 𝑆 = ∅ or 𝑆 = 𝐺 𝑖 𝜎, which is a closed subset of 𝐺, as claimed. 

Chapter 4

Generalized Free Products

The chapter introduces two equivalent types of free products of profinite groups. Both types have been independently introduced by the first author in [Har87] and by Oleg V. Melnikov in [Mel90].Î We adopt here the terminology of [Mel90]. The first type is written as ∗ 𝑡 ∈𝑇 𝐺 𝑡 , where 𝑇 is a compact 𝑇1 -topological space Î and the 𝐺 𝑡 ’s are closed subgroups of ∗ 𝑡 ∈𝑇 𝐺 𝑡 . We consider this type as an “inner free product” and refer to the second type as an “outer free product”. While the inner free product seems to be simpler in its description, the outer free product is technically easier to use. So, we prove properties of the outer free products and translate them to inner free products. Eventually we generalize Lemma 3.1.10 Î to inner free products: Thus, for ∗ 𝑡 ∈𝑇 𝐺 𝑡 as above, 𝐺 𝑟 ∩ 𝐺 𝑠𝑥 = 1 for all distinct Î 𝑔 𝑟, 𝑠 ∈ 𝑇 and 𝑥 ∈ ∗ 𝑡 ∈𝑇 𝐺 𝑡 ; moreover, 𝐺 𝑡 = 𝐺 𝑡 implies 𝑔 ∈ 𝐺 𝑡 for all 𝑡 ∈ 𝑇 and Î 𝑔 ∈ ∗ 𝑡 ∈𝑇 𝐺 𝑡 . Finally, we generalize Proposition 3.2.7 and prove that for every finite subgroup 𝐴 of 𝐺 there exist 𝑡 ∈ 𝑇 and 𝑥 ∈ 𝐺 such that 𝐴 ≤ 𝐺 𝑡𝑥 . Most of the material in Chapters 2 and 4 is covered in Chapter 5 of [Rib17]. The proofs may appear different, because we often employ the notion of étale compactness, dealt with in Chapter 1. Our treatment is intentionally less comprehensive, with the aim to obtain only those results that are needed in subsequent chapters, with occasional enrichments, that we have found interesting enough. However, note that when dealing with families of subgroups of a profinite group, indexed by a topological space, we do not limit ourselves to profinite spaces, since in Chapter 7 we deal with étale compact subsets of subgroups that are not necessarily profinite.

4.1 Inner Free Products We extend the notion of a free product of finitely many profinite groups to a notion of a free product of profinite groups indexed by a compact 𝑇1 -space. Definition 4.1.1 Let 𝐺 be a profinite group and 𝑇 a compact 𝑇1 -topological space and let G = {Γ𝑡 }𝑡 ∈𝑇 be an étale continuous family of closed subgroups of 𝐺 (Definition 2.1.1). Thus, Γ𝑠 ∩ Γ𝑡 = 1 if 𝑠, 𝑡 ∈ 𝑇 are distinct, and the map 𝑡 ↦→ Γ𝑡 from 𝑇 into Subgr(𝐺) is étale continuous. In particular, G is stellate (Definition 1.3.3). © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 D. Haran and M. Jarden, The Absolute Galois Group of a Semi-Local Field, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-030-89191-6_4

41

42

4 Generalized Free Products

We say that 𝐺 is an (inner) free product (in the sense of Melnikov) ofÎ the groups Î Γ𝑡 , with 𝑡 ranging over 𝑇, and write 𝐺 = ∗ 𝑡 ∈𝑇 Γ𝑡 (or, occasionally, 𝐺 = ∗ Γ∈ G Γ) if the following condition is satisfied: (4.1) Every continuous map 𝛾0 from 𝑈 (G) into a profinite group 𝐴, whose restriction to each of the groups Γ𝑡 is a homomorphism, uniquely extends to a homomorphism 𝛾 : 𝐺 → 𝐴. Occasionally, we drop the adjective “inner”, the clause “over 𝑇”, or the clause “in the sense of Melnikov”. Remark 4.1.2 Since every profinite group 𝐴 is an inverse limit of finite groups, the requirement on 𝛾 to be unique in Condition (4.1) implies that it suffices to demand that Condition (4.1) hold for finite 𝐴. Our definition of a free product over 𝑇 slightly generalizes that of Melnikov [Mel90, (1.2)], where 𝑇 is assumed to be profinite. The latter definition generalizes the notion of a free product of finitely many profinite groups introduced in Section 3.1: Lemma 4.1.3 Let 𝐺 be a profinite group and let 𝑇 be a finite discrete space. For each 𝑡 ∈ 𝑇 let Γ𝑡 be a closed subgroup of 𝐺. Then, 𝐺 is a free product of the groups Γ𝑡 , with 𝑡 ∈ 𝑇, in the sense of Melnikov if and only if 𝐺 is a free product of the groups Γ𝑡 in the sense of Proposition 3.1.1. Proof First we suppose that G = {Γ𝑡 | 𝑡 ∈ 𝑇 } is an étale continuous family that satisfies (4.1). Let (𝛾𝑡 : Γ𝑡 → 𝐴)𝑡 ∈𝑇 be a system of homomorphisms into a profinite group 𝐴. Since Γ𝑡 ∩ Γ𝑠 = 1 for 𝑠 ≠ 𝑡, there exists a unique map 𝛾0 : 𝑈 (G) = Ð |Γ𝑡 = 𝛾𝑡 for each 𝑡 ∈ 𝑇. For every open subset 𝑉 of 𝐴 the 𝑡 ∈𝑇 Γ𝑡 → 𝐴 with 𝛾0Ð inverse set 𝛾0−1 (𝑉) = 𝑡 ∈𝑇 𝛾𝑡−1 (𝑉) is open in 𝑈 (G), hence 𝛾0 is continuous. By (4.1), 𝛾0 uniquely extends to a homomorphism 𝛾 : 𝐺 → 𝐴. Hence, 𝐺 is a free product of the groups Γ𝑡 in the sense of Proposition 3.1.1. Conversely, we suppose that the conditions of Proposition 3.1.1 hold with 𝐼 = 𝑇. Since 𝑇 is discrete, the map 𝑡 ↦→ Γ𝑡 is étale continuous. By Lemma 3.1.10, Γ𝑠 ∩Γ𝑡 = 1 if 𝑠 ≠ 𝑡. Thus, {Γ𝑡 | 𝑡 ∈ 𝑇 } is an étale continuous family. Finally, Condition (4.1) follows from Condition (b) of Proposition 3.1.1. Thus, 𝐺 is a free product in the sense of Melnikov.  Remark 4.1.4 Let G = {Γ𝑡 }𝑡 ∈𝑇 be an étale continuous family of closed subgroups of · a profinite group 𝐺. Let 𝑇 0 = 𝑇 ∪{0} be the disjoint union of the topological spaces 𝑇 and the one-point space {0}, and let Γ0 = 1. Then, G 0 := {Γ𝑡 }𝑡 ∈𝑇 0 = G ∪ {1} 0 is also an étale continuous family of closed subgroups of 𝐺. Since 𝑈 (G) Î = 𝑈 (G ) 0 (except in the trivial Î case of G = ∅, in which 𝑈 (G), 𝑈 (G ) ⊆ 1), 𝐺 = ∗ 𝑡 ∈𝑇 Γ𝑡 if and only if 𝐺 = ∗ 𝑡 ∈𝑇 0 Γ𝑡 . In this sense we may assume without loss that 1 ∈ G.

4.2 Outer Free Products

43

Remark 4.1.5 Let 𝐺 be a profinite group and G a separated subset of Subgr(𝐺). Then, [Har87] defines 𝐺 to be a free product of the subgroups in G if (4.1) holds. This definition is equivalent to Definition 4.1.1, in which G is assumed to be étale continuous relative to a profinite space 𝑇 rather than separated. Indeed, by Remarks 4.1.4 and 2.1.6(c) we may assume that 1 ∈ G. It then follows from Propositions 2.1.8 and 2.1.7 that G is étale continuous relative to a profinite space 𝑇 if and only it is separated.

4.2 Outer Free Products We follow [Mel90, (1.14)] and represent inner free products of profinite groups as “outer free products of profinite groups”. Then, we use the new representation to prove more properties of inner free products of profinite groups. Definition 4.2.1 A free profinite product over a sheaf X is a sheaf-group structure (X, 𝜔, 𝐺), with the following universal property: (4.2) For every morphism 𝛼 of X into a profinite group 𝐴 there exists a unique homomorphism 𝜑 : 𝐺 → 𝐴 such that 𝜑 ◦ 𝜔 = 𝛼. The uniqueness part of Condition (4.2) and an inverse limit argument show that in order to prove (4.2) it suffices to prove it only for finite groups 𝐴. We loosely refer to the free profinite product over a sheaf also as the outer free product. Proposition 4.2.2 Over every sheaf X := (𝑋, 𝜏, 𝑇) of profinite groups there exists a free profinite product (X, 𝜔, 𝐺), unique up to isomorphism. Proof The uniqueness of the free profinite product over X follows from the universal property (4.2). Indeed, if another sheaf-group structure (X, 𝜔 0, 𝐺 0) satisfies (4.2), then there exist homomorphisms 𝜑 : 𝐺 → 𝐺 0 and 𝜑 0 : 𝐺 0 → 𝐺 such that 𝜑 ◦ 𝜔 = 𝜔 0 and 𝜑 0 ◦𝜔 0 = 𝜔. The uniqueness in (4.2) guarantees that 𝜑◦𝜑 0 = id𝐺0 and 𝜑 0 ◦𝜑 = id𝐺 . Hence, both 𝜑 and 𝜑 0 are isomorphisms. We prove the existence of the free profinite product over X by direct construction. The proof breaks up into three parts. Part A: Construction of 𝐺. We consider the abstract free product 𝐹 of the groups 𝑋𝑡 with Ð 𝑡 ranging over 𝑇. In particular, we consider every 𝑋𝑡 as a subgroup of 𝐹. Since 𝑋 = · 𝑡 ∈𝑇 𝑋𝑡 , there exists a unique map 𝜔0 : 𝑋 → 𝐹, the restriction of which to every 𝑋𝑡 is the identity map. For every normal subgroup 𝑁 of 𝐹 let 𝜋 𝑁 : 𝐹 → 𝐹/𝑁 be the quotient map. Let N be the collection of all normal subgroups 𝑁 of 𝐹 of finite index such that the map 𝜋 𝑁 ◦ 𝜔0 : 𝑋 → 𝐹/𝑁 is continuous. In other words, 𝜔−1 0 ( 𝑓 𝑁) is an open subset of 𝑋 for every 𝑓 ∈ 𝐹. If 𝑁1 , 𝑁2 ∈ N , then 𝑓 (𝑁1 ∩ 𝑁2 ) = 𝑓 𝑁1 ∩ 𝑓 𝑁2 , so 𝑁1 ∩ 𝑁2 ∈ N . Thus, we may consider the profinite group 𝐺 = lim 𝐹/𝑁 [FrJ08, ←−− 𝑁 ∈N p. 340, Section 17.2]. Let 𝜃 : 𝐹 → 𝐺 be the map defined by 𝜃 ( 𝑓 ) = ( 𝑓 𝑁)𝑁 ∈N and let 𝜔 = 𝜃 ◦ 𝜔0 .

44

4 Generalized Free Products

We prove that (X, 𝜔, 𝐺) is a free profinite product over X. Part B: The morphism 𝜔. For every 𝑁 ∈ N we denote the closure of 𝜃 (𝑁) in 𝐺 by ˆ By [FrJ08, p. 341, Lemma 17.2.1(a3), (b)] the family of all subsets 𝜃 ( 𝑓 ) 𝑁ˆ of 𝐺 𝑁. ˆ = 𝜔−1 ( 𝑓 𝑁) with 𝑓 ∈ 𝐹 and 𝑁 ∈ N is a basis for the topology on 𝐺 and 𝜔−1 (𝜃 ( 𝑓 ) 𝑁) 0 is open in 𝑋. Hence, 𝜔 is continuous. For every 𝑡 ∈ 𝑇 the map 𝜔0 maps 𝑋𝑡 identically onto its copy in 𝐹. In particular, 𝜔0 is a homomorphism on 𝑋𝑡 . Since 𝜃 is a homomorphism, 𝜔|𝑋𝑡 is a homomorphism. Therefore, 𝜔 : X → 𝐺 is a morphism (Definition 2.3.2). Part C: The universal property. We consider a profinite group 𝐴 and a morphism 𝛼 : X → 𝐴. Since 𝜔0 is the identity map on every 𝑋𝑡 , there exists a unique homomorphism 𝜑0 : 𝐹 → 𝐴 such that 𝛼 = 𝜑0 ◦ 𝜔0 . If 𝐴0 is an open normal subgroup of −1 −1 𝐴 and 𝑓 ∈ 𝐹, then 𝜔−1 0 ( 𝑓 𝜑0 ( 𝐴0 )) = 𝛼 (𝜑0 ( 𝑓 ) 𝐴0 ) is open (because 𝛼 : 𝑋 → 𝐴 is −1 continuous). Hence, 𝜑0 ( 𝐴0 ) ∈ N . By [FrJ08, p. 343, Lemma 17.2.2], there exists a unique homomorphism 𝜑 : 𝐺 → 𝐴 such that 𝜑0 = 𝜑 ◦ 𝜃. Hence, 𝜑 ◦ 𝜔 = 𝛼. Finally, if 𝜑 0 : 𝐺 → 𝐴 is an additional homomorphism that satisfies 𝜑 0 ◦ 𝜔 = 𝛼, then 𝜑00 := 𝜑 0 ◦ 𝜃 satisfies 𝜑00 ◦ 𝜔0 = 𝛼. : 𝐺O 𝜔 𝜃 𝜑0

Ð 𝑋 = · 𝑡 ∈𝑇 𝑋𝑡

𝜔0 𝛼

/𝐹 𝜑00

𝜋𝑁

! / 𝐹/𝑁

𝜑

𝜑0

$,   t 𝐴

Hence, by the uniqueness of 𝜑0 in the preceding paragraph, 𝜑00 = 𝜑0 , so 𝜑 0 ◦𝜃 = 𝜑◦𝜃. By the uniqueness of 𝜑 in the preceding paragraph, 𝜑 0 = 𝜑.  We use Corollary 2.3.3 to prove some basic properties of the outer free product over a profinite sheaf. Lemma 4.2.3 Let (X, 𝜔, 𝐺) be the free profinite product over a profinite sheaf X = (𝑋, 𝜏, 𝑇). For every 𝑡 ∈ 𝑇 consider the closed subgroup 𝐺 𝑡 = 𝜔(𝑋𝑡 ) of 𝐺. Then: (a) 𝜔 maps 𝑋𝑡 isomorphically onto 𝐺 𝑡 . (b) If 𝑠 and 𝑡 are distinct elements of 𝑇, then 𝐺 𝑠 ∩ 𝐺 𝑡 = 1. (c) 𝐺 = h𝐺 𝑡 i𝑡 ∈𝑇 . (d) (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) is the associated sheaf-group structure of {𝐺 𝑡 }𝑡 ∈𝑇 . (e) 𝜔 is injective on 𝑋 r 𝜔−1 (1). Proof Proof of (a) and (b). Let 𝑥 ∈ 𝑋𝑡 , 𝑥 ≠ 1. Then, 𝑋𝑡 has an open normal subgroup 𝑁 that does not contain 𝑥. Let 𝐴 = 𝑋𝑡 /𝑁 and let 𝛼𝑡 : 𝑋𝑡 → 𝐴 be the quotient map, so 𝛼𝑡 (𝑥) ≠ 1. By Corollary 2.3.3, 𝛼𝑡 extends to a morphism 𝛼 : X → 𝐴 such that 𝛼(𝑋𝑠 ) = 1. By (4.2), there exists a homomorphism 𝜑 : 𝐺 → 𝐴 such that 𝜑 ◦ 𝜔 = 𝛼. In particular, 1 ≠ 𝛼𝑡 (𝑥) = 𝛼(𝑥) = 𝜑(𝜔(𝑥)), hence 𝜔(𝑥) ≠ 1. Thus, 𝜔 maps 𝑋𝑡 injectively onto 𝐺 𝑡 , as claimed.

4.2 Outer Free Products

45

But 𝜑(𝐺 𝑠 ) = 𝜑(𝜔(𝑋𝑠 )) = 𝛼(𝑋𝑠 ) = 1, so 𝜔(𝑥) is not in 𝐺 𝑠 . This implies that 𝐺 𝑠 ∩ 𝐺 𝑡 = 1. Proof of (c). Let 𝐺 0 = h𝐺 𝑡 | 𝑡 ∈ 𝑇i and let 𝜔 0 : 𝑋 → 𝐺 0 be 𝜔, with restricted range. Then, (X, 𝜔 0, 𝐺 0) is also a free profinite product over X. Indeed, let 𝛼 be a morphism of X into a profinite group 𝐴. Then, there exists a homomorphism 𝜑 : 𝐺 → 𝐴 such that 𝜑 ◦ 𝜔 = 𝛼. Then, 𝜑 0 := 𝜑|𝐺0 also satisfies 𝜑 0 ◦ 𝜔 0 = 𝛼. Moreover, since 𝐺 0 = h𝜔 0 (𝑋𝑡 ) | 𝑡 ∈ 𝑇i, this 𝜑 0 is unique. By the uniqueness of the free profinite product (Proposition 4.2.2), there is an isomorphism 𝐺 0 → 𝐺 that maps every 𝐺 𝑡 identically onto itself. Hence, 𝐺 0 = 𝐺. Proof of (d). By Lemma 2.3.5(a), the map 𝑡 ↦→ 𝐺 𝑡 is étale continuous. By (a), 𝜔 is injective on 𝑋𝑡 , for every 𝑡 ∈ 𝑇. Hence, by Proposition 2.3.6, (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) is the associated sheaf-group structure. Proof of (e). Let 𝑥 and 𝑥 0 be elements of 𝑋 r 𝜔−1 (1) with 𝜔(𝑥) = 𝜔(𝑥 0). Then, 𝑥 ∈ 𝑋𝑡 and 𝑥 0 ∈ 𝑋𝑡 0 for some 𝑡, 𝑡 0 ∈ 𝑇. If 𝑡 = 𝑡 0, then 𝑥 = 𝑥 0, by (a). Otherwise, 𝑡 ≠ 𝑡 0 and 𝜔(𝑥) ∈ 𝐺 𝑡 ∩ 𝐺 𝑡 0 = 1 (by (b)), which contradicts the assumption on 𝑥.  Remark 4.2.4 Let X be the inverse limit of an inverse system (X𝑖 , 𝜋 𝑗𝑖 )𝑖 ≤ 𝑗 of profinite sheaves of profinite groups over a directed set (𝐼, ≤) (Remark 2.2.5). For every 𝑖 ∈ 𝐼 let (X𝑖 , 𝜔𝑖 , 𝐺 𝑖 ) be the free profinite product over X𝑖 . Then, for 𝑗 ≥ 𝑖, there exists a unique homomorphism 𝜋 0𝑗𝑖 such that the following diagram commutes: X𝑗

𝜔𝑗

/ 𝐺𝑗

𝜔𝑖

 / 𝐺𝑖 .

𝜋 0𝑗𝑖

𝜋 𝑗𝑖

 X𝑖

Then, 𝐺 = lim 𝐺 𝑖 is a profinite group, 𝜔 = lim 𝜔𝑖 : X → 𝐺 is a morphism, and for ←−− ←−− every 𝑖 ∈ 𝐼 the following diagram commutes: X

𝜔

/𝐺

𝜔𝑖

 / 𝐺𝑖 .

𝜋𝑖0

𝜋𝑖

 X𝑖

Here, 𝜋𝑖 , 𝜋𝑖0 are the projection maps of the inverse limits. In the notation of Remark 2.2.5, given 𝑡 ∈ 𝑇, we set 𝑡𝑖 = 𝜋𝑖 (𝑡). Then, 𝑋𝑡 = lim 𝑋𝑡𝑖 ←−− and the inverse limit of the isomorphisms 𝜔𝑖 |𝑋𝑖,𝑡𝑖 : 𝑋𝑖,𝑡𝑖 → 𝐺 𝑖,𝑡𝑖 is the isomorphism 𝜔|𝑋𝑡 : 𝑋𝑡 → 𝐺 𝑡 . Lemma 4.2.5 In the notation of Remark 4.2.4, (X, 𝜔, 𝐺) is the free profinite product over X. Proof We only have to prove that (X, 𝜔, 𝐺) satisfies the universal property (4.2) of Definition 4.2.1. Thus, given a profinite group 𝐴 and a morphism 𝛼 : X → 𝐴, we have to prove the existence of a unique homomorphism 𝜑 : 𝐺 → 𝐴 such that 𝜑 ◦ 𝜔 = 𝛼. By the statement that follows (4.2), we may assume that 𝐴 is finite.

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4 Generalized Free Products

By Remark 2.3.1, we may consider every profinite group as a sheaf (over a onepoint underlying space). Thus, by Lemma 2.2.6(a), there exist 𝑗 ∈ 𝐼 and a morphism 𝛼 𝑗 : X 𝑗 → 𝐴 such that 𝛼 𝑗 ◦ 𝜋 𝑗 = 𝛼. Since (X 𝑗 , 𝜔 𝑗 , 𝐺 𝑗 ) is the free profinite product over X 𝑗 , there exists a unique homomorphism 𝜑 𝑗 : 𝐺 𝑗 → 𝐴 such that 𝛼 𝑗 = 𝜑 𝑗 ◦ 𝜔 𝑗 . Then, the homomorphism 𝜑 = 𝜑 𝑗 ◦ 𝜋 0𝑗 of 𝐺 to 𝐴 satisfies 𝜑 ◦ 𝜔 = 𝛼. To show the uniqueness of 𝜑, let 𝜑 0 : 𝐺 → 𝐴 be another homomorphism with 𝛼 = 𝜑 0 ◦ 𝜔. As in the preceding paragraph, by Lemma 2.2.6(a), there are 𝑗 0 ∈ 𝐼 and a morphism 𝜑 0𝑗 0 : 𝐺 𝑗 0 → 𝐴 such that 𝜑 0𝑗 0 ◦ 𝜋 0𝑗 0 = 𝜑 0. By Lemma 2.2.6(b), we may replace both 𝑗 and 𝑗 0 by a 𝑘 ∈ 𝐽 with 𝑘 ≥ 𝑗, 𝑗 0. Hence, we may assume that 𝑗 0 = 𝑗, so 𝜑 0𝑗 : 𝐺 𝑗 → 𝐴 satisfies 𝜑 0𝑗 ◦ 𝜋 0𝑗 = 𝜑 0. Since 𝜋 0𝑗 ◦ 𝜔 = 𝜔 𝑗 ◦ 𝜋 𝑗 , we have 𝛼 𝑗 ◦ 𝜋 𝑗 = (𝜑 0𝑗 ◦ 𝜔 𝑗 ) ◦ 𝜋 𝑗 . Another application of Lemma 2.2.6(b) allows us to replace 𝑗 by some 𝑘 0 ≥ 𝑗 to assume that 𝛼 𝑗 = 𝜑 0𝑗 ◦ 𝜔 𝑗 . /𝐺

𝜔

X

𝜋 0𝑗

𝜋𝑗

 X𝑗 𝛼

 / 𝐺𝑗

𝜔𝑗 𝜑 0𝑗

𝛼𝑗

𝜑0

 ~ ,𝐴r By the uniqueness of 𝜑 𝑗 in the free profinite product (X 𝑗 , 𝜔 𝑗 , 𝐺 𝑗 ) over X 𝑗 we have 𝜑 0𝑗 = 𝜑 𝑗 . Thus, 𝜑 0 = 𝜑. 

4.3 Equivalence of Inner and Outer Free Products We prove in this section that the inner and the outer free products of profinite groups over profinite spaces are essentially the same and use this knowledge to prove that every such free product is an inverse limit of free products of finitely many profinite groups. Proposition 4.3.1 Let 𝐺 be a profinite group and 𝑇 a profinite space. Let {𝐺 𝑡 }𝑡 ∈𝑇 be Î a collection of closed subgroups of 𝐺. Then, 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 if and only if there exists a profinite sheaf X = (𝑋, 𝜏, 𝑇) of profinite groups and a morphism 𝜔 : X → 𝐺 such that (X, 𝜔, 𝐺) is the free profinite product over X and 𝜔 maps 𝑋𝑡 isomorphically onto 𝐺 𝑡 for every 𝑡 ∈ 𝑇. Î Proof Assume that 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 . In particular, {𝐺 𝑡 }𝑡 ∈𝑇 is an étale continuous family of closed subgroups of 𝐺 (Definition 4.1.1). Let (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) be the associated sheaf-group structure (Proposition 2.3.6) of {𝐺 𝑡 }𝑡 ∈𝑇 . Then, (𝑋, 𝜏, 𝑇) is a profinite sheaf and for each 𝑡 ∈ 𝑇, 𝜔 maps 𝑋𝑡 isomorphically onto 𝐺 𝑡 . Claim: (X, 𝜔, 𝐺) is the free profinite product over X. To this end we prove that 𝜔 satisfies the universal property Ð (4.2). Let 𝛼 be a morphism of X into a profinite group 𝐴. We define a map 𝜑0 : 𝑡 ∈𝑇 𝐺 𝑡 → 𝐴 in the following way: First of all we let 𝜑0 (1) = 1. Secondly, by Condition (2.1a), for every

4.3 Equivalence of Inner and Outer Free Products

47

Ð 𝑔 ∈ 𝑡 ∈𝑇 𝐺 𝑡 with 𝑔 ≠ 1 there exists a unique 𝑡 ∈ 𝑇 with 𝑔 ∈ 𝐺 𝑡 . By the preceding paragraph there is a unique 𝑥 ∈ 𝑋𝑡 such that 𝜔(𝑥) = 𝑔; we set 𝜑0 (𝑔) = 𝛼(𝑥). Since Ð 𝜔(𝑋𝑡 ) = 𝐺 𝑡 and · 𝑡 ∈𝑇 𝑋𝑡 = 𝑋, we have 𝜑0 ◦ 𝜔 = 𝛼. By Lemma 2.3.5(c), 𝜔 : 𝑋 → 𝐺 is a closed map. Hence, for every closed subset −1 −1 𝐴 Ð0 of 𝐴, the set 𝜑0 ( 𝐴0 ) = 𝜔(𝛼 ( 𝐴0 )) is closed in 𝐺, hence also in its subset 𝐺 . Therefore, 𝜑 is continuous. 𝑡 0 𝑡 ∈𝑇 By Condition (4.1), 𝜑0 extends to a unique homomorphism 𝜑 : 𝐺 → 𝐴. Since 𝜑0 ◦ 𝜔 = 𝛼, we have 𝜑 ◦ 𝜔 = 𝛼. If 𝜑 0 : 𝐺 → 𝐴 is another homomorphism that Ð satisfies 𝜑 0 ◦ 𝜔 = 𝛼, then 𝜑 0 coincides with 𝜑0 on 𝜔(𝑋) = 𝑡 ∈𝑇 𝐺 𝑡 , hence with 𝜑 on 𝐺. Conversely, let (X, 𝜔, 𝐺) be a free profinite product over a profinite sheaf X = (𝑋, 𝜏, 𝑇) such that 𝜔(𝑋𝑡 ) = 𝐺 𝑡 , for every 𝑡 ∈ 𝑇. By Lemma 2.3.5(a), the map 𝑡 ↦→ 𝐺 𝑡 is étale continuous. By Lemma 4.2.3(b), 𝐺 𝑠 ∩ 𝐺 𝑡 = 1 for distinct 𝑠, 𝑡 ∈ 𝑇. Hence, {𝐺 𝑡 }𝑡 ∈𝑇 is an étale continuous family and it remains to prove the universal property (4.1). Ð Let 𝐴 be a profinite group and let 𝜑0 : 𝑡 ∈𝑇 𝐺 𝑡 → 𝐴 be aÐcontinuous map whose restriction to every 𝐺 𝑡 is a homomorphism. Since 𝜔(𝑋) = 𝑡 ∈𝑇 𝐺 𝑡 , the map 𝜑0 ◦ 𝜔 : X → 𝐴 is a morphism. By (4.2), there exists Ð a unique homomorphism 𝜑 : 𝐺 → 𝐴 such that 𝜑0 ◦ 𝜔 = 𝜑 ◦ 𝜔. Since 𝜔(𝑋) = 𝑡 ∈𝑇 𝐺 𝑡 , the homomorphism 𝜑 extends 𝜑0 and it is unique with this property. Î Thus, 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 .  The maps defined in the last proposition between the two types of free products of profinite groups actually give an equivalence of categories. On the one hand we Î consider the inner free product 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 as a triple (𝑇, 𝐺, 𝑡 ↦→ 𝐺 𝑡 ), where 𝑇 is a profinite space, 𝐺 is a profinite group, and 𝑡 ↦→ 𝐺 𝑡 is an étale continuous map. A morphism 𝛼 : (𝑇, 𝐺, 𝑡 ↦→ 𝐺 𝑡 ) → (𝑇 0, 𝐺 0, 𝑡 0 ↦→ 𝐺 𝑡00 ) between two objects in this category consists of a continuous map 𝛼 : 𝑇 → 𝑇 0 of profinite spaces, and a homomorphism 𝛼 : 𝐺 → 𝐺 0 of profinite groups such that if 𝑡 ∈ 𝑇 and 𝑡 0 = 𝛼(𝑡), then 𝛼(𝐺 𝑡 ) ≤ 𝐺 𝑡00 . On the other hand, we consider the category of outer free products (𝑋, 𝜏, 𝑇, 𝜔, 𝐺), where (𝑋, 𝜏, 𝑇) is a profinite sheaf. Proposition 4.3.2 There exists an equivalence of categories Ö 𝐺 = ∗ 𝐺𝑡 (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) 𝑡 ∈𝑇

between the category of inner free products of profinite groups over profinite spaces (Definition 4.1.1) and the category of outer free products over profinite sheaves (Definition 4.2.1). This equivalence preserves inverse limits. Proof Let (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) be an outer free product.
By Lemma 4.2.3(a),(b), the continuous map 𝑋 → 𝑇×𝐺, given by 𝑥 ↦→ 𝜏(𝑥), 𝜔(𝑥) , is injective. By Lemma 2.3.5(b), it is closed. Hence, it is a homeomorphism onto its image. Therefore, we may assume that 𝑋 = {(𝑡, 𝑔) ∈ 𝑇 × 𝐺 | 𝑔 ∈ 𝐺 𝑡 } and 𝜏, 𝜔 are the coordinate projections 𝑋 → 𝑇 and 𝑋 → 𝐺, respectively. In what follows we restrict ourselves to outer free products of this kind and prove that, instead of an equivalence, we even have an isomorphism of categories:

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4 Generalized Free Products

Let 𝛼 : (𝑇, 𝐺, 𝑡 ↦→ 𝐺 𝑡 ) → (𝑇 0, 𝐺 0, 𝑡 0 ↦→ 𝐺 𝑡00 ) be a morphism of two inner free products. Let (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) and (𝑋 0, 𝜏 0, 𝑇 0, 𝜔 0, 𝐺 0) be their respective associated sheaf-group structures, of the above kind. By assumption, 𝛼(𝑔) ∈ 𝐺 𝛼(𝑡) for all 𝑡 ∈ 𝑇 and 𝑔 ∈ 𝐺 𝑡 . Hence 𝛼 : 𝐺 → 𝐺 0 and 𝛼 : 𝑇 → 𝑇 0 induce a unique continuous map 𝛼 : 𝑋 → 𝑋 0 such that 𝜔 𝜏 /𝑇 𝐺o 𝑋 𝛼

 𝐺0 o

𝛼 𝜔0

 𝑋0

𝛼 𝜏0

 / 𝑇0

commutes, namely, the map (𝑡, 𝑔) ↦→ (𝛼(𝑡), 𝛼(𝑔)). Then, 𝛼 : 𝑋 → 𝑋 0, together with 𝛼 : 𝑇 → 𝑇 0 and 𝛼 : 𝐺 → 𝐺 0, is a morphism of sheaf-group structures 𝛼 0 : (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) → (𝑋 0, 𝜏 0, 𝑇 0, 𝜔 0, 𝐺 0). If 𝛼 is the identity map of the inner free product (𝑇, 𝐺, 𝑡 ↦→ 𝐺 𝑡 ), then 𝛼 0 is the identity map of the corresponding outer free product (𝑋, 𝜏, 𝑇, 𝜔, 𝐺). In addition, the correspondence 𝛼 𝛼 0 commutes with compositions, hence this correspondence is a covariant functor F . Conversely, let (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) be an outer free product. As remarked earlier, we may assume that 𝑋 = {(𝑡, 𝑔) ∈ 𝑇 × 𝐺 | 𝑔 ∈ 𝐺 𝑡 } and 𝜏, 𝜔 are the coordinate Î projections. Put 𝐺 𝑡 = 𝜔(𝑋𝑡 ) for every 𝑡 ∈ 𝑇. By Proposition 4.3.1, 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 . Let 𝛼 0 : (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) ↦→ (𝑋 0, 𝜏 0, 𝑇 0, 𝜔 0, 𝐺 0) be a morphism of sheaf-group 0 0 0 structures,Îsay, given by maps Î 𝛼 : 𝑋 → 𝑋 , 𝛼 : 𝑇 → 𝑇 , and 𝛼 : 𝐺 → 𝐺 . 0 Let 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 and 𝐺 = ∗ 𝑡 0 ∈𝑇 0 𝐺 𝑡 0 be the corresponding inner free products. For
every 𝑡 ∈
𝑇 and 𝑡 0
= 𝛼(𝑡) ∈ 𝑇 0 we have 𝛼(𝑋𝑡 ) ≤ 𝑋𝑡00 , hence 𝛼 𝜔(𝑋𝑡 ) = 𝜔 0 𝛼(𝑋𝑡 ) ≤ 𝜔 0 𝑋𝑡00 , that is, 𝛼(𝐺 𝑡 ) ≤ 𝐺 𝑡00 . Thus, 𝛼 0 induces a morphism of these inner free products. Moreover, this correspondence preserves the identities and commutes with composition, hence it is a covariant functor F 0. The definitions of F and F 0 imply that F ◦ F 0 and F 0 ◦ F are identities. Finally, Î we consider an inverse system (𝐺 𝑖 , 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 , 𝑗 ≥𝑖 of inner free products 𝐺 𝑖 = ∗ 𝑡 ∈𝑇𝑖 𝐺 𝑖,𝑡 of profinite groups over profinite spaces 𝑇𝑖 . Let 𝐺 = lim 𝐺 𝑖 . For ←−− every 𝑖 ∈ 𝐼 let 𝑋𝑖 = {(𝑡, 𝑔) ∈ 𝑇𝑖 × 𝐺 𝑖 | 𝑔 ∈ 𝐺 𝑖,𝑡 }. Let 𝜏𝑖 and 𝜔𝑖 be the corresponding projections and let E𝑖 = (𝑋𝑖 , 𝜏𝑖 , 𝑇𝑖 , 𝜔𝑖 , 𝐺 𝑖 ) be the corresponding sheaf-group structure. Then, (E𝑖 , 𝜋 𝑗𝑖 ) 𝑗 ≥𝑖 , in which the 𝜋 𝑗𝑖 are the natural extensions of the original projections, is an inverse system of sheaf-group structures. Let E = (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) be the inverse Î limit of this system. By Lemma 4.2.5, it is an outer free product. Hence, 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 .  In view of the equivalence established in this section between the two categories, in the sequel every property proved about one of the categories will also be used for the other category.

4.4 Free Profinite Products in the Sense of Binz–Neukirch–Wenzel

49

4.4 Free Profinite Products in the Sense of Binz–Neukirch–Wenzel We show that the free product of profinite groups in the sense of Binz–Neukirch– Wenzel is a special case of the inner free product of profinite groups introduced in Section 4.1. Definition 4.4.1 Let 𝐼 be a set and for every 𝑖 ∈ 𝐼 let 𝐺 𝑖 be a profinite group. A family {𝛼𝑖 : 𝐺 𝑖 → 𝐴}𝑖 ∈𝐼 of homomorphisms into a profinite group 𝐴 is said to converge to 1 if for every open subgroup 𝐴0 of 𝐴, the set 𝐼0 = {𝑖 ∈ 𝐼 | 𝛼𝑖 (𝐺 𝑖 ) ≤ 𝐴0 } is cofinite in 𝐼 (that is, 𝐼 r 𝐼0 is finite). The free product in the sense of Binz–Neukirch–Wenzel [BNW71, p. 105] of {𝐺 𝑖 }𝑖 ∈𝐼 is defined to be a profinite group 𝐺 together with a family {𝜔𝑖 : 𝐺 𝑖 → 𝐺}𝑖 ∈𝐼 of homomorphisms that converges to 1 and has the following universal property: (4.3) For every profinite group 𝐴 and every family {𝛼𝑖 : 𝐺 𝑖 → 𝐴}𝑖 ∈𝐼 of homomorphisms that converges to 1 there exists a unique homomorphism 𝛼 : 𝐺 → 𝐴 such that 𝛼 ◦ 𝜔𝑖 = 𝛼𝑖 for every 𝑖 ∈ 𝐼. Î We write 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 𝑖 . Lemma 4.4.2 (One-point compactification of a union of profinite spaces) Let {𝑋𝑖 }𝑖 ∈𝐼 Ðbe a family of profinite spaces and let ∞ be an extra symbol. Set
· 𝑋 = · 𝑖 ∈𝐼 𝑋𝑖 ∪{∞}. Let T be the family of subsets 𝑈 of 𝑋 such that 𝑈 ∩ 𝑋𝑖 is open in 𝑋𝑖 for every 𝑖 ∈ 𝐼, and, if ∞ ∈ 𝑈, then 𝑋𝑖 ⊆ 𝑈 for almost all 𝑖 ∈ 𝐼, i.e. for all but finitely many 𝑖 ∈ 𝐼. Let B ⊆ T be the family of subsets 𝑈 of 𝑋 such that either Ð 𝑈 is an open-closed subset of 𝑋𝑖 for some 𝑖 ∈ 𝐼 or 𝑈 = {∞} ∪ 𝑖 ∈𝐼 0 𝑋𝑖 , for some cofinite 𝐼 0 ⊆ 𝐼. Then: (a) (b) (c) (d)

T is a topology on 𝑋. This topology induces on 𝑋𝑖 its original topology, for every 𝑖 ∈ 𝐼. B is a base for T and its elements are open-closed. 𝑋 is a profinite space.

Proof We leave the verification of Parts (a), (b), and (c) to the reader. To prove (d) we first consider Ð a point 𝑥 ∈ 𝑋𝑖 for some 𝑖 ∈ 𝐼. Then, ∞ belongs to the open neighborhood 𝑈 := 𝑖0 ≠𝑖 𝑋𝑖0 ∪ {∞} of ∞, which Ð is disjoint from the open neighborhood 𝑋𝑖 of 𝑥. Likewise, any two points 𝑥, 𝑥 0 of 𝑖 ∈𝐼 𝑋𝑖 can be separated by disjoint open subsets of 𝑋. Hence, 𝑋 is a Hausdorff space. Thus, in view of (c) and Definition 1.1.6(b), it suffices to prove that 𝑋 is compact. Let U be an open covering of 𝑋. There is a 𝑈 Ð ∈ U such that ∞ ∈ 𝑈. Then, there is a finite 𝐽 ⊆ 𝐼 such that 𝑋 r 𝑈 is contained in 𝑖 ∈𝐽 𝑋𝑖 . The latter set is compact, hence is covered by a finite subset U 0 of U. Thus, {𝑈} ∪ U 0 is a finite covering of 𝑋 which is contained in U.  Corollary
4.4.3 Let {𝐺 𝑖 }𝑖 ∈𝐼 be a family of profinite groups. Set 𝑋 = Ð · · 𝑖 ∈𝐼 𝐺 𝑖 ∪{∞} and equip 𝑋 with the profinite topology of Lemma 4.4.2Ðwith 𝐺 𝑖 replacing 𝑋𝑖 . Similarly, regarding 𝐼 as a union of finite discrete spaces, 𝐼 = · 𝑖 ∈𝐼 {𝑖},

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· set 𝑇 = 𝐼 ∪{∞} and equip 𝑇 with the profinite topology of Lemma 4.4.2 with {𝑖} replacing 𝑋𝑖 . Define 𝜏 : 𝑋 → 𝑇 by 𝜏(𝑔) = {𝑖} for each 𝑖 ∈ 𝐼 and every 𝑔 ∈ 𝐺 𝑖 , and 𝜏(∞) = ∞. We regard the subset {∞} of 𝑋 as the trivial group 1. Then:

(a) X = (𝑋, 𝜏, 𝑇) is a profinite sheaf. (b) Let 𝐺 be a profinite group and let {𝜔𝑖 : 𝐺 𝑖 → 𝐺}𝑖 ∈𝐼 be a family of homomorphisms. Define a map 𝜔 : 𝑋 → 𝐺 by 𝜔|𝐺𝑖 = 𝜔𝑖 , for every 𝑖 ∈ 𝐼, and 𝜔(∞) = 1. Then, 𝜔 is a morphism X → 𝐺 if and only if {𝜔𝑖 : 𝐺 𝑖 → 𝐺}𝑖 ∈𝐼 converges to 1 (Definition 4.4.1). (c) Let 𝐺, {𝜔𝑖 }𝑖 ∈𝐼 , and 𝜔 be as in (b). Then, 𝐺 is the free product of the family {𝜔𝑖 : 𝐺 𝑖 → 𝐺}𝑖 ∈𝐼 in the sense of Binz–Neukirch–Wenzel if and only if (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) is an outer free product in the sense of Melnikov.Î (d) Let 𝐺, {𝜔𝑖 }𝑖 ∈𝐼 , and 𝜔 be as in (b). Then, 𝐺 is the free product ∗ 𝑖 ∈𝐼 𝐺 𝑖 in the sense of Binz–Neukirch–Wenzel if and only if 𝐺 is isomorphic to the inner free Î product ( ∗ 𝑡 ∈𝑇 𝐺 𝑡 ) ∗ 1 in the sense of Melnikov. Proof Proof of (a). Set 𝑋𝑖 = 𝐺 𝑖 for each 𝑖 ∈ 𝐼. If 𝑉 ⊆ 𝑇, then ( 𝑋𝑖 𝑖 ∈ 𝑉 −1 𝑋𝑖 ∩ 𝜏 (𝑉) = and ∞ ∈ 𝑉 ⇐⇒ ∞ ∈ 𝜏 −1 (𝑉). ∅ 𝑖 ∉𝑉 Thus, if 𝑉 is open in 𝑇, that is, 𝑉 contains almost all 𝑖 ∈ 𝐼 if ∞ ∈ 𝑉, then 𝜏 −1 (𝑉) is open in 𝑋. Therefore, 𝜏 is continuous. The structure map 𝜇 : 𝑋 (2) → 𝑋, given by (𝑥 1 , 𝑥2 ) ↦→ 𝑥 1−1 𝑥2 , is continuous. Indeed, when restricted to the open subset 𝐺 𝑖(2) of 𝑋 (2) , it identifies with the continuous structure map 𝐺 𝑖(2) → 𝐺 𝑖 of 𝐺 𝑖 , for every 𝑖 ∈ 𝐼. Hence, 𝜇 is continuous at every (𝑥1Ð , 𝑥2 ) ≠ (∞, ∞). Consider a basic open neighborhood 𝑈 of ∞ in 𝑋, say, 𝑈 = {∞} ∪ 𝑖 ∈𝐼 0 𝑋𝑖 , for some cofinite 𝐼 0 ⊆ 𝐼. Then, 𝜇 maps the neighborhood 𝑈 (2) of (∞, ∞) into 𝑈, hence 𝜇 is continuous at (∞, ∞) as well. Proof of (b). By definition, 𝜔 is continuous if and only if for every open 𝑁 ⊳ 𝐺 and every 𝑎 ∈ 𝐺 the set 𝜔−1 (𝑎𝑁) is open in 𝑋. We examine the latter condition. Notice that for each 𝑖 ∈ 𝐼 the set 𝜔−1 (𝑎𝑁) ∩ 𝑋𝑖 = 𝜔−1 𝑖 (𝑎𝑁) is open in 𝑋𝑖 , because 𝜔𝑖 is continuous. Thus, if 𝑎 ∉ 𝑁, which implies ∞ ∉ 𝜔−1 (𝑎𝑁), the set 𝜔−1 (𝑎𝑁) is open in 𝑋. If 𝑎 ∈ 𝑁, which implies 𝑎𝑁 = 𝑁 and ∞ ∈ 𝜔−1 (𝑁), the set 𝜔−1 (𝑁) is open in 𝑋 if and only if 𝐺 𝑖 ⊆ 𝜔−1 (𝑁) for almost all 𝑖 ∈ 𝐼. Thus, 𝜔 is a morphism if and only if {𝜔𝑖 : 𝐺 𝑖 → 𝐺}𝑖 ∈𝐼 converges to 1. Proof of (c). Suppose 𝐺 is the free product in the sense of Binz–Neukirch–Wenzel and let 𝛼 : X → 𝐴 be a morphism into a profinite group 𝐴. For every 𝑖 ∈ 𝐼 let 𝛼𝑖 : 𝐺 𝑖 → 𝐴 be the restriction of 𝛼. By (b), {𝛼𝑖 }𝑖 ∈𝐼 converges to 1, hence there is a unique homomorphism 𝛽 : 𝐺 → 𝐴 such that 𝛽 ◦ 𝜔𝑖 = 𝛼𝑖 for every 𝑖 ∈ 𝐼. Thus, 𝛽 is the unique homomorphism 𝐺 → 𝐴 such that 𝛽 ◦ 𝜔 = 𝛼. Therefore, (X, 𝜔, 𝐺) is an outer free product. Conversely, suppose that (X, 𝜔, 𝐺) is an outer free product. Let {𝛼𝑖 : 𝐺 𝑖 → 𝐴}𝑖 ∈𝐼 be a family of homomorphisms into a profinite group 𝐴 that converges to 1. By (b),

4.4 Free Profinite Products in the Sense of Binz–Neukirch–Wenzel

51

this family defines a morphism 𝛼 : X → 𝐴. Hence, there is a unique homomorphism 𝛽 : 𝐺 → 𝐴 such that 𝛽 ◦ 𝜔 = 𝛼. Thus, 𝛽 is the unique homomorphism 𝐺 → 𝐴 such that 𝛽 ◦ 𝜔𝑖 = 𝛼𝑖 for every 𝑖 ∈ 𝐼. Therefore, 𝐺 is the free product in the sense of Binz–Neukirch–Wenzel. Proof of (d). Statement (d) is a consequence of (c) and Proposition 4.3.2.



Example 4.4.4 Let 𝐼 be a set and 𝐴 a profinite group. A map 𝜄 : 𝐼 → 𝐴 converges to 1 if 𝐼 r 𝜄−1 (𝐵) is a finite set for each open normal subgroup 𝐵 of 𝐴. A free profinite group is a profinite group 𝐺 and a map 𝜄 : 𝐼 → 𝐺 such that (4.4a) 𝜄 converges to 1, and (4.4b) for every map 𝜑0 of 𝐼 into a profinite group 𝐴 that converges to 1 there exists a unique homomorphism 𝜑 : 𝐺 → 𝐴 such that 𝜑 ◦ 𝜄 = 𝜑0 . Every map from a finite set converges to 1. Hence, the definition in Remark 3.1.6 is a special case of the latter definition. In particular, for distinct elements 𝑖, 𝑖 0 of 𝐼 we may consider a non-trivial profinite group 𝐴 and an element 𝑎 ∈ 𝐴 with 𝑎 ≠ 1. Then, define 𝜑0 : 𝐼 → 𝐴 by 𝜑0 (𝑖) = 𝑎 and 𝜑0 ( 𝑗) = 1 for all 𝑗 ∈ 𝐼 r {𝑖}. Let 𝜑 : 𝐺 → 𝐴 be the unique homomorphism with 𝜑 ◦ 𝜄 = 𝜑0 . Then, 𝜑(𝜄(𝑖)) = 𝜑0 (𝑖) = 𝑎 and 𝜑(𝜄(𝑖 0)) = 𝜑0 (𝑖 0) = 1. Thus, 𝜄 is injective. Consequently, we may identify 𝐼 with its image in 𝐺 and 𝜄 with the inclusion map. In this case, we say that 𝐼 is a base of the free profinite group 𝐺. Having done so, we take 𝐴 to be the additive group Zˆ := lim Z/𝑛Z [FrJ08, p. 12, ←−− §1.4]. Given an 𝑖 ∈ 𝐼, we define a map 𝜑0 : 𝐼 → Zˆ by 𝜑0 (𝑖) = 1 and 𝜑0 (𝑖 0) = 0 for each 𝑖 0 ≠ 𝑖. Let 𝜑 : 𝐺 → Zˆ be the unique homomorphism that extends 𝜑0 . Then, 𝜑 ˆ By [FrJ08, p. 360, Lemma 17.7.1], 𝐺 𝑖  Z. ˆ maps 𝐺 𝑖 := h𝑖i onto Z. It follows that the family {𝐺 𝑖 }𝑖 ∈𝐼 satisfies Î Condition (4.3). Hence, by Definition 4.4.1, 𝐺 is isomorphic to the free product ∗ 𝑖 ∈𝐼 Zˆ in the sense of Binz–Neukirch– Wenzel. It follows from Corollary 4.4.3 that 𝐺 is also isomorphic to the free profinite Î ˆ ∗ 1 in the sense of Melnikov. product ( ∗ 𝑖 ∈𝐼 Z) Writing 𝑚 for the cardinality of 𝐼, one usually denotes the free profinite group 𝐺 discussed above by 𝐹ˆ𝑚 . The importance of the group 𝐹ˆ𝑚 lies in the fact that for an infinite set 𝐼 it appears as the absolute Galois group of every function field 𝐸 of one variable over a separably closed field 𝐶 of cardinality 𝑚 [Jar11, p. 186, Cor. 9.4.9]. In the case where 𝑚 = ℵ0 , one usually denotes 𝐹ˆ𝑚 by 𝐹ˆ𝜔 , where 𝜔 is the ordinal number of the well ordered set {0, 1, 2, . . .} = {0} ∪· N. Example 4.4.5 Recall that a field 𝐾 is Hilbertian if Hilbert’s irreducibility theorem holds for 𝐾 [FrJ08, p. 219, Sec. 12.1]. Specifically, if 𝑓 ∈ 𝐾 (𝑇) [𝑋] is a separable irreducible polynomial and 𝑔 ∈ 𝐾 [𝑇] is non-zero, then the separable Hilbert subset 𝐻𝐾 ( 𝑓 , 𝑔) := {𝑎 ∈ 𝐾 | 𝑓 (𝑎, 𝑋) is defined and irreducible over 𝐾 and 𝑔(𝑎) ≠ 0} of 𝐾 is non-empty. Let 𝑒 be a positive integer and 𝐾 a countable Hilbertian field (e.g. 𝐾 is an infinite finitely generated extension of its prime field [FrJ08, p. 242, Thm. 13.4.2]), then the absolute Galois group of 𝐾sep (𝝈) is 𝐹ˆ𝑒 for almost all 𝝈 := (𝜎1 , . . . , 𝜎𝑒 ) ∈ Gal(𝐾) 𝑒 [FrJ08, p. 379, Thm. 18.5.6]. Here, 𝐾sep (𝝈) is the fixed field in 𝐾sep of 𝜎1 , . . . , 𝜎𝑒 .

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4 Generalized Free Products

Also, the clause “almost all” is meant here in the sense of the Haar measure of the profinite group Gal(𝐾) 𝑒 . See [FrJ08, Sections 18.1–18.4] for the definition and the basic properties of the Haar measure. In particular, we always take the Haar measure 𝜇 of Gal(𝐾) 𝑒 to be normalized, that is 𝜇(Gal(𝐾) 𝑒 ) = 1.

