Regularity and Substructures of Hom (Frontiers in Mathematics)

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Author: Friedrich Kasch | Adolf Mader

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Frontiers in Mathematics

Advisory Editorial Board Leonid Bunimovich (Georgia Institute of Technology, Atlanta, USA) Benoît Perthame (Ecole Normale Supérieure, Paris, France) Laurent Saloff-Coste (Cornell University, Rhodes Hall, USA) Igor Shparlinski (Macquarie University, New South Wales, Australia) Wolfgang Sprössig (TU Bergakademie, Freiberg, Germany) Cédric Villani (Ecole Normale Supérieure, Lyon, France)

Friedrich Kasch Adolf Mader

Regularity and

Substructures of

Hom

Birkhäuser Verlag Basel . Boston . Berlin

Authors: Friedrich Kasch Mathematisches Institut Universität München Theresienstrasse 39 D-80333 München e-mail: [email protected]

Adolf Mader Department of Mathematics University of Hawaii 2565 McCarthy Mall Honolulu, HI 96822 USA e-mail: [email protected]

2000 Mathematical Subject Classiﬁcation: 08A05, 08A35, 13A10, 13C10, 13C11, 13E99, 16D10, 16D40, 16D50, 16S50, 20K15, 20K20, 20K25

Library of Congress Control Number: 2008939516

Bibliographic information published by Die Deutsche Bibliothek Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliograﬁe; detailed bibliographic data is available in the Internet at .

ISBN 978-3-7643-9989-4 Birkhäuser Verlag AG, Basel · Boston · Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microﬁlms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained. © 2009 Birkhäuser Verlag AG Basel · Boston · Berlin P.O. Box 133, CH-4010 Basel, Switzerland Part of Springer Science+Business Media Cover design: Birgit Blohmann, Zürich, Switzerland Printed on acid-free paper produced from chlorine-free pulp. TCF ∞ Printed in Germany ISBN 978-3-7643-9989-4

e-ISBN 978-3-7643-9990-0

987654321

www.birkhauser.ch

Contents Preface

vii

I Notation and Background 1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Rings and Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Abelian Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II

III

Regular Homomorphisms 1 Deﬁnition and Characterization . . . . . . . . . . . . . . . . . . . 2 Partially Invertible Homomorphisms and Quasi-Inverses . . . . . 3 Regular Homomorphisms Generate Projective Direct Summands 4 Existence and Properties of Reg(A, M ) . . . . . . . . . . . . . . . 5 The Transfer Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Inherited Regularity . . . . . . . . . . . . . . . . . . . . . . . . . 7 Appendix: Various Formulas . . . . . . . . . . . . . . . . . . . . .

1 1 2 6

. . . . . . .

11 11 15 19 21 25 26 39

Indecomposable Modules 1 Reg(A, M ) = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Structure Theorems . . . . . . . . . . . . . . . . . . . . . . . . . .

41 41 42

IV Regularity in Modules 1 Fundamental Results . . . . . . . . . . . . . . . . . . . . . 2 Quasi-Inverses . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Basic Properties . . . . . . . . . . . . . . . . . . . 2.2 Partially Invertible Objects are Quasi-Inverses . . 3 Regular Elements Generate Projective Direct Summands . 4 Remarks on the Literature . . . . . . . . . . . . . . . . . . 5 The Transfer Rule . . . . . . . . . . . . . . . . . . . . . .

. . . . . . .

49 49 52 52 53 54 56 57

V Regularity in HomR (A, M ) as a One-sided Module 1 Iterated Endomorphism Rings . . . . . . . . . . . . . . . . . . . . . 2 Deﬁnitions and Characterizations . . . . . . . . . . . . . . . . . . .

59 59 60

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

vi

Contents 3 4

VI 1 2 3 4

Largest Regular Submodules . . . . . . . . . . . . . . . . . . . . . The Transfer Rule for S-Regularity . . . . . . . . . . . . . . . . . . Relative Regularity: U -Regularity and Semiregularity U -Regularity; Deﬁnition and Existence of U -Reg(A, M ) U -Regularity in Modules . . . . . . . . . . . . . . . . . . Semiregularity for Modules . . . . . . . . . . . . . . . . Semiregularity for Hom . . . . . . . . . . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

65 66 69 69 72 73 76

VII Reg(A, M ) and Other Substructures of Hom 1 Substructures of Hom . . . . . . . . . . . . . . . . . . . . . 2 Properties of Δ(A, M ) and ∇(A, M ) . . . . . . . . . . . . . 3 The Special Case HomR (R, M ) . . . . . . . . . . . . . . . . 4 Further Internal Properties of Δ(M ) . . . . . . . . . . . . . 5 Non-singular Modules . . . . . . . . . . . . . . . . . . . . . 6 A Correspondenc Between Submodules of HomR (A, M ) and of End(MR ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Correspondences for Modules . . . . . . . . . . . . . . . . .

81 . . . . 81 . . . . 85 . . . . 87 . . . . 90 . . . . 94 Ideals . . . . 95 . . . . 100

VIII 1 2 3 4

. . . .

Regularity in Homomorphism Groups of Abelian Groups Introduction . . . . . . . . . . . . . . . . . . . . . . . . Hom(A, M ) and Regularity . . . . . . . . . . . . . . . Mixed Groups . . . . . . . . . . . . . . . . . . . . . . . Regularity in Endomorphism Rings of Mixed Groups .

. . . .

. . . .

103 103 103 115 127

IX Regularity in Categories 1 Regularity in Preadditive Categories . . . . . . . . . . . . . . . . 2 Preadditive Categories . . . . . . . . . . . . . . . . . . . . . . . . 3 The Quasi-Isomorphism Category of Torsion-free Abelian Groups 4 Regularity in QA . . . . . . . . . . . . . . . . . . . . . . . . . . . √ 4.1 Realizing Q( p) . . . . . . . . . . . . . . . . . . . . . . . 4.2 Constructing the Group . . . . . . . . . . . . . . . . . . . 4.3 Computing the Quasi-Endomorphism Ring . . . . . . . . 5 Regularity in the Category of Groups . . . . . . . . . . . . . . .

. . . . . . . .

131 131 132 137 150 153 153 154 155

. . . .

. . . .

. . . .

. . . .

Bibliography

157

Index

161

Preface Regular rings were originally introduced by John von Neumann to clarify aspects of operator algebras ([33], [34], [9]). A continuous geometry is an indecomposable, continuous, complemented modular lattice that is not ﬁnite-dimensional ([8, page 155], [32, page V]). Von Neumann proved ([32, Theorem 14.1, page 208], [8, page 162]): Every continuous geometry is isomorphic to the lattice of right ideals of some regular ring. The book of K.R. Goodearl ([14]) gives an extensive account of various types of regular rings and there exist several papers studying modules over regular rings ([27], [31], [15]). In abelian group theory the interest lay in determining those groups whose endomorphism rings were regular or had related properties ([11, Section 112], [29], [30], [12], [13], [24]). An interesting feature was introduced by Brown and McCoy ([4]) who showed that every ring contains a unique largest ideal, all of whose elements are regular elements of the ring. In all these studies it was clear that regularity was intimately related to direct sum decompositions. Ware and Zelmanowitz ([35], [37]) deﬁned regularity in modules and studied the structure of regular modules. Nicholson ([26]) generalized the notion and theory of regular modules. In this purely algebraic monograph we study a generalization of regularity to the homomorphism group of two modules which was introduced by the ﬁrst author ([19]). Little background is needed and the text is accessible to students with an exposure to standard modern algebra. In the following, R is a ring with 1, and A, M are right unital R-modules. Let f ∈ HomR (A, M ). Then f is regular if there exists g ∈ HomR (M, A) such that f gf = f . If so, g is a quasi-inverse of f . The isomorphisms HomR (R, M ) ∼ = M and HomR (R, R) ∼ = R produce concepts of regularity in MR and R which are exactly those used by earlier authors. The basic theme of this monograph is to generalize earlier results on rings and modules to homomorphism groups and to obtain new results on regularity in homomorphism groups that can then be specialized to rings and modules. Chapter II contains basic properties of regular maps. Regular maps are associated with direct decompositions. Corollary II.1.3 (Characterization of Regularity). f ∈ HomR (A, M ) is regular if and only if Ker(f ) is a summand A and Im(f ) is a summand M .

viii

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This characterization already establishes that groups and rings of matrices over ﬁelds are regular since matrices are linear transformations and kernels and images of linear transformations are direct summands. Special quasi-inverses exist producing special decompositions. Proposition II.1.6. Let f ∈ HomR (A, M ) be regular. Then there exists g ∈ HomR (M, A) such that f gf = f and gf g = g. If so, then A = Ker(f ) ⊕ Im(g),

and M = Im(f ) ⊕ Ker(g).

Given modules AR and MR , let S = End(MR ) and T = End(AR ). Then HomR (A, M ) is an S-T -bimodule and this is the setting in which we operate. Regular maps produce projective summands. Theorem II.3.1. Let 0 = f ∈ HomR (A, M ) be regular. Then the following statements hold. 1) Sf is a nonzero S-projective direct summand of S HomR (A, M ) that is isomorphic to a cyclic left ideal of S that is a direct summand of S. 2) f T is a nonzero T -projective direct summand of HomR (A, M )T that is isomorphic to a cyclic right ideal of T that is a direct summand of T . There exists a largest regular S-T -submodule of H = HomR (A, M ), denoted by Reg(A, M ). Here “largest” means that any other regular S-T -submodule of H is contained in Reg(A, M ). Theorem II.4.3. Reg(A, M ) = {f ∈ HomR (A, M ) | Sf T is regular} is the largest regular S-T -submodule of HomR (A, M ). Previous authors studied regular rings and modules and modules over regular rings. We are now confronted with the more delicate problem of computing Reg(A, M ), and the corresponding specializations Reg(MR ), Reg(R), and Reg(A, A) = Reg(End(AR )). There are interesting results on the structure of Reg(A, M ). Theorem II.4.6. Every ﬁnitely or countably generated S-submodule of Reg(A, M ) is a direct sum of cyclic S-projective submodules that are isomorphic to left ideals of S and these ideals are direct summands of S. Every ﬁnitely generated S-submodule L of Reg(A, M ) is S-projective and a direct summand of S HomR (A, M ). The analogous results hold for Reg(A, M ) as a right T -module. Proposition II.2.5. Every epimorphic image of epimorphic image of Reg(A, M )T is T -ﬂat.

S

Reg(A, M ) is S-ﬂat, and every

Corollary II.4.8. Suppose that Reg(A, M ) contains no inﬁnite direct sums of Ssubmodules. Then Reg(A, M ) is the direct sum of ﬁnitely many simple projective

Preface

ix

S-modules, HomR (A, M ) = Reg(A, M ) ⊕ U for some S-submodule U and U contains no nonzero regular S-T -submodule. Every cyclic S-submodule of Reg(A, M ) is isomorphic to a left ideal of S that is a direct summand of S. Corollary II.4.9. Suppose that HomR (A, M ) is regular. Then every ﬁnitely generated S-submodule L of HomR (A, M ) is S-projective and a direct summand of S HomR (A, M ). Furthermore, L is the direct sum of ﬁnitely many cyclic S-projective submodules that are isomorphic to left ideals of S. There is a connection between regular elements in HomR (A, M ) and regular elements in HomR (M, A). If f ∈ HomR (A, M ) and g ∈ HomR (M, A) with f gf = f , then we call (f, g) a regular pair. Similarly, if h ∈ HomR (M, A) and k ∈ HomR (A, M ) with hkh = h, then (h, k) is a regular pair. If (f, g) is a regular pair, then also (gf g, f ) is a regular pair. We show that the so-called transfer (f, g) → (gf g, f ) produces all regular elements in HomR (M, A) from those in HomR (A, M ). We study “inherited regularity” when maps φ : A → A and μ : M → M are given and induce mappings on the homomorphism groups. The really important case of “inherited regularity” occurs when A = A1 ⊕ · · · ⊕ Am and M = M1 ⊕ · · · ⊕ Mn . We study the relationship between regular elements of HomR (A, M ) and those of its additive subgroups HomR (Ai , Mj ). It is convenient to identify HomR (A, M ), End(AR ) and End(MR ) with groups and rings of matrices. The results for m = n = 2 are as follows. Theorem II.6.14. Let A = A1 ⊕ A2 and M = M1 ⊕ M2 as before. We use the identiﬁcations ξ11 ξ21 ξ ∈ HomR (A1 , M1 ), ξ21 ∈ HomR (A2 , M1 ), , , 11 HomR (A, M ) = ξ12 ξ22 ξ12 ∈ HomR (A1 , M2 ), ξ22 ∈ HomR (A2 , M2 ). S = EndR (M ) = and

T = EndR (A) =

Then

ξ11 ξ12

μ11 μ12

μ21 μ22

α11 α12

α21 α22

ξ21 ξ22

| μij ∈ HomR (Mi , Mj )

| αij ∈ HomR (Ai , Aj ) .

∈ Reg(A, M )

if and only if for all μjt ∈ HomR (Mj , Mt ) and for all αsi ∈ HomR (As , Ai ) μjt ξij αsi ∈ Reg(As , Mt ). Theorem II.6.14 can be extended to arbitrary ﬁnite sums by induction. There is an interesting corollary.

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Preface

Corollary II.6.18. Let A = A1 ⊕ · · · ⊕ Am and M = M1 ⊕ · · · ⊕ Mn . Then HomR (A, M ) is regular, i.e., Reg(A, M ) = HomR (A, M ), if and only if ∀ i, j : Hom(Ai , Mj ) = Reg(Ai , Mj ). In Chapter III we consider the case that either A or M is indecomposable and Reg(A, M ) = 0. If this is the case much can be said about the structure of both modules. Theorem III.1.1. 1) Suppose that there is 0 = f ∈ HomR (A, M ) that is regular and μf is regular for all μ ∈ S = End(MR ). If M is directly indecomposable, then S is a division ring. 2) Suppose that there is 0 = f ∈ HomR (A, M ) that is regular and f α is regular for all α ∈ T = End(AR ). If A is directly indecomposable, then T is a division ring. Corollary III.2.3. Suppose that 0 = f ∈ Reg(A, M ), and M is indecomposable. Then End(MR ) is a division ring and either A = K ⊕ M1 ⊕ · · · ⊕ Mn where Mi ∼ = M , and HomR (K, M ) = 0 or for every i ∈ N there is a decomposition A = Ki ⊕ M1 ⊕ · · · ⊕ Mi with Mi ∼ = M and Ki = Ki+1 ⊕ Mi+1 . In the latter ∞ ∞ ∞ Mi = Mi and there is a homomorphism ρ : A → Mi such that case ∞ i=1

i=1

i=1

Mi ⊆ ρ(A) and Ker(ρ) =

∞ i=1

i=1

Ki .

There is a similar structure theorem when A is indecomposable (Corollary III.2.7). In Chapter IV we specialize the general theory to the case HomR (R, M ) ∼ = M . Then M is an S-R-bimodule where S = End(MR ). An element m ∈ M is regular with quasi-inverse ϕ ∈ HomR (M, R) if m = mϕ(m). This means that f ∈ HomR (R, M ) is regular in HomR (R, M ) if and only if f (1) ∈ M is regular in M. Regular modules are special as the following example shows. Theorem IV.1.2. Let M = 0 be a regular module over an integral domain R. Then R is a ﬁeld. A vector space over a division ring is regular. By means of the isomorphism HomR (R, M ) ∼ = M we can specialize our general results to immediately obtain numerous results on regularity in modules (Theorem IV.1.3, Theorem IV.1.4, Theorem IV.1.8, Theorem IV.3.1, Theorem IV.3.3, Theorem IV.3.4, Theorem IV.3.7). In Chapter V we consider H = HomR (A, M ) as a bimodule but also as a onesided module. Let S = End(MR ) and T = End(AR ). Then we have the bimodule S HT and the one-sided modules S H and HT and for each of these structures their concepts of regularity. Let f ∈ H = HomR (A, M ).

Preface

xi

• f is S-regular, i.e., f is regular as an element of exists σ ∈ HomS (H, S) such that (f σ)f = f .

S H,

if and only if there

• f is T -regular, i.e., f is regular as an element of HT , if and only if there exists τ ∈ HomT (H, T ) such that f (τ f ) = f . It turns out that the regularity of f ∈ HomR (A, M ) implies all other kinds of regularity. Lemma V.2.1. Let f ∈ HomR (A, M ) be regular and let g ∈ HomR (M, A) be such that f gf = f . Then the map σ : S HomR (A, M ) h → (h)σ = h ◦ g ∈ S End(MR ) = S S is a well-deﬁned S-homomorphism and f is S-regular with quasi-inverse σ. The map τ : HomR (M, A)T h → τ (h) = g ◦ h ∈ End(AR )T = TT is a well-deﬁned T -homomorphism and f is T -regular with quasi-inverse τ . There are numerous immediate corollaries to our results on regularity for modules. The existence of various largest regular submodules is an immediate consequence of Theorem II.4.3 and there are structure theorems that hold for the modules S H and HT . Theorem V.3.4. Let H = HomR (A, M ), let S = End(HT ) and suppose that N is a ﬁnitely generated S -submodule of Reg(HT ). Then N is an S -projective direct summand of H. Furthermore, N is the direct sum of ﬁnitely many cyclic projective submodules, each of which is isomorphic to a left ideal of S . The same is true for submodules of Reg(HT ) considered as a right T -module. In Chapter VI we introduce generalizations of regularity such as U -regularity and semiregularity. Semiregularity for modules was introduced by W.K. Nicholson ([26]). Zelmanowitz studied the class of regular modules, and Nicholson was interested in the wider class of semiregular modules. We list without proof some of his more striking results. Semiregularity is then deﬁned for Hom and there are three diﬀerent possibilities: semiregular, S-semiregular and T -semiregular. A sample result is as follows. Theorem VI.4.2. If f ∈ HomR (A, M ) is semiregular, then Sf lies over an Sprojective direct summand of S HomR (A, M ). In Chapter VII we consider additional substructures of H = HomR (A, M ) and study relations between them. The singular submodule of S HT is by deﬁnition the S-T -submodule Δ(A, M ) = {f ∈ HomR (A, M ) | Ker(f ) is large in A}. The cosingular submodule of H is the S-T -submodule ∇(A, M ) = {f ∈ HomR (A, M ) | Im(f ) is small in M }, and

xii

Preface

there are three diﬀerent concepts of radical in H: the radical of S H as an Smodule, and similarly the radical of HT as a T -module, and ﬁnally the most important radical for which there are two equivalent deﬁnitions: Rad(HomR (A, M )) = Rad(A, M ) = {f ∈ H | f HomR (M, A) ⊆ Rad(S)} = {f ∈ H | HomR (M, A)f ⊆ Rad(T )}. Proposition VII.2.4. If A or M is large restricted or injective, then Δ(A, M ) ⊆ Rad(A, M ). Corollary VII.3.1. 1) If RR or MR is injective, then Δ(MR ) ⊆ Rad(MR ). 2) If RR is injective, then Δ(RR ) ⊆ Rad(R). There is a correspondence between ideals of endomorphism rings and bisubmodules of Hom. Theorem VII.6.3. Let A and M be right R-modules, S = End(MR ), and T = End(AR ). Let L(S) denote the lattice of all two-sided ideals of S and L(H) the lattice of all S-T -submodules of H. Then Mdl : L(S) → L(H) : Mdl(I) = {f ∈ H | f HomR (M, A) ⊆ I} and Idl : L(H) → L(S) : Idl(K) =

g∈HomR (M,A)

Kg

are inclusion preserving maps with Mdl(Idl(Mdl(I))) = Mdl(I) and Idl(Mdl(Idl(K))) = Idl(K). In Chapter VIII we deal with regularity in homomorphism groups of abelian groups. The theory of abelian groups is a highly developed subject so that conclusive answers are possible using the available tools. We wish to compute the maximum regular bi-submodule of Hom(A, M ). A ﬁrst step consists in checking when Zf is regular for a regular homomorphism f ∈ Hom(A.M ). Proposition VIII.2.9. Let A be a group and let f ∈ Hom(A, M ). Then Zf = {nf | n ∈ Z} is a regular subgroup of Hom(A, M ) if and only if Ker(f ) is a direct summand of A, Im(f ) is a direct summand of M , and for all n ∈ Z, Im(f ) = n Im(f ) ⊕ Im(f )[n]. We can settle the torsion-free and torsion cases. Theorem VIII.2.13. Let A and M be torsion-free abelian groups. Then Reg(A, M ) = 0 unless A and M are both divisible. If both A and M are divisible, then Reg(A, M ) = Hom(A, M ).

Preface

xiii

Theorem VIII.2.14. Let A and M be p-primary groups. 1) Suppose that M is not reduced. Then Reg(A, M ) = 0. 2) Suppose that M is reduced. There are decompositions A = A1 ⊕ A2 and M = M1 ⊕ M2 such that A1 and M1 are elementary, and A2 and M2 have no direct summands of order p. Then i) Reg(A, M ) = 0 if M2 = 0, ii) Reg(A, M ) = 0 if A2 is not divisible, iii) Reg(A, M ) = Hom(A, M ) if M2 = 0 and A2 is divisible, i.e., if A is the direct sum of an elementary group and a divisible group and M is elementary. As usual the general case of a mixed group is much more diﬃcult. If f ∈ Reg(A, M ), then pf must be regular for every prime p. This fact alone has farreaching consequences. Theorem VIII.3.6. Let A and M be reduced abelian groups and assume that Reg(A, M ) = 0. Then there exists a non-void set of primes P(A, M ) such that ∀ p ∈ P(A, M ), A = pA ⊕ A[p], and

∀ p ∈ P(A, M ), M = M [p] ⊕ M ,

A[p] = 0, M [p] = 0.

The equations A = pA ⊕ A[p] clearly are very restrictive. Corollary VIII.3.10. Suppose that A is a reduced abelian group such that, for every prime p ∈ P, there is a decomposition A = pA ⊕ A[p]. Then A ∼ = G where G is a group with p∈P A[p] ⊆ G ⊆ p∈P A[p] such that G/ p∈P A[p] is divisible. If Reg(A, M ) = 0 we are dealing with groups of the kind in Corollary 3.10 in the best of situations. These groups appear to be very concrete and accessible yet numerous studies have shown that they are very elusive. Let t(G) denote the maximal torsion subgroup of G. p ∈ P, let Tp be a zero or nonzero elementary p-group, set P = For each T , T = p p∈P p∈P Tp = t(P ), and let P(T ) = {p ∈ P | Tp = 0}. A group A is a G ∗ (T )-group if T ⊆ A ⊆ P such that A/T is (torsion-free) and P(T )-divisible. A G ∗ (T )-group A will be called slim if Tp is cyclic or zero for every p ∈ P. Theorem VIII.3.16. Suppose that A ∈ G ∗ (T1 ) and M ∈ G ∗ (T2 ). Then t(Hom(A, M )) is a regular bi-submodule of Hom(A, M ). There are G ∗ (T )-groups A, M such that HomR (A, M ) contains non-regular homomorphisms. The situation is somewhat easier if A = M , which is the study of regularity in endomorphism rings of mixed groups. Despite numerous attempts ([29], [12], [30],

xiv

Preface

[13], [24]), it is not known which mixed abelian groups have regular endomorphism rings but there is one conclusive result. Theorem VIII.4.2. ([13, Theorem 4.1]) Let A be a slim G ∗ (T )-group such that A/T has ﬁnite rank and is divisible. Then End(A) is regular. The deﬁnition of regularity makes sense in any category and this is the topic of Chapter IX. In preadditive categories Reg(A, M ) exists as it does in module categories. Theorem IX.1.3. Let C be a preadditive category, A, M objects of C, S = C(M, M ) and T = C(A, A). Then Reg(A, M ) = {f ∈ C(A, M ) | Sf T is regular} is the largest regular S-T -submodule of C(A, M ). Preadditive categories are looked at in some detail, in particular, kernels, cokernels, images and coimages of morphisms in a category are needed for further discussions of regularity. Let A be an object in a category C and e = e2 ∈ C(A, A). Then the idempotent e splits in C if there exists an object M and mappings ι ∈ C(M, A), π ∈ C(A, M ) such that ιπ = e and πι = 1M . We say that idempotents split in C if all idempotents in C are splitting. The splitting of idempotents means that every idempotent determines a direct decomposition. By A ∈C A1 ⊕ A2 we mean that there exists a set of structural maps (insertions and projections) that make A the biproduct of A1 and A2 . If idempotents split in a preadditive category C, then we have the familiar decompositions associated with regular maps. Theorem IX.2.7. Let C be a preadditive category in which idempotents split and suppose that f ∈ C(A, M ) is regular and f gf = f for g ∈ C(M, A). Then the following statements hold. 1) e = f g ∈ C(M, M ) is an idempotent. 2) d = gf ∈ C(A, A) is an idempotent. 3) There are structural maps ιAi : Ai → A, and πAi : A → Ai such that d = ιA1 πA1 , πA1 ιA1 = 1A1 , 1A −d = ιA2 πA2 , πA2 ιA2 = 1A2 , and A ∈C A1 ⊕A2 . 4) There are structural maps ιMi : Mi → M , and πMi : M → Mi such that e = ιM1 πM1 , πM1 ιM1 = 1M1 , 1M − e = ιM2 πM2 , πM2 ιM2 = 1M2 , and M ∈C M1 ⊕ M2 . 5) πM1 f ιA1 : A1 → M1 is an isomorphism with inverse πA1 gιM1 . 6) ιA2 ∈ Ker(f ) and ιM2 ∈ Ker(f g). 7) ιM1 ∈ Im(f ) and ιA1 ∈ Im(gf ). In the preadditive category A of torsion-free abelian groups of ﬁnite rank there is only one group whose endomorphism ring is a division ring, namely Q,

Preface

xv

therefore regularity is not very interesting. Also there are highly non-unique direct decompositions in this category. To remedy this problem the quasi-isomorphism category QA was introduced by Bjarni Jonsson ([16], [17]). The objects of QA are the same as those of A, namely all torsion-free abelian groups of ﬁnite rank, which is the same as all additive subgroups of ﬁnite-dimensional Q-vector spaces. However, the morphism groups are now Q Hom(A, M ) = {rf | r ∈ Q, f ∈ Hom(A, M )} ⊆ Hom(QA, QM ) where QA and QM are the vector subspaces spanned by A and M in the Q-vector spaces in which they are contained as additive subgroups, and Hom(QA, QM ) is just the groups of linear transformations. The category QA is a Krull-Schmidt category, i.e., every object is “uniquely” the direct sum of indecomposable objects. Also this category contains an abundance of groups whose endomorphism rings are division rings. The regularity picture in the category QA is transparent but there are no conclusive simple theorems because the indecomposable objects are too numerous, complex, and unknown. Finally, we look at regularity in the category of (non-commutative) groups where semidirect products appear. Theorem IX.5.2. Let G and H be groups and f ∈ Hom(G, H). 1) Assume that f is regular. Let g ∈ Hom(H, G) such that f gf = f . Then e = gf ∈ End(G) is an idempotent, d = f g ∈ End(H) is an idempotent and G = Ker(f ) Im(e),

Im(e) ∼ = Im(f ),

and H = Ker(d) Im(f ).

2) Suppose that G = Ker(f ) K and H = N Im(f ). Then f is regular.

Chapter I

Notation and Background 1

Notation

Citations of numbered items belong to the chapter in which they appear, unless it is explicitly stated that the item is from a diﬀerent chapter (e.g., Theorem II.2.3). When the citation contains two index numbers (e.g., Theorem 1.1), it refers to the theorem in general. When the citation contains three index numbers (e.g., Corollary II.3.5.2), it refers to Item number 2 in Corollary II.3.5. In some statements ∀x : . . . the colon must be read as “it is true that”, and in statements ∃x : . . . the colon must be read as “such that”. We use := to mean “by deﬁnition equal”; e.g., frequently used abbreviations are H := HomR (A, M ), S := End(MR ), and T := End(AR ) where A and M are right R-modules. We mostly use the following notation to deﬁne functions. f : A a → f (a) ∈ B. In this way we avoid the awkward “where a is an element of A”. For X ⊆ B, the term f −1 (X) := {a ∈ A | f (a) ∈ X} denotes the preimage. • P denotes the set of all prime numbers. • Let P be a set of primes. A natural number n is a P -number if the prime factors of n all belong to P . • N0 := {0, 1, 2, . . .}, N := {1, 2, . . .} denote the set of natural numbers, one including 0 and the other excluding 0. • Q is the ﬁeld or group of rational numbers. Let A = B⊕C. Associated with such a decomposition we have the projections prB : A → B along C, and prC : A → C along B.

2

Chapter I. Notation and Background

These projections can also be viewed as endomorphisms of A: prB : A a → prB (a) ∈ A,

and

prC : A a → prC (a) ∈ A.

When viewed as endomorphisms we call the projections projectors. Let M be a module and K a submodule of M . A submodule L of M is a complement of K in M if L is maximal with respect to the condition that K ∩ L = 0. By Zorn’s Lemma every submodule K has complements. A submodule N of M is a supplement of K in M if N is minimal with respect to the property that K + N = M . Supplements may or may not exist. Let AR and MR be right R-modules. We apply maps of HomR (A, M ) on the left side of a ∈ A. If ψ ∈ HomR (A, M ), s ∈ S := End(MR ), t ∈ T := End(AR ), and a ∈ A, then (sψt)(a) = (s ◦ ψ ◦ t)(a) = s(ψ(t(a))) which shows that HomR (A, M ) is an S-T -bimodule. On the other hand, if R A and R M are left R-modules, then we apply maps of HomR (A, M ) on the right side of a ∈ A. If ψ ∈ HomR (A, M ), s ∈ S := End(R M ), t ∈ T := End(R A), and a ∈ A, then (a)(tψs) = (a)(t ◦ ψ ◦ s) = (((a)t)ψ)s which shows that HomR (A, M ) is a T -S-bimodule.

2

Rings and Modules

The notation K ⊆⊕ M means that K is a direct summand of the module M . A submodule K of MR is small or superﬂuous if whenever N + K = M for some submodule N of M , then N = M . We write K ⊆◦ M in this case. A submodule L of MR is large or essential if every nonzero submodule of M intersects L non-trivially. We write L ⊆∗ M in this case. The following is [20, Lemma 0.4]. Lemma 2.1. Let M = A ⊕ B be a direct decomposition of R-modules. Denote by C the set of all direct complements of A in M . Then Hom(B, A) ϕ → Bϕ := {ϕ(b) + b : b ∈ B} ∈ C deﬁnes a bijective mapping and ϕ + 1 : B b → ϕ(b) + b ∈ Bϕ is an isomorphism. Let MR be an R-module and denote by Rad(M ) the radical of R. Recall that the following are equivalent descriptions of Rad(M ) ([18]).

Rad(M ) = {A | A ⊆◦ M } = {B | B is maximal in M } = {Ker(φ) | φ ∈ HomR (M, N ), N semisimple}.

2. Rings and Modules

3

The radical Rad(R) of a ring R is Rad(RR ) which turns out to be the same as Rad(R R). Lemma 2.2. Let R be a ring, A and M right R-modules, S := End(MR ), and T := End(AR ). 1) Let K be a non-void subset of R that is closed under right multiplication by elements of R. Then K ⊆ Rad(R) if and only if for all a ∈ K, 1 − a is an invertible element in R. 2) Let f ∈ HomR (A, M ) and g ∈ HomR (M, A). Then 1S − f g is invertible in S if and only if 1T − gf is invertible in T . 3)

{f ∈ HomR (A, M ) | ∀ g ∈ HomR (M, A), gf ∈ Rad(T )} = {f ∈ HomR (A, M ) | ∀ g ∈ HomR (M, A), f g ∈ Rad(S)}.

Proof. 1) Assume ﬁrst that K ⊆ Rad(R). Then, for every a ∈ K, aR ⊆◦ RR , and hence (1 − a)R + aR = R implies that (1 − a)R = R. So there is s ∈ R such that (1−a)s = 1 and so 1+as = s. Since also a(−s) ∈ K, by specialization, there exists t ∈ R with (1 + as)t = st = 1. It follows that 1 − a = (1 − a)st = ((1 − a)s)t = t and further 1 = st = s(1 − a). Hence s is the inverse of 1 − a. Conversely, assume that for every a ∈ K, 1 − a is an invertible element in R. Denote by |K) the right ideal generated by K. We will show that |K) ⊆◦ RR which means that |K) ⊆ Rad(R). Assume that L is a right ideal of R with R = L + |K). Then (1) 1 = b + a1 + · · · + am , b ∈ L, ai ∈ K. If 1 = b ∈ L, then L = R and we are done. Therefore we can assume that am with m ≥ 1 really occurs. By assumption, 1 − am = b + a1 + · · · + am−1 is invertible, hence 1 = (b + a1 + · · · + am−1 )(1 − am )−1 .

(2)

Since b(1 − am )−1 ∈ L and ai (1 − am )−1 ∈ A for i = 1, . . . , m − 1, we have in (2) a representation analogous to (1) but with only m − 1 or no (for m = 1) summands from K. By induction we get that 1 ∈ L and |K) ⊆ Rad(R). 2) Suppose that 1S − f g is invertible in S. It is easy to check that (1T − gf )−1 = 1T + g(1S − f g)−1 f. Similarly, if 1T − gf is invertible in T , then (1S − f g)−1 = 1S + f (1T − gf )−1 g.

4

Chapter I. Notation and Background

3) Suppose that f ∈ HomR (A, M ) and for every g ∈ HomR (M, A), f g ∈ Rad(S). For all s ∈ S, t ∈ T , g ∈ HomR (M, A), also tgs ∈ HomR (M, A), and so f tgs ∈ Rad(S). Hence the right ideal f tgS is contained in Rad(S). By 1.) it follows that 1S −f tgs is invertible in S. Then, by 2.), T gsf is a left ideal contained in Rad(T ). Therefore f ∈ {f ∈ HomR (A, M ) | ∀ g ∈ HomR (M, A), gf ∈ Rad(T )}. The reverse containment follows in a similar fashion. The following result is the well-known Dual Basis Lemma for projective modules ([18, 5.4.3]). Lemma 2.3. The following properties of a module PR are equivalent. 1) P is projective. 2) Given a family {yi | i ∈ I} of generators of P , there exists a family of functionals {ϕi ∈ P ∗ := HomR (P, R) | i ∈ I} such that for every x ∈ P , all but a ﬁnite number of the ring elements ϕi (x), i ∈ I, are equal to 0 and x = i∈I yi ϕi (x). ∗ 3) There exist families {yi | i ∈ I} of generators of P and {ϕ i ∈ P := HomR (P, R) | i ∈ I} such that for every x ∈ P , we have x = i∈I yi ϕi (x) with ϕi (x) = 0 for all but a ﬁnite number of i.

