Direct Sum Decompositions of Torsion-Free Finite Rank Groups
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Direct Sum Decompositions of Torsion-Free Finite Rank Groups

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

EXECUTIVE EDITORS Earl J. Taft Rutgers University Piscataway, New Jersey

Zuhair Nashed University of Central Florida Orlando, Florida

EDITORIAL BOARD M. S. Baouendi University of California, San Diego Jane Cronin Rutgers University Jack K. Hale Georgia Institute of Technology S. Kobayashi University of California, Berkeley Marvin Marcus University of California, Santa Barbara W. S. Massey Yale University

Anil Nerode Cornell University Freddy van Oystaeyen University of Antwerp, Belgium Donald Passman University of Wisconsin, Madison Fred S. Roberts Rutgers University David L. Russell Virginia Polytechnic Institute and State University Walter Schempp Universität Siegen

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Direct Sum Decompositions of Torsion-Free Finite Rank Groups Theodore G. Faticoni Fordham University Bronx, New York, U.S.A.

Boca Raton London New York

Chapman & Hall/CRC is an imprint of the Taylor & Francis Group, an informa business

Chapman & Hall/CRC Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487‑2742 © 2007 by Taylor & Francis Group, LLC Chapman & Hall/CRC is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid‑free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number‑10: 1‑58488‑726‑5 (Hardcover) International Standard Book Number‑13: 978‑1‑58488‑726‑3 (Hardcover) This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. No part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright. com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC) 222 Rosewood Drive, Danvers, MA 01923, 978‑750‑8400. CCC is a not‑for‑profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging‑in‑Publication Data Faticoni, Theodore G. (Theodore Gerard), 1954‑ Direct sum decompositions of torsion‑free finite rank groups / Theodore G. Faticoni. p. cm. ‑‑ (Pure and applied mathematics) Includes bibliographical references and index. ISBN‑13: 978‑1‑58488‑726‑3 ISBN‑10: 1‑58488‑726‑5 1. Torsion free Abelian groups. 2. Direct sum decompositions. I. Title. QA180.F38 2006 512’.25‑‑dc22 Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

2006021040

To my parents who gave me life. To Adolph and Margaret Faticoni.

Contents Preface 1 Notation and Preliminary Results 1.1 Abelian Groups . . . . . . . . . . . 1.2 Associative Rings . . . . . . . . . . 1.3 Finite Dimensional Q-Algebras . . 1.4 Localization in Commutative Rings 1.5 Local-Global Remainder . . . . . . 1.6 Integrally Closed Rings . . . . . . 1.7 Semi-Perfect Rings . . . . . . . . . 1.8 Exercise . . . . . . . . . . . . . . .

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2 Motivation by Example 2.1 Some Well-Behaved Direct Sums . . . . 2.2 Some Badly-Behaved Direct Sums . . . 2.3 Corner’s Theorem . . . . . . . . . . . . 2.4 Arnold-Lady-Murley Theorem . . . . . . 2.4.1 Category Equivalence . . . . . . 2.4.2 Functor A(·) . . . . . . . . . . . 2.4.3 Elementary Uses of the Functors 2.5 Local Isomorphism . . . . . . . . . . . . 2.6 Exercises . . . . . . . . . . . . . . . . . 2.7 Questions for Future Research . . . . . .

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1 1 5 9 10 13 15 16 18

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19 20 25 27 29 30 31 32 35 44 45

3 Local Isomorphism Is Isomorphism 47 3.1 Integrally Closed Rings . . . . . . . . . . . . . . . . . . . 47 3.2 Conductor of an Rtffr Ring . . . . . . . . . . . . . . . . . 51 ix

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CONTENTS 3.3 3.4 3.5 3.6 3.7

Local Correspondence . . . . . Canonical Decomposition . . . Arnold’s Theorem . . . . . . . Exercises . . . . . . . . . . . . Questions for Future Research .

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4 Commuting Endomorphisms 4.1 Nilpotent Sets . . . . . . . . . . . . . . . . . . 4.2 Commutative Rtffr Rings . . . . . . . . . . . 4.2.1 Modules over Commutative Rings . . 4.2.2 Projectives over Commutative Rings . 4.2.3 Direct Sums over Commutative Rings 4.2.4 Commutative Endomorphism Rings . 4.3 E-Properties . . . . . . . . . . . . . . . . . . 4.4 Square-Free Ranks . . . . . . . . . . . . . . . 4.4.1 Rank Two Groups . . . . . . . . . . . 4.4.2 Groups of Rank ≥ 3 . . . . . . . . . . 4.5 Refinement and Square-Free Rank . . . . . . 4.6 Hereditary Endomorphism Rings . . . . . . . 4.7 Exercises . . . . . . . . . . . . . . . . . . . . 4.8 Questions for Future Research . . . . . . . . . 5 Refinement Revisited 5.1 Counting Isomorphism Classes . . . . . . . 5.1.1 Class Groups and Class Numbers . . 5.1.2 Class Group of the Integral Closure 5.1.3 Class Number and Refinement . . . 5.1.4 Quadratic Number Fields . . . . . . 5.1.5 Counting Theorems . . . . . . . . . 5.2 Integrally Closed Groups . . . . . . . . . . . 5.2.1 Review of Notation and Terminology 5.2.2 Locally Semi-Perfect Rings . . . . . 5.2.3 Semi-Primary Rtffr Groups . . . . . 5.2.4 Applications to Refinement . . . . . 5.3 Exercises . . . . . . . . . . . . . . . . . . . 5.4 Questions for Future Research . . . . . . . .

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55 59 62 64 66

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69 69 79 79 82 84 86 89 93 93 95 99 101 102 103

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107 108 108 110 113 116 117 122 123 124 127 130 134 135

6 Baer Splitting Property 139 6.1 Baer’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . 139 6.2 Splitting of Exact Sequences . . . . . . . . . . . . . . . . . 144 6.3 G-Compressed Projectives . . . . . . . . . . . . . . . . . . 146

CONTENTS 6.4 6.5 6.6

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Some Examples . . . . . . . . . . . . . . . . . . . . . . . . 151 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 Questions for Future Research . . . . . . . . . . . . . . . . 154

7 J -Groups, L-Groups, and S-Groups 7.1 Background on Ext . . . . . . . . . . 7.2 Finite Projective Properties . . . . . 7.3 Finitely Projective Groups . . . . . . 7.4 Finitely Faithful S-Groups . . . . . . 7.5 Isomorphism versus Local Isomorphism . . . . . . . . . . . . . 7.6 Analytic Number Theory . . . . . . 7.7 Eichler L-Groups Are J -Groups . . 7.8 Exercises . . . . . . . . . . . . . . . 7.9 Questions for Future Research . . . .

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155 155 157 159 162

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165 167 172 174 175

8 Gabriel Filters 8.1 Filters of Divisibility . . . . . . . . . . . . . . 8.1.1 Hereditary Torsion Classes . . . . . . 8.1.2 Gabriel Filters of Right Ideals . . . . . 8.2 Idempotent Ideals . . . . . . . . . . . . . . . 8.2.1 Traces of Covers . . . . . . . . . . . . 8.2.2 Bounded Gabriel Filters . . . . . . . . 8.2.3 Finite Filters of Divisibility . . . . . . 8.3 Gabriel Filters on Rtffr Rings . . . . . . . . . 8.3.1 Applications to Endomorphism Rings 8.3.2 Constructing Examples . . . . . . . . 8.3.3 Faithful Rings . . . . . . . . . . . . . 8.4 Gabriel Filters on QEnd(G) . . . . . . . . . . 8.4.1 Central Quasi-summands . . . . . . . 8.4.2 Q-Faithful Groups . . . . . . . . . . . 8.4.3 Q-Faithful E-Flat Groups . . . . . . . 8.5 Exercises . . . . . . . . . . . . . . . . . . . . 8.6 Questions for Future Research . . . . . . . . .

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177 177 179 181 185 185 190 192 196 196 199 201 204 204 208 211 213 216

9 Endomorphism Modules 9.1 Additive Structure of Rings . . . . . 9.2 E-Properties . . . . . . . . . . . . . 9.2.1 E-Rings . . . . . . . . . . . . 9.2.2 E-Finitely Generated Groups 9.2.3 E-Projective Groups . . . . .

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9.3

9.4 9.5 9.6

9.2.4 E-Generator Groups . . . . . . . . . . . . 9.2.5 E-Projective Rtffr Groups Characterized . 9.2.6 Noetherian Endomorphism Modules . . . Homological Dimensions . . . . . . . . . . . . . . 9.3.1 E-Projective Dimensions . . . . . . . . . 9.3.2 E-Flat Dimensions . . . . . . . . . . . . . 9.3.3 E-Injective Dimensions . . . . . . . . . . Self-Injective Rings . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . Questions for Future Research . . . . . . . . . . .

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234 236 239 242 243 246 248 251 252 255

A Pathological Direct Sums 257 A.1 Nonunique Direct Sums . . . . . . . . . . . . . . . . . . . 257 B ACD Groups 261 B.1 Example by Corner . . . . . . . . . . . . . . . . . . . . . . 261 C Power Cancellation 263 C.1 Failure of Power Cancellation . . . . . . . . . . . . . . . . 263 D Cancellation 265 D.1 Failure of Cancellation . . . . . . . . . . . . . . . . . . . . 265 E Corner Rings and Modules E.1 Topological Preliminaries . . . . . . . . . . . . . . . . . . E.2 The Construction of G . . . . . . . . . . . . . . . . . . . . E.3 Endomorphisms of G . . . . . . . . . . . . . . . . . . . . .

269 269 271 272

F Corner’s Theorem 277 F.1 Countable Endomorphism Rings . . . . . . . . . . . . . . 277 G Torsion Torsion-Free Groups 279 G.1 E-Torsion Groups . . . . . . . . . . . . . . . . . . . . . . 279 G.2 Self-Small Corner Modules . . . . . . . . . . . . . . . . . . 280 H E-Flat Groups 283 H.1 Ubiquity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 H.2 Unfaithful Groups . . . . . . . . . . . . . . . . . . . . . . 284 I

Zassenhaus and Butler 289 I.1 Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 I.2 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290

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J Countable E-Rings 295 J.1 Countable Torsion-Free E-Rings . . . . . . . . . . . . . . 295 K Dedekind E-Rings 301 K.1 Number Theoretic Preliminaries . . . . . . . . . . . . . . 301 K.2 Integrally Closed Rings . . . . . . . . . . . . . . . . . . . 302 Bibliography

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Index

311

Preface

This book is directed at mathematicians who wish to study advanced topics in direct sum decompositions of abelian groups and their variants. I assume that the reader has had experience with the ring theory covered in [6], and with the abelian group theory covered in the text that inspired us to write this text, [10]. A read through the abelian group text [46] will also be helpful. The use of reduced torsion-free finite rank abelian groups or of rtffr abelian groups can be traced back to solutions of linear equations with integer coefficients. We reinvent this point of view today when we study full free subgroups of a finite dimensional Q-vector space. Suppose that the rtffr group G has a direct sum decomposition (n1 )

G1

(nt )

⊕ · · · ⊕ Gt

(1)

for some integers t, n1 , . . . , nt > 0 and some indecomposable groups G1 , . . . , Gt . To eliminate redundancies we require that Gi ∼ = Gj ⇒ i = j. There are several theorems that give us conditions under which the integers n1 , . . . , nt and the isomorphism classes of the groups G1 , . . . , Gt are unique for G. Two of the most important results of this kind are the Azumaya-Krull-Schmidt Theorem and the Baer-Kulikov-Kaplansky Theorem. Each states that if the groups Gi satisfy some condition or other then the integers t, n1 , . . . , nt > 0 and the isomorphism classes (Gi ) of the groups Gi in (1) are unique to G. Thus one might expect this uniqueness to be common. This is not the case.

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For example, G is called completely decomposable if G has a direct sum decomposition (1) in which Gi is a rank one rtffr group for each i = 1, · · · , t. That is, Gi ⊂ Q. The Baer-Kulikov-Kaplansky Theorem states that if (1) is a direct sum of rank one groups G1 , . . . , Gt then the integers t, n1 , . . . , nt > 0 and the isomorphism classes of the groups G1 , . . . , Gt are unique to G. However, assume that t, n1 , . . . , nt > 1. Then several examples by B. J´onsson, A.L.S. Corner, and L. Fuchs and F. Loonstra (see [47] or the appendices) show that there are subgroups of finite index in G that possess direct sum decompositions whose indecomposable terms are not unique in any sensitive generalization of the term. Such a group H is called an almost completely decomposable group, or an acd group. Specifically, there are indecomposable acd groups G, H, A, B, C such that G(2) ∼ = H (2) while G and H are not isomorphic, and A ⊕ B ∼ = A ⊕ C while B and C are not isomorphic. Thus one can routinely construct subgroups H of finite index in G whose direct sum decompositions behave badly. Hence any theorems on direct sum decompositions of rtffr groups must respect this juxtaposition of uniqueness with nonuniqueness. B. J´onsson proved a theorem, considered by some to be the Fundamental Theorem of Torsion-Free Finite Rank Abelian Groups, that allows us to salvage a course uniqueness of the indecomposable terms in the direct sum (1). The idea is to substitute the weaker quasi-isomorphism for isomorphism in the Azumaya-Krull-Schmidt Theorem. Groups X and Y are quasi-isomorphic if there are maps f : X → Y and g : Y → X and . an integer n > 0 such that f g = n1Y and gf = n1X . We write X ∼ to Y . The group X is strongly = Y when X is quasi-isomorphic . ∼ indecomposable if X = Y implies that Y is indecomposable. THEOREM: [B. J´onsson] Let G be an rtffr group. 1. There are integers t, n1 , . . . , nt >. 0 and strongly indecomposable groups G1 , . . . , Gt such that Gi ∼ = Gj implies that i = j and such . n1 nt ∼ that G = G1 ⊕ · · · ⊕ Gt . .

∼ H m1 ⊕ · · · ⊕ H ms for some integers s, m1 , . . ., 2. Suppose that G = s 1 ms >. 0 and strongly indecomposable groups H1 , . . . , Hs such that Hi ∼ = Hj implies that i = j. Then s = t, ni = mi for each . i = 1, . . . , t, and Gi ∼ = Hi after a permutation of the subscripts. .

.

.

Thus. G(2) ∼ = H (2) implies that G ∼ = H, and A ⊕ B ∼ = A ⊕ C implies ∼ that B = C. While J´onsson’s Theorem accounts for some of the subgroup

PREFACE

xvii

structure of G, it misses the finer properties that a direct sum decomposition can have. For example, there are indecomposable groups that are quasi-isomorphic to groups that possess a nontrivial direct sum decomposition. Thus abelian groupists have replaced indecomposable rtffr groups with strongly indecomposable rtffr groups, direct sum decompositions of rtffr groups with quasi-direct sum decompositions of rtffr groups, and uniqueness up to isomorphism is replaced with uniqueness up to quasi-isomorphism. E. L. Lady [57] introduced another generalization of isomorphism called near isomorphism that is logically situated between isomorphism and quasi-isomorphism. Lady’s first applications of near isomorphism include the fact that if G(2) ∼ = H (2) then G and H are near isomorphic, ∼ and if A ⊕ B = A ⊕ C then B and C are near isomorphic. Near isomorphism shows up in direct sums of many types of groups. For example, if G is a completely decomposable rtffr group, if H ⊂ G is a subgroup, and if G/H is a finite p-group for some prime p ∈ Z, then a direct sum H ∼ = H1 ⊕ · · · ⊕ Hr of indecomposable groups Hi is unique up to near isomorphism. That is, if H ∼ = K1 ⊕ · · · ⊕ Ks for some indecomposable groups Kj then r = s and after a permutation of the subscripts Hi is near isomorphic to Ki for each i = 1, . . . , r. Near isomorphism is a group theoretic version of the genus class of lattices over orders. It is our point of view that near isomorphism and genus class are essentially the same measurement of algebraic properties, but of objects from different categories. Therefore, if two objects are near isomorphic or in the same genus class we will say that these objects are locally isomorphic. Following the publication of the papers [14] and [15], it became popular to study direct sum decompositions of G by studying the right ideal structure of End(G) with emphasis on the finitely generated projective right End(G)-modules. This book takes the point of view that a group G can be effectively studied by considering the finitely generated projective right End(G)-modules, the left End(G)-module G, and the ring E(G) = End(G)/N (End(G)). D. M. Arnold [10] introduced a functor A(·) to abelian groups that transforms direct summands of G into finitely generated projective right E(G)-modules. Our fundamental approach uses A(·) as the primary tool for understanding direct sum decompositions of rtffr groups. Thus A(·) enables us to treat the direct sum decompositions of the rtffr group G as

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a direct sum of finitely generated projective modules over the Noetherian semi-prime rtffr ring E(G). One advantage of considering E(G) over End(G) is that the ring and group structure of E(G) are simpler than that of End(G). Specifically, if S = center(E(G)) then E(G) is a finitely generated S-module. We will make extensive use of the localization theory of S to study E(G). There is a finite product E of classical maximal S-orders in which E(G) has finite index. We call E an integrally closed ring. Several applications of some Algebraic Number Theory to the ideal structure of E will give us good results on the group structure of G. For example, if h(G) is the number of isomorphism classes (H) of groups H that are locally isomorphic to G and if h(E) is the number of isomorphism classes (J) of nonzero fractional ideals J of E then h(E) divides h(G). While hereditary endomorphism rings have been scrutinized over the last 20 years, (see [54], and especially the papers of U. Albrecht), there seems to have been little attention paid to other pleasantly structured endomorphism rings. We will initiate an examination of the commutative property in E(G). The reader may be surprised at the number of groups G for which E(G) is commutative. Assuming that E(G) is a commutative ring, we can prove several results on the uniqueness and existence of direct sum decompositions (1) of G. Our investigations will show that G possesses some interesting properties when End(G) is commutative and End(G) ⊂ G ⊂ QEnd(G). Assuming that E(G) is commutative our work shows that for each integer n > 0, Gn has a locally unique decomposition. That is, 1. Gn = H1 ⊕ · · · ⊕ Hr where for each i = 1, . . . , r, Hi is an indecomposable group such that G is locally isomorphic to Hi ⊕ Hi0 for some group Hi0 , and 2. If Gn = K1 ⊕ · · · ⊕ Ks for some indecomposable groups Kj then r = s and after a permutation of the subscripts, Hi is locally isomorphic to Ki for i = 1, . . . , r. We show that Gn has a locally unique decomposition for each integer n > 0 if 1. G has a direct sum decomposition (1) in which the G1 , . . . , Gt are . strongly indecomposable rtffr groups such that Gi ∼ = Gj ⇒ i = j, and 2. E(Gi ) is commutative for each i = 1, . . . , t.

PREFACE

xix

Condition 2, that E(Gi ) is commutative, is true whenever G is a strongly indecomposable rtffr group with square-free rank rank(Gi ). Thus there are plenty of rtffr groups that satisfy these two conditions. This result is an extension of the Baer-Kulikov-Kaplansky Theorem. Let S and E be as above. There is a nonzero ideal τ ⊂ S called the conductor of G such that τ E ⊂ E(G) ⊂ E. It is possible to define a localization functor (·)τ in a manner that is consistent with the traditional localization of a commutative ring at a prime ideal. Localization theory shows us that if H ⊕H 0 ∼ = K⊕K 0 ∼ = G(n) for some integer n > 0 then H and K are locally isomorphic iff Aτ (H) and Aτ (K) are isomorphic right Eτ (G)-modules, where Aτ (x) = (A(x))τ is the localization of A(x) at τ . This gives us a different proof of Arnold’s Theorem: if G is locally isomorphic to H1 ⊕ H2 then G = G1 ⊕ G2 where Gi is locally isomorphic to Hi for i = 1, 2. An ideal P in a ring R is primary if R/P is a local ring with nilpotent Jacobson radical. That is, the Jacobson radical J of R/P is the unique maximal right ideal in R/P and J n = 0 for some integer n > 0. The group G is a semi-primary rtffr group if there is a group G with a direct sum decomposition (1) such that 1. E(Gi ) is a Dedekind domain for each integer i = 1, . . . , t and such that 2. for each i = 1, . . . , t there is a primary ideal N (End(Gi )) ⊂ Pi ⊂ (n ) (n ) End(Gi ) such that P1 G1 1 ⊕ · · · ⊕ Pt Gt t ⊂ G ⊂ G. Semi-primary groups are generalizations of almost completely decomposable groups with primary regulating quotient. They include certain strongly indecomposable groups G such that E(G) is commutative. Our results apply to subgroups of primary index in direct sums of strongly indecomposable bracket groups, strongly indecomposable strongly homogeneous groups, and strongly indecomposable Murley groups. It is natural to ask if rtffr groups G satisfy a splitting property like that of free groups. We will characterize in Chapter 6 those rtffr groups G that satisfy the Baer splitting property. That is, we examine those G for which a surjection π : G(c) −→ G splits whenever c is a cardinal. The main results are Theorems 6.3.1 and 6.3.2 where we relate, ´ a la U. Albrecht [3], the Baer splitting property for G to the vanishing of the tensor product M ⊗E G for nonzero right End(G)-modules M .

xx

PREFACE

A discussion of J. -groups, L-groups, and S-groups follows. The group G is a. J -group if G ∼ = H implies that G ∼ = H. The group G is an L-group ∼ that G is locally isomorphic to H. The group G is an if G = H implies . H implies that G generates H. While it is clear that J S-group if G ∼ = group ⇒ L-group ⇒ S-group the converses require some power. We use several deep results from Analytic Number Theory to show that the most finitely faithful S-groups are J -groups, and that most rtffr L-groups are J -groups. Over the past 30 years it has become fashionable to consider the structure of G as a left End(G)-module. R. S. Pierce introduced the following clever way of labelling module properties on groups that we will adopt as our own. Given a module theoretic property property we say that G is an E-property group if G satisfies property as a left End(G)module. Thus E-projective groups are projective left End(G)-modules, G is an E-finitely generated group if G is a finitely generated left End(G)module, an E-generator group is a generator as a left End(G)-module, and an E-Noetherian group is a Noetherian left End(G)-module. We will give a unified approach to rtffr groups G with these properties by showing that the group structure of G stems from a general structure theorem for finitely generated rtffr modules over an rtffr ring. Subsequently we show that several of these E-properties coincide. In the latter half of Chapter 9 we will examine the possible homological dimensions of the left End(G)module G. We show that for a given rtffr ring E and integer k less than the left global dimension of E there are rtffr groups Pk , Fk such that E = End(Pk ) = End(Fk ) and whose projective and flat dimensions, respectively, are k. Our techniques must be significantly modified if they are to give us broad results on the injective dimension. The text contains a number of exercises at the end of each chapter but the chapters themselves contain many statements like the reader will prove that . . . . Each of these is an intended unlisted exercise. The young abelian group theorist is advised to follow these directions. Also the author comes from the school of mathematics that emphasizes the use of examples. Examples guide our intuition and they steer us clear of falsehoods. Examples are used in this way throughout this text. Thus the reader should not be surprised at the number of examples used to motivate our discussions. I would like to thank Fordham University for giving me the time and the resources needed to write this book. I would also like to thank my colleagues for carefully reading this book in manuscript form and for their subsequent comments. These people include the group of mathematicians that I was introduced to during my formative years. They

PREFACE

xxi

are Ulrich Albrecht, Dave M. Arnold, R. A. Beaumont, Donna Beers, Anthony L. S. Corner, Manfred Dugas, Laszlo Fuchs, H. Pat Goeters, Anthony Giovanitti, Roger Hunter, E. Lee Lady, Ray Mines, DeAlberto Orsatti, Richard Pierce, K. M. Rangaswamy, James Reid, Fred Richman, Luigi Salce, Phil Schultz, Charles Vinsonhaler, Carol Walker, Elbert A. Walker, and William Wickless. Their encouragement over the years helped me forge this book. Since my research style produces more TEX files than paper files, I am both author and technical typist on this project. Thus, I put quite a lot of wear on my keyboard during the writing process. Any mathematical errors of commission, of omission, or of a typographical nature are my responsibility. Certain results are attributed to your author but give no publication reference. This is because these results do not appear elsewhere in the literature. Theodore G. Faticoni Department of Mathematics Fordham University Bronx, New York 10458 [email protected]

Chapter 1

Notation and Preliminary Results This text assumes that the reader is familiar with abelian groups and unital modules over associative rings with unity as contained in the texts by F.W. Anderson and K.R. Fuller [6], D.M. Arnold [10], and L. Fuchs [46]. We will reference but not prove those results that we feel fall outside of the line of thought of this book. I suggest that you use this chapter as a reference and nothing more. Skim through this chapter. Unless otherwise directed, do not attempt to plow through these results as though they were exercises.

1.1

Abelian Groups

Throughout, we let Z denote the ring of integers, Q denotes the field of rational numbers, and for a prime p ∈ Z, Zp is the localization of Z at p. We let E, R, S, T denote an associative ring with identity, hereafter referred to simply as a ring. Their modules are denoted by G, H, K, L, M, N , U, V , W . (The side will be specified.) The term group means abelian group, (i.e., a Z-module), and G, H, K denote groups. The group G is a torsion-free group if its torsion subgroup {x ∈ G nx = 0 for some integer n 6= 0} is zero. Thus G is torsion-free if x ∈ G and nx = 0 for some integer n 6= 0 implies that x = 0. We view Z and Q as groups or as rings, as the setting requires. If G and H are contained in a common Q-vector space V , then we say that G and H are quasi-equal if there are nonzero integers n, m such 1

2

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS

that nG ⊂ H and mH ⊂ G. In this case we write . G = H.

Let G be a group. Let I be a set, let c = card(I), and for each i ∈ I let Gi ∼ = G. The usual direct sum and direct product of c copies of G are denoted as G(c) = ⊕i∈I Gi Q Gc = i∈I Gi . We say that the group G is indecomposable if G = H ⊕K implies that H = 0 or K = 0. Z, Q, Zp , and Z/pk Z are indecomposable groups for primes p ∈ Z and integers k > 0. Given an integer n > 0 there is an indecomposable group Zn ⊂ G ⊂ Qn , so there are plenty of indecomposable torsion-free groups. Let G be a torsion-free group. We let QG denote the divisible hull of G. Since G is torsion-free QG is a Q-vector space spanned by G, or equivalently QG = Q ⊗Z G. The rank of G is the cardinality of a maximal Z-linearly independent subset of G, or equivalently

rank(G) = Q-dim(QG). With few exceptions, the torsion-free groups that we consider in this book have finite rank. T The group G is reduced if n>0 nG = 0, or equivalently if Hom(Q, G) = 0. An

rtffr group

1.1. ABELIAN GROUPS

3

is a reduced torsion-free finite rank group. As you read this book, pronounce rtffr as are-tee-ef-ef-are. A ring E is said to be an rtffr ring if its additive structure (E, +) is an rtffr group. A right E-module M is said to be an rtffr E-module if its additive structure (M, +) is an rtffr group. The reader should prove the following fact as an exercise. If G is an rtffr group and if H ⊂ G is a subgroup such that G/H is a bounded group (i.e., there is an integer n 6= 0 such that nG ⊂ H), then G/H is a finite group. T We say that G is p-reduced if k>0 pk G = 0, and we say that G is p-local if pG 6= G and qG = G for each prime q 6= p ∈ Z. If pG = G then we say that G is p-divisible. The group G is semi-local if there is a finite set of primes {p1 , . . . , ps } ⊂ Z such that pG = G for each p 6∈ {p1 , . . . , ps }. For example, Z and Q are rtffr groups, Z is reduced, and Q is divisible. The subgroup of Q na o a ∈ Z and n is a square-free integer n is reduced and torsion-free group of rank one. Moreover the subring of Q o na a, b ∈ Z and gcd(p, b) = 1 Zp = b is a reduced, p-reduced, torsion-free ring of rank one, but qZp = Zp for primes q 6= p ∈ Z. The ring of algebraic integers in an algebraic number field is an rtffr ring. The ring a 0 Z 0 = a, b, c ∈ Z Z Z b c is a reduced and torsion-free ring of rank three. Let G and H be groups. As usual, we let End(G) = the ring of group endomorphisms of G. Then End(G) is the set of group homomorphisms f : G −→ G. We let Hom(G, H) denote the group of group homomorphisms f : G −→ H. We consider G as a left End(G)-module by setting f · x = f (x) for f ∈ End(G) and x ∈ G. Similarly Hom(G, H) is a right End(G)-module if we set gf = g ◦ f for each g ∈ Hom(G, H) and f ∈ End(G). If G is an rtffr group, then End(G) is an rtffr ring.

4

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS Given groups G and H let

SG (H) =

P {f (G)

group maps f : G −→ H}.

We adopt the standard notation from [6]. Thus N ⊗R M denotes the tensor product of a right R-module N and a left R-module M . Since they are integral parts of our discussion we adopt a notation for a special tensor product and a special homset. TG (·) = · ⊗End(G) G and HG (·) = Hom(G, ·). There are natural transformations Θ Φ

: :

TG HG (·) −→ 1 1 −→ HG TG (·)

defined by ΘH (f ⊗ x) = f (x) ΦM (x)(·) = · ⊗ x. The reader should verify that these functors and associated natural transformations enjoy the following very useful identities. 1HG (H) = HG (ΘH ) ◦ ΦHG (H) 1TG (M ) = ΘTG (M ) ◦ TG (ΦM ) Let QHom(G, H) = Q ⊗ Hom(G, H) QEnd(G) = Q ⊗ End(G). An element f ∈ QHom(G, H) is called a quasi-homomorphism. The elements in QHom(G, H) are linear transformations f : QG → QH such . that f (G) = H. Observe that f is a quasi-homomorphism iff there is an integer n 6= 0 such that nf : G → H. We say that f ∈ QHom(G, H) is a quasi-isomorphism if there is a map g : QHom(H, G) such that f g = 1H

1.2. ASSOCIATIVE RINGS

5

and gf = 1G . Equivalently f : G → H is a quasi-isomorphism iff there is an integer n 6= 0 and a group homomorphism g : H → G such that f g = n1H and gf = n1G . We write .

G∼ =H when G is quasi-isomorphic to H. The following group is used in this text only as a source of illuminating examples. The ring of Z-adic integers is denoted by b = lim Z/nZ. Z ←n

Given a prime p ∈ Z the ring of p-adic integers b p = lim Z/pk Z Z ←k

b is a source for a number of important examples. To begin with, (1) Z b b b and Zp have uncountable rank, (2) for primes p ∈ Z, rp (Z) = rp (Zp ) = 1, b p = qZ bp. and (3) for primes q 6= p ∈ Z, Z

1.2

Associative Rings

Since some of our discussions deal with several rings at once we will need more than one symbol to denote a ring. For instance, E usually denotes an rtffr ring (a potential endomorphism ring of an rtffr group), and S denotes a commutative ring, often but not always S = the center of E = center(E) = {x ∈ E xy = yx for each y ∈ E}. Let E be a ring and let M be a right E-module. If x ∈ M then the right ideal annE (x) = {r ∈ E xr = 0} in E is called the annhilator of x in E and the ideal annE (M ) = {r ∈ E M r = 0} is called the annihilator of M in E. For example givenan integer n > 0, Z 0 1 0 annZ (Z/nZ) = nZ. If E = then let e11 = and let Z Z 0 0 0 0 0 0 N= . The reader can show that annE (e11 ) = and Z 0 Z Z 0 0 that annE (N ) = . Z Z

6

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS

In general, if M = E/I for some right ideal I ⊂ E then 1 + I ∈ E/I has annihilator annE (1 + I) = I in E, and annE (M ) is the largest ideal of E that is contained in I. If e2 = e ∈ E then annE (e) = (1 − e)E. If M is a simple right E-module (that is, the only E-submodules of M are 0 and M ), then annE (x) is a maximal right ideal in E for each x 6= 0 ∈ M . If E = End(G) then annE (G) = 0.

The Jacobson radical of E is the ideal J (E) defined as follows.

J (E) = ∩{M M ⊂ E is a maximal right ideal } = ∩{M M ⊂ E is a maximal left ideal } = ∩{annE (K) K is a simple left E-module } = ∩{annE (K) K is a simple right E-module } = {r ∈ E 1 + rx is a unit in E}

For example, given a prime p ∈ Z, J (Zp ) = pZp . Given n > 0 ∈ Z, let n = pn1 1 · · · pnt t be a product of powers n1 , . . . , nt > 0 of distinct primes p1 , . . . , pt . Let

Zn =

na m

o a, m ∈ Z, m is relatively prime to n .

Then Zn is a semi-local rtffr ring with Jacobson radical J (Zn ) = (p1 · · · pt )Zn . Evidently n ∈ J (Zn ). It is a healthy for exercise the reader to Z 0 0 0 show that J (Z) = 0 and that J = . Z Z Z 0

1.2. ASSOCIATIVE RINGS

7

Given an rtffr ring E then the ring

En =

nx m

x ∈ E and m ∈ Z is relatively prime to n

o

is a torsion-free finite rank ring that may not be reduced. Note that En ∼ = E ⊗Z Zn . LEMMA 1.2.1 Let E be a semi-local rtffr ring and assume that n = Q {primes p ∈ Z pE = 6 E}. Then n ∈ J (E). Proof: Let M be a maximal right ideal in E. We will prove in Lemma 4.2.3 that since E is reduced there is a prime p ∈ Z such that pE ⊂ M 6= E, and thus n ∈ M . Consequently n ∈ ∩{M M is a maximal ideal in E} = J (E). The ring R is local if any of the following equivalent properties hold. 1. R possesses a unique maximal right ideal M . 2. J (R) is the unique maximal right ideal of R. 3. u ∈ R is a unit of R iff u 6∈ J (R). b p are For example, fields are local rings. Given a prime p ∈ Z, Zp and Z local rings. The right ideal I ⊂ E is a nil right ideal if each x ∈ I is nilpotent. That is, for each x ∈ I there exists an integer n such that xn = 0. The nilradical of E is the ideal N (E) that is defined as follows. N (E) = {x ∈ E xE is a nil right ideal in E} = {x ∈ E Ex is a nil left ideal in E} X = {I I is a nil left ideal in E} = the largest nil ideal in E Let x ∈ E. Since xn = 0 implies that 1 − x is a unit in E (show that one, reader), 1 − xy is a unit of E for each x ∈ N (E) and y ∈ E. Thus N (E) ⊂ J (E). In the special case of an rtffr ring E we have the following relationships between N (E) and J (E).

8

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS

LEMMA 1.2.2 Let E be an rtffr ring. 1. N (QE) = J (QE) is the largest nilpotent ideal in QE. 2. N (E) = J (QE) ∩ E and QN (E) = J (QE). The reader may be surprised at the number of times we appeal to Nakayama’s Lemma in this text. LEMMA 1.2.3 [Nakayama] (See [6, page 169].) Let E be a ring, let M be a right E-module and let K ⊂ M be an E-submodule. 1. If M is finitely generated and if K + M J (E) = M then K = M . 2. If I is a nilpotent ideal in E such that K + M I = M then K = M . In particular if J ⊂ E is a right or a left ideal such that J +J (E) = E then J = E. An idempotent is an element e ∈ E such that e2 = e. Often we will avoid the term idempotent and just write e2 = e. Given a ring E and an e2 = e ∈ E then eEe = {exe x ∈ E}

is a ring with identity 1eEe = e. Although eEe ⊂ E, eEe is not a unital subring of E. LEMMA 1.2.4 Let E be a ring and let e2 = e ∈ E. Then EndE (eE) = eEe. Lifting theorems are a part of ideal theory. The following lifting results will be used throughout this text without fanfare. LEMMA 1.2.5 Let E be a ring. 1. If u + J (E) is a unit in E/J (E) then u is a unit in E. 2. If I is a nil ideal in E and if (f + I)2 = f + I ∈ E/I then there is an e2 = e ∈ E such that e − f ∈ I.

1.3. FINITE DIMENSIONAL Q-ALGEBRAS

1.3

9

Finite Dimensional Q-Algebras

Our source for structure results on rtffr rings is [10, Chapters 9-14]. The ring E is semi-prime if I is an ideal of E and if I 2 = 0 then I = 0. Equivalently, E is semi-prime iff N (E) = 0. Given a group G let E(G) = End(G)/N (End(G)). Notice that E(G) is in general a semi-prime ring. If G is an rtffr group then E(G) is a Noetherian semi-prime rtffr ring that is finitely generated by its center, S. See [10, Theorems 9.4, 9.9]. It is hard to overstate the importance of the semi-prime ring E(G) to our deliberations. Let G and H be rtffr groups. We say that H is a quasi-summand . of G if there is a group K such that G ∼ = H ⊕ K. Equivalently, H is a quasi-summand of G iff there is an integer n 6= 0 and maps f : G → H and g : H → G such. that f g = n1H . The rtffr group G is strongly indecomposable if G ∼ = H ⊕ K implies that either H = 0 or K = 0. Equivalently G is a strongly indecomposable group iff each subgroup of finite index in G is indecomposable. Each subgroup of Q is strongly b p are strongly indecomposable as is Z and for primes p ∈ Z, Zp and Z b p is strongly indecomposable indecomposable. Each pure subgroup of Z so there are plenty of strongly indecomposable groups around. A proof of the next lemma is found in [10, Theorem 9.10]. LEMMA 1.3.1 Suppose that G is an rtffr group such that (n1 )

G = G1

(nt )

⊕ · · · ⊕ Gt

for some integers t, n1 , . . . , nt >. 0 and some strongly indecomposable groups G1 , . . . , Gt such that Gi ∼ = Gj ⇒ i = j. Then E(G) ∼ = Matn1 (E(G1 )) × · · · × Matnt (E(Gt )).

(1.1)

The next result is [46, page 149, Lemma 92.2]. LEMMA 1.3.2 [J.D. Reid] Let G be an rtffr group. There is a bijection from the set of isomorphism classes of {(eQEnd(G)) e2 = e ∈ QEnd(G)} onto the set of quasi-isomorphism classes {[H] H is a quasi-summand of G}. The bijection is given by (eQEnd(G)) 7→ [eG]. The following classification of strongly indecomposable groups is due to J. D. Reid [46, page 149, Proposition 92.3].

10

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS

LEMMA 1.3.3 [J.D. Reid] Let G be an rtffr group. The following are equivalent. 1. G is strongly indecomposable. 2. QEnd(G) is a local ring. 3. An endomorphism f : G → G is either nilpotent or an injection. Some of the ideal structure of QE filters down to E. LEMMA 1.3.4 (See [10, Theorem 9.10].) Let E be an rtffr ring. The following are equivalent. 1. N (E) = 0. 2. There are integers t, n1 , . . . , nt > 0 and division Q-algebras D1 , . . ., Dt such that QE = Matn1 (D1 ) × · · · × Matnt (Dt )

(1.2)

as rings. 3. Each right ideal of QE is a direct summand of QE. LEMMA 1.3.5 (See [10, Corollary 10.14].) Let E be a semi-prime rtffr ring. Then 1. If I is a right ideal of E, there is an e2 = e ∈ QE and an integer n 6= 0 such that neE has finite index in I and nE ⊂ neE ⊕ n(1 − e)E ⊂ E ⊂ eE ⊕ (1 − e)E. 2. E is a Noetherian ring. 3. E is finitely generated as a module over its center center(E).

1.4

Localization in Commutative Rings

Let us review some ideas about localization in commutative rings. We will refer to these facts without fanfare. An element c ∈ S is a regular element if c is not a zero divisor in S. If I is an ideal in the commutative ring S we let Γ(I) = {c ∈ S c + I is a regular element in S/I}.

1.4. LOCALIZATION IN COMMUTATIVE RINGS

11

In general Γ(I) is a multiplicatively closed subset of S that contains 1. For example, if S/I is a finite ring then c ∈ Γ(I) iff c + I is a unit in S/I. For x ∈ S and c, d ∈ Γ(I) we define an equivalence relation ∼ on (x, c) ∈ S ×Γ(I) by (x, c) ∼ (y, d) iff xd = yc. Then the equivalence class of (1, c) is denoted by c−1 , and xc−1 is the equivalence class of (x, c). By setting K(I) = {x ∈ S xc = 0 for some c ∈ Γ(I)} then we can form the localization of S at I denoted by SI

= {xc−1 x ∈ S/K(I) and c ∈ Γ(I)}.

If S is an rtffr ring then S0 = QS, However, not every localization of S is formed by inverting a set of integers. Given an S-module X we write XI = X ⊗S SI . Given any injection f : X → Y the canonical map fI : XI −→ YI : xc−1 7−→ f (x)f (c)−1 is an injection. Thus the inclusion map J ⊂ S induces an injection JI ⊂ SI onto an ideal in SI . Indeed II ⊂ J (SI ) since for each x ∈ II and s ∈ SI , 1 + xs ≡ 1(modII ) and 1, being regular modulo I, maps onto a unit of SI . In particular, if S is an rtffr ring and if n ∈ Z then n ∈ J (SnS ). Note that SnS is different from Sn = S ⊗Z Zn . The following example will show us why Sn and SnS may be different rings. Let S = Z6 × Z15 . With n = 2 we have S2 = S ⊗Z Z2 = (Z6 × Z15 ) ⊗Z Z2 = Z2 × Q while with n = 3 we have S3 = S ⊗Z Z3 = Z3 × Z3 .

12

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS

But with I = 2S we have the following. Since (1, 0) maps to a unit in S/2S and since (1, 0) annihilates 0 × Z15 , we have K(2S) = 0 × Z15 = {x ∈ S xc = 0 for some c ∈ Γ(2S)}. Thus S2S = (S/K(2S))2S = Z2 . We observe that while 2 ∈ J (S2S ) we also have (1, 1)2 = (2, 2) 6∈ J (S2 ) = 2Z2 × 0. If I = 0 then Γ(0) is the set of regular elements in S and so S0 = QS is the classical ring of quotients of S. If I is a prime ideal in S then SI is a local ring with unique maximal ideal II . Furthermore, there is a natural isomorphism SI /II ∼ = (S/I)I , and SI /II is the field of fractions of the integral domain S/I. Let S be a commutative ring and let I ⊂ S be an ideal. If S/I is finite then Γ(I) maps to the set of units of S/I so that the canonical map S −→ SI : x 7→ x + K(I) induces an isomorphism S/I ∼ = SI /II ∼ = (S/I)I . Furthermore, if X is a finite S-module such that XI = 0 then the canonical map X −→ XI : x 7→ x1−1 is an isomorphism. For example, if p ∈ Z is a prime and if G is an abelian group such that pk G = 0 for some integer k then G ∼ = Gp ∼ = G ⊗Z Zp . Specifically if k k ∼ G is any group then G/p G = Gp /p Gp . LEMMA 1.4.1 Let I, J be ideals in a commutative integral domain S. Let X be an S-module. 1. (XI )J ∼ = (XJ )I . 2. If I ⊂ J and if S/I is finite then the natural surjection S/I → S/J takes units to units. Hence (SI )J ∼ = (SJ )I ∼ = SJ .

1.5. LOCAL-GLOBAL REMAINDER

1.5

13

Local-Global Remainder

Any theorem that describes a property of M in terms of some property of MI for each maximal ideal I ⊂ S is called a local-global property. We use several local-global properties in this text. THEOREM 1.5.1 [Local-Global Theorem] (See [72, page 105].) Let S be a commutative ring and let M be an S-module. Then M = 0 ⇔ MI = 0 for each maximal ideal I ⊂ E. Consequently f : M → N is a surjection iff the induced map fI : MI → NI is a surjection for each maximal ideal I ⊂ S. Dually f : M → N is an injection iff the induced map fI : MI → NI is an injection for each maximal ideal I ⊂ S. In particular, given N, K ⊂ M then K ⊂ N iff KI ⊂ NI as SI -submodules of MI for each maximal ideal I ⊂ S. Another type of Local-Global Theorem relates maps M → N to maps MI → NI for maximal ideals I in S. THEOREM 1.5.2 [Change of Rings] (See [72, page 106].) Let E be an S-algebra for some commutative ring S, and let M be a finitely presented right E-module. If C ⊂ S is a multiplicatively closed subset of S containing 1 then for each S-module N 1. HomE (M, N )[C −1 ] ∼ = HomE[C −1 ] (M [C −1 ], N [C −1 ]) as groups. 2. Ext1E (M, N )[C −1 ] ∼ = Ext1E[C −1 ] (M [C −1 ], N [C −1 ]) as groups. In particular if M is a finitely presented right E-module then EndE (M )I ∼ = EndEI (MI ). Furthermore, a finitely presented right Emodule M is a projective E-module iff MI is a projective right EI -module for each maximal ideal I in S. Beware: The Change of Rings Theorem is false unless M is finitely presented. The following congruence result is used in a variety of settings in the sequel. THEOREM 1.5.3 [Chinese Remainder Theorem] (See [30, page 46].) Let S be a commutative ring, let I1 , . . . , It be ideals in S such that Ii + Ij = S for 1 ≤ i 6= j ≤ t, and let M be a right S-module. Let M ∗ = M I1 ∩ · · · ∩ M It .

14

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS 1. Given x1 , . . . , xt ∈ M there is an x ∈ M such that x − xi ∈ M Ii

for each i = 1, . . . , t.

2. The diagonal map t

δ:

M M M −→ ∗ M M Ii i=1

such that δ(x + M ∗ ) = (x + I1 M, . . . , x + M It ) is an isomorphism. 3. M I1 ∩ · · · ∩ M It = M [I1 ∩ · · · ∩ It ]. Proof: See [30, Theorem 18.30-31] for a proof of parts 1 and 2. 3. Let I = I1 ∩ · · · ∩ It . By part 2 there is a canonical isomorphism t Y S ∼S = Ii I i=1

so there are isomorphisms M M∗

∼ =

t M M M Ii

∼ = M ⊗S

i=1

t Y S Ii i=1

S ∼ = M ⊗S I

∼ =

M . MI

Since these are canonical isomorphisms M ∗ = M I, as required by part 3. In particular, if t, n1 , . . . , nt > 0 are integers, if p1 , . . . , pt ∈ Z are distinct primes, and if n = pn1 1 · · · pnt t then given an abelian group G nG = pn1 1 G ∩ · · · ∩ pnt t G. Consequently, there is a natural isomorphism t

δ:

Y G G −→ nG pni i G i=1

such that δ(x + nG) = (x + pn1 1 G, . . . , x + pnt t G) of groups.

1.6. INTEGRALLY CLOSED RINGS

1.6

15

Integrally Closed Rings

See [10]. The commutative rtffr integral domain S is a Dedekind domain iff it satisfies one of the following equivalent conditions. 1. If S ⊂ S 0 ⊂ QS is a ring and if S 0 is a finitely generated S-module then S = S 0 . 2. If I ⊂ S is a nonzero ideal then I is invertible. That is, II ∗ = S where I ∗ = {q ∈ QS qI ⊂ S}. 3. Each nonzero ideal of S is a unique product of maximal ideals in S. 4. S is a hereditary Noetherian integral domain. 5. The localization SM is a discrete valuation domain for each maximal ideal M in S. That is, there is an element π ∈ SM such that each ideal of SM has the form π k SM for some integer k > 0. For example, Z is a Dedekind domain as is any pid. The ring of algebraic integers in an algebraic number field is a Dedekind domain. The rtffr ring E is integrally closed if whenever E ⊂ E 0 ⊂ QE is a ring such that E 0 /E is finite then E = E 0 . A classical maximal order is an integrally closed prime rtffr ring. A classical maximal order is finitely generated and torsion-free as a module over its center. The rtffr ring E is an integrally closed ring iff E = E1 × · · · × Et where each Ei is a classical maximal order. An E-lattice is a finitely generated right E-submodule of the right E-module QE (n) for some n ∈ Z. Dedekind domains are classical maximal orders. If E is a classical maximal order and if U is an E-lattice then Matn (E) and EndE (U ) are classical maximal orders. The next results are [69, Theorem 21.1], [10, Corollary 11.5], [10, Theorem 11.8], and [69, Theorems 18.10, 27.1]. LEMMA 1.6.1 Suppose that the rtffr ring E is a classical maximal order. 1. If I is a right ideal of finite index in E then O(I) = {q ∈ QE qI ⊂ I} is a classical maximal order.

16

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS 2. E is an hereditary Noetherian ring. That is, each one-sided ideal in E is a finitely generated projective E-module. 3. For each integer n ∈ Z, each right ideal in En is principal. 4. If U and V are E-lattices then U is locally isomorphic to V (see section 2.5) iff QU ∼ = QV as right QE-modules. (Warning: This is true only of lattices over maximal orders.)

Semi-prime rtffr rings are closely connected to integrally closed rings as the following result shows. The next two results follow from [10, page 127]. LEMMA 1.6.2 Let E be a semi-prime rtffr ring. There is a finite set of primes {p1 , . . . , ps } ⊂ Z such that Ep is a classical maximal order for all p 6∈ {p1 , . . . , ps }. LEMMA 1.6.3 Suppose that E is a semi-prime rtffr ring. There is an integrally closed ring E ⊂ QE and an integer n 6= 0 such that nE ⊂ E ⊂ E

1.7

Semi-Perfect Rings

The ring E is semi-perfect if 1. E/J (E) is semi-simple Artinian and 2. Given an e¯2 = e¯ ∈ E/J (E) there is an e2 = e ∈ E such that e¯ = e + J (E). That is, idempotents lift modulo J (E). See [6, Chapter 7 §27] for a complete discussion of semi-perfect rings and their modules. Fields, local rings, and Artinian rings are semi-perfect b p are semi-perfect rings. Z and Z6 are not semi-perfect but Zp and Z rings for primes p ∈ Z. Suppose that E is semi-perfect. Each indecomposable projective right E-module P has the form eE for some e2 = e ∈ E. Given e2 = e, f 2 = f ∈ E then eE/eJ (E) ∼ = f E/f J (E) ⇔ eE ∼ = f E. LEMMA 1.7.1 Let E be a semi-perfect ring and let P be a projective right E-module. Then

1.7. SEMI-PERFECT RINGS

17

1. P is indecomposable iff EndE (P ) is a local ring. 2. E is a direct sum of right E-modules with local endomorphism rings. Thus direct sum decompositions of projective modules over semiperfect rings are unique by the Azumaya-Krull-Schmidt Theorem 2.1.6. We will also find it necessary to use the following results on lifting idempotents in semi-perfect rings. LEMMA 1.7.2 Let E be a semi-perfect ring, and let e2 = e, f 2 = f ∈ E be such that eE ∼ = f E. There is a unit u ∈ E such that ueu−1 = f . Proof: Since e2 = e and f 2 = f we can write E = eE ⊕ (1 − e)E = f E ⊕ (1 − f )E and since E is semi-perfect (1 − e)E ∼ = (1 − f )E by The Azumaya-KrullSchmidt Theorem 2.1.6. Thus there is an isomorphism u : E → E such that ueu−1 ∈ u(eE) = f E −1

u(1 − e)u

and

∈ u((1 − e)E) = (1 − f )E.

Evidently u is multiplication by a unit of E so that 1 = u1u−1 = u(e + (1 − e))u−1 = ueu−1 ⊕ u(1 − e)u−1 ∈ f E ⊕ (1 − f )E. Since 1 can be written in exactly one way as an element in a direct sum we must have ueu−1 = f. This completes the proof of the lemma. LEMMA 1.7.3 Let E be semi-perfect and let e2 = e, f 2 = f ∈ E. If eE/eJ (E) ∼ = f E/f J (E) then eE ∼ = f E. Proof: Since e2 = e, eE is a projective right E-module so the isomorphism eE/eJ (E) ∼ = f E/f J (E) lifts to a map φ : eE −→ f E

18

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS

such that ker φ ⊂ eJ (E) and f E = φ(eE) + f J (E). By Nakayama’s Lemma 1.2.3, f E = φ(eE) and since f 2 = f ∈ E eE = U ⊕ ker φ where U ∼ = f E. Inasmuch as ker φ ⊂ eJ (E) another appeal to Nakayama’s Lemma 1.2.3 shows us that ker φ = 0, whence eE ∼ = f E. This completes the proof. LEMMA 1.7.4 Let E be a semi-perfect ring, and let J ⊂ J (R) be an ideal. Then idempotents lift modulo J in E. Proof: Let J ⊂ J (E) and let e ∈ E be such that e2 − e ∈ J ⊂ J (E). ¯ = x + J ∈ E. Let E = E/J and given x ∈ E let x Because E is semi-perfect there is an f 2 = f ∈ E such that e − f ∈ J (E). Then e, f¯ ∈ E are idempotents such that e−f¯ ∈ J (E) = J (E)/J. We will show that there is a unit u ∈ E such that uf¯u−1 = e. We have e + J (E) = f¯ + J (E) and e2 = e, f¯2 = f¯ ∈ E so that eE ∼ E E ∼ f¯E = (f¯ + J (E)) . = (e + J (E)) = ¯ eJ (E) J (E) J (E) f J (E) By Lemma 1.7.2 there is a unit u ∈ E such that e = uf¯u−1 . Since units lift modulo J ⊂ J (E), u = u + J for some unit u ∈ E and hence uf u−1 ≡ e(mod J). Thus idempotents lift modulo J in E.

1.8

Exercise

1. Verify each result mentioned in this chapter by proving it yourself or by finding it in the references.

Chapter 2

Motivation by Example The group G is an

rtffr group

if G is a reduced torsion-free finite abelian group. The topics that we will discuss in this text are properties of direct summands of rtffr groups G. These properties include the uniqueness of direct sum decompositions, as well as the properties of the left End(G)-module G. A direct sum decomposition G = G1 ⊕ · · · ⊕ G t

(2.1)

is said to be indecomposable if each Gi is indecomposable. If p ∈ Z is a prime and if k1 , . . . , kt > 0 are integers such that Gi ∼ = Z/pki Z for each i = 1, . . . , t then the decomposition (2.1) is indecomposable. If Gi ⊂ Q for each i = 1, . . . , t then the decomposition (2.1) is indecomposable. Suppose that (2.1) is an indecomposable decomposition of G and suppose that we can also write G∼ =H ⊕K ∼ = H ⊕ K 0.

(2.2)

We will pursue several broad questions on the nature of direct sums. For instance, is the direct sum (2.1) unique in some sense? What can be said about K and K 0 in (2.2)? Is H a direct sum of copies of the Gi ? Are the group structures of K and K 0 similar in some sense? Are there examples of badly behaved direct sum decompositions of G? Is there a simpler category in which the direct summands of G can be handled? 19

20

CHAPTER 2. MOTIVATION BY EXAMPLE

Suppose that we are given a surjection π : G(n) → G for some integer n > 0. If G is a free group then π is split. That is, is there a map φ : G → G(n) such that πφ = 1G , or equivalently, such that ker π a direct summand of G(n) . Under what conditions on G will π be split?

2.1

Some Well-Behaved Direct Sums

Before embarking on our journey we need a series of examples. The experience that these examples bring us will help us to develop intuition about the above questions. The constructions can be found in D. Arnold’s [10] or L. Fuchs’ [46]. Let G be a group. The purpose of this section is to show that under some conditions direct sum decompositions of G are well behaved, in a sense that we will make precise below. The group G is indecomposable if G = H ⊕ H 0 implies that H = 0 or H 0 = 0. For instance, Z is indecomposable, and given a prime p ∈ Z, Zp , Z(p∞ ), and Z/pk Z for integers k > 0 are indecomposable groups. Any subgroup of Q is indecomposable. The following example shows us that indecomposable rtffr groups can have finite index in a decomposable group. EXAMPLE 2.1.1 (See Lemma A.1.2.) Let X, Y ⊂ Q be groups such that Hom(X, Y ) = 0 = Hom(Y, X) and suppose that there is a prime p ∈ Z such that pX 6= X and pY 6= Y . Choose elements x ∈ X \ pX and y ∈ Y \ pY and define 1 G = (X ⊕ Y ) + (x + y)Z. p . Then G is an indecomposable group of rank two such that G = X ⊕ Y . The group G is strongly indecomposable if each subgroup of finite index in G is indecomposable. See Section 1.3. Given a prime p ∈ Z let

1 a Z = a, k ∈ Z . p pk

2.1. SOME WELL-BEHAVED DIRECT SUMS

21

EXAMPLE 2.1.2 (See Example A.1.1.) Let X, Y ⊂ Q be groups such that Hom(X, Y ) = 0 = Hom(Y, X) and suppose that there is a prime p ∈ Z such that pX 6= X and pY 6= Y . Choose elements x ∈ X \ pX and y ∈ Y \ pY and define 1 G = (X ⊕ Y ) + Z[ ](x + y). p Then G is a strongly indecomposable group of rank two. EXAMPLE 2.1.3 Let p ∈ Z be a prime and let G be a pure subgroup b p . Then G is strongly indecomposable. of Z . b p and suppose that G = Proof: Say G is pure in Z A ⊕ B. The reader ∼ b p ∩ G so that will show that G/pG = A/pA ⊕ B/pB. Then pG = pZ b p ∩ G) ∼ b p )/pZ bp ⊂ Z b p /pZ bp ∼ G/pG = G/(pZ = (G + pZ = Z/pZ. Thus G/pG is indecomposable, whence A/pA = 0 or B/pB = 0. But b p . Hence then A or B is a p-divisible subgroup of the p-reduced group Z A = 0 or B = 0, and therefore G is indecomposable. We say that G has a unique decomposition if 1. G has an indecomposable decomposition G ∼ = G1 ⊕ · · · ⊕ Gt and 2. Given an indecomposable decomposition G ∼ = G01 ⊕ · · · ⊕ G0s then s = t and after a permutation of the subscripts, Gi ∼ = G0i for each i = 1, . . . , t. In this case we call G1 ⊕ · · · ⊕ Gt the unique decomposition of G. The unique decomposition of an rtffr group G is necessarily indecomposable. Rtffr groups having unique decomposition are considered to be rare. EXAMPLE 2.1.4 1. The Fundamental Theorem of Abelian Groups states that if p ∈ Z is a prime and if G is a finite p-group then G has a unique decomposition G = Z/pn1 Z ⊕ · · · ⊕ Z/pnt Z for some integers 0 < n1 ≤ · · · ≤ nt . 2. Finitely generated free groups have unique decomposition as do the (possibly mixed) divisible groups.

22

CHAPTER 2. MOTIVATION BY EXAMPLE 3. A vector space over a field has a unique decomposition as a group.

EXAMPLE 2.1.5 Suppose End(G) is right Artinian. Because End(G) is then semi-perfect the indecomposable decomposition G = e1 G ⊕ · · · ⊕ et G is unique. An rtffr endomorphism ring End(G) contains a complete set of primitive orthogonal indecomposable idempotents {e1 , . . . , et } because QEnd(G) is Artinian. Then G has an indecomposable decomposition G = e1 G ⊕ · · · ⊕ et G. However this decomposition need not be unique. One of the more interesting properties associated with direct sums is the refinement property. The group G has the refinement property if 1. G possesses an indecomposable decomposition G = G1 ⊕ · · · ⊕ G t and 2. If G(n) ∼ = H ⊕ K for some integer n > 0 then there are integers h1 , . . . , ht ≥ 0 such that (h1 )

H = G1

(ht )

⊕ · · · ⊕ Gt

.

It is generally believed that rtffr groups with the refinement property are rare. A theorem on the uniqueness of indecomposable decompositions that will be essential to our investigations is the Azumaya-Krull-Schmidt Theorem, hereafter referred to as the AKS-Theorem. A proof can be found in [6]. The AKS Theorem represents the best conditions that one could hope for concerning the uniqueness of direct sum decompositions. THEOREM 2.1.6 [Azumaya-Krull-Schmidt] Let G1 , · · · , Gt be left E-modules such that EndE (Gi ) is a local ring for each i = 1, . . . , t and let G = G1 ⊕ · · · ⊕ Gt . 1. G1 ⊕ · · · ⊕ Gt is the unique decomposition for G. 2. G has the refinement property. A consequence of the AKS Theorem 2.1.6 is that if G is a direct sum of modules with local endomorphism rings then G ⊕ K ∼ = G ⊕ K 0 for 0 0 some modules K and K implies that K ∼ = K . See [30, Cancellation Theorem 21.2]. Some examples will help us see how to use the AKS Theorem 2.1.6.

2.1. SOME WELL-BEHAVED DIRECT SUMS

23

EXAMPLE 2.1.7 Let p1 , . . . , pt ∈ Z be a finite list of distinct primes and for each i = 1, . . . , t let Gi be the localization Gi = Zpi . Then End(Gi ) is a local commutative ring for each i and hence the AKS Theorem 2.1.6 states that G = G1 ⊕ · · · ⊕ Gt has a unique decomposition. The reader can show that End(G) is commutative in this case. EXAMPLE 2.1.8 If G1 , . . . , Gt are simple right modules over some ring R then EndR (Gi ) is a division ring for each i = 1, . . . , t. The AKS Theorem 2.1.6 states that G = G1 ⊕ · · · ⊕ Gt is the unique decomposition of G, and that G has the refinement property. EXAMPLE 2.1.9 An indecomposable projective module over a semiperfect ring has a local endomorphism ring. Thus the AKS Theorem 2.1.6 applies to direct sums of indecomposable projective modules over a semi-perfect ring. J.D. Reid’s Lemma 1.3.3 states that the rtffr group G is strongly indecomposable iff QEnd(G) is a local ring, so it is reasonable to expect that some form of the AKS Theorem 2.1.6 applies to direct sums of strongly indecomposable rtffr groups. The next theorem is referred to in the literature as J´onsson’s Theorem. It is generally accepted that this is the best possible existence and uniqueness result that we can expect for the general rtffr group G. See [46, page 150, Theorem 92.5]. THEOREM 2.1.10 [B. J´onsson] Let G be an rtffr group. There is a finite list of G1 , . . . , Gt strongly indecomposable groups such that .

1. G ∼ = G1 ⊕ · · · ⊕ G t . .

2. If G ∼ = G01 ⊕ · · · ⊕ G0s and if each G0i is strongly indecomposable . then s = t and after a permutation of the indices Gi ∼ = G0i for each i = 1, . . . , t. .

3. Given an integer n > 0 and a direct sum G(n) ∼ = H ⊕ H 0 then there is a subset I ⊂ {1, . . . , t} and integers hi > 0, i ∈ I, such that . (h ) H∼ = ⊕i∈I Gi i . .

.

4. If G ⊕ H ∼ = G ⊕ H 0 for some rtffr groups H and H 0 then H ∼ = H 0. Proof: We sketch a proof due to E.A. Walker [86]. The category of quasi-homomorphisms QAb is the category 1. whose objects are the torsion-free groups of finite rank, and

24

CHAPTER 2. MOTIVATION BY EXAMPLE 2. whose homsets are given by QHom(?, ?).

The rtffr group G is strongly indecomposable iff G is indecomposable in QAb, and G is strongly indecomposable iff QEnd(G) is a local Artinian ring, (Lemma 1.3.3). Thus each object in QAb is a direct sum of objects with local endomorphism rings. Then by a version of the AKS Theorem 2.1.6 for objects in additive categories in which idempotents split, each object in QAb has a unique decomposition G ∼ = G1 ⊕ · · · ⊕ Gt in QAb where each Gi is strongly indecomposable. Since two rtffr. groups are isomorphic in QAb iff they are quasi-isomorphic groups, G ∼ = G1 ⊕ · · · ⊕ Gt as groups. This completes the proof. We observe that the AKS Theorem 2.1.6 has been our only tool to this point for demonstrating the existence of a unique decomposition. The next result is a theorem on uniqueness and the refinement property of indecomposable decompositions of a certain type of group. Its proof is interesting in that it does not refer to the AKS Theorem 2.1.6. See [46, page 112, Proposition 86.1]. The group G is called completely decomposable if there are rank one groups G1 , . . . , Gt such that G = G1 ⊕ · · · ⊕ Gt . The group H has rank one if H ⊂ Q. THEOREM 2.1.11 [Baer-Kulikov-Kaplansky] Assume that G ∼ = G1 ⊕ · · · ⊕ Gt where each Gi is a rank one group. 1. G1 ⊕ · · · ⊕ Gt is the unique decomposition of G. 2. G has the refinement property. The group G is an acd group (almost completely decomposable group) if there is a completely decomposable group C and an integer n > 0 such that nC ⊂ G ⊂ C. See A. Mader’s text [60] for everything concerning acd groups. There is at least one type of acd group whose direct sum decompositions are well behaved. See [10, Theorem 2.3] for a proof. THEOREM 2.1.12 [R.A. Beaumont and R.S. Pierce] Suppose that . G = Gn1 1 ⊕ · · · ⊕ Gnt t for some integers t, n1 , . . . , nt > 0 and a chain G1 ⊂ . . . ⊂ Gt of rank one groups such that Gi ∼ = Gj iff i = j. Then G∼ = Gn1 1 ⊕ · · · ⊕ Gnt t . Theorem 2.2.1, however, will show us that acd groups can have badly behaved direct sum decompositions.

2.2. SOME BADLY-BEHAVED DIRECT SUMS

2.2

25

Some Badly-Behaved Direct Sums

Now that we have established some useful conditions under which direct sums are well behaved let us see what happens in more general situations. Every torsion-free group of finite rank has an indecomposable decomposition (i.e., the given group can be written as a direct sum of indecomposable groups) but this decomposition need not be unique. (See Theorem 2.2.1 below.) The purpose of this section is to give some examples that illustrate the unruly behavior of indecomposable decompositions of some rtffr groups. In the 1960s several examples came to light illustrating that within the class of acd groups there are examples of direct sum decompositions that are not unique in any sense of the word. Even though the completely decomposable groups possess a unique decomposition (Theorem 2.1.11), an indecomposable decomposition of an acd group may not be unique. THEOREM 2.2.1 [A.L.S. Corner] See Example B.1.1. Let n ≥ k ≥ 1 be integers. There is a group G = G(n) of rank n such that for each partition r1 + · · · + rk of n into k positive summands rj there is an indecomposable direct sum decomposition G = G1 ⊕ · · · ⊕ G k such that rj = rank(Gj ) for each j = 1, . . . , k. For instance, suppose that p, p1 , p2 , p3 are distinct primes, let A1 , A2 , A3 be rank one groups such that pi Ai 6= Ai 6= pAi for each i = 1, 2, 3, and let m = p1 p2 p3 . With n = 4 and k = 2 the above construction shows us that there is a group m(A21 ⊕ A2 ⊕ A3 ) ⊂ G ⊂ A21 ⊕ A2 ⊕ A3 that possesses indecomposable decompositions G = G1 ⊕ G2 = H1 ⊕ H2 such that rank(G1 ) = 3, rank(H1 ) = rank(H2 ) = 2, and rank(G2 ) = 1. Thus G does not possess a unique decomposition. The nonuniqueness of indecomposable direct sum decompositions of rtffr groups is also seen in the following examples.

26

CHAPTER 2. MOTIVATION BY EXAMPLE

THEOREM 2.2.2 [L. Fuchs and F. Loonstra] Examples C.1.2 and D.1.1. 1. There are acd groups G1 , G2 , G3 of ranks 1, 2, 2, respectively, such that G1 ⊕ G2 ∼ = G1 ⊕ G3 while G2 6∼ = G3 . 2. There are indecomposable acd groups G4 , G5 of rank 2 such that (2) (2) (2) G4 ∼ = G5 while G4 6∼ = G5 . Thus G = G4 does not possess a unique decomposition and does not satisfy the refinement property.

The next example illustrates that even if End(G1 ) = End(G2 ) = Z then indecomposable decompositions of G = G1 ⊕ G2 can fail to be unique.

THEOREM 2.2.3 [10, page 63, Example 6.9]. There are quasi-isomorphic indecomposable groups G1 , G2 , H1 , H2 such that G1 ⊕ G 2 ∼ = H 1 ⊕ H2 , and End(X) ∼ = Z for X = G1 , G2 , H1 , H2 , while H2 is not isomorphic to G1 , G2 , or H1 .

Thus in the AKS Theorem 2.1.6 the hypothesis that End(Gi ) is a local ring for each Gi cannot be weakened to the condition that End(Gi ) is a pid.

It is best if we use the above examples as guideposts to help us anticipate limitations in our results. For instance, while any theorem on the direct sum decompositions of acd groups might aspire to the conclusions of the AKS Theorem 2.1.6 or of the Baer-Kulikov-Kaplansky Theorem 2.1.11, it must also respect the badly behaved direct sum decompositions in the above section.

2.3. CORNER’S THEOREM

2.3

27

Corner’s Theorem

The fact that any rtffr ring E is the endomorphism ring of an rtffr group G will allow us to construct rtffr groups possessing a variety of left Emodule structures. These constructions are found in the Appendices. The group G is locally free if Gp is a free Zp -module for each prime p ∈ Z. M.C.R. Butler [21] constructs groups that have the same torsionfree rank as their endomorphism ring. See [10, Theorem 2.14]. THEOREM 2.3.1 [M.C.R. Butler] See Example I.1.1. Let E be an rtffr ring whose additive structure (E, +) is a locally free group. There is an rtffr group G such that E ⊂ G ⊂ QE as left E-modules and E∼ = End(G) as rings.

Let E be an rtffr ring. The left E-module M is called a Corner Emodule, or when E is understood simply a Corner module, if its additive structure (M, +) is a countable reduced torsion-free group. Similarly a Corner ring is an associative ring E with identity such that E is a Corner module. There is no confusion as to the orientation of M as all Corner modules are left modules. Call the group G a Corner group if G is a countable reduced torsion-free group. One of the questions that we will address is Is there a functorial means of characterizing the properties of G in terms of End(G)? Dually, we can ask Is there a functorial means that will allow us to characterize the properties of End(G) in terms of properties of G. Specifically, since G is a left End(G)-module we can ask about the module theoretic properties of G as a left End(G)-module. Corner’s Theorem provides a warehouse of flexible examples that will allow us to gauge how frequently certain properties occur. See Theorem F.1.1 where Corner’s Theorem follows from a more general result. See also [46, Theorem 110.1]. THEOREM 2.3.2 [A.L.S. Corner] Let E be a Corner ring. There is a group G such that E ∼ = End(G) as rings. If rank(E) = n then we can choose G so that rank(G) = 2n. If E is an rtffr ring then the group G constructed in Corner’s Theorem 2.3.2 is a pure and dense subgroup b E ⊂ G = hE, Eπi ⊂ E

28

CHAPTER 2. MOTIVATION BY EXAMPLE

b Then G fits into a short exact for some transcendental element π ∈ Z. sequence 0 → E −→ G −→ QE → 0 of left E-modules. The next two results will extend Corner’s Theorem by replacing E and QE with more general left E-modules while still retaining E = End(G). Given a Corner ring E and a Corner E-module M let O(M ) = {q ∈ QE qM ⊂ M }. The reader can show that under these hypotheses if annE (M ) = 0 then O(M ) is a Corner ring. See Theorem E.1.1 for a detailed proof. THEOREM 2.3.3 [36, T.G. Faticoni] Let E be a Corner ring and let M be a Corner module. There is a short exact sequence 0 → M −→ G −→ QC → 0

(2.3)

such that O(M ) ∼ = End(G) and where C ∼ = ⊕K K where K ranges over (ℵ ) o the cyclic E-submodules of M . For instance our choices for M in the above Theorem include any of the following left E-modules. 1. M = an E-submodule of a countably generated free E-module such that annE (M ) = 0. 2. M = K ⊕E where K is a Corner module. The summand E ensures that E = O(M ). We will find the following result useful in constructing rtffr groups G whose left End(G)-module structure is specified. The left E-module M is an rtffr E-module if its additive structure (M, +) is an rtffr group. See Theorem F.1.2 for a proof of the following result. THEOREM 2.3.4 [43, T.G. Faticoni and H.P. Goeters] Let E be an rtffr ring and let M be an rtffr E-module. There is an rtffr group G and a short exact sequence 0 → M −→ G −→ QE ⊕ QE → 0 such that O(M ) ∼ = End(G).

(2.4)

2.4. ARNOLD-LADY-MURLEY THEOREM

29

It is clear from (2.4) that rank(G) is finite. EXAMPLE 2.3.5 Let E be a Corner ring. 1. Given a Corner module M then E = O(M ⊕E). By Theorem 2.3.4 there is a short exact sequence 0 → M ⊕ E −→ G −→ QE ⊕ QE → 0 of left E-modules in which E ∼ = End(G). One might choose M ∼ = (r) E for some integer r > 0. 2. If we let M = E then E = O(M ) so there is a short exact sequence 0 → E −→ G −→ QE ⊕ QE → 0 of left E-modules such that E = End(G). For all practical modern purposes this is Corner’s Theorem 2.3.3. EXAMPLE 2.3.6 If we are willing to vary E but retain QE then we can use Theorem 2.3.4 as follows. Let A be a finite dimensional Q-algebra, let V be any finitely generated left A-module such that annA (V ) = 0, and let M be a free subgroup of V such that QM = V . Then E = O(M ) = {q ∈ A qM ⊂ M } is a subring of A such that (E, +) is a finitely generated free group and QE = A. An appeal to Theorem 2.3.4 produces a short exact sequence (2.4) in which E ∼ = End(G). Thus we should avoid conjectures like “every group has a proper nonzero direct summand,” “rank of G and rank of End(G) are related by some simple function,” and “indecomposable groups have local endomorphism rings.” Such buffoonery is exposed by Butler’s construction and Corner’s Theorem.

2.4

Arnold-Lady-Murley Theorem

While the examples in the previous sections give us an idea that characterizing module theoretic properties of G in terms of End(G) will not be easy, the Theorem of D.M. Arnold and E. L. Lady provides us with a friendly setting for studying direct sum decompositions of G. Finitely generated projective modules, at least on the surface, seem to be easier to work with than do torsion-free groups of finite rank.

30

2.4.1

CHAPTER 2. MOTIVATION BY EXAMPLE

Category Equivalence

Let G be an rtffr group. Recall the additive functors HG (·) and TG (·) and the associated natural transformations Θ and Φ from Chapter 1. Observe that HG (·) takes groups to right End(G)-modules, while TG (·) takes right End(G)-modules to groups. The first result classifies the direct summands of G(n) for integers n > 0 in terms of a finitely generated projective right End(G)-module. We will let Po (G) = {H G(n) ∼ = H ⊕ K for some integer n > 0 and some group K}. Po (End(G)) = the set of finitely generated projective right End(G)-modules. We consider Po (G) as a category of groups and we consider Po (End(G)) as a category of right End(G)-modules. THEOREM 2.4.1 [D. Arnold and L. Lady] See [10, page 47, Theorem 5.1]. If G is an rtffr group then the functors HG (·) : Po (G) −→ Po (End(G)) TG (·) : Po (End(G)) −→ Po (G) are inverse category equivalences. Proof: For the sake of the argument let E = End(G). Since Θ and Φ are natural transformations ΘH⊕K = ΘH ⊕ ΘK for groups H and K, and ΦM ⊕N = ΦM ⊕ ΦN for right End(G)-modules M and N . Moreover, since G is a group, for each integer n > 0 there is a group homomorphism ΘG(n) = ⊕n ΘG : TG HG (G(n) ) −→ G(n) . Thus given H ⊕ K ∼ = G(I) we can prove that ΘH is an isomorphism if we can prove that ΘG is an isomorphism. Similarly to show that ΦM is an isomorphism it suffices to show that ΦE is an isomorphism. Consider the map ΦE : E → HG TG (E). Notice that HG TG (E) ∼ = E with generator the map f : G → TG (E) such that f (x) = 1G ⊗ x for each x ∈ G. Then ΦE (1) = f which implies that ΦE is an isomorphism.

2.4. ARNOLD-LADY-MURLEY THEOREM

31

The reader proved in Chapter 1 that ΘTG (E) ◦ TG (ΦE ) = 1TG (E) . Then ΘTG (E) is an isomorphism since ΦE and TG (ΦE ) are isomorphisms. Inasmuch as G ∼ = TG (E), ΘG is an isomorphism. Given our reductions, the proof is complete. Let P(G) = {groups H H ⊕ H 0 ∼ = G(c) for some group H 0 and some cardinal c}. Let P(End(G)) be the category of projective right End(G)-modules. EXAMPLE 2.4.2 Let G = ⊕p Zp where p ranges over Q the primes in Z. Notice that G is not an rtffr group, that End(G) = p Zp is a semihereditary ring, and that G is a projective (=flat) left End(G)-module. Let I = G. Then I is a projective ideal in End(G) that is not finitely generated and such that TG (I) = IG = G. Inasmuch I ∼ 6 End(G) we have = shown that TG (·) : P(End(G)) → P(G) is not a category equivalence. EXAMPLE 2.4.3 Let G = Z(ℵo ) and let E = End(G). Then G is not an rtffr group but G is a cyclic projective left E-module, so TG (I) ∼ = IG for each right ideal I ⊂ End(G). If we let I = {f ∈ A f (G) has finite rank} then TG (I) = IG = G = TG (End(G)) but I 6∼ = End(G). Once again TG (·) : P(End(G)) → P(G) is not a category equivalence. See [32] for generalizations of Theorem 2.4.1 to self-small right Rmodules G over some ring R.

2.4.2

Functor A(·)

There is another functor and associated ring that we will single out. Let G be an rtffr group and let E(G) = End(G)/N (End(G)). While HG (·) takes direct summands of G to cyclic projective right End(G)-modules the ring End(G) can itself have complicated structure. One of the strengths of the functor A(·) below is that we can translate between direct summands of G and cyclic projective right E(G)-modules over the Noetherian semi-prime rtffr ring E(G). If E is an rtffr ring then two E-modules M and N are quasi-isomorphic if there is an integer m 6= 0 and E-module maps f : M → N , g : N → M such that f g = m1N and gf = m1M .

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CHAPTER 2. MOTIVATION BY EXAMPLE

THEOREM 2.4.4 Let G be an rtffr group. Let A(·) : Po (G) −→ Po (E(G)) be defined by A(·) = Hom(G, ·) ⊗End(G) End(G)/N (End(G)). 1. A(·) is an additive full functor that preserves direct sums. That is, given an E(G)-module map f : A(H) → A(H 0 ) there is a group map φ : H → H 0 such that A(φ) = f . 2. A(·) induces a bijection α : {(H) H ∈ Po (G)} −→ {(P ) P ∈ Po (E(G))} between the set of isomorphism classes (H) of H ∈ Po (G) and the set of isomorphism classes (P ) of P ∈ Po (A(G)). 3. A(·) induces a bijection between the set of quasi-isomorphism classes [H] of H ∈ Po (G) and the set of quasi-isomorphism classes [P ] of finitely generated projective right E(G)-modules. Proof: 1. Say f : A(H) → A(H 0 ) is an E(G)-module map. By the Arnold-Lady Theorem 2.4.1, HG (H) is a finitely generated projective End(G)-module so f lifts to a map g : HG (H) → HG (H 0 ). Because HG (·) is a category equivalence on Po (G) (Arnold-Lady Theorem 2.4.1), there is a group map φ : H → H 0 such that HG (φ) = g. The reader can verify that A(φ) = f . Parts 2 and 3 follow from [10, page 112, Corollary 9.6].

2.4.3

Elementary Uses of the Functors

The results in this subsection illustrate how one uses the above functors to translate properties of direct summands of G into properties of finitely generated projective modules over a ring. We will include the proofs since they are the first of their kind in this text. PROPOSITION 2.4.5 Let G be an rtffr group. The following are equivalent. 1. If G(n) = H ⊕ H 0 for some integer n > 0 then there is an integer m such that H ∼ = G(m) .

2.4. ARNOLD-LADY-MURLEY THEOREM

33

2. Each finitely generated projective right E(G)-module is free. Proof: Suppose that part 2 is true, let n > 0 be an integer, and let G(n) = H ⊕ K as groups. By Theorem 2.4.4 E(G)(n) ∼ = A(G(n) ) ∼ = A(H) ⊕ A(K) as projective right E(G)-modules. By part 2 A(H) ∼ = E(G)(m) for some integer m so that H ∼ = G(m) by Theorem 2.4.4. The converse is handled in an analogous manner so the proof is complete. The above work shows us that properties of direct sum decompositions of the rtffr group G are equivalent to properties of direct sum decompositions of free right End(G)-modules. This is one example where good results come from the Arnold-Lady Theorem 2.4.1. The general idea is that projective E(G)-modules are better behaved than direct summands of G(n) for integers n > 0. Since projective E(G)-modules are better behaved than those over End(G) it is natural to ask for an AKS-like Theorem for direct sums of groups whose endomorphism rings are pid’s. Recall the refinement property and unique decomposition from section 2.1. THEOREM 2.4.6 Suppose that G1 , . . . , Gt are indecomposable rtffr groups such that for each i 6= j, the composite of each pair of maps Gi → Gj → Gi is in N (EndR (Gi )). Assume that E(Gi ) is a pid for each i = 1, . . . , t. If we let G ∼ = G1 ⊕ · · · ⊕ Gt then 1. G has the refinement property. 2. G has a unique decomposition. ∼H ⊕K ∼ 3. If G(n) = = H ⊕ K 0 for some integer n > 0 and some rtffr groups K and K 0 then K ∼ = K 0. Proof: We begin the proof with some general comments about finitely generated projective right E(G)-modules. By Lemma 1.3.1 and Theorem 2.4.4 E(G) = E(G1 ) × · · · × E(Gt ) where E(Gi ) = A(Gi ) is indecomposable for each i = 1, . . . , t.

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CHAPTER 2. MOTIVATION BY EXAMPLE

Furthermore since E(Gi ) is a pid, given a finitely generated projective right E(G)-module P there are integers p1 , . . . , pt ≥ 0 such that P ∼ = E(G1 )(p1 ) ⊕ · · · ⊕ E(Gt )(pt ) . The uniqueness of the rank of free modules over a pid implies that the integers p1 , . . . , pt are unique to P . 1. Let H ∈ Po (G). Since A(H) ∈ Po (E(G)) we have A(H) ∼ = E(G1 )(h1 ) ⊕ · · · ⊕ E(Gt )(ht ) ∼ = A(G1 )(h1 ) ⊕ · · · ⊕ A(Gt )(ht ) (h ) (h ) ∼ = A(G1 1 ⊕ · · · ⊕ Gt t )

for some integers h1 , . . . , ht ≥ 0. Thus (h ) (h ) H∼ = G1 1 ⊕ · · · ⊕ Gt t

by Theorem 2.4.4, and hence G has the refinement property. 2. Suppose that G = G01 ⊕ · · · ⊕ G0s for some indecomposable right E-modules G01 , . . . , G0s . Then E(G) = E(G1 ) × · · · × E(Gt ) = A(G01 ) ⊕ · · · ⊕ A(G0s ). Consider the indecomposable central idempotent ei ∈ E(G) that corresponds to E(Gi ). Then ei A(G0j ) is a direct summand of A(G0j ), and since A(G0j ) is indecomposable, there is a permutation of the subscripts of A(G01 ), . . . , A(G0s ) such that ei A(G0i ) = A(G0i ) ⊂ E(Gi ) for i = 1, . . . , t. Since E(Gi ) is an indecomposable ideal in E(G) and since ei A(G0i ) is a direct summand of E(Gi ) we have ei A(G0i ) = E(Gi ) = A(Gi ). Then by Theorem 2.4.4, G0i ∼ = Gi . Since t is finite, s = t. 3. Cancellation follows from the uniqueness result in part 2 by counting indecomposable summands of K and K 0 . The reader should prove this if he/she has not already done so. This completes the proof. Given a subgroup H ⊂ Q, End(H) is a pid. Thus the Baer-KulikovKaplansky Theorem 2.1.11 is an immediate consequence of Theorem 2.4.6.

2.5. LOCAL ISOMORPHISM

35

We close this section with elementary translations of the refinement property and uniqueness of decomposition for G. PROPOSITION 2.4.7 Let G be an rtffr group. The following are equivalent. 1. G has the refinement property as a group. 2. End(G) has the refinement property as a right End(G)-module. 3. E(G) has the refinement property as a right E(G)-module. PROPOSITION 2.4.8 Let G be an rtffr group. The following are equivalent. 1. G has a unique decomposition. 2. End(G) has a unique decomposition as a right End(G)-module. 3. E(G) has a unique decomposition as a right E(G)-module.

2.5

Local Isomorphism

Let E be an rtffr ring and let M and N be rtffr E-modules. We will say that M and N are locally isomorphic if for each integer n > 0 there is an integer m 6= 0 and E-module maps fn : M → N , gn : N → M such that gcd(m, n) = 1, fn gn = m1N , and gn fn = m1M . Since we have not ruled out E = Z we have defined local isomorphism for rtffr groups G and H. Locally isomorphic rtffr groups are called nearly isomorphic in [10] and locally isomorphic E-lattices are said to be in the same genus class in [10, 69]. Our introduction of the new terminology is justified by the fact that we are using one term to describe two essentially identical concepts on the category of rtffr E-modules. The next lemma gives us the relationship between local isomorphism of rtffr groups and the local isomorphism of finitely generated projective modules. Its proof rests on the fact that an additive functor takes multiplication by an integer to multiplication by an integer. It is left as an exercise. LEMMA 2.5.1 Let G be an rtffr group, let H, K ∈ Po (G) be rtffr groups, and let E = End(G). The following are equivalent. 1. H and K are locally isomorphic as groups.

36

CHAPTER 2. MOTIVATION BY EXAMPLE 2. HG (H) and HG (K) are locally isomorphic as right E-modules.

Local isomorphism appears in several useful equivalent forms. We leave it to the reader to prove a similar result for finitely generated rtffr modules over an rtffr ring. Recall that Xn = X ⊗Z Zn for integers n 6= 0. LEMMA 2.5.2 Let G and H be rtffr groups. The following are equivalent. 1. G and H are locally isomorphic. 2. For each integer n 6= 0 there are fn ∈ HomE (G, H)n and gn ∈ HomE (H, G)n such that fn gn = 1H and gn fn = 1G . 3. For each prime p ∈ Z there are group maps fp ∈ HomE (G, H)p and gp ∈ HomE (H, G)p such that fp gp = 1H , and gp fp = 1G . 4. For each integer n 6= 0 there are there are fn ∈ HomE (G, H) and gn ∈ HomE (H, G) such that fn gn −1H ∈ nEnd(H) and gn fn −1G ∈ nEnd(G). Proof: 1 ⇒ 2 Given part 1 and an integer n 6= 0 there is an integer m 6= 0, and maps fn : G → H, and gn : H → G such that gcd(m, n) = 1, fn gn = m1H , gn fn = m1G . Since gcd(m, n) = 1, m is a unit in Zn , 1 1 1 fn ∈ HomE (G, H)n . Then ( m fn )gn = 1H and gn ( m fn ) = 1G , and so m which proves part 2. 2 ⇒ 3 is clear. 3 ⇒ 4 Let n = pn1 1 · · · pnt t 6= 0 be an integer where p1 , . . . , pt are the distinct prime divisors of n. Fix an i ∈ {1, . . . , t}. By part 3, there are maps fi ∈ HomE (G, H)pi and gi ∈ HomE (H, G)pi such that fi gi = 1H , and gi fi = 1G . Since X p ∼ X + pk X p = pk X p pk X p for p-reduced groups X and prime powers pk ∈ Z there are maps fi0 ∈ Hom(G, H) and gi0 ∈ Hom(H, G) such that fi0 − fi ∈ pni i HomE (G, H)pi gi0 − gi ∈ pni i HomE (H, G)pi . (X = Hom(G, H) in this case.) By the Chinese Remainder Theorem 1.5.3 there are f ∈ Hom(G, H) and g ∈ Hom(H, G) such that f − fi0 ∈ pni i HomE (G, H) g − gi0 ∈ pni i HomE (H, G)

2.5. LOCAL ISOMORPHISM

37

for each i = 1, · · · , t. Then gf − gfi0 ∈ pni i EndE (G) gfi0 − gi fi ∈ pni i EndE (G)pi so that gf − 1G = gf − gi fi = (gf − gfi0 ) + (gfi0 − gi0 fi0 ) + (gi0 fi0 − gi fi ) ∈ pni i EndE (G)pi . Inasmuch as gf and 1G ∈ End(G) gf − 1G ∈ pni i EndE (G)pi ∩ End(G) = pni i End(G). Since i ∈ {1, . . . , t} was arbitrarily selected, part 3 of the Chinese Remainder Theorem 1.5.3 shows us that gf − 1G ∈

t \

pni i EndE (G) = nEndE (G).

i=1

This proves part 4. 4 ⇒ 1 Choose an integer n 6= 0 such that gf − 1G ∈ nEnd(G) and f g−1H ∈ nEnd(H). Because G is an rtffr group we may choose n so large that End(G)n is a reduced ring. Then n ∈ J (End(G)n ) (try proving that fact, reader), and hence f g is a unit of End(G)n . Let u = (f g)−1 and choose an integer m such that gcd(m, n) = 1 and mu ∈ End(G). Then muf ∈ Hom(G, H) and (muf )g = m1G . Since QG and QH are finite dimensional vector spaces and since [g(uf )]2 = g(uf ) ∈ End(H) is an injection, g(uf ) = 1QH , and hence g(muf ) = m1H . This proves part 1 and completes the proof. The proof of the next result is an exercise. LEMMA 2.5.3 Let E be a semi-prime rtffr ring and let U and V be E-lattices. If E is a semi-local ring then U and V are locally isomorphic iff U ∼ =V. Proof: Suppose that U and V are locally isomorphic and say that there is an integer n 6= 0 such that pE = E for each prime p ∈ Z not dividing n. There is by definition an integer m and maps fn : U → V and gn : V → U such that gcd(m, n) = 1, fn gn = m1V , and gn fn = m1U . Then m is a unit in Zn so that fn and gn are isomorphisms. The converse is clear so the proof is complete.

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CHAPTER 2. MOTIVATION BY EXAMPLE

LEMMA 2.5.4 Let E be an rtffr ring. The following are equivalent for finitely generated projective rtffr E-modules M and N . 1. M and N are locally isomorphic. 2. Mp ∼ = Np as right Ep -modules for each prime p ∈ Z. 3. Mn ∼ = Nn as right En -modules for each integer n 6= 0 ∈ Z. 4. For each integer n 6= 0 there are injections fn : M → N , and gn : N → M such that fn (M ) + nN = N and gn (N ) + nM = M . Proof: 1 ⇒ 2 ⇒ 3 follow as they did in Lemma 2.5.2. 3 ⇒ 4 Assume part 3 and let n 6= 0 be an integer. There are fn ∈ HomE (Mn , Nn ) and gn ∈ HomE (Nn , Mn ) such that fn gn = 1Mn . Since M and N are finitely generated, we may choose an integer m such that gcd(m, n) = 1, mfn ∈ HomE (M, N ), mgn ∈ HomE (N, M ), and (mfn )(mgn ) = m2 1M . Inasmuch as m is a unit in Zn the isomorphism X ∼ Xn ∼ X + nXn = = nX nXn nXn and our choice of fn and gn show us that (mfn )(M ) + nN = N and (mgn )(N ) + nM = M. This proves part 4. 4 ⇒ 1 Assume part 4 and let n 6= 0 be an integer. Choose E-module maps f : M → N and g : N → M such that f (M ) + nN = N and g(N ) + nM = M . Since E is an rtffr ring we can assume without loss of generality that n is so large that the localization En is a reduced group. Then n ∈ J (En ) and since Mn and Nn are finitely generated right E-modules Nn = f (Mn ) + nNn = f (Mn ) by Nakayama’s Lemma 1.2.3. Similarly g(Nn ) = Mn , hence rank(Mn ) = rank(Nn ), whence f : Mn → Nn is an isomorphism. Let h : Nn → Mn be the inverse of f . Inasmuch as M and N are finitely generated there is an integer m 6= 0 such that gcd(m, n) = 1 and mh(N ) ⊂ M . Then f and mh are maps between M and N such that f (mh) = m1N and (mh)f = m1M . This completes the proof. An extreme case of the above lemma is found if E is a classical maximal order. Recall that an E-lattice is a finitely generated left E-module M of QE (n) for some integer n.

2.5. LOCAL ISOMORPHISM

39

LEMMA 2.5.5 Let E be an integrally closed semi-prime rtffr ring and let M, N be E-lattices. Then M and N are locally isomorphic iff QM ∼ = QN as right QE-modules. Proof: Since the integrally closed ring E equals E1 × · · · × Et for some classical maximal orders and since QM ∼ = QN Ei = QN iff QM Ei ∼ as QEi -modules for each i = 1, . . . , t, we have reduced the problem to the case where E is a classical maximal order. Given our reduction, suppose that QE is a prime Artinian ring. Then QM ∼ = QN iff M is locally isomorphic to N by [10, Corollary 12.4]. This completes the proof. The next two results relate local isomorphism to properties of direct sums. The argument is originally due to E. L. Lady. See [16]. LEMMA 2.5.6 Let E be an rtffr ring, and let G and H be rtffr Emodules. If G is locally isomorphic to H then G ⊕ G ∼ = H ⊕ K for some group K. Proof: Let n 6= 0 be any integer. Since H is locally isomorphic to G there is an integer m 6= 0 and group maps fn : G → H and gn : H → G such that gcd(m, n) = 1 and gn fn = m1G . Again there is an integer k 6= 0 and maps fm : G → H and gm : H → G such that gcd(k, m) = 1 and gm fm = k1G . Since gcd(k, m) = 1 there are integers a and b such that am + bk = 1. Consider the maps σ : G ⊕ G −→ H : x ⊕ y 7−→ afn (x) + bfm (y) : H −→ G ⊕ G :

z

7−→ gn (z) ⊕ gm (z).

Then σ(z) = afn gn (z) + bfm gm (z) = (am + bk)(z) = z and so G ⊕ G ∼ = H ⊕ ker σ. This completes the proof. The next result shows us that local isomorphism is tied to the cancellation property. THEOREM 2.5.7 Let H, K, and K 0 be rtffr groups. 1. If H⊕K is locally isomorphic to H⊕K 0 then K is locally isomorphic to K 0 . 2. If n > 0 is an integer and if H (n) ∼ = K (n) then H is locally isomorphic to K.

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CHAPTER 2. MOTIVATION BY EXAMPLE

Proof: 1. Suppose that H ⊕ K is locally isomorphic to H ⊕ K 0 . By Lemma 2.5.1 we may assume without loss of generality that H, K, K 0 are projective right modules over an rtffr ring E such that E = H ⊕ K. Let p ∈ Z be a prime. Since E is an rtffr ring we can choose an k integer multiple of p, say n 6= 0, such that ∩∞ k=1 n E = 0. Then En ∼ = Hn ⊕ Kn0 = Hn ⊕ Kn ∼ as right En -modules and consequently En ∼ Hn Kn ∼ Hn K0 ⊕ ⊕ n0 = = nEn nHn nKn nHn nKn as projective right En /nEn -modules. Since E is torsion-free of finite rank, En /nEn is finite, hence Artinian. Then an application of the AKS Theorem 2.1.6 implies that there is an isomorphism f¯ : Kn /nKn → Kn0 /nKn0 . There are natural isomorphisms K ∼ Kn f¯ K 0 ∼ Kn0 −→ . = = nK nKn nK 0 nKn0 Because K and K 0 are projective f¯ lifts to a map f : K → K 0 such that f (K) + nK 0 = K 0 . Similarly there is a map g : K 0 → K such that g(K 0 ) + nK = K. Then by Lemma 2.5.4(4), K is locally isomorphic to K 0 . Given our reductions we have completed the proof of part 1. Part 2 follows in a manner similar to that of part 1. This completes the proof of the lemma. There are at least two general conditions on G that imply a cancellation property for H ⊕ K. The group G is semi-local if there are at most finitely many primes p ∈ Z such that pG 6= G. COROLLARY 2.5.8 Let H and K be rtffr groups. 1. If K is semi-local then H ⊕ K ∼ = H ⊕ K0 ⇒ K ∼ = K 0. 2. If each right ideal of End(K) is principal then H ⊕ K ∼ = H ⊕ K0 ⇒ K ∼ = K 0. Proof: 1. Suppose that H ⊕ K ∼ = H ⊕ K 0 . Since K is semi-local there is an integer n 6= 0 such that nK 6= K and pK = K for each prime p ∈ Z such that gcd(p, n) = 1. By Theorem 2.5.7, K is locally isomorphic to K 0 so there is an integer m 6= 0 and injections fn : K → K 0 and gn : K 0 → K such that gcd(m, n) = 1, fn gn = m1K 0 , and such that

2.5. LOCAL ISOMORPHISM

41

1 gn fn = m1K . Inasmuch as mK = K we see that m gn : K 0 → K is a 1 1 map such that fn ( m gn ) = 1K 0 and ( m gn )fn = 1K . That is, K ∼ = K 0 as required. 2. Suppose that H ⊕K ∼ = H ⊕K 0 and that each right ideal of End(K) is principal. By Theorem 2.5.7, K is locally isomorphic to K 0 so there is an injection g : K 0 → K. The reader can show that Hom(K, K 0 )K = K 0 . Identify Hom(K, K 0 ) ∼ = gHom(K, K 0 ) with a right ideal of End(K). By the hypothesis on End(K) there is an f ∈ Hom(K, K 0 ) such that gHom(K, K 0 ) = f End(K). Thus when considered as subgroups of QK

g −1 f (K) = g −1 [f End(K)]K = g −1 [gHom(K, K 0 )]K = K 0 . Since H, K, K 0 have finite rank, rank(K) = rank(K 0 ), so that the surjection g −1 f : K → K 0 is an isomorphism. This completes the proof.

We can use local isomorphism to establish some kind of uniqueness for the direct sum decompositions of an rtffr group. Let E be an rtffr ring and let G be an rtffr left E-module. We say that G has a locally unique decomposition if 1. There is a direct sum decomposition G = G 1 ⊕ · · · ⊕ Gt

(2.5)

of indecomposable E-modules G1 , . . . , Gt , and 2. Given a direct sum decomposition G = G01 ⊕ · · · ⊕ G0s of indecomposable E-modules G01 , . . . , G0s then s = t and after a possible permutation of the indices G0i is locally isomorphic to Gi . In case G satisfies the above properties then G1 ⊕ · · · ⊕ Gt is called the locally unique decomposition of G. Observe that we do not require that the G1 , . . . , Gt are distinct groups in the above definition. Also, if E = Z then we have defined the locally unique decomposition of an rtffr group G. EXAMPLE 2.5.9 A locally unique decomposition that is not a unique decomposition is constructed as follows. Let E be a Dedekind domain

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CHAPTER 2. MOTIVATION BY EXAMPLE

with class number h > 1. There is an ideal I 6∼ = E such that I · · · I ∼ =E with h factors. Then M = I (h) ∼ = E (h−1) ⊕ (I · · · I) ∼ = E (h) . Thus M does not have a unique decomposition. However, the decomposition is locally unique since at each prime p ∈ Z, Ep is a pid (see [10] or [69]). Each direct summand of the projective Ep -module Mp is free. In particular any indecomposable direct summand U of Mp is isomorphic to Ep , whence Mp has a unique decomposition. Let E be an rtffr ring and let G be an rtffr left E-module. We say that G has the local refinement property if 1. There is a direct sum decomposition (2.5) for some indecomposable left E-modules G1 , . . . , Gt , and 2. Given H ∈ Po (G) then H = H1 ⊕ · · · ⊕ Hs for some indecomposable left E-modules H1 , . . . , Hs such that each Hi is locally isomorphic to some element of {G1 , . . . , Gt }. Observe that the Gi in the direct sum (2.5) need not be distinct. If E = Z then we have defined what we mean by the local refinement property for the rtffr group G. A large portion of the energy in this book is expended on the study of (locally) unique decompositions and the (local) refinement properties. Locally unique decomposition follows from the local refinement property. LEMMA 2.5.10 Let E be an rtffr ring and let G be an rtffr left Emodule. If G has the local refinement property then G has a locally unique decomposition. Proof: Suppose that G has the local refinement property. Then G has a direct sum decomposition (2.5) for some indecomposable left Emodules G1 , . . . , Gt . Suppose that G = G01 ⊕ · · · ⊕ G0s

(2.6)

for some indecomposable left E-modules G01 , . . . , G0s . Since G has the local refinement property, G01 is locally isomorphic to some element of {G1 , . . . , Gt }.

2.5. LOCAL ISOMORPHISM

43

∼ G0 for each i = 1, . . . , t after a We will show that s = t and that Gi = i permutation of the subscripts. We can assume without loss of generality that G01 is locally isomorphic to G1 and we know that G = G1 ⊕ (G2 ⊕ · · · ⊕ Gt ) and G1 ⊕ (G02 ⊕ · · · ⊕ G0s ) are (locally) isomorphic. Then by Theorem 2.5.7 G 2 ⊕ · · · ⊕ Gt

and G02 ⊕ · · · ⊕ G0s

are locally isomorphic. By a simple induction on rank(G), s − 1 = t − 1, s = t, and after a possible reindexing, Gi is locally isomorphic to G0i for each i = 2, . . . , t. This concludes the proof. We end this chapter by showing that at least one type of group has a unique decomposition. An idempotent e ∈ E is central if xe = ex for each x ∈ E. LEMMA 2.5.11 Let G = G1 ⊕ · · · ⊕ Gt be an rtffr group, for each integer i = 1, . . . , t let e2i = ei ∈ End(G) be central indecomposable idempotents such that ei (G) = Gi and such that 1 = e1 + ⊕ · · · ⊕ et . Then G1 ⊕ · · · ⊕ Gt is the unique decomposition of G. Proof: Let G = G1 ⊕ · · · ⊕ Gt be the given indecomposable decomposition and let {e1 , . . . , et } ⊂ End(G) be the associated complete set of primitive orthogonal indecomposable central idempotents. Suppose that G = G01 ⊕ · · · ⊕ G0s for some indecomposable groups G0i . Let {f1 , . . . , fs } be the complete set of orthogonal indecomposable idempotents corresponding to G01 ⊕ · · · ⊕ G0s . Because ei is central ei fj = fj ei for each i, j. It follows that e1 = e1 1G = e1 f1 ⊕ · · · ⊕ e1 fs is a decomposition of the indecomposable e1 into orthogonal idempotents, so that e1 = e1 fi for some i. In a similar manner, since G0i is indecomposable e1 fi = fi . That is e1 = fi . Reindexing if necessary we can assume that we have chosen f1 , . . . , ft such that ei = fi for each i = 1, . . . , t. Since e1 ⊕ · · · ⊕ et = 1G = f1 ⊕ · · · ⊕ ft ft+1 = . . . = fs = 0. Therefore Gi = ei G = fi G = G0i and the proof is complete. We will find that commuting endomorphisms of rtffr groups G occur with some regularity.

44

CHAPTER 2. MOTIVATION BY EXAMPLE

2.6

Exercises

p ∈ Z is a prime, n, m > 0 are integers, c and d are cardinals. G and H are groups, E is a ring, M is a right E-module, and f : G → H is a map. 1. If G is an rtffr group then for each index set I and each f : G → G(I) there is a finite subset J ⊂ I such that g(G) ⊂ G(J ) . 2. The linear transformation f : QG → QH is in QHom(G, H) iff . f (G) = H. 3. Hom(G, H) is a right End(G)-module. 4. Let p ∈ Z be prime. b p are indecomposable groups. (a) Z/pn Z, Z(p∞ ), and Z (b) Any subgroup of Q is indecomposable. 5. Let p ∈ Z be prime. Show that for each integer n > 0, End(Zp ) ∼ = Zp and End(Z/pn Z) ∼ = Z/pn Z are local rings. 6. Show that if G is an rtffr group then G/f (G) is finite for each injection f : G → G. 7. Show that if G is an rtffr group then a bounded quotient G/H is finite. 8. The rtffr group G is strongly indecomposable iff G is an indecomposable object in QAb. b p is strongly indecomposable. 9. A pure subgroup of Z 10. Let p ∈ Z be prime. Show that the subgroups of Z(p∞ ) form a bp. well ordered chain and show that End(Z(p∞ )) ∼ =Z 11. Let Q be the field of fraction of the ring Z[[x]] of power series. Determine the Z[[x]]-module Q/Z[[x]] and its Z[[x]]-endomorphism ring. 12. Let S ⊂ Q be a ring. Then each element of S is an integral multiple of a unit in S. 13. Let E be a Corner ring, let M be a Corner E-module, and let K be a right E-module. Let (2.4) be the short exact sequence constructed in Theorem 2.3.3. Show that TorkE (K, M ) ∼ = TorkE (K, G) for each integer k ≥ 1. Conclude that fdE (M ) = fdE (G).

2.7. QUESTIONS FOR FUTURE RESEARCH

45

14. Let E be a Corner ring, let M be a Corner E-module, and let N be a left E-module. Let (2.4) be the short exact sequence constructed in Theorem 2.3.3. Show that ExtkE (N, M ) ∼ = ExtkE (N, G) for each integer k ≥ 2. Conclude that pdE (M ) = pdE (G). 15. Show that Z(p∞ ) is an E-torsion group. 16. This more of a project than an exercise. Prove that there is a Corner group G such that IG 6= G for each finitely generated right ideal I ⊂ End(G) but such that K ⊗End(G) G = 0 for some nonzero finitely generated K. 17. Let G = G1 ⊕ · · · ⊕ Gt be an abelian group where E(Gi ) = End(Gi )/N (End(Gi )) is a pid for each i = 1, . . . t. Show that if H ⊕ H 0 ∼ = G(c) for some (h ) (h ) t cardinal c then H ∼ = G1 1 ⊕· · ·⊕Gt for some cardinals h1 , . . . , ht . 18. If G and H are locally isomorphic then Hom(G, H)G = H. 19. State and prove a result like Lemma 2.5.6 for locally isomorphic E-lattices G and H. 20. Prove part 2 of Lemma 2.5.7. If n > 0 is an integer and if G(n) ∼ = H (n) then G and H are locally isomorphic.

2.7

Questions for Future Research

In this section I will give some questions that came up during the writing process. These questions can be used to give the reader an eye toward future research in the area of reduced torsion-free finite rank abelian groups, better known as rtffr groups. Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. Extend the ideas of this book to self-small mixed groups. 2. Lift ideal and module theoretic properties from E/N (E) to E and conversely. 3. Lift group or ring theoretic properties from the Z-adic completion b of G to G. See [40]. G

46

CHAPTER 2. MOTIVATION BY EXAMPLE 4. Extend the Arnold-Lady-Murley Theorem to a larger class of groups (i.e., a class that contains Po (G)). See [32]. 5. Study the group theoretic of those G such that End(G) is an integrally closed ring. That is, End(G) is a product of classical maximal orders. 6. Classify the groups G such that End(G) is a semi-perfect ring. 7. In group theoretic terms, describe those groups G that have the Azumaya-Krull-Schmidt Property. For example, The Baer-KulikovKaplansky Theorem shows that the completely decomposable groups satisfy a strong version of this property. If End(G) is semiperfect then G has the Azumaya-Krull-Schmidt Property. 8. Characterize the radical ideals in End(G). For example, the Jacobson radical and the nil radical. 9. Give a correspondence between ideals of End(G) and the subgroups of G. Ideally, the subgroups of G should be the fully invariant Ggenerated subgroups of G.

10. Let E be an rtffr ring. Characterize those (maximal) right ideals I ⊂ E such that IG = G for some group G such that E = End(G). See [40].

Chapter 3

Local Isomorphism Is Isomorphism

rtffr means reduced torsion-free finite rank.

The functor A(·) strikes us as interesting. Thus we will make extensive use of the ideal structure of the semi-prime rtffr ring A(G) = E(G) = End(G)/N (End(G)). In fact A(·) becomes more important in this book than HG (·). This is different from the tone of the literature at the time of this writing. We will show that under a certain commutative localization functor Lτ (·) the local isomorphism of groups G and H translates into the isomorphism of right E(G)τ -modules.

3.1

Integrally Closed Rings

Recall that the ring E is integrally closed if given a ring E ⊂ E 0 ⊂ QE such that E 0 /E is finite then E = E 0 . A prime integrally closed rtffr ring is called a classical maximal order. By Lemma 1.6.3 if E is a semi-prime rtffr ring then QE is semi-simple Artinian. In this case there is an integrally closed ring E ⊂ QE and an integer n > 0 such that nE ⊂ E ⊂ E. 47

48

CHAPTER 3. LOCAL ISOMORPHISM IS ISOMORPHISM

The integrally closed ring E is a finite product E = E1 × · · · × Et

(3.1)

of classical maximal orders E i in the simple Artinian rings QE i . Let S = center(E), let S i = center(E i ) for each i = 1, . . . , t, and let S = S1 × · · · × St.

(3.2)

Then S = center(E), S i is a Dedekind domain for each i = 1, . . . , t, and nS ⊂ S ⊂ S. It follows that S and S are commutative Noetherian semi-prime rtffr rings such that S/S is finite and such that QS = QS. LEMMA 3.1.1 Let I ⊂ E be a right ideal that contains a regular element of E. Then E/I is a finite group. In particular at most finitely many (maximal) ideals of E contain I. Proof: Say that c ∈ I ⊂ E is regular. Then c is a unit in QE so it has an inverse d ∈ QE. There is an integer m 6= 0 such that md ∈ E so that c(md) = m1E ∈ I. Then E/I is bounded by m and since E is an rtffr ring E/I is finite. LEMMA 3.1.2 Let E be a semi-prime rtffr ring, let center(E) = S, and let I ⊂ S be an ideal. Then EI II ⊂ J (EI ). Proof: We know that II ⊂ J (SI ) and we know from Lemma 1.6.3 that because E is semi-prime, E is a finitely generated S-module. Nakayama’s Lemma 1.2.3 shows us that if J is a maximal right ideal J ⊂ EI such that EI II + J = EI then J = EI . (J is also an SI -submodule of EI .) Then EI II ⊂ ∩{J J is a maximal right ideal of E} = J (EI ). The next two results show us how to construct integrally closed rings E such that EI ⊂ E ⊂ E for ideals I ⊂ S. Lemma 1.6.1 states that if I is a right ideal of finite index in E then

O(I) = {q ∈ QE qI ⊂ I} is an integrally closed ring.

3.1. INTEGRALLY CLOSED RINGS

49

LEMMA 3.1.3 Let E be a semi-prime rtffr ring with center(E) = S. 0 0 Suppose that E ⊂ QE is an integrally closed ring such that IE ⊂ E for some ideal I ⊂ S of finite index in S. There is an integrally closed ring E ⊂ QE such that IE ⊂ E ⊂ E. Proof: Let

0

J = {q ∈ QE qE ⊂ E} ⊂ E. 0

0

Since 0 6= IE ⊂ E and since S/I is finite I ⊂ J is a right E -module of finite index in E. Write 0

0

0

E = E1 × · · · × Et 0

for some classical maximal orders E i . Then J = J1 ⊕ · · · ⊕ Jt 0

for some finitely generated right E i -modules Ji . Lemma 1.6.1 then implies that the ring E i = {q ∈ QE i qJi ⊂ Ji } is a classical maximal order in QE i so that E = E 1 × · · · × E t = O(J) is an integrally closed subring in QE. Since J is a left ideal in E, E ⊂ E, 0 and since I ⊂ IE ⊂ J, IE = EI ⊂ EJ = J ⊂ E which completes the proof. EXAMPLE 3.1.4 The following is an example of an rtffr Dedekind domain S and a subring S ⊂ S such that S/S is a finite group and the minimum number of generators for S is rank(S). Specify n > 1 ∈ Z, choose an algebraic number field k 6= Q, and let S denote the ring of algebraic integers in k. Then S is a Dedekind domain. A result from number theory (see, e.g., [58]) states that S is a free abelian group on [k : Q] = rank(S) generators. It follows that if p ∈ Z is a prime then Z/pZ- dim(S/pS) = rank(S) 6= 1

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CHAPTER 3. LOCAL ISOMORPHISM IS ISOMORPHISM

so that there is a proper subring S = Z + pS. Note that pS ⊂ S ⊂ S and that these inclusions are proper. Moreover, since S/pS ∼ = Z/pZ, S/pS, and, hence, S are generated by exactly rank(S) elements as S-modules. EXAMPLE 3.1.5 The ring S need not decompose as a ring. Let S1 and S2 be rtffr Dedekind domains. For i = 1, 2 choose nonzero proper ideals Ii ⊂ Si and construct a ring S as I1 × I2 ⊂ S = I1 × I2 + (1, 1)Z ⊂ S1 × S2 . Then (S1 × S2 )/S is finite but S is an indecomposable ring. (Observe that S does not contain the unique idempotents (1, 0), (0, 1) of S1 × S2 .) The next result is fundamental to our investigations. It originally appeared in [18]. A proof can be found in [10, Theorem 14.2]. THEOREM 3.1.6 [The Beaumont-Pierce-Wedderburn Theorem] Let E be an rtffr ring. There is a semi-prime Noetherian subring T ⊂ E . such that E = T ⊕ N (E). Using The Beaumont-Pierce-Wedderburn Theorem we can classify a number of modules up to quasi-isomorphism. EXAMPLE 3.1.7 This example shows that there can be little hope of classifying the additive structure of N (E). Let G be any abelian group and let n 0 E= n ∈ Z, x ∈ G . x n The reader will show that E is a commutative ring and that N (E) = 0 0 ∼ = G as groups. Then E is an rtffr ring iff G is an rtffr group. G 0 Since each semi-prime rtffr ring E has finite index in an integrally closed ring it is natural to ask if some similar kind of result is true of rtffr groups. Let us agree that G is an integrally closed group if E(G) is an integrally closed ring. LEMMA 3.1.8 If G is an rtffr group then there is a G-generated rtffr group G and an integer n = n(G) > 0 such that nG ⊂ G ⊂ G, E(G) is integrally closed, and E(G)/E(G) is finite.

3.2. CONDUCTOR OF AN RTFFR RING

51

Proof: Let E = End(G). By The Beaumont-Pierce-Wedderburn . Theorem 3.1.6 there is a semi-prime rtffr subring T ⊂ E such that E = T ⊕ N (E) and QE = QT ⊕ QN (E) as QT -modules. Let R = E(G) be the projected image of E into QT modulo QN (E). Then R is a semi. . prime Noetherian ring, T ⊂ R, R = T , and E = R ⊕ N (E). By Lemma 1.6.3 there is an integer n 6= 0 and an integrally closed ring T ⊂ QR such that nT ⊂ T ⊂ R ⊂ T . Let G = T G and observe that nG = (nT )G ⊂ T G = G ⊂ T G = G. Then G/G is finite, and . . . T ⊕ N (E) = R ⊕ N (E) = E = End(G). Hence T ⊂ E(G) and . . . T = R = E(G) = E(G). Since T is integrally closed E(G) = R ⊂ T = E(G) and since G/G is finite E(G)/E(G) is finite. This completes the proof. For instance G is an integrally closed group if End(G) is a Dedekind domain. Theorem 2.3.3 shows us that there are plenty of strongly indecomposable groups G that are not integrally closed groups. There is a strongly indecomposable rtffr group G such that End(G) is a hereditary domain but End(G) is not an integrally closed group. See Example 4.5.7.

3.2

Conductor of an Rtffr Ring

We will continue to use the notation introduced in the previous subsection. Specifically, E is a semi-prime rtffr ring and S = center(E). Using Lemma 1.6.3 we fix an integrally closed ring E ⊂ QE such that E ⊂ E and E/E is finite. Let S = center(E), and choose classical maximal orders E 1 , · · · , E t and Dedekind domains S 1 , · · · , S t as in (3.1) and (3.2).

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CHAPTER 3. LOCAL ISOMORPHISM IS ISOMORPHISM

An E-lattice is a finitely generated right E-submodule of QE (k) for some integer k > 0. The conductor of E into E is the largest ideal τ ⊂ S such that τ E ⊂ E ⊂ E.

Then τ = {q ∈ S qE ⊂ E}. There can be many integrally closed rings E such that E/E is finite, and for each E there is a conductor τ . Thus we say that an ideal τ ⊂ S is a conductor for E if there is an integrally closed ring E such that E/E is finite and such that τ is the conductor for E into E. Since E/E is finite τ contains an integer n 6= 0, so that S/τ and E/τ E are finite rings. Since S = center(E), τ S ⊂ S, so that (τ S)E = τ E ⊂ E. By the maximality of τ in S, τ = τ S is an ideal in S. We need to know that τ is unique to E in some sense. Recall that if I is an ideal in S then the radical of I is √ I = ∩{M M is a maximal right ideal such that I ⊂ M ⊂ S} 0

LEMMA 3.2.1 Let E be a semi-prime rtffr ring. Let E and E be integrally closed rtffr rings that contain E as a subring of finite index, 0 and let τ and τ 0 > 0 be the conductors for E and E into E, respectively. Then √ √ τ = τ 0. In particular given a maximal ideal M ⊂ S then τ ⊂ M ⊂ S iff EM is not integrally closed. Proof: Let I ⊂ S be the ideal defined by the finite intersection \ I = S ∩ {τN N is a maximal ideal in S such that EN 6= E N }. Choose a maximal ideal M of S such that EM 6= E M . Because localization commutes with finite intersection we have \ IM = SM ∩ {(τN )M EN 6= E N } = SM ∩ τM = τM

3.2. CONDUCTOR OF AN RTFFR RING

53

since (τN )M = QS for maximal ideals M 6= N in S. Thus for each maximal ideal M in S we have (IE)M = IM E M = τM E M ⊂ EM . By the Local-Global Theorem 1.5.1, IE ⊂ E, and by the maximality of τ , τ = I. Therefore τ ⊂ M ⊂ S iff EM is not integrally closed. √ √ It follows that if τ and τ 0 are conductors for E into E then τ = τ 0 . EXAMPLE 3.2.2 Let S be the ring of algebraic integers in an algebraic number field k 6= Q. Then S/pS ∼ = Z/pZ(n) where n = [k : Q]. Let S = Z + pS. Then pS ⊂ S ⊂ S and since S/pS ∼ = Z/pZ, pS is the conductor of S into S. The conductor of the semi-prime rtffr ring E gives a local condition that shows us when an E-lattice is a projective E-module. It also yields a product decomposition E = Eo × E1 where Eo is integrally closed. LEMMA 3.2.3 Let E be a semi-prime rtffr ring with conductor τ and let U be an E-lattice. Then U is a projective right E-module iff the localization Uτ is a projective right Eτ -module. Proof: Let U be an E-lattice. It is clear that if U is a projective right E-module then Uτ is a projective right Eτ -module. Conversely assume that Uτ is a projective Eτ -module. We will show that UM is a projective right EM -module for each maximal ideal M ⊂ S. Fix a maximal ideal M ⊂ S. There are two cases possible. If τ ⊂ M then because Uτ is projective, UM = (UM )τ = (Uτ )M is a projective right EM -module. Otherwise τ 6⊂ M so by Lemma 3.2.1, EM is an integrally closed ring. Then UM is an EM -lattice, hence a projective EM -module. Thus UM is a projective E-module for each maximal ideal M in S. Inasmuch as the E-lattice U is finitely presented over the Noetherian ring E, the Change of Rings Theorem 1.5.2 implies that U is a finitely generated projective right E-module. This completes the proof. Let E be a semi-prime rtffr ring with conductor τ ⊂ S and let

C = {c ∈ S c + τ is a unit in S/τ }.

(3.3)

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CHAPTER 3. LOCAL ISOMORPHISM IS ISOMORPHISM

Then C is a multiplicatively closed subset of S. Given an S-module X define X(τ ) = {x ∈ X xc = 0 for some c ∈ C}. Thus X(τ ) is the C-torsion submodule of X. THEOREM 3.2.4 [T.G. Faticoni] Let E be a semi-prime rtffr ring with conductor τ . 1. There are ideals S(τ ), S τ ⊂ S and E(τ ), E τ ⊂ E such that S = S(τ ) × S τ and E = E(τ ) × E τ , 2. Sτ = (S τ )τ and Eτ = (E τ )τ , 3. E(τ ) is an integrally closed ring. Proof: 1. and 2. One shows that S(τ ) is an ideal of S such that S/S(τ ) is a torsion-free group. Since S(τ ) is the kernel of localization at C, Sτ ∼ = (S/S(τ ))τ . Thus S/S(τ ) is a cyclic S-submodule of QS/QS(τ ), so that S/S(τ ) is an S-lattice such that Sτ = (S/S(τ ))τ is a projective Sτ -module. By Lemma 3.2.3, S/S(τ ) is a projective S-module, so that S = S(τ ) × S τ for some ideal S τ of S such that Sτ = (S τ )τ . In a similar manner E = E(τ ) ⊕ E τ and (E τ )τ ∼ = Eτ . Since E is a semi-prime rtffr ring and since E(τ ) is an ideal, E = E(τ ) × E τ . We will return to τ presently. 3. Let M be a maximal ideal in the ring S that contains S τ . For each c ∈ S(τ ) \ M the element (c, 0) is annihilated by S τ . Then SM ∼ = (S/S τ )M ∼ = S(τ )M and EM ∼ = (E/E τ )M ∼ = E(τ )M . Suppose for the sake of contradiction that τ ⊂ M . Let x ∈ S(τ ) \ M . Then x is annihilated by some element c ∈ C. See (3.3). However, since τ ⊂ M , x + M , a unit in S/M , is not annihilated by any of the units in S/τ . This contradiction shows us that τ 6⊂ M .

3.3. LOCAL CORRESPONDENCE

55

∼ E(τ )M is integrally closed for Therefore, by Lemma 3.2.1, EM = each maximal ideal τ 6⊂ M . For τ ⊂ M ⊂ S, E(τ )M = 0 since for each element x ∈ E(τ ) there is a c ∈ C such that xc = 0. Thus E(τ )M is integrally closed for each maximal ideal M ⊂ S, and hence E(τ ) is integrally closed. This completes the proof. For example let E = S = 2(Z6 ⊕ Z10 ) + Z(1 ⊕ 1). Then τ = 2(Z6 ⊕ Z10 ) and C = {(a, b) a − b ∈ τ }. Hence S(τ ) = 0, S τ = S, and Sτ = 2(Z2 ⊕ Z2 ) + Z2 (1 ⊕ 1).

3.3

Local Correspondence

The main result of this section shows how localization at the conductor can be used to translate local isomorphism into isomorphism. The proof of the theorem is partitioned into several lemmas that are interesting in their own right. For the proof of this theorem let τ be a conductor for E, and write E = E(τ ) × E τ as in Theorem 3.2.4. Given an E-lattice U let

U (τ ) = U S(τ ) = {x ∈ U xc = 0 for some c ∈ C}.

THEOREM 3.3.1 Let E be a semi-prime rtffr ring with conductor τ , and write E = E(τ ) × E τ as in Theorem 3.2.4. The localization functor (·)τ induces a bijection λτ : {[U ] U ∈ Po (E)} −→ {(W ) W ∈ Po (QE(τ ) × Eτ )}

from the set of local isomorphism classes [U ] of finitely generated projective right E-modules onto the set of isomorphism classes (W ) of finitely generated projective right QE(τ ) × Eτ -modules. LEMMA 3.3.2 Let E be a semi-prime rtffr ring with conductor τ and let U , V be E-lattices. Then U is locally isomorphic to V iff QU (τ ) ∼ = QV (τ ) and U τ is locally isomorphic to V τ . Proof: Write U = U (τ ) ⊕ U τ . It is clear that U is locally isomorphic to V iff U (τ ) is locally isomorphic to V (τ ) as E(τ )-lattices, and U τ is

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CHAPTER 3. LOCAL ISOMORPHISM IS ISOMORPHISM

locally isomorphic to V τ as E τ -lattices. By Theorem 3.2.4, E(τ ) is an integrally closed Noetherian semi-prime rtffr ring. Then by Lemma 1.6.1, U (τ ) is locally isomorphic to V (τ ) iff QU (τ ) ∼ = QV (τ ). This completes the proof. Begin the proof of Theorem 3.3.1: Define a functor as follows. Lτ (·) : Po (E) −→ Po (QE(τ ) × Eτ ) Lτ (U ) = QU (τ ) ⊕ Uτ . Since the localization functor is additive and exact, Lτ (·) is an additive exact functor. Define a function λτ by λτ ([U ]) = (Lτ (U )) = (QU (τ ) ⊕ Uτ ). We claim that λτ is a well defined function. Suppose that U , V ∈ Po (E) are locally isomorphic. By Lemma 3.3.2, QU (τ ) ∼ = QV (τ ) and U τ τ is locally isomorphic to V . Since τ contains an integer n 6= 0 ∈ τ we can choose an integer m and maps φn : U → V and ψn : V → U such that gcd(m, n) = 1, φn ψn = m1V , and ψn φn = m1U . Since gcd(m, n) = 1, m is a unit of En , and hence φn and ψn lift to isomorphisms between UnS ∼ = VnS . Inasmuch as nS ⊂ τ we have (XnS )τ = Xτ for E-lattices X. Thus Uτ ∼ = (Uτ )nS ∼ = (UnS )τ ∼ = (VnS )τ ∼ = Vτ . This proves the claim, and therefore λτ is a well defined function. We must show that λτ is a bijection. LEMMA 3.3.3 Let E be a semi-prime ring with conductor τ . Then for each finitely generated projective Eτ -module W there is a finitely generated projective E τ -module U such that Uτ ∼ = W. Proof: Let W ∈ Po (Eτ ) and write W = w1 Eτ + · · · + ws Eτ for some finite set {w1 , . . . , ws } ⊂ W . Let U be the E-lattice generated by w1 , · · · , ws . U = w1 E + · · · + ws E.

3.3. LOCAL CORRESPONDENCE

57

There is an E-module surjection π : ⊕si=1 xi E → U such that π(xi ) = wi for each i. Because localization is exact the image of πτ : ⊕si=1 xi Eτ → Uτ ⊂ W is Uτ = w1 Eτ + · · · + ws Eτ = W. Moreover Uτ = W is a finitely generated projective Eτ -module so by Lemma 3.2.3, U is a finitely generated projective E-module. The lemma is proved. COROLLARY 3.3.4 Let E be an rtffr ring with conductor τ . Then λτ is a surjection. Let X τ = XS τ for E-lattices X. Then X = X(τ ) ⊕ X τ . The following are technical lemmas that we need to prove that λτ is an injection. LEMMA 3.3.5 Let E be a semi-prime rtffr ring with conductor τ , and let n 6= 0 ∈ τ be an integer. 1. There is an ideal I ⊂ S such that (a) n ∈ I, (b) I + τ = S, and such that (c) c ∈ S is a unit modulo nS iff c is a unit modulo I ∩ τ . Consequently, UnS = UI∩τ for each finitely generated projective right E-module U . 2. (U τ )n = (U τ )nS for each finitely generated projective right Emodule U . Proof: 1. Let {M1 , . . . , Mr , T1 , . . . , Ts } be a complete set of the maximal ideals of S that contain nS, and assume that {T1 , . . . , Ts } is a complete set of the maximal ideals of S that contain τ . Let I = M 1 ∩ · · · ∩ Mr

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and let I ⊂ N ⊂ S be a maximal ideal. Using the Chinese Remainder Theorem 1.5.3 we see that S S ∼ S × ··· × = I M1 Mr and each S/Mi is a simple module. Thus S/N ∼ = S/Mi for some i = 1, · · · , r. Since S is commutative N = annS (1 + N ) = Mi , so that M1 , . . . , Mr is a complete list of the maximal ideals of S that contain I. In particular I is not contained in any of the maximal ideals T1 , . . . , Ts that contain τ , hence I +τ is not contained in any maximal ideal, whence I +τ = S. Another application of The Chinese Remainder Theorem 1.5.3 shows us that S ∼S S = × I ∩τ I τ as rings. We claim that {c ∈ S c + nS is a unit in S/nS} = {c ∈ S c + (I ∩ τ ) is a unit in S/(I ∩ τ )}. Since nS ⊂ I ∩ τ , and since units map to units, the elements c that are units modulo nS are units modulo I ∩ τ . Conversely, if c ∈ S maps to a unit in the finite ring S/(I ∩ τ ) then c 6∈ Mi or Tj for any Mi , Tj ∈ {M1 , . . . , Mr , T1 , . . . , Ts }. We have shown that {M1 , . . . , Mr , T1 , . . . , Ts } is a complete list of the maximal ideals that contain I ∩ τ . Thus c + nS is not in any of the maximal ideals of S/nS, and so c + nS is a unit of S/nS. As claimed, the set of units modulo nS is the set of units modulo I ∩ τ . Consequently, UnS = UI∩τ for each finitely generated projective right E-module. This proves part 1. 2. Since nS ⊂ τ , the ideal {x ∈ S xc = 0 for some c ∈ S such that c + nS ∈ S/nS is a unit} is contained in the ideal S(τ ). Thus S τ is torsion-free relative to the units modulo nS. Hence there is an inclusion S τ ⊂ (S τ )nS that lifts to one (S τ )n ⊂ (S τ )nS . To see that (S τ )n = (S τ )nS let N1 , . . . , Nr be a complete list of the maximal ideals of the commutative ring S that contain nS. For each N ∈ {N1 , . . . , Nr } ((S τ )n )N = (S τ )N = ((S τ )nS )N so that (S τ )n = (S τ )nS by the Local Global Theorem. (Prove this as an exercise, reader.) Then Unτ ∼ = (U τ )nS for each finitely generated projective S-module. This completes the proof.

3.4. CANONICAL DECOMPOSITION

59

COROLLARY 3.3.6 Let E be a semi-prime rtffr ring with conductor τ , and let n 6= 0 ∈ τ be an integer. Let U and V be finitely generated projective right E-modules such that QU (τ ) ∼ = Vτ . = QV (τ ) and Uτ ∼ V . Then Un ∼ = n Proof: By hypothesis, U (τ ) and V (τ ) are lattices over the integrally closed ring E(τ ) (Theorem 3.2.4), such that QU (τ ) ∼ = QV (τ ). Then U (τ ) and V (τ ) are locally isomorphic (Lemma 1.6.1(4)). Consequently, U (τ )n ∼ = V (τ )n , (Lemma 2.5.4(3)). By Lemma 3.3.5(1) UnS ∼ = UI∩τ ∼ = VI∩τ ∼ = VnS so (U τ )nS ∼ = (U S τ )nS ∼ = (V S τ )nS ∼ = (V τ )nS . Lemma 3.3.5(2) then shows us that (U τ )n ∼ = (U τ )nS ∼ = (V τ )nS ∼ = (V τ )n . Hence Un ∼ = U (τ )n ⊕ (U τ )n ∼ = V (τ )n ⊕ (V τ )n ∼ = Vn , which proves the Corollary. THEOREM 3.3.7 [T.G. Faticoni] Let E be a semi-prime rtffr ring with conductor τ . Then λτ is an injection. Specifically, if U , V ∈ Po (E) then U and V are locally isomorphic E-lattices iff QU (τ ) ∼ = QV (τ ) and ∼ Uτ = Vτ as Eτ -lattices. Proof: Suppose that QU (τ ) ∼ = QV (τ ) and that Uτ ∼ = Vτ . By Lemma 3.3.6, Un and Vn are isomorphic right E-modules for each integer n > 0. Thus U is locally isomorphic to V (Lemma 2.5.4), which completes the proof of the lemma. Proof of Theorem 3.3.1 completed: By Corollary 3.3.4 and Theorem 3.3.7, λτ is a bijection, which completes the proof.

3.4

Canonical Decomposition

Let us extend λτ to include rtffr groups G and the associated semi-prime rtffr ring A(G) = E(G) = End(G)/N (End(G)).

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Suppose that E(G) has conductor τ , and let Lτ (·) be the localization functor. Let E(G)(τ ) denote {x ∈ E(G) xc = 0 for some c ∈ S such that c is a unit modulo τ }. The reader will show that

E(G)(τ ) = E(G)S(τ ).

Recall Arnold’s Functor A(·) from Section 2.4.2. Let us define

Eτ (G) = QE(G)(τ ) × (E(G))τ Aτ (·) = Lτ (·) ◦ A(·) : Po (G) −→ Po (Eτ (G)).

We say that

G has conductor τ if τ is a conductor for E(G).

THEOREM 3.4.1 Let G be an rtffr group with conductor τ . The functor Aτ (·) induces a bijection αG (·) : {[H] H ∈ Po (G)} −→ {(W ) W ∈ Po (Eτ (G))} from the set of local isomorphism classes [H] of H ∈ Po (G) onto the set of isomorphism classes (W ) of W ∈ Po (Eτ (G)). Proof: By [10, Corollary 12.7], A(·) induces a bijection {[H] H ∈ Po (G)} → {[U ] U ∈ Po (E(G))} and Theorem 3.3.1 states that the functor Lτ (·) induces a bijection λτ : {[U ] U ∈ Po (E(G))} → {(W ) W ∈ Po (Eτ (G))}. The assignment [H] 7→ λτ [A(H)] defines the required bijection αG . The reader can prove the next two results as exercises.

3.4. CANONICAL DECOMPOSITION

61

THEOREM 3.4.2 Let G be an rtffr group and assume that E(G) is an integrally closed ring. The functor Q ⊗Z A(·) induces a bijection αG (·) : {[H] H ∈ Po (G)} −→ {(W ) W ∈ Po (QE(G))} from the set of local isomorphism classes [H] of H ∈ Po (G) onto the set of isomorphism classes (W ) of semi-simple right QE(G)-modules W . THEOREM 3.4.3 Let G be an rtffr group with conductor τ 6= S, and assume that QE(G) is a simple Artinian ring. The functor Aτ (·) induces a bijection αG (·) : {[H] H ∈ Po (G)} −→ {(W ) W ∈ Po ((E(G))τ )} from the set of local isomorphism classes [H] of H ∈ Po (G) onto the set of isomorphism classes (W ) of finitely generated projective right (E(G))τ modules W . Proof: Since QE(G) is simple, S is an integral domain and so S τ = S. Thus QE(G)(τ ) = 0. The reader can take it from here. These results lead us to a canonical direct sum decomposition of an rtffr group G. THEOREM 3.4.4 Let G be an rtffr group with conductor τ . Then G = G(τ ) ⊕ Gτ where E(G(τ )) = E(G)(τ ) is an integrally closed ring. Proof: By Theorem 3.2.4, E(G) = E(G)(τ ) × E(G)τ where E(G)(τ ) is integrally closed and E(G)τ (τ ) = 0. Let e2 = e ∈ S be such that E(G)e = E(G)(τ ) and E(G)(1 − e) = E(G)τ . Let G(τ ) = eG and Gτ = (1 − e)G. Thus G = G(τ ) ⊕ Gτ . By Theorem 2.4.4, E(G)(τ ) = A(G(τ )) and E(G)τ = A(Gτ ). Moreover, E(G(τ )) = E(G)(τ ) is integrally closed. This completes the proof.

If we combine the above theorem with Theorem 3.4.2 then we have shown that G can be directly decomposed as G = G(τ ) ⊕ Gτ where the direct sum decompositions of G(τ ) behave like modules over a semisimple Artinian ring, and where Gτ holds the interesting perhaps badly behaved direct sum decompositions.

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THEOREM 3.4.5 [T.G. Faticoni] Let G be an rtffr group with conductor τ . Then G = G(τ ) ⊕ Gτ where 1. Each H ∈ Po (G) can be written uniquely as H = H(τ )⊕H τ where H(τ ) ∈ Po (G(τ )) and H τ ∈ Po (Gτ ). 2. Direct sum decompositions of H(τ ) are functorially equivalent to direct sum decompositions of finitely generated semi-simple modules over a semi-simple Artinian ring. 3. Direct sum decompositions of H τ are functorially equivalent to direct sum decompositions of finitely generated projective modules over a semi-local semi-prime rtffr ring. Proof: Exercise.

3.5

Arnold’s Theorem

The power of Theorem 3.4.1 is that it translates problems on direct sum decompositions of an rtffr group G (which are not finitely generated) into problems on direct sum decompositions of finitely generated projective right modules over the semi-local semi-prime ring Eτ (G) = QE(G)(τ ) × (E(G))τ . Moreover the bijection translates local isomorphism into isomorphism and conversely. For this reason when discussing the local refinement property or locally unique decompositions of G we can instead study the refinement property or unique decompositions of the pleasantly structured ring Eτ (G). In this section we will put the above strategy to work. To begin with we will show that Theorem 3.3.1 can be used to give a short proof of an important result in the study of direct sum decompositions of rtffr groups. Part 2 of the following result is referred to in the literature as Arnold’s Theorem. THEOREM 3.5.1 Let G be an rtffr group with conductor τ , let Aτ (·) be the functor defined on page 60, and let H ∈ Po (G). 1. Aτ (H) ∼ = W1 ⊕ W2 as right Eτ (G)-modules iff H is locally isomorphic to H1 ⊕ H2 for some rtffr groups Hj such that Aτ (Hj ) ∼ = Wj for j = 1, 2.

3.5. ARNOLD’S THEOREM

63

2. [10, D. M. Arnold] H is locally isomorphic to G1 ⊕ G2 as rtffr groups iff H = H1 ⊕ H2 for some rtffr groups Hj such that Hj and Gj are locally isomorphic for j = 1, 2. Proof: 1. Suppose that Aτ (H) = W ∼ = W1 ⊕ W2 . By Theorem 2.4.4, A(H) = U is a finitely generated projective right E(G)-module such that Uτ ∼ = W . By the proof of Lemma 3.3.3, U ⊂ W . Let U1 = W1 ∩ U and let U2 = U/U1 . Because localization is exact (U2 )τ = (U/U1 )τ ∼ = W2 = W/W1 ∼ = Uτ /(U1 )τ ∼ so by Lemma 3.2.3, U2 is a projective right A(G)-module. Hence U ∼ = U1 ⊕ U2 . Furthermore, by Theorem 2.4.4 there are groups H1 , H2 ∈ Po (G) such that A(Hj ) ∼ = = Uj for j = 1, 2, and H = H1 ⊕ H2 . Then Aτ (Hj ) ∼ (Uj )τ ∼ W for j = 1, 2. The converse is clear. = j 2. Suppose that H is locally isomorphic to G1 ⊕ G2 . By Theorem 3.4.1, Aτ (H) ∼ = Aτ (G1 ⊕ G2 ) ∼ = Aτ (G1 ) ⊕ Aτ (G2 ). By part 1, H ∼ = H1 ⊕ H2 for some rtffr groups H1 and H2 such that Aτ (Gi ) ∼ = Aτ (Hi ) for i = 1, 2. Theorem 3.4.1 implies that Gi is locally isomorphic to Hi for i = 1, 2. The converse is evident so the proof is complete. COROLLARY 3.5.2 Let G be an rtffr group with conductor τ . The following are equivalent. 1. G is indecomposable. 2. If G is locally isomorphic to H then H is indecomposable. 3. E(G) is indecomposable as a right E(G)-module. 4. Eτ (G) is indecomposable as a right Eτ (G)-module. Recall locally unique decompositions and the local refinement property from Section 2.5. It follows from Theorem 3.4.1 and the AKS Theorem 2.1.6 that the rtffr group G has a locally unique decomposition if Eτ (G) is a semi-perfect ring. By the Baer-Kulikov-Kaplansky Theorem 2.1.11 completely decomposable groups have a locally unique decomposition and the reader should convince themselves that Eτ (G) is a semi-perfect ring in this case. The next two results show that the local refinement property can be treated as the refinement property.