4.5 Properties of Inner Free Products We use the equivalence of the two categories of free products of profinite groups in order to prove several properties of the inner free products of profinite groups. Î Setup 4.5.1 Let 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 be a free product over a profinite space 𝑇, fixed in this section. By Proposition 4.3.1 there exists a profinite sheaf X = (𝑋, 𝜏, 𝑇) and a morphism 𝜔 : X → 𝐺 such that (X, 𝜔, 𝐺) is the free profinite product over X. Moreover, for each 𝑡 ∈ 𝑇, 𝜔 maps the group 𝑋𝑡 = 𝜏 −1 (𝑡) isomorphically onto 𝐺 𝑡 . By Lemma 2.3.5(c), 𝜔 is closed. We fix the objects introduced in Setup 4.5.1 throughout the whole section. Lemma 4.5.2 Let 𝑟, 𝑠 ∈ 𝑇 be distinct and let 𝐴 be a finite group. Then, every homomorphism 𝛼𝑠 : 𝐺 𝑠 → 𝐴 extends to a homomorphism 𝛼 : 𝐺 → 𝐴 such that 𝛼(𝐺 𝑟 ) = 1. Proof Following Setup 4.5.1, we use Corollary 2.3.3 to extend 𝛼𝑠 ◦ 𝜔|𝑋𝑠 to a morphism 𝛼 0 : X → 𝐴 such that 𝛼 0 (𝑋𝑟 ) = 1. Let 𝛼 : 𝐺 → 𝐴 be the unique homomorphism that satisfies 𝛼 ◦ 𝜔 = 𝛼 0 (Definition 4.2.1). Then, 𝛼(𝐺 𝑟 ) = 𝛼(𝜔(𝑋𝑟 )) = 𝛼 0 (𝑋𝑟 ) = 1 and 𝛼(𝐺 𝑠 ) = 𝛼(𝜔(𝑋𝑠 )) = 𝛼 0 (𝑋𝑠 ) = 𝛼𝑠 (𝜔(𝑋𝑠 )) = 𝛼𝑠 (𝐺 𝑠 ).  Let 𝑆 be a subspace of 𝑇. We set 𝐺 (𝑆) = h𝐺 𝑡 i𝑡 ∈𝑆 . By definition, 𝐺 (𝑆) is a closed subgroup of 𝐺. Î Lemma 4.5.3 Let 𝑇 0 be a closed subset of 𝑇. Then, 𝐺 (𝑇 0) = ∗ 𝑡 ∈𝑇 0 𝐺 𝑡 . Proof In the notation of Setup 4.5.1, we set 𝑋 0 = 𝜏 −1 (𝑇 0) and 𝜔 0 = 𝜔|𝑋 0 . Then, X0 = (𝑋 0, 𝜏|𝑋 0 , 𝑇 0) is a closed subsheaf of X and 𝜔 0 (𝑋 0) = 𝐺 (𝑇 0) ≤ 𝐺, because 𝜔 is a closed map (Lemma 2.3.5(c)). We prove that Î (X0, 𝜔 0, 𝐺 (𝑇 0)) is a free profinite 0 0 product over X , so by Proposition 4.3.2, 𝐺 (𝑇 ) = ∗ 𝑡 ∈𝑇 0 𝐺 𝑡 . Let 𝛼 0 : X0 → 𝐴 be a morphism into a finite group 𝐴. We have to prove that there is a unique homomorphism 𝜑 0 : 𝐺 (𝑇 0) → 𝐴 such that 𝜑 0 ◦ 𝜔 0 = 𝛼 0 (Definition 4.2.1). By Lemma 2.2.9, 𝛼 0 extends to a morphism 𝛼 : X → 𝐴. By the universal property of outer free products, there exists a unique homomorphism 𝜑 : 𝐺 → 𝐴 such that 𝜑 ◦ 𝜔 = 𝛼. Let 𝜑 0 = 𝜑|𝐺 (𝑇 0 ) . Then, 𝜑 0 ◦ 𝜔 0 = 𝛼 0. This 𝜑 0 is unique, because 𝐺 (𝑇 0) = 𝜔 0 (𝑋 0).  The following lemma can be considered as an associative law for the inner free products over profinite spaces.

4.5 Properties of Inner Free Products

53

Ð Lemma 4.5.4 Let 𝑇 = · 𝑖 ∈𝐼 𝑇Î 𝑖 be a partition of 𝑇 intoÎfinitely many disjoint openclosed subspaces. Then, 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 (𝑇𝑖 ) and 𝐺 (𝑇𝑖 ) = ∗ 𝑡 ∈𝑇𝑖 𝐺 𝑡 for each 𝑖 ∈ 𝐼. Proof We set G = {𝐺 𝑡 }𝑡 ∈𝑇 and G𝑖 = {𝐺 𝑡 }𝑡 ∈𝑇𝑖 for each 𝑖 ∈ 𝐼. Since G is an étale continuous family (Definition 2.1.1), so are the G𝑖 . By Lemma 2.1.2, the 𝑈 (G𝑖 ) are closed in 𝐺. Since 𝐺 𝑠 ∩ 𝐺 𝑡 = 1 if 𝑠 ≠ 𝑡, we have 𝑈 (G𝑖 ) ∩ 𝑈 (G 𝑗 ) = {1} for 𝑖 ≠ 𝑗. Consider a profinite group 𝐴 and for each 𝑖 ∈ 𝐼 let 𝛼𝑖 : 𝐺 (𝑇𝑖 ) → 𝐴 be a homomorphism. Denote the restriction of 𝛼𝑖 to 𝑈 (G𝑖 ) by 𝛼𝑖0 . Since 𝑈 (G𝑖 ) ∩𝑈 (G 𝑗 ) = {1} and 𝛼𝑖0 (1) = 1 = 𝛼 𝑗0 (1) for 𝑖 ≠ 𝑗, there exists a unique map 𝛼0 : 𝑈 (G) → 𝐴 that extends each of the 𝛼𝑖0 ’s. Since the 𝛼𝑖0 ’s are continuous, so is 𝛼0 . By (4.1), 𝛼0 uniquely extends to a homomorphism 𝛼 : 𝐺 → 𝐴. Since 𝛼|𝑈 ( G𝑖 ) = 𝛼𝑖0 = 𝛼𝑖 |𝑈 ( G𝑖 ) , we have 𝛼|𝐺 (𝑇𝑖 ) = 𝛼𝑖 . Hence, 𝛼 uniquely extends the 𝛼𝑖 ’s. It follows Î from Proposition 3.1.1 that 𝐺 = ∗ 𝑖 ∈𝐼 𝐺 (𝑇𝑖 ). The last assertion of this lemma is Lemma 4.5.3.  · Definition 4.5.5 Let 𝑆 be an open-closed subset of 𝑇. Then, 𝑇 = 𝑆 ∪(𝑇 r 𝑆) is a partition of 𝑇 into two open-closed subspaces. By Lemma 4.5.4, 𝐺 = 𝐺 (𝑆)∗𝐺 (𝑇 r𝑆). Hence, by Proposition 3.1.1(b), there exists a unique epimorphism 𝜋 𝑆 : 𝐺 → 𝐺 (𝑆) such that 𝜋 𝑆 (𝑔) = 𝑔 if 𝑔 ∈ 𝐺 (𝑆) and 𝜋 𝑆 (𝑔) = 1 if 𝑔 ∈ 𝐺 (𝑇 r 𝑆). We call 𝜋 𝑆 the projection of 𝐺 onto 𝐺 (𝑆).

Lemma 4.5.6 Let 𝑈1 , . . . , 𝑈𝑛 be open-closed subsets of 𝑇. Then, 𝑛 𝑛 Ù Ù 𝐺 (𝑈𝑖 ) = 𝐺 ( 𝑈𝑖 ). 𝑖=1

𝑖=1

Proof By induction on 𝑛 it suffices to prove the lemma for 𝑛 = 2. To this end we consider the following partitions into open-closed subsets of 𝑇: · 𝑈1 = (𝑈1 r 𝑈2 ) ∪(𝑈 1 ∩ 𝑈2 ), · 𝑈2 = (𝑈1 ∩ 𝑈2 ) ∪(𝑈 2 r 𝑈1 ).

By Lemma 4.5.4, 𝐺 (𝑈1 ) = 𝐺 (𝑈1 r 𝑈2 ) ∗ 𝐺 (𝑈1 ∩ 𝑈2 ), 𝐺 (𝑈2 ) = 𝐺 (𝑈1 ∩ 𝑈2 ) ∗ 𝐺 (𝑈2 r 𝑈1 ). Hence, by Lemma 3.1.5,

𝐺 (𝑈1 ) ∩ 𝐺 (𝑈2 ) = 𝐺 (𝑈1 r 𝑈2 ) ∗ 𝐺 (𝑈1 ∩ 𝑈2 ) ∩ 𝐺 (𝑈1 ∩ 𝑈2 ) ∗ 𝐺 (𝑈2 r 𝑈1 ) = 𝐺 (𝑈1 ∩ 𝑈2 ), as claimed.



Lemma 4.5.7 Let {𝑈𝑖 }𝑖 ∈𝐼 be a family of open-closedÑsubsets of the Ñ profinite space 𝑇 which is closed under finite intersections. Then, 𝐺 ( 𝑖 ∈𝐼 𝑈𝑖 ) = 𝑖 ∈𝐼 𝐺 (𝑈𝑖 ). Ñ Ñ Proof First observe that 𝐺 ( 𝑖 ∈𝐼 𝑈𝑖 ) ⊆ Ñ𝑖 ∈𝐼 𝐺 (𝑈𝑖 ). For the converse inclusion it suffices, by Lemma 1.1.14, Ñ to prove that 𝑖 ∈𝐼 𝐺 (𝑈𝑖 ) ⊆ 𝐻 for every open subgroup 𝐻 of 𝐺 containing 𝐺 ( 𝑖 ∈𝐼 𝑈𝑖 ).

54

4 Generalized Free Products

Since {𝐺 𝑡 }𝑡 ∈𝑇 is an étale continuous family, 𝑈 = {𝑡 ∈ 𝑇 | 𝐺 𝑡 ≤ 𝐻} is an open Ñ subset of 𝑇 (Definition 2.1.1). Since 𝑇 is compact and 𝑖 ∈𝐼 𝑈𝑖 ⊆ 𝑈, there is a 𝑗 ∈ 𝐼 such that 𝑈 𝑗 ⊆ 𝑈 (Lemma 1.1.1). Thus, 𝐺 (𝑈 𝑗 ) = h𝐺 𝑡 i𝑡 ∈𝑈 𝑗 ≤ 𝐻. In particular, Ñ  𝑖 ∈𝐼 𝐺 (𝑈𝑖 ) ≤ 𝐻. Here is an improvement of Lemma 4.5.7. Ñ Lemma 4.5.8 Let {𝑆𝑖 }𝑖 ∈𝐼 be a collection of closed subsets of 𝑇. Then, 𝐺 ( 𝑖 ∈𝐼 𝑆𝑖 ) = Ñ 𝑖 ∈𝐼 𝐺 (𝑆 𝑖 ). Ñ Proof Set 𝑆 = 𝑖 ∈𝐼 𝑆𝑖 . For every 𝑖 ∈ 𝐼 let U𝑖 be the collection of all open-closed subsets of 𝑇 that contains 𝑆Ð𝑖 . Let W be the collection of all intersections Ñ 𝑈1 ∩· · ·∩𝑈𝑛 in which 𝑈1 , . . . , 𝑈𝑛 ∈ Ñ𝑖 ∈𝐼 U𝑖 . By Lemma 4.5.7, 𝐺 (𝑆𝑖 ) = 𝑈 ∈U Ð𝑖 𝐺 (𝑈) for every 𝑖 ∈ 𝐼 and 𝐺 (𝑆) = 𝑊 ∈W 𝐺 (𝑊). Moreover, if 𝑈1 , . . . , 𝑈𝑛 ∈ 𝑖 ∈𝐼 U𝑖 and 𝑊 = 𝑈1 ∩ · · · ∩𝑈𝑛 , then by Lemma 4.5.6, 𝐺 (𝑊) = 𝐺 (𝑈1 ) ∩ · · · ∩ 𝐺 (𝑈𝑛 ). Therefore, Ù Ù Ù Ù 𝐺 (𝑆) = 𝐺 (𝑊) = 𝐺 (𝑈) = 𝐺 (𝑆𝑖 ), 𝑊 ∈W

𝑖 ∈𝐼 𝑈 ∈U𝑖

𝑖 ∈𝐼

as claimed.



Remark 4.5.9 Every free product of profinite groups over a profinite space can be considered as an inverse limit of free products of finitely many profinite groups. Indeed, as a profinite space, 𝑇 is the inverse limit of an inverse system (𝑇𝑖 , 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 , 𝑗 ≥𝑖 of finite Ð discrete spaces. For every 𝑖 ∈ 𝐼 let 𝜋𝑖 : 𝑇 → 𝑇𝑖 be the projection. Then, 𝑇 = · 𝑡𝑖 ∈𝑇𝑖 𝜋𝑖−1 (𝑡 𝑖 ) is a partition into disjoint open-closed subsets. Î By Lemma 4.5.4, 𝐺 = ∗ 𝑡𝑖 ∈𝑇𝑖 𝐺 𝑡𝑖 , where 𝐺 𝑡𝑖 = h𝐺 𝑡 | 𝑡 ∈ 𝜋𝑖−1 (𝑡𝑖 )i. Î For every 𝑖 ∈ 𝐼 we set 𝐺 𝑖 = 𝐺. Then, 𝐺 𝑖 = ∗ 𝑡𝑖 ∈𝑇𝑖 𝐺 𝑡𝑖 is a free product of finitely many profinite groups and 𝐺 = lim 𝐺 𝑖 with respect to the identity maps. ←−− Applying the preceding observation to the situation of Remark 1.1.8 gives Ö 𝐺 = lim ∗ 𝐺 (𝑈), ←−− U ∈U 𝑈 ∈U

where U is the collection of all partitions of 𝑇 into disjoint open-closed subsets.

4.6 Further Properties of Inner Free Products We start with a lemma that strengthens the associative law (Lemma 4.5.4) for free products of profinite groups. Î Lemma 4.6.1 Let 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 be an inner free product of profinite groups over a profinite Î space 𝑇 and let Î𝛿 : 𝑇Î→ 𝑆 be a continuous map of profinite spaces. Then, 𝐺 = ∗ 𝑠 ∈𝑆 𝐺 (𝛿−1 (𝑠)) = ∗ 𝑠 ∈𝑆 ∗ 𝑡 ∈ 𝛿 −1 (𝑠) 𝐺 𝑡 . Proof We present 𝑆 as an inverse limit of an inverse system (𝑆𝑖 , 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 , 𝑗 ≥𝑖 of Ð finite discrete spaces. For every 𝑖 ∈ 𝐼 we have 𝑇 = · 𝑠𝑖 ∈𝑆𝑖 𝛿−1 (𝜋𝑖−1 (𝑠𝑖 )), where 𝜋𝑖 is the projection 𝑆 → 𝑆𝑖 . Hence, by Lemma 4.5.4,

4.6 Further Properties of Inner Free Products

(4.5)

𝐺=

55

Ö ∗ 𝐺 (𝛿−1 (𝜋𝑖−1 (𝑠𝑖 ))). 𝑠𝑖 ∈𝑆𝑖

Consider 𝑖, 𝑗 ∈ 𝐼, points 𝑠𝑖 ∈ 𝑆𝑖 and 𝑠 𝑗 ∈ 𝑆 𝑗 such that 𝑖 ≤ 𝑗 and 𝜋 𝑗𝑖 (𝑠 𝑗 ) = 𝑠𝑖 . Since −1 −1 −1 −1 −1 𝜋𝑖 = 𝜋 𝑗𝑖 ◦ 𝜋 𝑗 , we have 𝜋 −1 𝑗 (𝑠 𝑗 ) ⊆ 𝜋𝑖 (𝑠𝑖 ), so 𝛿 (𝜋 𝑗 (𝑠 𝑗 )) ⊆ 𝛿 (𝜋𝑖 (𝑠𝑖 )), and −1 −1 −1 −1 therefore 𝐺 (𝛿 (𝜋 𝑗 (𝑠 𝑗 ))) ≤ 𝐺 (𝛿 (𝜋𝑖 (𝑠𝑖 ))). It follows that, as 𝑖 ranges over 𝐼, the right-hand side of (4.5) ranges over an inverse system of profinite groups whose connecting maps are the identity maps. Going over to the inverse limit, we get by Lemma 4.2.5 and Proposition 4.3.2 that Ö (4.6) 𝐺 = ∗ lim 𝐺 (𝛿−1 (𝜋𝑖−1 (𝜋𝑖 (𝑠)))). ←−− 𝑠 ∈𝑆 𝑖 ∈𝐼

Since the connecting homomorphisms are inclusions and since Ù 𝜋𝑖−1 (𝜋𝑖 (𝑠)) = {𝑠} for every 𝑠 ∈ 𝑆, 𝑖 ∈𝐼

we get for every 𝑠 ∈ 𝑆 that lim 𝐺 (𝛿−1 (𝜋𝑖−1 (𝜋(𝑠)))) = ←−− 𝑖 ∈𝐼

Ù

𝐺 (𝛿−1 (𝜋𝑖−1 (𝜋𝑖 (𝑠))))

(Example 1.1.5)

𝑖 ∈𝐼

= 𝐺(

Ù

𝛿−1 (𝜋𝑖−1 (𝜋𝑖 (𝑠))))

(Lemma 4.5.8)

𝑖 ∈𝐼

= 𝐺 (𝛿−1 (

Ù

𝜋𝑖−1 (𝜋𝑖 (𝑠))))

𝑖 ∈𝐼 −1

= 𝐺 (𝛿 (𝑠)). Î Therefore, byÎ(4.6), 𝐺 = ∗ 𝑠 ∈𝑆 𝐺 (𝛿−1 (𝑠)). By Lemma 4.5.3, for each 𝑠 ∈ 𝑆 we have 𝐺 (𝛿−1 (𝑠)) = ∗ 𝑡 ∈ 𝛿 −1 (𝑠) 𝐺 𝑡 . Hence, Ö Ö Ö ∗ 𝐺𝑡 , 𝐺 = ∗ 𝐺 (𝛿−1 (𝑠)) = ∗ 𝑠 ∈𝑆

as claimed.

𝑠 ∈𝑆 𝑡 ∈ 𝛿 −1 (𝑠)



Here is a converse of Lemma 4.5.4: Lemma 4.6.2 Let 𝐼 be a finite set and 𝑇 a profinite space. Consider a partition, Ð 𝑇 = · 𝑖 ∈𝐼Î𝑇𝑖 , of 𝑇 into finitely subspaces. For each 𝑖 ∈ 𝐼 Î many disjoint open-closed Î let 𝐻𝑖 = ∗ 𝑡 ∈𝑇𝑖 𝐺 𝑡 . Let 𝐺 = ∗ 𝑖 ∈𝐼 𝐻𝑖 . Then, 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 . Ð Proof For each 𝑖 ∈ 𝐼 we set G𝑖 = {𝐺 𝑡 }𝑡 ∈𝑇𝑖 and let G = {𝐺 𝑡 }𝑡 ∈𝑇 = 𝑖 ∈𝐼 G𝑖 . We view G and the sets G𝑖 as subsets of Subgr(𝐺). Since 𝐻𝑖 ∩ 𝐻 𝑗 = 1 if 𝑖 ≠ 𝑗 (Lemma 3.1.10), and the G𝑖 are étale continuous families (Definition 2.1.1), so is G. Consider a profinite group 𝐴 and let 𝛼0 : 𝑈 (G) → 𝐴 be a continuous map whose restriction to 𝐺Î𝑡 , for every 𝑡 ∈ 𝑇, is a homomorphism. Since 𝐻𝑖 = ∗ 𝑡 ∈𝑇𝑖 𝐺 𝑡 , Condition (4.1) implies that for each 𝑖 ∈ 𝐼 the restriction of Î 𝛼0 to 𝑈 (G𝑖 ) uniquely extends to a homomorphism 𝛼𝑖 : 𝐻𝑖 → 𝐴. Since 𝐺 = ∗ 𝑖 ∈𝐼 𝐻𝑖 , the 𝛼𝑖 uniquely extend to a homomorphism 𝛼 : 𝐺 → 𝐴 (Proposition 3.1.1(b)). Thus, 𝛼0 uniquely extends to a homomorphism 𝛼 : 𝐺 → 𝐴. This proves Condition (4.1). Î Hence, by Definition 4.1.1, 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 , as claimed. 

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4 Generalized Free Products

Here is a commutative law for inner free products. Lemma 4.6.3 Let 𝐼 be a finite set and let 𝑇 be a profinite space. Let 𝐺 be a profinite group Î and for Î every (𝑖, 𝑡) ∈ 𝐼 × 𝑇Îlet 𝐺 𝑖𝑡Îbe a closed subgroup of 𝐺. Assume that 𝐺 = ∗ 𝑖 ∈𝐼 ( ∗ 𝑡 ∈𝑇 𝐺 𝑖𝑡 ). Then, 𝐺 = ∗ 𝑡 ∈𝑇 ( ∗ 𝑖 ∈𝐼 𝐺 𝑖𝑡 ). Ð Î Proof Since 𝐼 × 𝑇 = · 𝑖 ∈𝐼 {𝑖} × 𝑇, we have by Lemma 4.6.2 that 𝐺 = ∗ (𝑖,𝑡) ∈𝐼 ×𝑇 𝐺 𝑖𝑡 . Applying Lemma 4.6.1 to the projection 𝛿 : 𝐼 × 𝑇 → 𝑇 on the second component, Î 𝑡 ∈ 𝑇, Î we have by Lemma 4.5.3 that we get 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 (𝐼 ×Î{𝑡}). Now, for each Î 𝐺 (𝐼 × {𝑡}) = h𝐺 𝑖𝑡 i𝑖 ∈𝐼 = ∗ 𝑖 ∈𝐼 𝐺 𝑖𝑡 . Thus, 𝐺 = ∗ 𝑡 ∈𝑇 ( ∗ 𝑖 ∈𝐼 𝐺 𝑖𝑡 ), as claimed.  Let 𝐺 be a profinite group, let 𝑇 be a set, and for every 𝑡 ∈ 𝑇 let 𝐺 𝑡 be a closed subgroup of 𝐺. Then, we write 𝑔

[𝐺 𝑡 ]𝑡 ∈𝑇 = h𝐺 𝑡 i𝑡 ∈𝑇 , 𝑔 ∈𝐺 for the closed normal subgroup of 𝐺 generated by all of the subgroups 𝐺 𝑡 with 𝑡 ∈ 𝑇. Î Lemma 4.6.4 Let 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 be an inner free product of profinite groups over a profinite space 𝑇. For every 𝑡 ∈ 𝑇 let 𝐻𝑡 be a closed subgroup of 𝐺 𝑡 such that the Ð set 𝑡 ∈𝑇 𝐻𝑡 is closed in 𝐺. (a) We set 𝐻 = h𝐻𝑡 i𝑡 ∈𝑇 . Then, 𝐻 ∩ 𝐺 𝑡 = 𝐻𝑡 for every 𝑡 ∈ 𝑇. (b) Suppose that 𝐻𝑡 ⊳ 𝐺 𝑡 for every 𝑡 ∈ 𝑇 and set 𝐻 = [𝐻𝑡 ]𝑡 ∈𝑇 . Then, 𝐻 ∩ 𝐺 𝑡 = 𝐻𝑡 for every 𝑡 ∈ 𝑇. Proof By Setup 4.5.1, there exists a sheaf X = (𝑋, 𝜏, 𝑇) and a morphism 𝜔 : X → 𝐺 such that (X, 𝜔, 𝐺) is the free profinite product over X and 𝐺 𝑡 = 𝜔(𝑋𝑡 ), where Ð 𝑋𝑡 = 𝜏 −1 (𝑡) for every 𝑡 ∈ 𝑇. Thus, the subset 𝑌 = 𝜔−1 ( 𝑡 ∈𝑇 𝐻𝑡 ) of 𝑋 is closed, so 𝑌 is a profinite space (Remark 1.1.7(a)). We set 𝑌𝑡 = 𝑋𝑡 ∩ 𝑌 for each 𝑡 ∈ 𝑇. Claim: 𝐻𝑡 = 𝜔(𝑌𝑡 ) for each 𝑡 ∈ 𝑇. Indeed, consider 𝑡 ∈ 𝑇. By Lemma 4.2.3(a), 𝜔 maps 𝑋𝑡 isomorphically onto 𝐺 𝑡 . In particular 𝜔 maps the unit of 𝑋𝑡 onto the unit of 𝐺 𝑡 , which is also the unit of 𝐻𝑡 . Hence, the unit of 𝐻𝑡 belongs to 𝜔(𝑌𝑡 ). Next, consider a non-unit element ℎ of 𝐻𝑡 . Then, ℎ ∈ 𝐺 𝑡 , so there exists an 𝑥 ∈ 𝑋𝑡 such that 𝜔(𝑥) = ℎ. In addition, there exists a 𝑦 ∈ 𝑌 with 𝜔(𝑦) = ℎ. Since 𝜔 : 𝑋 → 𝐺 is injective on 𝑋 r 𝜔−1 (1) (Lemma 4.2.3(e)), we have 𝑥 = 𝑦, so 𝑦 ∈ 𝑋𝑡 ∩ 𝑌 = 𝑌𝑡 . Hence, ℎ = 𝜔(𝑦) ∈ 𝜔(𝑌𝑡 ). Thus, 𝐻𝑡 ⊆ 𝜔(𝑌𝑡 ). Conversely, let 𝑦 ∈ 𝑌𝑡 . If 𝜔(𝑦) is the unit of 𝐺 𝑡 , then 𝜔(𝑦) is also the unit of 𝐻𝑡 . Otherwise, 𝜔(𝑦) is a non-unit element of 𝐺 𝑡 . By the definition of 𝑌 there exists a 𝑡 0 ∈ 𝑇 with 𝜔(𝑦) ∈ 𝐻𝑡 0 , so also 𝜔(𝑦) ∈ 𝐺 𝑡 0 . It follows from Lemma 4.2.3(b) that 𝑡 0 = 𝑡, so 𝜔(𝑦) ∈ 𝐻𝑡 , so 𝜔(𝑌𝑡 ) ⊆ 𝐻𝑡 , and the proof of the Claim is complete. Having proved the Claim we consider 𝑡 ∈ 𝑇 and 𝑔 ∈ 𝐺 𝑡 r 𝐻𝑡 . In both cases it suffices to prove that 𝑔 ∉ 𝐻. To this end we consider the following commutative diagram.

4.6 Further Properties of Inner Free Products

57 𝜔

𝑋

/ Ð𝑡 ∈𝑇 𝐺 𝑡

𝐺

/ Ð𝑡 ∈𝑇 𝐻𝑡

𝐻

/ 𝐺𝑡

𝑋𝑡

𝑌

𝑌𝑡

/ 𝐻𝑡

Let 𝑥 ∈ 𝑋𝑡 be the unique element with 𝜔(𝑥) = 𝑔. By the Claim, 𝑥 ∈ 𝑋𝑡 r 𝑌𝑡 . Hence, there exists a homomorphism 𝜓𝑡 of 𝑋𝑡 onto a finite quotient 𝐴 of 𝐺 𝑡 such that 𝜓𝑡 (𝑥) ∉ 𝐵, where 𝐵 = 𝜓𝑡 (𝑌𝑡 ). If 𝐻𝑡 ⊳ 𝐺 𝑡 , then 𝐵 ⊳ 𝐴. By Corollary 2.3.3, 𝜓𝑡 extends to a morphism 𝜓 : X → 𝐴. In particular, (4.7)

𝜓(𝑥) ∉ 𝐵,

𝜓 −1 (𝐵)

is an open subset of 𝑋, and 𝜓 is a closed map (Fact 1.1.3(e)). We set 𝑆 = {𝑠 ∈ 𝑇 | 𝑌𝑠 ⊆ 𝜓 −1 (𝐵)} and observe that 𝑡 ∈ 𝑆, because 𝜓(𝑌𝑡 ) = 𝜓𝑡 (𝑌𝑡 ) = 𝐵. Since 𝜓 −1 (𝐵) is open in 𝑋, we have by Lemma 1.1.18 that 𝑆 is open in 𝑇. Hence, 𝑇 has an open-closed subset 𝑉 that contains 𝑡 and is contained in 𝑆. In particular, 𝜏 −1 (𝑉) is an open-closed subset of 𝑋. We may therefore define a morphism 𝜓 0 : 𝑋 → 𝐴 by 𝜓 0 (𝑧) = 𝜓(𝑧) for every 𝑧 ∈ 𝜏 −1 (𝑉) and 𝜓 0 (𝑧) = 1 for every 𝑧 ∈ 𝜏 −1 (𝑇 r 𝑉). In particular, since 𝜏(𝑥) = 𝑡 ∈ 𝑉, we have (4.8)

𝜓 0 (𝑥) = 𝜓(𝑥).

Let 𝜑 : 𝐺 → 𝐴 be the unique homomorphism with 𝜑 ◦ 𝜔 = 𝜓 0 given by (4.2) with replacing 𝛼. By the Claim, 𝜑(𝐻𝑠 ) = 𝜑(𝜔(𝑌𝑠 )) = 𝜓 0 (𝑌𝑠 ) ≤ 𝐵 for every 𝑠 ∈ 𝑇. Thus, if 𝐻 = h𝐻𝑠 i𝑠 ∈𝑇 , then 𝜑(𝐻) ≤ 𝐵. If 𝐻𝑠 ⊳ 𝐺 𝑠 for every 𝑠 ∈ 𝑇 and 𝐻 = [𝐻𝑠 ]𝑠 ∈𝑇 , then 𝐵 ⊳ 𝐴, so 𝜑(𝐻) ≤ 𝐵. But, by (4.8) and (4.7), 𝜑(𝑔) = 𝜑(𝜔(𝑥)) = 𝜓 0 (𝑥) = 𝜓(𝑥) ∉ 𝐵. This proves that 𝑔 ∉ 𝐻, as claimed.  𝜓0

Finally, we prove two results about quotients of free products. Î Lemma 4.6.5 Let 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 be an inner free product of profinite groups over a profinite space 𝑇. For every 𝑡 ∈ 𝑇 let 𝑁𝑡 be a closed normal subgroup of 𝐺 𝑡 and set Î 𝑁 = [𝑁𝑡 ]𝑡 ∈𝑇 . Then, 𝐺/𝑁 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 𝑁/𝑁. Proof By Definition 4.1.1, we have to prove Claims A and B below for the family G¯ := {𝐺 𝑡 𝑁/𝑁 }𝑡 ∈𝑇 of closed subgroups of 𝐺/𝑁. Claim A: The family G¯ is étale continuous. Indeed, by Setup 4.5.1 and Definition 4.1.1, the map 𝑡 ↦→ 𝐺 𝑡 of 𝑇 into Subgr(𝐺) is étale continuous. By Remark 1.2.1(a), the map Γ ↦→ Γ𝑁/𝑁 of Subgr(𝐺) into Subgr(𝐺/𝑁) is étale continuous. Hence, the map 𝑡 ↦→ 𝐺 𝑡 𝑁/𝑁 of 𝑇 into Subgr(𝐺/𝑁) is étale continuous. Thus, to complete the proof that G¯ is an étale continuous family of subgroups of 𝐺/𝑁, it suffices to prove that 𝐺 𝑠 𝑁/𝑁 ∩ 𝐺 𝑡 𝑁/𝑁 = 𝑁/𝑁 for all distinct 𝑠, 𝑡 ∈ 𝑇. This will be done as soon as we prove that if 𝑔 ∈ 𝐺 𝑠 r 𝑁, then 𝑔 ∉ 𝐺 𝑡 𝑁.

58

4 Generalized Free Products

To this end, we use the assumption that 𝑇 is a profinite space to choose a partition 𝑇 = 𝑇𝑠 ∪· 𝑇𝑡 of 𝑇 into open-closed sets such that 𝑠 ∈ 𝑇𝑠 and 𝑡 ∈ 𝑇𝑡 . By Lemma 4.5.4, Î Î 𝐺 = 𝐺 (𝑇𝑠 ) ∗ 𝐺 (𝑇𝑡 ), where 𝐺 (𝑇𝑠 ) = ∗ 𝑟 ∈𝑇𝑠 𝐺 𝑟 and 𝐺 (𝑇𝑡 ) = ∗ 𝑟 ∈𝑇𝑡 𝐺 𝑟 . Consider the homomorphism 𝛼 : 𝐺 → 𝐺 (𝑇𝑠 )/(𝐺 (𝑇𝑠 ) ∩ 𝑁) defined on 𝐺 (𝑇𝑠 ) as the quotient map and on 𝐺 (𝑇𝑡 ) as the trivial map. If 𝑟 ∈ 𝑇𝑠 , then 𝑁𝑟 ≤ 𝐺 (𝑇𝑠 ) ∩ 𝑁. If 𝑟 ∈ 𝑇𝑡 , then 𝑁𝑟 ≤ 𝐺 (𝑇𝑡 ). In each case 𝛼(𝑁𝑟 ) = 1. Hence, 𝛼(𝑁) = 1. In addition, 𝛼(𝐺 𝑡 ) = 1, because 𝐺 𝑡 ≤ 𝐺 (𝑇𝑡 ). Therefore, 𝛼(𝐺 𝑡 𝑁) = 1. On the other hand, 𝛼(𝑔) ≠ 1, because 𝑔 ∈ 𝐺 (𝑇𝑠 ) r (𝐺 (𝑇𝑠 ) ∩ 𝑁). It follows that 𝑔 ∉ 𝐺 𝑡 𝑁, as desired. Claim B: The family G¯ satisfies the universal property (4.1). be the quotient map and consider a continuous map Ø 𝛾0 : 𝐺 𝑡 𝑁/𝑁 → 𝐴

Let 𝜋 : 𝐺 → 𝐺/𝑁

𝑡 ∈𝑇

into a profinite group 𝐴 whoseÐrestriction to each of the groups 𝐺 𝑡 𝑁/𝑁 is a homomorphism. Then, 𝛾0 ◦ 𝜋 : 𝑡 ∈𝑇 𝐺 𝑡 → 𝐴 has the same properties. Hence, 𝛾0 ◦ 𝜋 extends to a homomorphism 𝛾˜ : 𝐺 → 𝐴. For every 𝑡 ∈ 𝑇 we have 𝛾(𝑁 ˜ 𝑡 ) = 𝛾0 (𝜋(𝑁𝑡 )) = 𝛾0 (1) = 1. Since 𝑁 = [𝑁𝑡 ]𝑡 ∈𝑇 , this implies that 𝛾(𝑁) ˜ = 1. It follows that there exists a homomorphism 𝛾 : 𝐺/𝑁 → 𝐴 such that 𝛾˜ = 𝛾 ◦ 𝜋. By our construction, 𝛾 extends 𝛾0 . If 𝛾 0 : Ð 𝐺/𝑁 → 𝐴 is another homomorphism that extends 𝛾0 , then 𝛾 0 ◦ 𝜋 extends 𝛾0 ◦ 𝜋 on 𝑡 ∈𝑇 𝐺 𝑡 . Hence, 𝛾 0 ◦ 𝜋 = 𝛾 ◦ 𝜋. Since 𝜋 is surjective, it follows that 𝛾 0 = 𝛾, as desired.  Î Lemma 4.6.6 Let 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 be an inner free product of profinite groups over a profinite space 𝑇. Let 𝑁 be a closed normal subgroup of 𝐺. For every 𝑡 ∈ 𝑇 we set Î 𝑁𝑡 = 𝐺 𝑡 ∩ 𝑁. Then, 𝐺/𝑁 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 𝑁/𝑁 if and only if 𝑁 = [𝑁𝑡 ]𝑡 ∈𝑇 . Î Proof If 𝑁 = [𝑁𝑡 ]𝑡 ∈𝑇 , then by Lemma Î 4.6.5, 𝐺/𝑁 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 𝑁/𝑁. Conversely, suppose that 𝐺/𝑁 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 𝑁/𝑁 and set 𝑁0 = [𝑁𝑡 ]𝑡 ∈𝑇 . Then, 𝑁0 is a closed normal subgroup of 𝐺 contained in 𝑁. Since 𝑁𝑡 ≤ 𝐺 𝑡 ∩ 𝑁0 ≤ 𝐺 𝑡 ∩ 𝑁 = 𝑁𝑡 , we have 𝐺 𝑡 ∩ 𝑁0 = 𝑁𝑡 , for every 𝑡 ∈ 𝑇. We have to show that 𝑁0 = 𝑁. We consider the quotient map 𝜋 : 𝐺/𝑁0 → 𝐺/𝑁. Its restriction Ø Ø 𝜋0 : 𝐺 𝑡 𝑁0 /𝑁0 → 𝐺 𝑡 𝑁/𝑁 𝑡 ∈𝑇

𝑡 ∈𝑇

is bijective. Indeed, for every 𝑡 ∈ 𝑇, 𝜋0 maps 𝐺 𝑡 𝑁0 /𝑁0 bijectively onto 𝐺 𝑡 𝑁/𝑁 (because 𝐺 𝑡 ∩ 𝑁0 = 𝑁𝑡 = 𝐺 𝑡 ∩ 𝑁)Îand, for distinct 𝑠, 𝑡 ∈ 𝑇, we have 𝐺 𝑠 𝑁/𝑁 ∩ 𝐺 𝑡 𝑁/𝑁 = 1 in 𝐺/𝑁, because 𝐺/𝑁 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 𝑁/𝑁. Î Using the universal property of ∗ 𝑡 ∈𝑇 𝐺 𝑡 𝑁/𝑁, the map 𝜋0−1 extends to a unique homomorphism 𝜋 0 : 𝐺/𝑁 → 𝐺/𝑁0 . Since 𝜋 ◦𝜋 0 : 𝐺/𝑁 → 𝐺/𝑁 extends the identity Ð map 𝜋 ◦ 𝜋0−1 of 𝑡 ∈𝑇 𝐺 𝑡 𝑁/𝑁, it is the identity of 𝐺/𝑁, by the uniqueness in Î the universal property of ∗ 𝑡 ∈𝑇 𝐺 𝑡 𝑁/𝑁. By Lemma 4.2.3(c), 𝐺 = h𝐺 𝑡 i𝑡 ∈𝑇 , and hence 𝐺/𝑁0 = h𝐺 𝑡 𝑁0 /𝑁0 i𝑡 ∈𝑇 . Therefore 𝜋 0 is surjective. It follows that 𝜋 is an isomorphism. Therefore, 𝑁0 = 𝑁, as claimed. 

4.7 Free Products as Semi-direct Products

59

4.7 Free Products as Semi-direct Products Let 𝐴 and 𝐵 be profinite groups, 𝜋 the projection of the free product 𝐴 ∗ 𝐵 onto 𝐵, given by 𝜋(𝑎) = 1 for each 𝑎 ∈ 𝐴 and 𝜋(𝑏) = 𝑏 for each 𝑏 ∈ 𝐵. Let 𝐵0 be a closed subgroup of 𝐵. We prove that 𝜋 −1 (𝐵0 ) is isomorphic to a semi-direct product of 𝐵0 with a free product, in the sense of Proposition 3.1.1, of a family of conjugates of 𝐴 in 𝐴 ∗ 𝐵.1 We start the proof of this statement with an alternative description of 𝐴 ∗ 𝐵. Setup 4.7.1 Consider the constant sheaf X = ( 𝐴 × 𝐵, pr, 𝐵) with the profinite space ˆ be the free profinite 𝐴 × 𝐵 and the projection pr on 𝐵 (Example 2.2.2). Let (X, 𝜔, 𝐴) product over this sheaf (Definition 4.2.1). For each 𝑏 ∈ 𝐵 the fiber pr−1 (𝑏) = 𝐴 × {𝑏} ˆ is a profinite Î group isomorphic to 𝐴; let 𝐴𝑏 = 𝜔( 𝐴 × {𝑏}). Then, 𝐴 is the inner free product ∗ 𝑏 ∈𝐵 𝐴𝑏 (Proposition 4.3.1). Moreover, the map 𝜔(𝑎, 1) ↦→ 𝑎 identifies 𝐴1 with 𝐴. Lemma 4.7.2 The group 𝐵 acts on 𝐴ˆ in the following way: (4.9)

𝜔(𝑎, 𝑏) 𝑥 = 𝜔(𝑎, 𝑏𝑥),

𝑎 ∈ 𝐴, 𝑏, 𝑥 ∈ 𝐵.

Proof The continuous action (𝑎, 𝑏) 𝑥 = (𝑎, 𝑏𝑥) of 𝐵 on 𝐴 × 𝐵 induces a right action ˆ Indeed, each 𝑥 ∈ 𝐵 defines a continuous map 𝜇 𝑥 : 𝐴 × 𝐵 → 𝐴 × 𝐵 by of 𝐵 on 𝐴. 𝜇 𝑥 (𝑎, 𝑏) := (𝑎, 𝑏) 𝑥 = (𝑎, 𝑏𝑥). Condition (4.2) gives a unique homomorphism 𝜔ˆ 𝑥 : 𝐴ˆ → 𝐴ˆ such that (4.10)

𝐴×𝐵

𝜔

/ 𝐴ˆ

𝜔

 / 𝐴ˆ

𝜇𝑥

 𝐴×𝐵

𝜔ˆ 𝑥

commutes. For each 𝑎ˆ ∈ 𝐴ˆ and each 𝑥 ∈ 𝐵 we define (4.11)

𝑎ˆ 𝑥 = 𝜔ˆ 𝑥 ( 𝑎). ˆ

If 𝑦 is another element of 𝐵, then 𝜇 𝑥 𝑦 = 𝜇 𝑦 ◦ 𝜇 𝑥 . By the uniqueness of (4.2), ˆ In addition, 𝜔ˆ 1 = id ˆ . 𝜔ˆ 𝑥 𝑦 = 𝜔ˆ 𝑦 ◦ 𝜔ˆ 𝑥 , that is, 𝑎ˆ 𝑥 𝑦 = ( 𝑎ˆ 𝑥 ) 𝑦 for every 𝑎ˆ ∈ 𝐴. 𝐴 ˆ Thus, (4.11) defines an action of 𝐵 on 𝐴 from the right. Property (4.9) follows from diagram (4.10): 𝜔(𝑎, 𝑏) 𝑥 = 𝜔ˆ 𝑥 (𝜔(𝑎, 𝑏)) = 𝜔(𝜇 𝑥 (𝑎, 𝑏)) = 𝜔(𝑎, 𝑏𝑥).  Definition 4.7.3 Let 𝐵 and 𝐶 be profinite groups with a right action of 𝐵 on 𝐶. The semi-direct product 𝐵 n 𝐶 is the group consisting of all pairs (𝑏, 𝑐) ∈ 𝐵 × 𝐶 with the 0 multiplication rule (𝑏, 𝑐) (𝑏 0, 𝑐 0) = (𝑏𝑏 0, 𝑐 𝑏 𝑐 0). Alternatively, 𝐵 n 𝐶 is a profinite group 𝐻 containing 𝐶 as a closed normal subgroup and 𝐵 as a closed subgroup such 𝜋0

that 𝐵 ∩ 𝐶 = 1 and 𝐵𝐶 = 𝐻. Finally, the sequence 1 −→ 𝐶 −→ 𝐵 n 𝐶 −→ 𝐵 −→ 1 with 𝜋 0 (𝑏, 𝑐) = 𝑏 is exact. 1 This section reproduces [HJP12, Section 2].

60

4 Generalized Free Products

Lemma 4.7.4 The homomorphism 𝛼 : 𝐴 ∗ 𝐵 → 𝐵 n 𝐴ˆ defined by 𝛼(𝑎) = 𝜔(𝑎, 1) for 𝑎 ∈ 𝐴 and 𝛼(𝑏) = 𝑏 for 𝑏 ∈ 𝐵 is an isomorphism. Its inverse 𝛼 0 : 𝐵 n 𝐴ˆ → 𝐴 ∗ 𝐵 is given by 𝛼 0 (𝑏) = 𝑏 on 𝐵 and by (4.12)

(4.13) on 𝐴ˆ = h𝜔( 𝐴 × 𝐵)i.

𝛼 0 (𝜔(𝑎, 𝑏)) = 𝑎 𝑏

Proof The proof has two parts. Part A: 𝛼 0 is well defined. The map 𝐴 × 𝐵 → 𝐴 ∗ 𝐵 defined by (𝑎, 𝑏) ↦→ 𝑎 𝑏 is continuous and its restriction to each fiber 𝐴 × {𝑏} is a homomorphism. By (4.2), there exists a unique homomorphism 𝛼 0 : 𝐴ˆ → 𝐴 ∗ 𝐵 such that 𝛼 0 (𝜔(𝑎, 𝑏)) = 𝑎 𝑏 . In ˆ we have to prove that order to extend 𝛼 0 from 𝐵 ∪ 𝐴ˆ to a homomorphism on 𝐵 n 𝐴, ˆ that is, that 𝛼 0 ( 𝑎ˆ 𝑥 ) = 𝛼 0 ( 𝑎) 𝛼 0 commutes with the action of 𝐵 on 𝐴, ˆ 𝑥 for all 𝑎ˆ ∈ 𝐴ˆ ˆ and 𝑥 ∈ 𝐵. But, since 𝜔( 𝐴 × 𝐵) generates 𝐴, it suffices to prove this for 𝑎ˆ = 𝜔(𝑎, 𝑏), for all 𝑎 ∈ 𝐴 and 𝑏 ∈ 𝐵. In this case, by (4.9), 𝛼 0 (𝜔(𝑎, 𝑏) 𝑥 ) = 𝛼 0 (𝜔(𝑎, 𝑏𝑥)) = 𝑎 𝑏𝑥 = (𝑎 𝑏 ) 𝑥 = (𝛼 0 (𝜔(𝑎, 𝑏))) 𝑥 . Part B: 𝛼 and 𝛼 0 are inverse to each other. By definition, both 𝛼 and 𝛼 0 are the identity map on 𝐵. For 𝑎 ∈ 𝐴 we have 𝛼 0 (𝛼(𝑎)) = 𝛼 0 (𝜔(𝑎, 1)) = 𝑎 1 = 𝑎. Conversely, in order to prove that 𝛼 ◦ 𝛼 0 | 𝐴ˆ = id 𝐴ˆ it suffices to prove 𝛼 ◦ 𝛼 0 (𝜔(𝑎, 𝑏)) = 𝜔(𝑎, 𝑏) for all (𝑎, 𝑏) ∈ 𝐴 × 𝐵. Indeed: (4.13)

(4.12)

(4.9)

𝛼(𝛼 0 (𝜔(𝑎, 𝑏))) = 𝛼(𝑎 𝑏 ) = 𝛼(𝑎) 𝛼(𝑏) = 𝜔(𝑎, 1) 𝑏 = 𝜔(𝑎, 1 · 𝑏) = 𝜔(𝑎, 𝑏), as needed.



We use the following result in the proof of Theorem 9.1.6. Lemma 4.7.5 Let 𝐴 and 𝐵 be profinite groups, let X = ( 𝐴 × 𝐵, pr, 𝐵) the sheaf, and ˆ be the free profinite product over X introduced in Setup 4.7.1. let (X, 𝜔, 𝐴) (a) The group 𝐵 n 𝐴ˆ is the free product of its subgroups 𝐴1 = 𝐴 and 𝐵. (b) Let 𝐾 = h𝐴𝑏 i𝑏 ∈𝐵Î≤ 𝐴 ∗ 𝐵. Let 𝜋 : 𝐴 ∗ 𝐵 → 𝐵 be the projection on the second factor. Then, 𝐾 = ∗ 𝑏 ∈𝐵 𝐴𝑏 = Ker(𝜋) and 𝐴 ∗ 𝐵 = 𝐵 n 𝐾. Proof Proof of (a). The isomorphism 𝛼 of Lemma 4.7.4 maps 𝐴 onto 𝐴1 and 𝐵 onto 𝐵. Moreover, Setup 4.7.1 identifies 𝐴1 with 𝐴. Î Proof of (b). By Setup 4.7.1, 𝐴ˆ = ∗ 𝑏 ∈𝐵 𝐴𝑏 . By (4.13), the isomorphism 𝛼 0 of Lemma 4.7.4 maps each 𝐴𝑏 = 𝜔( 𝐴 × {𝑏}) onto 𝐴𝑏 . Hence, 𝛼 0 maps 𝐴ˆ onto Î 𝐾, whence 𝐾 = ∗ 𝑏 ∈𝐵 𝐴𝑏 . Furthermore, 𝛼 0 maps 𝐵 identically onto itself, hence ˆ = 𝐵 n 𝐾. 𝐴 ∗ 𝐵 = 𝛼 0 (𝐵 n 𝐴)

4.7 Free Products as Semi-direct Products

61

Î ∗ 𝑏 ∈𝐵 𝐴𝑏

/ 𝐴∗𝐵

/ 𝐴ˆ O

/ 𝐵 n 𝐴ˆ O

1

𝛼

 /𝐾

1

𝛼0

𝛼

/𝐵

/1

/𝐵

/1

𝛼0

 / 𝐴∗𝐵

𝜋

Î ∗ 𝑏 ∈𝐵 𝐴𝑏 Since 𝜋 is the identity on 𝐵 and is trivial on 𝐾, 𝜋 ◦ 𝛼 0 = 𝜋 0, hence the isomorphism ˆ = 𝐾, so 𝐾 = Ker(𝜋).  𝛼 0 maps Ker(𝜋 0) onto Ker(𝜋). But Ker(𝜋 0) = 𝐴ˆ and 𝛼 0 ( 𝐴) If 𝐵0 is a closed subgroup of a profinite group 𝐵, then the quotient map 𝑏 ↦→ 𝑏𝐵0 is a continuous surjection 𝐵 → 𝐵/𝐵0 of profinite spaces. By Lemma 1.5.5, this map has a continuous section. Hence, by Fact 1.1.3(e), 𝐵 has a closed subset of representatives 𝑅 for the collection of left cosets 𝑏𝐵0 with 𝑏 ∈ 𝐵. This brings us to a result that we use in the proof of Theorem 9.2.3. In this result we start from 𝑅 and present 𝜋 −1 (𝐵0 ) as a free product of 𝐵0 and conjugates of 𝐴 with exponents ranging over 𝑅. Lemma 4.7.6 Let 𝐴 and 𝐵 be profinite groups, 𝐴 ∗ 𝐵 their free product, and 𝜋 : 𝐴 ∗ 𝐵 → 𝐵 the projection on the second factor. Consider a closed subgroup 𝐵0 of 𝐵 and let 𝑅 be a closed system of representatives of the left Î cosets 𝑏𝐵0 of 𝐵 modulo 𝐵0 . Let 𝐻 = 𝜋 −1 (𝐵0 ). Then, 𝐻 = 𝐵0 n Ker(𝜋) and 𝐻 = ( ∗ 𝑟 ∈𝑅 𝐴𝑟 ) ∗ 𝐵0 . Proof Let 𝐾 = Ker(𝜋). By Lemma 4.7.5(b), 𝐴 ∗ 𝐵 = 𝐵 n 𝐾. Since 𝜋 is injective on 𝐵0 , we have 𝐻 = 𝐵0 n 𝐾, where 𝐵0 acts on 𝐾 by conjugation in 𝐻. The map 𝑅 × 𝐵0 → 𝐵 given by (𝑟, 𝑏 0 ) ↦→ 𝑟𝑏 0 is a continuous bijection of profinite spaces, hence so is its inverse 𝛽 : 𝐵 → 𝑅 × 𝐵0 . Therefore, the composition of 𝛽 with the projection on 𝐵0 is a continuous surjection 𝛿 : 𝐵 → 𝐵0 ; it satisfies 𝛿−1 (𝑏 0 ) = 𝑅𝑏 0 for each 𝑏 0 ∈Î𝐵0 . By Lemma 4.7.5(b), 𝐾 = ∗ 𝑏 ∈𝐵 𝐴𝑏 . Hence, by Lemma 4.6.1, Ö Ö Ö Ö Ö Ö
𝑏

𝐾= ∗ ∗ 𝐴𝑏 = ∗ ∗ 𝐴 𝑟 𝑏0 = ∗ ∗ 𝐴𝑟 0 . 𝑏0 ∈𝐵0

𝑏 ∈𝑅𝑏0

𝑏0 ∈𝐵0

𝑟 ∈𝑅

𝑏0 ∈𝐵0

𝑟 ∈𝑅

Î By Lemma 4.7.5(b) (with ∗ 𝑟 ∈𝑅 𝐴𝑟 , 𝐵0 replacing 𝐴, 𝐵), Ö Ö
𝑏 Ö
𝐻 = 𝐵0 n 𝐾 = 𝐵0 n ∗ ∗ 𝐴𝑟 0 = ∗ 𝐴𝑟 ∗ 𝐵 0 . 𝑏0 ∈𝐵0

𝑟 ∈𝑅

𝑟 ∈𝑅

By the beginning of the proof, the action of 𝐵0 on 𝐾 is given by Ö 0 0 (𝑎 𝑏0 ) 𝑏0 = 𝑎 𝑏0 𝑏0 , for 𝑎 ∈ ∗ 𝐴𝑟 and 𝑏 0 , 𝑏 00 ∈ 𝐵0 . 𝑟 ∈𝑅



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4 Generalized Free Products

Construction 4.7.7 (Semi-constant sheaf) Let 𝑆 be a finite set and for each 𝔭 ∈ 𝑆 let 𝑇𝔭 be a profinite space and 𝐺 𝔭 a profinite group. We consider the profinite space Ð · 𝔭∈𝑆 𝑇𝔭 in which each 𝑇𝔭 is an open-closed subspace and the profinite space 𝑇 := Ð 𝑋 := · 𝔭∈𝑆 (𝑇𝔭 × 𝐺 𝔭 ) in which each 𝑇𝔭 × 𝐺 𝔭 is an open-closed subspace. Then, we define 𝜏 : 𝑋 → 𝑇 by 𝜏(𝑡, 𝑔) = 𝑡 and observe that X = (𝑋, 𝜏, 𝑇) is a sheaf (Definition 2.2.1), which we call semi-constant, and where 𝑋𝑡 = 𝜏 −1 (𝑡) = {𝑡} × 𝐺 𝔭 for each 𝑡 ∈ 𝑇𝔭 . If 𝑆 consists of one element, then X is a constant sheaf (Example 2.2.2). A semi-constant sheaf naturally arises in the following situation: Let 𝐺˜ be a profinite group. For each 𝔭 ∈ 𝑆 let 𝑅𝔭 be a closed subset of 𝐺˜ and 𝐺 𝔭 a closed ˜ We choose a homeomorphic copy 𝑇𝔭 of 𝑅𝔭 and a homeomorphism subgroup of 𝐺. Ð 𝜆 (𝑡) 𝜆𝔭 : 𝑇𝔭 → 𝑅𝔭 . As above we write 𝑇 = · 𝔭∈𝑆 𝑇𝔭 , and for each 𝑡 ∈ 𝑇𝔭 let 𝐺 𝑡 = 𝐺 𝔭 𝔭 be the conjugate of 𝐺 𝔭 by 𝜆𝔭 (𝑡). Then, the family (𝐺 𝑡 )𝑡 ∈𝑇 is étale continuous; in fact, ˜ given by 𝑡 ↦→ 𝐺 𝑡 is even strictly continuous. We let X be the the map 𝑇 → Subgr( 𝐺) semi-constant sheaf as in the preceding paragraph and define a map 𝜔 : 𝑋 → 𝐺˜ by 𝜔(𝑡, 𝑔) = 𝑔 𝜆𝔭 (𝑡) for 𝑡 ∈ 𝑇𝔭 , 𝑔 ∈ 𝐺 𝔭 , and 𝔭 ∈ 𝑆. Then, 𝜔 maps Ð each 𝑋𝜌𝑡 isomorphically onto 𝐺 𝑡 for each 𝑡 ∈ 𝑇. Thus, X is the associated sheaf of 𝔭∈𝑆 {𝐺 𝔭 }𝜌∈𝑅𝔭 . Î Î 𝜌 𝐺 is aÎ closed subgroup of 𝐺˜ such that 𝐺 = ∗ 𝔭∈𝑆 ∗ 𝜌∈𝑅𝔭 𝐺 𝔭 . Then, 𝐺 = Î Suppose Î ∗ 𝔭∈𝑆 ∗ 𝑡 ∈𝑇𝔭 𝐺 𝑡 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 (Lemma 4.5.4). By Definition 4.2.1, Proposition 4.2.2, and Lemma 4.2.3(d), (X, 𝜔, 𝐺) is, up to isomorphism, the outer free product over X. The importance of this observation lies in the fact that this free product is completely determined, up to isomorphism, by the data (𝑅𝔭 , 𝐺 𝔭 )𝔭∈𝑆 . In other words, Î Î if 𝐺 0 = ∗ 𝔭∈𝑆 ∗ 𝜌∈𝑅𝔭0 (𝐺 𝔭0 ) 𝜌 , the subset 𝑅𝔭0 of 𝐺 𝔭0 is homeomorphic to 𝑅𝔭 , and 𝐺 𝔭0  𝐺 𝔭 , for each 𝔭 ∈ 𝑆, then 𝐺 0  𝐺.