The following result on ﬂat modules can be found in [18, 10.5.2] or [1, 19.18]. Proposition 2.4. Let S P be a projective module over the ring S with 1, and let K be a submodule of P . Then P/K is ﬂat if and only if for every ﬁnitely generated right ideal I of R, it is true that K ∩ IM = IK. Lemma 2.5. Let M be a module and N a fully invariant subgroup of M . If M = K ⊕ L, then N = (N ∩ K) ⊕ (N ∩ L). Proof. Let prK and prL be the projectors belonging to the decomposition M = K ⊕ L. It is clear that N ∩ K ⊕ N ∩ L ⊆ N . To show equality let x ∈ N . Then x = prK (x) + prL (x) ∈ N ∩ K ⊕ N ∩ L where it is used that N is fully invariant in M . Remark. We have an S-S-homomorphism ρS : HomR (A, M ) ⊗T HomR (M, A) → S = End(MR ) given by ρS (f, g) = f ◦ g, and similarly, a T -T -homomorphism ρT : HomR (M, A) ⊗S HomR (A, M ) → T = End(AR ) given by ρT (g, f ) = g ◦ f. Proof. It is evident that the map ρS : HomR (A, M ) × HomR (M, A) → S satisﬁes ρS (f t, g) = ρA (f, tg) where t ∈ T and ρS (f1 + f2 , g) = ρS (f1 , g) + ρS (f2 , g) and ρS (f, g1 + g2 ) = ρS (f, g1 ) + ρS (f, g2 ) and hence induces a group homomorphism on the tensor product. Also ρS (s1 f, gs2 ) = s1 ρS (f, g)s2 . The argument for ρT is the same.

2. Rings and Modules

5

1) ρS is surjective if and only if there exist f1 , . . . , ft ∈ HomR (A, M ) and g1 , . . . gt ∈ HomR (M, A) such that f1 g1 + · · · + ft gt = 1M .

(3)

Proof. Suppose that ρS is surjective. Then there is an element in the tensor product that maps to 1M and an element in the tensor product is of the form f1 ⊗g1 +· · ·+ft ⊗gt . Conversely, if (3) exists, then clearly ρS is surjective. 2) If ρS is surjective, then SS is ﬁnitely generated. Proof. Suppose that ρS is surjective and (3) holds. Then S = f1 g1 S + · · · + ft gt S, so S is generated by the fi gi ∈ S. If ρS and ρT are both surjective, then A and M must have interacting properties. If A = R, then T = R, and if ρT = ρR is surjective, then MR is projective. Another interesting case occurs when A is simple. Question 2.6. Investigate situations where both ρS and ρT are surjective.

6

Chapter I. Notation and Background

3 Abelian Groups We now turn our attention to abelian groups. Notations and concepts on abelian groups are standard and can be found in [11]. • A torsion group is a group all of whose elements have ﬁnite order. A torsionfree group is a group all of whose nonzero elements have inﬁnite order. A group is mixed if it may have elements both of ﬁnite and inﬁnite order. The symbol t(A) denotes the maximal torsion subgroup of A, i.e., the subgroup consisting of all torsion elements of A. • A p-primary group or p-group is a torsion group all of whose elements have p-power order. A primary group is a group that is a p-group for some prime p. • Every torsion group T has a unique primary decomposition T = p∈P Tp where Tp is p-primary. The p-components Tp are fully invariant subgroups of T . For a general group G, we use Gp := t(G)p = {x ∈ G | ∃ n ∈ N : pn x = 0}. • Z(n) denotes the cyclic group of order n. • A group A is divisible if nA = A for every positive integer n. A group A is p-divisible if pA = A. Let P be a set of primes. A group D is P -divisible if pD = D for every p ∈ P . Every group contains a unique largest subgroup that is P -divisible. A group is divisible if and only if it is p-divisible for each p ∈ P. The divisible abelian groups are exactly the injective abelian groups. • Z(p∞ ) denotes the indecomposable divisible p-group; it is isomorphic to the ∞ p-primary component of Q/Z, i.e., Q/Z ∼ = p∈P Z(p ). It is known that ∞ ∼ ˆ ˆ End(Z(p )) = Zp where Zp is the ring of p-adic integers ([11, Volume I, page 181, Example 3]). Every divisible (= injective) abelian group is the direct sum of copies of the groups Q and Z(p∞ ), p ∈ P. • A group A is n-bounded if nA = 0 for some nonzero integer n. • A p-group A is p-elementary if and only if it is a direct sum of groups of order p. Equivalently, A is p-elementary if and only if A is p-bounded. A group A is elementary if it is the direct sum of p-elementary groups. An abelian group A is semisimple as a Z-module if and only if it is elementary. • Every group A has a decomposition A = B ⊕ C such that C is divisible (= injective) and B is reduced, i.e., B contains no nonzero divisible subgroup. • The p-socle of a group A is the subgroup A[p] := {x ∈ A : px = 0}. More generally, A[n] := {x ∈ A : nx = 0}. If A is a p-primary group, then n n A[p ] = A. • A subgroup B of a group A is pure if B ∩ nA = nB for all integers n. A bounded pure subgroup is a direct summand. The union of an ascending chain of pure subgroups is again pure.

3. Abelian Groups

7

• A subgroup B of a group A is p-pure if B ∩ pn A = pn B for all positive integers n. For a set of primes P0 a subgroup B of A is P0 -pure if B is p-pure for every p ∈ P0 . A subgroup B is pure in A if and only if it is p-pure for every prime p. Let A be a torsion-free abelian group. Then A ∼ = Z ⊗ A ⊆ Q ⊗ A shows that A is embedded in a Q-vector space. Hence, given a torsion-free abelian group A, it may always be assumed that A is an additive subgroup of a Q-vector space V , and doing so the Q-subspace QA spanned by A is the smallest divisible subgroup of V containing A. It can be shown easily that QA ∼ = Q ⊗ A. This means that QA is essentially unique and can, regardless of its containing vector space, always be thought of as QA = {ra | r ∈ Q, a ∈ A}. The group QA is the divisible hull of A. Let A and B be torsion-free groups and QA, QB their divisible hulls. Then Hom(QA, QB) = HomZ (QA, QB) = HomQ (QA, QB), that is to say that the group homomorphisms on the divisible hulls are exactly the linear transformations on QA to QB as vector spaces. Lemma 3.1. Let A be an abelian group and n ∈ Z. Then A has a maximal nbounded direct summand. Proof. An n-bounded direct summand is an n-bounded pure subgroup and nbounded pure subgroups are direct summands. It is well-known and easy to see that an ascending chain of pure subgroups is again pure. Also, an ascending chain of n-bounded subgroups is n-bounded. Therefore a straightforward application of Zorn’s Lemma establishes that there exist maximal n-bounded pure subgroups. In the context of regularity we will have to consider the following aspect of an abelian group. Proposition 3.2. Let A be any abelian group. Then for every p ∈ P there exists a direct decomposition A = Ap1 ⊕ Ap2 such that Ap1 is an elementary p-group and Ap2 has no direct summands of order p. Let π1p : A → Ap1 be the projection along Ap2 and deﬁne π1 : A x → π1 (x) := p∈P π1p (x) ∈ p∈P Ap1 . Then

p∈P

Ap1 ⊆ π1 (A) ⊆

p∈P

Ap1 ,

Ker(π1 ) =

p∈P

Ap2 ,

and Ker(π1 ) has no direct summands of prime order. Let P(A) := {p ∈ P | Ap1 = 0} and assume that A is reduced and ∀ p ∈ P(A) : Hom(Ap2 , Ap1 ) = 0.

8

Chapter I. Notation and Background

Then ∀ p ∈ P(A) : A = A[p] ⊕ pA, Ker(π1 ) is the maximal P(A)-divisible subgroup of A and A/ t(A) is P(A)-divisible. Proof. The decompositions exist by Lemma 3.1. The map π1 is evidently welldeﬁned, it acts as identity on p∈P Ap1 , and Ker(π1 ) is as claimed. Suppose that A is reduced, Ap1 = 0 and Hom(Ap2 , Ap1 ) = 0. Then Ap2 = pAp2 and (Ap2 )p is a pure subgroup of Ap2 , hence also p-divisible. But p-divisible p-groups are divisible and as A is reduced, (Ap2 )p = 0. It follows from A = Ap1 ⊕ Ap2 that A[p] = Ap1 and pA = pAp2 = Ap2 . Remark. In the decomposition A = Ap1 ⊕ Ap2 of Proposition 3.2, the summand Ap1 is unique if and only if Ap2 = pAp2 and the summand Ap1 is unique if and only if Ap2 [p] = 0 (Lemma 2.1). The following examples show that in general little can be said about the quotient π1 (A)/ t(π1 (A)) and Ker(π1 ). Example 3.3. Let X be a torsion-free group of rank ≤ 2ℵ0 . Then there is a group ∼ A such = X. In fact, ∀ p ∈ P, let Tp be cyclic of order p, let that π1 (A)/ t (π1 (A)) T := p∈P Tp and P := p∈P Tp . Then P/T is torsion-free divisible (a Q-vector space) of rank 2ℵ0 . Without loss of generality X ⊆ P/T and let A be the group with T ⊆ A ⊆ P and A/T = X. Then π1 (t(A)) = t(π1 (A)) = T and π1 (A) = A. Example 3.4. Let Tp be cyclic of order p, let T := p∈P Tp and let K be any p group that has no direct summand of prime order. Let A := T ⊕ K, A1 := Tp and p A2 := K ⊕ q =p Tq . Then π1 (A) = T and Ker(π1 ) = K. Let A be an abelian group and B a subgroup of A. A subgroup H of A containing B is called a pure hull of B in A if H is pure in A and if B ⊆ H ⊆ H and H is pure in A, then H = H. In general the questions of existence and uniqueness properties of pure hulls are diﬃcult and not completely resolved. However, in special cases, which includes the case of a torsion-free group A, there is an easy answer. Lemma 3.5. Let A be an abelian group, T the maximal torsion subgroup of A and B a subgroup of A containing T . Then there exists a unique smallest pure subgroup B∗A of A containing B. Furthermore, A/B∗A is torsion-free and B∗A /B is a torsion group. We call B∗A the puriﬁcation of B in A. Proof. Let B∗A be the unique subgroup of A such that B∗A /B = t(A/B). The claims are easily veriﬁed. Lemma 3.6. Let D ⊆ N be abelian groups, m a positive integer and assume that mD = D and m(N/D) = N/D. Then mN = N .

3. Abelian Groups

9

Proof. mN + D hyp. mN + mD mN N hyp. N = = m = = D D D D D so mN = N .

Proposition 3.7. Let G be an Abelian group and H a subgroup of G. Then H is a small subgroup of G if and only if 1) H ⊆ p∈P pG, 2) H has no nonzero divisible quotient group. Proof. (a) Suppose that H is small in G. If pG = G, then trivially H ⊆ pG. Suppose that pG = G. Then G/pG is a vector space over the prime ﬁeld Z/pZ and there is a family K of maximal subgroups of G containing pG whose intersection is pG. Let K ∈ K. If H is not contained in K, then H + K = G and K = G. As H is small we conclude that H ⊆ K and so H ⊆ (K) = pG. This, being true for every p, establishes 1). To prove 2) we assume to the contrary that there is a subgroup L of H such that H/L is nonzero divisible (= injective). Then G/L = H/L ⊕ K/L for some subgroup K of G and K = G since H/L = 0. Now G = H + K and G = K, contrary to the smallness of H. (b) Suppose now that 1.) and 2.) hold and that G = H +K for some subgroup K of G. By 1.), G = pG + K for any prime p, and hence p(G/K) = (pG + K)/K = G/K for every prime p which says that G/K is divisible. Therefore H/(H ∩ K) ∼ = (H + K)/K = G/K is divisible, and by assumption 2) H/(H ∩ K) = 0, so H ⊆ K and G = K showing that H is small. It is much easier to describe the large subgroups. Proposition 3.8. A subgroup L of a group G is large in G if and only if Soc(G) = p∈P G[p] ⊆ L and G/L is torsion. Proof. Suppose that L is large in G. If x is an element of prime order, then xZ ∩ L = 0 which says that Soc(G) ⊆ L. Let 0 = x ∈ G be arbitrary. Then xZ ∩ L = 0, so G/L is torsion. Conversely, Suppose L is a subgroup containing Soc(G) and such that G/L is torsion. Let x ∈ G. As G/L is torsion there is a least positive integer n such that xn ∈ L. If xn = 0 we have the desired nonzero intersection of xZ and L. So suppose that xn = 0 and n = 1. Then, for a prime p dividing n, the element x(n/p) is not in L by minimality of n and p(x(n/p)) = 0 which means that x(n/p) ∈ Soc(G) ⊆ L, a contradiction.

Chapter II

Regular Homomorphisms 1

Deﬁnition and Characterization

Let R be a ring with 1 ∈ R and denote by Mod-R the category of all unitary right R-modules. For arbitrary A, M ∈ Mod-R, let H := HomR (A, M ),

S := End(MR ),

T := End(AR ).

Then H is an S-T -bimodule. Deﬁnition 1.1. Let f ∈ HomR (A, M ). Then f is called regular if there exists g ∈ HomR (M, A) such that f gf = f. (1) If so, g is called a quasi-inverse of f (or r-inverse of f ). A subset X of HomR (A, M ) is regular if every element of X is regular. The deﬁnition of regular homomorphism used here includes the classical cases of regularity in a ring and in a module. Indeed the isomorphism HomR (R, M ) f → f (1) ∈ M

(2)

brings us to M . Similarly, the isomorphism HomR (R, R) ∼ = R brings us to the usual deﬁnition of regularity in a ring. An invertible element f is regular because f f −1 f = f . This means that “regular” is a generalization of invertible and motivates the term “quasi-inverse”. We will come back to quasi-inverses later. It follows from (1) that f (gf g)f = f,

(gf g)f (gf g) = gf g.

This shows that the quasi-inverse of f is not uniquely determined by f ; g and gf g are both quasi-inverses of f . We also see that f is a quasi-inverse of the regular element gf g.

12

Chapter II. Regular Homomorphisms

If (1) is multiplied by g on the left and also on the right, then we obtain two idempotents (3) e := f g = e2 ∈ S, d := gf = d2 ∈ T. By (1) and the deﬁnitions of e and d, it follows further that ef = f gf = f = f d.

(4)

If f = 0, then also e = 0 and d = 0. Since e and d are idempotents, also 1 − e and 1 − d are idempotents and M = e(M ) ⊕ (1 − e)(M ),

A = d(A) ⊕ (1 − d)(A).

(5)

The following theorem characterizes regular maps. Theorem 1.2. If f ∈ HomR (A, M ) is regular and f gf = f , then A = Ker(f ) ⊕ Im(gf ),

Im(gf ) ∼ = Im(f ) ∼ = Im(f g),

(6)

Ker(f g) = Im(1 − f g).

(7)

and M = Im(f ) ⊕ Ker(f g), Furthermore, the mapping f0 : Im(gf ) gf (a) → f gf (a) = f (a) ∈ Im(f ) is an isomorphism with inverse isomorphism g0 : Im(f ) f (a) → gf (a) ∈ Im(gf ). Proof. Set e = f g and d = gf . We have the decompositions (5). We show ﬁrst that (1 − d)(A) = Ker(f ). Indeed, by (4) f (1 − d) = f − f d = 0, so f (1 − d)(A) = 0, hence (1 − d)(A) ⊆ Ker(f ). Now let a ∈ Ker(f ). Then (1 − d)(a) = a − gf (a) = a, hence Ker(f ) ⊆ (1 − d)(A). Together we have Ker(f ) = (1 − d)(A). The complementary summand is d(A) = gf (A) and gf (A) ∼ = A/ Ker(f ) ∼ = Im(f ). Using (4) again, we get the chain Im(f ) = f (A) = ef (A) ⊆ e(M ) = f g(M ) ⊆ f (A), hence f (A) = e(M ) = f g(M ) which is a direct summand of M . The complementary summand is (1 − e)(A) = Ker(e) = Ker(f g).

1. Deﬁnition and Characterization

13

Since A = gf (A) ⊕ Ker(f ), we see that f0 is a monomorphism and f0 evidently surjective. Let x ∈ Im(gf ), say x = gf (a). Then g0 f0 (x) = g0 f0 (gf (a)) = gf gf (a) gf (a) = x so g0 f0 = 1. Let x ∈ Im(f ), say x = f (a). Then f0 g0 (x) = f gf (a) f (a) = x, so f0 g0 = 1.

is = =

We come to an important characterization of regularity. Corollary 1.3 (Characterization of Regularity). Let f ∈ HomR (A, M ). Then f is regular if and only if Ker(f ) ⊆⊕ A and Im(f ) ⊆⊕ M . Proof. Suppose f is regular. Apply Theorem 1.2. Conversely, assume that A = Ker(f ) ⊕ A0

and M = Im(f ) ⊕ M0 ,

(8)

and let π : M → Im(f ) be the projection of M onto Im(f ) along M0 , and let ι : A0 → A be the inclusion map. Deﬁne f0 : A0 a → f (a) ∈ Im(f ). By (8) f0 obviously is an isomorphism. Let g := ιf0−1 π. We will show that f gf = f . Let a ∈ A. Then, by (8), we have a = u + a0 , where u ∈ Ker(f ) and a0 ∈ A0 . It follows that f (a) = f (a0 )

and

f gf (a) = f gf (a0 ) = f f0−1 f (a0 ) = f (a0 ) = f (a),

hence f gf = f .

We remark that Corollary 1.3 already establishes that groups and rings of matrices over ﬁelds are regular since matrices are linear transformations and kernels and images of linear transformations are direct summands. By Corollary 1.5 it is easy to exhibit modules A and M such that HomR (A, M ) contains no nonzero regular homomorphisms. On the other hand, there are modules A and M such that every element in HomR (A, M ) is regular. Corollary 1.4. In the following three cases all elements of HomR (A, M ) are regular. 1) A and M are both semisimple. 2) A is semisimple and injective and M is arbitrary. 3) M is semisimple and projective and A is arbitrary. Proof. 1) Clear by Corollary 1.3. 2) Let 0 = f ∈ H. Then Ker(f ) is a submodule of A, and, A being semisimple, a direct summand of A: A = Ker(f ) ⊕ A0 . Since A is injective, A0 is also injective, and so is Im(f ) ∼ = A/ Ker(f ) ∼ = A0 . Thus Im(f ) ⊆⊕ M . By Corollary 1.3 it follows that f is regular.

14

Chapter II. Regular Homomorphisms

3) This case is dual to 2.). Since Im(f ) is a submodule of M and M is semisimple, Im(f ) is a direct summand of M and then also projective. Hence A/ Ker(f ) ∼ = Im(f ) is projective and this means that Ker(f ) is a direct summand of A. Again by Corollary 1.3, f is regular. Corollary 1.5. HomR (A, M ) contains a regular map = 0 if and only if there exist nonzero direct summands of A and of M that are isomorphic. Thus HomR (A, M ) contains nonzero regular maps if and only if HomR (M, A) contains nonzero regular maps. Proof. Let f = 0 be regular. The claim follows from Theorem 1.3 because A0 ∼ = A/ Ker(f ) ∼ = Im(f ) = 0. The proof of the converse is also easy. We include it for the convenience of the reader. Suppose that A = A1 ⊕ A0 ,

M = M1 ⊕ M0 ,

A0 ∼ = M1 = 0.

Let f0 : A0 → M1 be an isomorphism. Let • π0 be the projection of A onto A0 along A1 , • π1 the projection of M onto M1 along M0 , • ι0 the inclusion of A0 in A, and • ι1 the inclusion of M1 in M . Deﬁne f := ι1 f0 π0 ,

g := ι0 f0−1 π1 .

It is now easy to see that f = 0 and that f gf = f .

The proof shows that f0 is the restriction of the regular map f . The case of regular maps that are mutual quasi-inverses is particularly nice. Proposition 1.6. The following statements hold. 1) Let f ∈ HomR (A, M ) be regular. Then there exists a quasi-inverse g of f that is also regular and has quasi-inverse f , i.e., in addition to f gf = f we have that gf g = g. 2) Suppose that f ∈ HomR (A, M ) and g ∈ HomR (M, A) are such that f gf = f and gf g = g. Then A = Ker(f ) ⊕ Im(g),

and

M = Im(f ) ⊕ Ker(g).

3) Conversely, suppose that f ∈ HomR (A, M ) and g ∈ HomR (M, A), A = Ker(f )⊕Im(g) and M = Im(f )⊕Ker(g). Then there is α ∈ Aut(Im(f )) such that αf ∈ HomR (A, M ) and g are both regular and mutual quasi-inverses.

2. Partially Invertible Homomorphisms and Quasi-Inverses

15

4) Suppose that f ∈ HomR (A, M ). Let F := {h ∈ HomR (M, A) | f hf = f, hf h = h}. Then F is in bijective correspondence with Hom(M/ Im(f ), Im(f )). Proof. 1) Recall that f gf = f implies gf g f gf g = gf g,

f gf g f = f

saying that f is the quasi-inverse of the regular homomorphism gf g and f also has the quasi-inverse gf g. 2) We have decompositions A = Ker(f )⊕Im(gf ) and similarly M = Ker(g)⊕ Im(f g). We will show that Im(gf ) = Im(g) and Im(f g) = Im(f ). Obviously g(M ) ⊇ gf (A) ⊇ gf g(M ) = g(M ) which establishes that Im(gf ) = Im g. By symmetry Im(f g) = Im(f ). 3) As A = Ker(f ) ⊕ Im(g) and M = Im(f ) ⊕ Ker(g), the map f induces an isomorphism f0 : Im(g) → Im(f ) and g induces an isomorphism g0 : Im(f ) → Im(g). Let α := (f0 g0 )−1 ∈ Aut(Im(f )). Then (αf )g(αf ) = αf gg0−1 f0−1 f = αf,

and g(αf )g = gg0−1 f0−1 f g = g.

4) By 2), for any h ∈ F , we have M = Im(f ) ⊕ Ker(h). The assignment h → Ker(h) is clearly an injective assignment of h to a direct complement of Im(f ) in M . To show that the assignment is surjective, suppose that M = Im(f ) ⊕ C for some C. We deﬁne h : M → A to act as 0 on C and to agree with f −1 on Im(f ). It is routine to check that f hf = f and hf h = h. We have shown that F is in bijective correspondence with the direct complements of Im(f ) in M and by Lemma I.2.1 the set of complements in turn is in bijective correspondence with Hom(Ker(g), Im(f )) and Ker(g) ∼ = M/ Im(f ). As a matter of curiosity we consider two special cases. Remark. If f is regular and injective and f gf = f , then gf = 1A , M = Im(f ) ⊕ Ker(g) and Im(f ) ∼ = A = Im(g). If f is regular and surjective and f gf = f , then f g = 1M , A = Ker(f )⊕Im(g) and Im(f ) = M ∼ = Im(g). Question 1.7. Describe all pairs A, M such that all elements in H are regular.

2

Partially Invertible Homomorphisms and Quasi-Inverses

We have already seen that idempotents are very important for the study of regularity. An idempotent e is regular and is its own quasi-inverse since e3 = e. If

16

Chapter II. Regular Homomorphisms

f ∈ HomR (A, M ) is regular and f gf = f for g ∈ HomR (M, A), then we have the idempotents e := f g, d := gf and f and g are factors of these idempotents. Conversely, a factor of an idempotent is the quasi-inverse of some regular map as we will see next. Theorem 2.1. The following statements are equivalent for g ∈ HomR (M, A). 1) There exists h ∈ HomR (A, M ) such that e := hg = e2 = 0. 2) There exists k ∈ HomR (A, M ) such that d := gk = d2 = 0. 3) There exists f ∈ HomR (A, M ) such that f gf = f = 0. 4) There exist 0 = M0 ⊆⊕ M , and A0 ⊆⊕ A such that g0 : M0 x → g(x) ∈ A0

is an isomorphism.

Proof. 1) ⇒ 2): Suppose that e = hg = e2 = 0. Then for k := eh and d := gk = geh it follows that d2 = gehgeh = ge3 h = geh = d and

hdg = hgkg = hgehg = e3 = e = 0,

hence d = 0. 2) ⇒ 1): Similar to 1) ⇒ 2). Now we have (kdg)2 = kdgkdg = kd3 g = kdg. 1) ⇒ 3): Let f := eh. Then f gf = ehgeh = e3 h = eh = f and

ehg = e2 = e = 0, hence f = eh = 0. 3) ⇒ 1): Since f gf = f = 0, it follows that e = f g = 0. 3) ⇒ 4): This follows from Theorem 1.2 with M0 = e(M ) and A0 = d(A). 4) ⇒ 3): We choose f such that f gf = f = 0. By assumption we have M = M0 ⊕ M1 ,

A = A0 ⊕ A1 ,

and we have the isomorphism g0 : M0 x → g(x) ∈ A0 , g0 = 0. Now deﬁne f as

f : A0 → M0 : f = g0−1 ,

f : A1 → 0.

2. Partially Invertible Homomorphisms and Quasi-Inverses

17

Then, for x ∈ A, x = a0 + a1 where a0 ∈ A0 , a1 ∈ A1 , hence (f gf )(x) = f gf (a0 + a1 ) = f gf (a1 ) = f (a0 ) = f (x)

hence f gf = f .

Deﬁnition 2.2. We assume that the conditions of Theorem 2.1 are satisﬁed. Then 1) g is called partially invertible; 2) the total of A to M is deﬁned to be Tot(A, M ) := {f ∈ HomR (A, M ) | f is not partially invertible.}. The monograph ([20]) contains an extensive study of the total Tot(A, M ) of two R-modules. The total is in general not additively closed but it has the following multiplicative closure property. Lemma 2.3 ([20, Chapter II, Corollary 1.10]). If A, M, A , M are right R-modules and f ∈ HomR (A , A), g ∈ HomR (M, M ), then f Tot(A, M )g ⊆ Tot(A , M ). One of the major questions surrounding the total is to characterize the pairs A, M such that Tot(A, M ) is additively closed. The total is essential if M is a module with LE-decomposition, i.e., M = i∈I Mi where for each i, End(Mi ) is a local ring. In this case Tot(End(M )) is an ideal in End(M ) and ([20, Chapter IV, Corollary 2.4]) the quotient ring End(M )/ Tot(End(M )) is isomorphic to a product of endomorphism rings of vector spaces over division rings. By (3) every regular map is partially invertible. Thus we have the containments {f ∈ H | f is regular} ⊆ {f ∈ H | f is partially invertible} and Tot(A, M ) ⊆ {f ∈ HomR (A, M ) | f is NOT regular}. The following example shows, unsurprisingly, that a partially invertible element need not be regular. Example 2.4. Let p be a prime, and A = Za ⊕ Zb,

M = Zm ⊕ Zn

where a, b, m have order p and n has order p2 . Deﬁne f : A → M : f (a) = m, f (b) = pn,

g : M → A : g(m) = a, g(n) = 0.

These are well-deﬁned homomorphisms with the following properties. 1) f g(m) = m,

f g(n) = 0,

gf (a) = a,

gf (b) = 0.

2) e := f g is idempotent and is the projector of M onto Zm along Zn.

18

Chapter II. Regular Homomorphisms

3) d := gf is idempotent and is the projector of A onto Za along Zb. 4) gf g = g. 5) f0 : Za x → f (x) ∈ Zm is an isomorphism. 6) f is partially invertible by 2) or 3) but not regular as Im(f ) = Zm ⊕ Zpn is not a direct summand of M . The diﬀerence between a partially invertible and a regular map can be elucidated in general. Remark. Suppose that f ∈ HomR (A, M ), g ∈ HomR (M, A) and gf = e = e2 ∈ End(AR ). Then d := f gf g is an idempotent in End(MR ), and we have direct decompositions A = eA ⊕ (1 − e)A,

M = dM ⊕ (1 − d)M.

Now dM ⊆ Im(f ) and dM is a direct summand of M but this does not necessarily mean that Im(f ) is a direct summand of M , and the latter would have to happen if f were regular. Remark . We have seen that {f ∈ H | f is partially invertible} = {f ∈ H | f is a quasi-inverse}. Does this set have any algebraic properties? Let A = a be a cyclic group of order p and let M = m be a cyclic group of order p2 . Then Hom(A, M ) contains no partially invertible element, so {f ∈ H | f is partially invertible} = {f ∈ H | f is a quasi-inverse} = ∅. When does the regular map f have a unique quasi-inverse? This question is settled next. Theorem 2.5. Let f ∈ HomR (A, M ). There is a bijective correspondence between quasi-inverses of f and HomR (M/ Im(f ), A). Consequently, the regular homomorphism f has a unique quasi-inverse if and only if HomR (M/ Im(f ), A) = 0. Proof. We see from Theorem 1.2 that for any quasi-inverse g of f , the restriction of g to Im(f ) is the same map, namely f0−1 . Another quasi-inverse g can therefore only diﬀer from g on Ker(f g), and how it acts on Ker(f g) is without consequence as we can see from f = f g f

if and only if f (A) = f g f (A) = f g (f (A))

where only the action of g on f (A) enters. Now ﬁx a quasi-inverse g. We have the decomposition M = Im(f ) ⊕ Ker(f g). Let h : M → A be any other quasi-inverse of f . Then h Ker(f g)∈ HomR (Ker(f g), A) and the assignment h → h Ker(f g) is injective since h = f −1 on Im(f ). The preceding remark establishes that the assignment is also surjective. Since Ker(f g) ∼ = M/ Im(f ), the ﬁrst claim is proved and the second is a trivial consequence of the ﬁrst.

3. Regular Homomorphisms Generate Projective Direct Summands

19

It turns out that the quasi-inverse of f ∈ HomR (A, M ) is unique if End(MR ) is commutative. Proposition 2.6. Suppose that f ∈ HomR (A, M ) is regular and End(MR ) is commutative. Then f has a unique quasi-inverse. Proof. Let g1 and g2 be quasi-inverses of f . We have the decompositions M = Im(f )⊕Ker(f g1 ) = Im(f )⊕Ker(f g2 ). The direct summands of M are endomorphic images of M and hence fully invariant because End(MR ) is commutative. Therefore HomR (Ker(f g1 ), Im(f )) = 0 and this means that the complementary summand Ker(f g1 ) of Im(f ) is unique, so Ker(f g1 ) = Ker(f g2 ). Thus both g1 and g2 map Ker(f g1 ) = Ker(f g2 ) to 0 and on Im(f ) they both agree with f −1 . It follows that g1 = g2 . Corollary 2.7. Let f ∈ HomR (A, M ) be regular with quasi-inverse g ∈ HomR (M, A) and suppose that End(AR ) is commutative. Then f is the unique quasi-inverse of gf g. Proof. We have seen that (gf g)f (gf g) = gf g, so gf g is regular with quasi-inverse f that is unique by Proposition 2.6.

3

Regular Homomorphisms Generate Projective Direct Summands

In this short section we establish interesting properties of regular homomorphisms. Theorem 3.1. Let S := End(MR ), T := End(AR ) and assume that 0 = f ∈ HomR (A, M ) is regular. Then the following statements hold. 1) Sf is a nonzero S-projective direct summand of S HomR (A, M ) that is isomorphic to a cyclic left ideal of S, and this cyclic left ideal is a direct summand of S. 2) f T is a nonzero T -projective direct summand of HomR (A, M )T that is isomorphic to a cyclic right ideal of T and this cyclic right ideal of T is a direct summand of T . More precisely, let g ∈ HomR (M, A) be a quasi-inverse of f and set H := HomR (A, M ). Then: 1) S = Sf g ⊕ S(1 − f g), H = Sf ⊕ H(1 − gf ), and Sf ∼ = Sf g. 2) T = gf T ⊕ (1 − gf )T , H = f T ⊕ (1 − f g)H, and f T ∼ = gf T . Proof. 1) Since 0 = f ∈ H is regular and f gf = f for g ∈ HomR (M, A), the mapping e := f g is a nonzero idempotent in S, and S = Se ⊕ S(1 − e)

20

Chapter II. Regular Homomorphisms

where Se = 0 and Se is a projective left ideal of S. We claim that the Shomomorphism σ : Se se → sef = sf ∈ Sf is an isomorphism, showing that Sf is also projective. Obviously, σ is surjective, and σ is also injective. In fact, if sef = 0, then it follows that 0 = sef g = se2 = se. So se = 0 and σ is an isomorphism. Hence also Sf is S-projective. Furthermore d := gf = d2 ∈ T is an idempotent. For this idempotent we have Sf = Sf d ⊆ Hd = (Hg)f ⊆ Sf, where we used that Hg ⊆ S and Sf ⊆ H. Hence Sf = Hd and Hd is an S-direct summand of H. 2) The proof is similar to that of 1). First we have the epimorphism τ : dT dt → f dt = f t ∈ f T. If now f dt = 0, then it follows that 0 = gf dt = d2 t = dt, hence τ is also injective. Then together with dT , the T -module f T is also projective. Further we have eH = f gH ⊆ f T = ef T ⊆ eH, hence eH = f T ⊆⊕ HT .

Until now we considered H as an S-T -bimodule. We set S := End(HT ),

T := End(S H).

Then H is also a T -left and a S -right module. We ask how S and S are related. If s ∈ S, then the mapping s : H h → sh ∈ H

is in S . In particular, e ∈ S and e is again an idempotent. Since ef = f = 0, there exists a ∈ A with ef (a) = f (a) = 0, so also e = 0. As an element of S we denote e again by e. Similarly, we can assume that d ∈ T . Corollary 3.2. If 0 = f ∈ HomR (A, M ) is regular, then 1) S f = 0 is an S -projective direct summand of

S

HomR (A, M ).

2) f T = 0 is an T -projective direct summand of HomR (A, M )T . Proof. The same as the proof of Theorem 3.1.

Remark. Corollary 3.2 can also be proved using later results. If f is regular in H := HomR (A, M ), then (Lemma 2.1) f is a regular element in the modules S H and HT and the result follows from the module case (Theorem 3.1). We will return later to the rings S and T . They are important for the existence of certain substructures that we will consider later.

4. Existence and Properties of Reg(A, M )

4

21

Existence and Properties of Reg(A, M )

We will show that there exists a largest regular S-T -submodule of H = HomR (A, M ), denoted by Reg(A, M ). Here “largest” means that any other regular S-T -submodule of H is contained in Reg(A, M ). For the proof we need the following fact. Lemma 4.1. Let f ∈ HomR (A, M ) and g ∈ HomR (M, A). If f − f gf is regular, then f is regular. Proof. Since f − f gf is regular, there exists h ∈ HomR (M, A) such that (f − f gf )h(f − f gf ) = f − f gf.

(9)

Let k := h − gf h − hf g + gf hf g + g ∈ HomR (M, A) with g from (9). An easy computation shows that f kf = f . In fact, f kf = f hf − f gf hf − f hf gf + f gf hf gf + f gf = (f − f gf )h(f − f gf ) + f gf and by (9) f kf = f − f gf + f gf = f.

Recall that a subset X of HomR (A, M ) is regular if every element of X is regular. Deﬁnition 4.2. Let Reg(A, M ) := {f ∈ HomR (A, M ) | Sf T is regular}, where Sf T is the S-T -submodule of HomR (A, M ) generated by f . Theorem 4.3. Reg(A, M ) is the largest regular S-T -submodule of HomR (A, M ). Proof. The proof is achieved in four steps. (a) Let f ∈ H and assume that Sf T is regular. If ϕ ∈ Sf T , then SϕT ⊆ Sf T , hence also SϕT is regular and ϕ ∈ Reg(A, M ). Since ϕ was an arbitrary element of Sf T , it follows that Sf T ⊆ Reg(A, M ). (b) Since Reg(A, M ) is the sum of modules of the form Sf T , it is closed under left multiplication by S and right multiplication by T . (c) We show now that Reg(A, M ) is closed under addition. Let f1 , f2 ∈ Reg(A, M ). We have to show that S(f1 +f2 )T is regular. An element of S(f1 +f2 )T is of the form n

i=1 si (f1

+ f2 )ti =

n

i=1 si f1 ti

+

n

i=1 si f2 ti .