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THEOREM 3.5.3 [T.G. Faticoni] Let G be an rtffr group with conductor τ . The following are equivalent. 1. G has a locally unique decomposition. 2. E(G) has a locally unique decomposition. 3. Eτ (G) has a unique decomposition. Proof: 1 ⇔ 2 follows immediately from Theorems 2.4.1, 2.4.4 while 1 ⇔ 3 follows immediately from Theorem 3.5.1. The next result shows that a locally unique decomposition can be treated as a unique decomposition. THEOREM 3.5.4 [T.G. Faticoni] Let G be an rtffr group with conductor τ . The following are equivalent. 1. G has the local refinement property. 2. E(G) has the local refinement property. 3. Eτ (G) has the refinement property. Proof: Apply Theorems 2.4.4 and 3.5.1.

3.6

Exercises

Let E be a semi-prime rtffr ring with center S with conductor τ . Let E ⊂ QE be an integrally closed ring such that Eτ ⊂ E ⊂ E. Let S = center(E). 1. Let S be an Artinian ring. Show that S is local iff e2 = 2 ∈ S implies that e = 0 or 1. b p is actually left multi2. Show that any group endomorphism of Z b p . (Hint: Given a map f consider plication by some element of Z f (1).) 3. Let G be an rtffr. There is a bijection from the set of isomorphism classes of eQEnd(G) where e2 = e ∈ QEnd(G) onto the set of quasi-isomorphism classes [H] of quasi-summands H of G. The bijection is given by eQEnd(G) 7→ [eG].

3.6. EXERCISES

65

4. The group G is p-pure in H if G ⊂ H and if G/H is p-torsion-free. Show that (a) G is p-pure in H iff pG = pH ∩ G. b p is strongly indecomposable. (b) a p-pure subgroup G of Z b p . Show that 5. Let G be a p-pure subgroup of Z (a) G/pG ∼ = Z/pZ. bp. (b) End(G) is a pure subring of Z (c) End(G) is an integral domain. 6. Let x ∈ S and choose maximal ideals M1 , . . . , Mt of S such that (n ) M1n1 ∩ · · · ∩ Mt t ⊂ xS ⊂ M1 ∩ · · · ∩ Mt . Prove that each of the maximal ideals of S/xS are of the form Mi /xS. 7. Let H, K ∈ Po (G). Show that the following are equivalent. (a) H is locally isomorphic to H as groups. (b) A(H) is locally isomorphic to A(K) as right E(G)-modules. 8. Let {G1 , . . . , Gt } be a nilpotent set of rtffr groups. Then G = G1 ⊕ · · · ⊕ Gt has the refinement property iff each Gi has the refinement property. 9. (a) If E is a Noetherian semi-prime ring then QE is semi-simple. (b) If E ⊂ U ⊂ QE then EndE (U ) ⊂ QE. 10. If U is a generator for the ring E then the ring of EndE (U )-module endomorphisms of U is E. .

.

11. Let G and H be rtffr groups. Show that H ∼ = G iff A(H) ∼ = A(G). 12. Let S be the ring of algebraic integers in the algebraic number field k. Let n ∈ Z and let E = nZ + 1Z. Then S is generated as an E-module by exactly g elements where g is the composition length of the group S/nS. 13. Let I, J be ideals in the commutative ring S. (a) SI ⊗S SI ∼ = SJ ⊗S SI as S-modules. (b) (XI )J ∼ = (XJ )I naturally. (c) If I ⊂ J then the natural map S/I → S/J takes units to units.

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CHAPTER 3. LOCAL ISOMORPHISM IS ISOMORPHISM (d) (SI )J ∼ = SJ .

14. Let I ⊂ S be an ideal and let M be an S-module such that each element of M is annihilated by some finite power of I. Then M = 0 iff MN = 0 for each maximal ideal I ⊂ N ⊂ S. 15. Suppose that U and V are finitely generated projective right Emodules such that U/xU ∼ = V /xV . Show that there is an injection φ : U → V such that V = φ(U ) + V . 16. Suppose that U ⊂ W and that UI = W = W1 ⊕ W2 . Show that (W1 ∩ U )I = W1 . 17. Show that J (Sτ ) ⊂ J (Eτ ). 18. Let E be an rtffr ring and suppose that E is locally isomorphic to U1 ⊕ U2 for some finitely generated E-modules U1 and U2 . Then E = V1 ⊕ V2 for some cyclic right ideals V1 and V2 such that Ui is locally isomorphic to Vi for i = 1, 2. 19. If E is integrally closed then QE(τ ) = Eτ . Hint: 0 is a unit in the zero ring {0}. 20. Let G be an rtffr group. Show that the following are equivalent. (a) G is indecomposable. (b) G is locally isomorphic to some indecomposable group. (c) If G is locally isomorphic to H then H is indecomposable. (d) E(G) is indecomposable as a right E(G)-module. (e) Eτ (G) is indecomposable as a right Eτ (G)-module.

3.7

Questions for Future Research

Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. Is there some natural way of constructing the ring T ⊕ N (E) in the Beaumont-Pierce-Wedderburn Theorem? See Theorem 3.1.6. 2. Investigate the structure of the rtffr ring E as a module over S = center(E). How are the ideals of Z and S distributed in the ideal lattice of E?

3.7. QUESTIONS FOR FUTURE RESEARCH

67

3. Let G be an rtffr group. It is known that S is the ring of left End(G)-module homomorphisms φ : G −→ G. Describe functorially S in terms of G. See [32, 35]. 4. Describe a localization at (maximal) ideals in the rtffr ring E in a noncommutative setting. There is quite a large literature for this in ring theory. 5. Investigate the significance of the conductor τ of an rtffr ring. We know that E = E(τ ) × Eτ . See Theorem 3.2.4. Can we say more about τ or Eτ ? √ 6. Describe the rings E such that τ = τ . 7. Describe the rings E that have exactly one conductor τ . 8. Someone out there please come up with a better proof of Theorem 3.3.1. 9. Discuss the ideal theory of commutative rtffr rings. 10. Let G be an rtffr group. Realize the conductor τ of End(G) in terms of G. 11. Find applications for the correspondence given in Theorem 3.4.3. 12. Describe the direct sum decomposition G = G(τ ) ⊕ Gτ in group theoretic terms. For example, what is Gτ ? 13. Give a first principles proof of Arnold’s Theorem 3.5.1. 14. Classify the Jacobson radical of Sτ . See Theorem 3.2.4. 15. Investigate the existence of commutativity in rtffr rings E. 16. Let E be an rtffr ring. Investigate the existence of central idempotents in E/N (E). See [40].

Chapter 4

Commuting Endomorphisms Fix an rtffr group G with endomorphism ring End(G), and let E(G) = End(G)/N (End(G)). We will study direct sum decompositions of G in the presence of partial commutative conditions on rings associated with G.

4.1

Nilpotent Sets

Let G be an rtffr group. We begin with a short discussion of idempotents in End(G). The element e ∈ End(G) is an idempotent if e2 = e. We will use both e2 = e and idempotent to refer to the same property of e as the language permits. Let E be a ring and let {e1 , . . . , et } ⊂ E be a set of idempotents. We say that {e1 , . . . , et } is complete if e1 + · · · + et = 1E and we say that {e1 , . . . , et } is a set of orthogonal idempotents if ei ej = 0 for each 1 ≤ i 6= j ≤ t. If {e1 , e2 } is a set of orthogonal idempotents then we say that e1 and e2 are orthogonal idempotents. The reader should show that the sum of orthogonal idempotents is again an idempotent. The idempotent e ∈ E is primitive if given orthogonal idempotents e1 and e2 such that e = e1 + e2 69

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then either e1 = 0 or e2 = 0. The idempotent e is central if ex = xe for each x ∈ E. Idempotents in End(G) characterize the direct sum decompositions of G as follows. LEMMA 4.1.1 Let G be a group. 1. If e ∈ End(G) is an idempotent then G = eG ⊕ (1 − e)G. 2. If G = G1 ⊕ G2 then there are orthogonal idempotents e1 , e2 ∈ End(G) such that e1 + e2 = 1G , e1 G = G1 , and e2 G = G2 . 3. If e ∈ End(G) is an idempotent then e is primitive iff eG is indecomposable. 4. If {e1 , . . . , et } ⊂ End(G) is a complete set of primitive orthogonal idempotents then G = e1 G ⊕ · · · ⊕ et G is an indecomposable direct sum decomposition of G. 5. If G = G1 ⊕ · · · ⊕ Gt is an indecomposable decomposition of G then there is a complete set of primitive orthogonal idempotents {e1 , . . . , et } ⊂ End(G) such that Gi = ei G for each i = 1, . . . , t. Proof: 1. If e2 = e then each x ∈ G can be written as x = ex+(1−e)x so that G = eG+(1−e)G. If x ∈ eG∩(1−e)G then x = ex = (1−e)ex = 0 so that G = eG ⊕ (1 − e)G. 2. If G = G1 ⊕ G2 then let ei : G → Gi be the unique map defined as follows. Given x ∈ G there are unique x1 ∈ G1 and x2 ∈ G2 such that x = x1 ⊕ x2 . Then define ei ∈ End(G) by ei (x1 ⊕ x2 ) = xi

for i = 1, 2.

The reader will verify that e2i = ei , that ei ej = 0 for i 6= j, and that e1 + e2 = 1G . 3. Let e2 = e ∈ End(G). If e = e1 + e2 for some nonzero orthogonal idempotents e1 and e2 ∈ End(G) then as in part 1 we can write eG = e1 G ⊕ e2 G for some nonzero groups e1 G and (e − e1 )G. Thus eG is decomposable if e is not primitive. Conversely if eG = G1 ⊕ G2 for some nonzero groups then as in part 2 the canonical projections ei : G → Gi are nonzero orthogonal idempotents that satisfy e = e1 + e2 . Thus e is not primitive. This completes the proof of part 3.

4.1. NILPOTENT SETS

71

The proof of part 4 uses parts 1, 2, and 3 in a simple induction on t. This completes the proof of the lemma. If e2 = e and if H = eG then we say that H is the direct summand corresponding to e. If G = G1 ⊕ G2 then the natural projection e1 : G → G1 is an idempotent corresponding to G1 . The reader can prove the following lemma. LEMMA 4.1.2 Let G = G1 ⊕ G2 and for i = 1, 2 let e2i = ei ∈ End(G) correspond to Gi . 1. Hom(Gi , Gj ) = ej End(G)ei for each i, j ∈ {1, 2}. That is, each f : Gi → Gj satisfies f = ej f ei . 2. End(Gi ) = ei End(G)ei . Each indecomposable direct summand H of G corresponds to a primitive idempotent e ∈ End(G). We are thus motivated to consider complete sets of primitive orthogonal idempotents in End(G). Central indecomposable idempotents are of particular importance. LEMMA 4.1.3 Let E be any ring, let {e1 , . . . , et } ⊂ E be a complete set of primitive orthogonal idempotents, and suppose that ei is central for each i = 1, . . . , t. If e2 = e ∈ E is primitive then e ∈ {e1 , . . . , et }. Proof: Let e2 = e ∈ E be primitive. Because {e1 , . . . , et } is a complete set of central orthogonal idempotents (ei e)(ej e) = ei ej e = 0 for i 6= j. Then e = e1 e + · · · + et e is a sum of orthogonal idempotents. Since e is primitive there is a subscript, 1 say, such that e = e1 e. Since e1 is primitive and since e1 = e1 e + e1 (1 − e) is a sum of orthogonal idempotents we see that e1 (1 − e) = 0. That is e = e1 e = e1 . This completes the proof. The set {G1 , . . . , Gt } of groups is rigid if HomR (Gi , Gj ) = 0 for each 1 ≤ i 6= j ≤ t. B. Charles [22] calls {G1 , . . . , Gt } a semi-rigid set if HomR (Gi , Gj ) = 0 or HomR (Gj , Gi ) = 0 for each 1 ≤ i 6= j ≤ t. D. Arnold, R. Hunter, F. Richman [13] call {G1 , . . . , Gt } a pseudo-rigid set if for each 1 ≤ i 6= j ≤ t, each composition Gi → Gj → Gi

(4.1)

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of maps is zero. We will call {G1 , . . . , Gt } a nilpotent set if for each 1 ≤ i 6= j ≤ t each composition (4.1) of group maps is a nilpotent endomorphism of Gi . For instance {Zp primes p ∈ Z} is rigid. If G1 ⊂ G2 are rank one groups such that Hom(G2 , G1 ) = 0 then {G1 , G2 } is a semi-rigid set. If G1 and G2 are strongly indecomposable groups that are not quasi-isomorphic then we will show in Lemma 4.1.5 that {G1 , G2 } is a nilpotent set. For instance, let p ∈ Z be a prime and let G1 , G2 be pure b p of differing ranks. Then {G1 , G2 } is a nilpotent set. subgroups of Z The following classifies nilpotents sets in terms of a commutativity property. LEMMA 4.1.4 [34, T.G. Faticoni] Let G = G1 ⊕ · · · ⊕ Gt and for each i = 1, . . . , t let ei ∈ End(G) be the idempotent corresponding to Gi . The following are equivalent. 1. {G1 , . . . , Gt } is a nilpotent set. 2. Hom(Gi , Gj ) ⊂ N (End(G)) for each 1 ≤ i 6= j ≤ t. 3. ei is central modulo N (End(G)) for each i = 1, . . . , t. Proof: 1 ⇒ 2 Let i 6= j, let f : Gi → Gj , and let x : G → G. Then f = ej f ei . Since {G1 , . . . , Gt } is a nilpotent set ej f ei xej is a nilpotent endomorphism of Gj . For large m ∈ Z, !m t X (f x)m = ej f ei xek k=1

= (ej f ei xej )m = 0. The last expression is from the fact that (xek )(ej f ei ) = 0 for j 6= k and that ej f ei xej is nilpotent. Then f generates a nil right ideal of End(G), as required by part 2. 2 ⇒ 3 Let x ∈ End(G). Then for fixed i = 1, . . . , t P ei x = ei xei + i6=j ei xej and P xei = ei xei + i6=j ej xei . By part 2, ei xej ∈ N (End(G)) for each 1 ≤ i 6= j ≤ t, so that P P ei x − xei = i6=j ei xej − i6=j ej xei ∈ N (End(G)).

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73

Thus ei is central modulo N (End(G)), which proves part 3. 3 ⇒ 1 Let i 6= j and let f

g

Gi −→ Gj −→ Gi . Then gf = ei g(ej f ei ) and by part 3 0 = ej ei f ≡ ej f ei (mod N (End(G))). Thus gf ≡ 0(mod N (End(G))), hence gf is nilpotent, whence {G1 , . . ., Gt } is a nilpotent set. This completes the logical cycle. The following two results show us that it should be easy to determine when a set {G1 , . . . , Gt } of rtffr groups is nilpotent. LEMMA 4.1.5 Let G1 and G2 be strongly indecomposable rtffr groups. . ∼ Then {G1 , G2 } is not a nilpotent set iff G1 = G2 . .

Proof: Clearly {G1 , G2 } is not a nilpotent set if G1 ∼ = G2 . Conversely suppose that {G1 , G2 } is not nilpotent. There is a composition f

g

G1 −→ G2 −→ G1 such that gf is not nilpotent. By Lemma 1.3.3, QEnd(G1 ) is a local ring and since J (QEnd(G1 )) is a nilpotent ideal gf is a unit in QEnd(G1 ). Thus there is a map h ∈ End(G1 ) and an integer n 6= 0 such that (hg)f = . G1 ⊕ ker hg and since G2 is strongly indecomposable n1G1 . Then G2 ∼ = . ∼ G2 = G1 . This proves the lemma. LEMMA 4.1.6 Let {G1 , G2 } be a nilpotent set and for i = 1, 2 let Hi be a quasi-summand of Gi . Then {H1 , H2 } is a nilpotent set. Proof: Suppose that {G1 , G2 } is a nilpotent set and let H1 ⊂ G1 and H2 ⊂ G2 be quasi-summands. Consider a composition f

g

H1 −→ H2 −→ H1 of group maps. Since H2 is a quasi-summand of G2 , f : H1 → G2 . There is a map ρ : G2 → H2 and an integer m 6= 0 such that ρ(x) = mx for each x ∈ H2 . Let : H1 → G1 and π : G1 → H1 be the canonical injection/quasi-projection associated with the quasi-summand H1 such that π = n1H1 for some integer n. Because H2 is a quasi-summand of G2 (gρf )π : G1 → G1

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is an endomorphism of G1 that factors through G2 . Since {G1 , G2 } is a nilpotent set (gρf )π is a nilpotent endomorphism. Thus there is an integer k > 0 such that 0 = ((gρf )π)k+1 = (gρf π)k (gρf )π = (gρf n)k (gρf )π = nk (gρf )k+1 π. Inasmuch as π is a surjection and nk is an injection, (gρf )k = 0. Finally for each x ∈ H1 , gρf (x) = g(mf (x)) = gf (mx), so that (gf )k (mk x) = (gf )k−1 mk−1 (gρf )(x) = (gρf )k (x) = 0. Because H1 is torsion-free 0 = (gf )k (mk H1 ) = (gf )k (H1 ) which implies that gf is nilpotent. Hence {H1 , H2 } is a nilpotent set. One of the themes of this chapter is to study the direct sum decompositions of rtffr groups. With this in mind, Fitting’s Lemma leads us to an important result for rtffr groups. LEMMA 4.1.7 Let f : G → G be an endomorphism of the rtffr group G. There is an integer k > 0 such that . G = ker f k ⊕ image f k , f is nilpotent on ker f k , and f is idempotent on image f k . Proof: The given map f lifts to an endomorphism of the finite dimensional vector space QG so that by Fitting’s Lemma QG = ker f k ⊕ image f k where f is nilpotent on ker f k and f is idempotent on image f k . (See [30].) Then f k is an idempotent endomorphism of G, and hence . G = ker f k ⊕ image f k , as required by the lemma. The following result is anticipated by D.M. Arnold and E.L. Lady in [14] where they show that an rtffr group G = H ⊕ K enjoys an exchange property if H and K satisfy a property that is equivalent to assuming that {H, K} is a nilpotent set. THEOREM 4.1.8 Let {G1 , . . . , Gt } be a set of rtffr groups. The following are equivalent.

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75

1. {G1 , . . . , Gt } is a nilpotent set. 2. If 1 ≤ i 6= j ≤ t and if H is a quasi-summand of Gi and Gj then H = 0. 3. If 1 ≤ i 6= j ≤ t and if H is a strongly indecomposable quasisummand of Gi and Gj then H = 0. Proof: From the definition of nilpotent set it suffices to prove the theorem for t = 2. 1 ⇒ 2 is Lemma 4.1.6. 2 ⇒ 3 is clear. 3 ⇒ 1 Suppose that {G1 , G2 } is not a nilpotent set. There is a composition f

g

G1 −→ G2 −→ G1 such that gf is not nilpotent. By Fitting’s Lemma 4.1.7 there is an integer k > 0 such that . G1 = ker(gf )k ⊕ image (gf )k and (gf )k+1 = (gf )k . Then f (gf )k

g

G1 −−−−→ G2 −−−−→ G1 are maps such that gf (gf )k = (gf )k+1 is idempotent on image (gf )k 6= 0. But then f g image (gf )k −→ G2 −→ image (gf )k is an integer multiple of the identity map so that image (gf )k is a quasisummand of G2 . Thus part 3 is false and the proof is complete. THEOREM 4.1.9 Let G1 , . . . , Gt be strongly indecomposable . . ∼ groups such that Gi ∼ G ⇒ i = j. If G G ⊕ · · · ⊕ Gt then = j = 1 0 0 0 0 G = G1 ⊕· · ·⊕Gs for some nilpotent set {G1 , . . . , Gs } of indecomposable rtffr groups. Proof: Since G has finite rank we can write G = G01 ⊕ · · · ⊕ G0s for some indecomposable groups G01 , . . . , G0s . Suppose that i 6= j and let H be a strongly indecomposable quasi-summand of G0i and of G0j . Rewriting the summands of. G we can write H (2) as a quasi-summand of G. But by hypothesis Gi ∼ = Gj ⇒ i = j for each 1 ≤ i, j ≤ t. J´onsson’s Theorem 2.1.10 then implies that if H 6= 0 then the multiplicity of H in G1 ⊕ · · · ⊕ Gt is 1. Thus H = 0 so that {G01 , . . . , G0s } is a nilpotent set by Theorem 4.1.8.

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COROLLARY 4.1.10 Let {G1 , . . . , Gt } be a nilpotent set of rtffr (n ) (n ) groups. Then {G1 1 , . . . , Gt t } is a nilpotent set for any t-tuple (n1 , . . ., nt ) of positive integers. EXAMPLE 4.1.11 Let G be an rtffr group. J´onsson’s Theorem 2.1.10 states that there is a finite set of strongly indecomposable groups {G1 , . . . ., Gt } and integers n1 , . . . , nt > 0 such that Gi ∼ = Gj =⇒ i = j and .

(n1 )

G∼ = G1

(nt )

⊕ · · · ⊕ Gt

.

In this case Theorem 4.1.8 shows us that {G1 , . . . , Gt } is a nilpotent set. Thus each rtffr group G is quasi-isomorphic to a direct sum .

G∼ = G0 ⊕ H 0 where G0 = G1 ⊕ · · · ⊕ Gt . In particular Theorem 4.1.9 states that G01 ⊕ · · · ⊕ G0t is the direct sum of a nilpotent set {G01 , . . . , G0t }. Our interest in nilpotent sets stems from the fact that they enjoy a refinement and uniqueness property. Recall the functor A(·) from Section 2.4.2. THEOREM 4.1.12 Let {G1 , . . . , Gt } be a nilpotent set of indecomposable rtffr groups, and let G = G1 ⊕ · · · ⊕ Gt . Then 1. E(G) = E(G1 ) × · · · × E(Gt ) as rings. 2. G has a unique decomposition. Proof: 1. Let {e1 , . . . , et } ⊂ End(G) be a complete set of primitive orthogonal idempotents such that M ei (G) = Gi and ker ei = Gj . i6=j

By Lemma 4.1.4 the idempotents ei map to central idempotents e¯i in E(G) and by Theorem 2.4.4, E(G)¯ ei = A(Gi ) is a cyclic projective Emodule. Thus E(G) = E(G)¯ e1 ⊕ · · · ⊕ E(G)¯ et = A(G1 ) ⊕ · · · ⊕ A(Gt )

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77

as right ideals. Since the e¯i are central E(G) = A(G1 ) × · · · × A(Gt ). Thus A(Gi ) is a ring. The canonical map π : End(Gi ) → E(Gi ) taking f 7→ ei f ei has image in A(Gi ). Because {G1 , . . . , Gt } is a nilpotent set, each f ∈ A(Gi ) satisfies f ≡ ei f ei (mod N (End(G))). Thus π has image A(Gi ) and so E(Gi ) ∼ = A(Gi ). 2. Suppose that G = G01 ⊕ · · · ⊕ G0s for some indecomposable rtffr groups G0i . Let {e01 , . . . , e0s } ⊂ End(G) be a complete set of primitive orthogonal idempotents corresponding to G01 ⊕ · · · ⊕ G0s . The e0i map onto idempotents e¯0i ∈ E(G). Because e¯1 , . . . , e¯t is a complete set of primitive orthogonal central idempotents in E(G) e¯0i = e¯1 e¯0i + · · · + e¯t e¯0i is a sum of orthogonal idempotents. Since e¯01 is primitive e¯01 = e¯i e¯01 . We may assume without loss of generality that i = 1. Since e¯1 = e¯1 e¯01 + e¯1 (1 − e¯01 ) and since e¯1 is primitive we see that e¯01 = e¯1 . Iterating this process, we produce the identities e¯0 i = e¯i

for i = 1, . . . , s.

Since e¯01 + . . . + e¯0s = 1G , s = t. Therefore A(Gi ) ∼ ei = E(G)¯ e0i = A(G0i ) = E(G)¯ for each i = 1, . . . , t so that Gi ∼ = G0i by Theorem 2.4.4. This completes the proof.

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THEOREM 4.1.13 Let {G1 , . . . , Gt } be a nilpotent set of rtffr groups and let G = G1 ⊕ · · · ⊕ Gt . 1. If H ∈ Po (G) then H ∼ = H1 ⊕ · · · ⊕ Ht where for each, i = 1, . . . , t, Hi ∈ Po (Gi ). 2. If G = H ⊕ K ∼ = H ⊕ K 0 then K ∼ = K 0. Proof: 1. Given H ∈ Po (G) then A(H) is a finitely generated projective right E(G)-module. Since the e¯i are central in A(G) we can write A(H) ∼ e1 ⊕ · · · ⊕ A(H)¯ et = A(H)¯ as right E(G)-modules. By Theorem 2.4.4 there are H1 , . . . , Ht such that A(Hi ) ∼ ei = A(H)¯ for each i = 1, . . . , t so that A(H) ∼ = A(H1 ) ⊕ · · · ⊕ A(Ht ) ∼ = A(H1 ⊕ · · · ⊕ Ht ). By Theorem 2.4.4, H ∼ = H1 ⊕ · · · ⊕ Ht . Part 2 follows from Theorem 4.1.12 in the usual way, so the proof is complete. COROLLARY 4.1.14 Suppose that {G1 , . . . , Gt } is a nilpotent set of rtffr groups and suppose that (n1 )

H ⊕ H 0 = G1

(nt )

⊕ · · · ⊕ Gt

for some integers n1 , . . . , nt > 0. For each i = 1, . . . , t there are rtffr (n ) groups Hi and Hi0 such that Hi ⊕ Hi0 ∼ = Gi i and such that H ∼ = H1 ⊕ 0 0 0 · · · ⊕ Ht and H ∼ = H1 ⊕ · · · ⊕ Ht . The following result is somewhat of a surprise. THEOREM 4.1.15. Let G1 , . . . , Gt be strongly indecomposable rtffr . groups such that Gi ∼ = Gj ⇒ i = j, and let G = G1 ⊕ · · · ⊕ Gt . 1. G has a unique decomposition G = G01 ⊕ · · · ⊕ G0r . 2. If H ∈ Po (G) then H = H1 ⊕ · · · ⊕ Hs where for each i = 1, . . . , s, Hi ∈ Po (G0i ). Proof: Apply Theorems 4.1.9 and 4.1.12.

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4.2

79

Commutative Rtffr Rings

In this section we begin our investigation of the End(G)-module structure of rtffr groups G such that E(G) is a commutative ring. We will consider the finitely generated projective modules over rtffr commutative rings. Our techniques will produce good results on these rings.

4.2.1

Modules over Commutative Rings

Our first result gives us an easily verified condition that implies that the ring E(G) is commutative. LEMMA 4.2.1 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group, let E = End(G), let N = N (End(G)), and let k = QE/N . 1. The degree [k : Q] divides the integers (a) Q-dim(QE), (b) Q-dim(QN ), (c) Q-dim(QN G), (d) Q-dim(QG), and (e) Q-dim(QN k G/QN k+1 G) for each integer k ≥ 0. 2. If Q-dim(QE) or Q-dim(QG) is a square-free integer then k is a commutative field. Proof: 1. By the Beaumont-Pierce-Wedderburn Theorem 3.1.6 we can identify k with a subring of QE. Since QE and QG are then k-vector spaces and since QN and QN G are k-vector subspaces of QE and QG it follows that [k : Q] divides each of the integers in (a)-(e). 2. Since QE is a local ring QE/QN = k is a division ring. By part 1, the k-dimension of QE is a divisor of Q-dim(QE). Since the dimension of a division ring over its center is a perfect square (see [69]), and since Q-dim(QE) is assumed to be square-free k is commutative. A similar proof applies if Q-dim QG is square-free. This completes the proof. LEMMA 4.2.2 Let S be an Artinian commutative ring and let U be a finitely generated projective S-module. If annS (U ) = 0 then U ∼ = S ⊕ U0 0 for some S-module U .

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Proof: Let U be a finitely generated projective S-module. Since S is an Artinian commutative ring S is a finite product of local commutative rings say S = S1 × · · · × S t . Since annS (U ) = 0 U = U S1 ⊕ · · · ⊕ U St is a direct sum of nonzero projective modules. Since projective modules (n ) over local rings are free U Si ∼ = Si i for some integers n1 , . . . , nt > 0. Inasmuch as S = S1 × · · · × St , it follows that U ∼ = S ⊕ U 0 for some 0 S-module U . This completes the proof. The next result is of independent interest since it deals with the structure of right ideals in (not necessarily commutative) rtffr rings. A simple consequence of part 1 that we will not use in our deliberations is that each module of finite composition length over an rtffr ring is finitely presented. LEMMA 4.2.3 Let I be a right ideal in the rtffr ring E. 1. If I is a maximal right ideal in E then E/I is a finite p-group for some prime p ∈ Z. 2. If I ⊕ J has finite index in E for some right ideal J in E then I is finitely presented. Proof: 1. Let I be a maximal right ideal in E. Then N (E) ⊂ I and by the Beaumont-Pierce-Wedderburn Theorem 3.1.6 we have a group . decomposition E ∼ = E/N (E) ⊕ N (E). Hence E/N (E) is a reduced group. Thus we can assume without loss of generality that N (E) = 0. That is, QE is a semi-simple Artinian ring. Thus QI = eQE for some e2 = e ∈ QE. There is an integer m 6= 0 such that m(1 − e)E ⊂ E. Left multiplication by m(1 − e) induces an embedding of right E-modules E/(QI ∩ E) = E/(eQE ∩ E) ∼ = m(1 − e)E ⊂ E. Evidently, E/(QI ∩ E) is a right rtffr E-module. Furthermore since E/I is simple, either E/I is an elementary p-group for some prime p ∈ Z or E/I is divisible. Since E/I maps onto E/(QI ∩E), E/I is an elementary p-group, and since E has finite rank, E/I is finite. 2. There is an integer n 6= 0 such that n ∈ I ⊕ J. Thus (I ⊕ J)/nE is a finite right ideal in E/nE so that I ⊕ J is finitely generated. Suppose that π : P → I is a surjection where P is a finitely generated projective

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81

right E-module. Then π lifts to a surjection π : QP → QI which must split. Let f : QI → QP be the map such that πf = 1QI . Since I and P are finitely generated we may assume that f : I → P is such that . πf = m1I for some integer m 6= 0. Then P = f (I) ⊕ ker π and since P is finitely generated ker π is finitely generated. Hence I = π(P ) is a finitely presented right ideal of E which completes the proof. The following result anticipates Chapter 6 where we link the condition IG 6= G to the splitting of exact sequences involving G. The next theorem is proved in [10, Theorem 5.9] for ideals of finite index in commutative rings. We give a different proof. THEOREM 4.2.4 [10, D.M. Arnold, Theorem 5.9] Let G be an rtffr group with commutative endomorphism ring E = End(G). Then IG 6= G for each proper ideal I ⊂ E. Proof: It suffices to show that IG 6= G for each maximal proper ideal I ⊂ E. So choose a ideal I ⊂ E such that E/I is simple and such that IG = G. By Lemma 4.2.3(1) there is a prime p ∈ Z such that E/I is a finite p-group so that pE ⊂ I. Thus I/pE is a maximal ideal in the commutative Artinian ring E/pE. There is an idempotent e¯ ∈ E/pE such that I/pE = (¯ eE + J )/pE where J /pE = J (E/pE). Then G IG e¯G + J G e¯G + pG = = = pG pG pG pG by Nakayama’s Lemma 1.2.3. Since G/pG has annihilator pE in E, and since ¯1 − e¯ 6∈ pE annihilates e¯, e¯ = ¯1. Thus I/pE = E/pE and I = E. This completes the proof. The above result implies that if End(G) is commutative then Hom(G, IG) = I for each maximal ideal I ⊂ S. It does not state that Hom(G, IG) = I for each nonzero ideal I ⊂ S. It would be interesting to investigate the set {ideals I ⊂ S Hom(G, IG) = I}. LEMMA 4.2.5 Let S be an Artinian commutative ring and let M be an S-module such that annS (M ) = 0, let length(X) denote the composition length of X as a module over some subring R of S, and assume that length(S) < ∞. The following are equivalent. 1. annS (M/M I) = I for each ideal I ⊂ S and length(M ) ≤ length(S). 2. S ∼ = M as S-modules.

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Proof: Part 1 readily follows from part 2, so assume part 1 is true. Suppose that S is a local commutative Artinian ring. Since S is Artinian there is a simple ideal I 6= 0 such that length(I) = k = length(S/J (S)). By hypothesis annS (M/M I) = I. Furthermore, because M I 6= 0, k ≤ length(IM ), so that length(S/I) = length(S) − k ≥ length(M ) − length(M I) = length(M/M I). Induction on length(S) implies that S/I ∼ = M/M I. Since S is local I ⊂ J (S) so that M = xS + IM implies that M = xS by Nakayama’s Lemma 1.2.3. By hypothesis annS (x) = annS (M ) = 0 and so M ∼ = S. To complete the proof the Artinian commutative ring S satisfies S = S1 × · · · × S t for some local rings S1 , . . . , St . Then M = M S1 ⊕ · · · ⊕ M St where annSi (M Si ) = 0. If length(M ) = length(S) then length(M ) = length(M S1 ) + · · · + length(M St ). By the above argument length(M Si ) ≤ length(Si ) implies that M Si ∼ = Si . Then M ∼ S. This completes the proof. =

4.2.2

Projectives over Commutative Rings

Although the following is true for general commutative rings we will prove it for rtffr rings. We lift this proof from [75, Lemma VI.8.6]. LEMMA 4.2.6 Let S be a commutative rtffr ring and let I 2 = I be an ideal of S. Then I = eS for some e2 = e ∈ S. P Proof: Suppose that I = ni=1 xi S. Since I 2 = I for each xi there are yij ∈ I such that X xi = yij xj , j

which leads to the following system of equations. y12 x2 − ··· − y1n xn 0 = (1 − y11 )x1 − 0 = y21 x1 − (1 − y22 )x2 − · · · − y2n xn .. . 0 = yn1 x1 − yn2 x2 − · · · − (1 − ynn )xn

4.2. COMMUTATIVE RTFFR RINGS Thus the determinant det(M ) of the coefficient matrix (1 − y11 ) −y12 ··· −y1n y −(1 − y ) · · · −y2n 21 22 M = .. .. . . yn1 −yn2 · · · −(1 − ynn )

83

satisfies xi det(M ) = 0 for each i = 1, . . . , n. (See, e.g., [58, page 334335].) Thus I det(M ) = 0. However if we think of the determinant as a homogeneous polynomial of degree n then we see that it is of the form 1 − e for some e ∈ I. Then (1 − e)e ∈ (1 − e)I = 0 implies that e2 = e and so eS = I, as required to prove the lemma. Since the trace ideal I of a finitely generated projective module P is an idempotent ideal such that IP = P we have the following result. COROLLARY 4.2.7 Let S be a commutative ring and let P be a finitely generated projective S-module. There is an e2 = e ∈ S such that P = eP and (1 − e)P = 0. Proof: Let I be the trace ideal of the projective S-module P . Since = I, Lemma 4.2.6 implies that I = eS for some e2 = e ∈ S, and since IP = P , eP = eSP = IP = P . I2

COROLLARY 4.2.8 Let S be a commutative ring and let P be a finitely generated projective S-module. If annS (P ) = 0 then P is a generator over S. Proof: By Lemma 4.2.6, I = traceS (P ) = eS, and because annS (P ) = 0, (1 − e)P = 0 implies that e = 1. Thus I = S and hence P generates S. LEMMA 4.2.9 Let S be an rtffr commutative ring and let S ⊂ U ⊂ QS be a finitely generated S-module. Then U is a projective S-module iff U is an invertible fractional ideal for S. Proof: Suppose that S ⊂ U ⊂ QS is a projective S-module. By Lemma 4.2.6 there is an e2 = e ∈ S such that traceS (U ) = eS and (1 − e)U = 0. Since S ⊂ U , e = 1. That is, U generates S. Since HomS (U, S) = {q ∈ QS qU ⊂ S} = U ∗ , U U ∗ = S, whence U is invertible. Conversely suppose that S ⊂ U ⊂ QS and that U U ∗ = S. Then U generates S. Because S is commutative S ⊂ EndS (U ) ⊂ QS and it is

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known that U is then a finitely generated projective EndS (U )-module. Furthermore since S = U U ∗ EndS (U ) = EndS (U )S = EndS (U )U U ∗ = U U ∗ = S and hence U is a finitely generated projective S-module. This completes the proof. We will use the following result to construct a direct sum decomposition of a projective module. LEMMA 4.2.10 Suppose that S is an rtffr commutative ring and let U be a finitely generated projective S-module such that annS (U ) = 0. . 1. Given a nonzero ideal T = S there is a map f : U → S such that f (U ) + T = S. 2. Suppose that S is a semi-prime ring and let τ ⊂ S be a conductor . for S. If I = S is an ideal such that I +τ = S then I is an invertible ideal in S. Proof: 1. Since annS (U ) = 0, Lemma 4.2.9 implies that U (k) ∼ = S⊕U 0 for some integer k 6= 0 so that annS/T (U/U T ) = 0. Since U/U T is a finitely generated projective S/T -module Lemma 4.2.2 implies that U/U T ∼ = S/T ⊕ W for some S/T -module W . Thus there is a surjection f¯ : U → S/T. Since U is projective f¯ lifts to a map f : U → S such that f (U ) + T = S. 2. This is [10, Corollary 12.14]. We give a different proof. It suffices to prove this under the additional assumption that S is indecomposable as a ring. Localize at τ . Then Iτ + ττ = Sτ . Inasmuch as ττ ⊂ J (Sτ ), Iτ = Sτ . Hence IM ∼ = SM for each maximal ideal τ ⊂ M ⊂ S. For maximal ideals M ⊂ S such that τ 6⊂ M , SM is an integrally closed ring by Theorem 3.2.1. That is, SM is a finite product of local Dedekind . domains. Such a local ring is a pid. Since IM = SM , we see that IM ∼ = SM . Hence I is locally isomorphic to S so that I is a progenerator = invertible ideal in S. This proves the lemma.

4.2.3

Direct Sums over Commutative Rings

Now that we have disposed of the more general lemmas we will demonstrate that finitely generated projectives over commutative rtffr rings enjoy a natural direct sum decomposition.

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85

THEOREM 4.2.11 Suppose that S is an rtffr commutative semi-prime ring, and let U be a finitely generated projective S-module. Then U∼ = U1 ⊕ . . . ⊕ Us for some finitely generated projective ideals U1 , . . . , Us in S. Moreover, if annS (U ) = 0 then U1 can be chosen to be an invertible ideal in S. Proof: Let σ = traceS (U ). Then σ 2 = σ so by Lemma 4.2.6, σ = eS for some central e2 = e ∈ S and hence U is a projective generator over the ring eS. Thus we may assume without loss of generality that e = 1 and that annS (U ) = 0. Let R be an indecomposable factor of S and consider U R. Since U generates S, U R generates R. By Lemma 1.6.3 there is an integrally closed R and the associated conductor τ ⊂ R such that τ R ⊂ R ⊂ R ⊂ QR. By Theorem 3.2.4, R = R(τ )×Rτ , and since R is indecomposable R(τ ) = 0 and R = Rτ . Because R is an rtffr ring we can choose n so large that Rn is a reduced group. Then Lemma 3.3.5(2) implies that Rn = RnR and subsequently by Lemma 1.2.1, n ∈ J (Rn ). By Lemma 4.2.10 there is a map f : U → R such that f (U )+nR = R. Localize at n. By Lemma 1.2.1 f (U )n + nRn = Rn = f (U )n ⊃ R by Nakayama’s Lemma 1.2.3. Then mR ⊂ f (U ) ⊂ R for some integer m such that gcd(m, n) = 1. By Lemma 4.2.10(2), f (U ) is an invertible ideal in the semi-prime rtffr ring R. Thus U = f (U ) ⊕ U 0 = U 00 ⊕ U 0 for some invertible ideal U 00 in R. Since S = R1 × · · · × Rr for indecomposable factors Ri we see that U = U1 ⊕ · · · ⊕ Ur ⊕ U 00 for some projective S-module U 00 , where for each i = 1, . . . , t, Ui is an invertible ideal in Ri . An induction on the rank of U completes the proof. THEOREM 4.2.12 Let S be an indecomposable commutative rtffr ring and let U 6= 0 be a finitely generated projective S-module. Then U = U1 ⊕ · · · ⊕ Ut for some invertible ideals U1 , . . . , Ut in S. In this case U is locally isomorphic to S (r) for some integer r > 0.

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4.2.4

Commutative Endomorphism Rings

Now that we have the ring theoretic results out of the way we can apply them to structure theorems for rtffr groups with commutative endomorphism rings. Our approach will be to fix an rtffr group G and consider the semi-prime rtffr ring E(G) = End(G)/N (End(G)) giving general conditions under which E(G) is commutative. Recall the unique decomposition from Section 2.1 and the local refinement property from Section 2.5. THEOREM 4.2.13 [T.G. Faticoni] Let {G1 , . . . , Gt } be a nilpotent set of indecomposable rtffr groups such that E(Gi ) is a commutative ring for each i = 1, . . . , t, and let G = G1 ⊕ · · · ⊕ Gt . Then G has the local refinement property and G has a unique decomposition. Proof: Since {G1 , . . . , Gt } is a nilpotent set of indecomposable groups, Theorem 4.1.12(2) states that G = G1 ⊕ · · · ⊕ Gt has a unique decomposition. For each i = 1, . . . , t, A(Gi ) = E(Gi ) is indecomposable projective module. Then E(G) = E(G1 ) × · · · × E(Gt ) as rings by Theorem 4.1.12(1). Let H ∈ Po (G). By Theorem 2.4.4, A(H) is a finitely generated projective E(G)-module, and by Theorem 4.2.11 there are indecomposable projective ideals U1 , . . . , Us in E(G) such that A(H) ∼ = U1 ⊕ · · · ⊕ Us . Since U1 = U1 E(G1 ) ⊕ · · · ⊕ U1 E(Gt ) is indecomposable there is an index, say 1, such that U1 = U1 E(G1 ). Moreover, each Ui is locally isomorphic to E(Gi ). By Theorem 2.4.4, there are Hi ∈ Po (Gi ) such that Ui = A(Hi ), Hi is locally isomorphic to Gi , and H = H1 ⊕ · · · ⊕ Hs . Thus G has the local refinement property. This completes the proof. COROLLARY 4.2.14 Let G1 , . . . , Gt be strongly indecomposable rtffr . . ∼ groups such that Gi = Gj ⇒ i = j. Suppose that G = G1 ⊕ · · · ⊕ Gt . 1. If H ∈ Po (G) then there are groups Hi ∈ Po (Gi ) for each i = 1, . . . , t such that H = H1 ⊕ · · · ⊕ Ht . 2. G has a unique decomposition.

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THEOREM 4.2.15 . Let G1 , . . . , Gt be strongly indecomposable rtffr . groups such that Gi ∼ = Gj ⇒ i = j. Let G = G1 ⊕ · · · ⊕ Gt and assume that E(Gi ) is a commutative ring for each i = 1, . . . , t. Then G has the local refinement property and a unique decomposition. Proof: By Theorem 4.1.15, G = G01 ⊕ · · · ⊕ G0s for some nilpotent set of indecomposable groups. By hypothesis and by Theorem 4.1.12(1), QE(G) = QE(G1 ) × · · · × QE(Gt ) {G01 , . . . , G0s }

is a commutative ring, so E(G0i ) is an indecomposable commutative ring for each i = 1, . . . , t. Theorem 4.2.13 then states that G has the local refinement property and a unique decomposition. A good source of commutative rings that occur naturally is the class of rtffr E-rings. The ring E is an E-ring if the map λ : E → EndZ (E, +) sending r ∈ E to left multiplication by r, [λ(r)](x) = rx, is an isomorphism. Little is known about groups G such that End(G) is an E-ring. LEMMA 4.2.16 If E is an rtffr E-ring then E is commutative. Proof: The map λ : E → End(E, +) is an isomorphism so given a y ∈ E, the map x 7→ xy is also λ(z) for some z ∈ E. Then 1y = λ(z)(1) = z and so xy = yx for each x ∈ E. LEMMA 4.2.17 Let G be an rtffr group and let E = End(G). Suppose . that G ∼ = E. Then E is an E-ring. . . Proof: Since G = E we have End(E) = End(G) = E. The reader can show that E is an E-ring. LEMMA 4.2.18 Let G be an rtffr group with semi-prime commutative . endomorphism ring E = End(G) such that E ⊂ G ⊂ QE. Then Gp = Ep for each prime p ∈ Z. In particular G/nG ∼ = E/nE as groups for each integer n 6= 0. Proof: Since E is semi-prime Lemma 1.6.3 states that there is an integrally closed ring E such that E ⊂ E ⊂ QE and E/E is finite. Let . G = EG. Then G = G so that . . E ⊂ End(G) = E = E. Because E is integrally closed, E = End(G) and G is a flat E-module. Evidently we can assume without loss of generality that E = E and that

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G = G. Furthermore, since the integrally closed ring E is a finite product of Dedekind domains, we may further assume that E is a Dedekind domain. Let p ∈ Z be a prime and consider Ep ⊂ Gp ⊂ QE. Fix a maximal ideal pE ⊂ M ⊂ E. Since p ∈ M we see that (Ep )M = EM , and hence EM ⊂ GM ⊂ QE. Since E is a Dedekind domain, EM is a discrete valuation domain, so either GM = QE or GM = xEM for some x ∈ QE. Assume for the sake of contradiction that GM = QE. Since M is a maximal ideal in E G/M G ∼ = GM /MM GM = QE/MM QE = 0 so that G = M G. But E is commutative so by Theorem 4.2.4, G 6= M G for maximal ideals M ⊂ E. This contradiction shows us that GM 6= QE. Thus GM = xEM for some x ∈ QE so there is an integer k(M ) such that pk(M ) GM ⊂ EM . Since M = {maximal ideals M p ∈ M ⊂ E} is a finite set we can let k = sup{k(M ) M ∈ M} < ∞. Then pk GM ⊂ EM = (Ep )M

for each M ∈ M

so that pk Gp ⊂ ∩{pk GM p ∈ M ⊂ E is a maximal ideal} ⊂ ∩{EM p ∈ M ⊂ E is a maximal ideal} = Ep ⊂ Gp . by the Local-Global Theorem 1.5.1. Consequently, Gp = Ep . In particular G/pk G ∼ = E/pk E as groups for each integer k > 0. k1 Let n = p1 · · · pkt t be a product of powers of distinct primes. Then G/nG ∼ = G/pk11 G ⊕ · · · ⊕ G/pkt t G ∼ = E/nE as groups. This completes the proof.

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4.3

89

E-Properties

From these preliminary results we begin our investigation of the properties of G as a left E-module. The first E-property that we will discuss is E-flat. The group G is E-flat if G is a flat left E-module. Given a semi-prime commutative endomorphism ring we give a necessary and sufficient condition that the rtffr group G be E-flat. THEOREM 4.3.1 [T.G. Faticoni] Let G be an rtffr group such that E = End(G) is commutative and assume that E ⊂ G ⊂ QE. The following are equivalent. 1. Hom(G, IG) = I for each ideal I of finite index in E. 2. G is an E-flat group. Proof: 1 ⇒ 2 Fix a prime p. By hypothesis annE (G/IG) = I for each ideal I of finite index in E, and by Lemma 4.2.18 Gp /pk Gp ∼ = G/pk G ∼ = E/pk E ∼ = Ep /pk Ep as groups for each integer k > 0. Then by Lemma 4.2.5, G/pk G ∼ = E/pk E as E-modules. So there is an element x ∈ Gp such that Ep x + pk Gp = . Gp . By Lemma 4.2.18, Gp = Ep is a finitely generated Ep -module so that Gp ∼ = Ep x by Nakayama’s Lemma 1.2.3. Since Ep is commutative Gp = Ep x ∼ = Ep . Thus G is locally flat and hence G is E-flat. This proves part 2. 2 ⇒ 1 Suppose that G is E-flat and fix a prime p ∈ Z. By Lemma 4.2.18, Gp is an ideal of finite index in Ep . By hypothesis Ep is Noetherian and Gp is a flat left Ep -module, so Gp is a projective ideal of finite index in the commutative ring Ep . But then by Lemma 4.2.9, Gp is invertible, say Up Gp = Ep . Let I ⊂ E be an ideal of finite index in E and let J = Hom(G, IG). Then for each prime p ∈ Z Jp = JEp = JUp Gp = JGp Up = IGp Up = IEp = Ip so that J = I by the Local-Global Theorem 1.5.1. This proves part 1 and completes the proof. The examples will help illustrate the conditions in the above result. EXAMPLE 4.3.2 Let k be a field extension of Q such that [k : Q] = 3, and let O denote the ring of algebraic integers in k. There are rational

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primes p, q such that O/pO is a field of degree 3 over Z/pZ and qO = M M 0 M 00 for distinct maximal ideals M, M 0 , M 00 in O. Since the extension k/Q is minimal OM ∩ Op = E is an E-ring. (See Appendix K.) Let E = Z + pE ⊂ G ⊂ E where G is an E-submodule of E such that G/pE has dimension 2 over . Z/pZ. Since E ⊂ G ⊂ E, E ⊂ End(G) = E, and since E is a Dedekind domain E ⊂ End(G) ⊂ E. Because [End(G)/pE : E/pE] is then a proper divisor of [E/pE : E/pE] = 3, End(G) = E. However G is not flat as follows. Since we chose G such that G/pE has Z/pZ-dimension 2, G is generated by at least 2 elements as an Emodule. Thus Gp /pE p ∼ = G/pE is generated by two elements, so that Gp requires at least two generators over Ep . By our choice of E and p, Ep is a local ring with J (Ep ) = pE p . Finitely generated flat modules over the Noetherian domain Ep are projective, hence free Ep -modules. . Thus any projective module U = Ep satisfies U ∼ = Ep . In particular U is cyclic. Since Gp is not cyclic, it is not flat over Ep , whence G is not E-flat. EXAMPLE 4.3.3 Let E be a commutative semi-prime rtffr ring that is not hereditary. (There are plenty of these. Try to construct one as we did in the previous example.) There is an ideal I ⊂ E that is not projective, hence not flat, so that I ⊕ E is not a flat left E-module. Evidently E = O(I ⊕ E) = {q ∈ QE q(I ⊕ E) ⊂ I ⊕ E}. Then by Theorem 2.3.4 there is a short exact sequence 0 → I ⊕ E −→ G −→ QE ⊕ QE → 0 such that E ∼ = End(G). The fact that QE is a flat left E-module implies that there is an isomorphism of functors Tor1E (I ⊕ E, ·) ∼ = Tor1E (G, ·). Since I ⊕ E is not flat Tor1E (I ⊕ E, ·) 6= 0 so that G is not flat. Observe that I ⊕ E ⊂ G 6⊂ QE.

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EXAMPLE 4.3.4 Let E be any noncommutative rtffr ring whose additive structure is a free group. Butler’s Construction 2.3.1 shows us that there is a group G such that QG = QE = QEnd(G). Obviously End(G) is not commutative. The next theorem illustrates a theme that we will come back to several times in this book. We show that a number of the E-properties that we are examining agree on rtffr groups that have semi-prime commutative endomorphism rings. Notice that the endomorphism ring in the next result need not be commutative or semi-prime. THEOREM 4.3.5 Let G be an rtffr group with endomorphism ring E = End(G). If E ⊂ G ⊂ QE then the following are equivalent. .

1. G ∼ = E as groups. 2. G is an E-finitely generated group. 3. G is an E-finitely presented group. In this case End(G) is an E-ring. Proof: 3 ⇒ 2 is clear. 2 ⇒ 1 Say x1 , . . . , xn generate G as an E-module. Since E ⊂ G ⊂ QE there is an integer m 6= 0 such that mxi ∈ E so that mG ⊂ E ⊂ G. . 1 ⇒ 3 Since G = E, Lemma 4.2.3 states that G is finitely presented. This proves 3 and completes the logical cycle. We say that G is locally E-free iff Gp ∼ = Ep as E-modules for each p 6= 0 ∈ Z. The Local-Global Theorem 1.5.1 can be used to show that G is E-flat if G is locally E-free. The converse holds in the present situation. THEOREM 4.3.6 Let G be an rtffr group such that E = End(G) is commutative, and suppose that E ⊂ G ⊂ QE. Then G is an E-flat group iff G is locally E-free. Proof: If G is locally E-free then the above comment shows that G is E-flat. Conversely suppose that G is E-flat. Fix a prime p ∈ Z and choose by Lemma 3.3.5 a multiple n of p so large that En is a reduced group. By . Lemma 4.2.18, Gn = En , so by Lemma 4.2.3, Gn is a finitely presented ideal in En . Lemma 4.2.10 states that there is a map π : Gn → En such

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that π(Gn ) + nEn = En . By Lemma 1.2.1, n ∈ J (En ), so Nakayama’s Lemma 1.2.3 implies that π(Gn ) = En . Inasmuch as Gn has finite rank, π is an isomorphism. Thus G is locally E-free. The group G is E-projective if G is a projective left E-module, G is an E-generator if G generates the category of left E-modules, and G is an E-progenerator if G is E-finitely generated, E-projective, and an E-generator. The right ideal I ⊂ E is invertible if II ∗ = I ∗ I = E where I ∗ = {q ∈ QE qI ⊂ E}. These E-properties coincide when End(G) is a commutative semi-prime ring. THEOREM 4.3.7 Let G be an rtffr group with commutative semiprime endomorphism ring E = End(G). If E ⊂ G ⊂ QE then the following are equivalent. 1. G is an E-projective group. 2. G is an E-generator group. 3. G is an E-progenerator group. 4. G is an invertible ideal in End(G). In this case G is E-finitely presented and End(G) is an E-ring. Proof: 4 ⇒ 3 An invertible ideal in a commutative ring is a progenerator. 3 ⇒ 2 is clear. 2⇒ 1 If G is a generator over E then G is projective as a module over EndE (G). But EndE (G) = center(E) = E since E is commutative. Thus G is E-projective. 1 ⇒ 4 By hypothesis G is a projective E-module such that E ⊂ G ⊂ QE. There is a cardinal c and a direct sum E (c) ∼ = G ⊕ K of E-modules. Let {x1 , . . . , xn } ⊂ G be a maximal linearly independent subset of G. There is a finite direct sum E (m) that contains the xi so G ⊂ E (m) . Since E is semi-prime, E is Noetherian whence G is finitely generated. Since E ⊂ G ⊂ QE, G/E is a finitely generated torsion E-module, so that G/E is finite. Hence G is a finitely generated projective ideal of finite index in E. Thus G is invertible which proves part 4 and completes the logical cycle. EXAMPLE 4.3.8 Let G be the groups constructed in Example 4.3.2. Then End(G) is a commutative nonhereditary integral domain. Observe . that G = End(G) is finitely generated but that G is neither E-projective nor an E-generator.

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EXAMPLE 4.3.9 (See [43].) Let E be any ring whose additive structure (E, +) is a (locally) free group. Then E = End(G) for some group G such that E ⊂ G ⊂ QE. In fact the reader can prove that Gp ∼ = Ep for each prime p ∈ Z so that G is an E-flat group. But G is not E-finitely generated and not E-projective.

4.4

Square-Free Ranks

We investigate the strongly indecomposable rtffr groups whose rank is a square-free integer. We show that many E-properties coincide for these groups and that direct sums of nilpotent sets of these groups have the local refinement property.

4.4.1

Rank Two Groups

By a Theorem of J.D. Reid’s below [10, Theorem 3.3], a strongly indecomposable torsion-free group of rank two has a commutative endomorphism ring. Here is a ready-made interesting class of rtffr groups to which our techniques will provide good results. We will refine the work in the previous section by proving the following result. Given a subgroup X ⊂ Y , let X∗ be the pure subgroup of Y generated by X. THEOREM 4.4.1 [43, T.G. Faticoni and H.P. Goeters] A strongly indecomposable torsion-free group of rank two is E-flat. The proof of this theorem depends on the following classification of strongly indecomposable rank two groups. THEOREM 4.4.2 [J.D. Reid] Let G be a strongly indecomposable torsion-free group of rank two. Then QEnd(G) satisfies one of the following conditions. 1. QEnd(G) = Q. 2. QEnd(G) is a field extension of Q of degree 2. 3. QEnd(G) = Q[x]/(x2 ) where x is an indeterminant and where (x2 ) is the ideal generated by x2 . In any case End(G) is a commutative ring.

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Proof of Theorem 4.4.1: Let E = End(G). By Theorem 4.4.2, E is a commutative ring, so that QE is a local commutative ring and QG is a QE-module. If QE = Q then E is a pid and hence G is a flat left E-module. ∼ QG as QEIf QE is a field extension of Q of degree 2 then QE = modules since G has rank two. Thus we can assume without loss of generality that E ⊂ G ⊂ QE as E-modules. We will show that G is locally flat. Let M ⊂ E be a maximal ideal in E and let p be the unique prime integer in M . Since pEM ⊂ MM ⊂ EM and since EM /pEM has Zp /pZp dimension at most 2, either pEM = MM or pEM 6= MM ⊂ EM . Suppose that pEM = MM . Then EM ⊃ pEM ⊃ p2 EM ⊃ · · · is a composition series for EM so that EM is a discrete valuation domain. Hence GM is flat as an EM -module. . Otherwise pEM 6= MM ⊂ EM . By Lemma 4.2.18, GM = EM , so that GM /pGM ∼ = EM /pEM as groups. Since MM /pEM is the only proper nonzero ideal in the commutative ring EM /pEM , and since GM 6= MM GM (Nakayama’s Lemma 1.2.3), we see that MM /pEM = annEM /pEM (GM /MM GM ). Thus Lemma 4.2.5 states that GM /MM GM = EM (x+MM GM ) for some x ∈ GM . Another application of Nakayama’s Lemma 1.2.3 shows us that GM = EM x + MM GM = EM x. Since annEM (x) = annEM (GM ) = 0, GM ∼ = EM is a flat left E-module. The last case to consider is QE = Q[x]/(x2 ). Observe that QG is a QE-module of composition length 2. Since annQE (QG) = 0 Lemma 4.2.5 shows us that QE = QG so that E ⊂ G ⊂ QE. Let N = N (E)G and notice that since N (E)2 = 0, N (E) = Hom(G/N∗ , G). Since G/N∗ and N are rank one groups N = Hom(G/N∗ , G)G/N∗ is a pure subgroup of G. Thus given a prime p ∈ Z such that pG 6= G, we have Gp /Np ∼ = Zp , so that Gp = Ep x + Np = Ep x + N (E)p Gp . In the by now familiar manner Gp = Ep x is a flat E-module. In any case G is a locally flat E-module so that G is E-flat. This completes the proof. An appeal to the above theorem and Theorems 4.3.5 and 4.3.7 will prove the following result.

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THEOREM 4.4.3 Let G be a strongly indecomposable torsion-free group of rank two. The following are equivalent. .

1. G ∼ = End(G) as groups. 2. G is an E-finitely generated group. 3. G is an E-finitely presented group. 4. G is an E-projective group. 5. G is an E-generator group. 6. G is an E-progenerator group. 7. G is an invertible ideal in End(G). If one and hence all of the above conditions is true then End(G) is an E-ring and QEnd(G) is a field extension of Q of degree 2. The next example addresses the hypotheses of the above theorem. EXAMPLE 4.4.4 Let k = Q[α] be a field extension of Q of degree 2 and let E denote the ring of algebraic integers in k. Then (E, +) is a free group of rank 2. By a construction of M.C.R. Butler 2.3.1 there is a group E ⊂ G ⊂ k = QE such that E = End(G). We observe that E is not an E-ring, G and E are not quasi-isomorphic, but G is an E-flat group.

4.4.2

Groups of Rank ≥ 3

By Lemma 1.3.3, G is strongly indecomposable iff QEnd(G) is a local ring. In this case QE(G) = QEnd(G)/N (QEnd(G)) is a division ring. Because QE(G) is commutative if rank(G) is square-free (Lemma 4.2.1) and since we can pass from G to E(G) using the functor A(·) and Theorem 2.4.4, we can study G by applying some commutative ring theory to E(G). Our theorems can be applied to strongly indecomposable rtffr groups of rank three. Compare the next result to Reid’s Theorem 4.4.2. THEOREM 4.4.5 Let G be a strongly indecomposable torsion-free group of rank 3, let E = End(G), and let N = N (QEnd(G)). Then QE falls into one of the following classes. 1. N = 0 and QE is a field extension of Q of degree 1 or 3.

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CHAPTER 4. COMMUTING ENDOMORPHISMS 2. N 6= 0 and Q-dim(QE) = 3. Then QE/N = Q. Moreover (a) If Q-dim(QE) = 3 and N 2 6= 0 then QE ∼ = Q[x]/(x3 ). (b) If Q-dim(QE) = 3 and N 2 = 0 then N = Q ⊕ Q as ideals in QE. 3. N 2 6= 0, Q-dim(QE) ≥ 4, and QE/N = Q.

In any case QE/N is a field extension of Q of degree 1 or 3. Proof: Suppose that rank(G) = 3. Because G is strongly indecomposable of square-free rank, Theorem 1.3.3 states that QE/N = k is a field extension of Q whose degree [k : Q] divides QG and N QG. 1. Suppose that N = 0. Then k = QE is a field extension of Q whose degree [k : Q] divides Q-dim(QG) = 3. Thus [k : Q] = 1 or 3. 2. Suppose that N 6= 0 and that Q-dim(QE) = 3. Then [k : Q] < 3 is a proper divisor of Q-dim(QG) = 3 (Lemma 4.2.1), so that [k : Q] = 1. Thus k = Q. Consider the following cases. 2(a). Suppose that N 2 6= 0. Then 0 ⊂ N 2 ⊂ N ⊂ QE is a chain of distinct subspaces of QE which shows that N /N 2 ∼ = Q as Q-vector spaces. By Nakayama’s Lemma 1.2.3, N = QEy for some y ∈ QE such that y 2 6= 0 and y 3 = 0. Evidently there is a surjection f : Q[x] → Q[y] such that f (x) = y with ker f = (x3 ) = the ideal generated by x3 . Then a comparison of dimensions shows us that Q[x]/(x3 ) ∼ = Q[y]. 2 2(b). Assume that Q-dim(QE) = 3 and that N = 0. Then k ∼ =Q implies that the ideal structure of N is that of a two dimensional Qvector space, say N ∼ = Qx ⊕ Qy as ideals in QE. (Note: QE need not be commutative. See the example below.) 3. Assume that N 2 6= 0 and that Q-dim(QE) ≥ 4. By Lemma 4.2.1, [k : Q] divides Q-dim(QG) = 3 and Q-dim(N QG) < 3, so [k : Q] = 1. This completes the proof. EXAMPLE 4.4.6 Choose a commutative ring E whose additive structure is a free abelian group of rank 3 and such that QE is a local ring. For example, let E be the algebraic integers in a field extension k of degree 3 over Q. By Butler’s Construction in Theorem 2.3.1, E = End(G) for some group G such that E ⊂ G ⊂ QE. G is strongly indecomposable because QE is a local ring.

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EXAMPLE 4.4.7 We construct a strongly indecomposable rank 3 group G such that End(G) is a noncommutative ring of rank 3. Let E = R/I where 0 0 0 a 0 0 R = b a 0 a, b, c, d ∈ Z and I = 0 0 0 . d c a Z 0 0 Then E is a ring such that (E, +) is a free group of rank 3. By Butler’s Construction in Theorem 2.3.1, there is a group G such that QE = QEnd(G) and E ⊂ G ⊂ QE. Since QE is a local ring G is strongly indecomposable. EXAMPLE 4.4.8 We construct a strongly indecomposable rank 3 group G such that E = End(G) is a noncommutative ring of rank 4. Let M = R/I where 0 0 0 a 0 0 R = b a 0 a, b, c, d ∈ Z and I = 0 0 0 . 0 Z 0 d c a Then M = R/I is a left R-module such that annR (M ) = 0, and whose additive structure is a free group of rank 3. A slight variation of Butler’s Construction 2.3.1 produces a group G such that M ⊂ G ⊂ QM and such that QR ∼ = QEnd(G). Thus rank(R) = 4, rank(G) = 3, and because QR is a local ring, G is strongly indecomposable. For groups of rank 4 or 5 we have the following theorems. The ring of Hamiltonian quaternions over Q is the Q-vector space H with basis {1, i, j, k} where we define i2 = j 2 = k 2 = ijk = −1. Extend the multiplication linearly to give H a ring structure. Since ij = −ji the Hamiltonian Quaternions form a noncommutative domain. THEOREM 4.4.9 Let G be a strongly indecomposable torsion-free group of rank 4, let E = End(G) and let k = QE/N (QE). Then k is a field extension of Q of degree 1, 2, or 4. 1. If [k : Q] ≤ 2 then k is a commutative field. 2. If [k : Q] = 4 then either k is a commutative field or k is H = the noncommutative ring of Hamiltonian Quaternions over Q. Proof: The proof is a repeated application of Lemma 4.2.1 and is left as an exercise for the reader.

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EXAMPLE 4.4.10 Let E be a maximal Z-order in the ring of Hamiltonian Quaternions. Then (E, +) is a free group of rank 4 and E is a noncommutative domain. By Butler’s Construction in Theorem 2.3.1, E = End(G) for some group G such that E ⊂ G ⊂ QE. Compare to item 2 in the above theorem. THEOREM 4.4.11 Let G be a strongly indecomposable torsion-free group of rank 5, let E = End(G), and let k = QE/N (QE). 1. k is a commutative field of degree 1 or 5 over Q. 2. If N 6= 0 then k = Q is commutative.

THEOREM 4.4.12 Let G be a strongly indecomposable torsion-free group of rank p for some prime p ∈ Z, let E = End(G), and let k = QE/N (QE). Then 1. k is a commutative field of degree 1 or p over Q. 2. If N 6= 0 then k = Q is commutative. EXAMPLE 4.4.13 An example of End(G) for a strongly indecomposable rank 5 group. Let R be a noncommutative ring such that (R, +) is a free abelian group of rank 4 and let E be the ring (z, r) z ∈ Z, r ∈ R where (z, r)(z 0 , r0 ) = (zz 0 , rz 0 + zr0 + rr0 ). Then E is an associative ring with identity (1, 0). By Butler’s Construction in Theorem 2.3.1 there is a group E ⊂ G ⊂ QE such that QE ∼ = QEnd(G). Since QE is a local ring of Q-dimension 5, G is a strongly indecomposable group of rank 5. The indecomposable group G has the refinement property if given H ∈ Po (G) then H ∼ = Gm for some integer m > 0. THEOREM 4.4.14 A strongly indecomposable rtffr group having prime rank and having nonzero nilradical has the refinement property. Proof: Let H ∈ Po (G). Recall the functor A(·) from Theorem 2.4.4. Then A(H) is a finitely generated projective E(G)-module. By the above Theorem, QE(G) = Q when rank(G) = p is prime and N (End(G)) 6= 0, so that E(G) is a pid. Hence A(H) is a free E(G)-module of rank, say, m, whence H ∼ = G(m) . This completes the proof.

4.5. REFINEMENT AND SQUARE-FREE RANK

4.5

99

Refinement and Square-Free Rank

We have shown in Lemma 4.2.1 that E(G) is a commutative ring for strongly indecomposable groups of square-free rank. Then we can use the machinery that we have developed for direct sum decompositions of projective modules over commutative rtffr rings. The following results extend the Baer-Kulikov-Kaplansky Theorem 2.1.11 to direct sums of strongly indecomposable groups with square-free ranks. Recall unique decomposition from page 25 and the local refinement property from page 45. We give an extensive class of rtffr groups that have a locally unique decomposition. THEOREM 4.5.1 [T.G. Faticoni] Let G1 , . . . , Gt be strongly indecom. posable rtffr groups of square-free rank such that Gi ∼ = Gj ⇒ i = j. If . G = G1 ⊕ · · · ⊕ Gt then G has the local refinement property and G has a unique decomposition. Proof: By Lemma 4.2.1, E(Gi ) is commutative for each i = 1, . . . , t, so by Theorem 4.2.15, G has the local refinement property and a unique decomposition. Of course primes are square-free. COROLLARY 4.5.2 Let G1 , . . . , Gt be strongly indecomposable rtffr . . groups of prime ranks p1 , . . . , pt such that Gi ∼ = Gj ⇒ i = j. If G = G1 ⊕ · · · ⊕ Gt then G has the local refinement property and G has a unique decomposition. COROLLARY 4.5.3 Let G1 , . . . , Gt be strongly indecomposable rtffr . groups of differing prime ranks. If G = G1 ⊕ · · · ⊕ Gt then G has the local refinement property and G has a unique decomposition. Notice in the next result that although we cannot conclude that G has a unique decomposition we can say that G has a locally unique decomposition. THEOREM 4.5.4 Let G1 , . . . , Gt be strongly indecomposable . ∼ rtffr groups of square-free rank such that Gi = Gj ⇒ i = j, and let G = G1 ⊕ · · · ⊕ Gt .

(4.2)

(Note the equation here.) Then G has the local refinement property and a locally unique decomposition.

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Proof: Suppose that G has a direct sum decomposition (4.2). By Theorem 4.5.1, G1 ⊕· · ·⊕Gt has the local refinement property. The local refinement property implies a locally unique decomposition by Lemma 2.5.10. This completes the proof. THEOREM 4.5.5 Let H be a strongly indecomposable rtffr group of square-free rank, let n > 0 be an integer, and let G = H (n) . 1. Each K ∈ Po (G) is a direct sum K = K1 ⊕ · · · ⊕ Ks where each Ki is locally isomorphic to H. 2. H (n) is a locally unique decomposition of G. The following examples will show that in some sense our results are the best possible. EXAMPLE 4.5.6 There is a set {G1 , G2 } of quasi-isomorphic rank two groups such that G = G1 ⊕ G2 does not have the local refinement property. Theorem 2.2.1 states that there are acd groups G1 , G2 , G3 , G4 , such that 1. G = G1 ⊕ G2 ∼ = G3 ⊕ G4 2. rank(G1 ) = rank(G2 ) = 2, rank(G3 ) = 1, rank(G4 ) = 3. Evidently {G1 , G2 } is not a nilpotent set. Such a group G fails to have the local refinement property. Examination of the construction of G1 and G2 shows that E(Gi ) = Z is commutative for i = 1, 2. EXAMPLE 4.5.7 There is a strongly indecomposable torsion-free group of rank 4 that does not have the local refinement property. Proof: Let O denote a classical maximal Z-order in the ring of Hamil∼ tonian Quaterions over Q. Then O/3O = M2 (Z/3Z). Let 3O ⊂ E ⊂ O Z/3Z 0 be such that E/3O ∼ . It is known (see, e.g., [69]) that = Z/3Z Z/3Z E is a hereditary Noetherian primering. Let I be the ideal 3O ⊂ I ⊂ E 0 0 such that I/3O = . Then I 2 = I is a projective ideal. Z/3Z Z/3Z Z/3Z 0 Since I/3O does not generate , I is not a generator for 0 0 E. Therefore I is not locally isomorphic to E. It is clear from the above construction that I is generated by two elements so we can write E⊕E ∼ = I ⊕ J.

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By Corner’s Theorem 2.3.3, there is a group G such that E = End(G). Then by the Arnold-Lady-Murley Theorem 2.4.1, I ⊗E G ∼ = IG ∈ Po (G), G ⊕ G ∼ = IG ⊕ JG, and IG is not locally isomorphic to G. Thus G does not have a locally unique decomposition. The reader can show that End(IG) = EndO (I) = O. The above example illustrates the type of failure of local refinement that occurs whenever we are given an rtffr ring E and a finitely generated projective right ideal I that does not generate E. Such a projective right ideal I always exists in a hereditary rtffr ring E that is not integrally closed.

4.6

Hereditary Endomorphism Rings

There is another interesting class of rtffr groups G for which each H ∈ Po (G) is a direct sum of strongly indecomposables. THEOREM 4.6.1 Let {G1 , . . . , Gt } be a set of strongly indecompos. able rtffr groups such that Gi ∼ = Gj =⇒ i = j. Assume that E(Gi ) is a hereditary domain for each i = 1, . . . , t and let G = G 1 ⊕ · · · ⊕ Gt . Given H ∈ Po (G) there are integers m1 , . . . , mt ≥ 0 and indecomposable groups Hij ∈ Po (Gi ) such that .

Hij ∼ = Gi and such that

for each j = 1, . . . , mi

m1 mt M M H=[ H1j ] ⊕ · · · ⊕ [ Htj ]. j=1

j=1

Proof: Let H ∈ Po (G). Since {G1 , . . . , Gt } is a nilpotent set there are Hi ∈ Po (Gi ) such that H = H 1 ⊕ · · · ⊕ Ht so we may as well assume that G = G1 and that H = H1 . Under these reductions H ∈ Po (G) and G is strongly indecomposable. Since H ⊕. H 0 ∼ = G(n) for some integer n, J´onsson’s Theorem 2.1.10 (m) ∼ states that H = G for some integer m > 0. Then .

A(H) ∼ = E(G)(m)

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as right E(G)-modules. Since E(G) is a hereditary domain A(H) ∼ = U1 ⊕ · · · ⊕ Um .

where Ui ∼ = E(G) for each i = 1, . . . , m. By Theorem 2.4.4 we can write H∼ = H1 ⊕ · · · ⊕ Hm where A(Hi ) ∼ = Ui for each i = 1, .. . . , m. Finally since A(·). preserves quasi-isomorphism classes, A(Hi ) ∼ = G. This = E(G) implies that Hi ∼ completes the proof.

4.7

Exercises

Let E be an rtffr ring, let G, H be rtffr groups. 1. Let E be a ring and let P be a right E-module. Then P is finitely generated projective over E iff P is a generator in the category of left EndE (P )-modules. 2. Let M be a finitely presented right E-module. Show that M is a projective right E-module iff MI is a projective right EI -module for each maximal ideal I ⊂ S = center(E). √ 3. Let O denote the ring of algebraic integers in the field Q[ −5]. √ (a) Show by brute force that the ideal I = h2, 1 + −5i in O is a maximal nonprincipal ideal. (b) Show that 2O = I12 . (c) Conclude that I⊕I ∼ = O⊕O so that O has the local refinement property but not the refinement property. 4. Let E be the ring constructed in Example 4.5.7 and let I be the maximal ideal. (a) Show that E is hereditary. (b) Show that I is an ideal generated by two elements. (c) Show that I is not a generator. 5. Let E be a hereditary prime rtffr ring and let E be a classical maximal order such that E ⊂ E and E/E is finite. (a) Show that E is an ideal over E that is projective but not a generator over E.

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(b) Show that there is an idempotent ideal ∆ of finite index in E such that O(∆) = E. (c) Show that the hereditary prime rtffr ring E is a classical maximal order iff E and 0 are the only idempotent ideals in E. 6. Show that if E is an integral domain then E and 0 are the only idempotent ideals in E. 7. Construct an indecomposable rtffr subring E ⊂ Q × Q such that each finitely generated projective E-module is a generator for E and E ∼ = Z ⊕ Z as group. 8. Let S be a commutative rtffr ring and let U be a finitely generated ideal of finite index in S. Show that U is projective iff U is invertible iff Up ∼ = Sp for each prime p. 9. Let S be a commutative rtffr ring and let U be a finitely generated ideal of finite index in S. Show that U is invertible iff U is locally isomorphic to S. 10. Let S be a discrete valuation domain with field of fractions QS. If S ⊂ M ⊂ QS and if M 6= QS then M = Sx for some x ∈ QS. 11. Prove Fitting’s Lemma. If G is a left E-module of finite composition length and if f ∈ End(G) then there is an integer k > 0 such that G = ker f k ⊕ image f k where f is a nilpotent endomorphism of ker f k and f is an idempotent endomorphism of image f k . 12. If G is a torsion-free reduced p-local group of rank n and if rp (G) = (n) n then G ∼ = Zp . 13. Let E be a commutative ring and let M be a projective E-module. Then M generates E.

4.8

Questions for Future Research

Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. Give a general theory of nilpotent sets for infinite sets of groups or modules. See [34].

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2. Extend the theory in chapter 4 on the direct sum properties of . nilpotent sets G = G1 ⊕ · · · ⊕ Gt to direct sums in which Gi = Gj for each i 6= j. See [13]. 3. Find a condition on G, more general than square-free rank, that implies commutativity in QEnd(G) or QE(G). 4. Let S be a commutative rtffr ring. Discuss the localization of S at its maximal ideals and at multiplicatively closed sets contained in S. 5. Characterize the rtffr groups G that possess the (local) refinement property. 6. Characterize the rtffr groups G that possess the (locally) unique decomposition property. 7. Discuss the direct sum of strongly indecomposable rank two groups. These rank two groups have commutative endomorphism rings. 8. Just what is an E-flat group? 9. Discuss the structure of strongly indecomposable groups G of square-free rank. Extend that discussion to a large class of rtffr groups. 10. Discuss the local-global theory of (almost completely decomposable) rtffr groups. 11. Give a general ideal lattice for the nilradical of a strongly indecomposable rtffr group. Specifically, generalize Theorems 4.4.2,4.4.5, and 4.4.9 to larger ranks. 12. Find some kind of localization theory between the group G and S, the center of End(G). 13. Discuss the decomposition of groups G such that E(G) is hereditary. 14. Let G be an rtffr group. If E(G) is a hereditary ring then how does the ideal structure of E(G) affect G. For example, each right ideal is projective in E(G). There are at most finitely many isomorphism classes of right ideals in E(G). At most finitely many maximal right ideals are not progenerators for E(G). Discuss how these properties affect the properties of direct summands of G.

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15. Extend the results in chapter 4 to include rtffr groups G such that E(G) is integrally closed. That is, E(G) is a product of classical maximal orders.

Chapter 5

Refinement Revisited Recall that an indecomposable rtffr group G has the refinement property if given a direct sum G(n) ∼ = H ⊕ H 0 of groups then H ∼ = G(m) for some integer m > 0. We showed in Theorem 4.2.13 that if E(G) is a commutative ring then the indecomposable rtffr group G possesses the local refinement property. That is, given a direct sum G(n) ∼ = H ⊕ H 0 of groups then H ∼ = G1 ⊕ · · · ⊕ Gm for some groups G1 , . . . , Gm that are locally isomorphic to G. This brings two questions to mind.

1. If G has the local refinement property under what conditions does G have the refinement property?

2. If G has the local refinement property under what conditions does a subgroup H of finite index in G have the local refinement property?

We will investigate these problems in this chapter. For example, let G be a group such that E(G) has finite index in a Dedekind domain E. Then G has the local refinement property. But if E is not a pid then we will show that G has the local refinement property but it does not have the refinement property. By the Baer-Kulikov-Kaplansky Theorem 2.1.11, if G is a completely decomposable group then G has the refinement property. However Example 2.2.1 shows that it is not uncommon for a subgroup G of finite index in G to fail to possess the local refinement property. We will give conditions on G under which subgroups of finite index in G have the refinement property when G has the refinement property. 107

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CHAPTER 5. REFINEMENT REVISITED

Counting Isomorphism Classes Class Groups and Class Numbers

The integral closure E of the integral domain E is the largest subring E ⊂ QE such that E ⊂ E and such that E/E is finite. If E = E then E is an integrally closed ring. For example, each subring of Q is integrally closed. If i2 = −1 and if E = {a + 2bi a, b ∈ Z} then the reader can show that E = Z[i]. Given a Dedekind domain E and a nonzero ideal I ⊂ E then the integral closure of E = I + Z is E. It is known that the integral closure E of an rtffr integral domain E is a Dedekind domain, [10, Corollaries 10.12, 11.4].

For the remainder of this section we let E be an rtffr integral domain with integral closure E. The nonzero ideal I ⊂ E is invertible if I ∗ I = II ∗ = E where I ∗ = {q ∈ QE qI ⊂ E}.

By Lemma 4.2.9 a nonzero ideal I in the rtffr integral domain E is invertible iff it is projective. By Theorem 4.2.12, each finitely generated projective module over the commutative Noetherian integral domain E is a direct sum of invertible ideals. Fix a strongly indecomposable rtffr group G such that E(G) is a commutative ring, or equivalently an rtffr integral domain. This is the case, e.g., when rank(G) is a square-free integer. Theorem 4.2.13 shows us that G has the local refinement property. We define

Γ(G) = {isomorphism classes (H) G is locally isomorphic to H}

and we define the class number of G as

h(G) = card(Γ(G)).

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109

If E is an integral domain then a fractional ideal of E is aP nonzero Esubmodule of the field of fractions of E. If we define IJ = { finite xi yi xi ∈ I and yi ∈ J} for any fractional ideals I, J of E then γ(E) = the set of invertible fractional ideals of E is a group with I −1 = I ∗ and identity E. Let ρ(E) ⊂ γ(E) denote the set of principal fractional ideals of E. Evidently ρ(E) is a subgroup of γ(E). We define the class group of E to be Γ(E) = γ(E)/ρ(E). It is readily shown that fractional ideals I and J are isomorphic iff there is a q ∈ QE such that qI = J. Then the coset Iρ(E) in Γ(E) is the isomorphism class (I) of I. Thus Γ(E) = {isomorphism classes (I) I is an invertible ideal in E} and we define the class number of E to be h(E) = card(Γ(E)). The integral domain E is a Dedekind domain so each ideal in E is invertible. Thus Γ(E) is the set of isomorphism classes of nonzero ideals in E, and h(E) is the number of isomorphism classes of nonzero ideals in E. The next result shows us that h(G) is finite. The proof is due to E.L. Lady and can be found in [10, Theorem 11.11, page 137]. THEOREM 5.1.1 [E.L. Lady] Let G be an rtffr group. There are at most finitely many isomorphism classes of groups H that are locally isomorphic to G. Consequently h(G) < ∞. THEOREM 5.1.2 [Jordan-Zassenhaus] (See [10, Theorem 13.13, page 155].) Let E be an rtffr integral domain. There are at most finitely many isomorphism classes of nonzero ideals in E. Consequently h(E) < ∞.

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CHAPTER 5. REFINEMENT REVISITED Recall that the conductor of E into E is τ = the largest ideal τ in E such that τ E ⊂ E ⊂ E .

From section 3.2, τ is an ideal of E contained in E. Our investigations will relate the local refinement property in strongly indecomposable rtffr groups G to the class number of the integral closure E of E(G). LEMMA 5.1.3 Let G be an rtffr group. Then h(G) = h(E(G)). Proof: Given a group H that is locally isomorphic to G, Lemma 2.5.6 implies that G ⊕ G ∼ = H ⊕ H 0 for some group H 0 . Then H ∈ Po (G). By Lemma 2.5.1, A(H) is locally isomorphic to A(G). Then by Lemma 2.4.4, A(·) induces a functorial bijection α : Γ(G) → Γ(E(G)). Thus h(G) = h(E(G)). The reader will fill in the elementary details. Our investigation of h(G) for the rtffr group G is then translated into an investigation of h(E) for an rtffr integral domain E.

5.1.2

Class Group of the Integral Closure

The integral domain E is an rtffr integral domain if its additive structure (E, +) is an rtffr group. In this section E denotes an rtffr integral domain with integral closure E. We will prove the following result. THEOREM 5.1.4 [T.G. Faticoni] Let E be an rtffr integral domain. Then γ(E) is a free abelian group, and there is a group surjection α : γ(E) −→ γ(E). Thus γ(E) ∼ = γ(E) ⊕ ker α. Proof: Each fractional ideal I is uniquely the product of powers of finitely many maximal ideals of E, [69, page 47, Theorem 4.8]. Thus γ(E) is a free group on the set of maximal ideals of E. Define α(J) = JE for each J ∈ γ(E). Suppose that J is an invertible ideal in E. Because QE is commutative and because J ∗ J = E, J ∗ (JE) = E. Thus JE is an invertible ideal in E, hence α is well defined. Inasmuch as QE is commutative, we have IJE = IE · JE, for any ideals I, J ⊂ E, so that α is a group homomorphism. It remains to show that α is a surjection. The proof is a series of lemmas. LEMMA 5.1.5 Let T 6= 0 be an ideal in E and let I be an invertible ideal in E. There is an ideal I 0 ⊂ E such that I 0 ∼ = I and I 0 + T = E.

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111

. Proof: Since E is an integral domain T = E, and by Lemma 4.2.9, I is a projective E-module. Then by part 1 of Lemma 4.2.10 there is a map f : I → E such that f (I) + T = E. Since E is an integral domain f is an injection so letting I 0 = f (I) ∼ = I completes the proof. LEMMA 5.1.6 Let E be an rtffr integral domain with conductor τ and integral closure E. Let J ⊂ E be an ideal such that J + τ = E. 1. If τ ⊂ M ⊂ E is a maximal ideal then JM = EM . 2. If τ 6⊂ M ⊂ E is a maximal ideal then τM = EM = E M . Thus JM is a principal ideal in E M . 3. J is an invertible ideal in E. Proof: 1. Given a maximal ideal τ ⊂ M ⊂ E then JM + τM = EM and τM ⊂ MM = J (EM ). By Nakayama’s Lemma 1.2.3, EM = JM . 2. Given a maximal ideal M ⊂ E such that τ 6⊂ M then τ + M = E so that τM + MM = EM . Since MM = J (EM ), τM = EM , and so the inclusion τ E ⊂ E ⊂ E becomes E M = EM E M = τM E M ⊂ EM ⊂ E M . Thus EM = E M . It follows that JM is a projective ideal in the Dedekind domain E M . Since E M is then a local domain, JM is free, and since QJM ⊂ QE M , QJM ⊕ V = QE M for some vector space V over QE M . Inasmuch as the field QE M is indecomposable JM ∼ = EM . 3. In parts 1 and 2 we have shown that JM is a projective ideal in EM for each maximal ideal M ⊂ E. Then by the Local-Global Theorem 1.5.1, J is a projective ideal in E. By Lemma 4.2.9, J is an invertible ideal in E. This completes the proof of the lemma. LEMMA 5.1.7 Let E be an rtffr integral domain with integral closure E. Let J ⊂ E be an ideal such that J + τ = E. Then J ∩ E is an invertible ideal in E such that (J ∩ E) + τ = E. Proof: Because τ ⊂ E the modular law implies that (J + τ ) ∩ E = (J ∩ E) + τ = E. By part 3 of Lemma 5.1.6, J ∩ E is then an invertible ideal in E. LEMMA 5.1.8 Let E be an rtffr integral domain with conductor τ and with integral closure E. Let J ⊂ E be an ideal such that J + τ = E. Then [J ∩ E]E = J.

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Proof: Let J ⊂ E be an ideal such that J + τ = E and let J = J ∩ E. By the previous Lemma, J + τ = E. Given a maximal ideal τ ⊂ M ⊂ E we have JM = EM by Lemma 5.1.6(1). Furthermore because τM ⊂ MM = J (EM ) we have E M = J M + τM = J M + τM E M = J M by Nakayama’s Lemma 1.2.3. Then (JE)M = JM E M = EM E M = E M = J M . Given a maximal ideal τ 6⊂ M ⊂ E then Lemma 5.1.6(2) implies that JM is an ideal in EM = E M . Then (JE)M = JM E M = JM = J M ∩ EM = J M ∩ E M = J M so that JE = J by the Local-Global Theorem 1.5.1. Continuation of the Proof of Theorem 5.1.4: Let J be a frac0 0 tional ideal of E. There is a copy J of J such that J + τ = E (Lemma 0 5.1.5). Thus there is a q ∈ QE such that J = qJ . Moreover by Lemma 0 0 0 5.1.8, J ∩ E is an invertible ideal of E such that (J ∩ E)E = J . Then 0 0 J = q(J ∩ E) is an invertible ideal of E such that JE = qJ = J. Therefore α : γ(E) −→ γ(E) is a surjection, and hence γ(E) ∼ = γ(E) ⊕ ker α. This completes the proof of Theorem 5.1.4. It is clear that the principal fractional ideals J = aE of E are mapped to principal fractional ideals α(J) = aE of E. Because Γ(E) is formed by taking the quotient of γ(E) modulo the subgroup of principal fractional ideals, α induces a group surjection β : Γ(E) −→ Γ(E). THEOREM 5.1.9 Let E be an rtffr integral domain with integral closure E. Then β : Γ(E) −→ Γ(E) is a surjection of finite abelian groups. Proof: By the Jordan-Zassenhaus Theorem 5.1.2, Γ(E) and Γ(E) are finite abelian groups. Thus β is a group surjection of finite abelian groups. COROLLARY 5.1.10 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group such that E(G) is a commutative integral domain with integral closure E. Then h(E) divides h(G). Proof: By Theorem 5.1.9, there is a surjection β of finite groups. Then by Lagrange’s Theorem h(E(G)) = card(Γ(E(G))) is divisible by card(Γ(E)) = h(E). This completes the proof.

5.1. COUNTING ISOMORPHISM CLASSES

5.1.3

113

Class Number and Refinement

We continue to investigate h(G) when G is a strongly indecomposable rtffr group such that E(G) is a commutative integral domain with integral closure E. We say that G has the L-property if G is isomorphic to each group that is locally isomorphic to G. The L stands for E.L. Lady who first used local isomorphism on rtffr groups. See [56] where local isomorphism is called near isomorphism. THEOREM 5.1.11 Let G be a strongly indecomposable rtffr group such that E(G) is an integral domain. The following are equivalent. 1. h(G) = 1. 2. h(E(G)) = 1. 3. G has the L-property. 4. G has the refinement property. Proof: 1 ⇔ 2 is Lemma 5.1.3. 1 ⇔ 3 Let h(G) = 1 and suppose that G is locally isomorphic to H. Since h(G) = 1 implies that there is only one isomorphism class in Γ(G), (G) = (H) or equivalently G ∼ = H. The converse is proved in the same way. 2 ⇒ 4 By Theorem 2.4.7 we can prove that G has the refinement property if we prove that E(G) has the refinement property. Suppose that P ⊕ P 0 ∼ = E(G)(n) for some integer n. Since E(G) is an rtffr integral domain Theorem 4.2.15 states that E(G) has the local refinement property so that P = J1 ⊕ · · · ⊕ Jr for some invertible ideals J1 , . . . , Jr of E(G). Since h(E(G)) = 1, E(G) ∼ = Ji for each i = 1, . . . , r, and hence E(G) has the refinement property. 4 ⇒ 3 Let G be locally isomorphic to H. Since G is strongly indecomposable Theorem 3.5.1 implies that H is indecomposable. By Lemma 2.5.6 there is a direct sum G ⊕ G ∼ = H ⊕ H 0 of groups. Since G has the ∼ refinement property G = H, so that G has the L-property. This proves part 3 and completes the logical cycle. Let G be a strongly indecomposable rtffr group. We say that G has the proper local refinement property if G has the local refinement property but G does not have the refinement property. That is, there are isomorphic direct sums G(n) ∼ = H ⊕ K such that in no direct sum decomposition of H = G1 ⊕ · · · ⊕ Gn is each Gi isomorphic to G. The

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group G has the proper L-property if there is a group that is locally isomorphic to G but that is not isomorphic to G. THEOREM 5.1.12 Let G be a strongly indecomposable rtffr group. Assume that E(G) is an rtffr integral domain with integral closure E(G). 1. h(G) 6= 1. 2. h(E(G)) 6= 1. 3. G has the proper local refinement property. 4. G has the proper L-property. The next result shows us that the proper L-property and the proper local refinement property are inherited by subgroups of finite index. This kind of genetic behavior is uncommon in the study of rtffr groups. THEOREM 5.1.13 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group such that E(G) is a commutative integral domain. Let E be the integral closure of E(G). 1. If h(E) 6= 1 then G has the proper local refinement property. 2. h(E) = 1 if G has the refinement property. Proof: Part 2 is the contrapositive of part 1. Suppose that h(E) > 1. Since h(E) divides h(G), h(G) > 1, whence G does not have the refinement property, but G has the proper refinement property. This proves the Theorem. It is rare to find a property for an rtffr group G that is passed onto other groups in the quasi-equality class of G. COROLLARY 5.1.14 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group such that E(G) is a Dedekind domain, and assume that G has the proper local refinement property. If G is quasi-isomorphic to G then G has the proper local refinement property. Since E(G) is a commutative integral domain when G is a strongly indecomposable rtffr group of square-free rank we have proved the following corollary.

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115

COROLLARY 5.1.15 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group of square-free rank. Let E be the integral closure of E(G). 1. If h(E) 6= 1 then G has the proper local refinement property. 2. h(E) = 1 if G has the refinement property. Recall from Lemma 3.1.8 that if G is an rtffr group then there is an rtffr group G ⊂ G such that G/G is finite, E(G) is an integrally closed integral domain, E(G) ⊂ E(G), and E(G)/E(G) is finite. THEOREM 5.1.16 Let G be a strongly indecomposable integrally closed rtffr group of square-free rank. If G has the proper local refine. ment property and if G = G then G has the proper refinement property. Proof: Suppose that G has the proper refinement property. By Theorem 5.1.11, h(E(G)) 6= 1. Now E(G) is integrally closed and E(G)/E(G) is finite so E(G) is the integral closure of E(G). Hence Corollary 5.1.15 states that G has the proper refinement property. THEOREM 5.1.17 Let G1 , . . . , Gt be strongly indecomposable rtffr . groups of square-free ranks such that Gi ∼ = Gj ⇒ i = j. For each (n ) i = 1, . . . , t let E i be the integral closure of E(Gi ), and let G = G1 1 ⊕ (n ) · · · ⊕ Gt t for some integers n1 , . . . , nt > 0. 1. If some E i is not a pid then G has the proper local refinement property. 2. If G has the refinement property then E i is a pid for each i = 1, . . . , t. Proof: Part 1 is the contrapositive of part 2. 2. Suppose that G has the refinement property. One shows that Gi has the refinement property for each i = 1, . . . , t, so by Theorem 5.1.12, E i is a pid for each i = 1, . . . , t. The above theorems connect the refinement property for strongly indecomposable rtffr groups to the class number problem for Dedekind domains in algebraic number fields. Thus a classification of the strongly indecomposable rtffr groups that have the refinement property would naturally lead to a classification of the algebraic number fields whose

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class number is 1. The determination of the class number of an algebraic number field is generally regarded as a difficult problem. See [69]. Thus it would seem that the refinement property will be a fun one but a hard one to investigate.

5.1.4

Quadratic Number Fields

It is natural to ask for h(G) when QE(G) is a small field. Since there is a rather complete description of quadratic number fields k such that h(k) = 1 we will look there for a classification of strongly indecomposable rtffr groups G that have the refinement property. PROPOSITION 5.1.18 Let G be a strongly indecomposable rtffr group. 1. If QE(G) = Q then G has the refinement property. 2. If rank(G) is prime and if N (End(G)) 6= 0 then G has the refinement property. Proof: Part 1. follows from the fact that each subring of Q is a pid. 2. By hypothesis and Lemma 4.2.1, Q-dim(QE(G)) is a proper divisor of the prime Q-dim(QG) so that QE(G) = Q. Now use part 1. LEMMA 5.1.19 √ Let O be the ring of algebraic integers in a quadratic number field Q[ d] for some square-free integer d 6= 0. Let E be a subring of finite index in O. 1. Suppose that d < 0. Then h(E) = 1 if d ∈ {−1, −2, −3, −7, −11, −19, −43, −67, −163}. 2. Suppose that 1 < d < 100 is a square-free integer. Then h(E) 6= 1 iff d ∈ X = {10, 15, 26, 30, 34, 35, 39, 42, 51, 55, 58, 65, 66, 70, 74, 78, 79, 82, 85, 87, 91, 95}. Proof: 1. By [76, Theorem 10.5], the negative integers d ∈ {−1, −2, −3, −7, −11, −19, −43, −67, −163} are the only √ negative integers for which the ring O of algebraic integers in Q[ d] is a pid. Since O is the integral closure of E, Theorem 5.1.9 implies that h(O) 6= 1 divides h(E) 6= 1. 2. By [76, Theorem 10.5], the 22 values in X are exactly the squarefree integers 1 < d < 100 such that h(O) 6= 1. Now apply Theorem 5.1.12. This completes the proof.

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THEOREM 5.1.20 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group of rank 2. 1. If QEnd(G) = Q or N (End(G)) 6= 0 then G has the refinement property. √ 2. Otherwise, QEnd(G) = Q[ d] for some square-free integer d 6= 0. . Assume that End(G) = O where O is the ring of algebraic integers in QEnd(G). Then G has the proper refinement property provided that either (a) d 6∈ {−1, −2, −3, −7, −11, −19, −43, −67, −163} or (b) 1 < d < 100 and d ∈ {10, 15, 26, 30, 34, 35, 39, 42, 51, 55, 58, 65, 66, 70, 74, 78, 79, 82, 85, 87, 91, 95}. Proof: 1. Our assumptions amount to assuming that QE(G) = Q so that E(G) is a pid in this case. Thus the groups G in this case have the refinement property. . 2. Since E(G) = O, O is the integral closure of E(G). By the previous theorem h(O) 6= 1 in case d satisfies parts (a) or (b). By Corollary 5.1.15, G has the proper refinement property. Reader be warned: Suppose O is the ring of algebraic integers in any algebraic number field. Choose any nonzero ideal I ⊂ O. Then regardless of the class number of O, OI is a pid.

5.1.5

Counting Theorems

The following result is central to our discussion. THEOREM 5.1.21 [I. Kaplansky] (See [69]). Let E be a Noetherian integral domain and let I1 , . . . , Im , J1 , . . . , Jn be invertible ideals in E. Then I1 ⊕ · · · ⊕ Im ∼ = J1 ⊕ · · · ⊕ Jn ⇐⇒ m = n and I1 · · · Im ∼ = J1 · · · Jn . Proof: (⇐⇒) follows from [69, Theorem 38.13].

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Given an integer n > 0, a strongly indecomposable rtffr group G with square free rank, and an integral domain E let

νn (G) = card{unordered n-tuples {H1 , . . . , Hn } each Hi is an indecomposable rtffr group, and G(n) ∼ = H 1 ⊕ · · · ⊕ Hn } νn (E) = card{unordered n-tuples {I1 , . . . , In } each Ii is an invertible ideal in E, and E (n) ∼ = I1 ⊕ · · · ⊕ In }

and let e(G) = least integer e > 0 such that xe = 1 for all x ∈ Γ(E(G)) `(G) = least integer t > 0 such that Γ(E(G)) is a direct sum of t cyclic groups, say Γ(E(G)) = h(J1 )i × · · · × h(Jt )i

Observe that e(G), `(G) ≤ h(G) ≤ e(G)`(G). If Γ(E(G)) is a cyclic group then `(G) = 1 and e(G) = h(E(G)) = h(G). We will prove the following theorem. THEOREM 5.1.22 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group of square free rank, and let n > 0 be an integer. 1. If Γ(E(G)) is a cyclic group then νn (G) ≤ h(G)n−1 . 2. In general, νn (G) ≤ e(G)n`(G) ≤ h(G)n`(G) . The proof is a series of lemmas. Let

Kn (G) = {ordered m-tuples (I1 , · · · , In ) I1 , . . . , In are invertible ideals in E such that E (n) ∼ = I1 ⊕ · · · ⊕ In }.

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LEMMA 5.1.23 Let E be an rtffr integral domain, and let n > 0 be an integer. Then 1. There is a short exact sequence µ 0 −−−−→ Kn (E) −−−−→ Γ(E)(n) −−−−→ Γ(E) −−−−→ 0

(5.1)

of abelian groups where µ((I1 ), . . . , (In )) = (I1 · · · In ). 2. νn (E) ≤ card(Kn (E)) = h(E)n−1 . Proof: 1. Given (I1 , . . . , In ) ∈ ker µ then E · · · E = E ∼ = I1 · · · In so that E (n) ∼ = I1 ⊕ · · · ⊕ In by Theorem 5.1.21. Thus ker µ ⊂ Kn (E). Conversely given (I1 , . . . , In ) ∈ Kn (E) then E (n) ∼ = I1 ⊕ · · · ⊕ In so that E ∼ = I1 · · · In by Theorem 5.1.21. Thus (I1 , . . . , In ) ∈ ker µ, so that Kn (E) = ker µ. 2. By part 1 there is a short exact sequence (5.1) so that Lagrange’s Theorem shows us that h(E)n = card(Γ(E)(n) ) = h(E) · card(Kn (E)). That is, card(Kn (E)) = h(E)n−1 . Furthermore there is an obvious surjection from Kn (E) onto the set of unordered n-tuples {I1 , . . . , In } such that E (n) ∼ = I1 ⊕ · · · ⊕ In . Then νn (E) ≤ card(Kn (E)). This completes the proof of the lemma. Lemma 5.1.23(2) proves part 1 of Theorem 5.1.22 once we recall that h(G) = h(E(G)). LEMMA 5.1.24 Let E be a Noetherian integral domain, and let n > 0 be an integer. Then νn (E) ≤ e(E)n`(E) . Proof: Let t = `(E) and write the finite abelian group Γ(E) as Γ(E) = h(J1 )i × · · · × h(Jt )i for some fixed invertible ideals J1 , . . . , Jt ⊂ E. Then (Ji )e(G) = (E) for each i = 1, . . . , t so that each (I) ∈ Γ(E) can be written as I = J1e1 · · · Jtet

for some integers 0 ≤ ei < e(G).

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Form the set Mat(E) = {n × t matrices (eij )n×t

0 ≤ eij < e(G) for each

1 ≤ i ≤ n and 1 ≤ j ≤ t}. Since each element of Γ(E) is a product of powers of (J1 ), . . . , (Jt ), for each element (I1 , . . . , In ) ∈ Kn (E) there is an (eij )n×t ∈ Mat(E) such that Ii = J1ei1 · · · Jteit . There is then a (set theoretic) surjection Mat(E) −→ Kn (E) such that νn (E) = card(Kn (E)) ≤ card(Mat(E)) ≤ e(E)nt = e(E)n`(E) . This completes the proof of the lemma.

LEMMA 5.1.25 Let G be a strongly indecomposable rtffr group of square free rank. Then νn (G) = νn (E(G)). Proof: Let G be a strongly indecomposable rtffr group of square free rank. Suppose that G(n) = H1 ⊕ · · · ⊕ Hn for some indecomposable rtffr groups H1 , . . . , Hn . Then by Theorem 2.4.4 E(G)(n) = A(H1 ) ⊕ · · · ⊕ A(Hn ) where A(Hi ) is an indecomposable projective right E(G)-module. Since E(G) is a Noetherian integral domain, Theorem 4.2.12 states that A(Hi ) is an invertible ideal in E(G). Thus νn (G) ≤ νn (E(G)). Conversely suppose that E(G)(n) = I1 ⊕ · · · ⊕ In for some invertible ideals Ii in E(G). By Theorem 5.1.21, E(G) ∼ = I1 · · · In , and Theorem 2.4.4 implies that there are indecomposable groups Hi such that A(Hi ) = Ii and G(n) ∼ = H1 ⊕ · · · ⊕ Hn . Then νn (E(G)) ≤ νn (G) and it follows that νn (G) = νn (E(G)).

Proof of Theorem 5.1.22(2): By definition and Lemma 5.1.25 e(G) = e(E(G)), `(G) = `(E(G)), νn (G) = νn (E(G))

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121

so that by Lemma 5.1.24 νn (G) = νn (E(G)) ≤ e(E(G))n`(E(G)) = e(G)n`(G) . This completes the proof.

As an application of the above counting results we have a power cancellation result. THEOREM 5.1.26 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group of square free rank. If e(G) and n are relatively prime then G(n) ∼ = H (n) =⇒ G ∼ = H. Proof: Suppose that G(n) ∼ = H (n) . By Lemma 2.5.7, G and H are locally isomorphic. Theorem 2.4.4 implies that A(H) is an ideal in E(G) such that E(G)(n) ∼ = A(H)(n) . By Theorem 5.1.21, E(G) ∼ = A(H)n , so that the class (A(H)) ∈ Γ(E(G)) has order r dividing n. Since r must also divide e(G), and since e(G) and n are relatively prime, r = 1. Hence E(G) ∼ = A(H), and therefore G ∼ = H. THEOREM 5.1.27 Let G be an rtffr group such that Γ(E(G)) is a pgroup for some prime p. Then G(n) ∼ = H (n) =⇒ G ∼ = H for each integer n not divisible by p. Consider now some small values of n. Specifically we will show that these estimates are very good when n = 2. Let

hn (G) = card{x ∈ Γ(E(G)) xn = 1}.

THEOREM 5.1.28 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group of square free rank. 1. hn (G) = the number of isomorphism classes of groups H such that G(n) ∼ = H (n) . 2. h(G) − h2 (G) = the number of unordered pairs {(H), (K)} of distinct isomorphism classes of rtffr groups such that G ⊕ G ∼ = H ⊕ K.

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Proof: 1. By Theorem 2.4.4 it suffices to assume that G = E(G). So suppose that E(G)(n) ∼ = I (n) for some invertible ideal I ⊂ E(G). Theorem 5.1.21 states that E(G) ∼ = I · · · I (n factors), so that (I) ∈ {x ∈ Γ(E) xn = 1}. Hence the number of isomorphism classes of ideals I ⊂ E(G) such that E(G)(n) ∼ = I (n) is at most hn (G). The converse is proved by a similar argument. 2. Suppose that G ⊕ G ∼ 6 K. By = H ⊕ K as groups and that H ∼ = ∼ Theorem 2.4.4, E(G) ⊕ E(G) ∼ A(H) ⊕ A(K) and A(H) 6 A(K). Then = = ∼ by Theorem 5.1.21, A(H)A(K) = E(G), or in other words (A(H)) 6= (A(K)) and (A(H)) = (A(K))−1 . Hence card{(x, x−1 ) x−1 6= x ∈ Γ(E(G))} = # of x such that x2 6= 1 = h(E(G)) − h2 (G) = h(G) − h2 (G) by Lemma 5.1.25. This completes the proof. THEOREM 5.1.29 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group of square free rank. If h(G) is an odd integer then 1. G ⊕ G ∼ = H ⊕ H implies that G ∼ = H. 2. The number of sets {H, K} of nonisomorphic rtffr groups such that G⊕G∼ = H ⊕ K is h(G) − 1. Proof: 1. Suppose that G⊕G ∼ = H ⊕H. The group H corresponds to an element (I) ∈ Γ(E) such that I ·I ∼ = E. Since card(E) = h(E) = h(G) ∼ is odd, I = E. By Theorem 2.4.4, H ∼ = G. 2. Since h(G) is odd, h2 (G) = 1. Apply Theorem 5.1.28 to complete the proof.

5.2

Integrally Closed Groups

The rings in this section need not be commutative and the groups need not be strongly indecomposable. Recall that the rtffr group G is integrally closed group if E(G) is an integrally closed ring. We show in Lemma 3.1.8 that each rtffr group has finite index in an integrally closed group. We continue our theme that groups possessing the (local) refinement property and a locally unique decomposition are more common than once believed by identifying a naturally occurring class of subgroups of finite index in integrally closed groups that possess the local refinement property and a locally unique

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123

decomposition. Our goal should be contrasted with Example 2.2.1. Said example begins with a completely decomposable group G and constructs a subgroup G of finite index in G. By Theorem 2.1.11, G possesses a unique decomposition and it has the refinement property, but there are indecomposable decompositions G = G1 ⊕ G2 = H1 ⊕ H2 such that rank(G1 ) = rank(G2 ) = 2 and rank(H1 ) = 1. This juxtaposition of uniqueness with nonuniqueness invites us to study further. We can make some sense of this seemingly counterintuitive coincidence by examining direct sum decompositions of semi-primary groups. Definitions follow.

5.2.1

Review of Notation and Terminology

Fix a nilpotent set {G1 , · · · , Gt } of indecomposable rtffr groups, fix integers t, n1 , · · · , nt > 0, and let (n1 )

G = G1

(nt )

⊕ · · · ⊕ Gt

.

(5.2)

Nilpotency implies that Gi ∼ = Gj ⇒ i = j. We will make extensive use of the ideal structure of the semi-prime rtffr ring E(G), and the functor A(·) : Po (G) −→ Po (E(G)) from Theorem 2.4.4. Then because {G1 , . . . , Gt } is a nilpotent set (Theorem 4.1.8) we have E(G) = A(G1 )(n1 ) ⊕ · · · ⊕ A(Gt )(nt ) = Matn1 (E(G1 )) × · · · × Matnt (E(Gt )). There is at least one interesting class of rtffr ring that has the refinement property. PROPOSITION 5.2.1 The rtffr group G with conductor τ has the local refinement property and a locally unique decomposition if Eτ (G) is a semi-perfect ring. Proof: Given a semi-perfect ring Eτ (G), Lemma 1.7.1 and the AKS Theorem 2.1.6 state that Eτ (G) has the refinement property. Then Theorem 3.5.4 implies that G has the local refinement property. Lastly apply Lemma 2.5.10. Thus we are motivated to look for semi-perfect rtffr rings of the form Eτ (G). We will show that there are some fairly common rtffr groups G, called semi-primary groups, for which Eτ (G) is semi-perfect.

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We showed in Lemma 4.2.1 that E(G) is commutative if G is strongly indecomposable and if rank(G) is a square-free integer. Thus there are plenty of rtffr groups G with direct sum decomposition (5.2) such that QE(G) = Matn1 (k1 ) × · · · × Matnt (kt ) where ki is the field of fractions of the integral domain E(Gi ).

5.2.2

Locally Semi-Perfect Rings

Let E be a semi-prime rtffr ring with conductor τ . We give some fairly general conditions on τ and E that make Eτ semi-perfect. 5.2.2 Let S = center(E) and choose an integrally closed ring E ⊂ QE such that τ E ⊂ E ⊂ E. Then E = E1 × · · · × Et where each E i is a classical maximal order in the field ki . Let center(E) = S = S 1 × · · · × S t where S i = center(E i ) is a Dedekind domain for each i = 1, . . . , t. As in section 3.2, τ is an ideal in S so we can write

τ = τ1 ⊕ · · · ⊕ τt

where 0 6= τi ⊂ S i is an ideal of finite index.

Recall that an ideal I 6= 0 in a commutative ring S is a primary ideal if S/I is a local ring with nilpotent Jacobson radical. For instance, qZ is a primary ideal in Z iff q is a prime power. The primary ideals in a Dedekind domain are powers of maximal ideals. THEOREM 5.2.3 Suppose that E is a semi-prime rtffr ring with conductor τ = τ1 ⊕ · · · ⊕ τt as in (5.2.2). Then E has the local refinement property and a locally unique decomposition if E satisfies the following two conditions: 1. There are integers t, n1 , . . . , nt > 0 and fields k1 , . . . , kt such that QE = Matn1 (k1 ) × · · · × Matnt (kt ) and

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125

2. τi is a primary ideal in S i for each i = 1, . . . , t. Proof: The proof is a series of lemmas. The notations and conditions of the Theorem are used without fanfare. The first result is called the Change of Rings Theorem. I leave the proof as an exercise in the nature of long exact sequences and localizations. THEOREM 5.2.4 [72, page 107, Theorem 3.84] Let U be a finitely generated S-module and let I be an ideal in S. Then (EndS (U ))I ∼ = EndSI (UI ) as rings. LEMMA 5.2.5 Idempotents in Eτ /Eτ ττ lift to idempotents of Eτ . Proof: We first show that (E i )τ is a semi-perfect ring. Fix i ∈ {1, . . . , t}. By condition 1 of Theorem 5.2.3 QS i = center(QE i ) = center(Matni (ki )) = ki . Let Vi be a simple left module over the simple Artinian ring Matni (ki ) and let Ui = E i vi be a cyclic left E i -submodule of Vi . Then (n ) QUi = QE i vi = Vi ∼ = ki i

as ki -vector spaces. Since S i is a Dedekind domain, since QE i is a torsion-free Si -module, and since E i is a finitely generated torsion-free S i -module we see that Ui is a finitely generated projective S i -module. Thus EndS i (Ui ) is a finitely generated torsion-free S i -module. Furthermore (ni )

E i ⊂ EndS i (Ui ) ⊂ Endki (QUi ) = Endki (ki

) = Matni (ki )

and E i is a classical maximal order in Matni (ki ). Since EndS i (Ui )/E i is then a finitely generated torsion S i -module, E i = EndS i (Ui ), and since Ui is a finitely generated S i -module Theorem 5.2.4 implies that (E i )τ = (E i )τi ∼ = End(S i )τ (Uτi ). i

Condition 2 states that τi is a primary ideal in S i so that S i ∼ (S i )τi = τi (τi )τi

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is a local ring. Since (τi )τi ⊂ J ((Si )τi ), (S i )τi is a local ring. Moreover U is a finitely generated projective S i -module. Since Uτi is (n ) then a free module over the local ring (S i )τi Uτi ∼ = (S i )τi i . Then (E i )τ ∼ = End(S i )τ (Uτi ) ∼ = Matni ((S i )τi ) i

is a semi-perfect ring. As claimed E τ is semi-perfect. Now, let e ∈ Eτ be such that e2 − e ∈ ττ Eτ . Inasmuch as τ E ⊂ E ⊂ E, setting I = ττ2 E τ gives us I = ττ (ττ E τ ) ⊂ ττ Eτ ⊂ J (E τ ). Since idempotents lift modulo the nilpotent ideal ττ Eτ /I ⊂ Eτ /I, there is an (f + I)2 = f + I ∈ Eτ /I ⊂ E τ /I such that e − f ∈ ττ Eτ . By the above claim E τ is semi-perfect and I ⊂ J (E τ ), so by Lemma 1.7.4 there is a g 2 = g ∈ E τ such that f + I = g + I. Hence g ∈ f + I ⊂ Eτ , whence idempotents in Eτ /ττ Eτ lift to idempotents in Eτ . This completes the proof of the lemma. The rtffr ring E is semi-local iff there is some integer n 6= 0 such that each maximal ideal of E contains n. For example, if E is a semi-prime rtffr ring with conductor τ then Eτ is a semi-local ring. LEMMA 5.2.6 Let E be a semi-prime rtffr ring with conductor τ that satisfies conditions 1 and 2 of Theorem 5.2.3. Then Eτ is a semi-perfect ring. Proof: Suppose that E satisfies conditions 1 and 2. Since ττ Eτ ⊂ J (Eτ ) and since τ has finite index in S, Eτ /J (Eτ ) is a (finite =) semisimple Artinian ring. Furthermore if e2 = e ∈ Eτ /J (Eτ ) then e lifts modulo the nilpotent ideal J (Eτ )/ττ Eτ to an (e + ττ Eτ )2 = (e + ττ Eτ ) ∈ Eτ /ττ Eτ . By Lemma 5.2.5, there is an f 2 = f ∈ Eτ such that e − f ∈ ττ Eτ . Therefore Eτ is semi-perfect. Proof of Theorem 5.2.3: By Lemma 5.2.6, Eτ is semi-perfect, so Lemma 1.7.1 and the AKS Theorem 2.1.6 imply that Eτ has the refinement property. Using Theorem 3.5.3 we conclude that E has the local refinement property.

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127

In case QE = Matn (k) for some field k then S = center(E) is an integral domain with field of quotients k. If k has finite degree over Q then there is a unique Dedekind domain S ⊂ S ⊂ k called the integral closure of S such that S/S is finite. COROLLARY 5.2.7 Suppose that E is a prime rtffr ring with conductor τ such that QE ∼ = Matn (k) for some integer n > 0 and algebraic number field field k. Let S be the integral closure of S. If τ is a primary ideal in S then E has the local refinement property. EXAMPLE 5.2.8 Let k be an algebraic number field, let O be the ring of algebraic integers in k with class number h 6= 1, and let I 6= 0 be an ideal in O such that O ∼ 6 I. By Theorem 5.2.3 the finitely generated = projectives over E have the local refinement property. However since the class number h of O is not 1, I (h) ∼ = O(h) , while I 6∼ = O. Thus O has the local refinement property.

5.2.3

Semi-Primary Rtffr Groups

Let G ⊂ G be rtffr groups and suppose that G has the local refinement property. The results of this section will give conditions on G under which a subgroup of finite index in G has the local refinement property. This should be contrasted with examples due to Corner and Fuchs-Loonstra (see [10, 46]), that show that subgroups of finite index in completely decomposable rtffr groups can have badly behaved direct sum decompositions. The group G is a Dedekind group if E(G) is a Dedekind domain. If G is a Dedekind domain then E(G) is a commutative ring. We assume that we are given a finite nilpotent set {G1 , . . . , Gt } of Dedekind rtffr groups and integers n1 , . . . , nt > 0. Let (n1 )

G = G1

(nt )

⊕ · · · ⊕ Gt

.

(5.3)

For each i = 1, . . . , t choose a primary ideal Pi such that N (End(Gi )) ⊂ Pi ⊂ End(Gi ). Notice that because Pi is primary in End(Gi ) then End(Gi )/Pi is finite. (n ) The action Pi Gi i is unambiguous since Pi ⊂ End(Gi ). Then (n1 )

(P1 ⊕ · · · ⊕ Pt )G = P1 G1

(nt )

⊕ · · · ⊕ Pt Gt

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is uniquely defined. The rtffr groups we are interested in are those G such that (n1 )

P1 G 1

(nt )

⊕ · · · ⊕ Pt G t

⊂ G ⊂ G.

(5.4)

In this case we say that G has semi-primary index in G. Alternatively we say that G is a semi-primary rtffr group if there are integers t, n1 , . . . , nt > 0, a nilpotent set {G1 , . . . , Gt } of Dedekind rtffr groups, and primary ideals N (End(Gi )) ⊂ Pi ⊂ End(Gi ) such that (5.4) is true. Given a semi-primary rtffr group G we will use G and Pi in the way they are used in the above discussion. Evidently each Dedekind rtffr group is a semi-primary group so semiprimary groups exist in abundance. The following examples illustrate the settings that we are trying to generalize. EXAMPLE 5.2.9 Let p ∈ Z be a prime, let E = Mat2 (Z), and let E = 1Z + Mat2 (pZ). Then e2 = e ∈ E ⇒ e ∈ {0, 1}. Using Corner’s Theorem 2.3.3 there is an rtffr group G such that E = End(G). Let G = EG. One shows that E = End(G) and that pG ⊂ G ⊂ G. Then G is a semi-primary group, G is indecomposable, and G has a nontrivial direct sum decomposition. Moreover by Theorem 5.2.3, G and G have the local refinement property and unique decompositions. This example demonstrates that we are properly extending Theorem 2.4.6. EXAMPLE 5.2.10 Let G1 , . . . , Gt ⊂ Q be rank 1 groups such that Gi ∼ = Gj ⇒ i = j. The Baer-Kulikov-Kaplansky Theorem 2.1.11 states that the group G in (5.3) has unique decomposition. If q ∈ Z is a prime power and if G is a group such that qG ⊂ G ⊂ G then G is a semi-primary group. It is traditional [60] to call G an acd group with primary regulating quotient. EXAMPLE 5.2.11 Let {G1 , . . . , Gt } be a finite nilpotent set of rtffr groups such that End(Gi ) is a pid for each i = 1, . . . , t. Let G be the group in (5.3) and for each i = 1, . . . , t choose prime powers qi ∈ End(Gi ).

5.2. INTEGRALLY CLOSED GROUPS

129

Then each group G such that (n1 )

q 1 G1

(nt )

⊕ · · · ⊕ q t Gt

⊂ G ⊂ G

is a semi-primary rtffr group. EXAMPLE 5.2.12 Let G be a (strongly indecomposable) Dedekind rtffr group, fix an integer n > 0, and choose a primary ideal N (End(G)) ⊂ P ⊂ End(G). If G is any rtffr group such that PG

(n)

⊂G⊂G

(n)

then G is a semi-primary rtffr group. If n = 1 then we are looking at semi-primary rtffr groups G such that P G ⊂ G ⊂ G. The next result, the main result of this chapter, explains our interest in semi-primary rtffr groups. THEOREM 5.2.13 [T.G. Faticoni] Semi-primary rtffr groups have the local refinement property and a locally unique decomposition. Proof: The proof of this theorem consists of showing that E(G) satisfies conditions 1 and 2 in Theorem 5.2.3. Let G be a semi-primary rtffr group. There is a nilpotent set {G1 , . . ., Gt } of Dedekind rtffr groups, integers n1 , . . . , nt > 0, and primary ideals N (End(Gi )) ⊂ Pi ⊂ End(Gi ) such that (P1 ⊕ · · · ⊕ Pt )G ⊂ G ⊂ G where G is given in (5.3). Let P = P1 ⊕ · · · ⊕ Pt . For each i = 1, . . . , t, E(Gi ) is a Dedekind domain whose field of fractions QE(Gi ) is a number field. Since {G1 , . . . , Gt } is a nilpotent set Theorem 1.3.1 implies that E(G) = Matn1 (E(G1 )) × · · · × Matnt (E(Gt )).

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. . Since G = G, End(G) = End(G), so that QE(G) ∼ = QE(G) as rings. Then QE(G) = Matn1 (QE(G1 )) × · · · × Matnt (QE(Gt )) satisfies condition 1 of Theorem 5.2.3. As for condition 2, for each i = 1, . . . , t, Pi maps onto a nonzero primary ideal Ti ⊂ E(Gi ). Let T = T1 ⊕ · · · ⊕ Tt . Then T is an ideal of finite index in S. Since P G ⊂ G ⊂ G, P End(G) ⊂ End(G), and so P E(Gi ) = T E(Gi ) = Ti E(Gi ) ⊂ E(Gi ). Inasmuch as Ti ⊂ τi ⊂ E(Gi ) (where τ = τ1 ⊕ · · · ⊕ τt is the conductor of E(G)), and since Ti is a primary ideal in E(Gi ), τi is a primary ideal in E(Gi ). Thus E(G) satisfies condition 2. Then by Theorem 5.2.3, E(G) has the local refinement property and a locally unique decomposition. This completes the proof of the theorem.

5.2.4

Applications to Refinement

Some applications will show that the seemingly strong conditions in Theorem 5.2.3 are actually quite common. If H is a rank one group then End(H) ⊂ Q is a pid and a given prime p ∈ Z is either a unit in End(H) or a prime. Thus the rank one groups are Dedekind groups whose primary ideals are generated by powers of primes in Z. We say that G is homogeneous completely decomposable of type σ if G = H (n) for some rank one group H of type σ. If G is a completely decomposable rtffr group then G = G[σ1 ] ⊕ · · · ⊕ G[σt ] where G[σi ] is a homogeneous completely decomposable group of type σi , and the types σ1 , . . . , σt are distinct. Using this notation we can use Theorem 5.2.13 to study the uniqueness of direct sum decompositions of groups of semi-primary index in a completely decomposable rtffr group. The next two results should be contrasted with the acd groups constructed in Examples 2.2.1. THEOREM 5.2.14 [T.G. Faticoni] Let G = G[σ1 ] ⊕ · · · ⊕ G[σt ] be a completely decomposable rtffr group, where the groups G[σi ] are homogeneous completely decomposable of type σi , and where the types σ1 , . . . , σt are distinct. If G is a subgroup of finite index in G such that for each i = 1, . . . , t, (G + G[σi ])/G is a finite primary abelian group for each i = 1, . . . , t then G has a locally unique decomposition.

5.2. INTEGRALLY CLOSED GROUPS

131

(n )

Proof: By hypothesis G[σi ] = Gi i for some rank one group Gi , i = 1, . . . , t. The reader can show that {G1 , . . . , Gt } is a nilpotent set. Suppose that G has finite index in G and that for each i = 1, . . . , t, (G + G[σi ])/G is a finite pi -group. There are integers m1 , . . . , mt > 0 such that mt 1 pm 1 G[σ1 ] ⊕ · · · ⊕ pt G[σt ] ⊂ G ⊂ G[σ1 ] ⊕ · · · ⊕ G[σt ]

or in other words G has semi-primary index in G. By Theorem 5.2.13, G has the local refinement property. COROLLARY 5.2.15 [T.G. Faticoni and P. Schultz] (See [45].) Let G be a completely decomposable rtffr group and let q ∈ Z be a prime power. If G is a group such that qG ⊂ G ⊂ G then G has a locally unique decomposition. For the purposes of this discussion let H = H 1 ⊕ · · · ⊕ Hs where Hi ⊂ Q for each i = 1, . . . , s. A kernel group is a group G that fits into an exact sequence σ

0 → G −→ H −→ Q where σ is the unique map such that σ(x) = x for each x ∈ Hi and for each i = 1, . . . , s. Dually a cokernel group is a group G that fits into an exact sequence π 0 → X −→ H −→ G → 0 where X is a pure rank one subgroup of H. We call G a bracket group if G is either a kernel group of a cokernel group. If G is a strongly indecomposable bracket group then it is known that End(G) ⊂ Q is a pid. See [10]. Thus the strongly indecomposable bracket groups are Dedekind groups. A strongly homogeneous group is a group G such that for any pure rank one subgroups X, Y ⊂ G there is an automorphism α : G → G such that α(X) = Y . It is known that if G is a strongly indecomposable strongly homogeneous rtffr group then End(G) is a pid in which the primes p ∈ Z such that pG 6= G are primes in End(G). See [10]. Thus the strongly indecomposable strongly homogeneous rtffr groups are Dedekind groups. The group G is a Murley group if Z/pZ-dim(G/pG) ≤ 1 for each prime p ∈ Z. It is known that if G is a strongly indecomposable Murley

132

CHAPTER 5. REFINEMENT REVISITED

group then End(G) is a pid in which p ∈ Z is a prime in End(G) if b pG 6= G. Evidently G is a Murley group iff G is a pure subgroup of Z. See [62]. Direct sums of rtffr groups G such that End(G) is a pid are considered by D.M. Arnold, R. Hunter, and F. Richman in [13]. Certainly G is a Dedekind group if End(G) is a pid. We have shown that if G is a strongly indecomposable group of squarefree rank then E(G) is an rtffr integral domain. Lemma 3.1.8 implies that G has finite index in a Dedekind group, thus motivating the assumption that G is a Dedekind group. Let D denote the set of strongly indecomposable rtffr groups G such that G falls into at least one of the following categories. 1. Groups G such that E(G) is a pid, 2. Strongly indecomposable Murley groups, 3. Strongly indecomposable strongly homogeneous groups, 4. Strongly indecomposable bracket groups, 5. Rank one groups. THEOREM 5.2.16 [T.G. Faticoni] Let {G1 , . . . , Gt } be a nilpotent subset of D and let G be the direct sum (5.3) for some integers n1 , . . . , nt > 0. If qi ∈ E(Gi ) are powers of primes in the pid E(Gi ) and if G is a group such that (n1 )

q 1 G1

(nt )

⊕ · · · ⊕ q t Gt

⊂G⊂G

then G has the local refinement property. Proof: By the above discussion for each Gi ∈ D, E(Gi ) is a pid so each Gi is a Dedekind rtffr group. Thus G is a semi-primary group. An application of Theorem 5.2.13 shows us that G has the local refinement property. . In [13] it is shown that if End(G) is a pid and if H1 , . . . , Hr = G satisfy End(Hi ) = End(G) then H = H1 ⊕ · · · ⊕ Hr has a uniqueness of

5.2. INTEGRALLY CLOSED GROUPS

133

decomposition that is different from local uniqueness and that is a little too technical for this book. In general there is an integer q 6= 0 such that qG ⊂ Hi ⊂ G for each i = 1, . . . , r. Our techniques allow us to slightly expand the main theorem in [13] by restricting q and by relaxing the conditions on End(G) and End(Hi ). COROLLARY 5.2.17 Let {G1 , . . . , Gt } be a nilpotent subset of D such that E(G1 ) ∼ = E(Gi ) for each i = 1, . . . , t, and let (n1 )

G = G1

(nt )

⊕ · · · ⊕ Gt

.

(5.5)

If q ∈ E(G1 ) is a prime power then each group qG ⊂ G ⊂ G has the local refinement property. COROLLARY 5.2.18 Let G be a rtffr group such that E(G) is a pid, let n 6= 0 be an integer, and let q ∈ E(G) be a power of some prime (n) (n) in E(G). Then any group qG ⊂G⊂G has the local refinement property. COROLLARY 5.2.19 Let {G1 , . . . , Gt } be a nilpotent subset of D such that E(Gi ) ∼ = Z for each i = 1, . . . , t, and let G be a direct sum (5.5). If q ∈ Z is a prime power and if qG ⊂ G ⊂ G then G has the refinement property. We end this section with a result that shows that direct sums of semiprimary groups G are up to local isomorphism actually direct sums of semi-simple End(G)-modules. The reader will enjoy proving this result. THEOREM 5.2.20 [T.G. Faticoni] Let G be a semi-primary rtffr group. Let τ be the conductor of E(G) and assume that E(G)(τ ) = 0. Let (S, τ ) be the category of semi-simple End(G)-modules annihilated by τ . 1. There is a functor C(·) : Po (G) → (S, τ ) : H → A(H) ⊗E(G) E(G)τ /J (E(G)τ ). 2. C(·) induces a bijection from the set of local isomorphism classes (H) of H ∈ Po (G) onto the set of isomorphism classes (M ) of semi-simple right E(G)-modules such that M τ = 0.

134

5.3

CHAPTER 5. REFINEMENT REVISITED

Exercises

1. Show that J (Sτ ) ⊂ J (Eτ ). 2. Let E be a rtffr ring and suppose that E is locally isomorphic to U1 ⊕ U2 for some finitely generated E-modules U1 and U2 . Then E = V1 ⊕ V2 for some cyclic right ideals V1 and V2 such that Ui is locally isomorphic to Vi for i = 1, 2. 3. Let G be a rtffr group. Show that the following are equivalent. (a) G is indecomposable. (b) G is locally isomorphic to some indecomposable rtffr group. (c) If G is locally isomorphic to H then H is indecomposable. (d) E(G) is indecomposable as a right E(G)-module. (e) Eτ (G) is indecomposable as a right Eτ (G)-module. 4. Show that G has a locally unique decomposition if either (a) E(G) is a semi-perfect ring or (b) G is a completely decomposable rtffr group or (c) G = H (n) where E(H) is a Dedekind domain. 5. Suppose that S is a Noetherian integral domain and that its field of fractions k is a finitely generated S-module. Show that k = S. 6. Using the notation of Theorem 5.2.13 show that there is a ring theoretic embedding E(G) → E(G) that takes regular elements to regular elements. 7. Using the notation of Theorem 5.2.13 show that QE(G) = QE(G). 8. Let O be the ring of algebraic integers in the number field k. Show that (a) Matn (k) is the classical ring of quotients of Matn (O). (b) Matn (O) is a maximal order in Matn (k). 9. Let E be a semi-prime rtffr ring with conductor τ . Say E = E1 × E2 . (a) Show that τ = τ1 ⊕ τ2 where τi is the conductor of Ei for i = 1, 2.

5.4. QUESTIONS FOR FUTURE RESEARCH

135

(b) Show that Eτ = (E1 )τ × (E2 )τ = (E1 )τ1 × (E2 )τ2 . 10. Suppose that G is a strongly indecomposable rtffr group of squarefree rank. Suppose that t is an integer that is divisible by h(G). Given indecomposable groups H1 , . . . , Ht ∈ Po (G) then G(t) ∼ = H 1 ⊕ · · · ⊕ Ht . . 11. Suppose that G = G1 ⊕ · · · ⊕ Gt for some strongly indecomposable groups of square-free rank. Multiple appearances are allowed. Investigate the logical relations between the following statements. (a) h(G) = 1. (b) h(Gi ) = 1 for each i = 1, . . . , t. (c) Each Gi has the L-refinement property. (d) Each Gi has the refinement property.

5.4

Questions for Future Research

Let G be a rtffr group, let E be a rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. Use the theory of rtffr groups to advance the understanding of the class group of an algebraic number field. 2. Use the theory of the class group of an algebraic number field to advance the understanding of the theory of rtffr groups. 3. Give a (finite abelian) group structure to Γ(G), the set of isomorphism classes of groups locally isomorphic to G, and study the group. 4. Generalize the discussion given on class number h(G) from strongly indecomposable rtffr groups G to just plain rtffr groups G. 5. There are results from the literature that show that h(O), where O is the ring of algebraic integers in an algebraic number field, will grow in a certain way as the discriminant of O increases. 6. Make a study of the class group and class number of an algebraic number field and apply it to rtffr groups.

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CHAPTER 5. REFINEMENT REVISITED

7. Make a study of the class number h(G) of an rtffr group and apply it to the study of the class number of an algebraic number field. 8. See 5.1.9. Determine ker β. Specifically, determine if ker α is finitely generated. Determine if ker β is a direct summand of Γ(E). 9. Use the Mayer-Vietoris Sequence and the Kunneth Sequence to study Γ(E). See [69, Section 37]. 10. Discuss how

√

τ is related to h(G).

11. Nakayama’s Lemma and the Chinese Remainder Theorem are used continually in this text. Explain why this is so by using some other kind of theory. 12. Give a complete accounting of the groups G that have the Lproperty. 13. Give a complete accounting of the groups G that have the refinement property. 14. Let G be a strongly indecomposable rtffr group such that E(G) is a Noetherian integral domain. Give group theoretic conditions for h(G) = h(G0 ) for quasi-isomorphic groups G and G0 . 15. Give a more complete discussion of the integers νn (G), e(G), and `(G) given on page 118. 16. Improve the estimates in Theorem 5.1.22. 17. Do what is considered to be a great deal of fun in abelian groups. Relate the direct sum properties of G to integers associated with G. For example, G has the refinement property iff h(G) = 1. See also Theorem 5.1.26 and Theorem 5.1.28. 18. Characterize the groups G such that E(G) is a locally semi-perfect ring. 19. Characterize semi-primary rtffr groups. 20. Functorially realize the group G. 21. Extend the theory surrounding acd groups to semi-primary groups, or indicate where we cannot.

5.4. QUESTIONS FOR FUTURE RESEARCH

137

22. Let G be a strongly indecomposable rtffr group such that E(G) is a commutative rtffr integral domain. Let E be the integral closure of E(G). Investigate the converses of the statement “h(E) = 1 if G has the refinement property.” √ 23. Suppose that QEnd(G) = Q[ d] for some negative square-free integer d. If d ∈ {−1, −2, −3, −7, −11, −19, −43, −67, −163}, then discuss the refinement property in G.

Chapter 6

Baer Splitting Property The results in this section are from [38]. The short exact sequence π

0 → K −→ X −→ H → 0

(6.1)

of groups is said to be split if there is a map j : H → X such that πj = 1H . Recall that P(G) = { groups H H ⊕ H 0 ∼ = G(c) for some cardinal c and group H 0 }. We will consider the splitting of the short exact sequence (6.1) under different hypotheses on X and the groups H ∈ P(G). Closely related to the refinement property in rtffr groups is the Baer splitting property. The group G has the Baer splitting property if for each cardinal c, each surjection π : G(c) → G is split. Free groups have the Baer splitting property and we will show that if End(G) is commutative then G has the Baer splitting property. Baer’s Lemma 6.1.1 states that G has the Baer splitting property if IG 6= G for each right ideal I ⊂ End(G). The purpose of this chapter is to study the Baer splitting property in detail.

6.1

Baer’s Lemma

Given groups H and G we write SG (H) =

P {f (G) f ∈ Hom(G, H)}.

139

140

CHAPTER 6. BAER SPLITTING PROPERTY

We call SG (H) the G-socle in H, and we note that SG (H) is the largest G-generated subgroup of H. If H = SG (H) then we say that H is Ggenerated, and we say that H is finitely G-generated if H = f1 (G) + · · · + fn (G) for finitely many f1 , . . . , fn ∈ Hom(G, H). The next result, known as Baer’s Lemma, has assumed an important role in the study of the transfer of properties between the rtffr group G and its endomorphism ring End(G). One way of viewing this result is that it shows us that a group G behaves like a projective module over a ring if IG 6= G for each right ideal I ⊂ End(G). The version we use is found in [14]. LEMMA 6.1.1 [Baer’s Lemma] The following are equivalent for an rtffr group G. 1. IG 6= G for each right ideal I ⊂ EndS (G). 2. Each short exact sequence π

0 → K −→ X −→ G → 0

(6.2)

in which SG (X) + K = X is split exact. Proof: Assume part 1 and let I = πHom(G, X). Then I is a right ideal in the ring End(G). Let y ∈ G. There is an x ∈ X such that π(x) = y. Since X = SG (X) + K there are f1 , . . . , ft ∈ Hom(G, X), x1 , . . . , xt ∈ G, and z ∈ K such that ! t t X X y = π(x) = π fi (xi ) + z = πfi (xi ) ∈ IG. i=1

i=1

Thus G = IG. Inasmuch as JG 6= G for proper right ideals J ⊂ End(G), I = End(G). That is, there is a map f ∈ Hom(G, X) such that πf = 1G , so that (6.2) is split exact. This proves part 2. Conversely assume part 1 is false. There is a proper right ideal I ⊂ End(G) such that IG = G. Write I = {f1 , f2 , . . .} and let Gi ∼ = G for each i = 1, 2, . . .. There is a map π : ⊕∞ i=1 Gi −→ G such that for each x ∈ Gi , π(x) = fi (x). Because G has finite rank, given any map g : G −→ ⊕∞ i=1 Gi

6.1. BAER’S LEMMA

141

there is a finite set {1, . . . , n} such that g(G) ⊂ G1 ⊕ · · · ⊕ Gn so that there are maps gi : G → Gi such that g = g1 ⊕ · · · ⊕ gn . Then πg = f1 g1 + · · · + fn gn ∈ I 6= End(G). Specifically πg 6= 1G for any g so that π is not split. Thus part 2 is false, which completes the proof.

COROLLARY 6.1.2 The following are equivalent for an rtffr group G. 1. IG 6= G for each proper right ideal I ⊂ End(G). 2. For each cardinal c, each short exact sequence π

0 → K −→ G(c) −→ G → 0

(6.3)

is split exact. The rtffr group G is finitely faithful if IG 6= G for each proper ideal I of finite index in End(G). Of course faithful groups are finitely faithful. For rtffr groups the converse is true. For this reason we will use the term faithful and not finitely faithful throughout this chapter. LEMMA 6.1.3 Let G be an rtffr group. Then G is a faithful group iff G is a finitely faithful group. Proof: Suppose that IG 6= G for each right ideal of finite index in End(G), and let I ⊂ End(G). Then I is contained in a maximal right ideal M ⊂ End(G) which by Lemma 4.2.3 has finite index in End(G). By hypothesis IG ⊂ M G 6= G so that G is a faithful group. The converse is clear so the proof is complete. There is at least one interesting previously encountered class of group that consists of faithful groups. To prove the next result apply Theorems 4.2.4 and 6.1.1. THEOREM 6.1.4 [10, page 52, Theorem 5.9] Let G be an rtffr group such that End(G) is commutative. Then G is a faithful group.