4.8 Closed Subgroups of Free Products We generalize Lemma 3.1.10 to free profinite products over profinite spaces. We begin with a converse of Lemma 4.5.4: Lemma 4.8.1 Let {𝐺 𝑡 }𝑡 ∈𝑇 be an étale continuous family of closed subgroups of a profinite group 𝐺 over a profinite space 𝑇 (Definition 2.1.1). Assume that for every Î partition U into disjoint open-closed subsets of 𝑇 we have 𝐺 = ∗ 𝑈 ∈U 𝐺 (𝑈), where Î 𝐺 (𝑈) = h𝐺 𝑡 i𝑡 ∈𝑈 . Then, 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 . Ð Proof Set 𝑉 = 𝑡 ∈𝑇 𝐺 𝑡 . Let 𝐴 be a finite group and let 𝛾0 : 𝑉 → 𝐴 be a continuous map whose restriction to every 𝐺 𝑡 is a homomorphism. By Remark 4.1.2 it suffices to show that 𝛾0 extends to a unique homomorphism 𝛾 : 𝐺 → 𝐴. Since 𝐴 is discrete, every 𝑔 ∈ 𝑉 has an open neighborhood in 𝑉 on which 𝛾0 is constant. Thus, there is an open 𝐾 ⊳ 𝐺 such that 𝛾0 is constant on 𝑔𝐾 ∩ 𝑉. Since 𝑉 is closed (Lemma 2.1.2), hence compact (Fact 1.1.3(a)), we may assume that this 𝐾 is the same for all 𝑔 ∈ 𝑉. Thus, the map 𝛾 0 : 𝑉 𝐾 → 𝐴 defined by 𝛾 0 (𝑔𝑘) = 𝛾0 (𝑔) for each 𝑔 ∈ 𝑉 and all 𝑘 ∈ 𝐾 is well defined, hence also continuous. Moreover, the restriction of 𝛾 0 to 𝐺 𝑡 𝐾 is a homomorphism, for every 𝑡 ∈ 𝑇. Since the map 𝑡 ↦→ 𝐺 𝑡 is étale continuous, every 𝑠 ∈ 𝑇 has an open-closed neighborhood 𝑈 such that 𝐺 𝑡 ≤ 𝐺 𝑠 𝐾 for all 𝑡 ∈ 𝑈. By the compactness of 𝑇 there

4.8 Closed Subgroups of Free Products

63

is a partition U of 𝑇 into finitely many open-closed sets such that for every 𝑈 ∈ U there is a 𝑡𝑈 ∈ 𝑇 with 𝐺 𝑡 ≤ 𝐺 𝑡𝑈 𝐾 for all 𝑡 ∈ 𝑈. In particular, 𝐺 (𝑈) ≤ 𝐺 𝑡𝑈 𝐾 ⊆ 𝑉 𝐾. Thus, the restriction of 𝛾 0 to 𝐺 (𝑈) is a homomorphism 𝛾𝑈 : 𝐺 (𝑈) → 𝐴. Î Since 𝐺 = ∗ 𝑈 ∈U 𝐺 (𝑈), there is a homomorphism 𝛾 : 𝐺 → 𝐴 that extends 𝛾𝑈 for every 𝑈 ∈ U. In particular, 𝛾 extends 𝛾0 . This extension is unique, because by the assumption of the lemma applied to the trivial partition U = {𝑇 } of 𝑇 we have 𝐺 = 𝐺 (𝑇) = h𝐺 𝑡 i𝑡 ∈𝑇 .  Î Lemma 4.8.2 Let 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 be a free product of profinite groups over Ð a profinite space 𝑇. For each 𝑡 ∈ 𝑇 let 𝐻𝑡 be a closed subgroup of 𝐺 𝑡 such that Î𝑡 ∈𝑇 𝐻𝑡 is a closed subset of 𝐺. Then, the subgroup 𝐻 = h𝐻𝑡 i𝑡 ∈𝑇 of 𝐺 satisfies 𝐻 = ∗ 𝑡 ∈𝑇 𝐻𝑡 . Proof We break up the proof into two parts. Part A: We assume that 𝑇 is a finite discrete space. Let 𝐴 be a profinite group and for each 𝑡 ∈ 𝑇 let 𝜂𝑡 : 𝐻𝑡 → 𝐴 be a homomorphism. We have to find a homomorphism 𝜂 : 𝐻 → 𝐴 such that 𝜂| 𝐻𝑡 = 𝜂𝑡 for each 𝑡 ∈ 𝑇. Since the 𝐻𝑡 ’s generate 𝐻, the homomorphism 𝜂 will be unique with this property. Hence, by Remark 4.1.2, we may assume without loss that 𝐴 is finite. Then, for each 𝑡 ∈ 𝑇, the homomorphism 𝜂𝑡 extends to a homomorphism 𝜇𝑡 from an open subgroup 𝑈𝑡 of 𝐺 𝑡 that contains 𝐻𝑡 into 𝐴 [FrJ08, p. 8, Lemma 1.2.5(c)]. Claim: There is an open subgroup 𝑈 of 𝐺 containing 𝐻 such that 𝐺 𝑡 ∩ 𝑈 = 𝑈𝑡 for Î each 𝑡 ∈ 𝑇. Indeed, consider the cartesian product 𝐺¯ = 𝑡 ∈𝑇 𝐺 𝑡 and the cartesian Î map 𝜅 : 𝐺 → 𝐺¯ defined in (3.8). Since 𝑇 is finite, 𝑈¯ := 𝑡 ∈𝑇 𝑈𝑡 is an open subgroup ¯ Hence, 𝑈 = 𝜅 −1 (𝑈) ¯ is open in 𝐺 and of 𝐺. ¯ = h𝜅 −1 (𝑈𝑡 )i𝑡 ∈𝑇 ≥ h𝑈𝑡 i𝑡 ∈𝑇 ≥ h𝐻𝑡 i𝑡 ∈𝑇 = 𝐻. 𝑈 = 𝜅 −1 (𝑈) Now consider 𝑡 ∈ 𝑇. Then, 𝑈𝑡 ≤ 𝐺 𝑡 ∩ 𝑈. On the other hand, 𝜅(𝑈 ∩ 𝐺 𝑡 ) ≤ 𝑈¯ ∩ 𝜅(𝐺 𝑡 ) = 𝜅(𝑈𝑡 ) and 𝜅 is injective on 𝐺 𝑡 , so 𝑈 ∩ 𝐺 𝑡 ≤ 𝑈𝑡 . Thus, 𝐺 𝑡 ∩ 𝑈 = 𝑈𝑡 , as claimed. Î By Corollary 3.1.9, 𝐺 has a closed subgroup 𝑊 such that 𝑈 = 𝑊∗ ∗ 𝑡 ∈𝑇 𝑈𝑡 . Hence, there exists a homomorphism 𝜇 : 𝑈 → 𝐴 that extends 𝜇𝑡 for each 𝑡 ∈ 𝑇. The restriction 𝜂 of 𝜇 to 𝐻 is a homomorphism of 𝐻 into 𝐴 that satisfies 𝜂| 𝐻𝑡 = 𝜂𝑡 for each 𝑡 ∈ 𝑇, as desired. Part B: The general case. The map 𝑇 → Subgr(𝐻) given by 𝑡 ↦→ 𝐻𝑡 is étale continuous. Indeed, by Lemma 4.6.4(a), 𝐻𝑡 = 𝐻 ∩ 𝐺 𝑡 , for every 𝑡 ∈ 𝑇, hence the above map is the composition of the map 𝑇 → Subgr(𝐺) given by 𝑡 ↦→ 𝐺 𝑡 with the restriction Subgr(𝐺) → Subgr(𝐻) given by 𝐺 𝑡 ↦→ 𝐻 ∩ 𝐺 𝑡 . These two maps are étale continuous, by the assumption on 𝐺 and by Remark 1.2.1(c), respectively. Let U be a partition of 𝑇 into disjoint (hence, finitely many) open-closed subsets. Î By Lemma 4.5.4, 𝐺 = ∗ 𝑈 ∈U 𝐺 (𝑈), where 𝐺 (𝑈) = h𝐺 𝑡 i𝑡 ∈𝑈 . Since U is a finite Ð set, 𝑈Î ∈U 𝐻 (𝑈) is a closed subset of 𝐻, where 𝐻 (𝑈) = h𝐻𝑡 i𝑡 ∈𝑈 . Hence, by Part A, 𝐻 = ∗ 𝑈 ∈U 𝐻 (𝑈). Î It follows from Lemma 4.8.1 that 𝐻 = ∗ 𝑡 ∈𝑇 𝐻𝑡 , as claimed.  We are now in a position to prove the promised result.

64

4 Generalized Free Products

Î Proposition 4.8.3 Let 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 be a free product of profinite groups over a profinite space 𝑇. (a) If 𝑟, 𝑠 ∈ 𝑇 and 𝑟 ≠ 𝑠, then 𝐺 𝑟 ∩ 𝐺 𝑠𝑥 = 1 for every 𝑥 ∈ 𝐺. (b) Given 𝑡 ∈ 𝑇, we have 𝐺 𝑡 ∩ 𝐺 𝑡𝑥 = 1 for every 𝑥 ∈ 𝐺 r 𝐺 𝑡 . (c) If 𝑡 ∈ 𝑇 and 𝐺 𝑡 ≠ 1, then 𝑁𝐺 (𝐺 𝑡 ) = 𝐺 𝑡 . (d) Suppose that 𝐺 𝑡 ≠ 1 for each 𝑡 ∈ 𝑇. Then, G = {𝐺 𝑡 }𝑡 ∈𝑇 is a system of representatives for the 𝐺-orbits of Con(G)max . (e) For every finite group 𝐴 of 𝐺 there exist 𝑡 ∈ 𝑇 and 𝑥 ∈ 𝐺 such that 𝐴 ≤ 𝐺 𝑡𝑥 . (f) If 𝐺 𝑡 is torsion free for each 𝑡 ∈ 𝑇, then 𝐺 is also torsion free. Proof Proof of (a). Assume toward contradiction that there exist elements 𝑥 ∈ 𝐺 and 𝑔 ∈ 𝐺 𝑟 ∩ 𝐺 𝑠𝑥 with 𝑔 ≠ 1. We choose a homomorphism 𝛼𝑟 of 𝐺 𝑟 into a finite group 𝐴 such that 𝛼𝑟 (𝑔) ≠ 1. Then, we use Lemma 4.5.2 to extend 𝛼𝑟 to a homomorphism 𝛼 : 𝐺 → 𝐴 with 𝛼(𝐺 𝑠 ) = 1. Then, 𝛼(𝐺 𝑠𝑥 ) = 1. Therefore, on the one hand 𝛼(𝑔) = 𝛼𝑟 (𝑔) ≠ 1. On the other hand, 𝛼(𝑔) = 1 because 𝑔 ∈ 𝐺 𝑠𝑥 . This contradiction proves that 𝑥 and 𝑔 as above do not exist. Proof of (b). Since 𝐺 𝑡 is a closed subgroup of 𝐺, there exists an open normal subgroup 𝑁 of 𝐺 with 𝑥 ∉ 𝑁𝐺 𝑡 . Since the map 𝑡 ↦→ 𝐺 𝑡 from 𝑇 to Subgr(𝐺) is étale continuous (Definition 4.1.1), 𝑡 has an open-closed neighborhood 𝑈 in 𝑇 such that 𝐺 (𝑈) = h𝐺 𝑢 i𝑢 ∈𝑈 ≤ 𝑁𝐺 𝑡 . Hence, 𝑥 ∉ 𝐺 (𝑈). By Lemma 4.5.4, 𝐺 = 𝐺 (𝑈) ∗𝐺 (𝑇 r𝑈). By Lemma 3.1.10(b), 𝐺 (𝑈) ∩𝐺 (𝑈) 𝑥 = 1. Since 𝐺 𝑡 ≤ 𝐺 (𝑈), we have 𝐺 𝑡 ∩ 𝐺 𝑡𝑥 = 1, as claimed. Proof of (c). Statement (c) is a consequence of Statement (b). 𝑔

𝑔

Proof of (d). If 𝐺 𝑡 ≤ 𝐺 𝑠 for 𝑡, 𝑠 ∈ 𝑇 and 𝑔 ∈ 𝐺, then 𝐺 𝑡 ∩ 𝐺 𝑠 = 𝐺 𝑡 ≠ 1. 𝑔 Hence, by (a), 𝑠 = 𝑡, so by (c), 𝑔 ∈ 𝐺 𝑡 . This implies that 𝐺 𝑡 = 𝐺 𝑠 . It follows that 𝐺 𝑡 ∈ Con(G)max . Moreover, G is a system of representatives for the 𝐺-orbits of Con(G)max , as claimed. Proof of (e). Remark 4.5.9 gives an inverse system h𝐺 𝑖 , 𝑇𝑖 , 𝜋 𝑗𝑖 i𝑖, 𝑗 ∈𝐼 , 𝑗 ≥𝑖 of profinite groups 𝐺 𝑖 and finite discrete spaces 𝑇𝑖 such that for each 𝑖 ∈ 𝐼 the group Î 𝐺 𝑖 = ∗ 𝑡𝑖 ∈𝑇𝑖 𝐺 𝑡𝑖 is a free product of finitely many profinite groups and 𝐺 = lim 𝐺 𝑖 . ←−− Let 𝜋𝑖 : 𝐺 → 𝐺 𝑖 and 𝜋𝑖 : 𝑇 → 𝑇𝑖 be the projections on the 𝑖th components and consider the subgroup 𝐴𝑖 = 𝜋𝑖 ( 𝐴) of 𝐺 𝑖 . Then, 𝐴 = lim 𝐴𝑖 . ←−− By Corollary 3.2.8, for each 𝑖 ∈ 𝐼 and 𝑡 ∈ 𝑇𝑖 the subset 𝑌𝑖,𝑡 = {𝑥 ∈ 𝐺 𝑖 | 𝐴𝑖 ≤ 𝐺 𝑡𝑥 } of 𝐺 𝑖 is closed. Hence, for each 𝑖 ∈ 𝐼, the subset 𝑌𝑖 = {(𝑥, 𝑡) ∈ 𝐺 𝑖 × 𝑇𝑖 | 𝐴𝑖 ≤ 𝐺 𝑡𝑥 } =

Ø

(𝑌𝑖,𝑡 × {𝑡})

𝑡 ∈𝑇𝑖

of 𝐺 𝑖 ×𝑇𝑖 is also closed. In particular, 𝑌𝑖 is a compact Hausdorff space (Fact 1.1.3(a)). By Proposition 3.2.7, 𝑌𝑖 is non-empty. If 𝑗 ≥ 𝑖, then 𝜋 𝑗𝑖 (𝑌 𝑗 ) ⊆ 𝑌𝑖 . Hence, the inverse limit 𝑌 = lim 𝑌𝑖 is non-empty [FrJ08, p. 2, Lemma. 1.1.3]. Moreover, 𝑌 ⊆ 𝐺 × 𝑇. ←−− Each (𝑥, 𝑡) ∈ 𝑌 satisfies 𝐴 ≤ 𝐺 𝑡𝑥 , as desired. Proof of (f). Statement (f) is an immediate corollary of statement (e).



Chapter 5

Relative Embedding Problems

For a profinite group 𝐺 we introduce the definition of an “embedding problem relative to a subset G of Subgr(𝐺)” and accordingly the concepts “𝐺 is G-projective”, “𝐺 is properly G-projective”, and “𝐺 is strongly G-projective”. These definitions appear in [Pop95] and [HJP05]. We prove some variants of them, which will be useful later. We prove that free products of profinite groups over profinite spaces are strongly projective and that every closed subgroup of a strongly projective group is again strongly projective (Proposition 5.4.4). We also prove that, under some favorable conditions on a profinite group 𝐺 and a subset G of Subgr(𝐺), every finite proper G-embedding problem for 𝐺 has a proper “strong” solution that satisfies some local conditions (Proposition 5.6.8), given in advance.

5.1 Embedding Problems and Projectivity We introduce the notions of G-projectivity, as introduced in [Har87], [Pop95], and [HJP05]. See Remark 5.1.6 for a discussion on the differences among those notions. Let 𝐺 be a profinite group and let G be a subset of Subgr(𝐺). Definition 5.1.1 ([Pop95, p. 128, Sec. 2]) A G-embedding problem for 𝐺 is a triple (5.1)

(𝜑 : 𝐺 → 𝐴, 𝛼 : 𝐵 → 𝐴, B),

where (5.2a) 𝜑 is a homomorphism and 𝛼 is an epimorphism of profinite groups, (5.2b) B is an étale compact subset of Subgr(𝐵) which is closed under conjugation in 𝐵 and under taking subgroups, that is, B = Con(B) (thus, by Corollary 1.4.3 and Remark 1.2.7, B is strictly closed), and (5.2c) for each Γ ∈ G there exists a homomorphism 𝛾Γ : Γ → 𝐵 such that 𝛼 ◦ 𝛾Γ = 𝜑|Γ and 𝛾Γ (Γ) ∈ B; this implies, in particular, that (5.2d) 𝜑(G) ⊆ 𝛼(B). We say that the G-embedding problem (5.1) for 𝐺 is © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 D. Haran and M. Jarden, The Absolute Galois Group of a Semi-Local Field, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-030-89191-6_5

65

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5 Relative Embedding Problems

(5.3a) (5.3b) (5.3c) (5.3d)

finite if 𝐵 is finite; proper if 𝜑 is surjective; rigid if 𝛼 is B-rigid, i.e. 𝛼 is injective on each 𝐵0 ∈ B; and split if 𝛼 splits, i.e. there is a homomorphism 𝜆 : 𝐴 → 𝐵 such that 𝛼 ◦ 𝜆 = id 𝐴.

(5.4) A (proper) [strong] solution of (5.1) is a (surjective) homomorphism 𝛾 : 𝐺 → 𝐵 such that 𝛼 ◦ 𝛾 = 𝜑 [and 𝛾(G) ⊆ B]. (5.5) We say that 𝐺 is (properly) [strongly] G-projective if (a) G is an étale compact subset of Subgr(𝐺), and (b) every finite (proper) G-embedding problem (5.1) is (properly) [strongly] solvable. Remark 5.1.2 The parentheses and the brackets in (5.4) indicate that the definition actually contains four sentences: (5.6a) A solution of (5.1) is a homomorphism 𝛾 : 𝐺 → 𝐵 such that 𝛼 ◦ 𝛾 = 𝜑. (5.6b) A proper solution of (5.1) is a surjective homomorphism 𝛾 : 𝐺 → 𝐵 such that 𝛼 ◦ 𝛾 = 𝜑. (5.6c) A strong solution of (5.1) is a homomorphism 𝛾 : 𝐺 → 𝐵 such that 𝛼 ◦ 𝛾 = 𝜑 and 𝛾(G) ⊆ B. (5.6d) A proper strong solution of (5.1) is an epimorphism 𝛾 : 𝐺 → 𝐵 such that 𝛼 ◦ 𝛾 = 𝜑 and 𝛾(G) ⊆ B. A similar interpretation applies to (5.5) and to all occurrences of this pattern in the rest of this monograph. Occasionally, when 𝐺, 𝐴, and 𝐵 are known, we abbreviate (5.1) (and similar embedding problems) as (𝜑, 𝛼, B). Notice that if 𝛼 is B-rigid, (5.2d) implies (5.2c). Indeed, let Γ ∈ G. Then, there is a 𝐵0 ∈ B such that 𝛼 maps 𝐵0 isomorphically onto 𝜑(Γ); put 𝛾Γ = (𝛼| 𝐵0 ) −1 ◦ 𝜑|Γ . Then, 𝛼 ◦ 𝛾Γ = 𝜑|Γ . Notation 5.1.3 The fiber product of homomorphisms of profinite groups 𝛼 : 𝐵 → 𝐴 and 𝜑¯ : 𝐴ˆ → 𝐴 is the profinite group 𝐵ˆ := 𝐵 × 𝐴 𝐴ˆ := {(𝑏, 𝑎) ˆ ∈ 𝐵 × 𝐴ˆ | 𝛼(𝑏) = 𝜑( ¯ 𝑎)}. ˆ ˆ ˆ ˆ Let 𝛼ˆ : 𝐵 → 𝐴 and 𝛽 : 𝐵 → 𝐵 be the projections on the second and first coordinates of 𝐵ˆ (see Diagram (5.7) below). ˆ 𝛼, Then, ( 𝐵, ˆ 𝛽) is a triple consisting of a profinite group 𝐵ˆ with homomorphisms ˆ ˆ 𝛼ˆ : 𝐵 → 𝐴 and 𝛽 : 𝐵ˆ → 𝐵 such that 𝛼 ◦ 𝛽 = 𝜑¯ ◦ 𝛼. ˆ Moreover, if 𝛽∗ : 𝐵∗ → 𝐵 and 𝛼∗ : 𝐵∗ → 𝐴ˆ are homomorphisms of profinite groups that satisfy 𝛼 ◦ 𝛽∗ = 𝜑¯ ◦ 𝛼∗ , then there exists a unique homomorphism 𝛾 ∗ : 𝐵∗ → 𝐵ˆ such that 𝛽 ◦ 𝛾 ∗ = 𝛽∗ and 𝛼ˆ ◦ 𝛾 ∗ = 𝛼∗ . ˆ 𝛼, Thus, ( 𝐵, ˆ 𝛽) is unique with these properties [FrJ08, pp. 499–502]. We refer to the square in Diagram (5.7) as cartesian.

5.1 Embedding Problems and Projectivity

67

Lemma 5.1.4 Let 𝐺 be a profinite group and G an étale compact subset of Subgr(𝐺). Let (𝜑 : 𝐺 → 𝐴, 𝛼 : 𝐵 → 𝐴, B) be a finite G-embedding problem for 𝐺. Then, there is a 𝐾0 ∈ Open(𝐺) contained in Ker(𝜑) such that for every 𝐾ˆ ∈ OpenNormal(𝐺) contained in 𝐾0 there exists a commutative diagram (5.7)

𝐺 𝜑ˆ

𝐵ˆ

𝛼ˆ

 / 𝐴ˆ

 𝐵

𝛼

  /𝐴

𝜑 𝜑¯

𝛽

ˆ and a subset Bˆ of Subgr( 𝐵) ˆ is a finite proper ˆ such that ( 𝜑, with 𝐵ˆ = 𝐵 × 𝐴 𝐴, ˆ 𝛼, ˆ B) ˆ rigid G-embedding problem for 𝐺 with Ker( 𝜑) ˆ = 𝐾ˆ and 𝛽( B) ⊆ B. Moreover: (a) (b) (c) (d)

If 𝐺 is G-projective, then 𝛼ˆ splits. ˆ then 𝛾 := 𝛽 ◦ 𝛾ˆ is a solution of (5.1). If 𝛾ˆ is a solution of ( 𝜑, ˆ 𝛼, ˆ B), ˆ then 𝛾 := 𝛽 ◦ 𝛾ˆ is a strong solution of (5.1). If 𝛾ˆ is a strong solution of ( 𝜑, ˆ 𝛼, ˆ B), ˆ then 𝛾 := 𝛽 ◦ 𝛾ˆ is a If 𝜑 is surjective and 𝛾ˆ is a proper solution of ( 𝜑, ˆ 𝛼, ˆ B), proper solution of (5.1).

Proof We consider an open normal subgroup 𝐾ˆ of 𝐺 and break up the proof into five parts. ˆ let 𝜑ˆ : 𝐺 → 𝐴ˆ Part A: Diagram (5.7). Assume that 𝐾ˆ ≤ Ker(𝜑). Let 𝐴ˆ = 𝐺/𝐾, be the quotient map (in particular, 𝜑ˆ is surjective) and let 𝜑¯ : 𝐴ˆ → 𝐴 be the map induced by 𝜑. Then, 𝜑 = 𝜑¯ ◦ 𝜑. ˆ We consider the fiber product 𝐵ˆ = 𝐵 × 𝐴 𝐴ˆ with the projection maps 𝛼ˆ : 𝐵ˆ → 𝐴ˆ and 𝛽 : 𝐵ˆ → 𝐵 on the coordinates. Then, diagram (5.7) commutes. Since 𝛼 is surjective, so is 𝛼ˆ [FrJ08, p. 500, Lemma 22.2.3(b)]. By (5.2c), for each Γ ∈ G, there exists a homomorphism 𝛾Γ : Γ → 𝐵 such that 𝛼 ◦ 𝛾Γ = 𝜑|Γ and 𝛾Γ (Γ) ∈ B. Since 𝛼 ◦ 𝛾Γ = 𝜑¯ ◦ 𝜑| ˆ Γ , there is a unique homomorphism 𝛾ˆ Γ : Γ → 𝐵ˆ such that 𝛽 ◦ 𝛾ˆ Γ = 𝛾Γ and 𝛼ˆ ◦ 𝛾ˆ Γ = 𝜑| ˆ Γ . Let (5.8) Bˆ = Con({ 𝛾ˆ Γ (Γ) | Γ ∈ G}). ˆ is a proper G-embedding problem for 𝐺 such that 𝛽( B) ˆ ⊆ B. Then, ( 𝜑, ˆ 𝛼, ˆ B) Part B: We may choose 𝛾Γ so that (5.9) Γ ∩ 𝐾ˆ ≤ Ker(𝛾Γ ) for every Γ ∈ G. Indeed, let Γ ∈ G. Since 𝐵 and 𝐴 are finite, Ker(𝜑) is open in 𝐺 and Ker(𝛾Γ ) is open in Γ. Hence, there exists a 𝐾Γ ∈ OpenNormal(𝐺) such that 𝐾Γ ≤ Ker(𝜑) and Γ ∩ 𝐾Γ ≤ Ker(𝛾Γ ). It follows that 𝛾Γ extends to a homomorphism 𝛾Γ0 : Γ𝐾Γ → 𝐵 that satisfies 𝛾Γ0 (𝑔𝑘) = 𝛾Γ (𝑔) for all 𝑔 ∈ Γ and 𝑘 ∈ 𝐾Γ . Indeed, if 𝑔1 , 𝑔2 ∈ Γ and 𝑘 1 , 𝑘 2 ∈ 𝐾Γ satisfy 𝑔1 𝑘 1 = 𝑔2 𝑘 2 , then 𝑔2−1 𝑔1 = 𝑘 2 𝑘 1−1 ∈ Γ ∩ 𝐾Γ ≤ Ker(𝛾Γ ), so 𝛾Γ (𝑔2−1 𝑔1 ) = 1, hence 𝛾Γ (𝑔1 ) = 𝛾Γ (𝑔2 ). Therefore, 𝛾Γ0 is a well defined homomorphism. Moreover, 𝛼 ◦ 𝛾Γ0 = 𝜑|Γ𝐾Γ , 𝐾Γ ≤ Ker(𝛾Γ0 ), and 𝛾Γ0 (Γ𝐾Γ ) = 𝛾Γ (Γ) ∈ B.

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5 Relative Embedding Problems

Each of the subsets Subgr(Γ𝐾Γ ) is étale open in Subgr(𝐺) and their union covers G. Since G is étale compact, there exist Γ1 , . . . , Γ𝑚 ∈ G such that Ð𝑚 G ⊆ 𝑖=1 Subgr(Γ Ñ𝑚 𝑖 𝐾Γ𝑖 ). Set 𝐾0 = 𝑖=1 𝐾Γ𝑖 and assume that 𝐾ˆ ≤ 𝐾0 . For every Γ ∈ G we may choose an 𝑖 such that Γ ≤ Γ𝑖 𝐾Γ𝑖 and replace 𝛾Γ by the restriction of 𝛾Γ0 𝑖 to Γ. Then, 𝛼 ◦ 𝛾Γ = 𝜑|Γ and 𝛾Γ (Γ) ≤ 𝛾Γ0 𝑖 (Γ𝑖 𝐾Γ𝑖 ) ∈ B, whence 𝛾Γ (Γ) ∈ B. Since 𝐾ˆ ≤ 𝐾Γ𝑖 ≤ Ker(𝛾Γ0 𝑖 ), we have Γ ∩ 𝐾ˆ ≤ Γ ∩ Ker(𝛾Γ0 𝑖 ) = Ker(𝛾Γ ). ˆ Part C: If (5.9) holds, then 𝛼ˆ is B-rigid. Indeed, consider 𝐵ˆ 0 ∈ Bˆ and 𝑏 ∈ 𝐵ˆ 0 such that 𝛼(𝑏) ˆ = 1. By (5.8), there are Γ ∈ G, 𝑔 ∈ Γ, and 𝑏ˆ ∈ 𝐵ˆ such that ˆ ˆ ˆ 𝑏 𝛼( ˆ 𝑏) 𝑏 = 𝛾ˆ Γ (𝑔). Then, 𝛼( ˆ 𝛾ˆ Γ (𝑔)) = 𝛼(𝑏 ˆ 𝑏 ) = 𝛼(𝑏) ˆ = 1. In addition, by (5.9), ˆ 𝑔 ∈ Γ ∩ 𝐾 ≤ Ker(𝛾Γ ). Hence, 𝛽( 𝛾ˆ Γ (𝑔)) = 𝛾Γ (𝑔) = 1. Therefore, since the square in (5.7) is cartesian, 𝛾ˆ Γ (𝑔) = 1 [FrJ08, p. 500, Lemma 22.2.3(a)]. It follows that ˆ −1 ˆ 𝑏 = 𝛾ˆ Γ (𝑔) 𝑏 = 1. This means that 𝛼ˆ is B-rigid, as claimed. Part D: If 𝐺 is G-projective, we may choose 𝐾0 so that 𝛼ˆ splits. Indeed, in this case there exists a homomorphism 𝛾𝐺 : 𝐺 → 𝐵 such that 𝛼 ◦ 𝛾𝐺 = 𝜑. Hence, we may apply Parts A, B, and C to {𝐺} and Subgr(𝐵) rather than to G and B. It follows ˆ from the so modified Part C that 𝛼ˆ maps 𝛾ˆ 𝐺 (𝐺) isomorphically onto 𝜑(𝐺) ˆ = 𝐴, ˆ where 𝛾ˆ 𝐺 : 𝐺 → 𝐵 is the unique homomorphism with 𝛽 ◦ 𝛾ˆ 𝐺 = 𝛾𝐺 and 𝛼ˆ ◦ 𝛾ˆ 𝐺 = 𝜑. ˆ Therefore, 𝛼ˆ splits. Part E: Proof of (b), (c), and (d). If 𝛾ˆ : 𝐺 → 𝐵ˆ is a homomorphism such that 𝛼ˆ ◦ 𝛾ˆ = 𝜑, ˆ then 𝛼 ◦ 𝛾 = 𝜑, so 𝛾 is a solution of (5.1). Under the assumptions of (c), 𝛾(G) ˆ ⊆ Bˆ (by (5.4)). By the end of Part A, ˆ ⊆ B. Hence, 𝛾(G) = 𝛽( 𝛾(G)) 𝛽( B) ˆ ⊆ B. This means that 𝛾 is strong, as (c) claims. If 𝜑 is surjective, then so is 𝜑. ¯ Hence, by [FrJ08, p. 500, Lemma 22.2.3(b)], 𝛽 is surjective. Therefore, 𝛾 = 𝛽 ◦ 𝛾ˆ is surjective. In other words, 𝛾 is proper (by (5.4)), as claimed by (d).  In view of Definition 5.1.1, the following corollary is an immediate consequence of Lemma 5.1.4. Corollary 5.1.5 Let 𝐺 be a profinite group and let G be an étale compact subset of Subgr(𝐺). Then, the group 𝐺 is (properly) [strongly] G-projective if and only if every finite proper rigid G-embedding problem (5.1) for 𝐺 has a (proper) [strong] solution. If 𝐺 is G-projective, then 𝐺 is (properly) [strongly] G-projective if and only if every finite proper split G-embedding problem (5.1) for 𝐺 has a (proper) [strong] solution. Remark 5.1.6 We compare the terminology of [Pop95] and [HJP05]. In [Pop95, p. 128, Sec. 2], 𝐺 is “strongly G-projective” if every finite proper G-embedding problem has a strong solution. In view of Corollary 5.1.5, this is equivalent to our definition of strong G-projectivity. In [HJP05, p. 1966, Sec. 5], our G-embedding problem is called a “G-embedding problem with local data” and is assumed to be rigid. Then, [HJP05] calls a group 𝐺 “strongly G-projective” if every such finite problem has a solution which, in

5.2 Strongly G-Projective Groups

69

the terminology of this monograph, is strong. Again, by Corollary 5.1.5, this is equivalent to 𝐺 being strongly G-projective in our terms.

5.2 Strongly G-Projective Groups We give a few examples of strongly G-projective groups that will be needed later. Lemma Î𝑛 5.2.1 For 𝑖 = 1, Ð. . . , 𝑛 let 𝐺 𝑖 be a strongly G𝑖 -projective profinite group. Let 𝐺 = ∗ 𝑖=1 𝐺 𝑖 and G = 𝑖 G𝑖 . Then, 𝐺 is strongly G-projective. Proof By Remark 1.2.1(b) the inclusions Subgr(𝐺 𝑖 ) → Subgr(𝐺) are étale continuous, so by Fact 1.1.3(d) the images of G𝑖 in Subgr(𝐺) are also étale compact; hence so is their union G. Consider a finite G-embedding problem (5.1) for 𝐺. For each 𝑖 we have 𝜑(G𝑖 ) ⊆ 𝜑(G) ⊆ 𝛼(B), so there exists a homomorphism 𝛾𝑖 : 𝐺 𝑖 → 𝐵 such that 𝛼 ◦ 𝛾𝑖 = 𝜑|𝐺𝑖 and 𝛾𝑖 (G𝑖 ) ⊆ B. Since 𝐺 is a free product of the 𝐺 𝑖 ’s, there exists a unique homomorphism 𝛾 : 𝐺 → 𝐵 such that 𝛾|𝐺𝑖 = 𝛾𝑖 for each 𝑖. Thus, 𝛼 ◦ 𝛾 = 𝜑 and 𝛾(G) ⊆ B, as desired.  The next result generalizes Lemma 5.2.1 to inner free products over profinite spaces. Î Proposition 5.2.2 Let 𝐺 = ∗ 𝑡 ∈𝑇 Γ𝑡 be a free product of profinite groups over a profinite space 𝑇 and set G = {Γ𝑡 }𝑡 ∈𝑇 . Then, 𝐺 is strongly G-projective. Proof By Lemma 2.1.2, G is étale compact. Let (5.1) be a finite rigid G-embedding problem for 𝐺. In particular, 𝐴 is a finite group. For every 𝑡 ∈ 𝑇 let 𝐴𝑡 = 𝜑(Γ𝑡 ) ≤ 𝐴. By Remark 1.2.1(a), Ð𝑛 the map 𝑡 ↦→ 𝐴𝑡 is étale continuous. Hence, Lemma 2.1.3 gives a partition 𝑇 = · 𝑖=1 𝑇𝑖 into disjoint open-closed subsets and groups 𝐴1 , . . . , 𝐴𝑛 ∈ 𝜑(G) such that 𝜑(Γ𝑡 ) ≤ 𝐴𝑖 for every Ð 𝑡 ∈ 𝑇𝑖 and each 1 ≤ 𝑖 ≤ 𝑛. Put 𝑈 = 𝑈 (G) and let 𝑈𝑖 = 𝑡 ∈𝑇𝑖 Γ𝑡 for each 𝑖. By Definition 4.1.1, the map 𝑇 → Subgr(𝐺) given by 𝑡 ↦→ Γ𝑡 is étale continuous. Hence, by Fact 1.1.3(d), each of the sets {Γ𝑡 }𝑡 ∈𝑇𝑖 is étale compact. By Lemma 1.2.5, 𝑈 and each of the sets 𝑈𝑖 are closed in 𝐺. Let 𝐾 = Ker(𝜑). By (5.2d), 𝜑(G) ⊆ 𝛼(B) and by (5.3c), 𝛼 is injective on each 𝐵0 ∈ B. Hence, for each 𝑖 there is a 𝐵𝑖 ∈ B such that the restriction of 𝛼 to 𝐵𝑖 is an isomorphism 𝛼𝑖 : 𝐵𝑖 → 𝐴𝑖 . Define 𝛾 : 𝑈 → 𝐵 by 𝛾|Γ𝑡 = 𝛼𝑖−1 ◦ 𝜑|Γ𝑡 if 𝑡 ∈ 𝑇𝑖 . Then, 𝛾 is continuous on each 𝑈𝑖 and also on 𝐾 ∩ 𝑈, because 𝜑(𝐾) = 1. Since each 𝑈𝑖 is closed and 𝐾 is open-closed in 𝐺, 𝐾 ∩ 𝑈, 𝑈1 r 𝐾, . . . , 𝑈𝑛 r 𝐾 are closed subsets of 𝐺, and 𝑈 is their union. By Proposition 4.8.3(a), this union is disjoint. Hence, 𝛾 is continuous. Condition (4.1) yields a unique extension of 𝛾 : 𝑈 → 𝐵 to a homomorphism 𝛾 : 𝐺 → 𝐵. By definition, 𝛾(G) ⊆ B. Both 𝛼 ◦ 𝛾 and 𝜑 coincide on 𝑈, so, by the uniqueness, they are equal. Thus, 𝛾 solves (5.1). By Corollary 5.1.5, 𝐺 is strongly G-projective. 

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5 Relative Embedding Problems

Corollary 5.2.3 Let 𝐹 be a free profinite group of rank ≥ ℵ0 and, as in ProposiÎ tion 5.2.2, let 𝐺 = ∗ 𝑡 ∈𝑇 Γ𝑡 be the free product of profinite groups over a profinite space 𝑇. Set G = {Γ𝑡 }𝑡 ∈𝑇 . Then, 𝐹 ∗ 𝐺 is properly strongly G-projective. Proof By Corollary 5.1.5, we have to prove that every finite proper rigid G-embedding problem (𝜑 : 𝐹 ∗ 𝐺 → 𝐴, 𝛼 : 𝐵 → 𝐴, B) has a proper strong solution. Proposition 5.2.2 yields a homomorphism 𝛾𝐺 : 𝐺 → 𝐵 that satisfies 𝛼 ◦ 𝛾𝐺 = 𝜑|𝐺 and 𝛾𝐺 (G) ⊆ B. By [FrJ08, p. 616, Prop. 25.6.2], there is an epimorphism 𝛾𝐹 : 𝐹 → 𝛼−1 (𝜑(𝐹)) ≤ 𝐵 that satisfies 𝛼 ◦ 𝛾𝐹 = 𝜑|𝐹 . In particular, Ker(𝛼) ⊆ 𝛾𝐹 (𝐹).

(5.10)

Then, the unique homomorphism 𝛾 : 𝐹 ∗ 𝐺 → 𝐵 that extends both 𝛾𝐺 and 𝛾𝐹 satisfies 𝛾(G) ⊆ B. Its image 𝐵 0 satisfies 𝛼(𝐵 0) = 𝐴 (because 𝜑 is surjective) and Ker(𝛼) ≤ 𝐵 0 (by (5.10)). Hence, 𝐵 0 = 𝐵. Therefore, 𝛾 is an epimorphism, as needed.  Lemma 5.2.4 Let G be an étale compact subset of Subgr(𝐺). Then, 𝐺 is (properly) [strongly] G-projective if and only if it is (properly) [strongly] G 𝐺 -projective. Proof Assume that 𝐺 is (properly) [strongly] G 𝐺 -projective. Let (5.1) be a finite proper G-embedding problem for 𝐺. Then, (5.1) is a G 𝐺 -embedding problem for 𝐺 as well. Indeed, let Γ ∈ G and 𝜎 ∈ 𝐺. There is a homomorphism 𝛾Γ : Γ → 𝐵 such that 𝛼 ◦ 𝛾Γ = 𝜑|Γ and 𝛾Γ (Γ) ∈ B and there is a 𝑏 ∈ 𝐵 such that 𝛼(𝑏) = 𝑎 := 𝜑(𝜎). Let [𝜎 −1 ], [𝑏], [𝑎] be the conjugations by 𝜎 −1 , 𝑏, 𝑎 in 𝐺, 𝐵, 𝐴, respectively. Then, 𝛼 ◦ [𝑏] = [𝑎] ◦ 𝛼 and 𝜑 ◦ [𝜎 −1 ] = [𝑎] −1 ◦ 𝜑. Set 𝛿 = [𝑏] ◦ 𝛾Γ ◦ [𝜎 −1 ] |Γ 𝜎 . Then, 𝛿 : Γ 𝜎 → 𝐵 is a homomorphism such that 𝛿(Γ 𝜎 ) = [𝑏] (𝛾Γ (Γ)) ∈ B and 𝛼 ◦ 𝛿 = 𝛼 ◦ [𝑏] ◦ 𝛾Γ ◦ [𝜎 −1 ] |Γ 𝜎 = [𝑎] ◦ 𝛼 ◦ 𝛾Γ ◦ [𝜎 −1 ] |Γ 𝜎 = [𝑎] ◦ 𝜑|Γ ◦ [𝜎 −1 ] |Γ 𝜎 = [𝑎] ◦ [𝑎] −1 ◦ 𝜑|Γ 𝜎 = 𝜑|Γ 𝜎 . G𝐺 ,

Since G ⊆ a (proper) [strong] solution of (5.1) as a G 𝐺 -embedding problem also (properly) [strongly] solves (5.1) as a G-embedding problem. Conversely, assume that 𝐺 is (properly) [strongly] G-projective. Let (5.1) be a finite proper G 𝐺 -embedding problem for 𝐺. Since G ⊆ G 𝐺 , (5.1) is a G-embedding problem for 𝐺 as well. Its (proper) [strong] solution 𝛾 : 𝐺 → 𝐵 also (properly) [strongly] solves (5.1) as a G 𝐺 -embedding problem. Indeed, if 𝛾(G) ⊆ B, then 𝛾(G 𝐺 ) ⊆ B 𝐵 = B.  Remark 5.2.5 (a) Every profinite group 𝐺 is strongly {𝐺}-projective. (b) We observe that both statements “𝐺 is strongly ∅-projective” and “𝐺 is ∅-projective” are equivalent to “𝐺 is projective”, i.e., for every pair of epimorphisms of profinite groups (𝜑 : 𝐺 → 𝐴, 𝛼 : 𝐵 → 𝐴) with 𝐵 finite there exists a homomorphism 𝛾 : 𝐺 → 𝐵 such that 𝛼 ◦ 𝛾 = 𝜑.

5.3 Infinite Embedding Problems

71

5.3 Infinite Embedding Problems A classical result of Karl Gruenberg [FrJ08, p. 503, Lemma 22.3.2] says that if 𝐺 is projective, then 𝛾 as in Remark 5.2.5(b) exists even if 𝐵 and 𝐴 are arbitrary profinite groups. There are similar results for G-projective groups [Har87, p. 289, Lemma 7.4], if G is a separated subset of Subgr(𝐺). We are not aware of an analogue for arbitrary strongly G-projective groups. However, the following special case turns out to be useful. Lemma 5.3.1 ([HJP05, Lemma 5.3]) Let 𝐺 be strongly G-projective. Then, every rigid G-embedding problem (5.1), with 𝐴 finite and rank(𝐵) ≤ ℵ0 , is strongly solvable. Proof Choose a sequence Ker(𝛼) = 𝑁0 ≥ 𝑁1 ≥ 𝑁2 . . . in OpenNormal(𝐵) with Ñ𝑛 𝑁 𝑖=1 𝑖 = 1. For each 𝑖 ≥ 0 put 𝐵𝑖 = 𝐵/𝑁𝑖 , let 𝛽𝑖 : 𝐵 → 𝐵𝑖 and 𝛼𝑖 : 𝐵𝑖 → 𝐵𝑖−1 be the quotient maps, and let B𝑖 = 𝛽𝑖 (B). Then, B𝑖 is closed under conjugation and taking subgroups and 𝛼𝑖 (B𝑖 ) = B𝑖−1 for each 𝑖 ≥ 1. Identify 𝐴 with 𝐵0 and 𝛼 with 𝛽0 . Since (5.1) is rigid, 𝛼 is injective on each Γ ∈ B, so 𝛼𝑖 is injective on each Δ ∈ B𝑖 . Therefore, we may inductively construct a sequence of homomorphisms 𝛾𝑖 : 𝐺 → 𝐵𝑖 satisfying: 𝛾0 = 𝜑, 𝛾𝑖 (G) ⊆ B𝑖 , and 𝛼𝑖 ◦ 𝛾𝑖 = 𝛾𝑖−1 , 𝑖 = 1, 2, 3, . . . . The 𝛾𝑖 ’s define a homomorphism 𝛾 : 𝐺 → 𝐵 such that 𝛽𝑖 ◦ 𝛾 = 𝛾𝑖 for every 𝑖. By Condition (5.2b) of Definition 5.1.1, B = Con(B) is étale compact. Hence, by Corollary 1.4.3, B is strictly compact. By the third paragraph of Section 1.2, Subgr(𝐵) = lim Subgr(𝐵𝑖 ). Hence, by the first paragraph of this proof and [FrJ08, ←−− p. 3, Cor. 1.1.6], the inclusion map B → lim B𝑖 is surjective. Thus, B = lim B𝑖 . It ←−− ←−− follows that 𝛾(G) ⊆ B, so 𝛾 is a strong solution of (5.1).  Remark 5.3.2 Let 𝐺 be strongly G-projective and let 𝑁 ∈ OpenNormal(𝐺). Set 𝐴 = 𝐺/𝑁 and let 𝜑 : 𝐺 → 𝐴 be the quotient map. Let 𝐹 be the free profinite group on the set 𝐴, and let 𝐴1 , . . . , 𝐴𝑛 be the groups in the subset 𝜑(G) of Î𝑛Subgr( 𝐴). For 𝑖 = 1, . . . , 𝑛 let 𝐵𝑖 be an isomorphic copy of 𝐴𝑖 . We set 𝐵 = 𝐹∗ ∗ 𝑖=1 𝐵𝑖 and B = Con({𝐵1 , . . . , 𝐵𝑛 }), and choose an epimorphism 𝛼 : 𝐵 → 𝐴 that maps 𝐹 onto 𝐴 and each 𝐵𝑖 isomorphically onto its copy 𝐴𝑖 . Then, (𝜑, 𝛼, B) is an infinite rigid G-embedding problem for 𝐺 with rank(𝐵) < ℵ0 . We call (𝜑, 𝛼, B) the 𝑁-standard G-embedding problem for 𝐺. By Lemma 5.3.1, there is a homomorphism 𝛾 : 𝐺 → 𝐵 with 𝛼 ◦ 𝛾 = 𝜑 and 𝛾(G) ⊆ B. The following lemma deals with an infinite embedding problem and a weaker form of a solution. It is extracted from [Pop95, p. 132, Theorem 3]. Lemma 5.3.3 Let 𝐺 be a properly G-projective profinite group and let 𝐺 𝜑ˆ

𝛾 𝛼ˆ

𝐵ˆ

𝜓

𝛽

 𝐵

 / 𝐴ˆ

𝛼

  /𝐴

𝜑

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5 Relative Embedding Problems

be a commutative diagram of profinite groups such that 𝐵 is finite, 𝛼, 𝛽, 𝛼, ˆ 𝜓, 𝜑, ˆ 𝜑 are epimorphisms and 𝛾 has yet to be introduced. Let B and Bˆ be subfamilies of Subgr(𝐵) ˆ = B, ˆ respectively, and 𝛽( B) ˆ ⊆ B. ˆ such that Con(B) = B and Con( B) and Subgr( 𝐵) ˆ 𝛼ˆ : 𝐵ˆ → 𝐴, ˆ B) ˆ are Suppose that (𝜑 : 𝐺 → 𝐴, 𝛼 : 𝐵 → 𝐴, B) and ( 𝜑ˆ : 𝐺 → 𝐴, proper G-embedding problems for 𝐺. Then: (a) There exists a proper solution 𝛾 of (𝜑, 𝛼, B) and a strictly closed subset H of ˆ such that 𝛼(H Subgr( 𝐵) ˆ ) = StrictClosure( 𝜑(G)) ˆ and 𝛽(H ) = 𝛾(G). (b) Every Δ ∈ H is an inverse limit with surjective connecting maps of finite quotients of groups Γ𝐶 ∈ G such that all 𝜑(Γ ˆ 𝐶 ) are contained in a given strictly ˆ open neighborhood of 𝛼(Δ) ˆ ∈ Subgr( 𝐴). ˆ 𝛼ˆ : 𝐵ˆ → 𝐴, ˆ B) ˆ has a proper solution, say, Remark 5.3.4 Notice that if ( 𝜑ˆ : 𝐺 → 𝐴, ˆ then the assertions of the lemma follow with H = 𝛾(StrictClosure(G)) 𝛾ˆ : 𝐺 → 𝐵, ˆ and 𝛾 = 𝛽 ◦ 𝛾. ˆ Proof (Proof of the lemma) The proof breaks up into four parts. ˆ | 𝐶 ≤ Ker(𝛽)}. For every Part A: Reduction step. Let N = {𝐶 ∈ OpenNormal( 𝐵) ˆ 𝛼(𝐶), ˆ 𝐶 ∈ N let 𝐵𝐶 = 𝐵/𝐶 and 𝐴𝐶 = 𝐴/ ˆ let 𝛽𝐶 : 𝐵ˆ → 𝐵𝐶 and 𝜑𝐶 : 𝐴ˆ → 𝐴𝐶 be ˆ and let 𝛼𝐶 : 𝐵𝐶 → 𝐴𝐶 be the epimorphism the quotient maps, put B𝐶 = 𝛽𝐶 ( B), induced by 𝛼. ˆ Then, 𝜑𝐶 ◦ 𝛼ˆ = 𝛼𝐶 ◦ 𝛽𝐶 and (𝜑𝐶 ◦ 𝜑, ˆ 𝛼𝐶 , B𝐶 ) is a finite proper G-embedding problem for 𝐺. By assumption, both 𝛽 and 𝜑 are surjective. Hence, given a group 𝐶 ∈ N , there are epimorphisms 𝛽¯ : 𝐵𝐶 → 𝐵 and 𝜓¯ : 𝐴𝐶 → 𝐴 such that the diagram 𝐺 𝜑ˆ 𝛼ˆ

𝐵ˆ

𝛽

𝛼𝐶

𝐵𝐶   𝐵

𝜑

𝜑𝐶

𝛽𝐶



 / 𝐴ˆ

𝛽¯

 / 𝐴𝐶

𝜓

𝜓¯ 𝛼

 } /𝐴

commutes. Let 𝛾𝐶 be a proper solution of (𝜑𝐶 ◦ 𝜑, ˆ 𝛼𝐶 , B𝐶 ); that is, 𝛾𝐶 : 𝐺 → 𝐵𝐶 is an epimorphism such that 𝛼𝐶 ◦ 𝛾𝐶 = 𝜑𝐶 ◦ 𝜑ˆ (but not necessarily 𝛾𝐶 (G) ⊆ B𝐶 ). ˆ Then, 𝛾 = 𝛽¯ ◦ 𝛾𝐶 is a proper solution of (𝜑, 𝛼, B). Moreover, if H ⊆ Subgr( 𝐵) satisfies 𝛽𝐶 (H ) = 𝛾𝐶 (G), then 𝛽(H ) = 𝛾(G). Hence, we may replace (𝜑, 𝛼, B) with (𝜑𝐶 ◦ 𝜑, ˆ 𝛼𝐶 , B𝐶 ), for any 𝐶 ∈ N . Part B: An inverse system. Let 𝔅𝐶 be the collection of all families 𝛾𝐶 (G), where 𝛾𝐶 is a proper solution of (𝜑𝐶 ◦ 𝜑, ˆ 𝛼𝐶 , B𝐶 ). Since 𝐵𝐶 is finite, 𝔅𝐶 is a finite collection of subsets of Subgr(𝐵𝐶 ). Since 𝐺 is G-projective, 𝔅𝐶 ≠ ∅. If 𝐶 0 ∈ N contains 𝐶 ∈ N , then the corresponding quotient map 𝛽𝐶,𝐶 0 : 𝐵𝐶 → 𝐵𝐶 0 yields a map of 𝔅𝐶 into 𝔅𝐶 0 that we also denote by 𝛽𝐶,𝐶 0 . Thus, (𝔅𝐶 , 𝛽𝐶,𝐶 0 )𝐶,𝐶 0 is a

5.4 The Closed Subgroup Theorem

73

projective system of non-empty finite sets. Then, the following diagram, ignoring 𝛾𝐶 and 𝛾𝐶 0 , commutes. 𝐺 𝜑ˆ

𝛾𝐶 𝛾𝐶 0

𝐵ˆ

𝛼ˆ

 / 𝐴ˆ

𝛼𝐶

 / 𝐴𝐶

𝜑𝐶

𝛽𝐶

 𝐵𝐶

𝜑𝐶,𝐶 0

𝛽𝐶,𝐶 0

  𝐵𝐶 0

𝛼𝐶 0



/ 𝐴𝐶 0

By [FrJ08, p. 3, Cor. 1.1.4], the inverse limit of this system is non-empty. An element in this inverse limit is a family (𝛾𝐶 (G))𝐶 ∈N , where (5.11) 𝛾𝐶 : 𝐺 → 𝐵𝐶 is an epimorphism that satisfies 𝛼𝐶 ◦ 𝛾𝐶 = 𝜑𝐶 ◦ 𝜑ˆ (but not necessarily 𝛾𝐶 0 = 𝛽𝐶,𝐶 0 ◦ 𝛾𝐶 ) and 𝛽𝐶,𝐶 0 (𝛾𝐶 (G)) = 𝛾𝐶 0 (G) for all 𝐶 ≤ 𝐶 0 in N . Part C: Another inverse system. We fix one such family (𝛾𝐶 (G))𝐶 . Then, (𝛾𝐶 (G))𝐶 is itself a projective system of non-empty finite families of finite groups, with surjective connecting maps 𝛽𝐶,𝐶 0 . Its inverse limit is a strictly closed subfamily ˆ such that 𝛽𝐶 (H ) = 𝛾𝐶 (G) for every 𝐶 ∈ N . Then, H of Subgr( 𝐵) (5.11)

𝜑𝐶 ◦ 𝛼(H ˆ ) = 𝛼𝐶 ◦ 𝛽𝐶 (H ) = 𝛼𝐶 ◦ 𝛾𝐶 (G) = 𝜑𝐶 ◦ 𝜑(G) ˆ for every 𝐶 ∈ N , hence 𝛼(H ˆ ) = StrictClosure( 𝜑(G)), ˆ as claimed in (a). Part D: Proof of (b).