22

Chapter II. Regular Homomorphisms

n n Set ϕ1 = i=1 si f1 ti and ϕ2 = i=1 si f2 ti . Then ϕ1 ∈ Sf1 T is regular and there exists g ∈ HomR (M, A) such that ϕ1 gϕ1 = ϕ1 . Now consider (ϕ1 + ϕ2 ) − (ϕ1 + ϕ2 )g(ϕ1 + ϕ2 ) = ϕ2 − ϕ1 gϕ2 − ϕ2 gϕ1 − ϕ2 gϕ2 ,

(10)

where we used that ϕ1 gϕ1 − ϕ1 = 0. Since ϕ1 g ∈ S, gϕ1 , ϕ2 g ∈ T , the element in (10) is in Sϕ2 T . We now use that Sϕ2 T is regular which is true by (a). We can then apply Lemma 4.1 to the element in (10) with ϕ1 + ϕ2 in place of f , and obtain that ϕ1 + ϕ2 is regular. Since ϕ1 + ϕ2 was an arbitrary element of S(f1 + f2 )T , this submodule is regular. So S(f1 + f2 )T ⊆ Reg(A, M ) and in particular, f1 + f2 ∈ Reg(A, M ). (d) Finally we show that Reg(A, M ) is the largest regular S-T -submodule of H. Assume that Λ is a regular S-T -submodule of H and f ∈ Λ. Then Sf T is a regular S-T -submodule of H and by deﬁnition of Reg(A, M ) we have Sf T ⊆ Reg(A, M ), in particular f ∈ Reg(A, M ). This means that Λ ⊆ Reg(A, M ) and Reg(A, M ) is the largest regular S-T -submodule of H. From Corollary 1.4 we can get examples for the extreme case HomR (A, M ) = Reg(A, M ) but the conditions in Corollary 1.4 are by no means necessary. For example HomZ (Q, Q) = Reg(Q, Q) ∼ = Q and QZ is not semisimple. An example of the other extreme is Reg(Z, Z) = 0 in spite of the fact that HomZ (Z, Z) ∼ =Z contains the regular elements ±1. We state an immediate consequence of Corollary 1.5. Corollary 4.4. Let A and M be modules such that no nonzero summand of A is isomorphic with a summand of M , then Reg(A, M ) = 0. We now come to interesting properties of the bimodule Reg(A, M ). In order to avoid repetitions we abstract the essential step in the proofs of several theorems in a lemma. Lemma 4.5. Let H be a left S-module and N a submodule with the property that every cyclic submodule of N is a direct summand of H. Then the following statements hold. 1) Every ﬁnitely generated S-submodule U of N is a direct summand of H and a direct sum of cyclic submodules. 2) Every countably generated submodule of N is the direct sum of cyclic submodules. Proof. 1) The proof is by induction on the number n of generators f1 , . . . , fn ∈ N n of the submodule U := Sf i of N . The start n = 1 of the induction is i=1 n−1 hypothesis. Let n > 1 and set V := i=1 Sfi . Then U = V + Sfn and both V and U are submodules of N . By induction hypothesis V is a direct sum of cyclic submodules and a direct summand of H. Hence there is an idempotent

4. Existence and Properties of Reg(A, M )

23

d ∈ End(S H) such that Hd = V . We further have that H = Hd ⊕ H(1 − d) and fn = fn d + fn (1 − d). Since fn d ∈ Hd = V , it follows that U = V + Sfn = V + S(fn d + fn (1 − d)) = V ⊕ Sfn (1 − d). This shows that U is a direct sum of cyclic submodules. It remains to show that U is a direct summand of H. Now Sfn (1 − d) ⊆ Sfn + Sfn d ⊆ N + V = N and by hypothesis Sf (1 − d) is again a direct summand of H. Hence there exists an idempotent e ∈ End(S H) such that Sfn (1 − d) = He. Thus U = V ⊕ Sfn (1 − d) is the direct sum of the two modules V and He = Sfn (1−d) ⊆ H(1−d). Intersecting H = He⊕H(1−e) with H(1−d) we obtain H(1−d) = He⊕(H(1−d)∩H(1−e)). Consequently, H

= =

Hd ⊕ H(1 − d) = Hd ⊕ He ⊕ (H(1 − d) ∩ H(1 − e)) U ⊕ (H(1 − d) ∩ H(1 − e)).

So U is a direct summand of H, and the proof is complete. ∞ 2) Let U = n=1 Sfn ⊆ N . By the proof of 1), for n = 1, 2, . . ., we have Sf Sfn = (Sf1 + · · · + Sfn−1 ) ⊕ Cn for some submodule Cn . Clearly 1 + ··· + ∞ ∞ C = n=1 n n=1 Cn = U . Theorem 4.6. Every ﬁnitely or countably generated S-submodule of Reg(A, M ) is a direct sum of cyclic S-projective submodules that are isomorphic to left ideals of S that are direct summands of S. Every ﬁnitely generated S-submodule L of Reg(A, M ) is S-projective and a direct summand of S HomR (A, M ). The analogous results hold for Reg(A, M ) as a right T -module. Proof. By Theorem 3.1 every cyclic S-submodule of Reg(A, M ) is a projective direct summand of H = HomR (A, M ). Hence Lemma 4.5 applies and it is clear that the submodules in question are projective in this case. We will make use of a lemma for a general module S H. Lemma 4.7. Let S H be a left S-module and suppose that S H contains no inﬁnite direct sum of submodules and every cyclic submodule of H is a direct summand. Then H is ﬁnitely generated and semisimple. Proof. By Lemma 4.5 we have that every ﬁnitely generated submodule is a direct summand of H. We show next that every submodule of H is ﬁnitely generated. By way of contradiction assume that N is a submodule that is not ﬁnitely generated. Then N contains elements n1 , n2 , . . . such that Sn1 Sn1 + Sn2 · · ·

k i=1

Sni

k+1 i=1

Sni · · ·

24

Chapter II. Regular Homomorphisms

As every ﬁnitely generated submodule of H is a direct summand of H and hence of every submodule that contains it, we obtain decompositions Sn1 + Sn2 = Sn1 ⊕ K1 , Sn1 + Sn2 + Sn3 = (Sn1 + Sn2 ) ⊕ K2 , .. . Sn1 + · · · + Snk+1 = (Sn1 + · · · + Snk ) ⊕ Kk .. . ∞ So i=1 Sni = n1 R ⊕ K1 ⊕ K2 ⊕ · · · which is an inﬁnite direct sum contradicting the hypothesis. So every submodule is ﬁnitely generated and therefore a direct summand of H. In particular, H = Soc(H) ⊕ N for some submodule N where Soc(H) is the sum of all simple submodules of H. Suppose that Sm is a nonzero cyclic submodule of N . There is a maximal left ideal I of S containing AnnS (m) and Im ⊆ Sm is a cyclic submodule of H. Hence Sm = Im ⊕ K for some K and K∼ = Sm/Im ∼ = S/I is simple. But then K ⊆ Soc(H)∩N = 0, a contradiction. Corollary 4.8. Suppose that Reg(A, M ) contains no inﬁnite direct sums of Ssubmodules. Then Reg(A, M ) is the direct sum of ﬁnitely many simple projective S-modules, HomR (A, M ) = Reg(A, M ) ⊕ U for some S-submodule U and U contains no nonzero regular S-T -submodule. Every cyclic S-submodule of Reg(A, M ) is isomorphic to a left ideal of S that is a direct summand of S. There is an analogous result for HomR (A, M )T . Proof. By Theorem 4.6 and Lemma 4.7 the S-module Reg(A, M ) must be ﬁnitely generated and a ﬁnite direct sum of simple submodules. By Lemma 4.5 the submodule Reg(A, M ) is a direct summand of H. Finally, by deﬁnition of Reg(A, M ), the S-submodule U contains no nonzero regular S-T -submodule. Corollary 4.9. Suppose that HomR (A, M ) is regular. Then every ﬁnitely generated S-submodule L of HomR (A, M ) is S-projective and a direct summand of S HomR (A, M ). Furthermore, L is the direct sum of ﬁnitely many cyclic S-projective submodules that are isomorphic to left ideals of S. Under certain ﬁniteness conditions we can obtain much more speciﬁc information on Reg(A, M ). Theorem 4.10. Suppose that HomR (A, M ) is regular and contains no inﬁnite direct sums of S-submodules. Then HomR (A, M ) is a ﬁnitely generated semisimple S-submodule. The simple direct summands of HomR (A, M ) are S-projective and isomorphic to left ideals of S. Proof. It follows from the hypothesis that H := HomR (A, M ) is regular that every cyclic submodule is a direct summand. The fact that H contains no inﬁnite direct sums implies that every S-submodule of H is ﬁnitely generated. Hence every

5. The Transfer Rule

25

submodule of S H is a direct summand of H. It follows that Rad(H) = 0 and this together with the fact that every submodule of H is a direct summand implies that H is a direct sum of simple modules, i.e., H is semisimple. (See Chapter I). The following result generalizes part of [37, Theorem 1.11]. A ring R is left (right) perfect if every left (right) R-module has a projective cover. Theorem 4.11. Suppose that S := End(MR ) is left perfect. Then Reg(A, M ) is S-projective. If T := End(AR ) is right perfect, then Reg(A, M ) is T -projective. Proof. As S is left perfect, by [9, page 68, P(3’)] direct limits of S-projective modules are S-projective. The S-module Reg(A, M ) is the direct limit of its ﬁnitely generated submodules and by Theorem 4.6 these are all S-projective. Hence their direct limit Reg(A, M ) is S-projective.

5

The Transfer Rule

There is a connection between regular elements in HomR (A, M ) and regular elements in HomR (M, A) that is as surprising as it is simple. If f ∈ HomR (A, M ) and g ∈ HomR (M, A) with f gf = f , then we call (f, g) a regular pair. Similarly, if h ∈ HomR (M, A) and k ∈ HomR (A, M ) with hkh = h, then (h, k) is a regular pair. If (f, g) is a regular pair, then also (gf g, f ) is a regular pair because gf g · f · gf g = gf g but now with the regular homomorphism gf g ∈ HomR (M, A). We will show that with these formal rules we can obtain all regular elements in HomR (M, A) from those in HomR (A, M ). Deﬁnition 5.1. If (f, g) is a regular with pair f ∈ HomR (A, M ), then we call trf : (f, g) → (gf g, f ) the transfer rule or the transfer and we also write trf(f, g) = (gf g, f ). Applying trf twice to the pair (f, g) we obtain trf

trf

(f, g) → (gf g, f ) → (f gf gf, gf g) = (f, gf g). Hence applying trf twice to (f, g) brings us to (f, gf g). Thus we come back to f with which we started, but now paired with the quasi-inverse gf g in place of g. Note that transfer produced a regular pair (f, gf g) whose entries are mutual quasi-inverses. If f ∈ H is regular and g1 , g2 are quasi-inverses of f , then trf(f, g1 ) = (g1 f g1 , f ) and trf(f, g2 ) = (g2 f g2 , f ). The following example shows that g1 f g1 = g2 f g2 is possible.

26

Chapter II. Regular Homomorphisms

Example 5.2. Let A = Za ⊕ Zb where ord(a) = ∞, ord(b) = p, and let M = A. Then also M = Z(a + b) ⊕ Zb. Let f : A ra + sb → ra ∈ M , g1 : M ra + sb → ra ∈ A, and g2 : M ra + sb = r(a + b) + (s − r)b → r(a + b) ∈ A. It is easily checked that f g1 f = f and f g2 f = f but g1 f g1 = g2 f g2 , in fact g1 f g1 (a) = a while g2 f g2 (a) = a + b. Theorem 5.3. If trf is applied to all regular pairs (f, g) with f ∈ HomR (A, M ), then the set of ﬁrst entries in trf(f, g) = (gf g, f ) is the set of all regular elements in HomR (M, A). Proof. Let (h, k) be a regular pair with h ∈ HomR (M, A) and k ∈ HomR (A, M ). Then trf(h, k) = (khk, h) is again a regular pair with khk ∈ HomR (A, M ). Applying trf again, then we get (h, khk). Hence every regular element in HomR (M, A) is obtained by an application of trf to the element khk ∈ HomR (A, M ). Let (f, g) be a regular pair. Recall (Theorem 1.2) that we have decompositions A = Im(gf ) ⊕ Ker(f ),

M = Im(f ) ⊕ Ker(f g),

and

Im(f ) = Im(f g).

Using the regular pair (gf g, f ) obtained by transfer, one gets further relationships. To wit: 1) Im(gf ) = Im(gf g), Ker(gf g) = Ker(f g), 2) Im(g) = Im(gf ) ⊕ (Im(g) ∩ Ker(f )) ⊆ A. Proof. We have A = Ker(f ) ⊕ Im(gf ) = Ker(f ) ⊕ Im(gf g), with Im(gf g) ⊆ Im(gf ). It follows that Im(gf ) = Im(gf g). We have M = Im(f ) ⊕ Ker(f g) = Im(f ) ⊕ Ker(gf g), with Ker(gf ) ⊆ Ker(gf g). It follows that Ker(gf ) = Ker(gf g). We have Im(gf ) ⊆ Im(g) and A = Ker(f ) ⊕ Im(gf ), hence Im(g) = Im(gf ) ⊕ (Im(g) ∩ Ker(f )). Question 5.4. How does transfer act on Reg?

6

Inherited Regularity

We are interested in regularity in HomR (A, M ) for R-modules A, M and ask how regularity interacts with related homomorphism groups. Suppose that A and M are two more R-modules and that homomorphisms φ : A → A ,

μ : M → M

6. Inherited Regularity

27

are given. We then have induced maps on homomorphism groups as follows. φ∗ : HomR (A , M ) f μ∗ : HomR (A, M ) f φ∗ : HomR (M, A) f μ∗ : HomR (M , A) f

→ f ◦ φ ∈ HomR (A, M ), → μ ◦ f ∈ HomR (A, M ), → φ ◦ f ∈ HomR (M, A ), → f ◦ μ ∈ HomR (M, A).

Proposition 6.1. Let f ∈ HomR (A , M ) and assume the existence of the maps listed above. Then the following statements hold. 1) If φ is surjective and φ∗ (f ) is regular, then f is regular. 2) If μ is injective and μ∗ (f ) is regular, then f is regular. 3) If φ∗ is surjective and f is regular, then φ∗ (f ) is regular. 4) If μ∗ is surjective and f is regular, then μ∗ (f ) is regular. Proof. 1) Suppose that φ is surjective and φ∗ (f ) is regular. Then there exists g ∈ HomR (M, A) such that φ∗ (f )gφ∗ (f ) = φ∗ (f ). Equivalently, f φgf φ = f φ. Since φ is surjective, it can be canceled, and we obtain the regularity equation f (φg)f = f . 2) Suppose that μ is injective and μ∗ (f ) is regular. Then there is g ∈ HomR (M , A) such that μ∗ (f )gμ∗ (f ) = μ∗ (f ). Equivalently, μf gμf = μf . Since μ is injective, it can be canceled and we obtain the regularity equation f (gμ)f = f . 3) Suppose that φ∗ is surjective and f gf = f . Since φ∗ is surjective there exists h ∈ HomR (M, A) such that g = φ∗ (h) = φh. We conclude that f φhf φ = f φ, or, equivalently, φ∗ (f )hφ∗ (f ) = φ∗ (f ). 4) Suppose that μ∗ is surjective and f gf = f for a g ∈ HomR (M, A ). Since ∗ μ is surjective, there is h ∈ HomR (M , A) such that g = μ∗ (h) = hμ. We conclude that μf hμf = μf , or, equivalently, μ∗ (f )hμ∗ (f ) = μ∗ (f ). The claims of the following corollary are immediate consequences of Proposition 6.1.1 and 6.1.2. Corollary 6.2. 1) If φ ∈ HomR (A, A ) is surjective and HomR (A, M ) is regular, then HomR (A , M ) is regular. 2) If μ ∈ HomR (M, M ) is injective and HomR (A, M ) is regular, then HomR (A, M ) is regular. An interesting special case of Corollary 6.2 is as follows. Corollary 6.3. 1) If B ⊆ A and HomR (A, M ) is regular, then also HomR (A/B, M ) is regular. 2) If N ⊆ M , and HomR (A, M ) is regular, then HomR (A, N ) is regular.

28

Chapter II. Regular Homomorphisms

Proof. Apply Corollary 6.2 with φ : A A/B and μ : N → M .

We will now study the case of a map φ : A → A more closely and check what hypotheses are needed in order to get results on the relationships of Reg(A, M ) and Reg(A , M ). We have φ : A → A

induces

φ∗ : HomR (A , M ) → HomR (A, M ).

(11)

The abelian group H := HomR (A, M ) is an S-T -bimodule where S := End(MR ) and T := End(AR ) while H := HomR (A , M ) is an S-T -bimodule where S := End(MR ) as before and T := End(AR ). Clearly, the map φ∗ is an S-module morphism. Suppose that

or

∀ t ∈ End(AR ), ∃ t ∈ End(AR ) such that t φ = φt

(12)

∀ t ∈ End(AR ), ∃ t ∈ End(AR ) such that t φ = φt.

(13)

Then in either case ∀ f ∈ HomR (A , M ) : φ∗ (f )t = φ∗ (f t ). Proposition 6.4. Let φ : A → A and assume that ∀ g ∈ HomR (M, A ), ∃ h ∈ HomR (M, A) such that g = φh,

(14)

and assume (12). Then φ∗ (Reg(A , M )) ⊆ Reg(A, M ). Proof. Let f ∈ Reg(A , M ). Then f is regular and there is g ∈ HomR (M, A ) such that f gf = f . By (14) we get g = φh and hence φ∗ (f )hφ ∗ (f ) = f φhf φ = f gf φ = f φ = φ∗ (f ), showing that φ∗ (f ) ∈ HomR (A, M ) is regular. By deﬁnition of Reg the claim is established if we can show that Sφ∗ (Reg(A , M ))T ⊆ φ∗ (Reg(A , M )). Since φ∗ is an S-homomorphism, we have Sφ∗ (Reg(A , M )) = φ∗ (S Reg(A , M )) ⊆ φ∗ (Reg(A , M )). Now let f ∈ Reg(A , M ) and t ∈ End(AR ). By (12) there is t ∈ End(AR ) such that φt = tφ. Then φ∗ (f ) · t = f φt = f t φ = φ∗ (f t ) which is regular because f t ∈ Reg(A , M ). Proposition 6.5. Let φ : A → A and assume that φ is surjective and that (13) holds. Then −1 (φ∗ ) [Reg(A, M )] ⊆ Reg(A , M ). Proof. Let f ∈ HomR (A , M ) such that φ∗ (f ) ∈ Reg(A, M ). Then φ∗ (f ) is regular and there is g ∈ HomR (M, A) such that φ∗ (f )gφ∗ (f ) = φ∗ (f ) or equivalently, f φgf φ = f φ. Since φ is surjective, we obtain that f (φg)f = f showing that f is regular. By deﬁnition of Reg the claim is established if we can show that −1 −1 S (φ∗ ) [Reg(A , M )]T ⊆ (φ∗ ) [Reg(A , M )]. Since φ∗ is an S-homomorphism, −1 for s ∈ S we have φ∗ (sf ) = sφ∗ (f ) ∈ Reg(A, M )), hence sf ∈ (φ∗ ) [Reg(A , M )]. Now let t ∈ End(AR ). By (13) there is t ∈ End(AR ) such that φt = t φ. Then φ∗ (f t ) = f t φ = f φt = φ∗ (f )t ∈ Reg(A, M ).

6. Inherited Regularity

29

We now study the case of a map φ : M → M more closely and check what hypotheses are needed in order to get results on the relationships of Reg(A, M ) and Reg(A, M ). We have μ : M → M

induces

μ∗ : HomR (A, M ) → HomR (A, M ).

(15)

The H := HomR (A, M ) is an S-T -bimodule where S := End(MR ) and T := End(AR ) while H := HomR (A, M ) is an S -T -bimodule where S := End(MR ) and T = End(AR ). Clearly, the map μ∗ is an T -module morphism. Suppose that

or

∀ s ∈ End(MR ), ∃ s ∈ End(MR ) such that s μ = μs

(16)

∀ s ∈ End(MR ), ∃ s ∈ End(MR ) such that s μ = μs.

(17)

Then in either case ∀ f ∈ HomR (A, M ) : s μ∗ (f ) = μ∗ (sf ). Proposition 6.6. Let μ : M → M and assume that ∀ g ∈ HomR (M, A), ∃ h ∈ HomR (M , A) such that g = hμ,

(18)

and that (16) holds. Then μ∗ (Reg(A, M )) ⊆ Reg(A, M ). Proof. Let f ∈ Reg(A, M ). Then f is regular and there is g ∈ HomR (M, A) such that f gf = f . By (18) we get μ∗ (f gf ) = μf gf = μf hμf = μf , showing that μ∗ (f ) = μf ∈ HomR (A, M ) is regular. By deﬁnition of Reg the claim is established if we can show that S μ∗ (Reg(A, M ))T ⊆ μ∗ (Reg(A, M )). Since μ∗ is a T homomorphism, we have μ∗ (Reg(A, M ))T = μ∗ (Reg(A, M )T ) ⊆ μ∗ (Reg(A, M )). Now let f ∈ Reg(A, M ) and s ∈ S = End(MR ). By (17) there is s ∈ End(MR ) such that s μ = μs. Then s · μ∗ (f ) = s μf = μsf = μ∗ (sf ) which is regular because sf ∈ Reg(A, M ). Proposition 6.7. Let μ : M → M and assume that μ is injective and that (16) holds. Then −1 (μ∗ ) [Reg(A, M )] ⊆ Reg(A, M ). Proof. Let f ∈ HomR (A, M ) such that μ∗ (f ) ∈ Reg(A, M ). Then μ∗ (f ) is regular and there is g ∈ HomR (M , A) such that μ∗ (f )gμ∗ (f ) = μ∗ (f ) or equivalently, μf gμf = μf . Since μ is injective, we obtain that f (gμ)f = f showing that f is regular. By deﬁnition of Reg the claim is established if we can show that −1 −1 S (μ∗ ) [Reg(A, M )]T ⊆ (μ∗ ) [Reg(A, M )]. Since μ∗ is a T -homomorphism, −1 for t ∈ T we have μ∗ (f t) = μ∗ (f )t ∈ Reg(A, M )), hence f t ∈ (μ∗ ) [Reg(A, M )]. Now let s ∈ End(MR ). By (17) there is s ∈ End(MR ) such that s μ = μs. Then μ∗ (sf ) = μsf = s μf = s μ∗ (f ) ∈ Reg(A, M ).

30

Chapter II. Regular Homomorphisms We now apply our results to the short exact sequence ι

φ

B A A/B where ι is the insertion and φ is the natural epimorphism. We obtain the exact sequence φ∗

ι∗

HomR (A/B, M ) HomR (A, M ) → HomR (B, M ). Hence we have • φ∗ (Reg(A/B, M )) ⊆ HomR (A, M ), • ι∗ (Reg(A, M )) ⊆ HomR (B, M ), −1

• (φ∗ )

∗ −1

• (ι )

[Reg(A, M )] ⊆ HomR (A/B, M ),

[Reg(B, M )] ⊆ HomR (A, M ). ι

φ

Theorem 6.8. Let B A A/B be given. 1) Suppose that ∀ t ∈ End((A/B)R ), ∃ t ∈ End(AR ) : t φ = φt. Then −1

(φ∗ )

[Reg(A, M )] ⊆ Reg(A/B, M ).

2) Suppose that ι∗ : HomR (M, B) → HomR (M, A) is surjective and ∀ t ∈ End(BR ), ∃ t ∈ End(AR ) : t ι = ιt. Then −1

(ι∗ ) ι

[Reg(B, M )] ⊆ Reg(A, M ).

μ

Theorem 6.9. Let N M M/N be given. 1) Suppose that μ∗ : HomR (M/N, A) → Hom(M, A) is surjective and N is fully invariant in M . Then μ∗ (Reg(A, M )) ⊆ Reg(A, M/N ). 2) Suppose that ι∗ : HomR (M, A) → HomR (N, A) is surjective and ∀ s ∈ End(NR ), ∃ s ∈ End(MR ) : s ι = ιs. Then ι∗ (Reg(A, N )) ⊆ Reg(A, M ). We next consider the crucial case when A and M are direct sums. Let A and M be right R-modules and suppose that A = A1 ⊕ A2 ,

M = M1 ⊕ M2 .

(19)

6. Inherited Regularity

31

It is a standard fact that HomR (A, M ) ∼ = HomR (A1 , M1 ) ⊕ HomR (A1 , M2 ) ⊕ HomR (A2 , M1 ) ⊕ HomR (A2 , M2 ). We will study the relationship between regular elements of HomR (A, M ) and those of its additive subgroups HomR (Ai , Mj ). It is convenient to identify HomR (A, M ) with a group of matrices as follows. Lemma 6.10. Suppose that A = A1 ⊕ A2 and M = M1 ⊕ M2 . Then HomR (A, M ) may be identiﬁed with the additive group of all matrices ξ ∈ HomR (A1 , M1 ), ξ21 ∈ HomR (A2 , M1 ), ξ11 ξ21 , 11 ξ12 ξ22 ξ12 ∈ HomR (A1 , M2 ), ξ22 ∈ HomR (A2 , M2 ). A matrix acts on an element a1 + a2 , where ai ∈ Ai , according to ordinary matrix multiplication, a1 ξ11 (a1 ) + ξ21 (a2 ) ξ11 ξ21 . = ξ12 ξ22 a2 ξ12 (a1 ) + ξ22 (a2 ) The mapping HomR (A, M )

ξ11 ξ12

ξ21 ξ22

→ ξij ∈ HomR (Ai , Mj )

is a surjective homomorphism of additive groups. Composition of functions is matrix multiplication. Proof. Well-known and straightforward to check.

We ﬁrst relate regular elements in HomR (Ai , Mj ) with special regular matrices in HomR (A, M ). Theorem 6.11 (Up and Down Theorem). Suppose that A = A1 ⊕ A2 and M M1 ⊕ M2 . Then the following statements are true. ξ11 0 ∈ HomR (A, M ) 1) ξ11 ∈ HomR (A1 , M1 ) is regular if and only if 0 0 regular. 0 ξ21 2) ξ21 ∈ HomR (A2 , M1 ) is regular if and only if ∈ HomR (A, M ) 0 0 regular. 0 0 3) ξ12 ∈ HomR (A1 , M2 ) is regular if and only if ∈ HomR (A, M ) ξ12 0 regular.

=

is

is

is

32

Chapter II. Regular Homomorphisms

4) ξ22 ∈ HomR (A2 , M2 ) is regular if and only if

0 0

0 ξ22

∈ HomR (A, M ) is

regular. Proof. 1) Observe that for η11 ∈ HomR (M1 , A1 ), ξ11 0 η11 0 ξ11 ξ11 η11 ξ11 = ξ11 ⇒ 0 0 0 0 0 Conversely, if

ξ11 0

0 0

η11 η12

η21 η22

ξ11 0

0 0

0 0

=

then by (23) also ξ11 η11 ξ11 = ξ11 . 2) By (32) 0 ξ21 η11 η21 0 0 ξ21 = 0 0 η12 η22 0 0 0

ξ11 0

=

0 0

ξ11 0

0 0

.

,

ξ21 η12 ξ21 0

.

(20)

η11 = Hence if ξ21 η12 ξ21 = ξ21 for some R (M2 , A1 ), then choosing η21 ∈ Hom 0 ξ21 0 ξ21 is regular. Conversely, if η21 = η22 = 0 in (20), we see that 0 0 0 0 0 ξ21 and it follows that ξ21 η12 ξ21 = is regular, then the matrix in (20) equals 0 0 ξ21 . 3) By (29) η11 η21 0 0 0 0 0 0 = . (21) η12 η22 ξ12 0 ξ12 η21 ξ12 0 ξ12 0 η11 = Hence if ξ12 η21 ξ12 = ξ12 for some R (M1 , A2 ), then choosing η12 ∈ Hom 0 0 0 0 is regular. Conversely, if η12 = η22 = 0 in (21), we see that ξ21 0 ξ12 0 0 0 and it follows that ξ12 η21 ξ12 = is regular, then the matrix in (21) equals ξ12 0 ξ12 . 4) By (38) η11 η21 0 0 0 0 0 0 = . (22) 0 ξ22 0 ξ22 η22 ξ22 η12 η22 0 ξ22 η11 Hence if ξ22 η22 ξ22 = ξ22 for some η22 ∈ Hom R (M2 , A2 ), then choosing 0 0 0 0 η21 = η12 = 0 in (22), we see that is regular. Conversely, if 0 ξ22 0 ξ22 0 0 and it follows that ξ22 η22 ξ22 is regular, then the matrix in (22) equals 0 ξ22 ξ22 .

=

=

6. Inherited Regularity

33

Corollary 6.12. Suppose that A = A1 ⊕ A2 and M = M1 ⊕ M2 . If Reg(A, M ) = HomR (A, M ), then also Reg(Ai , Mj ) = HomR (Ai , Mj ) for all i, j ∈ {1, 2}. We will see later that the converse of Corollary 6.12 is also true (Corollary 6.18) A simple example shows that the restriction of a regular map to a direct summand need not be regular. Example 6.13. Let A = Zu ⊕ Zv and M = Zu ⊕ Zv be free groups. Then also M = Z(u +2v )⊕Zv . Let f ∈ Hom(A, M ) be the map given by f (u) = 2(u +2v ) and f (v) = u + 2v . Then Im(f ) = Z(u + 2v ) is a free direct summand of M and hence Ker(f ) is also a summand of A, so f is regular. On the other hand f (Zu) = 2 Im(f ) is not a direct summand of M and therefore f Zu is not regular. We now study regularity in HomR (A1 ⊕ A2 , M1 ⊕ M2 ) more generally. The following theorem generalizes a result of Goodearl ([14, Theorem 1.7]) and our proof owes much to the proof of Goodearl’s Lemma 1.6. Theorem 6.14. Let A = A1 ⊕ A2 and M = M1 ⊕ M2 as before. We identify HomR (A, M ) with a matrix group as in Lemma 6.10. Furthermore, we use the identiﬁcations μ11 μ21 | μij ∈ HomR (Mi , Mj ) S := EndR (M ) = μ12 μ22 and

α11 α12

T := EndR (A) = Then

α21 α22

| αij ∈ HomR (Ai , Aj ) .

ξ21 ξ22

ξ11 ξ12

∈ Reg(A, M )

if and only if for all μjt ∈ HomR (Mj , Mt ) and for all αsi ∈ HomR (As , Ai ) μjt ξij αsi ∈ Reg(As , Mt ). Proof. Suppose that T -bimodule. Hence μ11 0 ξ11 ξ12 0 0

ξ11 ξ12 ξ21 ξ22

ξ21 ξ22

∈ Reg(A, M ). Recall that Reg(A, M ) is an S-

α11 0

0 0

=

μ11 ξ11 α11 0

0 0

∈ Reg(A, M ).

By Theorem 6.11 it follows that μ11 ξ11 α11 ∈ HomR (A1 , M1 ) is regular. Similarly, 0 μ21 ξ11 ξ21 μ21 ξ22 α12 0 0 0 = ∈ Reg(A, M ). 0 0 ξ12 ξ22 0 0 α12 0

34

Chapter II. Regular Homomorphisms

By Theorem 6.11 it follows that μ21 ξ22 α12 ∈ HomR (A1 , M1 ) is regular. Similarly, 0 0 0 0 ξ11 ξ21 α11 0 ∈ Reg(A, M ). = 0 μ22 ξ12 ξ22 0 0 μ22 ξ12 α11 0 By Theorem 6.11 it follows that μ22 ξ12 α11 ∈ HomR (A1 , M2 ) is regular. The remaining cases work mutatis mutandis (see the Appendix to this section for the relevant matrix identities). We have shown that μ(μjt ξij αsi )α = (μμjt )ξij (αsi α) is regular for all μ ∈ End(Mt ) and all α ∈ End(As ). It is left to show that sums of such elements are (k) (k) regular. Consider k μjt ξij αsi . Using suitable formulas in the appendix we have matrices (k) (k) 0 μjt ξij αsi ∈ Reg(A, M ) 0 0 where the nonzero entry is in position (t, s) in the matrix (the displayed case is s = 2, t = 1). Hence (k) (k) (k) (k) 0 μjt ξij αsi 0 μjt ξij αsi k = ∈ Reg(A, M ); k 0 0 0 0 (k) (k) it follows that k μjt ξij αsi ∈ Reg(As , Mt ). Hence End(Mt )(μjt ξij αsi ) End(As ) is regular and therefore μjtξij αsi ∈ Reg(A s , Mt ). ξ11 ξ21 Conversely, let ξ = ∈ HomR (A, M ) and assume that for all ξ12 ξ22 μjt ∈ HomR (Mj , Mt ) and for all αsi ∈ HomR (As , Ai ), μjt ξij αsi ∈ Reg(As , Mt ). Assume ﬁrst that ξ12 = 0. By assumption there is η11 ∈ HomR (M1 , A1 ) such that ξ11 η11 ξ11 = ξ11 and there is η22 ∈ HomR (M2 , A2 ) such that ξ22 η22 ξ22 = ξ22 . Then η11 0 ξ11 ξ21 η11 0 ξ11 ξ21 ξ ξ = 0 ξ22 0 ξ22 0 η22 0 η22 ξ11 η11 ξ11 ξ11 η11 ξ21 + ξ21 η22 ξ22 = 0 ξ22 η22 ξ22 0 ξ11 η11 ξ21 + ξ21 η22 ξ22 − ξ21 = ξ+ 0 0 0 ξ21 = ξ+ 0 0 = ξ11 η11 ξ21 + ξ21 η22 ξ22 − ξ21 ∈ Reg(A2 , M1 ) by hypothesis. Hence there where ξ21 is η12 ∈ HomR (M1 , A2 ) such that ξ21 η12 ξ21 = ξ21 .