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CHAPTER 6. BAER SPLITTING PROPERTY

A faithfully E-flat group is a faithful group G that is flat as a left End(G)-module. EXAMPLE 6.1.5 Since each subring of Q is a pid, G is a faithfully E-flat group if End(G) ⊂ Q. Given a prime p ∈ Z, IZ(p∞ ) = Z(p∞ ) for b p . Thus Z(p∞ ) is not a faithful each nonzero ideal I in End(Z(p∞ )) ∼ =Z group even though End(G) is a pid. Evidently, Z(p∞ ) is not E-flat. The next example shows us that there are plenty of faithfully E-flat groups. EXAMPLE 6.1.6 The group constructed according to Corner’s Theorem 2.3.3 is faithfully flat. Proof: Let E be an rtffr group and let G be the group constructed according to Corner’s Theorem. There is a short exact sequence 0 → E −→ G −→ QE → 0 of left E-modules. Since E and QE are flat left E-modules, there is an exact sequence 0 = Tor1E (·, E) −→ Tor1E (·, G) −→ Tor1E (·, QE) = 0. Then Tor1E (·, G) = 0 and hence G is E-flat. Let I ⊂ End(G) be a right ideal. We may assume without loss of generality that I is a maximal right ideal. Lemma 4.2.3 states that E/I is a finite p-group for some prime p ∈ Z so pE ⊂ I ⊂ E. An application of E/I ⊗· to the above short exact sequence produces the exact sequence Tor1E (E/I, QE) → E/I → G/IG → QE/IQE. Since QE is a flat left E-module and since E/I is finite, Tor1E (E/I, QE) = 0 = QE/IQE, so that G/IG ∼ = E/I 6= 0. This completes the proof. We will construct rtffr groups whose properties are from the set {faithful, not faithful, E-flat, not E-flat}. EXAMPLE 6.1.7 There is an rtffr G that is E-flat but that is not faithful.

6.1. BAER’S LEMMA

143

Proof: Let k be an algebraic number field such that [k : Q] ≥ 2, and let O denote the ring of algebraic integers in k. It is known that Z/pZ-dim(O/pO) = [k : Q]. (See [58].) Let S = Z + pO 6= O, let A= and let M =

QS QHom(pO, S) QHom(S, pO) QO

=

k k k k

,

S pO

. Then let

E = {q ∈ A qM ⊂ M } S Hom(pO, S) = Hom(S, pO) O S pO . = pO O Furthermore I =

S pO pO pO

IM =

S pO pO pO

is a right ideal in E such that

S pO

=

S pO

= M.

Theorem 2.3.4 states that there is an rtffr group G and a short exact sequence 0 → M −→ G −→ A ⊕ A → 0

(6.4)

such that E ∼ = End(G). As in Example 6.1.6 we can prove that IM = M implies that IG = G. Thus G is not faithful. Since M is the first column of E, M is a projective (= flat) left E-module. Then as in Example 6.1.6, G is an E-flat group.

EXAMPLE 6.1.8 There is a faithful rtffr group that is not E-flat. Proof: In Example 4.3.2 we constructed a group G such that End(G) . is an integral domain but not a Dedekind domain and G = End(G) is not a flat left End(G). Since End(G) is commutative G is faithful. With these examples to guide us we will discuss faithful rtffr groups.

144

6.2

CHAPTER 6. BAER SPLITTING PROPERTY

Splitting of Exact Sequences

The groups in this section need not have finite rank. We continue our study of the Baer splitting property by considering a more general concept. Let G be a torsion-free group. 1. Let H ∈ P(G). Then (G, H) has the Baer splitting property if π the short exact sequence 0 → K −→ X −→ H → 0 is split exact whenever SG (X) + K = X. The group G has the Baer splitting property iff for each cardinal c and group H ∈ P(G), each short π exact sequence 0 → K −→ G(c) −→ H → 0 is split exact. 2. Let H ∈ P(G). Then (G, H) has the endlich Baer splitting property if for each positive integer c, the short exact sequence 0 → K −→ π G(c) −→ H → 0 is split exact. The group G has the endlich Baer splitting property iff for each positive integer c, each short exact π sequence 0 → K −→ G(c) −→ G → 0 is split exact. The term Baer splitting property was coined by U. Albrecht and H.P. Goeters [5] and the term endlich Baer splitting property first appears in [38]. Evidently G has the endlich Baer splitting property if G has the Baer splitting property. We will eventually prove that for an rtffr group G the endlich Baer splitting property implies the Baer splitting property.

The next two results show us that the set of groups H such that (G, H) has the Baer splitting property is closed under direct sums. LEMMA 6.2.1 Let G, H, H 0 be groups. If (G, H) and (G, H 0 ) have the Baer splitting property then (G, H ⊕ H 0 ) has the Baer splitting property. Proof: Suppose that (G, H) and (G, H 0 ) have the Baer splitting property and consider the short exact sequence π

0 → K −→ X −→ H ⊕ H 0 → 0

(6.5)

where SG (X) = X. We will show that (6.5) is split. Let ρ : H ⊕ H 0 −→ H be the canonical projection with ker ρ = H 0 . Since (G, H) has the Baer splitting property there is an injection g : H → X such that 1H = (ρπ)g = ρ(πg). Then H ⊕ H 0 = πg(H) ⊕ H 0

and X = g(H) ⊕ X 0

6.2. SPLITTING OF EXACT SEQUENCES

145

where X 0 = ker ρπ and where g(H) ∼ = H. Let g 00 : πg(H) ∼ = g(H) ⊂ X and notice that πg 00 (πg(H)) = πg(H). We can assume without loss of generality that πg 00 (x) = x for each x ∈ πg(H). The reader will show as an exercise that there is a short exact sequence π0

0 → K −→ X 0 −→ H 0 → 0 in which SG (X 0 ) = X 0 and π 0 is the restriction of π to X 0 . Since (G, H 0 ) has the Baer splitting property there is an injection g 0 : H 0 −→ X 0 such that π 0 g 0 = 1H 0 . It is then an easy task to show that π(g 00 ⊕ g 0 )(πg(x) ⊕ x0 ) = πg 00 (πg(x)) + πg 0 (x0 ) = πg(x) ⊕ x0 for each x ∈ H and x0 ∈ H 0 . Thus π(g 00 ⊕g 0 ) = 1H⊕H 0 so that (G, H ⊕H 0 ) has the Baer splitting property. This completes the proof. THEOREM 6.2.2 Let G, H be rtffr groups. 1. If (G, H) has the Baer splitting property then (G, H (n) ) has the Baer splitting property for each integer n > 0. 2. If (G, H) has the endlich Baer splitting property then (G, H (n) ) has the endlich Baer splitting property for each integer n > 0. Proof: Apply induction to n > 0 using Lemma 6.2.1 as the induction step. The above lemma is the induction step in a transfinite induction. THEOREM 6.2.3 [T.G. Faticoni] (See [38].) Let G, H be rtffr groups. Then (G, H) has the Baer splitting property iff (G, H (d) ) has the Baer splitting property for each cardinal d. Proof: Let H ∈ Po (G), let Hi ∼ = H for each i ∈ I, suppose that (G, H) has the Baer splitting property, let I be an index set, and let π

0 → K −→ X −→

M i∈I

Hi → 0

(6.6)

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CHAPTER 6. BAER SPLITTING PROPERTY

be a short exact sequence in which SG (X) + K = X. Given a subset J ⊂ I let M M ρJ : Hi −→ Hj i∈I

j∈J

be the canonical projection. Let B be the set of pairs (J, gJ ) such that J ⊂ I, gJ : ⊕j∈J Hj → X, and ρJ πgJ = 1Lj∈J Hj . Given i ∈ I, the Baer splitting property produces a map gi : Hi → X such that ρi πgi = 1Hi so B 6= ∅. Partially order B by the usual inclusion of sets and restriction of functions. An elementary argument shows that each chain in B contains an upper bound in B. Then by Zorn’s Lemma B contains a maximal element (Im , gm ). Assume for the sake of contradiction that Im 6= I. Let i ∈ I \ Im . The proof of Lemma 6.2.1 implies that there is a gi : Hi → X such that ρIm ∪{i} π(gm ⊕ gi ) = 1⊕`∈Im ∪{i} H` . Thus (Im ∪ {i}, gm ⊕ gi ) ∈ B. Since (Im , gm ) is maximal in B it must be that Im = I and ρI = 1⊕i∈I Hk = πgm . Thus (G, H (d) ) has the Baer splitting property for each cardinal d. COROLLARY 6.2.4 Let G be a group. 1. G has the endlich Baer splitting property iff (G, H) has the endlich Baer splitting property for each H ∈ Po (G). 2. G has the Baer splitting property iff (G, G) has the Baer splitting property. Proof: Parts 1 and 2 follow from Theorems 6.2.2 and 6.2.3.

6.3

G-Compressed Projectives

Baer’s Lemma 6.1.1 shows that we should be interested in simple right End(G)-modules M such that TG (M ) 6= 0. From this point of view, introduced by U. Albrecht [4], it is natural to consider the set of right End(G)-modules M such that TG (M ) = 0. Let M 6= 0 be a right End(G)-module. We say that M is Gcompressed if TG (M/N ) 6= 0 for each proper End(G)-submodule N ⊂ M .

6.3. G-COMPRESSED PROJECTIVES

147

The right End(G)-module K is finitely M -generated if there is an integer n and an isomorphism K ∼ = M (n) /N for some subgroup N ⊂ M (n) . The module M is finitely G-compressed TG (M/K) 6= 0 for each proper finitely generated submodule K of M . Let E = End(G). Given an H ∈ Po (G) the Arnold-Lady-Murley Theorem 2.4.1 shows that HG (H) is a finitely generated projective right End(G)-module. The Arnold-Lady-Murley Theorem 2.4.1 shows us that the dual module of HG (H) is HomE (HG (H), E) ∼ = HomE (HG (H), HG (G)) ∼ = Hom(H, G). Given a projective E = End(G)-module P the trace of P in E is the ideal ∆P =

P {f (P ) f ∈ HomE (P, E)}.

If H ∈ P(G) then the trace of H is the trace ideal of HG (H)

∆H = ∆HG (H) . It is known that if P is a projective right E-module then 1. ∆P is the smallest ideal in E such that P ∆P = P , and 2. ∆2P = ∆P . Notice that condition 6 in the following result is similar to the condition that IG 6= G for each right ideal I ⊂ End(G). THEOREM 6.3.1 [38, T.G. Faticoni] The following are equivalent for H ∈ Po (G). 1. HG (H)(n) is G-compressed for each integer n > 0. 2. HG (H) is G-compressed. 3. ∆H ⊂ I for each maximal right ideal I ⊂ End(G) such that IG = G. 4. IHom(H, G) = Hom(H, G) for each maximal right ideal I ⊂ End(G) such that IG = G.

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5. TG (M ) 6= 0 for each finitely HG (H)-generated right End(G)-modules M 6= 0. 6. N G 6= H for each proper End(G)-submodule N ⊂ HomR (G, H). 7. (G, H) has the Baer splitting property. 8. (G, H) has the endlich Baer splitting property. Proof: Throughout this proof let E = End(G), fix H ∈ Po (G), and let P = HG (H). Then P is a finitely generated projective right E-module. We will prove 1 ⇒ 2 ⇒ 3 ⇔ 4, 3 ⇒ 1 ⇔ 5 2⇔6⇒7⇒8⇒6 1 ⇒ 2 is clear. 2 ⇒ 3 We prove the contrapositive. Assume there is a maximal right ideal I ⊂ E such that IG = G but ∆P 6⊂ I. Let π : E → E/I be the natural projection map. There is a map f : P → E such that f (P ) 6⊂ I, and because I is a maximal right ideal of E, f (P ) + I = E. Hence πf : P → E/I is a surjection. Furthermore, because IG = G, TG (P/ ker πf ) ∼ = TG (E/I) ∼ = G/IG = 0 so that P is not G-compressed. 3 ⇔ 4 We have observed that P ∗ = Hom(H, G) and it is known that ∆P = ∆P ∗ in general. Assume part 3 is true. Given IG = G then ∆H ⊂ I. Then because ∆H = ∆HG (H) = ∆HG (H)∗ , IHG (H)∗ = HG (H)∗ . This proves part 4. Conversely assume part 4 is true. Let IG = G. Then IHG (H)∗ = IHom(H, G) = Hom(H, G) = HG (H)∗ implies that ∆H = ∆HG (H) = ∆HG (H)∗ ⊂ I. This is part 3. 3 ⇒ 1 We prove the contrapositive. Suppose there is an integer n > 0 and an E-submodule N ⊂ P n such that TG (P (n) /N ) = 0. Because P is finitely generated projective there is a maximal right E-submodule N ⊂ M ⊂ P n , and because P n /M is simple there is a maximal right ideal I ⊂ E such that P n /M ∼ = E/I. Then 0 = TG (P n /M ) = TG (E/I) = G/IG

6.3. G-COMPRESSED PROJECTIVES

149

so that P (n) is not G-compressed. The proof of 5 ⇔ 1 as an exercise. 2 ⇔ 6 Let N ⊂ HG (H) be a right E-submodule and apply the right exact functor TG (·) to the short exact sequence ı

π

0 → N −→ HG (H) −→ HG (H)/N → 0

(6.7)

to form the diagram TG (N )

TG (ı)-

TG HG (H)

- TG (HG (H)/N )

-0

ΘH ?

H of groups. Notice that image ΘH TG (ı) = N G. Because H ∈ Po (G) the Arnold-Lady-Murley Theorem 2.4.1 implies that ΘH is an isomorphism. Then N G = H iff TG (ı) is a surjection iff TG (HG (H)/N ) = 0. Then 2 ⇔ 6 follows immediately. 6 ⇒ 7 follows in the same way that we proved Baer’s Lemma 6.1.1, but in proving 6 ⇒ 7 use an End(G)-submodule N ⊂ HG (G) where we used a right ideal I ⊂ End(G). 7 ⇒ 8 is clear. 8 ⇒ 6 Suppose that N G = H for some proper right End(G)-submodule N ⊂ HG (H). Since HG (H) is finitely generated, there is a maximal right End(G)-submodule K ⊂ HG (H) that contains N . By Lemma 4.2.3 the simple module HG (H)/K is a finite p-group, so that pHG (H) ⊂ K ⊂ HG (H). Since H has finite rank K/pHG (H) ⊂ HG (H)/pHG (H) is finite and thus K is finitely generated by the generators for pHG (H) and representativesP of the cosets for K/pHG (H). Write K = ni=1 πi End(G) for some π1 , . . . , πn ∈ K. There is a short exact sequence 0

0 → H −→

n M

π

Gi −→ H → 0

(6.8)

i=1

where Gk ∼ = G for each i ∈ I and where π is the unique map such that π(x) = πi (x) for each i = 1, . . . , n each x ∈ Gi . Notice that Pand n π is a surjection since H = KG = i ). Furthermore, any i=1 πi (G P n map : H → ⊕ni=1 Gi can be written as = i=1 i for some maps i : H → Gi . Then for any map g : G → H we have ! n ! n n X X X πg = πi i g = πi i g. i=1

i=1

i=1

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CHAPTER 6. BAER SPLITTING PROPERTY

Since πi ∈ K and since i : H → Gi we have πg ∈ K for each g : G → H. Since K 6= HG (H) the reader will show that π 6= 1H for any : H → ⊕ni=1 Gi . Thus (6.8) does not split which proves that part 6 is false. This completes the logical circuit. The next result illustrates the connection between G-compressed modules, faithful groups, and Baer’s lemma. THEOREM 6.3.2 [38, T.G. Faticoni]. for an rtffr group G.

The following are equivalent

1. IG 6= G for each proper right ideal I ⊂ E. 2. IG 6= G for each finitely generated proper right ideal I ⊂ E. 3. TG (M ) 6= 0 for each finitely presented right End(G)-module M 6= 0. 4. TG (M ) 6= 0 for each finitely generated right End(G)-module M 6= 0. 5. G has the Baer splitting property. 6. G has the endlich Baer splitting property. Proof: 6 ⇔ 5 ⇔ 4 is Theorem 6.3.1(8), (7), (5). 4 ⇒ 3 is clear. 3 ⇒ 2 follows as in Theorem 6.3.1(5) ⇒ (4). 2 ⇒ 1 Given part 2 and a right ideal I ⊂ End(G) there is a maximal right ideal I ⊂ M ⊂ End(G). By Lemma 4.2.3, M is finitely presented so that IG ⊂ M G 6= G. This proves part 1. 1 ⇒ 5 is Baer’s Lemma 6.1.1. This completes the logical sequence. In the presence of a flatness hypothesis the above result can be improved a little. THEOREM 6.3.3 [3, U. Albrecht]. Let G be an E-flat rtffr group. Then G has the Baer splitting property iff TG (M ) 6= 0 for each right R-module M 6= 0. Proof: Say G has the Baer splitting property, and let M 6= 0 be a right End(G)-module. Let 0 6= Mo ⊂ M be any finitely generated End(G)-submodule. Since G is E-flat, TG (Mo ) ⊂ TG (M ), and by Theorem 6.3.2, TG (Mo ) 6= 0. Then TG (M ) 6= 0. The converse is Theorem 6.3.2 so the proof is complete.

6.4. SOME EXAMPLES

6.4

151

Some Examples

The examples in this section illustrate the abundance of the splitting properties, and they show that the Baer splitting property and the endlich Baer splitting property are in general different notions. EXAMPLE 6.4.1 1. Let G be rtffr group with commutative endomorphism ring. By Theorem 6.3.2 and Theorem 6.1.4, G has the Baer splitting property, so by Theorem 6.2.3, G(c) has the Baer splitting property for each cardinal c. However, let Io be the ideal of those f ∈ End(G(c) ) such that f (G) has finite rank. Then Io is a proper ideal of End(G(c) ) such that Io G(c) = G(c) . Thus the finite rank hypothesis in Theorem 6.3.2 is necessary. 2. G = Q has the Baer splitting property, and G = Z(p∞ ) does not have the endlich Baer splitting property. 3. D.M. Arnold and E.L. Lady [14] give an example of a completely decomposable abelian group G of rank 2 such that IG = G for some proper pure ideal I ⊂ End(G). 4. The group G constructed in Example 6.1.7 is one for which {right ideals I ⊂ End(G) IG = G} is a finite set. In [4], U. Albrecht proves that the self-small E-flat group G is faithful iff TG (K) 6= 0 for each right End(G)-module K 6= 0. The following example shows us that even if G is a faithful Corner group it may happen that TG (K) = 0 for some nonzero End(G)-module K. EXAMPLE 6.4.2 [36, T.G. Faticoni] There is a Corner group G such that 1. TG (N ) 6= 0 for each finite right End(G)-module N 6= 0. 2. There is a right End(G)-module K 6= 0 such that TG (K) = 0. Specifically G is faithful but TG (K) = 0 for some right E-module K 6= 0. Proof: Let p ∈ Z be a prime and let E be the localization of Z[x] at the maximal ideal (x, p). That is, E = Z[x](x,p) . Let M be the left E-module E E E ⊕ ⊕ ⊕ ··· (x) (x2 ) (x3 )

152

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where (xk ) is the ideal in E generated by xk . One shows that M is a reduced torsion-free left E-module. By Nakayama’s Lemma 1.2.3, J M 6= M where J = J (E). By Theorem 2.3.3 there is a short exact sequence 0 −→ M −→ G −→ QC −→ 0

(6.9)

∼ End(G). Since p ∈ J (E) = constructed in Theorem 2.3.3 such that E = J is the unique maximal ideal in E one proves in the now familiar way that G 6= JG because M 6= JM . Next let K be the left E-module K = (E/pE)bo ⊕ (E/pE)b1 ⊕ (E/pE)b2 ⊕ · · · where bj x = bj−1 for each j ≥ 1 and bo x = 0. An application of K ⊗E · to (6.9) produces the exact sequence K ⊗E M −→ K ⊗E G −→ K ⊗E QC → 0. Since QC is divisible and since K is a torsion group K ⊗E QC = 0. The reader will show that Kx = K. Since each element of M is annihilated by some power of x the usual argument shows us that K ⊗E M = 0. Therefore K ⊗E G = 0. Specifically, G is faithful but TG (K) = 0 for some right End(G)-module K 6= 0. This completes the proof. In the above example G is not self-small. The reader can produce a map f : G → G(ℵo ) whose image is not in any finite direct sum of copies of G. A few remarks are in order before we end this section. U. Albrecht [3, Theorem 2.1, Corollary 2.2] extends Baer’s Lemma by showing that the self-small right R-module G has the finite Baer splitting property iff IG 6= G for each proper right ideal I ⊂ E iff TG (M ) 6= 0 for each nonzero finitely generated right End(G)-module M 6= 0. Furthermore, he shows that if G is a self-small E-flat module, then G has the finite Baer splitting property iff TG (M ) 6= 0 for each right End(G)-module M 6= 0, [3, Corollary 2.3]. The faithful condition is central to the characterization of the rtffr groups G such that Ext1 (G, G) is a torsion-free abelian group, [11, 43]. D.M. Arnold and J. Hausen [12] use the endlich Baer splitting property to study modules G that have the summand intersection property, [41] and [12] use the endlich Baer splitting property to characterize those modules G such that End(G) is right semi-hereditary, and [3], [14], and [41] use the Baer splitting property to characterize those R-modules G such that End(G) is right Noetherian, right hereditary. K. Fuller [47] calls G completely faithful and U. Albrecht [3] calls G fully faithful if TG (M ) 6= 0 for each nonzero right End(G)-module M .

6.5. EXERCISES

6.5

153

Exercises

G, H, K are rtffr groups, p ∈ Z is a prime, E is an rtffr ring, M , N , are rtffr E-modules (left or right, depending on the setting). 1. Prove any of the results in this section left as exercises for the reader. 2. Let H ∈ P(G) and suppose that (G, H) has the endlich Baer splitting property. If H is finitely G-generated then H ∈ Po (G). 3. Refer to the proof of Theorem 6.2.3. Show that each chain in C contains an upper bound in C. 4. Let M be a right End(G)-module. Then M (n) is G-compressed for each integer n > 0 iff TG (N ) 6= 0 for each finitely M -generated right End(G)-module N 6= 0. 5. Let P be a projective right E-module, let P ∗ be the dual of P , and let I be a left ideal of E. Prove that IP = P iff ∆P ⊂ I and that ∆P = ∆P ∗ . 6. Let H ∈ Po (G). Show that Hom(G, H)∗ = Hom(H, G). 7. Prove that the following are equivalent. (a) IG 6= G for each proper right ideal I ⊂ E. (b) TG (M ) 6= 0 for each nonzero finitely generated M ∈ ModEndR (G) . 8. If IG 6= G for each proper finitely generated right ideal I ⊂ End(G) then G has the endlich Baer splitting property. 9. If H ∈ Po (G) and if N ⊂ HG (H) is such that N G = H then TG (HG (H)/N ) = 0. 10. Complete the proof of Theorem 6.2.3 by showing that π 6= 1H for any : H → G(c) . 11. If E is a reduced torsion-free ring of finite rank and if I is a maximal right ideal of E then E/I is finite. 12. Review the proof of Lemma 6.2.1. Show that the map g 00 : πg(H) → H 00 ⊂ X such that g 00 (πg(x)) = g(x) is a well defined injection such that πg 00 = 1πg(H) .

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13. Prove that the sequence 0 → K → X 0 → H 0 → 0 given in Lemma 6.2.1 is exact. 14. If (G, H) has the endlich Baer splitting property then (G, H (n) ) has the endlich Baer splitting property for each integer n > 0.

6.6

Questions for Future Research

Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. Given a ring E let Ω(E) denote the set of rtffr groups G such that E∼ = End(G). Some of the properties of E are seen in Ω(E). 1. Describe the rings E such that each G ∈ Ω(E) is a faithful group. Commutative and hereditary rings E satisfy this property. See [40] for more rings of this type. 2. In [43], it is shown that E is hereditary iff each group G ∈ Ω(E) is E-flat. Classify the rings E such that each group G ∈ Ω(E) is a faithfully E-flat module. 3. The group G has the Baer splitting property if for each cardinal c each surjection π : G(c) −→ G is split. Dualize the Baer splitting property in the obvious way. The group G has the dual-Baer splitting property if for each cardinal c, each injection G −→ Gc is split. See [5]. Characterize those rtffr groups G that satisfy the dualBaer splitting property. Develop a general theory as in Chapter 6. 4. Describe the rtffr rings E such that each G ∈ Ω(E) has the dualBaer splitting property. See [40]. 5. Characterize the G-compressed End(E)-modules. See Theorem 6.3.1. 6. Characterize the (rtffr) groups G such that TG (M ) 6= 0 for each nonzero right End(G)-module M . 7. Describe the rings E such that for each G ∈ Ω(E), TG (M ) 6= 0 for each nonzero right E-module M .

Chapter 7

J -Groups, L-Groups, and S-Groups The rtffr group G is called a J -group if G is isomorphic to each subgroup of finite index in G, G is called an L-group if G is locally isomorphic to each subgroup of finite index in G, and G is an S-group if each subgroup of finite index in G is G-generated. Obviously J -groups ⇒ L-groups ⇒ S-groups. In the first half of the chapter we will discuss how the torsion subgroup of Ext1Z (G, G) is related to (finitely) faithful S-groups. In the second half of the chapter we will scrutinize the converse relationships L-groups ⇒ J -groups, and finitely faithful S-groups ⇒ J -groups.

7.1

Background on Ext

Let G and H be groups. Call G an S-group if each subgroup of finite index in G is G-generated. S-groups are introduced by D.M. Arnold in [11]. Any group of the form Z ⊕ H is an S-group. If R is a pure subring b p for some prime p then R is an S-group. of Z Recall that G is finitely faithful if IG 6= G for each maximal right ideal of finite index in End(G). We investigate finitely faithful S-groups in an effort to find partial solutions to the question under what conditions on G are the groups quasi-isomorphic to G actually isomorphic to G? We will abbreviate notation as follows. Ext1Z (G, H) = Ext(G, H)

155

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CHAPTER 7. J -GROUPS, L-GROUPS, AND S-GROUPS

Then Ext(G, H) consists of the short exact sequences π

0 → H −→ X −→ G → 0

(7.1)

modulo an equivalence relation. See [46] for details. The split short exact sequences (7.1) correspond to 0 in Ext(G, H). The short exact sequence (7.1) is quasi-split if there is a map : G → X and an integer n 6= 0 such that π = n1G . Let tExt(G, H) = the torsion subgroup of Ext(G, H). C.L. Walker’s Theorem [46, Section 102] states that the quasi-split short exact sequences correspond to the torsion part of Ext(G, H). THEOREM 7.1.1 [85, C.L. Walker] The group tExt(G, H) is the set of equivalence classes of quasi-split short exact sequences (7.1). In particular given an equivalence class [E] ∈ tExt(G, H) of a short exact sequence E in (7.1) then n[E] = 0 iff there is a map : G → X such that π = n1G . Consequently Ext(G, H) is torsion-free iff each quasi-split short exact sequence (7.1) is split exact. Given a prime p ∈ Z and a torsion-free group G, G/pG is a Z/pZvector space. The p-rank of G is rp (G) = Z/pZ- dim(G/pG). Since Z/pZ-linearly independent elements in G/pG lift to Z-linearly independent elements in G, rp (G) ≤ rank(G) < ∞. For rtffr groups G and H, Ext(G, H) is a divisible group so there are integers {rp |p ∈ Z is a prime } such that as groups M Ext(G, H) ∼ Z(p∞ )(rp ) . (7.2) = Q(ℵ1 ) ⊕ primes p∈Z The numbers rp are called the p-rank of Ext(G, H). R.B. Warfield, Jr. [87] determines the values of the p-ranks rp of Ext(G, H). THEOREM 7.1.2 [R.B. Warfield, Jr.] (See [87].) Let G and H be torsion-free groups of finite p-rank for all primes p ∈ Z. The p-rank rp of Ext(G, H) is given by rp = rp (G)rp (H) − rp (Hom(G, H))

7.2. FINITE PROJECTIVE PROPERTIES

157

for all primes p ∈ Z. Consequently Ext(G, H) is p-torsion-free iff rp (G)rp (H) = rp (Hom(G, H)).

7.2

Finite Projective Properties

In this section we characterize the torsion-freeness of Ext(G, H) via a projective property. Let G be a group and consider the short exact sequence π

0 → K −→ H −→ H/K → 0.

(7.3)

1. Given a prime p ∈ Z, G is p-finitely H-projective if G is projective relative to (7.3) whenever H/K is a finite p-group. 2. G is finitely H-projective if G is projective relative to (7.3) whenever H/K is a finite group. 3. G is finitely projective if G is finitely G-projective. If p ∈ Z is prime, if pG = G, and if pH 6= H then G is trivially pfinitely H-projective. The reader can show that if Hom(G, H) = 0 then G is not p-finitely H-projective for any prime p ∈ Z such that pG 6= G and pH 6= H. EXAMPLE 7.2.1 Let G ⊂ H ⊂ Q. Then G is p-finitely H-projective. Proof: We may assume without loss of generality that pH 6= H and that G is p-dense in H. Given a map f¯ : G → H/pk H then f¯(pk G) = 0. Write H/pk H = hx + pk Hi. Since G is p-dense in H, H/pk H = (G + pk H)/pk H so we can assume that x ∈ G. Then f¯(x) = nx + pk H and so f¯ lifts to a map f : G → H over the inclusion map G ⊂ H. We demonstrate a relationship between G-projectiveness and quasiisomorphism. We will use without fanfare the fact that the torsion-free group G is (p)-finitely H-projective iff HG (·) is exact on the short exact sequence (7.3) whenever H/K is a finite p-group. The reader will prove that if H/H 0 is a finite group then Ext(G, H) ∼ = Ext(G, H 0 ). The Z-adic completion of G is b = lim G/nG G →n

(7.4)

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CHAPTER 7. J -GROUPS, L-GROUPS, AND S-GROUPS

and the p-adic completion of G is b p = lim G/pk G. G →k>0

b and G is p-pure and p-dense in G bp . The group G is pure and dense in G b b b That is, G ∩ nG = nG and G + nG = G for each integer n 6= 0. We will use the notation HG (·) = Hom(G, ·). THEOREM 7.2.2 [T.G. Faticoni and H.P. Goeters] (See [44].) Let G and H be torsion-free groups and let p ∈ Z be a prime. The following are equivalent. 1. G is (p-)finitely H-projective. 2. Ext(G, H) is (p-)torsion-free. b 3. Hom(G, H) is a (p-)pure and (p-)dense subgroup of Hom(G, H), b p )). (of Hom(G, H Proof: 1 ⇔ 2 An application of HG (·) to the short exact sequence p

π

0 → H −→ H −→ H/pH → 0 yields the exact sequence π∗

p

HG (H) −→ HG (H/pH) → Ext(G, H) −→ Ext(G, H) in which π ∗ = HG (π). Then Ext(G, H) is p-torsion-free iff multiplication by p is an injection iff π ∗ is a surjection iff G is p-finitely H-projective. b is pure-injective Ext(G, H) b = 2 ⇔ 3. Since G is torsion-free and H 0. Thus an application of HG (·) to the short exact sequence π b −→ b 0 → H −→ H H/H →0

yields the exact sequence ∗

∗

π b → b b = 0. HG (H) → HG (H) HG (H/H) → Ext(G, H) → Ext(G, H)

Then Ext(G, H) is p-torsion-free iff image π ∗ is p-divisible iff ker π ∗ = b image ∗ is p-pure and p-dense in HG (H). This completes the logical cycle.

7.3. FINITELY PROJECTIVE GROUPS

159

EXAMPLE 7.2.3 [87, R.B. Warfield, Jr.] Let G ⊂ H ⊂ Q be groups. We showed in Example 7.2.1 that G is finitely H-projective. Consequently Ext(G, H) is torsion-free.

7.3

Finitely Projective Groups

The group G is finitely faithful if IG 6= G for each right ideal I of finite index in End(G). Finitely faithful S-groups were introduced by D.M. Arnold [11]. We will continue to use the terminology finitely faithful Sgroup instead of faithful S-group since finitely faithful S-group seems to capture the spirit of our applications. Rank one groups are finitely faithful S-groups. Of course free groups of finite rank are finitely faithful S-groups, but G = Z(ℵo ) is an S-group without being faithful. If G is a finitely faithful S-group then G(n) is a finitely faithful S-group. (This follows from part 1 of Theorem 7.3.1 below.) We will approach the problem of classifying the finitely faithful Sgroups by considering the finitely projective groups. We begin by showing that the finitely faithful S-groups have finite p-rank for all primes p ∈ Z. Let π

0 → H −→ G −→ G/H → 0

(7.5)

be a short exact sequence in which G/H is finite. THEOREM 7.3.1 Let G be a reduced torsion-free group and let p ∈ Z be a prime. The following are equivalent. 1. G/pG is finite and G is projective relative to each short exact sequence (7.5) in which pG ⊂ H ⊂ G. 2. (a) If I is a right ideal of End(G) that contains some power of p then IG 6= G. (b) If H is a subgroup of G such that pG ⊂ H ⊂ G then H is G-generated. Proof: Let E = End(G). 1 ⇒ 2 Assume part 1, and let pE ⊂ I ⊂ E be a right ideal such that IG = G. By part 1, E/pE ∼ = HG (G/pG) ∼ = Hom(G/pG, G/pG)

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so that E/pE is a simple Artinian ring. Let I be a right ideal of E such that pE ⊂ I ⊂ E and IG = G. There is no loss of generality in assuming that I is maximal in E and thus that pE ⊂ I. Since E/pE is a simple Artinian ring there is an e2 = e ∈ E/pE such that e(E/pE) = I/pE. Then G/pG = IG/pG = (I/pE)(G/pG) = e(E/pE)(G/pG) = e(G/pG). Inasmuch as x(G/pG) 6= 0 for each nonzero x ∈ E/pE and since (1 − e)(G/pG) = 0 we conclude that e = 1. Thus I = E, as required by part (a). Let pG ⊂ H ⊂ G. For each x ∈ H \pG there is a map fx : G → H/pG such that x + pG ∈ fx (G). By hypothesis in part 1, fx lifts to a map gx : G → H such that x ∈ gx (G) + pG. Then H is G-generated, which proves part (b). This completes the proof of 1 ⇒ 2. 2 ⇒ 1. We first show that G/pG is finite. Let I ⊂ E be the ideal such that I/pE = {f ∈ E/pE Z/pZ- dim(f (G/pG)) is finite }. Given a subgroup pG ⊂ H ⊂ G such that H/pG ∼ = Z/pZ there is by part (b) a surjection fH : G → H/pG. Notice that fH ∈ I/pE and also that G/pG = (I/pE)(G/pG) = IG/pG. By part (a), I = E. Thus 1G has finite image in G/pG and hence G/pG is finite. Apply HG (·) to the short exact sequence (7.5). To show that π ∗ = HG (π) is a surjection we will prove the next lemma. LEMMA 7.3.2 Let G be a torsion-free group that satisfies Theorem 7.3.1(2). Then HG (Z/pZ) = Hom(G, Z/pZ) is a simple right E-module. Proof: For any nonzero map f : G → Z/pZ, ker f is a maximal subgroup of G that contains pG. Then pE ⊂ annE (f ) = HG (ker f ) ⊂ E. We claim that annE (f ) is a maximal right ideal of E. Suppose that annE (f ) ⊂ I ⊂ E and that annE (f ) 6= I. Then by part 2(b) we have pG ⊂ ker f = HG (ker f )G 6= IG ⊂ G.

7.3. FINITELY PROJECTIVE GROUPS

161

By the maximality of ker f in G we have IG = G and hence part 2(a) implies that I = E. As claimed for each f ∈ HG (Z/pZ), annE (f ) is a maximal right ideal in E. Now assume to the contrary that there are f 6= 0 6= g ∈ HG (Z/pZ) such that f E 6= gE are distinct simple right E-modules. Then f E ⊕ gE ⊂ HG (Z/pZ). By the above paragraph annE (f + g) is a maximal right ideal in E, whence annE (f ⊕ g) is a maximal right ideal of E such that annE (f ⊕ g) = annE (f ) ∩ annE (g) ⊂ annE (f ). Thus annE (f ⊕ g) = annE (f ) = annE (g). Since annE (f ⊕ g) = HG (ker(f ⊕ g)) and since (f ⊕ g)G is finite, part 2(b) implies that ker(f ⊕ g) is Ggenerated, so that ker(f ⊕ g) = HG (ker(f ⊕ g))G = annE (f ⊕ g)G = annE (f )G = ker f as abelian groups. Similarly ker(f ⊕ g) = ker g so g = uf for some automorphism u of Z/pZ. Since u is then multiplication by some integer, f E = gE. This contradiction to our choice of f E 6= gE shows us that HG (Z/pZ) = f E is a simple right E-module. This completes the proof of the Lemma. To continue the proof of Theorem 7.3.1: Choose a subgroup pG ⊂ H ⊂ G such that H/pG ∼ = Z/pZ. Consider the short exact sequence (7.5). By part 2(b) and because H/pG is simple there is a homomorphism f : G → H such that f (G) + pG = H. Then 0 6= πf ∈ HG (H/pG) ∼ = HG (Z/pZ) so by Lemma 7.3.2, πf E = HG (Z/pZ). Therefore image π ∗ = πHG (G) ⊃ HG (H/pG) for each such H. Inasmuch as G/pG is a direct sum of copies of Z/pZ, image π ∗ = HG (G/pG). That is, π ∗ is a surjection which proves part 1 and completes the proof.

162

7.4

CHAPTER 7. J -GROUPS, L-GROUPS, AND S-GROUPS

Finitely Faithful S-Groups

Using Theorem 7.3.1 we can characterize the finitely projective groups as being finitely faithful S-groups. THEOREM 7.4.1 [D.M. Arnold] (See [11].) The following are equivalent for a torsion-free group G for which rp (G) is finite for all primes p ∈ Z. 1. G is finitely projective. 2. Ext(G, G) is torsion-free. 3. G is a finitely faithful S-group. 4. rp (G)2 = rp (End(G)) for each prime p ∈ Z. 5. End(G)/pEnd(G) ∼ = Matrp (G) (Z/pZ) as rings for each prime p ∈ Z. 6. Each quasi-split short exact sequence 0 → G → X → G → 0 is split exact. Proof: 2 ⇔ 3 follow from Theorem 7.2.2 and 2 ⇔ 6 follows from Theorem 7.1.1. 1 ⇒ 5 Suppose that G is finitely projective. An application of HG (·) to the short exact sequence p

π

0 → G −→ G −→ G/pG → 0 yields the isomorphisms End(G)/pEnd(G) = image π ∗ = HG (G/pG) ∼ = Hom(G/pG, G/pG). Since G is rtffr Hom(G/pG, G/pG) ∼ = Matrp (G) (Zp /pZ) which proves 5. 5 ⇒ 4 By part 5 End(G)/pEnd(G) ∼ = Matrp (G) (Zp /pZ) ∼ = Hom(G/pG, G/pG) as rings. Then rp (End(G)) = Z/pZ- dim(Hom(G/pG, G/pG)) = rp (G)2 .

7.4. FINITELY FAITHFUL S-GROUPS

163

This proves part 4. 4 ⇒ 2 follows from Theorem 7.1.2. 2 ⇒ 1 follows from Theorem 7.2.2, which completes the logical cycle.

For instance, if G is an rtffr group then rp (G) is finite for all primes p ∈ Z. COROLLARY 7.4.2 Let G be an rtffr group. If G is a finitely faithful S-group then 1. G(k) is a finitely faithful S-group for each integer k > 0. 2. If G = H ⊕ H 0 then H is a finitely faithful S-group. . 3. If G = H then H is a finitely faithful S-group. Proof: 1. Since G is a finitely faithful S-group, Ext(G, G) is torsionfree (Theorem 7.4.1), which implies that Ext(G(k) , G(k) ) is torsion-free for each k > 0. Theorem 7.4.1 implies that G(k) is a finitely faithful S-group. 2. If G = H ⊕ H 0 then Ext(H, H) is a direct summand of the torsionfree group Ext(G, G). Now apply Theorem 7.4.1. . 3. If G = H then by Theorem 7.4.1, Ext(G, G) ∼ = Ext(H, H) are torsion-free groups. This completes the proof. COROLLARY 7.4.3 If G is a finitely faithful S-group then End(G) is a semi-prime Noetherian ring. Proof: For each prime p ∈ Z, End(G)/pEnd(G) is a simple ring, so N (End(G)) ⊂ pEnd(G) for all p. Thus pN (End(G)) = N (End(G)) is divisible. Since G is reduced N (End(G)) = 0. This completes the proof. EXAMPLE 7.4.4 This is an example of an S-group that is not faithful. Proof: Let 0 6= X ⊂ Q be a subgroup such that End(X) = Z, Hom(X, Z) = 0, and let G= Z ⊕ X. Clearly G is an S-group. Observe Z 0 that End(G) = is not semi-prime. Observe that for each Z Z Z 0 integer n, In = is a right ideal of End(G) such that In G = G. Z nZ Thus G is not faithful. In the above example Ext(G, G) is not torsion-free. The next example gives a nonsplit quasi-split short exact sequence.

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EXAMPLE 7.4.5 Let Z ⊂ X ⊂ Q be such that Hom(X, Z). Then G = Z ⊕ X is an S-group that is not faithful and so Ext(X, Z) has nonzero torsion subgroup. Thus there is a quasi-split, nonsplit extension 0 → Z → H → X → 0 of groups. The embedding sends Z to Z(p ⊕ 1) ∈ Z ⊕ X. Notice that Z(p ⊕ 1) is a pure subgroup, a quasi-summand, but not a direct summand of Z ⊕ X so that 0 → Z. → H → X → 0 is not split exact. Compare this to the fact that if G ∼ = H then H ∼ = Z ⊕ X. bp. An example of torsion-free Ext is found in Z b p be pure subgroups of Z b p . InasEXAMPLE 7.4.6 Let G ⊂ H ⊂ Z bp → Z b p we see much as each map f : G → H lifts to a unique map f : Z that b p = Hom(G, b H). b Z ⊂ Hom(G, H) ⊂ Z b p , Hom(G, H) is pure and Moreover since H is torsion-free and pure in Z b H). b Then by Theorem 7.2.2, Ext(G, H) is torsion-free. dense in Hom(G, The examples of S-groups so far either have nonzero nilradical or they have commutative endomorphism ring. For finitely faithful S-groups our examples thus far satisfy End(G)/pEnd(G) ∼ = Z/pZ. Note the next example. EXAMPLE 7.4.7 [11, D.M. Arnold] This is an example of a finitely faithful S-group such that rp (G) = 2 and End(G)/pEnd(G) ∼ = Mat2 (Z/pZ). Let E be a maximal Z-order in the Hamiltonian Quaternions over Q. If p 6= 2 ∈ Z is a prime then E/pE ∼ = Mat2×2 (Z/pZ) and the completion ∼ b b b Ep satisfies E = Mat2×2 (Z). (See [69].) Let M be a pure and dense rtffr E-submodule such that ! b 0 Z 0 Z ⊂M ⊂ b 0 Z 0 Z and using Theorem 2.3.4 construct a short exact sequence of left Emodules such that 0 → M −→ G −→ QE ⊕ QE → 0. (The reader can show that E = O(M ).) Since rp (E) = 4 = rp (G)2 , G is a finitely faithful S-group with noncommutative endomorphism ring E.

7.5. ISOMORPHISM VERSUS LOCAL ISOMORPHISM

7.5

165

Isomorphism versus Local Isomorphism

In this section we engage in a little alphabet soup. Let G be an rtffr group. 1. G is a J -group if G is isomorphic to each subgroup of finite index in G. J is for the author of J´ onsson’s Theorem 2.1.10. 2. G is an L-group if G is locally isomorphic to each subgroup of finite index in G. L is for E.L. Lady who first used local (= near) isomorphism on abelian groups. See [56]. 3. G is an S-group if each subgoup of finite index in G is G-generated. S is for G-socle, SG (H). See Section 6.1. Evidently G is an S-group if G is an L-group if G is a J -group. We will investigate the implications S-group ⇒ L-group ⇒ J -group. Examples of J -groups include the following. 1. Groups of the form Z⊕H, with H ⊂ Q, are J -groups. See Theorem 2.1.12. 2. Groups of the form Z ⊕ K, with K 6= 0, are S-groups. The reader can show that Z ⊕ K is a J -group if K is a J -group. 3. If X ⊂ Q then G = X (n) is a J -group for each integer n > 0. It is interesting to note that we did not give an example of an L-group that is not a J -group. Call G a Murley group if p-rank(G) ≤ 1 for each prime p ∈ Z. We will show that certain classes of J -groups are Murley groups. THEOREM 7.5.1 [H.P. Goeters] The following are equivalent for the rtffr group G. 1. G is a Murley group. 2. G is a J -group and End(G) is commutative. 3. G is an S-group and End(G) is commutative. Proof: 1 ⇒ 2 Let G be a Murley group and let H be a subgroup of finite index in G. There is a chain of subgroups of G such that H ⊂ H1 ⊂ · · · ⊂ Hk ⊂ G

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∼ Z/pZ. Because G/pG ∼ and such that Hi /Hi−1 = = Hk /pHk for quasiisomorphic groups, the groups Hi are Murley groups. But then pH1 ⊂ H ⊂ H1 and p-rank(H1 ) ≤ 1 so by induction H = pH1 ∼ = ··· ∼ = Hk ∼ = G. = H2 ∼ = H1 ∼ Hence G is a J -group. Since p-rank(G) ≤ 1 for each prime p ∈ Z, the isomorphisms G/pG ∼ = b Since Z b is pure injective each b ⊂ Z. Z/pZ lift to a pure imbedding G b → Z. b That is, End(G) imbeds in f : G → G lifts to a unique map fˆ : Z b the commutative ring Z. This proves part 2. 2 ⇒ 3 is clear. 3 ⇒ 1 Suppose that G is an S-group and that End(G) is commutative. Then End(G)/pEnd(G) is a commutative ring for each prime p ∈ Z. By Theorem 4.2.4 and part 3, G is a finitely faithful S-group so by Theorem 7.4.1 End(G)/pEnd(G) ∼ = Matrp (G)×rp (G) (Z/pZ) as rings for all primes p ∈ Z. Consequently rp (G) ≤ 1 for all primes p so that G is a Murley group. This completes the proof. EXAMPLE 7.5.2 This is an example of a strongly indecomposable J -group that is not a Murley group. This example is inspired by D.M. Arnold’s example in [11]. Choose a prime p 6= 2 ∈ Z and let E be a classical maximal Zp -order in the Hamiltonian Quaternions over Q. Then each right ideal I ⊂ E is principal, I = xE for some x ∈ E. bp satisfies E bp ∼ Since Ep /pEp ∼ = Mat2×2 (Z/pZ), the completion E = b Mat2×2 (Zp ). As in Example 7.4.7 construct a pure and dense rtffr Esubmodule M such that ! bp 0 Zp 0 Z ⊂M ⊂ bp 0 Zp 0 Z and using Theorem 2.3.4 construct a short exact sequence of left Emodules such that 0 → M −→ G −→ QE ⊕ QE → 0. (The reader can show that E = O(M ).) Since rp (E) = 4 = rp (G)2 , G is a finitely faithful S-group but not a Murley group. . In fact given H = G, there is a right ideal I ⊂ E such that H = IG. Since I ∼ = E, H ∼ = G, so that G is an indecomposable J -group.

7.6. ANALYTIC NUMBER THEORY

7.6

167

Analytic Number Theory

The next series of results will demonstrate a beautiful connection between J -groups, L-groups, and S-groups. However, they require some deep results on Analytic Number Theory whose proofs would take us too far afield. Thus we will state some results and reference their proofs. The following theorem gives a rather general condition under which a finitely faithful S-group is a J -group. Let i, j, k be such that −1 = i2 = j 2 = k 2 = ijk and let k be an algebraic number field. The algebra D = k1 ⊕ ki ⊕ kj ⊕ kk is called a totally definite quaternion algebra if each embedding of k into the complex numbers is an embedding of k into the reals. Observe that k = center(D), D is a division k-algebra, and D has dimension 4 over its center k. If we let k = Q then D is the Q-algebra of Hamiltonian Quaternions. Thus the classical Q-algebra of Hamiltonian Quaternions is a totally definite quaternion algebra. Let G be an rtffr group and write QE(G) = A1 × · · · × At for some nonzero simple Artinian Q-algebras A1 , . . . , At . We say that the rtffr group G is an Eichler group if Ai is not a totally definite quaterion algebra for each simple factor Ai of QE(G). Note that Eichler groups are rtffr groups. A proof of the next result is found in [44, Proposition III.6]. THEOREM 7.6.1 If the Eichler group G is a finitely faithful S-group then G is a J -group. The next result shows just how rare an exception the totally definite quaternion algebra is. COROLLARY 7.6.2 Write the rtffr group G as .

(n1 )

G∼ = G1

(nt )

⊕ · · · ⊕ Gt

(7.6)

for some strongly indecomposable groups G1 , . . . , Gt and integers t, n1 , . ∼ . . ., nt > 0 such that Gi = Gj ⇒ i = j. Suppose that G is a finitely faithful S-group. Then G is a J -group in either of the following cases. 1. ni 6= 1 for each i = 1, . . . , t. 2. rank(Gi ) is not divisible by 4 for each i = 1, . . . , t.

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Proof: Since we are assuming that G is a finitely faithful S-group, (n ) it suffices to show that for each i, QE(Gi i ) is not a totally definite quaternion algebra. (n ) 1. In case ni 6= 1 then QE(Gi i ) is a nontrivial ring of matrices so (n ) it is not a domain. Thus QE(Gi i ) is not a totally definite quaternion algebra. 2. By part 1 we can assume without loss of generality that ni = 1. Since rank(Gi ) is not divisible by 4, and since QE(Gi ) is a division ring, rank(E(Gi )) 6= 4. (See Lemma 4.2.1.) Then QE(Gi ) is not a totally definite quaternion algebra. The conditions ni 6= 1 and rank(Gi ) not divisible by 4 are satisfied in case 1. G = H (2) for some group H, or 2. For each i = 1, . . . , t, (ni )

QE(Gi

) = Matni (ki )

for some algebraic number field ki . Before we continue with our study of L-groups we will introduce a bit of power. THEOREM 7.6.3 [R.B. Warfield, Jr.] (See [88].) Let G and H be rtffr groups. Then G and H are locally isomorphic iff G(n) ∼ = H (n) for some integer n > 0. We need a pair of lemmas that are themselves interesting. The semiprime ring E is integrally closed if given a ring E ⊂ R ⊂ QE such that R/E is finite then E = R. Equivalently E is integrally closed iff E is a finite product of classical maximal orders. The following is a cancellation result for rtffr groups such that A(G) is integrally closed. D.M. Arnold [10] has proved a similar result. THEOREM 7.6.4 [T.G. Faticoni] Let G be an Eichler group such that E(G) is an integrally closed ring. If H is an rtffr group then G⊕G∼ = G ⊕ H =⇒ G ∼ = H. . . Proof: By J´onsson’s Theorem 2.1.10, G ∼ = H, so that E(G) = A(H). By Theorem 2.4.4 we have

E(G) ⊕ E(G) ∼ = E(G) ⊕ A(H)

7.6. ANALYTIC NUMBER THEORY

169

as right modules over the integrally closed ring E(G). There are classical maximal orders E i such that E(G) = E 1 × · · · × E t . Since E(G) ∼ = A(H)E i for each i = 1, . . . , t we have = A(H) iff E i ∼ reduced to the case where E(G) is a classical maximal order. So assume that E(G) is a classical maximal order with center S. Given a right ideal I ⊂ E(G) let nr(I) be the reduced norm of I in S. (See [69, page 214].) From E(G) ⊕ E(G) ∼ = E(G) ⊕ A(H) and [69, page 311, Corollary 35.11(ii)] we see that nr(E(G)) = nr(E(G)) · nr(E(G)) = nr(E(G)) · nr(A(H)) = nr(A(H)). Inasmuch as G is Eichler, [69, page 311, Corollary 35.11(ii)] implies that E(G) ∼ = A(H), so that by Theorem 2.4.4, G ∼ = H. COROLLARY 7.6.5 Let G be an L-group. Then E(G) is an integrally closed ring. . Proof: Let E = E(G). By Lemma 3.1.8 there is a group G = G such that E = E(G) is an integrally closed ring. Because G is an L-group G is (n) locally isomorphic to G and then Theorem 7.6.3 states that G(n) ∼ =G for some integer n > 0. Then Matn (End(G)) ∼ = End(G(n) ) ∼ = End(G

(n)

)∼ = Matn (End(G))

so that Matn (E) ∼ = Matn (E) is an integrally closed ring. Since maximal order is a Morita invariant property, [69], integrally closed is a Morita invariant property, so that E is integrally closed. This completes the proof. LEMMA 7.6.6 If G is an rtffr L-group and if G is locally isomorphic to H ⊕ K then H is an L-group. Proof: Let G be locally isomorphic to H ⊕ K and suppose that . H0 ∼ = H. Then G, H ⊕ K, and H 0 ⊕ K are quasi-isomorphic. Since G is assumed to be an L-group H ⊕ K and H 0 ⊕ K are locally isomorphic groups. Then by Theorem 2.5.7, H is locally isomorphic to H 0 , whence H is an L-group. This completes the proof.

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170

LEMMA 7.6.7 Let G be an rtffr L-group. There is a set {G1 , . . ., Gt } of strongly indecomposable groups and integers t, n1 , . . . , nt > 0 such that .

1. Gi ∼ = Gj ⇒ i = j, 2. G is locally isomorphic to (nt )

G0 = G t

(nt )

⊕ · · · ⊕ Gt

,

3. for each i = 1, . . . , t, Gi is an L-group. Proof: Suppose that G is an L-group. 1 and 2. By J´onsson’s Theorem 2.1.10 there are strongly indecomposable groups {G1 , . . . , Gt } and integers n1 , . . . , nt > 0 such that . Gi ∼ = Gj ⇒ i = j and such that G is quasi-isomorphic to G0 . Since G is an L-group G is locally isomorphic to G0 . Part 3 follows from Lemma 7.6.6, so the proof is complete. COROLLARY 7.6.8 Let G be an rtffr L-group. Then G = G 1 ⊕ · · · ⊕ Gr for some strongly indecomposable L-groups G1 , . . . , Gr . Proof : Apply Arnold’s Theorem 3.5.1(2) to the above lemma. LEMMA 7.6.9 Let G be an indecomposable rtffr L-group. Then End(G) is a classical maximal order in the division ring QEnd(G). Proof: Let E = End(G) and let N = N (E). By Lemma 7.6.7 an indecomposable L-group is strongly indecomposable so that QEnd(G) is a local ring. That is, each x ∈ QE \ QN is a unit in QE, Lemma 1.3.3. Since G is reduced there is an integer n 6= 0 such that ∩k>0 nk G = 0. Let I = nE + N . Then G and IG are locally isomorphic so that by Theorem 2.5.1, HG (G) = E and HG (IG) are locally isomorphic right Emodules. Lemma 2.5.4 states that the localizations En and HG (IG)n are isomorphic, so HG (IG)n = xEn for some x ∈ HG (IG)n . Since x 6∈ Nn is a nonzero-divisor in En , Nn ⊂ HG (IG)n = xEn implies that Nn = xNn = xk Nn for each integer k > 0. Moroeover there is an integer k > 0 such that N k = 0. Then (xEn )k ⊂ (nEn + Nn )k ⊂ nk En + nk−1 Nn + · · · + nNnk−1 ⊂ nEn

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171

hence Nn = xk Nn ⊂ (xE)k Nn ⊂ nNn ⊂ Nn whence nN = N . Subsequently N = ∩k nk G = 0, so that N = 0. Then by Corollary 7.6.5, E/N (E) = E is integrally closed. Inasmuch as the semi-prime local ring QE is a division ring E is a classical maximal order in a division ring. This completes the proof. We can use some of the above results to prove that an rtffr L-group is almost faithfully E-flat. LEMMA 7.6.10 Let G be an rtffr L-group and let E = End(G). If E is semi-prime then G is a faithful E-flat group. Proof: Suppose that E is semi-prime and let I be a right ideal of E. Apply I ⊗E · to the exact sequence 0 → G −→ QG −→ QG/G → 0 to produce the following exact sequence of groups. Tor1E (I, QG/G) −→ I ⊗E G −→ I ⊗E QG. By Corollary 7.6.5, E is integrally closed, hence hereditary, whence I is a projective (= flat) right E-module. Consequently Tor1E (I, ·) = 0 and so I ⊗E G −→ I ⊗E QG is an injection. Since QG is a direct summand of QE (n) as a left E-module and since QE is a flat E-module QG is a flat left E-module. Thus I ⊗E QG → E ⊗E QG ∼ = QG is an injection and so I ⊗E G → G is an injection. Then G is E-flat. Next suppose that I is a maximal right ideal of finite index in E such that IG = G. Since I is a finitely generated projective right EndR (G)module the Arnold-Lady-Murley Theorem 2.4.1 implies that I ∼ = E is principal, say I = xE for some x ∈ E. Then xG = xEG = IG = G. Since G has finite rank x is an automorphism of G. That is I = E so that G is faithful. This completes the proof.

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172

7.7

Eichler L-Groups Are J -Groups

To this point there have been three classes of groups that we have considered. We showed that the finitely faithful S-groups are J -groups except in one case, and we know that J -groups are L-groups. The main Theorems 7.7.2 and 7.7.3 will show us that Eichler L-groups are J -groups. LEMMA 7.7.1 Let G be an rtffr L-group. If G = H ⊕ K then H is an L-group. .

Proof: Suppose that G is an L-group and let H 0 ∼ = H. Then .

H0 ⊕ K ∼ =H ⊕K =G implies that H 0 ⊕ K, H ⊕ K, and G are locally isomorphic since G is an L-group. But then Theorem 2.5.7 implies that H 0 and H are locally isomorphic. Thus H is an L-group. Note the use of Analytic Number Theory in the proof of the next two theorems. THEOREM 7.7.2 [T.G. Faticoni] The following are equivalent for an indecomposable Eichler group G. 1. G is a J -group. 2. G is an L-group. 3. G is a finitely faithful S-group. Proof: We have commented on the implication 1 ⇒ 2. 2 ⇒ 3 Of course G is an S-group if G is a L-group. By Lemma 7.6.9, End(G) is a semi-prime ring so Lemma 7.6.10 implies that the indecomposable L-group is faithful. This proves part 3. 3 ⇒ 1 If G is a finitely faithful S-group then by Corollary 7.4.2, G(2) is a finitely faithful S-group, so that G(2) is a J -group by Corollary 7.6.2. By Lemma 7.7.1, G is then an L-group. . To show that G is a J -group suppose that G = H. Then H is locally isomorphic to G so that G ⊕ H ∼ = G ⊕ G (Lemma 2.5.6). Because G is indecomposable, End(G) is a classical maximal order in the division Qalgebra QEnd(G), Lemma 7.6.9. Then by Theorem 7.6.4, G ∼ = H. Hence G is a J -group and part 1 is proved. This completes the proof. THEOREM 7.7.3 [T.G. Faticoni] Eichler L-groups are J -groups.

7.7. EICHLER L-GROUPS ARE J -GROUPS

173

. Proof: Let G be an Eichler L-group and let H = G. Then H is locally isomorphic to G. By Corollary 7.6.8 G = G 1 ⊕ · · · ⊕ Gr for some strongly indecomposable L-groups G1 , . . . , Gr . The reader will prove as an exercise that each Gi is an Eichler group. By Arnold’s Theorem 3.5.1(2) H = H1 ⊕ · · · ⊕ H r where H1 , . . . , Hr are indecomposable groups such that Gi and Hi are . locally isomorphic. In particular, Gi ∼ = Hi . By Theorem 7.7.2, the indecomposable Eichler L-groups Gi are J -groups. Hence Gi ∼ = Hi for ∼ each i = 1, . . . , r, whence G = H, and therefore G is a J -group. COROLLARY 7.7.4 The Eichler group G is a J -group iff it is an L-group. There is one other interesting case where our groups overlap. THEOREM 7.7.5 The following are equivalent for an Eichler group G. 1. G is a finitely faithful S-group. 2. End(G) is semi-prime and G is a J -group. Proof: Assume part 1. Let G be a finitely faithful S-group. By Corollary 7.4.2, G(2) is a finitely faithful S-group, so by Theorem 7.6.1, G(2) is a J -group. By Lemma 7.7.1, G is then an L-group, so by Theorem 7.7.3, G is a J -group. Lemma 7.4.3 states that End(G) is a semi-prime ring which proves part 2. Conversely, assume that G is a J -group with semi-prime endomorphism ring. By Lemma 7.6.10, G is faithful. Since G is clearly an S-group we have proved part 1. EXAMPLE 7.7.6 An example due to Eichler ([69, midpage 305]) produces a totally definite quaternion algebra D with center k an algebraic number field, a maximal O-order E in D, where O is the ring of algebraic integers in k, and a nonprincipal right ideal E ⊂ L ⊂ D such that nr(L) = αO for some α ∈ k. Then L and E have the same reduced norm, but L ∼ 6 E. Compare this to the last lines of the proof of Theorem 7.6.4. = This explains the necessity of the Eichler hypothesis in Theorem 7.6.4.

174

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Exercises

G, H, K are rtffr groups, p ∈ Z is a prime, E is an rtffr ring, M , N , are rtffr E-modules (left or right, depending on the setting). 1. G is a J -group if G is a pure subgroup of Z. .

π 2. Let 0 → K → X −→ G → 0 be quasi-split. Then X ∼ = G ⊕ K.

3. Let G be a faithful rtffr group. Show that if I ⊂ End(G) is a maximal right ideal then IG 6= G iff I = Hom(G, IG). 4. Suppose that H/H 0 is finite. Then G is finitely H-projective iff G is finitely H 0 -projective. 5. If H/H 0 is a finitely group then the inclusion H 0 ⊂ H induces a canonical isomorphism Ext(G, H) ∼ = Ext(G, H 0 ) of groups. 6. Let G be a torsion-free group such that rp (G) is countably infinite. If G is p-finitely G-projective then End(G) is uncountable. 7. Let G and H be torsion-free groups and let p ∈ Z be a prime. If G is (p-)finitely H-projective then for each integer k > 0 (a) G(k) is (p-)finitely H-projective. (b) G is (p-)finitely H (k) -projective. 8. Let G and H be rtffr groups. If G is finitely H-projective then π each short exact sequence 0 → K −→ H −→ G → 0 is split exact. 9. Let G be an rtffr group and suppose that G/H is finite. Then G is a finitely faithful S-group iff H is a finitely faithful S-group. 10. Let G be an rtffr group. Then G is a finitely faithful S-group iff G is a finitely faithful L-group. 11. Let G and H be rtffr groups. Assume that H 0 has finite index in H. (a) Show that G is p-finitely H-projective iff G is p-finitely H 0 projective. (b) Show that G is finitely H-projective iff G is finitely H 0 -projective. 12. Let G and H be torsion-free groups and let p ∈ Z be a prime. If G is (p-)finitely H-projective then for each integer k > 0

7.9. QUESTIONS FOR FUTURE RESEARCH

175

(a) G(k) is (p-)finitely H-projective. (b) G is (p-)finitely H (k) -projective. b then G is a J -group. 13. [C. Murley] If G is a pure subgroup of Z 14. Show that an indecomposable L-group is strongly indecomposable. 15. Show that an indecomposable L-group is a finitely faithful S-group.

7.9

Questions for Future Research

Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. Dualize p-finitely H-projective groups and study the resulting class of groups. 2. If G is p-finitely projective then Hom(G, Z/pZ) is a simple End(G)module. Study the modules Hom(G, Z/pk Z). Study the module Hom(G, Q). Study other interesting classes of End(G)-modules in terms of Hom(G, ·). 3. Study the category of G-plexes when G is a p-finitely projective group. See [32] for relevant definitions. 4. Characterize group theoretically rtffr J -groups, N -groups, and especially S-groups. 5. Delete as much of the Analytic Number Theory used in our work on L-groups as is possible. 6. Describe the rtffr groups G such that G ⊕ G ∼ = G ⊕ H =⇒ G ∼ = H. 7. Characterize the endomorphism rings of rtffr J -groups.

Chapter 8

Gabriel Filters The material in this chapter comes from [40] where the filter of divisibility D(G) = {right ideals I ⊂ End(G) IG = G} is studied. This filter is nicely structured. We are most interested in the subfilters of D(M ) called Gabriel filters that correspond to a hereditary torsion class of right E-modules K such that K ⊗E G = 0. In particular, we will show that D(G) is in many interesting cases bounded by an idempotent ideal ∆. That is, there is an ideal ∆ of End(G) such that ∆2 = ∆ and such that IG = G implies ∆ ⊂ I.

8.1

Filters of Divisibility

Let E be an rtffr ring and let G be an rtffr group. Our work on the Baer Splitting Property in section 6.1 motivates us to investigate the set

DE (G) = D(G) = {right ideals I ⊂ End(G) IG = G}.

We call D(G) the filter of divisibility of G. Then G is a faithful right E-module iff D(G) = {End(G)}. We will consider faithful modules to be uninteresting in this section as they ignore any interesting structure there might be in D(G). Let the first example act as an intuition builder.

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EXAMPLE 8.1.1 Let X ⊂ Y be groups such that Hom(Y, X) = 0 and SX (Y ) = Y . Let G = X ⊕ Y , and let ∆ = Hom(X, G)Hom(G, X) ⊂ End(G). (For example, we could choose X = Z and nn o Y = n, m ∈ Z and m is square-free .) m Because Hom(Y, X) = 0, ∆ is a direct summand of End(G) such that ∆2 = Hom(X, G)(Hom(G, X)Hom(X, G))Hom(G, X) = Hom(X, G)Hom(X, X)Hom(G, X) = Hom(X, G)Hom(G, X) = ∆. Furthermore, because Hom(Y, X) = 0 and Y = SX (Y ) we have ∆G = Hom(X, G)Hom(G, X)(X ⊕ Y ) = Hom(X, G)Hom(X, X)X = Hom(X, G)X = SX (X) ⊕ SX (Y ) = X ⊕Y = G. Thus D(G) contains the infinite set {∆ + mEnd(G) m ∈ Z}. The reader will show that ∆ is the smallest right ideal in D(G). We will let E = End(G) = {q ∈ QEnd(G) qG ⊂ G}. Theorem 2.3.4 can be used to construct a short exact sequence π 0 −→ G −→ H −→ QE ⊕ QE −→ 0

(8.1)

in which E = End(H). Since IG = G and IQE = QE for each right ideal of the form I = ∆ + mE for some m 6= 0 ∈ Z, tensoring (8.1) with E/I shows that H = 0. IH

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179

Hence IH = H for each m ∈ Z. Thus there are infinitely many right ideals I ⊂ E such that E/I is finite and IH = H. But since JQE 6= QE for each proper right ideal J in QE, JH 6= H for each pure proper right ideal in E. Specifically, ∆H 6= H even though IH = H for each right ideal ∆ ⊂ I = ∆ + mE of finite index in E.

8.1.1

Hereditary Torsion Classes

A nonempty set S of right End(G)-modules is a hereditary torsion class if 1. S is closed under the formation of direct sums. That is, if Mk ∈ S, k ∈ I then ⊕k∈I Mk ∈ S. 2. S is closed under the formation of short exact sequences. That is, if 0 −→ K −→ M −→ N −→ 0 is a short exact sequence then K, N ∈ S iff M ∈ S. The prototypical examples of hereditary torsion classes are the set of groups, the set of torsion groups, and the set of torsion p-groups for some prime p ∈ Z. Our motivation for considering hereditary torsion classes is the set TorE (L) = {right E-modules K K ⊗E L = 0} where L is a left E-module. PROPOSITION 8.1.2 Let E be a ring and let L be a flat left Emodule. Then TorE (L) is a hereditary torsion class. Proof: Let Mk ∈ Tor(L), k ∈ I. Because ·⊗E L is an additive functor (⊕k∈I Mk ) ⊗E L ∼ = ⊕k∈I (Mk ⊗E L) = 0. Next consider the short exact sequence 0 −−−−→ K −−−−→ M −−−−→ N −−−−→ 0 of right E-modules. An application of the exact functor · ⊗E L yields the exact sequence α 0 −−−−→ K ⊗E L −−−−→ M ⊗E L −−−−→ N ⊗E L −−−−→ 0. It follows that K ⊗E L = N ⊗E L = 0 iff M ⊗E L = 0. Hence Tor(L) is a hereditary torsion class.

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EXAMPLE 8.1.3 TorE (L) is a hereditary torsion class if L is a prob p , so jective left E-module. If p ∈ Z is a prime then End(Z(p∞ )) = Z ∞ b p -modules of the form that TorE (Z(p )) is the set of Z H ⊕D b p -module. But where H is a torsion p-group and D is a divisible Z ∞ Tor(Z(p )) is not a hereditary torsion class since the quotient field Qp b p is in Tor(Z(p∞ )) and Z b p ⊂ Qp , but Z b p 6∈ Tor(Z(p∞ )). of Z An important example of hereditary torsion classes comes from idempotent ideals. PROPOSITION 8.1.4 Let E be a ring and let ∆ ⊂ E be an ideal such that ∆2 = ∆ and · ⊗E ∆ is an exact functor. The set {right E-modules K K∆ = 0} is a hereditary torsion class. We leave the proof as an exercise for the reader. There is a relationship between a hereditary torsion class and the cyclic submodules it contains. Let X be a set of right E-modules. The hereditary torsion class generated by X , denoted by TorE (X ), is the intersection of all of the hereditary torsion classes that contain X . Given a class T of right E-modules let CycE (T ) denote the cyclic modules contained in T . Given a hereditary torsion class T the reader can show that C ∈ CycE (T ) iff C is a cyclic submodule of some M ∈ T . LEMMA 8.1.5 TorE (CycE (T )) = T if T is a hereditary torsion class. Proof: Clearly TorE (CycE (T )) ⊂ T . Let M ∈ T . There is a submodule K ⊂ M such that ⊕x∈M xEnd(G) ∼ = M. K Since the hereditary torsion class TorE (CycE (T )) is closed under arbitrary direct sums and factor modules M ∈ TorE (CycE (T )). Thus TorE (CycE (T )) = T , and the proof is complete. Some examples will serve to illustrate these ideas. A module M is called semi-Artinian if each nonzero factor M/N of M has nonzero socle. The simple modules appearing in the socle of M/N are called composition factors of M . The reader is encouraged to prove these examples.

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181

EXAMPLE 8.1.6 Let E be a ring and let S be a set of simple right E-modules. The hereditary torsion class TorE (S) is the set of semiArtinian right E-modules M whose composition factors are are in S. EXAMPLE 8.1.7 Let G be a left E-module. Then TorE {E/I IG = G} = {M

for each x ∈ M , xE ⊂ E/I

for some I ⊂ E such that IG = G}. EXAMPLE 8.1.8 The group G in Example 6.4.2 satisfies TG (M ) 6= 0 for each nonzero finitely generated (cyclic) right End(G)-module M but there is a right End(G)-module K 6= 0 such that TG (K) = 0. That is, {E/I IG = G} = ∅ while {K K ⊗E G = 0} 6= ∅.

8.1.2

Gabriel Filters of Right Ideals

In this section we indirectly study sets of modules of the form CycE (T ) for some hereditary torsion class T . Given a right ideal I ⊂ E and x ∈ E let (I : x) = {a ∈ E xa ∈ I}. The reader can show that (I : x) = annE (x + I) where x + I ∈ E/I. A Gabriel filter on E is a set γ of right ideals of E such that 1. Given I, J ∈ γ and I ⊂ I 0 then I ∩ J, I 0 ∈ γ. 2. Given I ∈ γ and x ∈ E then (I : x) ∈ γ. 3. Given a right ideal J ⊂ E and I ∈ γ such that (J : x) ∈ γ for each x ∈ I then J ∈ γ. The relationship between Gabriel filters and hereditary torsion classes follows. LEMMA 8.1.9 Let E be a ring. 1. Let γ be a Gabriel filter on E. Then the hereditary torsion class E TorE I ∈ γ is the set of right E-modules K such that each I cyclic submodule of K has the form E/I for some I ∈ γ. 2. Let T be a hereditary torsion class of right E-modules. Then E right ideals I ⊂ E ∈ T is a Gabriel filter on E. I

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Proof: 1. Let TorE {E/I I ∈ γ} = TorE (γ). Since TorE (γ) is closed under direct sums and quotients, the set of right E-modules K such that each cyclic submodule of K has the form E/I for some I ∈ γ is contained in TorE (γ). On the other hand, if K ∈ TorE (γ) then xE ∈ {E/I I ∈ γ} for each x ∈ K. Thus TorE (γ) is contained in the set of right E-modules K such that each cyclic submodule of K has the form E/I for some I ∈ γ, and hence the two sets coincide. 2. Let T be a hereditary torsion class, and let E GabE (T ) = right ideals I ⊂ E ∈T . I One readily shows that if I, J ∈ GabE (T ) and if I ⊂ I 0 then E/(I ∩ J) ⊂ E/I ⊕ E/J, and E/I maps onto E/I 0 . Thus I ∩ J, I 0 ∈ GabE (T ). Next let I ∈ GabE (T ), let J ⊂ E be a right ideal, and suppose that (J : x) ∈ GabE (T ) for each x ∈ I. We have a short exact sequence 0 −−−−→

I +J E E −−−−→ −−−−→ −−−−→ 0. J J I +J

Since I ⊂ I + J and since the hereditary torsion class T is closed under quotients, E/(I + J) ∈ T . Let x ∈ I. Since I +J xE + J E ⊃ = ∈T J J (J : x) and since T is closed under direct sums and quotients, (I + J)/J ∈ T . Finally, since T is closed under extensions we see that E/J ∈ T . Thus J ∈ GabE (T ), from which we see that GabE (T ) is a Gabriel filter on E. This completes the proof. PROPOSITION 8.1.10 Let E be a ring, let Gab(E) be the set of Gabriel filters of E, and let Tor(E) be the set of hereditary torsion classes on E. There are inverse bijections GAB : Tor(E) −−−−→ Gab(E) TOR : Gab(E) −−−−→ Tor(E) given by GAB(T ) = GabE (T ) and TOR(γ) = TorE (γ). Proof: The reader can verify that GAB(TOR(γ)) = γ for each Gabriel filter γ and that TOR(GAB(T )) = T for each hereditary torsion class T .

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183

Therefore we can study hereditary torsion classes of right E-modules by studying Gabriel filters on E. The reader can verify the examples as exercises. EXAMPLE 8.1.11 Let T be the class of right E-modules such that for each x ∈ M ∈ T , xE is a bounded group (i.e., (xE)m = 0 for some integer m 6= 0). Then Tor(T ) is the hereditary torsion class of modules whose cyclic submodules are bounded groups. The associated Gabriel filter Gab(T ) is the set of right ideals I ⊂ E such that E/I is a torsion group. The following sets are important to our discussions in this Chapter. Let E be a ring and let M be a left E-module. DE (M ) = {right ideals I ⊂ E IM = M } maxE (M ) = {maximal right ideals I ⊂ E IM = M }. These sets are not necessarily Gabriel filters so we will look for Gabriel filters associated with them. Let δE (M ) = {right ideals I ⊂ E (I : x)M = M ∀x ∈ E} µE (M ) = the least Gabriel filter on E that contains maxE (M ).

Given I ∈ δE (M ) then I = (I : 1) so that δE (M ) ⊂ DE (M ). Since the ring E is understood we will drop it from our notation. The next result will allow us to say more about the right ideals in µ(M ) as well as being of independent interest. LEMMA 8.1.12 Let E be an rtffr ring. Then I ∈ µ(M ) iff E/I is a finite right E-module and each composition factor of E/I is isomorphic to some simple module in {E/J J ∈ maxE (M )}. Proof: Let µ0 (M ) be the set of right ideals I ⊂ E such that IM = M , E/I is finite, and each composition factor of E/I is isomorphic to E/J for some J ∈ maxE (M ). By Lemma 4.2.3, each I ∈ maxE (M ) has finite index in E, so that maxE (M ) ⊂ µ0 (M ). The reader will show that µ0 (M ) is a Gabriel filter on E. By definition as the least such filter, µ(M ) ⊂ µ0 (M ). Conversely, let I ∈ µ0 (M ). Then IM = M , E/I is finite and each composition factor of E/I is of the form E/J for some J ∈ maxE (M ) ⊂

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µ(M ). We observe that E/I is a factor module of the direct sum of cyclic modules M xE + I . I x∈E

Because E is an rtffr ring, torsion cyclic E-modules are bounded, hence finite groups. Thus for each x ∈ E, (xE + I)/I ∼ = E/(I : x) is finite. Let K be a simple submodule of E/(I : x). Then K ∼ = J/(I : x) for some (I : x) ⊂ J ∈ µ(M ). By a simple induction of the composition length of E/(I : x), each cyclic submodule of J/(I : x) is of the form E/L for some L ∈ µ(M ). Since µ(M ) is a Gabriel filter (I : x) ∈ µ(E) for each x ∈ E, and so I ∈ µ(M ). It follows that µ0 (M ) = µ(M ). An ideal ∆ in some ring is idempotent if ∆2 = ∆. Idempotent ideals will be important in the sequel. LEMMA 8.1.13 Suppose that ∆2 = ∆ is an ideal in E. Then D(∆) = {right ideals I ⊂ E I∆ = ∆} δ(∆) = the largest Gabriel filter contained in D(∆) max(∆) = {maximal right ideals I ∆ ⊂ I ⊂ E}. In particular ∆ is the unique minimal element in D(∆). Proof: This follows from the definitions using M = ∆. It follows that if ∆2 = ∆ then δ(∆) is generated by the set of cyclic right E-modules K such that K ⊗E ∆ = 0. There is an interesting characterization of the hereditary torsion class corresponding to δ(M ). LEMMA 8.1.14 Let M be a left E-module. Then K ∈ Tor(δ(M )) iff L ⊗E M = 0 for each cyclic submodule L ⊂ K. Proof: Suppose that x ∈ L ∈ Tor(δ(M )). Since δ(M ) is a Gabriel filter xE ∼ = E/I for some I ∈ δ(M ). Then E/I ⊗E M = 0 and since each submodule of L is a sum of cyclics K ⊗E M = 0 for each K ⊂ L. The converse is clear so the proof is complete. LEMMA 8.1.15 Let E be an rtffr ring and let M be a left E-module. 1. If γ ⊂ D(M ) is a Gabriel filter on E then γ ⊂ δ(M ). 2. If I ∈ D(M ) is an ideal then I ∈ δ(M ).

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185

Proof: 1. Say I ∈ γ ⊂ D(M ). Then for each x ∈ E, (I : x) ∈ γ ⊂ D(M ) so that (I : x)M = M . Thus I ∈ δ(M ) and hence γ ⊂ δ(M ). 2. Let I ∈ D(M ) be an ideal and let x ∈ E. Then I ⊂ (I : x) so that M = IM ⊂ (I : x)M ⊂ M which implies that (I : x) ∈ D(M ) for each x ∈ E. Hence I ∈ δ(M ). The next result shows that even though in general µ(M ) 6= δ(M ) 6= D(M ) they still share the same maximal right ideals. LEMMA 8.1.16 Let M be a left E-module. 1. maxE (M ) = max(δ(M )) = max(µ(M )). 2. µ(M ) = {right ideals I ⊂ E IG = G and E/I is semi-Artinian}. Proof: 1. Since δ(M ) ⊂ D(M ) and since maxE (M ) ⊂ µ(M ) we have max(δ(M )) ⊂ maxE (M ) ⊂ max(µ(M )). Let J ∈ max(µ(M )) and let x ∈ E. Inasmuch as E/J is simple, E/J ∼ = E/(J : x), so that xE ⊗E M = E/(J : x) ⊗E M ∼ = E/J ⊗E M ∼ = M/JM = 0. Then Lemma 8.1.14 states that J ∈ δ(M ). This proves part 1. 2. is left as an exercise.

8.2

Idempotent Ideals

As above E denotes an rtffr ring. Further restrict M to be an rtffr left E-module. We will show that the Gabriel filters δ(M ) and µ(M ) on E b are determined by an idempotent ideal ∆ ⊂ E.

8.2.1

Traces of Covers

The following concepts are found in [6]. Let M be a left E-module and let K ⊂ M be a submodule. We say that K is superfluous in M if K + N = M =⇒ N = M for each submodule N ⊂ M . For instance, Nakayama’s Lemma 1.2.3 implies that J(E)M is superfluous in each finitely generated left E-module M . However each proper submodule of Z(p∞ ) is superfluous in Z(p∞ ). In a ring E, the Jacobson

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radical J (E) is the largest superfluous E-submodule of E. No proper subgroup of Z is superfluous in Z, but for each prime p ∈ Z, pZp is superfluous in Zp . If N is a nilpotent ideal then N M is superfluous in M for each left E-module M . The reader can prove the following facts as exercises. 1. If N ⊂ K ⊂ L ⊂ M and if K is superfluous in L then K is superfluous in M and K/N is superfluous in M/N . 2. Suppose that P is a finitely generated projective left E-module. An E-submodule K ⊂ P is superfluous iff K ⊂ J(E)P . The left E-module M possesses a projective cover if there is a projective left E-module P and a surjection π : P → M such that ker π is superfluous in P . Our work in this section will show that projective covers are abundant when considering modules over rtffr rings. The trace of the projective cover π : P → M is ∆M = the trace ideal of P in E. Evidently ∆M is an idempotent ideal in D(M ). Since ∆M P = P we have ∆M M = M . A Gabriel filter γ is bounded if for each I ∈ γ there is a two sided ideal J ∈ γ such that J ⊂ I. We say that γ is bounded by (the idempotent ideal) ∆ if γ = { right ideals I ⊂ E ∆ ⊂ I}. In particular ∆ ∈ γ. For example, each Gabriel filter on a commutative ring is bounded. If ∆ is an idempotent ideal in E then by Lemma 8.1.13, ∆ is the unique minimal element in δ(∆). The filter of nonzero ideals on Z is bounded but is not bounded by a unique ideal. LEMMA 8.2.1 Let π : P → M be a projective cover of the left Emodule M . Then D(∆M ) = δ(M ) iff δ(M ) is bounded. Proof: If D(∆M ) = δ(M ) then by the above comment δ(M ) is bounded by ∆M . Conversely suppose that δ(M ) is bounded. By Lemma 8.1.13, δ(M ) contains every ideal in D(M ), and by the above comments, ∆M M = M . Then ∆M ∈ δ(M ). It remains to show that δ(M ) ⊂ D(∆M ). Let I ∈ δ(M ). There is an ideal J ⊂ I such that J ∈ δ(M ). Then JM = M so that JP + ker π = P . Because J is an ideal of E and because ker π is superfluous in P , JP = P . As usual ∆M ⊂ J ⊂ I so that I ∈ D(∆M ). Hence D(∆M ) = δ(M ), which completes the proof.

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187

EXAMPLE 8.2.2 1. Given a ring E then the canonical map E → E/J (E) is a projective cover. 2. Every module over a right Artinian ring possesses a projective cover, [6]. 3. The ring E is called semi-perfect if each finitely generated left Emodule possesses a projective cover. E is semi-perfect iff E/J(E) is semi-simple Artinian and idempotents in E/J(E) lift to idempotents of E. (See [6].) Local rings are semi-perfect. EXAMPLE 8.2.3 The following is an example of a finitely generated module M over a ring E such that E is not semi-perfect, M is not projective, but M possesses a finitelygenerated cover. projective 0 0 Z 0 ⊕ , let P = Let A be the ring of matrices Z Z Z Z 0 0 0 0 0 0 ⊕ , and let K = x ∈ Z . The reader x 0 x 0 Z Z 0 0 can show that J(A) = so that K ⊂ J(A)P is superfluous in Z 0 P . Then P/K is a doubly generated left E-module and the natural map π : P → P/K is a projective cover. An important source of projective covers is the following result. LEMMA 8.2.4 [40, T.G. Faticoni]. For each index i ∈ I let Ei be a ring, let Mi be a finitely generated left Ei -module, let πi : Pi → Mi be a projective cover of Mi over Ei , and let ∆i be the trace ideal of Pi in Ei . If there is an integer n such that each Mi is generated by at most n elements then Q Q 1. i∈I Mi is generated by at most n elements over i∈I Ei , Q Q 2. i∈I Pi is a finitely generated projective left i∈I Ei -module Q whose trace ideal satisfies ∆ = i∈I ∆i , and Q Q Q Q 3. i∈I πi : i∈I Pi → i∈I Mi is a projective cover over i∈I Ei . Proof: 1. By hypothesis, for each i > 0 there is a set of generators {xi1 , . . . , xin } ⊂ Mi . For each j = 1, . . . , n define a sequence Y xj = (x1j , x2j , . . .) ∈ Mi . i∈I

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Q We claim thatQ {x1 , . . . , xn } generates i Mi . Given y ∈ i Mi write y = (yi )i . By our choice of xij for each i there are rj = (rij )i , j = 1, . . . , n, such that yi =

n X

rij xij .

j=1

Then n n n n X X X X = y = (yi )i = rij xij (rij xij )i = (rij )i (xij )i = rj xj . j=1

j=1

i

j=1

Q

j=1

Q

As claimed {x1 , . . . , xn } generates M = i Mi over E = i Ei . 2. For each i ∈ I, Q Pi is generated by at most n elements over Ei , so that by part 1, P = i Pi is a finitely generated projective left E-module. Inasmuch as Y HomE (P, E) = HomEi (Pi , Ei ) i∈I

Q it follows that the trace ideal for P in E is the product ∆ = i∈I ∆i . 3. Suppose that Q for each i ∈ I, Ki is a superfluous Ei -submodule of Pi andQlet K = Q i Ki . Since Ki is superfluous in Pi , Ki ⊂ J (Ei )Pi . Since J ( i Ei ) = i J (Ei ) ! ! Y Y Y K= Ki ⊂ J Ei Pi = J (E)P. i

i

i

Since P is finitely generated, Nakayama’s Lemma 1.2.3 shows us that K is superfluous in P . Q The reader can show that because M = i Mi is finitely generated Y π= πi : P −−−−→ M i

Q is a surjection. Inasmuch as ker(π) = i (ker πi ), π is a projective cover of M over E. This completes the proof. The reader might ask where in the realm of rtffr groups will we have a chance to see an infinite product of rings. The next few results will answer this question. Given a prime p let pω G =

∞ \

pk G.

n=1

Recall that idempotents lift modulo nilpotent ideals. (See [6].)

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189

LEMMA 8.2.5 Let E be an rtffr ring. Then for each prime p the p-adic bp of E/pω E is a semi-perfect ring. completion E Proof: Let Ep be the localization of E at p. Because E has finite rank Ep /pk Ep ∼ = E/pk E is a finite (hence Artinian) ring for each integer bp /pk E bp . k > 0, so that idempotents lift module the Jacobson radical of E b Furthermore p ∈ J (Ep ). bp ). Because bp /J (E Let e¯2 = e¯ ∈ E bp )/pk E bp = J (E bp /pk E bp ) J (E and because bp bp /pk E pk−1 E bp /pk E bp for each integer k > 0 we can construct is a nilpotent ideal of E bp such that a sequence (ek )k in E bp ) = e¯, e1 + J (E

bp , ek−1 − ek ∈ pk−1 E

bp ). e2k − ek ∈ pk J (E

bp its limit exists. Since (ek )k clearly converges in the p-adic topology on E bp for each k. bp . Moreover e2 − e ≡ e2 − ek ∈ pk E That is, limk ek = e ∈ E k ω 2 bp = 0, e = e. That is, e is an idempotent lifting of e¯. Hence Since p E b Ep is semi-perfect. LEMMA 8.2.6 Let E be an rtffr ring, and let M be an rtffr right Eb p -module by at most rank(M ) cp is generated as a Z module. Then M b c is generated as a Zelements for each prime p ∈ Z. Furthermore M module by at most rank(M ) elements. Proof: Let p ∈ Z be a prime. Because Z/pZ-linearly independent subsets of M/pM lift to Z-independent subsets of M , M/pM has Z/pZcp /pM cp ∼ dimension at most rank(M ). Thus M = M/pM has dimension at rank(M ) b c cp . It follows that there most rank(M ), so that Zp maps onto Mp /pM b p -submodule K of M cp generated by at most rank(M ) elements is a free Z b p , and such that K + pM cp = M cp . Thus K is dense in M cp . Since K over Z cp . Thus M c is is finitely generated it is p-adically complete so that K = M Q pc c generated by at most rank(M ) elements. By Lemma 8.2.4, M = p Mp b=Q Z b p -module. is then generated by at most rank(M ) elements as a Z p

This proves the Lemma. THEOREM 8.2.7 [40, T.G. Faticoni] Let E be an rtffr ring and let c possesses a finitely generated M be an rtffr right E-module. Then M b projective cover over E.

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b = Q E b c Proof: Because E and M are rtffr groups, E p p and M = Q c Mp where p ranges over the primes in Z. Fix p ∈ Z. By Lemma 8.2.6, p

b p -module by rank(M ) elements. Consequently M cp is generated as a Z cp M bp (Lemma 8.2.5), possesses a projective cover over the semi-perfect ring E Q c c so that M = p Mp possesses a finitely generated projective cover over b (Lemma 8.2.4). This completes the proof. the ring E b possesses a COROLLARY 8.2.8 Let G be a rtffr group. Then G \ b is genfinitely generated projective cover over End(G). Moreover G \ erated by at most rank(G) elements as a left End(G)-module.

8.2.2

Bounded Gabriel Filters

In this section we examine the existence and the usefulness of idempotent ideals ∆ such that ∆M = M . Specifically we show that µ(M ) is finite iff it is bounded by an idempotent ideal. Let us agree that

c) = {right ideals I ⊂ E b (I : x)M c=M c∀x ∈ E}. b δ(M

c) denotes the largest Gabriel filter in D b (M c), the set of right Then δ(M E b such that I M c=M c. (See Lemma 8.1.15.) Furthermore let ideals I ⊂ E

c) = {right ideals I ⊂ E b E/I b is finite and I M c=M c}. µ(M

c) is a Gabriel filter on E. b We will show that Gabriel Observe that µ(M filters over p-adically complete torsion-free rings are especially nice. THEOREM 8.2.9 Let E be an rtffr ring and let M be an rtffr right E-module. c) is a bounded Gabriel filter on E. b 1. δ(M c) is a bounded Gabriel filter on E. b 2. µ(M

8.2. IDEMPOTENT IDEALS

191

b c). Since E b is a finitely generated Z-module Proof: 1. Let I ∈ δ(M b b b (Lemma 8.2.6) we can write E/I = Zx1 + · · · + Zxn for some integer n b and elements xi ∈ E/I. An elementary argument shows us that b (I : x1 ) ∩ · · · ∩ (I : xn ) ⊂ annEb (E/I) ⊂ I. c) is a Gabriel filter, (I : x1 ) ∩ · · · ∩ (I : xn ) ∈ δ(M c), from which Since δ(M b c). Thus δ(M c) is bounded. it follows that the ideal annEb (E/I) is in δ(M 2. Proceed as in part 1. This completes the proof. COROLLARY 8.2.10 Let E be an rtffr ring and let M be an rtffr right E-module. Then µ(M ) is a bounded Gabriel filter on E. Proof: Let I ∈ µ(M ). By Lemma 8.1.12, E/I is torsion so there is an integer m 6= 0 such that mE1 = E(m1) ⊂ I. Hence E/I is bounded, and since E is an rtffr, E/I is finite. Write {x1 , . . . , xn } = E/I. Then mE ⊂ (I : x1 ) ∩ · · · ∩ (I : xn ) ⊂ annE (E/I) ⊂ I so that annE (E/I) ⊂ I is an ideal in µ(M ). This completes the proof. THEOREM 8.2.11 Let E be an rtffr ring and let M be an rtffr right b E-module. There is a finitely generated idempotent ideal ∆M c ⊂ E such that c) = {right ideals I ⊂ E b ∆ c ⊂ I}. In particular, ∆ c ∈ δ(M c). 1. δ(M M

M

c) = {right ideals I ⊂ E b E/I b is finite and ∆ c ⊂ I}. 2. µ(M M c possesses a projective cover π : P → M c Proof: By Theorem 8.2.7, M b Let ∆ c be the trace ideal of P in E. b Then ∆ c is the idempotent over E. M M b such that ∆ cP = P , and IP = P iff ∆ c ⊂ I. ideal in E M M c = π(P ) = π(∆ cP ) = ∆ cM c, ∆ c ∈ δ(M c), and so {right 1. Since M M M M b c ideals ∆M c ⊂ I ⊂ E} ⊂ δ(M ). c). By Theorem 8.2.9, δ(M c) is bounded, so Conversely let I ∈ δ(M c c c=M c, there an ideal J ⊂ I such that J M = M . Because π(JP ) = J M JP + ker π = P . Inasmuch as ker π is superfluous in P , and because b JP is a left E-submodule of P , JP = P . Hence ∆M c ⊂ J and thus c b δ(M ) = {right ideals I ⊂ E ∆M c ⊂ I}. Prove part 2 in a manner similar to part 1. This completes the proof. The rtffr group H constructed at the end of Example 8.1.1 should dispel any notions that δ(M ) = µ(M ) is bounded by an idempotent ideal when M is an rtffr right E-module.

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EXAMPLE 8.2.12 This example shows that µ(M ) need not be finite. Let π : P −→M be the projective cover constructed in Example 8.2.3. 0 0 Then ∆M = , and Z Z δ(M ) =

nZ 0 Z Z

n∈Z

is bounded by the idempotent ideal ∆M . Furthermore, nZ 0 µ(M ) = n 6= 0 ∈ Z Z Z is not bounded by an idempotent ideal.

8.2.3

Finite Filters of Divisibility

In this section we study modules M for which µ(M ) and δ(M ) are finite sets. c) are identical The following Lemma shows us that µ(M ) and µ(M posets. LEMMA 8.2.13 Let E be an rtffr ring and let M be an rtffr right c) defined E-module. There are inverse poset isomorphisms µ(M ) ∼ = µ(M by I 7→ Ib and I 7→ I ∩ E. c) and let I = Ib ∩ E. Then E/ b Ib and E/I are Proof: Let Ib ∈ µ(M b Because I is dense finite. There is an integer m 6= 0 such that m ∈ I, I. b b c in I and because I ∈ µ(M ) c = (I + mI)(M b c) = IbM c=M c = M + mM c. IM + mM + mM Then by the Modular Law and because m ∈ I we have M

c) ∩ M = (M + mM c) ∩ M = (IM + mM c ∩ M) = IM + (mM = IM + mM = IM.

c) −→ µ(M ). Thus the assignment Ib → 7 Ib ∩ E defines a function µ(M

8.2. IDEMPOTENT IDEALS

193

Conversely let I ∈ µ(M ) and choose a nonzero integer m ∈ I. Then c = (I + mI)(M b c) = IM + mM c = M + mM c=M c IbM + mM c). Thus the assignment I 7→ Ib defines a function so that Ib ∈ µ(M c µ(M ) −→ µ(M ). Finally it is well known that Ib ∩ E = I for each right ideal I of finite 0 ∩ E = I 0 for each right ideal I 0 of finite index index in E, and that I\ b Thus the above maps are mutual inverses, which completes the in E. proof. Using the above correspondence we can characterize µ(M ) in terms of an idempotent ideal. THEOREM 8.2.14 Let E be an rtffr ring and let M be an rtffr right b E-module. There is a finitely generated idempotent ideal ∆M c ⊂ E such that b b µ(M ) = {right ideals I ⊂ E E/I is finite and ∆M c ⊂ I ⊂ E}. Moreover ∆M c is generated by at most rank(E) elements. Proof: The proof is an application of Lemma 8.2.13 and Theorem 8.2.11. We leave the details to the reader. We consider the Gabriel filters that are bounded by an idempotent ideal to be the most important examples of Gabriel filters for reasons that will be obvious in the next Theorem. Since each rtffr module M b the projective cover over E is associated with a projective cover π over E, brings with it the trace ideal ∆M c of the projective. The next result gives at least one case in which ∆M leads us to an idempotent ideal in E. This c is the case when µ(M ) is finite. THEOREM 8.2.15 [40, T.G. Faticoni] Let E be an rtffr ring and let M be an rtffr right E-module. The following are equivalent for M . 1. µ(M ) is finite. 2. µ(M ) = D(∆) for some idempotent ideal ∆ in E. b 3. ∆M c has finite index in E. If M satisfies one and hence all of the above conditions then ∆ = ∆M c ∩E.

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b Proof: 3 ⇒ 2 Suppose that ∆M c has finite index in E. Then by Theorem 8.2.11 b is finite and ∆ c ⊂ I} c) = {right ideals I ⊂ E b E/I µ(M M b = {right ideals I ⊂ E ∆M c ⊂ I}. If we let ∆ = ∆M c ∩ E then ∆ has finite index in E. The reader can show that ∆ is idempotent since ∆M c is idempotent. Lemma 8.2.13 then implies that b and ∆ ⊂ I} ∆ ∈ µ(M ) = {I ∩ E I is a right ideal in E b and ∆ ⊂ I ∩ E} = {I ∩ E I is a right ideal in E = {right ideals J ⊂ E ∆ ⊂ J}. This proves part 2. 2 ⇒ 1 follows immediately from Lemma 8.1.12. 1 ⇒ 3 Because µ(M ) = {I1 , · · · , In } is finite, and because µ(M ) is bounded, µ(M ) contains a unique minimal element ∆ = I1 ∩ · · · ∩ In ∈ µ(M ). b is the unique minimal element of Then Lemma 8.2.13 shows us that ∆ c b b µ(M ). We observe that ∆ has finite index in E. ∼ c b c To see that ∆M c = ∆ notice that µ(M ) = µ(M ) implies that µ(M ) is b We showed in Theorem finite, hence bounded by the idempotent ideal ∆. c 8.2.11 that ∆M c is the unique minimal element of µ(M ). This proves part 3 and completes the logical cycle. Let E be the ring and let M be the module constructed possessing a projective cover π : P → M in Example 8.2.3. We showed that ∆M = 0 0 c such that . We then constructed an rtffr group M ⊂ G ⊂ M Z Z each element of µ(G) has finite index in E. Specifically ∆M 6∈ µ(G). Furthermore, µ(G) is infinite and does not possess a unique minimal element. EXAMPLE 8.2.16 The reader will show that µ(G) is finite if G is the group constructed in Example 6.1.7. There is at least one class of rtffr ring in which each Gabriel filter is bounded by an idempotent ideal.

8.2. IDEMPOTENT IDEALS

195

THEOREM 8.2.17 Let E be an rtffr ring and let M be an rtffr left c) = 0. (For example, let M be any module E-module such that annEb (M such that E = End(M ).) If E is a semi-prime ring then µ(M ) is finite. Proof: Suppose that E is semi-prime and that annE (M ) = 0. Then b is semi-prime. Write E Y ∆M = ∆p c p

bp . Since where ∆p is an idempotent ideal in the semi-prime ring E c) = 0, ann b (∆p ) = 0. However, since E bp is semi-prime there is annEb (M Ep . b such that ∆p ⊕ I = E bp . Since then I∆p ⊂ ∆p ∩ I = 0, an ideal I in E . b I = 0 and so ∆p = Ep . Now for all but at most finitely many primes p, bp is an integrally closed ring. In such a ring, proper ideals are invertible, E bp . Then not idempotent. See [69]. Say ∆p J = E bp . bp ) = ∆p (∆p J) = ∆p J = E ∆p = ∆p (E b for all but at most finitely many primes p. Thus That is ∆p = E Q b p b E/∆M c= p Ep /∆p is finite, and hence by the previous Theorem, µ(M ) is finite. This completes the proof. Let E be a rtffr ring and let M be a rtffr right E-module. We say that M is faithful if IM 6= M for each right ideal I in E. Evidently M is faithful iff D(M ) = {E}. Since each right ideal in E is contained in a maximal right ideal and since each maximal right ideal I in E has finite index in E, we see that M is faithful iff IM 6= M for each right ideal I of finite index in E, Lemma 4.2.3. It is clear that M is a faithful left E-module iff µ(M ) = {E}. As we showed in Chapter 6, the faithful property is connected to the splitting property of short exact sequences. The next result characterizes the faithful left E-modules. THEOREM 8.2.18 Let E be an rtffr ring and let M be an rtffr right E-module. The following are equivalent for M . 1. D(M ) = {E}. 2. δ(M ) = {E}. 3. µ(M ) = {E}. c) = {E}. b 4. µ(M

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5. If I is a right ideal in E then there is an x ∈ E such that (I : x)M 6= M . b 6. ∆M c = E. 7. Let K be a torsion (as a group) right E-module such that K ⊗E M = 0. There exists a cyclic submodule N ⊂ K such that N ⊗E M 6= 0. Proof: 1 ⇒ 2⇒ 3 follows from the inclusions µ(M ) ⊂ δ(M ) ⊂ D(M ). 3 ⇒ 1 Suppose that part 1 is false. There is a right ideal I such that IM = M . We may assume that I is a maximal right ideal of E and hence that I ∈ max(M ) ⊂ µ(M ). This proves that part 3 is false. c) (Lemma 3 ⇔ 4 follows from the isomorphism of posets µ(M ) ∼ = µ(M 8.2.13). 2 ⇔ 5 follows from the definition of δ(M ). 3 ⇔ 6 is Theorem 8.2.14. 7 ⇒ 3 Suppose that µ(M ) 6= {E}. Let I ⊂ E be a maximal right ideal of E such that IM = M . Then E/I = (x + I)E ∼ = E/(I : x) for each x ∈ E. It follows that E/(I : x) = 0 for each x ∈ E and hence part 7 is false. 3 ⇒ 7 follows in a similar manner. This completes the logical cycle.

8.3

Gabriel Filters on Rtffr Rings

Let E denote a fixed rtffr ring and let Ω(E) = { rtffr groups G End(G) ∼ = E as rings} ∆Gb = the trace ideal of a projective cover b over E. b π:P →G

(See Theorem 8.2.7.) In this section we will investigate the rtffr rings E such that each G ∈ Ω(E) is faithful. We characterize those idempotent b such that ∆ = ∆ b for some G ∈ Ω(E). ideals ∆ ⊂ E G

8.3.1

Applications to Endomorphism Rings

Let G be an rtffr group. It is natural to investigate how the above Theorems can be used to explore the faithful property on G. In the next result we characterize those rtffr groups G for which δ(G) is a finite set.

8.3. GABRIEL FILTERS ON RTFFR RINGS

197

THEOREM 8.3.1 The following are equivalent for the rtffr group G. 1. δE (G) is finite. 2. µE (G) is finite. b is finite. 3. µEb (G) \ 4. ∆Gb has finite index in End(G). 5. There is an idempotent ideal ∆ of finite index in End(G) such that δE (G) = {right ideals I ∆ ⊂ I ⊂ End(G)}. In this case ∆ = ∆Gb ∩ End(G). Proof: 5 ⇒ 1 ⇒ 2 ⇒ 3 are clear, 3 ⇒ 4 ⇒ 5 is Theorem 8.2.15. This completes the logical cycle. The faithful rtffr groups are characterized in the following result. THEOREM 8.3.2 The following are equivalent for the rtffr group G. 1. D(G) = {End(G)}. 2. δ(G) = {End(G)}. 3. µ(G) = {End(G)}. \ b = {End(G)}. 4. µ(G) 5. If I is a right ideal in End(G) then there is an x ∈ End(G) such that (I : x)G 6= G. \ 6. ∆Gb = End(G). 7. Let K be a torsion (as a group) right End(G)-module such that K ⊗End(G) G = 0. There exists a cyclic submodule N ⊂ K such that N ⊗End(G) G 6= 0. The Proof follows from Theorem 8.2.18. The short exact sequence π 0 −−−−→ K −−−−→ X −−−−→ G −−−−→ 0

(8.2)

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is quasi-split if there is a map : G → X and. an integer n 6= 0 such that π = n1G . If (8.2) is quasi-split then X ∼ = G ⊕ K. The property δ(G) = µ(G) is characterized in terms of quasi-splitting. The faithful property is intimately associated with the splitting of short exact sequences. Thus we include this connection. THEOREM 8.3.3 The following are equivalent for an rtffr group G. 1. If IG = G for some right ideal I ⊂ End(G) then End(G)/I is finite. That is, δ(G) = µ(G). 2. Each short exact sequence (8.2) in which X = SG (X) + K is quasisplit. 3. Each short exact sequence (8.2) in which X ∼ = G(c) for some cardinal c is quasi-split. Proof: 1 ⇒ 2 is proved in the same manner that we proved Baer’s Lemma 6.1.1. 2 ⇒ 3 is clear. 3 ⇒ 1 Suppose that IG = G for Psome proper right ideal I in End(G)submodule of HG (H). Write I = k∈I πk End(G) for some index set I, let Gk = G for each k ∈ I, and let π : ⊕k∈I Gk −→ G be the unique map such that π(xk ) = πk (xk ) for each xP k ∈ Gk and each k ∈ I. Notice that π is a surjection since G = IG = k∈I πk (Gk ). Thus there is a short exact sequence (8.2) in which X ∼ = ⊕k∈I Gk . By part 5, (8.2) is quasisplit. There is a map : G → ⊕k∈I Gk and an integer n 6= 0 such that π = n1G . Since the rtffr group G is self-small we can write = ⊕k∈I k for some maps k : G → Gk . Then ! ! ! X X X n1G = π = πk k = πk k ∈ I k

k

k

by our choice of πk ∈ I. Then End(G)/I is bounded, hence finite, and part 1 follows. This completes the logical cycle. It is interesting to list the following result since it characterizes those rtffr groups G such that IG = G for some proper nonzero pure right ideal I ⊂ End(G) in terms of a splitting result. We make the following elementary observation. Suppose that E is an rtffr ring and let I ⊂ E be a right ideal such that E/I is a torsion group. There is an integer n 6= 0 such that n1G ∈ I so that nE ⊂ I ⊂ E. Since E has finite rank the bounded group E/I is finite.

8.3. GABRIEL FILTERS ON RTFFR RINGS

199

COROLLARY 8.3.4 The following are equivalent for the rtffr group G. 1. There is a proper nonzero pure right ideal I ⊂ End(G) such that IG = G. 2. There is a short exact sequence π 0 −−−−→ K −−−−→ G(c) −−−−→ G −−−−→ 0

(8.3)

that is not quasi-split. Proof: This theorem is the contrapositive of Theorem 8.3.3.

8.3.2

Constructing Examples

Given a torsion-free ring R and a left R-module M let OR (M ) = {r ∈ QR rM ⊂ M }.

LEMMA 8.3.5 Let E be an rtffr ring and let M be an rtffr left Emodule. Then c) = E. b OE (M ) = E iff OEb (M c) = E b and let r ∈ OE (M ). Then rM ⊂ Proof: Suppose that OEb (M c, rM c⊂ M and since M is a pure subgroup of the pure injective group M c. Thus M b ∩ QE = E r∈E and hence OE (M ) ⊂ E. The inclusion E ⊂ OE (M ) is obvious. b ⊂ O b (M c) and Conversely, assume that OE (M ) = E. Evidently E E c)/E b is torsion. It is a short exercise to show that OE (M ) is a pure OEb (M c). Then E = OE (M ) is pure and dense in E b and in subgroup of OEb (M c), so that E b = O b (M c). This completes the proof. OEb (M E b such that The next result characterizes the idempotent ideals ∆ in E ∆ = ∆G for some G ∈ Ω(E). THEOREM 8.3.6 Let E be an rtffr ring and let ∆ be an idempotent b The following are equivalent. ideal in E.

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1. ∆ = ∆Gb for some G ∈ Ω(E). b b = O b (∆). 2. ∆ is finitely generated as a left E-module and E E Proof: 1 ⇒ 2 Suppose that ∆ = ∆Gb for some G ∈ Ω(E). By Theorem b E b ⊂ O b (∆). 8.2.14, ∆ is finitely generated. Since ∆ is an ideal in E, E b=G b implies that On the other hand, by Lemma 8.3.5, ∆G b = E. b OEb (∆) ⊂ OEb (G) b and part 2 follows. Then OEb (∆) = E b 2 ⇒ 1 Since ∆ is a finitely generated torsion-free (=projective) Zmodule ∆ is a Z-adically complete group. b Let {x1 , . . . , xn } be a set of generators of ∆ as a left Z-module. Let M = (QEx1 + · · · + QExn ) ∩ ∆. b i for each i = 1, · · · , n, M is pure and dense in SinceP Zxi is dense in Zx b c = ∆. ∆ = i Exi . That is, M c possesses a projective cover π : P → M c over By Theorem 8.2.7, M 2 b c E. Since ∆ = ∆ = M c = ∆ = ∆∆ = ∆M c = π(∆P ). π(P ) = M

(8.4)

b and since ker π is Hence ∆P + ker π = P . Since ∆ is an ideal in E superfluous in P , ∆P = P . Then ∆P ⊂ ∆ where ∆P denotes the trace ideal of P . On the other hand, because ∆P and ∆ are idempotent ideals, ∆P ⊂ ∆ implies that ∆P ∆ = ∆P . By (8.4) ∆ = π(P ) = π(∆P P ) = ∆P π(P ) = ∆P ∆ = ∆P . Thus ∆ = ∆P . c) = O b (∆) = E b so that OE (M ) = E Finally by hypothesis OEb (M E (Lemma 8.3.5). Theorem 2.3.3 constructs a short exact sequence 0 −−−−→ M −−−−→ G −−−−→ QC −−−−→ 0 of left E-modules such that E ∼ = End(G) naturally. It follows that c b M = G so by the above paragraph ∆ = ∆P = ∆Gb . This completes the proof. Z 0 Z 0 EXAMPLE 8.3.7 Let E = and let ∆ = . Then Z Z Z 0 ∆ is a cyclic projective left E-module, ∆2 = ∆, and E = OE (∆). Conb is a finitely generated idempotent ideal of E b such that sequently ∆

8.3. GABRIEL FILTERS ON RTFFR RINGS

201

b = O b (∆). b E By Theorem 8.3.6 there is an rtffr group G such that E b = ∆G and E ∼ ∆⊂G⊂∆ = End(G). Furthermore G/∆ ∼ = QE ⊕ QE so that ∆G 6= G, while µ(G) = {right ideals I E/I is finite and ∆ ⊂ I ⊂ E} is infinite. The following is an interesting example of an idempotent ideal of finite index in the ring. EXAMPLE 8.3.8 Let O be a maximal order in the ring of Hamiltonian Quaternions H. (See [69].) It is known that for primes p 6= 2 ∈ Z, O/pO ∼ = Mat2×2 (Z/pZ). Let pO ⊂ ∆ ⊂ O be the right ideal in O such that 0 0 . ∆/pO = Z/pZ Z/pZ One shows that ∆ is a finitely generated projective idempotent ideal in OH (∆) where pO ⊂ OH (∆) ⊂ O is such that Z/pZ 0 . OH (∆)/pO = Z/pZ Z/pZ b = ∆ b for some group G such that OH (∆)) Theorem 8.3.6 states that ∆ G ∼ b = ∆, b ∆G = G, and = End(G). Since OH (∆)/∆ is finite and since G hence δ(G) = µ(G) is finite. Observe that G is strongly indecomposable. b b b = Q Z EXAMPLE 8.3.9 Let E = Z. Then E p p . Let ∆ = ⊕p Zp . 2 Then one readily verifies that ∆ = ∆ and that E = OE (∆). However ∆ 6= ∆Gb for any rtffr group G such that E ∼ = End(G) since ∆ is not a b finitely generated ideal in E.

8.3.3

Faithful Rings

The ring E is a faithful ring if given a right ideal I ⊂ E and a G ∈ Ω(E) such that IG = G then I = E. D.M. Arnold and E.L. Lady [14] and [41] show that the rtffr ring is a faithful ring if E is commutative, right hereditary, or local. More examples of faithful rings are given at the end of this subsection. The following theorem characterizes the faithful rtffr rings. THEOREM 8.3.10 Let E be an rtffr ring. The following are equivalent for E. 1. E is a faithful ring.

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b such that E b = 2. If ∆ is a finitely generated idempotent ideal of E b OEb (∆) then ∆ = E. b b= 3. If P is a finitely generated projective left E-module such that E b OEb (P ) then P generates E. Proof: 1 ⇒ 2 Assume part 2 is false. Then there is a proper finitely b such that O b (∆). By Theorem 8.3.6 generated idempotent ideal ∆ ⊂ E E b and ∆G b = G. b Then part 1 there is a G ∈ Ω(E) such that ∆ = ∆Gb 6= E is false. 2 ⇒ 3 Assume part 2. Let P be a finitely generated projective left b b = O b (P ), and let ∆ be the trace ideal of P in E-module such that E E b Then ∆ is a finitely generated idempotent ideal of E b and by Lemma E. b b 8.3.5, E = OEb (∆). By part 2, ∆ = E, which proves part 3. 3 ⇒ 1 Assume part 3 and let G ∈ Ω(E). By Theorem 8.2.7 there is b over E. b If we let ∆ be a finitely generated projective cover π : P → G the trace ideal of P then by Lemma 8.3.5 b ⊂ O b (P ) = O b (∆) ⊂ O b (G) b = E. b E E E E b so that by part 3, P generates E. b Hence ∆ = E b is Then OEb (P ) = E, b Inasmuch as ∆ = ∆ b , Theorem 8.3.2 implies the trace ideal of P in E. G that G is a faithful group. This proves part 1 and completes the logical cycle. The following are examples of faithful rings. The ring E is local if E possesses a unique maximal right ideal. In this case J (E) is the unique maximal right ideal of E and E/J (E) is a division ring. EXAMPLE 8.3.11 A local rtffr ring is faithful. Proof: Say E/J (E) is simple. It follows from Lemma 4.2.3 that E/J (E) is an elementary p-group for some prime p ∈ Z. Then pE 6= E b = E bp . One readily and qE = E for each prime q 6= p ∈ Z. Thus E \ bp ) so that E bp is a local ring. Since a projective proves that J (E)p = J (E bp -module is then free it generates E bp . Theorem 8.3.10 then implies E that each G ∈ Ω(E) is faithful. The ring E is subcommutative if each right ideal in E is an ideal in E. Commutative rings are subcommutative. If H is a maximal order in the ring of Hamiltonian Quaternions over Q then there is a unique prime ideal 2H ⊂ I ⊂ H such that H/I is a field. Then the localization H2 b 2 are noncommutative subcommutative rings. and the completion H

8.3. GABRIEL FILTERS ON RTFFR RINGS

203

EXAMPLE 8.3.12 Each subcommutative rtffr domain is faithful. Proof: Let E be an rtffr subcommutative domain, and let ∆ be a b such that O b (∆) = E. b By finitely generated idempotent ideal of E E b Theorem 8.3.10 it suffices to show that ∆ = E. Since E is a subcommutative domain, each right ideal has finite index in E, and hence for each integer n, each right ideal of E/nE is an ideal. b is subcommutative. By Lemma 8.2.5, E bp is a semi-perfect ring Thus E for each prime p ∈ Z so we can write bp = E1 ⊕ · · · ⊕ En E bp . This direct sum is one for some indecomposable (right) ideals Ei ⊂ E of indecomposable semi-perfect rings so each Ei is a local ring. Consequently ∆p = ∆1 ⊕ · · · ⊕ ∆n where ∆i ⊂ Ei . But by Nakayama’s Lemma the local Noetherian ring does not contain a proper nonzero ideal I 2 = I. Thus ∆i = Ei for each bp , whence ∆ = E. b Given our reductions and i = 1, . . . , n, hence ∆p = E Theorem 8.3.10, E is a faithful ring. We have shown in Theorem 8.2.17 that semi-prime rtffr rings are bounded rings. That is, each right ideal contains a nonzero ideal. We characterize the rtffr bounded rings by first characterizing those rtffr rings E such that δ(G) is finite for each G ∈ Ω(E). THEOREM 8.3.13 [40, T.G. Faticoni] The following are equivalent for the rtffr ring E. 1. µ(G) is finite for each G ∈ Ω(E). 2. δ(G) is finite for each G ∈ Ω(E). b is a finitely generated idempotent ideal such that E b = 3. If ∆ ⊂ E b OEb (∆) then E/∆ is finite. Proof: 1 ⇔ 2 is Theorem 8.3.2. 1 ⇒ 3 Assume part 1, and let ∆ be a finitely generated idempotent b such that E b = O b (∆). By Theorem 8.3.6 there is a group ideal in E E G ∈ Ω(E) such that ∆Gb = ∆. By Theorem 8.3.2 and part 1, ∆Gb has b This proves part 3. finite index in E. 3 ⇒ 1 Assume part 3, let G ∈ Ω(E), and consider µ(G). By Lemma b = { right ideals I E/I b is finite 8.2.13 and Theorem 8.3.2, µ(G) ∼ = µ(G)

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b It therefore suffices to show that ∆ b has finite index and ∆Gb ⊂ I ⊂ E}. G b in E. b is a finitely generated left module over Since G is an rtffr group, G b b b By Lemma E, and thus ∆Gb is a finitely generated Z-submodule of E. b so that by part 3, ∆ b has finite index in E. b Given 8.3.5, OEb (∆) = E, G our reductions µ(G) is finite. In Example 8.2.12 we constructed a group G such that δ(G) = µ(G) is infinite.

8.4

Gabriel Filters on QEnd(G)

In this section we characterize in terms of quasi-summands of G the Gabriel filter of right ideals I of QEnd(G) such that IQG = QG. Let Example 8.2.12 act as an intuition builder.

8.4.1

Central Quasi-summands

In the previous sections we discussed the right ideals I of finite index in End(G) such that IG = G. In this section we will investigate the pure right ideal I ⊂ End(G) such that G/IG is torsion, or equivalently such that (QI)QG = QG. This naturally leads us to consider D(QG) = {right ideals I of QEnd(G) I(QG) = QG}. δ(QG) = {right ideals I of QEnd(G) (I : x)(QG) = QG for each x ∈ QEnd(G)}.

Inasmuch as QEnd(G) is Artinian and since the Gabriel filter δ(QG) contains finite intersections there is an idempotent ideal ∆QG ⊂ QEnd(G) such that δ(QG) = D(∆QG ) = { right ideals I ⊂ QEnd(G) ∆QG ⊂ I}. EXAMPLE 8.4.1 Let X and Y be rtffr groups such that Y Hom(X, Y )(X), 0 = Hom(Y, X), and Z 0 End(X ⊕ Y ) = . Z Z

=

8.4. GABRIEL FILTERS ON QEND(G)

205

0 0 See, e.g., Example 8.1.1. Then ∆ = is such that ∆G = G for Z Z . any group G = X ⊕ Y . Thus Q∆QG = QG. The reader can show that Q∆ is the smallest element of δ(QG). . Given a quasi-direct sum decomposition G = H ⊕ H 0 ⊂ QG there are idempotents eH , eH 0 ∈ QEnd(G) such that . eH G = H,

. eH 0 G = K,

and eH ⊕ eH 0 = 1G .

Call H a nilpotent quasi-summand of G if each composition of maps H → H 0 → H is naturally a nilpotent endomorphism of G. In this case . we call G = H ⊕ H 0 a nilpotent quasi-decomposition of G. The reader can show that if H is a nilpotent quasi-summand of G and if K is a nilpotent quasi-summand of H then K is a nilpotent quasi-summand of . G. Evidently, if G = H ⊕ H 0 is a nilpotent quasi-decomposition of G then {H, H 0 } is a nilpotent set. The first lemma characterizes nilpotent quasi-summands in terms of idempotents in QEnd(G)/N (QEnd(G)). . LEMMA 8.4.2 Let G = H ⊕ H 0 and let eH : G −→ H be the canonical idempotent associated with H. Consider maps f : H −→ H 0 as endomorphisms of G in the natural way. The following are equivalent for H. 1. H is a nilpotent quasi-summand of G. 2. Hom(H, H 0 ) ⊂ N (QEnd(G)). 3. eH is central modulo N (QEnd(G)). Proof: 1 ⇒ 2 Let a : H −→ H 0 and x : G −→ G be elements of QEnd(G). Then a = eH 0 aeH and since H is a nilpotent quasi-summand eH xeH 0 aeH is a nilpotent endomorphism of G. Hence (xa)m+1 = (xa)(eE xeB aeE )m = 0 for large enough m. That is, a generates a nil left ideal of QEnd(G). 2 ⇒ 3 Let x ∈ QEnd(G). By part 2, eH xeH 0 , eH 0 xeH ∈ N (QEnd(G)). Then eH x − xeH

= (eH xeH + eH xeH 0 ) − (eH xeH + eH 0 xeH ) = eH xeH 0 − eH 0 xeH ∈ N (QEnd(G))

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so that eH is central modulo N (QEnd(G)). a

b

3 ⇒ 1 Let H 6= H 0 and let H −→ H 0 −→ H. Then ba = eH b(eH 0 aeH ). By part 3, 0 = eH (eH 0 a) ≡ (eH 0 a)eH (mod N (QEnd(G))) so that ba ≡ 0(mod N (QEnd(G))). Then ba is nilpotent, and hence H is a nilpotent quasi-summand of G. This completes the logical cycle. Given a quasi-summand H of G let ∆G,H

= QHom(H, G)QHom(G, H) = the ideal generated by compositions G −→ H −→ G.

Evidently ∆G,H is an ideal in QEnd(G). Because H is a quasi-summand of G, QHom(G, H)Hom(H, G) = QEnd(H) so that ∆2G,H

= (QHom(H, G)Hom(G, H))(QHom(H, G)Hom(G, H)) = QHom(H, G)QEnd(H)QHom(G, H) = QHom(H, G)Hom(G, H) = ∆G,H .

That is, ∆G,H is an idempotent ideal of QEnd(G). Let E be a ring. Recall that GAB(E) denotes the set of Gabriel filters of right ideals on E, and let IDEM(E) denote the set of idempotent ideals ∆ ⊂ E. Partially order GAB(QEnd(G)) and IDEM(QEnd(G)) by inclusion. Let Nil(G) denote the set of quasi-isomorphism classes [H] of nilpotent quasi-summands H of G, and define a partial order ≤ on Nil(G) by declaring that [K] ≤ [H] if K is quasi-isomorphic to a quasi-summand of H. Because G has finite rank ≤ is a partial order on Nil(G). The next result is a pair of bijections that characterize the nilpotent quasi-summands of QEnd(G) in terms of idempotent ideals of QEnd(G) and Gabriel filters of right ideals in QEnd(G). LEMMA 8.4.3 Define maps ∆ : GAB(QEnd(G)) −−−−→ IDEM(QEnd(G)) D : IDEM(QEnd(G)) −−−−→ Nil(G)

8.4. GABRIEL FILTERS ON QEND(G)

207

by ∆(γ) = the unique minimal element of γ D(∆) = [eG], where ∆ is generated by the idempotent e in QEnd(G). Then ∆ and D are isomorphisms of posets. Proof: Given a Gabriel filter γ on the Artinian ring QEnd(G) the intersection of all right ideals in γ is an ideal ∆ = ∆(γ) that is evidently the unique minimal element in γ. This ∆ is necessarily an idempotent ideal. Thus γ = D(∆) for some unique idempotent ideal ∆ of QEnd(G). The function ∆ is the one that sends γ 7→ ∆(γ). It is an easy exercise to show that the function ∆(·) is a bijection. Next given ∆2 = ∆ ⊂ QEnd(G) there is a unique e2 = e ∈ QEnd(G) such that e is central modulo N (QEnd(G)) and ∆ = QEnd(G)eQEnd(G) ([75, Corollary VIII.6.4]). Then by Lemma 8.4.2, eG is a nilpotent quasisummand of G. The function D is the one that sends ∆ 7→ [eG]. The reader can show that D : IDEM(QEnd(G)) −→ Nil(G) is a bijection. Q 0 the reader can show EXAMPLE 8.4.4 In the ring E = Q Q that IDEM(E)= Q 0 0 0 0 0 Q 0 0, , , , ,E . Q 0 0 Q Q Q 0 0 The Gabriel filters on E are those that contain one of the elements in IDEM(E). Given ∆ ∈ IDEM(E) let e2∆ = e∆ = e ∈ E be the unique idempotent such that ∆ = EeE. Then eG is the nilpotent quasi-summand of G that corresponds to ∆. Let H be a nilpotent quasi-summand of G. We say that H is a Qδ-quasi-summand of G if ∆G,H QG = QG, or equivalently if ∆G,H ∈ δ(QG). If H is a Qδ-quasi-summand of G and if no proper quasisummand of H is a Qδ-quasi-summand then we say that H is a minimal Qδ-quasi-summand. The lemma follows immediately from Lemma 8.4.3.

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PROPOSITION 8.4.5 Let G be an rtffr group and let H be a nilpotent quasi-summand of G. Then H is a minimal Qδ-quasi-summand of G iff ∆G,H is the unique minimal element ∆QG of δ(QG). LEMMA 8.4.6 Let G be an rtffr group. 1. A minimal Qδ-quasi-summand of G is unique up to quasi-isomorphism. 2. If H 0 is a (minimal) Qδ-quasi-summand of H and if H is a (minimal) Qδ-quasi-summand of G then H 0 is a minimal Qδ-quasisummand of G. Proof: 1. Suppose that H and H 0 are minimal Qδ-quasi-summands of G then by Proposition 8.4.5, ∆G,H and ∆G,H 0 equal the unique minimal . element ∆QG of δ(QG). By Lemma 8.4.2, H = H 0 . We leave the proof of part 2 as an exercise for the reader. The next result reduces the study of δ(G) to the groups G such that δ(G) = µ(G). That is, to those groups G such that each I ∈ δ(G) has finite index in End(G). . LEMMA 8.4.7 Let G be an rtffr group. Then G = H ⊕H 0 where H is a minimal Qδ-quasi-summand of G that is unique up to quasi-isomorphism and such that IH 6= H for each proper right ideal I ⊂ QEnd(H). Proof: By Lemma 8.4.6, the unique minimal element ∆QG ∈ δ(QG) corresponds to a minimal Qδ-quasi-summand H of G. Since ∆QG is unique H is unique up to quasi-isomorphism. Furthermore, the ideal ∆QH ⊂ QEnd(H) corresponds to a minimal Qδ-quasi-summand H 0 of H. Then H 0 is a Qδ-quasi-summand of G. The minimality of H shows us that H 0 = H, and hence that ∆QH = QEnd(H). That is, IQH 6= QH for each proper right ideal I ⊂ QEnd(H). This completes the proof.

8.4.2

Q-Faithful Groups

We say that G is a Q-faithful group if IQG 6= QG for each proper right ideal I ⊂ QEnd(G), or equivalently if D(QG) = {QEnd(G)}. Evidently G is a Q-faithful group iff IG = G ⇒ I has finite index in End(G). We will use a variation on the Baer splitting property to characterize Qfaithful groups. Theorem 8.3.3 implies that G is a Q-faithful rtffr group iff given a short exact sequence π 0 −−−−→ K −−−−→ X −−−−→ G −−−−→ 0

(8.5)

8.4. GABRIEL FILTERS ON QEND(G)

209

in which SG (X) + K = X, then (8.5) is quasi-split. That is, there is a map : G −→ X and an integer n 6= 0 such that π = n1G . Theorem 8.2.17 shows us that if E is semi-prime then each G ∈ Ω(E) is a Qfaithful rtffr group. Another important class of Q-faithful rtffr groups is the class of strongly indecomposable rtffr groups. THEOREM 8.4.8 If G 6= 0 is a strongly indecomposable rtffr group then IQG 6= QG for each proper right ideal I ⊂ QEnd(G). Proof: Let G be a strongly indecomposable rtffr group and suppose that IQG = QG for some proper right ideal I ⊂ QEnd(G). Because QEnd(G) is a local ring (see the comment preceding J´onsson Theorem 2.1.10), I ⊂ J (QEnd(G)) so that J (QEnd(G))QG = QG. Since QG is finitely generated Nakayama’s Lemma shows that QG = 0. This completes the proof. COROLLARY 8.4.9 Let G be a strongly indecomposable rtffr group. Then µ(G) = δ(G). THEOREM 8.4.10 The following are equivalent for an rtffr group G. 1. G is Q-faithful. . 2. If G = H ⊕ H 0 and if QHom(H, H 0 )H = H 0 6= 0 then some nonzero strongly indecomposable quasi-summand of H is isomorphic to a quasi-summand of H 0 . 3. Each short exact sequence (8.5) in which SG (X) + K = X is quasisplit. Proof: 1 ⇔ 2 G is a Q-faithful group iff IQG 6= QG for each proper right ideal I ⊂ QEnd(G) iff ∆QG = QEnd(G) iff G is a minimal Qδquasi-summand of G (Lemma 8.4.3), iff no proper quasi-summand of G is a Qδ-quasi-summand of G. The reader can show that this is equivalent to part 2. 1 ⇔ 3 Suppose that G is a Q-faithful group, and let (8.5) be a short exact sequence such that SG (X) + K = X. Then IG = G ⇒ End(G)/I is finite for each ideal I ⊂ End(G). Let I be the right ideal I = πHom(G, X) ⊂ End(G). Since SG (X) + K = X, G = πHom(G, X)G = IG. Then End(G)/I is finite. That is, there is an integer n 6= 0 and a map ∈ Hom(G, X) such that π = n1G . This proves part 3. Conversely assume part 3. Suppose that IG = G for some proper right ideal I of End(G). Let π1 , π2 , · · · denote the generators of I, let G =

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Gk for each k = 1, 2, . . ., and define π : ⊕∞ k=1 Gk → G to be the unique map such that π(x) = πk (x) for each x ∈ Gk and each k = 1, 2, . . .. Then ! ∞ X G = IG = πk End(G) G k=1

= =

=

∞ X k=1 ∞ X k=1 ∞ X

πk End(G)G πk (Gk ) ! π(Gk ) = π

M

Gk

.

k

k=1

By part 3 there is a map : G −→ ⊕k Gk and an integer P n 6= 0 such that π = n1G . Since G has finite rank we will write = m k=1 k for some finite list of maps k : G −→ Gk . Then k ∈ End(G) for each k = 1, . . . , m and hence n1G = π =

m X

πk k ∈ I.

k=1

Therefore End(G)/I is finite, QEnd(G) = QI, and thus part 1 is proved. This completes the logical cycle. For the next result, for each subgroup H of G let H∗ be the pure subgroup of G generated by H. We leave it as an exercise for the reader to show that if I ⊂ End(G) is a maximal right ideal then IG 6= G iff I = Hom(G, IG). Let N = N (End(G)). THEOREM 8.4.11 The following are equivalent for the rtffr group G. 1. G is a Q-faithful group. 2. I = Hom(G, IG∗ ) for each maximal pure right ideal I ⊂ End(G). 3. N (End(G)) = Hom(G, N G∗ ). 4. If π : P → QG is a projective cover over QEnd(G) then P generates QEnd(G). Proof: 1 ⇒ 2 Assume that G is Q-faithful and let I be a maximal pure right ideal in End(G). Evidently I ⊂ Hom(G, IG∗ ) and Hom(G, IG∗ ) is a pure right ideal in End(G). By our assumption on I, I = Hom(G, IG∗ ).

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211

2 ⇒ 3 Assume that I = Hom(G, IG∗ ) for each maximal pure right ideal in End(G). Since QN = J (QEnd(G)), QN is the intersection of the maximal right ideals of QEnd(G). Thus N is the intersection of the maximal pure right ideals of I. Then N

⊂ Hom(G, N G∗ ) ⊂ ∩{Hom(G, IG∗ ) I ⊂ End(G) is a maximal pure right ideal} = ∩{I I ⊂ End(G) is a maximal pure right ideal} by part 2 = N.

Thus N = Hom(G, N G∗ ), which proves part 3. 3 ⇒ 1 Suppose that G is not a Q-faithful group. By the comments beginning section 8.4.1 on page 204 there is a pure ideal ∆ ⊂ End(G) such that ∆QG = QG. Since QEnd(G)/N is a semi-simple Artinian ring there is an ideal N ⊂ I ⊂ QEnd(G) such that ∆ ∩ I = N and ∆ + I = QEnd(G). It follows that IQG = I(∆QG) ⊂ (I ∩ ∆)QG = N QG so that I is a nonnilpotent subset of Hom(G, N G∗ ). Hence N 6= Hom(G, N G∗ ) which completes the contrapositive. 1 ⇔ 4 In the notation of part 4, P generates QEnd(G) iff QEnd(G) is its trace ideal ∆QG iff IQG 6= QG for each proper right ideal I in QEnd(G) (see page 204), iff G is a Q-faithful group. This completes the logical cycle.

8.4.3

Q-Faithful E-Flat Groups

The results in this subsection show that when G is an E-flat Q-faithful group then the results in the previous section can be strengthened. The next result shows us that when G is an E-flat group then we lose very little information about δ(G) when we pass to δ(QG). LEMMA 8.4.12 Let G be an rtffr E-flat group. If QI ⊂ QEnd(G) and (QI)QG = QG then (QI ∩ End(G))G = G. In particular (∆QG ∩ End(G))G = G. Proof: Let I ∈ δ(QG). Since G is a flat left End(G)-module an application of · ⊗End(G) G to the inclusion End(G) QEnd(G) ⊂ (I ∩ End(G)) I

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yields the inclusions G (I ∩ End(G))G

∼ = ⊂ ∼ =

End(G) ⊗ G (I ∩ End(G)) End(G) QEnd(G) ⊗End(G) G I QG = 0. IQG

Thus G = (I ∩ End(G))G for each I ∈ δ(QG). This completes the proof. EXAMPLE 8.4.13 Let H and H 0 be strongly indecomposable rtffr groups such that Hom(H 0 , H) = 0, and such that SH (H 0 ) = Hom(H, H 0 )H is a full subgroup of H 0 , in which the factor group . H 0 /SH (H 0 ) is infinite. Let G = H ⊕ H 0 . Then H is a nilpotent quasisummand of G and ∆G,H = QHom(H, G)QHom(G, H) is in δ(QG). Observe that since Hom(H, H 0 )H 6= H 0 then G is not an E-flat group since otherwise G = (∆G,H ∩ End(G))G = Hom(H, G)Hom(G, H)G = Hom(H, G)H 6= G. This is an example of a quasi-isomorphism class (G) of rtffr groups in which no G0 ∈ (G) is an E-flat group. THEOREM 8.4.14 The following are equivalent for the E-flat rtffr group. 1. G is a Q-faithful group. 2. I = Hom(G, IG) for each maximal pure right ideal I ⊂ End(G). 3. N (End(G)) = Hom(G, N G). Proof: Given a pure right ideal I ⊂ End(G) there is an inclusion G ∼ End(G) QEnd(G) QG ⊗End(G) G ⊂ ⊗End(G) G = . = IG I QI QIQG Thus IG = IG∗ , whence the Theorem follows immediately from Theorem 8.4.11.

8.5. EXERCISES

8.5

213

Exercises

G, H, K are rtffr groups, p ∈ Z is a prime, E is an rtffr ring, M , N , are rtffr E-modules (left or right, depending on the setting). 1. Let H ∈ P(G) and suppose that (G, H) has the endlich Baer splitting property. If H is finitely G-generated then H ∈ Po (G). 2. Refer to the proof of Theorem 6.2.3. Show that each chain in C contains an upper bound in C. 3. Let M be a right End(G)-module. Then M (n) is G-compressed for each integer n > 0 iff TG (N ) 6= 0 for each finitely M -generated right End(G)-module N 6= 0. 4. Let P be a projective right E-module, let P ∗ be the dual of P , and let I be a left ideal of E. Prove that IP = P iff ∆P ⊂ I and that ∆P = ∆P ∗ . 5. Prove Proposition 8.1.4. 6. Let H ∈ Po (G). Show that Hom(G, H)∗ = Hom(H, G). 7. If IG 6= G for each proper finitely generated right ideal I ⊂ End(G) then G has the endlich Baer splitting property. 8. If H ∈ Po (G) and if N ⊂ HG (H) is such that N G = H then TG (HG (H)/N ) = 0. 9. Let G be a self-small right E-module. Then each H ∈ Po (G) is G-small. 10. Prove Example 8.1.7. 11. Prove Proposition 8.1.10. 12. A torsion-free group of finite rank is self-small. 13. If E is a reduced torsion-free ring of finite rank and if I is a maximal right ideal of E then E/I is finite. 14. If (G, H) has the endlich Baer splitting property then (G, H (n) ) has the endlich Baer splitting property for each integer n > 0. 15. Prove Theorem 8.2.14.

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16. Let G be a right E-module. Then Cyc(tor(G)) = {End(G)/I IG = G}. 17. Let X = Z, Y = Z[ p1 ], G = X ⊕ Y , and let ∆ = Hom(X, G)Hom(G, X). Show that G is a cyclic projective right End(G)-module, that ∆G = G, and that ∆ is a direct summand of End(G). 18. Let E be a ring, let Gab(E) be the set of Gabriel filters of E, and let Tor(E) be the set of hereditary torsion classes on E. There are inverse bijections Gab : Tor(E) −→ Gab(E) and Tor : Gab(E) −→ Tor(E). 19. Let ∆ be an idempotent ideal in E. Then Gab(∆) = {right ideals I ⊂ E ∆ ⊂ I} is a Gabriel filter on E. The hereditary torsion class corresponding to Gab(∆) is the set TorE (∆) of modules annihilated by ∆. 20. Show that I ∈ µ(M ) iff E/I is a semi-Artinian right E-module and each composition factor of E/I is isomorphic to some simple module in {E/J J ∈ maxE (M )}. 21. Let M be a right E-module. Then µ(M ) = δE (M ) ∩ µ(0). 22. Let M be a left E-module. If E/I is an Artinian module for each I ∈ δE (M ) then δE (M ) = µ(M ). 23. If N ⊂ K ⊂ L ⊂ M and if K is superfluous in M then L is superfluous in M and K/N is superfluous in M/N . 24. Suppose that P is a finitely generated projective left E-module. An E-submodule K ⊂ P is superfluous iff K ⊂ J(E)P . 25. If N is a nilpotent ideal then N M is superfluous in M for each left E-module M . 26. If ∆2 = ∆ ⊂ E then the hereditary torsion class corresponding to D(∆) is the set of right E-modules K such that K∆ = 0. 27. Show that µ(M ) equals the set of right ideals I ⊂ E such that E/I is a semi-Artinian E-module whose composition factors are isomorphic to E/J for some J ∈ maxE (M ). 28. Show that if π : P → M is a projective cover then M is finitely generated iff P is finitely generated.