Finally, every Δ ∈ H is an inverse limit lim 𝛾𝐶 (Γ𝐶 ), ←−−

𝐶 ∈N

where Γ𝐶 ∈ G for every 𝐶 ∈ N , with surjective connecting maps 𝛽𝐶,𝐶 0 . Since

𝜑𝐶 𝛼(Δ) ˆ = 𝛼𝐶 (𝛽𝐶 (Δ)) = 𝛼𝐶 (𝛾𝐶 (Γ𝐶 )) = 𝜑𝐶 𝜑(Γ ˆ 𝐶) for every 𝐶 ∈ N , there is a 𝐶 0 ∈ N such that if 𝐶 ≤ 𝐶 0, then 𝜑(Γ ˆ 𝐶 ) is contained in a given strictly open neighborhood of 𝛼(Δ). ˆ Replacing N by {𝐶 ∈ N | 𝐶 ≤ 𝐶 0 } we may assume that this holds for every 𝐶 ∈ N , as required by (b). 

5.4 The Closed Subgroup Theorem We follow a hint of Pop in [Pop99, paragraph 4 of page 10] for the proof of a closed subgroup theorem for strongly projective profinite groups, a variant of [Har87, Thm. 5.1].

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Lemma 5.4.1 Let 𝐺 be a profinite group and G an étale compact subset of Subgr(𝐺). Let 𝐻 be a closed subgroup of 𝐺 and let 𝑁 ∈ OpenNormal(𝐺). Then, there is an 𝐻 0 ∈ Open(𝐺) containing 𝐻 such that (5.12) for every Γ0 ∈ G there is a Γ ∈ G with Γ0 ∩ 𝐻 0 ≤ (Γ ∩ 𝐻)𝑁. Ñ Proof We set S = {𝐻 0 ∈ Open(𝐺) | 𝐻 ≤ 𝐻 0 }. Let Γ ∈ G. Since 𝐻 0 ∈S 𝐻 0 = 𝐻 Ñ (Lemma 1.1.14), we have 𝐻 0 ∈S Γ ∩ 𝐻 0 = Γ ∩ 𝐻, and since Γ ∩ 𝐻 is contained in (Γ ∩ 𝐻)𝑁 ∈ Open(𝐺), there is an 𝐻Γ ∈ S such that Γ ∩ 𝐻Γ ≤ (Γ ∩ 𝐻)𝑁 (Lemma 1.1.12(a)). Choose 𝑁Γ ∈ OpenNormal(𝐺) such that 𝑁Γ ≤ 𝐻Γ ∩ 𝑁. Then, Γ𝑁Γ ∩ 𝐻Γ = (Γ ∩ 𝐻Γ )𝑁Γ ≤ (Γ ∩ 𝐻)𝑁. Thus, if Γ0 ∈ Subgr(Γ𝑁Γ ), then Γ0 ∩ 𝐻Γ ≤ Γ𝑁Γ ∩Ð𝐻Γ ≤ (Γ ∩ 𝐻)𝑁. Since G ⊆ ) and G is étale compact, there Γ∈ G Subgr(Γ𝑁ΓÐ Ñ𝑛 are groups 𝑛 Γ1 , . . . , Γ𝑛 ∈ G such that G ⊆ 𝑖=1 Subgr(Γ𝑖 𝑁Γ𝑖 ). Let 𝐻 0 = 𝑖=1 𝐻Γ𝑖 . Then, 𝐻 0 ∈ S and for every Γ0 ∈ G there is an 𝑖 such that Γ0 ≤ Γ𝑖 𝑁Γ𝑖 , whence Γ0 ∩ 𝐻 0 ≤ Γ0 ∩ 𝐻Γ𝑖 ≤ (Γ𝑖 ∩ 𝐻)𝑁.  Proposition 5.4.2 Let 𝐺 be a strongly G-projective profinite group. Let 𝐻 be a closed subgroup of 𝐺 and set H = {Γ ∩ 𝐻 | Γ ∈ G 𝐺 }. Then, 𝐻 is strongly H -projective. Proof By Definition (5.5)(a), G is an étale compact subset of Subgr(𝐺), hence, by Corollary 1.4.3, so is G 𝐺 . By Lemma 5.2.4, we may assume that G = G 𝐺 . By Remark 1.2.1(c), H is étale compact. We have to solve a finite H -embedding problem for 𝐻 (5.13)

(𝜓 : 𝐻 → 𝐶, 𝛽 : 𝐷 → 𝐶, D).

The rest of the proof breaks up into several parts. We first assume that 𝐻 is open in 𝐺; only in the last part do we consider the general case. Part A: An infinite rigid G-embedding problem for 𝐺. Since Ker(𝜓) is open in 𝐻 and 𝐻 is open in 𝐺, Ker(𝜓) is open in 𝐺. Hence, there is an 𝑁 ∈ OpenNormal(𝐺) such that 𝑁 ≤ Ker(𝜓). Therefore, by Lemma 5.1.4 we may replace (5.13) by another finite H -embedding problem to assume that (5.13) is rigid, 𝜓 is surjective, and 𝑁 = Ker(𝜓) ∈ OpenNormal(𝐺). In particular, 𝑁 ≤ 𝐻. Let (𝜑 : 𝐺 → 𝐴, 𝛼 : 𝐵 → 𝐴, B) be the 𝑁-standard G-embedding problem for 𝐺 and let 𝛾 : 𝐺 → 𝐵 be its solution (Remark 5.3.2 with its notation: 𝐴 = 𝐺/𝑁, 𝜑(G) = { 𝐴1 , . . . , 𝐴𝑛 }, and B = Con(𝐵1 , . . . , 𝐵𝑛 )). Then, we may assume that 𝐶 ≤ 𝐴 and 𝜓 = 𝜑| 𝐻 . Part B: An infinite rigid H -embedding problem for 𝐻. Let 𝐷 0 = 𝛼−1 (𝐶). Since 𝛼(𝛾(𝐻)) = 𝜑(𝐻) ≤ 𝐶, we have 𝛾(𝐻) ≤ 𝛼−1 (𝐶) = 𝐷 0. Let 𝛿 : 𝐻 → 𝐷 0 be the restriction of 𝛾 to 𝐻 and 𝛽 0 : 𝐷 0 → 𝐶 the restriction of 𝛼 to 𝐷 0. This gives the following commutative diagram (where 𝜋 is yet to be introduced):

5.4 The Closed Subgroup Theorem

75

𝐺` /O 𝐻

𝛾 𝜑

𝐵`

y

 /𝐴 _

𝛼

𝜓

𝛿

0P y 𝐷0

/O /7 𝐶

𝛽0

𝜋

𝛽

𝐷 Let D 0 = {𝐵0 ∩ 𝐷 0 | 𝐵0 ∈ B} ⊆ Subgr(𝐷 0). Then, Con(D 0) = D 0, because Con(B) = B, Also, 𝛿(H ) ⊆ D 0. Indeed, let 𝐻0 ∈ H . By the definition of H , there is a 𝐺 0 ∈ G such that 𝐻0 = 𝐺 0 ∩ 𝐻. Since 𝛾(G) ⊆ B (Remark 5.3.2), there is a 𝐵0 ∈ B such that 𝛾(𝐺 0 ) = 𝐵0 . Hence, 𝛿(𝐻0 ) = 𝛾(𝐺 0 ∩ 𝐻) ≤ 𝛾(𝐺 0 ) ∩ 𝛾(𝐻) ≤ 𝐵0 ∩ 𝐷 0 ∈ D 0, whence 𝛿(𝐻0 ) ∈ D 0. Since 𝛼 : 𝐵 → 𝐴 is injective on every 𝐵0 ∈ B (Remark 5.3.2), the homomorphism 𝛽 0 : 𝐷 0 → 𝐶 is injective on every 𝐷 00 ∈ D 0. Since 𝛼 ◦ 𝛾 = 𝜑, we have 𝛽 0 ◦ 𝛿 = 𝜓. Thus, (𝜓, 𝛽 0, D 0) is a rigid H -embedding problem for 𝐻 and 𝛿 is a strong solution of this problem. Part C: (𝛽 0, 𝛽, D) is a finite
rigid D 0-embedding problem for 𝐷 0. We first claim that 𝛽 0 (D 0) ⊆ Con 𝜓(H ) . Indeed, by Part B, every 𝐷 00 ∈ D 0 is a subgroup of 𝐵𝑖𝑏 ∩ 𝐷 0 for some 1 ≤ 𝑖 ≤ 𝑛 and 𝑏 ∈ 𝐵. Since { 𝐴1 , . . . , 𝐴𝑛 } = 𝜑(G) is closed under conjugation in 𝐴 (by the assumption G = G 𝐺 made in the first paragraph of the proof), there is a 𝐺 0 ∈ G such that 𝜑(𝐺 0 ) = 𝐴𝑖𝛼(𝑏) . By Part A, the relation Ker(𝜑) = 𝑁 ≤ 𝐻 implies the equality 𝜑(𝐺 0 ) ∩ 𝜑(𝐻) = 𝜑(𝐺 0 ∩ 𝐻). Therefore, 𝛽 0 (𝐷 00 ) ≤ 𝛽 0 (𝐵𝑖𝑏 ∩ 𝐷 0) ≤ 𝛼(𝐵𝑖 ) 𝛼(𝑏) ∩ 𝛼(𝐷 0) ≤ 𝐴𝑖𝛼(𝑏) ∩ 𝐶 = 𝜑(𝐺 0 ) ∩ 𝜑(𝐻) = 𝜑(𝐺 0 ∩ 𝐻) = 𝜓(𝐺 0 ∩ 𝐻) ∈ 𝜓(H ). Since (5.13) is a rigid H -embedding problem, we have that 𝜓(H ) ⊆ 𝛽(D) and D = Con(D). Hence, by the preceding paragraph, 𝛽 0 (D 0) ⊆ Con(𝜓(H )) ⊆ 𝛽(D). Thus, (𝛽 0, 𝛽, D) is a finite rigid D 0-embedding problem for 𝐷 0. Part D: Application of the Kurosh subgroup theorem and a solution. Subgroup Theorem (Proposition 3.1.7), 𝑚 Ö 𝐷 0 = 𝐸 0 ∗ 𝐸 1 ∗ 𝐸 2 ∗ · · · ∗ 𝐸𝑟 ∗ ∗ 𝐷 0𝑗 ,

By the Kurosh

𝑗=1

where 𝐸 0 is a finitely generated free profinite group and, for 𝑖 = 1, . . . , 𝑟, 0 ∈ D 0 are representatives of 𝐸 𝑖 = 𝐹 𝑏𝑖 ∩ 𝐷 0 with some 𝑏 𝑖 ∈ 𝐵, and 𝐷 10 , . . . , 𝐷 𝑚 0 the conjugacy classes in D (Part B). Since 𝐹 is a finitely generated free profinite group, so is 𝐹 𝑏𝑖 . Since 𝐷 0 is open in 𝐵, 𝐸 𝑖 is open in 𝐹 𝑏𝑖 . Hence, by [FrJ08, p. 358, Prop. 17.6.2], 𝐸 𝑖 is a finitely generated free profinite group. It follows

76

5 Relative Embedding Problems

that 𝐸 := 𝐸 0 ∗ 𝐸 1 ∗ · · · ∗ 𝐸𝑟 is also a finitely generated free profinite group and Î 0 𝐷 0 = 𝐸∗ ∗ 𝑚 𝑗=1 𝐷 𝑗 . Recall that 𝐸 is projective [FrJ08, p. 507, Cor. 22.4.5]. Thus, by Remark 5.2.5, 𝐸 is strongly ∅-projective and 𝐷 0𝑗 is strongly {𝐷 0𝑗 }-projective for every 𝑗. Hence, by 0 }-projective. Whence, by Lemma 5.2.4, Lemma 5.2.1, 𝐷 0 is strongly {𝐷 10 , . . . , 𝐷 𝑚 𝐷 0 is strongly D 0-projective. Therefore, there is a homomorphism 𝜋 : 𝐷 0 → 𝐷 such that 𝛽 ◦ 𝜋 = 𝛽 0 and 𝜋(D 0) ⊆ D. Since, by Part B, 𝛿(H ) ⊆ D 0, we have that 𝜋 ◦ 𝛿 strongly solves (5.13). Part E: The general case. We now drop the assumption that 𝐻 is open. Since Ker(𝜓) is open in 𝐻, there exists an open normal subgroup 𝑁 of 𝐺 such that 𝑁 ∩ 𝐻 ≤ Ker(𝜓). Then, the homomorphism 𝜓0 : 𝑁 𝐻 → 𝐶 defined by 𝜓0 (𝑘 ℎ) = 𝜓(ℎ) for 𝑘 ∈ 𝑁 and ℎ ∈ 𝐻 extends 𝜓. By Lemma 5.4.1, there is an 𝐻 0 ∈ Open(𝐺) containing 𝐻 such that (5.12) holds. Without loss of generality 𝐻 0 ≤ 𝑁 𝐻. Let 𝜓 0 : 𝐻 0 → 𝐶 be the restriction of 𝜓0 . Then, 𝜓 0 extends 𝜓, hence is surjective, and 𝑁 ∩ 𝐻 0 ≤ Ker(𝜓 0). Moreover, 𝜓 0 maps H 0 := {Γ ∩ 𝐻 0 | Γ ∈ G} into 𝛽(D). Indeed, consider Γ0 ∈ G. Then, by (5.12), there is a Γ ∈ G with 𝜓 0 (Γ0 ∩ 𝐻 0) ≤ 𝜓(Γ ∩ 𝐻) ∈ 𝛽(D), and since D is closed under taking subgroups, 𝜓 0 (Γ0 ∩ 𝐻 0) ∈ 𝛽(D). Thus, (𝜓 0, 𝛽, D) is a finite H 0-embedding problem for 𝐻 0. By Part D, there exists a homomorphism 𝛾 0 : 𝐻 0 → 𝐷 such that 𝛽 ◦ 𝛾 0 = 𝜓 0 and 𝛾 0 (H 0) ⊆ D. Then, 𝛾 𝐻 = 𝛾 0 | 𝐻 is a homomorphism of 𝐻 into 𝐷 such that 𝛽 ◦ 𝛾 𝐻 = 𝜓 and 𝛾 𝐻 (H ) ⊆ D, as desired.  Proposition 5.4.2 yields the following generalization of Proposition 3.2.7 to strongly G-projective groups. Proposition 5.4.3 Let 𝐺 be a strongly G-projective profinite group and 𝐻 a finite subgroup of 𝐺. Suppose that G is non-empty and 𝐺-invariant. Then, there is a Γ ∈ G such that 𝐻 ≤ Γ. Proof Let H = {𝐻 ∩ Γ | Γ ∈ G}. Then, H is 𝐻-invariant. By Proposition 5.4.2, 𝐻 is strongly H -projective. Notice that 1 ∈ OpenNormal(𝐻). Consider the 1-standard H -embedding problem (𝜑 : 𝐻 → 𝐴, 𝛼 : 𝐵 → 𝐴, B) for 𝐻 (Remark 5.3.2 with Î𝑛 the notation: 𝜑 is an isomorphism, 𝐵 = 𝐹∗ ∗ 𝑖=1 𝐵𝑖 , and B = Con(𝐵1 , . . . , 𝐵𝑛 )). Since 𝐻 is strongly H -projective, the H -embedding problem has a strong solution 𝛾 : 𝐻 → 𝐵. If 𝛾(𝐻) = 1, then 𝜑(𝐻) = 𝛼(𝛾(𝐻)) = 1. Since 𝜑 is an isomorphism, 𝐻 = 1, and the assertion is trivial. So assume that 𝛾(𝐻) ≠ 1. Since 𝐻 is finite, so is 𝛾(𝐻). Hence, by Proposition 3.2.7, 𝛾(𝐻) is contained in a 𝐵-conjugate of one of the groups 𝐹, 𝐵1 , . . . , 𝐵𝑛 . This group cannot be 𝐹, because 𝐹 is projective [FrJ08, p. 507, Cor. 22.4.6], and as such torsion free [FrJ08, p. 507, Prop. 22.4.7]. Hence, there is an 𝑖 such that 𝐻 is contained in a 𝐵-conjugate of 𝐵𝑖 . It follows that there is a Γ ∈ G such that 𝜑(𝐻) = 𝛼(𝛾(𝐻)) is contained in 𝜑(𝐻 ∩ Γ), an 𝐴-conjugate of 𝛼(𝐵𝑖 ). But 𝜑 is an isomorphism, so 𝐻 is contained in 𝐻 ∩ Γ, that is, 𝐻 ≤ Γ.  For the sake of completeness we mention a predecessor of Proposition 5.4.2 that uses the notion of a “separated” (Definition 2.1.5) instead of an “étale compact” set

5.5 The Set Gmax

77

of subgroups. Note that “𝐺 is projective with respect to a family G of its subgroups” in [Har87] means “𝐺 is separated and strongly G-projective” in our sense. See also Lemma 2.1.12. Proposition 5.4.4 (The closed subgroup theorem [Har87, Thm. 5.1]) Let 𝐺 be a profinite group and G a 𝐺-invariant separated subset of Subgr(𝐺). Suppose that 𝐺 is strongly G-projective. Let 𝐻 be a closed subgroup of 𝐺 and set H = {Γ∩𝐻 | Γ ∈ G}. Then, 𝐻 is strongly H -projective.

5.5 The Set Gmax Let 𝐺 be a profinite group and G a subset of Subgr(𝐺). Recall that Gmax is the set of all maximal elements of G and that (Lemma 1.3.5) (5.14) if G is étale compact, then every Γ ∈ G is contained in some Γ0 ∈ Gmax . Lemma 5.5.1 Let 𝐺 be a profinite group and G an étale compact subset of Subgr(𝐺). Then, 𝐺 is (strongly) G-projective if and only if 𝐺 is (strongly) Gmax -projective. Proof By Lemma 1.3.6(a), Gmax is also étale compact. We consider a finite Gmax -embedding problem (Definition 5.1.1) (5.15)

(𝜑 : 𝐺 → 𝐴, 𝛼 : 𝐵 → 𝐴, B)

for 𝐺. Since every Γ0 ∈ G is contained in some Γ ∈ Gmax , (5.15) is also a G-embedding problem for 𝐺. A (strong) solution 𝛾 : 𝐺 → 𝐵 of the latter is also a (strong) solution of (5.15) as a Gmax -embedding problem, because Gmax ⊆ G. Conversely, suppose (5.15) is a finite G-embedding problem for 𝐺. Then, it is also a Gmax -embedding problem, because Gmax ⊆ G. A (strong) solution 𝛾 : 𝐺 → 𝐵 of the latter is also a (strong) solution of (5.15) as a G-embedding problem, because every Γ0 ∈ G is contained in some Γ ∈ Gmax .  Example 5.5.2 Let 𝐴 be a finite abelian group and let 𝐴1 , 𝐴2 ≤ 𝐴 be two distinct proper subgroups of 𝐴 such that 𝐴1 ∩ 𝐴2 ≠ 1. Let Zˆ = h𝜏i and let 𝐺 = 𝐴∗ h𝜏i. Observe that 𝐴 is strongly { 𝐴}-projective and Zˆ is strongly ∅-projective. Hence, by Lemma 5.2.1, 𝐺 is strongly { 𝐴}-projective. Therefore, by Lemma 5.5.1, 𝐺 is strongly { 𝐴, 𝐴1 , 𝐴2 }-projective. Since { 𝐴, 𝐴1 , 𝐴2 }𝐺 = { 𝐴, 𝐴1𝜏 , 𝐴2𝜏 }𝐺 , by Lemma 5.2.4, 𝐺 is strongly { 𝐴, 𝐴1𝜏 , 𝐴2𝜏 }-projective. Let G = { 𝐴, 𝐴1𝜏 , 𝐴2𝜏 }. Then, Gmax = G. Nevertheless, (5.16a) 𝐴1𝜏 ∩ 𝐴2𝜏 ≠ 1, because 𝐴1 ∩ 𝐴2 ≠ 1. (5.16b) 𝑁𝐺 ( 𝐴1𝜏 ) = 𝐴 𝜏 ≠ 𝐴1𝜏 . (5.16c) It is not true that (Gmax ) 𝐺 = (G 𝐺 )max , because 𝐴1𝜏 ∈ Gmax , but 𝐴1 is a proper subgroup of 𝐴 ∈ G. To avoid these problems we assume in the next result that G is 𝐺-invariant.

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5 Relative Embedding Problems

Proposition 5.5.3 ([HJP05, Prop. 5.5 and Prop. 5.8]) Let 𝐺 be a profinite group and G an étale compact 𝐺-invariant subset of Subgr(𝐺). Suppose that 𝐺 is strongly G-projective and let 𝑁0 ⊳ 𝐺 be open. Then: (a) Γ1 ∩ Γ2 = 1 for all distinct Γ1 , Γ2 ∈ Gmax . (b) 𝑁𝐺 (Γ) = Γ for every non-trivial Γ ∈ Gmax . (c) Gmax r Subgr(𝑁0 ) is étale profinite, so also separated. Proof By Lemma 1.3.6, Gmax is also étale compact and 𝐺-invariant. By Lemma 5.5.1, 𝐺 is strongly Gmax -projective. We may therefore replace G by Gmax to assume that G = Gmax . Then, we may assume that 1 ∉ G. Otherwise G = {1} and the assertions are trivial. Toward a proof of (a), (b), (c). Let Γ1 , Γ2 ∈ G be distinct, and let 𝑔 ∈ 𝐺 r Γ1 . Since G = Gmax , for every Δ ∈ G we have Γ1 ∪ Γ2 , Γ1 ∪ {𝑔} 6 ⊆ Δ. Hence, by Lemma 1.2.3, there is an open 𝑁 ⊳ 𝐺 such that (5.17)

Γ1 ∪ Γ2 , Γ1 ∪ {𝑔} 6 ⊆ Δ𝑁 for every Δ ∈ G.

Replacing 𝑁 by a smaller open normal subgroup of 𝐺, we may have, in addition to (5.17), that 𝑁 ≤ 𝑁0 and Γ1 6 ≤ 𝑁.

(5.18) Let

(𝜑 : 𝐺 → 𝐴, 𝛼 : 𝐵 → 𝐴, B)

(5.19)

be the 𝑁-standard G-embedding problem for 𝐺 – see Remark 5.3.2 with its notation: (5.20)

𝐴 = 𝐺/𝑁,

𝜑(G) = { 𝐴1 , . . . , 𝐴𝑛 },

and

B = Con{𝐵1 , . . . , 𝐵𝑛 }.

Set 𝐽 = {𝑖 | 𝐵𝑖 ≠ 1}. By Lemmas 3.1.11 and 3.1.10, Bmax = {𝐵 𝜏𝑗 | 𝑗 ∈ 𝐽, 𝜏 ∈ 𝐵} and (5.21)

for distinct 𝐵10 , 𝐵20 ∈ Bmax we have 𝐵10 ∩ 𝐵20 = 1 and 𝑁 𝐵 (𝐵10 ) = 𝐵10 .

By Lemma 5.3.1, embedding problem (5.19) has a strong solution 𝛾 : 𝐺 → 𝐵. In particular, 𝛾(Γ1 ), 𝛾(Γ2 ) ∈ B. Hence, (5.22)

there are 𝐵10 , 𝐵20 ∈ Bmax such that 𝛾(Γ1 ) ≤ 𝐵10 and 𝛾(Γ2 ) ≤ 𝐵20 .

Applying 𝛼 to the latter relations, we have 𝜑(Γ1 ) ≤ 𝛼(𝐵10 ) and 𝜑(Γ2 ) ≤ 𝛼(𝐵20 ). Since 𝐵10 is a conjugate of some 𝐵𝑖 ∈ Bmax and 𝜑(G) is closed under conjugation, (5.20) yields a Δ ∈ G with 𝜑(Δ) = 𝛼(𝐵10 ). Thus, we have (5.23)

𝛾(Γ1 ) ≤ 𝐵10 , 𝛾(Γ2 ) ≤ 𝐵20 , 𝜑(Γ1 ) ≤ 𝛼(𝐵10 ) = 𝜑(Δ), 𝜑(Γ2 ) ≤ 𝛼(𝐵20 ).

Claim A: 𝐵10 ≠ 𝐵20 . Otherwise, by (5.23), 𝛼(𝐵10 ) = 𝛼(𝐵20 ) = 𝜑(Δ), hence 𝜑(Γ1 ), 𝜑(Γ2 ) ≤ 𝜑(Δ). In particular, Γ1 , Γ2 ≤ 𝜑−1 (𝜑(Δ)) = Δ𝑁, a contradiction to (5.17). Claim B: 𝐵10 ≠ (𝐵10 ) 𝛾 (𝑔) . Otherwise, by (5.21), 𝛾(𝑔) ∈ 𝐵10 . Hence, applying 𝛼 and using (5.23), we get 𝜑(𝑔) ∈ 𝜑(Δ). In addition, by (5.23), 𝜑(Γ1 ) ≤ 𝜑(Δ). Therefore, Γ1 ∪ {𝑔} ⊆ 𝜑−1 (𝜑(Δ)) = Δ𝑁, a contradiction to (5.17). Proof of (a). It follows from Claim A and (5.21) that 𝐵10 ∩ 𝐵20 = 1, hence

5.5 The Set Gmax

79 (5.23)

𝛾(Γ1 ∩ Γ2 ) ⊆ 𝛾(Γ1 ) ∩ 𝛾(Γ2 ) ⊆ 𝐵10 ∩ 𝐵20 = 1. Apply 𝛼 to the latter relation to get that 𝜑(Γ1 ∩Γ2 ) = 1. Hence, Γ1 ∩Γ2 ≤ Ker(𝜑) = 𝑁. Since for every open normal subgroup 𝑁0 of 𝐺 there exists an open normal subgroup 𝑁 ≤ 𝑁0 with the latter relation, we have Γ1 ∩ Γ2 = 1. Proof of (b). Assume toward contradiction that there exists a 𝑔 ∈ 𝐺 r Γ such that Γ𝑔 = Γ. Apply the arguments of “Toward a proof of (a), (b), (c)” to Γ rather than to Γ1 . Then, by (5.23), (5.24)

𝛾(Γ) ≤ 𝐵10 and 𝛾(Γ) = 𝛾(Γ𝑔 ) = 𝛾(Γ) 𝛾 (𝑔) ≤ (𝐵10 ) 𝛾 (𝑔) .

By (5.18), Γ 6 ≤ 𝑁, so 𝛼(𝛾(Γ)) = 𝜑(Γ) ≠ 1, hence 𝛾(Γ) ≠ 1. Therefore, by (5.24), 𝐵10 ∩ (𝐵10 ) 𝛾 (𝑔) ≠ 1. On the other hand, by Claim B, 𝐵10 and (𝐵10 ) 𝛾 (𝑔) are distinct groups in Bmax . Hence, by (5.21), 𝐵10 ∩ (𝐵10 ) 𝛾 (𝑔) = 1. This is a contradiction. Proof of (c). By Remark 1.2.1(a), the map 𝛾∗ : G → B defined by 𝛾∗ (Γ) = 𝛾(Γ) is étale continuous. If Γ ∉ Subgr(𝑁0 ), then, by (5.18), Γ 6 ≤ 𝑁. Hence, by (5.19) and (5.20), 𝛼(𝛾(Γ)) = 𝜑(Γ) ≠ 1. Thus, 𝛾(Γ) 6 ≤ Ker(𝛼). Therefore, 𝛾∗ restricts to 𝛾∗,0 : G r Subgr(𝑁0 ) → B r Subgr(Ker(𝛼)). By Lemma 3.1.11(b), the map B r Subgr(Ker(𝛼)) → Bmax that maps Δ to the unique 𝐵 0 ∈ Bmax containing Δ is also étale continuous. Hence, by (5.22), its composition with 𝛾∗,0 is an étale continuous map 𝛾ˆ : G rSubgr(𝑁0 ) → Bmax . By (5.23), 𝛾(Γ ˆ 1 ) = 𝐵10 and 𝛾(Γ ˆ 2 ) = 𝐵20 . 0 0 By Lemma 3.1.11(a), Bmax is étale profinite. Since 𝐵1 , 𝐵2 ∈ Bmax are not equal, there are two disjoint étale open-closed subsets B10 , B20 of Bmax such that Bmax = B10 ∪· B20 and 𝐵𝑖0 ∈ B𝑖0, for 𝑖 = 1, 2. Let G𝑖 = 𝛾ˆ −1 (B𝑖0), for 𝑖 = 1, 2. Then, G r Subgr(𝑁0 ) is the disjoint union of the étale open-closed subsets G1 , G2 , and Γ𝑖 ∈ G𝑖 , for 𝑖 = 1, 2. It follows from Lemma 1.1.9 that G r Subgr(𝑁0 ) is étale profinite. Finally, by (a), G is stellate, hence so is G r Subgr(𝑁0 ). Therefore, by Corollary 2.1.10(d), G r Subgr(𝑁0 ) is also separated, as (c) claims.  Corollary 5.5.4 Let 𝐺 be a profinite group and G a 𝐺-invariant étale-compact subset of Subgr(𝐺) such that G = Gmax . Suppose that 𝐺 is strongly G-projective. Then, Γ ∩ Γ 𝜎 = 1 for every Γ ∈ G and 𝜎 ∈ 𝐺 r Γ. Proof By assumption, G = G 𝐺 . Let Γ ∈ G and 𝜎 ∈ 𝐺 r Γ. Then, by Proposition 5.5.3(b), Γ ≠ Γ 𝜎 or Γ = 1. Hence, by Proposition 5.5.3(a), Γ ∩ Γ 𝜎 = 1, as claimed. Corollary 5.5.5 Let 𝐺 be a profinite group and G an étale compact 𝐺-invariant ´ subset of Subgr(𝐺). Suppose 1 ∉ EtaleClosure(G) and 𝐺 is strongly G-projective. Then, Gmax is separated. Proof By Remark 1.3.1, there is an open 𝑁0 ⊳ 𝐺 such that G = G r Subgr(𝑁0 ). Hence, the assertion follows from Proposition 5.5.3(c).  The converse of Corollary 5.5.5 is not true: Corollary 5.5.6 Let 𝐺 be the free product in the sense of Binz–Neukirch–Wenzel (Definition 4.4.1) of non-trivial profinite groups 𝐺 𝑖 with 𝑖 ranging over an infinite set 𝐼. Then: (a) The set G = {𝐺 𝑖 }𝑖 ∈𝐼 is separated and 𝐺 is strongly G-projective. ´ (b) 1 ∈ EtaleClosure(G).

80

5 Relative Embedding Problems

· Proof Proof of (a). By Corollary 4.4.3(a),(d), 𝐼 ∗ = 𝐼 ∪{∞} is a profinite space Î and 𝐺 is the free profinite product 𝑖 ∈𝐼 ∗ 𝐺 𝑖 over 𝐼 ∗ with 𝐺 ∞ = 1. In particular, the family G ∗ = {𝐺 𝑖 }𝑖 ∈𝐼 ∗ is étale continuous (Definition 4.1.1). By Proposition 2.1.7, G ∗ is separated. Hence, by Remark 2.1.6(c), G is also separated. By Proposition 5.2.2, 𝐺 is strongly G ∗ -projective. Since 𝐺 ∞ = 1, 𝐺 is also strongly G-projective.

Proof of (b). Let 𝑁 be an open normal subgroup of 𝐺. By Definition 4.4.1, 𝑁 contains all but finitely many of the groups 𝐺 𝑖 . Since 𝐼 is infinite, 𝑁 contains ´ infinitely many of the groups 𝐺 𝑖 . Hence, by Remark 1.3.1, 1 ∈ EtaleClosure(G).  The following problem is related to Problem 2.1.11. Problem 5.5.7 Let 𝐺 be a profinite group and G an étale compact subset of Subgr(𝐺) closed under conjugation. Suppose that 𝐺 is strongly G-projective. Is Gmax separated? We give a partial solution to this problem: Proposition 5.5.8 Let 𝐺 be a profinite group of rank ≤ ℵ0 and let G be a subset of Subgr(𝐺). Assume that G is 𝐺-invariant and 𝐺 is strongly G-projective. Then, Gmax is separated. Proof By Lemmas 1.3.6 and 5.5.1, we may assume that G = Gmax . Since G is étale compact (by (5.5)), so is G r {1} (Remark 1.2.6). By Remark 2.1.6(c) it suffices to show that G r {1} is separated, so without loss of generality 1 ∉ G. Let Γ0, Γ00 ∈ G be distinct. Since rank(𝐺) ≤ ℵ0 , there Ñ is a sequence 𝑁1 ≥ 𝑁2 ≥ 𝑁3 ≥ · · · of open normal subgroups of 𝐺 such that ∞ 𝑖=1 𝑁𝑖 = 1. Set G𝑖 = G r Subgr(𝑁𝑖 ), for every 𝑖. By Proposition 5.5.3(c), G ⊆ G 1 2 ⊆ G3 ⊆ · · · is a sequence of étale profinite spaces. Ñ Since ∞ 𝑁 = 1, we have 𝑖 𝑖=1 ∞ Ø 𝑖=1

G𝑖 =

∞ Ø 𝑖=1

(G r Subgr(𝑁𝑖 )) = G r

∞ Ù

Subgr(𝑁𝑖 )

𝑖=1

= G r Subgr(

∞ Ù

𝑁𝑖 ) = G r {1} = G.

𝑖=1

Since Γ0, Γ00 ≠ 1, there is a 𝑘 such that Γ0, Γ00 6 ≤ 𝑁 𝑘 , hence Γ0, Γ00 ∈ G𝑖 for every 𝑖 ≥ 𝑘. Since G𝑘 is an étale profinite space, there are disjoint étale open-closed subsets G𝑘0 , G𝑘00 of G𝑘 such that G𝑘 = G𝑘0 ∪ G𝑘00, Γ0 ∈ G𝑘0 , and Γ00 ∈ G𝑘00. Applying induction, we conclude from Corollary 1.1.17(a) that for each 𝑙 ≥ 𝑘 there are disjoint étale openclosed subsets G𝑙0, G𝑙00 of G𝑙 (in particular étale compact) such that G𝑙 = G𝑙0 ∪ G𝑙00, 0 ∩ G , and G 00 = G 00 ∩ G . G𝑙0 = G𝑙+1 𝑙 𝑙+1 Ð 𝑙 Ð𝑙 ∞ ∞ Set G 0 = 𝑙=𝑘 G𝑙0 and G 00 = 𝑙=𝑘 G𝑙00. Then, G 0, G 00 are disjoint subsets of G 0 00 0 0 such that G = G ∪ G , G𝑙 = G ∩ G𝑙 , and G𝑙00 = G 00 ∩ G𝑙 , for every 𝑙 ≥ 𝑘. By definition, G𝑙0 = G 0 ∩ G𝑙 = G 0 ∩ (G r Subgr(𝑁𝑙 )) = G 0 r Subgr(𝑁𝑙 ). Similarly, G𝑙00 = G 00 r Subgr(𝑁𝑙 ). Hence, by Lemma 1.2.10, G 0, G 00 are étale compact. Therefore, by Lemma 1.2.5, both 𝑈 (G 0) and 𝑈 (G 00) are closed in 𝐺. By Proposition 5.5.3(a), G is stellate. Thus, G is separated (Definition 2.1.5). 

5.6 Prescribed Solutions

81

5.6 Prescribed Solutions Fix a strongly G-projective profinite group 𝐺 with G ⊆ (G 𝐺 )max . Then, G = Gmax . In particular, G is an étale compact subset of Subgr(𝐺) (Definition (5.5)). By Proposition 5.5.3, Γ1 ∩ Γ2 = 1 for all distinct Γ1 , Γ2 ∈ G. By Definition (5.5), every finite G-embedding problem (5.25)

(𝜑 : 𝐺 → 𝐴, 𝛼 : 𝐵 → 𝐴, B)

has a strong solution 𝛾. An interesting question is whether we can, to some extent, prescribe the solution. We consider G as a family {Γ𝑡 }𝑡 ∈𝑇 , where 𝑇 is G, with its étale topology, and 𝑡 ↦→ Γ𝑡 is the identity map G → G. In particular, the map 𝑡 → Γ𝑡 is étale continuous. It follows from the first paragraph that the family G is étale continuous (Definition 2.1.1). Claim: 𝑇 is a 𝑇1 -space. Indeed, consider distinct Γ, Γ0 ∈ G. Then, Γ 6 ⊆ Γ0. Hence, by Lemma 1.1.12(b), 𝐺 has an open normal subgroup 𝑁 such that Γ𝑁 6 ⊆ Γ0 𝑁. Thus, Subgr(Γ0 𝑁) ∩ G is an étale open neighborhood of Γ0 in G that does not contain Γ. Thus, 𝑇 is a 𝑇1 -space (see the paragraph preceding Lemma 1.1.23). Definition 5.6.1 In view of the Claim, we consider the sheaf X = (𝑋, 𝜏, 𝑇) and the sheaf-group structure (𝑋, 𝜏, 𝑇,Ð𝜔, 𝐺) that Proposition 2.3.6 associates with G. Then, Ð we set 𝑈 := 𝑈 (G) = 𝜔(𝑋) = 𝑡 ∈𝑇 𝜔(𝑋𝑡 ) = Γ∈ G Γ. Let 𝐵 be a profinite group. For any map 𝜓 : 𝑋 → 𝐵 and for 𝑏 ∈ 𝐵 let 𝜓 𝑏 : 𝑋 → 𝐵 be the map 𝑥 ↦→ 𝑏 −1 𝜓(𝑥)𝑏. If 𝜓 is continuous, then so is 𝜓 𝑏 . If 𝜓 is a morphism of sheaves X → 𝐵, then so is 𝜓 𝑏 . Two sheaf morphisms 𝜓, 𝜓 0 : X → 𝐵 are quasi-conjugate, written 𝜓 ∼ 𝜓 0, if for every 𝑡 ∈ 𝑇 there exists a 𝑏 ∈ 𝐵 such that 𝜓 𝑏 |𝑋𝑡 = 𝜓 0 |𝑋𝑡 . A G-conductor is a continuous map 𝛿 : 𝑈 → 𝐵 such that 𝛿|Γ : Γ → 𝐵 is a homomorphism, for every Γ ∈ G. Two G-conductors 𝛿, 𝛿 0 : 𝑈 → 𝐵 are quasi-conjugate, written 𝛿 ∼ 𝛿 0, if for every Γ ∈ G there exists a 𝑏 ∈ 𝐵 such that 𝛿 𝑏 |Γ = 𝛿 0 |Γ . Remark 5.6.2 If 𝛿 : 𝑈 → 𝐵 is a G-conductor, then 𝜓 = 𝛿 ◦ 𝜔 is a morphism of the sheaf X = (𝑋, 𝜏, 𝑇) into the group 𝐵 (Definition 2.3.2). Indeed, since 𝜔(𝑋) = 𝑈, the composition 𝛿 ◦ 𝜔 is well defined. Conversely, every morphism of sheaves 𝜓 : X → 𝐵 is of the form 𝛿 ◦ 𝜔, for a unique G-conductor 𝛿. Indeed, for every 𝑡 ∈ 𝑇, 𝜔 induces a group isomorphism 𝑋𝑡 → 𝜔(𝑋𝑡 ). Hence, there is a unique homomorphism 𝛿𝑡 : 𝜔(𝑋𝑡 ) → 𝐵 such that 𝜓 = 𝛿𝑡 ◦ 𝜔 on 𝑋𝑡 . By Proposition 5.5.3(a), the 𝛿𝑡 ’s extend to a well defined map 𝛿 : 𝑈 → 𝐵. Since 𝑋 is compact and 𝑈 is Hausdorff, 𝜔 : 𝑋 → 𝑈 is closed (Fact 1.1.3(e)). Note that for 𝛿−1 (𝐶) = 𝜔(𝜓 −1 (𝐶)) for every subset 𝐶 of 𝐵. Hence, if 𝐶 is closed, also 𝛿−1 (𝐶) is closed. Therefore, 𝛿 is continuous. It follows that 𝛿 is a G-conductor. By Definition 5.6.1, if 𝛿, 𝛿 0 : 𝑈 → 𝐵 are two G-conductors, then 𝛿 ∼ 𝛿 0 if and only if 𝛿 ◦ 𝜔 ∼ 𝛿 0 ◦ 𝜔.

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Lemma 5.6.3 Let 𝐵 be a finite group and 𝛿 : 𝑈 → 𝐵 a G-conductor. Then: ¯ an epimorphism 𝜋 : 𝐺 → 𝐺, ¯ and a map 𝛿1 : 𝜋(𝑈) → 𝐵 (a) There is a finite group 𝐺, such that 𝛿1 ◦ 𝜋|𝑈 = 𝛿. (b) The map G → Subgr(𝐵), given by Γ ↦→ 𝛿(Γ), is continuous with respect to the étale topologies on both G ⊆ Subgr(𝐺) and Subgr(𝐵). Proof Proof of (a). Apply Corollary 1.1.17(b) to extend 𝛿 to a continuous map 𝛿 0 : 𝐺 → 𝐵. Since 𝐺 is the inverse limit of its finite quotients, by Lemma 1.1.16(b) ¯ an epimorphism 𝜋 : 𝐺 → 𝐺, ¯ and a map 𝛿¯ : 𝐺¯ → 𝐵 such there is a finite group 𝐺, ¯ 𝜋 (𝑈 ) will satisfy 𝛿1 ◦ 𝜋|𝑈 = 𝛿, as desired. that 𝛿¯ ◦ 𝜋 = 𝛿 0. The map 𝛿1 := 𝛿| Proof of (b). Let 𝜋 and 𝛿 0 be as in the proof of (a) and let 𝐾 = Ker(𝜋). Then, 𝛿 0 is constant on the cosets of 𝐾 in 𝐺. Therefore, if Γ, Γ0 ∈ G and Γ0 ≤ Γ𝐾, then 𝛿(Γ0) ≤ 𝛿 0 (Γ𝐾) = 𝛿(Γ). This is the étale continuity of the map Γ ↦→ 𝛿(Γ).  Lemma 5.6.4 Let 𝐺 and G be as in the opening of the present section and set 𝑈 = 𝑈 (G) (Definition 5.6.1). Consider an epimorphism 𝛼 : 𝐵 → 𝐴 of finite groups, and let 𝛿 : 𝑈 → 𝐵 and 𝜑 : 𝑈 → 𝐴 be G-conductors such that 𝛼 ◦ 𝛿 ∼ 𝜑. Then, there is a G-conductor 𝛿 0 : 𝑈 → 𝐵 such that 𝛿 0 ∼ 𝛿 and 𝛼 ◦ 𝛿 0 = 𝜑. Proof Since 𝛿 and 𝜑 are continuous and 𝐵 is finite, there is an open 𝑁 ⊳ 𝐺 such that 𝛿(𝑈 ∩ 𝑁) = 1 and 𝜑(𝑈 ∩ 𝑁) = 1. By the first paragraph of the present section, G is an étale compact subset of Subgr(𝐺). By Lemma 5.2.4, 𝐺 is strongly G 𝐺 -projective. Hence, by Corollary 1.4.3, G 𝐺 is an étale compact subset of Subgr(𝐺), which, by definition, is 𝐺-invariant. By Lemma 5.2.4, 𝐺 is strongly G 𝐺 -projective. Hence, by Proposition 5.5.3(c), (G 𝐺 )max r Subgr(𝑁) is étale profinite. Since G ⊆ (G 𝐺 )max and G is étale closed in
Subgr(𝐺) (Fact 1.1.3(b)), the set G 0 := G r Subgr(𝑁) = G ∩ (G 𝐺 )max r Subgr(𝑁) is an étale closed subset of the étale profinite space (G 𝐺 )max r Subgr(𝑁). Therefore, G 0 is étale profinite (Remark 1.1.7(a)). Since 𝛼 ◦ 𝛿 ∼ 𝜑, for each Γ ∈ G 0 there is an 𝑎 ∈ 𝐴 such that (𝛼 ◦ 𝛿|Γ ) 𝑎 = 𝜑|Γ . Choose 𝑏 ∈ 𝐵 such that 𝛼(𝑏) = 𝑎. Then, 𝛼 ◦ 𝛿 𝑏 |Γ = 𝜑|Γ . By Corollary 1.1.17(c), there is an open-closed subset 𝑉 of 𝐺 containing Γ such that 𝛼 ◦ 𝛿 𝑏 and 𝜑 coincide on 𝑉 ∩ 𝑈 (G 0). By Corollary 1.1.13, there is an open 𝐾 ⊳ 𝐺 such that Γ𝐾 ⊆ 𝑉. Since G 0 is étale profinite, there is an étale open-closed neighborhood GΓ0 of Γ in G 0 contained in Subgr(Γ𝐾). If Δ ∈ GΓ0 , then Δ ⊆ Γ𝐾 ⊆ 𝑉 and Δ ⊆ 𝑈 (G 0), hence 𝛼 ◦ 𝛿 𝑏 |Δ = 𝜑|Δ . Since G 0 is compact, the open-closed covering {GΓ0 }Γ∈ G0 of G 0 has a finite subcovering, say, {G10 , . . . , G𝑛0 }. Applying Lemma 1.1.10, we may assume that G10 , . . . , G𝑛0 are disjoint. By construction, for each 𝑖 there is a 𝑏 𝑖 ∈ 𝐵 such that 𝛼 ◦ 𝛿 𝑏𝑖 |Γ = 𝜑|Γ for every Γ ∈ G𝑖0. Notice that if 𝑔 ∈ Γ ∩ 𝑁, then 𝛿(𝑔) = 1, so 𝛿 𝑏𝑖 (𝑔) = 𝑏 −1 𝑖 𝛿(𝑔)𝑏 𝑖 = 1. Define 𝛿 0 : 𝑈 → 𝐵 by 𝛿 0 |Γ = 𝛿 𝑏𝑖 |Γ for every Γ ∈ G𝑖0 and 𝛿 0 (𝑔) = 1 for every 𝑔 ∈ Γ ∈ G ∩ Subgr(𝑁). Since 𝑛 Ø · · G = (G r Subgr(𝑁)) ∪(G ∩ Subgr(𝑁)) = ( · G𝑖0) ∪(G ∩ Subgr(𝑁)), 𝑖=1

5.6 Prescribed Solutions

83

and G is stellate (Proposition 5.5.3(a)), 𝛿 0 is well defined. Also, 𝛿 0 ∼ 𝛿, 𝛼 ◦ 𝛿 0 = 𝜑. Moreover, 𝛿 0 is continuous on each 𝑈 (G𝑖0) r 𝑁, because 𝛿 0 is 𝛿 𝑏𝑖 there, and 𝛿 0 is continuous on 𝑈 ∩ 𝑁, because 𝛿 0 is constant there. In addition, 𝑛 Ø · 𝑈 = · (𝑈 (G𝑖0) r 𝑁) ∪(𝑈 ∩ 𝑁) 𝑖=1

is a partition into disjoint closed subsets. Therefore, 𝛿 0 is continuous on 𝑈. It follows that 𝛿 0 is a G-conductor, as desired.  Lemma 5.6.5 Let 𝐾 be an open normal subgroup of 𝐺 and let 𝛿 : 𝑈 → 𝐵 be a G-conductor into a finite group 𝐵. Then, there is an open 𝑁 ⊳ 𝐺 contained in 𝐾 such that for all 𝑔 ∈ 𝐺 and Γ ∈ G (*) either Γ𝑁 ∩ Γ𝑔 𝑁 ≤ 𝐾 or there is a Δ ∈ G with 𝛿(Γ) ≤ 𝛿(Δ) and 𝑔 ∈ Δ𝐾. Proof Let 𝜎 ∈ 𝐺 and Δ ∈ G. Claim: There is an open 𝑀 = 𝑀 (𝜎, Δ) ⊳ 𝐺 such that (*) holds for all open 𝑁 ⊳ 𝐺 contained in 𝐾 ∩ 𝑀, all 𝑔 ∈ 𝜎𝑀, and all Γ ∈ Subgr(Δ𝑀). We break up the proof into two cases: Case A: 𝜎 ∉ Δ. By Corollary 5.5.4, Δ ∩ Δ 𝜎 = 1 ≤ 𝐾. By Lemma 1.1.14, the intersection of all open normal subgroups of 𝐺 is 1. Hence, by Lemma 1.1.12(b), Ù Ù Ù (Δ𝑀 ∩ Δ 𝜎 𝑀) ≤ Δ𝑀 ∩ Δ 𝜎 𝑀 = Δ ∩ Δ 𝜎 = 1, 𝑀

𝑀

𝑀

where 𝑀 ranges over OpenNormal(𝐺). Therefore, by Lemma 1.1.12(a), there is an open 𝑀 ⊳ 𝐺 contained in 𝐾 such that Δ𝑀 ∩ Δ 𝜎 𝑀 ≤ 𝐾. If Γ ≤ Δ𝑀, then Γ𝑀 ≤ Δ𝑀 and Γ 𝜎 𝑀 ≤ Δ 𝜎 𝑀, hence Γ𝑀 ∩ Γ 𝜎 𝑀 ≤ 𝐾. If 𝑔 ∈ 𝜎𝑀, then Γ𝑔 𝑀 = Γ 𝜎 𝑀, hence Γ𝑀 ∩ Γ𝑔 𝑀 ≤ 𝐾. If 𝑁 ⊳ 𝐺 and 𝑁 ≤ 𝑀, then 𝑁 ≤ 𝐾 and Γ𝑁 ∩ Γ𝑔 𝑁 ≤ 𝐾. Thus, (*) holds. Case B: 𝜎 ∈ Δ. By Lemma 5.6.3(b), there is an open 𝑀 ⊳ 𝐺 such that 𝛿(Γ) ≤ 𝛿(Δ) for all Γ ∈ Subgr(Δ𝑀). Without loss of generality 𝑀 ≤ 𝐾. So if 𝑔 ∈ 𝜎𝑀, then 𝑔 ∈ Δ𝐾. Thus, the second disjunct of (*) holds. Since 𝐺 × G is compact, finitely many of the open neighborhoods 𝑛 𝜎𝑀 (𝜎, Δ) × Subgr(Δ𝑀 (𝜎, Δ)) cover Ñ𝑛𝐺 × G, say, {𝜎𝑖 𝑀𝑖 × Subgr(Δ𝑀𝑖 )}𝑖=1 . By the above claim, (*) holds with 𝑁 = 𝑖=1 𝑀𝑖 .  Remark 5.6.6 In the sequel we use Lemma 5.1.4 and its proof, with some additional care: Let (5.25) be a finite G-embedding problem and let 𝛿 : 𝑈 → 𝐵 be a G-conductor such that 𝛼 ◦ 𝛿 = 𝜑|𝑈 . Let 𝐾 be an open normal subgroup of 𝐺 contained in Ker(𝜑), ˆ Then, the map let 𝐴ˆ = 𝐺/𝐾, and consider the diagram (5.7) with 𝐵ˆ = 𝐵 × 𝐴 𝐴. ˆ𝛿 : 𝑈 → 𝐵ˆ defined by 𝛿(𝑢) ˆ = (𝛿(𝑢), 𝜑(𝑢)) ˆ for each 𝑢 ∈ 𝑈 is continuous and satisfies 𝛼ˆ ◦ 𝛿ˆ = 𝜑| ˆ 𝑈 and 𝛽 ◦ 𝛿ˆ = 𝛿. Moreover, 𝛿ˆ is the unique function with these properties. Note that for every Γ ∈ G, the restrictions 𝛿|Γ and 𝜑| ˆ Γ are group homomorphisms, ˆ Γ . Thus, 𝛿ˆ is a G-conductor. hence so is 𝛿| Now, for every Γ ∈ G, Part A of the proof of Lemma 5.1.4 establishes a homomorphism 𝛾Γ : Γ → 𝐵 such that 𝛼 ◦ 𝛾Γ = 𝜑|Γ and 𝛾Γ (Γ) ∈ B. In our situation we

84

5 Relative Embedding Problems

ˆ mentioned in may take 𝛾Γ to be 𝛿|Γ . Then, the unique homomorphism 𝛾ˆ Γ : Γ → 𝐵, Part A of the proof of Lemma 5.1.4, that satisfies 𝛽 ◦ 𝛾ˆ Γ = 𝛾Γ and 𝛼ˆ ◦ 𝛾ˆ Γ = 𝜑| ˆ Γ is ˆ Γ . Thus, by (5.8), just 𝛿| ˆ (5.26) Bˆ = Con({ 𝛾ˆ Γ (Γ) | Γ ∈ G}) = Con( 𝛿(G)). Lemma 5.6.7 Let 𝐺 be a (properly) strongly G-projective profinite group (Definition (5.5)). Assume that G is a set of representatives of the conjugacy classes of (G 𝐺 )max . Let (5.25) be a finite proper G-embedding problem and let 𝛿 : 𝑈 → 𝐵 be a G-conductor such that 𝛼 ◦ 𝛿 ∼ 𝜑|𝑈 . Assume that (a) B = Con(𝛿(G)) and (b) 𝛼 is injective on 𝛿(Γ) for each Γ ∈ G. Then, there exists a (proper) strong solution 𝛾 of (5.25) such that 𝛾|𝑈 ∼ 𝛿. Proof We set 𝐾 = Ker(𝜑) and assume by Lemma 5.6.4 that (5.27)

𝛼 ◦ 𝛿 = 𝜑|𝑈 .