6. Inherited Regularity Then

η11 ξ 0 = ξ+ ξ = ξ+ξ

0 η22 η11 0 η11 0

35

0 ξ21 0 0 0 ξ21 0 ξ21 =ξ+ ξ=ξ+ 0 0 0 0 η12 0 0 0 η11 0 0 0 0 ξ ξ−ξ ξ−ξ η22 0 η22 η12 0 0 0 0 η11 0 ξ−1 ξ − 1 ξ. η22 0 η22 η12 0

Using that ξXξ = ξ + ξY ξ if and only if ξ(X − Y )ξ = ξ, this shows that ξ is regular. ξ11 ξ21 Now consider a general element ξ = ∈ HomR (A, M ) such that ξ12 ξ22 for all μjt ∈ HomR (Mj , Mt ) and for all αsi ∈ HomR (As , Ai ) it is true that μjt ξij αsi ∈ Reg(As , Mt ). Choose η21 ∈ HomR (M2 , A1 ) such that ξ12 η21 ξ12 = ξ12 . Then 0 η21 ξ11 ξ21 ξ11 ξ21 0 η21 ξ ξ = 0 0 ξ12 ξ22 0 0 ξ12 ξ22 ξ11 ξ21 0 ξ11 η21 = 0 ξ12 η21 ξ12 ξ22 ξ11 η21 ξ12 ξ11 η21 ξ22 ξ11 η21 ξ12 ξ11 η21 ξ22 = . = ξ12 η21 ξ12 ξ12 η21 ξ22 ξ12 ξ12 η21 ξ22

Hence ξ

0 η21 0 0

ξ−ξ =

ξ11 η21 ξ12 − ξ11 0

ξ11 η21 ξ22 − ξ21 ξ12 η21 ξ22 − ξ22

which is regular by the special case treated ﬁrst. It follows that ξ is regular. We need to show further that every element in End(M )ξ End(A) is regular. Let α11 α21 μ11 μ21 α := ∈ End(A) and μ := ∈ End(M ). α12 α22 μ12 μ22 Then

μξα

⎡

⎤ μ11 ξ11 α11 + μ11 ξ21 α12 + μ21 ξ12 α11 + μ21 ξ22 α12 ⎢ μ11 ξ11 α21 + μ11 ξ21 α22 + μ21 ξ12 α21 + μ21 ξ22 α22 ⎥ ⎥. = ⎢ ⎣ μ12 ξ11 α11 + μ12 ξ21 α12 + μ22 ξ12 α11 + μ22 α12 ξ22 ⎦ μ12 ξ11 α21 + μ12 ξ21 α22 + μ22 ξ12 α21 + μ22 ξ22 α22

∈ HomR (At , Ai ) we have Then for μjs ∈ HomR (Mj , Ms ) and αti μ1s (μ11 ξ11 α11 + μ11 ξ21 α12 + μ21 ξ12 α11 + μ21 ξ22 α12 ) αt1 = (μ1s μ11 )ξ11 (α11 αt1 ) + (μ1s μ11 )ξ21 (α12 αt1 ) ) + (μ1s μ21 )ξ12 (α11 αt1 ) + (μ1s μ21 )ξ22 (α12 αt1 ∈ Reg(At , Ms ),

36

Chapter II. Regular Homomorphisms μ1s (μ11 ξ11 α21 + μ11 ξ21 α22 + μ21 ξ12 α21 + μ21 ξ22 α22 ) αt2 = (μ1s μ11 )ξ11 (α21 αt2 ) + (μ1s μ11 )ξ21 (α22 αt2 ) ) + (μ1s μ21 )ξ12 (α21 αt2 ) + (μ1s μ21 )ξ22 (α22 αt2 ∈ Reg(At , Ms ), μ2s (μ12 ξ11 α11 + μ12 ξ21 α12 + μ22 ξ12 α11 + μ22 ξ22 α12 ) αt1 = (μ2s μ12 )ξ11 (α11 αt1 ) + (μ2s μ12 )ξ21 (α12 αt1 ) ) + (μ2s μ22 )ξ12 (α11 αt1 ) + (μ2s μ22 )ξ22 (α12 αt1 ∈ Reg(At , Ms ),

μ2s (μ12 ξ11 α21 + μ12 ξ21 α22 + μ22 ξ12 α21 + μ22 ξ22 α22 ) αt2 = (μ2s μ12 )ξ11 (α21 αt2 ) + (μ2s μ12 )ξ21 (α22 αt2 ) ) + (μ2s μ22 )ξ12 (α21 αt2 ) + (μ2s μ22 )ξ22 (α22 αt2 ∈ Reg(At , Ms ).

By what has been shown, μξα is regular in HomR (A, M ). Finally, we must show that the sums k μ(k) ξα(k) are regular. The entry in position (s, t) of a matrix (k) (k) μ(k) ξα(k) are sums of elements of the form μjt ξij αsi and so are the entries of the (k) (k) matrix k μ ξα . These entries are in Reg(As , Mt ) and hence k μ(k) ξα(k) ∈ Reg(A, M ). This completes the proof that ξ ∈ Reg(A, M ). Using the characterization of regular maps we have a useful consequence. Corollary 6.15. Let A = A1 ⊕ A2 and M = M1 ⊕ M2 as before. Suppose that ξ11 ξ21 ∈ Reg(A, M ). ξ12 ξ22 Then for all μjt ∈ HomR (Mj , Mt ) and for all αsi ∈ HomR (As , Ai ), Ker(μjt ξij αsi ) ⊆⊕ As

and

Im(μjt ξij αsi ) ⊆⊕ Mt .

We will use later the following special case. Corollary 6.16. Let A and M be modules such that A = A1 ⊕A2 and M = M1 ⊕M2 and no nonzero summand of A2 is isomorphic to a summand of M1 and no nonzero summand of A1 is isomorphic to a summand of M2 . Suppose that ξ11 ξ21 ∈ Reg(A, M ). ξ12 ξ22 Then ξ21 = 0, ξ12 = 0, and for every μ21 ∈ Hom(M2 , M1 ), every α21 ∈ Hom(A2 , A1 ), every μ12 ∈ Hom(M1 , M2 ), and every α12 ∈ Hom(A1 , A2 ), μ21 ξ22 = 0,

ξ11 α21 = 0,

μ12 ξ11 = 0,

ξ22 α12 = 0,

6. Inherited Regularity

37

and μ21 ξ22 α12 ∈ Reg(A1 , M1 ),

μ12 ξ11 α21 ∈ Reg(A2 , M2 ).

Proof. By hypothesis and Corollary 4.4, Reg(A1 , M2 ) = 0 and Reg(A 2 , M1 ) = 0. μ11 μ21 ∈ By Theorem 6.14, ξ21 = 0 and ξ12 = 0. We also have that for all μ μ22 12 α11 α21 ∈ End(A), End(M ) and all α12 α22 μ11 ξ11 α11 + μ21 ξ22 α12 μ11 ξ11 α21 + μ21 ξ22 α22 μ12 ξ11 α11 + μ22 ξ22 α12 μ12 ξ11 α21 + μ22 ξ22 α22 ξ11 0 α11 α21 μ11 μ21 ∈ Reg(A, M ). = 0 ξ22 μ12 μ22 α12 α22 By Theorem 6.14 it follows that μ11 ξ11 α21 +μ21 ξ22 α22 ∈ Reg(A2 , M1 ) = 0. Choosing μ11 = 0 and α22 = 1, we ﬁnd that μ21 ξ22 = 0. The analogous claims follow similarly. Finally, by Theorem 6.14 we get μ11 ξ11 α11 ∈ Reg(A1 , M1 ) and μ12 ξ11 α21 ∈ Reg(A2 , M2 ). Theorem 6.14 can be extended to arbitrary ﬁnite sums by induction. Theorem 6.17. Let A = A1 ⊕ · · · ⊕ Am and ⎡ ξ11 · · · ξm1 ⎢ .. .. .. ξ=⎣ . . . ξ1n

M = M1 ⊕ · · · ⊕ Mn . Then ⎤ ⎥ ⎦ ∈ Reg(A, M )

· · · ξmn

if and only if for all μjt ∈ HomR (Mj , Mt ) and for all αsi ∈ HomR (As , Ai ), μjt ξij αsi ∈ Reg(As , Mt ). Proof. Set A1 := A1 , A2 := A2 ⊕ · · · ⊕ Am , M1 := M1 , and M2 := M2 ⊕ · · · ⊕ Mn . Then A = A1 ⊕ A2 and M = M1 ⊕ Mn . Then η11 η21 ξ= η12 η22 where ⎡

η11 = ξ11 , η21 =

ξ21

· · · ξn1

, η12

⎡ ⎤ ξ12 ξ22 ⎢ .. ⎥ ⎢ .. = ⎣ . ⎦ , η22 = ⎣ . ξ1m ξ2m

⎤ ξn2 .. ⎥ . . ⎦ · · · ξnm ··· .. .

We now wish to apply induction using Theorem 6.14. To do so we must consider homomorphisms induced by ξ on the subsums A1 and A2 of A to the subsums M1 and M2 of M . All such homomorphisms are matrices with entries ξij ∈ HomR (Ai , Mj ).

38

Chapter II. Regular Homomorphisms

We must then check whether the conditions of Theorem 6.14 prevail. To do so we must apply the endomorphisms of the subsums. All such endomorphisms are matrices with entries μij ∈ HomR (Mi , Mj ) and αij ∈ HomR (Ai , Aj ) for various sets of subscripts. Looking at products μηα we see that their entries are of the form (i,j) μjs ξij αti ∈ HomR (At , Ms ). In one direction we have that each summand is in some Reg, hence also the sum, and in the other direction we know that the sum is in some Reg for all choices of the μjs and αti and choosing all but one of the μ and all but one of the α to be 0, we get the necessity of the conditions μjs ξij αti ∈ Reg(At , Ms ). Corollary 6.18. Let A = A1 ⊕· · ·⊕Am and M = M1 ⊕· · ·⊕Mn . Then HomR (A, M ) is regular, i.e., Reg(A, M ) = HomR (A, M ), if and only if ∀ i, j : Hom(Ai , Mj ) = Reg(Ai , Mj ). Corollary 6.19. Let A = A1 ⊕ · · · ⊕ Am , M = M1 ⊕ · · · ⊕ Mn , and N be R-modules such that Ai ∼ =N ∼ = Mj for all i and j. Then the following statements hold. 1) Reg(A, M ) = 0 if and only if Reg(N, N ) = 0. 2) Reg(A, M ) = HomR (A, M ) if and only if Reg(N, N ) = End(NR ). 3) End(N k ) is regular if and only if End(N ) is regular. Proof. Select isomorphisms σi : Ai → N and τj : Mj → N . 1) Suppose that Reg(N, N ) = 0 and let 0 = η ∈ Reg(N, N ). Then τ1−1 ησ1 ∈ HomR (A1 , M1 ) and τ1−1 ησ1 = 0. Let ⎡ ⎢ ⎢ ξ := ⎢ ⎣

τ1−1 ησ1 0 .. .

0 0 .. .

0

0

⎤ ··· 0 ··· 0 ⎥ ⎥ .. ⎥ ∈ HomR (A, M ). ··· . ⎦ ··· 0

By Theorem 6.17, 0 = ξ ∈ Reg(A, M ) if and only if for all αi1 ∈ HomR (Ai , A1 ) and all μ1j it is true that μ1j (τ1−1 ησ1 )αi1 ∈ Reg(Ai , Mj ). But τj (μ1j τ1−1 ησ1 αi1 )σi−1 = (τj μ1j τ1−1 )η(σ1 αi1 σi−1 ) ∈ (End N )η(End N ) ⊆ Reg(N, N ) and it follows that μ1j (τ1−1 ησ1 )αi1 ∈ Reg(Ai , Mj ). Conversely, suppose that Reg(A, M ) = 0 and let ⎡

ξ11 ⎢ .. 0 = ξ = ⎣ . ξ1n

⎤ ξm1 .. ⎥ ∈ Reg(A, M ). . ⎦ · · · ξmn ··· .. .

Then there is an entry ξij = 0 in position (j, i). Let Ej1 ∈ End(M ) be the n × nmatrix with entry τ1 τj−1 in position (1, j) and 0 everywhere else, and let E1i ∈

7. Appendix: Various Formulas

39

End(A) be the m × m matrix with entry σi−1 σ1 in position (i, 1) and 0 everywhere else. Then ⎤ ⎡ τ1 τj−1 ξij σi σ1−1 · · · 0 ⎢ .. .. .. ⎥ ∈ Reg(A, M ). 0 = Ej1 ξE1i = ⎣ . . . ⎦ 0 ··· 0 It is now clear that 0 = τ1 τi−1 ξji σj σ1−1 ∈ Reg(A1 , M1 ). So Reg(A1 , M1 ) = 0, and hence also Reg(N, N ) = 0. 2) is an immediate consequence of Theorem 6.17 and 3) is a special case of 2). Corollary 6.20. The matrix ring Mn (R) is regular if and only if End(RR ) is regular.

7 Appendix: Various Formulas =

In general,

μ21 μ22

μ11 μ12

μ21 μ22

ξ11 ξ12

ξ21 ξ22

α11 α12

ξ11 α11 + ξ21 α12 ξ12 α11 + ξ22 α12

α21 α22

ξ11 α21 + ξ21 α22 ξ12 α21 + α22 ξ22

⎤ μ11 ξ11 α11 + μ11 ξ21 α12 + μ21 ξ12 α11 + μ21 ξ22 α12 ⎢ μ11 ξ11 α21 + μ11 ξ21 α22 + μ21 ξ12 α21 + μ21 ξ22 α22 ⎥ ⎥. ⎢ ⎦ ⎣ μ12 ξ11 α11 + μ12 ξ21 α12 + μ22 ξ12 α11 + μ22 ξ22 α12 μ12 ξ11 α21 + μ12 ξ21 α22 + μ22 ξ12 α21 + μ22 ξ22 α22 ⎡

=

μ11 μ12

[μij ][ξij ][αij ] = [ (i,j) μjs ξij αti ].

There are 16 possibilities for the maps μjt ξij αsi namely: ⎫ ⎫ μ11 ξ11 α11 ⎪ μ12 ξ11 α11 ⎪ ⎪ ⎪ ⎬ ⎬ μ21 ξ12 α11 μ22 ξ12 α11 ∈ HomR (A1 , M1 ), ∈ HomR (A1 , M2 ), μ11 ξ21 α12 ⎪ μ12 ξ21 α12 ⎪ ⎪ ⎪ ⎭ ⎭ μ21 ξ22 α12 μ22 ξ22 α12 ⎫ ⎫ μ12 ξ11 α21 ⎪ μ11 ξ11 α21 ⎪ ⎪ ⎪ ⎬ ⎬ μ21 ξ12 α21 μ22 ξ12 α21 ∈ HomR (A2 , M1 ), ∈ HomR (A2 , M2 ), μ11 ξ21 α22 ⎪ μ12 ξ21 α22 ⎪ ⎪ ⎪ ⎭ ⎭ μ21 ξ22 α22 μ22 ξ22 α22 and there are 16 corresponding matrix equations:

40

Chapter II. Regular Homomorphisms

μ11 ξ11 α11 0

0 0

μ21 ξ12 α11 0

0 0

μ11 ξ21 α12 0

0 0

μ21 ξ22 α12 0

0 0

0 μ12 ξ11 α11

0 0

0 μ22 ξ12 α11

0 0

0 μ12 ξ21 α12

0 0

0 μ22 ξ22 α12

0 0

0 0

μ11 ξ11 α21 0

0 μ21 ξ12 α21 0 0 0 μ11 ξ21 α22 0 0 0 0

μ22 ξ22 α22 0

0 0

0 μ12 ξ11 α21

0 0 0 μ22 ξ12 α21 0 0 0 μ12 ξ21 α22 0 0

0 μ22 ξ22 α22

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

μ11 0

0 0

0 μ21 0 0 μ11 0

0 0

0 μ21 0 0 0 μ12 0 0 0 μ12 0 0 μ11 0 0 0 μ11 0 0 0 0 μ12 0 0 0 μ12 0 0

0 0 0 μ22 0 0 0 μ22 0 0 μ21 0 0 0 μ21 0 0 0 0 μ22 0 0 0 μ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

ξ11 ξ12

ξ21 ξ22

α11 0

0 0

α11 0

0 0

0 α12

0 0

0 α12

0 0

α11 0

0 0

α11 0

0 0

0 α12

0 0

0 α12

0 0

0 α21 0 0 0 α21 0 0 0 0

0 α22

0 0

0 α22

0 α21 0 0 0 α21 0 0 0 0

0 α22

0 0

0 α22

(23) (24) (25) (26) (27) (28) (29) (30) (31) (32) (33) (34) (35) (36) (37) (38)

Chapter III

Indecomposable Modules 1

Reg(A, M ) = 0

A module M is directly indecomposable (or simply indecomposable) if and only if 0 and M are the only direct summands of M . This means that 0 and 1 are the only idempotents in End(MR ). We now study the situation that Reg(A, M ) = 0 and one of the modules A or M is indecomposable. It turns out that much can be said under assumptions weaker than Reg(A, M ) = 0. Theorem 1.1. 1) Suppose that there is 0 = f ∈ HomR (A, M ) that is regular and μf is regular for all μ ∈ S := End(MR ). If M is directly indecomposable, then S is a division ring. 2) Suppose that there is 0 = f ∈ HomR (A, M ) that is regular and f α is regular for all α ∈ T := End(AR ). If A is directly indecomposable, then T is a division ring. The proof of Theorem 1.1 requires a well-known lemma on division rings. For the sake of completeness we include it with proof. Lemma 1.2. Let R be a ring with 1 ∈ R and assume that every nonzero element of R has a right inverse. Then R is a division ring. The same is true for left inverses. Proof. Let 0 = r ∈ R and suppose that rs = 1 and st = 1. Then t = 1t = rst = r1 = r. Thus s is both right and left inverse of r and r is invertible. Proof of Theorem 1.1. 1) Let M be directly indecomposable, and assume that 0 = f ∈ HomR (A, M ) is regular. Hence Im(f ) ⊆⊕ M and M being indecomposable, it follows that Im(f ) = M . Let 0 = μ ∈ S. Then μf (A) = μM = 0 and therefore μf = 0. By assumption also μf is regular. Hence there exists g ∈ HomR (M, A) such that μf gμf = μf and then e := μf g = e2 = 0. Since M is indecomposable,

42

Chapter III. Indecomposable Modules

its only nonzero idempotent is 1, so e = 1 and μf g = μ(f g) = 1 with f g ∈ S, i.e., every 0 = μ ∈ S possesses a right inverse. By Lemma 1.2 it follows that S is a division ring. 2) Let A be indecomposable, and assume that 0 = f ∈ HomR (A, M ) is regular. Then Ker(f ) ⊆⊕ A and since f = 0 and A is indecomposable, Ker(f ) = 0. Let 0 = α ∈ T . Then there is a ∈ A such that α(a) = 0 and it follows that f (α(a)) = 0, so f α = 0. Since f α is regular by hypothesis, there exists g ∈ HomR (M, A) such that f αgf α = f α. Hence gf α is a nonzero idempotent of T and we get gf α = 1 and gf ∈ T is a left inverse of α. By Lemma 1.2 it follows that T is a division ring. If M is indecomposable and if S := End(MR ) is not a division ring then Reg(A, M ) = 0 for every A ∈ Mod-R. Similarly, if A is indecomposable and T := End(AR ) is not a division ring then Reg(A, M ) = 0 for every M ∈ Mod-R. An example of a module M that is indecomposable but End(MR ) is not a division ring is M = Z/p2 Z where p is a prime number. Simple modules are examples of modules whose endomorphism rings are division rings. This is Schur’s Lemma. Example 1.3. The abelian groups whose endomorphism rings are division rings are the groups Q with End(Q) = Q and the cyclic groups A of order p with End(A) = Z/pZ. Theorem 1.4. Let R be a commutative ring and MR a faithful R-module. Suppose that End(MR ) is a division ring. Then R is an integral domain, M = Q(R) and End(MR ) = Q(R) where Q(R) is the ﬁeld of quotients of R. Proof. Since R is commutative and M is faithful, R ⊆ S := End(MR ). Hence R is an integral domain and Q(R) ⊆ S. Hence M is a Q(R)-vector space. Let Q(R)m be a one-dimensional subspace of M . Then Q(R)m is an R-submodule of M and a direct summand. Since M is indecomposable, M = Q(R)m ∼ = Q(R) and the result follows.

2

Structure Theorems

Suppose that 0 = f ∈ Reg(A, M ) and M is indecomposable. Then A = Ker(f ) ⊕ M where M ∼ = M and End(MR ) is a division ring. Furthermore, for every s ∈ S and t ∈ T , the map sf t is regular (also sums of such). This being so, the module A must have a special structure. Similarly, if A is indecomposable and Reg(A, M ) = 0, then M must have a special structure. We ﬁrst settle the very special case when both M and A are indecomposable. Proposition 2.1. Suppose that A and M are indecomposable modules. 1) If 0 = f ∈ HomR (A, M ) is regular, then it is bijective and A ∼ = M.

2. Structure Theorems

43

2) If Reg(A, M ) = 0, then A ∼ = End(MR ) are division rings. = M and End(AR ) ∼ If so, then Reg(A, M ) = HomR (A, M ). Proof. 1) We know that A = Ker(f ) ⊕ Im(gf ) and M = Im(f ) ⊕ Ker(f g). As f = 0 and A and M are indecomposable, we have Ker(f ) = 0 and M = Im(f ). Hence f is bijective. 2) By 1) A ∼ = M and End(AR ) ∼ = End(MR ) are division rings by Theorem 1.1. If so every nonzero element of HomR (A, M ) is a bijection and hence every map in HomR (A, M ) is regular. The following lemma contains the essential step to the structure theorem. Lemma 2.2. Suppose that 0 = f ∈ Reg(A, M ), and M is indecomposable. Then End(MR ) is a division ring and A = Ker(f )⊕M where M ∼ = M . Suppose further that HomR (Ker(f ), M ) = 0. Then there is a decomposition Ker(f ) = K ⊕M with M ∼ = M. Proof. By Theorem 1.1 it follows that End(MR ) is a division ring. In particular f itself is regular and therefore we have a decomposition A = Ker(f ) ⊕ M where M ∼ = Im(f ) ⊆⊕ M . As f = 0 and M is indecomposable, we have Im(f ) = M . Suppose now that HomR (Ker(f ), M ) = 0. Then also HomR (Ker(f ), M ) = 0. Let 0 = ϕ ∈ HomR (Ker(f ), M ), let π : A → Ker(f ) be the projection along M , and let ι : M → A be the insertion. Then ιϕπ ∈ End(AR ) and f ιϕπ is a nonzero regular map in HomR (A, M ). Hence Im(f ιϕπ) = M and Ker(f ιϕπ) = Ker(ϕ) ⊕ M ⊆⊕ A. Then Ker(ϕ) ⊆⊕ Ker(f ) and Ker(f ) = Ker(ϕ) ⊕ M where M ∼ =

Ker(f ) Ker(ϕ)

∼ =

Ker(f )⊕M Ker(ϕ)⊕M

∼ =

A Ker(φ)⊕M

=

A Ker(f ιϕπ)

∼ = Im(f ιϕπ) ∼ = M.

Corollary 2.3. Suppose that 0 = f ∈ Reg(A, M ), and M is indecomposable. Then End(MR ) is a division ring and either A = K ⊕M1 ⊕· · ·⊕Mn where Mi ∼ = M , and HomR (K, M ) = 0, or for every i ∈ N there is a decomposition A = Ki ⊕M1 ⊕· · ·⊕ ∞ ∞ Mi with Mi ∼ = M and Ki = Ki+1 ⊕ Mi+1 . In the latter case i=1 Mi = i=1 Mi ∞ ∞ and there is a homomorphism ρ : A → i=1 Mi such that M ⊆ ρ(A) and i i=1 ∞ Ker(ρ) = i=1 Ki . Proof. The decompositions are obtained by repeated applications of Lemma 2.2 as follows. Since f is regular we have A = K1 ⊕ M1 where K1 = Ker(f ) and M1 ∼ = M . Suppose that Hom(K1 , M ) = 0. Then by Lemma 2.2 we have A = K2 ⊕ M2 ⊕ M1 , where K1 = K2 ⊕ M2 and M2 ∼ = M . Suppose that inductively, A = Kn ⊕ Mn ⊕ · · · ⊕ M2 ⊕ M1 where Kn ⊕ Mn = Kn−1 ⊆ Ker(f ), Mi ∼ = M , and f is injective on M1 . If HomR (Kn , M ) = 0, then the theorem is proved. So suppose that HomR (Kn , M ) = 0. Then HomR (Kn , M1 ) = 0. Let 0 = φ ∈ HomR (Kn , M1 ). Then ιφπ ∈ End(AR ) where π : A → M1 is the projection along K1 = Ker(f ) and ι : M1 → A is the insertion. Then f := f ιφπ : A → M is regular and f (M1 ) = f ιφπ(M1 ) = f φ(M1 ) = 0 because 0 = φ(M1 ) ⊆ M1 and f is injective on M1 . It follows that Im(f ) = M and Ker(f ) = Mn−1 ⊕ · · · ⊕ M2 ⊕ Ker(φ) ⊆⊕ A

44

Chapter III. Indecomposable Modules

and hence Kn+1 : = Ker(φ) ⊆⊕ Kn , so Kn = Kn+1 ⊕ Mn+1 for some submodule Mn+1 of A. Thus we have A = Kn+1 ⊕ Mn+1 ⊕ · · · ⊕ M2 ⊕ M1 and ﬁnally Mn+1 ∼ = A A ∼ = ) = M . Im(f = Kn+1 ⊕Mn ⊕···⊕M1 Ker(f ) This settles the ﬁrst casein which the process terminates. In the second case ∞ it is easy to see that the sum i=1 Mi is direct. The map ρ is obtained as follows. Let a ∈ A and write a = ki + m1 + · · · + mk according to the decomposition A = Ki ⊕ M1 ⊕ · · · ⊕ Mk the components m1 , . . . , mi , i ≤ k do not depend on the choice of k. This is because Ki = Ki+1 ⊕ Mi+1 . We therefore obtain a well-deﬁned map ∞ ρ(a) = (m , . . . , m , . . .) that is obviously homomorphic. Clearly, Ker(ρ) = 1 i i=1 Ki and ∞ i=1 Mi ⊆ Im(ρ). The converse of Corollary 2.3 need not hold. Example 2.4. There exist abelian groups A = A1 ⊕ A2 and M such that M ∼ = A2 are indecomposable groups whose endomorphism rings are division rings, and Hom(A1 , A2 ) = 0 but Reg(A, M ) = 0. Proof. Let A1 ∼ = Z(p∞ ), A2 ∼ = Q, M ∼ = Q, and set A := A1 ⊕ A2 . Then ∼ ∼ M = A2 = Q are indecomposable groups whose endomorphism ring is a division ring, namely Q. Also Hom(A1 , A2 ) = Hom(Z(p∞ ), Q) = 0 but Hom(A2 , A1 ) = Hom(Q, Z(p∞ )) contains an epimorphism φ. For the purpose of showing that Reg(A, M ) = 0, suppose that ξ = [ξ11 , ξ21 ] ∈ Reg(A, M ), ξ11 ∈ Hom(A1 , M ), ξ2 1 ∈ Hom(A2 , M ) ∼ = Hom(Q, Q) (where ξ11 = 0 actually). By Corollary II.6.15 we obtain that Ker(φξ21 ) ⊆⊕ M ∼ = Q. Suppose that ξ21 = 0. Then it is an isomorphism and φξ21 = 0. As M ∼ = Q is indecomposable, the kernel Ker(φξ21 ) is either 0 or M , neither of which is possible. We conclude that ξ = 0. By way of converse to Corollary 2.3 we are left with the hope that Reg(A, M ) = 0 if A = M1 ⊕ · · · ⊕ Mn , Mi ∼ = M , and End(MR ) is a division ring. The answer is yes. Theorem 2.5. Let M0 be an R-module whose endomorphism ring D := EndR (M0 ) is a division ring. Let A = A1 ⊕ · · · ⊕ Am and M = M1 ⊕ · · · ⊕ Mn with Ai ∼ = Mj ∼ = M0 . Then Reg(A, M ) = HomR (A, M ). Proof. We observe that D ∼ = HomR (M, M ) ∼ = HomR (Ai , Mj ) ∼ = EndR (Ai ) ∼ = EndR (Mj ) and all these groups are regular as in any of these all nonzero maps are isomorphisms. Now HomR (A, M ), End(AR ) and End(MR ) may be considered groups and rings of matrices with entries αsi ∈ Hom(As , Ai ), ξij ∈ HomR (Ai , Mj ), and μjt ∈ Hom(Mj , Mt ). Every map μjt ξij αsi ∈ HomR (As , Mt ) is regular hence every map in HomR (A, M ) is regular by Theorem II.6.17. We now consider the case that A is indecomposable and there is a certain abundance of regular maps.

2. Structure Theorems

45

Lemma 2.6. Suppose that 0 = f ∈ Hom(A, M ), for every μ ∈ End(MR ) the composite μf ∈ HomR (A, M ) is regular, and A is indecomposable. Then End(AR ) is a division ring and M = A ⊕L with A ∼ = A. Suppose further that HomR (A, L) = 0. Then there is a decomposition L = A ⊕ L with A ∼ = A. Proof. By Theorem 1.1 we have that End(AR ) is a division ring. In particular f itself is regular and therefore we have A = Ker(f ) ⊆⊕ A and since A is indecomposable Ker(f ) = 0. Also A ∼ = Im(f ) ⊆⊕ M , hence M = A ⊕ L for some L and (A, L) = 0, also HomR (A , L) = 0. Let 0 = ψ ∈ HomR (A , L) A ∼ A. As Hom = R and let π : M → A be the projection along L and let ι : L → M be the insertion. Then ιψπ ∈ End(MR ) and ιψπf is a regular map in HomR (A, M ). Note that ιψπf A = ιψπA = ιψA = ψA = 0. Hence ιψπf = 0 and therefore injective because A is indecomposable. We have further that A := (ιψπf )(A) ⊆⊕ M and as A ⊆ L, hence we get L = A ⊕ L for some L with A ∼ = A. Corollary 2.7. Suppose that 0 = f ∈ Reg(A, M ) is regular, and A is indecomposable. Then End(AR ) is a division ring, and either M = A1 ⊕ · · · ⊕ An ⊕ L where Ai ∼ = A and HomR (A, L) = 0, or for every i ∈ N there exist decompositions ⊕ · · · ⊕ Ai ⊕ Li where Ai ∼ M = A = A and Li = Ai+1 ⊕ Li+1 .In the latter case 1 ∞ ∞ ∞ A = A and there is a i i i=1 ∞ homomorphism ρ : M → i=1 Ai such that i=1 ∞ i=1 Ai ⊆ ρ(M ) and Ker(ρ) = i=1 Li . Proof. Lemma 2.6 establishes that End(AR ) is a division ring, and M = A1 ⊕ L1 for some A1 ∼ = A and some L1 . If HomR (A, L1 ) = 0, then we have the claimed decomposition of M . If HomR (A, L1 ) = 0, then Lemma 2.6 tells us that L1 = A2 ⊕ L2 for some A2 ∼ = A and some L2 . We assume by induction that M = A1 ⊕ · · · ⊕ Ai ⊕ Li with Ai ∼ = A. If HomR (A, Li ) = 0, then Lemma 2.6 tells us that Li = Ai+1 ⊕ Li+1 for some Ai+1 ∼ = A and some Li+1 . This shows that we have the two claimed possibilities. If the process does terminate, then we obtain an not ∞ ∞ . .. In this case i=1 Ai = i=1 Ai because uniqueness of the inﬁnite set A1 , A2 , . ∞ ∞ representation x = i=1 xi ∈ i=1 Ai has ﬁnite character. Suppose that M = A1 ⊕ · · · ⊕ Ai ⊕ Li where Ai ∼ = A and Li = Ai+1 ⊕ Li+1 for all i ∈ N. The map ρ is obtained as follows. Let m ∈ M and write m = a1 + · · · + ak + k according to the decomposition M = A1 ⊕ · · · ⊕ Ak ⊕ Lk , then the components m1 , . . . , mi , i ≤ k, do not depend on the choice of k. This is because Li = Ai+1 ⊕ Li+1 . We therefore obtain a well-deﬁned map ∞ ρ(m) = , . . . , a , . . .) that is obviously homomorphic. Clearly, Ker(ρ) = (a 1 i i=1 Li and ∞ A ⊆ Im(ρ). i i=1 The converse of Corollary 2.7 need not hold. Example 2.8. There exists an indecomposable abelian group A whose endomorphism ring is a division ring and there exists a group M = A ⊕L such that A ∼ =A and Hom(A, L) = 0, yet Reg(A, M ) = 0.

46

Chapter III. Indecomposable Modules

∼ Proof. Let A ∼ = Q, L ∼ = Z, and M = A ⊕L. Then = Q, A ∼ End(A) = Q is a ξ11 ∈ Reg(A, M ), ξ11 ∈ division ring and Hom(A , L) = 0. Suppose that ξ = ξ12 Hom(A, A ), ξ12 ∈ Hom(A, L) (actually ξ12 = 0). By way of contradiction assume that ξ11 ∈ Hom(A, A ) is not zero. Then ξ11 is invertible. Let 0 = φ ∈ Hom(L, A ), such maps being plentiful as L is free. Then 0 = ξ11 φ ∈ Hom(L, A ) is regular, so Z∼ =L∼ = ξ11 φL ⊆⊕ A ∼ = Q which is a contradiction. In the category of torsion-free abelian groups of ﬁnite rank every group is the direct sum of ﬁnitely many indecomposable groups, more generally, Noetherian modules are the direct sum of ﬁnitely many indecomposable submodules. This being fairly common, it is of interest to consider the case when both A and M are ﬁnite direct sums of indecomposable submodules. We denote by tp A the isomorphism class of the R-module A and call it ∼ ∼ the type of A. Thus A1 , A2 ∈ tp A means that A1 = A2 = A. Let T denote the class of all types of R-modules. Let A = i∈I Ai be a direct decomposition of A. Let Tcr (A) be the set of types of the direct summands of A. For each type τ ∈ Tcr (A) wecan then collect the direct summands into τ -homogeneous components Aτ := {Ai | Ai ∈ τ } and obtain the homogeneous decomposition A = ρ∈Tcr (A) Aρ . Remark. Note that Tcr (A) depends on the given decomposition of A and not just on A. If Mod-R happens to be a Remak-Krull-Schmidt category, i.e., a category with unique direct decompositions with indecomposable direct summands, then Tcr (A) is an invariant of A for each R-module. More generally, Tcr (A) is an invariant of A if the individual module A has a unique decomposition as a sum of indecomposable submodules. We ﬁrst deal with the case when both A and M are “homogeneous”. Lemma 2.9. Suppose A0 and M0 are directly indecomposable modules. Let A = A1 ⊕· · ·⊕Am , where Ai ∼ = A0 for all i, and let M = M1 ⊕· · ·⊕Mn , where Mj ∼ = M0 for all j. Suppose that Reg(A, M ) = 0. Then A0 ∼ = EndR (M0 ) = M0 , EndR (A0 ) ∼ are division rings and Reg(A, M ) = Hom(A, M ). Proof. Let

⎤ ξ11 · · · ξm1 ⎢ .. .. .. ⎥ ∈ Reg(A, M ). 0 = ξ = ⎣ . . . ⎦ ξ1n · · · ξmn ⎡

As ξ = 0, there is an entry ξij = 0, and μjt ξij αsi ∈ Reg(As , Mt ) for every μjt ∈ HomR (Mj , Mt ) and every αsi ∈ HomR (As , Ai ) (Theorem II.6.17). In particular, for t = j, s = i, μjj = 1Mj , αii = 1Ai we get that 0 = ξij ∈ Reg(Ai , Mj ). It follows from Proposition 2.1 that A0 ∼ = End(M0 ) = Ai ∼ = Mj ∼ = M0 and End(A0 ) ∼ are division rings. By Corollary II.6.19, Reg(A, M ) = HomR (A, M ).