8.5. EXERCISES

215

c) is a finitely Gabriel filter on E. b 29. Show that µ(M 30. Show that µ0 (M ) = {right ideals I ⊂ E E/I is a finite right Emodule such that IM = M and each composition factor of E/I is isomorphic to E/J for some J ∈ maxE (M )} is a Gabriel filter on the reduced torsion-free ring E of finite rank. 31. Show that Ib∩ E = I for each right ideal I of finite index in E, and 0 ∩ E = I 0 for each right ideal I 0 of finite index in E. b that I\ 32. Let E be a reduced torsion-free ring of finite rank and let M be a reduced torsion-free left E-module of finite rank. There is an b such that idempotent ideal ∆M ⊂ E b µ(M ) = {right ideals I ⊂ E E/I is finite and ∆M ⊂ Ib ⊂ E}. 33. Show that µ(G) is finite for the group G constructed in Example 6.1.7. 34. Let E be a Corner ring (not necessarily of finite rank) and let P be a finitely generated projective left E-module. Let ∆P be the trace of P in E. Then OE (∆M ) = OE (P ). c). 35. Show that OE (M ) is pure and dense in OEb (M b 1 + · · · + Ex b n=M c. Show that 36. P Let E be an rtffr ringPand let Ex n n b i=1 Exi . i=1 Exi is dense in 37. Let E be a local rtffr ring such that pE 6= E for some prime p ∈ Z. \ bp ). Prove that J (E)p = J (E 38. Let E be an rtffr ring. b is a semi-prime ring. (a) If E is a semi-prime ring then E b is a subcommutative (b) If E is a subcommutative ring then E ring. .

π 39. Let 0 → K → X −→ G → 0 be quasi-split . Then X ∼ = G ⊕ K.

40. Show that if H is a nilpotent quasi-summand of G and if H 0 is a nilpotent quasi-summand of H then H 0 is a nilpotent quasisummand of G. 41. Let G be a faithful rtffr group. Show that if I ⊂ End(G) is a maximal right ideal then I = Hom(G, IG) iff IG = 6 G.

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42. Let G be an rtffr group and suppose that G/H is finite. Then G is a finitely faithful S-group iff H is a finitely faithful S-group. 43. Let G be an rtffr group. Then G is a finitely faithful S-group iff G is a faithful L-group.

8.6

Questions for Future Research

Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. Give an example G in which {right ideals I ⊂ End(G) IG = G} is not a Gabriel filter. That is, find an rtffr group G such that D(G) 6= δ(G). 2. Give an internal characterization of the class Tor(G) of right E = End(G)-modules K such that K ⊗E G = 0. 3. Describe all Gabriel filters on the rtffr ring E. 4. Describe all hereditary torsion classes T of right E-modules. 5. Let E = End(G) and let I ⊂ E be a right maximal right ideal. If IG = G then HomE (I, E) = E = EndE (I). Describe the set of right ideals I ⊂ E such that HomE (I, E) = I. 6. Describe the idempotent ideals ∆ in E. 7. Describe the superfluous subgroups of G. 8. Describe the rtffr groups G that possess a projective cover over End(G). 9. Characterize the groups G such that End(G) is semi-perfect. 10. Further investigate δ(G) and µ(G). 11. Characterize those G such that δ(G) is finite. 12. Suppose δ(G) = D(∆) for some idempotent ideal ∆. Describe ∆ in terms of G. 13. Further characterize the faithful rings E.

8.6. QUESTIONS FOR FUTURE RESEARCH

217

14. Describe the Gabriel filters on QE and their associated hereditary torsion classes in terms of G. 15. Study (E-flat) Q-faithful rtffr groups.

Chapter 9

Endomorphism Modules In Chapter 4 we began a recurring theme, that certain E-properties coincide for rtffr groups. In this chapter we emphasize that coincidence. We will consider rtffr groups that are E-finitely generated, E-finitely presented, E-projective, E-generators, or E-flat. Some of these properties are classified up to quasi-isomorphism. We will also consider the homological dimensions of G as a left End(G)-module. The author’s research interests will of course greatly influence the list of properties that are examined here.

9.1

Additive Structure of Rings

A left E-module M is said to be a finitely generated rtffr E-module if M is a finitely generated left or right E-module whose additive structure (M, +) is an rtffr group. What follows is an extension of the BeaumontPierce-Wedderburn Theorem 3.1.6 from rings to finitely generated rtffr modules. Such a discussion is fundamental to our subsequent discussion of rtffr modules over rtffr rings. LEMMA 9.1.1 Let E be an rtffr ring and let T = E/N (E). Let M be an rtffr left E-module and suppose that there is an E-submodule K ⊂ M such that M/K is a finitely generated rtffr T -module. Then .

1. (M/K) ⊕ M 0 ∼ = T (n) for some finitely generated rtffr left T -module 0 M and some integer n, and .

2. M ∼ = (M/K) ⊕ K as groups. Proof: 1. Since M/K is a finitely generated rtffr left T -module QM/QK is a finitely generated projective left QT -module. There is 219

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a finitely generated projective QT -module P and an integer m such that (QM/QK) ⊕ P ∼ = QT (m) . Since P is finitely generated there is a finitely generated T -submodule M 0 ⊂ P such that QM 0 = P . Then (M/K) ⊕ M 0 ⊂ QT (m) and since both M/K and M 0 are finitely generated T -modules there is an integer n 6= 0 such that n(M/K ⊕ M 0 ) ⊂ T (m) . Furthermore since T (m) n(M/K ⊕ M 0 ) is a finitely generated T -module that is also a torsion group there is an integer n0 6= 0 such that n0 T (m) ⊂ n(M/K ⊕ M 0 ) ⊂ T (m) . This proves part 1. 2. Since M/K is a finitely generated left T -module there are x1 , . . . , xr ∈ M such that . M = T x1 + · · · + T xr + K. Thus there is a T -module mapping π : T (r) −→ M −→ M/K which lifts to a surjection π : QT (r) −→ QM/QK of projective QT -modules over the semi-simple ring QT . There is a QT module map f : QM/QK −→ QT (r) such that πf = 1QM/QK . Since M/K is a finitely generated T -module there is an integer n 6= 0 such that image nf ⊂ T (r) so that π(nf ) = n1M/K . It follows that . M∼ = (M/K) ⊕ K. This completes the proof. To classify the additive structure of rtffr semi-prime rings up to quasiisomorphism it suffices to classify the additive structure of a classical maximal order E over its center S. The next result follows immediately from [69, Theorem 39.14] which implies that a classical maximal order E is a projective S-module so that E ∼ = S (m−1) ⊕ J for some ideal J ⊂ S.

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221

LEMMA 9.1.2 Let E be a prime rtffr ring and let S = center(E). . There is an integer m 6= 0 such that E ∼ = S (m) as S-modules. Our classification of the additive structure of a semi-prime rtffr ring E is reduced to the case where E = S is a Dedekind domain. The ring E is a strongly indecomposable ring if (E, +) is strongly indecomposable. The next result is due to R. A. Beaumont and R.S. Pierce [18]. THEOREM 9.1.3 Let E be a Dedekind domain. There is a strongly . indecomposable Dedekind domain R ⊂ QE such that E = R(m) for some integer m. Moreover R ∼ = center(End(E, +)) as rings. Proof: Suppose that E is a Dedekind domain. Let . R = {Dedekind domains R ⊂ QE RE = E and RE is a finitely generated R-module }. Since E is an rtffr Dedekind domain E ∈ R 6= ∅ and since E has finite rank R contains a Dedekind domain R of minimal rank. Since a finitely generated torsion-free R-module is projective and since RE is a finitely . . generated torsion-free R-module E = RE = R(m) for some integer m 6= 0. We claim that R is strongly indecomposable as a group. Observe that R is a cyclic left End(R, +)-module. Given an ideal I 6= 0 ⊂ End(R, +) then IQR 6= 0 is an ideal in QR. Since QR is a field, IQR = QR, so that I 2 6= 0. Thus End(R, +) is a prime ring. By Lemma 9.1.2, End(R, +) is finitely generated over the integral domain S = center(End(R, +)). . Then R is a finitely generated rtffr S-module and thus R = S (`) for some integer ` > 0. By the minimality of R, S has finite index in R, and since S is a pure subgroup of End(R, +), S = R. . Finally write R = B (r) for some strongly indecomposable group B and integer r > 0. Since R = S commutes with the idempotent projections mapping R onto each copy of B, B is an ideal in R, hence R/B is . finite, whence R = B is strongly indecomposable.

9.2

E-Properties

In this section we will visit each of the E-properties mentioned in the introduction of this chapter.

222

9.2.1

CHAPTER 9. ENDOMORPHISM MODULES

E-Rings

Given a ring E with identity there is a natural embedding of rings λ : E → End(E, +) such that for f ∈ E, λ(f )(x) = f x for each x ∈ E. That is, λ(f ) is left multiplication by f on E. We ask When is λ an isomorphism? The ring E is called an E-ring if λ is an isomorphism. Examples of E-rings include Z, Q, Z/nZ for integers n. Other examples are hard to come by in the rtffr case. But once you find them they can be used to illuminate difficult points in our research. The existence of E-rings with a variety of ring structures is demonstrated in the following result. However, the constructed E-rings have countably infinite rank. See Theorem J.1.3 for a proof. THEOREM 9.2.1 [42, T.G. Faticoni] Each countable reduced torsionfree commutative ring is a pure subring of an E-ring. The countable hypothesis in the above theorem can be relaxed as M. Dugas, A. Mader, and C. Vinsonhaler [28] show us that each cotorsionfree commutative ring is a pure subring of an E-ring. Thus there are plenty of E-rings. For instance, if x1 , x2 , x3 , . . . are commuting indeterminants then Z[x1 , x2 , x3 , . . .]/(x21 , x32 , x43 , . . .) and Z[x1 , x2 , x3 , . . .] are contained in E-rings E1 and E2 respectively. These constructions show that in general the ring theoretic structure of an E-ring can be quite complicated. However the rtffr E-rings have a relatively uncomplicated ring structure. We will call the ring E a Dedekind E-ring if E is a Dedekind domain and an E-ring. See Appendix K. THEOREM 9.2.2 Let E be an rtffr integral domain. Then E is an E-ring iff E is a strongly indecomposable ring. In this case there is a unique Dedekind E-ring E such that E ⊂ E and E/E is finite. Proof: Suppose that E is not strongly indecomposable. Then there are nonzero subgroups A and B of E such that E/(A ⊕ B) is finite. There is a group map e : E → E such that e2 = ne for some integer . n 6= 0 and eE = A. Then e, n − e ∈ End(E) are such that e(n − e) = 0. Since E is an integral domain, End(E) 6= E, and then E is not an Ering. Consequently if the domain E is an rtffr E-ring then E is strongly indecomposable.

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223

Conversely, suppose that E is strongly indecomposable rtffr integral domain. By Theorem 9.1.3, E ∼ = S = center(End(E)). Since E is an integral domain End(E) is a prime ring so that QEnd(E) ∼ = Matn (D). 2 for some division ring D. Let e = e 6= 0 ∈ QEnd(E). Then E ∼ = . eE ⊕ (1 − e)E and since E is strongly indecomposable E = eE. That is, e = 1E . Then QE ⊂ QEnd(E) = D is a QS-vector space and QE = QS is a D-vector space. Since these rings are finite dimensional D = QS = QE and hence E = S = End(E). Therefore E is an E-ring, and the proof is complete. Our next result extends the Beaumont-Pierce Theorem 9.1.3 on the additive structure of rtffr prime rings. THEOREM 9.2.3 Let E be a prime rtffr ring. Then S = . (n) center(End(E)) is a strongly indecomposable E-ring and E = S as S-modules where S is a strongly indecomposable Dedekind E-ring such . that S = S. . (m) Proof: We proved in Theorems 9.1.2 and 9.1.3 that E = S where S is a strongly indecomposable integral domain. By Theorem 9.2.2, S . is an E-ring, and since S = S, S is a strongly indecomposable Dedekind E-ring. COROLLARY 9.2.4 Let E be a semi-prime rtffr ring. There are strongly indecomposable Dedekind E-rings R1 , · · · , Rt and integers . n1 , . . . , nt ≥ 0 such that Ri = Rj implies that i = j and such that . (n ) (n ) E = R1 1 ⊕ · · · ⊕ R t t as groups. The additive structure of rtffr E-rings is contained in the following theorem. A proof can be found in [10, Corollary 14.7] but try to prove it yourself without peeking. THEOREM 9.2.5 [73, R. Bowshell, P. Schultz] Let E be an rtffr ring. Then E is an E-ring iff there are strongly indecomposable Dedekind E-rings E 1 , . . . , E t such that 1. Hom(E i , E j ) = 0 for each 1 ≤ i 6= j ≤ t and 2. n(E 1 ⊕ · · · ⊕ E t ) ⊂ E ⊂ E 1 ⊕ · · · ⊕ E t for some integer n > 0.

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We have reduced the classification of rtffr E-rings to the case where the rtffr ring E is a Dedekind E-ring. The key to understanding Dedekind E-rings is an understanding of the distribution of the maximal ideals in the ring of algebraic integers in an algebraic number field. See Theorem K.2.4 for a proof. THEOREM 9.2.6 [T.G. Faticoni] Suppose that the rtffr ring E is a Dedekind domain and let O denote the ring of algebraic integers in QE. Then E is an E-ring iff given a proper subfield Q ⊂ k ⊂ QE there are maximal ideals M, M 0 ⊂ O such that M E 6= E, M 0 E = E, and M ∩ k = M 0 ∩ k. The field k is a minimal field extension of Q if Q and k are the only subfields of k. For minimal field extensions k of Q the classification of E-rings in k is a bit simpler. THEOREM 9.2.7 Let E be a Dedekind domain and let O be the ring of algebraic integers in QE. Further suppose that QE is a minimal field extension of Q. Then E is an E-ring if there is a prime p ∈ Z and maximal ideals M , M 0 ⊂ O such that M E 6= E, M 0 E = E, and p ∈ M ∩ M 0. Let p ∈ Z be a prime. The group G is p-local if pG 6= G and qG = G for each prime q 6= p ∈ Z. Theorem 9.2.6 and some Algebraic Number Theory will show that each algebraic number field k is the field of fractions of some Dedekind E-ring E. The question of which fields are of the form QE for some p-local Dedekind E-ring E is answered masterfully by R.S. Pierce [65] and R.S. Pierce and C. Vinsonhaler [66]. The number field k is said to be p-realizable if k = QE for some p-local Dedekind E-ring E. See [65, Theorem 2.1] for a proof of the following theorem. THEOREM 9.2.8 [R.S. Pierce] Let Γ be the Galois group Γ = Gal(k/Q) where k is a finite Galois extension of Q. If Γ has no nontrivial, normal, cyclic subgroups, and if Γ is not isomorphic to the symmetric group S4 or the alternating groups A4 , then k is p-realizable for all primes p ∈ Z.

9.2.2

E-Finitely Generated Groups

The group G is E-cyclic if G = End(G)x for some x ∈ G, and G is Efinitely generated if G is a finitely generated left End(G)-module. Each ring is an E-cyclic group, and in general if G is a finitely generated

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225

left module over some ring then G is E-finitely generated. Evidently Noetherian rtffr E-rings and their nonzero ideals are E-finitely generated groups. Furthermore if M is a left E-module then E ⊕ M is an E-cyclic group. Fortunately the class of E-finitely generated rtffr groups is closed under quasi-isomorphism. Given an integer m > 0 we have constructed in Example 4.3.2 a group G that requires exactly m ≥ 2 generators as a left End(G)-module. The group in question is quasi-equal to an E-cyclic group. Thus the number of generators for an E-finitely generated rtffr group can vary wildly in the quasi-isomorphism class of the group. We will present J.D. Reid’s [67] classification of the E-finitely generated rtffr groups. Our presentation differs from that in [67]. LEMMA 9.2.9 Let E be a semi-prime rtffr ring and let M and M 0 be torsion-free left E-modules with M 0 finitely generated. If π : M → M 0 . is a surjection then M ∼ = ker π ⊕ M 0 . Proof: Observe that π lifts to a surjection π : QM → QM 0 of left modules over the semi-simple ring QE. Since every QE-module is projective QM ∼ = ker π ⊕ QM 0 . Thus there is a QE-module map σ : QM 0 → QM such that πσ = 1QM 0 . Since M 0 is finitely generated there is an integer n 6= 0 such that nσ(M 0 ) ⊂ M . Inasmuch as e = σπ = (σπ)2 ∈ EndQT (QM ) and since ne ∈ EndT (M ), we have n1M = n(1 − e) ⊕ ne. Hence (n1M )M ⊂ [n(1 − e)](M ) ⊕ (ne)(M ) ⊂ ker(π|M ) ⊕ (nσ)(M 0 ) ⊂ M which completes the proof. LEMMA 9.2.10 Let E be a semi-prime rtffr and let M be a finitely generated torsion-free left E-module. There are integers t, n1 , . . . , nt > 0 and Dedekind rtffr E-rings R1 , . . . , Rt such that .

(n ) (n ) M∼ = R1 1 ⊕ · · · ⊕ Rt t .

Proof: Since M is finitely generated there is an integer n and a surjec. (n) (n) ∼ tion π : E → M . By Lemma 9.2.9, E = M ⊕ker π, and by Theorem 9.2.4, there are integers t, m1 , . . . , mt > 0 and strongly indecomposable Dedekind E-rings R1 , . . . , Rt such that .

(m ) (m ) E∼ = R1 1 ⊕ · · · ⊕ R t t .

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Then by J´onsson’s Theorem 2.1.10, there are integers t, n1 , . . . , nt such that . (n ) (n ) M∼ = R1 1 ⊕ · · · ⊕ Rt t which completes the proof. The next result is a classification of the group structure of a finitely generated left module over an rtffr ring. Recall that X∗ is the purification of X in a larger group. THEOREM 9.2.11 [T.G. Faticoni] Let E be an rtffr ring and let M . be an rtffr left E-module. If M is finitely generated then M = Mo ⊕ N where (n )

(n )

1. Mo = R1 1 ⊕ · · · ⊕ Rt t for some integers t, n1 , . . . , nt > 0 and Dedekind E-rings R1 , . . . , Rt , 2. N = (N (E)Mo )∗ = the purification of N (E)Mo in M , and 3. (N (E)Mo )∗ /N (E)Mo is finite. Proof: Let E be an rtffr ring and let M be a finitely generated rtffr left E-module. By the Beaumont-Pierce-Wedderburn Theorem 3.1.6 there is . a subring T ⊂ E such that E = T ⊕ N (E) as groups and it is clear that Mo = M/(N (E)M )∗ is a finitely generated left T -module. Then by Lemma 9.2.9 .

M∼ = Mo ⊕ (N (E)M )∗ as T -modules. Furthermore by Lemma 9.2.10 there are integers t, n1 , . . ., nt > 0 and Dedekind E-rings R1 , . . . , Rt such that (n1 )

Mo = R 1

(nt )

⊕ · · · ⊕ Rt

.

Evidently the module M/N (E)M is finitely generated over the Noetherian ring E/N (E) so (N (E)M )∗ /N (E)M is a finitely generated torsion E/N (E)-module. Since (N (E)M )∗ /N (E)M is a bounded quotient of an rtffr group, it is finite. This proves the theorem. THEOREM 9.2.12 [J.D. Reid] (See [67].) Let G be an rtffr group and let E = End(G). The following are equivalent. 1. G is E-finitely generated.

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227

2. There are integers t, n1 , . . . , nt > 0, strongly indecomposable Dedekind E-rings R1 , . . . , Rt , and a group N such that . (n ) (n ) G = R1 1 ⊕ · · · ⊕ Rt t ⊕ N and where N = SR1 ⊕···⊕Rt (N ). Proof: Assume part 1. Let E = End(G) and suppose that G is an E-finitely generated rtffr group. Let Go = G/(N (E)G)∗ . By Lemma 9.2.9 . G∼ = Go ⊕ (N (E)Go )∗ where (N (E)Go )∗ /N (E)Go is finite. Let N 0 = N (E)Go . Inasmuch as . (N (E)Go )∗ /N 0 is finite, and since N (E)Go = SGo (N 0 ), G ∼ = Go ⊕ N 0 for some group N 0 such that SGo (N 0 ) = N 0 . By Theorem 9.2.11 there are integers t, n1 , . . ., nt > 0 and strongly indecomposable Dedekind E-rings R1 , . . . , Rt such that .

(n ) (n ) Go ∼ = R1 1 ⊕ · · · ⊕ Rt t . .

∼ SR ⊕···⊕R (N 0 ) so we let N = SR ⊕···⊕R (N 0 ). Hence G = Then N 0 = t t 1 1 Go ⊕ N for some rtffr group N such that N = SR1 ⊕···⊕Rt (N ). This proves part 2. . Conversely, assume part 2. Suppose that G = Go ⊕ N where Go and N satisfy the conditions in part 2. If we let R = R1 × · · · × Rt then Go is a finitely generated R-module, and hence Go is an E-finitely generated group. Let x1 , . . . , xr ∈ Go be generators for Go over R. Inasmuch as R and Hom(Go , N ) embed naturally in QEnd(G) we have the following equations. . G = Go ⊕ SR (N ) . = Go ⊕ Hom(Go , N )Go = End(Go )x1 + · · · + End(Go )xr + Hom(Go , N )Go = End(Go )x1 + · · · + End(Go )xr + Hom(Go , N )End(Go )x1 + · · · + Hom(Go , N )End(Go )xr = End(Go )x1 + · · · + End(Go )xr + Hom(Go , N )x1 + · · · + Hom(Go , N )xr . = End(G)x1 + · · · + End(G)xr . Thus G, and hence any group quasi-isomorphic to G is E-finitely generated. This concludes the proof.

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Several of the E-properties imply the E-finitely generated property for rtffr groups. The group G is E-projective if G is a projective left End(G)-module, and G is an E-generator if G generates the category of left End(G)-modules. G is an E-progenerator if G is a finitely generated projective generator in the category of left End(G)-modules. Groups of the form Z ⊕ N are E-projective while E-rings are E-progenerator groups. COROLLARY 9.2.13 Let G be a strongly indecomposable E-finitely . generated rtffr group. Then G = E for some Dedekind E-ring E. Proof: If G is a strongly indecomposable E-finitely generated rtffr . group then by Theorem 9.2.12, G = R1 = E for some Dedekind E-ring E. COROLLARY 9.2.14 Let G be an rtffr group. Then G is E-finitely generated iff G is quasi-isomorphic to an E-cyclic group. Proof: Let E = End(G) and suppose that G is an E-finitely generated rtffr group. By Theorem 9.2.12 there are positive integers t, n1 , . . . , nt , strongly indecomposable E-rings R1 , . . . , Rt , and a group N such that N

= SGo (N ) (n )

(nt )

Go = R1 1 ⊕ · · · ⊕ Rt . G = Go ⊕ N.

, and

It suffices to show that Go ⊕N is an E-cyclic group. Let xi be a generator of Ri as an Ri -module and let x = x1 ⊕ · · · ⊕ xt . Since these are group homomorphisms, End(Go ⊕N )x contains each copy of Ri in Go . Thus End(Go ⊕ N )x ⊃ Go . Furthermore, by hypothesis N = SGo (N ) = Hom(Go , N )Go = Hom(Go , N )x so that End(Go ⊕ N )x ⊃ N . Therefore Go ⊕ N = End(Go ⊕ N )x. COROLLARY 9.2.15 An E-projective rtffr group is E-finitely generated.

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229

Proof: Let E = End(G), suppose that G is E-projective, and write G⊕P ∼ = E (I) for some index set I. Since G has finite rank there is a finitePsubset {x1 , . . . , xn } ⊂ G and a finite subset Io ⊂ I such that QG = ni=1 Qxi and {x1 , . . . , xn } ⊂ E (Io ) . Then G = E (I) ∩ QG ⊂ E (Io ) . The reader can show that because G is a direct summand of E (I) , G is a direct summand of E (Io ) , thus proving that G is finitely generated as a left E-module. This proves the lemma. COROLLARY 9.2.16 Let G be an E-generator rtffr group and let S = center(End(G)). Then G is E-finitely generated and G is a finitely generated projective S-module. Proof: Let E = End(G), suppose that G is an E-generator, and let S = EndE (G). Since G is a generator for E, G is a finitely generated projective right S = EndE (G)-module. But S ⊂ End(G) so that G is E-finitely generated. For strongly indecomposable rtffr groups we can prove a strong relationship between these E-properties. The rtffr group G is E-finitely presented if G is a finitely presented left E-module. THEOREM 9.2.17 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group. The following are equivalent. 1. G is quasi-isomorphic to a Dedekind E-ring. 2. G is quasi-isomorphic to an E-cyclic group. 3. G is an E-finitely generated group. 4. G is an E-finitely presented group. 5. G is quasi-isomorphic to an E-projective group. 6. G is quasi-isomorphic to an E-generator group. 7. G is quasi-isomorphic to an E-progenerator group. Proof: 1 ⇒ 2 ⇒ 3 is clear. . 3 ⇒ 4 By Lemma 9.2.13, G = R for some strongly indecomposable . Dedekind E-ring R, so that End(G) = R is a Noetherian integral domain. The E-finitely generated G is thus E-finitely presented.

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. 4 ⇒ 5 and 6 By Lemma 9.2.13, G = R for some strongly indecomposable Dedekind E-ring R. This proves part 5 and 6. 5 or 6 ⇒ 1 Suppose that G is an E-projective rtffr group or an Egenerator rtffr group. By Lemmas 9.2.15 and 9.2.16, G is E-finitely generated, and since G is a strongly indecomposable group, Lemma 9.2.13 . implies that G = R for some Dedekind E-ring R. This proves part 1. Inasmuch as 1 ⇒ 7 ⇒ 5 is clear the proof is complete. An example will show that some of these implications do not possess a converse. EXAMPLE 9.2.18 There is an E-cyclic rtffr group that is not Eprojective. Proof: Let k be a quadratic field extension of Q and let O denote the ring of algebraic integers in O. There is a prime p ∈ Z such that pO = M M 0 for some maximal ideals M 6= M 0 in O. Let \ S = OM ∩ {Oq q 6= p ∈ Z is a prime }. Since M OM 6= OM , M S 6= S, and since M 0 Ox = Ox for each x ∈ {I, primes q 6= p}, M 0 S = M 0 . By Theorem 9.2.7, S is an E-ring and one can show that O/qO ∼ = S/qS ∼ = (Z/qZ)(2) for each prime q 6= p. ∼ Furthermore S/pS = Z/pZ. Thus Z is a pure subring of S. By counting p-ranks of the rank one group N = S/Z we see that qN 6= N for each prime q 6= p while pN = N . Therefore G = S ⊕N is E-finitely generated. Note that N is not an S-module even though it is S-generated. Thus G is E-cyclic but not E-projective.

9.2.3

E-Projective Groups

. LEMMA 9.2.19 Let S be a commutative rtffr ring and let I = S be an ideal. If I is a projective S-module then I ⊕ I = S ⊕ K for some S-module K. Proof: By hypothesis S/I is a finite commutative ring so that S/I = T1 × · · · × Tr for some local rings T1 , . . . , Tr . Thus I/I 2 = L1 ⊕ · · · ⊕ Lr where Li is a Ti -module. Since the projective S-module I generates S, Li generates Ti , and since Ti is a local ring, there is a surjection Li → Ti . Therefore there is a surjection I → I/I 2 → S/I that lifts to a map f : I → S such that f (I) + I = S. It follows that I ⊕ I ∼ = S ⊕ ker(f ⊕ 1I ) where (f ⊕ 1I ) : I ⊕ I → S is the obvious surjection. This proves the lemma.

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231

Consider the problem of classifiying the E-projective rtffr groups. The investigation is complicated by the fact that if N is any rtffr group then Z ⊕ N is E-cyclic and E-projective. Another example of an Eprojective group is as follows. Let S be an E-ring, and let N be an S-module. We say that N is an S-linear module if HomS (S, N ) = Hom(S, N ). Our S-linear modules are referred to as E-modules by A. Mader and C. Vinsonhaler [61]. Any group is a Z-linear module, a Q-vector space is a Q-linear module, and one can prove that if I is an ideal of finite index in an E-ring S then I is an S-linear module. EXAMPLE 9.2.20 We give a general construction of E-projective rtffr . groups. Let S be an E-ring, let I = S be a projective nonprincipal ideal, and let N be an S-linear module. Then G=I ⊕N is an E-projective group that is generated by two elements as a left End(G)-module. Proof: Since S is an E-ring I is S-linear and since N is S-linear I = HomS (S, I) = Hom(S, I) and N

= HomS (S, N ) = Hom(S, N ).

The following chain of equations shows that G = I ⊕ N is S-linear. G = HomS (S, G) ∼ = HomS (S, I) ⊕ HomS (S, N ) ∼ = Hom(S, I) ⊕ Hom(S, N ) ∼ = Hom(S, I ⊕ N ) ∼ = Hom(S, G). The next series of equations shows us that G is direct summand of . End(G)(2) . Since I = S is a projective ideal Lemma 9.2.19 states that I ⊕I ∼ = S ⊕ K for some S-module K. Then because G is S-linear, End(G) ⊕ End(G) ∼ = Hom(I ⊕ I, G) ⊕ Hom(N ⊕ N, G) ∼ = Hom(S ⊕ K, G) ⊕ Hom(N ⊕ N, G) ∼ = Hom(S, G) ⊕ Hom(K ⊕ N ⊕ N, G) ∼ = G ⊕ Hom(K ⊕ N ⊕ N, G)

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as left EndS (G) = End(G)-modules. Thus G is a projective left End(G)module that is generated by at most two elements as a left End(G)module. EXAMPLE 9.2.21 Let S be an rtffr E-ring and let N be an rtffr S-linear module. By the above example G = S ⊕ N is an E-cyclic Eprojective rtffr group. Our work here will show that the above examples are the only kind of E-projective rtffr group. To motivate our discussion of E-cyclic Eprojective rtffr groups, notice that by Lemmas 9.2.14 and 9.2.15 every E-projective rtffr group is quasi-isomorphic to an E-cyclic E-projective rtffr group. THEOREM 9.2.22 [63, G.P. Niedzwecki, J.D. Reid] Let G be an rtffr group, let E = End(G), and let S = center(E). Then G is an E-cyclic E-projective group iff G ∼ = S ⊕ N for some S-linear module N . In this case S is an rtffr E-ring. Proof: Suppose that G = S ⊕ N where S = center(E) and where N is an S-linear module. We have shown in Example 9.2.21 that G is E-cyclic and E-projective. Conversely let E = End(G) and assume that G is an E-cyclic Eprojective rtffr group. There is a split surjection π:E→G of left E-modules so there is an e2 = e ∈ E such that G ∼ = Ee. Thus S = EndE (G) = EndE (Ee) ∼ = eEe by Lemma 1.2.4. If we let N = (1 − e)G then G = eG ⊕ (1 − e)G ∼ = eEe ⊕ N ∼ =S⊕N where N is an S-module. On the other hand because G ∼ = S ⊕ N each group homomorphism f : S → S commutes with each element of S so that S∼ = HomS (S, S) = Hom(S, S) is an E-ring. Similarly HomS (S, N ) = Hom(S, N )

9.2. E-PROPERTIES

233

so that N is an S-linear module. This completes the proof of the theorem.

We have more to say about E-projective groups once we have studied E-generator groups. See Theorem 9.2.28. It is known that P is a finitely generated projective right E-module iff P is a generator as a left EndE (P )-module. This observation connects E-projective groups with the generator property as follows. LEMMA 9.2.23 Let G be an rtffr group and let S = center(End(G)). Then G is an E-projective group iff G is a generator in the category of right S-modules. Proof: Observe that S = EndEnd(G) (G). Then by Lemma 9.2.15, G is E-projective iff G is E-finitely generated E-projective iff G is a generator as a right S-module. The module U is called quasi-projective if U is projective relative to each short exact sequence 0 −→ K −→ U −→ V −→ 0 of left E-modules. The group G is E-quasi-projective if G is a quasiprojective left End(G)-module. The reader can prove as an exercise that . if E is an rtffr E-ring and if E = I is a projective ideal then E and I are locally isomorphic as groups and as E-modules. PROPOSITION 9.2.24 [84, C. Vinsonhaler, W. Wickless] Let G be an rtffr group. Then G is E-quasi-projective iff G is an E-finitely generated E-projective group. Proof: Let G be a quasi-projective left E-module. Because G is an rtffr group there is an integer m 6= 0 such that ∩k>0 mk G = 0. Since G/mG is a finite group there is a finitely generated E-submodule F ⊂ G such that (F +mG)/mG = G/mG = M . The usual properties of relative projectivity (see [6, Proposition 16.12]) show us that the quasi-projective left E-module G is projective relative to each short exact sequence 0 → K −→ F −→ G/mG → 0 of left E-modules. Thus the natural projection G → G/mG lifts to a map f : G → F such that ker f ⊂ mG. Since G is torsion-free ker f is pure in G and so ker f = m(ker f ) = 0 by our choice of m. Thus f is a

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regular endomorphism of G, hence f (G) ⊂ F ⊂ G, whence G/F is finite. (Prove that one, reader.) Then G is a finitely generated quasi-projective left E-module. Conversely, suppose that G is E-projective and is generated by n elements. Because G has finite rank there is a finite set {x1 , . . . , xr } ⊂ G such that annE (x1 , . . . , xr ) = 0. Then x = x1 ⊕ · · · ⊕ xr ∈ G(r) satisfies annE (x) = 0 so that E ∼ = Ex ⊂ G(r) . That same theory of relative projectivity in [6] shows us that the quasi-projective E-module G is projective relative to each short exact sequence 0 → K −→ E (n) −→ G → 0. Then the identity 1G lifts to a map f : G → E (n) . That is, our sequence is split exact, and therefore G is a projective left E-module. This completes the proof.

9.2.4

E-Generator Groups

A left E-module U is a generator over E iff U is finitely generated projective as a right EndE (U )-module. Thus we are motivated to study the E-generator groups. That is, those groups G that are generators in the category of left End(G)-modules. Recall that G is called an E-progenerator if G is a finitely generated projective generator in the category of left End(G)-modules. THEOREM 9.2.25 [T.G. Faticoni] Let G be an rtffr group, let E = End(G), and let S = center(E). The following are equivalent. 1. G is an E-progenerator group. 2. G is an E-generator group. 3. G is a finitely generated projective S-module. 4. G ∼ = U1 ⊕ · · · ⊕ Us for some projective ideals U1 , . . . , Us ⊂ S. Proof: 1 ⇒ 2 is clear and 2 ⇒ 3 follows from the introductory motivational statement. 3 ⇒ 4 Since G is a finitely generated projective module over the commutative rtffr ring S, Theorem 4.2.11 states that G = U1 ⊕ · · · ⊕ Us for some projective ideals U1 , . . . , Us ⊂ S.

9.2. E-PROPERTIES

235

4 ⇒ 1 Since G is a finitely generated projective S-module G is a generator for EndS (G). The reader will recall that E = EndS (G) so that G is a generator as a left E-module. On the other hand we will show that G is a generator over S. Let ∆ be the trace ideal in S for the projective S-module G. Since the trace is an idempotent ideal in S, Lemma 4.2.6 states that ∆ = eS for some e2 = e ∈ S. Inasmuch as annS (G) = 0, and since eG = ∆G = G, e = 1G . That is G generates S. Then G is a finitely generated projective left module over EndS (G) = E. This proves 1 which completes the logical cycle. COROLLARY 9.2.26 Let G be an rtffr group and let S = center(End(G)). Then G is an E-generator iff G = U ⊕ U 0 where U is an invertible ideal of S and U 0 is some projective S-module. Proof: Let G = U ⊕U 0 for some invertible ideal U over S and some Smodule U 0 . Then G is finitely generated projective over S, which implies that G is E-projective. Conversely, suppose that G is an E-generator group. Since S = S1 × · · · × St is a finite product of indecomposable commutative rtffr rings S1 , . . . , St there is no loss of generality if we prove the result for G under the assumption that S is indecomposable. So, Theorem 9.2.25 states that G = U1 ⊕ · · · ⊕ Us for some indecomposable finitely generated projective ideals U1 , . . ., Us of S. Let ∆ = traceS (U1 ). By Lemma 4.2.6, there is an e2 = e ∈ S such that ∆ = eS. Since S is indecomposable e = 1G and hence U1 generates S. In other words U1 is an invertible ideal in S. This completes the proof. The group G is E-self-generating if G generates each End(G)-submodule of G(n) for each n > 0. The following result shows that the notions of an E-generating and E-self-generating rtffr are interchangeable. LEMMA 9.2.27 Let G be an rtffr group. Then G is an E-self-generating group iff G is an E-generator group. Proof: Let E = End(G). Let x1 , . . . , xn ∈ G be a maximal linearly independent subset of G. Then f (x1 , . . . , xn ) = 0 iff f (G) = 0 iff f = 0. Thus E∼ = E(x1 ⊕ · · · ⊕ xn ) ⊂ G(n) as left E-modules. Then G generates E so that G is a generator as a left E-module. The converse is clear, so the proof is complete.

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9.2.5

E-Projective Rtffr Groups Characterized

Lemma 9.2.15 leaves open the question of just how many generators an E-projective rtffr group requires. That issue is addressed in the famous paper [16] written at the University of Connecticut during its Special Year in Algebra 1981. The main theorem of [16] is our next result. Let us agree to call the module M a local direct summand of N if given an integer n 6= 0 there is an integer m and module maps fn : N → M and gn : M → N such that fn gn = m1M . THEOREM 9.2.28 [UConn ’81 Theorem] Let G be an rtffr group, let E = End(G), and let S = center(E). The following are equivalent. 1. G is an E-projective group and generated by two elements. 2. G is an E-projective group. 3. G is an E-quasi-projective group. 4. G is a local direct summand of E. 5. G is E-finitely generated and Gp is a cyclic projective left Ep module for each prime p ∈ Z. 6. There is an rtffr E-ring S, an invertible ideal I in S, and an S-linear module N such that G = I ⊕ N . Proof: We will prove the logical circuit 6 ⇒ 5 ⇒ 4 ⇒ 1 ⇒ 2 ⇒ 3 ⇒ 6. 6 ⇒ 5 Suppose that G ∼ = I ⊕ N as in part 6. Since I is invertible, QI is an invertible ideal in the Artinian commutative ring QS, so that . QI = QS. Hence I = S. Because I is a generator over S, there is an integer n and an S-module K such that I (n) ∼ = S ⊕ K. Subsequently HomS (I, G)(n) ∼ = HomS (I (n) , G) ∼ = HomS (S, G) ⊕ HomS (K, G) as left E-modules. Then G ∼ = HomS (S, G) is a direct summand of HomS (I, G)(n) ⊕ HomS (N, G)(n) ∼ = HomS (I ⊕ N, G)(n) ∼ = E (n) as left E-modules, and so G is E-finitely generated. Furthermore, let p ∈ Z be prime. Choose n ∈ Z large enough that . Sn is a reduced group. Since I = S is an invertible ideal in S, In is an invertible ideal in Sn . Since Sn /nSn is then generated by In /nIn and

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237

since the finite commutative ring Sn /nSn is a product of local rings, there is a map f : In → Sn such that f (In ) + nSn = Sn . By Lemma 1.2.1 and . our choice of n, n ∈ J (Sn ), and hence f (In ) = Sn . Since Sn = In have finite rank, In ∼ = Sn . Because S and I are finitely presented S-modules we can apply the Change of Rings Theorem 1.5.2 to show that HomS (I, G)n ∼ = HomSn (In , Gn ) ∼ = HomSn (Sn , Gn ) ∼ = HomS (S, G)n . Thus Gn = HomS (S, G)n is a direct summand of HomS (I ⊕ N, G)n ∼ = En as a left En -module. This proves part 5. 5 ⇒ 4 is proved along the lines of Lemma 2.5.2. 4 ⇒ 1 Suppose that G is a local summand of E. We claim that G is E-projective and generated as a left E-module by two elements. Let n 6= 0 ∈ Z. There is an integer m and E-module maps fn : E → G and gn : G → E such that gcd(m, n) = 1 and fn gn = m1G . By part 4, there is an integer k and maps fm : E → G and gm : G → E such that fm gm = k1G . Since gcd(k, m) = 1 there are integers a, b such that ak + bm = 1. Then the maps f : E ⊕ E −→ G : (x, y) 7→ afm (x) + bfn (y) g : G −→ E ⊕ E : z 7→ gm (z) ⊕ gn (z) satisfy f g(z) = afm gm (z) + bfn gn (z) = (ak + bm)z = z. ∼ G ⊕ ker f and hence G is projective and generated by two Thus E ⊕ E = elements. This proves part 1. 1 ⇒ 2 ⇒ 3 are clear. 3 ⇒ 6 By Proposition 9.2.24 the quasi-projective right E-module is a finitely generated projective left E-module, and by Lemma 9.2.23, G is a generator over S. We will proceed in a pair of lemmas. The first might be interesting in its own right. LEMMA 9.2.29 Let G be an E-finitely generated rtffr group and let S = center(End(G)). Then S is a Noetherian semi-prime commutative ring. Proof: Since G is E-finitely generated Theorem 9.2.12 states that there are integers t, n1 , . . . , nt > 0 and strongly indecomposable Dedekind E-rings R1 , . . . , Rt such that . (n ) (n ) G = R1 1 ⊕ · · · ⊕ Rt t ⊕ N

(9.1)

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where N is generated by R = R1 ⊕ · · · ⊕ Rt . .

We will assume that Ri ∼ = Rj ⇒ i = j. Observe that SRi (Rj ) = Iij is an ideal in Rj . Since Rj is a Dedekind domain (Iij )−1 Iij = Rj for nonzero ideals I in Rj . Consequently if there is a nonzero map Ri → Rj then we can assume that Rj is a direct summand of N . Thus we can asssume that in (9.1), Hom(Ri , Rj ) = 0 for each i 6= j, and hence by Theorem 9.2.5, R is an E-ring. In particular R is a semi-prime ring. Since R is an S-submodule of G, and since R generates a subgroup of finite index in G, xR 6= 0 for each x ∈ S. Thus S ⊂ End(R, +) ∼ =R and hence S is a Noetherian semi-prime commutative rtffr ring. This proves the lemma. LEMMA 9.2.30 Under the hypotheses of Theorem 9.2.28, there is a . map f : G → S such that f (G) = S is a progenerator S-module. Proof: Since S is a semi-prime rtffr ring Lemma 1.6.3 implies that there is an integer m 6= 0 large enough so that Sm is a reduced rtffr ring. Let U = G/N∗ and let M1 , . . . , Ms be a complete list of the maximal ideals of S that contain mS. Since U is a generator over S there are maps fi : U → S such that fi (U ) 6⊂ Mi . By the Chinese Remainder Theorem 1.5.3 there is a map f : U → S such that f (U ) 6⊂ Mi

for any i = 1, . . . , s.

Then f (U ) + mS is an ideal in S that is not included in any of the maximal ideals that contain mS, whence f (U ) + mS = S. Furthermore f (U )m + mSm = Sm and by Lemma 1.2.1, m ∈ J (Sm ), so that f (U )m = . Sm ⊃ S. Thus f (U ) = S and therefore by Lemma 4.2.10, f (U ) is a progenerator ideal in S. It follows that G maps onto the progenerator f (U ), which proves the lemma. We continue with the proof of Theorem 9.2.28. By the above lemma there is an S-module surjection G → G/N∗ → S whose image I is . a progenerator over S. Evidently I = S is an invertible ideal of S. Hence G∼ = I ⊕N for some S-module N . Since S is then a quasi-summand of G, Hom(S, G) = HomS (S, G) so that I and N are S-linear modules. This proves part 6 and completes the logical cycle.

9.2. E-PROPERTIES

9.2.6

239

Noetherian Endomorphism Modules

The group G is E-Noetherian if G is a Noetherian left End(G)-module. In this section we consider the quasi-direct sum decomposition of the ENoetherian rtffr groups. In the process we classify Noetherian modules over rtffr rings. LEMMA 9.2.31 Let E be an rtffr ring. Then E is a right Noetherian ring iff for each integer k > 0, N (E)k /N (E)k+1 is a finitely generated right E-module. Proof: Suppose that for each integer k > 0, N (E)k /N (E)k+1 is a finitely generated right E-module. Then N (E)/N (E)2 is finitely generated by the Noetherian ring E/N (E) so that N (E)/N (E)2 and E/N (E) are Noetherian right E-modules. Hence E/N (E)2 is a Noetherian right E-module. Because E is rtffr N (E)k = 0 for some integer k > 0. Continuing inductively on k we eventually show that E/N (E)k = E is right Noetherian. The converse is clear so the proof is complete. The following result can be viewed as a generalization of R.S. Pierce’s classification (Theorem 9.2.3) of the additive structure of a prime rtffr ring. THEOREM 9.2.32 Let E be an rtffr ring and let M be an rtffr left E-module. Then M is a Noetherian left E-module iff .

(n ) (n ) M∼ = R1 1 ⊕ · · · ⊕ Rt t

for some Dedekind E-rings R1 , . . . , Rt . Proof: Suppose that M is a Noetherian left rtffr E-module. Using . the Beaumont-Pierce-Wedderburn Theorem 3.1.6 write E = T ⊕ N (E) for some semi-prime subring T of E. By Theorem 9.1.2 .

(m ) (m ) T ∼ = R1 1 ⊕ · · · ⊕ Rt t

for some integers m1 , . . . , mt and some distinct Dedekind E-rings R1 , . . . , R t . Because M is a Noetherian E-module each E-submodule of each factor of M is Noetherian. In particular the following E/N (E)-modules are Noetherian. M1 =

M , N (E)M∗

M2 =

N (E)M∗ , N (E)2 M∗

···

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Since N (E) is nilpotent there is an integer k such that Mk+1 = 0. By our choice of T the modules M1 , M2 , . . . , Mk are Noetherian left T -modules. Thus an iterated application of Lemma 9.2.9 shows us that .

M∼ = M 1 ⊕ · · · ⊕ Mk as left T -modules. There is a finitely generated free T -module P and a surjection π : P → M of T -modules. There is a left T -submodule P 0 ⊂ P such that . . ker π ⊕ P 0 = P . Then P 0 = M and since P ∼ = T (m) for some integer m J´onsson’s Theorem 2.1.10 states that .

(n ) (n ) M∼ = R1 1 ⊕ · · · ⊕ Rt t

for some integers n1 , . . . , nt ≥ 0. The converse is an exercise for the reader. This completes the proof.

The proof of the main result of this section is an application of Lemma 9.2.32. THEOREM 9.2.33 [64, A. Paras]. Let E be an rtffr ring. The following are equivalent. 1. E is a right Noetherian ring. 2. E is a left Noetherian ring. .

(n )

(n )

3. E ∼ = R1 1 ⊕ · · · ⊕ Rt t for some integers n1 , . . . , nt and some distinct Dedekind E-rings. In the process of proving Theorem 9.2.32 we inadvertently proved the following result. COROLLARY 9.2.34 Let M be a Noetherian rtffr module over the rtffr ring E. Then .

M∼ =

M N (E)M∗ ⊕ ⊕ · · · ⊕ N (E)k M N (E)M∗ N (E)2 M∗

as left E/N (E)-modules for some integer k > 0. We are now in a position to characterize the E-Noetherian rtffr groups. The E-Noetherian rtffr groups are characterized up to quasiisomorphism by A. Paras [64] where she proves the following theorem.

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241

THEOREM 9.2.35 [64, A. Paras]. Let G be an rtffr group and let E = End(G). Then G is an E-Noetherian rtffr group iff . (n ) (n ) G = R 1 1 ⊕ · · · ⊕ Rt t

(9.2)

where t, n1 , . . . , nt > 0 are integers and where R1 , . . . , Rt are Dedekind E-rings. In this case End(G) is a left Noetherian rtffr ring. Proof: If G can be written as in (9.2) then G is E-Noetherian by Theorem 9.2.32. Conversely, suppose that G is E-Noetherian. Since G has finite rank, there is an integer n and an embedding E −→ G(n) of left E-modules. Then G, G(n) , and E are left Noetherian E-modules. That is, E is a Noetherian ring. By Theorem 9.2.33 there are Dedekind E-rings R1 , . . . , Rt and integers t, k1 , . . . , kt > 0 such that .

(k ) (k ) E∼ = R1 1 ⊕ · · · ⊕ R t t

and such that (9.2) is true. This completes the proof. The E-Noetherian property provides us with another example of overlapping E-properties for rtffr groups. Of course an E-Noetherian group is an E-finitely generated group. THEOREM 9.2.36 Let G be an rtffr group and suppose that E = End(G) is semi-prime. The following are equivalent for G. 1. G is an E-Noetherian group. 2. G is an E-finitely generated group. 3. G is quasi-isomorphic to an E-projective group. 4. G is quasi-isomorphic to an E-generator group. 5. G is quasi-isomorphic to an E-cyclic E-progenerator group. Proof: Let E = End(G) be semi-prime. 5 ⇒ 4 and 3 is clear. 4 or 3 ⇒ 2 follows from Corollaries 9.2.15 and 9.2.16. 2 ⇒ 1 Suppose that G is E-finitely generated. Since E is semi-prime, QE is a semi-simple ring, and since QG is a projective left QE-module

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there is an integer n > 0 and a split embedding f : QG → QE (n) of left QE-modules. Since G is finitely generated as a left E-module we may assume that f (G) is a quasi-summand of E (n) . By Theorem 9.2.3 there are integers t, n1 , . . . , nt > 0 and Dedekind E-rings R1 , . . . , Rt such that .

(n ) E∼ = R(n1 ) ⊕ · · · ⊕ Rt t

so by J´onsson’s Theorem 2.1.10, G has a quasi-direct sum decomposition (9.2) for some integers m1 , . . . , mt ≥ 0. Then G is E-Noetherian by Theorem 9.2.35. 1 ⇒ 5 Say G is E-Noetherian. Then G has a quasi-direct sum decomposition (m1 )

G = R1

(mt )

⊕ · · · ⊕ Rt

for some integers t, m1 , . . . , mt > 0 and Dedekind. E-rings R1 , . . . , Rt . We can assume without loss of generality that Ri ∼ = Rj ⇒ i = j for each 1 ≤ i, j ≤ t. Then the Lemma 4.1.4 characterizing nilpotent sets and the semi-prime hypothesis imply that Hom(Ri , Rj ) ⊂ N (E) = 0 for each i 6= j. Because End(Ri ) = Ri for each i = 1, . . . , t we have End(G) = Matn1 (R1 ) × · · · × Matnt (Rt ). As a left End(G)-module we have

R1 G = ... ⊕ · · · ⊕ R1 n 1

Rt .. . . Rt n t

The reader will show as an exercise that G is a cyclic progenerator as a left End(G)-module.

9.3

Homological Dimensions

Theorem 2.3.4 provides us with an effective method for constructing groups G whose structure as a left End(G)-module has certain properties. This kind of example is an indispensable tool for developing intuition in studying properties of rtffr groups. We will use Theorem 2.3.4 to construct groups with prescribed homological dimension.

9.3. HOMOLOGICAL DIMENSIONS

9.3.1

243

E-Projective Dimensions

Let E be an rtffr ring and let M be a left E-module. The projective dimension of M

pdE (M ) is the least integer k ≥ 0 for which there is a long exact sequence 0 −→ Pk −→ Pk−1 −→ · · · −→ P0 −→ M −→ 0

(9.3)

whose terms Pj are projective left E-modules for each j = 0, . . . , k. Equivalently pdE (M ) = k if k is the least positive integer such that Extk+j E (M, ·) = 0 for each integer j > 0. Then M is a projective left E-module iff pdE (M ) = 0 iff Ext1E (M, ·) = 0. (See [72].) One of the first examples of an rtffr group with prescribed projective dimension is found in H.M.K Angad-Gaur’s Thesis [7]. In it he shows that for each integer n > 0 there is an rtffr group G such that pdE (G) = n where E = End(G). C. Vinsonhaler and W. Wickless [81, Corollary 6] show that for each integer n > 0 there is a completely decomposable rtffr group G such that rank(G) = 2n + 1 and pdE (G) = n. We will use the construction in Theorem 2.3.4 to show that pdEnd(G) (G) can be almost any integer less than the left global dimension of End(G). LEMMA 9.3.1 Let E be an rtffr ring and let M be an rtffr left Emodule. There is a short exact sequence 0 → M ⊕ E −→ G −→ QE ⊕ QE → 0

(9.4)

of rtffr left E-modules such that E ∼ = End(G). Then 1. pdE (G) = pdE (M ) if pdE (M ) ≥ 1 and 2. pdE (G) = 1 if pdE (M ) = 0. Proof: 1 and 2. Let M be an rtffr left E-module, let pdE (M ) = k ≥ 1, and let N be any left E-module. By Theorem 2.3.4 there is a short exact sequence (9.4) such that E ∼ = End(G). Inasmuch as (9.4) is nonsplit,

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Ext1E (QE, ·) 6= 0. Furthermore think of ExtkE (QE, ·) as ker δ2 /image δ3 where δ

δ

3 2 HomE (QE, E (c3 ) ) −→ HomE (QE, E (c2 ) ) −→ HomE (QE, E (c1 ) )

for some cardinals c1 , c2 , c3 . Inasmuch as Hom(QE, E(c) ) = 0, ExtkE (QE, ·) = 0 for each integer k ≥ 2. Thus pdE (QE) = 1. Given a fixed integer k ≥ 0 an application of the contravariant functor HomE ( , ?) to (9.4) yields the long exact sequence k+j Extk+j E (QE ⊕ QE, ?) → ExtE (G, ?) k+j+1 → Extk+j (QE ⊕ QE, ?) E (M, ?) → ExtE

for each integer j > 0. Since pdE (QE) = 1 ≤ k we have k+j+1 (QE ⊕ QE, ?) = 0 Extk+j E (QE ⊕ QE, ?) = ExtE

for each integer j > 0 and so k+j ∼ Extk+j E (G, ?) = ExtE (M, ?)

(9.5)

for integers j > 0. In particular, if pdE (M ) = ∞ then ExtkE (M, ?) 6= 0 for all integers k > 0 so that by (9.5), pdE (G) = ∞. If 1 ≤ pdE (M ) = k < ∞ then Extk+j E (M, ?) = 0 for all integers j > 0 so that by (9.5), pdE (G) ≤ k. Since k ≥ 1, ExtkE (M, ?) 6= 0 and so pdE (G) = k by (9.5). j If pdE (M ) = 0 then ExtjE (G, ?) ∼ = ExtE (M, ?) = 0 for each j > 0 implies that pdE (G) ≤ 1. Assume to the contrary that pdE (G) = 0. Because G is then an E-projective rtffr group Corollary 9.2.15 states that G is E-finitely generated. Hence QE is finitely generated by E, a contradiction. Thus pdE (G) = 1 which completes the proof. With the aid of the above lemma we will construct groups whose projective dimensions over their endomorphism rings are prescribed values. Recall that gd(E) = 1 + {pdE (I) I ⊂ E is a left ideal }. THEOREM 9.3.2 Let E be an rtffr ring and suppose that there is a left ideal I ⊂ E such that pdE (I) = n ≤ ∞. There is an rtffr group G such that End(G) ∼ = E and pdE (G) = n.

9.3. HOMOLOGICAL DIMENSIONS

245

Proof: Suppose that I is a left ideal of E such that pdE (I) = n > 0 and let M = I ⊕ E. Then pdE (M ) = pdE (I) = n and by Theorem 2.3.4 there is a short exact sequence (9.4) of left E-modules such that E∼ = End(G). Lemma 9.3.1 states that pdE (G) = pdE (M ) = n. THEOREM 9.3.3 Let E be a Noetherian rtffr ring such that gd(E) = n + 1 < ∞. For each integer 0 < k ≤ n there is an rtffr group Gk such that E ∼ = End(Gk ) and pdE (Gk ) = k. Proof: Let 0 < k ≤ n be an integer. There is a left ideal I ⊂ E such that pdE (I) = n, so there is a long exact sequence δ

δ

δ

n 2 1 0 → Pn −→ · · · −→ P1 −→ P0 −→ I → 0

of left E-modules in which each Pi is a projective. Since E is Noetherian I is finitely presented so we can choose such a sequence in which each Pi is finitely generated projective. Let Mk = image δk . The reader will note that the induced long exact sequence δ

δ

k k · · · −→ Mk → 0 0 → Pk −→

has exactly n − k projective terms so that pdE (Mk ) = n − k. Using M = Mn−k we can use Theorem 2.3.4 to construct a short exact sequence (9.4) of left E-modules such that E ∼ = End(G). By Lemma 9.3.1, pdE (G) = pdE (Mn−k ) = k. We will state but will not prove a stronger result for countable reduced torsion-free groups whose ranks may not be finite. THEOREM 9.3.4 [36, T.G. Faticoni] Let E be a Corner ring and suppose that gd(E) = n + 1. Then for each integer 0 < k ≤ n there is a Corner group Gk such that E ∼ = End(Gk ) and pdE (Gk ). EXAMPLE 9.3.5 Let E = Z[x]/(x2 ), let I = (x)/(x2 ) ∼ = Z, and let M = I ⊕ E. By Theorem 2.3.4 there is a short exact sequence (9.4) such that E ∼ = End(G). By Lemma 9.3.1, pdE (G) = pdE (M ), and it is readily verified that x

x

x

· · · −→ E −→ E −→ I → 0

(9.6)

is a projective resolution of I such that ker x = I ⊂ E is not a direct summand of E. Thus pdE (M ) = pdE (I) = ∞. By Lemma 9.3.1 there is an rtffr group G such that E ∼ = End(G) and pdE (G) = pdE (I) = ∞.

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EXAMPLE 9.3.6 Let E be an rtffr ring and let G be a ring constructed according to Corner’s Theorem such that E ∼ = End(G). Then G fits into a short exact sequence 0 → E −→ G −→ QE → 0 of left E-modules. By Lemma 9.3.1, pdE (G) = 1.

9.3.2

E-Flat Dimensions

The flat dimension of M fdE (M ) is the least integer k > 0 for which there is a long exact sequence (9.3) in which each Pi is a flat left E-module for each i = 0, . . . , k. Equivalently, fdE (M ) = k iff k is the least positive integer such that Tork+j E (?, M ) = 0 for each integer j > 0. Then M is a flat left E-module iff fdE (M ) = 0 iff Tor1E (?, M ) = 0. Moreover M is a flat left E-module iff given a short exact sequence 0 → K −→ N −→ L → 0 of right E-modules then the induced sequence 0 → K ⊗E M −→ N ⊗E M −→ L ⊗E M → 0 of groups is exact. C. Vinsonhaler and W. Wickless [81, Theorem 5] show that for each integer n > 0 there is a completely decomposable rtffr group G such that rank(G) = 2n + 1 and fdE (G) = n. We address the question of which integers are the flat dimensions of rtffr groups. The lemma indicates how we will construct rtffr groups of prescribed flat dimension. Recall that O(M ) = {q ∈ QE qM ⊂ M }. LEMMA 9.3.7 Let E be an rtffr ring and let M be a left rtffr E-module such that E = O(M ). If there is a short exact sequence 0 → M −→ G −→ QE ⊕ QE → 0 of left E-modules then fdE (G) = fdE (M ).

(9.7)

9.3. HOMOLOGICAL DIMENSIONS

247

Proof: Given fdE (M ) = k an application of ? ⊗E ? to (9.7) yields the long exact sequence Tork+j−1 (?, QE ⊕ QE) → Tork+j (?, M ) → Tork+j (?, G) → Tork+j (?, QE ⊕ QE) for each integer j > 0. Because QE is a flat left E-module Tork+j−1 (?, QE ⊕ QE) = Tork+j (?, QE ⊕ QE) = 0 for each k + j − 1 > 0. Thus Tork+j (?, M ) ∼ = Tork+j (?, G)

(9.8)

for each j > 0. If fdE (M ) = ∞ then by (9.8), fdE (G) = ∞. If 1 ≤ fdE (M ) = k < ∞ then Tork (?, M ) 6= 0 and Tork+j (?, M ) = 0 for each integer j > 0. Then by (9.8), fdE (G) = k. If fdE (M ) = 0 then M and QE are flat left E-modules. It follows from the exactness of 0 = Tor1 (?, M ) → Tor1 (?, G) → Tor1 (?, QE ⊕ QE) = 0 that fdE (G) = 0. This completes the proof. We can apply Lemma 9.3.7 to the next result in the same manner that we used Lemma 9.3.1 to prove Theorem 9.3.2. In constructing (9.7) in the next proof choose M = I ⊕ E and then apply Theorem 2.3.4. THEOREM 9.3.8 [T.G. Faticoni] Let E be an rtffr ring and let I be a left ideal in E such that fdE (I) = n. There is an rtffr group G such that E ∼ = End(G) and fdE (G) = n. The following result is proved in the same manner that we proved Theorem 9.3.3, but use Tor(?, ?) instead of Ext(?, ?). THEOREM 9.3.9 Let E be a semi-prime rtffr ring and suppose that there is a left ideal I of E such that fdE (I) = n. Then for each integer 0 ≤ k ≤ n there is an rtffr group Gk such that E ∼ = End(Gk ) and fdE (Gk ) = k. EXAMPLE 9.3.10 Let E = Z[x]/(x2 ), let I = (x)/(x2 ), and let M = I ⊕E. Then the group in the center term of (9.4) has fdE (G) = fdE (I) = ∞ as there is an infinite sequence (9.6) of flat modules E such that no kernel ker x is a direct summand of E.

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EXAMPLE 9.3.11 Let E be an rtffr ring and let G be a group constructed according to Corner’s Theorem 2.3.3 such that E ∼ = End(G). There is a short exact sequence 0 → E −→ G −→ QE → 0 of left E-modules so G is a flat left E-module. That is, fdE (G) = 0. Compare to Example 9.3.6 where we proved that pdE (G) = 1.

9.3.3

E-Injective Dimensions

Define the injective dimension of M

idE (M ) to be the least integer k ≥ 0 for which there exists an exact sequence 0 −→ M −→ E0 −→ E1 · · · −→ Ek −→ 0

(9.9)

where each Ei is an injective left E-module. Equivalently idE (M ) = k iff k is the least positive integer such that Extk+j (?, M ) = 0 for each integer j > 0. The constructions illustrating the injective dimensions over an rtffr ring E turn out to be harder than the examples we gave about the projective and flat dimensions. For instance, while our techniques give us good results when we consider the injective dimension of QG, they say very little about the injective dimension of G. Thus our results are incomplete. LEMMA 9.3.12 Let E be an rtffr ring. There is a short exact sequence 0 → E −→ G −→ QE → 0

(9.10)

of left E-modules such that E = End(G) and idE (QG) = idE (QE). Proof: Let idE (QE) = k. Applying · ⊗ Q to 9.10 we have a split exact sequence 0 → QE −→ QG −→ QE → 0 Then idE (QE) = idE (QG).

9.3. HOMOLOGICAL DIMENSIONS

249

THEOREM 9.3.13 [T.G. Faticoni] Let E be an rtffr ring and let G be the group constructed for E according to Corner’s Theorem 2.3.3. If idE (QE) = n then idE (QG) = n. Proof: By Corner’s Theorem there is a short exact sequence (9.10) of left E-modules such that E ∼ = End(G). Then by Lemma 9.3.12, n = idE (QE) = idE (QG). Let E be an rtffr ring and let G be the group constructed according to Corner’s Theorem 2.3.3. Examples 9.3.6 and 9.3.11 show us that pdE (G) = 1, fdE (G) = 0, and idE (QG) = idE (QE). Of course idE (QE) = 0 if E is a semi-prime ring. So in the by now familiar manner the general condition idE (QE) = 0 leads us to the following theorem. THEOREM 9.3.14 [T.G. Faticoni] Let E be an rtffr ring such that idE (QE) = 0, let M be an rtffr left E-module such that E = O(M ). There is a short exact sequence 0 → M −→ G −→ QE ⊕ QE → 0

(9.11)

of left E-modules such that E ∼ = End(G) and such that idE (G) = idE (M ). Proof: Use Theorem 2.3.4 to construct the short exact sequence (9.11) in which E ∼ = End(G). The usual argument shows us that idE (G) = idE (M ). For example, if E is a semi-prime rtffr ring then idE (G) = idE (E) for any group G constructed according to Corner’s Theorem 2.3.3. EXAMPLE 9.3.15 For each integer n there is a torsion-free group G of rank n such that pdE (G) = fdE (G) = ∞ and idE (QG) = 0 where E = End(G). Proof: Let E = Z[x]/(xn ). Then E is a torsion-free ring of rank n whose additive structure is a free group. Butler’s Construction in Theorem 2.3.1 states that E = End(G) for some group G such that E ⊂ G ⊂ QE as left E-modules. The reader can show that QE = Q[x]/(xn ) is a self-injective ring so that idE (QG) = idE (QE) = 0. The rest follows from Lemmas 9.3.1 and 9.3.7.

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EXAMPLE 9.3.16 Let A be a finite dimensional Q-algebra and suppose that M is a finitely generated left A-module. If annA (M ) = 0 then there is a full subring E ⊂ A and an rtffr group G such that E = End(G) and idE (QG) = min{idE (QE), idE (M )}. Proof: Let F be a full free subgroup of M and let E = O(F ) = {q ∈ A qF ⊂ F }. It is an easy exercise to show that for each q ∈ A there is an n 6= 0 ∈ Z such that nq(F ) ⊂ F . Thus QE = A. Theorem 2.3.4 constructs a short exact sequence 0 → F −→ G −→ QE ⊕ QE → 0 of left E-modules such that E ∼ = End(G). So idE (QG) = max{idE (QE), idE (QM )}. Let us agree to call the rtffr group G an E-injective group if QG is an injective left End(G)-module. C. Vinsonhaler and W. Wickless [83] characterize the injective hull of a torsion-free group of finite rank while C. Vinsonhaler [79] characterizes certain E-injective rtffr groups.

Z 0 Z 0 and let I = . Then Z Z Z 0 I is a projective ideal of E such that annE (I) = 0 and QI is an injective left E-module. The reader can verify that E = {q ∈ QE qI ⊂ I}. Theorem 2.3.4 produces an exact sequence EXAMPLE 9.3.17 Let E =

0 −→ I −→ G −→ QE ⊕ QE −→ 0 of left E-modules such that E ∼ = End(G). Since QE is a hereditary ring, idE (M ) ≤ 1 for each left E-module M . It is an interesting exercise to show that Mat2 (Q) is the injective hull of QE so that idE (QE) = 1. Thus G and the group constructed according to Corner’s Theorem 2.3.3 satisfy pdE (G) = 1, fdE (G) = 0, and idE (G) = 1. The above constructions show us that any classification of the Ehomological dimension of an rtffr group G will require an approach that differs significantly from the techniques used in this book. See [32] for a characterization of homological dimensions of modules over their endomorphism rings.

9.4. SELF-INJECTIVE RINGS

9.4

251

Self-Injective Rings

The literature contains a fairly complete characterization of those rtffr groups for which End(G) is a left or right hereditary ring. See [4, 10, 12, 41, 35, 43, 53]. Let us consider the rtffr groups G such that QEnd(G) is a left self-injective ring. Because QEnd(G) is Artinian it is known (see [30]) that QEnd(G) is right self-injective iff QEnd(G) is left self-injective. We will need the following ideas. G-monomorphisms are introduced in [37]. A G-monomorphism is a group map ˆ : H → H 0 such that ˆ = TG (ı) for some injection ı : M → M 0 of right End(G)-modules. The group H is G-presented if H = G(n) /K for some integer n > 0 and some G-generated subgroup K ⊂ G(n) . It is easy to see that ˆ : K → G is a monomorphism in the category G-Pre of G-presented groups iff SG (ker ˆ) = 0. By [32, Lemma 5.1.1] a group map ˆ : K → G is a Gmonomorphism iff ˆ = TG (ı) for some injection ı : I −→ HG (G) of right End(G)-modules iff ˆ is a monomorphism in the category G-Pre. Our characterization of rtffr groups such that QEnd(G) is a right self-injective ring is in terms of a lifting property for G-monomorphisms. THEOREM 9.4.1 [T.G. Faticoni] Let G be an rtffr group and let E = End(G). The following are equivalent. 1. QE is self-injective. 2. If ˆ : K → G is a monomorphism in G-Pre and if f : K → G is a group map then there is a map g : G → G and an integer n such that nf = gˆ . Proof: Assume part 1. Then QE is right self-injective. Let ˆ : K → G be a monomorphism in G-Pre, and let h : K → G be a group map. By [32, Lemma 5.1.1], ˆ is a G-monomorphism, so ˆ = TG (ı) for some injection ı : I → E of right E-modules. Specifically TG (I) ∼ = K. By the adjoint isomorphism HomE (I, E) ∼ = HomE (I, HG (G)) ∼ = Hom(TG (I), G) = Hom(K, G) and the reader can verify that this isomorphism sends each map φ : I → E to TG (φ). Thus h = TG (φ) for some φ : I → E. Since QE is selfinjective there is a map γ : QE → QE such that φ = γı, and there is an integer n such that nγ(E) ⊂ E. Then nh = nTG (φ) = nTG (γı) = (ng)ˆ . This proves part 2.

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Conversely assume part 2, let ı : I → QE be an inclusion of right QE-modules, and let f : I → QE be a QE-module map. Write I = QJ for some finitely generated right ideal J ⊂ E so that there is an integer n 6= 0 such that nı(J) ⊂ E. By definition TG (nı) : TG (J) → TG (E) = G is a G-monomorphism. Choose an integer m 6= 0 such that mf (J) ⊂ E so that TG (mf ) : TG (J) → TG (E). By part 2 there is an integer m 6= 0 and a group map g : TG (E) → TG (E) such that TG (mf ) = gTG (nı). By the Arnold-Lady-Murley Theorem 2.4.1 we can write g = TG (h) for some h : E → E. Then TG (mf ) = TG (nhı) or in other words TG (mf − nhı) = 0 : TG (J) −→ TG (E). Since J ⊂ E the adjoint isomorphism implies that mf − nhı = 0 whence n f = (m h)ı. This completes the proof. COROLLARY 9.4.2 Let G be an rtffr group such that QE is selfinjective. If ˆ : K → G is a monomorphism in G-Pre (for instance, if ˆ is an injection) and if f : K → G is a group map then there is a map ψ : G → G and an integer n such that nf = ψˆ .

9.5

Exercises

Let E be an rtffr ring, let G, H be rtffr groups. 1. Let E = End(G) and S = center(E). Prove that (a) E = EndS (G). (b) S = EndE (G). (c) If G = A ⊕ B then A is an S-submodule of G. 2. Let M be a finitely generated right E-module whose additive structure is a torsion group. Show that mM = 0 for some integer m 6= 0. 3. A hard problem. Find a finitely generated right ideal in an rtffr ring that is not finitely presented.

9.5. EXERCISES

253

4. Show that the finite modules over an rtffr ring E are finitely presented. 5. Let R be any ring and let N be a left R-module. Then N is generated by R as a group. Hint: For fixed x ∈ N consider the map λx : R → N such that λx (r) = rx for each r ∈ R. 6. Show that a finitely presented left E-module M possesses a projective resolution · · · P2 → P1 → P0 → M → 0 such that the projective modules P0 , P1 , P2 , . . . are finitely generated. 7. Let E be an rtffr ring and let I be a maximal right ideal of E. If pdE (I) = n then for each integer 0 < k ≤ n there is a finite rank group Gk such that E = End(Gk ) and pdE (Gk ) = k. 8. Let G be an E-finitely generated rtffr group N = N (End(G)). . Show that if G0 is a left End(G)-submodule of G such that G = . 0 0 G + N G then G = G . 9. Let G be an rtffr group and let N (End(G)) = N . Construct rtffr G . and G0 such that G0 ⊂ G is a left End(G)-submodule, G = G0 +N G, but G0 is not quasi-isomorphic to G. 10. If E is an rtffr semi-prime ring and if P ⊂ E (m) is a right Esubmodule then any E-module surjection π : E (n) → P is quasisplit. That is, there is a map σ : P → E (n) and an integer k such that πσ = k1P . 11. Let E be a semi-prime rtffr ring and let M be an E-submodule of E (n) for some integer n 6= 0. Then M is an E-Noetherian rtffr group. . 12. Let E be an rtffr E-ring and let E = I be a projective ideal of E. (a) E is locally isomorphic to I as groups. (b) E is E-locally isomorphic to I as an E-module. 13. Let R be an rtffr integral domain. Prove that if I ⊂ R is a nonzero ideal then R/I is finite. 14. Let R be an rtffr integral domain and let M be a finitely generated R-module. Then the R-torsion submodule T of M is finite and . M/T ∼ = R(m) for some integer m > 0.

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15. Suppose that G is an rtffr group and that A ⊕ B is a subgroup of finite index in G. Show that there is a quasi-idempotent e ∈ End(G) corresponding to A. That is, there is an e ∈ End(G) such that e2 = ne for some integer n 6= 0, e(x) = nx for each x ∈ A, and e(B) = 0. 16. Show that if E is a ring then E (n) is a cyclic projective left Matn (E)module. 17. Let R be an E-ring. Then R(n) is a cyclic projective left End(R(n) )module. 18. Let E be an rtffr ring. (a) Find an rtffr left E-module M such that QM is the injective hull of the left E-module E. ∼ End(G) and (b) Find an rtffr group G such that E = idE (QG) = 0. (c) Let E be an rtffr ring and let M be an rtffr left E-module. Let lengthE (M ) denote the composition length of QM as a left QE-module. Show that for each integer n there is an rtffr group G such that lengthEnd(E) (G) = n. 19. Let E be a semi-prime rtffr ring and let M , N be left rtffr Emodules. Show that ExtnE (M, N ) is a bounded group for each integer n ≥ 1. 20. Let A be a self-injective finite dimensional Q-algebra and let M be any free subgroup of A such that QM = A. (a) Prove that there is a group G such that M ⊂ G ⊂ A and such that QEnd(G) = A. (b) Prove that G is an E-injective rtffr group and QEnd(G) is a self-injective ring. (c) By varying M produce a variety of properties on G. For instance, Show that under one choice of M , G has E-property while under another choice G does not have E-property. Be creative. (d) Is it possible that G is strongly indecomposable for one choice of M , and decomposable for another? 21. Find a better way of investigating the injective dimension of G as a left E-module and not of QG.

9.6. QUESTIONS FOR FUTURE RESEARCH

9.6

255

Questions for Future Research

Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. We have characterized the finitely generated rtffr E-modules M in terms of direct sums of E-rings. Characterize larger classes of Emodules. For instance, consider the injective E-modules. Dual the definition of finitely generated E-modules to arrive at the finitely cogenerated E-modules. See [6]. 2. Study The Beaumont-Pierce-Wedderburn Theorem for a ring E. Give a canonical or natural value for the semi-prime ring T ⊂ E. 3. Prove the converse of Theorem 9.2.11. 4. This question is due to R. Pierce. Let property be a module theoretic property, and say that G is an E-property group if G satisfies property. Pick property and characterize the E-property groups. 5. Fix property. Characterize all of the rtffr rings E such that each G ∈ Ω(E) satisfies property. 6. Fix property. Construct examples of indecomposable rtffr E-property groups. 7. Use Galois Theory to characterize the strongly indecomposable Erings. See [39, 65, 66]. 8. See [32]. Find an internal characterization of rtffr groups G that possess infinite flat, projective, or injective dimension over End(G). 9. Find an internal characterization of rtffr groups G that possess finite flat, projective, or injective dimension over End(G). 10. Characterize the rtffr groups G such that QG is an injective left End(G)-module. 11. Characterize the rtffr groups G such that QEnd(G) is an injective left or right End(G)-module.

Appendix A

Pathological Direct Sums We have sectioned these examples into the appendices because of the different techniques used. Reread the first few results of the chapters. You will see the use of ring and module theory, and of functors and functorial methods. These ideas were not widely used in abelian group theory during the period of the 1950s and 1960s. A.L.S. Corner [23] broke that mold with the publication of his famous result now called Corner’s Theorem F.1.1 wherein he proved that each countable reduced torsion-free ring is the group endomorphism ring of a torsion-free group. His technique was pure number theory and linear algebra over Z. Compare this to the Arnold-Lady Murley Theorem 2.4.1 that is pure category theory. That is why we have these appendices. The results of the chapters use modern techniques while the techniques in the appendices are all number theory. They represent the techniques used in the early years of abelian groups.

A.1

Nonunique Direct Sums

We have already claimed that direct sums of the general rtffr group are not unique in any sense. In this appendix we will give examples to justify our claim. We begin with the most elementary of examples. These examples come from [46, Sections 90]. EXAMPLE A.1.1 Let n ≥ 2 be an integer. There exists an rtffr group G such that G = A1 ⊕ · · · ⊕ An−1 ⊕ B = C ⊕ D where the groups Ai , B, C, D are indecomposable, rank(Ai ) = 1, rank(B) = rank(C) = rank(D) = n − 1. 257

258

APPENDIX A. PATHOLOGICAL DIRECT SUMS

Proof: Let n ≥ 3 be an integer, let p, q, p1 , · · · , pn−1 be different primes, and let {a1 , . . . , an−1 } and {b1 , . . . , bn−1 } be bases for the Q-vector space V . Define Aj

= hp−∞ aj i for each j = 1, · · · , n − 1 j

−∞ B = hp−∞ 1 b1 , . . . , pn−1 bn−1 ,

p−1 q −1 (b1 + b2 ), . . . , p−1 q −1 (b1 + bn−2 )i and let G = A1 ⊕ · · · ⊕ An−1 ⊕ B. LEMMA A.1.2 B is indecomposable. Proof: Suppose that B = U ⊕ W as abelian groups. We have a quasi-isomorphism . −∞ U ⊕ W = hp−∞ 1 b1 , . . . , pn−1 bn−1 i = Bo and the subgroups hp−∞ bj i are fully inavariant in Bo because the pj j are different primes. Since these subgroups have rank one, there is a complete set of central indecomposable idempotents e1 , . . . , en−1 ∈ EndZ (Bo ) such that ej (x) = x for each x ∈ hp−∞ bj i. j Thus each idempotent in EndZ (Bo ) is a sum of a subset of {ej 1, . . . , n − 1}. For example, given the canonical idempotent map

i =

e : Bo −→ U there are disjoint subsets σ, ψ of {1, . . . , n − 1} such that e = ⊕j∈σ ej . Suppose for the sake of contradiction that there are 1 ∈ ψ and j ∈ σ. Then e(p−1 q −1 (b1 ⊕ bj )) = p−1 q −1 (bj ). This is contrary to the fact that bj has p-height 0 in hp−∞ bj i. Thus j σ = ∅, hence U = 0, whence B is indecomposable.

A.1. NONUNIQUE DIRECT SUMS

259

Proof of Example A.1.1: We choose an ordered base a1 , . . . , an−1 , b1 , . . . , bn−1 for V and primes p 6= q. Choose integeres s and t such that ps − qt = 1 and then choose cj = paj + tbj and dj = qaj + sbj for each j = 1, . . . , n − 1. Let −∞ −1 −1 C = hp−∞ 1 c1 , . . . , pn−1 cn−1 , p (c1 + c2 ), . . . , p (c1 + cn−2 )i −∞ −1 −1 D = hp−∞ 1 d1 , . . . , pn−1 dn−1 , q (d1 + d2 ), . . . , q (d1 + dn−2 )i.

Then C ∩ D = 0, C, D ⊂ G, and by the above lemma C and D are indecomposable. By our choices of cj , dj , s, t we have aj

= scj − tdj

bj

= pdj − qcj

b1 + bj

= p(d1 + dj ) − q(c1 + cj ).

Thus G = C + D and therefore G = C ⊕ D. Because A1 ∼ 6 C, D in the above example, it follows that a direct sum = decomposition of an rtffr group G into indecomposables is not necessarily unique.

Appendix B

ACD Groups B.1

Example by Corner

The next example shows that rtffr groups can have many direct sum decompositions into indecomposables. EXAMPLE B.1.1 [Corner] (See [46].) Let n ≥ k ≥ 1 be integers. There is a group G = G(n) of rank n such that for each partition r1 + · · · + rk of n into k positive summands rj there is an indecomposable direct sum decomposition G = G1 ⊕ · · · ⊕ Gk such that rj = rank(Gj ) for each j = 1, . . . , k. Proof: Let p, p1 , . . . , pn−k , q1 , . . . , qn−k be a list of different primes, let V be a Q-vector space, let u1 , . . . , uk , x1 , . . . , xn−k be a basis for V , and let G = hp∞ u1 , . . . , p∞ uk , ∞ p∞ 1 x1 , . . . , pn−k xn−k , −1 q1−1 (u1 + x1 ), . . . , qn−k (u1 + xn−k )i.

We will show that G has the indicated property. Let n = r1 + · · · + rk be a partition of n with rj ≥ 1 for each j = 1, . . . , k. 261

262

APPENDIX B. ACD GROUPS Suppose there are s1 , . . . , sk ∈ Z such that s1 + · · · + sk = 1

and unknowns v1 , . . . , vk . The system of linear equations s1 v1 + s2 v2 + · · · + sk vk = u1 −v1 + v2 = u2 .. .. . . −v1 + vk = uk has determinant 1. Hence the system has a solution v1 , . . . , vk ∈ Z such that hv1 , . . . , vk i = hu1 , . . . , uk i. Let −1 Gj = hp∞ vj , p∞ i xi , qi (vj + xi ) i = tj + 1, . . . , tj+1 i

where t1 = 0 and tj = (r1 − 1) + · · · + (rj−1 − 1). Observe that u1 + xi = (vj + xi ) + s1 v1 + · · · + sj−1 vj−1 + (sj − 1)vj + · · · + sk vk .

Then in choosing the s1 , . . . , sk require that 1 (mod )qi for i = tj + 1, · · · , tj+1 sj = . 0 (mod )qj otherwise [Comment: For example, let Qj = q1 · · · qj−1 qj+1 · · · qn−k , P choose integers m` such that ` m` Q` = 1, and then choose sj = mti +1 Qti +1 + · · · + mti+1 Qti+1 .] Subsequently Gj ⊂ G for each j = 1, . . . , k, and p∞ u 1 , . . . , p ∞ u k , ∞ p∞ 1 x1 , . . . , pn−k xn−k , −1 q1−1 (u1 + x1 ), . . . , qn−k (u1 + xn−k ) ∈ G 1 + · · · + Gk . Thus G = G1 ⊕ · · · ⊕ Gk where by construction rank(Gj ) = rj and by Lemma A.1.2, Gj is indecomposable. This completes the construction.

Appendix C

Power Cancellation The group G satisfies power cancellation if given a group H and an integer n ≥ 1 such that G(n) ∼ = H (n) then G ∼ = H.

C.1

Failure of Power Cancellation

The next example shows that rtffr groups fail to satisfy the power cancellation property. EXAMPLE C.1.1 Let p ≥ 3 be a prime. There are nonisomorphic rtffr groups A and B of the same rank such that A(n) ∼ = B (n) iff p divides n. Proof: Let k be an algebraic number field of class number h such that p is a prime divisor of h. Let O = O(k) denote the ring of algebraic integers in k, and let Γ denote the ideal class group of O. There is a subgroup Γ(p) of the finite abelian group Γ of order p. Since p ≥ 3 there are at least two isomorphism classes (I) 6= (J) in Γ(p) of order p. Then for integers n > 0 In ∼ = J n iff p divides n. Because O is a Dedekind domain, Steinitz’ Theorem [69, Theorem 4.13] states that I (n) ∼ = J (n) iff I n ∼ = J n . So I (n) ∼ = J (n) iff p divides n. Using Corner’s Theorem F.1.3 there is an rtffr group G such that EndZ (G) ∼ = O. By the Arnold-Lady-Murley Theorem 2.4.1 there are rtffr groups A, B ∈ Po (G) such that HG (A) = I and HG (B) = J. Then HG (A(n) ) = I (n) ∼ = J (n) = HG (B (n) ) 263

264

APPENDIX C. POWER CANCELLATION

iff p divides n. By Theorem 2.4.1, A(n) ∼ = B (n) iff p divides n. This completes the proof. Fuchs and Loonstra give an example of nonisomorphic rtffr groups G and H such that Gn ∼ = H n for some integer n > 1. EXAMPLE C.1.2 [L. Fuchs and F. Loonstra] (See [46, Theorem 90.3].) Let m ≥ 2 be an integer. There are two torsion-free indecomposable rank 2 groups G and H such that G(n) ∼ = H (n) iff m divides n.

Appendix D

Cancellation The group G satisfies the cancellation property if for any groups K, L G⊕K ∼ = G ⊕ L implies K ∼ = L.

D.1

Failure of Cancellation

The next example shows that the category of rtffr groups does not enjoy the cancellation property. THEOREM D.1.1 [L. Fuchs and F. Loonstra] (See [46, Theorem 90.4].) Let m ≥ 1 be an integer. There are groups G, K, and pairwise nonisomorphic indecomposable groups H1 , · · · , Hm such that rank(B) = 1, rank(Hj ) = 2, and G∼ = B ⊕ Hj for each j = 1, . . . , m. Proof: Begin with two infinite disjoint sets of primes P and Q, and a prime po not in P ∪ Q. Let b, x, y 6= 0 ∈ Q. Construct rank one groups B = hp−1 b p ∈ P i,

X = hp−1 x p ∈ P i,

Y = hq −1 y q ∈ Qi.

By Lemma A.1.2, H1 = hX ⊕ Y, p−1 o (x ⊕ y)i is an indecomposable rank 2 group. Define G = B ⊕ H1 . Choose integers qi , ri , si , ti such that qi ti − ri si = 1 and let bi = qi b + si x

xi = r i b + t i x 265

266

APPENDIX D. CANCELLATION

for each i = 2, . . . , m. We let Bi = hp−1 bi p ∈ P i∗ and Xi = hp−1 i xi p ∈ P i∗ be isomorphic rank one subgroups of B ⊕ X. Then B ⊕ X = Bi ⊕ Xi . We will refine our choices of bi and xi so that for some integers 1 < ki < po , G = Bi ⊕ Hi with Hi = hp−1 o (ki xi + y), xi ∈ Xi , y ∈ Y i.

(D.1)

Notice that by Lemma A.1.2, in any choice of ki , Hi is indecomposable. From our choices of qi , ri , si , ti , and xi we see that ki xi + y = ki ri b + (ki ti − 1)x + (x + y) is divisible by p. Hence ki ri ≡ 0 mod p

and

ki ti ≡ 1 mod p.

(D.2)

Since 1 < k1 < p we can choose ri = p, qi = ki , and si , ti such that qi ti − ri si = 1. Then (D.2) is satisfied and hence (D.1) holds. It remains to prove the lemma. LEMMA D.1.2 Let H1 , . . . , Hm be groups as defined in (D.1). Then the Hi are quasi-isomorphic rank 2 acd groups such that Hi ∼ 6 Hj = for each 1 ≤ i 6= j ≤ m. Proof: Let φ : Hi −→ Hj be an isomorphism. Then φ(X) = X and φ(Y ) = Y and φ(xi ) = ±xi and φ(y) = ±y. Thus φ(ki xi + y) = ±(kj xj ± y). Preservation of divisibility implies that kj ≡ ±ki mod p. Consequently with k1 = 1, k2 = 2, · · · , km = m

D.1. FAILURE OF CANCELLATION

267

and p > 2m − 1 then kj 6≡ ±ki mod p. Thus no two of the H1 , . . . , Hm are isomorphic. This proves the lemma and completes the proof of the example. The above example shows that for each m ≥ 1 there exists an rtffr of rank 3 G with m inequivalent direct sum decompositions. It also shows that cancellation fails m different times in G. It is interesting to note that E.L. Lady [10, Theorem 11.11(a)] has shown that in any group direct sum decomposition G = B ⊕ H ∼ = B ⊕ K there are at most finitely many isomorphism classes of K.

Appendix E

Corner Rings and Modules E.1

Topological Preliminaries

Fix a Corner ring E. That is, the additive group (E, +) is a countable reduced torsion-free abelian group. A left E-module M is a Corner Emodule if (M, +) is a countable reduced torsion-free abelian group. A Corner E-module is always a left E-module. Fix a Corner E-module M and assume that annE (M ) = 0. Given a left E-module M Po (M ) is the set of finite subsets of M . We have a linear topology Γ(QE, QM ) = Γ(QM ) on QE whose neighborhoods of 0 are annQE (F ) over all finite subsets F ⊂ QM . This Haussdorf topology on QE is called the QM -topology. d is the completion of QE under the Γ(QM )-topology. QE Let d qM ⊂ M }. b ) = {q ∈ QE O(M

269

270

APPENDIX E. CORNER RINGS AND MODULES

b )-module. Then M is a left O(M

b Γ(M ) is the linear topology whose basis of open neighborhoods of zero is {annO(M (F ) F is a finite subset of QM }. b ) b ). This linear topology is called the M -topology on O(M We will prove the following. THEOREM E.1.1 [36, T.G. Faticoni]. Let E be a Corner ring and let M be a Corner E-module such that annE (M ) = 0. There is an exact sequence 0 −−−−→ M −−−−→ G −−−−→ QC −−−−→ 0

(E.1)

b )-modules such that of left O(M 1. C is a direct sum of the cyclic E-submodules of M (ℵo ) . b )∼ b ) 2. There is a topological isomorphism O(M = EndZ (G) where O(M b is endowed with the Γ(M )-topology and EndZ (G) is endowed with the finite topology. Proof: d LEMMA E.1.2 QM is a left QE-module. d Then r is the limit of a net Proof: Let x ∈ QM and let r ∈ QE. {rF F } ⊂ QE where ranges rF ranges over the finite subsets of QM . Choose a finite set F ⊂ QM such that x ∈ F and rF 0 − rF ∈ annE (x) for each finite set x ∈ F 0 ⊂ QM . Then (rF 0 − rF )x = 0 implies that d rF x = rF 0 x. Define rx = rF x. This makes QM a left QE-module.

b ) is complete in the M -topology. LEMMA E.1.3 O(M d so b ) is also a Cauchy net in QE Proof: A Cauchy net {rF }F in O(M d such that rbx = rF x ∈ M for the that it converges to an element rb ∈ QE b ). This completes subnet {rF x ∈ F }. Then rbM ⊂ M so that rb ∈ O(M the proof.

E.2. THE CONSTRUCTION OF G

E.2

271

The Construction of G

c denote the Z-adic completion of the CorE.2.1 In what follows let M b⊂E b b be an uncountable domain whose ner O(M )-module M . Let P ⊂ Z elements are rational multiples of units in P. (See [23, Section 2].) Let Π ⊂ P be a countable integral domain that satisfies the following lemma. LEMMA E.2.2 Let {λi i ∈ N} = L ⊂ P be a Π-linearly independent set. Let {xi i ∈ N} be an ordered subset of QM . P 1. If i∈N λi xi = 0 then xi = 0 for each i ∈ N. P 2. annO(M ( i∈N λi xi ) = annO(M ({xi i ∈ N}). b b ) ) c such that LEMMA E.2.3 There is a Corner submodule N ⊂ M 1. M ∩ N = 0. 2. For each finite set F ⊂ M there is a uF ∈ N such that annE (uF ) = annE (F ). 3. N is a direct sum of cyclic submodules of M (ℵo ) . Proof: Because M is countable, Po (M ) is countable, and because P is uncountable, there is a countable set L = {1, λF x F ∈ Po (M ) and x ∈ F } ⊂ P that is Π-linearly independent. Given F ∈ Po (M ) let X uF = λF x x

(E.2)

x∈F

and let N=

X finite

EuF .

(E.3)

F ⊂QM

The value of N in (E.3) does not change in this appendix. P Because L is Π-linearly independent, Lemma E.2.2(1) implies that finite F ⊂QM EuF is a direct sum and that M ∩N = 0. Thus N satisfies Lemma E.2.3(1). Fix a finite subset F ⊂ QM , and form the element uF in (E.2). Because L is Π-linearly independent, Lemma E.2.2(2) implies that annE (uF ) = annE (F ), so that Lemma E.2.3(2) is satisfied.

272

APPENDIX E. CORNER RINGS AND MODULES

Finally, given a finite subset F ⊂ QM , Lemma E.2.3(2) implies that there is an injection EuF → M (F ) such that uF → ⊕x∈F x where x ∈ Mx ∼ = M . Thus N embeds in M (ℵo ) as required by Lemma E.2.3(3). Inasmuch as M ⊕ N is countable and P is uncountable there is a countable set of units A = {mn m ∈ M, n ∈ N } ⊂ P

(E.4)

that is algebraically independent over Π. Let G be the pure subgroup of c generated by M ⊕ N and mn for each m ⊕ n ∈ M ⊕ N . M G = hM, N, Emn m ∈ M, n ∈ N i∗ X c ∩ Q M ⊕ N, = M {Emn m ∈ M, n ∈ N }

(E.5)

c. Then G is a Corner E-submodule of M PROPOSITION E.2.4 Let mn ∈ A be as in (E.4). P 1. {E(m + n)mn m ∈ M, n ∈ N } is a direct sum of cyclic Esubmodules of M (ℵo ) . b )-module. 2. G is a left O(M Proof: 1. The independence of the sum follows from Lemma E.2.2(1). Because mn is a unit in P there is a natural embedding E(m ⊕ n)mn ∼ = E(m ⊕ n) ⊂ M ⊕ N ⊂ M ⊕ M (ℵo ) . This is part 1. 2. Observe that QG is a left QE-module, so by Lemma E.1.2, QG d b )-module. Because M is a is a left QE-module. Thus QG is a left O(M b )-module, G = M c ∩ QG is a left O(M b )-module. This completes left O(M part 2.

E.3

Endomorphisms of G

The proof of the main result of this section is similar to the traditional proof of Corner’s Theorem [46, Theorem 110.1]. Since [46, Theorem 110.1] is the only publication of the proof of Corner’s Theorem in the last 30 years we will give a complete proof of that Theorem here.

E.3. ENDOMORPHISMS OF G

273

LEMMA E.3.1 Let η ∈ EndZ (G). For each m ∈ M and n ∈ N there is an rmn ∈ E such that η(m) = rmn m and η(n) = rmn n. c, η lifts to a map Proof: Let η ∈ EndZ (G). Because G is pure in M c −→ M c ηb : M and to a Q-linear homomorphism η : QG −→ QG. Let m ∈ M and n ∈ N . By the definition of A in (E.4), there is a finite subset F ⊂ M ⊕ N such that η((m ⊕ n)mn ) = ηb(m ⊕ n)mn X = x+ rm0 n0 (m0 ⊕ n0 )m0 n0

(E.6) (E.7)

m0 ⊕n0 ∈F

X

η(m ⊕ n) = y +

sm0 n0 (m0 ⊕ n0 )m0 n0

(E.8)

m0 ⊕n0 ∈F

where x, y ∈ Q(M ⊕ N ) and rm0 n0 , sm0 n0 ∈ E. As in [46, Theorem 110.1], substitute (E.7) and (E.8) into (E.6). Then use Lemma E.2.2(1) to compare coefficients in equations (E.6) and (E.7) and show that for each m0 ⊕ n0 ∈ F with m ⊕ n 6= m0 ⊕ n0 we have x = rm0 n0 (m0 ⊕ n0 ) = 0, sm0 ⊕n0 (m0 ⊕ n0 ) = 0, y = rmn (m ⊕ n) = η(m ⊕ n).

(E.9)

Finally, because m ⊕ n, 0 ⊕ n ∈ M ⊕ N there are rm , rn ∈ E such that η(m) = rm m and η(n) = rn n. Then rmn (m ⊕ n) = η(m ⊕ n) = η(m) ⊕ η(n) = rm m ⊕ rn n. Using (E.9) and (E.10) we arrive at rm m = rmn m and rn n = rmn n. This proves the lemma.

(E.10)

274

APPENDIX E. CORNER RINGS AND MODULES

LEMMA E.3.2 There is an isomorphism of rings b )∼ O(M = EndZ (G). b )-submodule of M c. InasProof: By Proposition E.2.4, G is an O(M c = 0 implies that r = 0 for r ∈ O(M b ), much as rG = 0 implies that rM b there is an embedding of rings O(M ) −→ EndZ (G). We claim that this embedding is an isomorphism. Let η ∈ EndZ (G) and let F ⊂ M be a finite set. By Lemma E.2.3(2), there is a uF ∈ N such that annE (uF ) = annE (F ). Lemma E.3.1 states that for each m ∈ F there is an rF m ∈ QE such that η(m) = rF m m and η(uF ) = rF m uF for each m ∈ F . Let rF be any one of the rF x with x ∈ F . Then rF m uF = η(uF ) = rF uF for each m ∈ F , so that rF − rF m ∈ annE (uF ) = annE (F ) ⊂ annE (m). Hence rF m = rF m m = η(m) for each m ∈ F . d under Subsequently, the set {rF F ∈ Po (M )} is a Cauchy net in QR d of the Cauchy net. By the QM -topology. Let rb denote the limit in QR (E.1.2), rbm = rF m = η(m) for each finite set F ⊂ QM and each m ∈ F . Thus [η − rb](M ) = 0. c, η = rb. Since M is pure and dense in M b Therefore the embedding O(M ) −→ EndZ (G) is an isomorphism. b )∼ LEMMA E.3.3 The isomorphism of rings O(M = EndZ (G) is a topob logical one, taking the M -topology on O(M ) onto the finite topology of EndZ (G).

E.3. ENDOMORPHISMS OF G

275

Proof: The proof, being natural, is left for the reader. Proof of Theorem E.1.1: Given the Corner Module M construct N and G as in (E.3) and (E.5). Let X C = N⊕ E(m ⊕ n)mn . m∈M,n∈N

c, Note that because M and G are pure and dense in M G/M ∼ = QC b )-modules. By Proposition E.2.4(1), C is a direct sum of as left O(M cyclic submodules of M (ℵo ) . Thus Theorem E.1.1(1) is satisfied. Furb )∼ thermore by Lemmas E.3.2 and E.3.3, O(M = EndZ (G) topologically, where EndZ (G) is endowed with the finite topology. This completes the proof of Theorem E.1.1.

Appendix F

Corner’s Theorem F.1

Countable Endomorphism Rings

THEOREM F.1.1 [23, A.L.S. Corner]. Each countable reduced torsion-free ring E is the endomorphism ring of a countable reduced torsionfree abelian group G. Furthermore, there is a short exact sequence (E.1) of left E-modules in which C is a direct sum of cyclic submodules of M (ℵ0 ) . Proof: Corner’s Theorem follows from Theorem E.1.1.

Let

O(M ) = {q ∈ QE qM ⊂ M }.

THEOREM F.1.2 [43, T.G. Faticoni, H.P. Goeters]. Let E be a reduced torsion-free finite rank ring and let M be a reduced torsion-free finite rank left E-module. Then O(M ) ∼ = EndZ (G) for some reduced torsion-free finite rank group G. Furthermore, there is a short exact sequence (E.1) of left O(M )-modules in which C ∼ = O(M ) ⊕ O(M ). Proof: When constructing G use a maximal linearly independent subset X of M instead of using each element m ∈ M . 277

278

APPENDIX F. CORNER’S THEOREM

THEOREM F.1.3 [23, A.L.S. Corner]. If E is a reduced torsion-free finite rank ring then E = EndZ (G) for some reduced torsion-free finite rank group G. Furthermore, there is a short exact sequence (E.1) of left E-modules in which C is a cyclic free left E-module.

Appendix G

Torsion Torsion-Free Groups G.1

E-Torsion Groups

Let E be an integral domain. A left E-module M is torsion if for each x ∈ M there is an r 6= 0 ∈ E such that xr = 0. We construct torsion-free groups G such that EndZ (G) is an integral domain, and such that G is a torsion left EndZ (G)-module. EXAMPLE G.1.1 There is a countable torsion-free group G such that EndZ (G) = Z[[x]] and such that G is a countable torsion left EndZ (G)module. Proof: Let E = Z[x] and let M = Z[x]/(x) ⊕ Z[x]/(x2 ) ⊕ Z[x]/(x3 ) ⊕ · · · . d = Q[[x]], and O(M b )= Then Γ(QM ) is the x-adic topology on QE, QE Z[[x]]. Theorem E.1.1 states that there is a short exact sequence 0 −→ M −→ G −→ QC −→ 0 of left Z[[x]]-modules and such that 1. EndZ (G) = Z[[x]], and 2. C is a direct sum of submodules of M (ℵo ) . Then M , G, and QC are torsion left Z[[x]]-modules. Thus G is a torsion left EndZ (G)-module. This proves the Example.

279

280

APPENDIX G. TORSION TORSION-FREE GROUPS

G.2

Self-Small Corner Modules

The groups in the above constructions are torsion modules over their endomorphism rings. The next construction produces E-torsion-free groups. The group G is self-small if for each cardinal c the natural mapping Hom(G, G)(c) −→ Hom(G, G(c) ) is an isomorphism of right EndZ (G)-modules. We will prove THEOREM G.2.1 Let E be a Corner ring and let M be a Corner E-module. There is a short exact sequence (E.1) in which 1. C is a free left O(M )-module, 2. O(M ) ∼ = EndZ (G), and 3. G is a self-small group. We require a lemma. LEMMA G.2.2 As in (E.2.1) choose a countable integral domain Π ⊂ b and an uncountable integral domain Π ⊂ P ⊂ Z. b Let M be a Corner Z c such that module such that annE (M ) = 0. There is an element u ∈ M annE (u) = 0. Proof: Write M = {m1 , m2 , m3 , · · ·} and since P is uncountable there is a countable set L = {λ1 , λ2 , λ3 , · · ·} ⊂ P that is algebraically independent over the integral domain Π. (See [23, Section 2].) There is a convergent sum u=

∞ X

mk pk

k=1

c. in the p-adic topology on M If r ∈ E and ru = 0 then ∞ X k=1

rmk pk = 0.

G.2. SELF-SMALL CORNER MODULES

281

By Lemma E.2.2(2), rmk pk = 0 = rmk for each integer k > 0. Hence c is the element that rM = 0 = r by our hypothesis on M . Thus u ∈ M we seek. Proof of Theorem G.2.1: By Lemma G.2.2 there is an element c such that annE (u) = 0. By taking λu for some λ ∈ P if necessary u∈M we can assume that M ∩ Eu = 0. c. Because M ⊕ Eu is countable there is an algeThen M ⊕ Eu ⊂ M braically independent set A = {m m ∈ M } ⊂ P over Π. Let C = Eu +

X

E(m ⊕ u)m

m∈M

and let c. G = Q(M ⊕ C) ∩ M

(G.1)

c. Thus there is a short exact Evidently G is a left E-submodule of M sequence 0 −−−−→ M −−−−→ G −−−−→ QC −−−−→ 0 of left E-modules. By our choice of u E∼ = Eu ∼ = E(m ⊕ u)m for each m ∈ M . Because A ∪ {1} is algebraically independent over Π, C is a free module. (See Lemma E.2.3(2).) Thus Theorem G.2.1(1) is satisfied. Because annO(M ) (G) = annO(M ) (M ) = 0 there is an embedding O(M ) −→ EndZ (G) of rings. We claim that this is an isomorphism. Using Lemma 2.3.3, for each m ∈ M there is an rm ∈ QE such that η(m) = rm m and η(u) = rn u. Let ro be one of the rm . Then for each m ∈ M , ro u = rm u. Since annE (u) = 0, ro = rm for all m ∈ M . Hence η(m) = ro m for all c) is contained in the m ∈ M . Thus [η − ro ](M ) = 0. Then [η − ro ](M c divisible subgroup of M , which in this case is 0. That is, η = ro and the proof of Theorem G.2.1(2) is complete. Lastly, G is a countable group in which the finite topology is discrete (u ∈ G.) Then by [15, Proposition 2.1], G is self-small. This proves Theorem G.2.1(3) and completes the proof.

Appendix H

E-Flat Groups Recall that G is faithful if IG 6= G for each proper right ideal I ⊂ EndZ (G), and that G is called fully faithful if K ⊗EndZ (G) G 6= 0 for each nonzero right EndZ (G)-module K. G is E-flat if G is a flat left E-module. G is faithfully E-flat if G is a faithful group and a flat left E-module.

H.1

Ubiquity

THEOREM H.1.1 The group constructed according to Corner’s Theorem F.1.1 is E-flat. Proof: Let E be a Corner ring. By Corner’s Theorem F.1.1 there is an exact sequence 0 −→ E −→ G −→ QC −→ 0 of left E-modules in which 1. E ∼ = EndZ (G), and 2. C is a free left E-module. Then there is an exact sequence of Tor groups Tor1E (·, E) −→ Tor1E (·, G) −→ Tor1E (·, QC). Since E is free, Tor1E (·, E) = 0, and since C is free, QC is flat. Thus Tor1E (·, QC) = 0. It follows that Tor1E (·, G) = 0, whence G is a flat left E-module. 283

284

APPENDIX H. E-FLAT GROUPS

THEOREM H.1.2 Let E be an rtffr ring, and let G be a group constructed according to Corner’s Theorem F.1.3 such that E = EndZ (G). Then G is faithfully E-flat. Proof: By the previous theorem G is a flat left E-module. To see that G is fully faithful (i.e., to see that K ⊗E G 6= 0 for nonzero right E-modules K), it suffices to show that IG 6= G for each maximal right ideal I ⊂ E. There is an exact sequence Tor1E (E/I, QC) −→ E/I ⊗E E −→ E/I ⊗E G −→ E/I ⊗E QC of groups. Since E is rtffr, E/I is finite (Lemma 4.2.3). Thus E/I ⊗E QC = 0. Furthermore since QC is a flat left E-module Tor1E (E/I, QC) = 0. Hence 0 6= E/I ∼ = G/IG whence G is faithful. This completes the proof.

H.2

Unfaithful Groups

THEOREM H.2.1 There is a countable torsion-free group G with endomorphism ring E such that 1. E is an integral domain, 2. IG 6= G for each nonzero ideal I ⊂ E, 3. There is a nonzero E-module K such that K ⊗E G = 0. That is, there is a countable torsion-free group G that is faithful but not fully faithful. Proof: 1. Let E = Z[x](x,2) be the localization of Z[x] at the ideal generated by x and 2. Let M=

∞ M

E/xk E.

k=1

Then the reader shows that O(M ) = E. By Theorem E.1.1 there is a short exact sequence 0 −→ M −→ G −→ QC −→ 0

285 of left E-modules such that C is a direct sum of submodules of M (ℵo ) . Then G is a torsion E-module. Choose the unique maximal right ideal I. Then E/I ∼ = Z/2Z so that E/I ⊗E QC = 0 = Tor1E (E/I, QC) and so 0 6= M/IM ∼ = G/IG. This proves part 2. For part 3 let c0 , c1 , c2 , · · · be elements such that ck 2 = 0 for all k ≥ 0 c0 x = 0 and ck+1 x = ck for all k ≥ 0. Let K=

∞ X

ck E.

k=0

Because QC is divisible and K is an elementary 2-group K ⊗E QC = 0. Because M is x-torsion while K is x-divisible, K ⊗E M = 0. Then K ⊗E G = 0. This proves part 3 and completes the proof. THEOREM H.2.2 There is a faithful group that is not E-flat. THEOREM H.2.3 There is an E-flat group that is not faithful. Z 0 and let M be the projective left E-module Proof: Let E = Z Z Z 0 Z . Evidently O(M ) = = E. M= Z Z Z A little moreeffort (using matrix units) shows that the trace ideal of Z 0 M in E is I = . Then IM = M so that (I + nE)M = M for Z 0 each integer n. By Theorem E.1.1 there is a short exact sequence 0 −−−−→ M −−−−→ G −−−−→ QC −−−−→ 0 in which C is a finitely generated free left E-module. Since M is projective as a left E-module, G is flat as a left E-module. Then E/(I + nE) ⊗E M = E/(I + nE) ⊗E QC = 0 for each integer n > 0. An application of E/(I + nE) ⊗E · to the short exact sequence shows us that (I + nE)G = G. Thus G is E-flat but not faithful.

286

APPENDIX H. E-FLAT GROUPS

THEOREM H.2.4 There is a countable torsion-free group that is not faithful and not E-flat. Proof: Let

and let

Z 0 0 E= Z Z 0 Z Z Z

Z 0 Z 0 0 0 M = Z Z 0 0 = Z Z . Z Z 0 Z Z 0

Since

Q 0 QM = Q Q Q 0

is not a projective left QE-module, M is not flat as a left E-module. This is because the indecomposable projective QE-modules are the columns of Q 0 0 QE = Q Q 0 . Q Q Q 0 0 0 0 0 0 does not contain a nonzero ideal of E, annE (M ) = Since 0 Z Z Z 0 0 0. Observe that I = Z Z 0 is a right ideal in E such that Z Z 0

Z 0 0 Z 0 Z 0 IM = Z Z 0 Z Z = Z Z = M. Z Z 0 Z 0 Z 0 Then JM = M for each proper right ideal Z 0 0 J = Z Z 0 . Z Z nZ We note that I ⊂ J ⊂ E and that E/J is finite. Now use Theorem F.1.2 to construct a short exact sequence 0 −−−−→ M −−−−→ G −−−−→ QC −−−−→ 0

287 of left E-modules such that C is a free E-module and E ∼ = EndZ (G). The by now familiar arguments show us that since JM = M , JG = G, and since M is not flat, G is not E-flat.

Appendix I

Zassenhaus and Butler I.1

Statement

A is a finite dimensional Q-algebra and V = A is a cyclic free left A-module with free basis . Identify A via left multiplication with the subring AL of EndQ (V ). Let Q = the set of primes p ∈ Z. For primes p ∈ Z and groups X we let Xp = Zp X ⊂ QX. Let O(X) = OA (X) = {a ∈ AL aX ⊂ X}. Let ∈ M ⊂ V be a full free abelian subgroup. Then QM = V and Mp is a free Zp -module for each p ∈ Q. Because M is a free abelian group we have the local-global relationships \ M= Mp , p∈Q

O(M )p = O(Mp ) for each p ∈ Q, and \ O(M ) = O(Mp ). p∈Q

We will prove the following theorem. 289

290

APPENDIX I. ZASSENHAUS AND BUTLER

THEOREM I.1.1 [T.G. Faticoni] Let A, V , and M be as above. There is a subgroup M ⊂ G ⊂ V such that 1. G is a locally free group, 2. O(M ) ⊂ EndZ (G) ⊂ A, 3. EndZ (G)p /O(M )p is a finite p-group for each prime p ∈ Z.

I.2

Proof

T We construct G as the intersection G = p Kp where the Kp are O(Mp ) = O(M )p -submodules of V . The Kp are chosen in a series of lemmas. Since V has finite Q-dimension we can enumerate the elements of [ (I.1) EndQ (V ) \ AL = {φp p ∈ Q}. LEMMA I.2.1 Let p ∈ Q, let φ = φp , and suppose that φ(Mp ) 6⊂ Mp . Then choose Kp = Mp .

(I.2)

In particular, Kp contains Mp , Kp is a left Ep -module, and Kp is a finitely generated free Zp -module. LEMMA I.2.2 Let 0 6= φ = φp for some p ∈ Q. Suppose that φ satisfies 1. φ(Mp ) ⊂ Mp , and 2. φ() = 0. There is a cyclic Ep -submodule Np ⊂ V such that φ(Np ) 6⊂ Np . Proof: Let p ∈ Q and let 0 6= φ = φp satisfy conditions 1 and 2 above. Let Ep = O(Mp ) and let Rp = Ep [φ] denote the subring of EndQ (V ) generated by Eq and φ. Assume, for the sake of contradiction that each cyclic Ep -submodule Np ⊂ V satisfies φ(Np ) ⊂ Np .

(I.3)

291 The contradiction that we are looking for is φ = 0. Since each Ep -submodule of V is a sum of cyclic Ep -submodules of V , condition (I.3) implies that each Ep -submodule of V is an Rp -module.

(I.4)

Arbitrarily choose x ∈ V . Since is a free basis for V , Ep is a cyclic free left Ep -module with basis . There is then a short exact sequence ρ

0 −→ L −→ Ep −→ Ep x −→ 0 of left Ep -modules in which ρ(s) = sx for each s ∈ Ep . Tensoring with the right Ep -module Rp yields the commutative diagram with exact rows Rp ⊗ L νL 0

? - L

- Rp ⊗ Ep 1 ⊗ ρ- Rp ⊗ Ep x

ν ? - Ep

-0

νx ρ

? - Ep x

-0

where the maps νY are scalar multiplication maps νY (r ⊗ y) = ry. By (I.4), L, Ep , and Ep x are left Rp -modules, so νL , ν , and νx are well defined Rp -module maps. Because ν , 1 ⊗ ρ, and νx are surjections, the commutativity of the diagram shows us that for each r ∈ Rp and y = s ∈ Ep ρ(ry) = ρ(ν (r ⊗ y)) = ρ(ν (r ⊗ s)) = νx (1 ⊗ ρ)(r ⊗ s) = νx (r ⊗ sx) = r(sx) = rρ(y) Hence ρ is an Rp -module map. Recall the hypothesis that φ() = 0. Then φ(x) = φ(ρ()) = ρ(φ()) = 0

292

APPENDIX I. ZASSENHAUS AND BUTLER

so that φ(x) = 0 for any x ∈ V . Thus φ(V ) = 0 which implies that φ = 0. This contradicts our choice of φ 6= 0 so condition (I.3) is false, that is, φNp 6⊂ Np some cyclic Ep -submodule of V . This proves the lemma. COROLLARY I.2.3 Let 0 6= φ = φp for some p ∈ Q, suppose that φ(Mp ) ⊂ Mp and that φ() = 0. There is an O(Mp )-submodule Mp ⊂ Kp ⊂ V such that Kp is a finitely generated free Zp -module, and φ(Kp ) 6⊂ Kp . Proof: Let O(Mp ) = Ep . By Lemma I.2.2, given φ such that φ(Mp ) ⊂ Mp and φ() = 0 there is a cyclic Ep -submodule Np ⊂ V such that φ(Np ) 6⊂ Np . Then Np is a torsion-free quotient of the finitely generated free Zp -module Ep , hence Np is a finitely generated free Zp -module, whence pk Mp + Np is a finitely generated free Zp -module for each integer k > 0. Let x ∈ Np be such that φ(x) 6∈ Np . Since the q-adic topology on the free Zp -module Mp + Np is discrete there is an integer k = kp > 0 such that φ(x) 6∈ pk Mp + Np . Choose Kp = Mp + p−k Np

(I.5)

and observe that Mp ⊂ Kp , that Kp is a finitely generated free Zp module, and that φ(Kp ) 6⊂ Kp . There is one case left in our choices of Kp . LEMMA I.2.4 Let 0 6= φ = φp for some p ∈ Q, suppose that φ(Mp ) ⊂ Mp , and suppose that φ() 6= 0. There is a finitely generated O(Mp )module Kp such that Mp ⊂ Kp ⊂ V and φ(Kp ) 6⊂ Kp . Proof: Let O(Mp ) = Ep . Since φ(Mp ) ⊂ Mp and since φ 6∈ AL , 0 6= mφ − r 6∈ Ep = O(Mp )

(I.6)

for any integer m 6= 0 and element r ∈ Ep . Because Mp and Ep are full subgroups of V we can write 0 6= φ(m) = r for some integer m 6= 0 and some r ∈ Ep . Because φ(Mp ) ⊂ Mp and r ∈ Ep we have (mφ − r)(Mp ) ⊂ Mp and (mφ − r) = 0. By (I.6), mφ − r 6= 0 so we can apply Corollary I.2.3 to show that there is a left Ep -submodule Mp ⊂ Kp ⊂ V such that Kp is a finitely generated

293 free Zp -module, and such that (mφ − r)(Kp ) 6⊂ Kp . Since r(Kp ) ⊂ Kp , mφ(Kp ) 6⊂ Kp , whence φ(Kp ) 6⊂ Kp . This proves the lemma. For φ = φp 6∈ AL such that φ(Mp ) ⊂ Mp , choose Mp ⊂ Kp ⊂ V

(I.7)

as in Lemma I.2.4. This takes into account all of the cases for φ ∈ EndQ (V ) \ AL . Proof of Theorem I.1.1: Let E = O(M ), let Kp denote the left Ep -modules chosen in (I.2), (I.5), and (I.7) and let \ G= Kp . (I.8) p∈Q

Each Kp is an Ep -module, each Kp is a finitely generated free Zp -module, and ∈ G is a unimodular element. Since localization commutes with . intersections, Gp = Kp = Mp for all primes p ∈ Z. In each of our choices, Kp is a finitely generated free Zp -module. Hence G is a locally free full subgroup of V T , as required by Theorem I.1.1(1). Since E = p∈Q Ep , G is a left E-module such that ψ() 6= 0 for each 0 6= ψ ∈ E ⊂ AL . Hence E ⊂ EndZ (G) ⊂ EndQ (V ). By our choices of φp and Kp , for each φ ∈ EndQ (V ) \ AL there is a p ∈ Q such that φ = φp and φ(Kp ) 6⊂ Kp . However for ψ ∈ EndZ (G), ψ(Kp ) = ψ(Gp ) ⊂ Gp = Kp for each p ∈ Q. Thus EndZ (G)

\

(EndQ (V ) \ AL ) = ∅

and hence EndZ (G) ⊂ AL . This proves condition Theorem I.1.1(2). Since Gp = Kp for all primes p ∈ Z it follows that EndZ (G)p ⊂ O(Kp ). By our construction of Kp in (I.2), (I.5), and (I.7) we see that Kp /Mp is a finite p-group so that O(Kp )/O(Mp )

294

APPENDIX I. ZASSENHAUS AND BUTLER

is a finite p-group. Thus EndZ (G)p /O(Mp ) is a finite p-group for each p ∈ Q. This is what is required in Theorem I.1.1(3). This completes the proof of the theorem. THEOREM I.2.5 [90, H. Zassenhaus]. Let E be a ring such that (E, +) is a finitely generated free ring. There is a group E ⊂ G ⊂ QE such that E = EndZ (G) where the action of E on G is the natural one as an E-submodule of QE. THEOREM I.2.6 [21, M.C.R. Butler]. Let E be a locally free rtffr ring. There is a group E ⊂ G ⊂ QE such that E = EndZ (G) where the action of E on G is the natural one as an E-submodule of QE.

Appendix J

Countable E-Rings J.1

Countable Torsion-Free E-Rings

An E-ring is a ring E for which the left representation embedding λ : E −→ EndZ (E) is an isomorphism of rings. A Dedekind E-ring is an E-ring R that is a Dedekind domain. THEOREM J.1.1 An E-ring is a commutative ring. R. Bowshell and P. Schultz [20] characterize the rtffr E-rings. THEOREM J.1.2 [20, R. Bowshell, P. Schultz]. The rtffr ring E is an E-ring iff . E = R1 ⊕ · · · ⊕ Rt for some integer t > 0 and some Dedekind E-rings R1 , . . . , Rt such that HomZ (Ri , Rj ) = 0 for each 1 ≤ i 6= j ≤ t. This structure theorem seemed to indicate that the structure of Erings was restricted as far as rtffr groups go. The following theorem shows that the countable torsion-free E-rings have a much more diverse additive structure. The group G is p-local if pG 6= G. THEOREM J.1.3 [42, T.G. Faticoni]. Let S be a countable reduced torsion-free commutative ring. There is a E-ring E and a pure and dense embedding of rings S −→ E.

295

296

APPENDIX J. COUNTABLE E-RINGS

Before proving this theorem an example or two along with a comparison to Theorem J.1.2 will illustrate the difference between rtffr E-rings and countable torsion-free E-rings. EXAMPLE J.1.4 Let S = Z[x1 , x2 , x3 , . . .]/I where x1 , x2 , x3 , . . . are commuting indeterminants, and where I is an ideal in S. By Theorem J.1.3, S is a pure and dense subring of an E-ring E. We point out that every commutative ring can be realized in this way. Specifically Z[x]/(xn ) embeds in an E-ring E. This E-ring contains a nonzero nilpotent element x. Rtffr E-rings do not contain nonzero nilpotent elements. Proof of Theorem J.1.3: As in the proof of Corner’s Theorem 2.3.3 we require an uncountable integral domain. The proof is left as an exercise for the reader. LEMMA J.1.5 Let K be a uncountable field extension of Q, and let B be a commutative K-algebra. Let S be a countable Q-subalgebra of B. There exists an uncountable set U ⊂ K that is algebraically independent over S. For groups G and primes p ∈ Z, let us define Gp =

G ⊗ Zp . div(G ⊗ Zp )

Then Gp is a reduced p-local group, and Gp is countable if G is countable. To reduce to the p-local case, observe that if S is a countable reduced torsion-free commutative ring then S is a pure and dense subring of Q S where p ranges over the primes of Z. Thus for each prime p ∈ Z p p we will embed each Sp as a pure and dense subring of an E-ring Q Ep . We can then embed S as a pure and dense subring of the E-ring p Ep . Let W ∪ {1} bp -module with basis W ∪ {1}. Since be a p-basis of S. Then Sb is a free Z bp of S is countable there is an algebraically independent subset U ⊂ Z units such that

297

card(U ) = card(W ) + 1. Write U = {d, uw w ∈ W }. We will show that E = hS[uw + dw w ∈ W ]i∗ b Certainly is the E-ring we seek, the purification taking place in S. S ⊂ E ⊂ Sb so that S is a pure and dense subring of E. It remains to prove that E is an E-ring. LEMMA J.1.6 Let S, W, uw , d, E be as above. Then 1. E ∩ dE = 0. 2. Let φ : E → Sb be a group homomorphism such that dφ(S)∩φ(S) = 0. Then f φ = 0 if f φ(1) = 0. Proof: 1. It suffices to show that {d, uw +dw w ∈ W } is algebraically independent over QS. Let X = {xd , xw w ∈ W } be a set of indeterminants and consider the evaluation maps α : QS[X] −→ QS[U ] and β : QS[X] −→ QS[U ] such that α(xd ) = β(xd ) = d α(xw ) = uw + dw β(xw ) = uw . Since U is algebraically independent and since W ⊂ S, β is a ring isomorphism. Let P (X) be a polynomial of minimial xd -degree such that α(P (X)) = 0. We can write

298

APPENDIX J. COUNTABLE E-RINGS

P (X) = xd Q(X) + R(X) where the xd -degree of R(X) is 0. Then α(P (X)) = 0 implies that α(R(X)) = −dα(Q(X)). Arbitrarily enumerate the terms of R(X) and let Rk (X) be the k-th term in R(X). For each integer k ≥ 1 and w ∈ W let e(w, k) = the xw -degree of Rk (X). Then ! Y

α(Rk (X)) = ak

e(w,k)

(uw + dw)

! = ak

Y

ue(w,k) w

+ dBk

w∈W

w∈W

for some ak ∈ QS and some Bk ∈ QS[U ]. Then Y ue(w,k) = β(Rk (X)) w w∈W

so that α(Rk (X)) = β(Rk (X)) + dBk . By setting B =

P

k

Bk then we have

α(R(X)) = β(R(X)) + bB = −dα(Q(X)). It follows that R(X) = xd β −1 [−B − α(Q(X))]. Since R(X) has xd -degree 0 it must be that R(X) = 0. That is P (X) = xd Q(X). bp and α is Z bp -linear so Now d is a unit in Z α(Q(X)) = d−1 α(xd Q(X)) = d−1 α(P (X)) = 0. This contradiction to the minimality of the xd -degree of P (X) shows that α : QS[X] −→ QS[U ] is an isomorphism. In particular {d, uw +dw dw ∈ W } is algebraically independent over QS. This proves part 1.

299 2. Assume that φ : E −→ Sb is a function such that φ(1) = 0 and bp -module map φ : E b −→ S, b φ(E) ∩ dφ(E) = 0. Since φ lifts to a Z bp . φ(u) = uφ(1) for each u ∈ Z Thus, for each w ∈ W , φ(uw + dw) = uw φ(1) + dφ(w) = dφ(w) ∈ φ(E) ∩ dφ(E) = 0 by part 1. Therefore 0 = φ(uw + dw) = dφ(w) = φ(w) for all w ∈ W . b=Z bp · W and since φ(1) = 0 by hypothesis, φ = 0. This Inasmuch as E proves part 2 and completes the proof of the lemma. Proof of Theorem J.1.3: Let S be a countable reduced torsion-free commutative ring. We have already observed that S ⊂ E is a pure and dense embedding of rings. Let φ : E −→ E be a group homomorphism, let u = φ(1), and consider φ − u. By Lemma J.1.6(1), E ∩ dE = 0, and by Lemma J.1.6(2), φ − u = 0. I.e. φ = u ∈ E, hence E is an E-ring. This concludes the proof of Theorem J.1.3.

Appendix K

Dedekind E-Rings Our references in this section for commutative ring theory are [6, 49]. The number theory we use can be found in [58, 66]. The ring R is a Dedekind E-ring if the group (R, +) is an E-ring and if the ring R is a Dedekind domain. We will show that the Dedekind domain R is an E-ring provided it is divisible by a certain set of primes in the ring of algebraic integers in algebraic number field QR.

K.1

Number Theoretic Preliminaries

At all times p ∈ Z is a prime number, E and F are algebraic number fields, OE denotes the ring of algebraic integers in E, and L(E) denotes the lattice of subrings of E containing OE . Let spec(E) denote the set of maximal ideals in a ring OE . For more general commutative rings R, spec(R) denotes the set of maximal ideals in R. Let E ⊂ F be a subfield, let P ∈ spec(E) and M ∈ spec(F ). We say that M lies over P if P ⊂ M . Equivalently, M lies over P if P ⊂ M ∩E. For unramified primes P ∈ E it is known that X [F : E] = {[OF /M : OE /P ] M ∈ spec(F ) and M lies over P .} If there are at least two different primes in OF lying over P then we say that P splits in F or in OF . If [F : E] primes in OF lie over P then we say that P splits completely in F . Thus P splits completely

301

302

APPENDIX K. DEDEKIND E-RINGS

in F iff OF /M ∼ = OE /P for each M ∈ spec(F ) lying over P . Complete splitting occurs often enough as [58, Theorem 6] states that infinitely many primes P ∈ spec(E) split completely in F .

K.2

Integrally Closed Rings

Let OF ⊂ R ⊂ F be a Dedekind domain. The support of R in F is the set σF (R) = {M ∈ spec(F ) M R 6= R} and the divisibility of R in F is

δF (R) = {M ∈ spec(F ) M R = R}. For a Dedekind domain OF ⊂ R ⊂ F , the sets σF (R) and δF (R) form a partition of spec(F ). Given a set σ ⊂ spec(F ) then let

Oσ = ∩{(OF )M M ∈ σ} where (OF )M is the classic localization of OF at the maximal ideal M in OF . K.2.1 Let σ ⊂ spec(F ). Then σF (Oσ ) = σ and δ(Oσ ) = σ 0 = the complement of σ in spec(F ). Moreover, if C(R) = {c ∈ OF cR = R} then by [49, page 73, exercise 7] R = OF [C −1 ]. Furthermore σF (R) = {M ∈ spec(F ) M ∩ C = ∅} and δF (R) = {M ∈ spec(F ) c ∈ M for some c ∈ C}.

303 Thus to study E-rings we need to study localizations Oσ for subsets σ ⊂ spec(F ). A classic result from commutative ring theory follows. Let S ⊂ R be rings. Then R is integral over S if R is finitely generated by S. LEMMA K.2.2 Let S ⊂ R ⊂ F be subrings of F . Then R is integral over S iff R = SOF . Proof: If R = SOF then R is finitely generated by S because OF is a finitely generated free abelian group. Hence R is integral over S. Conversely if R is integral over S then R is a finitely generated Smodule. Since R is integrally closed SOF ⊂ R and since R is finitely generated over the integrally closed domain SOF , SOF = R. THEOREM K.2.3 [10, Theorem 14.3] Suppose that OF ⊂ R ⊂ F and that R is a Dedekind domain. The following are equivalent. 1. (R, +) is strongly indecomposable. 2. F is the smallest subfield E ⊂ F such that R is integral over R ∩E. 3. F is the smallest subfield E ⊂ F such that R is finitely generated by R ∩ E. 4. R is an E-ring. Proof: The proof follows from the Lemma and from [10, Theorem 14.3].

THEOREM K.2.4 [39, Proposition 2.1] Let F be an algebraic number field, and let σ ⊂ spec(F ). The following are equivalent. 1. Oσ is an E-ring. 2. For each proper subfield E ⊂ F there are M, M 0 ∈ spec(F ) such that M ∩ E = M 0 ∩ E while M ∈ σ and M 0 6∈ σ. Proof: Assume that R = Oσ is not an E-ring. By Theorem K.2.3, there is a proper subfield E ⊂ F such that R = SOE where S = R ∩ E. Since S is a Dedekind domain with field of quotients E, S = OE [C −1 ] for some multiplicatively closed subset C ⊂ OE . Then OF [C −1 ] = R.

304

APPENDIX K. DEDEKIND E-RINGS

Fix M 0 ∈ spec(F ) such that M 0 R = R. There is a c ∈ C such that 1 · c = c ∈ M 0 . Thus M R = R for each maximal ideal M lying over M 0 ∩ S. This is the negation of part 2. Conversely, suppose that R = Oσ is an E-ring, let E ⊂ F be a proper subfield of F , and let S = R ∩ E. Since R is an E-ring, R is not integral over S. Since the integral property is a local-global property in F there is a prime P ∈ spec(E) such that RP is not integral over the discrete valuation domain SP . If RP = F then PP = PP R P ∩ S P = F ∩ S P = S P , contrary to the fact that the ring SP is never equal to its Jacobson radical PP . Thus RP 6= F . S and R are Dedekind domains and R is not finitely generated by S, so SP = (OE )P ⊂ SP OF = (OF )P 6= RP . Since RP 6= F there is an ideal M ∈ spec(F ) lying over P such that (MP )M = MM 6= RM = (RP )M 6= F. It follows that M RP 6= RP . RP is not integral over (OF )P , so there is an ideal M 0 ∈ spec(F ) lying over P such that (RP )M 0 = RM 0 is not finitely generated by the discrete valuation domain ((OF )P )M 0 = (OF )M 0 . Then RM 0 = F , and hence M 0 RP = RP . Localizing at primes Q ∈ spec(E) other than P yields M 0 RQ = RQ because there is an element c ∈ M 0 ∩ E \ Q. By the Local-Global Theorem, M 0 R = R. Since M and M 0 both lie over P we have proved part 2. THEOREM K.2.5 Let p ∈ Z be prime and let F be a minimal field extension of Q. Then F is the field of fractions of a p-local E-ring iff p splits in F . Examples of minimal field extensions are found by examining the finite lattice of subfields of the algebraic number field F . Thus quadratic number fields and in general fields F such that [F : Q] = p a prime are minimal field extensions. Another method is to take a Galois extension K/Q and then take a maximal subgroup G ⊂ Gal(K/Q). The fixed field F of G is minimal by the Galois correspondence. The degree [F : Q] is the degree [Gal(K/Q) : G].

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Index A A(·) 31, 59 acd group 24, 128, 143 Albrecht, U. 146, 150, 151 Albrecht, U. and Goeters, H.P. 144 almost completely decomposable group 24 Anderson, F.W. and Fuller, K.R. 1 Angad-Gaur 243 annhilator 5 Arnold, D.M. 1, 81, 159, 162, 164, 166, 168 Arnold’s Theorem 62 Arnold, D.M. and Hausen, J. 152 Arnold, D.M. , Hunter, R., and Richman, F. 71, 132 Arnold, D.M. and Lady, E.L. 74, 151, 201 Arnold-Lady-Murley Theorem 30 Artinian commutative ring 79, 81 associative ring 1 Azumaya-Krull-Schmidt Theorem 22 B Baer splitting property 139, 140, 144, 152, 177 Baer-Kulikov-Kaplansky Theorem 24, 26, 99 Baer’s Lemma 139, 146, 148, 151 Beaumont, R.A. and Pierce, R.S. 24, 50, 79, 221

Beaumont-Pierce-Wedderburn Theorem 50, 79, 219, 226, 239 bounded filter 186, 190 bounded ring 203 Bowshell, R. and Schultz, P. 223 bracket group 131 Butler’s Construction 91 Butler, M.C.R. 27, 289 C center 5, 48 central idempotents 48, 71 Change of Rings 13 Charles, B. 71 Chinese Remainder Theorem 13 classical maximal order 15, 38, 47, 51 class group of E 109 class number of G 108 class number of E 109 cokernel group 131 commutative ring 79, 81, 86 complete set of idempotents 71 completely decomposable 24, 243 completely faithful 152 composition factors 180 conductor of E 51 conductor of G 60 Corner, A.L.S. 25, 277 Corner E-module 27 Corner group 27 Corner module 27 Corner ring 27

311

312 Corner’s Theorem 27, 277 CycE (T ) 180 D D(QG) 204 D(G) 177 c) 190 DEb (M DE (M )183 δ(M ) 185 δE (G) 197 c) 190 δ(M δ(QG) 208 ∆Gb 196 ∆QG 204, 208 ∆G,H 206, 208 Dedekind group 127, 130 Dedekind E-ring 222, 226, 229, 240, 295 Dedekind ring 15 direct product 2 direct sum 2 discrete valuation domain 15 divisibility of R in F 302 dual module of 147 Dugas, M., Mader, A., and Vinsonhaler, C. 222 E Eτ (G) 61 E-properties 91 E-cyclic group 224, 229, 232 E-finitely generated 91, 95, 224, 226, 229, 241 E-finitely presented 91, 95, 229 E-flat 89 E-generator 92, 95, 226, 229, 234, 241 E-injective 250 E-lattice 15, 35, 38 E-modules 231 E-Noetherian 239, 241

INDEX E-progenerator 228, 229, 241 E-projective 92, 95, 228, 229, 232, 241 E-projective generator 92, 95, 229, 234 E-quasi-projective 233 E-ring 87, 91, 95, 295 E-self-generating 235 Eichler group 173 endlich Baer splitting property 144, 148, 151, 152 endomorphism ring 3 Ext(G, H) 156 F faithfully E-flat group 142 faithful group 141, 201 faithful ring 201 Faticoni-Schultz Theorem 131 Faticoni, T.G. 53, 59, 71, 79, 89, 99, 110, 114, 117, 121, 129, 168, 172, 189, 203, 222, 229, 234, 245, 249 Faticoni, T.G. and Goeters, H.P. 28, 93 filter of divisibility of 177 finite Baer splitting property 152 finite index 141 finitely faithful 141, 155, 159 finitely faithful S-group 155, 159, 162 finitely G-compressed 147, 162 finitely G-generated 140 finitely generated module 79 finitely generated projective 79, 85 finitely H-projective 157 finitely HG (H)-generated 148 finitely M -generated 147 finitely projective 157, 162 flat dimension 246 Fuchs, L. 1, 20

INDEX

313

Fuller, K.R. 152 fully faithful 152 J Fundamental Theorem of Abelian J -group 155, 165 Groups 21 J´onsson’s Theorem 23 Jacobson radical 6 G Jordan-Zassenhaus Theorem 109 G-compressed 146, 147 G-generated 140 G-monomorphism 251 K G-presented group 251 kernel group 131 G-socle 140, 165 GabE (T ) 182 L Gab(E) 182, 182 L-group 155, 165 GAB(T ) 182 Lady, E.L. 39, 109 Gabriel filter 177, 181, 190, 193 left self-injective ring 276 genus class 35 lies over 301 Goeters, H.P. 165 local ring 7 group 1 local refinement property 42, 61, 86 local-global property 13 H Local-global Theorem 13 h(G) 108 localization of S at I 11 h(E) 109 locally E-free 91 HG (·) 4, 30 locally isomorphic 35 Hamiltonian Quaternions 167 locally unique decomposition 62 hereditary torsion class 179, 198, 200, 203 M homogeneous completely c) 190 µ(M decomposable 130 µ(M ) 183, 185, 191 homomorphisms 3 µE (G) 197 µE (M ) 183 I Mader, A. 24 IDEM(E) 206 maxE (M )183 idempotent 8 maxE (M ) 183 idempotent ideal 177 minimal Qδ-quasi-summand 207 indecomposable 2, 20 module 1 injective dimension 248 Murley group 131, 165 integers 1 integral closure 122 N integral over 303 Nakayama’s Lemma 8, 82 integrally closed 15, 47, 87, 168 natural transformation 4 invertible ideal 92, 95

314 nearly isomorphic 34 Niedzwecki, G.P. and Reid, J.D. 232 nil right ideal 7 nilpotent quasi-decomposition 205 nilpotent quasi-summand 205 nilpotent set 72 nilradical 7 Noetherian endomorphism ring 240

O OR (M ) 199 Ω(E) 196, 201 orthogonal idempotents 71

P Φ4 P(G) 30 p-divisible 3 p-finitely H-projective 157 p-local 3, 224 p-rank 156 p-realizable 224 p-reduced 3 Pierce, R.S. 224 Pierce, R.S. and Vinsonhaler, C. 224 power cancellation 263 primary ideal 124 primary regulating quotient 131 projective cover 186, 189 projective dimension 243 pseudo-rigid 71 pure subgroup 93 Q

INDEX Q-faithful 208, 211 Qδ-quasi-summand 207 quasi-equal 1 quasi-homomorphism 5 quasi-isomorphism 5 quasi-projective 233 quasi-split 156, 209 quasi-summand 9 R rank one group 130 rank 2 rationals 1 reduced 2 reduced norm of 169 refinement property 22, 37, 97, 110, 119 regular element 10 Reid, J.D. 9, 23, 93, 225 Reid’s Theorem 93 rigid 80 ring 1 rtffr 2, 3 rtffr group 2, 19, 69 rtffr right E-modules 3, 35 rtffr ring 3, 35 S S-group 155, 165, 190 SG (H) 154 S-linear module 231 Schultz, P. 246, 323 self-injective ring 251 semi-Artinian 180 semi-local 3 semi-perfect ring 16, 62, 187 semi-prime 8 semi-primary rtffr group 128 semi-primary index 128 semi-rigid 71 simple module 80

INDEX split 20 split exact sequence 20 splits 301 splits completely 301 square-free rank 132 strongly homogeneous group 131 strongly indecomposable group 9, 20, 23, 88, 120, 146 strongly indecomposable ring 221 subcommutative ring 202 superfluous 185 support of R 302 T Θ4 tExt(G, H) 156 TG (·) 4, 30 TorE (X ) 180 Tor(E) 182 TOR(γ) 182 trace of H, of P 147 trace of projective cover 186 totally definite quaternion algebra 167, 170 U, V, W UConn ’81 Theorem, 236 unique decomposition 21 Vinsonhaler, C. and Wickless, W. 243, 246, 250 Vinsonhaler, C. 250 Walker, E.A. 23 Walker, C.L. 156 Walker’s Theorem, C.L. 156 Warfield, Jr., R.B. 156, 159 Warfield’s p-rank Theorem 156 Warfield’s Theorem 168 Z Zassenhaus, H. 294 Z-adic integers 5

315

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

EXECUTIVE EDITORS Earl J. Taft Rutgers University Piscataway, New Jersey

Zuhair Nashed University of Central Florida Orlando, Florida

EDITORIAL BOARD M. S. Baouendi University of California, San Diego Jane Cronin Rutgers University Jack K. Hale Georgia Institute of Technology S. Kobayashi University of California, Berkeley Marvin Marcus University of California, Santa Barbara W. S. Massey Yale University

Anil Nerode Cornell University Freddy van Oystaeyen University of Antwerp, Belgium Donald Passman University of Wisconsin, Madison Fred S. Roberts Rutgers University David L. Russell Virginia Polytechnic Institute and State University Walter Schempp Universität Siegen

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Direct Sum Decompositions of Torsion-Free Finite Rank Groups Theodore G. Faticoni Fordham University Bronx, New York, U.S.A.