Claim A: If 𝛾 is a strong solution of (5.25), then 𝛾(Γ) = 𝛿(Γ) = 1 for every Γ ∈ G contained in 𝐾. Indeed, let Γ be a group in G with Γ ≤ 𝐾. Since 𝐾 = Ker(𝜑), we have 𝜑(Γ) = 1. Since 𝛾 is a strong solution of (5.25), 𝛾(Γ) ∈ B. Since, by assumption, B = Con(𝛿(G)), we have 𝛿(Γ) ∈ B. By Assumptions (a) and (b), 𝛼 is injective on both 𝛾(Γ) and 𝛿(Γ). Since Γ ≤ 𝐾 = Ker(𝜑) and 𝛼 ◦ 𝛾 = 𝜑, we have 𝛼(𝛾(Γ)) = 𝜑(Γ) = 1, so 𝛾(Γ) = 1. By (5.27), 𝛼(𝛿(Γ)) = 𝜑(Γ) = 1. Hence, 𝛿(Γ) = 1, as claimed. Claim B: There is a (proper) strong solution 𝛾 of (5.25) such that for every Γ ∈ G there are Δ ∈ G and 𝑏 ∈ 𝐵 with 𝛿(Γ) ≤ 𝛿(Δ) and 𝛾(Γ) ≤ 𝛿(Δ) 𝑏 . By Definition (5.5), G is étale compact. Hence, by Proposition 5.5.3(c), Corollary 1.4.3, and Lemma 5.2.4, H := G 𝐺 r Subgr(𝐾) is étale profinite. Moreover, by Fact 1.1.3(a), G 0 := G r Subgr(𝐾) is étale compact. Therefore, G 0 is a profinite space which is étale closed in H . By Lemma 5.6.3(b), the map G 0 → Subgr(𝐵) given by Γ ↦→ 𝛿(Γ) is étale Ð𝑛 continuous. Hence, by Lemma 2.1.3, there is a partition G 0 = · 𝑖=1 G𝑖 into étale open-closed subsets and for every 𝑖 there is (5.28) a Γ𝑖 ∈ G𝑖 such that 𝛿(Γ) ≤ 𝛿(Γ𝑖 ) for every Γ ∈ G𝑖 . By the preceding paragraph and Example 1.1.11, H𝑖 := G𝑖𝐺 is étale closed in H for 𝑖 = 1, . . . , 𝑛. Since G is a set of representatives of the conjugacy classes of 0 𝐺 Ð𝑛 (G 𝐺 )max , we have G𝑖𝐺 ∩ G 𝐺 𝑗 = ∅ if 𝑖 ≠ 𝑗. Hence, H = (G ) = · 𝑖=1 H𝑖 . Therefore, each of the sets H𝑖 is also étale open in H . Thus, by Lemma 2.1.4, there is an open subgroup 𝑁 ⊳ 𝐺 such that if Γ, Γ0 ∈ H are not in the same H𝑖 , then Γ𝑁 ∩ Γ0 𝑁 ≤ 𝐾. Since 𝑁 ≤ Γ𝑁 ∩ Γ0 𝑁, we have 𝑁 ≤ 𝐾. ˆ with 𝑁 replacing We let 𝐴ˆ = 𝐺/𝑁 and consider the embedding problem ( 𝜑, ˆ 𝛼, ˆ B), ˆ 𝐾, appearing in Diagram (5.7) having the properties ensured by Lemma 5.1.4 and with B given by (5.26). In particular, 𝑁 = Ker( 𝜑). ˆ The preceding paragraph says that (5.29)

if Γ, Γ0 ∈ H are not in the same H𝑖 , then 𝜑(Γ) ˆ ∩ 𝜑(Γ ˆ 0) ≤ Ker( 𝜑). ¯

5.6 Prescribed Solutions

85

Since, by assumption, 𝐺 is (properly) strongly G-projective, there exists a (proper) ˆ Then, 𝛾 := 𝛽 ◦ 𝛾ˆ is a strong solution 𝛾ˆ : 𝐺 → 𝐵ˆ of embedding problem ( 𝜑, ˆ 𝛼, ˆ B). (proper) strong solution of (5.25). We show that each Γ ∈ G satisfies the conclusion of Claim B. Thus, (5.30)

we have to find Δ ∈ G and 𝑏 ∈ 𝐵 with 𝛿(Γ) ≤ 𝛿(Δ) and 𝛾(Γ) ≤ 𝛿(Δ) 𝑏 .

If Γ ≤ 𝐾, then by Claim A, 𝛾(Γ) = 𝛿(Γ) = 1, so Δ := Γ and 𝑏 := 1 satisfy the required condition. Thus, we may assume that 𝜑(Γ) ≠ 1, so there is 1 ≤ 𝑖 ≤ 𝑛 with Γ ∈ G𝑖 . In particular, by (5.28), 𝛿(Γ) ≤ 𝛿(Γ𝑖 ). ˆ Consider the conductor 𝛿 : 𝑈 → 𝐵ˆ given by Remark 5.6.6 satisfying (5.32) 𝛼ˆ ◦ 𝛿ˆ = 𝜑| ˆ 𝑈 and 𝛽 ◦ 𝛿ˆ = 𝛿. (5.31)

ˆ so relation (5.26) supplies a group Λ ∈ G Since 𝛾ˆ is strong, we have 𝛾(Γ) ˆ ∈ B, 𝑏ˆ . Apply 𝛽 and 𝛼 ˆ ˆ ˆ and an element 𝑏 ∈ 𝐵 such that 𝛾(Γ) ˆ ≤ 𝛿(Λ) ˆ to this equation and use (5.32) to get 𝛾(Γ) ≤ 𝛿(Λ) 𝑏 and 𝜑(Γ) ˆ ≤ 𝜑(Λ ˆ 𝑔 ), ˆ and 𝑔 ∈ 𝐺 satisfies 𝜑(𝑔) ˆ Then, apply 𝜑¯ on the second where 𝑏 = 𝛽( 𝑏) ˆ = 𝛼( ˆ 𝑏). inequality in (5.33) and use the relation 𝜑 = 𝜑¯ ◦ 𝜑ˆ (Diagram 5.7) to get

(5.33)

(5.34)

𝜑(Γ) ≤ 𝜑(Λ𝑔 ).

Since 𝜑(Γ) ≠ 1, (5.34) implies that 𝜑(Λ) ≠ 1, so Λ ∈ G 0. Hence, there is 1 ≤ 𝑗 ≤ 𝑛 such that Λ ∈ G 𝑗 . Then, Γ ∈ H𝑖 , because Γ ∈ G𝑖 , and Λ𝑔 ∈ H 𝑗 . In addition, by (5.33) and since 𝜑(Γ) ≠ 1, we have 𝜑(Γ) ˆ ∩ 𝜑(Λ ˆ 𝑔 ) = 𝜑(Γ) ˆ 6≤ Ker( 𝜑). ¯ Hence, by (5.29), 𝑖 = 𝑗. Therefore, by (5.28), 𝛿(Λ) ≤ 𝛿(Γ𝑖 ). By (5.33), 𝛾(Γ) ≤ 𝛿(Λ) 𝑏 ≤ 𝛿(Γ𝑖 ) 𝑏 . It follows that the group Δ := Γ𝑖 is in G and satisfies 𝛿(Γ) ≤ 𝛿(Δ) and 𝛾(Γ) ≤ 𝛿(Δ) 𝑏 , as required by (5.30). Claim C: There is a (proper) strong solution 𝛾 of the G-embedding problem (5.25) such that for every Γ ∈ G there are Δ ∈ G and 𝑏 ∈ Ker(𝛼) with 𝛿(Γ) ≤ 𝛿(Δ) and 𝛾(Γ) ≤ 𝛿(Δ) 𝑏 . Indeed, by Lemma 5.6.5, there is an open 𝑁 ⊳ 𝐺 contained in 𝐾 such that if Δ ∈ G and 𝑔 ∈ 𝐺 satisfy Δ𝑁 ∩ Δ𝑔 𝑁 6 ≤ 𝐾, then there is a Δ0 ∈ G with 𝛿(Δ) ≤ 𝛿(Δ0) and 𝑔 ∈ Δ0 𝐾. ˆ constructed above, Let 𝐴ˆ = 𝐺/𝑁 and consider the embedding problem ( 𝜑, ˆ 𝛼, ˆ B) with diagram (5.7). The preceding paragraph says that 𝑎ˆ 6 ≤ Ker( 𝜑), ˆ and 𝜑(Δ) (5.35) if Δ ∈ G, 𝑎ˆ ∈ 𝐴, ˆ ∩ 𝜑(Δ) ˆ ¯ then there is a Δ0 ∈ G with 0 0 𝛿(Δ) ≤ 𝛿(Δ ) and 𝜑( ¯ 𝑎) ˆ ∈ 𝜑(Δ ).

ˆ such that By Claim B there is a (proper) strong solution 𝛾ˆ : 𝐺 → 𝐵ˆ of ( 𝜑, ˆ 𝛼, ˆ B) ˆ ˆ for every Γ ∈ G there are Δ ∈ G and 𝑏 ∈ 𝐵 with ˆ ˆ ˆ 𝑏ˆ . 𝛿(Γ) ≤ 𝛿(Δ) and 𝛾(Γ) ˆ ≤ 𝛿(Δ) ˆ Then, 𝛾 is a (proper) strong solution of (5.25) with Let 𝛾 = 𝛽 ◦ 𝛾ˆ and 𝑏 = 𝛽( 𝑏).

(5.36)

86

(5.37)

5 Relative Embedding Problems

𝛿(Γ) ≤ 𝛿(Δ) and 𝛾(Γ) ≤ 𝛿(Δ) 𝑏 .

We complete the proof of Claim C by proving that 𝑏 can be chosen to be an element of Ker(𝛼). If Γ ≤ 𝐾, then, by Claim A, 𝛾(Γ) = 1, so 𝑏 can be chosen to be 1. Otherwise, Γ 6 ≤ 𝐾, that is, 𝜑( ¯ 𝜑(Γ)) ˆ = 𝜑(Γ) ≠ 1. Apply 𝛼ˆ to (5.36) and use ˆ ˆ 𝛼( ˆ 𝑏) 𝛼( ˆ 𝑏) (5.32) to get 𝜑(Γ) ˆ ≤ 𝜑(Δ) ˆ and 𝜑(Γ) ˆ ≤ 𝜑(Δ) ˆ . Thus, 𝜑(Δ) ˆ ∩ 𝜑(Δ) ˆ 6 ≤ Ker( 𝜑). ¯ 0 0 ˆ ∈ 𝜑(Δ0). Hence, by (5.35), there is a Δ ∈ G with 𝛿(Δ) ≤ 𝛿(Δ ) and 𝛼(𝑏) = 𝜑( ¯ 𝛼( ˆ 𝑏)) Replacing Δ by Δ0 in (5.37), we still have 𝛿(Γ) ≤ 𝛿(Δ), 𝛾(Γ) ≤ 𝛿(Δ) 𝑏 , and also 𝛼(𝑏) ∈ 𝜑(Δ). Since 𝛼 maps 𝛿(Δ) onto 𝜑(Δ) (by (5.27)), there is a 𝑑 ∈ 𝛿(Δ) such that −1 𝛼(𝑑) = 𝛼(𝑏). We have 𝑑 −1 𝑏 ∈ Ker(𝛼) and 𝛿(Δ) 𝑑 𝑏 = 𝛿(Δ) 𝑏 . Claim D: There exists a (proper) strong solution 𝛾 such that 𝛾|𝑈 ∼ 𝛿. Indeed, let 𝛾 : 𝐺 → 𝐵 be the solution of embedding problem (5.25) given by Claim C and consider Γ ∈ G. This claim supplies Δ ∈ G and 𝑏 ∈ Ker(𝛼) with 𝛿(Γ) ≤ 𝛿(Δ) and 𝛾(Γ) ≤ 𝛿(Δ) 𝑏 . Moreover, (5.27)

𝛼 ◦ 𝛿 𝑏 |Γ = (𝛼 ◦ 𝛿) 𝛼(𝑏) |Γ = 𝛼 ◦ 𝛿|Γ = 𝜑|Γ = 𝛼 ◦ 𝛾|Γ . Since, by (b), 𝛼 is injective on 𝛿(Δ) 𝑏 , we have 𝛿 𝑏 |Γ = 𝛾|Γ . By Definition 5.6.1, this means that 𝛾|𝑈 ∼ 𝛿, as desired.  Proposition 5.6.8 Let 𝐺 be a (properly) strongly G-projective profinite group. Assume that G is a set of representatives of the conjugacy classes of (G 𝐺 )max . Let (5.25) be a finite proper G-embedding problem and let 𝛿 : 𝑈 → 𝐵 be a G-conductor such that 𝛼 ◦ 𝛿 ∼ 𝜑|𝑈 . Assume that B = Con(𝛿(G)). Then, embedding problem (5.25) has a (proper) strong solution 𝛾 such that 𝛾|𝑈 ∼ 𝛿. Proof We reduce the proposition to the case where 𝛼| 𝛿 (Γ) is injective for every Γ ∈ G. As in the proof of Lemma 5.6.7, we set 𝐾 = Ker(𝜑) and replace 𝛿 by a quasiconjugate conductor, if necessary, to assume that 𝛼 ◦ 𝛿 = 𝜑|𝑈 . By Lemma 5.6.3(a), ˆ an epimorphism 𝜑ˆ : 𝐺 → 𝐴, ˆ and a map 𝛿1 : 𝜑(𝑈) there is a finite group 𝐴, ˆ → 𝐵 such that 𝛿1 ◦ 𝜑| ˆ 𝑈 = 𝛿. Thus, for every Γ ∈ G we have 𝛿1 | 𝜑ˆ (Γ) ◦ 𝜑| ˆ Γ = 𝛿|Γ .

(5.38)

In particular, 𝛿1 | 𝜑ˆ (Γ) : 𝜑(Γ) ˆ → 𝛿(Γ) is a group homomorphism. Without loss of generality, 𝐾ˆ = Ker( 𝜑) ˆ ≤ 𝐾. Consider diagram (5.7) of Lemma 5.1.4 together with the conductor 𝛿ˆ : 𝑈 → 𝐵ˆ in Remark 5.6.6 satisfying 𝛼ˆ ◦ 𝛿ˆ = 𝜑| ˆ 𝑈. 𝑈 𝐺 𝛿ˆ

𝜑ˆ

 𝜑(𝑈) ˆ

𝛾ˆ

!

𝐵ˆ

𝛼ˆ

 / 𝐴ˆ

𝛼

 /𝐴

𝛽 𝛿

𝛿1

!  0𝐵

𝜑ˆ

𝜑¯

5.6 Prescribed Solutions

87

ˆ Since 𝛼ˆ is B-rigid (Lemma 5.1.4), 𝛼| ˆ 𝛿ˆ (Γ) is injective for every Γ ∈ G. Since 𝜑¯ is surjective, so is 𝛽 [FrJ08, p. 500, Lemma 22.2.3(b)]. ˆ has a (proper) strong solution 𝛾ˆ By Lemma 5.6.7, embedding problem ( 𝜑, ˆ 𝛼, ˆ B) ˆ Then, 𝛾 : = 𝛽 ◦ 𝛾ˆ is a (proper) strong solution of embedding problem with 𝛾| ˆ 𝑈 ∼ 𝛿. (5.25). ˆ ˆ Γ . Applying 𝛽 on both sides, we Given Γ ∈ G, there exists a 𝑏ˆ ∈ 𝐵ˆ with 𝛿ˆ𝑏 |Γ = 𝛾| 𝑏 ˆ that 𝛿 |Γ = 𝛾|Γ . It follows that 𝛾|𝑈 ∼ 𝛿, as desired. have with 𝑏 = 𝛽( 𝑏) 

Chapter 6

Strong Proper Projectivity

Let P be a set of prime numbers, 𝐺 a profinite group, and G a subset of Subgr(𝐺). The main result, Proposition 6.4.8, of this chapter asserts that if 𝐺 is properly G-projective (equation (5.5) in Definition 5.1.1) and G is of P-type (Definition 6.4.3), then 𝐺 is strongly properly G-projective.

6.1 Prosolvable Subgroups of G-Projective Groups We investigate the nature of closed prosolvable subgroups of profinite groups 𝐺 which are strongly projective with respect to certain subsets of Subgr(𝐺). We start with a sufficient condition on a closed subgroup of a free product of finitely many finite groups to be non-solvable. To this end we denote the order of a profinite group 𝐺, as a “supernatural number” [FrJ08, pp. 520–522, Sec. 22.8], by #𝐺. Î Recall that a supernatural number is a formal product 𝑛 := 𝑝 𝑛 𝑝 , where 𝑝 ranges over all prime numbers andÎ 𝑛 𝑝 is either a non-negative number ∞. Given another Î or supernatural number 𝑚 := 𝑝 𝑚 𝑝 , we write lcm(𝑚, 𝑛) := 𝑝 max(𝑚 𝑝 ,𝑛 𝑝 ) . And for a profinite group 𝐺, we define the order to be #𝐺 = lcm{(𝐺 : 𝑈) | 𝑈 ∈ Open(𝐺)}. Note that if 𝐺 is finite, then #𝐺 = card(𝐺). The following result is a variant of [Pop95, Claim 1 of Theorem 1]. Lemma 6.1.1 Let 𝐺 1 and 𝐺 2 be finite groups and 𝐺 3 , . . . , 𝐺 𝑛 profinite groups. Let Î𝑛 𝐺 = ∗ 𝑖=1 𝐺 𝑖 and let 𝐷 be a closed subgroup of 𝐺. Suppose that there exist 𝜎1 , 𝜎2 ∈ 𝐺 and distinct prime numbers 𝑝 1 , 𝑝 2 such that 𝑝 1 |#(𝐺 1𝜎1 ∩ 𝐷) and 𝑝 2 |#(𝐺 2𝜎2 ∩ 𝐷). Then, 𝐷 is non-prosolvable. Proof Let 1 ≤ 𝑖 ≤ 2. By Cauchy’s Theorem, 𝐺 𝑖𝜎𝑖 ∩ 𝐷 has a subgroup Δ𝑖 of order 𝑝 𝑖 . It suffices to prove that the closed subgroup hΔ1 , Δ2 i of 𝐷 is not prosolvable. Hence, we may assume that 𝐷 = hΔ1 , Δ2 i. The rest of the proof breaks up into three parts. Part A: Assume first that 𝐺 𝑖𝜎𝑖 = Δ𝑖 , for 1, 2. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 D. Haran and M. Jarden, The Absolute Galois Group of a Semi-Local Field, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-030-89191-6_6

89

90

6 Strong Proper Projectivity

Then, the conjugate 𝐺 𝑖 of Δ𝑖 is of order 𝑝 𝑖 . Let 𝑔𝑖 be its generator, of order 𝑝 𝑖 . By [Jar94, Prop. 1.4], there exists a finite non-solvable group 𝑆 and elements 𝑦 𝑎1 , 𝑎2 ∈ 𝑆Î of orders 𝑝 1 , 𝑝 2 , respectively, such that 𝑆 = h𝑎 1𝑥 , 𝑎 2 i for all 𝑥, 𝑦 ∈ 𝑆. 𝑛 Since 𝐺 = ∗ 𝑖=1 𝐺 𝑖 , there is a homomorphism 𝜔 : 𝐺 → 𝑆 such that 𝜔(𝑔1 ) = 𝑎 1 , 𝜔(𝑔2 ) = 𝑎 2 , and 𝜔(𝐺 𝑘 ) = 1 for 𝑘 ≥ 3. Then, 𝑆 = h𝑎 1𝜔 ( 𝜎1 ) , 𝑎 2𝜔 ( 𝜎2 ) i = h𝜔(Δ1 ), 𝜔(Δ2 )i = 𝜔(𝐷) ≤ 𝑆, so 𝜔(𝐷) = 𝑆. Since 𝑆 is not solvable, 𝐷 is not prosolvable. Part B: 𝐺 𝑖𝜎𝑖 ∩ 𝐷 = Δ𝑖 , for 1, 2. Indeed, without loss of generality 𝑖 = 1. Let 𝜌 : 𝐺 → 𝐺 1 be an epimorphism that maps 𝐺 1 onto itself and 𝜌(𝐺 𝑘 ) = 1 for 𝑘 ≥ 2. Then, 𝜌(Δ2 ) ≤ 𝜌(𝐺 2𝜎2 ) = 1, so 𝜌(𝐷) = 𝜌(hΔ1 , Δ2 i) = 𝜌(Δ1 ) ≤ 𝜌(𝐺 1𝜎1 ∩ 𝐷) ≤ 𝜌(𝐷). Thus, 𝜌(𝐺 1𝜎1 ∩ 𝐷) = 𝜌(Δ1 ). But 𝜌|𝐺 𝜎1 is injective and maps Δ1 onto 𝜌(Δ1 ) as well, 1 so 𝐺 1𝜎1 ∩ 𝐷 = Δ1 . Part C: Replacing 𝐺 by an open subgroup. 𝑖 ∈ {1, 2}. By Lemma 1.1.14, Ñ Let H = {𝐻 ∈ Open(𝐺) | 𝐷 ≤Ñ𝐻} and consider 𝜎𝑖 𝐻 = 𝐷. Hence, by Part B, (𝐺 ∩ 𝐻) = 𝐺 𝑖𝜎𝑖 ∩ 𝐷 = Δ𝑖 . Since 𝐺 𝑖𝜎𝑖 𝐻 ∈H 𝐻 ∈H 𝑖 is finite and H is closed under finite intersections, there is an 𝐻𝑖0 ∈ H such that 𝐺 𝑖𝜎𝑖 ∩ 𝐻𝑖0 = Δ𝑖 . Then, 𝐻 := 𝐻10 ∩ 𝐻20 is an open subgroup of 𝐺 and 𝐺 𝑖𝜎𝑖 ∩ 𝐻 = Δ𝑖 , for 𝑖 = 1, 2. By Proposition 3.1.7, there are finite subsets 𝑆1 , . . . , 𝑆 𝑛 of 𝐺 and a free profinite group 𝐹 such that 𝑛 Ö Ö ∗ (𝐺 𝑘𝜏 ∩ 𝐻). (6.1) 𝐻=𝐹∗ ∗ 𝑘=1 𝜏 ∈𝑆𝑘

By Addendum 3.1.8, for each 1 ≤ 𝑖 ≤ 2 there are 𝜏𝑖 ∈ 𝑆𝑖 and 𝜂𝑖 ∈ 𝐻 such that Δ𝑖 = 𝐺 𝑖𝜎𝑖 ∩ 𝐻 = (𝐺 𝑖𝜏𝑖 ∩ 𝐻) 𝜂𝑖 . Î𝑛 Thus, by Part A, applied to (6.1) rather than to 𝐺 = ∗ 𝑖=1 𝐺 𝑖 , 𝐷 is non-solvable. Lemma 6.1.2 Let 𝐺 be a profinite group and let G be a 𝐺-invariant étale compact subset of Subgr(𝐺). Let Γ1 , Γ2 ∈ Gmax be non-conjugate in 𝐺. Then, there is an 𝑔 𝑔 𝑁 ∈ OpenNormal(𝐺) such that Γ1 1 ∪ Γ2 2 6 ⊆ Γ𝑁 for all 𝑔1 , 𝑔2 ∈ 𝐺 and all Γ ∈ G. 𝑔

𝑔

Proof Let (𝑔1 , 𝑔2 ) ∈ 𝐺 × 𝐺. Then, by Lemma 1.3.6(b), Γ1 1 , Γ2 2 are distinct groups 𝑔 𝑔 in Gmax . Hence, Γ1 1 ∪ Γ2 2 6 ⊆ Γ for all Γ ∈ G. By Lemma 1.2.3, there is an 𝑔 𝑔 𝑁 := 𝑁𝑔1 ,𝑔2 ∈ OpenNormal(𝐺) such that Γ1 1 ∪ Γ2 2 6 ⊆ Γ𝑁 for all Γ ∈ G. This remains true if we replace (𝑔1 , 𝑔2 ) by any (𝑔1 𝑛1 , 𝑔2 𝑛2 ) in the neighborhood 𝑔1 𝑁 × 𝑔2 𝑁 of (𝑔1 , 𝑔2 ) in 𝐺 × 𝐺. Indeed, let Γ ∈ G. By assumption there is 1 ≤ 𝑖 ≤ 2 and 𝜎 ∈ Γ𝑖 such that −1 𝑔 𝑖 𝜎 ∉ Γ𝑁. Since 𝑛𝑖 ∈ 𝑁 ≤ Γ𝑁, we have 𝜎 𝑔𝑖 ∉ (Γ𝑁) 𝑛𝑖 , hence 𝜎 𝑔𝑖 𝑛𝑖 ∉ Γ𝑁. Thus, 𝑔2 𝑛2 𝑔𝑖 𝑛𝑖 𝑔1 𝑛1 Γ𝑖 6 ⊆ Γ𝑁; in particular, Γ1 ∪ Γ2 6 ⊆ Γ𝑁. Ð Since 𝐺 × 𝐺 = (𝑔1 ,𝑔2 ) ∈𝐺×𝐺 𝑔1 𝑁𝑔1 ,𝑔2 × 𝑔2 𝑁𝑔1 ,𝑔2 and 𝐺 × 𝐺 is compact, there Ð is a finite subset 𝐽 of 𝐺 × 𝐺 Ñ such that 𝐺 × 𝐺 = (𝑔1 ,𝑔2 ) ∈𝐽 𝑔1 𝑁𝑔1 ,𝑔2 × 𝑔2 𝑁𝑔1 ,𝑔2 . The open normal subgroup 𝑁 = (𝑔1 ,𝑔2 ) ∈𝐽 𝑁𝑔1 ,𝑔2 of 𝐺 has the required property. 

6.1 Prosolvable Subgroups of G-Projective Groups

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The following result is a variant of [Jar94, Prop. 2.3]. Lemma 6.1.3 Let 𝐺 be a strongly G-projective profinite group with respect to a 𝐺-invariant étale compact subset G of Subgr(𝐺). Suppose that there exist nonconjugate Γ1 , Γ2 ∈ Gmax and distinct prime numbers 𝑝 1 , 𝑝 2 such that 𝑝 1 |#Γ1 and 𝑝 2 |#Γ2 . Then, 𝐺 is non-prosolvable. Proof By Lemma 1.3.6, Gmax is étale compact and 𝐺-invariant. By Lemma 5.5.1, 𝐺 is strongly Gmax -projective. Hence, we may assume that G = Gmax . 𝑔 𝑔 By Lemma 6.1.2, there is an 𝑁 ∈ OpenNormal(𝐺) such that Γ1 1 ∪ Γ2 2 6 ⊆ Γ𝑁 for all 𝑔1 , 𝑔2 ∈ 𝐺 and all Γ ∈ G. In particular, Γ1 𝑁 ≠ Γ2 𝑁, otherwise Γ1 ∪ Γ2 ⊆ Γ1 𝑁. Taking 𝑁 sufficiently small we may assume that 𝑝 𝑖 |#(Γ𝑖 𝑁/𝑁), for 𝑖 = 1, 2. Let (𝛼 : 𝐺 → 𝐵, 𝜑 : 𝐵 → 𝐴, B) be the 𝑁-standard G-embedding problem Î𝑛for 𝐺 (Remark 5.3.2 with its notation: 𝐴 = 𝐺/𝑁, 𝜑(G) = { 𝐴1 , . . . , 𝐴𝑛 }, 𝐵 = 𝐹∗ ∗ 𝑖=1 𝐵𝑖 , and B = Con(𝐵1 , . . . , 𝐵𝑛 )). Let 𝛾 : 𝐺 → 𝐵 be a strong solution of this problem. Since Γ1 𝑁 ≠ Γ2 𝑁, we have 𝜑(Γ1 ) ≠ 𝜑(Γ2 ). Hence, we may renumber 𝐴1 , . . . , 𝐴𝑛 to assume that 𝜑(Γ𝑖 ) = Γ𝑖 𝑁/𝑁 = 𝐴𝑖 for = 1, 2. Then, (6.2)

𝐴1𝑎1 ∪ 𝐴2𝑎2 6 ⊆ 𝐴 𝑗 for all 𝑎 1 , 𝑎 2 ∈ 𝐴 and all 𝑗 ∈ {1, . . . , 𝑛}.

Let 1 ≤ 𝑖 ≤ 2. Since 𝛾(G) ⊆ B, there exist 𝑏 𝑖 ∈ 𝐵 and 1 ≤ 𝑗 (𝑖) ≤ 𝑛 such that 𝑖 𝛾(Γ𝑖 ) ≤ 𝐵 𝑏𝑗 𝑖(𝑖) . Apply 𝛼 to get 𝐴𝑖 = 𝜑(Γ𝑖 ) = 𝛼(𝛾(Γ𝑖 )) ≤ 𝛼(𝐵 𝑏𝑗 𝑖(𝑖) ) = 𝐴 𝑎𝑗 (𝑖) , where 𝑎−1

𝑎 𝑖 = 𝛼(𝑏 𝑖 ). Thus, 𝐴𝑖 𝑖 ≤ 𝐴 𝑗 (𝑖) . By (6.2), 𝑗 (1) ≠ 𝑗 (2). Moreover, by the choice of 𝑁 we have 𝑝 𝑖 |#𝐴𝑖 . Since 𝐴𝑖 = 𝜑(Γ𝑖 ) = 𝛼(𝛾(Γ𝑖 )), also 𝑝 𝑖 |#𝛾(Γ𝑖 ). Since 𝛾(Γ𝑖 ) ≤ 𝐵 𝑏𝑗 𝑖(𝑖) , also 𝑝 𝑖 |#(𝐵 𝑏𝑗 𝑖(𝑖) ∩ 𝛾(𝐺)). Since 𝐵 = 𝐵 𝑗 (1) ∗ 𝐵 𝑗 (2) ∗ 𝐵 0, for some 𝐵 0, it follows from Lemma 6.1.1 that 𝛾(𝐺) is non-prosolvable. Hence, also 𝐺 is non-prosolvable, as claimed.  Lemma 6.1.4 Let 𝐺 be a prosolvable group and G a 𝐺-invariant étale compact family of closed subgroups of 𝐺. Suppose that 𝐺 is strongly G-projective. Then, one of the following conditions holds: (a) There exists a prime number 𝑙 such that all Γ ∈ Gmax are pro-𝑙 groups. (b) All Γ ∈ Gmax are finite and 𝐺-conjugate. (c) 𝐺 ∈ Gmax . Proof By Lemma 1.3.6 and Lemma 5.5.1 we may assume that G = Gmax . Let G0 be a system of representatives for the 𝐺-conjugacy classes of the groups in G. We distinguish between two cases: Case A: card(G0 ) ≥ 2. Since every Γ ∈ G0 is maximal, 1 ∉ G0 . By Lemma 6.1.3, for all distinct Γ1 , Γ2 ∈ G0 there are no distinct prime numbers 𝑝 1 , 𝑝 2 such that 𝑝 1 |#Γ1 and 𝑝 2 |#Γ2 ; i.e., there is a prime number 𝑙 such that both Γ1 and Γ2 are pro-𝑙 groups. Since both Γ1 and Γ2 are non-trivial, there is a prime number 𝑙 such that all Γ ∈ G0 are pro-𝑙 groups. Hence, also their conjugates are pro-𝑙 groups, so (a) holds. Case B: card(G0 ) = 1. Choose Γ ∈ G. Then, all groups in G are isomorphic to Γ. We may assume that they are infinite, otherwise (b) holds. If Γ = 𝐺, then (c) holds. Otherwise, there is an open subgroup 𝐻 of 𝐺 such that Γ ≤ 𝐻 ≠ 𝐺. Set H = {𝐻 ∩ Γ0 | Γ0 ∈ G} ⊆ Subgr(𝐻). By Remark 1.2.1(c), H is étale compact.

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Since G is 𝐺-invariant, H is 𝐻-invariant. Since Γ ∈ H and Γ ∈ G is maximal, we have Γ ∈ Hmax . Claim: There is a Δ ∈ Hmax not 𝐻-conjugate to Γ. Indeed, choose 𝑔 ∈ 𝐺 r 𝐻. Then, 𝐻 ∩ Γ𝑔 ∈ H . Since H is étale compact, Lemma 1.3.5 ensures the existence of Δ ∈ Hmax with 𝐻 ∩ Γ𝑔 ≤ Δ. Let ℎ ∈ 𝐻. Then, 𝑔ℎ−1 ∉ 𝐻, in particular, by Proposition 5.5.3(b), 𝑔ℎ−1 ∉ Γ = 𝑁𝐺 (Γ). Hence, −1 Γ𝑔ℎ ≠ Γ, so Γ𝑔 ≠ Γℎ . By Proposition 5.5.3(a), Γ𝑔 ∩ Γℎ = 1. Since Γ𝑔 is not finite, 1 ≠ 𝐻 ∩ Γ𝑔 ≤ Γ𝑔 . Hence, 𝐻 ∩ Γ𝑔 6 ≤ Γℎ . Since 𝐻 ∩ Γ𝑔 ≤ Δ, the latter relation implies that Δ 6 ≤ Γℎ , in particular, Δ ≠ Γℎ . This proves the Claim. By Proposition 5.4.2, 𝐻 is strongly H -projective. Since 𝐺 is prosolvable, so is 𝐻. Hence, Case A can be applied, with (𝐻, H ) replacing (𝐺, G). Thus, one of the conditions (a), (b), (c) holds for (𝐻, H ). By the Claim this condition cannot be (b). Also, (c) is impossible. Indeed, if 𝐻 ∈ Hmax , then Hmax = {𝐻}, in contrast to the Claim. It follows that Γ ∈ H is pro-𝑙 for some prime number 𝑙. Since every group in G is isomorphic to Γ, Condition (a) holds for the pair (𝐺, G). This concludes the proof of the lemma.  Proposition 6.1.5 Let 𝐺 be a profinite group and G a 𝐺-invariant étale compact subset of Subgr(𝐺). Suppose that 𝐺 is strongly G-projective. Let 𝐻 be a prosolvable closed subgroup of 𝐺 and set H = {𝐻 ∩ Γ | Γ ∈ G}. Then, one of the following conditions holds: (a) There exists a prime number 𝑙 such that all Δ ∈ Hmax are pro-𝑙 groups. (b) All Δ ∈ Hmax are finite and 𝐻-conjugate. (c) 𝐻 ∈ Hmax . Proof By Remark 1.2.1(c), H is étale compact. Since G is 𝐺-invariant, H is 𝐻-invariant. By Proposition 5.4.2, 𝐻 is strongly H -projective. Hence, by Lemma 6.1.4, one of the Conditions (a), (b), or (c) holds.  Î Corollary 6.1.6 Let 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 be a free product of profinite groups over a profinite space 𝑇 and set G = {𝐺 𝑡𝜎 | 𝑡 ∈ 𝑇, 𝜎 ∈ 𝐺}. Let 𝐻 be a prosolvable closed subgroup of 𝐺 and set H = {𝐻 ∩ Γ | Γ ∈ G}. Then, one of the following conditions holds: (a) There exists a prime number 𝑙 such that all Δ ∈ Hmax are pro-𝑙 groups. (b) All Δ ∈ Hmax are finite and 𝐻-conjugate. (c) 𝐻 ∈ Hmax . Proof Let G0 = {𝐺 𝑡 }𝑡 ∈𝑇 . By Definition 4.1.1 and Lemma 2.1.2, G0 is étale compact. By Proposition 5.2.2, 𝐺 is strongly G0 -projective. Hence, by Lemma 5.2.4, 𝐺 is strongly G-projective. Thus, our corollary is a special case of Proposition 6.1.5.  Using cohomology theory we get another characterization of prosolvable subgroups of strongly projective groups. (Actually, we use cohomology only as a notation.) Remark 6.1.7 Let 𝑝 be a prime number. Recall that the cohomological 𝑝-dimension cd 𝑝 (𝐹) of a profinite group 𝐹 is the smallest number 𝑛 such that 𝐻 𝑖 (𝐺, 𝐴) 𝑝 = 0 for all 𝑖 ≥ 𝑛 + 1 and all torsion 𝐺-modules 𝐴 [Rib70, p. 196, Def. 1.1].

6.2 Groups of Local p-Type

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Let 𝐹 𝑝 be a Sylow 𝑝-subgroup of 𝐹. By [Rib70, p. 207, Cor. 2.2] and [Rib70, p. 211, Prop. 3.1] cd 𝑝 (𝐹) ≥ 2 if and only if there is an epimorphism 𝛼 : 𝐵 → 𝐴 of finite groups and a homomorphism 𝜑 : 𝐹 𝑝 → 𝐴 such that Ker(𝛼) is a 𝑝-group and no homomorphism 𝛾 : 𝐹 𝑝 → 𝐵 satisfies 𝛼 ◦ 𝛾 = 𝜑. Proposition 6.1.8 Let 𝐺 be a strongly G-projective profinite group, where G is a 𝐺-invariant étale compact subset of Subgr(𝐺), and 𝐹 a closed prosolvable subgroup of 𝐺. Suppose that there exist distinct prime numbers 𝑝, 𝑝 0 such that (a) cd 𝑝 (𝐹), cd 𝑝0 (𝐹) ≥ 2, and (b) for no Γ ∈ G does 𝐹 ∩ Γ contain an element of order 𝑝. Then, there exists a Γ ∈ G such that 𝐹 ≤ Γ. Proof We choose a 𝑝-Sylow subgroup 𝐹 𝑝 of 𝐹 and set F𝑝 = {𝐹 𝑝 ∩ Γ | Γ ∈ G}. By Proposition 5.4.2, 𝐹 𝑝 is strongly F𝑝 -projective. Since cd 𝑝 (𝐹) ≥ 2, there is an epimorphism 𝛼 : 𝐵 → 𝐴 of finite groups and a homomorphism 𝜑 : 𝐹 𝑝 → 𝐴 such that there is no homomorphism 𝛾 : 𝐹 𝑝 → 𝐵 with 𝛼 ◦ 𝛾 = 𝜑. Since 𝐹 𝑝 is strongly F𝑝 -projective, this means that (𝜑, 𝛼, Subgr(𝐵)) is not an F𝑝 -embedding problem (condition (5.2c)). Thus, there is a Γ ∈ G such that for Φ 𝑝 := 𝐹 𝑝 ∩ Γ ∈ F𝑝 there is no homomorphism 𝛾 0 : Φ 𝑝 → 𝐵 with 𝛼 ◦ 𝛾 0 = 𝜑|Φ 𝑝 . In particular, Φ 𝑝 ≠ 1, so 𝑝|#Φ 𝑝 . Since Φ 𝑝 = 𝐹 𝑝 ∩ Γ ≤ 𝐹 ∩ Γ, also 𝑝|#(𝐹 ∩ Γ). By (b), 𝐹 ∩ Γ is infinite. Similarly, there is a Γ0 ∈ G such that 𝑝 0 |#(𝐹 ∩ Γ0). Now let F = {𝐹 ∩ Γ | Γ ∈ G}. Again, by Proposition 5.4.2, 𝐹 is strongly F -projective. By the preceding two paragraphs, there exists no prime number 𝑙 such that all Φ ∈ F are pro-𝑙 groups. As remarked earlier, not all Φ ∈ F are finite. Thus, statements (a) and (b) of Proposition 6.1.5 do not hold. Hence, statement (c) of Proposition 6.1.5 holds. In other words, 𝐹 ∈ F . Thus, there exists a Γ ∈ G such that 𝐹 = 𝐹 ∩ Γ ≤ Γ, as claimed. 

6.2 Groups of Local p-Type We prove that the absolute Galois group of a non-archimedean local field is of “local 𝑝-type” for every prime number 𝑝 different from the residue characteristic. Our treatment of this notion replaces the cohomological approach of [Pop95, Sec. 3] by an explicit use of the group of 3 × 3 unitriangular matrices over F 𝑝 (Notation 6.2.4). Given a profinite group 𝐺 and a prime number 𝑝, we set 𝐺 𝑝 for a 𝑝-Sylow group of 𝐺. Recall that 𝐺 𝑝 is unique up to conjugation. This implies that the statements below do not depend on the choice of 𝐺 𝑝 . Definition 6.2.1 (Local 𝑝-type) Let 𝑝 be a prime number. We fix an epimorphism 𝛼 𝑝 : 𝐵 𝑝 → 𝐴 𝑝 of finite 𝑝-groups. A profinite group 𝐺 is of local 𝑝-type (with respect to the epimorphism 𝛼 𝑝 : 𝐵 𝑝 → 𝐴 𝑝 ) if (a) there exists an epimorphism 𝐺 𝑝 → 𝐴 𝑝 ; and (b) there exists no pair (𝜑, 𝛾) comprising an epimorphism 𝜑 : 𝐺 𝑝 → 𝐴 𝑝 and a homomorphism 𝛾 : 𝐺 𝑝 → 𝐵 𝑝 such that 𝛼 𝑝 ◦ 𝛾 = 𝜑.

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We specify the epimorphism 𝛼 𝑝 : 𝐵 𝑝 → 𝐴 𝑝 in Setup 6.2.6 below. Remark 6.2.2 Let 𝐺 be of local 𝑝-type. Then, cd 𝑝 (𝐺) ≥ 2 (Remark 6.1.7). Lemma 6.2.3 Let 𝑝 be a prime number and 𝐺 a profinite group. Then, there is an open 𝑁0 ⊳ 𝐺 such that if 𝑁 ≤ 𝑁0 is normal in 𝐺, then 𝐺¯ = 𝐺/𝑁 satisfies (a) If there is an epimorphism 𝐺 𝑝 → 𝐴 𝑝 , then there is an epimorphism 𝐺¯ 𝑝 → 𝐴 𝑝 . (b) If 𝐺 is of local 𝑝-type, then 𝐺¯ is of local 𝑝-type. (c) Assume that 𝐺 = lim 𝐺 𝑖 , where (𝐺 𝑖 , 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 ,𝑖 ≤ 𝑗 is an inverse system of profinite ←−− groups with 𝜋 𝑗𝑖 : 𝐺 𝑗 → 𝐺 𝑖 surjective for all 𝑗 ≥ 𝑖. If the 𝐺 𝑖 are of local 𝑝-type, then so is 𝐺. Proof Proof of (a). Let 𝜑 : 𝐺 𝑝 → 𝐴 𝑝 be an epimorphism. Its kernel is open in 𝐺 𝑝 , hence there is an open 𝑁0 ⊳ 𝐺 such that 𝑁0 ∩ 𝐺 𝑝 ≤ Ker(𝜑). Let 𝑁 ⊳ 𝐺 be contained in 𝑁0 and put 𝐺¯ = 𝐺/𝑁. The quotient map 𝜋 : 𝐺 → 𝐺¯ maps 𝐺 𝑝 onto ¯ Thus, the kernel of its restriction 𝜋 𝑝 : 𝐺 𝑝 → 𝐺¯ 𝑝 to a Sylow 𝑝-subgroup 𝐺¯ 𝑝 of 𝐺. 𝐺 𝑝 is 𝑁 ∩ 𝐺 𝑝 ≤ 𝑁0 ∩ 𝐺 𝑝 ≤ Ker(𝜑). The first isomorphism theorem provides an epimorphism 𝜑¯ : 𝐺¯ 𝑝 → 𝐴 𝑝 such that 𝜑¯ ◦ 𝜋 𝑝 = 𝜑. Proof of (b). By (a) there is an epimorphism 𝜑¯ : 𝐺¯ 𝑝 → 𝐴 𝑝 . Then, 𝜑 = 𝜑¯ ◦ 𝜋 𝑝 is an epimorphism 𝐺 𝑝 → 𝐴 𝑝 . In addition, there is no homomorphism 𝛾¯ : 𝐺¯ → 𝐵 𝑝 such that 𝛼 𝑝 ◦ 𝛾¯ = 𝜑, ¯ otherwise 𝛾 = 𝛾¯ ◦ 𝜋 𝑝 would be a homomorphism 𝐺 𝑝 → 𝐵 𝑝 such that 𝛼 𝑝 ◦ 𝛾 = 𝜑, a contradiction to Definition 6.2.1(b). Proof of (c). For every 𝑖 ∈ 𝐼 let 𝜋𝑖 : 𝐺 → 𝐺 𝑖 be the projection of 𝐺 onto 𝐺 𝑖 . Since the 𝜋 𝑗𝑖 ’s are surjective, so are the 𝜋𝑖 ’s. Let 𝐺 𝑝 be a 𝑝-Sylow subgroup of 𝐺. Then, 𝐺 𝑖, 𝑝 = 𝜋𝑖 (𝐺 𝑝 ) is a 𝑝-Sylow subgroup of 𝐺 𝑖 and the restrictions 𝜋𝑖, 𝑝 : 𝐺 𝑝 → 𝐺 𝑖, 𝑝 of 𝜋𝑖 to 𝐺 𝑝 and 𝜋 𝑗,𝑖, 𝑝 : 𝐺 𝑗, 𝑝 → 𝐺 𝑖, 𝑝 of 𝜋 𝑗𝑖 to 𝐺 𝑗, 𝑝 , for every 𝑗 ≥ 𝑖, are all surjective. Fix 𝑖 ∈ 𝐼. Since 𝐺 𝑖 is of local 𝑝-type (Definition 6.2.1), there exists an epimorphism 𝜑𝑖 : 𝐺 𝑖, 𝑝 → 𝐴 𝑝 . Hence, 𝜑 = 𝜑𝑖 ◦ 𝜋𝑖, 𝑝 is an epimorphism of 𝐺 𝑝 onto 𝐴𝑝. Let 𝜑 : 𝐺 𝑝 → 𝐴 𝑝 be an epimorphism. Suppose there is a homomorphism 𝛾 : 𝐺 𝑝 → 𝐵 𝑝 such that 𝛼 𝑝 ◦ 𝛾 = 𝜑. Then, there is a 𝑘 ∈ 𝐼 such that Ker(𝜋 𝑘, 𝑝 ) ≤ Ker(𝛾) ≤ Ker(𝜑). The first isomorphism theorem provides a homomorphism 𝛾 𝑘 : 𝐺 𝑘, 𝑝 → 𝐵 𝑝 such that 𝛾 = 𝛾 𝑘 ◦ 𝜋 𝑘, 𝑝 , and an epimorphism 𝜑 𝑘 : 𝐺 𝑘, 𝑝 → 𝐴 𝑝 such that 𝜑 = 𝜑 𝑘 ◦ 𝜋 𝑘, 𝑝 . Since 𝛼 𝑝 ◦ 𝛾 = 𝜑 and 𝜋 𝑘, 𝑝 is surjective, we have 𝛼 𝑝 ◦ 𝛾 𝑘 = 𝜑 𝑘 , a contradiction.  We introduce a group 𝑈 𝑝 needed to specify 𝐴 𝑝 , 𝐵 𝑝 , and 𝛼 𝑝 . Notation 6.2.4 We write 𝑍 𝑛 for the additive cyclic group Z/𝑛Z of order 𝑛. For a prime number 𝑝 we consider the group 𝑀3 (F 𝑝 ) of 3 × 3 matrices with entries in F 𝑝 . Let 𝑈 𝑝 be the group of unitriangular matrices in 𝑀3 (F 𝑝 ), that is   1𝑎𝑐       © ª 𝑈 𝑝 = 𝑀 (𝑎, 𝑏, 𝑐) := ­0 1 𝑏 ® ∈ M3 (F 𝑝 ) | 𝑎, 𝑏, 𝑐 ∈ F 𝑝 .     «0 0 1¬   3 It is a group of order 𝑝 . In particular, one can show that 𝑈2 is the dihedral group of order 8. Let 𝛼 0𝑝 : 𝑈 𝑝 → 𝑍 𝑝 ⊕ 𝑍 𝑝 be the map 𝑀 (𝑎, 𝑏, 𝑐) ↦→ (𝑎, 𝑏).