2. Structure Theorems

47

We recall that an isomorphism class of modules is called a type and denoted by σ, τ, ρ, . . .. We will use End τ to denote the isomorphism class of the endomorphism ring of a group in τ . Theorem 2.10. Suppose that A = A1 ⊕ · · · ⊕ Am and M = M 1 ⊕ · · · ⊕ Mn are decompositions with indecomposable direct summands. Let A = ρ∈Tcr (A) Aρ and M = ρ∈Tcr (M ) Mρ be the associated homogeneous decompositions. Suppose that 0 = ξ = [ξστ ] ∈ Reg(A, M ) where ξστ ∈ HomR (Aσ , Mτ ). Suppose that ξστ = 0. Then the following statements hold. 1) σ = τ , 2) End(σ) = End(τ ) are division rings, 3) HomR (Aσ , Aσ ) = 0 whenever σ = σ, 4) HomR (Mτ , Mτ ) = 0 whenever τ = τ . Proof. 1), 2) By Theorem II.6.17 we have that μτ t ξστ αsσ ∈ Reg(As , Mt ) for all t, s, all μτ t ∈ HomR (Mτ , Mt ), and all αsσ ∈ HomR (As , Aσ ). In particular, for t = τ , s = σ, μτ τ = 1, ασσ = 1 we ﬁnd that 0 = ξστ ∈ Reg(Aσ , Mτ ). By Lemma 2.9 we conclude that σ = τ , End(σ) = End(τ ) is a division ring and Reg(Aσ , Mτ ) = HomR (Aσ , Mτ ). 3) We can write ξ = [ξij ] where ξij ∈ HomR (Ai , Mj ). As ξστ = ξσσ = 0 there exist i and j such that ξij = 0 and Ai , Mj ∈ σ. Also, by Theorem II.6.17, ξij ∈ Reg(Ai , Mj ) and this implies that ξij is bijective (Proposition 2.1). Now suppose that 0 = αsi ∈ HomR (As , Ai ). Then ξij αsi ∈ Reg(As , Mj ) and ξij αsi = 0. It follows again by Proposition 2.1 that As ∼ = Mj ∼ = Ai . This means that HomR (Aσ , Aσ ) = 0 whenever σ = σ. 4) Again we use ξ = [ξij ] where ξij ∈ HomR (Ai , Mj ) and assume that ξij = 0 where Ai , Mj ∈ τ , and hence ξij is bijective. Now suppose that 0 = μjt ∈ HomR (Mj , Mt ). Then μjt ξij ∈ Reg(Ai , Mt ) and μjt ξij = 0. It follows again by Proposition 2.1 that Mt ∼ = Ai ∼ = Mj . This means that HomR (Mτ , Mτ ) = 0 whenever τ = τ . We have the following converse. Theorem 2.11. Suppose that A = A1 ⊕ · · · ⊕ Am and M = M1⊕ · · · ⊕ Mn are decompositions with indecomposable direct summands. Let A = ρ∈Tcr (A) Aρ and M = ρ∈Tcr (M ) Mρ be the associated homogeneous decompositions. Let Σ ⊆ Tcr (A) ∩ Tcr (M ). Suppose that ξ = [ξστ ] ∈ HomR (A, M ) is such that 1) ξστ = 0 unless σ = τ , 2) ξσσ ∈ Reg(Aσ , Mσ ) for σ ∈ Σ, 3) ξσσ = 0 for σ ∈ / Σ, 4) whenever σ = σ, then HomR (Aσ , Aσ ) = 0,

48

Chapter III. Indecomposable Modules

5) whenever τ = τ , then HomR (Mτ , Mτ ) = 0. Then ξ ∈ Reg(A, M ). Proof. We need to show that μσσ ξσσ ασ σ ∈ Reg(Aσ , Mσ ) for all σ ∈ Σ, σ ∈ Tcr (A) and σ ∈ Tcr (M ). Let σ ∈ Σ be given. If σ = σ or σ = σ, then μσσ ξσσ ασ σ = 0 ∈ Reg(Aσ , Mσ ). So suppose that σ = σ = σ. Then μσσ ξσσ ασσ ∈ Reg(Aσ , Mσ ) because ξσσ ∈ Reg(Aσ , Mσ ) by hypothesis.

Chapter IV

Regularity in Modules 1 Fundamental Results A very interesting special case of our general concept brings us to the regularity of modules. In HomR (A, M ) we take A = RR . We then have T := End(RR ) = R where R acts by left multiplication on R and we obtain the S-R-bimodule HomR (R, M ) where S := End(MR ). Of course, M also is an S-R-bimodule. The ﬁrst basic observation that allows us to transfer our previous more general results to the module M is the routine fact that ρ : HomR (R, M ) f → f (1) ∈ M

(1)

is a bimodule isomorphism. We now come to the deﬁnition of regularity for modules. The map f ∈ HomR (R, M ) is regular if there exists ϕ ∈ HomR (M, R) such that f ϕf = f.

(2)

Applying ρ to f ϕf = f , i.e., evaluating at 1 ∈ R, we obtain m := f (1) = (f ϕf )(1) = f (ϕ(m)) = f (1)ϕ(m) = mϕ(m), so m = mϕ(m). Hence the previous deﬁnition reappears in the module setting as follows. Deﬁnition 1.1. An element m ∈ M is regular with quasi-inverse ϕ ∈ HomR (M, R) if m = mϕ(m). (3) This means that f ∈ HomR (R, M ) is regular in HomR (R, M ) if and only if f (1) ∈ M is regular in M . Several authors have studied regular modules ([37], [10], [35]). Regular modules are special as the following example shows.

50

Chapter IV. Regularity in Modules

Theorem 1.2. Let M = 0 be a regular module over an integral domain R. Then R is a ﬁeld. A vector space over a division ring is regular. Proof. Let R be an integral domain and M a nonzero regular R-module. Then for every 0 = a ∈ M there is ϕ ∈ HomR (M, R) such that aϕ(a) = a. Suppose that a is a torsion element, i.e., there is 0 = r ∈ R such that ar = 0. Then 0 = ϕ(ar) = ϕ(a)r and since R is an integral domain, ϕ(a) = 0 and therefore a = aϕ(a) = 0. So M contains no nonzero torsion elements, i.e., M is a torsionfree module. It now follows from aϕ(a) = a = 0 that a(ϕ(a) − 1) = 0 and therefore ϕ(a) = 1. Hence ϕ : M → R is surjective and since R is projective it follows that M = aR ⊕ K for every nonzero a ∈ M . Let 0 = s ∈ R. Then, as for a, we have M = asR ⊕ L for some L. As asR ⊆ aR we can intersect M = asR ⊕ L with aR to get aR = asR⊕(K∩aR). Then K∩aR ∼ = (aR/asR) s = 0 = aR/asR and (K∩aR)s ∼ and since M is torsion-free, we conclude that K ∩ aR = 0 and aR = asR. Hence there exists t ∈ R such that a = ast, so 1 = st, showing that s is invertible. Thus R is a ﬁeld. It is easy to see that vector spaces are regular modules. By means of the isomorphism (1) we can specialize our general results to immediately obtain numerous results on regularity in modules. Since we will have to work with f ∈ HomR (R, M ) we mention ﬁrst some simple properties of such an f . The image of f is Im(f ) = f (R) = f (1)R. Since f (1) ∈ M , we set m := f (1). Then Im(f ) = f (R) = mR. The kernel of f is Ker(f ) := {r ∈ R | f (r) = mr = 0} = AnnR (m), the annihilator of m in R. For f ∈ HomR (R, M ) and ϕ ∈ HomR (M, R) we have f ϕ ∈ S = End(MR ). With m := f (1) we have for all x ∈ M that f ϕ(x) = f (1 · ϕ(x)) = f (1)ϕ(x) = mϕ(x). This motivates the following deﬁnition of the product of a module element m ∈ M and a map ϕ ∈ HomR (M, R), mϕ : M x → mϕ(x) ∈ M. It is easily checked that mϕ ∈ S = End(MR ). In general, if f gf = f , then we have idempotents e := f g ∈ S and d := gf ∈ T . Hence for f ∈ HomR (R, M ) with f (1) = m and ϕ ∈ HomR (M, R) we have the idempotents e := mϕ ∈ S and d := ϕ(m) ∈ R.

1. Fundamental Results

51

Corresponding to Theorem II.1.2 we now have the following decomposition theorem for regular elements in a module. Theorem 1.3. If m ∈ M is regular with quasi-inverse ϕ ∈ HomR (M, R), then R = AnnR (m) ⊕ ϕ(m)R,

ϕ(m)R ∼ = mR = m(Im(ϕ)),

(4)

and M = mR ⊕ {x ∈ M | ϕ(x) ∈ AnnR (m)}. Furthermore, the mapping m0 : ϕ(m)R ϕ(m)r → mϕ(m)r = mr ∈ mR is an isomorphism with inverse isomorphism ϕ0 : mR mr → ϕ(m)r ∈ ϕ(m)R. Proof. This is just Theorem II.1.2 with Ker(f ) = AnnR (m), Im(f ) = mR, Im(ϕf ) = ϕf (R) = ϕ(mR) = ϕ(m)R, and Im(f ϕ) = f ϕ(M ) = m Im(ϕ). The following theorem is essentially a specialization of Corollary II.1.3. Theorem 1.4 (Characterization of Regularity). Let m ∈ M . Then the following statements are equivalent. 1) m is regular. 2) AnnR (m) ⊆⊕ R and mR ⊆⊕ M . 3) mR is an R-projective direct summand of M . Proof. The equivalence of 1) and 2) follows from Corollary II.1.3, and 3) follows from 2) because mR is isomorphic to the complementary direct summand of AnnR (m) in R. To ﬁnish, assume that mR is projective. We have the epimorphism R mR and since mR is projective, the kernel AnnR (m) is a direct summand of R. Hence 3) implies 2). Corresponding to Corollary II.1.5 we have: Corollary 1.5. The module M contains a regular element = 0 if and only if there exist nonzero direct summands of R and of M that are isomorphic. From Corollary II.1.4 we obtain the following cases of regular modules. Corollary 1.6. In the following three cases all elements of M are regular. 1) R and M are both semisimple. 2) R is semisimple and injective. 3) M is semisimple and projective.

52

Chapter IV. Regularity in Modules

Further examples of regular modules are obtained ([37, page 343]) as right ideals in regular rings. Rings of linear transformations are regular rings and the ring of linear transformations of a countable-dimensional vector space contains nonprojective right ideals. Now we will show that in every nonzero module there exists a largest regular submodule. Deﬁnition 1.7. Let MR be a given module. Let Reg(MR ) := {m ∈ M | SmR is regular}. Theorem 1.8. Reg(MR ) is the largest regular S-R-submodule of M . Proof. Theorem II.4.3 and ρ(Reg(R, M )) = Reg(MR ).

Question 1.9. Describe all modules MR with Reg(MR ) = M and Reg(MR ) = 0. Zelmanowitz’s paper is all about the case Reg(MR ) = M . The case Reg(MR ) = 0 may be unmanageable. What are the consequences of the assumption that Reg(MR ) = 0?

2

Quasi-Inverses

2.1

Basic Properties

In considering regularity in modules there is now a distinct asymmetry between M and HomR (M, R). To emphasize this asymmetry we set M ∗ := HomR (M, R) and use Greek letters ϕ, ψ, η, . . . for its elements. Note that M is an S-R-bimodule while M ∗ is an R-S-bimodule where S = End(MR ) and End(RR ) = R (by left multiplication). We observe that in addition to regular elements m ∈ M we also have regular maps ϕ ∈ M ∗ . 1) m ∈ M is regular with quasi-inverse ϕ ∈ M ∗ if and only if m = m ϕ(m). 2) ϕ ∈ M ∗ is regular with quasi-inverse m if and only if ϕ = ϕ(m)ϕ. In general, it follows from f gf = f that f (gf g)f = f,

(gf g)f (gf g) = gf g,

the interesting point being that gf g is a quasi-inverse of f that is again regular and has quasi-inverse f . Suppose now that f ∈ HomR (R, M ), f (1) = m and ϕ ∈ HomR (M, R) is a quasi-inverse of f or, equivalently, of m. Then (ϕf ϕ)(x) = ϕ(mϕ(x)) = ϕ(m)ϕ(x), i.e., ϕf ϕ = ϕ(m)ϕ. Hence ϕ(m)ϕ must be a quasi-inverse of m that is itself regular with quasi-inverse m. We will verify this directly. Suppose that mϕ(m) = m. Then applying ϕ to mϕ(m) = m we have that ϕ(m)2 = ϕ(m) and now m(ϕ(m)ϕ)(m) = mϕ(m)2 = mϕ(m) = m,

2. Quasi-Inverses and

53

(ϕ(m)ϕ)(m)(ϕ(m)ϕ) = ϕ(m)3 ϕ = ϕ(m)ϕ.

This shows that m and ϕ(m)ϕ are both regular and mutual quasi-inverses. The case of regular objects that are mutal quasi-inverses is particularly nice. Proposition 2.1. The following statements hold. 1) Let m ∈ M be regular. Then there exists a quasi-inverse ϕ of m that is also regular and has quasi-inverse m, i.e., in addition to mϕ(m) = m we have that ϕ(m)ϕ = ϕ. 2) Suppose that m ∈ M and ϕ ∈ M ∗ = HomR (M, R) are such that mϕ(m) = m and ϕ(m)ϕ = ϕ. Then R = AnnR (m) ⊕ Im(ϕ),

and

M = mR ⊕ Ker(ϕ).

Proof. Proposition II.1.6.

2.2

Partially Invertible Objects are Quasi-Inverses

If m ∈ M is regular and mϕ(m) = m for ϕ ∈ M ∗ = HomR (M, R), then we have the idempotents e := mϕ ∈ S, d := ϕ(m) ∈ R, and m and ϕ are “factors” of d = ϕm = ϕ(m) and of e = mϕ. Conversely, a “factor” of an idempotent is the quasi-inverse of some regular map as we will see next. The result that we specialize here is Theorem II.2.1. Due to the asymmetry inherent in our present set-up, Theorem II.2.1 splits into two parts. We ﬁrst deal with the case that ϕ ∈ M ∗ is a factor of an idempotent. Lemma 2.2. The following statements are equivalent for ϕ ∈ HomR (M, R). 1) There exists m ∈ M such that mϕ = (mϕ)2 = 0. 2) There exists n ∈ M such that ϕ(n) = ϕ(n)2 = 0. 3) There exists m ∈ M such that mϕ(m) = m = 0. 4) There exist 0 = M0 ⊆⊕ M , and R0 ⊆⊕ R such that ϕ0 : M0 x → ϕ(x) ∈ R0 is an isomorphism. Next we consider the case when m ∈ M is a factor of an idempotent. Lemma 2.3. The following statements are equivalent for m ∈ M . 1) There exists ψ ∈ M ∗ such that mψ = (mψ)2 = 0. 2) There exists η ∈ M ∗ such that η(m) = η(m)2 = 0.

54

Chapter IV. Regularity in Modules

3) There exists ϕ ∈ M ∗ such that mϕ(m) = m = 0. 4) There exist 0 = M0 ⊆⊕ M , and R0 ⊆⊕ R such that m0 : R0 x → mx ∈ M0 is an isomorphism. Deﬁnition 2.4. 1) ϕ ∈ M ∗ is partially invertible if the equivalent conditions of Lemma 2.2 are satisﬁed. 2) m ∈ M is partially invertible if the equivalent conditions of Lemma 2.3 are satisﬁed. 3) The total of ϕ = HomR (M, R) is deﬁned to be Tot(M, R) := {ϕ ∈ M ∗ | ϕ is not partially invertible.}. 4) The total of the right R-module M is deﬁned to be Tot(MR ) := {m ∈ M | m is not partially invertible.}. For an arbitrary module M , one of the following statements is true ([20, page 28]). 1) M = Tot(M ). n 2) M = U ⊕ i=1 mi R with n ≥ 1, U ⊆ Tot(M ), mi regular, and mi R projective for i = 1, . . . , n. 3) M contains a ﬁnitely direct summand of the form i∈N mi R where the mi R have the same properties as in 2). We mention two results on uniqueness of quasi-inverses that are the specializations of Proposition II.2.6 and Corollary II.2.7. Proposition 2.5. Suppose that m ∈ M is regular and End(MR ) is commutative. Then m has a unique quasi-inverse. Proposition 2.6. Let m ∈ M be regular with quasi-inverse ϕ ∈ HomR (M, R) and suppose that R is commutative. Then m is the unique quasi-inverse of ϕ(m)ϕ.

3

Regular Elements Generate Projective Direct Summands

Here again we obtain interesting results on regular module elements by specializing general results.

3. Regular Elements Generate Projective Direct Summands

55

Theorem 3.1. Let 0 = m ∈ M be regular and set S = End(MR ). Then: 1) Sm is a nonzero S-projective direct summand of S M that is isomorphic to a cyclic left ideal of S that is a direct summand of S. 2) mR is a nonzero R-projective direct summand of MR that is isomorphic to a cyclic right ideal of R that is a direct summand of R. More precisely, let ϕ ∈ HomR (M, R) be a quasi-inverse of m. Then: 1) M = Sm⊕M (1−ϕ(m)), S = Smϕ⊕S(1−mϕ), and Sm = M ϕ(m) ∼ = Smϕ. 2) R = ϕ(m)R ⊕ (1 − ϕ(m))R, M = mR ⊕ (1 − mϕ)M , and mR ∼ = ϕ(m)R. Proof. Apply ρ to the corresponding result on

S

HomR (R, M ) (Theorem II.3.1).

Until now we considered M as an S-R-bimodule. We set S := End(MR ),

E := End(S M ).

Then M is also an S-E-bimodule. Corollary 3.2. If 0 = m ∈ M is regular, then 1) Sm = 0 is an S-projective direct summand of S M . 2) mE = 0 is an E-projective direct summand of ME . Proof. 1) is nothing new. It coincides with Theorem 3.1.1. 2) follows from the more general Corollary II.3.2.

These results have implications for the structure of Reg(MR ). Theorem 3.3. Let N be a ﬁnitely generated S-submodule of Reg(MR ). Then N is an S-projective direct summand of M . Furthermore, N is the direct sum of ﬁnitely many cyclic projective submodules, each of which is isomorphic to a left ideal of S. The same is true for submodules of Reg(MR ) considered as a right R-module. Proof. This is obtained by either applying ρ to Theorem II.4.6 or as a corollary of Theorem 3.1 using Lemma II.4.5. Theorem 3.4. Suppose that Reg(MR ) contains no inﬁnite direct sums of S-submodules. Then Reg(MR ) is the direct sum of ﬁnitely many simple projective Smodules, M = Reg(MR ) ⊕ U for some S-submodule U and U contains no nonzero regular S-R-submodule. Every cyclic S-submodule of Reg(MR ) is isomorphic to a left ideal of S that is a direct summand of S. An analogous result holds when Reg(MR )R contains no inﬁnite direct sums of R-submodules. Proof. Corollary II.4.8.

56

Chapter IV. Regularity in Modules

Question 3.5. When is Reg(MR ) S-projective, when R-projective? Question 3.6. Zelmanowitz has the assumption that R is “left self injective”. What can be said about HomR (A, M ) if AR is injective or if MR is projective? The following result is an immediate corollary of Theorem II.4.11. Theorem 3.7. Suppose that S := End(MR ) is left perfect. Then Reg(MR ) is Sprojective. If R is right perfect, then Reg(MR ) is R-projective. Theorem 3.8. ([37, Theorem 1.11]) Suppose that R is ﬁnite-dimensional, i.e., RR contains no inﬁnite direct sums of right ideals. Then every regular R-module is projective. Proof. Let M be a regular R-module. By Lemma II.4.5 we have that every ﬁnitely generated submodule is a direct summand of M and every ﬁnitely or countably generated submodule of M is a direct sum of cyclic regular modules. By Zorn there is a maximal submodule that is the direct sum of cyclic submodules. Let N = a∈A Rma , ma ∈ M , be such a maximal submodule. We claim that N = M . Let 0 = m ∈ M . By Theorem 1.3 the cyclic module mR is isomorphic with a right ideal of R, so mR contains no inﬁnite direct sums. By Theorem 3.4 we conclude that mR = n1 R ⊕ · · · ⊕ nt R with t a natural number and ni R simple. But Rni is simple and N ∩ ni R = 0 implies that ni R ⊆ N for every i, hence m ∈ mR ⊆ N , and we have shown that M ⊆ N , so M = N .

4 Remarks on the Literature The paper [37] contains numerous results on regular modules. Many of these results have been generalized in this monograph in two ways: we deal with the bimodule S HomR (A, M )T instead of S MR and we do not assume that HomR (A, M ) is regular but instead make statements about the largest regular submodule Reg(A, M ). This is a generalization because HomR (A, M ) is regular if and only if HomR (A, M ) = Reg(A, M ). Zelmanowitz extensively discusses endomorphism rings of regular modules which is a topic speciﬁc to regular modules and we only cite a few sample results without proof. The endomorphism ring of a regular module need not be regular but ([37, (3.2)]) it is always semiprime. Theorem 4.1 ([37, Theorem 3.4]). If End(R M ) is regular.

RM

is a regular module, then the center of

Corollary 4.2 ([37, Corollary 3.5]). If J is a commutative regular left ideal of R, then End(R J) is a commutative regular ring. Theorem 4.3 ([37, Theorem 3.8]). Let R be a commutative ring and suppose that M is an R-module with the property that every cyclic submodule is contained in a projective direct summand of M . If End(R M ) is a regular ring, then M is a regular module.

5. The Transfer Rule

57

Corollary 4.4 ([37, Corollary 4.2], [35, Theorem 3.6]). If M is a ﬁnitely generated regular module, then End(M ) is a regular ring. Theorem 4.5 ([37, Theorem 3.8]). If R M is a regular R-module with homogeneous socle, then {α ∈ End(M ) | M α is ﬁnite-dimensional} is an ideal of End(M ) and a simple regular ring. Theorem 4.6 ([37, Theorem 3.8]). Let R M be a regular module. Then End(R M ) is semisimple with minimum condition if and only if M is ﬁnite-dimensional.

5

The Transfer Rule

The transfer rule can also be restricted to the module setting. If m ∈ M and ϕ ∈ HomR (M, R) with mϕ(m) = m, then we call (m, ϕ) a regular pair. Similarly, if ϕ ∈ HomR (M, R) and k ∈ M with ϕ(k)ϕ = ϕ, then (ϕ, k) is a regular pair. If (m, ϕ) is a regular pair, then also (ϕ(m)ϕ, m) is a regular pair. Deﬁnition 5.1. If (m, ϕ) is a regular pair with m ∈ M , then we call trf : (m, ϕ) → (ϕ(m)ϕ, m) the transfer rule or the transfer and we also write trf(m, ϕ) = (ϕ(m)ϕ, m). Theorem 5.2. If trf is applied to all regular pairs (m, ϕ) with m ∈ M , then the set of ﬁrst entries in trf(m, ϕ) = (ϕ(m)ϕ, m) is the set of all regular elements in HomR (M, R).

Chapter V

Regularity in HomR(A, M ) as a One-sided Module 1 Iterated Endomorphism Rings So far we discussed regularity in HomR (A, M ), and then by using HomR (R, M ) ∼ = M we arrived at regularity in modules. We can now look at module structures on HomR (A, M ) over certain rings and study regularity of HomR (A, M ) as a module over these rings. Explicitly, we have module structures on H := HomR (A, M ) as follows. •

SH

where S = End(MR ), ∀s ∈ S, ∀h ∈ H, s · h = s ◦ h,

• HT where T = End(AR ), ∀t ∈ T, ∀h ∈ H, h · t = h ◦ t, •

S H

where S = End(HT ), ∀s ∈ S , ∀h ∈ H, s · h = s (h),

• HT where T = End(S H), ∀t ∈ T , ∀h ∈ H, h · t = (h)t , •

S H

where S = End(HT ), ∀s ∈ S , ∀h ∈ H, s · h = s (h),

• HT where T = End(S H), ∀t ∈ T , ∀h ∈ H, h · t = (h)t . Associated with each of these there is a largest regular bi-submodule and one might think that there is an unending chain of such substructures of H. However, this is not so because the process of forming endomorphism rings comes to a standstill after three steps as we will explain next. This is well-known and can be found in [1, pages 60-61]. Given an R-module MR we have its endomorphism ring E1 := End(MR ), and M becomes a left E1 -module E1 M where left multiplication by s ∈ E1 is left action of s on M , i.e., for x ∈ M , s·x = s(x). Let E2 := End(E1 M ) and M become a right E2 -module where again right multiplication by t ∈ E2 is right action of t. Further we have E3 := End ME2 . We will show that E3 ∼ = E2 = E1 and E4 := End(E3 M ) ∼

60

Chapter V. Regularity in HomR (A, M ) as a One-sided Module

with identical actions on M so that nothing new is obtained by iterating the process further. Proposition 1.1. Let M be a right R-module. 1) Set E0 := R and set E1 := End(MR ) = End(ME0 ). For s ∈ E1 and x ∈ M , deﬁne s · x := s(x), making M into a bimodule E1 ME0 . 2) Set E2 := End(E1 M ). For t ∈ E2 and x ∈ M , deﬁne x · t := (x)t, making M into a bimodule E1 ME2 . 3) ρ : E0 → E2 : ∀ r ∈ R, ∀ x ∈ M, (x)ρ(r) = x · r determines a well-deﬁned ring homomorphism. 4) Set E3 := End(ME2 ). For u ∈ E3 and x ∈ M , deﬁne u · x := u(x), making M into a bimodule E3 ME2 . 5) λ : E1 → E3 : ∀ s ∈ E1 , ∀ x ∈ M, λ(s)(x) = s(x) determines a well-deﬁned ring isomorphism. 6) ∀ x ∈ M, ∀s ∈ E1 : s · x = λ(s) · x, and hence End(E3 M ) = E2 . 7) Set E4 := End(E3 M ). For v ∈ E4 and x ∈ M , deﬁne x · v := (x)v, making M into a bimodule E3 ME4 . 8) ρ : E2 → E4 : ∀ t ∈ E2 , ∀ x ∈ M, (x)ρ(t) = (x)t determines a well-deﬁned ring isomorphism. 9) ∀ x ∈ M, ∀t ∈ E2 : x · t = x · ρ(t), and hence End(ME4 ) = E3 . Proof. Most statements being evident and well-known, we check only 5) and 6). Clearly λ is a ring homomorphism. Suppose that λ(s) = 0. Then ∀ x ∈ M : s(x) = 0 which means that s = 0. Now let t ∈ E3 . We will see that t : M → M is an R-homomorphism which says that t ∈ E1 and so λ(t) = t. We have that ρ : R → E2 : ∀ r ∈ R, ∀ x ∈ M, (x)ρ(r) = x · r is a well-deﬁned ring homomorphism. Recall that t ∈ E3 . We compute that t(xr) = t(xρ(r)) = t(x)ρ(r) = t(x) · r showing that t is indeed an R-homomorphism. The proofs of 8) and 9) are similar.

2 Deﬁnitions and Characterizations We already worked with H := HomR (A, M ) as an S-T -bimodule where S := End(MR ),

T := End(AR )

and we have additionally S := End(HT ), T := End(S H), S := End(HT ), T := End(S H).

2. Deﬁnitions and Characterizations

61

Thus H is an S-T -, an S-T -, an S -T -, an S -T -, and an S -T -module. This is as far as we can go because, by Proposition 1.1, we have S := End(S H) = S ,

and T := End(HT ) = T .

We also have, assuming that AR is a faithful R-module, R ⊆ T ,

S ⊆ S ,

and T ⊆ T .

So H = HomR (A, M ) is a module in several ways and for each we have regularity concepts and regular substructures. Let f ∈ H := HomR (A, M ). • f is S-regular, i.e., f is regular as an element of exists σ ∈ HomS (H, S) such that (f σ)f = f .

S H,

if and only if there

• f is T -regular, i.e., f is regular as an element of HT , if and only if there exists τ ∈ HomT (H, T ) such that f (τ f ) = f . • f is S -regular, i.e., f is regular as an element of exists σ ∈ HomS (H, S ) such that (f σ )f = f .

S H,

if and only if there

• f is T -regular, i.e., f is regular as an element of HT , if and only if there exists τ ∈ HomT (H, T ) such that f (τ f ) = f . • f is S -regular, i.e., f is regular as an element of exists σ ∈ HomS (H, S ) such that (f σ )f = f .

S H,

if and only if there

• f is T -regular, i.e., f is regular as an element of HT , if and only if there exists τ ∈ HomT (H, T ) such that f (τ f ) = f . It turns out that the regularity of f ∈ HomR (A, M ) implies all other kinds of regularity. Lemma 2.1. Let f ∈ HomR (A, M ) be regular and let g ∈ HomR (M, A) be such that f gf = f . Then the map σ : S HomR (A, M ) h → (h)σ := h ◦ g ∈ S End(MR ) = S S is a well-deﬁned S-homomorphism and f is S-regular with quasi-inverse σ. The map τ : HomR (M, A)T h → τ (h) := g ◦ h ∈ End(AR )T = TT is a well-deﬁned T -homomorphism and f is T -regular with quasi-inverse τ . The map σ : S HomR (A, M ) → S End(HT ) = S S : ∀h, x ∈ HomR (M, A), ((h)σ )(x) := h ◦ g ◦ x is a well-deﬁned S -homomorphism and f is S -regular with quasi-inverse σ .

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Chapter V. Regularity in HomR (A, M ) as a One-sided Module

The map τ : HomR (M, A)T → End(S H)T = TT : ∀h, x ∈ HomR (M, A), (x)τ (h) := x ◦ g ◦ h is a well-deﬁned T -homomorphism and f is T -regular with quasi-inverse τ . The map σ : S HomR (A, M ) → S End(HT ) = S S : ∀h, x ∈ HomR (M, A), ((h)σ )(x) := h ◦ g ◦ x is a well-deﬁned S -homomorphism and f is S -regular with quasi-inverse σ . The map τ : HomR (M, A)T → End(S H)T = TT : ∀h, x ∈ HomR (M, A), (x)(τ (h)) := x ◦ g ◦ h is a well-deﬁned T -homomorphism and f is T -regular with quasi-inverse τ . Proof. Clearly σ, τ , σ , τ , σ , τ are all additive. For s ∈ S, we have (sh)σ = shg = s(h)σ. The equation f gf = f now reads (f σ)f = f showing that f is S-regular. For t ∈ T , we have τ (ht) = ght = τ (h)t. The equation f gf = f now reads f τ (f ) = f showing that f is T -regular. We ﬁrst check that σ maps into S . In fact, let h, x ∈ H and t ∈ T . Then (hσ )(xt) = h◦g◦x◦t = ((hσ )(x))t. Next we check that σ is an S -homomorphism. Let s ∈ S and h, x ∈ H. Then g ◦ x ∈ T and ((s h)σ )(x) = s (h) ◦ g ◦ x = s (h ◦ g ◦ x) = s ((hσ )(x)) = (s ◦ (hσ ))(x), hence (s h)σ = s (hσ ). Finally, (f σ )f = f gf = f . We ﬁrst check that τ maps into T . In fact, let h, x ∈ H and s ∈ S. Then (sx)(τ )(h) = sxgh = s((x)(τ (h))). Next we check that τ is a T -homomorphism. Let t ∈ T and h, x ∈ H. Then x ◦ g ∈ S and (x)(τ (ht )) = x ◦ g ◦ (h)t = ((x◦g)h)t = ((x)τ (h))t = (x)(τ (h)◦t ), hence τ (ht ) = τ (h)t . Finally, f τ (f ) = f gf = f . We ﬁrst check that σ maps into S . In fact, let h, x ∈ H and t ∈ T . Then h ◦ g ∈ S and (hσ )(xt ) = h ◦ g ◦ (x)t = (h ◦ g ◦ x)t = (hσ )(x)t . Next we check that σ is an S -homomorphism. Let s ∈ S and h, x ∈ H. Then g ◦ x ∈ T because ∀ s ∈ S, (sh)(g ◦ x) = s ◦ h ◦ g ◦ x = s ◦ (h ◦ g ◦ x) = s((h)(g ◦ x)) and hence ((s h)σ )(x) = ((s (h))σ )(x) = s (h) ◦ g ◦ x = s (h ◦ g ◦ x) = s ((hσ )(x)) = (s ◦ (hσ ))(x), hence (s h)σ = s (hσ ). Finally, (f σ )f = f gf = f . We ﬁrst check that τ maps into T . In fact, let h, x ∈ H and s ∈ S . Then (s x)(τ (h)) = (s (x))(τ )(h) = s (x) ◦ g ◦ h = s (x)(g ◦ h) = s (x ◦ g ◦ h) = s ((x)(τ (h))). Next we check that τ is a T -homomorphism. Let t ∈ T and h, x ∈ H. Then x ◦ g ∈ S because (s(x ◦ g))(h) = s ◦ x ◦ g ◦ h = s ◦ (x ◦ g ◦ h) = s(x ◦ g)(h) and (x)(τ (ht )) = (x)(τ ((h)t )) = x ◦ g ◦ (h)t = (x ◦ g)(h)t = ((x ◦ g) ◦ h)t = ((x)τ (h))t = (x)(τ (h) ◦ t ), hence τ (ht ) = τ (h)t . Finally, f τ (f ) = f gf = f .

2. Deﬁnitions and Characterizations

63

It is not clear how the other types of regularity are related. The following example shows that regular and S-regular need not be equivalent. Example 2.2. There exist Z-modules A and M such that HomR (A, M ) contains no nonzero regular elements while HomR (A, M ) contains nonzero End(AR )-regular elements. Proof. Let p be a prime number, A = aZ, ord(a) = p, and B = bZ, ord(b) = p2 . Suppose that f ∈ Hom(A, M ) is regular. Then M = Im(f ) ⊕ U for some subgroup U . Since M is indecomposable and | Im(f )| ≤ |aZ| = p, it is not possible that Im(f ) = M , so Im(f ) = 0, and f = 0. Note that T := End(aZ) ∼ = Z/pZ and S := End(bZ) ∼ = Z/p2 Z. Let f ∈ Hom(aZ, bZ) be the map with f (a) = pb and deﬁne for h ∈ Hom(aZ, bZ) with h(a) = pbk the functional ϕ : Hom(aZ, bZ) h → k + pZ ∈ Z/pZ = End(aZ). Then ϕ(f ) = 1 and f ϕ(f ) = f so that f is a nonzero T -regular element. Question 2.3. When does f regular follow from the fact that f is S, T , S , T regular? We will make use of a lemma of Ware ([35, Lemma 2.2]) that is a generalization of a result of Chase ([5, Proposition 2.1], and used by Zelmanowitz ([37, Lemma 2.4]). We include it with proof. Lemma 2.4. Let S P be a projective S-module and K a submodule of P . Then P/K is ﬂat if and only if for every x ∈ K there is a homomorphism αx : P → K with xαx = x. Proof. (a) We ﬁrst assume that P is free on a basis {bi | i ∈ I}. Assume that P/K is ﬂat, let x ∈ K, and write x = i∈I si bi , where si ∈ S and all but ﬁnitely many of the si are equal to 0. Let Ix := i∈I si S. Then Ix is a ﬁnitely generated right ideal of S.By Proposition I.2.4 we have K ∩ Ix P = Ix K which contains x and hence x = i∈I ti ki where ki ∈ K, ti ∈ Ix , and all but stipulating ﬁnitely many of the ti are equal to 0. Deﬁne αx : P → K by that b i αx = ki if si = 0 and bi αx = 0 otherwise. Then xαx = i∈I si bi αx = i∈I ti ki = x. Conversely, assume that for every x ∈ K there is αx : P → K such that xαx = x. Let U be a right ideal of S and let x ∈ K ∩ U P be given. Then u b , where u ∈ U . Applying α we get x = xα = x = i i i x x i∈I i∈I ui bi αx = i∈I ui (bi αx ) ∈ U K, showing that K ∩ U P = U K. By Proposition I.2.4 we see that P/K is ﬂat. (b) Now we consider the general case where P is projective but not necessarily free. Then there exists a free module F such that F = P ⊕ L for some module L. Suppose that P/K is ﬂat and let x ∈ K. Then P/K ⊕ Q is ﬂat hence, by (a), there exists g : F → K such that xg = x. Let αx := g P . Then αx : P → K with xαx = x.