Boca Raton London New York

Chapman & Hall/CRC is an imprint of the Taylor & Francis Group, an informa business

Chapman & Hall/CRC Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487‑2742 © 2007 by Taylor & Francis Group, LLC Chapman & Hall/CRC is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid‑free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number‑10: 1‑58488‑726‑5 (Hardcover) International Standard Book Number‑13: 978‑1‑58488‑726‑3 (Hardcover) This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. No part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright. com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC) 222 Rosewood Drive, Danvers, MA 01923, 978‑750‑8400. CCC is a not‑for‑profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging‑in‑Publication Data Faticoni, Theodore G. (Theodore Gerard), 1954‑ Direct sum decompositions of torsion‑free finite rank groups / Theodore G. Faticoni. p. cm. ‑‑ (Pure and applied mathematics) Includes bibliographical references and index. ISBN‑13: 978‑1‑58488‑726‑3 ISBN‑10: 1‑58488‑726‑5 1. Torsion free Abelian groups. 2. Direct sum decompositions. I. Title. QA180.F38 2006 512’.25‑‑dc22 Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

2006021040

To my parents who gave me life. To Adolph and Margaret Faticoni.

Contents Preface 1 Notation and Preliminary Results 1.1 Abelian Groups . . . . . . . . . . . 1.2 Associative Rings . . . . . . . . . . 1.3 Finite Dimensional Q-Algebras . . 1.4 Localization in Commutative Rings 1.5 Local-Global Remainder . . . . . . 1.6 Integrally Closed Rings . . . . . . 1.7 Semi-Perfect Rings . . . . . . . . . 1.8 Exercise . . . . . . . . . . . . . . .

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2 Motivation by Example 2.1 Some Well-Behaved Direct Sums . . . . 2.2 Some Badly-Behaved Direct Sums . . . 2.3 Corner’s Theorem . . . . . . . . . . . . 2.4 Arnold-Lady-Murley Theorem . . . . . . 2.4.1 Category Equivalence . . . . . . 2.4.2 Functor A(·) . . . . . . . . . . . 2.4.3 Elementary Uses of the Functors 2.5 Local Isomorphism . . . . . . . . . . . . 2.6 Exercises . . . . . . . . . . . . . . . . . 2.7 Questions for Future Research . . . . . .

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19 20 25 27 29 30 31 32 35 44 45

3 Local Isomorphism Is Isomorphism 47 3.1 Integrally Closed Rings . . . . . . . . . . . . . . . . . . . 47 3.2 Conductor of an Rtffr Ring . . . . . . . . . . . . . . . . . 51 ix

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CONTENTS 3.3 3.4 3.5 3.6 3.7

Local Correspondence . . . . . Canonical Decomposition . . . Arnold’s Theorem . . . . . . . Exercises . . . . . . . . . . . . Questions for Future Research .

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4 Commuting Endomorphisms 4.1 Nilpotent Sets . . . . . . . . . . . . . . . . . . 4.2 Commutative Rtffr Rings . . . . . . . . . . . 4.2.1 Modules over Commutative Rings . . 4.2.2 Projectives over Commutative Rings . 4.2.3 Direct Sums over Commutative Rings 4.2.4 Commutative Endomorphism Rings . 4.3 E-Properties . . . . . . . . . . . . . . . . . . 4.4 Square-Free Ranks . . . . . . . . . . . . . . . 4.4.1 Rank Two Groups . . . . . . . . . . . 4.4.2 Groups of Rank ≥ 3 . . . . . . . . . . 4.5 Refinement and Square-Free Rank . . . . . . 4.6 Hereditary Endomorphism Rings . . . . . . . 4.7 Exercises . . . . . . . . . . . . . . . . . . . . 4.8 Questions for Future Research . . . . . . . . . 5 Refinement Revisited 5.1 Counting Isomorphism Classes . . . . . . . 5.1.1 Class Groups and Class Numbers . . 5.1.2 Class Group of the Integral Closure 5.1.3 Class Number and Refinement . . . 5.1.4 Quadratic Number Fields . . . . . . 5.1.5 Counting Theorems . . . . . . . . . 5.2 Integrally Closed Groups . . . . . . . . . . . 5.2.1 Review of Notation and Terminology 5.2.2 Locally Semi-Perfect Rings . . . . . 5.2.3 Semi-Primary Rtffr Groups . . . . . 5.2.4 Applications to Refinement . . . . . 5.3 Exercises . . . . . . . . . . . . . . . . . . . 5.4 Questions for Future Research . . . . . . . .

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69 69 79 79 82 84 86 89 93 93 95 99 101 102 103

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107 108 108 110 113 116 117 122 123 124 127 130 134 135

6 Baer Splitting Property 139 6.1 Baer’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . 139 6.2 Splitting of Exact Sequences . . . . . . . . . . . . . . . . . 144 6.3 G-Compressed Projectives . . . . . . . . . . . . . . . . . . 146

CONTENTS 6.4 6.5 6.6

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Some Examples . . . . . . . . . . . . . . . . . . . . . . . . 151 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 Questions for Future Research . . . . . . . . . . . . . . . . 154

7 J -Groups, L-Groups, and S-Groups 7.1 Background on Ext . . . . . . . . . . 7.2 Finite Projective Properties . . . . . 7.3 Finitely Projective Groups . . . . . . 7.4 Finitely Faithful S-Groups . . . . . . 7.5 Isomorphism versus Local Isomorphism . . . . . . . . . . . . . 7.6 Analytic Number Theory . . . . . . 7.7 Eichler L-Groups Are J -Groups . . 7.8 Exercises . . . . . . . . . . . . . . . 7.9 Questions for Future Research . . . .

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165 167 172 174 175

8 Gabriel Filters 8.1 Filters of Divisibility . . . . . . . . . . . . . . 8.1.1 Hereditary Torsion Classes . . . . . . 8.1.2 Gabriel Filters of Right Ideals . . . . . 8.2 Idempotent Ideals . . . . . . . . . . . . . . . 8.2.1 Traces of Covers . . . . . . . . . . . . 8.2.2 Bounded Gabriel Filters . . . . . . . . 8.2.3 Finite Filters of Divisibility . . . . . . 8.3 Gabriel Filters on Rtffr Rings . . . . . . . . . 8.3.1 Applications to Endomorphism Rings 8.3.2 Constructing Examples . . . . . . . . 8.3.3 Faithful Rings . . . . . . . . . . . . . 8.4 Gabriel Filters on QEnd(G) . . . . . . . . . . 8.4.1 Central Quasi-summands . . . . . . . 8.4.2 Q-Faithful Groups . . . . . . . . . . . 8.4.3 Q-Faithful E-Flat Groups . . . . . . . 8.5 Exercises . . . . . . . . . . . . . . . . . . . . 8.6 Questions for Future Research . . . . . . . . .

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177 177 179 181 185 185 190 192 196 196 199 201 204 204 208 211 213 216

9 Endomorphism Modules 9.1 Additive Structure of Rings . . . . . 9.2 E-Properties . . . . . . . . . . . . . 9.2.1 E-Rings . . . . . . . . . . . . 9.2.2 E-Finitely Generated Groups 9.2.3 E-Projective Groups . . . . .

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9.4 9.5 9.6

9.2.4 E-Generator Groups . . . . . . . . . . . . 9.2.5 E-Projective Rtffr Groups Characterized . 9.2.6 Noetherian Endomorphism Modules . . . Homological Dimensions . . . . . . . . . . . . . . 9.3.1 E-Projective Dimensions . . . . . . . . . 9.3.2 E-Flat Dimensions . . . . . . . . . . . . . 9.3.3 E-Injective Dimensions . . . . . . . . . . Self-Injective Rings . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . Questions for Future Research . . . . . . . . . . .

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234 236 239 242 243 246 248 251 252 255

A Pathological Direct Sums 257 A.1 Nonunique Direct Sums . . . . . . . . . . . . . . . . . . . 257 B ACD Groups 261 B.1 Example by Corner . . . . . . . . . . . . . . . . . . . . . . 261 C Power Cancellation 263 C.1 Failure of Power Cancellation . . . . . . . . . . . . . . . . 263 D Cancellation 265 D.1 Failure of Cancellation . . . . . . . . . . . . . . . . . . . . 265 E Corner Rings and Modules E.1 Topological Preliminaries . . . . . . . . . . . . . . . . . . E.2 The Construction of G . . . . . . . . . . . . . . . . . . . . E.3 Endomorphisms of G . . . . . . . . . . . . . . . . . . . . .

269 269 271 272

F Corner’s Theorem 277 F.1 Countable Endomorphism Rings . . . . . . . . . . . . . . 277 G Torsion Torsion-Free Groups 279 G.1 E-Torsion Groups . . . . . . . . . . . . . . . . . . . . . . 279 G.2 Self-Small Corner Modules . . . . . . . . . . . . . . . . . . 280 H E-Flat Groups 283 H.1 Ubiquity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 H.2 Unfaithful Groups . . . . . . . . . . . . . . . . . . . . . . 284 I

Zassenhaus and Butler 289 I.1 Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 I.2 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290

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J Countable E-Rings 295 J.1 Countable Torsion-Free E-Rings . . . . . . . . . . . . . . 295 K Dedekind E-Rings 301 K.1 Number Theoretic Preliminaries . . . . . . . . . . . . . . 301 K.2 Integrally Closed Rings . . . . . . . . . . . . . . . . . . . 302 Bibliography

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Index

311

Preface

This book is directed at mathematicians who wish to study advanced topics in direct sum decompositions of abelian groups and their variants. I assume that the reader has had experience with the ring theory covered in [6], and with the abelian group theory covered in the text that inspired us to write this text, [10]. A read through the abelian group text [46] will also be helpful. The use of reduced torsion-free finite rank abelian groups or of rtffr abelian groups can be traced back to solutions of linear equations with integer coefficients. We reinvent this point of view today when we study full free subgroups of a finite dimensional Q-vector space. Suppose that the rtffr group G has a direct sum decomposition (n1 )

G1

(nt )

⊕ · · · ⊕ Gt

(1)

for some integers t, n1 , . . . , nt > 0 and some indecomposable groups G1 , . . . , Gt . To eliminate redundancies we require that Gi ∼ = Gj ⇒ i = j. There are several theorems that give us conditions under which the integers n1 , . . . , nt and the isomorphism classes of the groups G1 , . . . , Gt are unique for G. Two of the most important results of this kind are the Azumaya-Krull-Schmidt Theorem and the Baer-Kulikov-Kaplansky Theorem. Each states that if the groups Gi satisfy some condition or other then the integers t, n1 , . . . , nt > 0 and the isomorphism classes (Gi ) of the groups Gi in (1) are unique to G. Thus one might expect this uniqueness to be common. This is not the case.

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For example, G is called completely decomposable if G has a direct sum decomposition (1) in which Gi is a rank one rtffr group for each i = 1, · · · , t. That is, Gi ⊂ Q. The Baer-Kulikov-Kaplansky Theorem states that if (1) is a direct sum of rank one groups G1 , . . . , Gt then the integers t, n1 , . . . , nt > 0 and the isomorphism classes of the groups G1 , . . . , Gt are unique to G. However, assume that t, n1 , . . . , nt > 1. Then several examples by B. J´onsson, A.L.S. Corner, and L. Fuchs and F. Loonstra (see [47] or the appendices) show that there are subgroups of finite index in G that possess direct sum decompositions whose indecomposable terms are not unique in any sensitive generalization of the term. Such a group H is called an almost completely decomposable group, or an acd group. Specifically, there are indecomposable acd groups G, H, A, B, C such that G(2) ∼ = H (2) while G and H are not isomorphic, and A ⊕ B ∼ = A ⊕ C while B and C are not isomorphic. Thus one can routinely construct subgroups H of finite index in G whose direct sum decompositions behave badly. Hence any theorems on direct sum decompositions of rtffr groups must respect this juxtaposition of uniqueness with nonuniqueness. B. J´onsson proved a theorem, considered by some to be the Fundamental Theorem of Torsion-Free Finite Rank Abelian Groups, that allows us to salvage a course uniqueness of the indecomposable terms in the direct sum (1). The idea is to substitute the weaker quasi-isomorphism for isomorphism in the Azumaya-Krull-Schmidt Theorem. Groups X and Y are quasi-isomorphic if there are maps f : X → Y and g : Y → X and . an integer n > 0 such that f g = n1Y and gf = n1X . We write X ∼ to Y . The group X is strongly = Y when X is quasi-isomorphic . ∼ indecomposable if X = Y implies that Y is indecomposable. THEOREM: [B. J´onsson] Let G be an rtffr group. 1. There are integers t, n1 , . . . , nt >. 0 and strongly indecomposable groups G1 , . . . , Gt such that Gi ∼ = Gj implies that i = j and such . n1 nt ∼ that G = G1 ⊕ · · · ⊕ Gt . .

∼ H m1 ⊕ · · · ⊕ H ms for some integers s, m1 , . . ., 2. Suppose that G = s 1 ms >. 0 and strongly indecomposable groups H1 , . . . , Hs such that Hi ∼ = Hj implies that i = j. Then s = t, ni = mi for each . i = 1, . . . , t, and Gi ∼ = Hi after a permutation of the subscripts. .

.

.

Thus. G(2) ∼ = H (2) implies that G ∼ = H, and A ⊕ B ∼ = A ⊕ C implies ∼ that B = C. While J´onsson’s Theorem accounts for some of the subgroup

PREFACE

xvii

structure of G, it misses the finer properties that a direct sum decomposition can have. For example, there are indecomposable groups that are quasi-isomorphic to groups that possess a nontrivial direct sum decomposition. Thus abelian groupists have replaced indecomposable rtffr groups with strongly indecomposable rtffr groups, direct sum decompositions of rtffr groups with quasi-direct sum decompositions of rtffr groups, and uniqueness up to isomorphism is replaced with uniqueness up to quasi-isomorphism. E. L. Lady [57] introduced another generalization of isomorphism called near isomorphism that is logically situated between isomorphism and quasi-isomorphism. Lady’s first applications of near isomorphism include the fact that if G(2) ∼ = H (2) then G and H are near isomorphic, ∼ and if A ⊕ B = A ⊕ C then B and C are near isomorphic. Near isomorphism shows up in direct sums of many types of groups. For example, if G is a completely decomposable rtffr group, if H ⊂ G is a subgroup, and if G/H is a finite p-group for some prime p ∈ Z, then a direct sum H ∼ = H1 ⊕ · · · ⊕ Hr of indecomposable groups Hi is unique up to near isomorphism. That is, if H ∼ = K1 ⊕ · · · ⊕ Ks for some indecomposable groups Kj then r = s and after a permutation of the subscripts Hi is near isomorphic to Ki for each i = 1, . . . , r. Near isomorphism is a group theoretic version of the genus class of lattices over orders. It is our point of view that near isomorphism and genus class are essentially the same measurement of algebraic properties, but of objects from different categories. Therefore, if two objects are near isomorphic or in the same genus class we will say that these objects are locally isomorphic. Following the publication of the papers [14] and [15], it became popular to study direct sum decompositions of G by studying the right ideal structure of End(G) with emphasis on the finitely generated projective right End(G)-modules. This book takes the point of view that a group G can be effectively studied by considering the finitely generated projective right End(G)-modules, the left End(G)-module G, and the ring E(G) = End(G)/N (End(G)). D. M. Arnold [10] introduced a functor A(·) to abelian groups that transforms direct summands of G into finitely generated projective right E(G)-modules. Our fundamental approach uses A(·) as the primary tool for understanding direct sum decompositions of rtffr groups. Thus A(·) enables us to treat the direct sum decompositions of the rtffr group G as

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a direct sum of finitely generated projective modules over the Noetherian semi-prime rtffr ring E(G). One advantage of considering E(G) over End(G) is that the ring and group structure of E(G) are simpler than that of End(G). Specifically, if S = center(E(G)) then E(G) is a finitely generated S-module. We will make extensive use of the localization theory of S to study E(G). There is a finite product E of classical maximal S-orders in which E(G) has finite index. We call E an integrally closed ring. Several applications of some Algebraic Number Theory to the ideal structure of E will give us good results on the group structure of G. For example, if h(G) is the number of isomorphism classes (H) of groups H that are locally isomorphic to G and if h(E) is the number of isomorphism classes (J) of nonzero fractional ideals J of E then h(E) divides h(G). While hereditary endomorphism rings have been scrutinized over the last 20 years, (see [54], and especially the papers of U. Albrecht), there seems to have been little attention paid to other pleasantly structured endomorphism rings. We will initiate an examination of the commutative property in E(G). The reader may be surprised at the number of groups G for which E(G) is commutative. Assuming that E(G) is a commutative ring, we can prove several results on the uniqueness and existence of direct sum decompositions (1) of G. Our investigations will show that G possesses some interesting properties when End(G) is commutative and End(G) ⊂ G ⊂ QEnd(G). Assuming that E(G) is commutative our work shows that for each integer n > 0, Gn has a locally unique decomposition. That is, 1. Gn = H1 ⊕ · · · ⊕ Hr where for each i = 1, . . . , r, Hi is an indecomposable group such that G is locally isomorphic to Hi ⊕ Hi0 for some group Hi0 , and 2. If Gn = K1 ⊕ · · · ⊕ Ks for some indecomposable groups Kj then r = s and after a permutation of the subscripts, Hi is locally isomorphic to Ki for i = 1, . . . , r. We show that Gn has a locally unique decomposition for each integer n > 0 if 1. G has a direct sum decomposition (1) in which the G1 , . . . , Gt are . strongly indecomposable rtffr groups such that Gi ∼ = Gj ⇒ i = j, and 2. E(Gi ) is commutative for each i = 1, . . . , t.

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Condition 2, that E(Gi ) is commutative, is true whenever G is a strongly indecomposable rtffr group with square-free rank rank(Gi ). Thus there are plenty of rtffr groups that satisfy these two conditions. This result is an extension of the Baer-Kulikov-Kaplansky Theorem. Let S and E be as above. There is a nonzero ideal τ ⊂ S called the conductor of G such that τ E ⊂ E(G) ⊂ E. It is possible to define a localization functor (·)τ in a manner that is consistent with the traditional localization of a commutative ring at a prime ideal. Localization theory shows us that if H ⊕H 0 ∼ = K⊕K 0 ∼ = G(n) for some integer n > 0 then H and K are locally isomorphic iff Aτ (H) and Aτ (K) are isomorphic right Eτ (G)-modules, where Aτ (x) = (A(x))τ is the localization of A(x) at τ . This gives us a different proof of Arnold’s Theorem: if G is locally isomorphic to H1 ⊕ H2 then G = G1 ⊕ G2 where Gi is locally isomorphic to Hi for i = 1, 2. An ideal P in a ring R is primary if R/P is a local ring with nilpotent Jacobson radical. That is, the Jacobson radical J of R/P is the unique maximal right ideal in R/P and J n = 0 for some integer n > 0. The group G is a semi-primary rtffr group if there is a group G with a direct sum decomposition (1) such that 1. E(Gi ) is a Dedekind domain for each integer i = 1, . . . , t and such that 2. for each i = 1, . . . , t there is a primary ideal N (End(Gi )) ⊂ Pi ⊂ (n ) (n ) End(Gi ) such that P1 G1 1 ⊕ · · · ⊕ Pt Gt t ⊂ G ⊂ G. Semi-primary groups are generalizations of almost completely decomposable groups with primary regulating quotient. They include certain strongly indecomposable groups G such that E(G) is commutative. Our results apply to subgroups of primary index in direct sums of strongly indecomposable bracket groups, strongly indecomposable strongly homogeneous groups, and strongly indecomposable Murley groups. It is natural to ask if rtffr groups G satisfy a splitting property like that of free groups. We will characterize in Chapter 6 those rtffr groups G that satisfy the Baer splitting property. That is, we examine those G for which a surjection π : G(c) −→ G splits whenever c is a cardinal. The main results are Theorems 6.3.1 and 6.3.2 where we relate, ´ a la U. Albrecht [3], the Baer splitting property for G to the vanishing of the tensor product M ⊗E G for nonzero right End(G)-modules M .

xx

PREFACE

A discussion of J. -groups, L-groups, and S-groups follows. The group G is a. J -group if G ∼ = H implies that G ∼ = H. The group G is an L-group ∼ that G is locally isomorphic to H. The group G is an if G = H implies . H implies that G generates H. While it is clear that J S-group if G ∼ = group ⇒ L-group ⇒ S-group the converses require some power. We use several deep results from Analytic Number Theory to show that the most finitely faithful S-groups are J -groups, and that most rtffr L-groups are J -groups. Over the past 30 years it has become fashionable to consider the structure of G as a left End(G)-module. R. S. Pierce introduced the following clever way of labelling module properties on groups that we will adopt as our own. Given a module theoretic property property we say that G is an E-property group if G satisfies property as a left End(G)module. Thus E-projective groups are projective left End(G)-modules, G is an E-finitely generated group if G is a finitely generated left End(G)module, an E-generator group is a generator as a left End(G)-module, and an E-Noetherian group is a Noetherian left End(G)-module. We will give a unified approach to rtffr groups G with these properties by showing that the group structure of G stems from a general structure theorem for finitely generated rtffr modules over an rtffr ring. Subsequently we show that several of these E-properties coincide. In the latter half of Chapter 9 we will examine the possible homological dimensions of the left End(G)module G. We show that for a given rtffr ring E and integer k less than the left global dimension of E there are rtffr groups Pk , Fk such that E = End(Pk ) = End(Fk ) and whose projective and flat dimensions, respectively, are k. Our techniques must be significantly modified if they are to give us broad results on the injective dimension. The text contains a number of exercises at the end of each chapter but the chapters themselves contain many statements like the reader will prove that . . . . Each of these is an intended unlisted exercise. The young abelian group theorist is advised to follow these directions. Also the author comes from the school of mathematics that emphasizes the use of examples. Examples guide our intuition and they steer us clear of falsehoods. Examples are used in this way throughout this text. Thus the reader should not be surprised at the number of examples used to motivate our discussions. I would like to thank Fordham University for giving me the time and the resources needed to write this book. I would also like to thank my colleagues for carefully reading this book in manuscript form and for their subsequent comments. These people include the group of mathematicians that I was introduced to during my formative years. They

PREFACE

xxi

are Ulrich Albrecht, Dave M. Arnold, R. A. Beaumont, Donna Beers, Anthony L. S. Corner, Manfred Dugas, Laszlo Fuchs, H. Pat Goeters, Anthony Giovanitti, Roger Hunter, E. Lee Lady, Ray Mines, DeAlberto Orsatti, Richard Pierce, K. M. Rangaswamy, James Reid, Fred Richman, Luigi Salce, Phil Schultz, Charles Vinsonhaler, Carol Walker, Elbert A. Walker, and William Wickless. Their encouragement over the years helped me forge this book. Since my research style produces more TEX files than paper files, I am both author and technical typist on this project. Thus, I put quite a lot of wear on my keyboard during the writing process. Any mathematical errors of commission, of omission, or of a typographical nature are my responsibility. Certain results are attributed to your author but give no publication reference. This is because these results do not appear elsewhere in the literature. Theodore G. Faticoni Department of Mathematics Fordham University Bronx, New York 10458 [email protected]

Chapter 1

Notation and Preliminary Results This text assumes that the reader is familiar with abelian groups and unital modules over associative rings with unity as contained in the texts by F.W. Anderson and K.R. Fuller [6], D.M. Arnold [10], and L. Fuchs [46]. We will reference but not prove those results that we feel fall outside of the line of thought of this book. I suggest that you use this chapter as a reference and nothing more. Skim through this chapter. Unless otherwise directed, do not attempt to plow through these results as though they were exercises.

1.1

Abelian Groups

Throughout, we let Z denote the ring of integers, Q denotes the field of rational numbers, and for a prime p ∈ Z, Zp is the localization of Z at p. We let E, R, S, T denote an associative ring with identity, hereafter referred to simply as a ring. Their modules are denoted by G, H, K, L, M, N , U, V , W . (The side will be specified.) The term group means abelian group, (i.e., a Z-module), and G, H, K denote groups. The group G is a torsion-free group if its torsion subgroup {x ∈ G nx = 0 for some integer n 6= 0} is zero. Thus G is torsion-free if x ∈ G and nx = 0 for some integer n 6= 0 implies that x = 0. We view Z and Q as groups or as rings, as the setting requires. If G and H are contained in a common Q-vector space V , then we say that G and H are quasi-equal if there are nonzero integers n, m such 1

2

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS

that nG ⊂ H and mH ⊂ G. In this case we write . G = H.

Let G be a group. Let I be a set, let c = card(I), and for each i ∈ I let Gi ∼ = G. The usual direct sum and direct product of c copies of G are denoted as G(c) = ⊕i∈I Gi Q Gc = i∈I Gi . We say that the group G is indecomposable if G = H ⊕K implies that H = 0 or K = 0. Z, Q, Zp , and Z/pk Z are indecomposable groups for primes p ∈ Z and integers k > 0. Given an integer n > 0 there is an indecomposable group Zn ⊂ G ⊂ Qn , so there are plenty of indecomposable torsion-free groups. Let G be a torsion-free group. We let QG denote the divisible hull of G. Since G is torsion-free QG is a Q-vector space spanned by G, or equivalently QG = Q ⊗Z G. The rank of G is the cardinality of a maximal Z-linearly independent subset of G, or equivalently

rank(G) = Q-dim(QG). With few exceptions, the torsion-free groups that we consider in this book have finite rank. T The group G is reduced if n>0 nG = 0, or equivalently if Hom(Q, G) = 0. An

rtffr group

1.1. ABELIAN GROUPS

3

is a reduced torsion-free finite rank group. As you read this book, pronounce rtffr as are-tee-ef-ef-are. A ring E is said to be an rtffr ring if its additive structure (E, +) is an rtffr group. A right E-module M is said to be an rtffr E-module if its additive structure (M, +) is an rtffr group. The reader should prove the following fact as an exercise. If G is an rtffr group and if H ⊂ G is a subgroup such that G/H is a bounded group (i.e., there is an integer n 6= 0 such that nG ⊂ H), then G/H is a finite group. T We say that G is p-reduced if k>0 pk G = 0, and we say that G is p-local if pG 6= G and qG = G for each prime q 6= p ∈ Z. If pG = G then we say that G is p-divisible. The group G is semi-local if there is a finite set of primes {p1 , . . . , ps } ⊂ Z such that pG = G for each p 6∈ {p1 , . . . , ps }. For example, Z and Q are rtffr groups, Z is reduced, and Q is divisible. The subgroup of Q na o a ∈ Z and n is a square-free integer n is reduced and torsion-free group of rank one. Moreover the subring of Q o na a, b ∈ Z and gcd(p, b) = 1 Zp = b is a reduced, p-reduced, torsion-free ring of rank one, but qZp = Zp for primes q 6= p ∈ Z. The ring of algebraic integers in an algebraic number field is an rtffr ring. The ring a 0 Z 0 = a, b, c ∈ Z Z Z b c is a reduced and torsion-free ring of rank three. Let G and H be groups. As usual, we let End(G) = the ring of group endomorphisms of G. Then End(G) is the set of group homomorphisms f : G −→ G. We let Hom(G, H) denote the group of group homomorphisms f : G −→ H. We consider G as a left End(G)-module by setting f · x = f (x) for f ∈ End(G) and x ∈ G. Similarly Hom(G, H) is a right End(G)-module if we set gf = g ◦ f for each g ∈ Hom(G, H) and f ∈ End(G). If G is an rtffr group, then End(G) is an rtffr ring.

4

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS Given groups G and H let

SG (H) =

P {f (G)

group maps f : G −→ H}.

We adopt the standard notation from [6]. Thus N ⊗R M denotes the tensor product of a right R-module N and a left R-module M . Since they are integral parts of our discussion we adopt a notation for a special tensor product and a special homset. TG (·) = · ⊗End(G) G and HG (·) = Hom(G, ·). There are natural transformations Θ Φ

: :

TG HG (·) −→ 1 1 −→ HG TG (·)

defined by ΘH (f ⊗ x) = f (x) ΦM (x)(·) = · ⊗ x. The reader should verify that these functors and associated natural transformations enjoy the following very useful identities. 1HG (H) = HG (ΘH ) ◦ ΦHG (H) 1TG (M ) = ΘTG (M ) ◦ TG (ΦM ) Let QHom(G, H) = Q ⊗ Hom(G, H) QEnd(G) = Q ⊗ End(G). An element f ∈ QHom(G, H) is called a quasi-homomorphism. The elements in QHom(G, H) are linear transformations f : QG → QH such . that f (G) = H. Observe that f is a quasi-homomorphism iff there is an integer n 6= 0 such that nf : G → H. We say that f ∈ QHom(G, H) is a quasi-isomorphism if there is a map g : QHom(H, G) such that f g = 1H

1.2. ASSOCIATIVE RINGS

5

and gf = 1G . Equivalently f : G → H is a quasi-isomorphism iff there is an integer n 6= 0 and a group homomorphism g : H → G such that f g = n1H and gf = n1G . We write .

G∼ =H when G is quasi-isomorphic to H. The following group is used in this text only as a source of illuminating examples. The ring of Z-adic integers is denoted by b = lim Z/nZ. Z ←n

Given a prime p ∈ Z the ring of p-adic integers b p = lim Z/pk Z Z ←k

b is a source for a number of important examples. To begin with, (1) Z b b b and Zp have uncountable rank, (2) for primes p ∈ Z, rp (Z) = rp (Zp ) = 1, b p = qZ bp. and (3) for primes q 6= p ∈ Z, Z

1.2

Associative Rings

Since some of our discussions deal with several rings at once we will need more than one symbol to denote a ring. For instance, E usually denotes an rtffr ring (a potential endomorphism ring of an rtffr group), and S denotes a commutative ring, often but not always S = the center of E = center(E) = {x ∈ E xy = yx for each y ∈ E}. Let E be a ring and let M be a right E-module. If x ∈ M then the right ideal annE (x) = {r ∈ E xr = 0} in E is called the annhilator of x in E and the ideal annE (M ) = {r ∈ E M r = 0} is called the annihilator of M in E. For example givenan integer n > 0, Z 0 1 0 annZ (Z/nZ) = nZ. If E = then let e11 = and let Z Z 0 0 0 0 0 0 N= . The reader can show that annE (e11 ) = and Z 0 Z Z 0 0 that annE (N ) = . Z Z

6

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS

In general, if M = E/I for some right ideal I ⊂ E then 1 + I ∈ E/I has annihilator annE (1 + I) = I in E, and annE (M ) is the largest ideal of E that is contained in I. If e2 = e ∈ E then annE (e) = (1 − e)E. If M is a simple right E-module (that is, the only E-submodules of M are 0 and M ), then annE (x) is a maximal right ideal in E for each x 6= 0 ∈ M . If E = End(G) then annE (G) = 0.

The Jacobson radical of E is the ideal J (E) defined as follows.

J (E) = ∩{M M ⊂ E is a maximal right ideal } = ∩{M M ⊂ E is a maximal left ideal } = ∩{annE (K) K is a simple left E-module } = ∩{annE (K) K is a simple right E-module } = {r ∈ E 1 + rx is a unit in E}

For example, given a prime p ∈ Z, J (Zp ) = pZp . Given n > 0 ∈ Z, let n = pn1 1 · · · pnt t be a product of powers n1 , . . . , nt > 0 of distinct primes p1 , . . . , pt . Let

Zn =

na m

o a, m ∈ Z, m is relatively prime to n .

Then Zn is a semi-local rtffr ring with Jacobson radical J (Zn ) = (p1 · · · pt )Zn . Evidently n ∈ J (Zn ). It is a healthy for exercise the reader to Z 0 0 0 show that J (Z) = 0 and that J = . Z Z Z 0

1.2. ASSOCIATIVE RINGS

7

Given an rtffr ring E then the ring

En =

nx m

x ∈ E and m ∈ Z is relatively prime to n

o

is a torsion-free finite rank ring that may not be reduced. Note that En ∼ = E ⊗Z Zn . LEMMA 1.2.1 Let E be a semi-local rtffr ring and assume that n = Q {primes p ∈ Z pE = 6 E}. Then n ∈ J (E). Proof: Let M be a maximal right ideal in E. We will prove in Lemma 4.2.3 that since E is reduced there is a prime p ∈ Z such that pE ⊂ M 6= E, and thus n ∈ M . Consequently n ∈ ∩{M M is a maximal ideal in E} = J (E). The ring R is local if any of the following equivalent properties hold. 1. R possesses a unique maximal right ideal M . 2. J (R) is the unique maximal right ideal of R. 3. u ∈ R is a unit of R iff u 6∈ J (R). b p are For example, fields are local rings. Given a prime p ∈ Z, Zp and Z local rings. The right ideal I ⊂ E is a nil right ideal if each x ∈ I is nilpotent. That is, for each x ∈ I there exists an integer n such that xn = 0. The nilradical of E is the ideal N (E) that is defined as follows. N (E) = {x ∈ E xE is a nil right ideal in E} = {x ∈ E Ex is a nil left ideal in E} X = {I I is a nil left ideal in E} = the largest nil ideal in E Let x ∈ E. Since xn = 0 implies that 1 − x is a unit in E (show that one, reader), 1 − xy is a unit of E for each x ∈ N (E) and y ∈ E. Thus N (E) ⊂ J (E). In the special case of an rtffr ring E we have the following relationships between N (E) and J (E).

8

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS

LEMMA 1.2.2 Let E be an rtffr ring. 1. N (QE) = J (QE) is the largest nilpotent ideal in QE. 2. N (E) = J (QE) ∩ E and QN (E) = J (QE). The reader may be surprised at the number of times we appeal to Nakayama’s Lemma in this text. LEMMA 1.2.3 [Nakayama] (See [6, page 169].) Let E be a ring, let M be a right E-module and let K ⊂ M be an E-submodule. 1. If M is finitely generated and if K + M J (E) = M then K = M . 2. If I is a nilpotent ideal in E such that K + M I = M then K = M . In particular if J ⊂ E is a right or a left ideal such that J +J (E) = E then J = E. An idempotent is an element e ∈ E such that e2 = e. Often we will avoid the term idempotent and just write e2 = e. Given a ring E and an e2 = e ∈ E then eEe = {exe x ∈ E}

is a ring with identity 1eEe = e. Although eEe ⊂ E, eEe is not a unital subring of E. LEMMA 1.2.4 Let E be a ring and let e2 = e ∈ E. Then EndE (eE) = eEe. Lifting theorems are a part of ideal theory. The following lifting results will be used throughout this text without fanfare. LEMMA 1.2.5 Let E be a ring. 1. If u + J (E) is a unit in E/J (E) then u is a unit in E. 2. If I is a nil ideal in E and if (f + I)2 = f + I ∈ E/I then there is an e2 = e ∈ E such that e − f ∈ I.

1.3. FINITE DIMENSIONAL Q-ALGEBRAS

1.3

9

Finite Dimensional Q-Algebras

Our source for structure results on rtffr rings is [10, Chapters 9-14]. The ring E is semi-prime if I is an ideal of E and if I 2 = 0 then I = 0. Equivalently, E is semi-prime iff N (E) = 0. Given a group G let E(G) = End(G)/N (End(G)). Notice that E(G) is in general a semi-prime ring. If G is an rtffr group then E(G) is a Noetherian semi-prime rtffr ring that is finitely generated by its center, S. See [10, Theorems 9.4, 9.9]. It is hard to overstate the importance of the semi-prime ring E(G) to our deliberations. Let G and H be rtffr groups. We say that H is a quasi-summand . of G if there is a group K such that G ∼ = H ⊕ K. Equivalently, H is a quasi-summand of G iff there is an integer n 6= 0 and maps f : G → H and g : H → G such. that f g = n1H . The rtffr group G is strongly indecomposable if G ∼ = H ⊕ K implies that either H = 0 or K = 0. Equivalently G is a strongly indecomposable group iff each subgroup of finite index in G is indecomposable. Each subgroup of Q is strongly b p are strongly indecomposable as is Z and for primes p ∈ Z, Zp and Z b p is strongly indecomposable indecomposable. Each pure subgroup of Z so there are plenty of strongly indecomposable groups around. A proof of the next lemma is found in [10, Theorem 9.10]. LEMMA 1.3.1 Suppose that G is an rtffr group such that (n1 )

G = G1

(nt )

⊕ · · · ⊕ Gt

for some integers t, n1 , . . . , nt >. 0 and some strongly indecomposable groups G1 , . . . , Gt such that Gi ∼ = Gj ⇒ i = j. Then E(G) ∼ = Matn1 (E(G1 )) × · · · × Matnt (E(Gt )).

(1.1)

The next result is [46, page 149, Lemma 92.2]. LEMMA 1.3.2 [J.D. Reid] Let G be an rtffr group. There is a bijection from the set of isomorphism classes of {(eQEnd(G)) e2 = e ∈ QEnd(G)} onto the set of quasi-isomorphism classes {[H] H is a quasi-summand of G}. The bijection is given by (eQEnd(G)) 7→ [eG]. The following classification of strongly indecomposable groups is due to J. D. Reid [46, page 149, Proposition 92.3].

10

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS

LEMMA 1.3.3 [J.D. Reid] Let G be an rtffr group. The following are equivalent. 1. G is strongly indecomposable. 2. QEnd(G) is a local ring. 3. An endomorphism f : G → G is either nilpotent or an injection. Some of the ideal structure of QE filters down to E. LEMMA 1.3.4 (See [10, Theorem 9.10].) Let E be an rtffr ring. The following are equivalent. 1. N (E) = 0. 2. There are integers t, n1 , . . . , nt > 0 and division Q-algebras D1 , . . ., Dt such that QE = Matn1 (D1 ) × · · · × Matnt (Dt )

(1.2)

as rings. 3. Each right ideal of QE is a direct summand of QE. LEMMA 1.3.5 (See [10, Corollary 10.14].) Let E be a semi-prime rtffr ring. Then 1. If I is a right ideal of E, there is an e2 = e ∈ QE and an integer n 6= 0 such that neE has finite index in I and nE ⊂ neE ⊕ n(1 − e)E ⊂ E ⊂ eE ⊕ (1 − e)E. 2. E is a Noetherian ring. 3. E is finitely generated as a module over its center center(E).

1.4

Localization in Commutative Rings

Let us review some ideas about localization in commutative rings. We will refer to these facts without fanfare. An element c ∈ S is a regular element if c is not a zero divisor in S. If I is an ideal in the commutative ring S we let Γ(I) = {c ∈ S c + I is a regular element in S/I}.

1.4. LOCALIZATION IN COMMUTATIVE RINGS

11

In general Γ(I) is a multiplicatively closed subset of S that contains 1. For example, if S/I is a finite ring then c ∈ Γ(I) iff c + I is a unit in S/I. For x ∈ S and c, d ∈ Γ(I) we define an equivalence relation ∼ on (x, c) ∈ S ×Γ(I) by (x, c) ∼ (y, d) iff xd = yc. Then the equivalence class of (1, c) is denoted by c−1 , and xc−1 is the equivalence class of (x, c). By setting K(I) = {x ∈ S xc = 0 for some c ∈ Γ(I)} then we can form the localization of S at I denoted by SI

= {xc−1 x ∈ S/K(I) and c ∈ Γ(I)}.

If S is an rtffr ring then S0 = QS, However, not every localization of S is formed by inverting a set of integers. Given an S-module X we write XI = X ⊗S SI . Given any injection f : X → Y the canonical map fI : XI −→ YI : xc−1 7−→ f (x)f (c)−1 is an injection. Thus the inclusion map J ⊂ S induces an injection JI ⊂ SI onto an ideal in SI . Indeed II ⊂ J (SI ) since for each x ∈ II and s ∈ SI , 1 + xs ≡ 1(modII ) and 1, being regular modulo I, maps onto a unit of SI . In particular, if S is an rtffr ring and if n ∈ Z then n ∈ J (SnS ). Note that SnS is different from Sn = S ⊗Z Zn . The following example will show us why Sn and SnS may be different rings. Let S = Z6 × Z15 . With n = 2 we have S2 = S ⊗Z Z2 = (Z6 × Z15 ) ⊗Z Z2 = Z2 × Q while with n = 3 we have S3 = S ⊗Z Z3 = Z3 × Z3 .

12

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS

But with I = 2S we have the following. Since (1, 0) maps to a unit in S/2S and since (1, 0) annihilates 0 × Z15 , we have K(2S) = 0 × Z15 = {x ∈ S xc = 0 for some c ∈ Γ(2S)}. Thus S2S = (S/K(2S))2S = Z2 . We observe that while 2 ∈ J (S2S ) we also have (1, 1)2 = (2, 2) 6∈ J (S2 ) = 2Z2 × 0. If I = 0 then Γ(0) is the set of regular elements in S and so S0 = QS is the classical ring of quotients of S. If I is a prime ideal in S then SI is a local ring with unique maximal ideal II . Furthermore, there is a natural isomorphism SI /II ∼ = (S/I)I , and SI /II is the field of fractions of the integral domain S/I. Let S be a commutative ring and let I ⊂ S be an ideal. If S/I is finite then Γ(I) maps to the set of units of S/I so that the canonical map S −→ SI : x 7→ x + K(I) induces an isomorphism S/I ∼ = SI /II ∼ = (S/I)I . Furthermore, if X is a finite S-module such that XI = 0 then the canonical map X −→ XI : x 7→ x1−1 is an isomorphism. For example, if p ∈ Z is a prime and if G is an abelian group such that pk G = 0 for some integer k then G ∼ = Gp ∼ = G ⊗Z Zp . Specifically if k k ∼ G is any group then G/p G = Gp /p Gp . LEMMA 1.4.1 Let I, J be ideals in a commutative integral domain S. Let X be an S-module. 1. (XI )J ∼ = (XJ )I . 2. If I ⊂ J and if S/I is finite then the natural surjection S/I → S/J takes units to units. Hence (SI )J ∼ = (SJ )I ∼ = SJ .

1.5. LOCAL-GLOBAL REMAINDER

1.5

13

Local-Global Remainder

Any theorem that describes a property of M in terms of some property of MI for each maximal ideal I ⊂ S is called a local-global property. We use several local-global properties in this text. THEOREM 1.5.1 [Local-Global Theorem] (See [72, page 105].) Let S be a commutative ring and let M be an S-module. Then M = 0 ⇔ MI = 0 for each maximal ideal I ⊂ E. Consequently f : M → N is a surjection iff the induced map fI : MI → NI is a surjection for each maximal ideal I ⊂ S. Dually f : M → N is an injection iff the induced map fI : MI → NI is an injection for each maximal ideal I ⊂ S. In particular, given N, K ⊂ M then K ⊂ N iff KI ⊂ NI as SI -submodules of MI for each maximal ideal I ⊂ S. Another type of Local-Global Theorem relates maps M → N to maps MI → NI for maximal ideals I in S. THEOREM 1.5.2 [Change of Rings] (See [72, page 106].) Let E be an S-algebra for some commutative ring S, and let M be a finitely presented right E-module. If C ⊂ S is a multiplicatively closed subset of S containing 1 then for each S-module N 1. HomE (M, N )[C −1 ] ∼ = HomE[C −1 ] (M [C −1 ], N [C −1 ]) as groups. 2. Ext1E (M, N )[C −1 ] ∼ = Ext1E[C −1 ] (M [C −1 ], N [C −1 ]) as groups. In particular if M is a finitely presented right E-module then EndE (M )I ∼ = EndEI (MI ). Furthermore, a finitely presented right Emodule M is a projective E-module iff MI is a projective right EI -module for each maximal ideal I in S. Beware: The Change of Rings Theorem is false unless M is finitely presented. The following congruence result is used in a variety of settings in the sequel. THEOREM 1.5.3 [Chinese Remainder Theorem] (See [30, page 46].) Let S be a commutative ring, let I1 , . . . , It be ideals in S such that Ii + Ij = S for 1 ≤ i 6= j ≤ t, and let M be a right S-module. Let M ∗ = M I1 ∩ · · · ∩ M It .

14

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS 1. Given x1 , . . . , xt ∈ M there is an x ∈ M such that x − xi ∈ M Ii

for each i = 1, . . . , t.

2. The diagonal map t

δ:

M M M −→ ∗ M M Ii i=1

such that δ(x + M ∗ ) = (x + I1 M, . . . , x + M It ) is an isomorphism. 3. M I1 ∩ · · · ∩ M It = M [I1 ∩ · · · ∩ It ]. Proof: See [30, Theorem 18.30-31] for a proof of parts 1 and 2. 3. Let I = I1 ∩ · · · ∩ It . By part 2 there is a canonical isomorphism t Y S ∼S = Ii I i=1

so there are isomorphisms M M∗

∼ =

t M M M Ii

∼ = M ⊗S

i=1

t Y S Ii i=1

S ∼ = M ⊗S I

∼ =

M . MI

Since these are canonical isomorphisms M ∗ = M I, as required by part 3. In particular, if t, n1 , . . . , nt > 0 are integers, if p1 , . . . , pt ∈ Z are distinct primes, and if n = pn1 1 · · · pnt t then given an abelian group G nG = pn1 1 G ∩ · · · ∩ pnt t G. Consequently, there is a natural isomorphism t

δ:

Y G G −→ nG pni i G i=1

such that δ(x + nG) = (x + pn1 1 G, . . . , x + pnt t G) of groups.

1.6. INTEGRALLY CLOSED RINGS

1.6

15

Integrally Closed Rings

See [10]. The commutative rtffr integral domain S is a Dedekind domain iff it satisfies one of the following equivalent conditions. 1. If S ⊂ S 0 ⊂ QS is a ring and if S 0 is a finitely generated S-module then S = S 0 . 2. If I ⊂ S is a nonzero ideal then I is invertible. That is, II ∗ = S where I ∗ = {q ∈ QS qI ⊂ S}. 3. Each nonzero ideal of S is a unique product of maximal ideals in S. 4. S is a hereditary Noetherian integral domain. 5. The localization SM is a discrete valuation domain for each maximal ideal M in S. That is, there is an element π ∈ SM such that each ideal of SM has the form π k SM for some integer k > 0. For example, Z is a Dedekind domain as is any pid. The ring of algebraic integers in an algebraic number field is a Dedekind domain. The rtffr ring E is integrally closed if whenever E ⊂ E 0 ⊂ QE is a ring such that E 0 /E is finite then E = E 0 . A classical maximal order is an integrally closed prime rtffr ring. A classical maximal order is finitely generated and torsion-free as a module over its center. The rtffr ring E is an integrally closed ring iff E = E1 × · · · × Et where each Ei is a classical maximal order. An E-lattice is a finitely generated right E-submodule of the right E-module QE (n) for some n ∈ Z. Dedekind domains are classical maximal orders. If E is a classical maximal order and if U is an E-lattice then Matn (E) and EndE (U ) are classical maximal orders. The next results are [69, Theorem 21.1], [10, Corollary 11.5], [10, Theorem 11.8], and [69, Theorems 18.10, 27.1]. LEMMA 1.6.1 Suppose that the rtffr ring E is a classical maximal order. 1. If I is a right ideal of finite index in E then O(I) = {q ∈ QE qI ⊂ I} is a classical maximal order.

16

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS 2. E is an hereditary Noetherian ring. That is, each one-sided ideal in E is a finitely generated projective E-module. 3. For each integer n ∈ Z, each right ideal in En is principal. 4. If U and V are E-lattices then U is locally isomorphic to V (see section 2.5) iff QU ∼ = QV as right QE-modules. (Warning: This is true only of lattices over maximal orders.)

Semi-prime rtffr rings are closely connected to integrally closed rings as the following result shows. The next two results follow from [10, page 127]. LEMMA 1.6.2 Let E be a semi-prime rtffr ring. There is a finite set of primes {p1 , . . . , ps } ⊂ Z such that Ep is a classical maximal order for all p 6∈ {p1 , . . . , ps }. LEMMA 1.6.3 Suppose that E is a semi-prime rtffr ring. There is an integrally closed ring E ⊂ QE and an integer n 6= 0 such that nE ⊂ E ⊂ E

1.7

Semi-Perfect Rings

The ring E is semi-perfect if 1. E/J (E) is semi-simple Artinian and 2. Given an e¯2 = e¯ ∈ E/J (E) there is an e2 = e ∈ E such that e¯ = e + J (E). That is, idempotents lift modulo J (E). See [6, Chapter 7 §27] for a complete discussion of semi-perfect rings and their modules. Fields, local rings, and Artinian rings are semi-perfect b p are semi-perfect rings. Z and Z6 are not semi-perfect but Zp and Z rings for primes p ∈ Z. Suppose that E is semi-perfect. Each indecomposable projective right E-module P has the form eE for some e2 = e ∈ E. Given e2 = e, f 2 = f ∈ E then eE/eJ (E) ∼ = f E/f J (E) ⇔ eE ∼ = f E. LEMMA 1.7.1 Let E be a semi-perfect ring and let P be a projective right E-module. Then

1.7. SEMI-PERFECT RINGS

17

1. P is indecomposable iff EndE (P ) is a local ring. 2. E is a direct sum of right E-modules with local endomorphism rings. Thus direct sum decompositions of projective modules over semiperfect rings are unique by the Azumaya-Krull-Schmidt Theorem 2.1.6. We will also find it necessary to use the following results on lifting idempotents in semi-perfect rings. LEMMA 1.7.2 Let E be a semi-perfect ring, and let e2 = e, f 2 = f ∈ E be such that eE ∼ = f E. There is a unit u ∈ E such that ueu−1 = f . Proof: Since e2 = e and f 2 = f we can write E = eE ⊕ (1 − e)E = f E ⊕ (1 − f )E and since E is semi-perfect (1 − e)E ∼ = (1 − f )E by The Azumaya-KrullSchmidt Theorem 2.1.6. Thus there is an isomorphism u : E → E such that ueu−1 ∈ u(eE) = f E −1

u(1 − e)u

and

∈ u((1 − e)E) = (1 − f )E.

Evidently u is multiplication by a unit of E so that 1 = u1u−1 = u(e + (1 − e))u−1 = ueu−1 ⊕ u(1 − e)u−1 ∈ f E ⊕ (1 − f )E. Since 1 can be written in exactly one way as an element in a direct sum we must have ueu−1 = f. This completes the proof of the lemma. LEMMA 1.7.3 Let E be semi-perfect and let e2 = e, f 2 = f ∈ E. If eE/eJ (E) ∼ = f E/f J (E) then eE ∼ = f E. Proof: Since e2 = e, eE is a projective right E-module so the isomorphism eE/eJ (E) ∼ = f E/f J (E) lifts to a map φ : eE −→ f E

18

CHAPTER 1. NOTATION AND PRELIMINARY RESULTS

such that ker φ ⊂ eJ (E) and f E = φ(eE) + f J (E). By Nakayama’s Lemma 1.2.3, f E = φ(eE) and since f 2 = f ∈ E eE = U ⊕ ker φ where U ∼ = f E. Inasmuch as ker φ ⊂ eJ (E) another appeal to Nakayama’s Lemma 1.2.3 shows us that ker φ = 0, whence eE ∼ = f E. This completes the proof. LEMMA 1.7.4 Let E be a semi-perfect ring, and let J ⊂ J (R) be an ideal. Then idempotents lift modulo J in E. Proof: Let J ⊂ J (E) and let e ∈ E be such that e2 − e ∈ J ⊂ J (E). ¯ = x + J ∈ E. Let E = E/J and given x ∈ E let x Because E is semi-perfect there is an f 2 = f ∈ E such that e − f ∈ J (E). Then e, f¯ ∈ E are idempotents such that e−f¯ ∈ J (E) = J (E)/J. We will show that there is a unit u ∈ E such that uf¯u−1 = e. We have e + J (E) = f¯ + J (E) and e2 = e, f¯2 = f¯ ∈ E so that eE ∼ E E ∼ f¯E = (f¯ + J (E)) . = (e + J (E)) = ¯ eJ (E) J (E) J (E) f J (E) By Lemma 1.7.2 there is a unit u ∈ E such that e = uf¯u−1 . Since units lift modulo J ⊂ J (E), u = u + J for some unit u ∈ E and hence uf u−1 ≡ e(mod J). Thus idempotents lift modulo J in E.

1.8

Exercise

1. Verify each result mentioned in this chapter by proving it yourself or by finding it in the references.

Chapter 2

Motivation by Example The group G is an

rtffr group

if G is a reduced torsion-free finite abelian group. The topics that we will discuss in this text are properties of direct summands of rtffr groups G. These properties include the uniqueness of direct sum decompositions, as well as the properties of the left End(G)-module G. A direct sum decomposition G = G1 ⊕ · · · ⊕ G t

(2.1)

is said to be indecomposable if each Gi is indecomposable. If p ∈ Z is a prime and if k1 , . . . , kt > 0 are integers such that Gi ∼ = Z/pki Z for each i = 1, . . . , t then the decomposition (2.1) is indecomposable. If Gi ⊂ Q for each i = 1, . . . , t then the decomposition (2.1) is indecomposable. Suppose that (2.1) is an indecomposable decomposition of G and suppose that we can also write G∼ =H ⊕K ∼ = H ⊕ K 0.

(2.2)

We will pursue several broad questions on the nature of direct sums. For instance, is the direct sum (2.1) unique in some sense? What can be said about K and K 0 in (2.2)? Is H a direct sum of copies of the Gi ? Are the group structures of K and K 0 similar in some sense? Are there examples of badly behaved direct sum decompositions of G? Is there a simpler category in which the direct summands of G can be handled? 19

20

CHAPTER 2. MOTIVATION BY EXAMPLE

Suppose that we are given a surjection π : G(n) → G for some integer n > 0. If G is a free group then π is split. That is, is there a map φ : G → G(n) such that πφ = 1G , or equivalently, such that ker π a direct summand of G(n) . Under what conditions on G will π be split?

2.1

Some Well-Behaved Direct Sums

Before embarking on our journey we need a series of examples. The experience that these examples bring us will help us to develop intuition about the above questions. The constructions can be found in D. Arnold’s [10] or L. Fuchs’ [46]. Let G be a group. The purpose of this section is to show that under some conditions direct sum decompositions of G are well behaved, in a sense that we will make precise below. The group G is indecomposable if G = H ⊕ H 0 implies that H = 0 or H 0 = 0. For instance, Z is indecomposable, and given a prime p ∈ Z, Zp , Z(p∞ ), and Z/pk Z for integers k > 0 are indecomposable groups. Any subgroup of Q is indecomposable. The following example shows us that indecomposable rtffr groups can have finite index in a decomposable group. EXAMPLE 2.1.1 (See Lemma A.1.2.) Let X, Y ⊂ Q be groups such that Hom(X, Y ) = 0 = Hom(Y, X) and suppose that there is a prime p ∈ Z such that pX 6= X and pY 6= Y . Choose elements x ∈ X \ pX and y ∈ Y \ pY and define 1 G = (X ⊕ Y ) + (x + y)Z. p . Then G is an indecomposable group of rank two such that G = X ⊕ Y . The group G is strongly indecomposable if each subgroup of finite index in G is indecomposable. See Section 1.3. Given a prime p ∈ Z let

1 a Z = a, k ∈ Z . p pk

2.1. SOME WELL-BEHAVED DIRECT SUMS

21

EXAMPLE 2.1.2 (See Example A.1.1.) Let X, Y ⊂ Q be groups such that Hom(X, Y ) = 0 = Hom(Y, X) and suppose that there is a prime p ∈ Z such that pX 6= X and pY 6= Y . Choose elements x ∈ X \ pX and y ∈ Y \ pY and define 1 G = (X ⊕ Y ) + Z[ ](x + y). p Then G is a strongly indecomposable group of rank two. EXAMPLE 2.1.3 Let p ∈ Z be a prime and let G be a pure subgroup b p . Then G is strongly indecomposable. of Z . b p and suppose that G = Proof: Say G is pure in Z A ⊕ B. The reader ∼ b p ∩ G so that will show that G/pG = A/pA ⊕ B/pB. Then pG = pZ b p ∩ G) ∼ b p )/pZ bp ⊂ Z b p /pZ bp ∼ G/pG = G/(pZ = (G + pZ = Z/pZ. Thus G/pG is indecomposable, whence A/pA = 0 or B/pB = 0. But b p . Hence then A or B is a p-divisible subgroup of the p-reduced group Z A = 0 or B = 0, and therefore G is indecomposable. We say that G has a unique decomposition if 1. G has an indecomposable decomposition G ∼ = G1 ⊕ · · · ⊕ Gt and 2. Given an indecomposable decomposition G ∼ = G01 ⊕ · · · ⊕ G0s then s = t and after a permutation of the subscripts, Gi ∼ = G0i for each i = 1, . . . , t. In this case we call G1 ⊕ · · · ⊕ Gt the unique decomposition of G. The unique decomposition of an rtffr group G is necessarily indecomposable. Rtffr groups having unique decomposition are considered to be rare. EXAMPLE 2.1.4 1. The Fundamental Theorem of Abelian Groups states that if p ∈ Z is a prime and if G is a finite p-group then G has a unique decomposition G = Z/pn1 Z ⊕ · · · ⊕ Z/pnt Z for some integers 0 < n1 ≤ · · · ≤ nt . 2. Finitely generated free groups have unique decomposition as do the (possibly mixed) divisible groups.

22

CHAPTER 2. MOTIVATION BY EXAMPLE 3. A vector space over a field has a unique decomposition as a group.

EXAMPLE 2.1.5 Suppose End(G) is right Artinian. Because End(G) is then semi-perfect the indecomposable decomposition G = e1 G ⊕ · · · ⊕ et G is unique. An rtffr endomorphism ring End(G) contains a complete set of primitive orthogonal indecomposable idempotents {e1 , . . . , et } because QEnd(G) is Artinian. Then G has an indecomposable decomposition G = e1 G ⊕ · · · ⊕ et G. However this decomposition need not be unique. One of the more interesting properties associated with direct sums is the refinement property. The group G has the refinement property if 1. G possesses an indecomposable decomposition G = G1 ⊕ · · · ⊕ G t and 2. If G(n) ∼ = H ⊕ K for some integer n > 0 then there are integers h1 , . . . , ht ≥ 0 such that (h1 )

H = G1

(ht )

⊕ · · · ⊕ Gt

.

It is generally believed that rtffr groups with the refinement property are rare. A theorem on the uniqueness of indecomposable decompositions that will be essential to our investigations is the Azumaya-Krull-Schmidt Theorem, hereafter referred to as the AKS-Theorem. A proof can be found in [6]. The AKS Theorem represents the best conditions that one could hope for concerning the uniqueness of direct sum decompositions. THEOREM 2.1.6 [Azumaya-Krull-Schmidt] Let G1 , · · · , Gt be left E-modules such that EndE (Gi ) is a local ring for each i = 1, . . . , t and let G = G1 ⊕ · · · ⊕ Gt . 1. G1 ⊕ · · · ⊕ Gt is the unique decomposition for G. 2. G has the refinement property. A consequence of the AKS Theorem 2.1.6 is that if G is a direct sum of modules with local endomorphism rings then G ⊕ K ∼ = G ⊕ K 0 for 0 0 some modules K and K implies that K ∼ = K . See [30, Cancellation Theorem 21.2]. Some examples will help us see how to use the AKS Theorem 2.1.6.

2.1. SOME WELL-BEHAVED DIRECT SUMS

23

EXAMPLE 2.1.7 Let p1 , . . . , pt ∈ Z be a finite list of distinct primes and for each i = 1, . . . , t let Gi be the localization Gi = Zpi . Then End(Gi ) is a local commutative ring for each i and hence the AKS Theorem 2.1.6 states that G = G1 ⊕ · · · ⊕ Gt has a unique decomposition. The reader can show that End(G) is commutative in this case. EXAMPLE 2.1.8 If G1 , . . . , Gt are simple right modules over some ring R then EndR (Gi ) is a division ring for each i = 1, . . . , t. The AKS Theorem 2.1.6 states that G = G1 ⊕ · · · ⊕ Gt is the unique decomposition of G, and that G has the refinement property. EXAMPLE 2.1.9 An indecomposable projective module over a semiperfect ring has a local endomorphism ring. Thus the AKS Theorem 2.1.6 applies to direct sums of indecomposable projective modules over a semi-perfect ring. J.D. Reid’s Lemma 1.3.3 states that the rtffr group G is strongly indecomposable iff QEnd(G) is a local ring, so it is reasonable to expect that some form of the AKS Theorem 2.1.6 applies to direct sums of strongly indecomposable rtffr groups. The next theorem is referred to in the literature as J´onsson’s Theorem. It is generally accepted that this is the best possible existence and uniqueness result that we can expect for the general rtffr group G. See [46, page 150, Theorem 92.5]. THEOREM 2.1.10 [B. J´onsson] Let G be an rtffr group. There is a finite list of G1 , . . . , Gt strongly indecomposable groups such that .

1. G ∼ = G1 ⊕ · · · ⊕ G t . .

2. If G ∼ = G01 ⊕ · · · ⊕ G0s and if each G0i is strongly indecomposable . then s = t and after a permutation of the indices Gi ∼ = G0i for each i = 1, . . . , t. .

3. Given an integer n > 0 and a direct sum G(n) ∼ = H ⊕ H 0 then there is a subset I ⊂ {1, . . . , t} and integers hi > 0, i ∈ I, such that . (h ) H∼ = ⊕i∈I Gi i . .

.

4. If G ⊕ H ∼ = G ⊕ H 0 for some rtffr groups H and H 0 then H ∼ = H 0. Proof: We sketch a proof due to E.A. Walker [86]. The category of quasi-homomorphisms QAb is the category 1. whose objects are the torsion-free groups of finite rank, and

24

CHAPTER 2. MOTIVATION BY EXAMPLE 2. whose homsets are given by QHom(?, ?).

The rtffr group G is strongly indecomposable iff G is indecomposable in QAb, and G is strongly indecomposable iff QEnd(G) is a local Artinian ring, (Lemma 1.3.3). Thus each object in QAb is a direct sum of objects with local endomorphism rings. Then by a version of the AKS Theorem 2.1.6 for objects in additive categories in which idempotents split, each object in QAb has a unique decomposition G ∼ = G1 ⊕ · · · ⊕ Gt in QAb where each Gi is strongly indecomposable. Since two rtffr. groups are isomorphic in QAb iff they are quasi-isomorphic groups, G ∼ = G1 ⊕ · · · ⊕ Gt as groups. This completes the proof. We observe that the AKS Theorem 2.1.6 has been our only tool to this point for demonstrating the existence of a unique decomposition. The next result is a theorem on uniqueness and the refinement property of indecomposable decompositions of a certain type of group. Its proof is interesting in that it does not refer to the AKS Theorem 2.1.6. See [46, page 112, Proposition 86.1]. The group G is called completely decomposable if there are rank one groups G1 , . . . , Gt such that G = G1 ⊕ · · · ⊕ Gt . The group H has rank one if H ⊂ Q. THEOREM 2.1.11 [Baer-Kulikov-Kaplansky] Assume that G ∼ = G1 ⊕ · · · ⊕ Gt where each Gi is a rank one group. 1. G1 ⊕ · · · ⊕ Gt is the unique decomposition of G. 2. G has the refinement property. The group G is an acd group (almost completely decomposable group) if there is a completely decomposable group C and an integer n > 0 such that nC ⊂ G ⊂ C. See A. Mader’s text [60] for everything concerning acd groups. There is at least one type of acd group whose direct sum decompositions are well behaved. See [10, Theorem 2.3] for a proof. THEOREM 2.1.12 [R.A. Beaumont and R.S. Pierce] Suppose that . G = Gn1 1 ⊕ · · · ⊕ Gnt t for some integers t, n1 , . . . , nt > 0 and a chain G1 ⊂ . . . ⊂ Gt of rank one groups such that Gi ∼ = Gj iff i = j. Then G∼ = Gn1 1 ⊕ · · · ⊕ Gnt t . Theorem 2.2.1, however, will show us that acd groups can have badly behaved direct sum decompositions.

2.2. SOME BADLY-BEHAVED DIRECT SUMS

2.2

25

Some Badly-Behaved Direct Sums

Now that we have established some useful conditions under which direct sums are well behaved let us see what happens in more general situations. Every torsion-free group of finite rank has an indecomposable decomposition (i.e., the given group can be written as a direct sum of indecomposable groups) but this decomposition need not be unique. (See Theorem 2.2.1 below.) The purpose of this section is to give some examples that illustrate the unruly behavior of indecomposable decompositions of some rtffr groups. In the 1960s several examples came to light illustrating that within the class of acd groups there are examples of direct sum decompositions that are not unique in any sense of the word. Even though the completely decomposable groups possess a unique decomposition (Theorem 2.1.11), an indecomposable decomposition of an acd group may not be unique. THEOREM 2.2.1 [A.L.S. Corner] See Example B.1.1. Let n ≥ k ≥ 1 be integers. There is a group G = G(n) of rank n such that for each partition r1 + · · · + rk of n into k positive summands rj there is an indecomposable direct sum decomposition G = G1 ⊕ · · · ⊕ G k such that rj = rank(Gj ) for each j = 1, . . . , k. For instance, suppose that p, p1 , p2 , p3 are distinct primes, let A1 , A2 , A3 be rank one groups such that pi Ai 6= Ai 6= pAi for each i = 1, 2, 3, and let m = p1 p2 p3 . With n = 4 and k = 2 the above construction shows us that there is a group m(A21 ⊕ A2 ⊕ A3 ) ⊂ G ⊂ A21 ⊕ A2 ⊕ A3 that possesses indecomposable decompositions G = G1 ⊕ G2 = H1 ⊕ H2 such that rank(G1 ) = 3, rank(H1 ) = rank(H2 ) = 2, and rank(G2 ) = 1. Thus G does not possess a unique decomposition. The nonuniqueness of indecomposable direct sum decompositions of rtffr groups is also seen in the following examples.

26

CHAPTER 2. MOTIVATION BY EXAMPLE

THEOREM 2.2.2 [L. Fuchs and F. Loonstra] Examples C.1.2 and D.1.1. 1. There are acd groups G1 , G2 , G3 of ranks 1, 2, 2, respectively, such that G1 ⊕ G2 ∼ = G1 ⊕ G3 while G2 6∼ = G3 . 2. There are indecomposable acd groups G4 , G5 of rank 2 such that (2) (2) (2) G4 ∼ = G5 while G4 6∼ = G5 . Thus G = G4 does not possess a unique decomposition and does not satisfy the refinement property.

The next example illustrates that even if End(G1 ) = End(G2 ) = Z then indecomposable decompositions of G = G1 ⊕ G2 can fail to be unique.

THEOREM 2.2.3 [10, page 63, Example 6.9]. There are quasi-isomorphic indecomposable groups G1 , G2 , H1 , H2 such that G1 ⊕ G 2 ∼ = H 1 ⊕ H2 , and End(X) ∼ = Z for X = G1 , G2 , H1 , H2 , while H2 is not isomorphic to G1 , G2 , or H1 .

Thus in the AKS Theorem 2.1.6 the hypothesis that End(Gi ) is a local ring for each Gi cannot be weakened to the condition that End(Gi ) is a pid.

It is best if we use the above examples as guideposts to help us anticipate limitations in our results. For instance, while any theorem on the direct sum decompositions of acd groups might aspire to the conclusions of the AKS Theorem 2.1.6 or of the Baer-Kulikov-Kaplansky Theorem 2.1.11, it must also respect the badly behaved direct sum decompositions in the above section.

2.3. CORNER’S THEOREM

2.3

27

Corner’s Theorem

The fact that any rtffr ring E is the endomorphism ring of an rtffr group G will allow us to construct rtffr groups possessing a variety of left Emodule structures. These constructions are found in the Appendices. The group G is locally free if Gp is a free Zp -module for each prime p ∈ Z. M.C.R. Butler [21] constructs groups that have the same torsionfree rank as their endomorphism ring. See [10, Theorem 2.14]. THEOREM 2.3.1 [M.C.R. Butler] See Example I.1.1. Let E be an rtffr ring whose additive structure (E, +) is a locally free group. There is an rtffr group G such that E ⊂ G ⊂ QE as left E-modules and E∼ = End(G) as rings.

Let E be an rtffr ring. The left E-module M is called a Corner Emodule, or when E is understood simply a Corner module, if its additive structure (M, +) is a countable reduced torsion-free group. Similarly a Corner ring is an associative ring E with identity such that E is a Corner module. There is no confusion as to the orientation of M as all Corner modules are left modules. Call the group G a Corner group if G is a countable reduced torsion-free group. One of the questions that we will address is Is there a functorial means of characterizing the properties of G in terms of End(G)? Dually, we can ask Is there a functorial means that will allow us to characterize the properties of End(G) in terms of properties of G. Specifically, since G is a left End(G)-module we can ask about the module theoretic properties of G as a left End(G)-module. Corner’s Theorem provides a warehouse of flexible examples that will allow us to gauge how frequently certain properties occur. See Theorem F.1.1 where Corner’s Theorem follows from a more general result. See also [46, Theorem 110.1]. THEOREM 2.3.2 [A.L.S. Corner] Let E be a Corner ring. There is a group G such that E ∼ = End(G) as rings. If rank(E) = n then we can choose G so that rank(G) = 2n. If E is an rtffr ring then the group G constructed in Corner’s Theorem 2.3.2 is a pure and dense subgroup b E ⊂ G = hE, Eπi ⊂ E

28

CHAPTER 2. MOTIVATION BY EXAMPLE

b Then G fits into a short exact for some transcendental element π ∈ Z. sequence 0 → E −→ G −→ QE → 0 of left E-modules. The next two results will extend Corner’s Theorem by replacing E and QE with more general left E-modules while still retaining E = End(G). Given a Corner ring E and a Corner E-module M let O(M ) = {q ∈ QE qM ⊂ M }. The reader can show that under these hypotheses if annE (M ) = 0 then O(M ) is a Corner ring. See Theorem E.1.1 for a detailed proof. THEOREM 2.3.3 [36, T.G. Faticoni] Let E be a Corner ring and let M be a Corner module. There is a short exact sequence 0 → M −→ G −→ QC → 0

(2.3)

such that O(M ) ∼ = End(G) and where C ∼ = ⊕K K where K ranges over (ℵ ) o the cyclic E-submodules of M . For instance our choices for M in the above Theorem include any of the following left E-modules. 1. M = an E-submodule of a countably generated free E-module such that annE (M ) = 0. 2. M = K ⊕E where K is a Corner module. The summand E ensures that E = O(M ). We will find the following result useful in constructing rtffr groups G whose left End(G)-module structure is specified. The left E-module M is an rtffr E-module if its additive structure (M, +) is an rtffr group. See Theorem F.1.2 for a proof of the following result. THEOREM 2.3.4 [43, T.G. Faticoni and H.P. Goeters] Let E be an rtffr ring and let M be an rtffr E-module. There is an rtffr group G and a short exact sequence 0 → M −→ G −→ QE ⊕ QE → 0 such that O(M ) ∼ = End(G).

(2.4)

2.4. ARNOLD-LADY-MURLEY THEOREM

29

It is clear from (2.4) that rank(G) is finite. EXAMPLE 2.3.5 Let E be a Corner ring. 1. Given a Corner module M then E = O(M ⊕E). By Theorem 2.3.4 there is a short exact sequence 0 → M ⊕ E −→ G −→ QE ⊕ QE → 0 of left E-modules in which E ∼ = End(G). One might choose M ∼ = (r) E for some integer r > 0. 2. If we let M = E then E = O(M ) so there is a short exact sequence 0 → E −→ G −→ QE ⊕ QE → 0 of left E-modules such that E = End(G). For all practical modern purposes this is Corner’s Theorem 2.3.3. EXAMPLE 2.3.6 If we are willing to vary E but retain QE then we can use Theorem 2.3.4 as follows. Let A be a finite dimensional Q-algebra, let V be any finitely generated left A-module such that annA (V ) = 0, and let M be a free subgroup of V such that QM = V . Then E = O(M ) = {q ∈ A qM ⊂ M } is a subring of A such that (E, +) is a finitely generated free group and QE = A. An appeal to Theorem 2.3.4 produces a short exact sequence (2.4) in which E ∼ = End(G). Thus we should avoid conjectures like “every group has a proper nonzero direct summand,” “rank of G and rank of End(G) are related by some simple function,” and “indecomposable groups have local endomorphism rings.” Such buffoonery is exposed by Butler’s construction and Corner’s Theorem.

2.4

Arnold-Lady-Murley Theorem

While the examples in the previous sections give us an idea that characterizing module theoretic properties of G in terms of End(G) will not be easy, the Theorem of D.M. Arnold and E. L. Lady provides us with a friendly setting for studying direct sum decompositions of G. Finitely generated projective modules, at least on the surface, seem to be easier to work with than do torsion-free groups of finite rank.

30

2.4.1

CHAPTER 2. MOTIVATION BY EXAMPLE

Category Equivalence

Let G be an rtffr group. Recall the additive functors HG (·) and TG (·) and the associated natural transformations Θ and Φ from Chapter 1. Observe that HG (·) takes groups to right End(G)-modules, while TG (·) takes right End(G)-modules to groups. The first result classifies the direct summands of G(n) for integers n > 0 in terms of a finitely generated projective right End(G)-module. We will let Po (G) = {H G(n) ∼ = H ⊕ K for some integer n > 0 and some group K}. Po (End(G)) = the set of finitely generated projective right End(G)-modules. We consider Po (G) as a category of groups and we consider Po (End(G)) as a category of right End(G)-modules. THEOREM 2.4.1 [D. Arnold and L. Lady] See [10, page 47, Theorem 5.1]. If G is an rtffr group then the functors HG (·) : Po (G) −→ Po (End(G)) TG (·) : Po (End(G)) −→ Po (G) are inverse category equivalences. Proof: For the sake of the argument let E = End(G). Since Θ and Φ are natural transformations ΘH⊕K = ΘH ⊕ ΘK for groups H and K, and ΦM ⊕N = ΦM ⊕ ΦN for right End(G)-modules M and N . Moreover, since G is a group, for each integer n > 0 there is a group homomorphism ΘG(n) = ⊕n ΘG : TG HG (G(n) ) −→ G(n) . Thus given H ⊕ K ∼ = G(I) we can prove that ΘH is an isomorphism if we can prove that ΘG is an isomorphism. Similarly to show that ΦM is an isomorphism it suffices to show that ΦE is an isomorphism. Consider the map ΦE : E → HG TG (E). Notice that HG TG (E) ∼ = E with generator the map f : G → TG (E) such that f (x) = 1G ⊗ x for each x ∈ G. Then ΦE (1) = f which implies that ΦE is an isomorphism.

2.4. ARNOLD-LADY-MURLEY THEOREM

31

The reader proved in Chapter 1 that ΘTG (E) ◦ TG (ΦE ) = 1TG (E) . Then ΘTG (E) is an isomorphism since ΦE and TG (ΦE ) are isomorphisms. Inasmuch as G ∼ = TG (E), ΘG is an isomorphism. Given our reductions, the proof is complete. Let P(G) = {groups H H ⊕ H 0 ∼ = G(c) for some group H 0 and some cardinal c}. Let P(End(G)) be the category of projective right End(G)-modules. EXAMPLE 2.4.2 Let G = ⊕p Zp where p ranges over Q the primes in Z. Notice that G is not an rtffr group, that End(G) = p Zp is a semihereditary ring, and that G is a projective (=flat) left End(G)-module. Let I = G. Then I is a projective ideal in End(G) that is not finitely generated and such that TG (I) = IG = G. Inasmuch I ∼ 6 End(G) we have = shown that TG (·) : P(End(G)) → P(G) is not a category equivalence. EXAMPLE 2.4.3 Let G = Z(ℵo ) and let E = End(G). Then G is not an rtffr group but G is a cyclic projective left E-module, so TG (I) ∼ = IG for each right ideal I ⊂ End(G). If we let I = {f ∈ A f (G) has finite rank} then TG (I) = IG = G = TG (End(G)) but I 6∼ = End(G). Once again TG (·) : P(End(G)) → P(G) is not a category equivalence. See [32] for generalizations of Theorem 2.4.1 to self-small right Rmodules G over some ring R.

2.4.2

Functor A(·)

There is another functor and associated ring that we will single out. Let G be an rtffr group and let E(G) = End(G)/N (End(G)). While HG (·) takes direct summands of G to cyclic projective right End(G)-modules the ring End(G) can itself have complicated structure. One of the strengths of the functor A(·) below is that we can translate between direct summands of G and cyclic projective right E(G)-modules over the Noetherian semi-prime rtffr ring E(G). If E is an rtffr ring then two E-modules M and N are quasi-isomorphic if there is an integer m 6= 0 and E-module maps f : M → N , g : N → M such that f g = m1N and gf = m1M .

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CHAPTER 2. MOTIVATION BY EXAMPLE

THEOREM 2.4.4 Let G be an rtffr group. Let A(·) : Po (G) −→ Po (E(G)) be defined by A(·) = Hom(G, ·) ⊗End(G) End(G)/N (End(G)). 1. A(·) is an additive full functor that preserves direct sums. That is, given an E(G)-module map f : A(H) → A(H 0 ) there is a group map φ : H → H 0 such that A(φ) = f . 2. A(·) induces a bijection α : {(H) H ∈ Po (G)} −→ {(P ) P ∈ Po (E(G))} between the set of isomorphism classes (H) of H ∈ Po (G) and the set of isomorphism classes (P ) of P ∈ Po (A(G)). 3. A(·) induces a bijection between the set of quasi-isomorphism classes [H] of H ∈ Po (G) and the set of quasi-isomorphism classes [P ] of finitely generated projective right E(G)-modules. Proof: 1. Say f : A(H) → A(H 0 ) is an E(G)-module map. By the Arnold-Lady Theorem 2.4.1, HG (H) is a finitely generated projective End(G)-module so f lifts to a map g : HG (H) → HG (H 0 ). Because HG (·) is a category equivalence on Po (G) (Arnold-Lady Theorem 2.4.1), there is a group map φ : H → H 0 such that HG (φ) = g. The reader can verify that A(φ) = f . Parts 2 and 3 follow from [10, page 112, Corollary 9.6].

2.4.3

Elementary Uses of the Functors

The results in this subsection illustrate how one uses the above functors to translate properties of direct summands of G into properties of finitely generated projective modules over a ring. We will include the proofs since they are the first of their kind in this text. PROPOSITION 2.4.5 Let G be an rtffr group. The following are equivalent. 1. If G(n) = H ⊕ H 0 for some integer n > 0 then there is an integer m such that H ∼ = G(m) .

2.4. ARNOLD-LADY-MURLEY THEOREM

33

2. Each finitely generated projective right E(G)-module is free. Proof: Suppose that part 2 is true, let n > 0 be an integer, and let G(n) = H ⊕ K as groups. By Theorem 2.4.4 E(G)(n) ∼ = A(G(n) ) ∼ = A(H) ⊕ A(K) as projective right E(G)-modules. By part 2 A(H) ∼ = E(G)(m) for some integer m so that H ∼ = G(m) by Theorem 2.4.4. The converse is handled in an analogous manner so the proof is complete. The above work shows us that properties of direct sum decompositions of the rtffr group G are equivalent to properties of direct sum decompositions of free right End(G)-modules. This is one example where good results come from the Arnold-Lady Theorem 2.4.1. The general idea is that projective E(G)-modules are better behaved than direct summands of G(n) for integers n > 0. Since projective E(G)-modules are better behaved than those over End(G) it is natural to ask for an AKS-like Theorem for direct sums of groups whose endomorphism rings are pid’s. Recall the refinement property and unique decomposition from section 2.1. THEOREM 2.4.6 Suppose that G1 , . . . , Gt are indecomposable rtffr groups such that for each i 6= j, the composite of each pair of maps Gi → Gj → Gi is in N (EndR (Gi )). Assume that E(Gi ) is a pid for each i = 1, . . . , t. If we let G ∼ = G1 ⊕ · · · ⊕ Gt then 1. G has the refinement property. 2. G has a unique decomposition. ∼H ⊕K ∼ 3. If G(n) = = H ⊕ K 0 for some integer n > 0 and some rtffr groups K and K 0 then K ∼ = K 0. Proof: We begin the proof with some general comments about finitely generated projective right E(G)-modules. By Lemma 1.3.1 and Theorem 2.4.4 E(G) = E(G1 ) × · · · × E(Gt ) where E(Gi ) = A(Gi ) is indecomposable for each i = 1, . . . , t.

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CHAPTER 2. MOTIVATION BY EXAMPLE

Furthermore since E(Gi ) is a pid, given a finitely generated projective right E(G)-module P there are integers p1 , . . . , pt ≥ 0 such that P ∼ = E(G1 )(p1 ) ⊕ · · · ⊕ E(Gt )(pt ) . The uniqueness of the rank of free modules over a pid implies that the integers p1 , . . . , pt are unique to P . 1. Let H ∈ Po (G). Since A(H) ∈ Po (E(G)) we have A(H) ∼ = E(G1 )(h1 ) ⊕ · · · ⊕ E(Gt )(ht ) ∼ = A(G1 )(h1 ) ⊕ · · · ⊕ A(Gt )(ht ) (h ) (h ) ∼ = A(G1 1 ⊕ · · · ⊕ Gt t )

for some integers h1 , . . . , ht ≥ 0. Thus (h ) (h ) H∼ = G1 1 ⊕ · · · ⊕ Gt t

by Theorem 2.4.4, and hence G has the refinement property. 2. Suppose that G = G01 ⊕ · · · ⊕ G0s for some indecomposable right E-modules G01 , . . . , G0s . Then E(G) = E(G1 ) × · · · × E(Gt ) = A(G01 ) ⊕ · · · ⊕ A(G0s ). Consider the indecomposable central idempotent ei ∈ E(G) that corresponds to E(Gi ). Then ei A(G0j ) is a direct summand of A(G0j ), and since A(G0j ) is indecomposable, there is a permutation of the subscripts of A(G01 ), . . . , A(G0s ) such that ei A(G0i ) = A(G0i ) ⊂ E(Gi ) for i = 1, . . . , t. Since E(Gi ) is an indecomposable ideal in E(G) and since ei A(G0i ) is a direct summand of E(Gi ) we have ei A(G0i ) = E(Gi ) = A(Gi ). Then by Theorem 2.4.4, G0i ∼ = Gi . Since t is finite, s = t. 3. Cancellation follows from the uniqueness result in part 2 by counting indecomposable summands of K and K 0 . The reader should prove this if he/she has not already done so. This completes the proof. Given a subgroup H ⊂ Q, End(H) is a pid. Thus the Baer-KulikovKaplansky Theorem 2.1.11 is an immediate consequence of Theorem 2.4.6.

2.5. LOCAL ISOMORPHISM

35

We close this section with elementary translations of the refinement property and uniqueness of decomposition for G. PROPOSITION 2.4.7 Let G be an rtffr group. The following are equivalent. 1. G has the refinement property as a group. 2. End(G) has the refinement property as a right End(G)-module. 3. E(G) has the refinement property as a right E(G)-module. PROPOSITION 2.4.8 Let G be an rtffr group. The following are equivalent. 1. G has a unique decomposition. 2. End(G) has a unique decomposition as a right End(G)-module. 3. E(G) has a unique decomposition as a right E(G)-module.

2.5

Local Isomorphism

Let E be an rtffr ring and let M and N be rtffr E-modules. We will say that M and N are locally isomorphic if for each integer n > 0 there is an integer m 6= 0 and E-module maps fn : M → N , gn : N → M such that gcd(m, n) = 1, fn gn = m1N , and gn fn = m1M . Since we have not ruled out E = Z we have defined local isomorphism for rtffr groups G and H. Locally isomorphic rtffr groups are called nearly isomorphic in [10] and locally isomorphic E-lattices are said to be in the same genus class in [10, 69]. Our introduction of the new terminology is justified by the fact that we are using one term to describe two essentially identical concepts on the category of rtffr E-modules. The next lemma gives us the relationship between local isomorphism of rtffr groups and the local isomorphism of finitely generated projective modules. Its proof rests on the fact that an additive functor takes multiplication by an integer to multiplication by an integer. It is left as an exercise. LEMMA 2.5.1 Let G be an rtffr group, let H, K ∈ Po (G) be rtffr groups, and let E = End(G). The following are equivalent. 1. H and K are locally isomorphic as groups.

36

CHAPTER 2. MOTIVATION BY EXAMPLE 2. HG (H) and HG (K) are locally isomorphic as right E-modules.

Local isomorphism appears in several useful equivalent forms. We leave it to the reader to prove a similar result for finitely generated rtffr modules over an rtffr ring. Recall that Xn = X ⊗Z Zn for integers n 6= 0. LEMMA 2.5.2 Let G and H be rtffr groups. The following are equivalent. 1. G and H are locally isomorphic. 2. For each integer n 6= 0 there are fn ∈ HomE (G, H)n and gn ∈ HomE (H, G)n such that fn gn = 1H and gn fn = 1G . 3. For each prime p ∈ Z there are group maps fp ∈ HomE (G, H)p and gp ∈ HomE (H, G)p such that fp gp = 1H , and gp fp = 1G . 4. For each integer n 6= 0 there are there are fn ∈ HomE (G, H) and gn ∈ HomE (H, G) such that fn gn −1H ∈ nEnd(H) and gn fn −1G ∈ nEnd(G). Proof: 1 ⇒ 2 Given part 1 and an integer n 6= 0 there is an integer m 6= 0, and maps fn : G → H, and gn : H → G such that gcd(m, n) = 1, fn gn = m1H , gn fn = m1G . Since gcd(m, n) = 1, m is a unit in Zn , 1 1 1 fn ∈ HomE (G, H)n . Then ( m fn )gn = 1H and gn ( m fn ) = 1G , and so m which proves part 2. 2 ⇒ 3 is clear. 3 ⇒ 4 Let n = pn1 1 · · · pnt t 6= 0 be an integer where p1 , . . . , pt are the distinct prime divisors of n. Fix an i ∈ {1, . . . , t}. By part 3, there are maps fi ∈ HomE (G, H)pi and gi ∈ HomE (H, G)pi such that fi gi = 1H , and gi fi = 1G . Since X p ∼ X + pk X p = pk X p pk X p for p-reduced groups X and prime powers pk ∈ Z there are maps fi0 ∈ Hom(G, H) and gi0 ∈ Hom(H, G) such that fi0 − fi ∈ pni i HomE (G, H)pi gi0 − gi ∈ pni i HomE (H, G)pi . (X = Hom(G, H) in this case.) By the Chinese Remainder Theorem 1.5.3 there are f ∈ Hom(G, H) and g ∈ Hom(H, G) such that f − fi0 ∈ pni i HomE (G, H) g − gi0 ∈ pni i HomE (H, G)

2.5. LOCAL ISOMORPHISM

37

for each i = 1, · · · , t. Then gf − gfi0 ∈ pni i EndE (G) gfi0 − gi fi ∈ pni i EndE (G)pi so that gf − 1G = gf − gi fi = (gf − gfi0 ) + (gfi0 − gi0 fi0 ) + (gi0 fi0 − gi fi ) ∈ pni i EndE (G)pi . Inasmuch as gf and 1G ∈ End(G) gf − 1G ∈ pni i EndE (G)pi ∩ End(G) = pni i End(G). Since i ∈ {1, . . . , t} was arbitrarily selected, part 3 of the Chinese Remainder Theorem 1.5.3 shows us that gf − 1G ∈

t \

pni i EndE (G) = nEndE (G).

i=1

This proves part 4. 4 ⇒ 1 Choose an integer n 6= 0 such that gf − 1G ∈ nEnd(G) and f g−1H ∈ nEnd(H). Because G is an rtffr group we may choose n so large that End(G)n is a reduced ring. Then n ∈ J (End(G)n ) (try proving that fact, reader), and hence f g is a unit of End(G)n . Let u = (f g)−1 and choose an integer m such that gcd(m, n) = 1 and mu ∈ End(G). Then muf ∈ Hom(G, H) and (muf )g = m1G . Since QG and QH are finite dimensional vector spaces and since [g(uf )]2 = g(uf ) ∈ End(H) is an injection, g(uf ) = 1QH , and hence g(muf ) = m1H . This proves part 1 and completes the proof. The proof of the next result is an exercise. LEMMA 2.5.3 Let E be a semi-prime rtffr ring and let U and V be E-lattices. If E is a semi-local ring then U and V are locally isomorphic iff U ∼ =V. Proof: Suppose that U and V are locally isomorphic and say that there is an integer n 6= 0 such that pE = E for each prime p ∈ Z not dividing n. There is by definition an integer m and maps fn : U → V and gn : V → U such that gcd(m, n) = 1, fn gn = m1V , and gn fn = m1U . Then m is a unit in Zn so that fn and gn are isomorphisms. The converse is clear so the proof is complete.

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CHAPTER 2. MOTIVATION BY EXAMPLE

LEMMA 2.5.4 Let E be an rtffr ring. The following are equivalent for finitely generated projective rtffr E-modules M and N . 1. M and N are locally isomorphic. 2. Mp ∼ = Np as right Ep -modules for each prime p ∈ Z. 3. Mn ∼ = Nn as right En -modules for each integer n 6= 0 ∈ Z. 4. For each integer n 6= 0 there are injections fn : M → N , and gn : N → M such that fn (M ) + nN = N and gn (N ) + nM = M . Proof: 1 ⇒ 2 ⇒ 3 follow as they did in Lemma 2.5.2. 3 ⇒ 4 Assume part 3 and let n 6= 0 be an integer. There are fn ∈ HomE (Mn , Nn ) and gn ∈ HomE (Nn , Mn ) such that fn gn = 1Mn . Since M and N are finitely generated, we may choose an integer m such that gcd(m, n) = 1, mfn ∈ HomE (M, N ), mgn ∈ HomE (N, M ), and (mfn )(mgn ) = m2 1M . Inasmuch as m is a unit in Zn the isomorphism X ∼ Xn ∼ X + nXn = = nX nXn nXn and our choice of fn and gn show us that (mfn )(M ) + nN = N and (mgn )(N ) + nM = M. This proves part 4. 4 ⇒ 1 Assume part 4 and let n 6= 0 be an integer. Choose E-module maps f : M → N and g : N → M such that f (M ) + nN = N and g(N ) + nM = M . Since E is an rtffr ring we can assume without loss of generality that n is so large that the localization En is a reduced group. Then n ∈ J (En ) and since Mn and Nn are finitely generated right E-modules Nn = f (Mn ) + nNn = f (Mn ) by Nakayama’s Lemma 1.2.3. Similarly g(Nn ) = Mn , hence rank(Mn ) = rank(Nn ), whence f : Mn → Nn is an isomorphism. Let h : Nn → Mn be the inverse of f . Inasmuch as M and N are finitely generated there is an integer m 6= 0 such that gcd(m, n) = 1 and mh(N ) ⊂ M . Then f and mh are maps between M and N such that f (mh) = m1N and (mh)f = m1M . This completes the proof. An extreme case of the above lemma is found if E is a classical maximal order. Recall that an E-lattice is a finitely generated left E-module M of QE (n) for some integer n.

2.5. LOCAL ISOMORPHISM

39

LEMMA 2.5.5 Let E be an integrally closed semi-prime rtffr ring and let M, N be E-lattices. Then M and N are locally isomorphic iff QM ∼ = QN as right QE-modules. Proof: Since the integrally closed ring E equals E1 × · · · × Et for some classical maximal orders and since QM ∼ = QN Ei = QN iff QM Ei ∼ as QEi -modules for each i = 1, . . . , t, we have reduced the problem to the case where E is a classical maximal order. Given our reduction, suppose that QE is a prime Artinian ring. Then QM ∼ = QN iff M is locally isomorphic to N by [10, Corollary 12.4]. This completes the proof. The next two results relate local isomorphism to properties of direct sums. The argument is originally due to E. L. Lady. See [16]. LEMMA 2.5.6 Let E be an rtffr ring, and let G and H be rtffr Emodules. If G is locally isomorphic to H then G ⊕ G ∼ = H ⊕ K for some group K. Proof: Let n 6= 0 be any integer. Since H is locally isomorphic to G there is an integer m 6= 0 and group maps fn : G → H and gn : H → G such that gcd(m, n) = 1 and gn fn = m1G . Again there is an integer k 6= 0 and maps fm : G → H and gm : H → G such that gcd(k, m) = 1 and gm fm = k1G . Since gcd(k, m) = 1 there are integers a and b such that am + bk = 1. Consider the maps σ : G ⊕ G −→ H : x ⊕ y 7−→ afn (x) + bfm (y) : H −→ G ⊕ G :

z

7−→ gn (z) ⊕ gm (z).

Then σ(z) = afn gn (z) + bfm gm (z) = (am + bk)(z) = z and so G ⊕ G ∼ = H ⊕ ker σ. This completes the proof. The next result shows us that local isomorphism is tied to the cancellation property. THEOREM 2.5.7 Let H, K, and K 0 be rtffr groups. 1. If H⊕K is locally isomorphic to H⊕K 0 then K is locally isomorphic to K 0 . 2. If n > 0 is an integer and if H (n) ∼ = K (n) then H is locally isomorphic to K.

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CHAPTER 2. MOTIVATION BY EXAMPLE

Proof: 1. Suppose that H ⊕ K is locally isomorphic to H ⊕ K 0 . By Lemma 2.5.1 we may assume without loss of generality that H, K, K 0 are projective right modules over an rtffr ring E such that E = H ⊕ K. Let p ∈ Z be a prime. Since E is an rtffr ring we can choose an k integer multiple of p, say n 6= 0, such that ∩∞ k=1 n E = 0. Then En ∼ = Hn ⊕ Kn0 = Hn ⊕ Kn ∼ as right En -modules and consequently En ∼ Hn Kn ∼ Hn K0 ⊕ ⊕ n0 = = nEn nHn nKn nHn nKn as projective right En /nEn -modules. Since E is torsion-free of finite rank, En /nEn is finite, hence Artinian. Then an application of the AKS Theorem 2.1.6 implies that there is an isomorphism f¯ : Kn /nKn → Kn0 /nKn0 . There are natural isomorphisms K ∼ Kn f¯ K 0 ∼ Kn0 −→ . = = nK nKn nK 0 nKn0 Because K and K 0 are projective f¯ lifts to a map f : K → K 0 such that f (K) + nK 0 = K 0 . Similarly there is a map g : K 0 → K such that g(K 0 ) + nK = K. Then by Lemma 2.5.4(4), K is locally isomorphic to K 0 . Given our reductions we have completed the proof of part 1. Part 2 follows in a manner similar to that of part 1. This completes the proof of the lemma. There are at least two general conditions on G that imply a cancellation property for H ⊕ K. The group G is semi-local if there are at most finitely many primes p ∈ Z such that pG 6= G. COROLLARY 2.5.8 Let H and K be rtffr groups. 1. If K is semi-local then H ⊕ K ∼ = H ⊕ K0 ⇒ K ∼ = K 0. 2. If each right ideal of End(K) is principal then H ⊕ K ∼ = H ⊕ K0 ⇒ K ∼ = K 0. Proof: 1. Suppose that H ⊕ K ∼ = H ⊕ K 0 . Since K is semi-local there is an integer n 6= 0 such that nK 6= K and pK = K for each prime p ∈ Z such that gcd(p, n) = 1. By Theorem 2.5.7, K is locally isomorphic to K 0 so there is an integer m 6= 0 and injections fn : K → K 0 and gn : K 0 → K such that gcd(m, n) = 1, fn gn = m1K 0 , and such that

2.5. LOCAL ISOMORPHISM

41

1 gn fn = m1K . Inasmuch as mK = K we see that m gn : K 0 → K is a 1 1 map such that fn ( m gn ) = 1K 0 and ( m gn )fn = 1K . That is, K ∼ = K 0 as required. 2. Suppose that H ⊕K ∼ = H ⊕K 0 and that each right ideal of End(K) is principal. By Theorem 2.5.7, K is locally isomorphic to K 0 so there is an injection g : K 0 → K. The reader can show that Hom(K, K 0 )K = K 0 . Identify Hom(K, K 0 ) ∼ = gHom(K, K 0 ) with a right ideal of End(K). By the hypothesis on End(K) there is an f ∈ Hom(K, K 0 ) such that gHom(K, K 0 ) = f End(K). Thus when considered as subgroups of QK

g −1 f (K) = g −1 [f End(K)]K = g −1 [gHom(K, K 0 )]K = K 0 . Since H, K, K 0 have finite rank, rank(K) = rank(K 0 ), so that the surjection g −1 f : K → K 0 is an isomorphism. This completes the proof.

We can use local isomorphism to establish some kind of uniqueness for the direct sum decompositions of an rtffr group. Let E be an rtffr ring and let G be an rtffr left E-module. We say that G has a locally unique decomposition if 1. There is a direct sum decomposition G = G 1 ⊕ · · · ⊕ Gt

(2.5)

of indecomposable E-modules G1 , . . . , Gt , and 2. Given a direct sum decomposition G = G01 ⊕ · · · ⊕ G0s of indecomposable E-modules G01 , . . . , G0s then s = t and after a possible permutation of the indices G0i is locally isomorphic to Gi . In case G satisfies the above properties then G1 ⊕ · · · ⊕ Gt is called the locally unique decomposition of G. Observe that we do not require that the G1 , . . . , Gt are distinct groups in the above definition. Also, if E = Z then we have defined the locally unique decomposition of an rtffr group G. EXAMPLE 2.5.9 A locally unique decomposition that is not a unique decomposition is constructed as follows. Let E be a Dedekind domain

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CHAPTER 2. MOTIVATION BY EXAMPLE

with class number h > 1. There is an ideal I 6∼ = E such that I · · · I ∼ =E with h factors. Then M = I (h) ∼ = E (h−1) ⊕ (I · · · I) ∼ = E (h) . Thus M does not have a unique decomposition. However, the decomposition is locally unique since at each prime p ∈ Z, Ep is a pid (see [10] or [69]). Each direct summand of the projective Ep -module Mp is free. In particular any indecomposable direct summand U of Mp is isomorphic to Ep , whence Mp has a unique decomposition. Let E be an rtffr ring and let G be an rtffr left E-module. We say that G has the local refinement property if 1. There is a direct sum decomposition (2.5) for some indecomposable left E-modules G1 , . . . , Gt , and 2. Given H ∈ Po (G) then H = H1 ⊕ · · · ⊕ Hs for some indecomposable left E-modules H1 , . . . , Hs such that each Hi is locally isomorphic to some element of {G1 , . . . , Gt }. Observe that the Gi in the direct sum (2.5) need not be distinct. If E = Z then we have defined what we mean by the local refinement property for the rtffr group G. A large portion of the energy in this book is expended on the study of (locally) unique decompositions and the (local) refinement properties. Locally unique decomposition follows from the local refinement property. LEMMA 2.5.10 Let E be an rtffr ring and let G be an rtffr left Emodule. If G has the local refinement property then G has a locally unique decomposition. Proof: Suppose that G has the local refinement property. Then G has a direct sum decomposition (2.5) for some indecomposable left Emodules G1 , . . . , Gt . Suppose that G = G01 ⊕ · · · ⊕ G0s

(2.6)

for some indecomposable left E-modules G01 , . . . , G0s . Since G has the local refinement property, G01 is locally isomorphic to some element of {G1 , . . . , Gt }.

2.5. LOCAL ISOMORPHISM

43

∼ G0 for each i = 1, . . . , t after a We will show that s = t and that Gi = i permutation of the subscripts. We can assume without loss of generality that G01 is locally isomorphic to G1 and we know that G = G1 ⊕ (G2 ⊕ · · · ⊕ Gt ) and G1 ⊕ (G02 ⊕ · · · ⊕ G0s ) are (locally) isomorphic. Then by Theorem 2.5.7 G 2 ⊕ · · · ⊕ Gt

and G02 ⊕ · · · ⊕ G0s

are locally isomorphic. By a simple induction on rank(G), s − 1 = t − 1, s = t, and after a possible reindexing, Gi is locally isomorphic to G0i for each i = 2, . . . , t. This concludes the proof. We end this chapter by showing that at least one type of group has a unique decomposition. An idempotent e ∈ E is central if xe = ex for each x ∈ E. LEMMA 2.5.11 Let G = G1 ⊕ · · · ⊕ Gt be an rtffr group, for each integer i = 1, . . . , t let e2i = ei ∈ End(G) be central indecomposable idempotents such that ei (G) = Gi and such that 1 = e1 + ⊕ · · · ⊕ et . Then G1 ⊕ · · · ⊕ Gt is the unique decomposition of G. Proof: Let G = G1 ⊕ · · · ⊕ Gt be the given indecomposable decomposition and let {e1 , . . . , et } ⊂ End(G) be the associated complete set of primitive orthogonal indecomposable central idempotents. Suppose that G = G01 ⊕ · · · ⊕ G0s for some indecomposable groups G0i . Let {f1 , . . . , fs } be the complete set of orthogonal indecomposable idempotents corresponding to G01 ⊕ · · · ⊕ G0s . Because ei is central ei fj = fj ei for each i, j. It follows that e1 = e1 1G = e1 f1 ⊕ · · · ⊕ e1 fs is a decomposition of the indecomposable e1 into orthogonal idempotents, so that e1 = e1 fi for some i. In a similar manner, since G0i is indecomposable e1 fi = fi . That is e1 = fi . Reindexing if necessary we can assume that we have chosen f1 , . . . , ft such that ei = fi for each i = 1, . . . , t. Since e1 ⊕ · · · ⊕ et = 1G = f1 ⊕ · · · ⊕ ft ft+1 = . . . = fs = 0. Therefore Gi = ei G = fi G = G0i and the proof is complete. We will find that commuting endomorphisms of rtffr groups G occur with some regularity.

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CHAPTER 2. MOTIVATION BY EXAMPLE

2.6

Exercises

p ∈ Z is a prime, n, m > 0 are integers, c and d are cardinals. G and H are groups, E is a ring, M is a right E-module, and f : G → H is a map. 1. If G is an rtffr group then for each index set I and each f : G → G(I) there is a finite subset J ⊂ I such that g(G) ⊂ G(J ) . 2. The linear transformation f : QG → QH is in QHom(G, H) iff . f (G) = H. 3. Hom(G, H) is a right End(G)-module. 4. Let p ∈ Z be prime. b p are indecomposable groups. (a) Z/pn Z, Z(p∞ ), and Z (b) Any subgroup of Q is indecomposable. 5. Let p ∈ Z be prime. Show that for each integer n > 0, End(Zp ) ∼ = Zp and End(Z/pn Z) ∼ = Z/pn Z are local rings. 6. Show that if G is an rtffr group then G/f (G) is finite for each injection f : G → G. 7. Show that if G is an rtffr group then a bounded quotient G/H is finite. 8. The rtffr group G is strongly indecomposable iff G is an indecomposable object in QAb. b p is strongly indecomposable. 9. A pure subgroup of Z 10. Let p ∈ Z be prime. Show that the subgroups of Z(p∞ ) form a bp. well ordered chain and show that End(Z(p∞ )) ∼ =Z 11. Let Q be the field of fraction of the ring Z[[x]] of power series. Determine the Z[[x]]-module Q/Z[[x]] and its Z[[x]]-endomorphism ring. 12. Let S ⊂ Q be a ring. Then each element of S is an integral multiple of a unit in S. 13. Let E be a Corner ring, let M be a Corner E-module, and let K be a right E-module. Let (2.4) be the short exact sequence constructed in Theorem 2.3.3. Show that TorkE (K, M ) ∼ = TorkE (K, G) for each integer k ≥ 1. Conclude that fdE (M ) = fdE (G).

2.7. QUESTIONS FOR FUTURE RESEARCH

45

14. Let E be a Corner ring, let M be a Corner E-module, and let N be a left E-module. Let (2.4) be the short exact sequence constructed in Theorem 2.3.3. Show that ExtkE (N, M ) ∼ = ExtkE (N, G) for each integer k ≥ 2. Conclude that pdE (M ) = pdE (G). 15. Show that Z(p∞ ) is an E-torsion group. 16. This more of a project than an exercise. Prove that there is a Corner group G such that IG 6= G for each finitely generated right ideal I ⊂ End(G) but such that K ⊗End(G) G = 0 for some nonzero finitely generated K. 17. Let G = G1 ⊕ · · · ⊕ Gt be an abelian group where E(Gi ) = End(Gi )/N (End(Gi )) is a pid for each i = 1, . . . t. Show that if H ⊕ H 0 ∼ = G(c) for some (h ) (h ) t cardinal c then H ∼ = G1 1 ⊕· · ·⊕Gt for some cardinals h1 , . . . , ht . 18. If G and H are locally isomorphic then Hom(G, H)G = H. 19. State and prove a result like Lemma 2.5.6 for locally isomorphic E-lattices G and H. 20. Prove part 2 of Lemma 2.5.7. If n > 0 is an integer and if G(n) ∼ = H (n) then G and H are locally isomorphic.

2.7

Questions for Future Research

In this section I will give some questions that came up during the writing process. These questions can be used to give the reader an eye toward future research in the area of reduced torsion-free finite rank abelian groups, better known as rtffr groups. Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. Extend the ideas of this book to self-small mixed groups. 2. Lift ideal and module theoretic properties from E/N (E) to E and conversely. 3. Lift group or ring theoretic properties from the Z-adic completion b of G to G. See [40]. G

46

CHAPTER 2. MOTIVATION BY EXAMPLE 4. Extend the Arnold-Lady-Murley Theorem to a larger class of groups (i.e., a class that contains Po (G)). See [32]. 5. Study the group theoretic of those G such that End(G) is an integrally closed ring. That is, End(G) is a product of classical maximal orders. 6. Classify the groups G such that End(G) is a semi-perfect ring. 7. In group theoretic terms, describe those groups G that have the Azumaya-Krull-Schmidt Property. For example, The Baer-KulikovKaplansky Theorem shows that the completely decomposable groups satisfy a strong version of this property. If End(G) is semiperfect then G has the Azumaya-Krull-Schmidt Property. 8. Characterize the radical ideals in End(G). For example, the Jacobson radical and the nil radical. 9. Give a correspondence between ideals of End(G) and the subgroups of G. Ideally, the subgroups of G should be the fully invariant Ggenerated subgroups of G.

10. Let E be an rtffr ring. Characterize those (maximal) right ideals I ⊂ E such that IG = G for some group G such that E = End(G). See [40].

Chapter 3

Local Isomorphism Is Isomorphism

rtffr means reduced torsion-free finite rank.

The functor A(·) strikes us as interesting. Thus we will make extensive use of the ideal structure of the semi-prime rtffr ring A(G) = E(G) = End(G)/N (End(G)). In fact A(·) becomes more important in this book than HG (·). This is different from the tone of the literature at the time of this writing. We will show that under a certain commutative localization functor Lτ (·) the local isomorphism of groups G and H translates into the isomorphism of right E(G)τ -modules.

3.1

Integrally Closed Rings

Recall that the ring E is integrally closed if given a ring E ⊂ E 0 ⊂ QE such that E 0 /E is finite then E = E 0 . A prime integrally closed rtffr ring is called a classical maximal order. By Lemma 1.6.3 if E is a semi-prime rtffr ring then QE is semi-simple Artinian. In this case there is an integrally closed ring E ⊂ QE and an integer n > 0 such that nE ⊂ E ⊂ E. 47

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CHAPTER 3. LOCAL ISOMORPHISM IS ISOMORPHISM

The integrally closed ring E is a finite product E = E1 × · · · × Et

(3.1)

of classical maximal orders E i in the simple Artinian rings QE i . Let S = center(E), let S i = center(E i ) for each i = 1, . . . , t, and let S = S1 × · · · × St.

(3.2)

Then S = center(E), S i is a Dedekind domain for each i = 1, . . . , t, and nS ⊂ S ⊂ S. It follows that S and S are commutative Noetherian semi-prime rtffr rings such that S/S is finite and such that QS = QS. LEMMA 3.1.1 Let I ⊂ E be a right ideal that contains a regular element of E. Then E/I is a finite group. In particular at most finitely many (maximal) ideals of E contain I. Proof: Say that c ∈ I ⊂ E is regular. Then c is a unit in QE so it has an inverse d ∈ QE. There is an integer m 6= 0 such that md ∈ E so that c(md) = m1E ∈ I. Then E/I is bounded by m and since E is an rtffr ring E/I is finite. LEMMA 3.1.2 Let E be a semi-prime rtffr ring, let center(E) = S, and let I ⊂ S be an ideal. Then EI II ⊂ J (EI ). Proof: We know that II ⊂ J (SI ) and we know from Lemma 1.6.3 that because E is semi-prime, E is a finitely generated S-module. Nakayama’s Lemma 1.2.3 shows us that if J is a maximal right ideal J ⊂ EI such that EI II + J = EI then J = EI . (J is also an SI -submodule of EI .) Then EI II ⊂ ∩{J J is a maximal right ideal of E} = J (EI ). The next two results show us how to construct integrally closed rings E such that EI ⊂ E ⊂ E for ideals I ⊂ S. Lemma 1.6.1 states that if I is a right ideal of finite index in E then

O(I) = {q ∈ QE qI ⊂ I} is an integrally closed ring.

3.1. INTEGRALLY CLOSED RINGS

49

LEMMA 3.1.3 Let E be a semi-prime rtffr ring with center(E) = S. 0 0 Suppose that E ⊂ QE is an integrally closed ring such that IE ⊂ E for some ideal I ⊂ S of finite index in S. There is an integrally closed ring E ⊂ QE such that IE ⊂ E ⊂ E. Proof: Let

0

J = {q ∈ QE qE ⊂ E} ⊂ E. 0

0

Since 0 6= IE ⊂ E and since S/I is finite I ⊂ J is a right E -module of finite index in E. Write 0

0

0

E = E1 × · · · × Et 0

for some classical maximal orders E i . Then J = J1 ⊕ · · · ⊕ Jt 0

for some finitely generated right E i -modules Ji . Lemma 1.6.1 then implies that the ring E i = {q ∈ QE i qJi ⊂ Ji } is a classical maximal order in QE i so that E = E 1 × · · · × E t = O(J) is an integrally closed subring in QE. Since J is a left ideal in E, E ⊂ E, 0 and since I ⊂ IE ⊂ J, IE = EI ⊂ EJ = J ⊂ E which completes the proof. EXAMPLE 3.1.4 The following is an example of an rtffr Dedekind domain S and a subring S ⊂ S such that S/S is a finite group and the minimum number of generators for S is rank(S). Specify n > 1 ∈ Z, choose an algebraic number field k 6= Q, and let S denote the ring of algebraic integers in k. Then S is a Dedekind domain. A result from number theory (see, e.g., [58]) states that S is a free abelian group on [k : Q] = rank(S) generators. It follows that if p ∈ Z is a prime then Z/pZ- dim(S/pS) = rank(S) 6= 1

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CHAPTER 3. LOCAL ISOMORPHISM IS ISOMORPHISM

so that there is a proper subring S = Z + pS. Note that pS ⊂ S ⊂ S and that these inclusions are proper. Moreover, since S/pS ∼ = Z/pZ, S/pS, and, hence, S are generated by exactly rank(S) elements as S-modules. EXAMPLE 3.1.5 The ring S need not decompose as a ring. Let S1 and S2 be rtffr Dedekind domains. For i = 1, 2 choose nonzero proper ideals Ii ⊂ Si and construct a ring S as I1 × I2 ⊂ S = I1 × I2 + (1, 1)Z ⊂ S1 × S2 . Then (S1 × S2 )/S is finite but S is an indecomposable ring. (Observe that S does not contain the unique idempotents (1, 0), (0, 1) of S1 × S2 .) The next result is fundamental to our investigations. It originally appeared in [18]. A proof can be found in [10, Theorem 14.2]. THEOREM 3.1.6 [The Beaumont-Pierce-Wedderburn Theorem] Let E be an rtffr ring. There is a semi-prime Noetherian subring T ⊂ E . such that E = T ⊕ N (E). Using The Beaumont-Pierce-Wedderburn Theorem we can classify a number of modules up to quasi-isomorphism. EXAMPLE 3.1.7 This example shows that there can be little hope of classifying the additive structure of N (E). Let G be any abelian group and let n 0 E= n ∈ Z, x ∈ G . x n The reader will show that E is a commutative ring and that N (E) = 0 0 ∼ = G as groups. Then E is an rtffr ring iff G is an rtffr group. G 0 Since each semi-prime rtffr ring E has finite index in an integrally closed ring it is natural to ask if some similar kind of result is true of rtffr groups. Let us agree that G is an integrally closed group if E(G) is an integrally closed ring. LEMMA 3.1.8 If G is an rtffr group then there is a G-generated rtffr group G and an integer n = n(G) > 0 such that nG ⊂ G ⊂ G, E(G) is integrally closed, and E(G)/E(G) is finite.

3.2. CONDUCTOR OF AN RTFFR RING

51

Proof: Let E = End(G). By The Beaumont-Pierce-Wedderburn . Theorem 3.1.6 there is a semi-prime rtffr subring T ⊂ E such that E = T ⊕ N (E) and QE = QT ⊕ QN (E) as QT -modules. Let R = E(G) be the projected image of E into QT modulo QN (E). Then R is a semi. . prime Noetherian ring, T ⊂ R, R = T , and E = R ⊕ N (E). By Lemma 1.6.3 there is an integer n 6= 0 and an integrally closed ring T ⊂ QR such that nT ⊂ T ⊂ R ⊂ T . Let G = T G and observe that nG = (nT )G ⊂ T G = G ⊂ T G = G. Then G/G is finite, and . . . T ⊕ N (E) = R ⊕ N (E) = E = End(G). Hence T ⊂ E(G) and . . . T = R = E(G) = E(G). Since T is integrally closed E(G) = R ⊂ T = E(G) and since G/G is finite E(G)/E(G) is finite. This completes the proof. For instance G is an integrally closed group if End(G) is a Dedekind domain. Theorem 2.3.3 shows us that there are plenty of strongly indecomposable groups G that are not integrally closed groups. There is a strongly indecomposable rtffr group G such that End(G) is a hereditary domain but End(G) is not an integrally closed group. See Example 4.5.7.

3.2

Conductor of an Rtffr Ring

We will continue to use the notation introduced in the previous subsection. Specifically, E is a semi-prime rtffr ring and S = center(E). Using Lemma 1.6.3 we fix an integrally closed ring E ⊂ QE such that E ⊂ E and E/E is finite. Let S = center(E), and choose classical maximal orders E 1 , · · · , E t and Dedekind domains S 1 , · · · , S t as in (3.1) and (3.2).

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An E-lattice is a finitely generated right E-submodule of QE (k) for some integer k > 0. The conductor of E into E is the largest ideal τ ⊂ S such that τ E ⊂ E ⊂ E.

Then τ = {q ∈ S qE ⊂ E}. There can be many integrally closed rings E such that E/E is finite, and for each E there is a conductor τ . Thus we say that an ideal τ ⊂ S is a conductor for E if there is an integrally closed ring E such that E/E is finite and such that τ is the conductor for E into E. Since E/E is finite τ contains an integer n 6= 0, so that S/τ and E/τ E are finite rings. Since S = center(E), τ S ⊂ S, so that (τ S)E = τ E ⊂ E. By the maximality of τ in S, τ = τ S is an ideal in S. We need to know that τ is unique to E in some sense. Recall that if I is an ideal in S then the radical of I is √ I = ∩{M M is a maximal right ideal such that I ⊂ M ⊂ S} 0

LEMMA 3.2.1 Let E be a semi-prime rtffr ring. Let E and E be integrally closed rtffr rings that contain E as a subring of finite index, 0 and let τ and τ 0 > 0 be the conductors for E and E into E, respectively. Then √ √ τ = τ 0. In particular given a maximal ideal M ⊂ S then τ ⊂ M ⊂ S iff EM is not integrally closed. Proof: Let I ⊂ S be the ideal defined by the finite intersection \ I = S ∩ {τN N is a maximal ideal in S such that EN 6= E N }. Choose a maximal ideal M of S such that EM 6= E M . Because localization commutes with finite intersection we have \ IM = SM ∩ {(τN )M EN 6= E N } = SM ∩ τM = τM

3.2. CONDUCTOR OF AN RTFFR RING

53

since (τN )M = QS for maximal ideals M 6= N in S. Thus for each maximal ideal M in S we have (IE)M = IM E M = τM E M ⊂ EM . By the Local-Global Theorem 1.5.1, IE ⊂ E, and by the maximality of τ , τ = I. Therefore τ ⊂ M ⊂ S iff EM is not integrally closed. √ √ It follows that if τ and τ 0 are conductors for E into E then τ = τ 0 . EXAMPLE 3.2.2 Let S be the ring of algebraic integers in an algebraic number field k 6= Q. Then S/pS ∼ = Z/pZ(n) where n = [k : Q]. Let S = Z + pS. Then pS ⊂ S ⊂ S and since S/pS ∼ = Z/pZ, pS is the conductor of S into S. The conductor of the semi-prime rtffr ring E gives a local condition that shows us when an E-lattice is a projective E-module. It also yields a product decomposition E = Eo × E1 where Eo is integrally closed. LEMMA 3.2.3 Let E be a semi-prime rtffr ring with conductor τ and let U be an E-lattice. Then U is a projective right E-module iff the localization Uτ is a projective right Eτ -module. Proof: Let U be an E-lattice. It is clear that if U is a projective right E-module then Uτ is a projective right Eτ -module. Conversely assume that Uτ is a projective Eτ -module. We will show that UM is a projective right EM -module for each maximal ideal M ⊂ S. Fix a maximal ideal M ⊂ S. There are two cases possible. If τ ⊂ M then because Uτ is projective, UM = (UM )τ = (Uτ )M is a projective right EM -module. Otherwise τ 6⊂ M so by Lemma 3.2.1, EM is an integrally closed ring. Then UM is an EM -lattice, hence a projective EM -module. Thus UM is a projective E-module for each maximal ideal M in S. Inasmuch as the E-lattice U is finitely presented over the Noetherian ring E, the Change of Rings Theorem 1.5.2 implies that U is a finitely generated projective right E-module. This completes the proof. Let E be a semi-prime rtffr ring with conductor τ ⊂ S and let

C = {c ∈ S c + τ is a unit in S/τ }.

(3.3)

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CHAPTER 3. LOCAL ISOMORPHISM IS ISOMORPHISM

Then C is a multiplicatively closed subset of S. Given an S-module X define X(τ ) = {x ∈ X xc = 0 for some c ∈ C}. Thus X(τ ) is the C-torsion submodule of X. THEOREM 3.2.4 [T.G. Faticoni] Let E be a semi-prime rtffr ring with conductor τ . 1. There are ideals S(τ ), S τ ⊂ S and E(τ ), E τ ⊂ E such that S = S(τ ) × S τ and E = E(τ ) × E τ , 2. Sτ = (S τ )τ and Eτ = (E τ )τ , 3. E(τ ) is an integrally closed ring. Proof: 1. and 2. One shows that S(τ ) is an ideal of S such that S/S(τ ) is a torsion-free group. Since S(τ ) is the kernel of localization at C, Sτ ∼ = (S/S(τ ))τ . Thus S/S(τ ) is a cyclic S-submodule of QS/QS(τ ), so that S/S(τ ) is an S-lattice such that Sτ = (S/S(τ ))τ is a projective Sτ -module. By Lemma 3.2.3, S/S(τ ) is a projective S-module, so that S = S(τ ) × S τ for some ideal S τ of S such that Sτ = (S τ )τ . In a similar manner E = E(τ ) ⊕ E τ and (E τ )τ ∼ = Eτ . Since E is a semi-prime rtffr ring and since E(τ ) is an ideal, E = E(τ ) × E τ . We will return to τ presently. 3. Let M be a maximal ideal in the ring S that contains S τ . For each c ∈ S(τ ) \ M the element (c, 0) is annihilated by S τ . Then SM ∼ = (S/S τ )M ∼ = S(τ )M and EM ∼ = (E/E τ )M ∼ = E(τ )M . Suppose for the sake of contradiction that τ ⊂ M . Let x ∈ S(τ ) \ M . Then x is annihilated by some element c ∈ C. See (3.3). However, since τ ⊂ M , x + M , a unit in S/M , is not annihilated by any of the units in S/τ . This contradiction shows us that τ 6⊂ M .

3.3. LOCAL CORRESPONDENCE

55

∼ E(τ )M is integrally closed for Therefore, by Lemma 3.2.1, EM = each maximal ideal τ 6⊂ M . For τ ⊂ M ⊂ S, E(τ )M = 0 since for each element x ∈ E(τ ) there is a c ∈ C such that xc = 0. Thus E(τ )M is integrally closed for each maximal ideal M ⊂ S, and hence E(τ ) is integrally closed. This completes the proof. For example let E = S = 2(Z6 ⊕ Z10 ) + Z(1 ⊕ 1). Then τ = 2(Z6 ⊕ Z10 ) and C = {(a, b) a − b ∈ τ }. Hence S(τ ) = 0, S τ = S, and Sτ = 2(Z2 ⊕ Z2 ) + Z2 (1 ⊕ 1).

3.3

Local Correspondence

The main result of this section shows how localization at the conductor can be used to translate local isomorphism into isomorphism. The proof of the theorem is partitioned into several lemmas that are interesting in their own right. For the proof of this theorem let τ be a conductor for E, and write E = E(τ ) × E τ as in Theorem 3.2.4. Given an E-lattice U let

U (τ ) = U S(τ ) = {x ∈ U xc = 0 for some c ∈ C}.

THEOREM 3.3.1 Let E be a semi-prime rtffr ring with conductor τ , and write E = E(τ ) × E τ as in Theorem 3.2.4. The localization functor (·)τ induces a bijection λτ : {[U ] U ∈ Po (E)} −→ {(W ) W ∈ Po (QE(τ ) × Eτ )}

from the set of local isomorphism classes [U ] of finitely generated projective right E-modules onto the set of isomorphism classes (W ) of finitely generated projective right QE(τ ) × Eτ -modules. LEMMA 3.3.2 Let E be a semi-prime rtffr ring with conductor τ and let U , V be E-lattices. Then U is locally isomorphic to V iff QU (τ ) ∼ = QV (τ ) and U τ is locally isomorphic to V τ . Proof: Write U = U (τ ) ⊕ U τ . It is clear that U is locally isomorphic to V iff U (τ ) is locally isomorphic to V (τ ) as E(τ )-lattices, and U τ is

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CHAPTER 3. LOCAL ISOMORPHISM IS ISOMORPHISM

locally isomorphic to V τ as E τ -lattices. By Theorem 3.2.4, E(τ ) is an integrally closed Noetherian semi-prime rtffr ring. Then by Lemma 1.6.1, U (τ ) is locally isomorphic to V (τ ) iff QU (τ ) ∼ = QV (τ ). This completes the proof. Begin the proof of Theorem 3.3.1: Define a functor as follows. Lτ (·) : Po (E) −→ Po (QE(τ ) × Eτ ) Lτ (U ) = QU (τ ) ⊕ Uτ . Since the localization functor is additive and exact, Lτ (·) is an additive exact functor. Define a function λτ by λτ ([U ]) = (Lτ (U )) = (QU (τ ) ⊕ Uτ ). We claim that λτ is a well defined function. Suppose that U , V ∈ Po (E) are locally isomorphic. By Lemma 3.3.2, QU (τ ) ∼ = QV (τ ) and U τ τ is locally isomorphic to V . Since τ contains an integer n 6= 0 ∈ τ we can choose an integer m and maps φn : U → V and ψn : V → U such that gcd(m, n) = 1, φn ψn = m1V , and ψn φn = m1U . Since gcd(m, n) = 1, m is a unit of En , and hence φn and ψn lift to isomorphisms between UnS ∼ = VnS . Inasmuch as nS ⊂ τ we have (XnS )τ = Xτ for E-lattices X. Thus Uτ ∼ = (Uτ )nS ∼ = (UnS )τ ∼ = (VnS )τ ∼ = Vτ . This proves the claim, and therefore λτ is a well defined function. We must show that λτ is a bijection. LEMMA 3.3.3 Let E be a semi-prime ring with conductor τ . Then for each finitely generated projective Eτ -module W there is a finitely generated projective E τ -module U such that Uτ ∼ = W. Proof: Let W ∈ Po (Eτ ) and write W = w1 Eτ + · · · + ws Eτ for some finite set {w1 , . . . , ws } ⊂ W . Let U be the E-lattice generated by w1 , · · · , ws . U = w1 E + · · · + ws E.

3.3. LOCAL CORRESPONDENCE

57

There is an E-module surjection π : ⊕si=1 xi E → U such that π(xi ) = wi for each i. Because localization is exact the image of πτ : ⊕si=1 xi Eτ → Uτ ⊂ W is Uτ = w1 Eτ + · · · + ws Eτ = W. Moreover Uτ = W is a finitely generated projective Eτ -module so by Lemma 3.2.3, U is a finitely generated projective E-module. The lemma is proved. COROLLARY 3.3.4 Let E be an rtffr ring with conductor τ . Then λτ is a surjection. Let X τ = XS τ for E-lattices X. Then X = X(τ ) ⊕ X τ . The following are technical lemmas that we need to prove that λτ is an injection. LEMMA 3.3.5 Let E be a semi-prime rtffr ring with conductor τ , and let n 6= 0 ∈ τ be an integer. 1. There is an ideal I ⊂ S such that (a) n ∈ I, (b) I + τ = S, and such that (c) c ∈ S is a unit modulo nS iff c is a unit modulo I ∩ τ . Consequently, UnS = UI∩τ for each finitely generated projective right E-module U . 2. (U τ )n = (U τ )nS for each finitely generated projective right Emodule U . Proof: 1. Let {M1 , . . . , Mr , T1 , . . . , Ts } be a complete set of the maximal ideals of S that contain nS, and assume that {T1 , . . . , Ts } is a complete set of the maximal ideals of S that contain τ . Let I = M 1 ∩ · · · ∩ Mr

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and let I ⊂ N ⊂ S be a maximal ideal. Using the Chinese Remainder Theorem 1.5.3 we see that S S ∼ S × ··· × = I M1 Mr and each S/Mi is a simple module. Thus S/N ∼ = S/Mi for some i = 1, · · · , r. Since S is commutative N = annS (1 + N ) = Mi , so that M1 , . . . , Mr is a complete list of the maximal ideals of S that contain I. In particular I is not contained in any of the maximal ideals T1 , . . . , Ts that contain τ , hence I +τ is not contained in any maximal ideal, whence I +τ = S. Another application of The Chinese Remainder Theorem 1.5.3 shows us that S ∼S S = × I ∩τ I τ as rings. We claim that {c ∈ S c + nS is a unit in S/nS} = {c ∈ S c + (I ∩ τ ) is a unit in S/(I ∩ τ )}. Since nS ⊂ I ∩ τ , and since units map to units, the elements c that are units modulo nS are units modulo I ∩ τ . Conversely, if c ∈ S maps to a unit in the finite ring S/(I ∩ τ ) then c 6∈ Mi or Tj for any Mi , Tj ∈ {M1 , . . . , Mr , T1 , . . . , Ts }. We have shown that {M1 , . . . , Mr , T1 , . . . , Ts } is a complete list of the maximal ideals that contain I ∩ τ . Thus c + nS is not in any of the maximal ideals of S/nS, and so c + nS is a unit of S/nS. As claimed, the set of units modulo nS is the set of units modulo I ∩ τ . Consequently, UnS = UI∩τ for each finitely generated projective right E-module. This proves part 1. 2. Since nS ⊂ τ , the ideal {x ∈ S xc = 0 for some c ∈ S such that c + nS ∈ S/nS is a unit} is contained in the ideal S(τ ). Thus S τ is torsion-free relative to the units modulo nS. Hence there is an inclusion S τ ⊂ (S τ )nS that lifts to one (S τ )n ⊂ (S τ )nS . To see that (S τ )n = (S τ )nS let N1 , . . . , Nr be a complete list of the maximal ideals of the commutative ring S that contain nS. For each N ∈ {N1 , . . . , Nr } ((S τ )n )N = (S τ )N = ((S τ )nS )N so that (S τ )n = (S τ )nS by the Local Global Theorem. (Prove this as an exercise, reader.) Then Unτ ∼ = (U τ )nS for each finitely generated projective S-module. This completes the proof.

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59

COROLLARY 3.3.6 Let E be a semi-prime rtffr ring with conductor τ , and let n 6= 0 ∈ τ be an integer. Let U and V be finitely generated projective right E-modules such that QU (τ ) ∼ = Vτ . = QV (τ ) and Uτ ∼ V . Then Un ∼ = n Proof: By hypothesis, U (τ ) and V (τ ) are lattices over the integrally closed ring E(τ ) (Theorem 3.2.4), such that QU (τ ) ∼ = QV (τ ). Then U (τ ) and V (τ ) are locally isomorphic (Lemma 1.6.1(4)). Consequently, U (τ )n ∼ = V (τ )n , (Lemma 2.5.4(3)). By Lemma 3.3.5(1) UnS ∼ = UI∩τ ∼ = VI∩τ ∼ = VnS so (U τ )nS ∼ = (U S τ )nS ∼ = (V S τ )nS ∼ = (V τ )nS . Lemma 3.3.5(2) then shows us that (U τ )n ∼ = (U τ )nS ∼ = (V τ )nS ∼ = (V τ )n . Hence Un ∼ = U (τ )n ⊕ (U τ )n ∼ = V (τ )n ⊕ (V τ )n ∼ = Vn , which proves the Corollary. THEOREM 3.3.7 [T.G. Faticoni] Let E be a semi-prime rtffr ring with conductor τ . Then λτ is an injection. Specifically, if U , V ∈ Po (E) then U and V are locally isomorphic E-lattices iff QU (τ ) ∼ = QV (τ ) and ∼ Uτ = Vτ as Eτ -lattices. Proof: Suppose that QU (τ ) ∼ = QV (τ ) and that Uτ ∼ = Vτ . By Lemma 3.3.6, Un and Vn are isomorphic right E-modules for each integer n > 0. Thus U is locally isomorphic to V (Lemma 2.5.4), which completes the proof of the lemma. Proof of Theorem 3.3.1 completed: By Corollary 3.3.4 and Theorem 3.3.7, λτ is a bijection, which completes the proof.

3.4

Canonical Decomposition

Let us extend λτ to include rtffr groups G and the associated semi-prime rtffr ring A(G) = E(G) = End(G)/N (End(G)).

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Suppose that E(G) has conductor τ , and let Lτ (·) be the localization functor. Let E(G)(τ ) denote {x ∈ E(G) xc = 0 for some c ∈ S such that c is a unit modulo τ }. The reader will show that

E(G)(τ ) = E(G)S(τ ).

Recall Arnold’s Functor A(·) from Section 2.4.2. Let us define

Eτ (G) = QE(G)(τ ) × (E(G))τ Aτ (·) = Lτ (·) ◦ A(·) : Po (G) −→ Po (Eτ (G)).

We say that

G has conductor τ if τ is a conductor for E(G).

THEOREM 3.4.1 Let G be an rtffr group with conductor τ . The functor Aτ (·) induces a bijection αG (·) : {[H] H ∈ Po (G)} −→ {(W ) W ∈ Po (Eτ (G))} from the set of local isomorphism classes [H] of H ∈ Po (G) onto the set of isomorphism classes (W ) of W ∈ Po (Eτ (G)). Proof: By [10, Corollary 12.7], A(·) induces a bijection {[H] H ∈ Po (G)} → {[U ] U ∈ Po (E(G))} and Theorem 3.3.1 states that the functor Lτ (·) induces a bijection λτ : {[U ] U ∈ Po (E(G))} → {(W ) W ∈ Po (Eτ (G))}. The assignment [H] 7→ λτ [A(H)] defines the required bijection αG . The reader can prove the next two results as exercises.

3.4. CANONICAL DECOMPOSITION

61

THEOREM 3.4.2 Let G be an rtffr group and assume that E(G) is an integrally closed ring. The functor Q ⊗Z A(·) induces a bijection αG (·) : {[H] H ∈ Po (G)} −→ {(W ) W ∈ Po (QE(G))} from the set of local isomorphism classes [H] of H ∈ Po (G) onto the set of isomorphism classes (W ) of semi-simple right QE(G)-modules W . THEOREM 3.4.3 Let G be an rtffr group with conductor τ 6= S, and assume that QE(G) is a simple Artinian ring. The functor Aτ (·) induces a bijection αG (·) : {[H] H ∈ Po (G)} −→ {(W ) W ∈ Po ((E(G))τ )} from the set of local isomorphism classes [H] of H ∈ Po (G) onto the set of isomorphism classes (W ) of finitely generated projective right (E(G))τ modules W . Proof: Since QE(G) is simple, S is an integral domain and so S τ = S. Thus QE(G)(τ ) = 0. The reader can take it from here. These results lead us to a canonical direct sum decomposition of an rtffr group G. THEOREM 3.4.4 Let G be an rtffr group with conductor τ . Then G = G(τ ) ⊕ Gτ where E(G(τ )) = E(G)(τ ) is an integrally closed ring. Proof: By Theorem 3.2.4, E(G) = E(G)(τ ) × E(G)τ where E(G)(τ ) is integrally closed and E(G)τ (τ ) = 0. Let e2 = e ∈ S be such that E(G)e = E(G)(τ ) and E(G)(1 − e) = E(G)τ . Let G(τ ) = eG and Gτ = (1 − e)G. Thus G = G(τ ) ⊕ Gτ . By Theorem 2.4.4, E(G)(τ ) = A(G(τ )) and E(G)τ = A(Gτ ). Moreover, E(G(τ )) = E(G)(τ ) is integrally closed. This completes the proof.

If we combine the above theorem with Theorem 3.4.2 then we have shown that G can be directly decomposed as G = G(τ ) ⊕ Gτ where the direct sum decompositions of G(τ ) behave like modules over a semisimple Artinian ring, and where Gτ holds the interesting perhaps badly behaved direct sum decompositions.

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THEOREM 3.4.5 [T.G. Faticoni] Let G be an rtffr group with conductor τ . Then G = G(τ ) ⊕ Gτ where 1. Each H ∈ Po (G) can be written uniquely as H = H(τ )⊕H τ where H(τ ) ∈ Po (G(τ )) and H τ ∈ Po (Gτ ). 2. Direct sum decompositions of H(τ ) are functorially equivalent to direct sum decompositions of finitely generated semi-simple modules over a semi-simple Artinian ring. 3. Direct sum decompositions of H τ are functorially equivalent to direct sum decompositions of finitely generated projective modules over a semi-local semi-prime rtffr ring. Proof: Exercise.

3.5

Arnold’s Theorem

The power of Theorem 3.4.1 is that it translates problems on direct sum decompositions of an rtffr group G (which are not finitely generated) into problems on direct sum decompositions of finitely generated projective right modules over the semi-local semi-prime ring Eτ (G) = QE(G)(τ ) × (E(G))τ . Moreover the bijection translates local isomorphism into isomorphism and conversely. For this reason when discussing the local refinement property or locally unique decompositions of G we can instead study the refinement property or unique decompositions of the pleasantly structured ring Eτ (G). In this section we will put the above strategy to work. To begin with we will show that Theorem 3.3.1 can be used to give a short proof of an important result in the study of direct sum decompositions of rtffr groups. Part 2 of the following result is referred to in the literature as Arnold’s Theorem. THEOREM 3.5.1 Let G be an rtffr group with conductor τ , let Aτ (·) be the functor defined on page 60, and let H ∈ Po (G). 1. Aτ (H) ∼ = W1 ⊕ W2 as right Eτ (G)-modules iff H is locally isomorphic to H1 ⊕ H2 for some rtffr groups Hj such that Aτ (Hj ) ∼ = Wj for j = 1, 2.

3.5. ARNOLD’S THEOREM

63

2. [10, D. M. Arnold] H is locally isomorphic to G1 ⊕ G2 as rtffr groups iff H = H1 ⊕ H2 for some rtffr groups Hj such that Hj and Gj are locally isomorphic for j = 1, 2. Proof: 1. Suppose that Aτ (H) = W ∼ = W1 ⊕ W2 . By Theorem 2.4.4, A(H) = U is a finitely generated projective right E(G)-module such that Uτ ∼ = W . By the proof of Lemma 3.3.3, U ⊂ W . Let U1 = W1 ∩ U and let U2 = U/U1 . Because localization is exact (U2 )τ = (U/U1 )τ ∼ = W2 = W/W1 ∼ = Uτ /(U1 )τ ∼ so by Lemma 3.2.3, U2 is a projective right A(G)-module. Hence U ∼ = U1 ⊕ U2 . Furthermore, by Theorem 2.4.4 there are groups H1 , H2 ∈ Po (G) such that A(Hj ) ∼ = = Uj for j = 1, 2, and H = H1 ⊕ H2 . Then Aτ (Hj ) ∼ (Uj )τ ∼ W for j = 1, 2. The converse is clear. = j 2. Suppose that H is locally isomorphic to G1 ⊕ G2 . By Theorem 3.4.1, Aτ (H) ∼ = Aτ (G1 ⊕ G2 ) ∼ = Aτ (G1 ) ⊕ Aτ (G2 ). By part 1, H ∼ = H1 ⊕ H2 for some rtffr groups H1 and H2 such that Aτ (Gi ) ∼ = Aτ (Hi ) for i = 1, 2. Theorem 3.4.1 implies that Gi is locally isomorphic to Hi for i = 1, 2. The converse is evident so the proof is complete. COROLLARY 3.5.2 Let G be an rtffr group with conductor τ . The following are equivalent. 1. G is indecomposable. 2. If G is locally isomorphic to H then H is indecomposable. 3. E(G) is indecomposable as a right E(G)-module. 4. Eτ (G) is indecomposable as a right Eτ (G)-module. Recall locally unique decompositions and the local refinement property from Section 2.5. It follows from Theorem 3.4.1 and the AKS Theorem 2.1.6 that the rtffr group G has a locally unique decomposition if Eτ (G) is a semi-perfect ring. By the Baer-Kulikov-Kaplansky Theorem 2.1.11 completely decomposable groups have a locally unique decomposition and the reader should convince themselves that Eτ (G) is a semi-perfect ring in this case. The next two results show that the local refinement property can be treated as the refinement property.