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95

We list some elementary properties of 𝑈 𝑝 . Lemma 6.2.5 (a) 𝛼 0𝑝 is an epimorphism. (b) 𝛼 0𝑝 maps no abelian subgroup of 𝑈 𝑝 onto 𝑍 𝑝 ⊕ 𝑍 𝑝 .
(c) 𝑀 (𝑎, 𝑏, 𝑐) 𝑘 = 𝑀 (𝑘𝑎, 𝑘 𝑏, 𝑘𝑐 + 𝑘2 𝑎𝑏) for all 𝑎, 𝑏, 𝑐 ∈ F 𝑝 and 𝑘 ∈ N. (d) If 𝑝 ≠ 2, then every 1 ≠ 𝑀 ∈ 𝑈 𝑝 is of order 𝑝. Every 1 ≠ 𝑀 ∈ 𝑈2 is of order 2, except 𝑀 (1, 1, 0) and 𝑀 (1, 1, 1), which are of order 4. Proof Proof of (a). Clear. Proof of (b). Let 𝐷 ≤ 𝑈 𝑝 such that 𝛼 0𝑝 (𝐷) = 𝑍 𝑝 ⊕ 𝑍 𝑝 . Then, one can lift (1, 0), (0, 1) ∈ 𝑍 𝑝 ⊕ 𝑍 𝑝 to elements of 𝑈 𝑝 , say, 𝑀 (1, 0, 𝑐), 𝑀 (0, 1, 𝑐 0) for some 𝑐, 𝑐 0 ∈ F 𝑝 . But, 𝑀 (1, 0, 𝑐) 𝑀 (0, 1, 𝑐 0) = 𝑀 (1, 1, 1 + 𝑐 + 𝑐 0) ≠ 𝑀 (1, 1, 𝑐 + 𝑐 0) = 𝑀 (0, 1, 𝑐 0)𝑀 (1, 0, 𝑐), so 𝐷 is not abelian. Proof of (c). Multiplication of matrices yields 𝑀 (𝑎, 𝑏, 𝑐) 𝑀 (𝑎 0, 𝑏 0, 𝑐 0) = 𝑀 (𝑎 + 𝑎 0, 𝑏 + 𝑏 0, 𝑎𝑏 0 + 𝑐 + 𝑐 0)
for all 𝑎, 𝑏, 𝑐 ∈ F 𝑝 . In particular, (c) holds for 𝑘 = 1, because 12 = 0. Proceeding by induction, assume that (c) holds for some 𝑘 ≥ 2. Then, 𝑀 (𝑎, 𝑏, 𝑐) 𝑘+1 = 𝑀 (𝑎, 𝑏, 𝑐) 𝑘 𝑀 (𝑎, 𝑏, 𝑐)   𝑘 = 𝑀 (𝑘𝑎, 𝑘 𝑏, 𝑘𝑐 + 𝑎𝑏) 𝑀 (𝑎, 𝑏, 𝑐) 2   𝑘 = 𝑀 (𝑘𝑎 + 𝑎, 𝑘 𝑏 + 𝑏, 𝑘𝑐 + 𝑎𝑏 + 𝑘𝑎𝑏 + 𝑐) 2   𝑘 +1 = 𝑀 ((𝑘 + 1)𝑎, (𝑘 + 1)𝑏, (𝑘 + 1)𝑐 + 𝑎𝑏). 2 Hence, (c) holds for all 𝑘 ≥ 1.
Proof of (d). Let 𝑀 = 𝑀 (𝑎, 𝑏, 𝑐). If 𝑝 ≠ 2, then 𝑝 | 𝑝2 . Hence, by (c), 𝑀 (𝑎, 𝑏, 𝑐) 𝑝 = 𝑀 (0, 0, 0) = 1. This is also true if 𝑝 = 2 and either 𝑎 = 0 or 𝑏 = 0, because 𝑀 (𝑎, 𝑏, 𝑐) 2 = 𝑀 (2𝑎, 2𝑏, 2𝑐 + 𝑎𝑏). Moreover, if 𝑝 = 2, then, by (c), 𝑀 (1, 1,
𝑐) 4 = 𝑀 (4𝑎, 4𝑏, 4𝑐 + 6𝑎𝑏) = 𝑀 (0, 0, 0) = 1. But, 𝑀 (1, 1, 𝑐) 2 = 𝑀 (2, 2, 2𝑐 + 22 1 · 1) = 𝑀 (0, 0, 1) ≠ 1. Hence, 𝑀 (1, 1, 𝑐) is of order 4, for 𝑐 = 0, 1. We fix 𝐴 𝑝 , 𝐵 𝑝 , and 𝛼 𝑝 for the rest of this monograph: Setup 6.2.6 We set 𝐴 𝑝 = 𝑍 𝑝 ⊕ 𝑍 𝑝 . It is a group of order 𝑝 2 . If 𝑝 ≠ 2, let 𝐵 𝑝 = 𝑈 𝑝 and 𝛼 𝑝 = 𝛼 0𝑝 . If 𝑝 = 2, let 𝛼200 : 𝑍4 ⊕ 𝑍2 → 𝐴2 be the epimorphism defined by 𝛼200 (1, 0) = (1, 1) and 𝛼200 (0, 1) = (1, 0). Put 𝐵2 = 𝑈2 × 𝐴2 (𝑍4 ⊕ 𝑍2 ) = {(𝑏 0, 𝑏 00) ∈ 𝑈2 × (𝑍4 ⊕ 𝑍2 ) | 𝛼20 (𝑏 0) = 𝛼200 (𝑏 00)}

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and let 𝛼2 : 𝐵2 → 𝐴2 be the epimorphism (𝑏 0, 𝑏 00) ↦→ 𝛼20 (𝑏 0) = 𝛼200 (𝑏 00). Furthermore, let 𝜋 0 : 𝐵2 → 𝑈2 and 𝜋 00 : 𝐵2 → 𝑍4 ⊕ 𝑍2 be the coordinate projections. 𝜋 00

𝐵2 𝜋0

 𝑈2

/ 𝑍4 ⊕ 𝑍2

𝛼2

𝛼20

𝛼200

#  / 𝐴2

6.3 Fields of Local p-Type Proposition 6.3.5 below gives sufficient conditions for a prime number 𝑝 and the absolute Galois group of a Henselian field to be of local 𝑝-type. Remark 6.3.1 (Valuation rings) Consider a valuation 𝑣 of a field 𝐾, let 𝑂 𝑣 = {𝑥 ∈ 𝐾 | 𝑣(𝑥) ≥ 0} be the valuation ring of 𝑣, let 𝔪 𝑣 = {𝑥 ∈ 𝐾 | 𝑣(𝑥) > 0} ¯ if 𝑣 is clear from the context) be the unique maximal ideal of 𝑂 𝑣 , and let 𝐾¯ 𝑣 (or 𝐾, be the residue field 𝑂 𝑣 /𝔪 𝑣 of 𝑣 [Efr06, Sections 3.1 and 3.2]. Now let 𝐿 be a Galois extension of 𝐾 and consider an extension 𝑤 of 𝑣 to 𝐿. Then, the decomposition group of 𝑤/𝑣 is 𝜎 𝐷 𝑤/𝑣 = {𝜎 ∈ Gal(𝐿/𝐾) | 𝑂 𝑤 = 𝑂 𝑤 }.

We denote the fixed field of 𝐷 𝑤/𝑣 in 𝐿 by 𝐿 𝑤/𝑣 and call 𝐿 𝑤/𝑣 the decomposition field of 𝑤/𝑣. By definition, 𝑤 is the unique valuation of 𝐿 that lies over the restriction of 𝑤 to 𝐿 𝑤/𝑣 [Efr06, p. 134, Prop. 15.1.2(b)]. The residue field of 𝐿 𝑤/𝑣 with respect to 𝑤 is equal to 𝐾¯ 𝑣 [Efr06, p. 134, Thm. 15.2.2]. Moreover, reduction modulo 𝔪 𝑤 maps the valuation ring of 𝐿 𝑤/𝑣 onto 𝐾¯ 𝑣 . If 𝐿¯ 𝑤 /𝐾¯ 𝑣 is separable, then 𝐿¯ 𝑤 /𝐾¯ 𝑣 is a Galois extension and reduction modulo 𝔪 𝑤 gives rise to an epimorphism Gal(𝐿/𝐾) → Gal( 𝐿¯ 𝑤 /𝐾¯ 𝑣 ). The latter is an isomorphism if the extension 𝑤/𝑣 is unramified, that is, 𝔪 𝑤 = 𝔪 𝑣 𝑂 𝑤 . If 𝑤 0 is another extension of 𝑣 to 𝐿, then there exists a 𝜌 ∈ Gal(𝐿/𝐾) such that −1 𝜌 𝑤 0 (𝑥) = 𝑤(𝑥 𝜌 ) for each 𝑥 ∈ 𝑂 𝑤 0 [Efr06, p. 134, Prop. 15.1.2(a)]. Thus, 𝑂 𝑤 0 = 𝑂 𝑤 . Remark 6.3.2 (Henselian fields) The special case where the field 𝐿 in Remark 6.3.1 is the separable algebraic closure 𝐾sep of 𝐾 plays a central role in the rest of this monograph. In this case the decomposition field of 𝑤/𝑣 is a Henselization of 𝐾 with respect to 𝑣 [Efr06, p. 139, Examples 15.3.8(1)]. We denote this field by 𝐾 𝑣 and rename both 𝑤 and 𝑤|𝐾𝑣 as 𝑣. By Remark 6.3.1, 𝑣 has a unique extension from 𝐾 𝑣 to 𝐾sep , hence a unique extension to every separable algebraic extension of 𝐾 𝑣 . Then, (𝐾 𝑣 , 𝑣) is a Henselian field. Thus, among other things, (𝐾 𝑣 , 𝑣) satisfies the Hensel–Rychlik lemma [Efr06, p. 162, (f)]: Let 𝑓 ∈ 𝐾 𝑣 [𝑋] be a polynomial and 𝑎 an element of 𝑂 𝑣 such that 𝑣( 𝑓 (𝑎)) > 2𝑣( 𝑓 0 (𝑎)), where 𝑓 0 is the derivative of 𝑓 . Then, there exists a unique element 𝑥 ∈ 𝑂 𝑣 such that 𝑓 (𝑥) = 0 and 𝑣(𝑥 − 𝑎) > 𝑣( 𝑓 0 (𝑎)). Finally, note that (𝐾 𝑣 , 𝑣) is unique up to conjugation by elements of Gal(𝐾) (Remark 6.3.1).

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Remark 6.3.3 (Ostrowski’s theorem) Let (𝐾, 𝑣) be a Henselian field and 𝐿 be a finite extension of 𝐾. By a theorem of Ostrowski, [𝐿 : 𝐾] = 𝑑 (𝐿/𝐾)𝑒(𝐿/𝐾) 𝑓 (𝐿/𝐾), ¯ if char( 𝐾) ¯ > 0 and 1 where 𝑑 (𝐿/𝐾) is the defect of 𝐿/𝐾. It is a power of char( 𝐾) × × otherwise. Moreover, 𝑒(𝐿/𝐾) = (𝑣(𝐿 ) : 𝑣(𝐾 )) is the ramification index of 𝐿/𝐾 ¯ is the residue degree of 𝐿/𝐾. By [Art67, p. 62, Thm. 10] or and 𝑓 (𝐿/𝐾) = [ 𝐿¯ : 𝐾] [Efr06, p. 154, Thm. 17.2.1 and the first paragraph of p. 155], each of the factors on the right-hand side of (6.3) is an integer. (6.3)

Lemma 6.3.4 Let (𝐾, 𝑣) be a Henselian field, 𝑝 a prime number, and 𝐾 𝑝 the fixed field in 𝐾sep of a 𝑝-Sylow subgroup of Gal(𝐾). Suppose that the group 𝑣(𝐾 ×𝑝 ) is 𝑝-divisible. Then, so is 𝑣(𝐾 × ). Proof Consider 𝑥 ∈ 𝐾 × . By assumption, there exists a 𝑦 ∈ 𝐾 ×𝑝 with 𝑣(𝑥) = 𝑝𝑣(𝑦). Set 𝐿 = 𝐾 (𝑦). Since Gal(𝐾sep /𝐾 𝑝 ) is a 𝑝-Sylow group of Gal(𝐾), we have 𝑝 - [𝐿 : 𝐾]. Hence, by (6.3), 𝑝 - (𝑣(𝐿 × ) : 𝑣(𝐾 × )). Since 𝑣(𝑦) ∈ 𝑣(𝐿 × ) and 𝑝𝑣(𝑦) = 𝑣(𝑥) ∈ 𝑣(𝐾 × ), we see that 𝑣(𝑦) ∈ 𝑣(𝐾 × ). Thus, 𝑣(𝑥) = 𝑝𝑣(𝑦) ∈ 𝑝𝑣(𝐾 × ). This implies that 𝑣(𝐾 × ) is 𝑝-divisible, as claimed.  ¯ is an algebraic Proposition 6.3.5 Let (𝐾, 𝑣) be a valued field whose residue field, 𝐾, ¯ be a prime number such that 𝑝|# Gal( 𝐾) ¯ extension of a finite field. Let 𝑝 ≠ char( 𝐾) and 𝑣(𝐾 × ) is not 𝑝-divisible. Let 𝐺 be the decomposition group of an extension of 𝑣 to 𝐾sep . Then, (a) 𝐺 𝑝  𝑍 n 𝑇 where 𝑇 = Z𝑚 𝑝 with 𝑚 ≥ 1 and 𝑍 = Z 𝑝 = h𝜎i, and there is an 𝑟 ∈ 1 + 𝑝Z 𝑝 ≤ Z×𝑝 such that 𝜏 𝜎 = 𝜏𝑟 , for every 𝜏 ∈ 𝑇. (b) 𝐺 is a group of local 𝑝-type (Definition 6.2.1) with respect to the epimorphism 𝛼 𝑝 : 𝐵 𝑝 → 𝐴 𝑝 introduced in Setup 6.2.6. Proof Proof of (a). The proof naturally breaks up into three parts. Part A: Reduction to a Henselian field. The fixed field 𝐿 of 𝐺 is the Henselian closure of (𝐾, 𝑣) [Efr06, p. 137, Theorem 15.3.5]. Since 𝑣(𝐿 × ) = 𝑣(𝐾 × ) and 𝐿¯ = 𝐾¯ [Efr06, p. 134, Thm. 15.2.2], we may replace 𝐾 by 𝐿. Thus, we may assume that 𝐾 is Henselian and 𝐺 = Gal(𝐾). Part B: Reduction to a pro-𝑝 absolute Galois group. Under this assumption, let 𝐾 𝑝 be the fixed field in 𝐾sep of a 𝑝-Sylow subgroup of Gal(𝐾). Then, 𝐾 𝑝 is Henselian and by Lemma 6.3.4, 𝑣(𝐾 ×𝑝 ) is not 𝑝-divisible. ¯ | [𝐾 𝑝 : 𝐾] and [ 𝐾¯ sep : 𝐾 𝑝 ] | [𝐾sep : 𝐾 𝑝 ], we In addition, since by (6.3), [𝐾 𝑝 : 𝐾] ¯ By assumption 𝑝|# Gal( 𝐾). ¯ have that Gal(𝐾 𝑝 ) is the 𝑝-Sylow subgroup of Gal( 𝐾). Hence, 𝑝|# Gal(𝐾 𝑝 ). It follows that we may replace 𝐾 by 𝐾 𝑝 to assume that 𝐺 = Gal(𝐾) = 𝐺 𝑝 and ¯ is a non-trivial pro-𝑝 group. Gal( 𝐾) Part C: End of the proof of (a). Since 𝐾¯ is an algebraic extension of a finite field, ¯ ≤ Zˆ [FrJ08, p. 15, Section 1.5]. Since Z 𝑝 is the 𝑝-Sylow subgroup of Zˆ Gal( 𝐾) ¯ ≤ Z 𝑝 . Since Gal( 𝐾) ¯ is non-trivial, we [FrJ08, p. 15, Lemma 1.4.5], we have Gal( 𝐾) ¯ have by [FrJ08, p. 13, Lemma 1.4.2(e)] that Gal( 𝐾)  Z 𝑝 .

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6 Strong Proper Projectivity

Let 𝜁 𝑝 be a root of unity of order 𝑝. Then, [𝐾 (𝜁 𝑝 ) : 𝐾] divides 𝑝 − 1 and is a power of 𝑝, so [𝐾 (𝜁 𝑝 ) : 𝐾] = 1, hence, 𝜁 𝑝 ∈ 𝐾. By [Efr06, p. 202, Example 22.1.6, × × ¯ n 𝑇, where 𝑇  Z𝑚 (22.1.1)], 𝐺 = Gal( 𝐾) 𝑝 , with 𝑚 = dimF 𝑝 (𝑣(𝐾 )/𝑝𝑣(𝐾 )), × ¯ and 𝑚 ≥ 1, because 𝑣(𝐾 ) is not 𝑝-divisible. Moreover, the action of Gal( 𝐾) on ¯ and 𝜏 ∈ 𝑇, then 𝜏 𝜎 = 𝜏𝑟 , where 𝑇 is as follows: If 𝜎 is a generator of Gal( 𝐾) × 𝑟 ∈ 1 + 𝑝Z 𝑝 ≤ Z 𝑝 , as (a) claims. Proof of (b). There are epimorphisms 𝜑1 : 𝑇 → 𝑍 𝑝 and 𝜑2 : 𝑍 → 𝑍 𝑝 . Then, the above action of 𝑍 on 𝑇 ensures that the map 𝜑 : 𝐺 𝑝 → 𝐴 𝑝 , given by (𝜏, 𝜎 𝑘 ) ↦→ (𝜑1 (𝜏), 𝜑2 (𝜎 𝑘 )), is an epimorphism. The rest of the proof breaks up into two cases. In each case we will consider the epimorphism 𝛼 𝑝 : 𝐵 𝑝 → 𝐴 𝑝 and a homomorphism 𝛾 : 𝐺 𝑝 → 𝐵 𝑝 and prove that 𝛼 𝑝 ◦ 𝛾 ≠ 𝜑. This will imply that 𝐺 is of local 𝑝-type (Definition 6.2.1). Case A: 𝑝 ≠ 2. If 𝜏 ∈ 𝑇, then, by Lemma 6.2.5(d), 𝛾(𝜏) is of order 1 or 𝑝. Let 𝜎 and 𝑟 be as in (a). Then, since 𝑟 ≡ 1 (mod 𝑝), we have 𝛾(𝜏) 𝛾 ( 𝜎) = 𝛾(𝜏) 𝑟 = 𝛾(𝜏). Hence, 𝛾(𝐺 𝑝 ) = 𝛾(𝑇)𝛾(𝑍) is abelian. Therefore, by Lemma 6.2.5(b), 𝛼 𝑝 (𝛾(𝐺 𝑝 )) ≠ 𝐴 𝑝 . It follows that 𝛼 𝑝 ◦ 𝛾 ≠ 𝜑. Case B: 𝑝 = 2. Assume toward contradiction that 𝛼2 ◦ 𝛾 = 𝜑. Put 𝛾 0 = 𝜋 0 ◦ 𝛾 and 𝛾 00 = 𝜋 00 ◦ 𝛾. Then, 𝛼20 ◦ 𝛾 0 = 𝜑 and 𝛼200 ◦ 𝛾 00 = 𝜑 (Setup 6.2.6). 𝐺 2 = Z2𝑚 o Z2 = 𝑇 o h𝜎i 𝛾0

𝛾

{

𝛾00

𝜋 00

𝐵2 𝜋0

 𝑈2

 / 𝑍4 ⊕ 𝑍2

𝛼2 𝛼20

𝜑

𝛼200

#  | / 𝑍 2 ⊕ 𝑍 2 = 𝐴2

Since 𝛼20 (𝛾 0 (𝐺 2 )) = 𝜑(𝐺 2 ) = 𝑍2 ⊕ 𝑍2 and 𝛼20 maps no abelian subgroup of 𝑈2 onto 𝑍2 ⊕ 𝑍2 (Lemma 6.2.5(b)), we have that 𝛾 0 (𝐺 2 ) is non-abelian. Since 0 𝛾 0 (𝐺 2 ) = 𝛾 0 (𝑇) o h𝛾 0 (𝜎)i with 𝛾 0 (𝑇) ⊳ 𝛾 0 (𝐺 2 ), and 𝛾 0 (𝜏) 𝛾 ( 𝜎) = 𝛾 0 (𝜏) 𝑟 for every 0 𝜏 ∈ 𝑇, this means that there is a 𝜏 ∈ 𝑇 such that 𝛾 (𝜏) is of order 4 and 𝑟 ≡ −1 (mod 4). In particular, by Lemma 6.2.5(d), 𝛾 0 (𝜏) = 𝑀 (1, 1, 0) or 𝛾 0 (𝜏) = 𝑀 (1, 1, 1). In both cases 𝜑(𝜏) = 𝛼20 (𝛾 0 (𝜏)) = (1, 1). This implies that 𝛾 00 (𝜏), which belongs to (𝛼200) −1 (1, 1), is also of order 4. Hence, 00 𝛾 00 (𝜏) 𝛾 ( 𝜎) = 𝛾 00 (𝜏) −1 ≠ 𝛾 00 (𝜏). This contradicts 𝑍4 ⊕ 𝑍2 being abelian.  Corollary 6.3.6 Let (𝐾, 𝑣) be a Henselian discrete valued field with a finite residue field 𝐾¯ 𝑣 and let 𝑝 ≠ char( 𝐾¯ 𝑣 ) be a prime number. Then, Gal(𝐾) is of local 𝑝-type (Definition 6.2.1). Proof In the notation of Proposition 6.3.5, 𝐺 = Gal(𝐾). Further, 𝐾¯ 𝑣 is a finite field, so Gal( 𝐾¯ 𝑣 )  Zˆ [FrJ08, p. 15, Sec. 1.5], and 𝑣(𝐾 × )  Z. Thus, 𝑝|# Gal( 𝐾¯ 𝑣 ) and 𝑣(𝐾 × ) is not 𝑝-divisible for every prime number 𝑝. It follows from Proposition 6.3.5 that Gal(𝐾) is of local 𝑝-type. 

6.4 Groups of P-Type and G-Projectivity

99

6.4 Groups of P-Type and G-Projectivity Fix a set of primes P, a profinite group 𝐺, and a set G ⊆ Subgr(𝐺). Building on Definition 6.2.1, we introduce the notion “G is of P-type” and give an example related to local fields. The main result of this section is that if G is of P-type and 𝐺 is G-projective, then 𝐺 is strongly G-projective (Proposition 6.4.8). Lemma 6.4.1 The absolute Galois group of a field 𝐾 of positive characteristic is torsion free. Proof Otherwise, Gal(𝐾) has a non-trivial finite subgroup 𝐺. Let 𝐿 be its fixed field. By [Lan93, p. 264, Thm. 1.8], 1 < [ 𝐾˜ : 𝐿] = card(𝐺) < ∞. By [Lan93, p. 299, Cor. 9.3], char(𝐿) = 0, a contradiction.  Lemma 6.4.2 For every prime number 𝑝 each of the groups Gal(F 𝑝 ((𝑡))) and Gal(Q 𝑝 ) is torsion free. Proof By Lemma 6.4.1, Gal(F 𝑝 ((𝑡))) is torsion free, so we consider the group Gal(Q 𝑝 ). Note that 𝑎 𝑝 − 𝑎 − 1 = −1 for every 𝑎 ∈ F 𝑝 , so 𝑣 𝑝 (𝑥 𝑝 − 𝑥 − 1) = 0 for each 𝑥 ∈ Z 𝑝 . Hence, the polynomial 𝑋 𝑝 − 𝑋 − 1 has no roots in Q 𝑝 , so Q 𝑝 is not algebraically closed. Now consider the polynomial 𝑓 (𝑋) := 𝑋 2 + 𝑝 3 −1 and its derivative, 𝑓 0 (𝑋) = 2𝑋. They satisfy 𝑣 𝑝 ( 𝑓 (1)) = 3 > 2𝑣 𝑝 (2) = 2𝑣 𝑝 ( 𝑓 0 (1)). Hence, by Remark 6.3.2, there is an 𝑥 ∈ Q 𝑝 with 𝑥 2 + 𝑝 3 − 1 = 0. Thus, −𝑥 2 ≠ 1 and −𝑥 2 = 𝑝 3 − 1 = 12 + · · · + 12 (𝑝 3 − 1 times) is a sum of squares. By [Lan93, p. 451, first paragraph of §2], this implies that Q 𝑝 is not a real field. If Gal(Q 𝑝 ) has a non-unit element of finite order, then Q 𝑝 has an algebraic ˜ 𝑝 : 𝐹] < ∞. By [Lan93, p. 299, Cor. 9.3 and p. 452, extension 𝐹 such that 1 < [ Q Prop. 2.4], 𝐹 is a real closed field. Hence, Q 𝑝 is real, in contradiction to the conclusion of the preceding paragraph. We conclude that Gal(Q 𝑝 ) is torsion free, as claimed. Definition 6.4.3 Let P be a set of prime numbers. For each profinite group Γ and every prime number 𝑝 we choose a 𝑝-Sylow subgroup Γ 𝑝 of Γ. Let 𝐴 𝑝 (and 𝐵 𝑝 ) be the groups specified in Setup 6.2.6. We set P (Γ) = {𝑝 ∈ P | 𝐴 𝑝 is a quotient of Γ 𝑝 }, L (Γ) = {𝑝 ∈ P | Γ is of local 𝑝-type}. A set G of closed subgroups of a profinite group 𝐺 is said to be of P-type if there exist non-negative integers 𝑛0 , 𝑛1 , 𝑛∞ , and strictly closed subsets GS , G∞ of Subgr(𝐺) such that G = GS ∪ G∞ (in particular, G is strictly closed, hence also étale compact, by Remark 1.2.7) and the following conditions hold: (a) Every Γ ∈ G∞ is finite of order bounded by 𝑛∞ . (b) Every Γ ∈ GS is prosolvable with the following properties: (b1) card(P (Γ)) ≥ 𝑛0 + 𝑛1 + 2. (b2) All but at most 𝑛1 elements 𝑝 of P satisfy the following condition: If 𝑝 ∈ P (Γ), then 𝑝 ∈ L (Γ). (b3) There are at most 𝑛0 + 1 elements 𝑝 ∈ P such that Γ contains an element of order 𝑝.

100

6 Strong Proper Projectivity

Remark 6.4.4 Let Γ0 → Γ be an epimorphism of profinite groups. Then, it follows from Definitions 6.2.1 and 6.4.3 that: (a) P (Γ0) ⊇ P (Γ) and L (Γ0) ⊇ L (Γ). (b) P (Γ) ∩ L (Γ0) ⊆ L (Γ). Example 6.4.5 Let P be a set of prime numbers with card(P) ≥ 3 and let 𝐺 be a profinite group. We set (6.4)

𝑛0 = 0, 𝑛1 = 1, and 𝑛∞ = 2.

Let G∞ and GS be strictly closed subsets of Subgr(𝐺) such that each Γ ∈ G∞ is isomorphic to Gal(R) and every Γ ∈ GS is isomorphic to the absolute Galois group of a non-archimedean classically local field 𝐹. Thus, 𝐹 is a finite extension of F 𝑝 ((𝑡)) or of Q 𝑝 for some prime number 𝑝. Claim: G := GS ∪ G∞ is of P-type. We prove the Claim by verifying Conditions (a) and (b) of Definition 6.4.3. (a) The order of every Γ ∈ G∞ is 2, and 𝑛∞ = 2. Next, consider Γ ∈ GS . Then, Γ ∈ GS is the absolute Galois group of a nonarchimedean classically local field 𝐹. Hence, (b) By [CaF67, p. 31, last statement of Cor. 1], Γ is prosolvable. (b1) Let 𝑝 ∈ P. If 𝑝 is different from the residue characteristic of 𝐹, then, by Corollary 6.3.6, Gal(𝐹) is of local 𝑝-type (Definition 6.2.1). In particular, 𝐴 𝑝 is a quotient of Gal(𝐹) 𝑝 , so 𝑝 ∈ P (Γ) (Definition 6.4.3). If 𝑝 is equal to the residue characteristic of 𝐹, then 𝐹 has an unramified extension 𝐹 0 of degree 𝑝 [Ser79, p. 54, Cor. 1]. Moreover, 𝐹 also has a totally ramified cyclic extension of degree 𝑝, namely the splitting field 𝐹 00 of the polynomial 𝑋 𝑝 − 𝑋 − 𝑎 −1 , where 𝑎 is a prime element of the ring of integers of 𝐹 [FrJ08, p. 29, Example 2.3.9, Case A]. Since 𝐹 0 ∩ 𝐹 00 = 𝐹, we have Gal(𝐹 0 𝐹 00/𝐹)  𝑍 𝑝 ⊕ 𝑍 𝑝 . Let 𝐹 𝑝 be the fixed field of a 𝑝-Sylow subgroup of Gal(𝐹). Since 𝑝 - [𝐹 𝑝 : 𝐹], we have Gal(𝐹 𝑝 𝐹 0 𝐹 00/𝐹 𝑝 )  Gal(𝐹 0 𝐹 00/𝐹)  𝑍 𝑝 ⊕ 𝑍 𝑝 , so 𝐴 𝑝 = 𝑍 𝑝 ⊕ 𝑍 𝑝 (Setup 6.2.6) is a quotient of the 𝑝-Sylow subgroup Gal(𝐹 𝑝 ) of Γ = Gal(𝐹). Thus, P (Γ) = P (Definition 6.4.3). Finally, by (6.4), 𝑛0 + 𝑛1 + 2 = 3 ≤ card(P). (b2) We have 𝑛0 + 1 = 1. By Corollary 6.3.6, Gal(𝐹) is of local 𝑞-type for every 𝑞 ∈ P different from the residue characteristic of 𝐹. (b3) By Lemma 6.4.2, Γ = Gal(𝐹) is torsion-free Thus, all of the requirements of Definition 6.4.3 have been established. Lemma 6.4.6 Let 𝐺 be a profinite group, let G be a strictly closed subset of Subgr(𝐺), and let 𝑛 ∈ N. Suppose that card(P (Γ)) > 𝑛 for every Γ ∈ G. Then, there exists an 𝑁0 ∈ OpenNormal(𝐺) such that for every homomorphism 𝜑 : 𝐺 → 𝐴 with Ker(𝜑) ≤ 𝑁0 we have: card(P (𝜑(Γ))) > 𝑛 for every Γ ∈ G. Proof Let Γ ∈ G. Since Γ = lim Γ𝑁/𝑁 and hence ←−− 𝑁 ∈OpenNormal(𝐺) Ð P (Γ) = 𝑁 P (Γ𝑁/𝑁) = lim P (Γ𝑁/𝑁), and card(P (Γ)) > 𝑛, there is an 𝑁 −−→ 𝑁 such that card(P (Γ𝑁/𝑁)) > 𝑛.

6.4 Groups of P-Type and G-Projectivity

101

If Γ0 ∈ 𝜈(Γ, 𝑁) (see (1.3)), that is, Γ0 𝑁 = Γ𝑁, then card(P (Γ0 𝑁/𝑁)) > 𝑛. By the strict compactness of G there are Γ1 , . . . , Γ 𝑘 ∈ G and 𝑁1 , . . . , 𝑁 𝑘 ∈ Ð𝑘 OpenNormal(𝐺) Ñ such that G ⊆ 𝑖=1 𝜈(Γ𝑖 , 𝑁𝑖 ) and card(P (Γ𝑖 𝑁𝑖 /𝑁𝑖 )) > 𝑛 for every 𝑖. Put 𝑁0 = 𝑖 𝑁𝑖 . Let 𝜑 : 𝐺 → 𝐴 be a homomorphism with Ker(𝜑) ≤ 𝑁0 and let Γ ∈ G. Then, there is an 𝑖 such that Γ𝑁𝑖 = Γ𝑖 𝑁𝑖 . Since Ker(𝜑) ≤ 𝑁𝑖 , there is an epimorphism 𝜑(Γ)  Γ Ker(𝜑)/Ker(𝜑) → Γ𝑁𝑖 /𝑁𝑖 = Γ𝑖 𝑁𝑖 /𝑁𝑖 . Since card(P (Γ𝑁𝑖 /𝑁𝑖 )) > 𝑛, Remark 6.4.4(a) implies that card(P (𝜑(Γ))) > 𝑛, as claimed.  Lemma 6.4.7 Let 𝐺 be a profinite group and G a subset of Subgr(𝐺) of P-type such that 𝐺 is strongly G-projective. Let 𝑛0 , 𝑛1 , 𝑛∞ and G∞ , GS be as in Definition 6.4.3. Let Δ be a closed subgroup of 𝐺 that satisfies the following conditions: (a) Δ is the inverse limit of an inverse system (Δ𝑖 , 𝜋 𝑗𝑖 )𝑖, 𝑗 ∈𝐼 , 𝑗 ≥𝑖 of finite quotients Δ𝑖 of groups belonging to G = GS ∪ G∞ such that all 𝜋 𝑗𝑖 are surjective. (b) If Δ is not finite, then card(P (Δ)) ≥ 𝑛0 + 𝑛1 + 2. (c) If Δ is not finite, then there are at most 𝑛0 + 1 elements 𝑝 ∈ P such that Δ contains an element of order 𝑝. Then, there are Γ ∈ G and 𝑔 ∈ 𝐺 such that Δ ≤ Γ𝑔 . Proof If GS is strictly closed in Subgr(𝐺), then so is GS𝐺 (Example 1.1.11). By Lemma 5.2.4, we may assume that G is 𝐺-invariant. If Δ is finite, it is contained in some Γ ∈ G (Proposition 5.4.3). Thus, we may assume that Δ is infinite. By (a), Δ = lim Δ𝑖 , where the Δ𝑖 are finite quotients of groups in GS ∪ G∞ . ←−−𝑖 ∈𝐼 By [Rib70, p. 39, Proposition 4.4(c)], there is a 𝑘 ∈ 𝐼 such that card(Δ 𝑘 ) > 𝑛∞ . Replacing 𝐼 by {𝑖 ∈ 𝐼 | 𝑖 ≥ 𝑘 } we may assume, for every 𝑖 ∈ 𝐼, that card(Δ𝑖 ) > 𝑛∞ and hence that Δ𝑖 is a quotient of some Γ𝑖 ∈ GS . In particular, by Definition 6.4.3(b), Δ𝑖 is a solvable group. Hence, (6.5)

Δ is a prosolvable group. P0

By (b), there is a ⊆ P (Δ) (Definition 6.4.3) of cardinality 𝑛0 + 𝑛1 + 2. Thus, by Lemma 6.2.3(a), there is a 𝑘 ∈ 𝐼 such that P 0 ⊆ P (Δ 𝑘 ). By Remark 6.4.4(a), P 0 ⊆ P (Δ𝑖 ) and P 0 ⊆ P (Γ𝑖 ) for every 𝑖 ≥ 𝑘. By Definition 6.4.3(b2), for every 𝑖 ≥ 𝑘 there are at most 𝑛1 elements 𝑝 ∈ P 0 such that Γ𝑖 is not of local 𝑝-type (Definition 6.2.1). Thus, P 0 ∩ L (Γ𝑖 ) has at least 𝑛0 + 2 elements. By Remark 6.4.4(b), P 0 ∩ L (Γ𝑖 ) ⊆ P 0 ∩ L (Δ𝑖 ), hence |P 0 ∩ L (Δ𝑖 )| ≥ 𝑛0 + 2. Moreover, if 𝑘 ≤ 𝑖 ≤ 𝑗, then Δ𝑖 isÑa quotient of Δ 𝑗 , so by Remark 6.4.4(b), P 0 ∩ L (Δ 𝑗 ) ⊆ P 0 ∩ L (Δ𝑖 ). Thus, P 00 := 𝑖 ≥𝑘 P 0 ∩ L (Δ𝑖 ) has at least 𝑛0 + 2 elements. By Lemma 6.2.3(c), P 00 ⊆ L (Δ). By Remark 6.2.2, cd 𝑝 (Δ) ≥ 2 for all 𝑝 ∈ P 00 .

(6.6)

By (c) there is a 𝑝 ∈ P 00 such that (6.7) Let

𝑝0

(6.8)

Δ contains no element of order 𝑝. ∈

P 00

be distinct from 𝑝. Then, cd 𝑝 (Δ), cd 𝑝0 (Δ ≥ 2.

Hence, by (6.7), (6.8), and Proposition 6.1.8, there is a Γ ∈ G = G 𝐺 such that Δ ≤ Γ. 

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6 Strong Proper Projectivity

We conclude this chapter with a proposition that will play an essential role in the proof of the main result of this monograph. Proposition 6.4.8 Let 𝐺 be a profinite group and let G ⊆ Subgr(𝐺) be of P-type (Definition 6.4.3). Assume that 𝐺 is properly G-projective. Then, 𝐺 is properly strongly G-projective. Proof By Corollary 5.1.5, it suffices to prove that every finite proper G-embedding problem (𝜑 : 𝐺 → 𝐴, 𝛼 : 𝐵 → 𝐴, B)

(6.9)

has a proper strong solution. Let G∞ , GS and 𝑛0 , 𝑛1 , 𝑛∞ be as in Definition 6.4.3. In particular, G = GS ∪ G∞ and both G∞ and GS are strictly closed in Subgr(𝐺). By Lemma 5.1.4 we may assume that Ker(𝜑) is contained in a prescribed open normal subgroup 𝑁0 of 𝐺. Hence, by Lemma 6.4.6, we may assume that (6.10)

card(P (𝜑(Γ))) ≥ 𝑛0 + 𝑛1 + 2 for every Γ ∈ GS .

Let 𝑇 = G. Since G is strictly closed in Subgr(𝐺), it is a profinite space and the identity map 𝑇 → G is strictly continuous. Thus, by Proposition 2.3.6, there exists a profinite sheaf X = (𝑋, 𝜏, 𝑇) and a morphism 𝜔 : X → 𝐺 that maps 𝑋𝑡 isomorphically onto 𝐺 𝑡 for every 𝑡 ∈ 𝑇. By Lemma 2.3.5(e), 𝜏 : 𝑋 → 𝑇 is open. Let 𝜀 : X → 𝐹 be the free profinite product over X, given by Proposition 4.2.2, and let 𝛼ˆ 1 : 𝐹 → 𝐺 be the unique homomorphism, given by Definition 4.2.1, such that 𝛼ˆ 1 ◦ 𝜀 = 𝜔. By Lemma 2.3.5(d), the map 𝑡 ↦→ 𝜀(𝑋𝑡 ) is strictly continuous. Hence, by Fact 1.1.3(e), F := {Φ𝑡 := 𝜀(𝑋𝑡 ) | 𝑡 ∈ 𝑇 } is strictly closed in Subgr(𝐹) and 𝛼ˆ 1 maps F onto G. Moreover, since 𝜔 maps 𝑋𝑡 isomorphically onto 𝐺 𝑡 and, by Lemma 4.2.3(a), 𝜀 maps 𝑋𝑡 isomorphically onto Φ𝑡 , we see that (6.11)

𝛼ˆ 1 maps Φ𝑡 isomorphically onto 𝐺 𝑡 , for every 𝑡 ∈ 𝑇 .

Thus, FS := 𝛼ˆ 1−1 (GS ) ∩ F such that F = FS ∪ F∞ . It

and F∞ := 𝛼ˆ 1−1 (G∞ ) ∩ F are strictly closed subsets of F follows that

(6.12) F is of P-type with respect to (FS , F∞ ) and with the same non-negative integers 𝑛0 , 𝑛1 , 𝑛∞ , given in Definition 6.4.3, as those associated with G. Now, let 𝐷 be a free profinite group of rank ≥ max(rank(𝐺), ℵ0 ). Then, there is an epimorphism 𝛼ˆ 2 : 𝐷 → 𝐺 [FrJ08, p. 349, Prop. 17.4.8]. Set 𝐵ˆ = 𝐷 ∗ 𝐹. ˆ By the universal property of 𝐷 ∗ 𝐹, there exists a unique Then, F ⊆ Subgr( 𝐵). ˆ epimorphism 𝛼ˆ : 𝐵 → 𝐺 thatÎextends both 𝛼ˆ 1 and 𝛼ˆ 2 . By (6.11), 𝛼ˆ maps F onto G. By Proposition 4.3.1, 𝐹  ∗ 𝑡 ∈𝑇 Φ𝑡 . Therefore, by Proposition 5.2.2, 𝐵ˆ is properly strongly F -projective, hence there exists an epimorphism 𝛽 : 𝐵ˆ → 𝐵 that maps F into B such that the following diagram commutes. 𝐵ˆ

𝛼ˆ

/𝐺

𝛼

 /𝐴

𝜑

𝛽

 𝐵 ˆ Then, Put Bˆ = Con(F ) ⊆ Subgr( 𝐵).

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103

ˆ = Con(G) and 𝛽( B) ˆ ⊆ B. 𝛼( ˆ B) By Lemma 5.3.3(a) (with 𝐴ˆ = 𝐺 and 𝜑ˆ : 𝐺 → 𝐴ˆ the identity map), there exists a proper solution 𝛾 of (𝜑, 𝛼, B). Moreover, since G is strictly closed, Lemma 5.3.3 ˆ such that gives a strictly closed subset H of Subgr( 𝐵) (6.13)

(6.14)

𝛼(H ˆ ) = G,

𝛽(H ) = 𝛾(G),

and (6.15)

every Δ ∈ H is an inverse limit of finite quotients of groups in G with surjective connecting maps.

In addition, we may even assume that these groups in G lie in a given strictly open neighborhood of 𝛼(Δ) ˆ in Subgr(𝐺). Claim: For each Δ ∈ H , there are Φ ∈ F and 𝑓 ∈ 𝐹 such that Δ ≤ Φ 𝑓 . ˆ F ) replacing (𝐺, G). We verify the conditions of Lemma 6.4.7 with ( 𝐵, Verification of (a): This is done in (6.15). Verification of (b): Here we have to prove that if Δ is infinite, then card(P (Δ)) ≥ 𝑛0 + 𝑛1 + 2. Indeed, by (6.14), there is a Γ ∈ G such that 𝛽(Δ) = 𝛾(Γ). In particular, Γ ∈ GS . Otherwise, since GS is strictly closed in Subgr(𝐺), there would be a strictly open neighborhood U of Γ in Subgr(𝐺) disjoint from GS . Hence, U ⊆ G∞ . By Definition 6.4.3(a), each group in U is of order ≤ 𝑛∞ . But since Δ is an inverse limit of finite quotients of groups in U for every Δ ∈ H and every strictly open neighborhood U of 𝛼(Δ) ˆ in Subgr(𝐺) the group Δ is an inverse limit of finite quotients of groups in U with surjective connecting maps, U is finite, a contradiction. Thus, 𝛼(𝛽(Δ)) = 𝜑(Γ), and by (6.10) and Remark 6.4.4(a), we have card(P (Δ)) ≥ card(𝜑(Γ)) ≥ 𝑛0 + 𝑛1 + 2, as needed. Verification of (c): Again, we have to prove that if Δ is infinite, then there are at most 𝑛0 + 1 elements 𝑝 ∈ P such that Δ contains an element of order 𝑝. Indeed, by (6.14), Γ := 𝛼(Δ) ˆ ∈ G. As in the Verification of (b), Γ ∈ GS . Let 𝑝 ∈ P such that Δ contains an element ℎ of order 𝑝. Then, hℎi is a finite ˆ By Proposition 3.2.7, hℎi is contained in a conjugate of 𝐷 or of 𝐹. subgroup of 𝐵. The first option is impossible, since 𝐷 is a free profinite group, hence torsion free [FrJ08, p. 507, Cor. Î 22.4.6 and Prop. 22.4.7]. Thus, hℎi is contained in a conjugate of 𝐹. But since 𝐹  ∗ 𝑡 ∈𝑇 Φ𝑡 , it follows from Proposition 4.8.3(e) that hℎi is contained in a conjugate of some Φ𝑡 ∈ F . The latter is mapped, by (6.11), injectively by 𝛼ˆ onto its image. Thus, Γ = 𝛼(Δ) ˆ contains an element of order 𝑝 as well. By Definition 6.4.3(b3), there are at most 𝑛0 + 1 elements 𝑝 ∈ P such that Δ contains an element of order 𝑝. This concludes the verification of the last condition of Lemma 6.4.7. We conclude from Lemma 6.4.7 that there are Φ ∈ F and 𝑓 ∈ 𝐹 such that Δ ≤ Φ 𝑓 , as claimed. ˆ Hence, by (6.14) and (6.13), End of proof: By the Claim, H ⊆ Con(F ) = B. ˆ 𝛾(G) = 𝛽(H ) ⊆ 𝛽( B) ⊆ B. This means that 𝛾 is a strong solution of (𝜑, 𝛼, B). It follows that 𝐺 is properly strongly G-projective, as claimed. 

Chapter 7

A Free Profinite Product over an Étale Compact Subset of Subgr(𝑮)

Iwasawa characterizes the free profinite group 𝐹ˆ𝜔 as a profinite group of rank ≤ ℵ0 for which every finite embedding problem is properly solvable [Iwa53] (see also [FrJ08, p. 581, Thm. 24.8.1]). The main result of this chapter follows [Pop96] and generalizes this characterization: Let 𝐺 be a profinite group of rank ≤ ℵ0 endowed with an étale compact subset G of Subgr(𝐺). Suppose that every Îfinite G-embedding problem for 𝐺 has a proper strong solution. Then, 𝐺  𝐹ˆ𝜔 ∗ ∗ Γ∈ G0 Γ, where G0 is a set of representatives of the distinct conjugacy classes in (G 𝐺 )max (Proposition 7.2.3).

7.1 A System of Representatives We prove that if the rank of a profinite group 𝐺 is at most ℵ0 , then every 𝐺-invariant étale compact subset G of Subgr(𝐺) such that G = Gmax and 𝐺 is strongly G-projective has an étale compact set of representatives for its 𝐺-orbits. A profinite 𝐺-space is a profinite space 𝑋 together with a continuous action of 𝐺 (from the right) on 𝑋. A morphism 𝜑 : 𝑋 → 𝑌 of profinite 𝐺-spaces is a continuous map 𝜑 : 𝑋 → 𝑌 which is 𝐺-invariant, i.e. 𝜑(𝑥 𝜎 ) = 𝜑(𝑥) 𝜎 for all 𝑥 ∈ 𝐺 and 𝜎 ∈ 𝐺. Given a profinite space 𝑋, we denote the set of all partitions of 𝑋 by Partitions(𝑋). Thus, an element of Partitions(𝑋) is a finite collection {𝑋1 , . . . , 𝑋𝑛 } of disjoint nonÐ𝑛 empty open-closed subsets of 𝑋 such that 𝑋 = · 𝑖=1 𝑋𝑖 (Remark 1.1.8). If 𝑋 is a profinite 𝐺-space, then a 𝐺-partition of 𝑋 is a partition {𝑋1 , . . . , 𝑋𝑛 } of 𝑋 which is 𝐺-invariant, i.e. {𝑋1𝜎 , . . . , 𝑋𝑛𝜎 } = {𝑋1 , . . . , 𝑋𝑛 } for all 𝜎 ∈ 𝐺. If P and P0 are 𝐺-partitions of 𝑋, then P00 = {𝑉 ∩ 𝑉 0 | 𝑉 ∈ P, 𝑉 0 ∈ P0, 𝑉 ∩ 𝑉 0 ≠ ∅} is a 𝐺-partition of 𝑋 which is finer than both P and P0. If P0 is finer than P, we define a map 𝜋P0 ,P from P0 onto P that maps each 𝑉 0 ∈ P0 onto the unique set 𝑉 ∈ P that contains 𝑉 0. Note that 𝜋P0 ,P is compatible with the actions of 𝐺 on P0 and P, and 𝜋P0 ,P ◦ 𝜋P00 ,P0 = 𝜋P00 ,P if P00 is finer than P0. Thus, we may consider the inverse limit lim P of the 𝐺-partitions of 𝑋. ←−− © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 D. Haran and M. Jarden, The Absolute Galois Group of a Semi-Local Field, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-030-89191-6_7

105

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7 A Free Profinite Product over an Étale Compact Subset of Subgr(𝐺)

Lemma 7.1.1 (After [HaJ85, Prop. 1.5]) Every profinite 𝐺-space 𝑋 is isomorphic to the inverse limit lim P, where P ranges over all 𝐺-partitions of 𝑋. ←−− Proof The proof depends on two claims. Claim A: Let 𝑉 be an open-closed subset of 𝑋. Then, 𝐺 has an open normal subgroup 𝑁 such that 𝑉 𝜎 = 𝑉 for all 𝜎 ∈ 𝑁. Indeed, given 𝑥 ∈ 𝑉, the continuity of the 𝐺-action 𝑋 × 𝐺 → 𝑋 at (𝑥, 1) yields an open-closed neighborhood 𝑈 𝑥 of 𝑥 in 𝑋 and an open normal subgroup 𝑁 𝑥 of 𝐺 such that 𝑈 𝑥𝜎 ⊆ 𝑉 for all 𝜎 ∈ 𝑁 𝑥 . In particular, 𝑈 𝑥 ⊆ 𝑉. Since 𝑉 is compact, Ð𝑘 Ñ𝑘 there are 𝑥1 , . . . , 𝑥 𝑘 ∈ 𝑉 such that 𝑉 = 𝑖=1 𝑈 𝑥𝑖 . We set 𝑁 = 𝑖=1 𝑁 𝑥𝑖 . Then, Ð −1 𝑘 𝑉 𝜎 = 𝑖=1 𝑈 𝑥𝜎𝑖 ⊆ 𝑉 for all 𝜎 ∈ 𝑁. Hence, 𝑉 𝜎 ⊆ 𝑉 for all 𝜎 ∈ 𝑁. Therefore, 𝑉 𝜎 = 𝑉 for all 𝜎 ∈ 𝑁, as claimed. Claim B: For every partition P of 𝑋 there exists a 𝐺-partition P0 of 𝑋 finer than P. Indeed, let P = {𝑉1 , . . . , 𝑉𝑚 }. By Claim A, there exists an open normal subgroup 𝑁 of 𝐺 such that 𝑉𝑖𝜎 = 𝑉𝑖 for 𝑖 = 1, . . . , 𝑚 and all 𝜎 ∈ 𝑁. Let 𝜎1 , . . . , 𝜎𝑛 be representatives of the right cosets of 𝐺 modulo 𝑁. For every 𝜎𝑛 𝜎1 ∩ · · · ∩ 𝑉 𝛼(𝑛) . Then, function 𝛼 : {1, . . . , 𝑛} → {1, . . . , 𝑚} we set 𝑉 𝛼 = 𝑉 𝛼(1) P0 = {𝑉 𝛼 | 𝑉 𝛼 ≠ ∅} is a 𝐺-partition of 𝑋 finer than P. (See also Lemma 1.1.10.) End of proof: For every 𝐺-partition P of 𝑋 we define a surjective map 𝜋P : 𝑋 → P that maps each 𝑥 ∈ 𝑋 onto the unique element 𝑉 in P that contains 𝑥. The inverse image of 𝑉 considered as an element of P is the open subset 𝑉 of 𝑋. Hence, 𝜋P is continuous. Our definitions imply that 𝜋P0 ,P ◦ 𝜋P0 = 𝜋P , if P0 is finer than P. Hence, the collection of all 𝜋P ’s yields a continuous surjective map 𝜋 : 𝑋 → lim P whose ←−− composition with the projection onto the P’th component is the map 𝜋P [FrJ08, p. 3, Cor. 1.1.6]. Since each of the maps 𝜋P0 ,P is compatible with the action of 𝐺, the map 𝜋P is a morphism of profinite 𝐺-spaces. If 𝑥 and 𝑥 0 are distinct elements of 𝑋, then there exist disjoint open-closed subsets 𝑉 and 𝑉 0 of 𝑋 such that 𝑥 ∈ 𝑉, 𝑥 0 ∈ 𝑉 0, and 𝑋 = 𝑉 ∪· 𝑉 0. Claim B yields a 𝐺-partition P of 𝑋 which is finer than the partition {𝑉, 𝑉 0 }. Then, 𝜋P (𝑥) ≠ 𝜋P (𝑥 0), so 𝜋(𝑥) ≠ 𝜋(𝑥 0). We conclude that 𝜋 is bijective. Since both 𝑋 and lim P are profinite spaces, 𝜋 is a homeomorphism (Fact 1.1.3(e)), ←−− as claimed.  Lemma 7.1.2 Let 𝐺 be a profinite group, 𝐼 a countable directed partially ordered set, and 𝑋 = lim 𝑋𝑖 an inverse limit of finite discrete 𝐺-spaces, with 𝑖 ranging over ←−− 𝐼. Let 𝑌 be a closed 𝐺-subspace of 𝑋 and let 𝑌 0 be a closed set of representatives for the 𝐺-orbits of 𝑌 . Then, 𝑋 has a closed set of representatives 𝑋 0 for its 𝐺-orbits such that 𝑋 0 ∩ 𝑌 = 𝑌 0. Proof We may assume that 𝑋 is the inverse limit of a sequence 𝑋1 o

𝜑1

𝑋2 o

𝜑2

𝑋3 o

𝜑3

···

of finite discrete 𝐺-spaces, where the 𝜑𝑖 ’s are surjective morphisms of 𝐺-spaces (Remark 1.1.7(d)). For each positive integer 𝑖 we let 𝜋𝑖 : 𝑋 → 𝑋𝑖 be the projection, so 𝜋𝑖 = 𝜑𝑖 ◦ 𝜋𝑖+1 . Set 𝑌𝑖 = 𝜋𝑖 (𝑌 ) and 𝑍𝑖 = 𝑋𝑖 r 𝑌𝑖 . Then, 𝑌𝑖𝐺 = 𝜋𝑖 (𝑌 ) 𝐺 = 𝜋𝑖 (𝑌 𝐺 ) =

7.1 A System of Representatives

107

𝜋𝑖 (𝑌 ) = 𝑌𝑖 . Hence, 𝑍𝑖𝐺 = 𝑍𝑖 . Moreover, 𝑌𝑖0 := 𝜋𝑖 (𝑌 0) ⊆ 𝜋𝑖 (𝑌 ) = 𝑌𝑖 ⊆ 𝑋𝑖 . By 0 ) = 𝜑 (𝜋 0 0 0 assumption, (𝑌 0) 𝐺 = 𝑌 , so (𝑌𝑖0) 𝐺 = 𝑌𝑖 and 𝜑𝑖 (𝑌𝑖+1 𝑖 𝑖+1 (𝑌 )) = 𝜋𝑖 (𝑌 ) = 𝑌𝑖 . (However, 𝑌𝑖0 need not be a set of representatives for the 𝐺-orbits of 𝑌𝑖 .) Since 𝑌 0 is closed in 𝑋, we have 𝑌 0 = lim 𝑌𝑖0. ←−− Construction: We construct for each positive integer 𝑖 a set 𝑍𝑖0 of representatives for 0 ) = 𝑋 0. the 𝐺-orbits of 𝑍𝑖 such that 𝑋𝑖0 := 𝑌𝑖0 ∪· 𝑍𝑖0 satisfies 𝜑𝑖 (𝑋𝑖+1 𝑖 The construction is carried out by induction on 𝑖. For 𝑖 = 1 we choose any set of representatives 𝑍10 for the 𝐺-orbits of 𝑍1 and let 𝑋10 = 𝑌10 ∪· 𝑍10 . Next let 𝑖 ≥ 2 and assume that we have constructed a set 𝑍𝑖0 of representatives for 0 . In particular, (𝑋 0 ) 𝐺 = (𝑌 0 ) 𝐺 ∪ (𝑍 0 ) 𝐺 = the 𝐺-orbits of 𝑍𝑖 such that 𝜑𝑖−1 (𝑋𝑖0) = 𝑋𝑖−1 𝑖 𝑖 𝑖 0 0 0 𝑌𝑖 ∪ 𝑍𝑖 = 𝑋𝑖 . Again, we set 𝑋𝑖 = 𝑌𝑖 ∪· 𝑍𝑖 . 0 ) = 𝑌 0 and 𝜑 (𝑋 ) = 𝑋 , we have 𝑌 0 ⊆ 𝜑−1 (𝑌 0 ) ⊆ 𝜑−1 (𝑋 ) = Since 𝜑𝑖 (𝑌𝑖+1 𝑖 𝑖+1 𝑖 𝑖 𝑖 𝑖 𝑖
𝐺 𝑖+1−1 𝑖 0 𝐺 −1 0 𝑋𝑖+1 . It follows from (𝑋𝑖 ) = 𝑋𝑖 that 𝜑𝑖 (𝑋𝑖 ) = 𝜑𝑖 (𝑋𝑖 ) = 𝑋𝑖+1 . Moreover, 𝜎 =𝑌 𝑌𝑖+1 𝑖+1 for every 𝜎 ∈ 𝐺. Hence,
𝐺 0 𝜑−1 = 𝑋𝑖+1 r 𝑌𝑖+1 = 𝑍𝑖+1 . 𝑖 (𝑋𝑖 ) r 𝑌𝑖+1 0 for the 𝐺-orbits of 𝑍 Thus, we can choose a set of representatives 𝑍𝑖+1 𝑖+1 such that 0 −1 0 0 𝑍𝑖+1 ⊆ 𝜑𝑖 (𝑋𝑖 ) r 𝑌𝑖+1 . In particular, 𝜑𝑖 (𝑍𝑖+1 ) ⊆ 𝑋𝑖0. 0 = 𝑌0 ∪ 0 𝐺 = (𝑌 0 ) 𝐺 ∪ (𝑍 0 ) 𝐺 = 𝑌 · 0 We let 𝑋𝑖+1 𝑖+1 ∪ 𝑍 𝑖+1 = 𝑖+1 𝑍 𝑖+1 . Then, (𝑋𝑖+1 ) 𝑖+1 𝑖+1 𝑋𝑖+1 . 0 ) = 𝑋 0. 0 ) = 𝑌 0 ⊆ 𝑋 0 and 𝜑 (𝑍 0 ) ⊆ 𝑋 0 = Claim A: 𝜑𝑖 (𝑋𝑖+1 Indeed, 𝜑𝑖 (𝑌𝑖+1 𝑖 𝑖 𝑖 𝑖 𝑖 𝑖+1 0 0 0 ) ⊇ 𝑍 0. · 𝑌𝑖 ∪ 𝑍𝑖 . Hence, it suffices to show that 𝜑𝑖 (𝑍𝑖+1 𝑖 To this end let 𝑧 ∈ 𝑍𝑖0. Then, there is an 𝑥 ∈ 𝑋𝑖+1 such that 𝜑𝑖 (𝑥) = 𝑧. Since 0 0 𝑍𝑖 ⊆ 𝑍𝑖 = 𝑋𝑖 r 𝑌𝑖 , we have 𝑧 ∉ 𝑌𝑖 . Hence, 𝑥 ∉ 𝑌𝑖+1 . Therefore, 𝑥 ∈ 𝜑−1 𝑖 (𝑋𝑖 ) r 𝑌𝑖+1 ⊆ 0 0 0 𝑍𝑖+1 . By the choice of 𝑍𝑖+1 , there are 𝑥 ∈ 𝑍𝑖+1 and 𝜎 ∈ 𝐺 such that 𝑥 0 = 𝑥 𝜎 . 0 ) ⊆ 𝑋 0 . But, 𝑧 ∉ 𝑌 = 𝑌 𝜎 −1 . Hence, Then, 𝑧 𝜎 = 𝜑𝑖 (𝑥 𝜎 ) = 𝜑𝑖 (𝑥 0) ∈ 𝜑𝑖 (𝑍𝑖+1 𝑖 𝑖 𝑖 𝑧 𝜎 ∈ 𝑋𝑖0 r 𝑌𝑖 = (𝑌𝑖0 ∪· 𝑍𝑖0) r 𝑌𝑖 = 𝑍𝑖0. Since 𝑍𝑖0 is a set of representatives for the 𝐺-orbits of 𝑍𝑖 and 𝑧, 𝑧 𝜎 ∈ 𝑍𝑖0, we have 𝑧 𝜎 = 𝑧. Hence, 𝜑𝑖 (𝑥 0) = 𝜑𝑖 (𝑥 𝜎 ) = 𝑧 𝜎 = 𝑧. 0 ). Thus, 𝑧 = 𝜑𝑖 (𝑥 0) ∈ 𝜑𝑖 (𝑍𝑖+1 This concludes the induction step. Next observe that the closed subset 𝑋 0 := lim 𝑋𝑖0 of 𝑋 contains 𝑌 0 := lim 𝑌𝑖0, ←−− ←−− because 𝑌𝑖0 ⊆ 𝑋𝑖0 for each 𝑖 ∈ 𝐼.