64

Chapter V. Regularity in HomR (A, M ) as a One-sided Module

Assume that for every x ∈ K there is a homomorphism αx : P → K with xαx = x. Let x ∈ K. Then there exists αx : P → K such that xαx = x. Extend αx to g : F → K by deﬁning Lg = 0. Then xg = x and by (a) (P ⊕ L)/K is ﬂat. Therefore P/K is ﬂat. Proposition 2.5. Every epimorphic image of morphic image of Reg(A, M )T is T -ﬂat.

S

Reg(A, M ) is S-ﬂat, and every epi-

Proof. Suppose that K is an S-submodule of Reg(A, M ). We will show that Reg(A, M )/K is ﬂat. Since direct limits of ﬂat modules are ﬂat, it suﬃces to show that every ﬁnitely generated submodule of Reg(A, M )/K is ﬂat. A ﬁnitely generated submodule of Reg(A, M )/K is of the form (N + K)/K where N is a ﬁnitely generated submodule of Reg(A, M ). The module N is regular and projective by Theorem II.4.6. We wish to apply Lemma 2.4 to conclude that (N + K)/K ∼ = N/(N ∩ K) is ﬂat. We have that N is a projective module and we need to show that for every x ∈ N ∩ K there is an S-homomorphism αx : N → N ∩ K such that xαx = x. Let x ∈ N ∩ K. As N is regular as a subset of HomR (A, M ), it is also S-regular (see Lemma 2.1), and hence there exists ϕx : N → S such that (xϕx )x = x. Deﬁne αx : N n → (nϕx )x which is an S-homomorphism. Then xαx = (xϕx )x = x. By Lemma 2.4 we obtain that (N + K)/K ∼ = N/(N ∩ K) is ﬂat. We now have numerous immediate corollaries to our results on regularity for modules. Corollary 2.6 (Characterization of Regularity). Let P be any one of the rings S, S , or S . Let f ∈ P HomR (A, M ).Then the following statements are equivalent. 1) f is P -regular. 2) AnnP (f ) ⊆⊕ P and P f ⊆⊕ HomR (A, M ). 3) P f is a P -projective direct summand of HomR (A, M ). Let P be any one of the rings T , T , or T . Let f ∈ HomR (A, M )P .Then the following statements are equivalent. 1) f is P -regular. 2) AnnP (f ) ⊆⊕ P and f P ⊆⊕ HomR (A, M ). 3) f P is a P -projective direct summand of HomR (A, M ). Proof. Theorem IV.1.4.

This also says that all elements of Reg(A, M ) are S-, T -, S -, T -, S - and T -regular. Later we will come back to this fact. The next and some later results we only formulate for S HomR (A, M ) but there are ﬁve more results corresponding to the other ﬁve module structures on H.

3. Largest Regular Submodules

65

Proposition 2.7. The following statements hold. 1) Let f ∈ H = HomR (A, M ) be S-regular. Then there exists a quasi-inverse ϕ ∈ HomS (H, S) of f that is also regular and has quasi-inverse f , i.e., in addition to f ◦ (f )ϕ = f we have that (f )ϕ ◦ ϕ = ϕ. 2) Suppose that f ∈ H and ϕ ∈ H ∗ = HomS (H, S) are such that f ◦ (f )ϕ = f and ϕ ◦ (f )ϕ = ϕ. Then S = AnnS (f ) ⊕ Im(ϕ),

and

HomR (A, M ) = Sf ⊕ Ker(ϕ).

Proof. Lemma 2.1.

Proposition 2.8. Suppose that f ∈ HomR (A, M ) is S-regular and T = End(S H) is commutative. Then f has a unique quasi-inverse.

Proof. Proposition IV.2.5.

Corollary 2.9. Let f ∈ HomR (A, M ) be S-regular with quasi-inverse ϕ ∈ HomS (H, S) and suppose that S is commutative. Then f is the unique quasiinverse of (f )ϕ ◦ ϕ.

Proof. Proposition IV.2.6.

3

Largest Regular Submodules

The existence of various largest regular submodules is an immediate consequence of Theorem IV.1.8. Deﬁnition 3.1. Let H = HomR (A, M ), S, T , etc. be as before. We deﬁne 1) Reg(S H) := {f ∈ H | Sf T is S-regular}. 2) Reg(HT ) := {f ∈ H | S f T is T -regular}. 3) Reg(S H) := {f ∈ H | S f T is S -regular}. 4) Reg(HT ) := {f ∈ H | S f T is T -regular}. 5) Reg(S H) := {f ∈ H | S f T is S -regular}. 6) Reg(HT ) := {f ∈ H | S f T is T -regular}. Here it is used that S = S and T = T . Theorem 3.2. 1) Reg(S H) is the largest S-regular S-T -submodule of S HT . 2) Reg(HT ) is the largest T -regular S -T -submodule of

S HT .

3) Reg(S H) is the largest S -regular S -T -submodule of

S HT .

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Chapter V. Regularity in HomR (A, M ) as a One-sided Module

4) Reg(HT ) is the largest T -regular S -T -submodule of

S HT .

5) Reg(S H) is the largest S -regular S -T -submodule of

6) Reg(HT ) is the largest T -regular S -T -submodule of

S HT . S HT .

Proof. Theorem IV.1.8.

We have already mentioned that f regular implies that f is S-, T -, S -, T -, S -, and T -regular. Hence Reg(A, M ) is S-, T -, etc. regular. But we do not know whether Reg(A, M ) is contained in Reg(S H), Reg(HT ), Reg(S H), Reg(HT ), Reg(S H), or Reg(HT ). We have that f ∈ Reg(A, M ) if and only if Sf T is regular; however this does not necessarily mean that Sf T is S-regular. Similarly for the other cases.

Question 3.3. For which pairs A, M does Sf T regular imply Sf T is S-regular? Same question for other combinations of rings. The modules Reg(S H), Reg(HT ), etc. have the properties that Reg(MR ) has for a general module MR and which were derived from the properties of Reg(A, M ) with A = R. We list these properties for Reg(S H) only; it is easy to formulate them for Reg(S H), Reg(HT ) and the other module structures on H. Theorem 3.4. Let N be a ﬁnitely generated S -submodule of Reg(HT ). Then N is an S -projective direct summand of H. Furthermore, N is the direct sum of ﬁnitely many cyclic projective submodules, each of which is isomorphic to a left ideal of S . The same is true for submodules of Reg(HT ) considered as a right T -module. Proof. Theorem IV.3.3.

Theorem 3.5. Suppose that Reg(HT ) contains no inﬁnite direct sums of S -submodules. Then Reg(HT ) is the direct sum of ﬁnitely many simple projective S modules, HomR (A, M ) = Reg(HT ) ⊕ U for some S -submodule U and U contains no nonzero regular S -T -submodule. Every cyclic S -submodule of Reg(HT ) is isomorphic to a left ideal of S that is a direct summand of S . Analogous results hold for the other module structures on HomR (A, M ). Proof. Theorem IV.3.4.

Theorem 3.6. Suppose that S is left perfect. Then Reg(HT ) is S -projective. If T is right perfect, then Reg(HT ) is T -projective. Proof. Theorem IV.3.7.

4

The Transfer Rule for S-Regularity

Here also we repeat the transfer rule of the plain regular case but we enter situations where regularity was not applied before.

4. The Transfer Rule for S-Regularity

67

Let f be S-regular and f σf = f with σ ∈ HomS (H, S). Then we get the idempotent e := f σ = e2 ∈ S, and ef = f. Further σ(f σ) ∈ HomS (H, S) where σ(f σ) : H h → (hσ)(f σ) ∈ S. As before, we call (f, σ) a regular pair and deﬁne trf(f, σ) = (σf σ, f ) where σf σ = σe ∈ HomS (H, S). Further, (σf σ, f ) is again a regular pair because σf σf σf σ = σe3 = σe = σf σ with σf σ = σe ∈ HomS (H, S). It follows further that trf(σf σ, f ) = (f σf σf, σf σ) = (f, σe). Corollary 4.1. If trf is applied to all S-regular pairs (f, σ) with f ∈ HomR (A, M ), then the set of all entries σf σ in trf(f, σ) = (σf σ, f ) is the set of all S-regular elements in HomS (H, S). Proof. Same as that of Theorem II.5.3.

Chapter VI

Relative Regularity: U -Regularity and Semiregularity 1

U -Regularity; Deﬁnition and Existence of U -Reg(A, M )

The paper [19] on U -regularity was the ﬁrst step in the study of regular substructures of Hom. We give here the main properties of U -regularity. Deﬁnition 1.1. 1) Let U be an S-T -submodule of S HT = S Hom(A, M )T . Then f ∈ H is called U -regular if and only if there exists g ∈ HomR (M, A) such that f gf − f ∈ U.

(1)

2) U-Reg(A, M ) := {f ∈ H | Sf T is U -regular} where Sf T is the S-T -submodule of H generated by f and “Sf T is U -regular” means that every element of Sf T is U -regular. We add some comments that will serve to familiarize the reader with the new notions. 1) For U = 0, U -regular = regular. 2) U ⊆ U-Reg(A, M ) because U is an S-T -submodule, and for u ∈ U we have u · 0 · u − u = −u ∈ U , hence u is in U-Reg(A, M ). 3) If U1 and U2 are S-T -submodules of H, and U1 ⊆ U2 , then U1 -Reg(A, M ) ⊆ U2 -Reg(A, M ).

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Chapter VI. Relative Regularity: U -Regularity and Semiregularity

4) If f ∈ H is U -regular with (1) and v ∈ U , then also f + v is U -regular, since, by (1), setting u := f gf − f ∈ U , we get (f + v)g(f + v) − (f + v) = u1 ∈ U where u1 = u + v + f gv + vgf . Here we used the fact that f g ∈ S, gf ∈ T and that U is an S-T -module. This implies also that S(f + v)T is U -regular if and only if Sf T is U -regular. For U -regularity we have the same important theorem that we proved earlier for other regularities. Theorem 1.2. The set U-Reg(A, M ) is the unique largest U -regular S-T -submodule of HomR (A, M ). The proof is more or less the same as in the previous case (namely Theorem II.4.3), yet we will give it in detail to reinforce the arguments. For the proof we need the following lemma. Lemma 1.3. Let U be an S-T -submodule of HomR (A, M ), let f ∈ HomR (A, M ) and let g ∈ HomR (M, A). If f − f gf is U -regular, then f is U -regular. Proof. Since f − f gf is U -regular, there exists h ∈ HomR (M, A) such that (f − f gf )h(f − f gf ) − (f − f gf ) ∈ U.

(2)

Let k := h − gf h − hf g + gf hf g + g ∈ HomR (M, A) with h from (2). An easy computation shows that f kf − f ∈ U . In fact, f kf

= f hf − f gf hf − f hf gf + f gf hf gf + f gf = (f − f gf )h(f − f gf ) + f gf

and by (2) f − f kf = (f − f gf ) − (f − f gf )h(f − f gf ) ∈ U.

Proof of Theorem 1.2. This proof is achieved in four steps. (a) Let f ∈ H and assume that Sf T is U -regular. If ϕ ∈ Sf T , then SϕT ⊆ Sf T , hence also SϕT is U -regular and ϕ ∈ U-Reg(A, M ). Since ϕ was an arbitrary element of Sf T , it follows that Sf T ⊆ U-Reg(A, M ). (b) Since U-Reg(A, M ) is the sum of modules of the form Sf T , it is closed under left multiplication by S and right multiplication by T . (c) We show now that U-Reg(A, M ) is closed under addition. Let f1 , f2 ∈ U-Reg(A, M ). We have to show that S(f1 + f2 )T is U -regular. An element of S(f1 + f2 )T is of the form n n n i=1 si (f1 + f2 )ti = i=1 si f1 ti + i=1 si f2 ti . n n Set ϕ1 = i=1 si f1 ti and ϕ2 = i=1 si f2 ti . Then ϕ1 ∈ Sf1 T is U -regular and there exists g ∈ HomR (M, A) such that ϕ1 gϕ1 − ϕ1 ∈ U . Now consider (ϕ1 +ϕ2 )−(ϕ1 +ϕ2 )g(ϕ1 +ϕ2 ) = (ϕ1 −ϕ1 gϕ1 )+(ϕ2 −ϕ1 gϕ2 −ϕ2 gϕ1 −ϕ2 gϕ2 ). (3)

1. U -Regularity; Deﬁnition and Existence of U -Reg(A, M )

71

Here ϕ1 gϕ1 − ϕ1 ∈ U and since ϕ1 g ∈ S, gϕ1 , ϕ2 g ∈ T , the term ϕ2 − ϕ1 gϕ2 − ϕ2 gϕ1 −ϕ2 gϕ2 in (3) is in Sϕ2 T . We now use that Sϕ2 T is U -regular which is true by (a). Therefore the element in (3) is U -regular and we can apply Lemma 1.3 with ϕ1 + ϕ2 in place of f , and obtain that ϕ1 + ϕ2 is U -regular. Since ϕ1 + ϕ2 was an arbitrary element of S(f1 + f2 )T , this submodule is U -regular. So S(f1 + f2 )T ⊆ U-Reg(A, M ) and in particular, f1 + f2 ∈ U-Reg(A, M ). (d) Finally we show that U-Reg(A, M ) is the largest U -regular S-T -submodule of H. Assume that Λ is a U -regular S-T -submodule of H and f ∈ Λ. Then Sf T is a U -regular S-T -submodule of H and by deﬁnition of U-Reg(A, M ) we have Sf T ⊆ U-Reg(A, M ), in particular f ∈ U-Reg(A, M ). This means that Λ ⊆ U-Reg(A, M ) and U-Reg(A, M ) is the largest regular S-T -submodule of H. In the following A and M will be ﬁxed and we will therefore write U-Reg instead of U-Reg(A, M ). As U-Reg is again an S-T submodule of H, U-Reg can be used in place of U in the deﬁnition above, i.e., we may consider (U-Reg)- Reg. Theorem 1.4. Let U be an S-T -submodule of H = S Hom(A, M )T . Then (U-Reg(A, M ))- Reg(A, M ) = U-Reg(A, M ). Proof. Since U ⊆ U-Reg, it follows that U-Reg ⊆ (U-Reg)- Reg . We have to prove the converse inclusion. Let f ∈ (U-Reg)- Reg. Then there exist g ∈ HomR (M, A) and w ∈ U-Reg such that f gf − f = w.

(4)

Since w ∈ U-Reg, there exists h ∈ HomR (M, A) with whw − w = u ∈ U.

(5)

w = f (gf − 1T ) = (f g − 1S )f

(6)

It follows from (4) that

and using (4), (5) and (6) together we obtain f

= f gf − w = f gf − whw + u = f gf − f (gf − 1T )h(f g − 1S )f + u = f [g − (gf − 1T )h(f g − 1S )]f + u.

The element in the bracket is in HomR (M, A). Hence the equation ﬁts the deﬁnition for f to be U -regular. We have now established that every (U-Reg)-regular element is U -regular.

72

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Chapter VI. Relative Regularity: U -Regularity and Semiregularity

U -Regularity in Modules

As before we can use the isomorphism ρ : HomR (R, M ) f → f (1) ∈ M to specialize U -regularity. As always S := End(MR ) and R = End(RR ). Then HomR (R, M ) and M are naturally S-R-modules and ρ becomes a bimodule isomorphism. Let U be an S-R-submodule of M and U = ρ−1 (U ). Then f ∈ HomR (R, M ) is U -regular if and only if there exists ϕ ∈ M ∗ := HomR (M, R) such that f ϕf − f ∈ U . Setting m = f (1) and applying ρ we obtain that mϕ(m) − m ∈ U . We deﬁne U -regularity in M in such a way that m is U -regular in M if and only if ρ−1 (m) is ρ−1 (U )-regular in HomR (R, M ). Deﬁnition 2.1. 1) Let U be an S-R-submodule of M . Then m ∈ M is called U -regular if and only if there exists ϕ ∈ HomR (M, R) such that mϕ(m) − m ∈ U. 2) U-Reg(MR ) := {m ∈ M | SmR is U -regular} where SmR is the S-Rsubmodule of M generated by m and “SmR is U -regular” means that every element of SmR is U -regular. As in the general case we have the following facts. 1) For U = 0, U -regular = regular. 2) U ⊆ U-Reg(MR ). 3) If U1 and U2 are S-R-submodules of M , and U1 ⊆ U2 , then U1 -Reg(MR ) ⊆ U2 -Reg(MR ). 4) If m ∈ M is U -regular and v ∈ U , then also m+v is U -regular and S(m+v)R is U -regular if and only if SmR is U -regular. Theorem 2.2. The set U-Reg(MR ) is the unique largest U -regular S-R-submodule of M . Theorem 2.3. Let U be an S-R-submodule of S MR . Then (U-Reg(MR ))- Reg(MR ) = U-Reg(MR ). We will now compare U -regularity with regularity in M/U . Proposition 2.4. Let U be an S-R-submodule of MR , and let : M m → m := m + U ∈ M/U = M be the natural R-epimorphism. Let m ∈ M . 1) If m is regular in M , then m is U -regular in M .

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73

2) If HomR (U, R) = 0 and m is regular in M , then m is U -regular in M . Proof. 1) Note that any element of M is of the form m for some m ∈ M . Let m be regular in M . Then there is ψ ∈ HomR (M , R) such that m(ψ(m)) = m which says that m((ψ ◦ )(m)) − m ∈ U where ψ ◦ ∈ HomR (M, R). 2) Assume that HomR (U, R) = 0, let m be U -regular in M , and let ϕ ∈ Hom(M, R) be such that mϕ(m)−m ∈ U . The hypothesis HomR (U, R) = 0 implies that ϕ factors through ∈ HomR (M/U, R), i.e., there is ψ ∈ HomR (M/U, R) such that ϕ = ψ ◦ . We conclude that m(ψ(m)) − m = mφ(m) − m = 0. We now look at some examples. Example 2.5. Let MZ be an abelian group. Then: 1) 0 = m ∈ M is regular if and only if M = mZ ⊕ N for some subgroup N and mZ ∼ = Z. However Reg(MZ ) = 0. 2) Let p a prime, and suppose that Hom(pM, Z) = 0. Then pM is a fully invariant subgroup of M , (M/pM )Z contains no nonzero regular elements although M/pM is regular as a module over the ﬁeld Z/pZ. We have pM − Reg(M ) = pM . 3) t(M ) − Reg(MR ) = t(M ) Proof. 1) Suppose that 0 = m ∈ M is regular. Then there is ϕ ∈ Hom(M, Z) such that mϕ(m) = m = 0. Hence ϕ = 0. By Theorem IV.1.3 we get Z = AnnZ (m) ⊕ ϕ(m)Z and hence Z = ϕ(m)Z, M = mZ ⊕ Ker(ϕ), and mZ ∼ = ϕ(m)Z = Z. Conversely, if M = mZ ⊕ N and mZ ∼ Z, then the map ϕ ∈ Hom(M, Z) with = ϕ(mk) = k and ϕ(N ) = 0 clearly is a quasi-inverse for m. By Theorem 1.2 we have Reg(MZ ) = 0. 2) We observe ﬁrst that pM is a fully invariant submodule of M and hence is an S-Z-submodule of M where S := End(M ). Observe that M/pM is a semisimple Z/pZ-module, and hence is regular. However as an abelian group it contains no nonzero regular elements whatsoever. Thus if m ∈ M is pM -regular, then m = 0, i.e., m ∈ pM . This shows that pM − Reg(MZ ) = pM . 3) Again t(M ) is fully invariant in M , and Hom(t(M ), Z) = 0. In this case (M/ t(M ))Z may well contain nonzero regular elements, but if m = 0 is such an element, then mk where 0 = k ∈ Z cannot be regular. So km is not t(M )-regular in M while k ∈ End(M ). The claim follows.

3

Semiregularity for Modules

Another form of regularity is semiregularity. This was introduced and studied for modules by W.K. Nicholson ([26]). Deﬁnition 3.1. (Nicholson) Let MR be an R-module. An element m ∈ M is semiregular if there exists a regular element n ∈ M such that n − m ∈ Rad(MR ).

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Chapter VI. Relative Regularity: U -Regularity and Semiregularity

Nicholson supplies the following equivalent characterizations of semiregularity. To state them, we need the notion of “lying over”. Deﬁnition 3.2. A submodule C ⊆ M lies over a summand of M if and only if M = A ⊕ B with A ⊆ C and C ∩ B is small. Proposition 3.3 ([26, Proposition 1.3]). Let M be a right R-module and m ∈ M . The following conditions on m are equivalent. 1) m is semiregular. 2) There is n ∈ M such that n is regular, and n − m ∈ Rad(M ). 3) There is e2 = e ∈ End(MR ) such that e(M ) ⊆ mR, e(M ) is projective and m − e(m) ∈ Rad(M ). 4) mR lies over a projective direct summand of M . 5) There exists ϕ ∈ M ∗ := HomR (M, R) such that ϕ(m)2 = ϕ(m) and m − mϕ(m) ∈ Rad(M ). 6) There is n ∈ mR such that n is regular, n − m ∈ Rad(M ), and mR = nR ⊕ (m − n)R. Proof. 1) ⇔ 2) is the deﬁnition. 2) ⇒ 3): By hypothesis we have a regular element n ∈ M such that n − m ∈ Rad(M ). As n is regular we further have ϕ ∈ M ∗ such that nϕ(n) = n, so also ϕ(n)2 = ϕ(n).

(7)

Assume for the moment that n ∈ mR. Let e : M x → e(x) = nϕ(x) ∈ M . Then e2 = (nϕ)(nϕ) = nϕ(n)ϕ(n) = nϕ(n) = e, with e(M ) ⊆ nR ⊆ mR. As n is regular, by Theorem IV.1.3, the submodule nR is a projective direct summand of M and hence e(M ), being a direct summand of nR, is also projective. Finally, m − e(m) = m − nϕ(m) = m − n + n − nϕ(m) = (m − n) + nϕ(n) − nϕ(m) = (m − n) + e(n − m) ∈ Rad(M ). We now drop the assumption that n ∈ mR and show that mϕ(n) is regular with m − mϕ(n) ∈ Rad(M ). Therefore mϕ(n) ∈ mR can be used in place of n showing that there was no loss of generality in assuming n ∈ mR. Note that ϕ(n) − ϕ(m) = ϕ(n − m) ∈ Rad(R), and therefore there is s ∈ R such that s(1 − ϕ(n) + ϕ(m)) = 1. Then, using (7), ϕ(n) = s(1 − ϕ(n) + ϕ(m))ϕ(n) = sϕ(m)ϕ(n). We have sϕ ∈ M ∗ and ﬁnd that (mϕ(n)) (sϕ)(mϕ(n))

= (8)

=

(mϕ(n)) (sϕ(m)ϕ(n)) (7)

(mϕ(n)) ϕ(n) = mϕ(n),

(8)

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75

showing that mϕ(n) is regular with quasi-inverse sϕ. Furthermore, m − mϕ(n) = m−n+n−mϕ(n) = (m−n)+nϕ(n)−mϕ(n) = (m−n)+(n−m)ϕ(n) ∈ Rad(M ). 3) ⇒ 4): Obvious. 4) ⇒ 5): By hypothesis there is a decomposition M = K ⊕ L such that K ⊆ mR, K is projective, and L ∩ mR is small. Hence mR = K ⊕ (L ∩ mR) and it follows that K = kR where m = k + for some ∈ L ∩ mR. As K is projective, by the Basis Lemma applied to the generating set {k}, we have ϕ ∈ K ∗ such that for every x ∈ K, x = kϕ(x). Extend ϕ to all of M by stipulating that ϕ(L) = 0. Then ϕ ∈ M ∗ and ϕ(m)2 = ϕ(k)2 = ϕ(kϕ(k)) = ϕ(k) = ϕ(m). Finally, m − mϕ(m) = k + − mϕ(k) = k + − kϕ(k) − ϕ(k) = − ϕ(k) ∈ L ∩ mR ⊆ Rad(M ). 5) ⇒ 6) By hypothesis we have ϕ ∈ M ∗ , ϕ(m)2 = ϕ(m), and m − mϕ(m) ∈ Rad(M ). Set n := mϕ(m). Then m − n ∈ Rad(M ), and mR = mϕ(m)R ⊕ m(1−ϕ(m))R = nR⊕(m−n)R. Hence (m−n)R is cyclic and contained in the radical, so small. It remains to show that n is regular. As nϕ(n) = mϕ(m)ϕ(mϕ(m)) = mϕ(m)ϕ(m)ϕ(m) = mϕ(m) = n this is indeed the case. 6) ⇒ 2) is again obvious.

An important characterization of semiregular modules, also due to Nicholson, follows. Theorem 3.4 ([26, Theorem 1.6]). For a module MR the following statements are equivalent. 1) M is semiregular. 2) For every ﬁnitely generated submodule N of M , there is e2 = e ∈ End(MR ) such that e(M ) ⊆ N , e(M ) is projective and N (1 − e) ⊆ Rad(M ). 3) Every ﬁnitely generated submodule N of M lies over a projective direct summand of M . Proof. 1) ⇒ 2): We induct on the number n of generators of N . For n = 1 the claim follows from Proposition 3.3.3). Suppose that n > 1 and N = a1 R ⊕ · · · ⊕ an R. Choose e2 = e ∈ End(MR ) such that e(M ) ⊆ an R, e(M ) is projective, and (1 − e)(an R) ⊆ Rad(M ). Set K := (1 − e)(a1 )R ⊕ · · · ⊕ (1 − e)(an−1 )R, and, by induction hypothesis, choose d2 = d ∈ End(MR ) such that d(M ) ⊆ K, d(M ) is projective and (1 − d)(K) ⊆ Rad(M ). Then e(K) = 0, hence (ed)(M ) = 0 which means that ed = 0. We have M = e(M ) ⊕ (1 − e)(M ) = d(M ) ⊕ (1 − d)(M ) and ed = 0. Now set k := e + d − de. Then k2

= =

(e + d − de)(e + d − de) = e2 + de + d2 − d2 e − de2 e + de + d − de − de = e + d − de = k

and we easily ﬁnd that ke = e2 + de − de2 = e,

and kd = d2 = d.

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Chapter VI. Relative Regularity: U -Regularity and Semiregularity

Further, N = K + an R ⊇ d(M ) + e(M ) and d(M ) + e(M ) = d(M ) ⊕ e(M ) because ed = 0. Also k(M ) = (e + d − de)(M ) ⊆ e(M ) + d(M ) = ke(M ) + kd(M ) ⊆ k(M ), so k(M ) = d(M ) ⊕ e(M ) is projective. Finally, (1 − k)(N ) = (1 − d − e + de)(N ) = (1 − d)(1 − e)(N ) ⊆ Rad(M ). 2) ⇒ 3): We have M = e(M )⊕(1−e)(M ) and (1−e)(N ) = N ∩(1−e)(M ) ⊆ Rad(M ). Since (1 − e)(N ) is ﬁnitely generated, it is in fact small. Hence N lies over a projective summand of M . ular.

3) ⇒ 1): By hypothesis and Proposition 3.3 every element of M is semireg

Zelmanowitz studied the class of regular modules, and Nicholson was interested in the wider class of semiregular modules. We list some of Nicholson’s more striking results without proof. Corollary 3.5 ([26, Corollary 1.7]). A projective module M is semiregular if and only if M/N has a projective cover for every ﬁnitely generated submodule N of M. Theorem 3.6 ([26, Theorem 1.12]). Let M be a countably ∞generated semiregular module and suppose that Rad(M ) is small. Then M ∼ = i=1 ei R where e2i = ei ∈ R. In particular, M is projective. Corollary 3.7 ([26, Corollary 1.18]). A ﬁnitely generated projective module M is semiregular if and only if it satisﬁes the following conditions. 1) Every ﬁnitely generated submodule of M/ Rad(M ) is a direct summand. 2) Direct decompositions of M/ Rad(M ) can be lifted to M .

4

Semiregularity for Hom

We now consider H := HomR (A, M ). We saw in Chapter IV, “Regularity in Modules” that H can naturally be considered a module over six possibly diﬀerent rings, in particular the rings S := End(MR ) and T := End(AR ). For each of these there is the concept of semiregular according to Nicholson. We will consider another natural deﬁnition. For the deﬁnition of Rad(A, M ) see Lemma I.2.2. Deﬁnition 4.1. The map f ∈ HomR (A, M ) is semiregular or S-semiregular or T -semiregular if there exists a regular or S-regular or T -regular map h ∈ H, respectively, such that h − f ∈ Rad(A, M ). (9) This deﬁnition diﬀers from Nicholson’s because Rad(A, M ), Rad(S H), and Rad(HT ) may diﬀer. However, it is known ([20, Corollary 2.3]) that Rad(S H) ∪ Rad(HT ) ⊆ Rad(A, M ).

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77

This means that there are potentially more semiregular, S-, and T -semiregular maps according to Deﬁnition 4.1 than there are according to Nicholson’s deﬁnition. On the other hand, we have also seen (Lemma 2.1) that f ∈ H regular is stronger than either S-regular or T -regular. We verify a remark for a semiregular map f and give a characterization of an S-regular map f . We work with S-regularity which is forced more or less by the fact that we have to apply the dual basis lemma for projective modules. Remark. Let f ∈ HomR (A, M ) and assume that f is semiregular. Then 1) f is Rad(A, M )-regular; 2) f is partially invertible. Proof. 1) By assumption we have h−f = u ∈ Rad(A, M ) where h is regular. Then h = f + u and there exists g ∈ HomR (M, A) such that h = hgh and as h = f + u, f + u = (f + u)g(f + u) = f gf + ug(f + u) + (f + u)gu. This implies that f − f gf = −u + ug(f + u) + (f + u)gu ∈ Rad(A, M ), and hence f is Rad(A, M )-regular. 2) Again we have a regular map h = f + u where u ∈ Rad(A, M ). By way of contradiction assume that f is not partially invertible. Then f ∈ Tot(A, M ) and h = f + u ∈ Tot(A, M ) + Rad(A, M ) = Tot(A, M )

(10)

by [20, Theorem 2.4]. However, h = hgh for some g ∈ HomR (M, A) and e := hg = e2 hence h is partially invertible contrary to (10). We now prove an interesting theorem that connects semiregularity with other standard notions. Theorem 4.2. If f ∈ HomR (A, M ) is semiregular, then Sf lies over an S-projective direct summand of S HomR (A, M ). Although f semiregular implies that f is S-semiregular, Nicholson’s result cannot be applied here because Rad(A, M ) is not known to be contained in Rad(S HomR (A, M )) and therefore f is not known to be semiregular in S HomR (A, M ) in the sense of Nicholson. Proof. By assumption we have h − f ∈ Rad(A, M ) with hgh = h for some g ∈ HomR (M, A). Since h − f ∈ Rad(A, M ) it follows that (h − f )g ∈ Rad(S). Then there exists s0 ∈ S such that (1 − (h − f )g)s0 = 1. This implies that f gs0 = 1 − (1 − hg)s0 .

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Chapter VI. Relative Regularity: U -Regularity and Semiregularity

Multiplying by hg on the left and using that (hg)2 = hg, we get hgf gs0 = hg. Now multiplying by hgf on the right, we get hgf gs0 hgf = hghgf = hgf, and this shows that hgf is regular. Hence hgf has all the good properties of regular homomorphisms. In particular, Shgf is a projective direct summand of S H. Since hg is an idempotent in S, we have that Shgf ⊆ Sf . This is the ﬁrst of the conditions that we had to satisfy. Now we set h1 := hgf,

g1 := gs0 ,

e1 := g1 h1 ∈ T.

It follows that h1 e1 = h1 g1 h1 = h1

(11)

and further Sh1 = Sh1 e1 ⊆ He1 = Hg1 h1 ⊆ Sh1 , hence He1 = Sh1 . It now follows with the idempotent e1 that SH

= He1 ⊕ H(1 − e1 ) = Sh1 ⊕ H(1 − e1 ).

We still have to show that U := Sf ∩ S H(1 − e1 ) ⊆ Rad(A, M ). We ﬁrst observe that U = Lf for some left ideal L of S. Then we show that f − h1 =: v ∈ Rad(A, M ). To do so, write f = h − u with u ∈ Rad(A, M ). Then v = f − h1

=

f − hgf = h − u − hg(h − u)

=

h − hgh − u + hgu = −u + hgu ∈ Rad(A, M ).

Since h1 (1 − e1 ) = 0 by (11), U = U (1 − e1 ) and the theorem is proved.

= Lf (1 − e1 ) = L(h1 + v)(1 − e1 ) = Lv(1 − e1 ) ⊆ Rad(A, M ),

Theorem 4.2 raises the immediate question whether the converse is true. We can use the results of Nicholson to show that f is S-semiregular, but it is open whether this also means that f is semiregular as a map.

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Theorem 4.3. Let f ∈ HomR (A, M ). If Sf lies over an S-projective S-direct summand of S HomR (A, M ), then f is S-semiregular. Proof. By Proposition 3.3.4, f is semiregular as a element of the module S H, hence there is a map g ∈ H that is regular as an element of the module S H, i.e., S-regular, and f − g ∈ Rad(S H). Since Rad(S H) ⊆ Rad(A, M ), the map f is S-semiregular in the sense of Deﬁnition 4.1. Question 4.4. If f ∈ Rad(A, M ), then by deﬁnition of Rad(A, M ) for every g ∈ HomR (A, M ) it follows that f g ∈ Rad(S) and, equivalently, gf ∈ Rad(T ). If σ ∈ HomS (S H, S) and f ∈ Rad(A, M ), is f σ ∈ Rad(S)? We conjecture that this is true.

Chapter VII

Reg(A, M ) and Other Substructures of Hom 1

Substructures of Hom

We turn to connections between Reg(A, M ) and other substructures of H := HomR (A, M ). By X ⊆∗ A we mean that X is large (= essential) in A, and by X ⊆◦ A we mean that X is small (= superﬂuous) in A. Interesting substructures of Hom are the following. The singular submodule of S HT is by deﬁnition the S-T -submodule Δ(A, M ) := {f ∈ HomR (A, M ) | Ker(f ) ⊆∗ A}. The cosingular submodule of H is the S-T -submodule ∇(A, M ) := {f ∈ HomR (A, M ) | Im(f ) ⊆◦ M }. We also call f ∈ HomR (A, M ) singular if Ker(f ) ⊆∗ A, and cosingular if Im(f ) ⊆◦ M. Δ(A, M ) and ∇(A, M ) are not only S-T -bimodules but “ideals” in Mod-R having the following properties. Lemma 1.1. 1) If f1 , f2 ∈ Δ(A, M ), then f1 + f2 ∈ Δ(A, M ). 2) If X, Y ∈ Mod-R, f ∈ Δ(A, M ), g ∈ HomR (M, Y ), and h ∈ HomR (X, A), then gf h ∈ Δ(X, Y ). 3) If f1 , f2 ∈ ∇(A, M ), then f1 + f2 ∈ ∇(A, M ).