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THEOREM 3.5.3 [T.G. Faticoni] Let G be an rtffr group with conductor τ . The following are equivalent. 1. G has a locally unique decomposition. 2. E(G) has a locally unique decomposition. 3. Eτ (G) has a unique decomposition. Proof: 1 ⇔ 2 follows immediately from Theorems 2.4.1, 2.4.4 while 1 ⇔ 3 follows immediately from Theorem 3.5.1. The next result shows that a locally unique decomposition can be treated as a unique decomposition. THEOREM 3.5.4 [T.G. Faticoni] Let G be an rtffr group with conductor τ . The following are equivalent. 1. G has the local refinement property. 2. E(G) has the local refinement property. 3. Eτ (G) has the refinement property. Proof: Apply Theorems 2.4.4 and 3.5.1.

3.6

Exercises

Let E be a semi-prime rtffr ring with center S with conductor τ . Let E ⊂ QE be an integrally closed ring such that Eτ ⊂ E ⊂ E. Let S = center(E). 1. Let S be an Artinian ring. Show that S is local iff e2 = 2 ∈ S implies that e = 0 or 1. b p is actually left multi2. Show that any group endomorphism of Z b p . (Hint: Given a map f consider plication by some element of Z f (1).) 3. Let G be an rtffr. There is a bijection from the set of isomorphism classes of eQEnd(G) where e2 = e ∈ QEnd(G) onto the set of quasi-isomorphism classes [H] of quasi-summands H of G. The bijection is given by eQEnd(G) 7→ [eG].

3.6. EXERCISES

65

4. The group G is p-pure in H if G ⊂ H and if G/H is p-torsion-free. Show that (a) G is p-pure in H iff pG = pH ∩ G. b p is strongly indecomposable. (b) a p-pure subgroup G of Z b p . Show that 5. Let G be a p-pure subgroup of Z (a) G/pG ∼ = Z/pZ. bp. (b) End(G) is a pure subring of Z (c) End(G) is an integral domain. 6. Let x ∈ S and choose maximal ideals M1 , . . . , Mt of S such that (n ) M1n1 ∩ · · · ∩ Mt t ⊂ xS ⊂ M1 ∩ · · · ∩ Mt . Prove that each of the maximal ideals of S/xS are of the form Mi /xS. 7. Let H, K ∈ Po (G). Show that the following are equivalent. (a) H is locally isomorphic to H as groups. (b) A(H) is locally isomorphic to A(K) as right E(G)-modules. 8. Let {G1 , . . . , Gt } be a nilpotent set of rtffr groups. Then G = G1 ⊕ · · · ⊕ Gt has the refinement property iff each Gi has the refinement property. 9. (a) If E is a Noetherian semi-prime ring then QE is semi-simple. (b) If E ⊂ U ⊂ QE then EndE (U ) ⊂ QE. 10. If U is a generator for the ring E then the ring of EndE (U )-module endomorphisms of U is E. .

.

11. Let G and H be rtffr groups. Show that H ∼ = G iff A(H) ∼ = A(G). 12. Let S be the ring of algebraic integers in the algebraic number field k. Let n ∈ Z and let E = nZ + 1Z. Then S is generated as an E-module by exactly g elements where g is the composition length of the group S/nS. 13. Let I, J be ideals in the commutative ring S. (a) SI ⊗S SI ∼ = SJ ⊗S SI as S-modules. (b) (XI )J ∼ = (XJ )I naturally. (c) If I ⊂ J then the natural map S/I → S/J takes units to units.

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CHAPTER 3. LOCAL ISOMORPHISM IS ISOMORPHISM (d) (SI )J ∼ = SJ .

14. Let I ⊂ S be an ideal and let M be an S-module such that each element of M is annihilated by some finite power of I. Then M = 0 iff MN = 0 for each maximal ideal I ⊂ N ⊂ S. 15. Suppose that U and V are finitely generated projective right Emodules such that U/xU ∼ = V /xV . Show that there is an injection φ : U → V such that V = φ(U ) + V . 16. Suppose that U ⊂ W and that UI = W = W1 ⊕ W2 . Show that (W1 ∩ U )I = W1 . 17. Show that J (Sτ ) ⊂ J (Eτ ). 18. Let E be an rtffr ring and suppose that E is locally isomorphic to U1 ⊕ U2 for some finitely generated E-modules U1 and U2 . Then E = V1 ⊕ V2 for some cyclic right ideals V1 and V2 such that Ui is locally isomorphic to Vi for i = 1, 2. 19. If E is integrally closed then QE(τ ) = Eτ . Hint: 0 is a unit in the zero ring {0}. 20. Let G be an rtffr group. Show that the following are equivalent. (a) G is indecomposable. (b) G is locally isomorphic to some indecomposable group. (c) If G is locally isomorphic to H then H is indecomposable. (d) E(G) is indecomposable as a right E(G)-module. (e) Eτ (G) is indecomposable as a right Eτ (G)-module.

3.7

Questions for Future Research

Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. Is there some natural way of constructing the ring T ⊕ N (E) in the Beaumont-Pierce-Wedderburn Theorem? See Theorem 3.1.6. 2. Investigate the structure of the rtffr ring E as a module over S = center(E). How are the ideals of Z and S distributed in the ideal lattice of E?

3.7. QUESTIONS FOR FUTURE RESEARCH

67

3. Let G be an rtffr group. It is known that S is the ring of left End(G)-module homomorphisms φ : G −→ G. Describe functorially S in terms of G. See [32, 35]. 4. Describe a localization at (maximal) ideals in the rtffr ring E in a noncommutative setting. There is quite a large literature for this in ring theory. 5. Investigate the significance of the conductor τ of an rtffr ring. We know that E = E(τ ) × Eτ . See Theorem 3.2.4. Can we say more about τ or Eτ ? √ 6. Describe the rings E such that τ = τ . 7. Describe the rings E that have exactly one conductor τ . 8. Someone out there please come up with a better proof of Theorem 3.3.1. 9. Discuss the ideal theory of commutative rtffr rings. 10. Let G be an rtffr group. Realize the conductor τ of End(G) in terms of G. 11. Find applications for the correspondence given in Theorem 3.4.3. 12. Describe the direct sum decomposition G = G(τ ) ⊕ Gτ in group theoretic terms. For example, what is Gτ ? 13. Give a first principles proof of Arnold’s Theorem 3.5.1. 14. Classify the Jacobson radical of Sτ . See Theorem 3.2.4. 15. Investigate the existence of commutativity in rtffr rings E. 16. Let E be an rtffr ring. Investigate the existence of central idempotents in E/N (E). See [40].

Chapter 4

Commuting Endomorphisms Fix an rtffr group G with endomorphism ring End(G), and let E(G) = End(G)/N (End(G)). We will study direct sum decompositions of G in the presence of partial commutative conditions on rings associated with G.

4.1

Nilpotent Sets

Let G be an rtffr group. We begin with a short discussion of idempotents in End(G). The element e ∈ End(G) is an idempotent if e2 = e. We will use both e2 = e and idempotent to refer to the same property of e as the language permits. Let E be a ring and let {e1 , . . . , et } ⊂ E be a set of idempotents. We say that {e1 , . . . , et } is complete if e1 + · · · + et = 1E and we say that {e1 , . . . , et } is a set of orthogonal idempotents if ei ej = 0 for each 1 ≤ i 6= j ≤ t. If {e1 , e2 } is a set of orthogonal idempotents then we say that e1 and e2 are orthogonal idempotents. The reader should show that the sum of orthogonal idempotents is again an idempotent. The idempotent e ∈ E is primitive if given orthogonal idempotents e1 and e2 such that e = e1 + e2 69

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then either e1 = 0 or e2 = 0. The idempotent e is central if ex = xe for each x ∈ E. Idempotents in End(G) characterize the direct sum decompositions of G as follows. LEMMA 4.1.1 Let G be a group. 1. If e ∈ End(G) is an idempotent then G = eG ⊕ (1 − e)G. 2. If G = G1 ⊕ G2 then there are orthogonal idempotents e1 , e2 ∈ End(G) such that e1 + e2 = 1G , e1 G = G1 , and e2 G = G2 . 3. If e ∈ End(G) is an idempotent then e is primitive iff eG is indecomposable. 4. If {e1 , . . . , et } ⊂ End(G) is a complete set of primitive orthogonal idempotents then G = e1 G ⊕ · · · ⊕ et G is an indecomposable direct sum decomposition of G. 5. If G = G1 ⊕ · · · ⊕ Gt is an indecomposable decomposition of G then there is a complete set of primitive orthogonal idempotents {e1 , . . . , et } ⊂ End(G) such that Gi = ei G for each i = 1, . . . , t. Proof: 1. If e2 = e then each x ∈ G can be written as x = ex+(1−e)x so that G = eG+(1−e)G. If x ∈ eG∩(1−e)G then x = ex = (1−e)ex = 0 so that G = eG ⊕ (1 − e)G. 2. If G = G1 ⊕ G2 then let ei : G → Gi be the unique map defined as follows. Given x ∈ G there are unique x1 ∈ G1 and x2 ∈ G2 such that x = x1 ⊕ x2 . Then define ei ∈ End(G) by ei (x1 ⊕ x2 ) = xi

for i = 1, 2.

The reader will verify that e2i = ei , that ei ej = 0 for i 6= j, and that e1 + e2 = 1G . 3. Let e2 = e ∈ End(G). If e = e1 + e2 for some nonzero orthogonal idempotents e1 and e2 ∈ End(G) then as in part 1 we can write eG = e1 G ⊕ e2 G for some nonzero groups e1 G and (e − e1 )G. Thus eG is decomposable if e is not primitive. Conversely if eG = G1 ⊕ G2 for some nonzero groups then as in part 2 the canonical projections ei : G → Gi are nonzero orthogonal idempotents that satisfy e = e1 + e2 . Thus e is not primitive. This completes the proof of part 3.

4.1. NILPOTENT SETS

71

The proof of part 4 uses parts 1, 2, and 3 in a simple induction on t. This completes the proof of the lemma. If e2 = e and if H = eG then we say that H is the direct summand corresponding to e. If G = G1 ⊕ G2 then the natural projection e1 : G → G1 is an idempotent corresponding to G1 . The reader can prove the following lemma. LEMMA 4.1.2 Let G = G1 ⊕ G2 and for i = 1, 2 let e2i = ei ∈ End(G) correspond to Gi . 1. Hom(Gi , Gj ) = ej End(G)ei for each i, j ∈ {1, 2}. That is, each f : Gi → Gj satisfies f = ej f ei . 2. End(Gi ) = ei End(G)ei . Each indecomposable direct summand H of G corresponds to a primitive idempotent e ∈ End(G). We are thus motivated to consider complete sets of primitive orthogonal idempotents in End(G). Central indecomposable idempotents are of particular importance. LEMMA 4.1.3 Let E be any ring, let {e1 , . . . , et } ⊂ E be a complete set of primitive orthogonal idempotents, and suppose that ei is central for each i = 1, . . . , t. If e2 = e ∈ E is primitive then e ∈ {e1 , . . . , et }. Proof: Let e2 = e ∈ E be primitive. Because {e1 , . . . , et } is a complete set of central orthogonal idempotents (ei e)(ej e) = ei ej e = 0 for i 6= j. Then e = e1 e + · · · + et e is a sum of orthogonal idempotents. Since e is primitive there is a subscript, 1 say, such that e = e1 e. Since e1 is primitive and since e1 = e1 e + e1 (1 − e) is a sum of orthogonal idempotents we see that e1 (1 − e) = 0. That is e = e1 e = e1 . This completes the proof. The set {G1 , . . . , Gt } of groups is rigid if HomR (Gi , Gj ) = 0 for each 1 ≤ i 6= j ≤ t. B. Charles [22] calls {G1 , . . . , Gt } a semi-rigid set if HomR (Gi , Gj ) = 0 or HomR (Gj , Gi ) = 0 for each 1 ≤ i 6= j ≤ t. D. Arnold, R. Hunter, F. Richman [13] call {G1 , . . . , Gt } a pseudo-rigid set if for each 1 ≤ i 6= j ≤ t, each composition Gi → Gj → Gi

(4.1)

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of maps is zero. We will call {G1 , . . . , Gt } a nilpotent set if for each 1 ≤ i 6= j ≤ t each composition (4.1) of group maps is a nilpotent endomorphism of Gi . For instance {Zp primes p ∈ Z} is rigid. If G1 ⊂ G2 are rank one groups such that Hom(G2 , G1 ) = 0 then {G1 , G2 } is a semi-rigid set. If G1 and G2 are strongly indecomposable groups that are not quasi-isomorphic then we will show in Lemma 4.1.5 that {G1 , G2 } is a nilpotent set. For instance, let p ∈ Z be a prime and let G1 , G2 be pure b p of differing ranks. Then {G1 , G2 } is a nilpotent set. subgroups of Z The following classifies nilpotents sets in terms of a commutativity property. LEMMA 4.1.4 [34, T.G. Faticoni] Let G = G1 ⊕ · · · ⊕ Gt and for each i = 1, . . . , t let ei ∈ End(G) be the idempotent corresponding to Gi . The following are equivalent. 1. {G1 , . . . , Gt } is a nilpotent set. 2. Hom(Gi , Gj ) ⊂ N (End(G)) for each 1 ≤ i 6= j ≤ t. 3. ei is central modulo N (End(G)) for each i = 1, . . . , t. Proof: 1 ⇒ 2 Let i 6= j, let f : Gi → Gj , and let x : G → G. Then f = ej f ei . Since {G1 , . . . , Gt } is a nilpotent set ej f ei xej is a nilpotent endomorphism of Gj . For large m ∈ Z, !m t X (f x)m = ej f ei xek k=1

= (ej f ei xej )m = 0. The last expression is from the fact that (xek )(ej f ei ) = 0 for j 6= k and that ej f ei xej is nilpotent. Then f generates a nil right ideal of End(G), as required by part 2. 2 ⇒ 3 Let x ∈ End(G). Then for fixed i = 1, . . . , t P ei x = ei xei + i6=j ei xej and P xei = ei xei + i6=j ej xei . By part 2, ei xej ∈ N (End(G)) for each 1 ≤ i 6= j ≤ t, so that P P ei x − xei = i6=j ei xej − i6=j ej xei ∈ N (End(G)).

4.1. NILPOTENT SETS

73

Thus ei is central modulo N (End(G)), which proves part 3. 3 ⇒ 1 Let i 6= j and let f

g

Gi −→ Gj −→ Gi . Then gf = ei g(ej f ei ) and by part 3 0 = ej ei f ≡ ej f ei (mod N (End(G))). Thus gf ≡ 0(mod N (End(G))), hence gf is nilpotent, whence {G1 , . . ., Gt } is a nilpotent set. This completes the logical cycle. The following two results show us that it should be easy to determine when a set {G1 , . . . , Gt } of rtffr groups is nilpotent. LEMMA 4.1.5 Let G1 and G2 be strongly indecomposable rtffr groups. . ∼ Then {G1 , G2 } is not a nilpotent set iff G1 = G2 . .

Proof: Clearly {G1 , G2 } is not a nilpotent set if G1 ∼ = G2 . Conversely suppose that {G1 , G2 } is not nilpotent. There is a composition f

g

G1 −→ G2 −→ G1 such that gf is not nilpotent. By Lemma 1.3.3, QEnd(G1 ) is a local ring and since J (QEnd(G1 )) is a nilpotent ideal gf is a unit in QEnd(G1 ). Thus there is a map h ∈ End(G1 ) and an integer n 6= 0 such that (hg)f = . G1 ⊕ ker hg and since G2 is strongly indecomposable n1G1 . Then G2 ∼ = . ∼ G2 = G1 . This proves the lemma. LEMMA 4.1.6 Let {G1 , G2 } be a nilpotent set and for i = 1, 2 let Hi be a quasi-summand of Gi . Then {H1 , H2 } is a nilpotent set. Proof: Suppose that {G1 , G2 } is a nilpotent set and let H1 ⊂ G1 and H2 ⊂ G2 be quasi-summands. Consider a composition f

g

H1 −→ H2 −→ H1 of group maps. Since H2 is a quasi-summand of G2 , f : H1 → G2 . There is a map ρ : G2 → H2 and an integer m 6= 0 such that ρ(x) = mx for each x ∈ H2 . Let : H1 → G1 and π : G1 → H1 be the canonical injection/quasi-projection associated with the quasi-summand H1 such that π = n1H1 for some integer n. Because H2 is a quasi-summand of G2 (gρf )π : G1 → G1

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is an endomorphism of G1 that factors through G2 . Since {G1 , G2 } is a nilpotent set (gρf )π is a nilpotent endomorphism. Thus there is an integer k > 0 such that 0 = ((gρf )π)k+1 = (gρf π)k (gρf )π = (gρf n)k (gρf )π = nk (gρf )k+1 π. Inasmuch as π is a surjection and nk is an injection, (gρf )k = 0. Finally for each x ∈ H1 , gρf (x) = g(mf (x)) = gf (mx), so that (gf )k (mk x) = (gf )k−1 mk−1 (gρf )(x) = (gρf )k (x) = 0. Because H1 is torsion-free 0 = (gf )k (mk H1 ) = (gf )k (H1 ) which implies that gf is nilpotent. Hence {H1 , H2 } is a nilpotent set. One of the themes of this chapter is to study the direct sum decompositions of rtffr groups. With this in mind, Fitting’s Lemma leads us to an important result for rtffr groups. LEMMA 4.1.7 Let f : G → G be an endomorphism of the rtffr group G. There is an integer k > 0 such that . G = ker f k ⊕ image f k , f is nilpotent on ker f k , and f is idempotent on image f k . Proof: The given map f lifts to an endomorphism of the finite dimensional vector space QG so that by Fitting’s Lemma QG = ker f k ⊕ image f k where f is nilpotent on ker f k and f is idempotent on image f k . (See [30].) Then f k is an idempotent endomorphism of G, and hence . G = ker f k ⊕ image f k , as required by the lemma. The following result is anticipated by D.M. Arnold and E.L. Lady in [14] where they show that an rtffr group G = H ⊕ K enjoys an exchange property if H and K satisfy a property that is equivalent to assuming that {H, K} is a nilpotent set. THEOREM 4.1.8 Let {G1 , . . . , Gt } be a set of rtffr groups. The following are equivalent.

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75

1. {G1 , . . . , Gt } is a nilpotent set. 2. If 1 ≤ i 6= j ≤ t and if H is a quasi-summand of Gi and Gj then H = 0. 3. If 1 ≤ i 6= j ≤ t and if H is a strongly indecomposable quasisummand of Gi and Gj then H = 0. Proof: From the definition of nilpotent set it suffices to prove the theorem for t = 2. 1 ⇒ 2 is Lemma 4.1.6. 2 ⇒ 3 is clear. 3 ⇒ 1 Suppose that {G1 , G2 } is not a nilpotent set. There is a composition f

g

G1 −→ G2 −→ G1 such that gf is not nilpotent. By Fitting’s Lemma 4.1.7 there is an integer k > 0 such that . G1 = ker(gf )k ⊕ image (gf )k and (gf )k+1 = (gf )k . Then f (gf )k

g

G1 −−−−→ G2 −−−−→ G1 are maps such that gf (gf )k = (gf )k+1 is idempotent on image (gf )k 6= 0. But then f g image (gf )k −→ G2 −→ image (gf )k is an integer multiple of the identity map so that image (gf )k is a quasisummand of G2 . Thus part 3 is false and the proof is complete. THEOREM 4.1.9 Let G1 , . . . , Gt be strongly indecomposable . . ∼ groups such that Gi ∼ G ⇒ i = j. If G G ⊕ · · · ⊕ Gt then = j = 1 0 0 0 0 G = G1 ⊕· · ·⊕Gs for some nilpotent set {G1 , . . . , Gs } of indecomposable rtffr groups. Proof: Since G has finite rank we can write G = G01 ⊕ · · · ⊕ G0s for some indecomposable groups G01 , . . . , G0s . Suppose that i 6= j and let H be a strongly indecomposable quasi-summand of G0i and of G0j . Rewriting the summands of. G we can write H (2) as a quasi-summand of G. But by hypothesis Gi ∼ = Gj ⇒ i = j for each 1 ≤ i, j ≤ t. J´onsson’s Theorem 2.1.10 then implies that if H 6= 0 then the multiplicity of H in G1 ⊕ · · · ⊕ Gt is 1. Thus H = 0 so that {G01 , . . . , G0s } is a nilpotent set by Theorem 4.1.8.

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COROLLARY 4.1.10 Let {G1 , . . . , Gt } be a nilpotent set of rtffr (n ) (n ) groups. Then {G1 1 , . . . , Gt t } is a nilpotent set for any t-tuple (n1 , . . ., nt ) of positive integers. EXAMPLE 4.1.11 Let G be an rtffr group. J´onsson’s Theorem 2.1.10 states that there is a finite set of strongly indecomposable groups {G1 , . . . ., Gt } and integers n1 , . . . , nt > 0 such that Gi ∼ = Gj =⇒ i = j and .

(n1 )

G∼ = G1

(nt )

⊕ · · · ⊕ Gt

.

In this case Theorem 4.1.8 shows us that {G1 , . . . , Gt } is a nilpotent set. Thus each rtffr group G is quasi-isomorphic to a direct sum .

G∼ = G0 ⊕ H 0 where G0 = G1 ⊕ · · · ⊕ Gt . In particular Theorem 4.1.9 states that G01 ⊕ · · · ⊕ G0t is the direct sum of a nilpotent set {G01 , . . . , G0t }. Our interest in nilpotent sets stems from the fact that they enjoy a refinement and uniqueness property. Recall the functor A(·) from Section 2.4.2. THEOREM 4.1.12 Let {G1 , . . . , Gt } be a nilpotent set of indecomposable rtffr groups, and let G = G1 ⊕ · · · ⊕ Gt . Then 1. E(G) = E(G1 ) × · · · × E(Gt ) as rings. 2. G has a unique decomposition. Proof: 1. Let {e1 , . . . , et } ⊂ End(G) be a complete set of primitive orthogonal idempotents such that M ei (G) = Gi and ker ei = Gj . i6=j

By Lemma 4.1.4 the idempotents ei map to central idempotents e¯i in E(G) and by Theorem 2.4.4, E(G)¯ ei = A(Gi ) is a cyclic projective Emodule. Thus E(G) = E(G)¯ e1 ⊕ · · · ⊕ E(G)¯ et = A(G1 ) ⊕ · · · ⊕ A(Gt )

4.1. NILPOTENT SETS

77

as right ideals. Since the e¯i are central E(G) = A(G1 ) × · · · × A(Gt ). Thus A(Gi ) is a ring. The canonical map π : End(Gi ) → E(Gi ) taking f 7→ ei f ei has image in A(Gi ). Because {G1 , . . . , Gt } is a nilpotent set, each f ∈ A(Gi ) satisfies f ≡ ei f ei (mod N (End(G))). Thus π has image A(Gi ) and so E(Gi ) ∼ = A(Gi ). 2. Suppose that G = G01 ⊕ · · · ⊕ G0s for some indecomposable rtffr groups G0i . Let {e01 , . . . , e0s } ⊂ End(G) be a complete set of primitive orthogonal idempotents corresponding to G01 ⊕ · · · ⊕ G0s . The e0i map onto idempotents e¯0i ∈ E(G). Because e¯1 , . . . , e¯t is a complete set of primitive orthogonal central idempotents in E(G) e¯0i = e¯1 e¯0i + · · · + e¯t e¯0i is a sum of orthogonal idempotents. Since e¯01 is primitive e¯01 = e¯i e¯01 . We may assume without loss of generality that i = 1. Since e¯1 = e¯1 e¯01 + e¯1 (1 − e¯01 ) and since e¯1 is primitive we see that e¯01 = e¯1 . Iterating this process, we produce the identities e¯0 i = e¯i

for i = 1, . . . , s.

Since e¯01 + . . . + e¯0s = 1G , s = t. Therefore A(Gi ) ∼ ei = E(G)¯ e0i = A(G0i ) = E(G)¯ for each i = 1, . . . , t so that Gi ∼ = G0i by Theorem 2.4.4. This completes the proof.

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THEOREM 4.1.13 Let {G1 , . . . , Gt } be a nilpotent set of rtffr groups and let G = G1 ⊕ · · · ⊕ Gt . 1. If H ∈ Po (G) then H ∼ = H1 ⊕ · · · ⊕ Ht where for each, i = 1, . . . , t, Hi ∈ Po (Gi ). 2. If G = H ⊕ K ∼ = H ⊕ K 0 then K ∼ = K 0. Proof: 1. Given H ∈ Po (G) then A(H) is a finitely generated projective right E(G)-module. Since the e¯i are central in A(G) we can write A(H) ∼ e1 ⊕ · · · ⊕ A(H)¯ et = A(H)¯ as right E(G)-modules. By Theorem 2.4.4 there are H1 , . . . , Ht such that A(Hi ) ∼ ei = A(H)¯ for each i = 1, . . . , t so that A(H) ∼ = A(H1 ) ⊕ · · · ⊕ A(Ht ) ∼ = A(H1 ⊕ · · · ⊕ Ht ). By Theorem 2.4.4, H ∼ = H1 ⊕ · · · ⊕ Ht . Part 2 follows from Theorem 4.1.12 in the usual way, so the proof is complete. COROLLARY 4.1.14 Suppose that {G1 , . . . , Gt } is a nilpotent set of rtffr groups and suppose that (n1 )

H ⊕ H 0 = G1

(nt )

⊕ · · · ⊕ Gt

for some integers n1 , . . . , nt > 0. For each i = 1, . . . , t there are rtffr (n ) groups Hi and Hi0 such that Hi ⊕ Hi0 ∼ = Gi i and such that H ∼ = H1 ⊕ 0 0 0 · · · ⊕ Ht and H ∼ = H1 ⊕ · · · ⊕ Ht . The following result is somewhat of a surprise. THEOREM 4.1.15. Let G1 , . . . , Gt be strongly indecomposable rtffr . groups such that Gi ∼ = Gj ⇒ i = j, and let G = G1 ⊕ · · · ⊕ Gt . 1. G has a unique decomposition G = G01 ⊕ · · · ⊕ G0r . 2. If H ∈ Po (G) then H = H1 ⊕ · · · ⊕ Hs where for each i = 1, . . . , s, Hi ∈ Po (G0i ). Proof: Apply Theorems 4.1.9 and 4.1.12.

4.2. COMMUTATIVE RTFFR RINGS

4.2

79

Commutative Rtffr Rings

In this section we begin our investigation of the End(G)-module structure of rtffr groups G such that E(G) is a commutative ring. We will consider the finitely generated projective modules over rtffr commutative rings. Our techniques will produce good results on these rings.

4.2.1

Modules over Commutative Rings

Our first result gives us an easily verified condition that implies that the ring E(G) is commutative. LEMMA 4.2.1 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group, let E = End(G), let N = N (End(G)), and let k = QE/N . 1. The degree [k : Q] divides the integers (a) Q-dim(QE), (b) Q-dim(QN ), (c) Q-dim(QN G), (d) Q-dim(QG), and (e) Q-dim(QN k G/QN k+1 G) for each integer k ≥ 0. 2. If Q-dim(QE) or Q-dim(QG) is a square-free integer then k is a commutative field. Proof: 1. By the Beaumont-Pierce-Wedderburn Theorem 3.1.6 we can identify k with a subring of QE. Since QE and QG are then k-vector spaces and since QN and QN G are k-vector subspaces of QE and QG it follows that [k : Q] divides each of the integers in (a)-(e). 2. Since QE is a local ring QE/QN = k is a division ring. By part 1, the k-dimension of QE is a divisor of Q-dim(QE). Since the dimension of a division ring over its center is a perfect square (see [69]), and since Q-dim(QE) is assumed to be square-free k is commutative. A similar proof applies if Q-dim QG is square-free. This completes the proof. LEMMA 4.2.2 Let S be an Artinian commutative ring and let U be a finitely generated projective S-module. If annS (U ) = 0 then U ∼ = S ⊕ U0 0 for some S-module U .

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Proof: Let U be a finitely generated projective S-module. Since S is an Artinian commutative ring S is a finite product of local commutative rings say S = S1 × · · · × S t . Since annS (U ) = 0 U = U S1 ⊕ · · · ⊕ U St is a direct sum of nonzero projective modules. Since projective modules (n ) over local rings are free U Si ∼ = Si i for some integers n1 , . . . , nt > 0. Inasmuch as S = S1 × · · · × St , it follows that U ∼ = S ⊕ U 0 for some 0 S-module U . This completes the proof. The next result is of independent interest since it deals with the structure of right ideals in (not necessarily commutative) rtffr rings. A simple consequence of part 1 that we will not use in our deliberations is that each module of finite composition length over an rtffr ring is finitely presented. LEMMA 4.2.3 Let I be a right ideal in the rtffr ring E. 1. If I is a maximal right ideal in E then E/I is a finite p-group for some prime p ∈ Z. 2. If I ⊕ J has finite index in E for some right ideal J in E then I is finitely presented. Proof: 1. Let I be a maximal right ideal in E. Then N (E) ⊂ I and by the Beaumont-Pierce-Wedderburn Theorem 3.1.6 we have a group . decomposition E ∼ = E/N (E) ⊕ N (E). Hence E/N (E) is a reduced group. Thus we can assume without loss of generality that N (E) = 0. That is, QE is a semi-simple Artinian ring. Thus QI = eQE for some e2 = e ∈ QE. There is an integer m 6= 0 such that m(1 − e)E ⊂ E. Left multiplication by m(1 − e) induces an embedding of right E-modules E/(QI ∩ E) = E/(eQE ∩ E) ∼ = m(1 − e)E ⊂ E. Evidently, E/(QI ∩ E) is a right rtffr E-module. Furthermore since E/I is simple, either E/I is an elementary p-group for some prime p ∈ Z or E/I is divisible. Since E/I maps onto E/(QI ∩E), E/I is an elementary p-group, and since E has finite rank, E/I is finite. 2. There is an integer n 6= 0 such that n ∈ I ⊕ J. Thus (I ⊕ J)/nE is a finite right ideal in E/nE so that I ⊕ J is finitely generated. Suppose that π : P → I is a surjection where P is a finitely generated projective

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81

right E-module. Then π lifts to a surjection π : QP → QI which must split. Let f : QI → QP be the map such that πf = 1QI . Since I and P are finitely generated we may assume that f : I → P is such that . πf = m1I for some integer m 6= 0. Then P = f (I) ⊕ ker π and since P is finitely generated ker π is finitely generated. Hence I = π(P ) is a finitely presented right ideal of E which completes the proof. The following result anticipates Chapter 6 where we link the condition IG 6= G to the splitting of exact sequences involving G. The next theorem is proved in [10, Theorem 5.9] for ideals of finite index in commutative rings. We give a different proof. THEOREM 4.2.4 [10, D.M. Arnold, Theorem 5.9] Let G be an rtffr group with commutative endomorphism ring E = End(G). Then IG 6= G for each proper ideal I ⊂ E. Proof: It suffices to show that IG 6= G for each maximal proper ideal I ⊂ E. So choose a ideal I ⊂ E such that E/I is simple and such that IG = G. By Lemma 4.2.3(1) there is a prime p ∈ Z such that E/I is a finite p-group so that pE ⊂ I. Thus I/pE is a maximal ideal in the commutative Artinian ring E/pE. There is an idempotent e¯ ∈ E/pE such that I/pE = (¯ eE + J )/pE where J /pE = J (E/pE). Then G IG e¯G + J G e¯G + pG = = = pG pG pG pG by Nakayama’s Lemma 1.2.3. Since G/pG has annihilator pE in E, and since ¯1 − e¯ 6∈ pE annihilates e¯, e¯ = ¯1. Thus I/pE = E/pE and I = E. This completes the proof. The above result implies that if End(G) is commutative then Hom(G, IG) = I for each maximal ideal I ⊂ S. It does not state that Hom(G, IG) = I for each nonzero ideal I ⊂ S. It would be interesting to investigate the set {ideals I ⊂ S Hom(G, IG) = I}. LEMMA 4.2.5 Let S be an Artinian commutative ring and let M be an S-module such that annS (M ) = 0, let length(X) denote the composition length of X as a module over some subring R of S, and assume that length(S) < ∞. The following are equivalent. 1. annS (M/M I) = I for each ideal I ⊂ S and length(M ) ≤ length(S). 2. S ∼ = M as S-modules.

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Proof: Part 1 readily follows from part 2, so assume part 1 is true. Suppose that S is a local commutative Artinian ring. Since S is Artinian there is a simple ideal I 6= 0 such that length(I) = k = length(S/J (S)). By hypothesis annS (M/M I) = I. Furthermore, because M I 6= 0, k ≤ length(IM ), so that length(S/I) = length(S) − k ≥ length(M ) − length(M I) = length(M/M I). Induction on length(S) implies that S/I ∼ = M/M I. Since S is local I ⊂ J (S) so that M = xS + IM implies that M = xS by Nakayama’s Lemma 1.2.3. By hypothesis annS (x) = annS (M ) = 0 and so M ∼ = S. To complete the proof the Artinian commutative ring S satisfies S = S1 × · · · × S t for some local rings S1 , . . . , St . Then M = M S1 ⊕ · · · ⊕ M St where annSi (M Si ) = 0. If length(M ) = length(S) then length(M ) = length(M S1 ) + · · · + length(M St ). By the above argument length(M Si ) ≤ length(Si ) implies that M Si ∼ = Si . Then M ∼ S. This completes the proof. =

4.2.2

Projectives over Commutative Rings

Although the following is true for general commutative rings we will prove it for rtffr rings. We lift this proof from [75, Lemma VI.8.6]. LEMMA 4.2.6 Let S be a commutative rtffr ring and let I 2 = I be an ideal of S. Then I = eS for some e2 = e ∈ S. P Proof: Suppose that I = ni=1 xi S. Since I 2 = I for each xi there are yij ∈ I such that X xi = yij xj , j

which leads to the following system of equations. y12 x2 − ··· − y1n xn 0 = (1 − y11 )x1 − 0 = y21 x1 − (1 − y22 )x2 − · · · − y2n xn .. . 0 = yn1 x1 − yn2 x2 − · · · − (1 − ynn )xn

4.2. COMMUTATIVE RTFFR RINGS Thus the determinant det(M ) of the coefficient matrix (1 − y11 ) −y12 ··· −y1n y −(1 − y ) · · · −y2n 21 22 M = .. .. . . yn1 −yn2 · · · −(1 − ynn )

83

satisfies xi det(M ) = 0 for each i = 1, . . . , n. (See, e.g., [58, page 334335].) Thus I det(M ) = 0. However if we think of the determinant as a homogeneous polynomial of degree n then we see that it is of the form 1 − e for some e ∈ I. Then (1 − e)e ∈ (1 − e)I = 0 implies that e2 = e and so eS = I, as required to prove the lemma. Since the trace ideal I of a finitely generated projective module P is an idempotent ideal such that IP = P we have the following result. COROLLARY 4.2.7 Let S be a commutative ring and let P be a finitely generated projective S-module. There is an e2 = e ∈ S such that P = eP and (1 − e)P = 0. Proof: Let I be the trace ideal of the projective S-module P . Since = I, Lemma 4.2.6 implies that I = eS for some e2 = e ∈ S, and since IP = P , eP = eSP = IP = P . I2

COROLLARY 4.2.8 Let S be a commutative ring and let P be a finitely generated projective S-module. If annS (P ) = 0 then P is a generator over S. Proof: By Lemma 4.2.6, I = traceS (P ) = eS, and because annS (P ) = 0, (1 − e)P = 0 implies that e = 1. Thus I = S and hence P generates S. LEMMA 4.2.9 Let S be an rtffr commutative ring and let S ⊂ U ⊂ QS be a finitely generated S-module. Then U is a projective S-module iff U is an invertible fractional ideal for S. Proof: Suppose that S ⊂ U ⊂ QS is a projective S-module. By Lemma 4.2.6 there is an e2 = e ∈ S such that traceS (U ) = eS and (1 − e)U = 0. Since S ⊂ U , e = 1. That is, U generates S. Since HomS (U, S) = {q ∈ QS qU ⊂ S} = U ∗ , U U ∗ = S, whence U is invertible. Conversely suppose that S ⊂ U ⊂ QS and that U U ∗ = S. Then U generates S. Because S is commutative S ⊂ EndS (U ) ⊂ QS and it is

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known that U is then a finitely generated projective EndS (U )-module. Furthermore since S = U U ∗ EndS (U ) = EndS (U )S = EndS (U )U U ∗ = U U ∗ = S and hence U is a finitely generated projective S-module. This completes the proof. We will use the following result to construct a direct sum decomposition of a projective module. LEMMA 4.2.10 Suppose that S is an rtffr commutative ring and let U be a finitely generated projective S-module such that annS (U ) = 0. . 1. Given a nonzero ideal T = S there is a map f : U → S such that f (U ) + T = S. 2. Suppose that S is a semi-prime ring and let τ ⊂ S be a conductor . for S. If I = S is an ideal such that I +τ = S then I is an invertible ideal in S. Proof: 1. Since annS (U ) = 0, Lemma 4.2.9 implies that U (k) ∼ = S⊕U 0 for some integer k 6= 0 so that annS/T (U/U T ) = 0. Since U/U T is a finitely generated projective S/T -module Lemma 4.2.2 implies that U/U T ∼ = S/T ⊕ W for some S/T -module W . Thus there is a surjection f¯ : U → S/T. Since U is projective f¯ lifts to a map f : U → S such that f (U ) + T = S. 2. This is [10, Corollary 12.14]. We give a different proof. It suffices to prove this under the additional assumption that S is indecomposable as a ring. Localize at τ . Then Iτ + ττ = Sτ . Inasmuch as ττ ⊂ J (Sτ ), Iτ = Sτ . Hence IM ∼ = SM for each maximal ideal τ ⊂ M ⊂ S. For maximal ideals M ⊂ S such that τ 6⊂ M , SM is an integrally closed ring by Theorem 3.2.1. That is, SM is a finite product of local Dedekind . domains. Such a local ring is a pid. Since IM = SM , we see that IM ∼ = SM . Hence I is locally isomorphic to S so that I is a progenerator = invertible ideal in S. This proves the lemma.

4.2.3

Direct Sums over Commutative Rings

Now that we have disposed of the more general lemmas we will demonstrate that finitely generated projectives over commutative rtffr rings enjoy a natural direct sum decomposition.

4.2. COMMUTATIVE RTFFR RINGS

85

THEOREM 4.2.11 Suppose that S is an rtffr commutative semi-prime ring, and let U be a finitely generated projective S-module. Then U∼ = U1 ⊕ . . . ⊕ Us for some finitely generated projective ideals U1 , . . . , Us in S. Moreover, if annS (U ) = 0 then U1 can be chosen to be an invertible ideal in S. Proof: Let σ = traceS (U ). Then σ 2 = σ so by Lemma 4.2.6, σ = eS for some central e2 = e ∈ S and hence U is a projective generator over the ring eS. Thus we may assume without loss of generality that e = 1 and that annS (U ) = 0. Let R be an indecomposable factor of S and consider U R. Since U generates S, U R generates R. By Lemma 1.6.3 there is an integrally closed R and the associated conductor τ ⊂ R such that τ R ⊂ R ⊂ R ⊂ QR. By Theorem 3.2.4, R = R(τ )×Rτ , and since R is indecomposable R(τ ) = 0 and R = Rτ . Because R is an rtffr ring we can choose n so large that Rn is a reduced group. Then Lemma 3.3.5(2) implies that Rn = RnR and subsequently by Lemma 1.2.1, n ∈ J (Rn ). By Lemma 4.2.10 there is a map f : U → R such that f (U )+nR = R. Localize at n. By Lemma 1.2.1 f (U )n + nRn = Rn = f (U )n ⊃ R by Nakayama’s Lemma 1.2.3. Then mR ⊂ f (U ) ⊂ R for some integer m such that gcd(m, n) = 1. By Lemma 4.2.10(2), f (U ) is an invertible ideal in the semi-prime rtffr ring R. Thus U = f (U ) ⊕ U 0 = U 00 ⊕ U 0 for some invertible ideal U 00 in R. Since S = R1 × · · · × Rr for indecomposable factors Ri we see that U = U1 ⊕ · · · ⊕ Ur ⊕ U 00 for some projective S-module U 00 , where for each i = 1, . . . , t, Ui is an invertible ideal in Ri . An induction on the rank of U completes the proof. THEOREM 4.2.12 Let S be an indecomposable commutative rtffr ring and let U 6= 0 be a finitely generated projective S-module. Then U = U1 ⊕ · · · ⊕ Ut for some invertible ideals U1 , . . . , Ut in S. In this case U is locally isomorphic to S (r) for some integer r > 0.

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4.2.4

Commutative Endomorphism Rings

Now that we have the ring theoretic results out of the way we can apply them to structure theorems for rtffr groups with commutative endomorphism rings. Our approach will be to fix an rtffr group G and consider the semi-prime rtffr ring E(G) = End(G)/N (End(G)) giving general conditions under which E(G) is commutative. Recall the unique decomposition from Section 2.1 and the local refinement property from Section 2.5. THEOREM 4.2.13 [T.G. Faticoni] Let {G1 , . . . , Gt } be a nilpotent set of indecomposable rtffr groups such that E(Gi ) is a commutative ring for each i = 1, . . . , t, and let G = G1 ⊕ · · · ⊕ Gt . Then G has the local refinement property and G has a unique decomposition. Proof: Since {G1 , . . . , Gt } is a nilpotent set of indecomposable groups, Theorem 4.1.12(2) states that G = G1 ⊕ · · · ⊕ Gt has a unique decomposition. For each i = 1, . . . , t, A(Gi ) = E(Gi ) is indecomposable projective module. Then E(G) = E(G1 ) × · · · × E(Gt ) as rings by Theorem 4.1.12(1). Let H ∈ Po (G). By Theorem 2.4.4, A(H) is a finitely generated projective E(G)-module, and by Theorem 4.2.11 there are indecomposable projective ideals U1 , . . . , Us in E(G) such that A(H) ∼ = U1 ⊕ · · · ⊕ Us . Since U1 = U1 E(G1 ) ⊕ · · · ⊕ U1 E(Gt ) is indecomposable there is an index, say 1, such that U1 = U1 E(G1 ). Moreover, each Ui is locally isomorphic to E(Gi ). By Theorem 2.4.4, there are Hi ∈ Po (Gi ) such that Ui = A(Hi ), Hi is locally isomorphic to Gi , and H = H1 ⊕ · · · ⊕ Hs . Thus G has the local refinement property. This completes the proof. COROLLARY 4.2.14 Let G1 , . . . , Gt be strongly indecomposable rtffr . . ∼ groups such that Gi = Gj ⇒ i = j. Suppose that G = G1 ⊕ · · · ⊕ Gt . 1. If H ∈ Po (G) then there are groups Hi ∈ Po (Gi ) for each i = 1, . . . , t such that H = H1 ⊕ · · · ⊕ Ht . 2. G has a unique decomposition.

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87

THEOREM 4.2.15 . Let G1 , . . . , Gt be strongly indecomposable rtffr . groups such that Gi ∼ = Gj ⇒ i = j. Let G = G1 ⊕ · · · ⊕ Gt and assume that E(Gi ) is a commutative ring for each i = 1, . . . , t. Then G has the local refinement property and a unique decomposition. Proof: By Theorem 4.1.15, G = G01 ⊕ · · · ⊕ G0s for some nilpotent set of indecomposable groups. By hypothesis and by Theorem 4.1.12(1), QE(G) = QE(G1 ) × · · · × QE(Gt ) {G01 , . . . , G0s }

is a commutative ring, so E(G0i ) is an indecomposable commutative ring for each i = 1, . . . , t. Theorem 4.2.13 then states that G has the local refinement property and a unique decomposition. A good source of commutative rings that occur naturally is the class of rtffr E-rings. The ring E is an E-ring if the map λ : E → EndZ (E, +) sending r ∈ E to left multiplication by r, [λ(r)](x) = rx, is an isomorphism. Little is known about groups G such that End(G) is an E-ring. LEMMA 4.2.16 If E is an rtffr E-ring then E is commutative. Proof: The map λ : E → End(E, +) is an isomorphism so given a y ∈ E, the map x 7→ xy is also λ(z) for some z ∈ E. Then 1y = λ(z)(1) = z and so xy = yx for each x ∈ E. LEMMA 4.2.17 Let G be an rtffr group and let E = End(G). Suppose . that G ∼ = E. Then E is an E-ring. . . Proof: Since G = E we have End(E) = End(G) = E. The reader can show that E is an E-ring. LEMMA 4.2.18 Let G be an rtffr group with semi-prime commutative . endomorphism ring E = End(G) such that E ⊂ G ⊂ QE. Then Gp = Ep for each prime p ∈ Z. In particular G/nG ∼ = E/nE as groups for each integer n 6= 0. Proof: Since E is semi-prime Lemma 1.6.3 states that there is an integrally closed ring E such that E ⊂ E ⊂ QE and E/E is finite. Let . G = EG. Then G = G so that . . E ⊂ End(G) = E = E. Because E is integrally closed, E = End(G) and G is a flat E-module. Evidently we can assume without loss of generality that E = E and that

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G = G. Furthermore, since the integrally closed ring E is a finite product of Dedekind domains, we may further assume that E is a Dedekind domain. Let p ∈ Z be a prime and consider Ep ⊂ Gp ⊂ QE. Fix a maximal ideal pE ⊂ M ⊂ E. Since p ∈ M we see that (Ep )M = EM , and hence EM ⊂ GM ⊂ QE. Since E is a Dedekind domain, EM is a discrete valuation domain, so either GM = QE or GM = xEM for some x ∈ QE. Assume for the sake of contradiction that GM = QE. Since M is a maximal ideal in E G/M G ∼ = GM /MM GM = QE/MM QE = 0 so that G = M G. But E is commutative so by Theorem 4.2.4, G 6= M G for maximal ideals M ⊂ E. This contradiction shows us that GM 6= QE. Thus GM = xEM for some x ∈ QE so there is an integer k(M ) such that pk(M ) GM ⊂ EM . Since M = {maximal ideals M p ∈ M ⊂ E} is a finite set we can let k = sup{k(M ) M ∈ M} < ∞. Then pk GM ⊂ EM = (Ep )M

for each M ∈ M

so that pk Gp ⊂ ∩{pk GM p ∈ M ⊂ E is a maximal ideal} ⊂ ∩{EM p ∈ M ⊂ E is a maximal ideal} = Ep ⊂ Gp . by the Local-Global Theorem 1.5.1. Consequently, Gp = Ep . In particular G/pk G ∼ = E/pk E as groups for each integer k > 0. k1 Let n = p1 · · · pkt t be a product of powers of distinct primes. Then G/nG ∼ = G/pk11 G ⊕ · · · ⊕ G/pkt t G ∼ = E/nE as groups. This completes the proof.

4.3. E-PROPERTIES

4.3

89

E-Properties

From these preliminary results we begin our investigation of the properties of G as a left E-module. The first E-property that we will discuss is E-flat. The group G is E-flat if G is a flat left E-module. Given a semi-prime commutative endomorphism ring we give a necessary and sufficient condition that the rtffr group G be E-flat. THEOREM 4.3.1 [T.G. Faticoni] Let G be an rtffr group such that E = End(G) is commutative and assume that E ⊂ G ⊂ QE. The following are equivalent. 1. Hom(G, IG) = I for each ideal I of finite index in E. 2. G is an E-flat group. Proof: 1 ⇒ 2 Fix a prime p. By hypothesis annE (G/IG) = I for each ideal I of finite index in E, and by Lemma 4.2.18 Gp /pk Gp ∼ = G/pk G ∼ = E/pk E ∼ = Ep /pk Ep as groups for each integer k > 0. Then by Lemma 4.2.5, G/pk G ∼ = E/pk E as E-modules. So there is an element x ∈ Gp such that Ep x + pk Gp = . Gp . By Lemma 4.2.18, Gp = Ep is a finitely generated Ep -module so that Gp ∼ = Ep x by Nakayama’s Lemma 1.2.3. Since Ep is commutative Gp = Ep x ∼ = Ep . Thus G is locally flat and hence G is E-flat. This proves part 2. 2 ⇒ 1 Suppose that G is E-flat and fix a prime p ∈ Z. By Lemma 4.2.18, Gp is an ideal of finite index in Ep . By hypothesis Ep is Noetherian and Gp is a flat left Ep -module, so Gp is a projective ideal of finite index in the commutative ring Ep . But then by Lemma 4.2.9, Gp is invertible, say Up Gp = Ep . Let I ⊂ E be an ideal of finite index in E and let J = Hom(G, IG). Then for each prime p ∈ Z Jp = JEp = JUp Gp = JGp Up = IGp Up = IEp = Ip so that J = I by the Local-Global Theorem 1.5.1. This proves part 1 and completes the proof. The examples will help illustrate the conditions in the above result. EXAMPLE 4.3.2 Let k be a field extension of Q such that [k : Q] = 3, and let O denote the ring of algebraic integers in k. There are rational

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primes p, q such that O/pO is a field of degree 3 over Z/pZ and qO = M M 0 M 00 for distinct maximal ideals M, M 0 , M 00 in O. Since the extension k/Q is minimal OM ∩ Op = E is an E-ring. (See Appendix K.) Let E = Z + pE ⊂ G ⊂ E where G is an E-submodule of E such that G/pE has dimension 2 over . Z/pZ. Since E ⊂ G ⊂ E, E ⊂ End(G) = E, and since E is a Dedekind domain E ⊂ End(G) ⊂ E. Because [End(G)/pE : E/pE] is then a proper divisor of [E/pE : E/pE] = 3, End(G) = E. However G is not flat as follows. Since we chose G such that G/pE has Z/pZ-dimension 2, G is generated by at least 2 elements as an Emodule. Thus Gp /pE p ∼ = G/pE is generated by two elements, so that Gp requires at least two generators over Ep . By our choice of E and p, Ep is a local ring with J (Ep ) = pE p . Finitely generated flat modules over the Noetherian domain Ep are projective, hence free Ep -modules. . Thus any projective module U = Ep satisfies U ∼ = Ep . In particular U is cyclic. Since Gp is not cyclic, it is not flat over Ep , whence G is not E-flat. EXAMPLE 4.3.3 Let E be a commutative semi-prime rtffr ring that is not hereditary. (There are plenty of these. Try to construct one as we did in the previous example.) There is an ideal I ⊂ E that is not projective, hence not flat, so that I ⊕ E is not a flat left E-module. Evidently E = O(I ⊕ E) = {q ∈ QE q(I ⊕ E) ⊂ I ⊕ E}. Then by Theorem 2.3.4 there is a short exact sequence 0 → I ⊕ E −→ G −→ QE ⊕ QE → 0 such that E ∼ = End(G). The fact that QE is a flat left E-module implies that there is an isomorphism of functors Tor1E (I ⊕ E, ·) ∼ = Tor1E (G, ·). Since I ⊕ E is not flat Tor1E (I ⊕ E, ·) 6= 0 so that G is not flat. Observe that I ⊕ E ⊂ G 6⊂ QE.

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EXAMPLE 4.3.4 Let E be any noncommutative rtffr ring whose additive structure is a free group. Butler’s Construction 2.3.1 shows us that there is a group G such that QG = QE = QEnd(G). Obviously End(G) is not commutative. The next theorem illustrates a theme that we will come back to several times in this book. We show that a number of the E-properties that we are examining agree on rtffr groups that have semi-prime commutative endomorphism rings. Notice that the endomorphism ring in the next result need not be commutative or semi-prime. THEOREM 4.3.5 Let G be an rtffr group with endomorphism ring E = End(G). If E ⊂ G ⊂ QE then the following are equivalent. .

1. G ∼ = E as groups. 2. G is an E-finitely generated group. 3. G is an E-finitely presented group. In this case End(G) is an E-ring. Proof: 3 ⇒ 2 is clear. 2 ⇒ 1 Say x1 , . . . , xn generate G as an E-module. Since E ⊂ G ⊂ QE there is an integer m 6= 0 such that mxi ∈ E so that mG ⊂ E ⊂ G. . 1 ⇒ 3 Since G = E, Lemma 4.2.3 states that G is finitely presented. This proves 3 and completes the logical cycle. We say that G is locally E-free iff Gp ∼ = Ep as E-modules for each p 6= 0 ∈ Z. The Local-Global Theorem 1.5.1 can be used to show that G is E-flat if G is locally E-free. The converse holds in the present situation. THEOREM 4.3.6 Let G be an rtffr group such that E = End(G) is commutative, and suppose that E ⊂ G ⊂ QE. Then G is an E-flat group iff G is locally E-free. Proof: If G is locally E-free then the above comment shows that G is E-flat. Conversely suppose that G is E-flat. Fix a prime p ∈ Z and choose by Lemma 3.3.5 a multiple n of p so large that En is a reduced group. By . Lemma 4.2.18, Gn = En , so by Lemma 4.2.3, Gn is a finitely presented ideal in En . Lemma 4.2.10 states that there is a map π : Gn → En such

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that π(Gn ) + nEn = En . By Lemma 1.2.1, n ∈ J (En ), so Nakayama’s Lemma 1.2.3 implies that π(Gn ) = En . Inasmuch as Gn has finite rank, π is an isomorphism. Thus G is locally E-free. The group G is E-projective if G is a projective left E-module, G is an E-generator if G generates the category of left E-modules, and G is an E-progenerator if G is E-finitely generated, E-projective, and an E-generator. The right ideal I ⊂ E is invertible if II ∗ = I ∗ I = E where I ∗ = {q ∈ QE qI ⊂ E}. These E-properties coincide when End(G) is a commutative semi-prime ring. THEOREM 4.3.7 Let G be an rtffr group with commutative semiprime endomorphism ring E = End(G). If E ⊂ G ⊂ QE then the following are equivalent. 1. G is an E-projective group. 2. G is an E-generator group. 3. G is an E-progenerator group. 4. G is an invertible ideal in End(G). In this case G is E-finitely presented and End(G) is an E-ring. Proof: 4 ⇒ 3 An invertible ideal in a commutative ring is a progenerator. 3 ⇒ 2 is clear. 2⇒ 1 If G is a generator over E then G is projective as a module over EndE (G). But EndE (G) = center(E) = E since E is commutative. Thus G is E-projective. 1 ⇒ 4 By hypothesis G is a projective E-module such that E ⊂ G ⊂ QE. There is a cardinal c and a direct sum E (c) ∼ = G ⊕ K of E-modules. Let {x1 , . . . , xn } ⊂ G be a maximal linearly independent subset of G. There is a finite direct sum E (m) that contains the xi so G ⊂ E (m) . Since E is semi-prime, E is Noetherian whence G is finitely generated. Since E ⊂ G ⊂ QE, G/E is a finitely generated torsion E-module, so that G/E is finite. Hence G is a finitely generated projective ideal of finite index in E. Thus G is invertible which proves part 4 and completes the logical cycle. EXAMPLE 4.3.8 Let G be the groups constructed in Example 4.3.2. Then End(G) is a commutative nonhereditary integral domain. Observe . that G = End(G) is finitely generated but that G is neither E-projective nor an E-generator.

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EXAMPLE 4.3.9 (See [43].) Let E be any ring whose additive structure (E, +) is a (locally) free group. Then E = End(G) for some group G such that E ⊂ G ⊂ QE. In fact the reader can prove that Gp ∼ = Ep for each prime p ∈ Z so that G is an E-flat group. But G is not E-finitely generated and not E-projective.

4.4

Square-Free Ranks

We investigate the strongly indecomposable rtffr groups whose rank is a square-free integer. We show that many E-properties coincide for these groups and that direct sums of nilpotent sets of these groups have the local refinement property.

4.4.1

Rank Two Groups

By a Theorem of J.D. Reid’s below [10, Theorem 3.3], a strongly indecomposable torsion-free group of rank two has a commutative endomorphism ring. Here is a ready-made interesting class of rtffr groups to which our techniques will provide good results. We will refine the work in the previous section by proving the following result. Given a subgroup X ⊂ Y , let X∗ be the pure subgroup of Y generated by X. THEOREM 4.4.1 [43, T.G. Faticoni and H.P. Goeters] A strongly indecomposable torsion-free group of rank two is E-flat. The proof of this theorem depends on the following classification of strongly indecomposable rank two groups. THEOREM 4.4.2 [J.D. Reid] Let G be a strongly indecomposable torsion-free group of rank two. Then QEnd(G) satisfies one of the following conditions. 1. QEnd(G) = Q. 2. QEnd(G) is a field extension of Q of degree 2. 3. QEnd(G) = Q[x]/(x2 ) where x is an indeterminant and where (x2 ) is the ideal generated by x2 . In any case End(G) is a commutative ring.

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Proof of Theorem 4.4.1: Let E = End(G). By Theorem 4.4.2, E is a commutative ring, so that QE is a local commutative ring and QG is a QE-module. If QE = Q then E is a pid and hence G is a flat left E-module. ∼ QG as QEIf QE is a field extension of Q of degree 2 then QE = modules since G has rank two. Thus we can assume without loss of generality that E ⊂ G ⊂ QE as E-modules. We will show that G is locally flat. Let M ⊂ E be a maximal ideal in E and let p be the unique prime integer in M . Since pEM ⊂ MM ⊂ EM and since EM /pEM has Zp /pZp dimension at most 2, either pEM = MM or pEM 6= MM ⊂ EM . Suppose that pEM = MM . Then EM ⊃ pEM ⊃ p2 EM ⊃ · · · is a composition series for EM so that EM is a discrete valuation domain. Hence GM is flat as an EM -module. . Otherwise pEM 6= MM ⊂ EM . By Lemma 4.2.18, GM = EM , so that GM /pGM ∼ = EM /pEM as groups. Since MM /pEM is the only proper nonzero ideal in the commutative ring EM /pEM , and since GM 6= MM GM (Nakayama’s Lemma 1.2.3), we see that MM /pEM = annEM /pEM (GM /MM GM ). Thus Lemma 4.2.5 states that GM /MM GM = EM (x+MM GM ) for some x ∈ GM . Another application of Nakayama’s Lemma 1.2.3 shows us that GM = EM x + MM GM = EM x. Since annEM (x) = annEM (GM ) = 0, GM ∼ = EM is a flat left E-module. The last case to consider is QE = Q[x]/(x2 ). Observe that QG is a QE-module of composition length 2. Since annQE (QG) = 0 Lemma 4.2.5 shows us that QE = QG so that E ⊂ G ⊂ QE. Let N = N (E)G and notice that since N (E)2 = 0, N (E) = Hom(G/N∗ , G). Since G/N∗ and N are rank one groups N = Hom(G/N∗ , G)G/N∗ is a pure subgroup of G. Thus given a prime p ∈ Z such that pG 6= G, we have Gp /Np ∼ = Zp , so that Gp = Ep x + Np = Ep x + N (E)p Gp . In the by now familiar manner Gp = Ep x is a flat E-module. In any case G is a locally flat E-module so that G is E-flat. This completes the proof. An appeal to the above theorem and Theorems 4.3.5 and 4.3.7 will prove the following result.

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THEOREM 4.4.3 Let G be a strongly indecomposable torsion-free group of rank two. The following are equivalent. .

1. G ∼ = End(G) as groups. 2. G is an E-finitely generated group. 3. G is an E-finitely presented group. 4. G is an E-projective group. 5. G is an E-generator group. 6. G is an E-progenerator group. 7. G is an invertible ideal in End(G). If one and hence all of the above conditions is true then End(G) is an E-ring and QEnd(G) is a field extension of Q of degree 2. The next example addresses the hypotheses of the above theorem. EXAMPLE 4.4.4 Let k = Q[α] be a field extension of Q of degree 2 and let E denote the ring of algebraic integers in k. Then (E, +) is a free group of rank 2. By a construction of M.C.R. Butler 2.3.1 there is a group E ⊂ G ⊂ k = QE such that E = End(G). We observe that E is not an E-ring, G and E are not quasi-isomorphic, but G is an E-flat group.

4.4.2

Groups of Rank ≥ 3

By Lemma 1.3.3, G is strongly indecomposable iff QEnd(G) is a local ring. In this case QE(G) = QEnd(G)/N (QEnd(G)) is a division ring. Because QE(G) is commutative if rank(G) is square-free (Lemma 4.2.1) and since we can pass from G to E(G) using the functor A(·) and Theorem 2.4.4, we can study G by applying some commutative ring theory to E(G). Our theorems can be applied to strongly indecomposable rtffr groups of rank three. Compare the next result to Reid’s Theorem 4.4.2. THEOREM 4.4.5 Let G be a strongly indecomposable torsion-free group of rank 3, let E = End(G), and let N = N (QEnd(G)). Then QE falls into one of the following classes. 1. N = 0 and QE is a field extension of Q of degree 1 or 3.

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CHAPTER 4. COMMUTING ENDOMORPHISMS 2. N 6= 0 and Q-dim(QE) = 3. Then QE/N = Q. Moreover (a) If Q-dim(QE) = 3 and N 2 6= 0 then QE ∼ = Q[x]/(x3 ). (b) If Q-dim(QE) = 3 and N 2 = 0 then N = Q ⊕ Q as ideals in QE. 3. N 2 6= 0, Q-dim(QE) ≥ 4, and QE/N = Q.

In any case QE/N is a field extension of Q of degree 1 or 3. Proof: Suppose that rank(G) = 3. Because G is strongly indecomposable of square-free rank, Theorem 1.3.3 states that QE/N = k is a field extension of Q whose degree [k : Q] divides QG and N QG. 1. Suppose that N = 0. Then k = QE is a field extension of Q whose degree [k : Q] divides Q-dim(QG) = 3. Thus [k : Q] = 1 or 3. 2. Suppose that N 6= 0 and that Q-dim(QE) = 3. Then [k : Q] < 3 is a proper divisor of Q-dim(QG) = 3 (Lemma 4.2.1), so that [k : Q] = 1. Thus k = Q. Consider the following cases. 2(a). Suppose that N 2 6= 0. Then 0 ⊂ N 2 ⊂ N ⊂ QE is a chain of distinct subspaces of QE which shows that N /N 2 ∼ = Q as Q-vector spaces. By Nakayama’s Lemma 1.2.3, N = QEy for some y ∈ QE such that y 2 6= 0 and y 3 = 0. Evidently there is a surjection f : Q[x] → Q[y] such that f (x) = y with ker f = (x3 ) = the ideal generated by x3 . Then a comparison of dimensions shows us that Q[x]/(x3 ) ∼ = Q[y]. 2 2(b). Assume that Q-dim(QE) = 3 and that N = 0. Then k ∼ =Q implies that the ideal structure of N is that of a two dimensional Qvector space, say N ∼ = Qx ⊕ Qy as ideals in QE. (Note: QE need not be commutative. See the example below.) 3. Assume that N 2 6= 0 and that Q-dim(QE) ≥ 4. By Lemma 4.2.1, [k : Q] divides Q-dim(QG) = 3 and Q-dim(N QG) < 3, so [k : Q] = 1. This completes the proof. EXAMPLE 4.4.6 Choose a commutative ring E whose additive structure is a free abelian group of rank 3 and such that QE is a local ring. For example, let E be the algebraic integers in a field extension k of degree 3 over Q. By Butler’s Construction in Theorem 2.3.1, E = End(G) for some group G such that E ⊂ G ⊂ QE. G is strongly indecomposable because QE is a local ring.

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EXAMPLE 4.4.7 We construct a strongly indecomposable rank 3 group G such that End(G) is a noncommutative ring of rank 3. Let E = R/I where 0 0 0 a 0 0 R = b a 0 a, b, c, d ∈ Z and I = 0 0 0 . d c a Z 0 0 Then E is a ring such that (E, +) is a free group of rank 3. By Butler’s Construction in Theorem 2.3.1, there is a group G such that QE = QEnd(G) and E ⊂ G ⊂ QE. Since QE is a local ring G is strongly indecomposable. EXAMPLE 4.4.8 We construct a strongly indecomposable rank 3 group G such that E = End(G) is a noncommutative ring of rank 4. Let M = R/I where 0 0 0 a 0 0 R = b a 0 a, b, c, d ∈ Z and I = 0 0 0 . 0 Z 0 d c a Then M = R/I is a left R-module such that annR (M ) = 0, and whose additive structure is a free group of rank 3. A slight variation of Butler’s Construction 2.3.1 produces a group G such that M ⊂ G ⊂ QM and such that QR ∼ = QEnd(G). Thus rank(R) = 4, rank(G) = 3, and because QR is a local ring, G is strongly indecomposable. For groups of rank 4 or 5 we have the following theorems. The ring of Hamiltonian quaternions over Q is the Q-vector space H with basis {1, i, j, k} where we define i2 = j 2 = k 2 = ijk = −1. Extend the multiplication linearly to give H a ring structure. Since ij = −ji the Hamiltonian Quaternions form a noncommutative domain. THEOREM 4.4.9 Let G be a strongly indecomposable torsion-free group of rank 4, let E = End(G) and let k = QE/N (QE). Then k is a field extension of Q of degree 1, 2, or 4. 1. If [k : Q] ≤ 2 then k is a commutative field. 2. If [k : Q] = 4 then either k is a commutative field or k is H = the noncommutative ring of Hamiltonian Quaternions over Q. Proof: The proof is a repeated application of Lemma 4.2.1 and is left as an exercise for the reader.

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EXAMPLE 4.4.10 Let E be a maximal Z-order in the ring of Hamiltonian Quaternions. Then (E, +) is a free group of rank 4 and E is a noncommutative domain. By Butler’s Construction in Theorem 2.3.1, E = End(G) for some group G such that E ⊂ G ⊂ QE. Compare to item 2 in the above theorem. THEOREM 4.4.11 Let G be a strongly indecomposable torsion-free group of rank 5, let E = End(G), and let k = QE/N (QE). 1. k is a commutative field of degree 1 or 5 over Q. 2. If N 6= 0 then k = Q is commutative.

THEOREM 4.4.12 Let G be a strongly indecomposable torsion-free group of rank p for some prime p ∈ Z, let E = End(G), and let k = QE/N (QE). Then 1. k is a commutative field of degree 1 or p over Q. 2. If N 6= 0 then k = Q is commutative. EXAMPLE 4.4.13 An example of End(G) for a strongly indecomposable rank 5 group. Let R be a noncommutative ring such that (R, +) is a free abelian group of rank 4 and let E be the ring (z, r) z ∈ Z, r ∈ R where (z, r)(z 0 , r0 ) = (zz 0 , rz 0 + zr0 + rr0 ). Then E is an associative ring with identity (1, 0). By Butler’s Construction in Theorem 2.3.1 there is a group E ⊂ G ⊂ QE such that QE ∼ = QEnd(G). Since QE is a local ring of Q-dimension 5, G is a strongly indecomposable group of rank 5. The indecomposable group G has the refinement property if given H ∈ Po (G) then H ∼ = Gm for some integer m > 0. THEOREM 4.4.14 A strongly indecomposable rtffr group having prime rank and having nonzero nilradical has the refinement property. Proof: Let H ∈ Po (G). Recall the functor A(·) from Theorem 2.4.4. Then A(H) is a finitely generated projective E(G)-module. By the above Theorem, QE(G) = Q when rank(G) = p is prime and N (End(G)) 6= 0, so that E(G) is a pid. Hence A(H) is a free E(G)-module of rank, say, m, whence H ∼ = G(m) . This completes the proof.

4.5. REFINEMENT AND SQUARE-FREE RANK

4.5

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Refinement and Square-Free Rank

We have shown in Lemma 4.2.1 that E(G) is a commutative ring for strongly indecomposable groups of square-free rank. Then we can use the machinery that we have developed for direct sum decompositions of projective modules over commutative rtffr rings. The following results extend the Baer-Kulikov-Kaplansky Theorem 2.1.11 to direct sums of strongly indecomposable groups with square-free ranks. Recall unique decomposition from page 25 and the local refinement property from page 45. We give an extensive class of rtffr groups that have a locally unique decomposition. THEOREM 4.5.1 [T.G. Faticoni] Let G1 , . . . , Gt be strongly indecom. posable rtffr groups of square-free rank such that Gi ∼ = Gj ⇒ i = j. If . G = G1 ⊕ · · · ⊕ Gt then G has the local refinement property and G has a unique decomposition. Proof: By Lemma 4.2.1, E(Gi ) is commutative for each i = 1, . . . , t, so by Theorem 4.2.15, G has the local refinement property and a unique decomposition. Of course primes are square-free. COROLLARY 4.5.2 Let G1 , . . . , Gt be strongly indecomposable rtffr . . groups of prime ranks p1 , . . . , pt such that Gi ∼ = Gj ⇒ i = j. If G = G1 ⊕ · · · ⊕ Gt then G has the local refinement property and G has a unique decomposition. COROLLARY 4.5.3 Let G1 , . . . , Gt be strongly indecomposable rtffr . groups of differing prime ranks. If G = G1 ⊕ · · · ⊕ Gt then G has the local refinement property and G has a unique decomposition. Notice in the next result that although we cannot conclude that G has a unique decomposition we can say that G has a locally unique decomposition. THEOREM 4.5.4 Let G1 , . . . , Gt be strongly indecomposable . ∼ rtffr groups of square-free rank such that Gi = Gj ⇒ i = j, and let G = G1 ⊕ · · · ⊕ Gt .

(4.2)

(Note the equation here.) Then G has the local refinement property and a locally unique decomposition.

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Proof: Suppose that G has a direct sum decomposition (4.2). By Theorem 4.5.1, G1 ⊕· · ·⊕Gt has the local refinement property. The local refinement property implies a locally unique decomposition by Lemma 2.5.10. This completes the proof. THEOREM 4.5.5 Let H be a strongly indecomposable rtffr group of square-free rank, let n > 0 be an integer, and let G = H (n) . 1. Each K ∈ Po (G) is a direct sum K = K1 ⊕ · · · ⊕ Ks where each Ki is locally isomorphic to H. 2. H (n) is a locally unique decomposition of G. The following examples will show that in some sense our results are the best possible. EXAMPLE 4.5.6 There is a set {G1 , G2 } of quasi-isomorphic rank two groups such that G = G1 ⊕ G2 does not have the local refinement property. Theorem 2.2.1 states that there are acd groups G1 , G2 , G3 , G4 , such that 1. G = G1 ⊕ G2 ∼ = G3 ⊕ G4 2. rank(G1 ) = rank(G2 ) = 2, rank(G3 ) = 1, rank(G4 ) = 3. Evidently {G1 , G2 } is not a nilpotent set. Such a group G fails to have the local refinement property. Examination of the construction of G1 and G2 shows that E(Gi ) = Z is commutative for i = 1, 2. EXAMPLE 4.5.7 There is a strongly indecomposable torsion-free group of rank 4 that does not have the local refinement property. Proof: Let O denote a classical maximal Z-order in the ring of Hamil∼ tonian Quaterions over Q. Then O/3O = M2 (Z/3Z). Let 3O ⊂ E ⊂ O Z/3Z 0 be such that E/3O ∼ . It is known (see, e.g., [69]) that = Z/3Z Z/3Z E is a hereditary Noetherian primering. Let I be the ideal 3O ⊂ I ⊂ E 0 0 such that I/3O = . Then I 2 = I is a projective ideal. Z/3Z Z/3Z Z/3Z 0 Since I/3O does not generate , I is not a generator for 0 0 E. Therefore I is not locally isomorphic to E. It is clear from the above construction that I is generated by two elements so we can write E⊕E ∼ = I ⊕ J.

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101

By Corner’s Theorem 2.3.3, there is a group G such that E = End(G). Then by the Arnold-Lady-Murley Theorem 2.4.1, I ⊗E G ∼ = IG ∈ Po (G), G ⊕ G ∼ = IG ⊕ JG, and IG is not locally isomorphic to G. Thus G does not have a locally unique decomposition. The reader can show that End(IG) = EndO (I) = O. The above example illustrates the type of failure of local refinement that occurs whenever we are given an rtffr ring E and a finitely generated projective right ideal I that does not generate E. Such a projective right ideal I always exists in a hereditary rtffr ring E that is not integrally closed.

4.6

Hereditary Endomorphism Rings

There is another interesting class of rtffr groups G for which each H ∈ Po (G) is a direct sum of strongly indecomposables. THEOREM 4.6.1 Let {G1 , . . . , Gt } be a set of strongly indecompos. able rtffr groups such that Gi ∼ = Gj =⇒ i = j. Assume that E(Gi ) is a hereditary domain for each i = 1, . . . , t and let G = G 1 ⊕ · · · ⊕ Gt . Given H ∈ Po (G) there are integers m1 , . . . , mt ≥ 0 and indecomposable groups Hij ∈ Po (Gi ) such that .

Hij ∼ = Gi and such that

for each j = 1, . . . , mi

m1 mt M M H=[ H1j ] ⊕ · · · ⊕ [ Htj ]. j=1

j=1

Proof: Let H ∈ Po (G). Since {G1 , . . . , Gt } is a nilpotent set there are Hi ∈ Po (Gi ) such that H = H 1 ⊕ · · · ⊕ Ht so we may as well assume that G = G1 and that H = H1 . Under these reductions H ∈ Po (G) and G is strongly indecomposable. Since H ⊕. H 0 ∼ = G(n) for some integer n, J´onsson’s Theorem 2.1.10 (m) ∼ states that H = G for some integer m > 0. Then .

A(H) ∼ = E(G)(m)

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as right E(G)-modules. Since E(G) is a hereditary domain A(H) ∼ = U1 ⊕ · · · ⊕ Um .

where Ui ∼ = E(G) for each i = 1, . . . , m. By Theorem 2.4.4 we can write H∼ = H1 ⊕ · · · ⊕ Hm where A(Hi ) ∼ = Ui for each i = 1, .. . . , m. Finally since A(·). preserves quasi-isomorphism classes, A(Hi ) ∼ = G. This = E(G) implies that Hi ∼ completes the proof.

4.7

Exercises

Let E be an rtffr ring, let G, H be rtffr groups. 1. Let E be a ring and let P be a right E-module. Then P is finitely generated projective over E iff P is a generator in the category of left EndE (P )-modules. 2. Let M be a finitely presented right E-module. Show that M is a projective right E-module iff MI is a projective right EI -module for each maximal ideal I ⊂ S = center(E). √ 3. Let O denote the ring of algebraic integers in the field Q[ −5]. √ (a) Show by brute force that the ideal I = h2, 1 + −5i in O is a maximal nonprincipal ideal. (b) Show that 2O = I12 . (c) Conclude that I⊕I ∼ = O⊕O so that O has the local refinement property but not the refinement property. 4. Let E be the ring constructed in Example 4.5.7 and let I be the maximal ideal. (a) Show that E is hereditary. (b) Show that I is an ideal generated by two elements. (c) Show that I is not a generator. 5. Let E be a hereditary prime rtffr ring and let E be a classical maximal order such that E ⊂ E and E/E is finite. (a) Show that E is an ideal over E that is projective but not a generator over E.

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(b) Show that there is an idempotent ideal ∆ of finite index in E such that O(∆) = E. (c) Show that the hereditary prime rtffr ring E is a classical maximal order iff E and 0 are the only idempotent ideals in E. 6. Show that if E is an integral domain then E and 0 are the only idempotent ideals in E. 7. Construct an indecomposable rtffr subring E ⊂ Q × Q such that each finitely generated projective E-module is a generator for E and E ∼ = Z ⊕ Z as group. 8. Let S be a commutative rtffr ring and let U be a finitely generated ideal of finite index in S. Show that U is projective iff U is invertible iff Up ∼ = Sp for each prime p. 9. Let S be a commutative rtffr ring and let U be a finitely generated ideal of finite index in S. Show that U is invertible iff U is locally isomorphic to S. 10. Let S be a discrete valuation domain with field of fractions QS. If S ⊂ M ⊂ QS and if M 6= QS then M = Sx for some x ∈ QS. 11. Prove Fitting’s Lemma. If G is a left E-module of finite composition length and if f ∈ End(G) then there is an integer k > 0 such that G = ker f k ⊕ image f k where f is a nilpotent endomorphism of ker f k and f is an idempotent endomorphism of image f k . 12. If G is a torsion-free reduced p-local group of rank n and if rp (G) = (n) n then G ∼ = Zp . 13. Let E be a commutative ring and let M be a projective E-module. Then M generates E.

4.8

Questions for Future Research

Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. Give a general theory of nilpotent sets for infinite sets of groups or modules. See [34].

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2. Extend the theory in chapter 4 on the direct sum properties of . nilpotent sets G = G1 ⊕ · · · ⊕ Gt to direct sums in which Gi = Gj for each i 6= j. See [13]. 3. Find a condition on G, more general than square-free rank, that implies commutativity in QEnd(G) or QE(G). 4. Let S be a commutative rtffr ring. Discuss the localization of S at its maximal ideals and at multiplicatively closed sets contained in S. 5. Characterize the rtffr groups G that possess the (local) refinement property. 6. Characterize the rtffr groups G that possess the (locally) unique decomposition property. 7. Discuss the direct sum of strongly indecomposable rank two groups. These rank two groups have commutative endomorphism rings. 8. Just what is an E-flat group? 9. Discuss the structure of strongly indecomposable groups G of square-free rank. Extend that discussion to a large class of rtffr groups. 10. Discuss the local-global theory of (almost completely decomposable) rtffr groups. 11. Give a general ideal lattice for the nilradical of a strongly indecomposable rtffr group. Specifically, generalize Theorems 4.4.2,4.4.5, and 4.4.9 to larger ranks. 12. Find some kind of localization theory between the group G and S, the center of End(G). 13. Discuss the decomposition of groups G such that E(G) is hereditary. 14. Let G be an rtffr group. If E(G) is a hereditary ring then how does the ideal structure of E(G) affect G. For example, each right ideal is projective in E(G). There are at most finitely many isomorphism classes of right ideals in E(G). At most finitely many maximal right ideals are not progenerators for E(G). Discuss how these properties affect the properties of direct summands of G.

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15. Extend the results in chapter 4 to include rtffr groups G such that E(G) is integrally closed. That is, E(G) is a product of classical maximal orders.

Chapter 5

Refinement Revisited Recall that an indecomposable rtffr group G has the refinement property if given a direct sum G(n) ∼ = H ⊕ H 0 of groups then H ∼ = G(m) for some integer m > 0. We showed in Theorem 4.2.13 that if E(G) is a commutative ring then the indecomposable rtffr group G possesses the local refinement property. That is, given a direct sum G(n) ∼ = H ⊕ H 0 of groups then H ∼ = G1 ⊕ · · · ⊕ Gm for some groups G1 , . . . , Gm that are locally isomorphic to G. This brings two questions to mind.

1. If G has the local refinement property under what conditions does G have the refinement property?

2. If G has the local refinement property under what conditions does a subgroup H of finite index in G have the local refinement property?

We will investigate these problems in this chapter. For example, let G be a group such that E(G) has finite index in a Dedekind domain E. Then G has the local refinement property. But if E is not a pid then we will show that G has the local refinement property but it does not have the refinement property. By the Baer-Kulikov-Kaplansky Theorem 2.1.11, if G is a completely decomposable group then G has the refinement property. However Example 2.2.1 shows that it is not uncommon for a subgroup G of finite index in G to fail to possess the local refinement property. We will give conditions on G under which subgroups of finite index in G have the refinement property when G has the refinement property. 107

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CHAPTER 5. REFINEMENT REVISITED

Counting Isomorphism Classes Class Groups and Class Numbers

The integral closure E of the integral domain E is the largest subring E ⊂ QE such that E ⊂ E and such that E/E is finite. If E = E then E is an integrally closed ring. For example, each subring of Q is integrally closed. If i2 = −1 and if E = {a + 2bi a, b ∈ Z} then the reader can show that E = Z[i]. Given a Dedekind domain E and a nonzero ideal I ⊂ E then the integral closure of E = I + Z is E. It is known that the integral closure E of an rtffr integral domain E is a Dedekind domain, [10, Corollaries 10.12, 11.4].

For the remainder of this section we let E be an rtffr integral domain with integral closure E. The nonzero ideal I ⊂ E is invertible if I ∗ I = II ∗ = E where I ∗ = {q ∈ QE qI ⊂ E}.

By Lemma 4.2.9 a nonzero ideal I in the rtffr integral domain E is invertible iff it is projective. By Theorem 4.2.12, each finitely generated projective module over the commutative Noetherian integral domain E is a direct sum of invertible ideals. Fix a strongly indecomposable rtffr group G such that E(G) is a commutative ring, or equivalently an rtffr integral domain. This is the case, e.g., when rank(G) is a square-free integer. Theorem 4.2.13 shows us that G has the local refinement property. We define

Γ(G) = {isomorphism classes (H) G is locally isomorphic to H}

and we define the class number of G as

h(G) = card(Γ(G)).

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109

If E is an integral domain then a fractional ideal of E is aP nonzero Esubmodule of the field of fractions of E. If we define IJ = { finite xi yi xi ∈ I and yi ∈ J} for any fractional ideals I, J of E then γ(E) = the set of invertible fractional ideals of E is a group with I −1 = I ∗ and identity E. Let ρ(E) ⊂ γ(E) denote the set of principal fractional ideals of E. Evidently ρ(E) is a subgroup of γ(E). We define the class group of E to be Γ(E) = γ(E)/ρ(E). It is readily shown that fractional ideals I and J are isomorphic iff there is a q ∈ QE such that qI = J. Then the coset Iρ(E) in Γ(E) is the isomorphism class (I) of I. Thus Γ(E) = {isomorphism classes (I) I is an invertible ideal in E} and we define the class number of E to be h(E) = card(Γ(E)). The integral domain E is a Dedekind domain so each ideal in E is invertible. Thus Γ(E) is the set of isomorphism classes of nonzero ideals in E, and h(E) is the number of isomorphism classes of nonzero ideals in E. The next result shows us that h(G) is finite. The proof is due to E.L. Lady and can be found in [10, Theorem 11.11, page 137]. THEOREM 5.1.1 [E.L. Lady] Let G be an rtffr group. There are at most finitely many isomorphism classes of groups H that are locally isomorphic to G. Consequently h(G) < ∞. THEOREM 5.1.2 [Jordan-Zassenhaus] (See [10, Theorem 13.13, page 155].) Let E be an rtffr integral domain. There are at most finitely many isomorphism classes of nonzero ideals in E. Consequently h(E) < ∞.

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CHAPTER 5. REFINEMENT REVISITED Recall that the conductor of E into E is τ = the largest ideal τ in E such that τ E ⊂ E ⊂ E .

From section 3.2, τ is an ideal of E contained in E. Our investigations will relate the local refinement property in strongly indecomposable rtffr groups G to the class number of the integral closure E of E(G). LEMMA 5.1.3 Let G be an rtffr group. Then h(G) = h(E(G)). Proof: Given a group H that is locally isomorphic to G, Lemma 2.5.6 implies that G ⊕ G ∼ = H ⊕ H 0 for some group H 0 . Then H ∈ Po (G). By Lemma 2.5.1, A(H) is locally isomorphic to A(G). Then by Lemma 2.4.4, A(·) induces a functorial bijection α : Γ(G) → Γ(E(G)). Thus h(G) = h(E(G)). The reader will fill in the elementary details. Our investigation of h(G) for the rtffr group G is then translated into an investigation of h(E) for an rtffr integral domain E.

5.1.2

Class Group of the Integral Closure

The integral domain E is an rtffr integral domain if its additive structure (E, +) is an rtffr group. In this section E denotes an rtffr integral domain with integral closure E. We will prove the following result. THEOREM 5.1.4 [T.G. Faticoni] Let E be an rtffr integral domain. Then γ(E) is a free abelian group, and there is a group surjection α : γ(E) −→ γ(E). Thus γ(E) ∼ = γ(E) ⊕ ker α. Proof: Each fractional ideal I is uniquely the product of powers of finitely many maximal ideals of E, [69, page 47, Theorem 4.8]. Thus γ(E) is a free group on the set of maximal ideals of E. Define α(J) = JE for each J ∈ γ(E). Suppose that J is an invertible ideal in E. Because QE is commutative and because J ∗ J = E, J ∗ (JE) = E. Thus JE is an invertible ideal in E, hence α is well defined. Inasmuch as QE is commutative, we have IJE = IE · JE, for any ideals I, J ⊂ E, so that α is a group homomorphism. It remains to show that α is a surjection. The proof is a series of lemmas. LEMMA 5.1.5 Let T 6= 0 be an ideal in E and let I be an invertible ideal in E. There is an ideal I 0 ⊂ E such that I 0 ∼ = I and I 0 + T = E.

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111

. Proof: Since E is an integral domain T = E, and by Lemma 4.2.9, I is a projective E-module. Then by part 1 of Lemma 4.2.10 there is a map f : I → E such that f (I) + T = E. Since E is an integral domain f is an injection so letting I 0 = f (I) ∼ = I completes the proof. LEMMA 5.1.6 Let E be an rtffr integral domain with conductor τ and integral closure E. Let J ⊂ E be an ideal such that J + τ = E. 1. If τ ⊂ M ⊂ E is a maximal ideal then JM = EM . 2. If τ 6⊂ M ⊂ E is a maximal ideal then τM = EM = E M . Thus JM is a principal ideal in E M . 3. J is an invertible ideal in E. Proof: 1. Given a maximal ideal τ ⊂ M ⊂ E then JM + τM = EM and τM ⊂ MM = J (EM ). By Nakayama’s Lemma 1.2.3, EM = JM . 2. Given a maximal ideal M ⊂ E such that τ 6⊂ M then τ + M = E so that τM + MM = EM . Since MM = J (EM ), τM = EM , and so the inclusion τ E ⊂ E ⊂ E becomes E M = EM E M = τM E M ⊂ EM ⊂ E M . Thus EM = E M . It follows that JM is a projective ideal in the Dedekind domain E M . Since E M is then a local domain, JM is free, and since QJM ⊂ QE M , QJM ⊕ V = QE M for some vector space V over QE M . Inasmuch as the field QE M is indecomposable JM ∼ = EM . 3. In parts 1 and 2 we have shown that JM is a projective ideal in EM for each maximal ideal M ⊂ E. Then by the Local-Global Theorem 1.5.1, J is a projective ideal in E. By Lemma 4.2.9, J is an invertible ideal in E. This completes the proof of the lemma. LEMMA 5.1.7 Let E be an rtffr integral domain with integral closure E. Let J ⊂ E be an ideal such that J + τ = E. Then J ∩ E is an invertible ideal in E such that (J ∩ E) + τ = E. Proof: Because τ ⊂ E the modular law implies that (J + τ ) ∩ E = (J ∩ E) + τ = E. By part 3 of Lemma 5.1.6, J ∩ E is then an invertible ideal in E. LEMMA 5.1.8 Let E be an rtffr integral domain with conductor τ and with integral closure E. Let J ⊂ E be an ideal such that J + τ = E. Then [J ∩ E]E = J.

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Proof: Let J ⊂ E be an ideal such that J + τ = E and let J = J ∩ E. By the previous Lemma, J + τ = E. Given a maximal ideal τ ⊂ M ⊂ E we have JM = EM by Lemma 5.1.6(1). Furthermore because τM ⊂ MM = J (EM ) we have E M = J M + τM = J M + τM E M = J M by Nakayama’s Lemma 1.2.3. Then (JE)M = JM E M = EM E M = E M = J M . Given a maximal ideal τ 6⊂ M ⊂ E then Lemma 5.1.6(2) implies that JM is an ideal in EM = E M . Then (JE)M = JM E M = JM = J M ∩ EM = J M ∩ E M = J M so that JE = J by the Local-Global Theorem 1.5.1. Continuation of the Proof of Theorem 5.1.4: Let J be a frac0 0 tional ideal of E. There is a copy J of J such that J + τ = E (Lemma 0 5.1.5). Thus there is a q ∈ QE such that J = qJ . Moreover by Lemma 0 0 0 5.1.8, J ∩ E is an invertible ideal of E such that (J ∩ E)E = J . Then 0 0 J = q(J ∩ E) is an invertible ideal of E such that JE = qJ = J. Therefore α : γ(E) −→ γ(E) is a surjection, and hence γ(E) ∼ = γ(E) ⊕ ker α. This completes the proof of Theorem 5.1.4. It is clear that the principal fractional ideals J = aE of E are mapped to principal fractional ideals α(J) = aE of E. Because Γ(E) is formed by taking the quotient of γ(E) modulo the subgroup of principal fractional ideals, α induces a group surjection β : Γ(E) −→ Γ(E). THEOREM 5.1.9 Let E be an rtffr integral domain with integral closure E. Then β : Γ(E) −→ Γ(E) is a surjection of finite abelian groups. Proof: By the Jordan-Zassenhaus Theorem 5.1.2, Γ(E) and Γ(E) are finite abelian groups. Thus β is a group surjection of finite abelian groups. COROLLARY 5.1.10 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group such that E(G) is a commutative integral domain with integral closure E. Then h(E) divides h(G). Proof: By Theorem 5.1.9, there is a surjection β of finite groups. Then by Lagrange’s Theorem h(E(G)) = card(Γ(E(G))) is divisible by card(Γ(E)) = h(E). This completes the proof.

5.1. COUNTING ISOMORPHISM CLASSES

5.1.3

113

Class Number and Refinement

We continue to investigate h(G) when G is a strongly indecomposable rtffr group such that E(G) is a commutative integral domain with integral closure E. We say that G has the L-property if G is isomorphic to each group that is locally isomorphic to G. The L stands for E.L. Lady who first used local isomorphism on rtffr groups. See [56] where local isomorphism is called near isomorphism. THEOREM 5.1.11 Let G be a strongly indecomposable rtffr group such that E(G) is an integral domain. The following are equivalent. 1. h(G) = 1. 2. h(E(G)) = 1. 3. G has the L-property. 4. G has the refinement property. Proof: 1 ⇔ 2 is Lemma 5.1.3. 1 ⇔ 3 Let h(G) = 1 and suppose that G is locally isomorphic to H. Since h(G) = 1 implies that there is only one isomorphism class in Γ(G), (G) = (H) or equivalently G ∼ = H. The converse is proved in the same way. 2 ⇒ 4 By Theorem 2.4.7 we can prove that G has the refinement property if we prove that E(G) has the refinement property. Suppose that P ⊕ P 0 ∼ = E(G)(n) for some integer n. Since E(G) is an rtffr integral domain Theorem 4.2.15 states that E(G) has the local refinement property so that P = J1 ⊕ · · · ⊕ Jr for some invertible ideals J1 , . . . , Jr of E(G). Since h(E(G)) = 1, E(G) ∼ = Ji for each i = 1, . . . , r, and hence E(G) has the refinement property. 4 ⇒ 3 Let G be locally isomorphic to H. Since G is strongly indecomposable Theorem 3.5.1 implies that H is indecomposable. By Lemma 2.5.6 there is a direct sum G ⊕ G ∼ = H ⊕ H 0 of groups. Since G has the ∼ refinement property G = H, so that G has the L-property. This proves part 3 and completes the logical cycle. Let G be a strongly indecomposable rtffr group. We say that G has the proper local refinement property if G has the local refinement property but G does not have the refinement property. That is, there are isomorphic direct sums G(n) ∼ = H ⊕ K such that in no direct sum decomposition of H = G1 ⊕ · · · ⊕ Gn is each Gi isomorphic to G. The

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group G has the proper L-property if there is a group that is locally isomorphic to G but that is not isomorphic to G. THEOREM 5.1.12 Let G be a strongly indecomposable rtffr group. Assume that E(G) is an rtffr integral domain with integral closure E(G). 1. h(G) 6= 1. 2. h(E(G)) 6= 1. 3. G has the proper local refinement property. 4. G has the proper L-property. The next result shows us that the proper L-property and the proper local refinement property are inherited by subgroups of finite index. This kind of genetic behavior is uncommon in the study of rtffr groups. THEOREM 5.1.13 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group such that E(G) is a commutative integral domain. Let E be the integral closure of E(G). 1. If h(E) 6= 1 then G has the proper local refinement property. 2. h(E) = 1 if G has the refinement property. Proof: Part 2 is the contrapositive of part 1. Suppose that h(E) > 1. Since h(E) divides h(G), h(G) > 1, whence G does not have the refinement property, but G has the proper refinement property. This proves the Theorem. It is rare to find a property for an rtffr group G that is passed onto other groups in the quasi-equality class of G. COROLLARY 5.1.14 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group such that E(G) is a Dedekind domain, and assume that G has the proper local refinement property. If G is quasi-isomorphic to G then G has the proper local refinement property. Since E(G) is a commutative integral domain when G is a strongly indecomposable rtffr group of square-free rank we have proved the following corollary.

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115

COROLLARY 5.1.15 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group of square-free rank. Let E be the integral closure of E(G). 1. If h(E) 6= 1 then G has the proper local refinement property. 2. h(E) = 1 if G has the refinement property. Recall from Lemma 3.1.8 that if G is an rtffr group then there is an rtffr group G ⊂ G such that G/G is finite, E(G) is an integrally closed integral domain, E(G) ⊂ E(G), and E(G)/E(G) is finite. THEOREM 5.1.16 Let G be a strongly indecomposable integrally closed rtffr group of square-free rank. If G has the proper local refine. ment property and if G = G then G has the proper refinement property. Proof: Suppose that G has the proper refinement property. By Theorem 5.1.11, h(E(G)) 6= 1. Now E(G) is integrally closed and E(G)/E(G) is finite so E(G) is the integral closure of E(G). Hence Corollary 5.1.15 states that G has the proper refinement property. THEOREM 5.1.17 Let G1 , . . . , Gt be strongly indecomposable rtffr . groups of square-free ranks such that Gi ∼ = Gj ⇒ i = j. For each (n ) i = 1, . . . , t let E i be the integral closure of E(Gi ), and let G = G1 1 ⊕ (n ) · · · ⊕ Gt t for some integers n1 , . . . , nt > 0. 1. If some E i is not a pid then G has the proper local refinement property. 2. If G has the refinement property then E i is a pid for each i = 1, . . . , t. Proof: Part 1 is the contrapositive of part 2. 2. Suppose that G has the refinement property. One shows that Gi has the refinement property for each i = 1, . . . , t, so by Theorem 5.1.12, E i is a pid for each i = 1, . . . , t. The above theorems connect the refinement property for strongly indecomposable rtffr groups to the class number problem for Dedekind domains in algebraic number fields. Thus a classification of the strongly indecomposable rtffr groups that have the refinement property would naturally lead to a classification of the algebraic number fields whose

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class number is 1. The determination of the class number of an algebraic number field is generally regarded as a difficult problem. See [69]. Thus it would seem that the refinement property will be a fun one but a hard one to investigate.

5.1.4

Quadratic Number Fields

It is natural to ask for h(G) when QE(G) is a small field. Since there is a rather complete description of quadratic number fields k such that h(k) = 1 we will look there for a classification of strongly indecomposable rtffr groups G that have the refinement property. PROPOSITION 5.1.18 Let G be a strongly indecomposable rtffr group. 1. If QE(G) = Q then G has the refinement property. 2. If rank(G) is prime and if N (End(G)) 6= 0 then G has the refinement property. Proof: Part 1. follows from the fact that each subring of Q is a pid. 2. By hypothesis and Lemma 4.2.1, Q-dim(QE(G)) is a proper divisor of the prime Q-dim(QG) so that QE(G) = Q. Now use part 1. LEMMA 5.1.19 √ Let O be the ring of algebraic integers in a quadratic number field Q[ d] for some square-free integer d 6= 0. Let E be a subring of finite index in O. 1. Suppose that d < 0. Then h(E) = 1 if d ∈ {−1, −2, −3, −7, −11, −19, −43, −67, −163}. 2. Suppose that 1 < d < 100 is a square-free integer. Then h(E) 6= 1 iff d ∈ X = {10, 15, 26, 30, 34, 35, 39, 42, 51, 55, 58, 65, 66, 70, 74, 78, 79, 82, 85, 87, 91, 95}. Proof: 1. By [76, Theorem 10.5], the negative integers d ∈ {−1, −2, −3, −7, −11, −19, −43, −67, −163} are the only √ negative integers for which the ring O of algebraic integers in Q[ d] is a pid. Since O is the integral closure of E, Theorem 5.1.9 implies that h(O) 6= 1 divides h(E) 6= 1. 2. By [76, Theorem 10.5], the 22 values in X are exactly the squarefree integers 1 < d < 100 such that h(O) 6= 1. Now apply Theorem 5.1.12. This completes the proof.

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THEOREM 5.1.20 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group of rank 2. 1. If QEnd(G) = Q or N (End(G)) 6= 0 then G has the refinement property. √ 2. Otherwise, QEnd(G) = Q[ d] for some square-free integer d 6= 0. . Assume that End(G) = O where O is the ring of algebraic integers in QEnd(G). Then G has the proper refinement property provided that either (a) d 6∈ {−1, −2, −3, −7, −11, −19, −43, −67, −163} or (b) 1 < d < 100 and d ∈ {10, 15, 26, 30, 34, 35, 39, 42, 51, 55, 58, 65, 66, 70, 74, 78, 79, 82, 85, 87, 91, 95}. Proof: 1. Our assumptions amount to assuming that QE(G) = Q so that E(G) is a pid in this case. Thus the groups G in this case have the refinement property. . 2. Since E(G) = O, O is the integral closure of E(G). By the previous theorem h(O) 6= 1 in case d satisfies parts (a) or (b). By Corollary 5.1.15, G has the proper refinement property. Reader be warned: Suppose O is the ring of algebraic integers in any algebraic number field. Choose any nonzero ideal I ⊂ O. Then regardless of the class number of O, OI is a pid.

5.1.5

Counting Theorems

The following result is central to our discussion. THEOREM 5.1.21 [I. Kaplansky] (See [69]). Let E be a Noetherian integral domain and let I1 , . . . , Im , J1 , . . . , Jn be invertible ideals in E. Then I1 ⊕ · · · ⊕ Im ∼ = J1 ⊕ · · · ⊕ Jn ⇐⇒ m = n and I1 · · · Im ∼ = J1 · · · Jn . Proof: (⇐⇒) follows from [69, Theorem 38.13].

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Given an integer n > 0, a strongly indecomposable rtffr group G with square free rank, and an integral domain E let

νn (G) = card{unordered n-tuples {H1 , . . . , Hn } each Hi is an indecomposable rtffr group, and G(n) ∼ = H 1 ⊕ · · · ⊕ Hn } νn (E) = card{unordered n-tuples {I1 , . . . , In } each Ii is an invertible ideal in E, and E (n) ∼ = I1 ⊕ · · · ⊕ In }

and let e(G) = least integer e > 0 such that xe = 1 for all x ∈ Γ(E(G)) `(G) = least integer t > 0 such that Γ(E(G)) is a direct sum of t cyclic groups, say Γ(E(G)) = h(J1 )i × · · · × h(Jt )i

Observe that e(G), `(G) ≤ h(G) ≤ e(G)`(G). If Γ(E(G)) is a cyclic group then `(G) = 1 and e(G) = h(E(G)) = h(G). We will prove the following theorem. THEOREM 5.1.22 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group of square free rank, and let n > 0 be an integer. 1. If Γ(E(G)) is a cyclic group then νn (G) ≤ h(G)n−1 . 2. In general, νn (G) ≤ e(G)n`(G) ≤ h(G)n`(G) . The proof is a series of lemmas. Let

Kn (G) = {ordered m-tuples (I1 , · · · , In ) I1 , . . . , In are invertible ideals in E such that E (n) ∼ = I1 ⊕ · · · ⊕ In }.

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LEMMA 5.1.23 Let E be an rtffr integral domain, and let n > 0 be an integer. Then 1. There is a short exact sequence µ 0 −−−−→ Kn (E) −−−−→ Γ(E)(n) −−−−→ Γ(E) −−−−→ 0

(5.1)

of abelian groups where µ((I1 ), . . . , (In )) = (I1 · · · In ). 2. νn (E) ≤ card(Kn (E)) = h(E)n−1 . Proof: 1. Given (I1 , . . . , In ) ∈ ker µ then E · · · E = E ∼ = I1 · · · In so that E (n) ∼ = I1 ⊕ · · · ⊕ In by Theorem 5.1.21. Thus ker µ ⊂ Kn (E). Conversely given (I1 , . . . , In ) ∈ Kn (E) then E (n) ∼ = I1 ⊕ · · · ⊕ In so that E ∼ = I1 · · · In by Theorem 5.1.21. Thus (I1 , . . . , In ) ∈ ker µ, so that Kn (E) = ker µ. 2. By part 1 there is a short exact sequence (5.1) so that Lagrange’s Theorem shows us that h(E)n = card(Γ(E)(n) ) = h(E) · card(Kn (E)). That is, card(Kn (E)) = h(E)n−1 . Furthermore there is an obvious surjection from Kn (E) onto the set of unordered n-tuples {I1 , . . . , In } such that E (n) ∼ = I1 ⊕ · · · ⊕ In . Then νn (E) ≤ card(Kn (E)). This completes the proof of the lemma. Lemma 5.1.23(2) proves part 1 of Theorem 5.1.22 once we recall that h(G) = h(E(G)). LEMMA 5.1.24 Let E be a Noetherian integral domain, and let n > 0 be an integer. Then νn (E) ≤ e(E)n`(E) . Proof: Let t = `(E) and write the finite abelian group Γ(E) as Γ(E) = h(J1 )i × · · · × h(Jt )i for some fixed invertible ideals J1 , . . . , Jt ⊂ E. Then (Ji )e(G) = (E) for each i = 1, . . . , t so that each (I) ∈ Γ(E) can be written as I = J1e1 · · · Jtet

for some integers 0 ≤ ei < e(G).

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Form the set Mat(E) = {n × t matrices (eij )n×t

0 ≤ eij < e(G) for each

1 ≤ i ≤ n and 1 ≤ j ≤ t}. Since each element of Γ(E) is a product of powers of (J1 ), . . . , (Jt ), for each element (I1 , . . . , In ) ∈ Kn (E) there is an (eij )n×t ∈ Mat(E) such that Ii = J1ei1 · · · Jteit . There is then a (set theoretic) surjection Mat(E) −→ Kn (E) such that νn (E) = card(Kn (E)) ≤ card(Mat(E)) ≤ e(E)nt = e(E)n`(E) . This completes the proof of the lemma.

LEMMA 5.1.25 Let G be a strongly indecomposable rtffr group of square free rank. Then νn (G) = νn (E(G)). Proof: Let G be a strongly indecomposable rtffr group of square free rank. Suppose that G(n) = H1 ⊕ · · · ⊕ Hn for some indecomposable rtffr groups H1 , . . . , Hn . Then by Theorem 2.4.4 E(G)(n) = A(H1 ) ⊕ · · · ⊕ A(Hn ) where A(Hi ) is an indecomposable projective right E(G)-module. Since E(G) is a Noetherian integral domain, Theorem 4.2.12 states that A(Hi ) is an invertible ideal in E(G). Thus νn (G) ≤ νn (E(G)). Conversely suppose that E(G)(n) = I1 ⊕ · · · ⊕ In for some invertible ideals Ii in E(G). By Theorem 5.1.21, E(G) ∼ = I1 · · · In , and Theorem 2.4.4 implies that there are indecomposable groups Hi such that A(Hi ) = Ii and G(n) ∼ = H1 ⊕ · · · ⊕ Hn . Then νn (E(G)) ≤ νn (G) and it follows that νn (G) = νn (E(G)).

Proof of Theorem 5.1.22(2): By definition and Lemma 5.1.25 e(G) = e(E(G)), `(G) = `(E(G)), νn (G) = νn (E(G))

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121

so that by Lemma 5.1.24 νn (G) = νn (E(G)) ≤ e(E(G))n`(E(G)) = e(G)n`(G) . This completes the proof.

As an application of the above counting results we have a power cancellation result. THEOREM 5.1.26 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group of square free rank. If e(G) and n are relatively prime then G(n) ∼ = H (n) =⇒ G ∼ = H. Proof: Suppose that G(n) ∼ = H (n) . By Lemma 2.5.7, G and H are locally isomorphic. Theorem 2.4.4 implies that A(H) is an ideal in E(G) such that E(G)(n) ∼ = A(H)(n) . By Theorem 5.1.21, E(G) ∼ = A(H)n , so that the class (A(H)) ∈ Γ(E(G)) has order r dividing n. Since r must also divide e(G), and since e(G) and n are relatively prime, r = 1. Hence E(G) ∼ = A(H), and therefore G ∼ = H. THEOREM 5.1.27 Let G be an rtffr group such that Γ(E(G)) is a pgroup for some prime p. Then G(n) ∼ = H (n) =⇒ G ∼ = H for each integer n not divisible by p. Consider now some small values of n. Specifically we will show that these estimates are very good when n = 2. Let

hn (G) = card{x ∈ Γ(E(G)) xn = 1}.

THEOREM 5.1.28 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group of square free rank. 1. hn (G) = the number of isomorphism classes of groups H such that G(n) ∼ = H (n) . 2. h(G) − h2 (G) = the number of unordered pairs {(H), (K)} of distinct isomorphism classes of rtffr groups such that G ⊕ G ∼ = H ⊕ K.

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Proof: 1. By Theorem 2.4.4 it suffices to assume that G = E(G). So suppose that E(G)(n) ∼ = I (n) for some invertible ideal I ⊂ E(G). Theorem 5.1.21 states that E(G) ∼ = I · · · I (n factors), so that (I) ∈ {x ∈ Γ(E) xn = 1}. Hence the number of isomorphism classes of ideals I ⊂ E(G) such that E(G)(n) ∼ = I (n) is at most hn (G). The converse is proved by a similar argument. 2. Suppose that G ⊕ G ∼ 6 K. By = H ⊕ K as groups and that H ∼ = ∼ Theorem 2.4.4, E(G) ⊕ E(G) ∼ A(H) ⊕ A(K) and A(H) 6 A(K). Then = = ∼ by Theorem 5.1.21, A(H)A(K) = E(G), or in other words (A(H)) 6= (A(K)) and (A(H)) = (A(K))−1 . Hence card{(x, x−1 ) x−1 6= x ∈ Γ(E(G))} = # of x such that x2 6= 1 = h(E(G)) − h2 (G) = h(G) − h2 (G) by Lemma 5.1.25. This completes the proof. THEOREM 5.1.29 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group of square free rank. If h(G) is an odd integer then 1. G ⊕ G ∼ = H ⊕ H implies that G ∼ = H. 2. The number of sets {H, K} of nonisomorphic rtffr groups such that G⊕G∼ = H ⊕ K is h(G) − 1. Proof: 1. Suppose that G⊕G ∼ = H ⊕H. The group H corresponds to an element (I) ∈ Γ(E) such that I ·I ∼ = E. Since card(E) = h(E) = h(G) ∼ is odd, I = E. By Theorem 2.4.4, H ∼ = G. 2. Since h(G) is odd, h2 (G) = 1. Apply Theorem 5.1.28 to complete the proof.

5.2

Integrally Closed Groups

The rings in this section need not be commutative and the groups need not be strongly indecomposable. Recall that the rtffr group G is integrally closed group if E(G) is an integrally closed ring. We show in Lemma 3.1.8 that each rtffr group has finite index in an integrally closed group. We continue our theme that groups possessing the (local) refinement property and a locally unique decomposition are more common than once believed by identifying a naturally occurring class of subgroups of finite index in integrally closed groups that possess the local refinement property and a locally unique

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123

decomposition. Our goal should be contrasted with Example 2.2.1. Said example begins with a completely decomposable group G and constructs a subgroup G of finite index in G. By Theorem 2.1.11, G possesses a unique decomposition and it has the refinement property, but there are indecomposable decompositions G = G1 ⊕ G2 = H1 ⊕ H2 such that rank(G1 ) = rank(G2 ) = 2 and rank(H1 ) = 1. This juxtaposition of uniqueness with nonuniqueness invites us to study further. We can make some sense of this seemingly counterintuitive coincidence by examining direct sum decompositions of semi-primary groups. Definitions follow.

5.2.1

Review of Notation and Terminology

Fix a nilpotent set {G1 , · · · , Gt } of indecomposable rtffr groups, fix integers t, n1 , · · · , nt > 0, and let (n1 )

G = G1

(nt )

⊕ · · · ⊕ Gt

.

(5.2)

Nilpotency implies that Gi ∼ = Gj ⇒ i = j. We will make extensive use of the ideal structure of the semi-prime rtffr ring E(G), and the functor A(·) : Po (G) −→ Po (E(G)) from Theorem 2.4.4. Then because {G1 , . . . , Gt } is a nilpotent set (Theorem 4.1.8) we have E(G) = A(G1 )(n1 ) ⊕ · · · ⊕ A(Gt )(nt ) = Matn1 (E(G1 )) × · · · × Matnt (E(Gt )). There is at least one interesting class of rtffr ring that has the refinement property. PROPOSITION 5.2.1 The rtffr group G with conductor τ has the local refinement property and a locally unique decomposition if Eτ (G) is a semi-perfect ring. Proof: Given a semi-perfect ring Eτ (G), Lemma 1.7.1 and the AKS Theorem 2.1.6 state that Eτ (G) has the refinement property. Then Theorem 3.5.4 implies that G has the local refinement property. Lastly apply Lemma 2.5.10. Thus we are motivated to look for semi-perfect rtffr rings of the form Eτ (G). We will show that there are some fairly common rtffr groups G, called semi-primary groups, for which Eτ (G) is semi-perfect.

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We showed in Lemma 4.2.1 that E(G) is commutative if G is strongly indecomposable and if rank(G) is a square-free integer. Thus there are plenty of rtffr groups G with direct sum decomposition (5.2) such that QE(G) = Matn1 (k1 ) × · · · × Matnt (kt ) where ki is the field of fractions of the integral domain E(Gi ).

5.2.2

Locally Semi-Perfect Rings

Let E be a semi-prime rtffr ring with conductor τ . We give some fairly general conditions on τ and E that make Eτ semi-perfect. 5.2.2 Let S = center(E) and choose an integrally closed ring E ⊂ QE such that τ E ⊂ E ⊂ E. Then E = E1 × · · · × Et where each E i is a classical maximal order in the field ki . Let center(E) = S = S 1 × · · · × S t where S i = center(E i ) is a Dedekind domain for each i = 1, . . . , t. As in section 3.2, τ is an ideal in S so we can write

τ = τ1 ⊕ · · · ⊕ τt

where 0 6= τi ⊂ S i is an ideal of finite index.

Recall that an ideal I 6= 0 in a commutative ring S is a primary ideal if S/I is a local ring with nilpotent Jacobson radical. For instance, qZ is a primary ideal in Z iff q is a prime power. The primary ideals in a Dedekind domain are powers of maximal ideals. THEOREM 5.2.3 Suppose that E is a semi-prime rtffr ring with conductor τ = τ1 ⊕ · · · ⊕ τt as in (5.2.2). Then E has the local refinement property and a locally unique decomposition if E satisfies the following two conditions: 1. There are integers t, n1 , . . . , nt > 0 and fields k1 , . . . , kt such that QE = Matn1 (k1 ) × · · · × Matnt (kt ) and

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125

2. τi is a primary ideal in S i for each i = 1, . . . , t. Proof: The proof is a series of lemmas. The notations and conditions of the Theorem are used without fanfare. The first result is called the Change of Rings Theorem. I leave the proof as an exercise in the nature of long exact sequences and localizations. THEOREM 5.2.4 [72, page 107, Theorem 3.84] Let U be a finitely generated S-module and let I be an ideal in S. Then (EndS (U ))I ∼ = EndSI (UI ) as rings. LEMMA 5.2.5 Idempotents in Eτ /Eτ ττ lift to idempotents of Eτ . Proof: We first show that (E i )τ is a semi-perfect ring. Fix i ∈ {1, . . . , t}. By condition 1 of Theorem 5.2.3 QS i = center(QE i ) = center(Matni (ki )) = ki . Let Vi be a simple left module over the simple Artinian ring Matni (ki ) and let Ui = E i vi be a cyclic left E i -submodule of Vi . Then (n ) QUi = QE i vi = Vi ∼ = ki i

as ki -vector spaces. Since S i is a Dedekind domain, since QE i is a torsion-free Si -module, and since E i is a finitely generated torsion-free S i -module we see that Ui is a finitely generated projective S i -module. Thus EndS i (Ui ) is a finitely generated torsion-free S i -module. Furthermore (ni )

E i ⊂ EndS i (Ui ) ⊂ Endki (QUi ) = Endki (ki

) = Matni (ki )

and E i is a classical maximal order in Matni (ki ). Since EndS i (Ui )/E i is then a finitely generated torsion S i -module, E i = EndS i (Ui ), and since Ui is a finitely generated S i -module Theorem 5.2.4 implies that (E i )τ = (E i )τi ∼ = End(S i )τ (Uτi ). i

Condition 2 states that τi is a primary ideal in S i so that S i ∼ (S i )τi = τi (τi )τi

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is a local ring. Since (τi )τi ⊂ J ((Si )τi ), (S i )τi is a local ring. Moreover U is a finitely generated projective S i -module. Since Uτi is (n ) then a free module over the local ring (S i )τi Uτi ∼ = (S i )τi i . Then (E i )τ ∼ = End(S i )τ (Uτi ) ∼ = Matni ((S i )τi ) i

is a semi-perfect ring. As claimed E τ is semi-perfect. Now, let e ∈ Eτ be such that e2 − e ∈ ττ Eτ . Inasmuch as τ E ⊂ E ⊂ E, setting I = ττ2 E τ gives us I = ττ (ττ E τ ) ⊂ ττ Eτ ⊂ J (E τ ). Since idempotents lift modulo the nilpotent ideal ττ Eτ /I ⊂ Eτ /I, there is an (f + I)2 = f + I ∈ Eτ /I ⊂ E τ /I such that e − f ∈ ττ Eτ . By the above claim E τ is semi-perfect and I ⊂ J (E τ ), so by Lemma 1.7.4 there is a g 2 = g ∈ E τ such that f + I = g + I. Hence g ∈ f + I ⊂ Eτ , whence idempotents in Eτ /ττ Eτ lift to idempotents in Eτ . This completes the proof of the lemma. The rtffr ring E is semi-local iff there is some integer n 6= 0 such that each maximal ideal of E contains n. For example, if E is a semi-prime rtffr ring with conductor τ then Eτ is a semi-local ring. LEMMA 5.2.6 Let E be a semi-prime rtffr ring with conductor τ that satisfies conditions 1 and 2 of Theorem 5.2.3. Then Eτ is a semi-perfect ring. Proof: Suppose that E satisfies conditions 1 and 2. Since ττ Eτ ⊂ J (Eτ ) and since τ has finite index in S, Eτ /J (Eτ ) is a (finite =) semisimple Artinian ring. Furthermore if e2 = e ∈ Eτ /J (Eτ ) then e lifts modulo the nilpotent ideal J (Eτ )/ττ Eτ to an (e + ττ Eτ )2 = (e + ττ Eτ ) ∈ Eτ /ττ Eτ . By Lemma 5.2.5, there is an f 2 = f ∈ Eτ such that e − f ∈ ττ Eτ . Therefore Eτ is semi-perfect. Proof of Theorem 5.2.3: By Lemma 5.2.6, Eτ is semi-perfect, so Lemma 1.7.1 and the AKS Theorem 2.1.6 imply that Eτ has the refinement property. Using Theorem 3.5.3 we conclude that E has the local refinement property.

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127

In case QE = Matn (k) for some field k then S = center(E) is an integral domain with field of quotients k. If k has finite degree over Q then there is a unique Dedekind domain S ⊂ S ⊂ k called the integral closure of S such that S/S is finite. COROLLARY 5.2.7 Suppose that E is a prime rtffr ring with conductor τ such that QE ∼ = Matn (k) for some integer n > 0 and algebraic number field field k. Let S be the integral closure of S. If τ is a primary ideal in S then E has the local refinement property. EXAMPLE 5.2.8 Let k be an algebraic number field, let O be the ring of algebraic integers in k with class number h 6= 1, and let I 6= 0 be an ideal in O such that O ∼ 6 I. By Theorem 5.2.3 the finitely generated = projectives over E have the local refinement property. However since the class number h of O is not 1, I (h) ∼ = O(h) , while I 6∼ = O. Thus O has the local refinement property.

5.2.3

Semi-Primary Rtffr Groups

Let G ⊂ G be rtffr groups and suppose that G has the local refinement property. The results of this section will give conditions on G under which a subgroup of finite index in G has the local refinement property. This should be contrasted with examples due to Corner and Fuchs-Loonstra (see [10, 46]), that show that subgroups of finite index in completely decomposable rtffr groups can have badly behaved direct sum decompositions. The group G is a Dedekind group if E(G) is a Dedekind domain. If G is a Dedekind domain then E(G) is a commutative ring. We assume that we are given a finite nilpotent set {G1 , . . . , Gt } of Dedekind rtffr groups and integers n1 , . . . , nt > 0. Let (n1 )

G = G1

(nt )

⊕ · · · ⊕ Gt

.

(5.3)

For each i = 1, . . . , t choose a primary ideal Pi such that N (End(Gi )) ⊂ Pi ⊂ End(Gi ). Notice that because Pi is primary in End(Gi ) then End(Gi )/Pi is finite. (n ) The action Pi Gi i is unambiguous since Pi ⊂ End(Gi ). Then (n1 )

(P1 ⊕ · · · ⊕ Pt )G = P1 G1

(nt )

⊕ · · · ⊕ Pt Gt

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is uniquely defined. The rtffr groups we are interested in are those G such that (n1 )

P1 G 1

(nt )

⊕ · · · ⊕ Pt G t

⊂ G ⊂ G.

(5.4)

In this case we say that G has semi-primary index in G. Alternatively we say that G is a semi-primary rtffr group if there are integers t, n1 , . . . , nt > 0, a nilpotent set {G1 , . . . , Gt } of Dedekind rtffr groups, and primary ideals N (End(Gi )) ⊂ Pi ⊂ End(Gi ) such that (5.4) is true. Given a semi-primary rtffr group G we will use G and Pi in the way they are used in the above discussion. Evidently each Dedekind rtffr group is a semi-primary group so semiprimary groups exist in abundance. The following examples illustrate the settings that we are trying to generalize. EXAMPLE 5.2.9 Let p ∈ Z be a prime, let E = Mat2 (Z), and let E = 1Z + Mat2 (pZ). Then e2 = e ∈ E ⇒ e ∈ {0, 1}. Using Corner’s Theorem 2.3.3 there is an rtffr group G such that E = End(G). Let G = EG. One shows that E = End(G) and that pG ⊂ G ⊂ G. Then G is a semi-primary group, G is indecomposable, and G has a nontrivial direct sum decomposition. Moreover by Theorem 5.2.3, G and G have the local refinement property and unique decompositions. This example demonstrates that we are properly extending Theorem 2.4.6. EXAMPLE 5.2.10 Let G1 , . . . , Gt ⊂ Q be rank 1 groups such that Gi ∼ = Gj ⇒ i = j. The Baer-Kulikov-Kaplansky Theorem 2.1.11 states that the group G in (5.3) has unique decomposition. If q ∈ Z is a prime power and if G is a group such that qG ⊂ G ⊂ G then G is a semi-primary group. It is traditional [60] to call G an acd group with primary regulating quotient. EXAMPLE 5.2.11 Let {G1 , . . . , Gt } be a finite nilpotent set of rtffr groups such that End(Gi ) is a pid for each i = 1, . . . , t. Let G be the group in (5.3) and for each i = 1, . . . , t choose prime powers qi ∈ End(Gi ).

5.2. INTEGRALLY CLOSED GROUPS

129

Then each group G such that (n1 )

q 1 G1

(nt )

⊕ · · · ⊕ q t Gt

⊂ G ⊂ G

is a semi-primary rtffr group. EXAMPLE 5.2.12 Let G be a (strongly indecomposable) Dedekind rtffr group, fix an integer n > 0, and choose a primary ideal N (End(G)) ⊂ P ⊂ End(G). If G is any rtffr group such that PG

(n)

⊂G⊂G

(n)

then G is a semi-primary rtffr group. If n = 1 then we are looking at semi-primary rtffr groups G such that P G ⊂ G ⊂ G. The next result, the main result of this chapter, explains our interest in semi-primary rtffr groups. THEOREM 5.2.13 [T.G. Faticoni] Semi-primary rtffr groups have the local refinement property and a locally unique decomposition. Proof: The proof of this theorem consists of showing that E(G) satisfies conditions 1 and 2 in Theorem 5.2.3. Let G be a semi-primary rtffr group. There is a nilpotent set {G1 , . . ., Gt } of Dedekind rtffr groups, integers n1 , . . . , nt > 0, and primary ideals N (End(Gi )) ⊂ Pi ⊂ End(Gi ) such that (P1 ⊕ · · · ⊕ Pt )G ⊂ G ⊂ G where G is given in (5.3). Let P = P1 ⊕ · · · ⊕ Pt . For each i = 1, . . . , t, E(Gi ) is a Dedekind domain whose field of fractions QE(Gi ) is a number field. Since {G1 , . . . , Gt } is a nilpotent set Theorem 1.3.1 implies that E(G) = Matn1 (E(G1 )) × · · · × Matnt (E(Gt )).

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. . Since G = G, End(G) = End(G), so that QE(G) ∼ = QE(G) as rings. Then QE(G) = Matn1 (QE(G1 )) × · · · × Matnt (QE(Gt )) satisfies condition 1 of Theorem 5.2.3. As for condition 2, for each i = 1, . . . , t, Pi maps onto a nonzero primary ideal Ti ⊂ E(Gi ). Let T = T1 ⊕ · · · ⊕ Tt . Then T is an ideal of finite index in S. Since P G ⊂ G ⊂ G, P End(G) ⊂ End(G), and so P E(Gi ) = T E(Gi ) = Ti E(Gi ) ⊂ E(Gi ). Inasmuch as Ti ⊂ τi ⊂ E(Gi ) (where τ = τ1 ⊕ · · · ⊕ τt is the conductor of E(G)), and since Ti is a primary ideal in E(Gi ), τi is a primary ideal in E(Gi ). Thus E(G) satisfies condition 2. Then by Theorem 5.2.3, E(G) has the local refinement property and a locally unique decomposition. This completes the proof of the theorem.

5.2.4

Applications to Refinement

Some applications will show that the seemingly strong conditions in Theorem 5.2.3 are actually quite common. If H is a rank one group then End(H) ⊂ Q is a pid and a given prime p ∈ Z is either a unit in End(H) or a prime. Thus the rank one groups are Dedekind groups whose primary ideals are generated by powers of primes in Z. We say that G is homogeneous completely decomposable of type σ if G = H (n) for some rank one group H of type σ. If G is a completely decomposable rtffr group then G = G[σ1 ] ⊕ · · · ⊕ G[σt ] where G[σi ] is a homogeneous completely decomposable group of type σi , and the types σ1 , . . . , σt are distinct. Using this notation we can use Theorem 5.2.13 to study the uniqueness of direct sum decompositions of groups of semi-primary index in a completely decomposable rtffr group. The next two results should be contrasted with the acd groups constructed in Examples 2.2.1. THEOREM 5.2.14 [T.G. Faticoni] Let G = G[σ1 ] ⊕ · · · ⊕ G[σt ] be a completely decomposable rtffr group, where the groups G[σi ] are homogeneous completely decomposable of type σi , and where the types σ1 , . . . , σt are distinct. If G is a subgroup of finite index in G such that for each i = 1, . . . , t, (G + G[σi ])/G is a finite primary abelian group for each i = 1, . . . , t then G has a locally unique decomposition.

5.2. INTEGRALLY CLOSED GROUPS

131

(n )

Proof: By hypothesis G[σi ] = Gi i for some rank one group Gi , i = 1, . . . , t. The reader can show that {G1 , . . . , Gt } is a nilpotent set. Suppose that G has finite index in G and that for each i = 1, . . . , t, (G + G[σi ])/G is a finite pi -group. There are integers m1 , . . . , mt > 0 such that mt 1 pm 1 G[σ1 ] ⊕ · · · ⊕ pt G[σt ] ⊂ G ⊂ G[σ1 ] ⊕ · · · ⊕ G[σt ]

or in other words G has semi-primary index in G. By Theorem 5.2.13, G has the local refinement property. COROLLARY 5.2.15 [T.G. Faticoni and P. Schultz] (See [45].) Let G be a completely decomposable rtffr group and let q ∈ Z be a prime power. If G is a group such that qG ⊂ G ⊂ G then G has a locally unique decomposition. For the purposes of this discussion let H = H 1 ⊕ · · · ⊕ Hs where Hi ⊂ Q for each i = 1, . . . , s. A kernel group is a group G that fits into an exact sequence σ

0 → G −→ H −→ Q where σ is the unique map such that σ(x) = x for each x ∈ Hi and for each i = 1, . . . , s. Dually a cokernel group is a group G that fits into an exact sequence π 0 → X −→ H −→ G → 0 where X is a pure rank one subgroup of H. We call G a bracket group if G is either a kernel group of a cokernel group. If G is a strongly indecomposable bracket group then it is known that End(G) ⊂ Q is a pid. See [10]. Thus the strongly indecomposable bracket groups are Dedekind groups. A strongly homogeneous group is a group G such that for any pure rank one subgroups X, Y ⊂ G there is an automorphism α : G → G such that α(X) = Y . It is known that if G is a strongly indecomposable strongly homogeneous rtffr group then End(G) is a pid in which the primes p ∈ Z such that pG 6= G are primes in End(G). See [10]. Thus the strongly indecomposable strongly homogeneous rtffr groups are Dedekind groups. The group G is a Murley group if Z/pZ-dim(G/pG) ≤ 1 for each prime p ∈ Z. It is known that if G is a strongly indecomposable Murley

132

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group then End(G) is a pid in which p ∈ Z is a prime in End(G) if b pG 6= G. Evidently G is a Murley group iff G is a pure subgroup of Z. See [62]. Direct sums of rtffr groups G such that End(G) is a pid are considered by D.M. Arnold, R. Hunter, and F. Richman in [13]. Certainly G is a Dedekind group if End(G) is a pid. We have shown that if G is a strongly indecomposable group of squarefree rank then E(G) is an rtffr integral domain. Lemma 3.1.8 implies that G has finite index in a Dedekind group, thus motivating the assumption that G is a Dedekind group. Let D denote the set of strongly indecomposable rtffr groups G such that G falls into at least one of the following categories. 1. Groups G such that E(G) is a pid, 2. Strongly indecomposable Murley groups, 3. Strongly indecomposable strongly homogeneous groups, 4. Strongly indecomposable bracket groups, 5. Rank one groups. THEOREM 5.2.16 [T.G. Faticoni] Let {G1 , . . . , Gt } be a nilpotent subset of D and let G be the direct sum (5.3) for some integers n1 , . . . , nt > 0. If qi ∈ E(Gi ) are powers of primes in the pid E(Gi ) and if G is a group such that (n1 )

q 1 G1

(nt )

⊕ · · · ⊕ q t Gt

⊂G⊂G

then G has the local refinement property. Proof: By the above discussion for each Gi ∈ D, E(Gi ) is a pid so each Gi is a Dedekind rtffr group. Thus G is a semi-primary group. An application of Theorem 5.2.13 shows us that G has the local refinement property. . In [13] it is shown that if End(G) is a pid and if H1 , . . . , Hr = G satisfy End(Hi ) = End(G) then H = H1 ⊕ · · · ⊕ Hr has a uniqueness of

5.2. INTEGRALLY CLOSED GROUPS

133

decomposition that is different from local uniqueness and that is a little too technical for this book. In general there is an integer q 6= 0 such that qG ⊂ Hi ⊂ G for each i = 1, . . . , r. Our techniques allow us to slightly expand the main theorem in [13] by restricting q and by relaxing the conditions on End(G) and End(Hi ). COROLLARY 5.2.17 Let {G1 , . . . , Gt } be a nilpotent subset of D such that E(G1 ) ∼ = E(Gi ) for each i = 1, . . . , t, and let (n1 )

G = G1

(nt )

⊕ · · · ⊕ Gt

.

(5.5)

If q ∈ E(G1 ) is a prime power then each group qG ⊂ G ⊂ G has the local refinement property. COROLLARY 5.2.18 Let G be a rtffr group such that E(G) is a pid, let n 6= 0 be an integer, and let q ∈ E(G) be a power of some prime (n) (n) in E(G). Then any group qG ⊂G⊂G has the local refinement property. COROLLARY 5.2.19 Let {G1 , . . . , Gt } be a nilpotent subset of D such that E(Gi ) ∼ = Z for each i = 1, . . . , t, and let G be a direct sum (5.5). If q ∈ Z is a prime power and if qG ⊂ G ⊂ G then G has the refinement property. We end this section with a result that shows that direct sums of semiprimary groups G are up to local isomorphism actually direct sums of semi-simple End(G)-modules. The reader will enjoy proving this result. THEOREM 5.2.20 [T.G. Faticoni] Let G be a semi-primary rtffr group. Let τ be the conductor of E(G) and assume that E(G)(τ ) = 0. Let (S, τ ) be the category of semi-simple End(G)-modules annihilated by τ . 1. There is a functor C(·) : Po (G) → (S, τ ) : H → A(H) ⊗E(G) E(G)τ /J (E(G)τ ). 2. C(·) induces a bijection from the set of local isomorphism classes (H) of H ∈ Po (G) onto the set of isomorphism classes (M ) of semi-simple right E(G)-modules such that M τ = 0.

134

5.3

CHAPTER 5. REFINEMENT REVISITED

Exercises

1. Show that J (Sτ ) ⊂ J (Eτ ). 2. Let E be a rtffr ring and suppose that E is locally isomorphic to U1 ⊕ U2 for some finitely generated E-modules U1 and U2 . Then E = V1 ⊕ V2 for some cyclic right ideals V1 and V2 such that Ui is locally isomorphic to Vi for i = 1, 2. 3. Let G be a rtffr group. Show that the following are equivalent. (a) G is indecomposable. (b) G is locally isomorphic to some indecomposable rtffr group. (c) If G is locally isomorphic to H then H is indecomposable. (d) E(G) is indecomposable as a right E(G)-module. (e) Eτ (G) is indecomposable as a right Eτ (G)-module. 4. Show that G has a locally unique decomposition if either (a) E(G) is a semi-perfect ring or (b) G is a completely decomposable rtffr group or (c) G = H (n) where E(H) is a Dedekind domain. 5. Suppose that S is a Noetherian integral domain and that its field of fractions k is a finitely generated S-module. Show that k = S. 6. Using the notation of Theorem 5.2.13 show that there is a ring theoretic embedding E(G) → E(G) that takes regular elements to regular elements. 7. Using the notation of Theorem 5.2.13 show that QE(G) = QE(G). 8. Let O be the ring of algebraic integers in the number field k. Show that (a) Matn (k) is the classical ring of quotients of Matn (O). (b) Matn (O) is a maximal order in Matn (k). 9. Let E be a semi-prime rtffr ring with conductor τ . Say E = E1 × E2 . (a) Show that τ = τ1 ⊕ τ2 where τi is the conductor of Ei for i = 1, 2.

5.4. QUESTIONS FOR FUTURE RESEARCH

135

(b) Show that Eτ = (E1 )τ × (E2 )τ = (E1 )τ1 × (E2 )τ2 . 10. Suppose that G is a strongly indecomposable rtffr group of squarefree rank. Suppose that t is an integer that is divisible by h(G). Given indecomposable groups H1 , . . . , Ht ∈ Po (G) then G(t) ∼ = H 1 ⊕ · · · ⊕ Ht . . 11. Suppose that G = G1 ⊕ · · · ⊕ Gt for some strongly indecomposable groups of square-free rank. Multiple appearances are allowed. Investigate the logical relations between the following statements. (a) h(G) = 1. (b) h(Gi ) = 1 for each i = 1, . . . , t. (c) Each Gi has the L-refinement property. (d) Each Gi has the refinement property.

5.4

Questions for Future Research

Let G be a rtffr group, let E be a rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. Use the theory of rtffr groups to advance the understanding of the class group of an algebraic number field. 2. Use the theory of the class group of an algebraic number field to advance the understanding of the theory of rtffr groups. 3. Give a (finite abelian) group structure to Γ(G), the set of isomorphism classes of groups locally isomorphic to G, and study the group. 4. Generalize the discussion given on class number h(G) from strongly indecomposable rtffr groups G to just plain rtffr groups G. 5. There are results from the literature that show that h(O), where O is the ring of algebraic integers in an algebraic number field, will grow in a certain way as the discriminant of O increases. 6. Make a study of the class group and class number of an algebraic number field and apply it to rtffr groups.

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CHAPTER 5. REFINEMENT REVISITED

7. Make a study of the class number h(G) of an rtffr group and apply it to the study of the class number of an algebraic number field. 8. See 5.1.9. Determine ker β. Specifically, determine if ker α is finitely generated. Determine if ker β is a direct summand of Γ(E). 9. Use the Mayer-Vietoris Sequence and the Kunneth Sequence to study Γ(E). See [69, Section 37]. 10. Discuss how

√

τ is related to h(G).

11. Nakayama’s Lemma and the Chinese Remainder Theorem are used continually in this text. Explain why this is so by using some other kind of theory. 12. Give a complete accounting of the groups G that have the Lproperty. 13. Give a complete accounting of the groups G that have the refinement property. 14. Let G be a strongly indecomposable rtffr group such that E(G) is a Noetherian integral domain. Give group theoretic conditions for h(G) = h(G0 ) for quasi-isomorphic groups G and G0 . 15. Give a more complete discussion of the integers νn (G), e(G), and `(G) given on page 118. 16. Improve the estimates in Theorem 5.1.22. 17. Do what is considered to be a great deal of fun in abelian groups. Relate the direct sum properties of G to integers associated with G. For example, G has the refinement property iff h(G) = 1. See also Theorem 5.1.26 and Theorem 5.1.28. 18. Characterize the groups G such that E(G) is a locally semi-perfect ring. 19. Characterize semi-primary rtffr groups. 20. Functorially realize the group G. 21. Extend the theory surrounding acd groups to semi-primary groups, or indicate where we cannot.

5.4. QUESTIONS FOR FUTURE RESEARCH

137

22. Let G be a strongly indecomposable rtffr group such that E(G) is a commutative rtffr integral domain. Let E be the integral closure of E(G). Investigate the converses of the statement “h(E) = 1 if G has the refinement property.” √ 23. Suppose that QEnd(G) = Q[ d] for some negative square-free integer d. If d ∈ {−1, −2, −3, −7, −11, −19, −43, −67, −163}, then discuss the refinement property in G.

Chapter 6

Baer Splitting Property The results in this section are from [38]. The short exact sequence π

0 → K −→ X −→ H → 0

(6.1)

of groups is said to be split if there is a map j : H → X such that πj = 1H . Recall that P(G) = { groups H H ⊕ H 0 ∼ = G(c) for some cardinal c and group H 0 }. We will consider the splitting of the short exact sequence (6.1) under different hypotheses on X and the groups H ∈ P(G). Closely related to the refinement property in rtffr groups is the Baer splitting property. The group G has the Baer splitting property if for each cardinal c, each surjection π : G(c) → G is split. Free groups have the Baer splitting property and we will show that if End(G) is commutative then G has the Baer splitting property. Baer’s Lemma 6.1.1 states that G has the Baer splitting property if IG 6= G for each right ideal I ⊂ End(G). The purpose of this chapter is to study the Baer splitting property in detail.

6.1

Baer’s Lemma

Given groups H and G we write SG (H) =

P {f (G) f ∈ Hom(G, H)}.

139

140

CHAPTER 6. BAER SPLITTING PROPERTY

We call SG (H) the G-socle in H, and we note that SG (H) is the largest G-generated subgroup of H. If H = SG (H) then we say that H is Ggenerated, and we say that H is finitely G-generated if H = f1 (G) + · · · + fn (G) for finitely many f1 , . . . , fn ∈ Hom(G, H). The next result, known as Baer’s Lemma, has assumed an important role in the study of the transfer of properties between the rtffr group G and its endomorphism ring End(G). One way of viewing this result is that it shows us that a group G behaves like a projective module over a ring if IG 6= G for each right ideal I ⊂ End(G). The version we use is found in [14]. LEMMA 6.1.1 [Baer’s Lemma] The following are equivalent for an rtffr group G. 1. IG 6= G for each right ideal I ⊂ EndS (G). 2. Each short exact sequence π

0 → K −→ X −→ G → 0

(6.2)

in which SG (X) + K = X is split exact. Proof: Assume part 1 and let I = πHom(G, X). Then I is a right ideal in the ring End(G). Let y ∈ G. There is an x ∈ X such that π(x) = y. Since X = SG (X) + K there are f1 , . . . , ft ∈ Hom(G, X), x1 , . . . , xt ∈ G, and z ∈ K such that ! t t X X y = π(x) = π fi (xi ) + z = πfi (xi ) ∈ IG. i=1

i=1

Thus G = IG. Inasmuch as JG 6= G for proper right ideals J ⊂ End(G), I = End(G). That is, there is a map f ∈ Hom(G, X) such that πf = 1G , so that (6.2) is split exact. This proves part 2. Conversely assume part 1 is false. There is a proper right ideal I ⊂ End(G) such that IG = G. Write I = {f1 , f2 , . . .} and let Gi ∼ = G for each i = 1, 2, . . .. There is a map π : ⊕∞ i=1 Gi −→ G such that for each x ∈ Gi , π(x) = fi (x). Because G has finite rank, given any map g : G −→ ⊕∞ i=1 Gi

6.1. BAER’S LEMMA

141

there is a finite set {1, . . . , n} such that g(G) ⊂ G1 ⊕ · · · ⊕ Gn so that there are maps gi : G → Gi such that g = g1 ⊕ · · · ⊕ gn . Then πg = f1 g1 + · · · + fn gn ∈ I 6= End(G). Specifically πg 6= 1G for any g so that π is not split. Thus part 2 is false, which completes the proof.

COROLLARY 6.1.2 The following are equivalent for an rtffr group G. 1. IG 6= G for each proper right ideal I ⊂ End(G). 2. For each cardinal c, each short exact sequence π

0 → K −→ G(c) −→ G → 0

(6.3)

is split exact. The rtffr group G is finitely faithful if IG 6= G for each proper ideal I of finite index in End(G). Of course faithful groups are finitely faithful. For rtffr groups the converse is true. For this reason we will use the term faithful and not finitely faithful throughout this chapter. LEMMA 6.1.3 Let G be an rtffr group. Then G is a faithful group iff G is a finitely faithful group. Proof: Suppose that IG 6= G for each right ideal of finite index in End(G), and let I ⊂ End(G). Then I is contained in a maximal right ideal M ⊂ End(G) which by Lemma 4.2.3 has finite index in End(G). By hypothesis IG ⊂ M G 6= G so that G is a faithful group. The converse is clear so the proof is complete. There is at least one interesting previously encountered class of group that consists of faithful groups. To prove the next result apply Theorems 4.2.4 and 6.1.1. THEOREM 6.1.4 [10, page 52, Theorem 5.9] Let G be an rtffr group such that End(G) is commutative. Then G is a faithful group.