Claim B: 𝑋 0 ∩𝑌 = 𝑌 0. Let 𝑥 ∈ 𝑋 0 ∩𝑌 . Then, 𝜋𝑖 (𝑥) ∈ 𝑋𝑖0 ∩𝑌𝑖 = (𝑌𝑖0 ∪· 𝑍𝑖0) ∩𝑌𝑖 = 𝑌𝑖0, for each 𝑖. Hence, 𝑥 ∈ lim 𝑌𝑖0 = 𝑌 0. The converse inclusion is trivial. ←−− 0 Claim C: 𝑋 is a set of representatives for the 𝐺-orbits of 𝑋. Since 𝑋𝑖 = (𝑋𝑖0) 𝐺 for each 𝑖, we have 𝑋 = (𝑋 0) 𝐺 . Let 𝑥, 𝑦 ∈ 𝑋 0 and 𝜎 ∈ 𝐺 such that 𝑥 𝜎 = 𝑦. If 𝜋𝑖 (𝑥) ∈ 𝑌𝑖 for each 𝑖, then 𝑥 ∈ lim 𝑌𝑖 = 𝑌 , hence also 𝑦 = 𝑥 𝜎 ∈ 𝑌 𝜎 = 𝑌 . By ←−− Claim B, 𝑥, 𝑦 ∈ 𝑋 0 ∩ 𝑌 = 𝑌 0. Since 𝑌 0 is a set of representatives for the 𝐺-orbits in 𝑌 , we have 𝑥 = 𝑦. Otherwise, there is an 𝑖 such that 𝜋𝑖 (𝑥) ∉ 𝑌𝑖 . Then, 𝜋𝑖 (𝑦) = 𝜋𝑖 (𝑥) 𝜎 ∉ 𝑌𝑖𝜎 = 𝑌𝑖 as well. Hence, 𝜋𝑖 (𝑥), 𝜋𝑖 (𝑦) ∈ 𝑋𝑖0 r 𝑌𝑖 = 𝑍𝑖0, by construction. Since 𝑍𝑖0 is a set of representatives for the 𝐺-orbits of 𝑍𝑖 , we have 𝜋𝑖 (𝑥) = 𝜋𝑖 (𝑦). Without loss of generality this holds for each 𝑖, hence 𝑥 = 𝑦.

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7 A Free Profinite Product over an Étale Compact Subset of Subgr(𝐺)

Thus, 𝑋 0 is a set of representatives for the 𝐺-orbits of 𝑋. This concludes the proof of the lemma.



Here is a useful consequence of Lemma 7.1.2. Lemma 7.1.3 Let 𝐺 be a profinite group of rank ≤ ℵ0 and let G be a 𝐺-invariant étale compact subset of Subgr(𝐺) such that G = Gmax and 𝐺 is strongly G-projective. Then, G has an étale compact subset of representatives for its 𝐺-orbits. Proof If 1 ∈ G, then G = {1}, because G = Gmax . Hence, G 0 = {1} is the desired subset of G. Thus, we may assume that 1 ∉ G. By [FrJ08, p. 340, Example 17.1.7(a)], there is a sequence 𝐺 = 𝑁0 ≥ 𝑁1 ≥ 𝑁2 ≥ · · · Ñ of open normal subgroups of 𝐺 such that ∞ 𝑖=0 𝑁𝑖 = 1. Let 𝑖 ≥ 0. By Proposition 5.5.3(c), G = G r Subgr(𝑁𝑖 ) is étale profinite. Also, 𝑖 Ð G𝑖𝐺 = G𝑖 , G𝑖 ⊆ G𝑖+1 , and ∞ G = G. 𝑖=0 𝑖 By Lemma 7.1.1, G𝑖 is isomorphic to the inverse limit of its 𝐺-invariant partitions. The base {Subgr(𝐻) | 𝐻 ∈ Open(𝐺)} for the étale topology on Subgr(𝐺) is countable. It induces a countable base for the étale topology on G𝑖 . Therefore, G𝑖 is isomorphic to the inverse limit over a countable set of finite 𝐺-spaces. Applying induction on 𝑖, we use Lemma 7.1.2 to find for each 𝑖 an étale closed 0 ∩ G . By subset G𝑖0 of representatives of the 𝐺-orbits ofÐG𝑖 such that G𝑖0 = G𝑖+1 𝑖 ∞ 0 0 0 Fact 1.1.3(a), G𝑖 is étale compact. Let G = 𝑖=0 G𝑖 . Then, for each 𝑖 we have G𝑖0 = G 0 ∩ G𝑖 . Hence, G𝑖0 = G 0 ∩ (G r Subgr(𝑁𝑖 )) = G 0 r Subgr(𝑁𝑖 ). Therefore, G 0 is a set of representatives for the 𝐺-orbits of G. By Lemma 1.2.10, G 0 is étale compact, as claimed.  Remark 7.1.4 Lemma 7.1.3 is all that we need to know about set of representatives to prove the main result of this monograph. However, Fact 2.2 of [Pop96] claims that the following result holds: Let 𝐺 be a countably generated profinite group and G an étale compact subset of Subgr(𝐺). Then, Con(G)max has an étale compact set of representatives for its 𝐺-orbits. Lemma 7.1.3 proves Pop’s claim under the assumption that G is strongly G-projective. We have not been able to prove the claim in general.

7.2 The Generalized Iwasawa Theorem We state and prove Pop’s generalization [Pop96, Theorem 2.7] of Iwasawa’s theorem. Lemma 7.2.1 Let 𝐷 be a closed subset of a profinite group 𝐺 such that 𝐺 = h𝐷i. (a) If rank(𝐺) ≤ ℵ0 , then the topology of 𝐷 has a countable base. (b) If the topology of 𝐷 has a countable base, then rank(𝐺) ≤ ℵ0 .

7.2 The Generalized Iwasawa Theorem

109

Proof Proof of (a). If rank(𝐺) ≤ ℵ0 , then card(OpenNormal(𝐺)) ≤ ℵ0 , hence {𝑁𝑔 | 𝑁 ∈ OpenNormal(𝐺), 𝑔 ∈ 𝐺} is a countable base for the topology of 𝐺. Thus, {𝑁𝑔 ∩ 𝐷 | 𝑁 ∈ OpenNormal(𝐺), 𝑔 ∈ 𝐺} is a countable base for the topology of 𝐷. Proof of (b). Let C be the family of open-closed subsets of 𝐷. Since 𝐷 is compact, every 𝐶 ∈ C is a union of finitely many basic sets, hence card(C) ≤ ℵ0 . Let 𝐴 be a finite group. Since 𝐺 = h𝐷i, every epimorphism 𝜑 : 𝐺 → 𝐴 is determined by its restriction to 𝐷. The latter is determined by a partition (𝜑−1 (𝑎) | 𝑎 ∈ 𝐴) of 𝐷 into disjoint open-closed subsets. Since there are only countably many partitions and there are countably many finite groups, up to isomorphism, there are only countably many epimorphisms of 𝐺 onto finite groups. In particular, there are only countably many kernels of such epimorphisms, hence card(OpenNormal(𝐺)) ≤ ℵ0 . Thus, rank(𝐺) ≤ ℵ0 .  The generalized Iwasawa theorem depends on Proposition 5.6.8: Proposition 7.2.2 ([Pop96, Thm. 4.5]) Let 𝐺 be a profinite group and G a subset of Subgr(𝐺). Suppose that 𝐺 and G satisfy the following conditions: (a1) rank(𝐺) ≤ ℵ0 . (a2) G is a set of representatives of the distinct conjugacy classes in (G 𝐺 )max . (a3) 𝐺 is properly strongly G-projective. Let 𝐺 0 be a profinite group and G 0 a subset of Subgr(𝐺 0) that also satisfy the above conditions. Assume that (b) there is a homeomorphism 𝜇 : 𝑈 (G) → 𝑈 (G 0) that satisfies the following condition: for every Γ ∈ G there is a Γ0 ∈ G 0 such that 𝜇|Γ : Γ → Γ0 is an isomorphism of groups. 0

Then, there is an isomorphism 𝜃 : 𝐺 0 → 𝐺 such that 𝜃 ((G 0) 𝐺 ) = G 𝐺 . 0 0 ∞ Proof By (a1) we can write Open(𝐺) = {𝑀 𝑗 }∞ 𝑗=1 and Open(𝐺 ) = {𝑀 𝑗 } 𝑗=1 . Let 0 𝜇 : 𝑈 (G) → 𝑈 (G ) be as in (b). The rest of the proof breaks up into three parts.

Part A: Induction.

We construct by induction

(7.1) sequences 𝐺 = 𝑁0 ≥ 𝑁1 ≥ 𝑁2 ≥ · · · and 𝐺 0 = 𝑁00 ≥ 𝑁10 ≥ 𝑁20 ≥ · · · of open normal subgroups of 𝐺 and 𝐺 0, respectively, and, for every 𝑗, isomorphisms 𝜃 𝑗 : 𝐺 0/𝑁 0𝑗 → 𝐺/𝑁 𝑗 , such that (7.2) 𝑁 𝑗 ≤

Ñ𝑗 𝑘=1

𝑀𝑘 for every even 𝑗 and 𝑁 0𝑗 ≤

Ñ𝑗 𝑘=1

𝑀𝑘0 for every odd 𝑗,

and in Diagram (7.4) in which 𝜄, 𝜄0 are inclusions and the other vertical maps are the quotient maps, (7.3a) the bottom rectangle commutes, i.e. 𝜃 𝑗 ◦ 𝜋 0𝑗 = 𝜋 𝑗 ◦ 𝜃 𝑗+1 , (7.3b) the outer rectangle quasi-commutes, that is, 𝜃 𝑗 ◦ 𝜑 0𝑗 ◦ 𝜄0 ∼ 𝜑 𝑗 ◦ 𝜄 ◦ 𝜇−1 ; equivalently, 𝜑 0𝑗 ◦ 𝜄0 ◦ 𝜇 ∼ 𝜃 −1 𝑗 ◦ 𝜑 𝑗 ◦ 𝜄, and

110

7 A Free Profinite Product over an Étale Compact Subset of Subgr(𝐺) 0 0 0 0 (7.3c) 𝜃 −1 𝑗 (𝜑 𝑗 (G)) ⊆ Con(𝜑 𝑗 (G )) for even 𝑗 and 𝜃 𝑗 (𝜑 𝑗 (G )) ⊆ Con(𝜑 𝑗 (G)) for odd 𝑗.

The groups 𝑁 0𝑗+1 and 𝑁 𝑗+1 as well as the dotted arrows in the diagram are objects yet to be defined in the induction step. 𝑈 (G _ 0) o

(7.4)

𝜇

𝑈 (G) _

𝜄0

 𝐺0

𝜑 0𝑗+1

𝜑 0𝑗

  y 𝐺 0/𝑁 0𝑗+1 o 𝜋 0𝑗

$  𝐺 0/𝑁 0𝑗

𝜄

 𝐺

𝛿 𝛾

𝜑 𝑗+1 𝜃 𝑗+1 𝛾¯ 𝛼

 / 𝐺/𝑁 𝑗+1

𝜑𝑗

𝜋𝑗 𝜃𝑗

$  z / 𝐺/𝑁 𝑗

For 𝑗 = 0 we have 𝐺/𝑁0  1  𝐺/𝑁00 , so 𝜃 00 , 𝜑00 , and 𝜑0 are taken to be the trivial maps. The construction of 𝑁10 , 𝑁1 , and 𝜃 1 is included in the following induction step. Part B: The induction step. Let 𝑗 ≥ 0 be an even integer. Assume that 𝑁0 , 𝑁1 , . . . , 𝑁 𝑗 , 𝑁00 , 𝑁10 , . . . , 𝑁 0𝑗 , and 𝜃 0 , 𝜃 1 , . . . , 𝜃 𝑗 have already been constructed such that Conditions (7.2), (7.3a), (7.3b), and (7.3c) hold with 𝑖 replacing 𝑗 for each 𝑖 between 0 and 𝑗 − 1, and Conditions (7.2), (7.3b), and (7.3c) hold also for 𝑗. Ñ 𝑗+1 Choose an open 𝑁 0𝑗+1 ⊳ 𝐺 0 such that 𝑁 0𝑗+1 ≤ 𝑁 0𝑗 ∩ 𝑘=1 𝑀𝑘0 and let 𝜑 0𝑗+1 : 𝐺 0 → 𝐺 0/𝑁 0𝑗+1 and 𝜋 0𝑗 : 𝐺 0/𝑁 0𝑗+1 → 𝐺 0/𝑁 0𝑗 be the quotient maps, so that 𝜑 0𝑗 = 𝜋 0𝑗 ◦ 𝜑 0𝑗+1 . Since 𝜇 is a G-conductor (Definition 5.6.1), so is 𝛿 := 𝜑 0𝑗+1 ◦𝜄0◦𝜇 : 𝑈 (G) → 𝐺 0/𝑁 0𝑗+1 . We set 𝛼 = 𝜃 𝑗 ◦ 𝜋 0𝑗 : 𝐺 0/𝑁 0𝑗+1 → 𝐺/𝑁 𝑗 . Then, (7.3b)

𝛼 ◦ 𝛿 = 𝜃 𝑗 ◦ 𝜋 0𝑗 ◦ 𝜑 0𝑗+1 ◦ 𝜄0 ◦ 𝜇 = 𝜃 𝑗 ◦ 𝜑 0𝑗 ◦ 𝜄0 ◦ 𝜇 ∼ 𝜑 𝑗 ◦ 𝜄 ◦ 𝜇−1 ◦ 𝜇 = 𝜑 𝑗 ◦ 𝜄. Hence, 𝛼◦𝛿 ∼ 𝜑 𝑗 ◦𝜄. By Proposition 5.6.8, there is an epimorphism 𝛾 : 𝐺 → 𝐺 0/𝑁 0𝑗+1 such that 𝛼 ◦ 𝛾 = 𝜑 𝑗 , 𝛾 ◦ 𝜄 ∼ 𝛿, and 𝛾(G) ⊆ Con(𝛿(G)). Let 𝑁 𝑗+1 = Ker(𝛾). Then, 𝑁 𝑗+1 is an open normal subgroup of 𝐺 contained in Ker(𝜑 𝑗 ). Let 𝜑 𝑗+1 : 𝐺 → 𝐺/𝑁 𝑗+1 be the quotient map. By the first isomorphism theorem, there is an isomorphism 𝛾¯ : 𝐺/𝑁 𝑗+1 → 𝐺 0/𝑁 0𝑗+1 such that 𝛾 = 𝛾¯ ◦ 𝜑 𝑗+1 . In particular, 𝛾(𝜑 ¯ 𝑗+1 (G)) ⊆ Con(𝛿(G)) = Con(𝜑 0𝑗+1 ◦ 𝜄0 ◦ 𝜇(G)) = Con(𝜑 0𝑗+1 (G 0)). Then, 𝜃 𝑗+1 := 𝛾¯ −1 satisfies 𝜃 𝑗+1 ◦ 𝛾 = 𝜑 𝑗+1 and 0 0 𝜃 −1 𝑗+1 (𝜑 𝑗+1 (G)) ⊆ Con(𝜑 𝑗+1 (G )).

We have 𝜃 𝑗 ◦ 𝜋 0𝑗 ◦ 𝛾 = 𝛼 ◦ 𝛾 = 𝜑 𝑗 = 𝜋 𝑗 ◦ 𝜑 𝑗+1 = 𝜋 𝑗 ◦ 𝜃 𝑗+1 ◦ 𝛾. Since 𝛾 is surjective, 𝜃 𝑗 ◦ 𝜋 0𝑗 = 𝜋 𝑗 ◦ 𝜃 𝑗+1 . From 𝜑 𝑗+1 = 𝜃 𝑗+1 ◦ 𝛾 and 𝛾 ◦ 𝜄 ∼ 𝛿 = 𝜑 0𝑗+1 ◦ 𝜄0 ◦ 𝜇 we deduce that 𝜑 𝑗+1 ◦ 𝜄 = 𝜃 𝑗+1 ◦ 𝛾 ◦ 𝜄 ∼ 𝜃 𝑗+1 ◦ 𝜑 0𝑗+1 ◦ 𝜄0 ◦ 𝜇. This completes the induction step for 𝑗 even. If 𝑗 is odd, apply the same proof, switching between 𝐺, G and 𝐺 0, G 0 and replacing −1 𝜃 𝑗 , 𝜃 𝑗+1 with 𝜃 −1 𝑗 , 𝜃 𝑗+1 .

7.2 The Generalized Iwasawa Theorem

111

Ñ End of proof: Having completed the induction, it follows from (7.2) that ∞ 𝑗=1 𝑁 𝑗 = 1 Ñ∞ 0 0 0 0 and 𝑗=1 𝑁 𝑗 = 1. Hence, 𝐺 = lim 𝐺/𝑁𝑖 , 𝐺 = lim 𝐺 /𝑁𝑖 , and the compatible ←−−𝑖 ←−−𝑖 0 isomorphisms 𝜃 𝑗 yield an isomorphism 𝜃 : 𝐺 0 → 𝐺 such that 𝜃 ((G 0) 𝐺 ) = G 𝐺 (by (7.3c)).  Proposition 7.2.3 Let 𝐺 be a profinite group of rank ≤ ℵ0 equipped with a subset G of Subgr(𝐺). Suppose that (a) 𝐺 is properly strongly G-projective (in particular, G is étale compact), and (b) G is a set of representatives for the distinct conjugacy classes in (G 𝐺 )max . Î Î Then, 𝐺  𝐹ˆ𝜔 ∗ ∗ Γ∈ G Γ, where ∗ Γ∈ G Γ is an inner free product of the groups Γ in G. Proof We break up the proof into four parts. Part A: Inner free product. By Proposition 5.5.8, (G 𝐺 )max is separated, hence G is separated. By Proposition 2.1.8, there exist a profinite space 𝑇 and an étale continuous family {𝐺 𝑡 }𝑡 ∈𝑇 of closed subgroups of 𝐺 such that {𝐺 𝑡 }𝑡 ∈𝑇 is either G or G ∪ {1}. Thus, by Proposition 2.3.6, there is, up to isomorphism, a unique sheaf-group structure (𝑋, 𝜏, 𝑇, 𝜔, 𝐺) with a profinite sheaf X = (𝑋, 𝜏, 𝑇) such that 𝜔 maps 𝑋𝑡 =: 𝜏 −1 (𝑡) isomorphically onto 𝐺 𝑡 for each 𝑡 ∈ 𝑇. By Proposition 4.2.2, there ˜ over X. Hence, exists a unique, up to isomorphism, free profinite product (X, 𝜌, 𝐺) Î Î ˜ by Proposition 4.3.1, 𝐺 = ∗ 𝑡 ∈𝑇 𝐺 𝑡 = ∗ Γ∈ G Γ is an inner free product over 𝑇. Moreover, for every 𝑡 ∈ 𝑇, 𝜌 maps 𝑋𝑡 isomorphically onto 𝐺 𝑡 (Lemma 4.2.3(a)) and 𝐺˜ = h𝐺 𝑡 i𝑡 ∈𝑇 (Lemma 4.2.3(c)). ˜ Part B: The free product 𝐹 := 𝐹ˆ𝜔 ∗ 𝐺. We consider 𝐹ˆ𝜔 and 𝐺˜ as closed subgroups of the free product 𝐹. Then, F =: {𝜌(𝑋𝑡 )}𝑡 ∈𝑇 is a subset of Subgr(𝐹). By Corollary 5.2.3, 𝐹 is properly strongly F -projective. In particular, by (5.5), F is étale compact. There is an alternative way to view 𝐹: Let 𝑇 0 be the disjoint union of 𝑇 with a · one-point-space {𝑡0 } and let 𝑋 0 = 𝑋 ∪{𝐺 𝑡0 }, with 𝐺 𝑡0 := 𝐹ˆ 𝜔 . Then, extend 𝜏 to 𝜏 0 : 𝑋 0 → 𝑇 0 by mapping 𝐹ˆ𝜔 onto {𝑡 0 }. By Lemma 4.6.2, 𝐹 is the inner free product on the sheaf X0 = (𝑋 0, 𝜏 0, 𝑇 0). By Proposition 4.8.3(a), it follows from this presentation that F is a set of representatives for the 𝐹-conjugacy classes of (F 𝐹 )max . Part C: A homeomorphism 𝑈 (F ) → 𝑈 (G). Since F and G are étale compact, 𝑈 (FÐ ) and 𝑈 (G) are profinite spaces (Lemma 1.2.5 and Remark 1.1.7(a)). Since 𝑋 = · 𝑡 ∈𝑇 𝑋𝑡 and 𝜌 maps each 𝑋𝑡 isomorphically onto 𝐺 𝑡 ⊆ 𝑈 (G), there is a unique map 𝜇 : 𝑈 (F ) = 𝜌(𝑋) → 𝑈 (G) which is the identity on each 𝜌(𝑋𝑡 ) = 𝐺 𝑡 such that 𝜇 ◦ 𝜌 = 𝜔. In particular, 𝜇 is a bijection. In addition, 𝜇 is continuous. Indeed, if 𝐶 is closed in 𝑈 (G), then, since 𝜔 is continuous, 𝜔−1 (𝐶) is closed in 𝑋; since 𝜌 : 𝑋 → 𝑈 (F ) is a continuous map of profinite spaces, and hence closed (Fact 1.1.3(e)), 𝜇−1 (𝐶) = 𝜌(𝜌 −1 (𝜇−1 (𝐶))) = 𝜌(𝜔−1 (𝐶)) is closed in 𝑈 (F ). By Fact 1.1.3(e), 𝜇 is a homeomorphism.

112

7 A Free Profinite Product over an Étale Compact Subset of Subgr(𝐺)

Part D: End of proof. Since rank(𝐺), rank( 𝐹ˆ𝜔 ) ≤ ℵ0 , by Lemma 7.2.1, 𝑈 (G) (hence also 𝑈 (F )) and 𝐹ˆ𝜔 have countable bases for their topologies. Therefore, so has the closed subset 𝑈 = 𝑈 (F ) ∪ 𝐹ˆ𝜔 of 𝐹. Again, by Lemma 7.2.1, rank(𝐹) ≤ ℵ0 . By Proposition 7.2.2, 𝐺  𝐹, as claimed. 

Chapter 8

Fundamental Result

We introduce the facts on classically local fields needed to prove the fundamental result of this monograph: Let 𝐹 be a countable ample Hilbertian field and let 𝑆 be a finite set of classically local primes of 𝐹. Suppose that Gal(𝐹) is GGal(𝐹 ),𝑆 -projective. Then, GGal(𝐹 ),𝑆,max has an étale profinite system R of representatives for its Gal(𝐹)-orbits and Gal(𝐹)  Î 𝐹ˆ𝜔 ∗ ∗ Γ∈R Γ (Proposition 8.4.3).

8.1 Classical Closures of a Field For each prime 𝔭 of a finite set 𝑆 of “classically local primes of a field 𝐾”, we consider the algebraic part 𝐾𝔭 of the completion of 𝐾 with respect to 𝔭. We prove that the set G𝐾 ,𝑆 := {Gal(𝐾𝔭 ) 𝜎 | 𝔭 ∈ 𝑆 and 𝜎 ∈ Gal(𝐾)} coincides with G𝐾 ,𝑆,max and is étale profinite. Definition 8.1.1 We consider a field 𝐾 and an equivalence class 𝔭 of discrete absolute values of 𝐾 [CaF67, p. 44, Def.] such that the completion 𝐾ˆ 𝔭 (which is defined up to 𝐾-isomorphism) is a classically local field. As introduced in Example 6.4.5, 𝐾ˆ 𝔭 is a finite extension of Q 𝑝 , a finite extension of F 𝑝 ((𝑡)), or R. In the first two cases we say that 𝔭 is a finite prime and in the third case we say that 𝔭 is an infinite prime. In both cases we also refer to 𝔭 as a classically local prime. Then, we embed 𝐾˜ in the algebraic closure of 𝐾ˆ 𝔭 and consider the 𝔭-closure 𝐾𝔭 = 𝐾sep ∩ 𝐾ˆ 𝔭 of 𝐾 at 𝔭. If 𝔭 is finite, 𝐾𝔭 is a Henselian closure of 𝐾 at each of the discrete valuations that belongs to 𝔭 [Efr06, p. 167, Cor. 18.3.3(b)]. If 𝔭 is infinite, then 𝐾𝔭 is a real closure of 𝐾 with respect to every ordering that belongs to 𝔭 [Pre84, p. 32, Lemma 3.13]. Setup 8.1.2 We consider a field 𝐾 and a finite set 𝑆 of classically local primes of 𝐾. We set Ù Ù 𝐾tot,𝑆 := 𝐾𝔭 . 𝔭∈𝑆 𝜎 ∈Gal(𝐾 )

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 D. Haran and M. Jarden, The Absolute Galois Group of a Semi-Local Field, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-030-89191-6_8

113

114

8 Fundamental Result

This field is a Galois extension of 𝐾 whose absolute Galois group will be described later. For each 𝔭 ∈ 𝑆, the set of all Henselian (resp. real) closures of 𝐾 at 𝔭 is {𝐾𝔭𝜎 } 𝜎 ∈Gal(𝐾 ) . See [EnP05, p. 121, paragraph preceding Thm. 5.2.2] for finite 𝔭 and [Lan93, p. 455, Thm. 2.9 and p. 456, Thm. 2.11] for infinite 𝔭. Hence, by Example 1.1.11, (8.1)

G𝐾 ,𝔭 := {Gal(𝐾𝔭 ) 𝜎 } 𝜎 ∈Gal(𝐾 ) = {Gal(𝐾𝔭𝜎 )} 𝜎 ∈Gal(𝐾 )

is a strictly closed subset of Subgr(Gal(𝐾)). Therefore, Ø (8.2) G𝐾 ,𝑆 := · G𝐾 ,𝔭 = {Gal(𝐾𝔭 ) 𝜎 | 𝔭 ∈ 𝑆, 𝜎 ∈ Gal(𝐾)} 𝔭∈𝑆

is also a strictly closed subset of Subgr(Gal(𝐾)). Observe that if 𝑀 is an extension of 𝐾 in 𝐾tot,𝑆 , then Gal(𝑀) acts on G𝐾 ,𝑆 by conjugation. In addition, we fix a set of prime numbers P of cardinality at least 3. By Example 6.4.5, G𝐾 ,𝑆 is of P-type. Remark 8.1.3 Note that we have not included C in the set of classically local fields. Accordingly, by our definition, none of the classically local primes 𝔭 is complex. In particular, 𝐾𝔭 is not separably closed, that is, Gal(𝐾𝔭 ) ≠ 1. It follows that 1 ∉ G𝐾 ,𝑆 . Lemma 8.1.4 Let 𝔭 be a classically local prime of 𝐾. Then, the restriction map (8.3) res𝔭 : Gal( 𝐾ˆ 𝔭 ) → Gal(𝐾𝔭 ) is an isomorphism. Proof By definition, 𝐾𝔭 = 𝐾sep ∩ 𝐾ˆ 𝔭 . First suppose that 𝔭 is an infinite prime. Then, 𝐾ˆ 𝔭 = R, hence 𝐾˜ 𝐾ˆ 𝔭 properly contains 𝐾ˆ 𝔭 , so 𝐾˜ 𝐾ˆ 𝔭 = C. Thus, both groups Gal(𝐾𝔭 ) and Gal( 𝐾ˆ 𝔭 ) are of order 2, so res𝔭 is an isomorphism. Next, consider the case where 𝔭 is a finite prime. By [Efr06, p. 172, Cor. 18.5.3], 𝐾sep 𝐾ˆ 𝔭 = 𝐾ˆ 𝔭,sep . Again, this implies that res𝔭 is an isomorphism.  Remark 8.1.5 The following statements hold for distinct classically local primes 𝔭 and 𝔮 of 𝐾. (8.4a) 𝐾𝔭 ≠ 𝐾sep . (8.4b) 𝐾𝔭𝜎 𝐾𝔭 = 𝐾sep if 𝜎 ∈ Gal(𝐾) r Gal(𝐾𝔭 ). (8.4c) 𝐾𝔭𝜎 𝐾𝔮 = 𝐾sep for all 𝜎 ∈ Gal(𝐾). Equivalently, (8.5a) Gal(𝐾𝔭 ) ≠ 1. (8.5b) Gal(𝐾𝔭 ) 𝜎 ∩ Gal(𝐾𝔭 ) = 1 if 𝜎 ∈ Gal(𝐾) r Gal(𝐾𝔭 ); in particular Aut(Gal(𝐾𝔭 )/𝐾) = 1. (8.5c) Gal(𝐾𝔭 ) 𝜎 ∩ Gal(𝐾𝔮 ) = 1 for all 𝜎 ∈ Gal(𝐾). These statements go back to Artin–Schreier in the infinite prime case, see [Lan93, p. 455, Thm. 2.9]. For finite primes the statements are due to F. K. Schmidt [Sch33], see [Efr06, p. 194, Cor. 21.1.4 and p. 195, Cor. 21.2.3].

8.2 Sections over Henselian Fields

115

Lemma 8.1.6 Let 𝑆 be a finite set of classically local primes of a field 𝐾. Then, G𝐾 ,𝑆 is strictly closed and G𝐾 ,𝑆 = G𝐾 ,𝑆,max . Moreover, the strict topology of G𝐾 ,𝑆 coincides with its étale topology. In particular, G𝐾 ,𝑆 is étale profinite. Proof We start with a proof of the equality G𝐾 ,𝑆 = G𝐾 ,𝑆,max . Assume toward contradiction that there exist 𝔭, 𝔭0 ∈ 𝑆 and 𝜎 ∈ Gal(𝐾) such that 𝐾𝔭 ⊂ (𝐾𝔭0 ) 𝜎 . Then, by Remark 8.1.5, (𝐾𝔭0 ) 𝜎 = 𝐾𝔭 𝐾𝔭𝜎0 = 𝐾sep , in contradiction to Remark 8.1.3. By Remark 8.1.5, G𝐾 ,𝑆,max = G𝐾 ,𝑆 . By Setup 8.1.2, G𝐾 ,𝑆 is strictly closed in Subgr(Gal(𝐾)). Since Subgr(Gal(𝐾)) is strictly profinite (Section 1.2), so is G𝐾 ,𝑆 . By Lemma 1.3.7, the strict topology of Subgr(Gal(𝐾)) coincides with its étale topology. Hence, G𝐾 ,𝑆 is also étale profinite, as claimed.  Remark 8.1.7 Having proved that the strict topology of G𝐾 ,𝑆 coincides with its étale topology, we drop the attributes “strict” and “étale” of G𝐾 ,𝑆 . In the case where 𝑆 is a singleton we apply the same convention for the sets G𝐾 ,𝔭 with 𝔭 ∈ 𝑆. We apply Remark 8.1.5 to give a crucial example of a Cantor space. Lemma 8.1.8 Let 𝐾 be a countable field. Let 𝑀 be an infinite extension of 𝐾 in 𝐾tot,𝑆 and R an étale compact set of representatives for the Gal(𝑀)-orbits of G𝐾 ,𝑆 . Then: Ð (a) R = · 𝔭∈𝑆 R 𝔭 , where each R 𝔭 is a closed set of representatives for the Gal(𝑀)-orbits of G𝐾 ,𝔭 (equation (8.1)). (b) R is a Cantor space. (c) For each 𝔭 ∈ 𝑆 there exists a closed subset 𝑅𝔭 of Gal(𝐾) such that R 𝔭 = {Gal(𝐾𝔭 ) 𝜌 | 𝜌 ∈ 𝑅𝔭 } and the map 𝜌 ↦→ Gal(𝐾𝔭 ) 𝜌 is a homeomorphism of 𝑅𝔭 onto R 𝔭 . Proof Our lemma is a special case of Lemma 1.5.8. Indeed, let 𝐺 = Gal(𝐾), 𝐻 = Gal(𝑀), 𝑁 = Gal(𝐾tot,𝑆 ), and {𝐽1 , . . . , 𝐽𝑛 } = {Gal(𝐾𝔭 )}𝔭∈𝑆 . Then, rank(𝐺) ≤ ℵ0 , because 𝐾 is countable, 𝑁 is a closed normal subgroup of 𝐺, because 𝐾tot,𝑆 is a Galois extension of 𝐾 (Setup 8.1.2), G := G𝐾 ,𝑆 is a closed subset of Subgr(𝐺) (Setup 8.1.2), and R is a closed subset of representatives for the 𝐻-orbits of G (by assumption). Then, (8.5a), (8.5b), and (8.5c) translate into Conditions (a1) and (a2) of Lemma 1.5.8. It follows that (b1), (b2), and (b3) of this lemma hold true. Those conditions translate for their part to statements (a), (b), and (c) of our lemma. 

8.2 Sections over Henselian Fields For an arbitrary valuation field (𝐹, 𝑣) we denote the residue field of 𝐹 under 𝑣 by 𝐹¯𝑣 ¯ if 𝑣 is clear from the context. Following [HJP07, Sec. 7], we consider a or also 𝐹, ¯ 𝐹¯ is a normal extension. Galois extension (𝑁, 𝑣)/(𝐹, 𝑣) of Henselian fields. Then, 𝑁/ ¯ ¯ For each 𝜎 ∈ Gal(𝑁/𝐹) define 𝜎 ¯ ∈ Aut( 𝑁/𝐹) by this rule: 𝜎 ¯ 𝑥¯ = 𝜎𝑥 for 𝑥 ∈ 𝑁

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¯ 𝐹). ¯ with 𝑣(𝑥) ≥ 0. The map 𝜎 ↦→ 𝜎 ¯ is an epimorphism 𝜌 : Gal(𝑁/𝐹) → Aut( 𝑁/ Its kernel is the inertia group: 𝐺 0 (𝑁/𝐹) = {𝜎 ∈ Gal(𝑁/𝐹) | 𝑣(𝜎𝑥 − 𝑥) > 0 for each 𝑥 ∈ 𝑁 with 𝑣(𝑥) ≥ 0}. Denote the fixed field in 𝑁 of 𝐺 0 (𝑁/𝐹) by 𝑁0 . Then, 𝑁¯ 0 is the maximal separable extension of 𝐹¯ in 𝑁¯ [End72, p. 152, Thm. 19.12]. Hence, 𝑁¯ 0 /𝐹¯ is Galois and the ¯ 𝐹) ¯ isomorphically onto Gal( 𝑁¯ 0 /𝐹). ¯ By [End72, p. 152, restriction maps Aut( 𝑁/ Thm. 19.12] or [Efr06, p. 141, Thm. 16.1.1], there is a short exact sequence 𝜌

¯ −→ 1. 1 −→ Gal(𝑁/𝑁0 ) −→ Gal(𝑁/𝐹) −→ Gal( 𝑁¯ 0 /𝐹) ¯ 𝐹) ¯ with its restriction to 𝑁¯ 0 . In addition, Here we have identified each 𝜎 ¯ ∈ Aut( 𝑁/ × × 𝑣(𝑁0 ) = 𝑣(𝐹 ) [End72, p. 153, Cor. 19.14]. Hence, 𝑁0 /𝐹 is an unramified extension. The ramification group of Gal(𝑁/𝐹) is  𝜎𝑥  𝐺 1 (𝑁/𝐹) = {𝜎 ∈ Gal(𝑁/𝐹) | 𝑣 − 1 > 0 for each 𝑥 ∈ 𝑁 × }. 𝑥 It is a normal subgroup of Gal(𝑁/𝐹) which is contained in 𝐺 0 (𝑁/𝐹) [End72, p. 159, ¯ > 0, Cor. 20.6]. Denote the fixed field of 𝐺 1 (𝑁/𝐹) in 𝑁 by 𝑁1 . When 𝑝 = char( 𝐹) ¯ Gal(𝑁/𝑁1 ) is the unique 𝑝-Sylow subgroup of Gal(𝑁/𝑁0 ). When char( 𝐹) = 0, Gal(𝑁/𝑁1 ) is trivial. See [End72, p. 167, Thm. 20.18] or [Efr06, p. 146, 2nd ¯ does not divide [𝑁1 : 𝑁0 ]. paragraph]. So, in both cases, char( 𝐹) Suppose now that 𝑁 = 𝐹sep . Then, 𝑁0 = 𝐹𝑢 is the inertia field and 𝑁1 = 𝐹𝑟 is the ramification field of 𝐹. In this case (8.6) becomes the short exact sequence (8.6)

𝜌

(8.7)

¯ −→ 1. 1 −→ Gal(𝐹𝑢 ) −→ Gal(𝐹) −→ Gal( 𝐹)

Also, 𝐹 ⊆ 𝐹𝑢 ⊆ 𝐹𝑟 ⊆ 𝐹sep , 𝐹𝑢 /𝐹 and 𝐹𝑟 /𝐹 are Galois extensions, and ¯ - [𝐹𝑟 : 𝐹𝑢 ]. In addition, Gal(𝐹𝑟 ) is a pro-𝑝 group if 𝑝 = char( 𝐹) ¯ ≠ 0 char( 𝐹) ¯ and Gal(𝐹𝑟 ) = 1 if char( 𝐹) = 0. Consider now a finite extension (𝐿, 𝑣)/(𝐹,
𝑣) of Henselian fields. As in Remark 6.3.3, let 𝑒 = 𝑒(𝐿/𝐹) = 𝑣(𝐿 × ) : 𝑣(𝐹 × ) be the ramification index and let 𝑑 be ¯ = 𝑝 > 0 and 𝑑 = 1 if the defect of 𝐿/𝐹. Recall that 𝑑 is a power of 𝑝 if char( 𝐹) ¯ char( 𝐹) = 0. When 𝑑 = 1, we say that 𝐿/𝐹 is defectless. In each case, ¯ (8.8) [𝐿 : 𝐹] = 𝑑 · (𝑣(𝐿 × ) : 𝑣(𝐹 × )) [ 𝐿¯ : 𝐹]. An arbitrary algebraic extension 𝑀/𝐹 is defectless if each finite subextension is ¯ - [𝑀 : 𝐹]. For example, 𝐹𝑟 /𝐹𝑢 is defectless. This is the case when char( 𝐹) defectless. In addition, by (8.7), ¯ for each finite subextension 𝐿/𝐹 of 𝐹𝑢 /𝐹. (8.9) [𝐿 : 𝐹] = [ 𝐿¯ : 𝐹] Hence, 𝐹𝑢 /𝐹 is unramified and defectless. Consequently, 𝐹𝑟 /𝐹 is defectless. Lemma 8.2.1 Let (𝐹, 𝑣) be a Henselian valued field. Use the above notation. (a) There is a field 𝐹 0 with 𝐹𝑢 𝐹 0 = 𝐹𝑟 and 𝐹𝑢 ∩ 𝐹 0 = 𝐹. (b) The short exact sequence 1 → Gal(𝐹𝑟 /𝐹𝑢 ) → Gal(𝐹𝑟 /𝐹) → Gal(𝐹𝑢 /𝐹) → 1 splits.

8.2 Sections over Henselian Fields

117

Proof Statement (b) is a Galois-theoretic interpretation of (a), hence it suffices to prove (a). ¯ Zorn’s lemma gives a maximal extension 𝐹 0 of 𝐹 in 𝐹𝑟 with residue field 𝐹. ¯ the value group of 𝐹 0 is Claim: For each prime number 𝑙 ≠ char( 𝐹) 𝑙-divisible. √ Otherwise, there is an 𝑎 ∈ 𝐹 0 with 𝑣(𝑎) ∉ 𝑙𝑣((𝐹 0) × ). Put 𝐿 = 𝐹 0 ( 𝑙 𝑎). In particular, 𝑎 ∉ (𝐹 0) 𝑙 , so by [Lan93, p. 297, Thm. 9.1], [𝐿 : 𝐹 0] = 𝑙. Also, 𝑣(𝐿 × )/𝑣((𝐹 0) × ) contains an element of order 𝑙, so, by (8.8), 𝑙 ≤ (𝑣(𝐿 × ) : 𝑣((𝐹 0) × )) ≤ [𝐿 : 𝐹 0] = 𝑙. Hence, (𝑣(𝐿 × ) : 𝑣((𝐹 0) × )) = 𝑙, so again by (8.8), [ 𝐿¯ : 𝐹 0] = 1. Now observe that 𝐹𝑟 /𝐹 is a Galois extension, so 𝐹𝑟 /𝐹 0 is also a Galois extension. If 𝐿 6 ⊆ 𝐹𝑟 , then 𝑙 = [𝐿 : 𝐹 0] = [𝐹𝑟 𝐿 : 𝐹𝑟 ]. This contradicts the fact that Gal(𝐹𝑟 ) is a pro-𝑝 group if ¯ = 𝑝 > 0 and trivial if char( 𝐹) ¯ = 0. It follows from this contradiction that char( 𝐹) 𝐿 ⊆ 𝐹𝑟 . Again, this contradicts the maximality of 𝐹 0. The latter contradiction proves our claim. Having proved the Claim, we consider a finite extension 𝑀 of 𝐹 in 𝐹 0. Then, ¯ ¯ so by (8.9), 𝐹𝑢 ∩ 𝑀 = 𝐹. 𝑀 = 𝐾¯ (by the definition of 𝐹 0), hence 𝐹𝑢 ∩ 𝑀 = 𝐾, 0 Therefore, 𝐹𝑢 ∩ 𝐹 = 𝐹. ¯ Since 𝐸/𝐹 0 is an algebraic Let 𝐸 = 𝐹𝑢 𝐹 0. Consider a prime number 𝑙 ≠ char( 𝐹). extension, 𝑣(𝐸 × ) is contained in the divisible hull of 𝑣(𝐹 0). Since 𝑣((𝐹 0) × ) is 𝑙-divisible, so is 𝑣(𝐸 × ). Since 𝐹𝑢 ⊆ 𝐸 ⊆ 𝐹𝑟 ⊆ 𝐹sep and 𝐹¯𝑢 = 𝐹¯sep , we have 𝐸¯ = 𝐹¯𝑟 . ¯ = 1 and Hence, for every finite extension 𝐸 0 of 𝐸 in 𝐹𝑟 we have 𝑒(𝐸 0/𝐸) = [𝐸 0 : 𝐸] 0 0 0 therefore [𝐸 : 𝐸] = 1 (because 𝐸 /𝐸 is defectless). Consequently, 𝐹𝑢 𝐹 = 𝐸 = 𝐹𝑟 , as claimed.  The following result is due to Kuhlmann–Pank–Roquette. Lemma 8.2.2 ([KPR86, Thm. 2.2]) Let (𝐹, 𝑣) be a Henselian field. (a) There is a field 𝐹 0 with 𝐹𝑟 ∩ 𝐹 0 = 𝐹 and 𝐹𝑟 𝐹 0 = 𝐹sep . (b) The short exact sequence 1 → Gal(𝐹𝑟 ) → Gal(𝐹) → Gal(𝐹𝑟 /𝐹) → 1 splits. Proof Statement (a) is a Galois-theoretic interpretation of (b). So, we prove (b). Let ¯ If 𝑝 = 0, then 𝐹𝑟 = 𝐹sep and we may take 𝐹 0 = 𝐹. Suppose 𝑝 ≠ 0. 𝑝 = char( 𝐹). ¯ By a theorem of Witt, cd 𝑝 (Gal( 𝐹)) ¯ ≤ 1 [Rib70, By (8.7), Gal(𝐹𝑢 /𝐹)  Gal( 𝐹). p. 256, Thm. 3.3]. Hence, cd 𝑝 (Gal(𝐹𝑢 /𝐹)) ≤ 1. Since 𝑝 - [𝐹𝑟 : 𝐹𝑢 ] (by the statement after (8.7)), cd 𝑝 (Gal(𝐹𝑟 /𝐹𝑢 )) = 0 [Rib70, p. 208, Cor. 2.3]. Hence, by [Rib70, p. 209, Prop. 2.6], cd 𝑝 (Gal(𝐹𝑟 /𝐹)) ≤ 1. Since Gal(𝐹𝑟 ) is a pro-𝑝 group, the short sequence in (b) splits [Rib70, p. 211, Prop. 3.1(iii)’].  Lemma 8.2.3 Let (𝐹, 𝑣) be a valued field. (a) Suppose (𝐹, 𝑣) is Henselian. Then, the epimorphism ¯ 𝜌 : Gal(𝐹) → Gal( 𝐹) induced by reduction at 𝑣 splits. ¯ is isomorphic to a subgroup of Gal(𝐹). (b) Each subgroup of Gal( 𝐹)

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Proof Proof of (a). The map 𝜌 decomposes as 𝜌¯

res res ¯ Gal(𝐹) −→ Gal(𝐹𝑟 /𝐹) −→ Gal(𝐹𝑢 /𝐹) −→ Gal( 𝐹).