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Chapter VII. Reg(A, M ) and Other Substructures of Hom

4) If X, Y ∈ Mod-R, f ∈ ∇(A, M ), g ∈ HomR (M, Y ), and h ∈ HomR (X, A), then gf h ∈ ∇(X, Y ). Proof. 1) Ker(f1 ) ∩ Ker(f2 ) ⊆ Ker(f1 + f2 ). As the intersection of two large submodules is large, also Ker(f1 ) ∩ Ker(f2 ) is large and so Ker(f1 + f2 ) is large. Hence f1 + f2 ∈ Δ(A, M ). 2) Suppose that f ∈ Δ(A, M ), i.e., f ∈ HomR (A, M ) and Ker(f ) is large. Let g ∈ HomR (M, Y ). As Ker(f ) ⊆ Ker(gf ) it follows that Ker(gf ) is large and gf ∈ Δ(A, Y ). Let h ∈ HomR (X, A). To show that Ker(f h) is large, let 0 = B ⊆ X be given. In case B ⊆ Ker(h) we have B ⊆ Ker(f h) which shows that Ker(f h) intersects B non-trivially. In case B ⊆ Ker(h), we have that h(B) = 0. Since Ker(f ) ⊆∗ A, it follows that Ker(f )∩h(B) = 0. Let b ∈ B with 0 = h(b) ∈ Ker(f ). Then f h(b) = 0, hence B ∩ Ker(f h) = 0. Together we have that Ker(f h) ⊆∗ X in all cases. 3) Im(f1 + f2 ) ⊆ Im(f1 ) + Im(f2 ) and the sum of two small submodules is clearly small. So Im(f1 + f2 ) is also small and f1 + f2 ∈ ∇(A, M ). 4) Suppose that f ∈ ∇(A, M ), i.e., f ∈ HomR (A, M ) and Im(f ) is small. Let h ∈ HomR (X, A). Then Im(f h) ⊆ Im(f ) ⊆◦ M so Im(f h) ⊆◦ M . Let g ∈ HomR (M, Y ) and suppose that Y1 + Im(gf h) = Y . We will show that Y1 = Y . Let x ∈ M , then g(x) = y1 + gf h(x1 ) for some y1 ∈ Y1 and some x1 ∈ X. Then y1 = g(x − f h(x1 )), so x = (x − f h(x1 )) + f h(x1 ) ∈ g −1 (Y1 ) + Im(f h). As Im(f h) is small in M , we have g −1 (Y1 ) = M , so Y1 ⊆ g(M ) and therefore Y = Y1 + Im(gf h) = Y1 . There are three diﬀerent concepts of radical in H. There are the radicals of H as left S-module, and of H as right T -module, namely Rad(S H) := {f ∈ H | Sf ⊆◦ S H}, Rad(HT ) := {f ∈ H | f T ⊆◦ HT }. The most important radical is the following for which there are two equivalent deﬁnitions (Lemma Ion radicals). Rad(H)

= Rad(HomR (A, M )) = Rad(A, M ) := {f ∈ H | f HomR (M, A) ⊆ Rad(S)} = {f ∈ H | HomR (M, A)f ⊆ Rad(T )}.

(1)

It is easy to see that Rad(S H), Rad(HT ) ⊆ Rad(H), and that for any A, M , Δ(A, M ), ∇(A, M ), Rad(A, M ) ⊆ Tot(A, M ). A natural question arises: For which pairs A, M does the total equal one or more of the other structures? The sub-bimodules Δ(A, M ), ∇(A, M ), and Rad(A, M ) all have intersection {0} with Reg(A, M ).

1. Substructures of Hom

83

Proposition 1.2. 1) Reg(A, M ) ∩ Δ(A, M ) = {0}. 2) Reg(A, M ) ∩ ∇(A, M ) = {0}. 3) Reg(A, M ) ∩ Rad(A, M ) = {0} Proof. 1) Indeed, if 0 = f ∈ Reg(A, M ), then Ker(f ) = A and Ker(f ) ⊆⊕ A, so Ker(f ) is not large in A and f ∈ Δ(A, M ). 2) If 0 = f is regular, then 0 = Im(f ) ⊆⊕ M , hence Im(f ) is not small in M and f ∈ ∇(A, M ). 3) If 0 = f is regular with f gf = f , then e := f g = e2 = 0 and e cannot be in Rad(S). Reg(A, M ) and Tot(A, M ) are on opposite substructures. Proposition 1.3. 1) Reg(A, M ) ∩ Tot(A, M ) = 0. 2) Tot(S) Reg(A, M ) = Reg(A, M ) Tot(T ) = 0. Proof. 1) This follows because every nonzero regular homomorphism is partially invertible. 2) Let s ∈ Tot(S), and f ∈ Reg(A, M ). Since Reg(A, M ) is a left S-module, we have sf ∈ Reg(A, M ). Assume sf = 0. Then since sf is regular, it is partially invertible. Then by Lemma 2.3.1, s must be partially invertible, a contradiction. Question 1.4. For which pairs A, M is HomR (A, M ) = Reg(A, M ) ⊕ Tot(A, M )? What can be said about Σ Tot(A, M ), the additive closure of Tot(A, M )? As Rad(A, M ) ⊆ Tot(A, M ) ([20, Theorem 2.4]) the new Proposition 1.3.1 generalizes Proposition 1.2.3. Example 1.5. Suppose that A and M are such that HomR (M, A) = 0. Then it is easily seen that • Rad(HomR (A, M ) = HomR (A, M ); • Reg(A, M ) = 0; • Tot(A, M ) = HomR (A, M ). We provide some simple examples from abelian group theory. Example 1.6. Let A = Z⊕Z and M = Q. Then Hom(M, A) = 0 and Hom(A, M ) ∼ = Q ⊕ Q.

84

Chapter VII. Reg(A, M ) and Other Substructures of Hom

1) For any f ∈ Hom(A, M ), we have A/ Ker(f ) ∼ = Im(f ) ⊆ Q. So Ker(f ) is not large in A (Proposition I.3.8) unless f = 0. We conclude that Δ(A, M ) = 0. 2) For any f ∈ Hom(A, M ), the image Im(f ) is ﬁnitely generated and therefore has no divisible epimorphic image. Applying Proposition I.3.7 we see that Im(f ) is small in M = Q and we conclude that ∇(A, M ) = Hom(A, M ). Example 1.7. Let A = Z(p2 ) and M = Z(p∞ ). Then Hom(M, A) = 0 and Hom(A, M ) ∼ = Z(p2 ). 1) For any f ∈ Hom(A, M ), we have that Ker(f ) is 0 or pZ(p2 ) or Z(p2 ). This means that the kernel is large if and only if pZ(p2 ) ⊆ Ker(f ). So the homomorphisms with large kernel are exactly those that factor through Z(p2 )/pZ(p2 ). We conclude that Δ(A, M ) ∼ = Hom(Z(p2 )/pZ(p2 ), Z(p∞ ) ∼ = Z(p). 2) For any f ∈ Hom(A, M ), the image Im(f ) ⊆ Z(p∞ ) is always small (Proposition I.3.7) and we conclude that ∇(A, M ) = Hom(A, M ). Example 1.8. Let A = Z and M = Z(p). Then Hom(M, A) = 0 and Hom(A, M ) ∼ = Z(p). 1) For any f ∈ Hom(A, M ), we have that Ker(f ) is large. We conclude that Δ(A, M ) = Hom(A, M ). 2) For any f ∈ Hom(A, M ), the image Im(f ) ⊆ Z(p) is not small unless Im(f ) = 0. We conclude that ∇(A, M ) = 0. Example 1.9. Let p and q be diﬀerent prime numbers, let A = Z(p) ⊕ Z(q 2 ) and M = A. Then Hom(A, M ) ∼ = Hom(Z(p), Z(p)) ⊕ Hom(Z(q 2 ), Z(q 2 )) ∼ = Z(p) ⊕ Z(q 2 ), while

Reg(A, M ) ∼ = Hom(Z(p), Z(p)) ∼ = Z(p),

so 0 = Reg(A, M ) = Hom(A, M ). Proof. Because of orders Hom(Z(p), Z(q 2 )) = 0 and Hom(Z(q 2 ), Z(p)) = 0. Since the cyclic group Z(p) of order p is simple, Reg(Z(p), Z(p)) = Hom(Z(p), Z(p)) while Reg(Z(q 2 ), Z(q 2 )) = 0 because Z(q 2 ) is indecomposable but its endomorphism ring is not a division ring. Example 1.10. Let A = p∈P Z(p) and M = p∈P Z(p). Then A is equal to the maximal torsion subgroup of M , and Z(p), Z(p)) Hom(A, M ) = Hom(A, A) = Hom( ∼ =

p∈P

Hom(Z(p),

p∈P

p∈P

Z(p)) ∼ =

p∈P

p∈P

Hom(Z(p), Z(p)),

2. Properties of Δ(A, M ) and ∇(A, M ) while Reg(A, M ) =

p∈P

85

Hom(Z(p), Z(p)).

Again 0 = Reg(A, M ) = Hom(A, M ). Proof. Note that A is elementary, i.e., semisimple, ∼ End(A) = p∈P Hom(Z(p), p∈P Z(p)) = p∈P End(Z(p)), and similarly,

End(M ) ∼ = p∈P End(Z(p)).

A map f ∈ Hom(A, M ) is regular if and only if Im(f ) is a direct summand of M and this is the case if and only if Im(f ) is ﬁnite. The reason for this is that for any f ∈ Hom(A,M ) = Hom(A, A), we have Im(f ) = p∈P Z(p) for some set P of primes and p∈P Z(p)/ p∈P Z(p) is divisible and therefore p∈P Z(p) Z(p) and hence not of Z(p) unless cannot be a direct summand of p∈P p∈P Z(p)/ Z(p) = 0, which is equivalent to P being ﬁnite. The maps in p∈P p∈P Hom(A, M ) = p∈P Z(p) with ﬁnite image are exactly those in p∈P Z(p). This subgroup of regular elements is evidently an S-T -submodule hence is the largest regular bi-submodule of Hom(A, M ).

2

Properties of Δ(A, M ) and ∇(A, M )

Let R be a ring and denote by Rad(R) the radical of R. The radical has the following well-known property: For a left or right ideal A of R it is true that A ⊆ Rad(R) ⇔ ∀a ∈ A, 1 − a is an invertible element of R.

(2)

Now consider H := HomR (A, M ), S := End(MR ), T := End(AR ). Then H is an S-T -bimodule. For f ∈ H and g ∈ HomR (M, A) we have gf ∈ T , f g ∈ S. Thus HomR (M, A)f is a left ideal in T and f HomR (M, A) is a right ideal in S. We ask whether (2) can be applied with gf or f g in place of a but with weaker or diﬀerent conditions replacing invertibility. We need further notions (see [20]). Deﬁnition 2.1. 1) A modules V is called locally injective if and only if for every submodule A ⊆ V that is not large (= essential) in V , there exists a nonzero injective submodule Q of V with A ∩ Q = 0. 2) A modules W is called locally projective if and only if for every submodule B ⊆ W that is not small in W , there exists a nonzero projective direct summand P of W with P ⊆ B. 3) A module N is called large restricted if and only if every monomorphism f : N → N with Im(f ) ⊆∗ N is an automorphism, i.e., Im(f ) = N .

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Chapter VII. Reg(A, M ) and Other Substructures of Hom

4) A module N is called small restricted if and only if every epimorphism f : N → N with Ker(f ) ⊆◦ N is an automorphism, i.e., Ker(f ) = 0. The following results are easily veriﬁed (see [20]). • Every injective module is locally injective. • Every projective and semiperfect module is locally projective. • Every injective module is large restricted. • Every projective module is small restricted. Remark. ([20]) If RR is not Noetherian, then there exist inﬁnite direct sums of injective R-modules that are not injective. But these are locally injective! We would like to have an example of a ring R such that RR is locally injective but not injective. Also, we know no example of a locally projective ring that is not projective. Theorem 2.2 ([20]). 1) The module A is locally injective if and only if for all M ∈ Mod-R, Δ(A, M ) = Tot(A, M ). 2) The module M is locally projective if and only if for all A ∈ Mod-R, ∇(A, M ) = Tot(A, M ). 3) If A is large restricted, then Δ(A, M ) ⊆ Rad(A, M ) for all M ∈ Mod-R. 4) If M is small restricted, then ∇(A, M ) ⊆ Rad(A, M ) for all A ∈ Mod-R. Combining these properties we obtain the following corollary. Corollary 2.3 ([20]). 1) If A is locally injective and M is locally projective, then Δ(A, M ) = ∇(A, M ) = Tot(A, M ). 2) If A is injective and M is locally projective or if A is locally injective and M is projective and semi-perfect, then Δ(A, M ) = ∇(A, M ) = Rad(A, M ) = Tot(A, M ). We include a variant of such results with proof. Proposition 2.4. If A or M is large restricted or injective, then Δ(A, M ) ⊆ Rad(A, M ).

3. The Special Case HomR (R, M )

87

Proof. Let f ∈ Δ(A, M ), i.e., Ker(f ) ⊆∗ A and let g ∈ HomR (M, A). Then also Ker(gf ) ⊆∗ A since Ker(f ) ⊆ Ker(gf ). But also Ker(f g) ⊆∗ M . To see this, let 0 = U ⊆ M . If g(U ) = 0, then U ⊆ Ker(f g) and all is well. So suppose that g(U ) = 0. Then Ker(f ) ∩ g(U ) = 0 since Ker(f ) ⊆∗ A. Hence there exists 0 = u ∈ U such that g(u) = 0 but f g(u) = 0. This means that Ker(f g) ⊆∗ M . Consider a ∈ Ker(1 − gf ) ∩ Ker(gf ). Then 0 = (1 − gf )a = a − gf (a) = a, hence Ker(1 − gf ) ∩ Ker(gf ) = 0. Since Ker(gf ) ⊆∗ A, this implies Ker(1 − gf ) = 0, hence 1 − gf is a monomorphism. It follows similarly that 1 − f g is a monomorphism. Further we have Ker(gf ) ⊆ Im(1 − gf ) which implies that Im(1 − gf ) ⊆∗ A. If A is large restricted, then it follows that 1 − gf is also an epimorphism, so altogether an automorphism. If A is injective, then it is large restricted. It follows similarly that 1 − f g is an automorphism if M is large restricted, and in particular if M is injective. Now apply (2) and consider the deﬁnition of Rad(A, M ). This is again a result connecting two substructures of Hom(A, M ). Next we consider ∇(A, M ). Here for f ∈ ∇(A, M ), Im(f ) ⊆◦ M and for g ∈ HomR (M, A), it follows that also Im(gf ) ⊆◦ A, Im(f g) ⊆◦ M . Then from Im(1 − gf ) + Im(gf ) = A,

Im(1 − f g) + Im(f g) = M

it follows that Im(1 − gf ) = A, Im(1 − f g) = M , hence 1 − gf , 1 − f g are epimorphisms. Proposition 2.5. Let f ∈ ∇(A, M ). 1) If A is projective, then for all g ∈ HomR (M, A), the map 1 − gf is regular. 2) If M is projective, then for all g ∈ HomR (M, A) the map 1 − f g is regular. Proof. 1) We know that 1 − gf is an epimorphism on the projective module A, and therefore splits. Thus Im(1 − gf ) and Ker(1 − gf ) are direct summands which implies that 1 − gf is regular. 2) Same as 1).

3

The Special Case HomR (R, M )

Until now the singular submodules did not occur explicitly. We show now that they are already included as a special case in our general considerations.

88

Chapter VII. Reg(A, M ) and Other Substructures of Hom

Finally we consider two special cases of this result; ﬁrst the case A = R, M arbitrary, and then the case A = M = R. Here we apply the isomorphism ρ : HomR (R, M ) f → f (1) ∈ M from HomR (R, M ) to M , and from HomR (R, R) to R. Then Δ(MR ) := ρ(Δ(R, M ) = {m ∈ M | AnnR (m) ⊆∗ RR }, that is, Δ(MR ) is properly the singular submodule of M . Here S := End(MR ) as before, but T = End(RR ) ∼ = R. Corollary 3.1. 1) If RR or MR is injective, then Δ(MR ) ⊆ Rad(MR ). 2) If RR is injective, then Δ(RR ) ⊆ Rad(R). We again consider the special case A = R and translate using ρ. Then f ∈ ∇(R, M ) ⇔ f (1)R ⊆◦ M ⇔ m := f (1) ∈ Rad(M ), i.e., ∇(MR ) := ρ(∇(R, M )) = Rad(M ). It follows for all g ∈ HomR (M, R) that g(m) ∈ Rad(R) and from this we get the result that 1 − g(m) is not only regular but invertible. For projective M , the map 1 − mg is again regular. If we assume ﬁnally that A = M = R, then m ∈ Rad(R) and the action of g ∈ HomR (R, R) is the left multiplication in R by an element of R. Now gm and mg are in Rad(R) and 1 − gm and 1 − mg are invertible. Question 3.2. If RR is injective, is it true that for all M ∈ Mod-R we have Rad(M ) = Tot(M )? Question 3.3. For ρ : HomR (R, M ) f → f (1) ∈ M , does it follow that ρ (Rad(R, M )) = Rad(M )? We found that ρ(∇(R, M )) = Rad(M ). What is now ρ(Rad(R, M ))? By deﬁnition, f ∈ Rad(R, M ) if and only if for all ϕ ∈ M ∗ := HomR (M, R), (ϕf )(r) = ϕ(f (1)r) = ϕ(f (1))r ∈ Rad(R), hence ϕ(f (1)) ∈ Rad(R). This implies that ρ(Rad(R, M )) = {m ∈ M | ∀ ϕ ∈ M ∗ , ϕ(m) ∈ Rad(R)}. Since for any homomorphism ϕ, ϕ(Rad(M )) ⊆ Rad(R), it follows that Rad(M ) ⊆ ρ(Rad(R, M )). Finally, we have to describe ρ(Tot(R, M )). Recall that Tot(MR ) := {m ∈ M | m is not partially invertible.}.

3. The Special Case HomR (R, M ) Proposition 3.4.

89

ρ(Tot(R, M )) = Tot(MR ).

Proof. Recall that f ∈ HomR (R, M ) is partially invertible if and only if there exists ϕ ∈ M ∗ such that ϕ(f (r)) = ϕ(f (1))r is left multiplication in R by the idempotent ϕ(f (1)). Hence ρ(Tot(R, M )) = {m ∈ M | ∀ ϕ ∈ M ∗ , ϕ(m) is not an idempotent = 0}. By Lemma IV.2.3, f ∈ HomR (R, M ) is partially invertible if and only if f (1) is partially invertible. Hence f is not partially invertible if and only if f (1) is not partially invertible. This is the claim. We now specialize the results of Theorem 2.2 and Corollary 2.3 to the case HomR (R, M ) and translate them by means of ρ into results about M . Corollary 3.5. Let R be a ring. 1) The module RR is locally injective if and only if ∀ M ∈ Mod-R : Δ(M ) = Tot(M ). 2) If the module RR is injective, then ∀ M ∈ Mod-R : Δ(M ) = Rad(M ) = Tot(M ). 3) If RR is injective and W is locally projective, then Δ(W ) = Rad(W ) = Tot(W ). 4) If RR is locally injective and W is locally projective, then Δ(W ) = Rad(W ) = Tot(W ). 5) If W is locally projective, then Rad(W ) = Tot(W ). 6) If RR is large restricted, then ∀ M ∈ Mod-R : Δ(M ) ⊆ ρ(Rad(R, M ). We now show with an example that the equalities that have been established under certain assumptions, are not always true.

90

Chapter VII. Reg(A, M ) and Other Substructures of Hom

Example 3.6. Let R be a ring that is not right-Kasch. This means that there exists a simple right R-module B for which there does not exist an isomorphic right ideal of R. Then the only homomorphism from B to R is the 0-mapping. Hence Rad(B) = 0 but ρ(Rad(R, B)) = Δ(B) = Tot(B) = B. Proof. To prove Δ(B) = B, let 0 = b ∈ B. Then bR = B and the mapping R r → br ∈ bR = B is an epimorphism, hence R/ AnnR (b) ∼ = B. Therefore C := AnnR (b) is a maximal right ideal of R. We will show that C is large in RR . Assume that 0 = U ⊆ RR and C ∩ U = 0. Since C is maximal it follows that C ⊕ U = R. But then B ∼ = R/C ∼ = U , a contradiction to the choice of B. Since C := AnnR (b) is large, b is singular, hence Δ(B) = B. What is now Tot(B)? Since {0} = HomR (B, R), no element in B is partially invertible, hence Tot(B) = B.

4

Further Internal Properties of Δ(M )

The advantage of identifying Δ(M ) with ρ(Δ(R, M )) rests on the fact that the general results about Δ(A, M ) can be applied to this special case. We now consider more internal properties. A module M is singular if Δ(M ) = M . It is easy to ﬁnd examples of singular modules, e.g., by employing the following remark. Remark. If A ⊆∗ M , then M/A is singular. Proof. Set m = m + A ∈ M/A. We have to show: If 0 = U is a right ideal of R, then AnnR (m) ∩ U = 0. Since U is a right ideal of R, mU is a submodule of M . Case mU = 0. Then U ⊆ AnnR (m) ⊆ AnnR (m). Case mU = 0. Choose u ∈ U such that mu = 0. Then muR = 0. Since muR ⊆ M and A ⊆∗ M , it follows that A ∩ muR = 0. Choose 0 = a = mur0 ∈ A ∩ muR. Then 0 = ur0 ∈ AnnR (m) ∩ U as needed. There are situations in which the Remark is used in the following form: If M/A is not singular, the A is not large in M . A module is nonsingular if its only singular element is 0. Note that in contrast “not singular” means that at least one element is not singular. By Zorn’s Lemma it follows immediately that every module contains a maximal nonsingular submodule. Lemma 4.1. 1) If A and B with A ∩ B = 0 are nonsingular submodules of a module M , then A + B = A ⊕ B is a nonsingular submodule of M . 2) If A ⊆∗ M and A is nonsingular, then M is nonsingular. 3) If A ⊆∗ M and A is singular, then M does not contain a nonsingular submodule diﬀerent from 0.

4. Further Internal Properties of Δ(M )

91

4) If C is a maximal nonsingular submodule of M , then Δ(M ) ∩ C = 0,

Δ(M ) + C = Δ(M ) ⊕ C ⊆∗ M,

and C is a complement of Δ(C) in M . 5) Let C be a maximal nonsingular submodule of M . If A is a complement of C in M with Δ(M ) ⊆ A, then Δ(M ) ⊆∗ A and A + C = A ⊕ C ⊆∗ M . Proof. 1) Let 0 = a ∈ A, b ∈ B, and assume that (a + b)r = ar + br = 0 where r ∈ R. Then ar = −br ∈ A∩B = 0, hence ar = 0. This implies that AnnR (a+b) ⊆ AnnR (a). Since by assumption AnnR (a) is not large, also AnnR (a + b) is not large and hence a + b is not singular. 2) The assumption that Δ(M ) = 0 would imply that A ∩ Δ(M ) = 0, and A would not be nonsingular contrary to hypothesis. 3) Assume by way of contradiction that B is a nonsingular submodule of M . Then A ∩ B = 0, hence A contains elements that are non-singular contrary to hypothesis. 4) By deﬁnition of Δ(M ) we have that Δ(M ) ∩ C = 0. Suppose 0 = U ⊆ M . Assume that (Δ(M ) + C) ∩ U = 0. Then U cannot contain a nonzero singular element since this would also belong to Δ(M ). Hence U is nonsingular. Since C ∩ U = 0, it follows by 1) that C + U is nonsingular contrary to the maximality of C. Therefore (Δ(M ) + C) ∩ U = 0. Now let B be a complement of Δ(M ) with C ⊆ B. Then B cannot contain a nonzero singular element since such an element would also belong to Δ(M ). Hence B is nonsingular and equal to the maximal nonsingular submodule C. 5) This statement is true in general for complements but in our case there is a shorter proof. Assume that 0 = U ⊆ A and Δ(M ) ∩ U = 0. Then U is nonsingular. Since C ∩ A = 0, also C ∩ U = 0. By 1) C + U is nonsingular which contradicts the maximality of C. By 4) it follows that A + C ⊆∗ M . We would like to apply our results also to abelian groups, that is, Z-modules, but it is not more work to do this not only for Z-modules but for (right) uniform rings. A ring R is called (right) uniform if and only if every nonzero right ideal of R is large in RR , or, what means the same thing, any two nonzero right ideals have nonzero intersection. It is obvious that any commutative ring without zero divisors is uniform. If now M is a module over a uniform ring R and if for some m ∈ M , AnnR (m) = 0, then AnnR (m) is a large right ideal, hence m is singular. This means: If m is torsion, then m is singular. Hence the maximal torsion submodule of M is equal to Δ(M ) and M is torsion–free means that M is nonsingular. If m is not torsion = singular, then mR is nonsingular; this we will use in the following. Lemma 4.2. Let RR be uniform. Then Δ(M ) is a complement of any maximal nonsingular submodule C of M .

92

Chapter VII. Reg(A, M ) and Other Substructures of Hom

Proof. Let C be a maximal nonsingular submodule of M and let A be a complement of C in M with Δ(M ) ⊆ A. Assume that a ∈ A, a ∈ / Δ(M ). Then aR is nonsingular. But since C ∩ A = 0, also C ∩ aR = 0, hence by Lemma 4.1.1, C + aR is nonsingular in contradiction to the maximality of C. Hence Δ(M ) = A. We need a deﬁnition from [25]. A module MR is quasi-continuous if every complement in M is a direct summand, and if A and B are direct summands of M and A ∩ B = 0, then A ⊕ B is a direct summand of M . Theorem 4.3. 1) Let C be a maximal nonsingular submodule of M and A a complement of C in M containing Δ(M ). If M is quasi-continuous, then A ⊕ C = M. 2) If M is quasi-continuous, RR is uniform, and C is a maximal nonsingular submodule of M , then Δ(M ) ⊕ C = M. Proof. 1) Since A and C are complements in a quasi-continuous module, they are direct summands. Since also A ∩ C = 0, it follows that A ⊕ C is a direct summand of M . But A ⊕ C ⊆∗ M by Lemma 4.1.5, so A ⊕ C = M . 2) By Lemma 4.2 we know that A = Δ(M ); hence A ⊕ C = M implies that Δ(M ) ⊕ C = M . We return to the general situation making no assumptions on R or M . By now we have seen many interesting relations between Δ(M ), the radical and the total, and also between Δ(M ) and maximal nonsingular submodules. Yet, it is still diﬃcult to ﬁnd, for a module M , explicit properties of Δ(M ) and a complement C of Δ(M ) in M . In the following we exhibit some results concerning this topic. We will have to use the following characterization of the socle of a module (see [18, 9.1.1]): Soc(M ) = B⊆∗ M B. (3) In general, Soc(M ) is not large in M , but if M is Artinian, then Soc(M ) ⊆∗ M since it is the intersection of ﬁnitely many large submodules. For a module MR and a right ideal I of R we set AnnM (I) := {m ∈ M | mI = 0}. Theorem 4.4. 1) For any module M , Δ(M ) ⊆ AnnM (Soc(RR )). 2) If Soc(RR ) is large in RR , then Δ(M ) ⊆∗ AnnM (Soc(RR )). 3) If Soc(RR ) is large in RR and if Soc(RR ) = Soc(R R), then Rad(R) ⊆ Δ(RR ).

4. Further Internal Properties of Δ(M )

93

Proof. 1) If m ∈ Δ(M ), then it follows by (3) that m Soc(RR ) = 0, hence m ∈ AnnM (Soc(RR )). 2) If m ∈ AnnM (Soc(RR )) and if Soc(RR ) is large in M , then m ∈ Δ(M ). 3) If U is a left R-module, then Rad(R)U ⊆ Rad(U ). If U is semisimple, then Rad(U ) = 0, hence Rad(R) Soc(R R) = 0. It follows that Rad(R) Soc(RR ) = 0 hence Rad(R) ⊆ Δ(RR ). In Corollary 3.5.2 we have seen that if the ring RR is injective, then Δ(M ) = Rad(M ) = Tot(M )

(4)

for all M ∈ Mod-R. This includes, for RR injective, the identities Δ(RR ) = Rad(R) = Tot(R). In Rad(R) and Tot(R) we did not indicate whether the ring is considered a right or left R-module. This is because the concepts are independent of the side of the R-action. For Rad(R) this is well-known. For Tot(R) it follows from the fact that “partially invertible” is independent of side. There is another situation where (4) is valid. First we need the following lemma which requires another deﬁnition from [25]. A module MR is continuous if every complement in M is a direct summand, and every submodule isomorphic to a direct summand is itself a direct summand. Lemma 4.5. ([25, Proposition 3.5]) If MR is continuous and S := End(MR ), then Rad(S) = Δ(M, M ) = {f ∈ S | Ker(f ) ⊆∗ M }. Theorem 4.6. If RR is continuous, then Δ(RR ) = Rad(R) = Tot(R). Proof. We apply Lemma 4.5 with RR in place of MR . Then S is now the ring of left multiplications on R by elements of R which we identify with R. Then we obtain by Lemma 4.5 that Δ(RR ) = Rad(R). It remains to show that Tot(R) ⊆ Δ(RR ). Let b ∈ R with AnnR (b) not large in RR . Then there exists a complement K of AnnR (b). Since RR is continuous, the submodule K is a direct summand of RR . Since AnnR (b) ∩ K = 0, the mapping K x → bx ∈ bK is an isomorphism. Since RR is continuous and K is a direct summand, also bK is a direct summand. By Theorem II.2.1.4 it follows that b is partially invertible. We have now shown: If b is not singular, then it is partially invertible. Hence Tot(R) ⊆ Δ(RR ).

94

5

Chapter VII. Reg(A, M ) and Other Substructures of Hom

Non-singular Modules

We have seen that any module contains a maximal nonsingular submodule. But what does it look like? Can we describe at least some nonsingular submodule? We give some information in this direction and start with simple modules. Lemma 5.1. Let U be a simple module. Then the following statements hold. 1) U is nonsingular if and only if U is projective. 2) If U ⊆ M and U ⊆ Rad(M ), then U ⊆⊕ M . 3) If U ⊆ M , U is nonsingular, and if C is a maximal nonsingular submodule of M , then U ⊆ C. Consequently, M contains a unique largest nonsingular submodule. Proof. 1) If 0 = u ∈ U , then uR = U since U is simple. The homomorphism g : R r → ur ∈ U is an isomorphism and Ker(g) = AnnR (u). Then R/ AnnR (u) ∼ = U and since U is simple, AnnR (u) is a maximal right ideal of R. Since U is nonsingular, AnnR (u) is not a large right ideal of R. Hence there exists 0 = B ⊆ RR such that B ∩ AnnR (u) = 0. Only if: Since AnnR (u) is maximal and B = 0 it follows that R = B + AnnR (u) = B ⊕ AnnR (u). Then B is projective and as B ∼ = R/ AnnR (u) ∼ = U , so is U . If: Since now U is projective, the epimorphism g splits: RR = uR ⊕ AnnR (u),

uR = 0.

Since uR ∩ AnnR (u) = 0, AnnR (u) is not large in RR , hence u is not singular. 2) By hypothesis U ⊆ Rad(M ). Therefore U is not small in M . Hence there exists M0 M such that M = M0 + U. Since M0 = M it follows that U ⊆ M0 and since U is simple M0 ∩ U = 0, hence M = M0 ⊕ U . 3) If U is a simple nonsingular submodule of M , then either U ⊆ C or U ∩C = 0. By Lemma 4.1.1 the second case would imply that C ⊕U is nonsingular contradicting the maximality of C. We denote by Doc(M ) the sum of all simple submodules of M that are not contained in Rad(M ). Then Doc(M ) is a direct summand of Soc(M ), the socle of M: Soc(M ) = Doc(M ) ⊕ (Soc(M ) ∩ Rad(M )) .

6. A Correspondence

95

Theorem 5.2. If M is projective or locally projective, then Doc(M ) is a nonsingular and projective submodule of M . If C is a maximal nonsingular submodule of M , then Doc(M ) ⊆ C. Proof. Let U be a simple submodule of M such that U ⊆ Rad(M ). Then by Lemma 5.1.2 U ⊆⊕ M . If M is projective, then so is U . If M is locally projective, then, since U is not small in M , we know that U contains a nonzero projective direct summand: P ⊆⊕ M . Since U is simple U = P . In either case U is nonsingular by Lemma 5.1.1. By Lemma 4.1.1 also Doc(M ) is nonsingular. This is clear if Doc(M ) is the sum of ﬁnitely many simple summands. If Doc(M ) is the sum of inﬁnitely many summands, and some element of Doc(M ) is singular, then this element is contained in a ﬁnite subsum that is nonsingular. Hence the singular element must be 0. That Doc(M ) is projective is clear. It follows from Lemma 5.1.3 that Doc(M ) ⊆ C. We mention two special cases of the theorem. Corollary 5.3. 1) If RR is a ring with 1 ∈ R, then Doc(RR ) is nonsingular and projective. 2) If M is a projective or locally projective module with Rad(M ) = 0, then Soc(M ) is nonsingular and projective.

6

A Correspondence Between Submodules of HomR (A, M ) and Ideals of End(MR )

In (1) we deﬁned the bi-submodule Rad(A, M ) by using Rad(S). But the reverse is also possible. Using Rad(A, M ) we can deﬁne a two-sided ideal in S contained in Rad(S). Let Idl(Rad(A, M )) :=

g∈HomR (M,A)

Rad(A, M )g.

Then this is a two-sided ideal in S since Rad(A, M ) is a left S-module and HomR (M, A) is a T -S-bimodule. That Idl(Rad(A, M )) ⊆ Rad(S) follows from (1). We generalize these constructions to arrive at a correspondence between the lattice of two-sided ideals of S and S-T -submodules of H = HomR (A, M ). Deﬁnition 6.1. Let I be a two-sided ideal in S, and K an S-T -submodule of H. Let Mdl(I) := {f ∈ H | f HomR (M, A) ⊆ I}, and Idl(K) :=

g∈HomR (M,A)

Kg.