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CHAPTER 6. BAER SPLITTING PROPERTY

A faithfully E-flat group is a faithful group G that is flat as a left End(G)-module. EXAMPLE 6.1.5 Since each subring of Q is a pid, G is a faithfully E-flat group if End(G) ⊂ Q. Given a prime p ∈ Z, IZ(p∞ ) = Z(p∞ ) for b p . Thus Z(p∞ ) is not a faithful each nonzero ideal I in End(Z(p∞ )) ∼ =Z group even though End(G) is a pid. Evidently, Z(p∞ ) is not E-flat. The next example shows us that there are plenty of faithfully E-flat groups. EXAMPLE 6.1.6 The group constructed according to Corner’s Theorem 2.3.3 is faithfully flat. Proof: Let E be an rtffr group and let G be the group constructed according to Corner’s Theorem. There is a short exact sequence 0 → E −→ G −→ QE → 0 of left E-modules. Since E and QE are flat left E-modules, there is an exact sequence 0 = Tor1E (·, E) −→ Tor1E (·, G) −→ Tor1E (·, QE) = 0. Then Tor1E (·, G) = 0 and hence G is E-flat. Let I ⊂ End(G) be a right ideal. We may assume without loss of generality that I is a maximal right ideal. Lemma 4.2.3 states that E/I is a finite p-group for some prime p ∈ Z so pE ⊂ I ⊂ E. An application of E/I ⊗· to the above short exact sequence produces the exact sequence Tor1E (E/I, QE) → E/I → G/IG → QE/IQE. Since QE is a flat left E-module and since E/I is finite, Tor1E (E/I, QE) = 0 = QE/IQE, so that G/IG ∼ = E/I 6= 0. This completes the proof. We will construct rtffr groups whose properties are from the set {faithful, not faithful, E-flat, not E-flat}. EXAMPLE 6.1.7 There is an rtffr G that is E-flat but that is not faithful.

6.1. BAER’S LEMMA

143

Proof: Let k be an algebraic number field such that [k : Q] ≥ 2, and let O denote the ring of algebraic integers in k. It is known that Z/pZ-dim(O/pO) = [k : Q]. (See [58].) Let S = Z + pO 6= O, let A= and let M =

QS QHom(pO, S) QHom(S, pO) QO

=

k k k k

,

S pO

. Then let

E = {q ∈ A qM ⊂ M } S Hom(pO, S) = Hom(S, pO) O S pO . = pO O Furthermore I =

S pO pO pO

IM =

S pO pO pO

is a right ideal in E such that

S pO

=

S pO

= M.

Theorem 2.3.4 states that there is an rtffr group G and a short exact sequence 0 → M −→ G −→ A ⊕ A → 0

(6.4)

such that E ∼ = End(G). As in Example 6.1.6 we can prove that IM = M implies that IG = G. Thus G is not faithful. Since M is the first column of E, M is a projective (= flat) left E-module. Then as in Example 6.1.6, G is an E-flat group.

EXAMPLE 6.1.8 There is a faithful rtffr group that is not E-flat. Proof: In Example 4.3.2 we constructed a group G such that End(G) . is an integral domain but not a Dedekind domain and G = End(G) is not a flat left End(G). Since End(G) is commutative G is faithful. With these examples to guide us we will discuss faithful rtffr groups.

144

6.2

CHAPTER 6. BAER SPLITTING PROPERTY

Splitting of Exact Sequences

The groups in this section need not have finite rank. We continue our study of the Baer splitting property by considering a more general concept. Let G be a torsion-free group. 1. Let H ∈ P(G). Then (G, H) has the Baer splitting property if π the short exact sequence 0 → K −→ X −→ H → 0 is split exact whenever SG (X) + K = X. The group G has the Baer splitting property iff for each cardinal c and group H ∈ P(G), each short π exact sequence 0 → K −→ G(c) −→ H → 0 is split exact. 2. Let H ∈ P(G). Then (G, H) has the endlich Baer splitting property if for each positive integer c, the short exact sequence 0 → K −→ π G(c) −→ H → 0 is split exact. The group G has the endlich Baer splitting property iff for each positive integer c, each short exact π sequence 0 → K −→ G(c) −→ G → 0 is split exact. The term Baer splitting property was coined by U. Albrecht and H.P. Goeters [5] and the term endlich Baer splitting property first appears in [38]. Evidently G has the endlich Baer splitting property if G has the Baer splitting property. We will eventually prove that for an rtffr group G the endlich Baer splitting property implies the Baer splitting property.

The next two results show us that the set of groups H such that (G, H) has the Baer splitting property is closed under direct sums. LEMMA 6.2.1 Let G, H, H 0 be groups. If (G, H) and (G, H 0 ) have the Baer splitting property then (G, H ⊕ H 0 ) has the Baer splitting property. Proof: Suppose that (G, H) and (G, H 0 ) have the Baer splitting property and consider the short exact sequence π

0 → K −→ X −→ H ⊕ H 0 → 0

(6.5)

where SG (X) = X. We will show that (6.5) is split. Let ρ : H ⊕ H 0 −→ H be the canonical projection with ker ρ = H 0 . Since (G, H) has the Baer splitting property there is an injection g : H → X such that 1H = (ρπ)g = ρ(πg). Then H ⊕ H 0 = πg(H) ⊕ H 0

and X = g(H) ⊕ X 0

6.2. SPLITTING OF EXACT SEQUENCES

145

where X 0 = ker ρπ and where g(H) ∼ = H. Let g 00 : πg(H) ∼ = g(H) ⊂ X and notice that πg 00 (πg(H)) = πg(H). We can assume without loss of generality that πg 00 (x) = x for each x ∈ πg(H). The reader will show as an exercise that there is a short exact sequence π0

0 → K −→ X 0 −→ H 0 → 0 in which SG (X 0 ) = X 0 and π 0 is the restriction of π to X 0 . Since (G, H 0 ) has the Baer splitting property there is an injection g 0 : H 0 −→ X 0 such that π 0 g 0 = 1H 0 . It is then an easy task to show that π(g 00 ⊕ g 0 )(πg(x) ⊕ x0 ) = πg 00 (πg(x)) + πg 0 (x0 ) = πg(x) ⊕ x0 for each x ∈ H and x0 ∈ H 0 . Thus π(g 00 ⊕g 0 ) = 1H⊕H 0 so that (G, H ⊕H 0 ) has the Baer splitting property. This completes the proof. THEOREM 6.2.2 Let G, H be rtffr groups. 1. If (G, H) has the Baer splitting property then (G, H (n) ) has the Baer splitting property for each integer n > 0. 2. If (G, H) has the endlich Baer splitting property then (G, H (n) ) has the endlich Baer splitting property for each integer n > 0. Proof: Apply induction to n > 0 using Lemma 6.2.1 as the induction step. The above lemma is the induction step in a transfinite induction. THEOREM 6.2.3 [T.G. Faticoni] (See [38].) Let G, H be rtffr groups. Then (G, H) has the Baer splitting property iff (G, H (d) ) has the Baer splitting property for each cardinal d. Proof: Let H ∈ Po (G), let Hi ∼ = H for each i ∈ I, suppose that (G, H) has the Baer splitting property, let I be an index set, and let π

0 → K −→ X −→

M i∈I

Hi → 0

(6.6)

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CHAPTER 6. BAER SPLITTING PROPERTY

be a short exact sequence in which SG (X) + K = X. Given a subset J ⊂ I let M M ρJ : Hi −→ Hj i∈I

j∈J

be the canonical projection. Let B be the set of pairs (J, gJ ) such that J ⊂ I, gJ : ⊕j∈J Hj → X, and ρJ πgJ = 1Lj∈J Hj . Given i ∈ I, the Baer splitting property produces a map gi : Hi → X such that ρi πgi = 1Hi so B 6= ∅. Partially order B by the usual inclusion of sets and restriction of functions. An elementary argument shows that each chain in B contains an upper bound in B. Then by Zorn’s Lemma B contains a maximal element (Im , gm ). Assume for the sake of contradiction that Im 6= I. Let i ∈ I \ Im . The proof of Lemma 6.2.1 implies that there is a gi : Hi → X such that ρIm ∪{i} π(gm ⊕ gi ) = 1⊕`∈Im ∪{i} H` . Thus (Im ∪ {i}, gm ⊕ gi ) ∈ B. Since (Im , gm ) is maximal in B it must be that Im = I and ρI = 1⊕i∈I Hk = πgm . Thus (G, H (d) ) has the Baer splitting property for each cardinal d. COROLLARY 6.2.4 Let G be a group. 1. G has the endlich Baer splitting property iff (G, H) has the endlich Baer splitting property for each H ∈ Po (G). 2. G has the Baer splitting property iff (G, G) has the Baer splitting property. Proof: Parts 1 and 2 follow from Theorems 6.2.2 and 6.2.3.

6.3

G-Compressed Projectives

Baer’s Lemma 6.1.1 shows that we should be interested in simple right End(G)-modules M such that TG (M ) 6= 0. From this point of view, introduced by U. Albrecht [4], it is natural to consider the set of right End(G)-modules M such that TG (M ) = 0. Let M 6= 0 be a right End(G)-module. We say that M is Gcompressed if TG (M/N ) 6= 0 for each proper End(G)-submodule N ⊂ M .

6.3. G-COMPRESSED PROJECTIVES

147

The right End(G)-module K is finitely M -generated if there is an integer n and an isomorphism K ∼ = M (n) /N for some subgroup N ⊂ M (n) . The module M is finitely G-compressed TG (M/K) 6= 0 for each proper finitely generated submodule K of M . Let E = End(G). Given an H ∈ Po (G) the Arnold-Lady-Murley Theorem 2.4.1 shows that HG (H) is a finitely generated projective right End(G)-module. The Arnold-Lady-Murley Theorem 2.4.1 shows us that the dual module of HG (H) is HomE (HG (H), E) ∼ = HomE (HG (H), HG (G)) ∼ = Hom(H, G). Given a projective E = End(G)-module P the trace of P in E is the ideal ∆P =

P {f (P ) f ∈ HomE (P, E)}.

If H ∈ P(G) then the trace of H is the trace ideal of HG (H)

∆H = ∆HG (H) . It is known that if P is a projective right E-module then 1. ∆P is the smallest ideal in E such that P ∆P = P , and 2. ∆2P = ∆P . Notice that condition 6 in the following result is similar to the condition that IG 6= G for each right ideal I ⊂ End(G). THEOREM 6.3.1 [38, T.G. Faticoni] The following are equivalent for H ∈ Po (G). 1. HG (H)(n) is G-compressed for each integer n > 0. 2. HG (H) is G-compressed. 3. ∆H ⊂ I for each maximal right ideal I ⊂ End(G) such that IG = G. 4. IHom(H, G) = Hom(H, G) for each maximal right ideal I ⊂ End(G) such that IG = G.

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5. TG (M ) 6= 0 for each finitely HG (H)-generated right End(G)-modules M 6= 0. 6. N G 6= H for each proper End(G)-submodule N ⊂ HomR (G, H). 7. (G, H) has the Baer splitting property. 8. (G, H) has the endlich Baer splitting property. Proof: Throughout this proof let E = End(G), fix H ∈ Po (G), and let P = HG (H). Then P is a finitely generated projective right E-module. We will prove 1 ⇒ 2 ⇒ 3 ⇔ 4, 3 ⇒ 1 ⇔ 5 2⇔6⇒7⇒8⇒6 1 ⇒ 2 is clear. 2 ⇒ 3 We prove the contrapositive. Assume there is a maximal right ideal I ⊂ E such that IG = G but ∆P 6⊂ I. Let π : E → E/I be the natural projection map. There is a map f : P → E such that f (P ) 6⊂ I, and because I is a maximal right ideal of E, f (P ) + I = E. Hence πf : P → E/I is a surjection. Furthermore, because IG = G, TG (P/ ker πf ) ∼ = TG (E/I) ∼ = G/IG = 0 so that P is not G-compressed. 3 ⇔ 4 We have observed that P ∗ = Hom(H, G) and it is known that ∆P = ∆P ∗ in general. Assume part 3 is true. Given IG = G then ∆H ⊂ I. Then because ∆H = ∆HG (H) = ∆HG (H)∗ , IHG (H)∗ = HG (H)∗ . This proves part 4. Conversely assume part 4 is true. Let IG = G. Then IHG (H)∗ = IHom(H, G) = Hom(H, G) = HG (H)∗ implies that ∆H = ∆HG (H) = ∆HG (H)∗ ⊂ I. This is part 3. 3 ⇒ 1 We prove the contrapositive. Suppose there is an integer n > 0 and an E-submodule N ⊂ P n such that TG (P (n) /N ) = 0. Because P is finitely generated projective there is a maximal right E-submodule N ⊂ M ⊂ P n , and because P n /M is simple there is a maximal right ideal I ⊂ E such that P n /M ∼ = E/I. Then 0 = TG (P n /M ) = TG (E/I) = G/IG

6.3. G-COMPRESSED PROJECTIVES

149

so that P (n) is not G-compressed. The proof of 5 ⇔ 1 as an exercise. 2 ⇔ 6 Let N ⊂ HG (H) be a right E-submodule and apply the right exact functor TG (·) to the short exact sequence ı

π

0 → N −→ HG (H) −→ HG (H)/N → 0

(6.7)

to form the diagram TG (N )

TG (ı)-

TG HG (H)

- TG (HG (H)/N )

-0

ΘH ?

H of groups. Notice that image ΘH TG (ı) = N G. Because H ∈ Po (G) the Arnold-Lady-Murley Theorem 2.4.1 implies that ΘH is an isomorphism. Then N G = H iff TG (ı) is a surjection iff TG (HG (H)/N ) = 0. Then 2 ⇔ 6 follows immediately. 6 ⇒ 7 follows in the same way that we proved Baer’s Lemma 6.1.1, but in proving 6 ⇒ 7 use an End(G)-submodule N ⊂ HG (G) where we used a right ideal I ⊂ End(G). 7 ⇒ 8 is clear. 8 ⇒ 6 Suppose that N G = H for some proper right End(G)-submodule N ⊂ HG (H). Since HG (H) is finitely generated, there is a maximal right End(G)-submodule K ⊂ HG (H) that contains N . By Lemma 4.2.3 the simple module HG (H)/K is a finite p-group, so that pHG (H) ⊂ K ⊂ HG (H). Since H has finite rank K/pHG (H) ⊂ HG (H)/pHG (H) is finite and thus K is finitely generated by the generators for pHG (H) and representativesP of the cosets for K/pHG (H). Write K = ni=1 πi End(G) for some π1 , . . . , πn ∈ K. There is a short exact sequence 0

0 → H −→

n M

π

Gi −→ H → 0

(6.8)

i=1

where Gk ∼ = G for each i ∈ I and where π is the unique map such that π(x) = πi (x) for each i = 1, . . . , n each x ∈ Gi . Notice that Pand n π is a surjection since H = KG = i ). Furthermore, any i=1 πi (G P n map : H → ⊕ni=1 Gi can be written as = i=1 i for some maps i : H → Gi . Then for any map g : G → H we have ! n ! n n X X X πg = πi i g = πi i g. i=1

i=1

i=1

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CHAPTER 6. BAER SPLITTING PROPERTY

Since πi ∈ K and since i : H → Gi we have πg ∈ K for each g : G → H. Since K 6= HG (H) the reader will show that π 6= 1H for any : H → ⊕ni=1 Gi . Thus (6.8) does not split which proves that part 6 is false. This completes the logical circuit. The next result illustrates the connection between G-compressed modules, faithful groups, and Baer’s lemma. THEOREM 6.3.2 [38, T.G. Faticoni]. for an rtffr group G.

The following are equivalent

1. IG 6= G for each proper right ideal I ⊂ E. 2. IG 6= G for each finitely generated proper right ideal I ⊂ E. 3. TG (M ) 6= 0 for each finitely presented right End(G)-module M 6= 0. 4. TG (M ) 6= 0 for each finitely generated right End(G)-module M 6= 0. 5. G has the Baer splitting property. 6. G has the endlich Baer splitting property. Proof: 6 ⇔ 5 ⇔ 4 is Theorem 6.3.1(8), (7), (5). 4 ⇒ 3 is clear. 3 ⇒ 2 follows as in Theorem 6.3.1(5) ⇒ (4). 2 ⇒ 1 Given part 2 and a right ideal I ⊂ End(G) there is a maximal right ideal I ⊂ M ⊂ End(G). By Lemma 4.2.3, M is finitely presented so that IG ⊂ M G 6= G. This proves part 1. 1 ⇒ 5 is Baer’s Lemma 6.1.1. This completes the logical sequence. In the presence of a flatness hypothesis the above result can be improved a little. THEOREM 6.3.3 [3, U. Albrecht]. Let G be an E-flat rtffr group. Then G has the Baer splitting property iff TG (M ) 6= 0 for each right R-module M 6= 0. Proof: Say G has the Baer splitting property, and let M 6= 0 be a right End(G)-module. Let 0 6= Mo ⊂ M be any finitely generated End(G)-submodule. Since G is E-flat, TG (Mo ) ⊂ TG (M ), and by Theorem 6.3.2, TG (Mo ) 6= 0. Then TG (M ) 6= 0. The converse is Theorem 6.3.2 so the proof is complete.

6.4. SOME EXAMPLES

6.4

151

Some Examples

The examples in this section illustrate the abundance of the splitting properties, and they show that the Baer splitting property and the endlich Baer splitting property are in general different notions. EXAMPLE 6.4.1 1. Let G be rtffr group with commutative endomorphism ring. By Theorem 6.3.2 and Theorem 6.1.4, G has the Baer splitting property, so by Theorem 6.2.3, G(c) has the Baer splitting property for each cardinal c. However, let Io be the ideal of those f ∈ End(G(c) ) such that f (G) has finite rank. Then Io is a proper ideal of End(G(c) ) such that Io G(c) = G(c) . Thus the finite rank hypothesis in Theorem 6.3.2 is necessary. 2. G = Q has the Baer splitting property, and G = Z(p∞ ) does not have the endlich Baer splitting property. 3. D.M. Arnold and E.L. Lady [14] give an example of a completely decomposable abelian group G of rank 2 such that IG = G for some proper pure ideal I ⊂ End(G). 4. The group G constructed in Example 6.1.7 is one for which {right ideals I ⊂ End(G) IG = G} is a finite set. In [4], U. Albrecht proves that the self-small E-flat group G is faithful iff TG (K) 6= 0 for each right End(G)-module K 6= 0. The following example shows us that even if G is a faithful Corner group it may happen that TG (K) = 0 for some nonzero End(G)-module K. EXAMPLE 6.4.2 [36, T.G. Faticoni] There is a Corner group G such that 1. TG (N ) 6= 0 for each finite right End(G)-module N 6= 0. 2. There is a right End(G)-module K 6= 0 such that TG (K) = 0. Specifically G is faithful but TG (K) = 0 for some right E-module K 6= 0. Proof: Let p ∈ Z be a prime and let E be the localization of Z[x] at the maximal ideal (x, p). That is, E = Z[x](x,p) . Let M be the left E-module E E E ⊕ ⊕ ⊕ ··· (x) (x2 ) (x3 )

152

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where (xk ) is the ideal in E generated by xk . One shows that M is a reduced torsion-free left E-module. By Nakayama’s Lemma 1.2.3, J M 6= M where J = J (E). By Theorem 2.3.3 there is a short exact sequence 0 −→ M −→ G −→ QC −→ 0

(6.9)

∼ End(G). Since p ∈ J (E) = constructed in Theorem 2.3.3 such that E = J is the unique maximal ideal in E one proves in the now familiar way that G 6= JG because M 6= JM . Next let K be the left E-module K = (E/pE)bo ⊕ (E/pE)b1 ⊕ (E/pE)b2 ⊕ · · · where bj x = bj−1 for each j ≥ 1 and bo x = 0. An application of K ⊗E · to (6.9) produces the exact sequence K ⊗E M −→ K ⊗E G −→ K ⊗E QC → 0. Since QC is divisible and since K is a torsion group K ⊗E QC = 0. The reader will show that Kx = K. Since each element of M is annihilated by some power of x the usual argument shows us that K ⊗E M = 0. Therefore K ⊗E G = 0. Specifically, G is faithful but TG (K) = 0 for some right End(G)-module K 6= 0. This completes the proof. In the above example G is not self-small. The reader can produce a map f : G → G(ℵo ) whose image is not in any finite direct sum of copies of G. A few remarks are in order before we end this section. U. Albrecht [3, Theorem 2.1, Corollary 2.2] extends Baer’s Lemma by showing that the self-small right R-module G has the finite Baer splitting property iff IG 6= G for each proper right ideal I ⊂ E iff TG (M ) 6= 0 for each nonzero finitely generated right End(G)-module M 6= 0. Furthermore, he shows that if G is a self-small E-flat module, then G has the finite Baer splitting property iff TG (M ) 6= 0 for each right End(G)-module M 6= 0, [3, Corollary 2.3]. The faithful condition is central to the characterization of the rtffr groups G such that Ext1 (G, G) is a torsion-free abelian group, [11, 43]. D.M. Arnold and J. Hausen [12] use the endlich Baer splitting property to study modules G that have the summand intersection property, [41] and [12] use the endlich Baer splitting property to characterize those modules G such that End(G) is right semi-hereditary, and [3], [14], and [41] use the Baer splitting property to characterize those R-modules G such that End(G) is right Noetherian, right hereditary. K. Fuller [47] calls G completely faithful and U. Albrecht [3] calls G fully faithful if TG (M ) 6= 0 for each nonzero right End(G)-module M .

6.5. EXERCISES

6.5

153

Exercises

G, H, K are rtffr groups, p ∈ Z is a prime, E is an rtffr ring, M , N , are rtffr E-modules (left or right, depending on the setting). 1. Prove any of the results in this section left as exercises for the reader. 2. Let H ∈ P(G) and suppose that (G, H) has the endlich Baer splitting property. If H is finitely G-generated then H ∈ Po (G). 3. Refer to the proof of Theorem 6.2.3. Show that each chain in C contains an upper bound in C. 4. Let M be a right End(G)-module. Then M (n) is G-compressed for each integer n > 0 iff TG (N ) 6= 0 for each finitely M -generated right End(G)-module N 6= 0. 5. Let P be a projective right E-module, let P ∗ be the dual of P , and let I be a left ideal of E. Prove that IP = P iff ∆P ⊂ I and that ∆P = ∆P ∗ . 6. Let H ∈ Po (G). Show that Hom(G, H)∗ = Hom(H, G). 7. Prove that the following are equivalent. (a) IG 6= G for each proper right ideal I ⊂ E. (b) TG (M ) 6= 0 for each nonzero finitely generated M ∈ ModEndR (G) . 8. If IG 6= G for each proper finitely generated right ideal I ⊂ End(G) then G has the endlich Baer splitting property. 9. If H ∈ Po (G) and if N ⊂ HG (H) is such that N G = H then TG (HG (H)/N ) = 0. 10. Complete the proof of Theorem 6.2.3 by showing that π 6= 1H for any : H → G(c) . 11. If E is a reduced torsion-free ring of finite rank and if I is a maximal right ideal of E then E/I is finite. 12. Review the proof of Lemma 6.2.1. Show that the map g 00 : πg(H) → H 00 ⊂ X such that g 00 (πg(x)) = g(x) is a well defined injection such that πg 00 = 1πg(H) .

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13. Prove that the sequence 0 → K → X 0 → H 0 → 0 given in Lemma 6.2.1 is exact. 14. If (G, H) has the endlich Baer splitting property then (G, H (n) ) has the endlich Baer splitting property for each integer n > 0.

6.6

Questions for Future Research

Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. Given a ring E let Ω(E) denote the set of rtffr groups G such that E∼ = End(G). Some of the properties of E are seen in Ω(E). 1. Describe the rings E such that each G ∈ Ω(E) is a faithful group. Commutative and hereditary rings E satisfy this property. See [40] for more rings of this type. 2. In [43], it is shown that E is hereditary iff each group G ∈ Ω(E) is E-flat. Classify the rings E such that each group G ∈ Ω(E) is a faithfully E-flat module. 3. The group G has the Baer splitting property if for each cardinal c each surjection π : G(c) −→ G is split. Dualize the Baer splitting property in the obvious way. The group G has the dual-Baer splitting property if for each cardinal c, each injection G −→ Gc is split. See [5]. Characterize those rtffr groups G that satisfy the dualBaer splitting property. Develop a general theory as in Chapter 6. 4. Describe the rtffr rings E such that each G ∈ Ω(E) has the dualBaer splitting property. See [40]. 5. Characterize the G-compressed End(E)-modules. See Theorem 6.3.1. 6. Characterize the (rtffr) groups G such that TG (M ) 6= 0 for each nonzero right End(G)-module M . 7. Describe the rings E such that for each G ∈ Ω(E), TG (M ) 6= 0 for each nonzero right E-module M .

Chapter 7

J -Groups, L-Groups, and S-Groups The rtffr group G is called a J -group if G is isomorphic to each subgroup of finite index in G, G is called an L-group if G is locally isomorphic to each subgroup of finite index in G, and G is an S-group if each subgroup of finite index in G is G-generated. Obviously J -groups ⇒ L-groups ⇒ S-groups. In the first half of the chapter we will discuss how the torsion subgroup of Ext1Z (G, G) is related to (finitely) faithful S-groups. In the second half of the chapter we will scrutinize the converse relationships L-groups ⇒ J -groups, and finitely faithful S-groups ⇒ J -groups.

7.1

Background on Ext

Let G and H be groups. Call G an S-group if each subgroup of finite index in G is G-generated. S-groups are introduced by D.M. Arnold in [11]. Any group of the form Z ⊕ H is an S-group. If R is a pure subring b p for some prime p then R is an S-group. of Z Recall that G is finitely faithful if IG 6= G for each maximal right ideal of finite index in End(G). We investigate finitely faithful S-groups in an effort to find partial solutions to the question under what conditions on G are the groups quasi-isomorphic to G actually isomorphic to G? We will abbreviate notation as follows. Ext1Z (G, H) = Ext(G, H)

155

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CHAPTER 7. J -GROUPS, L-GROUPS, AND S-GROUPS

Then Ext(G, H) consists of the short exact sequences π

0 → H −→ X −→ G → 0

(7.1)

modulo an equivalence relation. See [46] for details. The split short exact sequences (7.1) correspond to 0 in Ext(G, H). The short exact sequence (7.1) is quasi-split if there is a map : G → X and an integer n 6= 0 such that π = n1G . Let tExt(G, H) = the torsion subgroup of Ext(G, H). C.L. Walker’s Theorem [46, Section 102] states that the quasi-split short exact sequences correspond to the torsion part of Ext(G, H). THEOREM 7.1.1 [85, C.L. Walker] The group tExt(G, H) is the set of equivalence classes of quasi-split short exact sequences (7.1). In particular given an equivalence class [E] ∈ tExt(G, H) of a short exact sequence E in (7.1) then n[E] = 0 iff there is a map : G → X such that π = n1G . Consequently Ext(G, H) is torsion-free iff each quasi-split short exact sequence (7.1) is split exact. Given a prime p ∈ Z and a torsion-free group G, G/pG is a Z/pZvector space. The p-rank of G is rp (G) = Z/pZ- dim(G/pG). Since Z/pZ-linearly independent elements in G/pG lift to Z-linearly independent elements in G, rp (G) ≤ rank(G) < ∞. For rtffr groups G and H, Ext(G, H) is a divisible group so there are integers {rp |p ∈ Z is a prime } such that as groups M Ext(G, H) ∼ Z(p∞ )(rp ) . (7.2) = Q(ℵ1 ) ⊕ primes p∈Z The numbers rp are called the p-rank of Ext(G, H). R.B. Warfield, Jr. [87] determines the values of the p-ranks rp of Ext(G, H). THEOREM 7.1.2 [R.B. Warfield, Jr.] (See [87].) Let G and H be torsion-free groups of finite p-rank for all primes p ∈ Z. The p-rank rp of Ext(G, H) is given by rp = rp (G)rp (H) − rp (Hom(G, H))

7.2. FINITE PROJECTIVE PROPERTIES

157

for all primes p ∈ Z. Consequently Ext(G, H) is p-torsion-free iff rp (G)rp (H) = rp (Hom(G, H)).

7.2

Finite Projective Properties

In this section we characterize the torsion-freeness of Ext(G, H) via a projective property. Let G be a group and consider the short exact sequence π

0 → K −→ H −→ H/K → 0.

(7.3)

1. Given a prime p ∈ Z, G is p-finitely H-projective if G is projective relative to (7.3) whenever H/K is a finite p-group. 2. G is finitely H-projective if G is projective relative to (7.3) whenever H/K is a finite group. 3. G is finitely projective if G is finitely G-projective. If p ∈ Z is prime, if pG = G, and if pH 6= H then G is trivially pfinitely H-projective. The reader can show that if Hom(G, H) = 0 then G is not p-finitely H-projective for any prime p ∈ Z such that pG 6= G and pH 6= H. EXAMPLE 7.2.1 Let G ⊂ H ⊂ Q. Then G is p-finitely H-projective. Proof: We may assume without loss of generality that pH 6= H and that G is p-dense in H. Given a map f¯ : G → H/pk H then f¯(pk G) = 0. Write H/pk H = hx + pk Hi. Since G is p-dense in H, H/pk H = (G + pk H)/pk H so we can assume that x ∈ G. Then f¯(x) = nx + pk H and so f¯ lifts to a map f : G → H over the inclusion map G ⊂ H. We demonstrate a relationship between G-projectiveness and quasiisomorphism. We will use without fanfare the fact that the torsion-free group G is (p)-finitely H-projective iff HG (·) is exact on the short exact sequence (7.3) whenever H/K is a finite p-group. The reader will prove that if H/H 0 is a finite group then Ext(G, H) ∼ = Ext(G, H 0 ). The Z-adic completion of G is b = lim G/nG G →n

(7.4)

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CHAPTER 7. J -GROUPS, L-GROUPS, AND S-GROUPS

and the p-adic completion of G is b p = lim G/pk G. G →k>0

b and G is p-pure and p-dense in G bp . The group G is pure and dense in G b b b That is, G ∩ nG = nG and G + nG = G for each integer n 6= 0. We will use the notation HG (·) = Hom(G, ·). THEOREM 7.2.2 [T.G. Faticoni and H.P. Goeters] (See [44].) Let G and H be torsion-free groups and let p ∈ Z be a prime. The following are equivalent. 1. G is (p-)finitely H-projective. 2. Ext(G, H) is (p-)torsion-free. b 3. Hom(G, H) is a (p-)pure and (p-)dense subgroup of Hom(G, H), b p )). (of Hom(G, H Proof: 1 ⇔ 2 An application of HG (·) to the short exact sequence p

π

0 → H −→ H −→ H/pH → 0 yields the exact sequence π∗

p

HG (H) −→ HG (H/pH) → Ext(G, H) −→ Ext(G, H) in which π ∗ = HG (π). Then Ext(G, H) is p-torsion-free iff multiplication by p is an injection iff π ∗ is a surjection iff G is p-finitely H-projective. b is pure-injective Ext(G, H) b = 2 ⇔ 3. Since G is torsion-free and H 0. Thus an application of HG (·) to the short exact sequence π b −→ b 0 → H −→ H H/H →0

yields the exact sequence ∗

∗

π b → b b = 0. HG (H) → HG (H) HG (H/H) → Ext(G, H) → Ext(G, H)

Then Ext(G, H) is p-torsion-free iff image π ∗ is p-divisible iff ker π ∗ = b image ∗ is p-pure and p-dense in HG (H). This completes the logical cycle.

7.3. FINITELY PROJECTIVE GROUPS

159

EXAMPLE 7.2.3 [87, R.B. Warfield, Jr.] Let G ⊂ H ⊂ Q be groups. We showed in Example 7.2.1 that G is finitely H-projective. Consequently Ext(G, H) is torsion-free.

7.3

Finitely Projective Groups

The group G is finitely faithful if IG 6= G for each right ideal I of finite index in End(G). Finitely faithful S-groups were introduced by D.M. Arnold [11]. We will continue to use the terminology finitely faithful Sgroup instead of faithful S-group since finitely faithful S-group seems to capture the spirit of our applications. Rank one groups are finitely faithful S-groups. Of course free groups of finite rank are finitely faithful S-groups, but G = Z(ℵo ) is an S-group without being faithful. If G is a finitely faithful S-group then G(n) is a finitely faithful S-group. (This follows from part 1 of Theorem 7.3.1 below.) We will approach the problem of classifying the finitely faithful Sgroups by considering the finitely projective groups. We begin by showing that the finitely faithful S-groups have finite p-rank for all primes p ∈ Z. Let π

0 → H −→ G −→ G/H → 0

(7.5)

be a short exact sequence in which G/H is finite. THEOREM 7.3.1 Let G be a reduced torsion-free group and let p ∈ Z be a prime. The following are equivalent. 1. G/pG is finite and G is projective relative to each short exact sequence (7.5) in which pG ⊂ H ⊂ G. 2. (a) If I is a right ideal of End(G) that contains some power of p then IG 6= G. (b) If H is a subgroup of G such that pG ⊂ H ⊂ G then H is G-generated. Proof: Let E = End(G). 1 ⇒ 2 Assume part 1, and let pE ⊂ I ⊂ E be a right ideal such that IG = G. By part 1, E/pE ∼ = HG (G/pG) ∼ = Hom(G/pG, G/pG)

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so that E/pE is a simple Artinian ring. Let I be a right ideal of E such that pE ⊂ I ⊂ E and IG = G. There is no loss of generality in assuming that I is maximal in E and thus that pE ⊂ I. Since E/pE is a simple Artinian ring there is an e2 = e ∈ E/pE such that e(E/pE) = I/pE. Then G/pG = IG/pG = (I/pE)(G/pG) = e(E/pE)(G/pG) = e(G/pG). Inasmuch as x(G/pG) 6= 0 for each nonzero x ∈ E/pE and since (1 − e)(G/pG) = 0 we conclude that e = 1. Thus I = E, as required by part (a). Let pG ⊂ H ⊂ G. For each x ∈ H \pG there is a map fx : G → H/pG such that x + pG ∈ fx (G). By hypothesis in part 1, fx lifts to a map gx : G → H such that x ∈ gx (G) + pG. Then H is G-generated, which proves part (b). This completes the proof of 1 ⇒ 2. 2 ⇒ 1. We first show that G/pG is finite. Let I ⊂ E be the ideal such that I/pE = {f ∈ E/pE Z/pZ- dim(f (G/pG)) is finite }. Given a subgroup pG ⊂ H ⊂ G such that H/pG ∼ = Z/pZ there is by part (b) a surjection fH : G → H/pG. Notice that fH ∈ I/pE and also that G/pG = (I/pE)(G/pG) = IG/pG. By part (a), I = E. Thus 1G has finite image in G/pG and hence G/pG is finite. Apply HG (·) to the short exact sequence (7.5). To show that π ∗ = HG (π) is a surjection we will prove the next lemma. LEMMA 7.3.2 Let G be a torsion-free group that satisfies Theorem 7.3.1(2). Then HG (Z/pZ) = Hom(G, Z/pZ) is a simple right E-module. Proof: For any nonzero map f : G → Z/pZ, ker f is a maximal subgroup of G that contains pG. Then pE ⊂ annE (f ) = HG (ker f ) ⊂ E. We claim that annE (f ) is a maximal right ideal of E. Suppose that annE (f ) ⊂ I ⊂ E and that annE (f ) 6= I. Then by part 2(b) we have pG ⊂ ker f = HG (ker f )G 6= IG ⊂ G.

7.3. FINITELY PROJECTIVE GROUPS

161

By the maximality of ker f in G we have IG = G and hence part 2(a) implies that I = E. As claimed for each f ∈ HG (Z/pZ), annE (f ) is a maximal right ideal in E. Now assume to the contrary that there are f 6= 0 6= g ∈ HG (Z/pZ) such that f E 6= gE are distinct simple right E-modules. Then f E ⊕ gE ⊂ HG (Z/pZ). By the above paragraph annE (f + g) is a maximal right ideal in E, whence annE (f ⊕ g) is a maximal right ideal of E such that annE (f ⊕ g) = annE (f ) ∩ annE (g) ⊂ annE (f ). Thus annE (f ⊕ g) = annE (f ) = annE (g). Since annE (f ⊕ g) = HG (ker(f ⊕ g)) and since (f ⊕ g)G is finite, part 2(b) implies that ker(f ⊕ g) is Ggenerated, so that ker(f ⊕ g) = HG (ker(f ⊕ g))G = annE (f ⊕ g)G = annE (f )G = ker f as abelian groups. Similarly ker(f ⊕ g) = ker g so g = uf for some automorphism u of Z/pZ. Since u is then multiplication by some integer, f E = gE. This contradiction to our choice of f E 6= gE shows us that HG (Z/pZ) = f E is a simple right E-module. This completes the proof of the Lemma. To continue the proof of Theorem 7.3.1: Choose a subgroup pG ⊂ H ⊂ G such that H/pG ∼ = Z/pZ. Consider the short exact sequence (7.5). By part 2(b) and because H/pG is simple there is a homomorphism f : G → H such that f (G) + pG = H. Then 0 6= πf ∈ HG (H/pG) ∼ = HG (Z/pZ) so by Lemma 7.3.2, πf E = HG (Z/pZ). Therefore image π ∗ = πHG (G) ⊃ HG (H/pG) for each such H. Inasmuch as G/pG is a direct sum of copies of Z/pZ, image π ∗ = HG (G/pG). That is, π ∗ is a surjection which proves part 1 and completes the proof.

162

7.4

CHAPTER 7. J -GROUPS, L-GROUPS, AND S-GROUPS

Finitely Faithful S-Groups

Using Theorem 7.3.1 we can characterize the finitely projective groups as being finitely faithful S-groups. THEOREM 7.4.1 [D.M. Arnold] (See [11].) The following are equivalent for a torsion-free group G for which rp (G) is finite for all primes p ∈ Z. 1. G is finitely projective. 2. Ext(G, G) is torsion-free. 3. G is a finitely faithful S-group. 4. rp (G)2 = rp (End(G)) for each prime p ∈ Z. 5. End(G)/pEnd(G) ∼ = Matrp (G) (Z/pZ) as rings for each prime p ∈ Z. 6. Each quasi-split short exact sequence 0 → G → X → G → 0 is split exact. Proof: 2 ⇔ 3 follow from Theorem 7.2.2 and 2 ⇔ 6 follows from Theorem 7.1.1. 1 ⇒ 5 Suppose that G is finitely projective. An application of HG (·) to the short exact sequence p

π

0 → G −→ G −→ G/pG → 0 yields the isomorphisms End(G)/pEnd(G) = image π ∗ = HG (G/pG) ∼ = Hom(G/pG, G/pG). Since G is rtffr Hom(G/pG, G/pG) ∼ = Matrp (G) (Zp /pZ) which proves 5. 5 ⇒ 4 By part 5 End(G)/pEnd(G) ∼ = Matrp (G) (Zp /pZ) ∼ = Hom(G/pG, G/pG) as rings. Then rp (End(G)) = Z/pZ- dim(Hom(G/pG, G/pG)) = rp (G)2 .

7.4. FINITELY FAITHFUL S-GROUPS

163

This proves part 4. 4 ⇒ 2 follows from Theorem 7.1.2. 2 ⇒ 1 follows from Theorem 7.2.2, which completes the logical cycle.

For instance, if G is an rtffr group then rp (G) is finite for all primes p ∈ Z. COROLLARY 7.4.2 Let G be an rtffr group. If G is a finitely faithful S-group then 1. G(k) is a finitely faithful S-group for each integer k > 0. 2. If G = H ⊕ H 0 then H is a finitely faithful S-group. . 3. If G = H then H is a finitely faithful S-group. Proof: 1. Since G is a finitely faithful S-group, Ext(G, G) is torsionfree (Theorem 7.4.1), which implies that Ext(G(k) , G(k) ) is torsion-free for each k > 0. Theorem 7.4.1 implies that G(k) is a finitely faithful S-group. 2. If G = H ⊕ H 0 then Ext(H, H) is a direct summand of the torsionfree group Ext(G, G). Now apply Theorem 7.4.1. . 3. If G = H then by Theorem 7.4.1, Ext(G, G) ∼ = Ext(H, H) are torsion-free groups. This completes the proof. COROLLARY 7.4.3 If G is a finitely faithful S-group then End(G) is a semi-prime Noetherian ring. Proof: For each prime p ∈ Z, End(G)/pEnd(G) is a simple ring, so N (End(G)) ⊂ pEnd(G) for all p. Thus pN (End(G)) = N (End(G)) is divisible. Since G is reduced N (End(G)) = 0. This completes the proof. EXAMPLE 7.4.4 This is an example of an S-group that is not faithful. Proof: Let 0 6= X ⊂ Q be a subgroup such that End(X) = Z, Hom(X, Z) = 0, and let G= Z ⊕ X. Clearly G is an S-group. Observe Z 0 that End(G) = is not semi-prime. Observe that for each Z Z Z 0 integer n, In = is a right ideal of End(G) such that In G = G. Z nZ Thus G is not faithful. In the above example Ext(G, G) is not torsion-free. The next example gives a nonsplit quasi-split short exact sequence.

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EXAMPLE 7.4.5 Let Z ⊂ X ⊂ Q be such that Hom(X, Z). Then G = Z ⊕ X is an S-group that is not faithful and so Ext(X, Z) has nonzero torsion subgroup. Thus there is a quasi-split, nonsplit extension 0 → Z → H → X → 0 of groups. The embedding sends Z to Z(p ⊕ 1) ∈ Z ⊕ X. Notice that Z(p ⊕ 1) is a pure subgroup, a quasi-summand, but not a direct summand of Z ⊕ X so that 0 → Z. → H → X → 0 is not split exact. Compare this to the fact that if G ∼ = H then H ∼ = Z ⊕ X. bp. An example of torsion-free Ext is found in Z b p be pure subgroups of Z b p . InasEXAMPLE 7.4.6 Let G ⊂ H ⊂ Z bp → Z b p we see much as each map f : G → H lifts to a unique map f : Z that b p = Hom(G, b H). b Z ⊂ Hom(G, H) ⊂ Z b p , Hom(G, H) is pure and Moreover since H is torsion-free and pure in Z b H). b Then by Theorem 7.2.2, Ext(G, H) is torsion-free. dense in Hom(G, The examples of S-groups so far either have nonzero nilradical or they have commutative endomorphism ring. For finitely faithful S-groups our examples thus far satisfy End(G)/pEnd(G) ∼ = Z/pZ. Note the next example. EXAMPLE 7.4.7 [11, D.M. Arnold] This is an example of a finitely faithful S-group such that rp (G) = 2 and End(G)/pEnd(G) ∼ = Mat2 (Z/pZ). Let E be a maximal Z-order in the Hamiltonian Quaternions over Q. If p 6= 2 ∈ Z is a prime then E/pE ∼ = Mat2×2 (Z/pZ) and the completion ∼ b b b Ep satisfies E = Mat2×2 (Z). (See [69].) Let M be a pure and dense rtffr E-submodule such that ! b 0 Z 0 Z ⊂M ⊂ b 0 Z 0 Z and using Theorem 2.3.4 construct a short exact sequence of left Emodules such that 0 → M −→ G −→ QE ⊕ QE → 0. (The reader can show that E = O(M ).) Since rp (E) = 4 = rp (G)2 , G is a finitely faithful S-group with noncommutative endomorphism ring E.

7.5. ISOMORPHISM VERSUS LOCAL ISOMORPHISM

7.5

165

Isomorphism versus Local Isomorphism

In this section we engage in a little alphabet soup. Let G be an rtffr group. 1. G is a J -group if G is isomorphic to each subgroup of finite index in G. J is for the author of J´ onsson’s Theorem 2.1.10. 2. G is an L-group if G is locally isomorphic to each subgroup of finite index in G. L is for E.L. Lady who first used local (= near) isomorphism on abelian groups. See [56]. 3. G is an S-group if each subgoup of finite index in G is G-generated. S is for G-socle, SG (H). See Section 6.1. Evidently G is an S-group if G is an L-group if G is a J -group. We will investigate the implications S-group ⇒ L-group ⇒ J -group. Examples of J -groups include the following. 1. Groups of the form Z⊕H, with H ⊂ Q, are J -groups. See Theorem 2.1.12. 2. Groups of the form Z ⊕ K, with K 6= 0, are S-groups. The reader can show that Z ⊕ K is a J -group if K is a J -group. 3. If X ⊂ Q then G = X (n) is a J -group for each integer n > 0. It is interesting to note that we did not give an example of an L-group that is not a J -group. Call G a Murley group if p-rank(G) ≤ 1 for each prime p ∈ Z. We will show that certain classes of J -groups are Murley groups. THEOREM 7.5.1 [H.P. Goeters] The following are equivalent for the rtffr group G. 1. G is a Murley group. 2. G is a J -group and End(G) is commutative. 3. G is an S-group and End(G) is commutative. Proof: 1 ⇒ 2 Let G be a Murley group and let H be a subgroup of finite index in G. There is a chain of subgroups of G such that H ⊂ H1 ⊂ · · · ⊂ Hk ⊂ G

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∼ Z/pZ. Because G/pG ∼ and such that Hi /Hi−1 = = Hk /pHk for quasiisomorphic groups, the groups Hi are Murley groups. But then pH1 ⊂ H ⊂ H1 and p-rank(H1 ) ≤ 1 so by induction H = pH1 ∼ = ··· ∼ = Hk ∼ = G. = H2 ∼ = H1 ∼ Hence G is a J -group. Since p-rank(G) ≤ 1 for each prime p ∈ Z, the isomorphisms G/pG ∼ = b Since Z b is pure injective each b ⊂ Z. Z/pZ lift to a pure imbedding G b → Z. b That is, End(G) imbeds in f : G → G lifts to a unique map fˆ : Z b the commutative ring Z. This proves part 2. 2 ⇒ 3 is clear. 3 ⇒ 1 Suppose that G is an S-group and that End(G) is commutative. Then End(G)/pEnd(G) is a commutative ring for each prime p ∈ Z. By Theorem 4.2.4 and part 3, G is a finitely faithful S-group so by Theorem 7.4.1 End(G)/pEnd(G) ∼ = Matrp (G)×rp (G) (Z/pZ) as rings for all primes p ∈ Z. Consequently rp (G) ≤ 1 for all primes p so that G is a Murley group. This completes the proof. EXAMPLE 7.5.2 This is an example of a strongly indecomposable J -group that is not a Murley group. This example is inspired by D.M. Arnold’s example in [11]. Choose a prime p 6= 2 ∈ Z and let E be a classical maximal Zp -order in the Hamiltonian Quaternions over Q. Then each right ideal I ⊂ E is principal, I = xE for some x ∈ E. bp satisfies E bp ∼ Since Ep /pEp ∼ = Mat2×2 (Z/pZ), the completion E = b Mat2×2 (Zp ). As in Example 7.4.7 construct a pure and dense rtffr Esubmodule M such that ! bp 0 Zp 0 Z ⊂M ⊂ bp 0 Zp 0 Z and using Theorem 2.3.4 construct a short exact sequence of left Emodules such that 0 → M −→ G −→ QE ⊕ QE → 0. (The reader can show that E = O(M ).) Since rp (E) = 4 = rp (G)2 , G is a finitely faithful S-group but not a Murley group. . In fact given H = G, there is a right ideal I ⊂ E such that H = IG. Since I ∼ = E, H ∼ = G, so that G is an indecomposable J -group.

7.6. ANALYTIC NUMBER THEORY

7.6

167

Analytic Number Theory

The next series of results will demonstrate a beautiful connection between J -groups, L-groups, and S-groups. However, they require some deep results on Analytic Number Theory whose proofs would take us too far afield. Thus we will state some results and reference their proofs. The following theorem gives a rather general condition under which a finitely faithful S-group is a J -group. Let i, j, k be such that −1 = i2 = j 2 = k 2 = ijk and let k be an algebraic number field. The algebra D = k1 ⊕ ki ⊕ kj ⊕ kk is called a totally definite quaternion algebra if each embedding of k into the complex numbers is an embedding of k into the reals. Observe that k = center(D), D is a division k-algebra, and D has dimension 4 over its center k. If we let k = Q then D is the Q-algebra of Hamiltonian Quaternions. Thus the classical Q-algebra of Hamiltonian Quaternions is a totally definite quaternion algebra. Let G be an rtffr group and write QE(G) = A1 × · · · × At for some nonzero simple Artinian Q-algebras A1 , . . . , At . We say that the rtffr group G is an Eichler group if Ai is not a totally definite quaterion algebra for each simple factor Ai of QE(G). Note that Eichler groups are rtffr groups. A proof of the next result is found in [44, Proposition III.6]. THEOREM 7.6.1 If the Eichler group G is a finitely faithful S-group then G is a J -group. The next result shows just how rare an exception the totally definite quaternion algebra is. COROLLARY 7.6.2 Write the rtffr group G as .

(n1 )

G∼ = G1

(nt )

⊕ · · · ⊕ Gt

(7.6)

for some strongly indecomposable groups G1 , . . . , Gt and integers t, n1 , . ∼ . . ., nt > 0 such that Gi = Gj ⇒ i = j. Suppose that G is a finitely faithful S-group. Then G is a J -group in either of the following cases. 1. ni 6= 1 for each i = 1, . . . , t. 2. rank(Gi ) is not divisible by 4 for each i = 1, . . . , t.

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Proof: Since we are assuming that G is a finitely faithful S-group, (n ) it suffices to show that for each i, QE(Gi i ) is not a totally definite quaternion algebra. (n ) 1. In case ni 6= 1 then QE(Gi i ) is a nontrivial ring of matrices so (n ) it is not a domain. Thus QE(Gi i ) is not a totally definite quaternion algebra. 2. By part 1 we can assume without loss of generality that ni = 1. Since rank(Gi ) is not divisible by 4, and since QE(Gi ) is a division ring, rank(E(Gi )) 6= 4. (See Lemma 4.2.1.) Then QE(Gi ) is not a totally definite quaternion algebra. The conditions ni 6= 1 and rank(Gi ) not divisible by 4 are satisfied in case 1. G = H (2) for some group H, or 2. For each i = 1, . . . , t, (ni )

QE(Gi

) = Matni (ki )

for some algebraic number field ki . Before we continue with our study of L-groups we will introduce a bit of power. THEOREM 7.6.3 [R.B. Warfield, Jr.] (See [88].) Let G and H be rtffr groups. Then G and H are locally isomorphic iff G(n) ∼ = H (n) for some integer n > 0. We need a pair of lemmas that are themselves interesting. The semiprime ring E is integrally closed if given a ring E ⊂ R ⊂ QE such that R/E is finite then E = R. Equivalently E is integrally closed iff E is a finite product of classical maximal orders. The following is a cancellation result for rtffr groups such that A(G) is integrally closed. D.M. Arnold [10] has proved a similar result. THEOREM 7.6.4 [T.G. Faticoni] Let G be an Eichler group such that E(G) is an integrally closed ring. If H is an rtffr group then G⊕G∼ = G ⊕ H =⇒ G ∼ = H. . . Proof: By J´onsson’s Theorem 2.1.10, G ∼ = H, so that E(G) = A(H). By Theorem 2.4.4 we have

E(G) ⊕ E(G) ∼ = E(G) ⊕ A(H)

7.6. ANALYTIC NUMBER THEORY

169

as right modules over the integrally closed ring E(G). There are classical maximal orders E i such that E(G) = E 1 × · · · × E t . Since E(G) ∼ = A(H)E i for each i = 1, . . . , t we have = A(H) iff E i ∼ reduced to the case where E(G) is a classical maximal order. So assume that E(G) is a classical maximal order with center S. Given a right ideal I ⊂ E(G) let nr(I) be the reduced norm of I in S. (See [69, page 214].) From E(G) ⊕ E(G) ∼ = E(G) ⊕ A(H) and [69, page 311, Corollary 35.11(ii)] we see that nr(E(G)) = nr(E(G)) · nr(E(G)) = nr(E(G)) · nr(A(H)) = nr(A(H)). Inasmuch as G is Eichler, [69, page 311, Corollary 35.11(ii)] implies that E(G) ∼ = A(H), so that by Theorem 2.4.4, G ∼ = H. COROLLARY 7.6.5 Let G be an L-group. Then E(G) is an integrally closed ring. . Proof: Let E = E(G). By Lemma 3.1.8 there is a group G = G such that E = E(G) is an integrally closed ring. Because G is an L-group G is (n) locally isomorphic to G and then Theorem 7.6.3 states that G(n) ∼ =G for some integer n > 0. Then Matn (End(G)) ∼ = End(G(n) ) ∼ = End(G

(n)

)∼ = Matn (End(G))

so that Matn (E) ∼ = Matn (E) is an integrally closed ring. Since maximal order is a Morita invariant property, [69], integrally closed is a Morita invariant property, so that E is integrally closed. This completes the proof. LEMMA 7.6.6 If G is an rtffr L-group and if G is locally isomorphic to H ⊕ K then H is an L-group. Proof: Let G be locally isomorphic to H ⊕ K and suppose that . H0 ∼ = H. Then G, H ⊕ K, and H 0 ⊕ K are quasi-isomorphic. Since G is assumed to be an L-group H ⊕ K and H 0 ⊕ K are locally isomorphic groups. Then by Theorem 2.5.7, H is locally isomorphic to H 0 , whence H is an L-group. This completes the proof.

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170

LEMMA 7.6.7 Let G be an rtffr L-group. There is a set {G1 , . . ., Gt } of strongly indecomposable groups and integers t, n1 , . . . , nt > 0 such that .

1. Gi ∼ = Gj ⇒ i = j, 2. G is locally isomorphic to (nt )

G0 = G t

(nt )

⊕ · · · ⊕ Gt

,

3. for each i = 1, . . . , t, Gi is an L-group. Proof: Suppose that G is an L-group. 1 and 2. By J´onsson’s Theorem 2.1.10 there are strongly indecomposable groups {G1 , . . . , Gt } and integers n1 , . . . , nt > 0 such that . Gi ∼ = Gj ⇒ i = j and such that G is quasi-isomorphic to G0 . Since G is an L-group G is locally isomorphic to G0 . Part 3 follows from Lemma 7.6.6, so the proof is complete. COROLLARY 7.6.8 Let G be an rtffr L-group. Then G = G 1 ⊕ · · · ⊕ Gr for some strongly indecomposable L-groups G1 , . . . , Gr . Proof : Apply Arnold’s Theorem 3.5.1(2) to the above lemma. LEMMA 7.6.9 Let G be an indecomposable rtffr L-group. Then End(G) is a classical maximal order in the division ring QEnd(G). Proof: Let E = End(G) and let N = N (E). By Lemma 7.6.7 an indecomposable L-group is strongly indecomposable so that QEnd(G) is a local ring. That is, each x ∈ QE \ QN is a unit in QE, Lemma 1.3.3. Since G is reduced there is an integer n 6= 0 such that ∩k>0 nk G = 0. Let I = nE + N . Then G and IG are locally isomorphic so that by Theorem 2.5.1, HG (G) = E and HG (IG) are locally isomorphic right Emodules. Lemma 2.5.4 states that the localizations En and HG (IG)n are isomorphic, so HG (IG)n = xEn for some x ∈ HG (IG)n . Since x 6∈ Nn is a nonzero-divisor in En , Nn ⊂ HG (IG)n = xEn implies that Nn = xNn = xk Nn for each integer k > 0. Moroeover there is an integer k > 0 such that N k = 0. Then (xEn )k ⊂ (nEn + Nn )k ⊂ nk En + nk−1 Nn + · · · + nNnk−1 ⊂ nEn

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171

hence Nn = xk Nn ⊂ (xE)k Nn ⊂ nNn ⊂ Nn whence nN = N . Subsequently N = ∩k nk G = 0, so that N = 0. Then by Corollary 7.6.5, E/N (E) = E is integrally closed. Inasmuch as the semi-prime local ring QE is a division ring E is a classical maximal order in a division ring. This completes the proof. We can use some of the above results to prove that an rtffr L-group is almost faithfully E-flat. LEMMA 7.6.10 Let G be an rtffr L-group and let E = End(G). If E is semi-prime then G is a faithful E-flat group. Proof: Suppose that E is semi-prime and let I be a right ideal of E. Apply I ⊗E · to the exact sequence 0 → G −→ QG −→ QG/G → 0 to produce the following exact sequence of groups. Tor1E (I, QG/G) −→ I ⊗E G −→ I ⊗E QG. By Corollary 7.6.5, E is integrally closed, hence hereditary, whence I is a projective (= flat) right E-module. Consequently Tor1E (I, ·) = 0 and so I ⊗E G −→ I ⊗E QG is an injection. Since QG is a direct summand of QE (n) as a left E-module and since QE is a flat E-module QG is a flat left E-module. Thus I ⊗E QG → E ⊗E QG ∼ = QG is an injection and so I ⊗E G → G is an injection. Then G is E-flat. Next suppose that I is a maximal right ideal of finite index in E such that IG = G. Since I is a finitely generated projective right EndR (G)module the Arnold-Lady-Murley Theorem 2.4.1 implies that I ∼ = E is principal, say I = xE for some x ∈ E. Then xG = xEG = IG = G. Since G has finite rank x is an automorphism of G. That is I = E so that G is faithful. This completes the proof.

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172

7.7

Eichler L-Groups Are J -Groups

To this point there have been three classes of groups that we have considered. We showed that the finitely faithful S-groups are J -groups except in one case, and we know that J -groups are L-groups. The main Theorems 7.7.2 and 7.7.3 will show us that Eichler L-groups are J -groups. LEMMA 7.7.1 Let G be an rtffr L-group. If G = H ⊕ K then H is an L-group. .

Proof: Suppose that G is an L-group and let H 0 ∼ = H. Then .

H0 ⊕ K ∼ =H ⊕K =G implies that H 0 ⊕ K, H ⊕ K, and G are locally isomorphic since G is an L-group. But then Theorem 2.5.7 implies that H 0 and H are locally isomorphic. Thus H is an L-group. Note the use of Analytic Number Theory in the proof of the next two theorems. THEOREM 7.7.2 [T.G. Faticoni] The following are equivalent for an indecomposable Eichler group G. 1. G is a J -group. 2. G is an L-group. 3. G is a finitely faithful S-group. Proof: We have commented on the implication 1 ⇒ 2. 2 ⇒ 3 Of course G is an S-group if G is a L-group. By Lemma 7.6.9, End(G) is a semi-prime ring so Lemma 7.6.10 implies that the indecomposable L-group is faithful. This proves part 3. 3 ⇒ 1 If G is a finitely faithful S-group then by Corollary 7.4.2, G(2) is a finitely faithful S-group, so that G(2) is a J -group by Corollary 7.6.2. By Lemma 7.7.1, G is then an L-group. . To show that G is a J -group suppose that G = H. Then H is locally isomorphic to G so that G ⊕ H ∼ = G ⊕ G (Lemma 2.5.6). Because G is indecomposable, End(G) is a classical maximal order in the division Qalgebra QEnd(G), Lemma 7.6.9. Then by Theorem 7.6.4, G ∼ = H. Hence G is a J -group and part 1 is proved. This completes the proof. THEOREM 7.7.3 [T.G. Faticoni] Eichler L-groups are J -groups.

7.7. EICHLER L-GROUPS ARE J -GROUPS

173

. Proof: Let G be an Eichler L-group and let H = G. Then H is locally isomorphic to G. By Corollary 7.6.8 G = G 1 ⊕ · · · ⊕ Gr for some strongly indecomposable L-groups G1 , . . . , Gr . The reader will prove as an exercise that each Gi is an Eichler group. By Arnold’s Theorem 3.5.1(2) H = H1 ⊕ · · · ⊕ H r where H1 , . . . , Hr are indecomposable groups such that Gi and Hi are . locally isomorphic. In particular, Gi ∼ = Hi . By Theorem 7.7.2, the indecomposable Eichler L-groups Gi are J -groups. Hence Gi ∼ = Hi for ∼ each i = 1, . . . , r, whence G = H, and therefore G is a J -group. COROLLARY 7.7.4 The Eichler group G is a J -group iff it is an L-group. There is one other interesting case where our groups overlap. THEOREM 7.7.5 The following are equivalent for an Eichler group G. 1. G is a finitely faithful S-group. 2. End(G) is semi-prime and G is a J -group. Proof: Assume part 1. Let G be a finitely faithful S-group. By Corollary 7.4.2, G(2) is a finitely faithful S-group, so by Theorem 7.6.1, G(2) is a J -group. By Lemma 7.7.1, G is then an L-group, so by Theorem 7.7.3, G is a J -group. Lemma 7.4.3 states that End(G) is a semi-prime ring which proves part 2. Conversely, assume that G is a J -group with semi-prime endomorphism ring. By Lemma 7.6.10, G is faithful. Since G is clearly an S-group we have proved part 1. EXAMPLE 7.7.6 An example due to Eichler ([69, midpage 305]) produces a totally definite quaternion algebra D with center k an algebraic number field, a maximal O-order E in D, where O is the ring of algebraic integers in k, and a nonprincipal right ideal E ⊂ L ⊂ D such that nr(L) = αO for some α ∈ k. Then L and E have the same reduced norm, but L ∼ 6 E. Compare this to the last lines of the proof of Theorem 7.6.4. = This explains the necessity of the Eichler hypothesis in Theorem 7.6.4.

174

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Exercises

G, H, K are rtffr groups, p ∈ Z is a prime, E is an rtffr ring, M , N , are rtffr E-modules (left or right, depending on the setting). 1. G is a J -group if G is a pure subgroup of Z. .

π 2. Let 0 → K → X −→ G → 0 be quasi-split. Then X ∼ = G ⊕ K.

3. Let G be a faithful rtffr group. Show that if I ⊂ End(G) is a maximal right ideal then IG 6= G iff I = Hom(G, IG). 4. Suppose that H/H 0 is finite. Then G is finitely H-projective iff G is finitely H 0 -projective. 5. If H/H 0 is a finitely group then the inclusion H 0 ⊂ H induces a canonical isomorphism Ext(G, H) ∼ = Ext(G, H 0 ) of groups. 6. Let G be a torsion-free group such that rp (G) is countably infinite. If G is p-finitely G-projective then End(G) is uncountable. 7. Let G and H be torsion-free groups and let p ∈ Z be a prime. If G is (p-)finitely H-projective then for each integer k > 0 (a) G(k) is (p-)finitely H-projective. (b) G is (p-)finitely H (k) -projective. 8. Let G and H be rtffr groups. If G is finitely H-projective then π each short exact sequence 0 → K −→ H −→ G → 0 is split exact. 9. Let G be an rtffr group and suppose that G/H is finite. Then G is a finitely faithful S-group iff H is a finitely faithful S-group. 10. Let G be an rtffr group. Then G is a finitely faithful S-group iff G is a finitely faithful L-group. 11. Let G and H be rtffr groups. Assume that H 0 has finite index in H. (a) Show that G is p-finitely H-projective iff G is p-finitely H 0 projective. (b) Show that G is finitely H-projective iff G is finitely H 0 -projective. 12. Let G and H be torsion-free groups and let p ∈ Z be a prime. If G is (p-)finitely H-projective then for each integer k > 0

7.9. QUESTIONS FOR FUTURE RESEARCH

175

(a) G(k) is (p-)finitely H-projective. (b) G is (p-)finitely H (k) -projective. b then G is a J -group. 13. [C. Murley] If G is a pure subgroup of Z 14. Show that an indecomposable L-group is strongly indecomposable. 15. Show that an indecomposable L-group is a finitely faithful S-group.

7.9

Questions for Future Research

Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. Dualize p-finitely H-projective groups and study the resulting class of groups. 2. If G is p-finitely projective then Hom(G, Z/pZ) is a simple End(G)module. Study the modules Hom(G, Z/pk Z). Study the module Hom(G, Q). Study other interesting classes of End(G)-modules in terms of Hom(G, ·). 3. Study the category of G-plexes when G is a p-finitely projective group. See [32] for relevant definitions. 4. Characterize group theoretically rtffr J -groups, N -groups, and especially S-groups. 5. Delete as much of the Analytic Number Theory used in our work on L-groups as is possible. 6. Describe the rtffr groups G such that G ⊕ G ∼ = G ⊕ H =⇒ G ∼ = H. 7. Characterize the endomorphism rings of rtffr J -groups.

Chapter 8

Gabriel Filters The material in this chapter comes from [40] where the filter of divisibility D(G) = {right ideals I ⊂ End(G) IG = G} is studied. This filter is nicely structured. We are most interested in the subfilters of D(M ) called Gabriel filters that correspond to a hereditary torsion class of right E-modules K such that K ⊗E G = 0. In particular, we will show that D(G) is in many interesting cases bounded by an idempotent ideal ∆. That is, there is an ideal ∆ of End(G) such that ∆2 = ∆ and such that IG = G implies ∆ ⊂ I.

8.1

Filters of Divisibility

Let E be an rtffr ring and let G be an rtffr group. Our work on the Baer Splitting Property in section 6.1 motivates us to investigate the set

DE (G) = D(G) = {right ideals I ⊂ End(G) IG = G}.

We call D(G) the filter of divisibility of G. Then G is a faithful right E-module iff D(G) = {End(G)}. We will consider faithful modules to be uninteresting in this section as they ignore any interesting structure there might be in D(G). Let the first example act as an intuition builder.

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EXAMPLE 8.1.1 Let X ⊂ Y be groups such that Hom(Y, X) = 0 and SX (Y ) = Y . Let G = X ⊕ Y , and let ∆ = Hom(X, G)Hom(G, X) ⊂ End(G). (For example, we could choose X = Z and nn o Y = n, m ∈ Z and m is square-free .) m Because Hom(Y, X) = 0, ∆ is a direct summand of End(G) such that ∆2 = Hom(X, G)(Hom(G, X)Hom(X, G))Hom(G, X) = Hom(X, G)Hom(X, X)Hom(G, X) = Hom(X, G)Hom(G, X) = ∆. Furthermore, because Hom(Y, X) = 0 and Y = SX (Y ) we have ∆G = Hom(X, G)Hom(G, X)(X ⊕ Y ) = Hom(X, G)Hom(X, X)X = Hom(X, G)X = SX (X) ⊕ SX (Y ) = X ⊕Y = G. Thus D(G) contains the infinite set {∆ + mEnd(G) m ∈ Z}. The reader will show that ∆ is the smallest right ideal in D(G). We will let E = End(G) = {q ∈ QEnd(G) qG ⊂ G}. Theorem 2.3.4 can be used to construct a short exact sequence π 0 −→ G −→ H −→ QE ⊕ QE −→ 0

(8.1)

in which E = End(H). Since IG = G and IQE = QE for each right ideal of the form I = ∆ + mE for some m 6= 0 ∈ Z, tensoring (8.1) with E/I shows that H = 0. IH

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179

Hence IH = H for each m ∈ Z. Thus there are infinitely many right ideals I ⊂ E such that E/I is finite and IH = H. But since JQE 6= QE for each proper right ideal J in QE, JH 6= H for each pure proper right ideal in E. Specifically, ∆H 6= H even though IH = H for each right ideal ∆ ⊂ I = ∆ + mE of finite index in E.

8.1.1

Hereditary Torsion Classes

A nonempty set S of right End(G)-modules is a hereditary torsion class if 1. S is closed under the formation of direct sums. That is, if Mk ∈ S, k ∈ I then ⊕k∈I Mk ∈ S. 2. S is closed under the formation of short exact sequences. That is, if 0 −→ K −→ M −→ N −→ 0 is a short exact sequence then K, N ∈ S iff M ∈ S. The prototypical examples of hereditary torsion classes are the set of groups, the set of torsion groups, and the set of torsion p-groups for some prime p ∈ Z. Our motivation for considering hereditary torsion classes is the set TorE (L) = {right E-modules K K ⊗E L = 0} where L is a left E-module. PROPOSITION 8.1.2 Let E be a ring and let L be a flat left Emodule. Then TorE (L) is a hereditary torsion class. Proof: Let Mk ∈ Tor(L), k ∈ I. Because ·⊗E L is an additive functor (⊕k∈I Mk ) ⊗E L ∼ = ⊕k∈I (Mk ⊗E L) = 0. Next consider the short exact sequence 0 −−−−→ K −−−−→ M −−−−→ N −−−−→ 0 of right E-modules. An application of the exact functor · ⊗E L yields the exact sequence α 0 −−−−→ K ⊗E L −−−−→ M ⊗E L −−−−→ N ⊗E L −−−−→ 0. It follows that K ⊗E L = N ⊗E L = 0 iff M ⊗E L = 0. Hence Tor(L) is a hereditary torsion class.

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EXAMPLE 8.1.3 TorE (L) is a hereditary torsion class if L is a prob p , so jective left E-module. If p ∈ Z is a prime then End(Z(p∞ )) = Z ∞ b p -modules of the form that TorE (Z(p )) is the set of Z H ⊕D b p -module. But where H is a torsion p-group and D is a divisible Z ∞ Tor(Z(p )) is not a hereditary torsion class since the quotient field Qp b p is in Tor(Z(p∞ )) and Z b p ⊂ Qp , but Z b p 6∈ Tor(Z(p∞ )). of Z An important example of hereditary torsion classes comes from idempotent ideals. PROPOSITION 8.1.4 Let E be a ring and let ∆ ⊂ E be an ideal such that ∆2 = ∆ and · ⊗E ∆ is an exact functor. The set {right E-modules K K∆ = 0} is a hereditary torsion class. We leave the proof as an exercise for the reader. There is a relationship between a hereditary torsion class and the cyclic submodules it contains. Let X be a set of right E-modules. The hereditary torsion class generated by X , denoted by TorE (X ), is the intersection of all of the hereditary torsion classes that contain X . Given a class T of right E-modules let CycE (T ) denote the cyclic modules contained in T . Given a hereditary torsion class T the reader can show that C ∈ CycE (T ) iff C is a cyclic submodule of some M ∈ T . LEMMA 8.1.5 TorE (CycE (T )) = T if T is a hereditary torsion class. Proof: Clearly TorE (CycE (T )) ⊂ T . Let M ∈ T . There is a submodule K ⊂ M such that ⊕x∈M xEnd(G) ∼ = M. K Since the hereditary torsion class TorE (CycE (T )) is closed under arbitrary direct sums and factor modules M ∈ TorE (CycE (T )). Thus TorE (CycE (T )) = T , and the proof is complete. Some examples will serve to illustrate these ideas. A module M is called semi-Artinian if each nonzero factor M/N of M has nonzero socle. The simple modules appearing in the socle of M/N are called composition factors of M . The reader is encouraged to prove these examples.

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181

EXAMPLE 8.1.6 Let E be a ring and let S be a set of simple right E-modules. The hereditary torsion class TorE (S) is the set of semiArtinian right E-modules M whose composition factors are are in S. EXAMPLE 8.1.7 Let G be a left E-module. Then TorE {E/I IG = G} = {M

for each x ∈ M , xE ⊂ E/I

for some I ⊂ E such that IG = G}. EXAMPLE 8.1.8 The group G in Example 6.4.2 satisfies TG (M ) 6= 0 for each nonzero finitely generated (cyclic) right End(G)-module M but there is a right End(G)-module K 6= 0 such that TG (K) = 0. That is, {E/I IG = G} = ∅ while {K K ⊗E G = 0} 6= ∅.

8.1.2

Gabriel Filters of Right Ideals

In this section we indirectly study sets of modules of the form CycE (T ) for some hereditary torsion class T . Given a right ideal I ⊂ E and x ∈ E let (I : x) = {a ∈ E xa ∈ I}. The reader can show that (I : x) = annE (x + I) where x + I ∈ E/I. A Gabriel filter on E is a set γ of right ideals of E such that 1. Given I, J ∈ γ and I ⊂ I 0 then I ∩ J, I 0 ∈ γ. 2. Given I ∈ γ and x ∈ E then (I : x) ∈ γ. 3. Given a right ideal J ⊂ E and I ∈ γ such that (J : x) ∈ γ for each x ∈ I then J ∈ γ. The relationship between Gabriel filters and hereditary torsion classes follows. LEMMA 8.1.9 Let E be a ring. 1. Let γ be a Gabriel filter on E. Then the hereditary torsion class E TorE I ∈ γ is the set of right E-modules K such that each I cyclic submodule of K has the form E/I for some I ∈ γ. 2. Let T be a hereditary torsion class of right E-modules. Then E right ideals I ⊂ E ∈ T is a Gabriel filter on E. I

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Proof: 1. Let TorE {E/I I ∈ γ} = TorE (γ). Since TorE (γ) is closed under direct sums and quotients, the set of right E-modules K such that each cyclic submodule of K has the form E/I for some I ∈ γ is contained in TorE (γ). On the other hand, if K ∈ TorE (γ) then xE ∈ {E/I I ∈ γ} for each x ∈ K. Thus TorE (γ) is contained in the set of right E-modules K such that each cyclic submodule of K has the form E/I for some I ∈ γ, and hence the two sets coincide. 2. Let T be a hereditary torsion class, and let E GabE (T ) = right ideals I ⊂ E ∈T . I One readily shows that if I, J ∈ GabE (T ) and if I ⊂ I 0 then E/(I ∩ J) ⊂ E/I ⊕ E/J, and E/I maps onto E/I 0 . Thus I ∩ J, I 0 ∈ GabE (T ). Next let I ∈ GabE (T ), let J ⊂ E be a right ideal, and suppose that (J : x) ∈ GabE (T ) for each x ∈ I. We have a short exact sequence 0 −−−−→

I +J E E −−−−→ −−−−→ −−−−→ 0. J J I +J

Since I ⊂ I + J and since the hereditary torsion class T is closed under quotients, E/(I + J) ∈ T . Let x ∈ I. Since I +J xE + J E ⊃ = ∈T J J (J : x) and since T is closed under direct sums and quotients, (I + J)/J ∈ T . Finally, since T is closed under extensions we see that E/J ∈ T . Thus J ∈ GabE (T ), from which we see that GabE (T ) is a Gabriel filter on E. This completes the proof. PROPOSITION 8.1.10 Let E be a ring, let Gab(E) be the set of Gabriel filters of E, and let Tor(E) be the set of hereditary torsion classes on E. There are inverse bijections GAB : Tor(E) −−−−→ Gab(E) TOR : Gab(E) −−−−→ Tor(E) given by GAB(T ) = GabE (T ) and TOR(γ) = TorE (γ). Proof: The reader can verify that GAB(TOR(γ)) = γ for each Gabriel filter γ and that TOR(GAB(T )) = T for each hereditary torsion class T .

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183

Therefore we can study hereditary torsion classes of right E-modules by studying Gabriel filters on E. The reader can verify the examples as exercises. EXAMPLE 8.1.11 Let T be the class of right E-modules such that for each x ∈ M ∈ T , xE is a bounded group (i.e., (xE)m = 0 for some integer m 6= 0). Then Tor(T ) is the hereditary torsion class of modules whose cyclic submodules are bounded groups. The associated Gabriel filter Gab(T ) is the set of right ideals I ⊂ E such that E/I is a torsion group. The following sets are important to our discussions in this Chapter. Let E be a ring and let M be a left E-module. DE (M ) = {right ideals I ⊂ E IM = M } maxE (M ) = {maximal right ideals I ⊂ E IM = M }. These sets are not necessarily Gabriel filters so we will look for Gabriel filters associated with them. Let δE (M ) = {right ideals I ⊂ E (I : x)M = M ∀x ∈ E} µE (M ) = the least Gabriel filter on E that contains maxE (M ).

Given I ∈ δE (M ) then I = (I : 1) so that δE (M ) ⊂ DE (M ). Since the ring E is understood we will drop it from our notation. The next result will allow us to say more about the right ideals in µ(M ) as well as being of independent interest. LEMMA 8.1.12 Let E be an rtffr ring. Then I ∈ µ(M ) iff E/I is a finite right E-module and each composition factor of E/I is isomorphic to some simple module in {E/J J ∈ maxE (M )}. Proof: Let µ0 (M ) be the set of right ideals I ⊂ E such that IM = M , E/I is finite, and each composition factor of E/I is isomorphic to E/J for some J ∈ maxE (M ). By Lemma 4.2.3, each I ∈ maxE (M ) has finite index in E, so that maxE (M ) ⊂ µ0 (M ). The reader will show that µ0 (M ) is a Gabriel filter on E. By definition as the least such filter, µ(M ) ⊂ µ0 (M ). Conversely, let I ∈ µ0 (M ). Then IM = M , E/I is finite and each composition factor of E/I is of the form E/J for some J ∈ maxE (M ) ⊂

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µ(M ). We observe that E/I is a factor module of the direct sum of cyclic modules M xE + I . I x∈E

Because E is an rtffr ring, torsion cyclic E-modules are bounded, hence finite groups. Thus for each x ∈ E, (xE + I)/I ∼ = E/(I : x) is finite. Let K be a simple submodule of E/(I : x). Then K ∼ = J/(I : x) for some (I : x) ⊂ J ∈ µ(M ). By a simple induction of the composition length of E/(I : x), each cyclic submodule of J/(I : x) is of the form E/L for some L ∈ µ(M ). Since µ(M ) is a Gabriel filter (I : x) ∈ µ(E) for each x ∈ E, and so I ∈ µ(M ). It follows that µ0 (M ) = µ(M ). An ideal ∆ in some ring is idempotent if ∆2 = ∆. Idempotent ideals will be important in the sequel. LEMMA 8.1.13 Suppose that ∆2 = ∆ is an ideal in E. Then D(∆) = {right ideals I ⊂ E I∆ = ∆} δ(∆) = the largest Gabriel filter contained in D(∆) max(∆) = {maximal right ideals I ∆ ⊂ I ⊂ E}. In particular ∆ is the unique minimal element in D(∆). Proof: This follows from the definitions using M = ∆. It follows that if ∆2 = ∆ then δ(∆) is generated by the set of cyclic right E-modules K such that K ⊗E ∆ = 0. There is an interesting characterization of the hereditary torsion class corresponding to δ(M ). LEMMA 8.1.14 Let M be a left E-module. Then K ∈ Tor(δ(M )) iff L ⊗E M = 0 for each cyclic submodule L ⊂ K. Proof: Suppose that x ∈ L ∈ Tor(δ(M )). Since δ(M ) is a Gabriel filter xE ∼ = E/I for some I ∈ δ(M ). Then E/I ⊗E M = 0 and since each submodule of L is a sum of cyclics K ⊗E M = 0 for each K ⊂ L. The converse is clear so the proof is complete. LEMMA 8.1.15 Let E be an rtffr ring and let M be a left E-module. 1. If γ ⊂ D(M ) is a Gabriel filter on E then γ ⊂ δ(M ). 2. If I ∈ D(M ) is an ideal then I ∈ δ(M ).

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185

Proof: 1. Say I ∈ γ ⊂ D(M ). Then for each x ∈ E, (I : x) ∈ γ ⊂ D(M ) so that (I : x)M = M . Thus I ∈ δ(M ) and hence γ ⊂ δ(M ). 2. Let I ∈ D(M ) be an ideal and let x ∈ E. Then I ⊂ (I : x) so that M = IM ⊂ (I : x)M ⊂ M which implies that (I : x) ∈ D(M ) for each x ∈ E. Hence I ∈ δ(M ). The next result shows that even though in general µ(M ) 6= δ(M ) 6= D(M ) they still share the same maximal right ideals. LEMMA 8.1.16 Let M be a left E-module. 1. maxE (M ) = max(δ(M )) = max(µ(M )). 2. µ(M ) = {right ideals I ⊂ E IG = G and E/I is semi-Artinian}. Proof: 1. Since δ(M ) ⊂ D(M ) and since maxE (M ) ⊂ µ(M ) we have max(δ(M )) ⊂ maxE (M ) ⊂ max(µ(M )). Let J ∈ max(µ(M )) and let x ∈ E. Inasmuch as E/J is simple, E/J ∼ = E/(J : x), so that xE ⊗E M = E/(J : x) ⊗E M ∼ = E/J ⊗E M ∼ = M/JM = 0. Then Lemma 8.1.14 states that J ∈ δ(M ). This proves part 1. 2. is left as an exercise.

8.2

Idempotent Ideals

As above E denotes an rtffr ring. Further restrict M to be an rtffr left E-module. We will show that the Gabriel filters δ(M ) and µ(M ) on E b are determined by an idempotent ideal ∆ ⊂ E.

8.2.1

Traces of Covers

The following concepts are found in [6]. Let M be a left E-module and let K ⊂ M be a submodule. We say that K is superfluous in M if K + N = M =⇒ N = M for each submodule N ⊂ M . For instance, Nakayama’s Lemma 1.2.3 implies that J(E)M is superfluous in each finitely generated left E-module M . However each proper submodule of Z(p∞ ) is superfluous in Z(p∞ ). In a ring E, the Jacobson

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radical J (E) is the largest superfluous E-submodule of E. No proper subgroup of Z is superfluous in Z, but for each prime p ∈ Z, pZp is superfluous in Zp . If N is a nilpotent ideal then N M is superfluous in M for each left E-module M . The reader can prove the following facts as exercises. 1. If N ⊂ K ⊂ L ⊂ M and if K is superfluous in L then K is superfluous in M and K/N is superfluous in M/N . 2. Suppose that P is a finitely generated projective left E-module. An E-submodule K ⊂ P is superfluous iff K ⊂ J(E)P . The left E-module M possesses a projective cover if there is a projective left E-module P and a surjection π : P → M such that ker π is superfluous in P . Our work in this section will show that projective covers are abundant when considering modules over rtffr rings. The trace of the projective cover π : P → M is ∆M = the trace ideal of P in E. Evidently ∆M is an idempotent ideal in D(M ). Since ∆M P = P we have ∆M M = M . A Gabriel filter γ is bounded if for each I ∈ γ there is a two sided ideal J ∈ γ such that J ⊂ I. We say that γ is bounded by (the idempotent ideal) ∆ if γ = { right ideals I ⊂ E ∆ ⊂ I}. In particular ∆ ∈ γ. For example, each Gabriel filter on a commutative ring is bounded. If ∆ is an idempotent ideal in E then by Lemma 8.1.13, ∆ is the unique minimal element in δ(∆). The filter of nonzero ideals on Z is bounded but is not bounded by a unique ideal. LEMMA 8.2.1 Let π : P → M be a projective cover of the left Emodule M . Then D(∆M ) = δ(M ) iff δ(M ) is bounded. Proof: If D(∆M ) = δ(M ) then by the above comment δ(M ) is bounded by ∆M . Conversely suppose that δ(M ) is bounded. By Lemma 8.1.13, δ(M ) contains every ideal in D(M ), and by the above comments, ∆M M = M . Then ∆M ∈ δ(M ). It remains to show that δ(M ) ⊂ D(∆M ). Let I ∈ δ(M ). There is an ideal J ⊂ I such that J ∈ δ(M ). Then JM = M so that JP + ker π = P . Because J is an ideal of E and because ker π is superfluous in P , JP = P . As usual ∆M ⊂ J ⊂ I so that I ∈ D(∆M ). Hence D(∆M ) = δ(M ), which completes the proof.

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187

EXAMPLE 8.2.2 1. Given a ring E then the canonical map E → E/J (E) is a projective cover. 2. Every module over a right Artinian ring possesses a projective cover, [6]. 3. The ring E is called semi-perfect if each finitely generated left Emodule possesses a projective cover. E is semi-perfect iff E/J(E) is semi-simple Artinian and idempotents in E/J(E) lift to idempotents of E. (See [6].) Local rings are semi-perfect. EXAMPLE 8.2.3 The following is an example of a finitely generated module M over a ring E such that E is not semi-perfect, M is not projective, but M possesses a finitelygenerated cover. projective 0 0 Z 0 ⊕ , let P = Let A be the ring of matrices Z Z Z Z 0 0 0 0 0 0 ⊕ , and let K = x ∈ Z . The reader x 0 x 0 Z Z 0 0 can show that J(A) = so that K ⊂ J(A)P is superfluous in Z 0 P . Then P/K is a doubly generated left E-module and the natural map π : P → P/K is a projective cover. An important source of projective covers is the following result. LEMMA 8.2.4 [40, T.G. Faticoni]. For each index i ∈ I let Ei be a ring, let Mi be a finitely generated left Ei -module, let πi : Pi → Mi be a projective cover of Mi over Ei , and let ∆i be the trace ideal of Pi in Ei . If there is an integer n such that each Mi is generated by at most n elements then Q Q 1. i∈I Mi is generated by at most n elements over i∈I Ei , Q Q 2. i∈I Pi is a finitely generated projective left i∈I Ei -module Q whose trace ideal satisfies ∆ = i∈I ∆i , and Q Q Q Q 3. i∈I πi : i∈I Pi → i∈I Mi is a projective cover over i∈I Ei . Proof: 1. By hypothesis, for each i > 0 there is a set of generators {xi1 , . . . , xin } ⊂ Mi . For each j = 1, . . . , n define a sequence Y xj = (x1j , x2j , . . .) ∈ Mi . i∈I

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Q We claim thatQ {x1 , . . . , xn } generates i Mi . Given y ∈ i Mi write y = (yi )i . By our choice of xij for each i there are rj = (rij )i , j = 1, . . . , n, such that yi =

n X

rij xij .

j=1

Then n n n n X X X X = y = (yi )i = rij xij (rij xij )i = (rij )i (xij )i = rj xj . j=1

j=1

i

j=1

Q

j=1

Q

As claimed {x1 , . . . , xn } generates M = i Mi over E = i Ei . 2. For each i ∈ I, Q Pi is generated by at most n elements over Ei , so that by part 1, P = i Pi is a finitely generated projective left E-module. Inasmuch as Y HomE (P, E) = HomEi (Pi , Ei ) i∈I

Q it follows that the trace ideal for P in E is the product ∆ = i∈I ∆i . 3. Suppose that Q for each i ∈ I, Ki is a superfluous Ei -submodule of Pi andQlet K = Q i Ki . Since Ki is superfluous in Pi , Ki ⊂ J (Ei )Pi . Since J ( i Ei ) = i J (Ei ) ! ! Y Y Y K= Ki ⊂ J Ei Pi = J (E)P. i

i

i

Since P is finitely generated, Nakayama’s Lemma 1.2.3 shows us that K is superfluous in P . Q The reader can show that because M = i Mi is finitely generated Y π= πi : P −−−−→ M i

Q is a surjection. Inasmuch as ker(π) = i (ker πi ), π is a projective cover of M over E. This completes the proof. The reader might ask where in the realm of rtffr groups will we have a chance to see an infinite product of rings. The next few results will answer this question. Given a prime p let pω G =

∞ \

pk G.

n=1

Recall that idempotents lift modulo nilpotent ideals. (See [6].)

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189

LEMMA 8.2.5 Let E be an rtffr ring. Then for each prime p the p-adic bp of E/pω E is a semi-perfect ring. completion E Proof: Let Ep be the localization of E at p. Because E has finite rank Ep /pk Ep ∼ = E/pk E is a finite (hence Artinian) ring for each integer bp /pk E bp . k > 0, so that idempotents lift module the Jacobson radical of E b Furthermore p ∈ J (Ep ). bp ). Because bp /J (E Let e¯2 = e¯ ∈ E bp )/pk E bp = J (E bp /pk E bp ) J (E and because bp bp /pk E pk−1 E bp /pk E bp for each integer k > 0 we can construct is a nilpotent ideal of E bp such that a sequence (ek )k in E bp ) = e¯, e1 + J (E

bp , ek−1 − ek ∈ pk−1 E

bp ). e2k − ek ∈ pk J (E

bp its limit exists. Since (ek )k clearly converges in the p-adic topology on E bp for each k. bp . Moreover e2 − e ≡ e2 − ek ∈ pk E That is, limk ek = e ∈ E k ω 2 bp = 0, e = e. That is, e is an idempotent lifting of e¯. Hence Since p E b Ep is semi-perfect. LEMMA 8.2.6 Let E be an rtffr ring, and let M be an rtffr right Eb p -module by at most rank(M ) cp is generated as a Z module. Then M b c is generated as a Zelements for each prime p ∈ Z. Furthermore M module by at most rank(M ) elements. Proof: Let p ∈ Z be a prime. Because Z/pZ-linearly independent subsets of M/pM lift to Z-independent subsets of M , M/pM has Z/pZcp /pM cp ∼ dimension at most rank(M ). Thus M = M/pM has dimension at rank(M ) b c cp . It follows that there most rank(M ), so that Zp maps onto Mp /pM b p -submodule K of M cp generated by at most rank(M ) elements is a free Z b p , and such that K + pM cp = M cp . Thus K is dense in M cp . Since K over Z cp . Thus M c is is finitely generated it is p-adically complete so that K = M Q pc c generated by at most rank(M ) elements. By Lemma 8.2.4, M = p Mp b=Q Z b p -module. is then generated by at most rank(M ) elements as a Z p

This proves the Lemma. THEOREM 8.2.7 [40, T.G. Faticoni] Let E be an rtffr ring and let c possesses a finitely generated M be an rtffr right E-module. Then M b projective cover over E.

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b = Q E b c Proof: Because E and M are rtffr groups, E p p and M = Q c Mp where p ranges over the primes in Z. Fix p ∈ Z. By Lemma 8.2.6, p

b p -module by rank(M ) elements. Consequently M cp is generated as a Z cp M bp (Lemma 8.2.5), possesses a projective cover over the semi-perfect ring E Q c c so that M = p Mp possesses a finitely generated projective cover over b (Lemma 8.2.4). This completes the proof. the ring E b possesses a COROLLARY 8.2.8 Let G be a rtffr group. Then G \ b is genfinitely generated projective cover over End(G). Moreover G \ erated by at most rank(G) elements as a left End(G)-module.

8.2.2

Bounded Gabriel Filters

In this section we examine the existence and the usefulness of idempotent ideals ∆ such that ∆M = M . Specifically we show that µ(M ) is finite iff it is bounded by an idempotent ideal. Let us agree that

c) = {right ideals I ⊂ E b (I : x)M c=M c∀x ∈ E}. b δ(M

c) denotes the largest Gabriel filter in D b (M c), the set of right Then δ(M E b such that I M c=M c. (See Lemma 8.1.15.) Furthermore let ideals I ⊂ E

c) = {right ideals I ⊂ E b E/I b is finite and I M c=M c}. µ(M

c) is a Gabriel filter on E. b We will show that Gabriel Observe that µ(M filters over p-adically complete torsion-free rings are especially nice. THEOREM 8.2.9 Let E be an rtffr ring and let M be an rtffr right E-module. c) is a bounded Gabriel filter on E. b 1. δ(M c) is a bounded Gabriel filter on E. b 2. µ(M

8.2. IDEMPOTENT IDEALS

191

b c). Since E b is a finitely generated Z-module Proof: 1. Let I ∈ δ(M b b b (Lemma 8.2.6) we can write E/I = Zx1 + · · · + Zxn for some integer n b and elements xi ∈ E/I. An elementary argument shows us that b (I : x1 ) ∩ · · · ∩ (I : xn ) ⊂ annEb (E/I) ⊂ I. c) is a Gabriel filter, (I : x1 ) ∩ · · · ∩ (I : xn ) ∈ δ(M c), from which Since δ(M b c). Thus δ(M c) is bounded. it follows that the ideal annEb (E/I) is in δ(M 2. Proceed as in part 1. This completes the proof. COROLLARY 8.2.10 Let E be an rtffr ring and let M be an rtffr right E-module. Then µ(M ) is a bounded Gabriel filter on E. Proof: Let I ∈ µ(M ). By Lemma 8.1.12, E/I is torsion so there is an integer m 6= 0 such that mE1 = E(m1) ⊂ I. Hence E/I is bounded, and since E is an rtffr, E/I is finite. Write {x1 , . . . , xn } = E/I. Then mE ⊂ (I : x1 ) ∩ · · · ∩ (I : xn ) ⊂ annE (E/I) ⊂ I so that annE (E/I) ⊂ I is an ideal in µ(M ). This completes the proof. THEOREM 8.2.11 Let E be an rtffr ring and let M be an rtffr right b E-module. There is a finitely generated idempotent ideal ∆M c ⊂ E such that c) = {right ideals I ⊂ E b ∆ c ⊂ I}. In particular, ∆ c ∈ δ(M c). 1. δ(M M

M

c) = {right ideals I ⊂ E b E/I b is finite and ∆ c ⊂ I}. 2. µ(M M c possesses a projective cover π : P → M c Proof: By Theorem 8.2.7, M b Let ∆ c be the trace ideal of P in E. b Then ∆ c is the idempotent over E. M M b such that ∆ cP = P , and IP = P iff ∆ c ⊂ I. ideal in E M M c = π(P ) = π(∆ cP ) = ∆ cM c, ∆ c ∈ δ(M c), and so {right 1. Since M M M M b c ideals ∆M c ⊂ I ⊂ E} ⊂ δ(M ). c). By Theorem 8.2.9, δ(M c) is bounded, so Conversely let I ∈ δ(M c c c=M c, there an ideal J ⊂ I such that J M = M . Because π(JP ) = J M JP + ker π = P . Inasmuch as ker π is superfluous in P , and because b JP is a left E-submodule of P , JP = P . Hence ∆M c ⊂ J and thus c b δ(M ) = {right ideals I ⊂ E ∆M c ⊂ I}. Prove part 2 in a manner similar to part 1. This completes the proof. The rtffr group H constructed at the end of Example 8.1.1 should dispel any notions that δ(M ) = µ(M ) is bounded by an idempotent ideal when M is an rtffr right E-module.

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EXAMPLE 8.2.12 This example shows that µ(M ) need not be finite. Let π : P −→M be the projective cover constructed in Example 8.2.3. 0 0 Then ∆M = , and Z Z δ(M ) =

nZ 0 Z Z

n∈Z

is bounded by the idempotent ideal ∆M . Furthermore, nZ 0 µ(M ) = n 6= 0 ∈ Z Z Z is not bounded by an idempotent ideal.

8.2.3

Finite Filters of Divisibility

In this section we study modules M for which µ(M ) and δ(M ) are finite sets. c) are identical The following Lemma shows us that µ(M ) and µ(M posets. LEMMA 8.2.13 Let E be an rtffr ring and let M be an rtffr right c) defined E-module. There are inverse poset isomorphisms µ(M ) ∼ = µ(M by I 7→ Ib and I 7→ I ∩ E. c) and let I = Ib ∩ E. Then E/ b Ib and E/I are Proof: Let Ib ∈ µ(M b Because I is dense finite. There is an integer m 6= 0 such that m ∈ I, I. b b c in I and because I ∈ µ(M ) c = (I + mI)(M b c) = IbM c=M c = M + mM c. IM + mM + mM Then by the Modular Law and because m ∈ I we have M

c) ∩ M = (M + mM c) ∩ M = (IM + mM c ∩ M) = IM + (mM = IM + mM = IM.

c) −→ µ(M ). Thus the assignment Ib → 7 Ib ∩ E defines a function µ(M

8.2. IDEMPOTENT IDEALS

193

Conversely let I ∈ µ(M ) and choose a nonzero integer m ∈ I. Then c = (I + mI)(M b c) = IM + mM c = M + mM c=M c IbM + mM c). Thus the assignment I 7→ Ib defines a function so that Ib ∈ µ(M c µ(M ) −→ µ(M ). Finally it is well known that Ib ∩ E = I for each right ideal I of finite 0 ∩ E = I 0 for each right ideal I 0 of finite index index in E, and that I\ b Thus the above maps are mutual inverses, which completes the in E. proof. Using the above correspondence we can characterize µ(M ) in terms of an idempotent ideal. THEOREM 8.2.14 Let E be an rtffr ring and let M be an rtffr right b E-module. There is a finitely generated idempotent ideal ∆M c ⊂ E such that b b µ(M ) = {right ideals I ⊂ E E/I is finite and ∆M c ⊂ I ⊂ E}. Moreover ∆M c is generated by at most rank(E) elements. Proof: The proof is an application of Lemma 8.2.13 and Theorem 8.2.11. We leave the details to the reader. We consider the Gabriel filters that are bounded by an idempotent ideal to be the most important examples of Gabriel filters for reasons that will be obvious in the next Theorem. Since each rtffr module M b the projective cover over E is associated with a projective cover π over E, brings with it the trace ideal ∆M c of the projective. The next result gives at least one case in which ∆M leads us to an idempotent ideal in E. This c is the case when µ(M ) is finite. THEOREM 8.2.15 [40, T.G. Faticoni] Let E be an rtffr ring and let M be an rtffr right E-module. The following are equivalent for M . 1. µ(M ) is finite. 2. µ(M ) = D(∆) for some idempotent ideal ∆ in E. b 3. ∆M c has finite index in E. If M satisfies one and hence all of the above conditions then ∆ = ∆M c ∩E.

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b Proof: 3 ⇒ 2 Suppose that ∆M c has finite index in E. Then by Theorem 8.2.11 b is finite and ∆ c ⊂ I} c) = {right ideals I ⊂ E b E/I µ(M M b = {right ideals I ⊂ E ∆M c ⊂ I}. If we let ∆ = ∆M c ∩ E then ∆ has finite index in E. The reader can show that ∆ is idempotent since ∆M c is idempotent. Lemma 8.2.13 then implies that b and ∆ ⊂ I} ∆ ∈ µ(M ) = {I ∩ E I is a right ideal in E b and ∆ ⊂ I ∩ E} = {I ∩ E I is a right ideal in E = {right ideals J ⊂ E ∆ ⊂ J}. This proves part 2. 2 ⇒ 1 follows immediately from Lemma 8.1.12. 1 ⇒ 3 Because µ(M ) = {I1 , · · · , In } is finite, and because µ(M ) is bounded, µ(M ) contains a unique minimal element ∆ = I1 ∩ · · · ∩ In ∈ µ(M ). b is the unique minimal element of Then Lemma 8.2.13 shows us that ∆ c b b µ(M ). We observe that ∆ has finite index in E. ∼ c b c To see that ∆M c = ∆ notice that µ(M ) = µ(M ) implies that µ(M ) is b We showed in Theorem finite, hence bounded by the idempotent ideal ∆. c 8.2.11 that ∆M c is the unique minimal element of µ(M ). This proves part 3 and completes the logical cycle. Let E be the ring and let M be the module constructed possessing a projective cover π : P → M in Example 8.2.3. We showed that ∆M = 0 0 c such that . We then constructed an rtffr group M ⊂ G ⊂ M Z Z each element of µ(G) has finite index in E. Specifically ∆M 6∈ µ(G). Furthermore, µ(G) is infinite and does not possess a unique minimal element. EXAMPLE 8.2.16 The reader will show that µ(G) is finite if G is the group constructed in Example 6.1.7. There is at least one class of rtffr ring in which each Gabriel filter is bounded by an idempotent ideal.

8.2. IDEMPOTENT IDEALS

195

THEOREM 8.2.17 Let E be an rtffr ring and let M be an rtffr left c) = 0. (For example, let M be any module E-module such that annEb (M such that E = End(M ).) If E is a semi-prime ring then µ(M ) is finite. Proof: Suppose that E is semi-prime and that annE (M ) = 0. Then b is semi-prime. Write E Y ∆M = ∆p c p

bp . Since where ∆p is an idempotent ideal in the semi-prime ring E c) = 0, ann b (∆p ) = 0. However, since E bp is semi-prime there is annEb (M Ep . b such that ∆p ⊕ I = E bp . Since then I∆p ⊂ ∆p ∩ I = 0, an ideal I in E . b I = 0 and so ∆p = Ep . Now for all but at most finitely many primes p, bp is an integrally closed ring. In such a ring, proper ideals are invertible, E bp . Then not idempotent. See [69]. Say ∆p J = E bp . bp ) = ∆p (∆p J) = ∆p J = E ∆p = ∆p (E b for all but at most finitely many primes p. Thus That is ∆p = E Q b p b E/∆M c= p Ep /∆p is finite, and hence by the previous Theorem, µ(M ) is finite. This completes the proof. Let E be a rtffr ring and let M be a rtffr right E-module. We say that M is faithful if IM 6= M for each right ideal I in E. Evidently M is faithful iff D(M ) = {E}. Since each right ideal in E is contained in a maximal right ideal and since each maximal right ideal I in E has finite index in E, we see that M is faithful iff IM 6= M for each right ideal I of finite index in E, Lemma 4.2.3. It is clear that M is a faithful left E-module iff µ(M ) = {E}. As we showed in Chapter 6, the faithful property is connected to the splitting property of short exact sequences. The next result characterizes the faithful left E-modules. THEOREM 8.2.18 Let E be an rtffr ring and let M be an rtffr right E-module. The following are equivalent for M . 1. D(M ) = {E}. 2. δ(M ) = {E}. 3. µ(M ) = {E}. c) = {E}. b 4. µ(M

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5. If I is a right ideal in E then there is an x ∈ E such that (I : x)M 6= M . b 6. ∆M c = E. 7. Let K be a torsion (as a group) right E-module such that K ⊗E M = 0. There exists a cyclic submodule N ⊂ K such that N ⊗E M 6= 0. Proof: 1 ⇒ 2⇒ 3 follows from the inclusions µ(M ) ⊂ δ(M ) ⊂ D(M ). 3 ⇒ 1 Suppose that part 1 is false. There is a right ideal I such that IM = M . We may assume that I is a maximal right ideal of E and hence that I ∈ max(M ) ⊂ µ(M ). This proves that part 3 is false. c) (Lemma 3 ⇔ 4 follows from the isomorphism of posets µ(M ) ∼ = µ(M 8.2.13). 2 ⇔ 5 follows from the definition of δ(M ). 3 ⇔ 6 is Theorem 8.2.14. 7 ⇒ 3 Suppose that µ(M ) 6= {E}. Let I ⊂ E be a maximal right ideal of E such that IM = M . Then E/I = (x + I)E ∼ = E/(I : x) for each x ∈ E. It follows that E/(I : x) = 0 for each x ∈ E and hence part 7 is false. 3 ⇒ 7 follows in a similar manner. This completes the logical cycle.

8.3

Gabriel Filters on Rtffr Rings

Let E denote a fixed rtffr ring and let Ω(E) = { rtffr groups G End(G) ∼ = E as rings} ∆Gb = the trace ideal of a projective cover b over E. b π:P →G

(See Theorem 8.2.7.) In this section we will investigate the rtffr rings E such that each G ∈ Ω(E) is faithful. We characterize those idempotent b such that ∆ = ∆ b for some G ∈ Ω(E). ideals ∆ ⊂ E G

8.3.1

Applications to Endomorphism Rings

Let G be an rtffr group. It is natural to investigate how the above Theorems can be used to explore the faithful property on G. In the next result we characterize those rtffr groups G for which δ(G) is a finite set.

8.3. GABRIEL FILTERS ON RTFFR RINGS

197

THEOREM 8.3.1 The following are equivalent for the rtffr group G. 1. δE (G) is finite. 2. µE (G) is finite. b is finite. 3. µEb (G) \ 4. ∆Gb has finite index in End(G). 5. There is an idempotent ideal ∆ of finite index in End(G) such that δE (G) = {right ideals I ∆ ⊂ I ⊂ End(G)}. In this case ∆ = ∆Gb ∩ End(G). Proof: 5 ⇒ 1 ⇒ 2 ⇒ 3 are clear, 3 ⇒ 4 ⇒ 5 is Theorem 8.2.15. This completes the logical cycle. The faithful rtffr groups are characterized in the following result. THEOREM 8.3.2 The following are equivalent for the rtffr group G. 1. D(G) = {End(G)}. 2. δ(G) = {End(G)}. 3. µ(G) = {End(G)}. \ b = {End(G)}. 4. µ(G) 5. If I is a right ideal in End(G) then there is an x ∈ End(G) such that (I : x)G 6= G. \ 6. ∆Gb = End(G). 7. Let K be a torsion (as a group) right End(G)-module such that K ⊗End(G) G = 0. There exists a cyclic submodule N ⊂ K such that N ⊗End(G) G 6= 0. The Proof follows from Theorem 8.2.18. The short exact sequence π 0 −−−−→ K −−−−→ X −−−−→ G −−−−→ 0

(8.2)

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is quasi-split if there is a map : G → X and. an integer n 6= 0 such that π = n1G . If (8.2) is quasi-split then X ∼ = G ⊕ K. The property δ(G) = µ(G) is characterized in terms of quasi-splitting. The faithful property is intimately associated with the splitting of short exact sequences. Thus we include this connection. THEOREM 8.3.3 The following are equivalent for an rtffr group G. 1. If IG = G for some right ideal I ⊂ End(G) then End(G)/I is finite. That is, δ(G) = µ(G). 2. Each short exact sequence (8.2) in which X = SG (X) + K is quasisplit. 3. Each short exact sequence (8.2) in which X ∼ = G(c) for some cardinal c is quasi-split. Proof: 1 ⇒ 2 is proved in the same manner that we proved Baer’s Lemma 6.1.1. 2 ⇒ 3 is clear. 3 ⇒ 1 Suppose that IG = G for Psome proper right ideal I in End(G)submodule of HG (H). Write I = k∈I πk End(G) for some index set I, let Gk = G for each k ∈ I, and let π : ⊕k∈I Gk −→ G be the unique map such that π(xk ) = πk (xk ) for each xP k ∈ Gk and each k ∈ I. Notice that π is a surjection since G = IG = k∈I πk (Gk ). Thus there is a short exact sequence (8.2) in which X ∼ = ⊕k∈I Gk . By part 5, (8.2) is quasisplit. There is a map : G → ⊕k∈I Gk and an integer n 6= 0 such that π = n1G . Since the rtffr group G is self-small we can write = ⊕k∈I k for some maps k : G → Gk . Then ! ! ! X X X n1G = π = πk k = πk k ∈ I k

k

k

by our choice of πk ∈ I. Then End(G)/I is bounded, hence finite, and part 1 follows. This completes the logical cycle. It is interesting to list the following result since it characterizes those rtffr groups G such that IG = G for some proper nonzero pure right ideal I ⊂ End(G) in terms of a splitting result. We make the following elementary observation. Suppose that E is an rtffr ring and let I ⊂ E be a right ideal such that E/I is a torsion group. There is an integer n 6= 0 such that n1G ∈ I so that nE ⊂ I ⊂ E. Since E has finite rank the bounded group E/I is finite.

8.3. GABRIEL FILTERS ON RTFFR RINGS

199

COROLLARY 8.3.4 The following are equivalent for the rtffr group G. 1. There is a proper nonzero pure right ideal I ⊂ End(G) such that IG = G. 2. There is a short exact sequence π 0 −−−−→ K −−−−→ G(c) −−−−→ G −−−−→ 0

(8.3)

that is not quasi-split. Proof: This theorem is the contrapositive of Theorem 8.3.3.

8.3.2

Constructing Examples

Given a torsion-free ring R and a left R-module M let OR (M ) = {r ∈ QR rM ⊂ M }.

LEMMA 8.3.5 Let E be an rtffr ring and let M be an rtffr left Emodule. Then c) = E. b OE (M ) = E iff OEb (M c) = E b and let r ∈ OE (M ). Then rM ⊂ Proof: Suppose that OEb (M c, rM c⊂ M and since M is a pure subgroup of the pure injective group M c. Thus M b ∩ QE = E r∈E and hence OE (M ) ⊂ E. The inclusion E ⊂ OE (M ) is obvious. b ⊂ O b (M c) and Conversely, assume that OE (M ) = E. Evidently E E c)/E b is torsion. It is a short exercise to show that OE (M ) is a pure OEb (M c). Then E = OE (M ) is pure and dense in E b and in subgroup of OEb (M c), so that E b = O b (M c). This completes the proof. OEb (M E b such that The next result characterizes the idempotent ideals ∆ in E ∆ = ∆G for some G ∈ Ω(E). THEOREM 8.3.6 Let E be an rtffr ring and let ∆ be an idempotent b The following are equivalent. ideal in E.

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1. ∆ = ∆Gb for some G ∈ Ω(E). b b = O b (∆). 2. ∆ is finitely generated as a left E-module and E E Proof: 1 ⇒ 2 Suppose that ∆ = ∆Gb for some G ∈ Ω(E). By Theorem b E b ⊂ O b (∆). 8.2.14, ∆ is finitely generated. Since ∆ is an ideal in E, E b=G b implies that On the other hand, by Lemma 8.3.5, ∆G b = E. b OEb (∆) ⊂ OEb (G) b and part 2 follows. Then OEb (∆) = E b 2 ⇒ 1 Since ∆ is a finitely generated torsion-free (=projective) Zmodule ∆ is a Z-adically complete group. b Let {x1 , . . . , xn } be a set of generators of ∆ as a left Z-module. Let M = (QEx1 + · · · + QExn ) ∩ ∆. b i for each i = 1, · · · , n, M is pure and dense in SinceP Zxi is dense in Zx b c = ∆. ∆ = i Exi . That is, M c possesses a projective cover π : P → M c over By Theorem 8.2.7, M 2 b c E. Since ∆ = ∆ = M c = ∆ = ∆∆ = ∆M c = π(∆P ). π(P ) = M

(8.4)

b and since ker π is Hence ∆P + ker π = P . Since ∆ is an ideal in E superfluous in P , ∆P = P . Then ∆P ⊂ ∆ where ∆P denotes the trace ideal of P . On the other hand, because ∆P and ∆ are idempotent ideals, ∆P ⊂ ∆ implies that ∆P ∆ = ∆P . By (8.4) ∆ = π(P ) = π(∆P P ) = ∆P π(P ) = ∆P ∆ = ∆P . Thus ∆ = ∆P . c) = O b (∆) = E b so that OE (M ) = E Finally by hypothesis OEb (M E (Lemma 8.3.5). Theorem 2.3.3 constructs a short exact sequence 0 −−−−→ M −−−−→ G −−−−→ QC −−−−→ 0 of left E-modules such that E ∼ = End(G) naturally. It follows that c b M = G so by the above paragraph ∆ = ∆P = ∆Gb . This completes the proof. Z 0 Z 0 EXAMPLE 8.3.7 Let E = and let ∆ = . Then Z Z Z 0 ∆ is a cyclic projective left E-module, ∆2 = ∆, and E = OE (∆). Conb is a finitely generated idempotent ideal of E b such that sequently ∆

8.3. GABRIEL FILTERS ON RTFFR RINGS

201

b = O b (∆). b E By Theorem 8.3.6 there is an rtffr group G such that E b = ∆G and E ∼ ∆⊂G⊂∆ = End(G). Furthermore G/∆ ∼ = QE ⊕ QE so that ∆G 6= G, while µ(G) = {right ideals I E/I is finite and ∆ ⊂ I ⊂ E} is infinite. The following is an interesting example of an idempotent ideal of finite index in the ring. EXAMPLE 8.3.8 Let O be a maximal order in the ring of Hamiltonian Quaternions H. (See [69].) It is known that for primes p 6= 2 ∈ Z, O/pO ∼ = Mat2×2 (Z/pZ). Let pO ⊂ ∆ ⊂ O be the right ideal in O such that 0 0 . ∆/pO = Z/pZ Z/pZ One shows that ∆ is a finitely generated projective idempotent ideal in OH (∆) where pO ⊂ OH (∆) ⊂ O is such that Z/pZ 0 . OH (∆)/pO = Z/pZ Z/pZ b = ∆ b for some group G such that OH (∆)) Theorem 8.3.6 states that ∆ G ∼ b = ∆, b ∆G = G, and = End(G). Since OH (∆)/∆ is finite and since G hence δ(G) = µ(G) is finite. Observe that G is strongly indecomposable. b b b = Q Z EXAMPLE 8.3.9 Let E = Z. Then E p p . Let ∆ = ⊕p Zp . 2 Then one readily verifies that ∆ = ∆ and that E = OE (∆). However ∆ 6= ∆Gb for any rtffr group G such that E ∼ = End(G) since ∆ is not a b finitely generated ideal in E.

8.3.3

Faithful Rings

The ring E is a faithful ring if given a right ideal I ⊂ E and a G ∈ Ω(E) such that IG = G then I = E. D.M. Arnold and E.L. Lady [14] and [41] show that the rtffr ring is a faithful ring if E is commutative, right hereditary, or local. More examples of faithful rings are given at the end of this subsection. The following theorem characterizes the faithful rtffr rings. THEOREM 8.3.10 Let E be an rtffr ring. The following are equivalent for E. 1. E is a faithful ring.

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b such that E b = 2. If ∆ is a finitely generated idempotent ideal of E b OEb (∆) then ∆ = E. b b= 3. If P is a finitely generated projective left E-module such that E b OEb (P ) then P generates E. Proof: 1 ⇒ 2 Assume part 2 is false. Then there is a proper finitely b such that O b (∆). By Theorem 8.3.6 generated idempotent ideal ∆ ⊂ E E b and ∆G b = G. b Then part 1 there is a G ∈ Ω(E) such that ∆ = ∆Gb 6= E is false. 2 ⇒ 3 Assume part 2. Let P be a finitely generated projective left b b = O b (P ), and let ∆ be the trace ideal of P in E-module such that E E b Then ∆ is a finitely generated idempotent ideal of E b and by Lemma E. b b 8.3.5, E = OEb (∆). By part 2, ∆ = E, which proves part 3. 3 ⇒ 1 Assume part 3 and let G ∈ Ω(E). By Theorem 8.2.7 there is b over E. b If we let ∆ be a finitely generated projective cover π : P → G the trace ideal of P then by Lemma 8.3.5 b ⊂ O b (P ) = O b (∆) ⊂ O b (G) b = E. b E E E E b so that by part 3, P generates E. b Hence ∆ = E b is Then OEb (P ) = E, b Inasmuch as ∆ = ∆ b , Theorem 8.3.2 implies the trace ideal of P in E. G that G is a faithful group. This proves part 1 and completes the logical cycle. The following are examples of faithful rings. The ring E is local if E possesses a unique maximal right ideal. In this case J (E) is the unique maximal right ideal of E and E/J (E) is a division ring. EXAMPLE 8.3.11 A local rtffr ring is faithful. Proof: Say E/J (E) is simple. It follows from Lemma 4.2.3 that E/J (E) is an elementary p-group for some prime p ∈ Z. Then pE 6= E b = E bp . One readily and qE = E for each prime q 6= p ∈ Z. Thus E \ bp ) so that E bp is a local ring. Since a projective proves that J (E)p = J (E bp -module is then free it generates E bp . Theorem 8.3.10 then implies E that each G ∈ Ω(E) is faithful. The ring E is subcommutative if each right ideal in E is an ideal in E. Commutative rings are subcommutative. If H is a maximal order in the ring of Hamiltonian Quaternions over Q then there is a unique prime ideal 2H ⊂ I ⊂ H such that H/I is a field. Then the localization H2 b 2 are noncommutative subcommutative rings. and the completion H

8.3. GABRIEL FILTERS ON RTFFR RINGS

203

EXAMPLE 8.3.12 Each subcommutative rtffr domain is faithful. Proof: Let E be an rtffr subcommutative domain, and let ∆ be a b such that O b (∆) = E. b By finitely generated idempotent ideal of E E b Theorem 8.3.10 it suffices to show that ∆ = E. Since E is a subcommutative domain, each right ideal has finite index in E, and hence for each integer n, each right ideal of E/nE is an ideal. b is subcommutative. By Lemma 8.2.5, E bp is a semi-perfect ring Thus E for each prime p ∈ Z so we can write bp = E1 ⊕ · · · ⊕ En E bp . This direct sum is one for some indecomposable (right) ideals Ei ⊂ E of indecomposable semi-perfect rings so each Ei is a local ring. Consequently ∆p = ∆1 ⊕ · · · ⊕ ∆n where ∆i ⊂ Ei . But by Nakayama’s Lemma the local Noetherian ring does not contain a proper nonzero ideal I 2 = I. Thus ∆i = Ei for each bp , whence ∆ = E. b Given our reductions and i = 1, . . . , n, hence ∆p = E Theorem 8.3.10, E is a faithful ring. We have shown in Theorem 8.2.17 that semi-prime rtffr rings are bounded rings. That is, each right ideal contains a nonzero ideal. We characterize the rtffr bounded rings by first characterizing those rtffr rings E such that δ(G) is finite for each G ∈ Ω(E). THEOREM 8.3.13 [40, T.G. Faticoni] The following are equivalent for the rtffr ring E. 1. µ(G) is finite for each G ∈ Ω(E). 2. δ(G) is finite for each G ∈ Ω(E). b is a finitely generated idempotent ideal such that E b = 3. If ∆ ⊂ E b OEb (∆) then E/∆ is finite. Proof: 1 ⇔ 2 is Theorem 8.3.2. 1 ⇒ 3 Assume part 1, and let ∆ be a finitely generated idempotent b such that E b = O b (∆). By Theorem 8.3.6 there is a group ideal in E E G ∈ Ω(E) such that ∆Gb = ∆. By Theorem 8.3.2 and part 1, ∆Gb has b This proves part 3. finite index in E. 3 ⇒ 1 Assume part 3, let G ∈ Ω(E), and consider µ(G). By Lemma b = { right ideals I E/I b is finite 8.2.13 and Theorem 8.3.2, µ(G) ∼ = µ(G)

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b It therefore suffices to show that ∆ b has finite index and ∆Gb ⊂ I ⊂ E}. G b in E. b is a finitely generated left module over Since G is an rtffr group, G b b b By Lemma E, and thus ∆Gb is a finitely generated Z-submodule of E. b so that by part 3, ∆ b has finite index in E. b Given 8.3.5, OEb (∆) = E, G our reductions µ(G) is finite. In Example 8.2.12 we constructed a group G such that δ(G) = µ(G) is infinite.

8.4

Gabriel Filters on QEnd(G)

In this section we characterize in terms of quasi-summands of G the Gabriel filter of right ideals I of QEnd(G) such that IQG = QG. Let Example 8.2.12 act as an intuition builder.

8.4.1

Central Quasi-summands

In the previous sections we discussed the right ideals I of finite index in End(G) such that IG = G. In this section we will investigate the pure right ideal I ⊂ End(G) such that G/IG is torsion, or equivalently such that (QI)QG = QG. This naturally leads us to consider D(QG) = {right ideals I of QEnd(G) I(QG) = QG}. δ(QG) = {right ideals I of QEnd(G) (I : x)(QG) = QG for each x ∈ QEnd(G)}.

Inasmuch as QEnd(G) is Artinian and since the Gabriel filter δ(QG) contains finite intersections there is an idempotent ideal ∆QG ⊂ QEnd(G) such that δ(QG) = D(∆QG ) = { right ideals I ⊂ QEnd(G) ∆QG ⊂ I}. EXAMPLE 8.4.1 Let X and Y be rtffr groups such that Y Hom(X, Y )(X), 0 = Hom(Y, X), and Z 0 End(X ⊕ Y ) = . Z Z

=

8.4. GABRIEL FILTERS ON QEND(G)

205

0 0 See, e.g., Example 8.1.1. Then ∆ = is such that ∆G = G for Z Z . any group G = X ⊕ Y . Thus Q∆QG = QG. The reader can show that Q∆ is the smallest element of δ(QG). . Given a quasi-direct sum decomposition G = H ⊕ H 0 ⊂ QG there are idempotents eH , eH 0 ∈ QEnd(G) such that . eH G = H,

. eH 0 G = K,

and eH ⊕ eH 0 = 1G .

Call H a nilpotent quasi-summand of G if each composition of maps H → H 0 → H is naturally a nilpotent endomorphism of G. In this case . we call G = H ⊕ H 0 a nilpotent quasi-decomposition of G. The reader can show that if H is a nilpotent quasi-summand of G and if K is a nilpotent quasi-summand of H then K is a nilpotent quasi-summand of . G. Evidently, if G = H ⊕ H 0 is a nilpotent quasi-decomposition of G then {H, H 0 } is a nilpotent set. The first lemma characterizes nilpotent quasi-summands in terms of idempotents in QEnd(G)/N (QEnd(G)). . LEMMA 8.4.2 Let G = H ⊕ H 0 and let eH : G −→ H be the canonical idempotent associated with H. Consider maps f : H −→ H 0 as endomorphisms of G in the natural way. The following are equivalent for H. 1. H is a nilpotent quasi-summand of G. 2. Hom(H, H 0 ) ⊂ N (QEnd(G)). 3. eH is central modulo N (QEnd(G)). Proof: 1 ⇒ 2 Let a : H −→ H 0 and x : G −→ G be elements of QEnd(G). Then a = eH 0 aeH and since H is a nilpotent quasi-summand eH xeH 0 aeH is a nilpotent endomorphism of G. Hence (xa)m+1 = (xa)(eE xeB aeE )m = 0 for large enough m. That is, a generates a nil left ideal of QEnd(G). 2 ⇒ 3 Let x ∈ QEnd(G). By part 2, eH xeH 0 , eH 0 xeH ∈ N (QEnd(G)). Then eH x − xeH

= (eH xeH + eH xeH 0 ) − (eH xeH + eH 0 xeH ) = eH xeH 0 − eH 0 xeH ∈ N (QEnd(G))

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so that eH is central modulo N (QEnd(G)). a

b

3 ⇒ 1 Let H 6= H 0 and let H −→ H 0 −→ H. Then ba = eH b(eH 0 aeH ). By part 3, 0 = eH (eH 0 a) ≡ (eH 0 a)eH (mod N (QEnd(G))) so that ba ≡ 0(mod N (QEnd(G))). Then ba is nilpotent, and hence H is a nilpotent quasi-summand of G. This completes the logical cycle. Given a quasi-summand H of G let ∆G,H

= QHom(H, G)QHom(G, H) = the ideal generated by compositions G −→ H −→ G.

Evidently ∆G,H is an ideal in QEnd(G). Because H is a quasi-summand of G, QHom(G, H)Hom(H, G) = QEnd(H) so that ∆2G,H

= (QHom(H, G)Hom(G, H))(QHom(H, G)Hom(G, H)) = QHom(H, G)QEnd(H)QHom(G, H) = QHom(H, G)Hom(G, H) = ∆G,H .

That is, ∆G,H is an idempotent ideal of QEnd(G). Let E be a ring. Recall that GAB(E) denotes the set of Gabriel filters of right ideals on E, and let IDEM(E) denote the set of idempotent ideals ∆ ⊂ E. Partially order GAB(QEnd(G)) and IDEM(QEnd(G)) by inclusion. Let Nil(G) denote the set of quasi-isomorphism classes [H] of nilpotent quasi-summands H of G, and define a partial order ≤ on Nil(G) by declaring that [K] ≤ [H] if K is quasi-isomorphic to a quasi-summand of H. Because G has finite rank ≤ is a partial order on Nil(G). The next result is a pair of bijections that characterize the nilpotent quasi-summands of QEnd(G) in terms of idempotent ideals of QEnd(G) and Gabriel filters of right ideals in QEnd(G). LEMMA 8.4.3 Define maps ∆ : GAB(QEnd(G)) −−−−→ IDEM(QEnd(G)) D : IDEM(QEnd(G)) −−−−→ Nil(G)

8.4. GABRIEL FILTERS ON QEND(G)

207

by ∆(γ) = the unique minimal element of γ D(∆) = [eG], where ∆ is generated by the idempotent e in QEnd(G). Then ∆ and D are isomorphisms of posets. Proof: Given a Gabriel filter γ on the Artinian ring QEnd(G) the intersection of all right ideals in γ is an ideal ∆ = ∆(γ) that is evidently the unique minimal element in γ. This ∆ is necessarily an idempotent ideal. Thus γ = D(∆) for some unique idempotent ideal ∆ of QEnd(G). The function ∆ is the one that sends γ 7→ ∆(γ). It is an easy exercise to show that the function ∆(·) is a bijection. Next given ∆2 = ∆ ⊂ QEnd(G) there is a unique e2 = e ∈ QEnd(G) such that e is central modulo N (QEnd(G)) and ∆ = QEnd(G)eQEnd(G) ([75, Corollary VIII.6.4]). Then by Lemma 8.4.2, eG is a nilpotent quasisummand of G. The function D is the one that sends ∆ 7→ [eG]. The reader can show that D : IDEM(QEnd(G)) −→ Nil(G) is a bijection. Q 0 the reader can show EXAMPLE 8.4.4 In the ring E = Q Q that IDEM(E)= Q 0 0 0 0 0 Q 0 0, , , , ,E . Q 0 0 Q Q Q 0 0 The Gabriel filters on E are those that contain one of the elements in IDEM(E). Given ∆ ∈ IDEM(E) let e2∆ = e∆ = e ∈ E be the unique idempotent such that ∆ = EeE. Then eG is the nilpotent quasi-summand of G that corresponds to ∆. Let H be a nilpotent quasi-summand of G. We say that H is a Qδ-quasi-summand of G if ∆G,H QG = QG, or equivalently if ∆G,H ∈ δ(QG). If H is a Qδ-quasi-summand of G and if no proper quasisummand of H is a Qδ-quasi-summand then we say that H is a minimal Qδ-quasi-summand. The lemma follows immediately from Lemma 8.4.3.

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PROPOSITION 8.4.5 Let G be an rtffr group and let H be a nilpotent quasi-summand of G. Then H is a minimal Qδ-quasi-summand of G iff ∆G,H is the unique minimal element ∆QG of δ(QG). LEMMA 8.4.6 Let G be an rtffr group. 1. A minimal Qδ-quasi-summand of G is unique up to quasi-isomorphism. 2. If H 0 is a (minimal) Qδ-quasi-summand of H and if H is a (minimal) Qδ-quasi-summand of G then H 0 is a minimal Qδ-quasisummand of G. Proof: 1. Suppose that H and H 0 are minimal Qδ-quasi-summands of G then by Proposition 8.4.5, ∆G,H and ∆G,H 0 equal the unique minimal . element ∆QG of δ(QG). By Lemma 8.4.2, H = H 0 . We leave the proof of part 2 as an exercise for the reader. The next result reduces the study of δ(G) to the groups G such that δ(G) = µ(G). That is, to those groups G such that each I ∈ δ(G) has finite index in End(G). . LEMMA 8.4.7 Let G be an rtffr group. Then G = H ⊕H 0 where H is a minimal Qδ-quasi-summand of G that is unique up to quasi-isomorphism and such that IH 6= H for each proper right ideal I ⊂ QEnd(H). Proof: By Lemma 8.4.6, the unique minimal element ∆QG ∈ δ(QG) corresponds to a minimal Qδ-quasi-summand H of G. Since ∆QG is unique H is unique up to quasi-isomorphism. Furthermore, the ideal ∆QH ⊂ QEnd(H) corresponds to a minimal Qδ-quasi-summand H 0 of H. Then H 0 is a Qδ-quasi-summand of G. The minimality of H shows us that H 0 = H, and hence that ∆QH = QEnd(H). That is, IQH 6= QH for each proper right ideal I ⊂ QEnd(H). This completes the proof.

8.4.2

Q-Faithful Groups

We say that G is a Q-faithful group if IQG 6= QG for each proper right ideal I ⊂ QEnd(G), or equivalently if D(QG) = {QEnd(G)}. Evidently G is a Q-faithful group iff IG = G ⇒ I has finite index in End(G). We will use a variation on the Baer splitting property to characterize Qfaithful groups. Theorem 8.3.3 implies that G is a Q-faithful rtffr group iff given a short exact sequence π 0 −−−−→ K −−−−→ X −−−−→ G −−−−→ 0

(8.5)

8.4. GABRIEL FILTERS ON QEND(G)

209

in which SG (X) + K = X, then (8.5) is quasi-split. That is, there is a map : G −→ X and an integer n 6= 0 such that π = n1G . Theorem 8.2.17 shows us that if E is semi-prime then each G ∈ Ω(E) is a Qfaithful rtffr group. Another important class of Q-faithful rtffr groups is the class of strongly indecomposable rtffr groups. THEOREM 8.4.8 If G 6= 0 is a strongly indecomposable rtffr group then IQG 6= QG for each proper right ideal I ⊂ QEnd(G). Proof: Let G be a strongly indecomposable rtffr group and suppose that IQG = QG for some proper right ideal I ⊂ QEnd(G). Because QEnd(G) is a local ring (see the comment preceding J´onsson Theorem 2.1.10), I ⊂ J (QEnd(G)) so that J (QEnd(G))QG = QG. Since QG is finitely generated Nakayama’s Lemma shows that QG = 0. This completes the proof. COROLLARY 8.4.9 Let G be a strongly indecomposable rtffr group. Then µ(G) = δ(G). THEOREM 8.4.10 The following are equivalent for an rtffr group G. 1. G is Q-faithful. . 2. If G = H ⊕ H 0 and if QHom(H, H 0 )H = H 0 6= 0 then some nonzero strongly indecomposable quasi-summand of H is isomorphic to a quasi-summand of H 0 . 3. Each short exact sequence (8.5) in which SG (X) + K = X is quasisplit. Proof: 1 ⇔ 2 G is a Q-faithful group iff IQG 6= QG for each proper right ideal I ⊂ QEnd(G) iff ∆QG = QEnd(G) iff G is a minimal Qδquasi-summand of G (Lemma 8.4.3), iff no proper quasi-summand of G is a Qδ-quasi-summand of G. The reader can show that this is equivalent to part 2. 1 ⇔ 3 Suppose that G is a Q-faithful group, and let (8.5) be a short exact sequence such that SG (X) + K = X. Then IG = G ⇒ End(G)/I is finite for each ideal I ⊂ End(G). Let I be the right ideal I = πHom(G, X) ⊂ End(G). Since SG (X) + K = X, G = πHom(G, X)G = IG. Then End(G)/I is finite. That is, there is an integer n 6= 0 and a map ∈ Hom(G, X) such that π = n1G . This proves part 3. Conversely assume part 3. Suppose that IG = G for some proper right ideal I of End(G). Let π1 , π2 , · · · denote the generators of I, let G =

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Gk for each k = 1, 2, . . ., and define π : ⊕∞ k=1 Gk → G to be the unique map such that π(x) = πk (x) for each x ∈ Gk and each k = 1, 2, . . .. Then ! ∞ X G = IG = πk End(G) G k=1

= =

=

∞ X k=1 ∞ X k=1 ∞ X

πk End(G)G πk (Gk ) ! π(Gk ) = π

M

Gk

.

k

k=1

By part 3 there is a map : G −→ ⊕k Gk and an integer P n 6= 0 such that π = n1G . Since G has finite rank we will write = m k=1 k for some finite list of maps k : G −→ Gk . Then k ∈ End(G) for each k = 1, . . . , m and hence n1G = π =

m X

πk k ∈ I.

k=1

Therefore End(G)/I is finite, QEnd(G) = QI, and thus part 1 is proved. This completes the logical cycle. For the next result, for each subgroup H of G let H∗ be the pure subgroup of G generated by H. We leave it as an exercise for the reader to show that if I ⊂ End(G) is a maximal right ideal then IG 6= G iff I = Hom(G, IG). Let N = N (End(G)). THEOREM 8.4.11 The following are equivalent for the rtffr group G. 1. G is a Q-faithful group. 2. I = Hom(G, IG∗ ) for each maximal pure right ideal I ⊂ End(G). 3. N (End(G)) = Hom(G, N G∗ ). 4. If π : P → QG is a projective cover over QEnd(G) then P generates QEnd(G). Proof: 1 ⇒ 2 Assume that G is Q-faithful and let I be a maximal pure right ideal in End(G). Evidently I ⊂ Hom(G, IG∗ ) and Hom(G, IG∗ ) is a pure right ideal in End(G). By our assumption on I, I = Hom(G, IG∗ ).

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211

2 ⇒ 3 Assume that I = Hom(G, IG∗ ) for each maximal pure right ideal in End(G). Since QN = J (QEnd(G)), QN is the intersection of the maximal right ideals of QEnd(G). Thus N is the intersection of the maximal pure right ideals of I. Then N

⊂ Hom(G, N G∗ ) ⊂ ∩{Hom(G, IG∗ ) I ⊂ End(G) is a maximal pure right ideal} = ∩{I I ⊂ End(G) is a maximal pure right ideal} by part 2 = N.

Thus N = Hom(G, N G∗ ), which proves part 3. 3 ⇒ 1 Suppose that G is not a Q-faithful group. By the comments beginning section 8.4.1 on page 204 there is a pure ideal ∆ ⊂ End(G) such that ∆QG = QG. Since QEnd(G)/N is a semi-simple Artinian ring there is an ideal N ⊂ I ⊂ QEnd(G) such that ∆ ∩ I = N and ∆ + I = QEnd(G). It follows that IQG = I(∆QG) ⊂ (I ∩ ∆)QG = N QG so that I is a nonnilpotent subset of Hom(G, N G∗ ). Hence N 6= Hom(G, N G∗ ) which completes the contrapositive. 1 ⇔ 4 In the notation of part 4, P generates QEnd(G) iff QEnd(G) is its trace ideal ∆QG iff IQG 6= QG for each proper right ideal I in QEnd(G) (see page 204), iff G is a Q-faithful group. This completes the logical cycle.

8.4.3

Q-Faithful E-Flat Groups

The results in this subsection show that when G is an E-flat Q-faithful group then the results in the previous section can be strengthened. The next result shows us that when G is an E-flat group then we lose very little information about δ(G) when we pass to δ(QG). LEMMA 8.4.12 Let G be an rtffr E-flat group. If QI ⊂ QEnd(G) and (QI)QG = QG then (QI ∩ End(G))G = G. In particular (∆QG ∩ End(G))G = G. Proof: Let I ∈ δ(QG). Since G is a flat left End(G)-module an application of · ⊗End(G) G to the inclusion End(G) QEnd(G) ⊂ (I ∩ End(G)) I

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yields the inclusions G (I ∩ End(G))G

∼ = ⊂ ∼ =

End(G) ⊗ G (I ∩ End(G)) End(G) QEnd(G) ⊗End(G) G I QG = 0. IQG

Thus G = (I ∩ End(G))G for each I ∈ δ(QG). This completes the proof. EXAMPLE 8.4.13 Let H and H 0 be strongly indecomposable rtffr groups such that Hom(H 0 , H) = 0, and such that SH (H 0 ) = Hom(H, H 0 )H is a full subgroup of H 0 , in which the factor group . H 0 /SH (H 0 ) is infinite. Let G = H ⊕ H 0 . Then H is a nilpotent quasisummand of G and ∆G,H = QHom(H, G)QHom(G, H) is in δ(QG). Observe that since Hom(H, H 0 )H 6= H 0 then G is not an E-flat group since otherwise G = (∆G,H ∩ End(G))G = Hom(H, G)Hom(G, H)G = Hom(H, G)H 6= G. This is an example of a quasi-isomorphism class (G) of rtffr groups in which no G0 ∈ (G) is an E-flat group. THEOREM 8.4.14 The following are equivalent for the E-flat rtffr group. 1. G is a Q-faithful group. 2. I = Hom(G, IG) for each maximal pure right ideal I ⊂ End(G). 3. N (End(G)) = Hom(G, N G). Proof: Given a pure right ideal I ⊂ End(G) there is an inclusion G ∼ End(G) QEnd(G) QG ⊗End(G) G ⊂ ⊗End(G) G = . = IG I QI QIQG Thus IG = IG∗ , whence the Theorem follows immediately from Theorem 8.4.11.

8.5. EXERCISES

8.5

213

Exercises

G, H, K are rtffr groups, p ∈ Z is a prime, E is an rtffr ring, M , N , are rtffr E-modules (left or right, depending on the setting). 1. Let H ∈ P(G) and suppose that (G, H) has the endlich Baer splitting property. If H is finitely G-generated then H ∈ Po (G). 2. Refer to the proof of Theorem 6.2.3. Show that each chain in C contains an upper bound in C. 3. Let M be a right End(G)-module. Then M (n) is G-compressed for each integer n > 0 iff TG (N ) 6= 0 for each finitely M -generated right End(G)-module N 6= 0. 4. Let P be a projective right E-module, let P ∗ be the dual of P , and let I be a left ideal of E. Prove that IP = P iff ∆P ⊂ I and that ∆P = ∆P ∗ . 5. Prove Proposition 8.1.4. 6. Let H ∈ Po (G). Show that Hom(G, H)∗ = Hom(H, G). 7. If IG 6= G for each proper finitely generated right ideal I ⊂ End(G) then G has the endlich Baer splitting property. 8. If H ∈ Po (G) and if N ⊂ HG (H) is such that N G = H then TG (HG (H)/N ) = 0. 9. Let G be a self-small right E-module. Then each H ∈ Po (G) is G-small. 10. Prove Example 8.1.7. 11. Prove Proposition 8.1.10. 12. A torsion-free group of finite rank is self-small. 13. If E is a reduced torsion-free ring of finite rank and if I is a maximal right ideal of E then E/I is finite. 14. If (G, H) has the endlich Baer splitting property then (G, H (n) ) has the endlich Baer splitting property for each integer n > 0. 15. Prove Theorem 8.2.14.

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16. Let G be a right E-module. Then Cyc(tor(G)) = {End(G)/I IG = G}. 17. Let X = Z, Y = Z[ p1 ], G = X ⊕ Y , and let ∆ = Hom(X, G)Hom(G, X). Show that G is a cyclic projective right End(G)-module, that ∆G = G, and that ∆ is a direct summand of End(G). 18. Let E be a ring, let Gab(E) be the set of Gabriel filters of E, and let Tor(E) be the set of hereditary torsion classes on E. There are inverse bijections Gab : Tor(E) −→ Gab(E) and Tor : Gab(E) −→ Tor(E). 19. Let ∆ be an idempotent ideal in E. Then Gab(∆) = {right ideals I ⊂ E ∆ ⊂ I} is a Gabriel filter on E. The hereditary torsion class corresponding to Gab(∆) is the set TorE (∆) of modules annihilated by ∆. 20. Show that I ∈ µ(M ) iff E/I is a semi-Artinian right E-module and each composition factor of E/I is isomorphic to some simple module in {E/J J ∈ maxE (M )}. 21. Let M be a right E-module. Then µ(M ) = δE (M ) ∩ µ(0). 22. Let M be a left E-module. If E/I is an Artinian module for each I ∈ δE (M ) then δE (M ) = µ(M ). 23. If N ⊂ K ⊂ L ⊂ M and if K is superfluous in M then L is superfluous in M and K/N is superfluous in M/N . 24. Suppose that P is a finitely generated projective left E-module. An E-submodule K ⊂ P is superfluous iff K ⊂ J(E)P . 25. If N is a nilpotent ideal then N M is superfluous in M for each left E-module M . 26. If ∆2 = ∆ ⊂ E then the hereditary torsion class corresponding to D(∆) is the set of right E-modules K such that K∆ = 0. 27. Show that µ(M ) equals the set of right ideals I ⊂ E such that E/I is a semi-Artinian E-module whose composition factors are isomorphic to E/J for some J ∈ maxE (M ). 28. Show that if π : P → M is a projective cover then M is finitely generated iff P is finitely generated.