The map 𝜌, ¯ which is also induced by reduction, is an isomorphism (by (8.7)). By Lemmas 8.2.1 and 8.2.2, each of the restriction maps splits. Hence, 𝜌 splits. ¯ By (a), each Proof of (b). Let (𝐹 0, 𝑣) be the Henselization of (𝐹, 𝑣). Then, 𝐹 0 = 𝐹. ¯ is isomorphic to a subgroup of Gal(𝐹 0), hence of Gal(𝐹).  subgroup of Gal( 𝐹) Lemma 8.2.4 Let 𝐹/𝐾 be an extension of fields. Suppose 𝑣 is a valuation of 𝐹 which is trivial on 𝐾 and 𝐹¯ = 𝐾. Then: (a) res : Gal(𝐹) → Gal(𝐾) is an epimorphism which splits. If, in addition, (𝐹, 𝑣) is Henselian and 𝑣 is extended to 𝐹sep such that 𝑎¯ = 𝑎 for each 𝑎 ∈ 𝐾sep , then res is the epimorphism induced by reduction at 𝑣. (b) (𝐹, 𝑣) has a separable algebraic Henselian extension (𝐹 0, 𝑣) such that res : Gal(𝐹 0) → Gal(𝐾) is an isomorphism and 𝐹 0 is a purely inseparable extension of 𝐾. (c) Suppose 𝐾 is perfect. Then, (𝐹, 𝑣) has an algebraic Henselian extension (𝐹 00, 𝑣) such that 𝐹 00 is perfect, 𝐹 00 = 𝐾, and res : Gal(𝐹 00) → Gal(𝐾) is an isomorphism. Proof Replace (𝐹, 𝑣) by a Henselian closure, if necessary, to assume (𝐹, 𝑣) is Henselian. Let 𝜌 : Gal(𝐹) → Gal(𝐾) be the epimorphism induced by reduction at ¯ 𝑎¯ = 𝜎𝑎 ¯ = 𝑣. Then, for each 𝑎 ∈ 𝐾sep and each 𝜎 ∈ Gal(𝐹) we have, 𝜎𝑎 = 𝜎𝑎 = 𝜎 𝜌(𝜎)𝑎. Thus, res : Gal(𝐹) → Gal(𝐾) coincides with 𝜌. Lemma 8.2.3(a) gives a section 𝜌 0 : Gal(𝐾) → Gal(𝐹) of 𝜌. Let 𝐹 0 be the fixed field of 𝜌 0 (Gal(𝐾)) in 𝐹sep . Then, Gal(𝐹 0) → Gal(𝐾) is an isomorphism. Also, for all 𝑢 ∈ 𝐹 0 with 𝑣(𝑢) ≥ 0 and 𝜎 ∈ Gal(𝐹 0) we have 𝜎 ¯ 𝑢¯ = 𝜎𝑢 = 𝑢. ¯ Hence, 𝐹 0 is a purely inseparable extension of 𝐾. This concludes the proof of (a) and (b). 0 satisfies (c). When 𝐾 is perfect, 𝐹 00 = 𝐹ins 

8.3 G-Projectivity If 𝐾 is a PAC field, then Gal(𝐾) is projective [FrJ08, p. 207, Thm. 11.6.2]. Likewise, we prove in this section that if 𝐾 is PXC for a set X of separable field extensions of 𝐾 and G is the set of absolute Galois groups of those extensions, then Gal(𝐾) is G-projective. Lemma 8.3.1 Let 𝐿/𝐾 be a finite Galois extension. Let 𝜓 : 𝐵 → Gal(𝐿/𝐾) be an epimorphism of finite groups. Then, there exists a finitely generated regular extension 𝐸 of 𝐾 and a finite Galois extension 𝐹 of 𝐸 containing 𝐿 such that 𝐵 = Gal(𝐹/𝐸) and 𝜓 is the restriction res𝐹 /𝐿 : Gal(𝐹/𝐸) → Gal(𝐿/𝐾). Moreover, let 𝐾 ⊆ 𝐿 0 ⊆ 𝐿 and 𝐸 ⊆ 𝐹0 ⊆ 𝐹 be fields with 𝐿 0 ⊆ 𝐹0 . Suppose 𝜓 : Gal(𝐹/𝐹0 ) → Gal(𝐿/𝐿 0 ) is an isomorphism. Then, 𝐹0 is a purely transcendental extension of 𝐿 0 of transcendence degree card(𝐵).

8.3 G-Projectivity

119

Proof Let 𝑥 𝛽 , with 𝛽 ranging over 𝐵, be algebraically independent elements over 𝐾. 0 0 0 0 Define a faithful action of 𝐵 on 𝐹 = 𝐿(𝑥 𝛽 )𝛽 ∈𝐵 by (𝑥 𝛽 ) 𝛽 = 𝑥 𝛽𝛽 and 𝑎 𝛽 = 𝑎 𝜓 (𝛽 ) for 𝑎 ∈ 𝐿 and 𝛽 0 ∈ 𝐵. Denote the fixed field of 𝐵 in 𝐹 by 𝐸. Then, 𝐹/𝐸 is a Galois extension with Galois group 𝐵. By construction, 𝐹/𝐾 is a finitely generated separable extension. Hence, by [Lan64, p. 64, Prop. 6], 𝐸/𝐾 is also a finitely generated separable extension. In addition, res : Gal(𝐹/𝐸) → Gal(𝐿/𝐾) coincides with 𝜓 : 𝐵 → Gal(𝐿/𝐾). Hence, 𝐸 ∩ 𝐾˜ = 𝐸 ∩ 𝐹 ∩ 𝐾˜ = 𝐸 ∩ 𝐿 = 𝐾. Therefore, 𝐸/𝐾 is regular. Now let 𝐿 0 and 𝐹0 be as in the second paragraph of the lemma. Set 𝐵0 = Gal(𝐹/𝐹0 ) and choose a set 𝑅 of representatives for the left cosets of 𝐵 modulo 𝐵0 . Let 𝑤 1 , . . . , 𝑤 𝑚 be a basis for 𝐿/𝐿 0 . By assumption, 𝑚 = card(𝐵0 ). Consider 𝜌 ∈ 𝑅 and let Õ 𝛽 𝑡𝜌 𝑗 = 𝑤 𝑗 𝑥 𝜌𝛽 , 𝑗 = 1, . . . , 𝑚. 𝛽 ∈𝐵0 𝛽

Since det(𝑤 𝑗 ) ≠ 0, each 𝑥 𝜌𝛽 is a linear combination of 𝑡 𝜌 𝑗 with coefficients in 𝐿. Put t = (𝑡 𝜌 𝑗 | 𝜌 ∈ 𝑅, 𝑗 = 1, . . . , 𝑚), x = (𝑥 𝛽 | 𝛽 ∈ 𝐵). Both tuples contain exactly card(𝐵) elements and 𝐿(t) = 𝐿 (x) = 𝐹. So, 𝐿 0 (t) is a purely transcendental extension of 𝐿 0 . Each 𝑡 𝜌 𝑗 is fixed by 𝐵0 . Hence, 𝐿 0 (t) ⊆ 𝐹0 . Moreover, we have 𝑚 = [𝐿 : 𝐿 0 ] = [𝐿 (t) : 𝐿 0 (t)] = [𝐹 : 𝐿 0 (t)] ≥ [𝐹 : 𝐹0 ] = card(𝐵0 ) = 𝑚. Consequently, 𝐹0 = 𝐿 0 (t) and 𝐹0 /𝐿 0 is purely transcendental.  Definition 8.3.2 Let 𝐾 be a field and X a set of extensions of 𝐾 in 𝐾sep . We say that 𝐾 is PXC if every geometrically integral algebraic variety over 𝐾 with a simple 𝐾 0-rational point for each 𝐾 0 in X has a 𝐾-rational point. The following result is a special case of Corollary A2 of [JaR80]. Lemma 8.3.3 Let 𝑉 be a geometrically integral algebraic variety over a field 𝐾 with function field 𝐸. Suppose that 𝑉 has a simple 𝐾-rational point. Then, 𝐸 has a valuation which is trivial on 𝐾 and with 𝐾 as its residue field. We also need the following elementary result: Lemma 8.3.4 Let 𝑉 be a geometrically integral variety over an infinite field 𝑁. Suppose that the function field of 𝑉 is contained in a purely transcendental extension of 𝑁. Then, 𝑉 (𝑁) ≠ ∅. Proof Let (𝑥1 , . . . , 𝑥 𝑛 ) be a generic point of 𝑉. By assumption, 𝑁 (𝑥1 , . . . , 𝑥 𝑛 ) ⊆ 𝑁 (𝑡1 , . . . , 𝑡𝑟 ), with 𝑡1 , . . . , 𝑡𝑟 algebraically independent over 𝑁. Thus, 𝑓𝑖 (𝑡1 , . . . , 𝑡𝑟 ) 𝑥𝑖 = for 𝑖 = 1, . . . , 𝑛, 𝑓0 (𝑡 1 , . . . , 𝑡𝑟 ) where 𝑓0 , 𝑓1 , . . . , 𝑓𝑟 are polynomials with coefficients in 𝑁 and 𝑓0 ≠ 0. If 𝑟 = 0, then (𝑥1 , . . . , 𝑥 𝑛 ) is an 𝑁-rational point of 𝑉. If 𝑟 = 1, then 𝑓0 has at most finitely many roots in 𝑁. Since 𝑁 is infinite, induction on 𝑟 gives 𝑎 1 , . . . , 𝑎𝑟 ∈ 𝑁 such that 𝑓0 (𝑎 1 , . . . , 𝑎𝑟 ) ≠ 0. For 𝑖 = 1, . . . , 𝑛 let 𝑏 𝑖 = 𝑓𝑖 (𝑎 1 , . . . , 𝑎 𝑟 )/ 𝑓0 (𝑎 1 , . . . , 𝑎𝑟 ). Then, (𝑏 1 , . . . , 𝑏 𝑛 ) ∈ 𝑉 (𝑁), as desired. 

120

8 Fundamental Result

Lemma 8.3.5 Let 𝑀 be an infinite field and X a set of extensions of 𝑀 in 𝑀sep . Put G = {Gal(𝑀 0)} 𝑀 0 ∈X . Suppose that G is an étale compact subset of Gal(𝑀) and 𝑀 is PXC . Then, Gal(𝑀) is G-projective. Proof By Corollary 1.4.3, G Gal( 𝑀 ) is étale compact. If we prove that Gal(𝑀) is G Gal( 𝑀 ) -projective, it will follow from Lemma 5.2.4 that Gal(𝑀) is G-projective. Also, if a geometrically integral variety 𝑉 over 𝑀 has an 𝑀 0-rational point x for an extension 𝑀 0 of 𝑀 in 𝑀sep , then 𝑉 has an (𝑀 0) 𝜎 -rational point x 𝜎 for every 𝜎 ∈ Gal(𝑀). We may therefore assume that G is Gal(𝑀)-invariant. By Corollary 5.1.5, it suffices to solve every finite rigid G-embedding problem (8.10)

(𝜑 : Gal(𝑀) → 𝐴, 𝛼 : 𝐵 → 𝐴, B),

where 𝜑 is an epimorphism. Let 𝑁 be the fixed field of Ker(𝜑) in 𝑀sep . Then, identify 𝐴 with Gal(𝑁/𝑀) and 𝜑 with res 𝑀sep /𝑁 . Next, use Lemma 8.3.1 to construct a finitely generated regular extension 𝐸 of 𝑀 and a finite Galois extension 𝐹 of 𝐸 containing 𝑁 with these properties: (8.11a) 𝐵 = Gal(𝐹/𝐸) and 𝛼 is the restriction res𝐹 /𝑁 : Gal(𝐹/𝐸) → Gal(𝑁/𝑀). (8.11b) Let 𝑁0 be a field between 𝑀 and 𝑁 and 𝐹0 a field between 𝐸 and 𝐹 which contains 𝑁0 . Suppose res𝐹 /𝑁 : Gal(𝐹/𝐹0 ) → Gal(𝑁/𝑁0 ) is an isomorphism. Then, 𝐹0 is a purely transcendental extension of 𝑁0 . Since 𝐸/𝑀 is finitely generated and regular, one may view 𝐸 as the function field of a geometrically integral affine variety 𝑉 over 𝑀 [FrJ08, p. 175, Cor. 10.2.2]. Replacing 𝑉 by a Zariski-open non-empty subvariety, we may even assume that 𝑉 is affine and smooth. Now let {𝑁𝑖 }𝑖 ∈𝐼 = {𝑀 0 ∩ 𝑁 } 𝑀 0 ∈X . By rigidity, choose for each 𝑖 ∈ 𝐼 a field 𝐹𝑖 between 𝐸 and 𝐹 containing 𝑁𝑖 such that res𝐹 /𝑁 : Gal(𝐹/𝐹𝑖 ) → Gal(𝑁/𝑁𝑖 ) is an isomorphism. By (8.11b), 𝐹𝑖 is a purely transcendental extension of 𝑁𝑖 . Since 𝑀 is infinite, so is 𝑁𝑖 . Since 𝐸 𝑁𝑖 ⊆ 𝐹𝑖 , Lemma 8.3.4 guarantees that 𝑉 (𝑁𝑖 ) ≠ ∅. Therefore, 𝑉 (𝑀 0) ≠ ∅ for each 𝑀 0 ∈ X. Since 𝑀 is PXC, 𝑉 has an 𝑀-rational point, which by assumption is simple. By Lemma 8.3.3, 𝐸 has a valuation which is trivial on 𝑀 and with 𝑀 as its residue field. Lemma 8.2.4(b) gives an algebraic extension 𝐸 0 of 𝐸 such that 0 0 /𝑀 res𝐸sep sep : Gal(𝐸 ) → Gal(𝑀) 0 /𝐹 ◦ 𝛾 0 solves (8.10), is an isomorphism. Denote its inverse by 𝛾 0. Then, 𝛾 = res𝐸sep as desired. 

8.4 Semi-Local Fields We prove the fundamental result of this monograph. Definition 8.4.1 We say that a field 𝑀 is ample if 𝑉 (𝑀) is Zariski-dense in 𝑉 for every geometrically integral algebraic variety 𝑉 over 𝑀 with a simple 𝑀-rational point [Jar11, p. 68, Def. 5.3.2].

8.4 Semi-Local Fields

121

By [Pop96, Prop. 1.2] or [Jar11, p. 72, Lemma 5.5.1], every algebraic extension of an ample field is ample. The following result can be considered as a local-global principle for proper G-embedding problems. Lemma 8.4.2 Let 𝑀 be an ample Hilbertian field and let G be a strictly closed Gal(𝑀)-invariant subset of Subgr(Gal(𝑀)) of P-type (Setup 8.1.2). Suppose that Gal(𝑀) is G-projective. Then, every finite proper G-embedding problem for Gal(𝑀) has a proper strong solution. Proof By Remark 1.2.7, G is étale compact. Hence, by Preposition 6.4.8, it suffices to prove that every finite proper G-embedding problem (8.12)

(𝜑 : Gal(𝑀) → 𝐴, 𝛼 : 𝐵 → 𝐴, B)

has a proper solution. By Lemma 5.1.5, we may assume that this problem splits. Then, by [Pop96, Main Theorem B] or [Jar11, p. 89, Thm. 5.10.2(a)], embedding problem (8.12) is properly solvable. Note that the latter source uses “solution” rather than “proper solution”.  We are now in a position to prove the fundamental result of this monograph, due to Pop [Pop96, Thm. 2.8]. Proposition 8.4.3 As in Setup 8.1.2, let 𝑆 be a finite set of classically local primes of a countable field 𝐾. Consider an extension 𝑀 of 𝐾 in 𝐾tot,𝑆 which is ample and Hilbertian. Set G = G𝐾 ,𝑆 and suppose that Gal(𝑀) is G-projective. Then, G = Gmax and G has an étale profinite system R of representatives for its Gal(𝑀)-orbits such Î that Gal(𝑀)  𝐹ˆ𝜔 ∗ ∗ Γ∈R Γ. Proof By Setup 8.1.2, G is a Gal(𝑀)-invariant subset of Subgr(Gal(𝑀)). By Lemma 8.1.6, G = Gmax is strictly closed and étale profinite. In particular, G is étale compact. By Setup 8.1.2, G is of P-type. Since Gal(𝑀) is G-projective, it follows from Lemma 8.4.2 that every finite proper G-embedding problem for Gal(𝑀) has a proper strong solution. Since 𝑀 is countable, rank(Gal(𝑀)) ≤ ℵ0 . It follows from Lemma 7.1.3 that G has an étale compact system R of representatives for its Gal(𝑀)-orbits. Since G is étale profinite, also R is étale profinite. Every Γ ∈ G is Gal(𝑀)-conjugate to a subgroup of some Γ0 ∈ R. Hence, every finite R-embedding problem for Gal(𝑀) has a proper strong solution. Thus, Gal(𝑀) is properly strongly R-projective. By definition, R is also a set of representatives 𝑀)) for the distinct conjugacy classes of (R Gal(Î max . It follows from Proposition 7.2.3, with R replacing G, that Gal(𝑀)  𝐹ˆ𝜔 ∗ ∗ Γ∈R Γ, as claimed. 

Chapter 9

Main Result

As in Section 8.1 we fix a countable Hilbertian field 𝐾, a finite set 𝑆 of classically local primes of 𝐾, a positive integer 𝑒, and for each 𝔭 ∈ 𝑆 consider the 𝔭-topology of 𝐾. The intersections of basic 𝔭-open subsets of 𝐾 with 𝔭 ranging over 𝑆 form a basis for the 𝑆-adic topology of 𝐾. Let 𝐾𝔭 and 𝐾tot,𝑆 be as in Setup 8.1.2. For each 𝝈 = (𝜎1 , . . . , 𝜎𝑒 ) ∈ Gal(𝐾) 𝑒 we consider the fixed field 𝐾tot,𝑆 (𝝈) = {𝑥 ∈ 𝐾tot,𝑆 | 𝑥 𝜎𝑖 = 𝑥 for 𝑖 = 1, . . . , 𝑒} of 𝝈 in 𝐾tot,𝑆 . Let 𝐾tot,𝑆 [𝝈] be the maximal Galois extension of 𝐾 in 𝐾tot,𝑆 (𝝈). We prove in this chapter that for almost all 𝝈 ∈ Gal(𝐾) 𝑒 the absolute Galois group Gal(𝐾tot,𝑆 [𝝈]) is the free product of 𝐹ˆ𝜔 with the free product of some of 𝜌 the groups Gal(𝐾𝔭 ), where 𝔭 ∈ 𝑆 and 𝜌 ∈ Gal(𝐾). The latter factor depends, up to isomorphism, only on 𝑆. Likewise, we prove that the absolute Galois group Gal(𝐾tot,𝑆 (𝝈)) is the free 𝜌 product of 𝐹ˆ𝑒 with the free product of some of the groups Gal(𝐾𝔭 ).

9.1 The Groups Gal(𝑲tot,𝑺 [𝝈]) This section contains the main result of this monograph mentioned above. Definition 9.1.1 A field extension 𝐸 of 𝐾 in 𝐾tot,𝑆 is said to be P𝑆C (pseudo 𝑆-closed) 𝜌 if every geometrically integral variety 𝑉 over 𝐸 with a simple 𝐾𝔭 -point for all 𝔭 ∈ 𝑆 and 𝜌 ∈ Gal(𝐾) has an 𝐸-rational point. Note that 𝐸 is P𝑆C if and only if 𝐸 is PXC 𝜌 with respect to the set X = {𝐾𝔭 } {𝔭∈𝑆, 𝜌∈Gal(𝐾 ) } (Definition 8.3.2). By [Pop96, Prop. 3.1] or [Jar11, p. 75, Example 5.6.4], every P𝑆C field is ample (Definition 8.4.1). We also note that if 𝐸 is a P𝑆C-extension of 𝐾 in 𝐾tot,𝑆 and 𝑀 is an extension of 𝐸 in 𝐾tot,𝑆 , then 𝑀 is also P𝑆C. This statement is a special case of [Jar91a, Lemma 7.2] that we do not use in this monograph.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 D. Haran and M. Jarden, The Absolute Galois Group of a Semi-Local Field, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-030-89191-6_9

123

124

9 Main Result

Remark 9.1.2 (Cantor free products) We consider the special case of Construction 4.7.7, where 𝑆 is the finite set of prime divisors of 𝐾 that we have fixed in this section and 𝐺˜ = Gal(𝐾). Moreover, for each 𝔭 ∈ 𝑆 the closed subset 𝑅𝔭 of 𝐺˜ that we consider is now a Cantor space and 𝐺 𝔭 = Gal(𝐾𝔭 ). Then, the corresponding semi-constant sheaf Ø Ø X = ( · 𝑅𝔭 × Gal(𝐾𝔭 ), 𝜏, · 𝑅𝔭 ) 𝔭∈𝑆

𝔭∈𝑆

depends, up to isomorphism, only on 𝑆. We call X a Cantor semi-constant sheaf. If a closed subgroup 𝐺 of Gal(𝐾) is isomorphic to Ö Ö ∗ ∗ Gal(𝐾𝔭 ) 𝜌 , 𝔭∈𝑆 𝜌∈𝑅𝔭

then 𝐺 depends, up to isomorphism, only on 𝑆 (Construction 4.7.7). We say that 𝐺 is a Cantor free product over 𝑆. Theorem 27.4.8 of [FrJ08] asserts for a countable Hilbertian field 𝐾 and a positive integer 𝑒 that for almost all 𝝈 ∈ Gal(𝐾) 𝑒 the field 𝐾sep [𝝈] is Hilbertian. A surprising result of Lior Bary-Soroker and Arno Fehm [BSF13, Thm. 1.1] immensely generalizes this assertion. They consider a Galois extension 𝑁 of a field 𝐾 and an 𝑒-tuple 𝝈 := (𝜎1 , . . . , 𝜎𝑒 ) ∈ Gal(𝐾) 𝑒 . Let 𝑁 (𝝈) = 𝑁 ∩ 𝐾sep (𝝈) and let 𝑁 [𝝈] = 𝑁 ∩ 𝐾sep [𝝈] be the maximal Galois extension of 𝐾 in 𝑁 (𝝈). Then: Proposition 9.1.3 Let 𝐾 be a countable Hilbertian field, 𝑁 a Galois extension of 𝐾, and 𝑒 a positive integer. Then, for almost all 𝝈 ∈ Gal(𝐾) 𝑒 the field 𝑁 [𝝈] is Hilbertian. Remark 9.1.4 It is well known that the field 𝑁 appearing in Proposition 9.1.3 need not be Hilbertian. For example, the maximal solvable extension Qsolv of Q is not Hilbertian, because it has no quadratic extension. More subtle is the main result of [BSF14] saying that if 𝐾 and 𝑆 are as in the introduction to this chapter, then 𝐾tot,𝑆 is not Hilbertian Lemma 9.1.5 Let 𝐾 be a Hilbertian field, 𝑆 a finite set of classically local primes of 𝐾, and 𝑒 a positive integer. Then, for almost all 𝝈 ∈ Gal(𝐾) 𝑒 the field 𝐾tot,𝑆 [𝝈] is an infinite extension of 𝐾. Proof Proceeding by induction, assume that there are linearly disjoint separable quadratic extensions 𝐿 1 , . . . , 𝐿 𝑛 of 𝐾 contained in 𝐾tot,𝑆 . Consider the absolutely irreducible polynomial 𝑓 (𝑇, 𝑋) := 𝑋 2 − 𝑋 −𝑇 over 𝐾 and note that 𝑓 (𝑎, 𝑋) is a separable polynomial for each 𝑎 ∈ 𝐾. By [Jar91b, Prop. 12.3] in the non-archimedean case and by [Jar91b, Prop. 16.7] in the real case, for each 𝔭 ∈ 𝑆 there exists a 𝔭-open neighborhood 𝑈𝔭 of 0 in 𝐾 such that if 𝑎 ∈ 𝑈𝔭 , then the splitting field of 𝑓 (𝑎, 𝑋) over 𝐾 is contained in 𝐾𝔭 . By [FrJ08, p. 224, Cor. 12.2.3], 𝐾 has a separable Hilbert subset 𝐻 such that if 𝑎 ∈ 𝐻, then 𝑓 (𝑎, 𝑋) is irreducible over 𝐿. By Geyer, the Hilbert subsets of 𝐾 are dense Ñ in the 𝑆-topology [Gey78, Lemma 3.4]. Hence, there exists an 𝑎 ∈ 𝐻 ∩ 𝔭∈𝑆 𝑈𝔭 . Let ˜ Then, 𝐿 𝑛+1 := 𝐾 (𝑏) is a separable quadratic extension 𝑏 be a root of 𝑓 (𝑎, 𝑋) in 𝐾. of 𝐾 in 𝐾tot,𝑆 which is linearly disjoint from 𝐿.

9.1 The Groups Gal(𝐾tot,𝑆 [𝝈 ])

125

We have therefore constructed a sequence 𝐿 1 , 𝐿 2 , 𝐿 3 , . . . of quadratic separable extensions of 𝐾 in 𝐾tot,𝑆 . By [FrJ08, p. 378, Lemma 18.5.3], for almost all 𝝈 ∈ Gal(𝐾) 𝑒 the field 𝐾tot,𝑆 (𝝈) contains the compositum 𝑁 of infinitely many of the 𝐿 𝑖 . Since the 𝐿 𝑖 ’s are linearly disjoint over 𝐾, the field 𝑁 is an infinite Galois extension of 𝐾 in 𝐾tot,𝑆 (𝝈). Hence, 𝑁 ⊆ 𝐾tot,𝑆 [𝝈]. It follows that [𝐾tot,𝑆 [𝝈] : 𝐾] = ∞, as claimed.  With this we arrive at the main result of our monograph: Theorem 9.1.6 Let 𝐾 be a countable Hilbertian field and 𝑒 ≥ 1 an integer. Then, for almost all 𝝈 ∈ Gal(𝐾) 𝑒 the field 𝑀 := 𝐾tot,𝑆 [𝝈] is Hilbertian, P𝑆C, and ample. Moreover, there exists an étale profinite set of representatives R for the Gal(𝑀)-orbits of G𝐾 ,𝑆 such that R is a Cantor space, and there exists a commutative diagram of profinite groups 𝜋 / 𝐹ˆ𝜔 ∗ Î / Ker(𝜋) / 𝐹ˆ𝜔 /1 (9.1) 1 ∗O Γ∈R Γ O O 𝛼

1

/ Gal(𝐾tot,𝑆 )

/ Gal(𝑀)

res

/ Gal(𝐾tot,𝑆 /𝑀)

/ 1,

where 𝜋 is the projection on the first factor, the horizontal sequences are exact, and the vertical arrows are isomorphisms. Also, Ö Ö Ö ∗ Γ= ∗ ∗ Gal(𝐾𝔭 ) 𝜌 Γ∈R

𝔭∈𝑆 𝜌∈𝑅𝔭

is a Cantor free product over 𝑆 (Remark 9.1.2). Proof By [GeJ02, Thm. A], for almost all 𝝈 ∈ Gal(𝐾) 𝑒 the field 𝑀 := 𝐾tot,𝑆 [𝝈] is P𝑆C, hence by Definition 9.1.1, 𝑀 is also ample. By Proposition 9.1.3 and Lemma 9.1.5, for almost all 𝝈 ∈ Gal(𝐾) 𝑒 the field 𝑀 is in addition Hilbertian and [𝑀 : 𝐾] = ∞. By Setup 8.1.2, G𝐾 ,𝑆 is Gal(𝐾)-invariant and strictly closed, hence an étale compact subset of Subgr(Gal(𝐾)) (Remark 1.2.7). Since 𝑀 ⊆ 𝐾tot,𝑆 , the set G𝐾 ,𝑆 is also an étale compact subset of Subgr(Gal(𝑀)). By the preceding paragraph 𝑀 is P𝑆C. Hence, by Lemma 8.3.5, Gal(𝑀) is G𝐾 ,𝑆 -projective. By Setup 8.1.2, G𝐾 ,𝑆 is of P-type. Hence, by Lemma 8.4.2, Gal(𝑀) is properly strongly G𝐾 ,𝑆 -projective. It follows from Proposition 8.4.3 that G𝐾 ,𝑆 = G𝐾 ,𝑆,max and there exists an étale profinite system of representatives R for the Gal(𝑀)-orbits of G𝐾 ,𝑆 . Also, by that proposition, there exists an isomorphism Ö 𝛼 : Gal(𝑀) → 𝐹ˆ𝜔 ∗ ∗ Γ. Γ∈R

Moreover, 𝛼 maps the family Î R as a subset of Subgr(Gal(𝑀)) Ð onto the family R as a subset of Subgr( 𝐹ˆ𝜔 ∗ ∗ Γ∈R Γ). By Lemma 8.1.8, R = · 𝔭∈𝑆 R 𝔭 is a Cantor space, where each R 𝔭 is a closed set of representatives for the Gal(𝑀)-orbits of G𝐾 ,𝔭 . In addition, still by Lemma 8.1.8, Gal(𝐾) has a closed subset 𝑅𝔭 such that R 𝔭 = {Gal(𝐾𝔭 ) 𝜌 }𝜌∈𝑅𝔭 and the map 𝜌 ↦→ Gal(𝐾𝔭 ) 𝜌 is a homeomorphism of 𝑅𝔭 onto R 𝔭 . By Lemma 4.5.4,

126

9 Main Result

Ö Ö Ö Ö Ö ∗ Gal(𝐾𝔭 ) 𝜌 . ∗ Γ= ∗ ∗ Γ= ∗ Γ∈R

𝔭∈𝑆 Γ∈R 𝔭

𝔭∈𝑆 𝜌∈𝑅𝔭

Î Thus, ∗ Γ∈R Γ is a Cantor free product over 𝑆. Since 𝑀 ⊆ 𝐾tot,𝑆 and R is a system of representatives for the Gal(𝑀)-orbits of G𝐾 ,𝑆 , the closed normal subgroup of Gal(𝑀) generated by R is Gal(𝐾tot,𝑆 ). Î By Lemma 4.7.5(b), the closed normal subgroup of 𝐹ˆ𝜔 ∗ ∗ Γ∈R Γ generated by R is Ker(𝜋). Hence, 𝛼(Gal(𝐾tot,𝑆 )) = Ker(𝜋), so 𝛼 induces an isomorphism of Gal(𝐾tot,𝑆 /𝑀) onto 𝐹ˆ𝜔 such that diagram (9.1) commutes.  Remark 9.1.7 When [HJP12] was published, [BSF13] was not yet available, so it was not yet known that almost all fields 𝐾tot,𝑆 [𝝈] are Hilbertian. Instead, we considered an (𝑒 + 1)-tuple (𝝈, 𝜏) taken at random in Gal(𝐾) 𝑒+1 and chose a proper finite extension 𝑀 of 𝐾tot,𝑆 [𝝈, 𝜏] in 𝐾tot,𝑆 [𝝈]. By a theorem of Weissauer, 𝑀 is Hilbertian [FrJ08, p. 262, Thm. 13.9.1]. In addition, as an algebraic extension of a P𝑆C field, 𝑀 is also P𝑆C. This allowed us, as in the proof of Theorem 9.1.6, to prove that Gal(𝑀) fits into a diagram like (9.1). Finally, we used Lemma 4.7.5 to conclude that Gal(𝐾tot,𝑆 ) is a Cantor free product over 𝑆.

9.2 Finitely Generated Groups We continue to consider a countable Hilbertian field 𝐾, a finite set 𝑆 of classically local primes of 𝐾, a positive integer 𝑒, and set 𝑁 = 𝐾tot,𝑆 . Building on our previous results, we prove in this section that for almost all 𝝈 ∈ Gal(𝐾) 𝑒 the group Gal(𝑁 (𝝈)) is isomorphic to the free product of 𝐹ˆ𝑒 and a Cantor free product over 𝑆. Lemma 9.2.1 Let 𝐺 and 𝐻 be closed subgroups of Gal(𝐾). Suppose that 𝐺 is a Cantor free product Îover 𝑆 and let 𝑅 be either a finite set or a Cantor set in Gal(𝐾). Suppose that 𝐻 := ∗ 𝜌∈𝑅 𝐺 𝜌 is a free product. Then, 𝐻 is a Cantor free product over 𝑆. Proof By Remark 9.1.2, for each 𝔭 ∈ 𝑆 there exists a closed Cantor subspace 𝑅𝔭 of Gal(𝐾) such that Ö Ö (9.2) 𝐺= ∗ ∗ Gal(𝐾𝔭 ) 𝜌 . 𝔭∈𝑆 𝜌∈𝑅𝔭

In particular, 𝑅𝔭 is a Cantor space for each 𝔭 ∈ 𝑆. The finiteness of 𝑆 allows us to use the commutativity and associativity of the free products (Lemmas 4.6.3 and 4.5.4) and to rewrite 𝐻 in the following way: Ö Ö Ö Ö 0 0 𝐻 = ∗ 𝐺𝜌 = ∗ ∗ ∗ Gal(𝐾𝔭 ) 𝜌𝜌 𝜌0 ∈𝑅

(9.3)

𝜌0 ∈𝑅 𝔭∈𝑆 𝜌∈𝑅𝔭

Ö Ö Ö 0 = ∗ ∗ ∗ Gal(𝐾𝔭 ) 𝜌𝜌 𝔭∈𝑆 𝜌0 ∈𝑅 𝜌∈𝑅𝔭

Ö = ∗

Ö ∗

𝔭∈𝑆 (𝜌,𝜌0 ) ∈𝑅𝔭 ×𝑅

0

Gal(𝐾𝔭 ) 𝜌𝜌 .

9.2 Finitely Generated Groups

127

Since non-trivial free factors in a free product of profinite groups are distinct (by Proposition 4.8.3(a)), the map 𝑅𝔭 ×𝑅 → 𝑅𝔭 𝑅 given by (𝜌, 𝜌 0) ↦→ 𝜌𝜌 0 is a continuous bijection of profinite spaces, so it is a homeomorphism (Remark 1.1.7(c)). Since 𝑅𝔭 is a Cantor space and 𝑅 is either finite or a Cantor space, 𝑅𝔭 × 𝑅 is a Cantor space (Corollary 1.5.4). Therefore, 𝑅𝔭 𝑅 is a Cantor space and the right-hand side of (9.3) can be rewritten as a Cantor free product over 𝑆: Ö Ö 𝐻= ∗ ∗ Gal(𝐾𝔭 ) 𝜇 , 𝔭∈𝑆 𝜇 ∈𝑅𝔭 𝑅

as claimed.



Lemma 9.2.2 For almost all 𝝈 ∈ Gal(𝐾) 𝑒 we have Gal(𝑁/𝑁 (𝝈))  𝐹ˆ𝑒 . Proof For each 𝝈 ∈ Gal(𝐾) 𝑒 the group Gal(𝑁/𝑁 (𝝈)) is generated by 𝜎1 | 𝑁 , . . . , 𝜎𝑒 | 𝑁 . Thus, by [FrJ08, p. 360, Lemma 17.7.1], it suffices to prove that for each finite group 𝐵 which is generated by 𝑒 elements and for almost all 𝝈 ∈ Gal(𝐾) 𝑒 , the field 𝑁 (𝝈) has a Galois extension in 𝑁 with Galois group 𝐵. To this end, consider 𝐵 as a subgroup of 𝑆 𝑛 , where 𝑛 = card(𝐵). Let 𝑐 0 , . . . , 𝑐 𝑛−1 be integers such that 𝑋 𝑛 + 𝑐 𝑛−1 𝑋 𝑛−1 + · · · + 𝑐 0 = (𝑋 − 1) (𝑋 − 2) · · · (𝑋 − 𝑛). By [Jar91b, Propositions 12.3 and 16.7], c = (𝑐 0 , . . . , 𝑐 𝑛−1 ) has an open 𝑆-adic neighborhood 𝐴 in 𝐾 𝑛 such that if (𝑎 0 , . . . , 𝑎 𝑛−1 ) ∈ 𝐴, then 𝑋 𝑛 + 𝑎 𝑛−1 𝑋 𝑛−1 + · · · + 𝑎 0 splits into linear factors over each 𝐾𝔭 with 𝔭 ∈ 𝑆, hence it splits over 𝑁. Now consider the general polynomial 𝑓 (T, 𝑋) = 𝑋 𝑛 + 𝑇𝑛−1 𝑋 𝑛−1 + · · · + 𝑇0 of degree 𝑛, having 𝑆 𝑛 as its Galois group over 𝐿 (T) for each field extension 𝐿 of 𝐾. Suppose we have inductively constructed linearly disjoint finite Galois extensions 𝐿 1 , . . . , 𝐿 𝑘 of 𝐾 in 𝑁 with Galois group 𝑆 𝑛 . Set 𝐿 = 𝐿 1 · · · 𝐿 𝑘 . By [FrJ08, p. 224, Cor. 12.2.3 and p. 231, Lemma 13.1.1], 𝐾 𝑛 has a separable Hilbert subset 𝐻 such that Gal( 𝑓 (a, 𝑋), 𝐾)  Gal( 𝑓 (a, 𝑋), 𝐿)  𝑆 𝑛 for each a ∈ 𝐻. By [Gey78, Lemma 3.4], there exists an a ∈ 𝐻 ∩ 𝐴. The splitting field 𝐿 𝑘+1 of 𝑓 (a, 𝑋) over 𝐾 has Galois group 𝑆 𝑛 , is linearly disjoint from 𝐿 over 𝐾, and is contained in 𝑁. This completes the induction. By construction, for each 𝑘 there are elements 𝜎𝑘,1 , . . . , 𝜎𝑘,𝑒 ∈ Gal(𝐿 𝑘 /𝐾) such that h𝜎𝑘,1 , . . . , 𝜎𝑘,𝑒 i  𝐵. By Borel–Cantelli [FrJ08, p. 372, Lemma 18.3.5], for almost all 𝝈 ∈ Gal(𝐾) 𝑒 there exists a 𝑘 such that 𝜎𝑖 | 𝐿𝑘 = 𝜎𝑘,𝑖 , 𝑖 = 1, . . . , 𝑒. Therefore, 𝑁 (𝝈)𝐿 𝑘 ⊆ 𝑁 and Gal(𝑁 (𝝈)𝐿 𝑘 /𝑁 (𝝈))  Gal(𝐿 𝑘 /𝐿 𝑘 (𝝈𝑘 ))  𝐵, as claimed.



Theorem 9.2.3 Let 𝐾 be a countable Hilbertian field, 𝑆 a finite set of classically local primes of 𝐾, and 𝑒 a positive integer. Then, for almost all 𝝈 ∈ Gal(𝐾) 𝑒 there is a Cantor free product 𝐷 over 𝑆 such that Gal(𝐾tot,𝑆 (𝝈))  𝐹ˆ𝑒 ∗ 𝐷. Proof By Theorem 9.1.6, for almost all 𝝈 ∈ Gal(𝐾) 𝑒 , there is an isomorphism Gal(𝐾tot,𝑆 [𝝈])  𝐹ˆ𝜔 ∗ 𝐶, where 𝐶 is a Cantor free product over 𝑆. Moreover,

128

9 Main Result

Diagram (9.1) of the proof of that theorem gives the upper two rows of the following commutative diagram, they are short exact sequences, where the upper vertical arrows are isomorphism. 1

/ Ker(𝜋) O

/ 𝐹ˆ𝜔 ∗ 𝐶 O

1

/ Gal(𝐾tot,𝑆 )

/ Gal(𝐾tot,𝑆 [𝝈]) O

1

/ Gal(𝐾tot,𝑆 )

/ Gal(𝐾tot,𝑆 (𝝈))

𝜋

res

/ 𝐹ˆ𝜔 O

/1

/ Gal(𝐾tot,𝑆 /𝐾tot,𝑆 [𝝈]) O

/1

/ Gal(𝐾tot,𝑆 /𝐾tot,𝑆 (𝝈))

/ 1.

The bottom row of the diagram is also a short exact sequence and the two bottom vertical arrows are inclusions, so the whole diagram is commutative. In addition, by Lemma 9.2.2, Gal(𝐾tot,𝑆 /𝐾tot,𝑆 (𝝈))  𝐹ˆ𝑒 for almost all 𝝈 ∈ Gal(𝐾) 𝑒 . Note that Gal(𝐾tot,𝑆 (𝝈)) = res−1 (Gal(𝐾tot,𝑆 /𝐾tot,𝑆 (𝝈))). Therefore, by Lemma 4.7.6, Ö (9.4) Gal(𝐾tot,𝑆 (𝝈))  Gal(𝐾tot,𝑆 /𝐾tot,𝑆 (𝝈)) ∗ ∗ 𝐶 𝜌 , 𝜌∈𝑅

where 𝑅 is a closed system of representatives for Gal(𝐾tot,𝑆 /𝐾tot,𝑆 [𝝈]) modulo Gal(𝐾tot,𝑆 /𝐾tot,𝑆 (𝝈)). By [FrJ08, p. 358, Prop. 17.6.2], every open subgroup of 𝐹ˆ𝜔 is isomorphic to 𝐹ˆ𝜔 . Hence, Gal(𝐾tot,𝑆 /𝐾tot,𝑆 (𝝈)) is of infinite index in Gal(𝐾tot,𝑆 /𝐾tot,𝑆 [𝝈]). Therefore, 𝑅 is infinite. This implies, by Lemma 1.5.6, that 𝑅 Î is a Cantor subspace of Gal(𝐾tot,𝑆 [𝝈]). By Lemma 9.2.1, 𝐷 = ∗ 𝜌∈𝑅 𝐶 𝜌 is a Cantor free product over 𝑆. In addition, Gal(𝐾tot,𝑆 (𝝈))  𝐹ˆ𝑒 ∗ 𝐷, as claimed.  Pop’s result [Pop96, Thm. 3] is now a consequence of Theorem 9.1.6. Remark 9.2.4 (The absolute Galois group of 𝐾tot,𝑆 ) Pop [Pop96, Thm. 3] proves (in our terminology) that the group Gal(𝐾tot,𝑆 ) is a Cantor free product over 𝑆. This is also an easy consequence of our results. Indeed, in the proof of Theorem 9.1.6, 𝑀 := 𝐾tot,𝑆 [𝝈] is an extension of 𝐾 in 𝑁 := 𝐾tot,𝑆 such that Gal(𝑀)  𝐺 ∗ 𝐹ˆ𝜔 , where 𝐺 is a Cantor free product over 𝑆. Then, Gal(𝐾tot,𝑆 ) is isomorphic to the kernel Î of the projection 𝜋 : 𝐺 ∗ 𝐹ˆ𝜔 → 𝐹ˆ𝜔 . Hence, by Lemma 4.7.5(b), Gal(𝐾tot,𝑆 )  ∗ 𝑓 ∈ 𝐹ˆ 𝐺 𝑓 . It follows 𝜔 from Lemma 9.2.1 that Gal(𝐾tot,𝑆 ) is a Cantor free product over 𝑆.

References

[Art67]

E. Artin, Algebraic Numbers and Algebraic Functions, Gordon and Breach, New York, 1967. [Ax67] J. Ax, Solving diophantine problems modulo every prime, Annals of Mathematics 85 (1967), 161–183. [BSF13] L. Bary-Soroker and A. Fehm, Random Galois extensions of Hilbertian fields, Journal de théorie des nombres de Bordeaux 25 (2013), 31–42. [BSF14] L. Bary-Soroker and A. Fehm, On fields of totally S-adic numbers, with an appendix by Florian Pop, Valuation theory in interaction, EMS Series of Congress Reports, European Mathematical Society, Zurich, 2014, 11–15. [BNW71] E. Binz, J. Neukirch, and G.H. Wenzel, A subgroup theorem for free products of profinite groups, Journal of Algebra 19 (1971), 104–109. [Bou89] N. Bourbaki, General Topology, Chapters 1–4, Springer, Berlin, 1989. [CaF67] J.W.S. Cassels and A. Fröhlich, Algebraic Number Theory, Academic Press, London, 1967. [Die84] V. Diekert, Über die absolute Galoisgruppe dyadischer Zahlkörper, Journal für die reine und angewandte Mathematik 350 (1984), 152–172. [Efr06] I. Efrat, Valuations, Orderings, and Milnor 𝐾-Theory, Mathematical Surveys and Monographs 124, Americal Mathematical Society, 2006. [End72] O. Endler, Valuation Theory, Springer, Berlin, 1972. [EnP05] A.J. Engler and A. Prestel, Valued Fields, Springer, Berlin, Heidelberg, 2005. [FrJ08] M.D. Fried and M. Jarden, Field Arithmetic, third edition, revised by Moshe Jarden, Ergebnisse der Mathematik (3) 11, Springer, Heidelberg, 2008. [Gey78] W.-D. Geyer, Galois groups of intersections of local fields, Israel Journal of Mathematics 30 (1978), 382–396. [GeJ02] W.-D. Geyer and M. Jarden, PSC Galois extensions of Hilbertian fields, Mathematische Nachrichten 236 (2002), 119–160. [Gim95] R. Gill-more, On free products of profinite groups, M.Sc. Thesis, Tel Aviv University 1995. [HJP05] D. Haran, M. Jarden, and F. Pop, P-adically projective groups as absolute Galois groups, International Mathematics Research Notices 32 (2005), 1957–1995.

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 D. Haran and M. Jarden, The Absolute Galois Group of a Semi-Local Field, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-030-89191-6

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130

[HJP07] [HJP09] [HJP12] [Har87] [HaJ85] [Hrb95] [HeR85] [HeR63] [HoY61] [Iwa53] [JaW82] [Jar91a] [Jar91b] [Jar94] [Jar11] [JaR80] [Koc67] [KPR86] [Lan64] [Lan93] [MaM99]

References

D. Haran, M. Jarden, and F. Pop, Projective group structures as absolute Galois structures with block approximation, Memoirs of AMS 189 (2007), 1–56, D. Haran, M. Jarden, and F. Pop, The absolute Galois group of the field of totally 𝑆-adic numbers, Nagoya Mathematical Journal 194 (2009), 91–147. D. Haran, M. Jarden, and F. Pop, The absolute Galois group of subfields of the fields of totally 𝑆-adic numbers, Functiones et Approximatio Commentarii Mathematici 46 (2012), 205–223. D. Haran, On closed subgroups of free products of profinite groups, Proceedings of the London Mathematical Society 55 (1987), 266–298. D. Haran and M. Jarden, The absolute Galois group of a pseudo real closed field, Annali della Scuola Normale Superiore di Pisa 12 (1985), 449–489. D. Harbater, Fundamental groups and embedding problems in characteristic 𝑝, Contemporary Mathematics 186 (1995), 353–369. W. Herfort and L. Ribes, Torsion elements and centralizers in free products of profinite groups, Journal für die reine und angewandte Mathematik 358 (1985), 155–161. E. Hewitt and K.A. Ross, Abstract Harmonic Analysis I, Die Grundlehren der Mathematischen Wissenschaften 115, Springer-Verlag, Berlin, 1963. J.G. Hocking and G.S. Young, Topology, Addison-Wesley, Reading, 1961. K. Iwasawa, On solvable extensions of algebraic number fields, Annals of Mathematics 58 (1953), 548–572. U. Jannsen and K. Wingberg, Die Stuktur der absoluten Galoisgruppe 𝔭-adischer Zahlkörper, Inventiones Mathematicae 70 (1982), 71–98. M. Jarden, Algebraic realization of 𝑝-adically projective groups, Compositio Mathematica 79 (1991), 21–62. M. Jarden, Intersection of local algebraic extensions of a Hilbertian field (A. Barlotti et al., eds), NATO ASI Series C 333 (1991), 343–405, Kluwer, Dordrecht. M. Jarden, Prosolvable subgroups of free products of profinite groups, Communications of Algebra 22 (1994), 1467–1494. M. Jarden, Algebraic Patching, Springer Monographs in Mathematics, Springer 2011. M. Jarden and P. Roquette, The Nullstellensatz over 𝑝-adically closed fields, Journal of the Mathematical Society of Japan 32 (1980), 425– 460. H. Koch, Über die Galoissche Gruppe der algebraischen Abschließung eines Potenzreihenkörpers mit endlichem Konstantenkörper, Mathematische Nachrichten 35, 1967. F.-V. Kuhlmann, M. Pank, and P. Roquette, Immediate and purely wild extensions of valued fields, manuscripta mathematicae 55 (1986), 39–67. S. Lang, Introduction to Algebraic Geometry, Interscience Publishers, New York, 1964. S. Lang, Algebra, third edition, Addison-Wesley, Reading, 1993. G. Malle, B.H. Matzat, Inverse Galois Theory, Springer Monographs in Mathematics, Springer-Verlag, Berlin 1999.

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Symbol Index

𝐵 n 𝐶, 59 ℭ, 15 cd 𝑝 , 92 𝐶 𝐺 := {𝑐 𝜎 | 𝑐 ∈ 𝐶 and 𝜎 ∈ 𝐺}, 4 Con(G), 14 Env(G), 13 ´ EtaleClosure, 10

N -topology, 34 𝜈(Δ, 𝑁), 8 Open(𝐺), 8 OpenNormal(𝐺), 8 Partitions(𝑋), 105 P (Γ), 99 rank(𝐺), 18

G 𝐻 , 14 #𝐺, 89 Gmax , 12 𝐺 (𝑆), 52 𝐾sep (𝝈), 51 𝐾tot,𝑆 , 114

𝑆-adic topology, 123 StrictClosure, 10 Subgr(𝐺), 1, 8 𝑇1 (topological space), 7

L (Γ), 99

𝑈 (G), 10 𝑈 𝑝 , 94

𝑁 [𝝈], 124

𝑍 𝑛 , 94

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 D. Haran and M. Jarden, The Absolute Galois Group of a Semi-Local Field, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-030-89191-6

133

Subject Index

absolute Galois group (of a field), xv abstract group, xv almost all (elements in a set), 49 ample (field), 120 associated sheaf, 32 associated sheaf-group structure, 32 Bary-Soroker, 124 base (of a free profinite group), 51 Binz–Neukirch–Wenzel, 35, 49–51, 79 Borel–Cantelli, 127 Cantor free product (over 𝑆), 124 Cantor middle-third set, 15 Cantor semi-constant sheaf, 124 Cantor space, 15 cartesian (square), 66 cartesian kernel, 38 cartesian map, 38 classically local field, 100 classically local prime, 113 closed (map), 2 closed subset of 𝑋/𝐺, 6 closed subsheaf, 26 cofinite (subset), 49 cohomological 𝑝-dimension, 92 compact (topological space), 1 conductor, 81 connecting maps, 2 constant sheaf, 26

contains (sheaf contains a subsheaf), 26 continuous action, 4 continuous free product (of profinite groups), 33 converge to 1 (family of homomorphisms), 49 converges to 1 (map), 51 decomposition field (of extensions of valuations), 96 decomposition ring, 96 defect (of a finite extension of Henselian fields), 97 defectless (algebraic extension of a Henselian field), 116 directed family (of subgroups), 34 disjoint union, xv envelope (of a subset G), 13 étale closed (subset), 8 étale closure, 10 étale compact (subset), 8 étale continuous (family), 21 étale continuous (function), 9 étale Hausdorff (subset), 8 étale open (subset), 8 étale profinite, 11 étale topology (of Subgr(𝐺)), 8 Fehm, 124 fiber product, 66

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 D. Haran and M. Jarden, The Absolute Galois Group of a Semi-Local Field, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-030-89191-6

135

136

finer (Σ-partition), 24 finer (one topology than the other), 9 finer partitions, 3 finite (G-embedding problem), 66 finite (sheaf), 26 finite prime, 113 free product (of finitely many profinite groups), 33 free product in the sense of Binz–Neukirch–Wenzel, 49 free profinite group, 51 free profinite product over a sheaf, 43 group generated, xv Gruenberg, 71 Hausdorff (space), 2 Hensel–Rychlik lemma, 96 Henselian field, 96–98 Henselization (of a valued field), 96 Herfort, 36, 37 Herfort–Ribes, 33 Hilbertian (field), 51 inertia field, 116 inertia group, 116 infinite prime, 113 inner free product (in the sense of Melnikov), 42 inverse limit, 2 inverse limit (of profinite sheaves), 27 isolated (point in a topological space), 15 isomorphism of sheaf-group structures, 30 Iwasawa, viii Krasner, viii Kuhlmann–Pank–Roquette, 117 Kurosh decomposition (of a subgroup), 35 Kurosh Subgroup Theorem, 75 Kurosh subgroup theorem, 35 local P-type, 99 local 𝑝-type (profinite group), 93

Subject Index

Melnikov, 41, 42 morphism (of a sheaf into a group), 30 morphism (of profinite 𝐺-spaces), 105 morphism (of sheaves), 27 morphism of sheaf-group structures, 30 normalized (Haar measure), 52 open subset of 𝑋/𝐺, 6 open-closed subsheaf, 26 order (of a profinite group), 89 Ostrowski, 97 Ostrowski’s theorem, 97 outer free product, 43 partition, 22 𝐺-partition, 105 Σ-partition, 24 Pop, vii, xi, 128 procyclic (group), ix product topology, 2 profinite, 3 profinite 𝐺-space, 105 profinite (sheaf), 26 profinite group, 4 projection (of a free product), 53 projection (of an inverse limit on its 𝑖th component), 2 proper (G-embedding problem), 66 proper solution (of a G-embedding problem), 66 proper solution (of a G-embedding problem), 66 proper strong solution (of a G-embedding problem), 66 properly G-projective, 66 pseudo-𝑆-closed, P𝑆C, 123 PXC , 119 quasi-conjugate (morphisms of a sheaf into a profinite group), 81 ramification field, 116 ramification group, 116

Subject Index

ramification index (of a finite extension of Henselian fields), 97 rank (of a profinite group), 18 real field, 99 residue degree (of a finite extension of Henselian fields), 97 residue field, 96 Ribes, 36, 37 rigid (embedding problem), 66 section (of a surjective map), 18 semi-constant (sheaf of profinite groups), 62 semi-direct product, 59 separated (family of subgroups), 22 sheaf (of profinite groups), 26 sheaf-group structure, 30 solution (of a G-embedding problem), 66 split (embedding problem), 66 𝑁-standard G-embedding problem for 𝐺, 71 stellate, 11

137

strict closure, 10 strict topology, 1 strictly closed (subset of Subgr(𝐺)), 8 strictly compact (subset of Subgr(𝐺)), 8 strictly continuous (function), 8 strictly Hausdorff (subset of Subgr(𝐺)), 8 strictly open (subset of Subgr(𝐺)), 8 strong solution (of a G-embedding problem), 66 strongly G-projective, 66 structure map (of a sheaf), 26 supernatural number, 89 Tychonoff’s theorem, 2 uniformly continuous (group operations), 26 unitriangular matrix, 94 unramified (extension of valuations), 96 valuation ring, 96