8.5. EXERCISES

215

c) is a finitely Gabriel filter on E. b 29. Show that µ(M 30. Show that µ0 (M ) = {right ideals I ⊂ E E/I is a finite right Emodule such that IM = M and each composition factor of E/I is isomorphic to E/J for some J ∈ maxE (M )} is a Gabriel filter on the reduced torsion-free ring E of finite rank. 31. Show that Ib∩ E = I for each right ideal I of finite index in E, and 0 ∩ E = I 0 for each right ideal I 0 of finite index in E. b that I\ 32. Let E be a reduced torsion-free ring of finite rank and let M be a reduced torsion-free left E-module of finite rank. There is an b such that idempotent ideal ∆M ⊂ E b µ(M ) = {right ideals I ⊂ E E/I is finite and ∆M ⊂ Ib ⊂ E}. 33. Show that µ(G) is finite for the group G constructed in Example 6.1.7. 34. Let E be a Corner ring (not necessarily of finite rank) and let P be a finitely generated projective left E-module. Let ∆P be the trace of P in E. Then OE (∆M ) = OE (P ). c). 35. Show that OE (M ) is pure and dense in OEb (M b 1 + · · · + Ex b n=M c. Show that 36. P Let E be an rtffr ringPand let Ex n n b i=1 Exi . i=1 Exi is dense in 37. Let E be a local rtffr ring such that pE 6= E for some prime p ∈ Z. \ bp ). Prove that J (E)p = J (E 38. Let E be an rtffr ring. b is a semi-prime ring. (a) If E is a semi-prime ring then E b is a subcommutative (b) If E is a subcommutative ring then E ring. .

π 39. Let 0 → K → X −→ G → 0 be quasi-split . Then X ∼ = G ⊕ K.

40. Show that if H is a nilpotent quasi-summand of G and if H 0 is a nilpotent quasi-summand of H then H 0 is a nilpotent quasisummand of G. 41. Let G be a faithful rtffr group. Show that if I ⊂ End(G) is a maximal right ideal then I = Hom(G, IG) iff IG = 6 G.

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42. Let G be an rtffr group and suppose that G/H is finite. Then G is a finitely faithful S-group iff H is a finitely faithful S-group. 43. Let G be an rtffr group. Then G is a finitely faithful S-group iff G is a faithful L-group.

8.6

Questions for Future Research

Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. Give an example G in which {right ideals I ⊂ End(G) IG = G} is not a Gabriel filter. That is, find an rtffr group G such that D(G) 6= δ(G). 2. Give an internal characterization of the class Tor(G) of right E = End(G)-modules K such that K ⊗E G = 0. 3. Describe all Gabriel filters on the rtffr ring E. 4. Describe all hereditary torsion classes T of right E-modules. 5. Let E = End(G) and let I ⊂ E be a right maximal right ideal. If IG = G then HomE (I, E) = E = EndE (I). Describe the set of right ideals I ⊂ E such that HomE (I, E) = I. 6. Describe the idempotent ideals ∆ in E. 7. Describe the superfluous subgroups of G. 8. Describe the rtffr groups G that possess a projective cover over End(G). 9. Characterize the groups G such that End(G) is semi-perfect. 10. Further investigate δ(G) and µ(G). 11. Characterize those G such that δ(G) is finite. 12. Suppose δ(G) = D(∆) for some idempotent ideal ∆. Describe ∆ in terms of G. 13. Further characterize the faithful rings E.

8.6. QUESTIONS FOR FUTURE RESEARCH

217

14. Describe the Gabriel filters on QE and their associated hereditary torsion classes in terms of G. 15. Study (E-flat) Q-faithful rtffr groups.

Chapter 9

Endomorphism Modules In Chapter 4 we began a recurring theme, that certain E-properties coincide for rtffr groups. In this chapter we emphasize that coincidence. We will consider rtffr groups that are E-finitely generated, E-finitely presented, E-projective, E-generators, or E-flat. Some of these properties are classified up to quasi-isomorphism. We will also consider the homological dimensions of G as a left End(G)-module. The author’s research interests will of course greatly influence the list of properties that are examined here.

9.1

Additive Structure of Rings

A left E-module M is said to be a finitely generated rtffr E-module if M is a finitely generated left or right E-module whose additive structure (M, +) is an rtffr group. What follows is an extension of the BeaumontPierce-Wedderburn Theorem 3.1.6 from rings to finitely generated rtffr modules. Such a discussion is fundamental to our subsequent discussion of rtffr modules over rtffr rings. LEMMA 9.1.1 Let E be an rtffr ring and let T = E/N (E). Let M be an rtffr left E-module and suppose that there is an E-submodule K ⊂ M such that M/K is a finitely generated rtffr T -module. Then .

1. (M/K) ⊕ M 0 ∼ = T (n) for some finitely generated rtffr left T -module 0 M and some integer n, and .

2. M ∼ = (M/K) ⊕ K as groups. Proof: 1. Since M/K is a finitely generated rtffr left T -module QM/QK is a finitely generated projective left QT -module. There is 219

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a finitely generated projective QT -module P and an integer m such that (QM/QK) ⊕ P ∼ = QT (m) . Since P is finitely generated there is a finitely generated T -submodule M 0 ⊂ P such that QM 0 = P . Then (M/K) ⊕ M 0 ⊂ QT (m) and since both M/K and M 0 are finitely generated T -modules there is an integer n 6= 0 such that n(M/K ⊕ M 0 ) ⊂ T (m) . Furthermore since T (m) n(M/K ⊕ M 0 ) is a finitely generated T -module that is also a torsion group there is an integer n0 6= 0 such that n0 T (m) ⊂ n(M/K ⊕ M 0 ) ⊂ T (m) . This proves part 1. 2. Since M/K is a finitely generated left T -module there are x1 , . . . , xr ∈ M such that . M = T x1 + · · · + T xr + K. Thus there is a T -module mapping π : T (r) −→ M −→ M/K which lifts to a surjection π : QT (r) −→ QM/QK of projective QT -modules over the semi-simple ring QT . There is a QT module map f : QM/QK −→ QT (r) such that πf = 1QM/QK . Since M/K is a finitely generated T -module there is an integer n 6= 0 such that image nf ⊂ T (r) so that π(nf ) = n1M/K . It follows that . M∼ = (M/K) ⊕ K. This completes the proof. To classify the additive structure of rtffr semi-prime rings up to quasiisomorphism it suffices to classify the additive structure of a classical maximal order E over its center S. The next result follows immediately from [69, Theorem 39.14] which implies that a classical maximal order E is a projective S-module so that E ∼ = S (m−1) ⊕ J for some ideal J ⊂ S.

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221

LEMMA 9.1.2 Let E be a prime rtffr ring and let S = center(E). . There is an integer m 6= 0 such that E ∼ = S (m) as S-modules. Our classification of the additive structure of a semi-prime rtffr ring E is reduced to the case where E = S is a Dedekind domain. The ring E is a strongly indecomposable ring if (E, +) is strongly indecomposable. The next result is due to R. A. Beaumont and R.S. Pierce [18]. THEOREM 9.1.3 Let E be a Dedekind domain. There is a strongly . indecomposable Dedekind domain R ⊂ QE such that E = R(m) for some integer m. Moreover R ∼ = center(End(E, +)) as rings. Proof: Suppose that E is a Dedekind domain. Let . R = {Dedekind domains R ⊂ QE RE = E and RE is a finitely generated R-module }. Since E is an rtffr Dedekind domain E ∈ R 6= ∅ and since E has finite rank R contains a Dedekind domain R of minimal rank. Since a finitely generated torsion-free R-module is projective and since RE is a finitely . . generated torsion-free R-module E = RE = R(m) for some integer m 6= 0. We claim that R is strongly indecomposable as a group. Observe that R is a cyclic left End(R, +)-module. Given an ideal I 6= 0 ⊂ End(R, +) then IQR 6= 0 is an ideal in QR. Since QR is a field, IQR = QR, so that I 2 6= 0. Thus End(R, +) is a prime ring. By Lemma 9.1.2, End(R, +) is finitely generated over the integral domain S = center(End(R, +)). . Then R is a finitely generated rtffr S-module and thus R = S (`) for some integer ` > 0. By the minimality of R, S has finite index in R, and since S is a pure subgroup of End(R, +), S = R. . Finally write R = B (r) for some strongly indecomposable group B and integer r > 0. Since R = S commutes with the idempotent projections mapping R onto each copy of B, B is an ideal in R, hence R/B is . finite, whence R = B is strongly indecomposable.

9.2

E-Properties

In this section we will visit each of the E-properties mentioned in the introduction of this chapter.

222

9.2.1

CHAPTER 9. ENDOMORPHISM MODULES

E-Rings

Given a ring E with identity there is a natural embedding of rings λ : E → End(E, +) such that for f ∈ E, λ(f )(x) = f x for each x ∈ E. That is, λ(f ) is left multiplication by f on E. We ask When is λ an isomorphism? The ring E is called an E-ring if λ is an isomorphism. Examples of E-rings include Z, Q, Z/nZ for integers n. Other examples are hard to come by in the rtffr case. But once you find them they can be used to illuminate difficult points in our research. The existence of E-rings with a variety of ring structures is demonstrated in the following result. However, the constructed E-rings have countably infinite rank. See Theorem J.1.3 for a proof. THEOREM 9.2.1 [42, T.G. Faticoni] Each countable reduced torsionfree commutative ring is a pure subring of an E-ring. The countable hypothesis in the above theorem can be relaxed as M. Dugas, A. Mader, and C. Vinsonhaler [28] show us that each cotorsionfree commutative ring is a pure subring of an E-ring. Thus there are plenty of E-rings. For instance, if x1 , x2 , x3 , . . . are commuting indeterminants then Z[x1 , x2 , x3 , . . .]/(x21 , x32 , x43 , . . .) and Z[x1 , x2 , x3 , . . .] are contained in E-rings E1 and E2 respectively. These constructions show that in general the ring theoretic structure of an E-ring can be quite complicated. However the rtffr E-rings have a relatively uncomplicated ring structure. We will call the ring E a Dedekind E-ring if E is a Dedekind domain and an E-ring. See Appendix K. THEOREM 9.2.2 Let E be an rtffr integral domain. Then E is an E-ring iff E is a strongly indecomposable ring. In this case there is a unique Dedekind E-ring E such that E ⊂ E and E/E is finite. Proof: Suppose that E is not strongly indecomposable. Then there are nonzero subgroups A and B of E such that E/(A ⊕ B) is finite. There is a group map e : E → E such that e2 = ne for some integer . n 6= 0 and eE = A. Then e, n − e ∈ End(E) are such that e(n − e) = 0. Since E is an integral domain, End(E) 6= E, and then E is not an Ering. Consequently if the domain E is an rtffr E-ring then E is strongly indecomposable.

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223

Conversely, suppose that E is strongly indecomposable rtffr integral domain. By Theorem 9.1.3, E ∼ = S = center(End(E)). Since E is an integral domain End(E) is a prime ring so that QEnd(E) ∼ = Matn (D). 2 for some division ring D. Let e = e 6= 0 ∈ QEnd(E). Then E ∼ = . eE ⊕ (1 − e)E and since E is strongly indecomposable E = eE. That is, e = 1E . Then QE ⊂ QEnd(E) = D is a QS-vector space and QE = QS is a D-vector space. Since these rings are finite dimensional D = QS = QE and hence E = S = End(E). Therefore E is an E-ring, and the proof is complete. Our next result extends the Beaumont-Pierce Theorem 9.1.3 on the additive structure of rtffr prime rings. THEOREM 9.2.3 Let E be a prime rtffr ring. Then S = . (n) center(End(E)) is a strongly indecomposable E-ring and E = S as S-modules where S is a strongly indecomposable Dedekind E-ring such . that S = S. . (m) Proof: We proved in Theorems 9.1.2 and 9.1.3 that E = S where S is a strongly indecomposable integral domain. By Theorem 9.2.2, S . is an E-ring, and since S = S, S is a strongly indecomposable Dedekind E-ring. COROLLARY 9.2.4 Let E be a semi-prime rtffr ring. There are strongly indecomposable Dedekind E-rings R1 , · · · , Rt and integers . n1 , . . . , nt ≥ 0 such that Ri = Rj implies that i = j and such that . (n ) (n ) E = R1 1 ⊕ · · · ⊕ R t t as groups. The additive structure of rtffr E-rings is contained in the following theorem. A proof can be found in [10, Corollary 14.7] but try to prove it yourself without peeking. THEOREM 9.2.5 [73, R. Bowshell, P. Schultz] Let E be an rtffr ring. Then E is an E-ring iff there are strongly indecomposable Dedekind E-rings E 1 , . . . , E t such that 1. Hom(E i , E j ) = 0 for each 1 ≤ i 6= j ≤ t and 2. n(E 1 ⊕ · · · ⊕ E t ) ⊂ E ⊂ E 1 ⊕ · · · ⊕ E t for some integer n > 0.

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We have reduced the classification of rtffr E-rings to the case where the rtffr ring E is a Dedekind E-ring. The key to understanding Dedekind E-rings is an understanding of the distribution of the maximal ideals in the ring of algebraic integers in an algebraic number field. See Theorem K.2.4 for a proof. THEOREM 9.2.6 [T.G. Faticoni] Suppose that the rtffr ring E is a Dedekind domain and let O denote the ring of algebraic integers in QE. Then E is an E-ring iff given a proper subfield Q ⊂ k ⊂ QE there are maximal ideals M, M 0 ⊂ O such that M E 6= E, M 0 E = E, and M ∩ k = M 0 ∩ k. The field k is a minimal field extension of Q if Q and k are the only subfields of k. For minimal field extensions k of Q the classification of E-rings in k is a bit simpler. THEOREM 9.2.7 Let E be a Dedekind domain and let O be the ring of algebraic integers in QE. Further suppose that QE is a minimal field extension of Q. Then E is an E-ring if there is a prime p ∈ Z and maximal ideals M , M 0 ⊂ O such that M E 6= E, M 0 E = E, and p ∈ M ∩ M 0. Let p ∈ Z be a prime. The group G is p-local if pG 6= G and qG = G for each prime q 6= p ∈ Z. Theorem 9.2.6 and some Algebraic Number Theory will show that each algebraic number field k is the field of fractions of some Dedekind E-ring E. The question of which fields are of the form QE for some p-local Dedekind E-ring E is answered masterfully by R.S. Pierce [65] and R.S. Pierce and C. Vinsonhaler [66]. The number field k is said to be p-realizable if k = QE for some p-local Dedekind E-ring E. See [65, Theorem 2.1] for a proof of the following theorem. THEOREM 9.2.8 [R.S. Pierce] Let Γ be the Galois group Γ = Gal(k/Q) where k is a finite Galois extension of Q. If Γ has no nontrivial, normal, cyclic subgroups, and if Γ is not isomorphic to the symmetric group S4 or the alternating groups A4 , then k is p-realizable for all primes p ∈ Z.

9.2.2

E-Finitely Generated Groups

The group G is E-cyclic if G = End(G)x for some x ∈ G, and G is Efinitely generated if G is a finitely generated left End(G)-module. Each ring is an E-cyclic group, and in general if G is a finitely generated

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225

left module over some ring then G is E-finitely generated. Evidently Noetherian rtffr E-rings and their nonzero ideals are E-finitely generated groups. Furthermore if M is a left E-module then E ⊕ M is an E-cyclic group. Fortunately the class of E-finitely generated rtffr groups is closed under quasi-isomorphism. Given an integer m > 0 we have constructed in Example 4.3.2 a group G that requires exactly m ≥ 2 generators as a left End(G)-module. The group in question is quasi-equal to an E-cyclic group. Thus the number of generators for an E-finitely generated rtffr group can vary wildly in the quasi-isomorphism class of the group. We will present J.D. Reid’s [67] classification of the E-finitely generated rtffr groups. Our presentation differs from that in [67]. LEMMA 9.2.9 Let E be a semi-prime rtffr ring and let M and M 0 be torsion-free left E-modules with M 0 finitely generated. If π : M → M 0 . is a surjection then M ∼ = ker π ⊕ M 0 . Proof: Observe that π lifts to a surjection π : QM → QM 0 of left modules over the semi-simple ring QE. Since every QE-module is projective QM ∼ = ker π ⊕ QM 0 . Thus there is a QE-module map σ : QM 0 → QM such that πσ = 1QM 0 . Since M 0 is finitely generated there is an integer n 6= 0 such that nσ(M 0 ) ⊂ M . Inasmuch as e = σπ = (σπ)2 ∈ EndQT (QM ) and since ne ∈ EndT (M ), we have n1M = n(1 − e) ⊕ ne. Hence (n1M )M ⊂ [n(1 − e)](M ) ⊕ (ne)(M ) ⊂ ker(π|M ) ⊕ (nσ)(M 0 ) ⊂ M which completes the proof. LEMMA 9.2.10 Let E be a semi-prime rtffr and let M be a finitely generated torsion-free left E-module. There are integers t, n1 , . . . , nt > 0 and Dedekind rtffr E-rings R1 , . . . , Rt such that .

(n ) (n ) M∼ = R1 1 ⊕ · · · ⊕ Rt t .

Proof: Since M is finitely generated there is an integer n and a surjec. (n) (n) ∼ tion π : E → M . By Lemma 9.2.9, E = M ⊕ker π, and by Theorem 9.2.4, there are integers t, m1 , . . . , mt > 0 and strongly indecomposable Dedekind E-rings R1 , . . . , Rt such that .

(m ) (m ) E∼ = R1 1 ⊕ · · · ⊕ R t t .

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Then by J´onsson’s Theorem 2.1.10, there are integers t, n1 , . . . , nt such that . (n ) (n ) M∼ = R1 1 ⊕ · · · ⊕ Rt t which completes the proof. The next result is a classification of the group structure of a finitely generated left module over an rtffr ring. Recall that X∗ is the purification of X in a larger group. THEOREM 9.2.11 [T.G. Faticoni] Let E be an rtffr ring and let M . be an rtffr left E-module. If M is finitely generated then M = Mo ⊕ N where (n )

(n )

1. Mo = R1 1 ⊕ · · · ⊕ Rt t for some integers t, n1 , . . . , nt > 0 and Dedekind E-rings R1 , . . . , Rt , 2. N = (N (E)Mo )∗ = the purification of N (E)Mo in M , and 3. (N (E)Mo )∗ /N (E)Mo is finite. Proof: Let E be an rtffr ring and let M be a finitely generated rtffr left E-module. By the Beaumont-Pierce-Wedderburn Theorem 3.1.6 there is . a subring T ⊂ E such that E = T ⊕ N (E) as groups and it is clear that Mo = M/(N (E)M )∗ is a finitely generated left T -module. Then by Lemma 9.2.9 .

M∼ = Mo ⊕ (N (E)M )∗ as T -modules. Furthermore by Lemma 9.2.10 there are integers t, n1 , . . ., nt > 0 and Dedekind E-rings R1 , . . . , Rt such that (n1 )

Mo = R 1

(nt )

⊕ · · · ⊕ Rt

.

Evidently the module M/N (E)M is finitely generated over the Noetherian ring E/N (E) so (N (E)M )∗ /N (E)M is a finitely generated torsion E/N (E)-module. Since (N (E)M )∗ /N (E)M is a bounded quotient of an rtffr group, it is finite. This proves the theorem. THEOREM 9.2.12 [J.D. Reid] (See [67].) Let G be an rtffr group and let E = End(G). The following are equivalent. 1. G is E-finitely generated.

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227

2. There are integers t, n1 , . . . , nt > 0, strongly indecomposable Dedekind E-rings R1 , . . . , Rt , and a group N such that . (n ) (n ) G = R1 1 ⊕ · · · ⊕ Rt t ⊕ N and where N = SR1 ⊕···⊕Rt (N ). Proof: Assume part 1. Let E = End(G) and suppose that G is an E-finitely generated rtffr group. Let Go = G/(N (E)G)∗ . By Lemma 9.2.9 . G∼ = Go ⊕ (N (E)Go )∗ where (N (E)Go )∗ /N (E)Go is finite. Let N 0 = N (E)Go . Inasmuch as . (N (E)Go )∗ /N 0 is finite, and since N (E)Go = SGo (N 0 ), G ∼ = Go ⊕ N 0 for some group N 0 such that SGo (N 0 ) = N 0 . By Theorem 9.2.11 there are integers t, n1 , . . ., nt > 0 and strongly indecomposable Dedekind E-rings R1 , . . . , Rt such that .

(n ) (n ) Go ∼ = R1 1 ⊕ · · · ⊕ Rt t . .

∼ SR ⊕···⊕R (N 0 ) so we let N = SR ⊕···⊕R (N 0 ). Hence G = Then N 0 = t t 1 1 Go ⊕ N for some rtffr group N such that N = SR1 ⊕···⊕Rt (N ). This proves part 2. . Conversely, assume part 2. Suppose that G = Go ⊕ N where Go and N satisfy the conditions in part 2. If we let R = R1 × · · · × Rt then Go is a finitely generated R-module, and hence Go is an E-finitely generated group. Let x1 , . . . , xr ∈ Go be generators for Go over R. Inasmuch as R and Hom(Go , N ) embed naturally in QEnd(G) we have the following equations. . G = Go ⊕ SR (N ) . = Go ⊕ Hom(Go , N )Go = End(Go )x1 + · · · + End(Go )xr + Hom(Go , N )Go = End(Go )x1 + · · · + End(Go )xr + Hom(Go , N )End(Go )x1 + · · · + Hom(Go , N )End(Go )xr = End(Go )x1 + · · · + End(Go )xr + Hom(Go , N )x1 + · · · + Hom(Go , N )xr . = End(G)x1 + · · · + End(G)xr . Thus G, and hence any group quasi-isomorphic to G is E-finitely generated. This concludes the proof.

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Several of the E-properties imply the E-finitely generated property for rtffr groups. The group G is E-projective if G is a projective left End(G)-module, and G is an E-generator if G generates the category of left End(G)-modules. G is an E-progenerator if G is a finitely generated projective generator in the category of left End(G)-modules. Groups of the form Z ⊕ N are E-projective while E-rings are E-progenerator groups. COROLLARY 9.2.13 Let G be a strongly indecomposable E-finitely . generated rtffr group. Then G = E for some Dedekind E-ring E. Proof: If G is a strongly indecomposable E-finitely generated rtffr . group then by Theorem 9.2.12, G = R1 = E for some Dedekind E-ring E. COROLLARY 9.2.14 Let G be an rtffr group. Then G is E-finitely generated iff G is quasi-isomorphic to an E-cyclic group. Proof: Let E = End(G) and suppose that G is an E-finitely generated rtffr group. By Theorem 9.2.12 there are positive integers t, n1 , . . . , nt , strongly indecomposable E-rings R1 , . . . , Rt , and a group N such that N

= SGo (N ) (n )

(nt )

Go = R1 1 ⊕ · · · ⊕ Rt . G = Go ⊕ N.

, and

It suffices to show that Go ⊕N is an E-cyclic group. Let xi be a generator of Ri as an Ri -module and let x = x1 ⊕ · · · ⊕ xt . Since these are group homomorphisms, End(Go ⊕N )x contains each copy of Ri in Go . Thus End(Go ⊕ N )x ⊃ Go . Furthermore, by hypothesis N = SGo (N ) = Hom(Go , N )Go = Hom(Go , N )x so that End(Go ⊕ N )x ⊃ N . Therefore Go ⊕ N = End(Go ⊕ N )x. COROLLARY 9.2.15 An E-projective rtffr group is E-finitely generated.

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229

Proof: Let E = End(G), suppose that G is E-projective, and write G⊕P ∼ = E (I) for some index set I. Since G has finite rank there is a finitePsubset {x1 , . . . , xn } ⊂ G and a finite subset Io ⊂ I such that QG = ni=1 Qxi and {x1 , . . . , xn } ⊂ E (Io ) . Then G = E (I) ∩ QG ⊂ E (Io ) . The reader can show that because G is a direct summand of E (I) , G is a direct summand of E (Io ) , thus proving that G is finitely generated as a left E-module. This proves the lemma. COROLLARY 9.2.16 Let G be an E-generator rtffr group and let S = center(End(G)). Then G is E-finitely generated and G is a finitely generated projective S-module. Proof: Let E = End(G), suppose that G is an E-generator, and let S = EndE (G). Since G is a generator for E, G is a finitely generated projective right S = EndE (G)-module. But S ⊂ End(G) so that G is E-finitely generated. For strongly indecomposable rtffr groups we can prove a strong relationship between these E-properties. The rtffr group G is E-finitely presented if G is a finitely presented left E-module. THEOREM 9.2.17 [T.G. Faticoni] Let G be a strongly indecomposable rtffr group. The following are equivalent. 1. G is quasi-isomorphic to a Dedekind E-ring. 2. G is quasi-isomorphic to an E-cyclic group. 3. G is an E-finitely generated group. 4. G is an E-finitely presented group. 5. G is quasi-isomorphic to an E-projective group. 6. G is quasi-isomorphic to an E-generator group. 7. G is quasi-isomorphic to an E-progenerator group. Proof: 1 ⇒ 2 ⇒ 3 is clear. . 3 ⇒ 4 By Lemma 9.2.13, G = R for some strongly indecomposable . Dedekind E-ring R, so that End(G) = R is a Noetherian integral domain. The E-finitely generated G is thus E-finitely presented.

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. 4 ⇒ 5 and 6 By Lemma 9.2.13, G = R for some strongly indecomposable Dedekind E-ring R. This proves part 5 and 6. 5 or 6 ⇒ 1 Suppose that G is an E-projective rtffr group or an Egenerator rtffr group. By Lemmas 9.2.15 and 9.2.16, G is E-finitely generated, and since G is a strongly indecomposable group, Lemma 9.2.13 . implies that G = R for some Dedekind E-ring R. This proves part 1. Inasmuch as 1 ⇒ 7 ⇒ 5 is clear the proof is complete. An example will show that some of these implications do not possess a converse. EXAMPLE 9.2.18 There is an E-cyclic rtffr group that is not Eprojective. Proof: Let k be a quadratic field extension of Q and let O denote the ring of algebraic integers in O. There is a prime p ∈ Z such that pO = M M 0 for some maximal ideals M 6= M 0 in O. Let \ S = OM ∩ {Oq q 6= p ∈ Z is a prime }. Since M OM 6= OM , M S 6= S, and since M 0 Ox = Ox for each x ∈ {I, primes q 6= p}, M 0 S = M 0 . By Theorem 9.2.7, S is an E-ring and one can show that O/qO ∼ = S/qS ∼ = (Z/qZ)(2) for each prime q 6= p. ∼ Furthermore S/pS = Z/pZ. Thus Z is a pure subring of S. By counting p-ranks of the rank one group N = S/Z we see that qN 6= N for each prime q 6= p while pN = N . Therefore G = S ⊕N is E-finitely generated. Note that N is not an S-module even though it is S-generated. Thus G is E-cyclic but not E-projective.

9.2.3

E-Projective Groups

. LEMMA 9.2.19 Let S be a commutative rtffr ring and let I = S be an ideal. If I is a projective S-module then I ⊕ I = S ⊕ K for some S-module K. Proof: By hypothesis S/I is a finite commutative ring so that S/I = T1 × · · · × Tr for some local rings T1 , . . . , Tr . Thus I/I 2 = L1 ⊕ · · · ⊕ Lr where Li is a Ti -module. Since the projective S-module I generates S, Li generates Ti , and since Ti is a local ring, there is a surjection Li → Ti . Therefore there is a surjection I → I/I 2 → S/I that lifts to a map f : I → S such that f (I) + I = S. It follows that I ⊕ I ∼ = S ⊕ ker(f ⊕ 1I ) where (f ⊕ 1I ) : I ⊕ I → S is the obvious surjection. This proves the lemma.

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231

Consider the problem of classifiying the E-projective rtffr groups. The investigation is complicated by the fact that if N is any rtffr group then Z ⊕ N is E-cyclic and E-projective. Another example of an Eprojective group is as follows. Let S be an E-ring, and let N be an S-module. We say that N is an S-linear module if HomS (S, N ) = Hom(S, N ). Our S-linear modules are referred to as E-modules by A. Mader and C. Vinsonhaler [61]. Any group is a Z-linear module, a Q-vector space is a Q-linear module, and one can prove that if I is an ideal of finite index in an E-ring S then I is an S-linear module. EXAMPLE 9.2.20 We give a general construction of E-projective rtffr . groups. Let S be an E-ring, let I = S be a projective nonprincipal ideal, and let N be an S-linear module. Then G=I ⊕N is an E-projective group that is generated by two elements as a left End(G)-module. Proof: Since S is an E-ring I is S-linear and since N is S-linear I = HomS (S, I) = Hom(S, I) and N

= HomS (S, N ) = Hom(S, N ).

The following chain of equations shows that G = I ⊕ N is S-linear. G = HomS (S, G) ∼ = HomS (S, I) ⊕ HomS (S, N ) ∼ = Hom(S, I) ⊕ Hom(S, N ) ∼ = Hom(S, I ⊕ N ) ∼ = Hom(S, G). The next series of equations shows us that G is direct summand of . End(G)(2) . Since I = S is a projective ideal Lemma 9.2.19 states that I ⊕I ∼ = S ⊕ K for some S-module K. Then because G is S-linear, End(G) ⊕ End(G) ∼ = Hom(I ⊕ I, G) ⊕ Hom(N ⊕ N, G) ∼ = Hom(S ⊕ K, G) ⊕ Hom(N ⊕ N, G) ∼ = Hom(S, G) ⊕ Hom(K ⊕ N ⊕ N, G) ∼ = G ⊕ Hom(K ⊕ N ⊕ N, G)

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as left EndS (G) = End(G)-modules. Thus G is a projective left End(G)module that is generated by at most two elements as a left End(G)module. EXAMPLE 9.2.21 Let S be an rtffr E-ring and let N be an rtffr S-linear module. By the above example G = S ⊕ N is an E-cyclic Eprojective rtffr group. Our work here will show that the above examples are the only kind of E-projective rtffr group. To motivate our discussion of E-cyclic Eprojective rtffr groups, notice that by Lemmas 9.2.14 and 9.2.15 every E-projective rtffr group is quasi-isomorphic to an E-cyclic E-projective rtffr group. THEOREM 9.2.22 [63, G.P. Niedzwecki, J.D. Reid] Let G be an rtffr group, let E = End(G), and let S = center(E). Then G is an E-cyclic E-projective group iff G ∼ = S ⊕ N for some S-linear module N . In this case S is an rtffr E-ring. Proof: Suppose that G = S ⊕ N where S = center(E) and where N is an S-linear module. We have shown in Example 9.2.21 that G is E-cyclic and E-projective. Conversely let E = End(G) and assume that G is an E-cyclic Eprojective rtffr group. There is a split surjection π:E→G of left E-modules so there is an e2 = e ∈ E such that G ∼ = Ee. Thus S = EndE (G) = EndE (Ee) ∼ = eEe by Lemma 1.2.4. If we let N = (1 − e)G then G = eG ⊕ (1 − e)G ∼ = eEe ⊕ N ∼ =S⊕N where N is an S-module. On the other hand because G ∼ = S ⊕ N each group homomorphism f : S → S commutes with each element of S so that S∼ = HomS (S, S) = Hom(S, S) is an E-ring. Similarly HomS (S, N ) = Hom(S, N )

9.2. E-PROPERTIES

233

so that N is an S-linear module. This completes the proof of the theorem.

We have more to say about E-projective groups once we have studied E-generator groups. See Theorem 9.2.28. It is known that P is a finitely generated projective right E-module iff P is a generator as a left EndE (P )-module. This observation connects E-projective groups with the generator property as follows. LEMMA 9.2.23 Let G be an rtffr group and let S = center(End(G)). Then G is an E-projective group iff G is a generator in the category of right S-modules. Proof: Observe that S = EndEnd(G) (G). Then by Lemma 9.2.15, G is E-projective iff G is E-finitely generated E-projective iff G is a generator as a right S-module. The module U is called quasi-projective if U is projective relative to each short exact sequence 0 −→ K −→ U −→ V −→ 0 of left E-modules. The group G is E-quasi-projective if G is a quasiprojective left End(G)-module. The reader can prove as an exercise that . if E is an rtffr E-ring and if E = I is a projective ideal then E and I are locally isomorphic as groups and as E-modules. PROPOSITION 9.2.24 [84, C. Vinsonhaler, W. Wickless] Let G be an rtffr group. Then G is E-quasi-projective iff G is an E-finitely generated E-projective group. Proof: Let G be a quasi-projective left E-module. Because G is an rtffr group there is an integer m 6= 0 such that ∩k>0 mk G = 0. Since G/mG is a finite group there is a finitely generated E-submodule F ⊂ G such that (F +mG)/mG = G/mG = M . The usual properties of relative projectivity (see [6, Proposition 16.12]) show us that the quasi-projective left E-module G is projective relative to each short exact sequence 0 → K −→ F −→ G/mG → 0 of left E-modules. Thus the natural projection G → G/mG lifts to a map f : G → F such that ker f ⊂ mG. Since G is torsion-free ker f is pure in G and so ker f = m(ker f ) = 0 by our choice of m. Thus f is a

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regular endomorphism of G, hence f (G) ⊂ F ⊂ G, whence G/F is finite. (Prove that one, reader.) Then G is a finitely generated quasi-projective left E-module. Conversely, suppose that G is E-projective and is generated by n elements. Because G has finite rank there is a finite set {x1 , . . . , xr } ⊂ G such that annE (x1 , . . . , xr ) = 0. Then x = x1 ⊕ · · · ⊕ xr ∈ G(r) satisfies annE (x) = 0 so that E ∼ = Ex ⊂ G(r) . That same theory of relative projectivity in [6] shows us that the quasi-projective E-module G is projective relative to each short exact sequence 0 → K −→ E (n) −→ G → 0. Then the identity 1G lifts to a map f : G → E (n) . That is, our sequence is split exact, and therefore G is a projective left E-module. This completes the proof.

9.2.4

E-Generator Groups

A left E-module U is a generator over E iff U is finitely generated projective as a right EndE (U )-module. Thus we are motivated to study the E-generator groups. That is, those groups G that are generators in the category of left End(G)-modules. Recall that G is called an E-progenerator if G is a finitely generated projective generator in the category of left End(G)-modules. THEOREM 9.2.25 [T.G. Faticoni] Let G be an rtffr group, let E = End(G), and let S = center(E). The following are equivalent. 1. G is an E-progenerator group. 2. G is an E-generator group. 3. G is a finitely generated projective S-module. 4. G ∼ = U1 ⊕ · · · ⊕ Us for some projective ideals U1 , . . . , Us ⊂ S. Proof: 1 ⇒ 2 is clear and 2 ⇒ 3 follows from the introductory motivational statement. 3 ⇒ 4 Since G is a finitely generated projective module over the commutative rtffr ring S, Theorem 4.2.11 states that G = U1 ⊕ · · · ⊕ Us for some projective ideals U1 , . . . , Us ⊂ S.

9.2. E-PROPERTIES

235

4 ⇒ 1 Since G is a finitely generated projective S-module G is a generator for EndS (G). The reader will recall that E = EndS (G) so that G is a generator as a left E-module. On the other hand we will show that G is a generator over S. Let ∆ be the trace ideal in S for the projective S-module G. Since the trace is an idempotent ideal in S, Lemma 4.2.6 states that ∆ = eS for some e2 = e ∈ S. Inasmuch as annS (G) = 0, and since eG = ∆G = G, e = 1G . That is G generates S. Then G is a finitely generated projective left module over EndS (G) = E. This proves 1 which completes the logical cycle. COROLLARY 9.2.26 Let G be an rtffr group and let S = center(End(G)). Then G is an E-generator iff G = U ⊕ U 0 where U is an invertible ideal of S and U 0 is some projective S-module. Proof: Let G = U ⊕U 0 for some invertible ideal U over S and some Smodule U 0 . Then G is finitely generated projective over S, which implies that G is E-projective. Conversely, suppose that G is an E-generator group. Since S = S1 × · · · × St is a finite product of indecomposable commutative rtffr rings S1 , . . . , St there is no loss of generality if we prove the result for G under the assumption that S is indecomposable. So, Theorem 9.2.25 states that G = U1 ⊕ · · · ⊕ Us for some indecomposable finitely generated projective ideals U1 , . . ., Us of S. Let ∆ = traceS (U1 ). By Lemma 4.2.6, there is an e2 = e ∈ S such that ∆ = eS. Since S is indecomposable e = 1G and hence U1 generates S. In other words U1 is an invertible ideal in S. This completes the proof. The group G is E-self-generating if G generates each End(G)-submodule of G(n) for each n > 0. The following result shows that the notions of an E-generating and E-self-generating rtffr are interchangeable. LEMMA 9.2.27 Let G be an rtffr group. Then G is an E-self-generating group iff G is an E-generator group. Proof: Let E = End(G). Let x1 , . . . , xn ∈ G be a maximal linearly independent subset of G. Then f (x1 , . . . , xn ) = 0 iff f (G) = 0 iff f = 0. Thus E∼ = E(x1 ⊕ · · · ⊕ xn ) ⊂ G(n) as left E-modules. Then G generates E so that G is a generator as a left E-module. The converse is clear, so the proof is complete.

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9.2.5

E-Projective Rtffr Groups Characterized

Lemma 9.2.15 leaves open the question of just how many generators an E-projective rtffr group requires. That issue is addressed in the famous paper [16] written at the University of Connecticut during its Special Year in Algebra 1981. The main theorem of [16] is our next result. Let us agree to call the module M a local direct summand of N if given an integer n 6= 0 there is an integer m and module maps fn : N → M and gn : M → N such that fn gn = m1M . THEOREM 9.2.28 [UConn ’81 Theorem] Let G be an rtffr group, let E = End(G), and let S = center(E). The following are equivalent. 1. G is an E-projective group and generated by two elements. 2. G is an E-projective group. 3. G is an E-quasi-projective group. 4. G is a local direct summand of E. 5. G is E-finitely generated and Gp is a cyclic projective left Ep module for each prime p ∈ Z. 6. There is an rtffr E-ring S, an invertible ideal I in S, and an S-linear module N such that G = I ⊕ N . Proof: We will prove the logical circuit 6 ⇒ 5 ⇒ 4 ⇒ 1 ⇒ 2 ⇒ 3 ⇒ 6. 6 ⇒ 5 Suppose that G ∼ = I ⊕ N as in part 6. Since I is invertible, QI is an invertible ideal in the Artinian commutative ring QS, so that . QI = QS. Hence I = S. Because I is a generator over S, there is an integer n and an S-module K such that I (n) ∼ = S ⊕ K. Subsequently HomS (I, G)(n) ∼ = HomS (I (n) , G) ∼ = HomS (S, G) ⊕ HomS (K, G) as left E-modules. Then G ∼ = HomS (S, G) is a direct summand of HomS (I, G)(n) ⊕ HomS (N, G)(n) ∼ = HomS (I ⊕ N, G)(n) ∼ = E (n) as left E-modules, and so G is E-finitely generated. Furthermore, let p ∈ Z be prime. Choose n ∈ Z large enough that . Sn is a reduced group. Since I = S is an invertible ideal in S, In is an invertible ideal in Sn . Since Sn /nSn is then generated by In /nIn and

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237

since the finite commutative ring Sn /nSn is a product of local rings, there is a map f : In → Sn such that f (In ) + nSn = Sn . By Lemma 1.2.1 and . our choice of n, n ∈ J (Sn ), and hence f (In ) = Sn . Since Sn = In have finite rank, In ∼ = Sn . Because S and I are finitely presented S-modules we can apply the Change of Rings Theorem 1.5.2 to show that HomS (I, G)n ∼ = HomSn (In , Gn ) ∼ = HomSn (Sn , Gn ) ∼ = HomS (S, G)n . Thus Gn = HomS (S, G)n is a direct summand of HomS (I ⊕ N, G)n ∼ = En as a left En -module. This proves part 5. 5 ⇒ 4 is proved along the lines of Lemma 2.5.2. 4 ⇒ 1 Suppose that G is a local summand of E. We claim that G is E-projective and generated as a left E-module by two elements. Let n 6= 0 ∈ Z. There is an integer m and E-module maps fn : E → G and gn : G → E such that gcd(m, n) = 1 and fn gn = m1G . By part 4, there is an integer k and maps fm : E → G and gm : G → E such that fm gm = k1G . Since gcd(k, m) = 1 there are integers a, b such that ak + bm = 1. Then the maps f : E ⊕ E −→ G : (x, y) 7→ afm (x) + bfn (y) g : G −→ E ⊕ E : z 7→ gm (z) ⊕ gn (z) satisfy f g(z) = afm gm (z) + bfn gn (z) = (ak + bm)z = z. ∼ G ⊕ ker f and hence G is projective and generated by two Thus E ⊕ E = elements. This proves part 1. 1 ⇒ 2 ⇒ 3 are clear. 3 ⇒ 6 By Proposition 9.2.24 the quasi-projective right E-module is a finitely generated projective left E-module, and by Lemma 9.2.23, G is a generator over S. We will proceed in a pair of lemmas. The first might be interesting in its own right. LEMMA 9.2.29 Let G be an E-finitely generated rtffr group and let S = center(End(G)). Then S is a Noetherian semi-prime commutative ring. Proof: Since G is E-finitely generated Theorem 9.2.12 states that there are integers t, n1 , . . . , nt > 0 and strongly indecomposable Dedekind E-rings R1 , . . . , Rt such that . (n ) (n ) G = R1 1 ⊕ · · · ⊕ Rt t ⊕ N

(9.1)

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where N is generated by R = R1 ⊕ · · · ⊕ Rt . .

We will assume that Ri ∼ = Rj ⇒ i = j. Observe that SRi (Rj ) = Iij is an ideal in Rj . Since Rj is a Dedekind domain (Iij )−1 Iij = Rj for nonzero ideals I in Rj . Consequently if there is a nonzero map Ri → Rj then we can assume that Rj is a direct summand of N . Thus we can asssume that in (9.1), Hom(Ri , Rj ) = 0 for each i 6= j, and hence by Theorem 9.2.5, R is an E-ring. In particular R is a semi-prime ring. Since R is an S-submodule of G, and since R generates a subgroup of finite index in G, xR 6= 0 for each x ∈ S. Thus S ⊂ End(R, +) ∼ =R and hence S is a Noetherian semi-prime commutative rtffr ring. This proves the lemma. LEMMA 9.2.30 Under the hypotheses of Theorem 9.2.28, there is a . map f : G → S such that f (G) = S is a progenerator S-module. Proof: Since S is a semi-prime rtffr ring Lemma 1.6.3 implies that there is an integer m 6= 0 large enough so that Sm is a reduced rtffr ring. Let U = G/N∗ and let M1 , . . . , Ms be a complete list of the maximal ideals of S that contain mS. Since U is a generator over S there are maps fi : U → S such that fi (U ) 6⊂ Mi . By the Chinese Remainder Theorem 1.5.3 there is a map f : U → S such that f (U ) 6⊂ Mi

for any i = 1, . . . , s.

Then f (U ) + mS is an ideal in S that is not included in any of the maximal ideals that contain mS, whence f (U ) + mS = S. Furthermore f (U )m + mSm = Sm and by Lemma 1.2.1, m ∈ J (Sm ), so that f (U )m = . Sm ⊃ S. Thus f (U ) = S and therefore by Lemma 4.2.10, f (U ) is a progenerator ideal in S. It follows that G maps onto the progenerator f (U ), which proves the lemma. We continue with the proof of Theorem 9.2.28. By the above lemma there is an S-module surjection G → G/N∗ → S whose image I is . a progenerator over S. Evidently I = S is an invertible ideal of S. Hence G∼ = I ⊕N for some S-module N . Since S is then a quasi-summand of G, Hom(S, G) = HomS (S, G) so that I and N are S-linear modules. This proves part 6 and completes the logical cycle.

9.2. E-PROPERTIES

9.2.6

239

Noetherian Endomorphism Modules

The group G is E-Noetherian if G is a Noetherian left End(G)-module. In this section we consider the quasi-direct sum decomposition of the ENoetherian rtffr groups. In the process we classify Noetherian modules over rtffr rings. LEMMA 9.2.31 Let E be an rtffr ring. Then E is a right Noetherian ring iff for each integer k > 0, N (E)k /N (E)k+1 is a finitely generated right E-module. Proof: Suppose that for each integer k > 0, N (E)k /N (E)k+1 is a finitely generated right E-module. Then N (E)/N (E)2 is finitely generated by the Noetherian ring E/N (E) so that N (E)/N (E)2 and E/N (E) are Noetherian right E-modules. Hence E/N (E)2 is a Noetherian right E-module. Because E is rtffr N (E)k = 0 for some integer k > 0. Continuing inductively on k we eventually show that E/N (E)k = E is right Noetherian. The converse is clear so the proof is complete. The following result can be viewed as a generalization of R.S. Pierce’s classification (Theorem 9.2.3) of the additive structure of a prime rtffr ring. THEOREM 9.2.32 Let E be an rtffr ring and let M be an rtffr left E-module. Then M is a Noetherian left E-module iff .

(n ) (n ) M∼ = R1 1 ⊕ · · · ⊕ Rt t

for some Dedekind E-rings R1 , . . . , Rt . Proof: Suppose that M is a Noetherian left rtffr E-module. Using . the Beaumont-Pierce-Wedderburn Theorem 3.1.6 write E = T ⊕ N (E) for some semi-prime subring T of E. By Theorem 9.1.2 .

(m ) (m ) T ∼ = R1 1 ⊕ · · · ⊕ Rt t

for some integers m1 , . . . , mt and some distinct Dedekind E-rings R1 , . . . , R t . Because M is a Noetherian E-module each E-submodule of each factor of M is Noetherian. In particular the following E/N (E)-modules are Noetherian. M1 =

M , N (E)M∗

M2 =

N (E)M∗ , N (E)2 M∗

···

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Since N (E) is nilpotent there is an integer k such that Mk+1 = 0. By our choice of T the modules M1 , M2 , . . . , Mk are Noetherian left T -modules. Thus an iterated application of Lemma 9.2.9 shows us that .

M∼ = M 1 ⊕ · · · ⊕ Mk as left T -modules. There is a finitely generated free T -module P and a surjection π : P → M of T -modules. There is a left T -submodule P 0 ⊂ P such that . . ker π ⊕ P 0 = P . Then P 0 = M and since P ∼ = T (m) for some integer m J´onsson’s Theorem 2.1.10 states that .

(n ) (n ) M∼ = R1 1 ⊕ · · · ⊕ Rt t

for some integers n1 , . . . , nt ≥ 0. The converse is an exercise for the reader. This completes the proof.

The proof of the main result of this section is an application of Lemma 9.2.32. THEOREM 9.2.33 [64, A. Paras]. Let E be an rtffr ring. The following are equivalent. 1. E is a right Noetherian ring. 2. E is a left Noetherian ring. .

(n )

(n )

3. E ∼ = R1 1 ⊕ · · · ⊕ Rt t for some integers n1 , . . . , nt and some distinct Dedekind E-rings. In the process of proving Theorem 9.2.32 we inadvertently proved the following result. COROLLARY 9.2.34 Let M be a Noetherian rtffr module over the rtffr ring E. Then .

M∼ =

M N (E)M∗ ⊕ ⊕ · · · ⊕ N (E)k M N (E)M∗ N (E)2 M∗

as left E/N (E)-modules for some integer k > 0. We are now in a position to characterize the E-Noetherian rtffr groups. The E-Noetherian rtffr groups are characterized up to quasiisomorphism by A. Paras [64] where she proves the following theorem.

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241

THEOREM 9.2.35 [64, A. Paras]. Let G be an rtffr group and let E = End(G). Then G is an E-Noetherian rtffr group iff . (n ) (n ) G = R 1 1 ⊕ · · · ⊕ Rt t

(9.2)

where t, n1 , . . . , nt > 0 are integers and where R1 , . . . , Rt are Dedekind E-rings. In this case End(G) is a left Noetherian rtffr ring. Proof: If G can be written as in (9.2) then G is E-Noetherian by Theorem 9.2.32. Conversely, suppose that G is E-Noetherian. Since G has finite rank, there is an integer n and an embedding E −→ G(n) of left E-modules. Then G, G(n) , and E are left Noetherian E-modules. That is, E is a Noetherian ring. By Theorem 9.2.33 there are Dedekind E-rings R1 , . . . , Rt and integers t, k1 , . . . , kt > 0 such that .

(k ) (k ) E∼ = R1 1 ⊕ · · · ⊕ R t t

and such that (9.2) is true. This completes the proof. The E-Noetherian property provides us with another example of overlapping E-properties for rtffr groups. Of course an E-Noetherian group is an E-finitely generated group. THEOREM 9.2.36 Let G be an rtffr group and suppose that E = End(G) is semi-prime. The following are equivalent for G. 1. G is an E-Noetherian group. 2. G is an E-finitely generated group. 3. G is quasi-isomorphic to an E-projective group. 4. G is quasi-isomorphic to an E-generator group. 5. G is quasi-isomorphic to an E-cyclic E-progenerator group. Proof: Let E = End(G) be semi-prime. 5 ⇒ 4 and 3 is clear. 4 or 3 ⇒ 2 follows from Corollaries 9.2.15 and 9.2.16. 2 ⇒ 1 Suppose that G is E-finitely generated. Since E is semi-prime, QE is a semi-simple ring, and since QG is a projective left QE-module

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there is an integer n > 0 and a split embedding f : QG → QE (n) of left QE-modules. Since G is finitely generated as a left E-module we may assume that f (G) is a quasi-summand of E (n) . By Theorem 9.2.3 there are integers t, n1 , . . . , nt > 0 and Dedekind E-rings R1 , . . . , Rt such that .

(n ) E∼ = R(n1 ) ⊕ · · · ⊕ Rt t

so by J´onsson’s Theorem 2.1.10, G has a quasi-direct sum decomposition (9.2) for some integers m1 , . . . , mt ≥ 0. Then G is E-Noetherian by Theorem 9.2.35. 1 ⇒ 5 Say G is E-Noetherian. Then G has a quasi-direct sum decomposition (m1 )

G = R1

(mt )

⊕ · · · ⊕ Rt

for some integers t, m1 , . . . , mt > 0 and Dedekind. E-rings R1 , . . . , Rt . We can assume without loss of generality that Ri ∼ = Rj ⇒ i = j for each 1 ≤ i, j ≤ t. Then the Lemma 4.1.4 characterizing nilpotent sets and the semi-prime hypothesis imply that Hom(Ri , Rj ) ⊂ N (E) = 0 for each i 6= j. Because End(Ri ) = Ri for each i = 1, . . . , t we have End(G) = Matn1 (R1 ) × · · · × Matnt (Rt ). As a left End(G)-module we have

R1 G = ... ⊕ · · · ⊕ R1 n 1

Rt .. . . Rt n t

The reader will show as an exercise that G is a cyclic progenerator as a left End(G)-module.

9.3

Homological Dimensions

Theorem 2.3.4 provides us with an effective method for constructing groups G whose structure as a left End(G)-module has certain properties. This kind of example is an indispensable tool for developing intuition in studying properties of rtffr groups. We will use Theorem 2.3.4 to construct groups with prescribed homological dimension.

9.3. HOMOLOGICAL DIMENSIONS

9.3.1

243

E-Projective Dimensions

Let E be an rtffr ring and let M be a left E-module. The projective dimension of M

pdE (M ) is the least integer k ≥ 0 for which there is a long exact sequence 0 −→ Pk −→ Pk−1 −→ · · · −→ P0 −→ M −→ 0

(9.3)

whose terms Pj are projective left E-modules for each j = 0, . . . , k. Equivalently pdE (M ) = k if k is the least positive integer such that Extk+j E (M, ·) = 0 for each integer j > 0. Then M is a projective left E-module iff pdE (M ) = 0 iff Ext1E (M, ·) = 0. (See [72].) One of the first examples of an rtffr group with prescribed projective dimension is found in H.M.K Angad-Gaur’s Thesis [7]. In it he shows that for each integer n > 0 there is an rtffr group G such that pdE (G) = n where E = End(G). C. Vinsonhaler and W. Wickless [81, Corollary 6] show that for each integer n > 0 there is a completely decomposable rtffr group G such that rank(G) = 2n + 1 and pdE (G) = n. We will use the construction in Theorem 2.3.4 to show that pdEnd(G) (G) can be almost any integer less than the left global dimension of End(G). LEMMA 9.3.1 Let E be an rtffr ring and let M be an rtffr left Emodule. There is a short exact sequence 0 → M ⊕ E −→ G −→ QE ⊕ QE → 0

(9.4)

of rtffr left E-modules such that E ∼ = End(G). Then 1. pdE (G) = pdE (M ) if pdE (M ) ≥ 1 and 2. pdE (G) = 1 if pdE (M ) = 0. Proof: 1 and 2. Let M be an rtffr left E-module, let pdE (M ) = k ≥ 1, and let N be any left E-module. By Theorem 2.3.4 there is a short exact sequence (9.4) such that E ∼ = End(G). Inasmuch as (9.4) is nonsplit,

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Ext1E (QE, ·) 6= 0. Furthermore think of ExtkE (QE, ·) as ker δ2 /image δ3 where δ

δ

3 2 HomE (QE, E (c3 ) ) −→ HomE (QE, E (c2 ) ) −→ HomE (QE, E (c1 ) )

for some cardinals c1 , c2 , c3 . Inasmuch as Hom(QE, E(c) ) = 0, ExtkE (QE, ·) = 0 for each integer k ≥ 2. Thus pdE (QE) = 1. Given a fixed integer k ≥ 0 an application of the contravariant functor HomE ( , ?) to (9.4) yields the long exact sequence k+j Extk+j E (QE ⊕ QE, ?) → ExtE (G, ?) k+j+1 → Extk+j (QE ⊕ QE, ?) E (M, ?) → ExtE

for each integer j > 0. Since pdE (QE) = 1 ≤ k we have k+j+1 (QE ⊕ QE, ?) = 0 Extk+j E (QE ⊕ QE, ?) = ExtE

for each integer j > 0 and so k+j ∼ Extk+j E (G, ?) = ExtE (M, ?)

(9.5)

for integers j > 0. In particular, if pdE (M ) = ∞ then ExtkE (M, ?) 6= 0 for all integers k > 0 so that by (9.5), pdE (G) = ∞. If 1 ≤ pdE (M ) = k < ∞ then Extk+j E (M, ?) = 0 for all integers j > 0 so that by (9.5), pdE (G) ≤ k. Since k ≥ 1, ExtkE (M, ?) 6= 0 and so pdE (G) = k by (9.5). j If pdE (M ) = 0 then ExtjE (G, ?) ∼ = ExtE (M, ?) = 0 for each j > 0 implies that pdE (G) ≤ 1. Assume to the contrary that pdE (G) = 0. Because G is then an E-projective rtffr group Corollary 9.2.15 states that G is E-finitely generated. Hence QE is finitely generated by E, a contradiction. Thus pdE (G) = 1 which completes the proof. With the aid of the above lemma we will construct groups whose projective dimensions over their endomorphism rings are prescribed values. Recall that gd(E) = 1 + {pdE (I) I ⊂ E is a left ideal }. THEOREM 9.3.2 Let E be an rtffr ring and suppose that there is a left ideal I ⊂ E such that pdE (I) = n ≤ ∞. There is an rtffr group G such that End(G) ∼ = E and pdE (G) = n.

9.3. HOMOLOGICAL DIMENSIONS

245

Proof: Suppose that I is a left ideal of E such that pdE (I) = n > 0 and let M = I ⊕ E. Then pdE (M ) = pdE (I) = n and by Theorem 2.3.4 there is a short exact sequence (9.4) of left E-modules such that E∼ = End(G). Lemma 9.3.1 states that pdE (G) = pdE (M ) = n. THEOREM 9.3.3 Let E be a Noetherian rtffr ring such that gd(E) = n + 1 < ∞. For each integer 0 < k ≤ n there is an rtffr group Gk such that E ∼ = End(Gk ) and pdE (Gk ) = k. Proof: Let 0 < k ≤ n be an integer. There is a left ideal I ⊂ E such that pdE (I) = n, so there is a long exact sequence δ

δ

δ

n 2 1 0 → Pn −→ · · · −→ P1 −→ P0 −→ I → 0

of left E-modules in which each Pi is a projective. Since E is Noetherian I is finitely presented so we can choose such a sequence in which each Pi is finitely generated projective. Let Mk = image δk . The reader will note that the induced long exact sequence δ

δ

k k · · · −→ Mk → 0 0 → Pk −→

has exactly n − k projective terms so that pdE (Mk ) = n − k. Using M = Mn−k we can use Theorem 2.3.4 to construct a short exact sequence (9.4) of left E-modules such that E ∼ = End(G). By Lemma 9.3.1, pdE (G) = pdE (Mn−k ) = k. We will state but will not prove a stronger result for countable reduced torsion-free groups whose ranks may not be finite. THEOREM 9.3.4 [36, T.G. Faticoni] Let E be a Corner ring and suppose that gd(E) = n + 1. Then for each integer 0 < k ≤ n there is a Corner group Gk such that E ∼ = End(Gk ) and pdE (Gk ). EXAMPLE 9.3.5 Let E = Z[x]/(x2 ), let I = (x)/(x2 ) ∼ = Z, and let M = I ⊕ E. By Theorem 2.3.4 there is a short exact sequence (9.4) such that E ∼ = End(G). By Lemma 9.3.1, pdE (G) = pdE (M ), and it is readily verified that x

x

x

· · · −→ E −→ E −→ I → 0

(9.6)

is a projective resolution of I such that ker x = I ⊂ E is not a direct summand of E. Thus pdE (M ) = pdE (I) = ∞. By Lemma 9.3.1 there is an rtffr group G such that E ∼ = End(G) and pdE (G) = pdE (I) = ∞.

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EXAMPLE 9.3.6 Let E be an rtffr ring and let G be a ring constructed according to Corner’s Theorem such that E ∼ = End(G). Then G fits into a short exact sequence 0 → E −→ G −→ QE → 0 of left E-modules. By Lemma 9.3.1, pdE (G) = 1.

9.3.2

E-Flat Dimensions

The flat dimension of M fdE (M ) is the least integer k > 0 for which there is a long exact sequence (9.3) in which each Pi is a flat left E-module for each i = 0, . . . , k. Equivalently, fdE (M ) = k iff k is the least positive integer such that Tork+j E (?, M ) = 0 for each integer j > 0. Then M is a flat left E-module iff fdE (M ) = 0 iff Tor1E (?, M ) = 0. Moreover M is a flat left E-module iff given a short exact sequence 0 → K −→ N −→ L → 0 of right E-modules then the induced sequence 0 → K ⊗E M −→ N ⊗E M −→ L ⊗E M → 0 of groups is exact. C. Vinsonhaler and W. Wickless [81, Theorem 5] show that for each integer n > 0 there is a completely decomposable rtffr group G such that rank(G) = 2n + 1 and fdE (G) = n. We address the question of which integers are the flat dimensions of rtffr groups. The lemma indicates how we will construct rtffr groups of prescribed flat dimension. Recall that O(M ) = {q ∈ QE qM ⊂ M }. LEMMA 9.3.7 Let E be an rtffr ring and let M be a left rtffr E-module such that E = O(M ). If there is a short exact sequence 0 → M −→ G −→ QE ⊕ QE → 0 of left E-modules then fdE (G) = fdE (M ).

(9.7)

9.3. HOMOLOGICAL DIMENSIONS

247

Proof: Given fdE (M ) = k an application of ? ⊗E ? to (9.7) yields the long exact sequence Tork+j−1 (?, QE ⊕ QE) → Tork+j (?, M ) → Tork+j (?, G) → Tork+j (?, QE ⊕ QE) for each integer j > 0. Because QE is a flat left E-module Tork+j−1 (?, QE ⊕ QE) = Tork+j (?, QE ⊕ QE) = 0 for each k + j − 1 > 0. Thus Tork+j (?, M ) ∼ = Tork+j (?, G)

(9.8)

for each j > 0. If fdE (M ) = ∞ then by (9.8), fdE (G) = ∞. If 1 ≤ fdE (M ) = k < ∞ then Tork (?, M ) 6= 0 and Tork+j (?, M ) = 0 for each integer j > 0. Then by (9.8), fdE (G) = k. If fdE (M ) = 0 then M and QE are flat left E-modules. It follows from the exactness of 0 = Tor1 (?, M ) → Tor1 (?, G) → Tor1 (?, QE ⊕ QE) = 0 that fdE (G) = 0. This completes the proof. We can apply Lemma 9.3.7 to the next result in the same manner that we used Lemma 9.3.1 to prove Theorem 9.3.2. In constructing (9.7) in the next proof choose M = I ⊕ E and then apply Theorem 2.3.4. THEOREM 9.3.8 [T.G. Faticoni] Let E be an rtffr ring and let I be a left ideal in E such that fdE (I) = n. There is an rtffr group G such that E ∼ = End(G) and fdE (G) = n. The following result is proved in the same manner that we proved Theorem 9.3.3, but use Tor(?, ?) instead of Ext(?, ?). THEOREM 9.3.9 Let E be a semi-prime rtffr ring and suppose that there is a left ideal I of E such that fdE (I) = n. Then for each integer 0 ≤ k ≤ n there is an rtffr group Gk such that E ∼ = End(Gk ) and fdE (Gk ) = k. EXAMPLE 9.3.10 Let E = Z[x]/(x2 ), let I = (x)/(x2 ), and let M = I ⊕E. Then the group in the center term of (9.4) has fdE (G) = fdE (I) = ∞ as there is an infinite sequence (9.6) of flat modules E such that no kernel ker x is a direct summand of E.

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EXAMPLE 9.3.11 Let E be an rtffr ring and let G be a group constructed according to Corner’s Theorem 2.3.3 such that E ∼ = End(G). There is a short exact sequence 0 → E −→ G −→ QE → 0 of left E-modules so G is a flat left E-module. That is, fdE (G) = 0. Compare to Example 9.3.6 where we proved that pdE (G) = 1.

9.3.3

E-Injective Dimensions

Define the injective dimension of M

idE (M ) to be the least integer k ≥ 0 for which there exists an exact sequence 0 −→ M −→ E0 −→ E1 · · · −→ Ek −→ 0

(9.9)

where each Ei is an injective left E-module. Equivalently idE (M ) = k iff k is the least positive integer such that Extk+j (?, M ) = 0 for each integer j > 0. The constructions illustrating the injective dimensions over an rtffr ring E turn out to be harder than the examples we gave about the projective and flat dimensions. For instance, while our techniques give us good results when we consider the injective dimension of QG, they say very little about the injective dimension of G. Thus our results are incomplete. LEMMA 9.3.12 Let E be an rtffr ring. There is a short exact sequence 0 → E −→ G −→ QE → 0

(9.10)

of left E-modules such that E = End(G) and idE (QG) = idE (QE). Proof: Let idE (QE) = k. Applying · ⊗ Q to 9.10 we have a split exact sequence 0 → QE −→ QG −→ QE → 0 Then idE (QE) = idE (QG).

9.3. HOMOLOGICAL DIMENSIONS

249

THEOREM 9.3.13 [T.G. Faticoni] Let E be an rtffr ring and let G be the group constructed for E according to Corner’s Theorem 2.3.3. If idE (QE) = n then idE (QG) = n. Proof: By Corner’s Theorem there is a short exact sequence (9.10) of left E-modules such that E ∼ = End(G). Then by Lemma 9.3.12, n = idE (QE) = idE (QG). Let E be an rtffr ring and let G be the group constructed according to Corner’s Theorem 2.3.3. Examples 9.3.6 and 9.3.11 show us that pdE (G) = 1, fdE (G) = 0, and idE (QG) = idE (QE). Of course idE (QE) = 0 if E is a semi-prime ring. So in the by now familiar manner the general condition idE (QE) = 0 leads us to the following theorem. THEOREM 9.3.14 [T.G. Faticoni] Let E be an rtffr ring such that idE (QE) = 0, let M be an rtffr left E-module such that E = O(M ). There is a short exact sequence 0 → M −→ G −→ QE ⊕ QE → 0

(9.11)

of left E-modules such that E ∼ = End(G) and such that idE (G) = idE (M ). Proof: Use Theorem 2.3.4 to construct the short exact sequence (9.11) in which E ∼ = End(G). The usual argument shows us that idE (G) = idE (M ). For example, if E is a semi-prime rtffr ring then idE (G) = idE (E) for any group G constructed according to Corner’s Theorem 2.3.3. EXAMPLE 9.3.15 For each integer n there is a torsion-free group G of rank n such that pdE (G) = fdE (G) = ∞ and idE (QG) = 0 where E = End(G). Proof: Let E = Z[x]/(xn ). Then E is a torsion-free ring of rank n whose additive structure is a free group. Butler’s Construction in Theorem 2.3.1 states that E = End(G) for some group G such that E ⊂ G ⊂ QE as left E-modules. The reader can show that QE = Q[x]/(xn ) is a self-injective ring so that idE (QG) = idE (QE) = 0. The rest follows from Lemmas 9.3.1 and 9.3.7.

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EXAMPLE 9.3.16 Let A be a finite dimensional Q-algebra and suppose that M is a finitely generated left A-module. If annA (M ) = 0 then there is a full subring E ⊂ A and an rtffr group G such that E = End(G) and idE (QG) = min{idE (QE), idE (M )}. Proof: Let F be a full free subgroup of M and let E = O(F ) = {q ∈ A qF ⊂ F }. It is an easy exercise to show that for each q ∈ A there is an n 6= 0 ∈ Z such that nq(F ) ⊂ F . Thus QE = A. Theorem 2.3.4 constructs a short exact sequence 0 → F −→ G −→ QE ⊕ QE → 0 of left E-modules such that E ∼ = End(G). So idE (QG) = max{idE (QE), idE (QM )}. Let us agree to call the rtffr group G an E-injective group if QG is an injective left End(G)-module. C. Vinsonhaler and W. Wickless [83] characterize the injective hull of a torsion-free group of finite rank while C. Vinsonhaler [79] characterizes certain E-injective rtffr groups.

Z 0 Z 0 and let I = . Then Z Z Z 0 I is a projective ideal of E such that annE (I) = 0 and QI is an injective left E-module. The reader can verify that E = {q ∈ QE qI ⊂ I}. Theorem 2.3.4 produces an exact sequence EXAMPLE 9.3.17 Let E =

0 −→ I −→ G −→ QE ⊕ QE −→ 0 of left E-modules such that E ∼ = End(G). Since QE is a hereditary ring, idE (M ) ≤ 1 for each left E-module M . It is an interesting exercise to show that Mat2 (Q) is the injective hull of QE so that idE (QE) = 1. Thus G and the group constructed according to Corner’s Theorem 2.3.3 satisfy pdE (G) = 1, fdE (G) = 0, and idE (G) = 1. The above constructions show us that any classification of the Ehomological dimension of an rtffr group G will require an approach that differs significantly from the techniques used in this book. See [32] for a characterization of homological dimensions of modules over their endomorphism rings.

9.4. SELF-INJECTIVE RINGS

9.4

251

Self-Injective Rings

The literature contains a fairly complete characterization of those rtffr groups for which End(G) is a left or right hereditary ring. See [4, 10, 12, 41, 35, 43, 53]. Let us consider the rtffr groups G such that QEnd(G) is a left self-injective ring. Because QEnd(G) is Artinian it is known (see [30]) that QEnd(G) is right self-injective iff QEnd(G) is left self-injective. We will need the following ideas. G-monomorphisms are introduced in [37]. A G-monomorphism is a group map ˆ : H → H 0 such that ˆ = TG (ı) for some injection ı : M → M 0 of right End(G)-modules. The group H is G-presented if H = G(n) /K for some integer n > 0 and some G-generated subgroup K ⊂ G(n) . It is easy to see that ˆ : K → G is a monomorphism in the category G-Pre of G-presented groups iff SG (ker ˆ) = 0. By [32, Lemma 5.1.1] a group map ˆ : K → G is a Gmonomorphism iff ˆ = TG (ı) for some injection ı : I −→ HG (G) of right End(G)-modules iff ˆ is a monomorphism in the category G-Pre. Our characterization of rtffr groups such that QEnd(G) is a right self-injective ring is in terms of a lifting property for G-monomorphisms. THEOREM 9.4.1 [T.G. Faticoni] Let G be an rtffr group and let E = End(G). The following are equivalent. 1. QE is self-injective. 2. If ˆ : K → G is a monomorphism in G-Pre and if f : K → G is a group map then there is a map g : G → G and an integer n such that nf = gˆ . Proof: Assume part 1. Then QE is right self-injective. Let ˆ : K → G be a monomorphism in G-Pre, and let h : K → G be a group map. By [32, Lemma 5.1.1], ˆ is a G-monomorphism, so ˆ = TG (ı) for some injection ı : I → E of right E-modules. Specifically TG (I) ∼ = K. By the adjoint isomorphism HomE (I, E) ∼ = HomE (I, HG (G)) ∼ = Hom(TG (I), G) = Hom(K, G) and the reader can verify that this isomorphism sends each map φ : I → E to TG (φ). Thus h = TG (φ) for some φ : I → E. Since QE is selfinjective there is a map γ : QE → QE such that φ = γı, and there is an integer n such that nγ(E) ⊂ E. Then nh = nTG (φ) = nTG (γı) = (ng)ˆ . This proves part 2.

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Conversely assume part 2, let ı : I → QE be an inclusion of right QE-modules, and let f : I → QE be a QE-module map. Write I = QJ for some finitely generated right ideal J ⊂ E so that there is an integer n 6= 0 such that nı(J) ⊂ E. By definition TG (nı) : TG (J) → TG (E) = G is a G-monomorphism. Choose an integer m 6= 0 such that mf (J) ⊂ E so that TG (mf ) : TG (J) → TG (E). By part 2 there is an integer m 6= 0 and a group map g : TG (E) → TG (E) such that TG (mf ) = gTG (nı). By the Arnold-Lady-Murley Theorem 2.4.1 we can write g = TG (h) for some h : E → E. Then TG (mf ) = TG (nhı) or in other words TG (mf − nhı) = 0 : TG (J) −→ TG (E). Since J ⊂ E the adjoint isomorphism implies that mf − nhı = 0 whence n f = (m h)ı. This completes the proof. COROLLARY 9.4.2 Let G be an rtffr group such that QE is selfinjective. If ˆ : K → G is a monomorphism in G-Pre (for instance, if ˆ is an injection) and if f : K → G is a group map then there is a map ψ : G → G and an integer n such that nf = ψˆ .

9.5

Exercises

Let E be an rtffr ring, let G, H be rtffr groups. 1. Let E = End(G) and S = center(E). Prove that (a) E = EndS (G). (b) S = EndE (G). (c) If G = A ⊕ B then A is an S-submodule of G. 2. Let M be a finitely generated right E-module whose additive structure is a torsion group. Show that mM = 0 for some integer m 6= 0. 3. A hard problem. Find a finitely generated right ideal in an rtffr ring that is not finitely presented.

9.5. EXERCISES

253

4. Show that the finite modules over an rtffr ring E are finitely presented. 5. Let R be any ring and let N be a left R-module. Then N is generated by R as a group. Hint: For fixed x ∈ N consider the map λx : R → N such that λx (r) = rx for each r ∈ R. 6. Show that a finitely presented left E-module M possesses a projective resolution · · · P2 → P1 → P0 → M → 0 such that the projective modules P0 , P1 , P2 , . . . are finitely generated. 7. Let E be an rtffr ring and let I be a maximal right ideal of E. If pdE (I) = n then for each integer 0 < k ≤ n there is a finite rank group Gk such that E = End(Gk ) and pdE (Gk ) = k. 8. Let G be an E-finitely generated rtffr group N = N (End(G)). . Show that if G0 is a left End(G)-submodule of G such that G = . 0 0 G + N G then G = G . 9. Let G be an rtffr group and let N (End(G)) = N . Construct rtffr G . and G0 such that G0 ⊂ G is a left End(G)-submodule, G = G0 +N G, but G0 is not quasi-isomorphic to G. 10. If E is an rtffr semi-prime ring and if P ⊂ E (m) is a right Esubmodule then any E-module surjection π : E (n) → P is quasisplit. That is, there is a map σ : P → E (n) and an integer k such that πσ = k1P . 11. Let E be a semi-prime rtffr ring and let M be an E-submodule of E (n) for some integer n 6= 0. Then M is an E-Noetherian rtffr group. . 12. Let E be an rtffr E-ring and let E = I be a projective ideal of E. (a) E is locally isomorphic to I as groups. (b) E is E-locally isomorphic to I as an E-module. 13. Let R be an rtffr integral domain. Prove that if I ⊂ R is a nonzero ideal then R/I is finite. 14. Let R be an rtffr integral domain and let M be a finitely generated R-module. Then the R-torsion submodule T of M is finite and . M/T ∼ = R(m) for some integer m > 0.

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15. Suppose that G is an rtffr group and that A ⊕ B is a subgroup of finite index in G. Show that there is a quasi-idempotent e ∈ End(G) corresponding to A. That is, there is an e ∈ End(G) such that e2 = ne for some integer n 6= 0, e(x) = nx for each x ∈ A, and e(B) = 0. 16. Show that if E is a ring then E (n) is a cyclic projective left Matn (E)module. 17. Let R be an E-ring. Then R(n) is a cyclic projective left End(R(n) )module. 18. Let E be an rtffr ring. (a) Find an rtffr left E-module M such that QM is the injective hull of the left E-module E. ∼ End(G) and (b) Find an rtffr group G such that E = idE (QG) = 0. (c) Let E be an rtffr ring and let M be an rtffr left E-module. Let lengthE (M ) denote the composition length of QM as a left QE-module. Show that for each integer n there is an rtffr group G such that lengthEnd(E) (G) = n. 19. Let E be a semi-prime rtffr ring and let M , N be left rtffr Emodules. Show that ExtnE (M, N ) is a bounded group for each integer n ≥ 1. 20. Let A be a self-injective finite dimensional Q-algebra and let M be any free subgroup of A such that QM = A. (a) Prove that there is a group G such that M ⊂ G ⊂ A and such that QEnd(G) = A. (b) Prove that G is an E-injective rtffr group and QEnd(G) is a self-injective ring. (c) By varying M produce a variety of properties on G. For instance, Show that under one choice of M , G has E-property while under another choice G does not have E-property. Be creative. (d) Is it possible that G is strongly indecomposable for one choice of M , and decomposable for another? 21. Find a better way of investigating the injective dimension of G as a left E-module and not of QG.

9.6. QUESTIONS FOR FUTURE RESEARCH

9.6

255

Questions for Future Research

Let G be an rtffr group, let E be an rtffr ring, and let S = center(E) be the center of E. Let τ be the conductor of G. There is no loss of generality in assuming that G is strongly indecomposable. 1. We have characterized the finitely generated rtffr E-modules M in terms of direct sums of E-rings. Characterize larger classes of Emodules. For instance, consider the injective E-modules. Dual the definition of finitely generated E-modules to arrive at the finitely cogenerated E-modules. See [6]. 2. Study The Beaumont-Pierce-Wedderburn Theorem for a ring E. Give a canonical or natural value for the semi-prime ring T ⊂ E. 3. Prove the converse of Theorem 9.2.11. 4. This question is due to R. Pierce. Let property be a module theoretic property, and say that G is an E-property group if G satisfies property. Pick property and characterize the E-property groups. 5. Fix property. Characterize all of the rtffr rings E such that each G ∈ Ω(E) satisfies property. 6. Fix property. Construct examples of indecomposable rtffr E-property groups. 7. Use Galois Theory to characterize the strongly indecomposable Erings. See [39, 65, 66]. 8. See [32]. Find an internal characterization of rtffr groups G that possess infinite flat, projective, or injective dimension over End(G). 9. Find an internal characterization of rtffr groups G that possess finite flat, projective, or injective dimension over End(G). 10. Characterize the rtffr groups G such that QG is an injective left End(G)-module. 11. Characterize the rtffr groups G such that QEnd(G) is an injective left or right End(G)-module.

Appendix A

Pathological Direct Sums We have sectioned these examples into the appendices because of the different techniques used. Reread the first few results of the chapters. You will see the use of ring and module theory, and of functors and functorial methods. These ideas were not widely used in abelian group theory during the period of the 1950s and 1960s. A.L.S. Corner [23] broke that mold with the publication of his famous result now called Corner’s Theorem F.1.1 wherein he proved that each countable reduced torsion-free ring is the group endomorphism ring of a torsion-free group. His technique was pure number theory and linear algebra over Z. Compare this to the Arnold-Lady Murley Theorem 2.4.1 that is pure category theory. That is why we have these appendices. The results of the chapters use modern techniques while the techniques in the appendices are all number theory. They represent the techniques used in the early years of abelian groups.

A.1

Nonunique Direct Sums

We have already claimed that direct sums of the general rtffr group are not unique in any sense. In this appendix we will give examples to justify our claim. We begin with the most elementary of examples. These examples come from [46, Sections 90]. EXAMPLE A.1.1 Let n ≥ 2 be an integer. There exists an rtffr group G such that G = A1 ⊕ · · · ⊕ An−1 ⊕ B = C ⊕ D where the groups Ai , B, C, D are indecomposable, rank(Ai ) = 1, rank(B) = rank(C) = rank(D) = n − 1. 257

258

APPENDIX A. PATHOLOGICAL DIRECT SUMS

Proof: Let n ≥ 3 be an integer, let p, q, p1 , · · · , pn−1 be different primes, and let {a1 , . . . , an−1 } and {b1 , . . . , bn−1 } be bases for the Q-vector space V . Define Aj

= hp−∞ aj i for each j = 1, · · · , n − 1 j

−∞ B = hp−∞ 1 b1 , . . . , pn−1 bn−1 ,

p−1 q −1 (b1 + b2 ), . . . , p−1 q −1 (b1 + bn−2 )i and let G = A1 ⊕ · · · ⊕ An−1 ⊕ B. LEMMA A.1.2 B is indecomposable. Proof: Suppose that B = U ⊕ W as abelian groups. We have a quasi-isomorphism . −∞ U ⊕ W = hp−∞ 1 b1 , . . . , pn−1 bn−1 i = Bo and the subgroups hp−∞ bj i are fully inavariant in Bo because the pj j are different primes. Since these subgroups have rank one, there is a complete set of central indecomposable idempotents e1 , . . . , en−1 ∈ EndZ (Bo ) such that ej (x) = x for each x ∈ hp−∞ bj i. j Thus each idempotent in EndZ (Bo ) is a sum of a subset of {ej 1, . . . , n − 1}. For example, given the canonical idempotent map

i =

e : Bo −→ U there are disjoint subsets σ, ψ of {1, . . . , n − 1} such that e = ⊕j∈σ ej . Suppose for the sake of contradiction that there are 1 ∈ ψ and j ∈ σ. Then e(p−1 q −1 (b1 ⊕ bj )) = p−1 q −1 (bj ). This is contrary to the fact that bj has p-height 0 in hp−∞ bj i. Thus j σ = ∅, hence U = 0, whence B is indecomposable.

A.1. NONUNIQUE DIRECT SUMS

259

Proof of Example A.1.1: We choose an ordered base a1 , . . . , an−1 , b1 , . . . , bn−1 for V and primes p 6= q. Choose integeres s and t such that ps − qt = 1 and then choose cj = paj + tbj and dj = qaj + sbj for each j = 1, . . . , n − 1. Let −∞ −1 −1 C = hp−∞ 1 c1 , . . . , pn−1 cn−1 , p (c1 + c2 ), . . . , p (c1 + cn−2 )i −∞ −1 −1 D = hp−∞ 1 d1 , . . . , pn−1 dn−1 , q (d1 + d2 ), . . . , q (d1 + dn−2 )i.

Then C ∩ D = 0, C, D ⊂ G, and by the above lemma C and D are indecomposable. By our choices of cj , dj , s, t we have aj

= scj − tdj

bj

= pdj − qcj

b1 + bj

= p(d1 + dj ) − q(c1 + cj ).

Thus G = C + D and therefore G = C ⊕ D. Because A1 ∼ 6 C, D in the above example, it follows that a direct sum = decomposition of an rtffr group G into indecomposables is not necessarily unique.

Appendix B

ACD Groups B.1

Example by Corner

The next example shows that rtffr groups can have many direct sum decompositions into indecomposables. EXAMPLE B.1.1 [Corner] (See [46].) Let n ≥ k ≥ 1 be integers. There is a group G = G(n) of rank n such that for each partition r1 + · · · + rk of n into k positive summands rj there is an indecomposable direct sum decomposition G = G1 ⊕ · · · ⊕ Gk such that rj = rank(Gj ) for each j = 1, . . . , k. Proof: Let p, p1 , . . . , pn−k , q1 , . . . , qn−k be a list of different primes, let V be a Q-vector space, let u1 , . . . , uk , x1 , . . . , xn−k be a basis for V , and let G = hp∞ u1 , . . . , p∞ uk , ∞ p∞ 1 x1 , . . . , pn−k xn−k , −1 q1−1 (u1 + x1 ), . . . , qn−k (u1 + xn−k )i.

We will show that G has the indicated property. Let n = r1 + · · · + rk be a partition of n with rj ≥ 1 for each j = 1, . . . , k. 261

262

APPENDIX B. ACD GROUPS Suppose there are s1 , . . . , sk ∈ Z such that s1 + · · · + sk = 1

and unknowns v1 , . . . , vk . The system of linear equations s1 v1 + s2 v2 + · · · + sk vk = u1 −v1 + v2 = u2 .. .. . . −v1 + vk = uk has determinant 1. Hence the system has a solution v1 , . . . , vk ∈ Z such that hv1 , . . . , vk i = hu1 , . . . , uk i. Let −1 Gj = hp∞ vj , p∞ i xi , qi (vj + xi ) i = tj + 1, . . . , tj+1 i

where t1 = 0 and tj = (r1 − 1) + · · · + (rj−1 − 1). Observe that u1 + xi = (vj + xi ) + s1 v1 + · · · + sj−1 vj−1 + (sj − 1)vj + · · · + sk vk .

Then in choosing the s1 , . . . , sk require that 1 (mod )qi for i = tj + 1, · · · , tj+1 sj = . 0 (mod )qj otherwise [Comment: For example, let Qj = q1 · · · qj−1 qj+1 · · · qn−k , P choose integers m` such that ` m` Q` = 1, and then choose sj = mti +1 Qti +1 + · · · + mti+1 Qti+1 .] Subsequently Gj ⊂ G for each j = 1, . . . , k, and p∞ u 1 , . . . , p ∞ u k , ∞ p∞ 1 x1 , . . . , pn−k xn−k , −1 q1−1 (u1 + x1 ), . . . , qn−k (u1 + xn−k ) ∈ G 1 + · · · + Gk . Thus G = G1 ⊕ · · · ⊕ Gk where by construction rank(Gj ) = rj and by Lemma A.1.2, Gj is indecomposable. This completes the construction.

Appendix C

Power Cancellation The group G satisfies power cancellation if given a group H and an integer n ≥ 1 such that G(n) ∼ = H (n) then G ∼ = H.

C.1

Failure of Power Cancellation

The next example shows that rtffr groups fail to satisfy the power cancellation property. EXAMPLE C.1.1 Let p ≥ 3 be a prime. There are nonisomorphic rtffr groups A and B of the same rank such that A(n) ∼ = B (n) iff p divides n. Proof: Let k be an algebraic number field of class number h such that p is a prime divisor of h. Let O = O(k) denote the ring of algebraic integers in k, and let Γ denote the ideal class group of O. There is a subgroup Γ(p) of the finite abelian group Γ of order p. Since p ≥ 3 there are at least two isomorphism classes (I) 6= (J) in Γ(p) of order p. Then for integers n > 0 In ∼ = J n iff p divides n. Because O is a Dedekind domain, Steinitz’ Theorem [69, Theorem 4.13] states that I (n) ∼ = J (n) iff I n ∼ = J n . So I (n) ∼ = J (n) iff p divides n. Using Corner’s Theorem F.1.3 there is an rtffr group G such that EndZ (G) ∼ = O. By the Arnold-Lady-Murley Theorem 2.4.1 there are rtffr groups A, B ∈ Po (G) such that HG (A) = I and HG (B) = J. Then HG (A(n) ) = I (n) ∼ = J (n) = HG (B (n) ) 263

264

APPENDIX C. POWER CANCELLATION

iff p divides n. By Theorem 2.4.1, A(n) ∼ = B (n) iff p divides n. This completes the proof. Fuchs and Loonstra give an example of nonisomorphic rtffr groups G and H such that Gn ∼ = H n for some integer n > 1. EXAMPLE C.1.2 [L. Fuchs and F. Loonstra] (See [46, Theorem 90.3].) Let m ≥ 2 be an integer. There are two torsion-free indecomposable rank 2 groups G and H such that G(n) ∼ = H (n) iff m divides n.

Appendix D

Cancellation The group G satisfies the cancellation property if for any groups K, L G⊕K ∼ = G ⊕ L implies K ∼ = L.

D.1

Failure of Cancellation

The next example shows that the category of rtffr groups does not enjoy the cancellation property. THEOREM D.1.1 [L. Fuchs and F. Loonstra] (See [46, Theorem 90.4].) Let m ≥ 1 be an integer. There are groups G, K, and pairwise nonisomorphic indecomposable groups H1 , · · · , Hm such that rank(B) = 1, rank(Hj ) = 2, and G∼ = B ⊕ Hj for each j = 1, . . . , m. Proof: Begin with two infinite disjoint sets of primes P and Q, and a prime po not in P ∪ Q. Let b, x, y 6= 0 ∈ Q. Construct rank one groups B = hp−1 b p ∈ P i,

X = hp−1 x p ∈ P i,

Y = hq −1 y q ∈ Qi.

By Lemma A.1.2, H1 = hX ⊕ Y, p−1 o (x ⊕ y)i is an indecomposable rank 2 group. Define G = B ⊕ H1 . Choose integers qi , ri , si , ti such that qi ti − ri si = 1 and let bi = qi b + si x

xi = r i b + t i x 265

266

APPENDIX D. CANCELLATION

for each i = 2, . . . , m. We let Bi = hp−1 bi p ∈ P i∗ and Xi = hp−1 i xi p ∈ P i∗ be isomorphic rank one subgroups of B ⊕ X. Then B ⊕ X = Bi ⊕ Xi . We will refine our choices of bi and xi so that for some integers 1 < ki < po , G = Bi ⊕ Hi with Hi = hp−1 o (ki xi + y), xi ∈ Xi , y ∈ Y i.

(D.1)

Notice that by Lemma A.1.2, in any choice of ki , Hi is indecomposable. From our choices of qi , ri , si , ti , and xi we see that ki xi + y = ki ri b + (ki ti − 1)x + (x + y) is divisible by p. Hence ki ri ≡ 0 mod p

and

ki ti ≡ 1 mod p.

(D.2)

Since 1 < k1 < p we can choose ri = p, qi = ki , and si , ti such that qi ti − ri si = 1. Then (D.2) is satisfied and hence (D.1) holds. It remains to prove the lemma. LEMMA D.1.2 Let H1 , . . . , Hm be groups as defined in (D.1). Then the Hi are quasi-isomorphic rank 2 acd groups such that Hi ∼ 6 Hj = for each 1 ≤ i 6= j ≤ m. Proof: Let φ : Hi −→ Hj be an isomorphism. Then φ(X) = X and φ(Y ) = Y and φ(xi ) = ±xi and φ(y) = ±y. Thus φ(ki xi + y) = ±(kj xj ± y). Preservation of divisibility implies that kj ≡ ±ki mod p. Consequently with k1 = 1, k2 = 2, · · · , km = m

D.1. FAILURE OF CANCELLATION

267

and p > 2m − 1 then kj 6≡ ±ki mod p. Thus no two of the H1 , . . . , Hm are isomorphic. This proves the lemma and completes the proof of the example. The above example shows that for each m ≥ 1 there exists an rtffr of rank 3 G with m inequivalent direct sum decompositions. It also shows that cancellation fails m different times in G. It is interesting to note that E.L. Lady [10, Theorem 11.11(a)] has shown that in any group direct sum decomposition G = B ⊕ H ∼ = B ⊕ K there are at most finitely many isomorphism classes of K.

Appendix E

Corner Rings and Modules E.1

Topological Preliminaries

Fix a Corner ring E. That is, the additive group (E, +) is a countable reduced torsion-free abelian group. A left E-module M is a Corner Emodule if (M, +) is a countable reduced torsion-free abelian group. A Corner E-module is always a left E-module. Fix a Corner E-module M and assume that annE (M ) = 0. Given a left E-module M Po (M ) is the set of finite subsets of M . We have a linear topology Γ(QE, QM ) = Γ(QM ) on QE whose neighborhoods of 0 are annQE (F ) over all finite subsets F ⊂ QM . This Haussdorf topology on QE is called the QM -topology. d is the completion of QE under the Γ(QM )-topology. QE Let d qM ⊂ M }. b ) = {q ∈ QE O(M

269

270

APPENDIX E. CORNER RINGS AND MODULES

b )-module. Then M is a left O(M

b Γ(M ) is the linear topology whose basis of open neighborhoods of zero is {annO(M (F ) F is a finite subset of QM }. b ) b ). This linear topology is called the M -topology on O(M We will prove the following. THEOREM E.1.1 [36, T.G. Faticoni]. Let E be a Corner ring and let M be a Corner E-module such that annE (M ) = 0. There is an exact sequence 0 −−−−→ M −−−−→ G −−−−→ QC −−−−→ 0

(E.1)

b )-modules such that of left O(M 1. C is a direct sum of the cyclic E-submodules of M (ℵo ) . b )∼ b ) 2. There is a topological isomorphism O(M = EndZ (G) where O(M b is endowed with the Γ(M )-topology and EndZ (G) is endowed with the finite topology. Proof: d LEMMA E.1.2 QM is a left QE-module. d Then r is the limit of a net Proof: Let x ∈ QM and let r ∈ QE. {rF F } ⊂ QE where ranges rF ranges over the finite subsets of QM . Choose a finite set F ⊂ QM such that x ∈ F and rF 0 − rF ∈ annE (x) for each finite set x ∈ F 0 ⊂ QM . Then (rF 0 − rF )x = 0 implies that d rF x = rF 0 x. Define rx = rF x. This makes QM a left QE-module.

b ) is complete in the M -topology. LEMMA E.1.3 O(M d so b ) is also a Cauchy net in QE Proof: A Cauchy net {rF }F in O(M d such that rbx = rF x ∈ M for the that it converges to an element rb ∈ QE b ). This completes subnet {rF x ∈ F }. Then rbM ⊂ M so that rb ∈ O(M the proof.

E.2. THE CONSTRUCTION OF G

E.2

271

The Construction of G

c denote the Z-adic completion of the CorE.2.1 In what follows let M b⊂E b b be an uncountable domain whose ner O(M )-module M . Let P ⊂ Z elements are rational multiples of units in P. (See [23, Section 2].) Let Π ⊂ P be a countable integral domain that satisfies the following lemma. LEMMA E.2.2 Let {λi i ∈ N} = L ⊂ P be a Π-linearly independent set. Let {xi i ∈ N} be an ordered subset of QM . P 1. If i∈N λi xi = 0 then xi = 0 for each i ∈ N. P 2. annO(M ( i∈N λi xi ) = annO(M ({xi i ∈ N}). b b ) ) c such that LEMMA E.2.3 There is a Corner submodule N ⊂ M 1. M ∩ N = 0. 2. For each finite set F ⊂ M there is a uF ∈ N such that annE (uF ) = annE (F ). 3. N is a direct sum of cyclic submodules of M (ℵo ) . Proof: Because M is countable, Po (M ) is countable, and because P is uncountable, there is a countable set L = {1, λF x F ∈ Po (M ) and x ∈ F } ⊂ P that is Π-linearly independent. Given F ∈ Po (M ) let X uF = λF x x

(E.2)

x∈F

and let N=

X finite

EuF .

(E.3)

F ⊂QM

The value of N in (E.3) does not change in this appendix. P Because L is Π-linearly independent, Lemma E.2.2(1) implies that finite F ⊂QM EuF is a direct sum and that M ∩N = 0. Thus N satisfies Lemma E.2.3(1). Fix a finite subset F ⊂ QM , and form the element uF in (E.2). Because L is Π-linearly independent, Lemma E.2.2(2) implies that annE (uF ) = annE (F ), so that Lemma E.2.3(2) is satisfied.

272

APPENDIX E. CORNER RINGS AND MODULES

Finally, given a finite subset F ⊂ QM , Lemma E.2.3(2) implies that there is an injection EuF → M (F ) such that uF → ⊕x∈F x where x ∈ Mx ∼ = M . Thus N embeds in M (ℵo ) as required by Lemma E.2.3(3). Inasmuch as M ⊕ N is countable and P is uncountable there is a countable set of units A = {mn m ∈ M, n ∈ N } ⊂ P

(E.4)

that is algebraically independent over Π. Let G be the pure subgroup of c generated by M ⊕ N and mn for each m ⊕ n ∈ M ⊕ N . M G = hM, N, Emn m ∈ M, n ∈ N i∗ X c ∩ Q M ⊕ N, = M {Emn m ∈ M, n ∈ N }

(E.5)

c. Then G is a Corner E-submodule of M PROPOSITION E.2.4 Let mn ∈ A be as in (E.4). P 1. {E(m + n)mn m ∈ M, n ∈ N } is a direct sum of cyclic Esubmodules of M (ℵo ) . b )-module. 2. G is a left O(M Proof: 1. The independence of the sum follows from Lemma E.2.2(1). Because mn is a unit in P there is a natural embedding E(m ⊕ n)mn ∼ = E(m ⊕ n) ⊂ M ⊕ N ⊂ M ⊕ M (ℵo ) . This is part 1. 2. Observe that QG is a left QE-module, so by Lemma E.1.2, QG d b )-module. Because M is a is a left QE-module. Thus QG is a left O(M b )-module, G = M c ∩ QG is a left O(M b )-module. This completes left O(M part 2.

E.3

Endomorphisms of G

The proof of the main result of this section is similar to the traditional proof of Corner’s Theorem [46, Theorem 110.1]. Since [46, Theorem 110.1] is the only publication of the proof of Corner’s Theorem in the last 30 years we will give a complete proof of that Theorem here.

E.3. ENDOMORPHISMS OF G

273

LEMMA E.3.1 Let η ∈ EndZ (G). For each m ∈ M and n ∈ N there is an rmn ∈ E such that η(m) = rmn m and η(n) = rmn n. c, η lifts to a map Proof: Let η ∈ EndZ (G). Because G is pure in M c −→ M c ηb : M and to a Q-linear homomorphism η : QG −→ QG. Let m ∈ M and n ∈ N . By the definition of A in (E.4), there is a finite subset F ⊂ M ⊕ N such that η((m ⊕ n)mn ) = ηb(m ⊕ n)mn X = x+ rm0 n0 (m0 ⊕ n0 )m0 n0

(E.6) (E.7)

m0 ⊕n0 ∈F

X

η(m ⊕ n) = y +

sm0 n0 (m0 ⊕ n0 )m0 n0

(E.8)

m0 ⊕n0 ∈F

where x, y ∈ Q(M ⊕ N ) and rm0 n0 , sm0 n0 ∈ E. As in [46, Theorem 110.1], substitute (E.7) and (E.8) into (E.6). Then use Lemma E.2.2(1) to compare coefficients in equations (E.6) and (E.7) and show that for each m0 ⊕ n0 ∈ F with m ⊕ n 6= m0 ⊕ n0 we have x = rm0 n0 (m0 ⊕ n0 ) = 0, sm0 ⊕n0 (m0 ⊕ n0 ) = 0, y = rmn (m ⊕ n) = η(m ⊕ n).

(E.9)

Finally, because m ⊕ n, 0 ⊕ n ∈ M ⊕ N there are rm , rn ∈ E such that η(m) = rm m and η(n) = rn n. Then rmn (m ⊕ n) = η(m ⊕ n) = η(m) ⊕ η(n) = rm m ⊕ rn n. Using (E.9) and (E.10) we arrive at rm m = rmn m and rn n = rmn n. This proves the lemma.

(E.10)

274

APPENDIX E. CORNER RINGS AND MODULES

LEMMA E.3.2 There is an isomorphism of rings b )∼ O(M = EndZ (G). b )-submodule of M c. InasProof: By Proposition E.2.4, G is an O(M c = 0 implies that r = 0 for r ∈ O(M b ), much as rG = 0 implies that rM b there is an embedding of rings O(M ) −→ EndZ (G). We claim that this embedding is an isomorphism. Let η ∈ EndZ (G) and let F ⊂ M be a finite set. By Lemma E.2.3(2), there is a uF ∈ N such that annE (uF ) = annE (F ). Lemma E.3.1 states that for each m ∈ F there is an rF m ∈ QE such that η(m) = rF m m and η(uF ) = rF m uF for each m ∈ F . Let rF be any one of the rF x with x ∈ F . Then rF m uF = η(uF ) = rF uF for each m ∈ F , so that rF − rF m ∈ annE (uF ) = annE (F ) ⊂ annE (m). Hence rF m = rF m m = η(m) for each m ∈ F . d under Subsequently, the set {rF F ∈ Po (M )} is a Cauchy net in QR d of the Cauchy net. By the QM -topology. Let rb denote the limit in QR (E.1.2), rbm = rF m = η(m) for each finite set F ⊂ QM and each m ∈ F . Thus [η − rb](M ) = 0. c, η = rb. Since M is pure and dense in M b Therefore the embedding O(M ) −→ EndZ (G) is an isomorphism. b )∼ LEMMA E.3.3 The isomorphism of rings O(M = EndZ (G) is a topob logical one, taking the M -topology on O(M ) onto the finite topology of EndZ (G).

E.3. ENDOMORPHISMS OF G

275

Proof: The proof, being natural, is left for the reader. Proof of Theorem E.1.1: Given the Corner Module M construct N and G as in (E.3) and (E.5). Let X C = N⊕ E(m ⊕ n)mn . m∈M,n∈N

c, Note that because M and G are pure and dense in M G/M ∼ = QC b )-modules. By Proposition E.2.4(1), C is a direct sum of as left O(M cyclic submodules of M (ℵo ) . Thus Theorem E.1.1(1) is satisfied. Furb )∼ thermore by Lemmas E.3.2 and E.3.3, O(M = EndZ (G) topologically, where EndZ (G) is endowed with the finite topology. This completes the proof of Theorem E.1.1.

Appendix F

Corner’s Theorem F.1

Countable Endomorphism Rings

THEOREM F.1.1 [23, A.L.S. Corner]. Each countable reduced torsion-free ring E is the endomorphism ring of a countable reduced torsionfree abelian group G. Furthermore, there is a short exact sequence (E.1) of left E-modules in which C is a direct sum of cyclic submodules of M (ℵ0 ) . Proof: Corner’s Theorem follows from Theorem E.1.1.

Let

O(M ) = {q ∈ QE qM ⊂ M }.

THEOREM F.1.2 [43, T.G. Faticoni, H.P. Goeters]. Let E be a reduced torsion-free finite rank ring and let M be a reduced torsion-free finite rank left E-module. Then O(M ) ∼ = EndZ (G) for some reduced torsion-free finite rank group G. Furthermore, there is a short exact sequence (E.1) of left O(M )-modules in which C ∼ = O(M ) ⊕ O(M ). Proof: When constructing G use a maximal linearly independent subset X of M instead of using each element m ∈ M . 277

278

APPENDIX F. CORNER’S THEOREM

THEOREM F.1.3 [23, A.L.S. Corner]. If E is a reduced torsion-free finite rank ring then E = EndZ (G) for some reduced torsion-free finite rank group G. Furthermore, there is a short exact sequence (E.1) of left E-modules in which C is a cyclic free left E-module.

Appendix G

Torsion Torsion-Free Groups G.1

E-Torsion Groups

Let E be an integral domain. A left E-module M is torsion if for each x ∈ M there is an r 6= 0 ∈ E such that xr = 0. We construct torsion-free groups G such that EndZ (G) is an integral domain, and such that G is a torsion left EndZ (G)-module. EXAMPLE G.1.1 There is a countable torsion-free group G such that EndZ (G) = Z[[x]] and such that G is a countable torsion left EndZ (G)module. Proof: Let E = Z[x] and let M = Z[x]/(x) ⊕ Z[x]/(x2 ) ⊕ Z[x]/(x3 ) ⊕ · · · . d = Q[[x]], and O(M b )= Then Γ(QM ) is the x-adic topology on QE, QE Z[[x]]. Theorem E.1.1 states that there is a short exact sequence 0 −→ M −→ G −→ QC −→ 0 of left Z[[x]]-modules and such that 1. EndZ (G) = Z[[x]], and 2. C is a direct sum of submodules of M (ℵo ) . Then M , G, and QC are torsion left Z[[x]]-modules. Thus G is a torsion left EndZ (G)-module. This proves the Example.

279

280

APPENDIX G. TORSION TORSION-FREE GROUPS

G.2

Self-Small Corner Modules

The groups in the above constructions are torsion modules over their endomorphism rings. The next construction produces E-torsion-free groups. The group G is self-small if for each cardinal c the natural mapping Hom(G, G)(c) −→ Hom(G, G(c) ) is an isomorphism of right EndZ (G)-modules. We will prove THEOREM G.2.1 Let E be a Corner ring and let M be a Corner E-module. There is a short exact sequence (E.1) in which 1. C is a free left O(M )-module, 2. O(M ) ∼ = EndZ (G), and 3. G is a self-small group. We require a lemma. LEMMA G.2.2 As in (E.2.1) choose a countable integral domain Π ⊂ b and an uncountable integral domain Π ⊂ P ⊂ Z. b Let M be a Corner Z c such that module such that annE (M ) = 0. There is an element u ∈ M annE (u) = 0. Proof: Write M = {m1 , m2 , m3 , · · ·} and since P is uncountable there is a countable set L = {λ1 , λ2 , λ3 , · · ·} ⊂ P that is algebraically independent over the integral domain Π. (See [23, Section 2].) There is a convergent sum u=

∞ X

mk pk

k=1

c. in the p-adic topology on M If r ∈ E and ru = 0 then ∞ X k=1

rmk pk = 0.

G.2. SELF-SMALL CORNER MODULES

281

By Lemma E.2.2(2), rmk pk = 0 = rmk for each integer k > 0. Hence c is the element that rM = 0 = r by our hypothesis on M . Thus u ∈ M we seek. Proof of Theorem G.2.1: By Lemma G.2.2 there is an element c such that annE (u) = 0. By taking λu for some λ ∈ P if necessary u∈M we can assume that M ∩ Eu = 0. c. Because M ⊕ Eu is countable there is an algeThen M ⊕ Eu ⊂ M braically independent set A = {m m ∈ M } ⊂ P over Π. Let C = Eu +

X

E(m ⊕ u)m

m∈M

and let c. G = Q(M ⊕ C) ∩ M

(G.1)

c. Thus there is a short exact Evidently G is a left E-submodule of M sequence 0 −−−−→ M −−−−→ G −−−−→ QC −−−−→ 0 of left E-modules. By our choice of u E∼ = Eu ∼ = E(m ⊕ u)m for each m ∈ M . Because A ∪ {1} is algebraically independent over Π, C is a free module. (See Lemma E.2.3(2).) Thus Theorem G.2.1(1) is satisfied. Because annO(M ) (G) = annO(M ) (M ) = 0 there is an embedding O(M ) −→ EndZ (G) of rings. We claim that this is an isomorphism. Using Lemma 2.3.3, for each m ∈ M there is an rm ∈ QE such that η(m) = rm m and η(u) = rn u. Let ro be one of the rm . Then for each m ∈ M , ro u = rm u. Since annE (u) = 0, ro = rm for all m ∈ M . Hence η(m) = ro m for all c) is contained in the m ∈ M . Thus [η − ro ](M ) = 0. Then [η − ro ](M c divisible subgroup of M , which in this case is 0. That is, η = ro and the proof of Theorem G.2.1(2) is complete. Lastly, G is a countable group in which the finite topology is discrete (u ∈ G.) Then by [15, Proposition 2.1], G is self-small. This proves Theorem G.2.1(3) and completes the proof.

Appendix H

E-Flat Groups Recall that G is faithful if IG 6= G for each proper right ideal I ⊂ EndZ (G), and that G is called fully faithful if K ⊗EndZ (G) G 6= 0 for each nonzero right EndZ (G)-module K. G is E-flat if G is a flat left E-module. G is faithfully E-flat if G is a faithful group and a flat left E-module.

H.1

Ubiquity

THEOREM H.1.1 The group constructed according to Corner’s Theorem F.1.1 is E-flat. Proof: Let E be a Corner ring. By Corner’s Theorem F.1.1 there is an exact sequence 0 −→ E −→ G −→ QC −→ 0 of left E-modules in which 1. E ∼ = EndZ (G), and 2. C is a free left E-module. Then there is an exact sequence of Tor groups Tor1E (·, E) −→ Tor1E (·, G) −→ Tor1E (·, QC). Since E is free, Tor1E (·, E) = 0, and since C is free, QC is flat. Thus Tor1E (·, QC) = 0. It follows that Tor1E (·, G) = 0, whence G is a flat left E-module. 283

284

APPENDIX H. E-FLAT GROUPS

THEOREM H.1.2 Let E be an rtffr ring, and let G be a group constructed according to Corner’s Theorem F.1.3 such that E = EndZ (G). Then G is faithfully E-flat. Proof: By the previous theorem G is a flat left E-module. To see that G is fully faithful (i.e., to see that K ⊗E G 6= 0 for nonzero right E-modules K), it suffices to show that IG 6= G for each maximal right ideal I ⊂ E. There is an exact sequence Tor1E (E/I, QC) −→ E/I ⊗E E −→ E/I ⊗E G −→ E/I ⊗E QC of groups. Since E is rtffr, E/I is finite (Lemma 4.2.3). Thus E/I ⊗E QC = 0. Furthermore since QC is a flat left E-module Tor1E (E/I, QC) = 0. Hence 0 6= E/I ∼ = G/IG whence G is faithful. This completes the proof.

H.2

Unfaithful Groups

THEOREM H.2.1 There is a countable torsion-free group G with endomorphism ring E such that 1. E is an integral domain, 2. IG 6= G for each nonzero ideal I ⊂ E, 3. There is a nonzero E-module K such that K ⊗E G = 0. That is, there is a countable torsion-free group G that is faithful but not fully faithful. Proof: 1. Let E = Z[x](x,2) be the localization of Z[x] at the ideal generated by x and 2. Let M=

∞ M

E/xk E.

k=1

Then the reader shows that O(M ) = E. By Theorem E.1.1 there is a short exact sequence 0 −→ M −→ G −→ QC −→ 0

285 of left E-modules such that C is a direct sum of submodules of M (ℵo ) . Then G is a torsion E-module. Choose the unique maximal right ideal I. Then E/I ∼ = Z/2Z so that E/I ⊗E QC = 0 = Tor1E (E/I, QC) and so 0 6= M/IM ∼ = G/IG. This proves part 2. For part 3 let c0 , c1 , c2 , · · · be elements such that ck 2 = 0 for all k ≥ 0 c0 x = 0 and ck+1 x = ck for all k ≥ 0. Let K=

∞ X

ck E.

k=0

Because QC is divisible and K is an elementary 2-group K ⊗E QC = 0. Because M is x-torsion while K is x-divisible, K ⊗E M = 0. Then K ⊗E G = 0. This proves part 3 and completes the proof. THEOREM H.2.2 There is a faithful group that is not E-flat. THEOREM H.2.3 There is an E-flat group that is not faithful. Z 0 and let M be the projective left E-module Proof: Let E = Z Z Z 0 Z . Evidently O(M ) = = E. M= Z Z Z A little moreeffort (using matrix units) shows that the trace ideal of Z 0 M in E is I = . Then IM = M so that (I + nE)M = M for Z 0 each integer n. By Theorem E.1.1 there is a short exact sequence 0 −−−−→ M −−−−→ G −−−−→ QC −−−−→ 0 in which C is a finitely generated free left E-module. Since M is projective as a left E-module, G is flat as a left E-module. Then E/(I + nE) ⊗E M = E/(I + nE) ⊗E QC = 0 for each integer n > 0. An application of E/(I + nE) ⊗E · to the short exact sequence shows us that (I + nE)G = G. Thus G is E-flat but not faithful.

286

APPENDIX H. E-FLAT GROUPS

THEOREM H.2.4 There is a countable torsion-free group that is not faithful and not E-flat. Proof: Let

and let

Z 0 0 E= Z Z 0 Z Z Z

Z 0 Z 0 0 0 M = Z Z 0 0 = Z Z . Z Z 0 Z Z 0

Since

Q 0 QM = Q Q Q 0

is not a projective left QE-module, M is not flat as a left E-module. This is because the indecomposable projective QE-modules are the columns of Q 0 0 QE = Q Q 0 . Q Q Q 0 0 0 0 0 0 does not contain a nonzero ideal of E, annE (M ) = Since 0 Z Z Z 0 0 0. Observe that I = Z Z 0 is a right ideal in E such that Z Z 0

Z 0 0 Z 0 Z 0 IM = Z Z 0 Z Z = Z Z = M. Z Z 0 Z 0 Z 0 Then JM = M for each proper right ideal Z 0 0 J = Z Z 0 . Z Z nZ We note that I ⊂ J ⊂ E and that E/J is finite. Now use Theorem F.1.2 to construct a short exact sequence 0 −−−−→ M −−−−→ G −−−−→ QC −−−−→ 0

287 of left E-modules such that C is a free E-module and E ∼ = EndZ (G). The by now familiar arguments show us that since JM = M , JG = G, and since M is not flat, G is not E-flat.

Appendix I

Zassenhaus and Butler I.1

Statement

A is a finite dimensional Q-algebra and V = A is a cyclic free left A-module with free basis . Identify A via left multiplication with the subring AL of EndQ (V ). Let Q = the set of primes p ∈ Z. For primes p ∈ Z and groups X we let Xp = Zp X ⊂ QX. Let O(X) = OA (X) = {a ∈ AL aX ⊂ X}. Let ∈ M ⊂ V be a full free abelian subgroup. Then QM = V and Mp is a free Zp -module for each p ∈ Q. Because M is a free abelian group we have the local-global relationships \ M= Mp , p∈Q

O(M )p = O(Mp ) for each p ∈ Q, and \ O(M ) = O(Mp ). p∈Q

We will prove the following theorem. 289

290

APPENDIX I. ZASSENHAUS AND BUTLER

THEOREM I.1.1 [T.G. Faticoni] Let A, V , and M be as above. There is a subgroup M ⊂ G ⊂ V such that 1. G is a locally free group, 2. O(M ) ⊂ EndZ (G) ⊂ A, 3. EndZ (G)p /O(M )p is a finite p-group for each prime p ∈ Z.

I.2

Proof

T We construct G as the intersection G = p Kp where the Kp are O(Mp ) = O(M )p -submodules of V . The Kp are chosen in a series of lemmas. Since V has finite Q-dimension we can enumerate the elements of [ (I.1) EndQ (V ) \ AL = {φp p ∈ Q}. LEMMA I.2.1 Let p ∈ Q, let φ = φp , and suppose that φ(Mp ) 6⊂ Mp . Then choose Kp = Mp .

(I.2)

In particular, Kp contains Mp , Kp is a left Ep -module, and Kp is a finitely generated free Zp -module. LEMMA I.2.2 Let 0 6= φ = φp for some p ∈ Q. Suppose that φ satisfies 1. φ(Mp ) ⊂ Mp , and 2. φ() = 0. There is a cyclic Ep -submodule Np ⊂ V such that φ(Np ) 6⊂ Np . Proof: Let p ∈ Q and let 0 6= φ = φp satisfy conditions 1 and 2 above. Let Ep = O(Mp ) and let Rp = Ep [φ] denote the subring of EndQ (V ) generated by Eq and φ. Assume, for the sake of contradiction that each cyclic Ep -submodule Np ⊂ V satisfies φ(Np ) ⊂ Np .

(I.3)

291 The contradiction that we are looking for is φ = 0. Since each Ep -submodule of V is a sum of cyclic Ep -submodules of V , condition (I.3) implies that each Ep -submodule of V is an Rp -module.

(I.4)

Arbitrarily choose x ∈ V . Since is a free basis for V , Ep is a cyclic free left Ep -module with basis . There is then a short exact sequence ρ

0 −→ L −→ Ep −→ Ep x −→ 0 of left Ep -modules in which ρ(s) = sx for each s ∈ Ep . Tensoring with the right Ep -module Rp yields the commutative diagram with exact rows Rp ⊗ L νL 0

? - L

- Rp ⊗ Ep 1 ⊗ ρ- Rp ⊗ Ep x

ν ? - Ep

-0

νx ρ

? - Ep x

-0

where the maps νY are scalar multiplication maps νY (r ⊗ y) = ry. By (I.4), L, Ep , and Ep x are left Rp -modules, so νL , ν , and νx are well defined Rp -module maps. Because ν , 1 ⊗ ρ, and νx are surjections, the commutativity of the diagram shows us that for each r ∈ Rp and y = s ∈ Ep ρ(ry) = ρ(ν (r ⊗ y)) = ρ(ν (r ⊗ s)) = νx (1 ⊗ ρ)(r ⊗ s) = νx (r ⊗ sx) = r(sx) = rρ(y) Hence ρ is an Rp -module map. Recall the hypothesis that φ() = 0. Then φ(x) = φ(ρ()) = ρ(φ()) = 0

292

APPENDIX I. ZASSENHAUS AND BUTLER

so that φ(x) = 0 for any x ∈ V . Thus φ(V ) = 0 which implies that φ = 0. This contradicts our choice of φ 6= 0 so condition (I.3) is false, that is, φNp 6⊂ Np some cyclic Ep -submodule of V . This proves the lemma. COROLLARY I.2.3 Let 0 6= φ = φp for some p ∈ Q, suppose that φ(Mp ) ⊂ Mp and that φ() = 0. There is an O(Mp )-submodule Mp ⊂ Kp ⊂ V such that Kp is a finitely generated free Zp -module, and φ(Kp ) 6⊂ Kp . Proof: Let O(Mp ) = Ep . By Lemma I.2.2, given φ such that φ(Mp ) ⊂ Mp and φ() = 0 there is a cyclic Ep -submodule Np ⊂ V such that φ(Np ) 6⊂ Np . Then Np is a torsion-free quotient of the finitely generated free Zp -module Ep , hence Np is a finitely generated free Zp -module, whence pk Mp + Np is a finitely generated free Zp -module for each integer k > 0. Let x ∈ Np be such that φ(x) 6∈ Np . Since the q-adic topology on the free Zp -module Mp + Np is discrete there is an integer k = kp > 0 such that φ(x) 6∈ pk Mp + Np . Choose Kp = Mp + p−k Np

(I.5)

and observe that Mp ⊂ Kp , that Kp is a finitely generated free Zp module, and that φ(Kp ) 6⊂ Kp . There is one case left in our choices of Kp . LEMMA I.2.4 Let 0 6= φ = φp for some p ∈ Q, suppose that φ(Mp ) ⊂ Mp , and suppose that φ() 6= 0. There is a finitely generated O(Mp )module Kp such that Mp ⊂ Kp ⊂ V and φ(Kp ) 6⊂ Kp . Proof: Let O(Mp ) = Ep . Since φ(Mp ) ⊂ Mp and since φ 6∈ AL , 0 6= mφ − r 6∈ Ep = O(Mp )

(I.6)

for any integer m 6= 0 and element r ∈ Ep . Because Mp and Ep are full subgroups of V we can write 0 6= φ(m) = r for some integer m 6= 0 and some r ∈ Ep . Because φ(Mp ) ⊂ Mp and r ∈ Ep we have (mφ − r)(Mp ) ⊂ Mp and (mφ − r) = 0. By (I.6), mφ − r 6= 0 so we can apply Corollary I.2.3 to show that there is a left Ep -submodule Mp ⊂ Kp ⊂ V such that Kp is a finitely generated

293 free Zp -module, and such that (mφ − r)(Kp ) 6⊂ Kp . Since r(Kp ) ⊂ Kp , mφ(Kp ) 6⊂ Kp , whence φ(Kp ) 6⊂ Kp . This proves the lemma. For φ = φp 6∈ AL such that φ(Mp ) ⊂ Mp , choose Mp ⊂ Kp ⊂ V

(I.7)

as in Lemma I.2.4. This takes into account all of the cases for φ ∈ EndQ (V ) \ AL . Proof of Theorem I.1.1: Let E = O(M ), let Kp denote the left Ep -modules chosen in (I.2), (I.5), and (I.7) and let \ G= Kp . (I.8) p∈Q

Each Kp is an Ep -module, each Kp is a finitely generated free Zp -module, and ∈ G is a unimodular element. Since localization commutes with . intersections, Gp = Kp = Mp for all primes p ∈ Z. In each of our choices, Kp is a finitely generated free Zp -module. Hence G is a locally free full subgroup of V T , as required by Theorem I.1.1(1). Since E = p∈Q Ep , G is a left E-module such that ψ() 6= 0 for each 0 6= ψ ∈ E ⊂ AL . Hence E ⊂ EndZ (G) ⊂ EndQ (V ). By our choices of φp and Kp , for each φ ∈ EndQ (V ) \ AL there is a p ∈ Q such that φ = φp and φ(Kp ) 6⊂ Kp . However for ψ ∈ EndZ (G), ψ(Kp ) = ψ(Gp ) ⊂ Gp = Kp for each p ∈ Q. Thus EndZ (G)

\

(EndQ (V ) \ AL ) = ∅

and hence EndZ (G) ⊂ AL . This proves condition Theorem I.1.1(2). Since Gp = Kp for all primes p ∈ Z it follows that EndZ (G)p ⊂ O(Kp ). By our construction of Kp in (I.2), (I.5), and (I.7) we see that Kp /Mp is a finite p-group so that O(Kp )/O(Mp )

294

APPENDIX I. ZASSENHAUS AND BUTLER

is a finite p-group. Thus EndZ (G)p /O(Mp ) is a finite p-group for each p ∈ Q. This is what is required in Theorem I.1.1(3). This completes the proof of the theorem. THEOREM I.2.5 [90, H. Zassenhaus]. Let E be a ring such that (E, +) is a finitely generated free ring. There is a group E ⊂ G ⊂ QE such that E = EndZ (G) where the action of E on G is the natural one as an E-submodule of QE. THEOREM I.2.6 [21, M.C.R. Butler]. Let E be a locally free rtffr ring. There is a group E ⊂ G ⊂ QE such that E = EndZ (G) where the action of E on G is the natural one as an E-submodule of QE.

Appendix J

Countable E-Rings J.1

Countable Torsion-Free E-Rings

An E-ring is a ring E for which the left representation embedding λ : E −→ EndZ (E) is an isomorphism of rings. A Dedekind E-ring is an E-ring R that is a Dedekind domain. THEOREM J.1.1 An E-ring is a commutative ring. R. Bowshell and P. Schultz [20] characterize the rtffr E-rings. THEOREM J.1.2 [20, R. Bowshell, P. Schultz]. The rtffr ring E is an E-ring iff . E = R1 ⊕ · · · ⊕ Rt for some integer t > 0 and some Dedekind E-rings R1 , . . . , Rt such that HomZ (Ri , Rj ) = 0 for each 1 ≤ i 6= j ≤ t. This structure theorem seemed to indicate that the structure of Erings was restricted as far as rtffr groups go. The following theorem shows that the countable torsion-free E-rings have a much more diverse additive structure. The group G is p-local if pG 6= G. THEOREM J.1.3 [42, T.G. Faticoni]. Let S be a countable reduced torsion-free commutative ring. There is a E-ring E and a pure and dense embedding of rings S −→ E.

295

296

APPENDIX J. COUNTABLE E-RINGS

Before proving this theorem an example or two along with a comparison to Theorem J.1.2 will illustrate the difference between rtffr E-rings and countable torsion-free E-rings. EXAMPLE J.1.4 Let S = Z[x1 , x2 , x3 , . . .]/I where x1 , x2 , x3 , . . . are commuting indeterminants, and where I is an ideal in S. By Theorem J.1.3, S is a pure and dense subring of an E-ring E. We point out that every commutative ring can be realized in this way. Specifically Z[x]/(xn ) embeds in an E-ring E. This E-ring contains a nonzero nilpotent element x. Rtffr E-rings do not contain nonzero nilpotent elements. Proof of Theorem J.1.3: As in the proof of Corner’s Theorem 2.3.3 we require an uncountable integral domain. The proof is left as an exercise for the reader. LEMMA J.1.5 Let K be a uncountable field extension of Q, and let B be a commutative K-algebra. Let S be a countable Q-subalgebra of B. There exists an uncountable set U ⊂ K that is algebraically independent over S. For groups G and primes p ∈ Z, let us define Gp =

G ⊗ Zp . div(G ⊗ Zp )

Then Gp is a reduced p-local group, and Gp is countable if G is countable. To reduce to the p-local case, observe that if S is a countable reduced torsion-free commutative ring then S is a pure and dense subring of Q S where p ranges over the primes of Z. Thus for each prime p ∈ Z p p we will embed each Sp as a pure and dense subring of an E-ring Q Ep . We can then embed S as a pure and dense subring of the E-ring p Ep . Let W ∪ {1} bp -module with basis W ∪ {1}. Since be a p-basis of S. Then Sb is a free Z bp of S is countable there is an algebraically independent subset U ⊂ Z units such that

297

card(U ) = card(W ) + 1. Write U = {d, uw w ∈ W }. We will show that E = hS[uw + dw w ∈ W ]i∗ b Certainly is the E-ring we seek, the purification taking place in S. S ⊂ E ⊂ Sb so that S is a pure and dense subring of E. It remains to prove that E is an E-ring. LEMMA J.1.6 Let S, W, uw , d, E be as above. Then 1. E ∩ dE = 0. 2. Let φ : E → Sb be a group homomorphism such that dφ(S)∩φ(S) = 0. Then f φ = 0 if f φ(1) = 0. Proof: 1. It suffices to show that {d, uw +dw w ∈ W } is algebraically independent over QS. Let X = {xd , xw w ∈ W } be a set of indeterminants and consider the evaluation maps α : QS[X] −→ QS[U ] and β : QS[X] −→ QS[U ] such that α(xd ) = β(xd ) = d α(xw ) = uw + dw β(xw ) = uw . Since U is algebraically independent and since W ⊂ S, β is a ring isomorphism. Let P (X) be a polynomial of minimial xd -degree such that α(P (X)) = 0. We can write

298

APPENDIX J. COUNTABLE E-RINGS

P (X) = xd Q(X) + R(X) where the xd -degree of R(X) is 0. Then α(P (X)) = 0 implies that α(R(X)) = −dα(Q(X)). Arbitrarily enumerate the terms of R(X) and let Rk (X) be the k-th term in R(X). For each integer k ≥ 1 and w ∈ W let e(w, k) = the xw -degree of Rk (X). Then ! Y

α(Rk (X)) = ak

e(w,k)

(uw + dw)

! = ak

Y

ue(w,k) w

+ dBk

w∈W

w∈W

for some ak ∈ QS and some Bk ∈ QS[U ]. Then Y ue(w,k) = β(Rk (X)) w w∈W

so that α(Rk (X)) = β(Rk (X)) + dBk . By setting B =

P

k

Bk then we have

α(R(X)) = β(R(X)) + bB = −dα(Q(X)). It follows that R(X) = xd β −1 [−B − α(Q(X))]. Since R(X) has xd -degree 0 it must be that R(X) = 0. That is P (X) = xd Q(X). bp and α is Z bp -linear so Now d is a unit in Z α(Q(X)) = d−1 α(xd Q(X)) = d−1 α(P (X)) = 0. This contradiction to the minimality of the xd -degree of P (X) shows that α : QS[X] −→ QS[U ] is an isomorphism. In particular {d, uw +dw dw ∈ W } is algebraically independent over QS. This proves part 1.

299 2. Assume that φ : E −→ Sb is a function such that φ(1) = 0 and bp -module map φ : E b −→ S, b φ(E) ∩ dφ(E) = 0. Since φ lifts to a Z bp . φ(u) = uφ(1) for each u ∈ Z Thus, for each w ∈ W , φ(uw + dw) = uw φ(1) + dφ(w) = dφ(w) ∈ φ(E) ∩ dφ(E) = 0 by part 1. Therefore 0 = φ(uw + dw) = dφ(w) = φ(w) for all w ∈ W . b=Z bp · W and since φ(1) = 0 by hypothesis, φ = 0. This Inasmuch as E proves part 2 and completes the proof of the lemma. Proof of Theorem J.1.3: Let S be a countable reduced torsion-free commutative ring. We have already observed that S ⊂ E is a pure and dense embedding of rings. Let φ : E −→ E be a group homomorphism, let u = φ(1), and consider φ − u. By Lemma J.1.6(1), E ∩ dE = 0, and by Lemma J.1.6(2), φ − u = 0. I.e. φ = u ∈ E, hence E is an E-ring. This concludes the proof of Theorem J.1.3.

Appendix K

Dedekind E-Rings Our references in this section for commutative ring theory are [6, 49]. The number theory we use can be found in [58, 66]. The ring R is a Dedekind E-ring if the group (R, +) is an E-ring and if the ring R is a Dedekind domain. We will show that the Dedekind domain R is an E-ring provided it is divisible by a certain set of primes in the ring of algebraic integers in algebraic number field QR.

K.1

Number Theoretic Preliminaries

At all times p ∈ Z is a prime number, E and F are algebraic number fields, OE denotes the ring of algebraic integers in E, and L(E) denotes the lattice of subrings of E containing OE . Let spec(E) denote the set of maximal ideals in a ring OE . For more general commutative rings R, spec(R) denotes the set of maximal ideals in R. Let E ⊂ F be a subfield, let P ∈ spec(E) and M ∈ spec(F ). We say that M lies over P if P ⊂ M . Equivalently, M lies over P if P ⊂ M ∩E. For unramified primes P ∈ E it is known that X [F : E] = {[OF /M : OE /P ] M ∈ spec(F ) and M lies over P .} If there are at least two different primes in OF lying over P then we say that P splits in F or in OF . If [F : E] primes in OF lie over P then we say that P splits completely in F . Thus P splits completely

301

302

APPENDIX K. DEDEKIND E-RINGS

in F iff OF /M ∼ = OE /P for each M ∈ spec(F ) lying over P . Complete splitting occurs often enough as [58, Theorem 6] states that infinitely many primes P ∈ spec(E) split completely in F .

K.2

Integrally Closed Rings

Let OF ⊂ R ⊂ F be a Dedekind domain. The support of R in F is the set σF (R) = {M ∈ spec(F ) M R 6= R} and the divisibility of R in F is

δF (R) = {M ∈ spec(F ) M R = R}. For a Dedekind domain OF ⊂ R ⊂ F , the sets σF (R) and δF (R) form a partition of spec(F ). Given a set σ ⊂ spec(F ) then let

Oσ = ∩{(OF )M M ∈ σ} where (OF )M is the classic localization of OF at the maximal ideal M in OF . K.2.1 Let σ ⊂ spec(F ). Then σF (Oσ ) = σ and δ(Oσ ) = σ 0 = the complement of σ in spec(F ). Moreover, if C(R) = {c ∈ OF cR = R} then by [49, page 73, exercise 7] R = OF [C −1 ]. Furthermore σF (R) = {M ∈ spec(F ) M ∩ C = ∅} and δF (R) = {M ∈ spec(F ) c ∈ M for some c ∈ C}.

303 Thus to study E-rings we need to study localizations Oσ for subsets σ ⊂ spec(F ). A classic result from commutative ring theory follows. Let S ⊂ R be rings. Then R is integral over S if R is finitely generated by S. LEMMA K.2.2 Let S ⊂ R ⊂ F be subrings of F . Then R is integral over S iff R = SOF . Proof: If R = SOF then R is finitely generated by S because OF is a finitely generated free abelian group. Hence R is integral over S. Conversely if R is integral over S then R is a finitely generated Smodule. Since R is integrally closed SOF ⊂ R and since R is finitely generated over the integrally closed domain SOF , SOF = R. THEOREM K.2.3 [10, Theorem 14.3] Suppose that OF ⊂ R ⊂ F and that R is a Dedekind domain. The following are equivalent. 1. (R, +) is strongly indecomposable. 2. F is the smallest subfield E ⊂ F such that R is integral over R ∩E. 3. F is the smallest subfield E ⊂ F such that R is finitely generated by R ∩ E. 4. R is an E-ring. Proof: The proof follows from the Lemma and from [10, Theorem 14.3].

THEOREM K.2.4 [39, Proposition 2.1] Let F be an algebraic number field, and let σ ⊂ spec(F ). The following are equivalent. 1. Oσ is an E-ring. 2. For each proper subfield E ⊂ F there are M, M 0 ∈ spec(F ) such that M ∩ E = M 0 ∩ E while M ∈ σ and M 0 6∈ σ. Proof: Assume that R = Oσ is not an E-ring. By Theorem K.2.3, there is a proper subfield E ⊂ F such that R = SOE where S = R ∩ E. Since S is a Dedekind domain with field of quotients E, S = OE [C −1 ] for some multiplicatively closed subset C ⊂ OE . Then OF [C −1 ] = R.

304

APPENDIX K. DEDEKIND E-RINGS

Fix M 0 ∈ spec(F ) such that M 0 R = R. There is a c ∈ C such that 1 · c = c ∈ M 0 . Thus M R = R for each maximal ideal M lying over M 0 ∩ S. This is the negation of part 2. Conversely, suppose that R = Oσ is an E-ring, let E ⊂ F be a proper subfield of F , and let S = R ∩ E. Since R is an E-ring, R is not integral over S. Since the integral property is a local-global property in F there is a prime P ∈ spec(E) such that RP is not integral over the discrete valuation domain SP . If RP = F then PP = PP R P ∩ S P = F ∩ S P = S P , contrary to the fact that the ring SP is never equal to its Jacobson radical PP . Thus RP 6= F . S and R are Dedekind domains and R is not finitely generated by S, so SP = (OE )P ⊂ SP OF = (OF )P 6= RP . Since RP 6= F there is an ideal M ∈ spec(F ) lying over P such that (MP )M = MM 6= RM = (RP )M 6= F. It follows that M RP 6= RP . RP is not integral over (OF )P , so there is an ideal M 0 ∈ spec(F ) lying over P such that (RP )M 0 = RM 0 is not finitely generated by the discrete valuation domain ((OF )P )M 0 = (OF )M 0 . Then RM 0 = F , and hence M 0 RP = RP . Localizing at primes Q ∈ spec(E) other than P yields M 0 RQ = RQ because there is an element c ∈ M 0 ∩ E \ Q. By the Local-Global Theorem, M 0 R = R. Since M and M 0 both lie over P we have proved part 2. THEOREM K.2.5 Let p ∈ Z be prime and let F be a minimal field extension of Q. Then F is the field of fractions of a p-local E-ring iff p splits in F . Examples of minimal field extensions are found by examining the finite lattice of subfields of the algebraic number field F . Thus quadratic number fields and in general fields F such that [F : Q] = p a prime are minimal field extensions. Another method is to take a Galois extension K/Q and then take a maximal subgroup G ⊂ Gal(K/Q). The fixed field F of G is minimal by the Galois correspondence. The degree [F : Q] is the degree [Gal(K/Q) : G].

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Index A A(·) 31, 59 acd group 24, 128, 143 Albrecht, U. 146, 150, 151 Albrecht, U. and Goeters, H.P. 144 almost completely decomposable group 24 Anderson, F.W. and Fuller, K.R. 1 Angad-Gaur 243 annhilator 5 Arnold, D.M. 1, 81, 159, 162, 164, 166, 168 Arnold’s Theorem 62 Arnold, D.M. and Hausen, J. 152 Arnold, D.M. , Hunter, R., and Richman, F. 71, 132 Arnold, D.M. and Lady, E.L. 74, 151, 201 Arnold-Lady-Murley Theorem 30 Artinian commutative ring 79, 81 associative ring 1 Azumaya-Krull-Schmidt Theorem 22 B Baer splitting property 139, 140, 144, 152, 177 Baer-Kulikov-Kaplansky Theorem 24, 26, 99 Baer’s Lemma 139, 146, 148, 151 Beaumont, R.A. and Pierce, R.S. 24, 50, 79, 221

Beaumont-Pierce-Wedderburn Theorem 50, 79, 219, 226, 239 bounded filter 186, 190 bounded ring 203 Bowshell, R. and Schultz, P. 223 bracket group 131 Butler’s Construction 91 Butler, M.C.R. 27, 289 C center 5, 48 central idempotents 48, 71 Change of Rings 13 Charles, B. 71 Chinese Remainder Theorem 13 classical maximal order 15, 38, 47, 51 class group of E 109 class number of G 108 class number of E 109 cokernel group 131 commutative ring 79, 81, 86 complete set of idempotents 71 completely decomposable 24, 243 completely faithful 152 composition factors 180 conductor of E 51 conductor of G 60 Corner, A.L.S. 25, 277 Corner E-module 27 Corner group 27 Corner module 27 Corner ring 27

311

312 Corner’s Theorem 27, 277 CycE (T ) 180 D D(QG) 204 D(G) 177 c) 190 DEb (M DE (M )183 δ(M ) 185 δE (G) 197 c) 190 δ(M δ(QG) 208 ∆Gb 196 ∆QG 204, 208 ∆G,H 206, 208 Dedekind group 127, 130 Dedekind E-ring 222, 226, 229, 240, 295 Dedekind ring 15 direct product 2 direct sum 2 discrete valuation domain 15 divisibility of R in F 302 dual module of 147 Dugas, M., Mader, A., and Vinsonhaler, C. 222 E Eτ (G) 61 E-properties 91 E-cyclic group 224, 229, 232 E-finitely generated 91, 95, 224, 226, 229, 241 E-finitely presented 91, 95, 229 E-flat 89 E-generator 92, 95, 226, 229, 234, 241 E-injective 250 E-lattice 15, 35, 38 E-modules 231 E-Noetherian 239, 241

INDEX E-progenerator 228, 229, 241 E-projective 92, 95, 228, 229, 232, 241 E-projective generator 92, 95, 229, 234 E-quasi-projective 233 E-ring 87, 91, 95, 295 E-self-generating 235 Eichler group 173 endlich Baer splitting property 144, 148, 151, 152 endomorphism ring 3 Ext(G, H) 156 F faithfully E-flat group 142 faithful group 141, 201 faithful ring 201 Faticoni-Schultz Theorem 131 Faticoni, T.G. 53, 59, 71, 79, 89, 99, 110, 114, 117, 121, 129, 168, 172, 189, 203, 222, 229, 234, 245, 249 Faticoni, T.G. and Goeters, H.P. 28, 93 filter of divisibility of 177 finite Baer splitting property 152 finite index 141 finitely faithful 141, 155, 159 finitely faithful S-group 155, 159, 162 finitely G-compressed 147, 162 finitely G-generated 140 finitely generated module 79 finitely generated projective 79, 85 finitely H-projective 157 finitely HG (H)-generated 148 finitely M -generated 147 finitely projective 157, 162 flat dimension 246 Fuchs, L. 1, 20

INDEX

313

Fuller, K.R. 152 fully faithful 152 J Fundamental Theorem of Abelian J -group 155, 165 Groups 21 J´onsson’s Theorem 23 Jacobson radical 6 G Jordan-Zassenhaus Theorem 109 G-compressed 146, 147 G-generated 140 G-monomorphism 251 K G-presented group 251 kernel group 131 G-socle 140, 165 GabE (T ) 182 L Gab(E) 182, 182 L-group 155, 165 GAB(T ) 182 Lady, E.L. 39, 109 Gabriel filter 177, 181, 190, 193 left self-injective ring 276 genus class 35 lies over 301 Goeters, H.P. 165 local ring 7 group 1 local refinement property 42, 61, 86 local-global property 13 H Local-global Theorem 13 h(G) 108 localization of S at I 11 h(E) 109 locally E-free 91 HG (·) 4, 30 locally isomorphic 35 Hamiltonian Quaternions 167 locally unique decomposition 62 hereditary torsion class 179, 198, 200, 203 M homogeneous completely c) 190 µ(M decomposable 130 µ(M ) 183, 185, 191 homomorphisms 3 µE (G) 197 µE (M ) 183 I Mader, A. 24 IDEM(E) 206 maxE (M )183 idempotent 8 maxE (M ) 183 idempotent ideal 177 minimal Qδ-quasi-summand 207 indecomposable 2, 20 module 1 injective dimension 248 Murley group 131, 165 integers 1 integral closure 122 N integral over 303 Nakayama’s Lemma 8, 82 integrally closed 15, 47, 87, 168 natural transformation 4 invertible ideal 92, 95

314 nearly isomorphic 34 Niedzwecki, G.P. and Reid, J.D. 232 nil right ideal 7 nilpotent quasi-decomposition 205 nilpotent quasi-summand 205 nilpotent set 72 nilradical 7 Noetherian endomorphism ring 240

O OR (M ) 199 Ω(E) 196, 201 orthogonal idempotents 71

P Φ4 P(G) 30 p-divisible 3 p-finitely H-projective 157 p-local 3, 224 p-rank 156 p-realizable 224 p-reduced 3 Pierce, R.S. 224 Pierce, R.S. and Vinsonhaler, C. 224 power cancellation 263 primary ideal 124 primary regulating quotient 131 projective cover 186, 189 projective dimension 243 pseudo-rigid 71 pure subgroup 93 Q

INDEX Q-faithful 208, 211 Qδ-quasi-summand 207 quasi-equal 1 quasi-homomorphism 5 quasi-isomorphism 5 quasi-projective 233 quasi-split 156, 209 quasi-summand 9 R rank one group 130 rank 2 rationals 1 reduced 2 reduced norm of 169 refinement property 22, 37, 97, 110, 119 regular element 10 Reid, J.D. 9, 23, 93, 225 Reid’s Theorem 93 rigid 80 ring 1 rtffr 2, 3 rtffr group 2, 19, 69 rtffr right E-modules 3, 35 rtffr ring 3, 35 S S-group 155, 165, 190 SG (H) 154 S-linear module 231 Schultz, P. 246, 323 self-injective ring 251 semi-Artinian 180 semi-local 3 semi-perfect ring 16, 62, 187 semi-prime 8 semi-primary rtffr group 128 semi-primary index 128 semi-rigid 71 simple module 80

INDEX split 20 split exact sequence 20 splits 301 splits completely 301 square-free rank 132 strongly homogeneous group 131 strongly indecomposable group 9, 20, 23, 88, 120, 146 strongly indecomposable ring 221 subcommutative ring 202 superfluous 185 support of R 302 T Θ4 tExt(G, H) 156 TG (·) 4, 30 TorE (X ) 180 Tor(E) 182 TOR(γ) 182 trace of H, of P 147 trace of projective cover 186 totally definite quaternion algebra 167, 170 U, V, W UConn ’81 Theorem, 236 unique decomposition 21 Vinsonhaler, C. and Wickless, W. 243, 246, 250 Vinsonhaler, C. 250 Walker, E.A. 23 Walker, C.L. 156 Walker’s Theorem, C.L. 156 Warfield, Jr., R.B. 156, 159 Warfield’s p-rank Theorem 156 Warfield’s Theorem 168 Z Zassenhaus, H. 294 Z-adic integers 5

315

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