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Chapter VII. Reg(A, M ) and Other Substructures of Hom

We will see that Mdl(I) is an S-T -submodule of H, and Idl(K) is a two-sided ideal of S and that the maps Mdl and Idl have properties that are reminiscent of a Galois connection. Lemma 6.2. In the above context the following statements are valid. 1) Mdl(I) is an S-T -submodule of HomR (A, M ) and Idl(K) is a two-sided ideal of S; 2) if I1 ⊆ I2 , then Mdl(I1 ) ⊆ Mdl(I2 ) and if K1 ⊆ K2 , then Idl(K1 ) ⊆ Idl(K2 ); 3) I HomR (A, M ) ⊆ Mdl(I), 4) Idl(Mdl(I)) ⊆ I,

Idl(K) HomR (A, M ) ⊆ K;

K ⊆ Mdl(Idl(K));

5) Mdl(Idl(Mdl(I))) = Mdl(I), and Idl(Mdl(Idl(K))) = Idl(K); 6) for bi-submodules K1 , K2 of HomR (A, M ), Idl(K1 + K2 ) = Idl(K1 ) + Idl(K2 ); 7) for ideals I1 , I2 of End(MR ), Mdl(I1 ∩ I2 ) = Mdl(I1 ) ∩ Mdl(I2 ). Proof. 1) If f1 g, f2 g ∈ I, then (f1 + f2 )g ∈ I and Mdl(I) is additively closed. As I is a left ideal in S, if f g ∈ I, then also s(f g) = (sf )g ∈ I so Mdl(I) is a left S-module. Since HomR (M, A) is a T -S-module, it follows that tg ∈ HomR (M, A) for any t ∈ T and any g ∈ HomR (M, A). If f g ∈ I for all g, then for all t and all g it follows that f (tg) = (f t)g ∈ I, hence Mdl(I) is also a right T -module. It is similarly easy to see that Idl(K) is a two-sided ideal in S. 2) The deﬁnitions of Mdl(I) and Idl(K) make it clear that these constructions preserve inclusions. 3) For s ∈ I, f ∈ H, and g ∈ HomR (M, A) we have (sf )g = s(f g), and since f g ∈ S and I is an ideal, it follows that s(f ng) ∈ I, hence sf ∈ Mdl(I). This says that IH ⊆ Mdl(I). Now suppose that i=1 hi gi ∈ Idl(K) where hi ∈ Since gi f ∈ T K, gi ∈ HomR (M, A), and let f ∈ H. Then (hi gi )f = hi (gi f ). n (g f ) ∈ K, hence ( and K is a right T -module, it follows that h i i i=1 hi gi ) f = n h (g f ) ∈ K. i=1 i i 4) Here we need to deal with Idl(Mdl(I)) =

g∈HomR (M,A)

Mdl(I)g.

If f ∈ Mdl(I), then f g ∈ I, hence Mdl(I)g ⊆ I, and so Idl(Mdl(I)) ⊆ I. For the second inclusion, let h ∈ K, and g ∈ HomR (M, A). Then hg ∈ Idl(K) by deﬁnition of Idl(K), hence h ∈ Mdl(Idl(K)).

6. A Correspondence

97

5) We use 4) and 2). From Idl(Mdl(I)) ⊆ I it follows that Mdl(Idl(Mdl(I))) ⊆ Mdl(I).

(5)

In K ⊆ Mdl(Idl(K)) we take K = Mdl(I) to obtain Mdl(I) ⊆ Mdl(Idl(Mdl(I))). This together with (5) establishes the ﬁrst claim. From K ⊆ Mdl(Idl(K)) it follows that Idl(K) ⊆ Idl(Mdl(Idl(K))).

(6)

Taking I = Idl(K) in Idl(Mdl(I)) ⊆ I we get Idl(Mdl(Idl(K))) ⊆ Idl(K) which together with (6) establishes the second claim. 6) Obviously,

Idl(K1 + K2 ) = (K1 + K2 )g g∈HomR (M,A)

=

g∈HomR (M,A)

=

K1 g +

K2 g

g∈HomR (M,A)

Idl(K1 ) + Idl(K2 ).

7) Equally simply, Mdl(I1 ∩ I2 ) = {f ∈ H | f HomR (M, A) ⊆ I1 ∩ I2 } = Mdl(I1 ) ∩ Mdl(I2 ). Remark. In Lemma 6.2 we can rephrase 3) in terms of annihilators. The containment IH ⊆ Mdl(I) means that I ⊆ AnnS (H/ Mdl(I))

(7)

Idl(K) ⊆ AnnS (H/K).

(8)

and Idl(K)H ⊆ K means

We reformulate (some of) our results. Theorem 6.3. Let A and M be right R-modules, H := HomR (A, M ), S := End(MR ), and T := End(AR ). Let L(S) denote the lattice of all two-sided ideals of S and L(H) the lattice of all S-T -submodules of H. Then Mdl : L(S) → L(H),

Idl : L(H) → L(S)

are inclusion preserving maps with Mdl(Idl(Mdl(I))) = Mdl(I) and Idl(Mdl(Idl(K))) = Idl(K).

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Chapter VII. Reg(A, M ) and Other Substructures of Hom

Proof. Lemma 6.2. Remark. With the standard notation T := End(AR ) we also have Mdl : L(T ) → L(H) : Mdl(I) = {f ∈ H | Hom(M, A)f ⊆ I} where I is an ideal of T , and Idl : L(H) → L(T ) : Idl(K) =

g∈HomR (M,A)

gK

where K is an S-T -submodule of H, with properties as for the previous operators Mdl and Idl. If two-sided ideals of S and two-sided ideals of T are considered simultaneously we diﬀerentiate between Idl(K) = IdlS (K) and IdlT (K) := g∈HomR (M,A) gK ⊆ T . Question 6.4. Natural questions arise: 1) For which A, M , I, and K is it true that Mdl(Idl(K)) = K or Idl(Mdl(I)) = I? 2) When is it true that I HomR (A, M ) = Mdl(I) or Idl(K) HomR (A, M ) = K? 3) For which A and M is Idl(Rad(A, M )) = Rad(S)? Example 6.5. Let p be a prime, A := Z(p) and M := Z(p∞ ). Then • T := End(A) = Z/pZ, Rad(T ) = 0,

p (the ring of p-adic integers), Rad(S) = pZ

p, • S := End(M ) = Z • H := Hom(A, M ) = Hom(A, M [p]) = Z(p), • Hom(M, A) = 0, • Rad(H) = Rad(A, M ) = {f ∈ H | f Hom(M, A) ⊆ Rad(S)} = H, • Idl(Rad(H)) := g∈HomR (M,A) Rad(A, M )g = 0. Hence in this case Idl(Rad(H)) Rad(S). Of particular interest is the equality I = Idl(Mdl(I)) because it will turn Mdl Idl into a closure operator as we will see below. We have Idl(Mdl(I)) = g∈HomR (M,A) Mdl(I)g ⊆ I. To get equality it is necessary and suﬃcient to ﬁnd for each s ∈ I mappings gi ∈ HomR (M, A) and fi ∈ Mdl(I) such that s = f1 g1 + · · · + fn gn . There is an intriguing result pointing in this direction. We ask when can every endomorphism of M be factored through A? The answer follows.

6. A Correspondence

99

Lemma 6.6. Let A, M , and S = End(MR ) be as before. Then for any s ∈ S there exist f ∈ HomR (A, M ) and g ∈ HomR (M, A) such that s = f g if and only if A has a direct summand isomorphic to M . Proof. Suppose that A = M ⊕ N and σ : M → M is an isomorphism. Let s ∈ S. Then let g : M → A : g = σs and let f : A → M be such that f = σ −1 on M and f = 0 on N . Then f g = s. Conversely, assume that S 1 = f g for some f ∈ HomR (A, M ) and some g ∈ HomR (M, A). Then f is surjective and A = M ⊕ Ker(f ) with Im(g) = M ∼ = M. Lemma 6.6 suggests specializing to modules A that have a direct summand isomorphic with M . Lemma 6.7. Let M be an R-module and A = M ⊕ N with M ∼ = M . Let I be an ideal of S := End(MR ). Then, given s ∈ I, there is σ ∈ HomR (M, A) and f ∈ Mdl(I) such that s = f σ. Proof. By hypothesis there is an isomorphism σ : M → M and we let π : A → M be the projection along N . Let s ∈ I. Let f : A → M : f := sσ −1 π. Then s = f σ and we claim that f ∈ Mdl(I). This requires that for all g ∈ HomR (M, A) the composite f g ∈ I. So let g ∈ HomR (M, A). Then f g = sσ −1 πg with σ −1 πg ∈ S. Since I is an ideal and s ∈ I, we get that f g ∈ I. Note that by Lemma 6.7 the crucial hypothesis (9) of the following theorem is satisﬁed if A has a direct summand isomorphic to M . Theorem 6.8. Let A and M be R-modules, S := End(MR ), H := HomR (A, M ), and assume that Idl Mdl(I) = I for all ideals I of S. (9) Let L(S) be the set of all ideals of S. Then L(S) is a complete lattice under set inclusion where the least upper bounds are the sums and the greatest lower bounds are the intersections. Let L(H)cl := {K ⊆ S HT | K = Mdl Idl(K)}, partially ordered by set inclusion. Then Mdl(I) ∈ L(H)cl for every I ∈ L(S) and L(H)cl is a complete lattice where the greatest lower bounds are the intersections. The functions Mdl : L(S) → L(H)cl and Idl : L(H)cl → L(S) are inverse lattice isomorphisms. Proof. The lattice properties of L(S) are well-known and routinely veriﬁed. To show that L(H)cl is a complete lattice, we ﬁrst note that H ∈ L(H)cl because by Lemma 6.2.4), H ⊆ Mdl Idl(H) ⊆ H. We show next that L(H)cl is closed under arbitrary intersections which establishes the existence of greatest lower cl be a family of modules in L(H) . Then K ⊆ bounds. Let K i Kj implies that i Idl( Ki ) ⊆ j Idl(Kj ) and further j Kj ⊆ Mdl Idl( j Kj ) ⊆ j Mdl Idl(Kj ) = cl j Kj . To show the existence of least upper bounds in L(H) , let Ki be a family

100

Chapter VII. Reg(A, M ) and Other Substructures of Hom

of modules in L(H)cl . Let U := {K ∈ L(H)cl | ∀ i : Ki ⊆ K}. Then H ∈ U and L = U ∈ L(H)cl because L(H)cl is closed under intersections, and L is the least upper bound of the family by the way it is deﬁned. It follows from Lemma 6.2.6) that Mdl(I) ∈ L(H)cl . By what has been shown, it is clear the the operators Mdl and Idl are inverse lattice isomorphisms.

7

Correspondences for Modules

We consider now the special case HomR (R, M ) ∼ = M . For readers that are only interested in this case, we repeat the main notions and results. Let us consider an R-module MR , and S := End(MR ). Then M is an S-Rbimodule. Deﬁnition 7.1. Let K ⊆ S MR , I ⊆ S SS , and J ⊆ R RR . In particular, I and J are two-sided ideals in S and R respectively. For m ∈ M and ψ ∈ HomR (M, R), we have mψ : M → M : (mψ)(x) = m · ψ(x), while for m ∈ M and ϕ ∈ HomR (M, R), we have ϕm : R → R : (ϕm)(x) = ϕ(m) · x. Hence mψ ∈ S and ϕm ∈ End(RR ) = R. By specializing the general concepts we arrive at the following deﬁnitions. 1) IdlS (K) := ψ∈HomR (M,R) Kψ ⊆ S, and IdlR (K) := ϕ∈HomR (M,R) ϕK. 2) Mdl(I) := {m ∈ M | m HomS (M, S) ⊆ I}, and Mdl(J) := {m ∈ M | HomR (M, R)m ⊆ J}. Remark. The map mψ : M → M : (mψ)(x) = m · ψ(x) is an S-homomorphism because for s ∈ S and x ∈ M , ((sm)ψ)(x)

= s(m)ψ(x) = s(mψ(x)) = s((mψ)(x)) = (s ◦ (mψ))(x) = (s(mψ))(x),

so (sm)ψ = s(mψ). Here we used that s · m = s(m). Corollary 7.2. Let L(S) denote the lattice of all two-sided ideals of S and L(M ) the lattice of all S-R-submodules of M . Then Mdl : L(S) → L(M ),

Idl : L(M ) → L(S),

Mdl : L(R) → L(M ),

Idl : L(M ) → L(R)

are well-deﬁned inclusion preserving maps with the following additional properties where K ⊆ S MR , I is a two-sided ideal of S, and J is a two-sided ideal of R. 1) K ⊆ Mdl(IdlS (K)), K ⊆ Mdl(IdlR (K)), IdlS (Mdl(I)) ⊆ I, IdlR (Mdl(J)) ⊆ J. 2) IM ⊆ Mdl(I), and M J ⊆ Mdl(J).

7. Correspondences for Modules

101

3) Mdl(IdlS (Mdl(I))) = Mdl(I), Mdl(IdlR (Mdl(J))) = Mdl(J). 4) IdlS (Mdl(IdlS (K))) = IdlS (K), IdlR (Mdl(IdlR (K))) = IdlR (K). 5) For S-R-submodules K1 , K2 of M , Idl(K1 + K2 ) = Idl(K1 ) + Idl(K2 ). 6) For ideals I1 , I2 of S, Mdl(I1 ∩ I2 ) = Mdl(I1 ) ∩ Mdl(I2 ). Proof. All claims are special cases of the general results, in particular Lemma 6.2 but can also be derived analogously to the general results. As an example we prove that IM ⊆ Mdl(I). Let s ∈ I, m ∈ M , ϕ ∈ HomR (M, R). Then (sm)ϕ = s(mϕ) ∈ I, since mϕ ∈ R and I is a right ideal in R, hence sm ∈ Mdl(I).

We specialize Lemma 6.7. Lemma 7.3. Let M be an R-module and R = M ⊕ N with M ∼ = M . Let I be an ideal of S := End(MR ). Then, given s ∈ I, there is σ ∈ HomR (M, R) and f ∈ Mdl(I) such that s = f σ. This shows that the assumption (10) is satisﬁed if M is isomorphic to a direct summand of R. Theorem 7.4. Let M be an R-module, S := End(MR ), and assume that IdlS Mdl(I) = I for all ideals I of S.

(10)

Let L(S) be the set of all two-sided ideals of S. Then L(S) is a complete lattice under set inclusion where the least upper bounds are the sums and the greatest lower bounds are the intersections. Let L(M )cl := {K ⊆ S MR | K = Mdl IdlS (K)}, partially ordered by set inclusion. Then Mdl(I) ∈ L(H)cl for every I ∈ L(S) and L(M )cl is a complete lattice where the greatest lower bounds are the intersections. The functions Mdl : L(S) → L(M )cl

and

IdlS : L(M )cl → L(S)

are inverse lattice isomorphisms. Proof. This is just a special case of Theorem 6.8. Similarly we have the following result.

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Chapter VII. Reg(A, M ) and Other Substructures of Hom

Theorem 7.5. Let M be an R-module, S := End(MR ), and assume that IdlR Mdl(J) = J for all ideals J of R.

(11)

Let L(R) be the set of all two-sided ideals of R. Then L(R) is a complete lattice under set inclusion where the least upper bounds are the sums and the greatest lower bounds are the intersections. Let L(M )cl := {K ⊆ S MR | K = Mdl IdlR (K)}, partially ordered by set inclusion. Then Mdl(J) ∈ L(H)cl for every J ∈ L(R) and L(M )cl is a complete lattice where the greatest lower bounds are the intersections. The functions Mdl : L(R) → L(M )cl are inverse lattice isomorphisms.

and

IdlR : L(M )cl → L(R)

Chapter VIII

Regularity in Homomorphism Groups of Abelian Groups 1

Introduction

We are concerned in this chapter with regularity in homomorphism groups of abelian groups. Previous authors [29], [12], [30], [13] dealt with those groups that have regular endomorphism rings. Cognizant of the existence of the largest regular ideal Reg(A, A) in the endomorphism ring of the group A, their results have been generalized in [24] to computing Reg(A, A) = Reg(End(A)). Here we study Hom(A, M ) as an End(M )-End(A)-bimodule in view of regularity. In this chapter “group” will mean “abelian group” and p will always stand for a prime number. The theory of abelian groups is a highly developed subject so that conclusive answers are possible using the available tools. It is hoped that the results obtained and the ideas employed will be helpful in other contexts. It is routine to generalize the results of this section to modules over a principal ideal domain.

2

Hom(A, M ) and Regularity

We begin by considering indecomposable groups. The indecomposable torsion groups are completely known. They are • the cyclic primary groups Z(pn ) of order pn , n ∈ N, • the divisible Pr¨ ufer groups Z(p∞ ). In addition there are very many indecomposable torsion-free groups, among them the rank-one groups, i.e., the additive subgroups of Q, in particular Q itself. There

104

Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

are 2ℵ0 non-isomorphic rank-one groups, and there are indecomposable torsionfree groups of any cardinality less than the ﬁrst strongly inaccessible cardinal ([11, Theorem 89.2]). Proposition 2.1. 1) Reg(Z(p), Z(p)) = Hom(Z(p), Z(p)) ∼ = Z(p). 2) For 2 ≤ n ≤ ∞ and 1 ≤ m ≤ ∞, Reg(Z(pm ), Z(pn )) = 0 and Reg(Z(pn ), Z(pm )) = 0. 3) Reg(Q, Q) = Hom(Q, Q) ∼ = Q. 4) If A is torsion-free and indecomposable but A = Q, then for any group M , Reg(A, M ) = 0 and Reg(M, A) = 0. Proof. 1) is true because Z(p) is simple and 3) is true because every nonzero map in Hom(Q, Q) is an isomorphism. ˆ p , the ring It is well-known that End(Z(pn )) ∼ = Z/pn Z, and End(Z(p∞ )) ∼ =Z of p-adic integers, and for n ≥ 2 these endomorphism rings are not division rings. Hence Theorem III.1.1 establishes 2). 4) Suppose that Reg(A, M ) = 0 or Reg(M, A) = 0. Then it follows from Theorem III.1.1 that End(A) is a division ring. Since (multiplication by) n ∈ N is an endomorphism of A, we also have 1/n ∈ End(A) and this means that A is divisible. The only divisible indecomposable group is Q (up to isomorphism). We established the contrapositive of the claim. In some cases the absence of any nonzero regular maps is easily seen. Lemma 2.2. Let A and M be groups such that one of these is torsion and the other is torsion-free. Then Hom(A, M ) contains no regular elements except for the zero map. Proof. A direct summand of a torsion group is torsion and a direct summand of a torsion-free group is torsion-free. Hence there is no nonzero summand of A that is isomorphic to a summand of M . By Corollary II.1.5 there are no nonzero regular elements. The following lemma says that in the study of torsion groups one can – as is usually done – restrict oneself to primary groups. Lemma 2.3. Let A and M be torsion groups and let A = p∈P Ap and M = M be the primary decompositions. Then p p∈P Hom(A, M ) ∼ =

p∈P

Hom(Ap , Mp ), and Reg(A, M ) ∼ =

p∈P

Reg(Ap , Mp ).

2. Hom(A, M ) and Regularity

105

Proof. It is well-known and easy to see (ordersof elements!) that Hom(A, M ) ∼ = ∼ ∼ Hom(A , M ). Similarly, End(A) End(A ) and End(M ) = = p p p p∈P p∈P p∈P End(Mp ). The actions of (αp ) ∈ End(A) and of (μp ) ∈ End(M ) on (fp ) ∈ Hom(A, M ) is given by (μp )(fp )(αp ) = (μp fp αp ), i.e., the action is componentwise, and the claim is now clear.

Proposition 2.4. Let A and M be divisible torsion groups. Then Reg(A, M ) = 0. Proof. By Lemma 2.3 we assume without loss of generality that A and M are pgroups. Divisible p-groups are direct sums of groups isomorphic to Z(p∞ ). Suppose that 0 = f ∈ Reg(A, M ). Then there are decompositions A = A1 ⊕ A2 and M = M1 ⊕ M2 such that A1 ∼ = M1 ∼ = Z(p∞ ) and 0 = f (A1 ) ⊆ M1 . By Theorem II.6.14 the map f induces a map 0 = f11 ∈ Reg(A1 , M1 ) which is a contradiction by Proposition 2.1. It is well-known that every group A contains a unique largest divisible subgroup DA and since the divisible abelian groups are exactly the injective groups, we have a direct decomposition A = DA ⊕ RA , where RA contains no nonzero divisible subgroup. Groups that contain no nonzero divisible subgroups are called reduced. Lemma 2.5. Let A and M be groups such that one of these is divisible and the other is reduced. Then Hom(A, M ) contains no nonzero regular maps. Proof. A direct summand of a reduced group is reduced and a direct summand of a divisible group is divisible. Hence there is no nonzero summand of A that is isomorphic to a summand of M . By Corollary II.1.5 the group Hom(A, M ) contains no nonzero regular element. We show next that for the purpose of computing Reg(A, M ) we may assume that both A and M are divisible or that both are reduced. tf Lemma 2.6. Let A and M be any groups and consider decompositions A = DA ⊕ t tf t t t DA ⊕ RA and M = DM ⊕ DM ⊕ RM where DA , DM are divisible torsion groups, tf tf , DM are divisible torsion-free groups, and RA , RM are reduced groups. Such DA decompositions exist. Let f ∈ Hom(A, M ) and represent f as ⎤ ⎡ f11 f21 f31 f = ⎣ f12 f22 f32 ⎦ where f13 f23 f33 tf tf t tf tf , DM ), f21 ∈ Hom(DA , DM ), f31 ∈ Hom(RA , DM ), f11 ∈ Hom(DA tf t t t t ), f12 ∈ Hom(DA , DM ), f22 ∈ Hom(DA , DM ), f32 ∈ Hom(RA , DM tf t , RM ), f23 ∈ Hom(DA , RM ), f33 ∈ Hom(RA , RM ). f13 ∈ Hom(DA

Then the following statements are true.

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Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

1) If f ∈ Reg(A, M ), then f21 = 0, f31 = 0, f12 = 0, f22 = 0, f32 = 0, f13 = 0, and f23 = 0. t = 0, then f11 = 0. 2) If f ∈ Reg(A, M ) and DM

3) If f ∈ Reg(A, M ) and RA / t(RA ) = 0, then f11 = 0. t = 0 and RA / t(RA ) = 0. Then 4) Assume that DM

⎡

f11 f =⎣ 0 0

0 0 0

⎤ 0 0 ⎦ ∈ Reg(A, M ). 0

t )p = 0} ⊆ t(RM ). 5) If f ∈ Reg(A, M ), then Im(f33 ) ⊆ {(RM )p | (Dm t 6) Assume that Im(f33 ) ⊆ {(RM )p | (Dm )p = 0}, and f33 ∈ Reg(RA , RM ). Then ⎤ ⎡ 0 0 0 f = ⎣ 0 0 0 ⎦ ∈ Reg(A, M ). 0 0 f33 t = 0, RA / t(RA ) = 0, Im(f33 ) ⊆ 7) Assume that DM and f33 ∈ Reg(RA , RM ). Then

⎡

f11 f =⎣ 0 0

t {(RM )p | (Dm )p = 0},

⎤ 0 0 0 0 ⎦ ∈ Reg(A, M ). 0 f33

Proof. 1) Assume that f ∈ Reg(A, M ). Then f22 = 0 by Theorem II.6.14 and Proposition 2.4, f21 = 0 because there are no nonzero maps on a torsion group to a torsion-free group, f13 = 0 and f23 = 0 because there are no nonzero maps of a divisible group to a reduced group; the other statements are true by Theorem II.6.14, Lemma 2.2 and Lemma 2.5. tf t , DM ) = 0, for every map 2) By Theorem II.6.14 we have μf11 ∈ Reg(DA tf t tf and there μ ∈ Hom(DM , DM ). Suppose that f11 = 0. Then 0 = Im(f11 ) ⊆ DM tf t is μ ∈ Hom(DM , DM ) such that μf11 = 0, a contradiction. We conclude that f11 = 0. tf ) = 0, for every map 3) By Theorem II.6.14 we have f11 α ∈ Reg(RA , DM tf tf and, as α ∈ Hom(RA , DM ). Suppose that f11 = 0. Then 0 = Im(f11 ) ⊆ DM tf RA / t(RA ) = 0, there is α ∈ Hom(RA , DM ) such that f11 α = 0, a contradiction. We conclude that f11 = 0.

4) By Theorem II.6.14 we have to show that all possible maps of the form tf tf and α has codomain DA lie in the appropriate μf11 α where μ has domain DM

2. Hom(A, M ) and Regularity

107

Reg. The possibilities are depicted in the following diagram. tf DA t DA

−→

tf f11 DA −→

tf DM

tf DM

RA

t −→ DM RM

t tf tf We have Hom(DA , DA ) = 0, and Hom(DM , RM ) = 0 which settles aﬃrmatively tf ) = 0 and ﬁve of the nine possible combinations μf11 α. But also Hom(RA , DM tf t Hom(DM , DM ) = 0 by our assumptions. This settles three more cases aﬃrmaf11

tf tf tf tf → DA −→ DM → DM which tively, and leaves the single composite μf11 α : DA tf tf tf tf tf tf is in Reg(DA , DM ) because Reg(DA , DM ) = Hom(DA , DM ). t 5) Suppose that f ∈ Reg(A, M ) and Im(f33 ) ⊆ {(RM )p | (Dm )p = 0}. t t Then there is μ : RM → D M such that 0t = μf33 ∈ Reg(RA , DM ) = 0, a contra)p = 0}. diction. Hence Im(f33 ) ⊆ {(RM )p | (Dm 6) By Theorem II.6.14 we have to show that all possible maps of the form μf33 α where μ has domain RM and α has codomain RA lie in the appropriate Reg. The possibilities are depicted in the following diagram. tf DA t DA

f33

−→ RA −→ RM

RA

tf DM

t −→ DM RM

tf t We have Hom(DA , RA ) = 0, and Hom(DM , RA ) = 0 which settles aﬃrmatively six of the nine possible combinations μf33 α. But also the composite μf11 α : RA → f33 tf is equal to 0 because Im(f33 ) ⊆ t(RM ) by assumption. RA −→ RM → DM f33

t Further the composite μf11 α : RA → RA −→ RM → DM is equal to 0 because t of the assumption that Im(f33 ) ⊆ {(RM )p | (Dm )p = 0}. This leaves the case f33

μf11 α : RA → RA −→ RM → RM . In this case μf11 α ∈ Reg(RA , RM ) because f33 ∈ Reg(RA , RM ) by assumption and μ ∈ End(RM ) and α ∈ End(RA ). 7) is the combination of 4) and 6). We wish to extend these results by computing the maximum regular bisubmodule of Hom(A, M ) in general. A ﬁrst step consists in checking when pf is regular for a regular homomorphism f ∈ Hom(A.M ). Lemma 2.7. Let A and M be abelian groups and suppose that f ∈ Hom(A, M ) is regular. Then pf is also regular if and only if Im(f ) = p Im(f ) ⊕ Im(f )[p].

(1)

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Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

Proof. As f is regular we have decompositions A = Ker(f ) ⊕ A1 ,

M = Im(f ) ⊕ M1 ,

A1 ∼ = Im(f ).

Regarding pf , it is easily seen that Im(pf ) = p Im(f ),

and

Ker(pf ) = Ker(f ) ⊕ A1 [p].

Let D denote the maximal divisible subgroup of Im(f ). Assume ﬁrst that pf is also regular. Then Im(pf ) = p Im(f ) ⊆⊕ M , hence Im(f ) = p Im(f ) ⊕ C for some C, and pC ∼ = p (Im(f )/p Im(f )) = 0, so C ⊆ Im(f )[p]. Therefore p Im(f ) = p(p Im(f )). Also D = pD ⊆ p Im(f ), and hence p Im(f ) = D ⊕ B for some subgroup B. Here pB = B as B is a summand of a p-divisible group, but B is reduced by the choice of D. This means that the pprimary component Bp of B must be 0 because p-divisible p-groups are divisible. So (Im(f ))[p] = D[p] ⊕ C. (2) At this point we have A1 ∼ = Im(f ) = D ⊕ B ⊕ C. Therefore A1 = D ⊕ B ⊕ C ∼ ∼ with D = D, B = B, and C ∼ = C. As Ker(pf ) = Ker(f ) ⊕ A1 [p] is a summand of A, so is A1 [p] = D [p] ⊕ B [p] ⊕ C [p]. This makes D [p] a summand of A and it follows from the structure theory of divisible abelian groups that D [p] = 0. Thus D[p] = 0 and by (2), C = (Im(f ))[p]. Altogether we have Im(f ) = D ⊕ B ⊕ C = p Im(f ) ⊕ Im(f )[p] as claimed. Assume now that Im(f ) = p Im(f ) ⊕ Im(f )[p]. We must show that pf is regular and this is the case if Im(pf ) ⊆⊕ M and Ker(pf ) ⊆⊕ A. As Im(pf ) = p Im(f ), it follows from the assumptions and the fact that a summand of a summand is a summand that Im(pf ) ⊆⊕ M . From A1 ∼ = Im(f ) it follows from the assumptions that A1 = A1 [p] ⊕ pA1 and then from A = Ker(f ) ⊕ A1 that A = Ker(f ) ⊕ A1 [p] ⊕ pA1 . Since Ker(pf ) = Ker(f ) ⊕ A1 [p] as observed above, we have A = Ker(pf ) ⊕ pA1 as needed. Corollary 2.8. Suppose that f ∈ Hom(A, M ) and pf are both regular. Then pk f is regular for any integer k ≥ 0. Proof. By Lemma 2.7 we have A = Ker(f ) ⊕ A1 , M = Im(f ) ⊕ M1 and Im(f ) = p Im(f ) ⊕ Im(f )[p]. We will show that Im(pk f ) ⊆⊕ M and Ker(pk f ) ⊆⊕ A which says that pk f is regular. The group p Im(f ) is p-divisible and hence for any k ≥ 1, we have Im(pk f ) = pk−1 (p Im(f )) = p Im(f ) and this is a direct summand of M . It is easily seen that Ker(pk f ) = Ker(f ) ⊕ A1 [pk ]. Since (p Im(f ))p = 0, we have that Im(f )[pk ] = Im(f )[p] which is a direct summand of Im(f ). As A1 ∼ = Im(f ), also A1 [pk ] = A1 [p] is a direct summand of A1 and hence Ker(pk f ) = Ker(f ) ⊕ A1 [pk ] is a direct summand of A which concludes the proof. In the same vein we characterize regular cyclic subgroups of Hom(A, M ).

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109

Proposition 2.9. Let A be a group and let f ∈ Hom(A, M ). Then Zf := {nf | n ∈ Z} is a regular subgroup of Hom(A, M ) if and only if Ker(f ) ⊆⊕ A, and for all n ∈ Z,

and

Im(f ) ⊆⊕ M

Im(f ) = n Im(f ) ⊕ Im(f )[n].

(3)

(4)

Proof. Suppose that Zf is regular. Then f is regular and there exist decompositions (5) A = Ker(f ) ⊕ A1 , M = Im(f ) ⊕ M1 . We will use several times the induced isomorphism f˜ := f A1 : A1 → Im(f ). Let n ∈ Z be given. Then nf is also regular. We ﬁrst observe that Im(nf ) = n Im(f ) ⊆ Im(f ), and Ker(nf ) = Ker(f ) ⊕ A1 [n].

(6)

The ﬁrst of these is obvious. To verify the second, let x ∈ A. Using (5) write x = y + z with y ∈ Ker(f ) and z ∈ A1 . Then x ∈ Ker(nf )

⇔ (nf )(y + z) = nf (z) = 0 ⇔ f (z) ∈ Im(f )[n] ⇔ z ∈ A1 [n],

where we applied the isomorphism (f˜)−1 . We verify next that Im(f ) = n Im(f ) + Im(f )[n].

(7)

In fact, as Im(nf ) = n Im(f ) ⊆⊕ M , there is a direct decomposition Im(f ) = n Im(f ) ⊕ M2 , and nM2 ⊆ n Im(f ) ∩ M2 = 0, showing that M2 ⊆ Im(f )[n] and so Im(f ) = n Im(f ) ⊕ M2 ⊆ n Im(f ) + Im(f )[n] ⊆ Im(f ) and (7) is established. We will use next that A1 [n] ⊆⊕ Ker(nf ) ⊆⊕ A,

hence

A1 [n] ⊆⊕ A1 .

Let A1 = A1 [n] ⊕ A2 . Then Im(f ) = f˜(A1 ) = Im(f )[n] ⊕ f˜(A2 ).

(8)

We will now show that f˜(A2 ) = n Im(f ) which will establish (4). Let x ∈ f (A2 ). By (7) we can write x = ny + z where y ∈ Im(f ) and z ∈ Im(f )[n]. It follows that nx = n2 y ∈ f (A2 ) ∩ n2 Im(f ) = n2 f (A2 ) because f (A2 ) is pure (even a summand) in Im(f ). Hence there is x ∈ f (A2 ) such that nx = n2 x , so n(x − nx ) = 0. Therefore x − nx ∈ f˜(A2 ) ∩ Im(f )[n] = 0 by (8), x = nx ∈ nf (A2 ), and we have shown that f˜(A2 ) = nf˜(A2 ) and further A2 = nA2 . From A1 = A1 [n]⊕A2 we now

110

Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

ﬁnd that A2 = nA2 = nA1 and so A1 = A1 [n] ⊕ nA1 . Applying the isomorphism (f˜)−1 we get Im(f ) = Im(f )[n] ⊕ n Im(f ). For the converse assume that (4) and (3) hold. Then f is regular by (3) and Corollary II.1.3. Let n ∈ Z. By (4) Im(nf ) = n Im(f ) ⊆⊕ Im(f ) ⊆⊕ M , so Im(nf ) ⊆⊕ M . Also A = Ker(f ) ⊕ A1 for some A1 . Furthermore, A1 ∼ = Im(f ) = Im(f )[n] ⊕ n Im(f ) which says that A1 [n] ⊆⊕ A1 and we see that Ker(nf ) = Ker(f ) ⊕ A1 [n] is a summand of A. By Corollary II.1.3 nf is regular. Corollary 2.10. 1) Suppose that M has no elementary direct summands and either A or M is reduced. Then Reg(A, M ) = 0. 2) Suppose that A has no elementary direct summands and either A or M is reduced. Then Reg(A, M ) = 0. Proof. 1) Suppose that f ∈ Reg(A, M ) and let p ∈ P. Then pf is regular. By Lemma 2.7 we have Im(f ) = p Im(f )⊕Im(f )[p]. Also Im(f ) ⊆⊕ M , so Im(f )[p] ⊆⊕ M , and by hypothesis Im(f )[p] = 0. So Im(f ) = p Im(f ) which makes the Im(f ) p-divisible for every prime p which means that Im(f ) is divisible. If M is reduced, it follows that Im(f ) = 0 and so f = 0. Also A ∼ = Ker(f ) ⊕ Im(f ) (Theorem II.1.2) and if A is reduced then again Im(f ) = 0, f = 0. 2) We have A ∼ = Ker(f ) ⊕ Im(f ) and for every prime p, Im(f ) = p Im(f ) ⊕ Im(f )[p]. Since A has no elementary direct summands, Im(f ) is divisible as before. Therefore Im(f ) = 0, i.e., f = 0, if either A or M is reduced. The following useful lemma is just a specialization of Theorem II.6.14 and Corollary II.1.5. Lemma 2.11. Suppose that A = A1 ⊕ A2 and M = M1 ⊕ M2 and no nonzero summand of A1 is isomorphic to a summand of M2 . Suppose that f11 f21 ∈ Reg(A, M ). f= f12 f22 Then f21 = 0, f12 = 0, and for every μij ∈ Hom(Mi , Mj ), and every αij ∈ Hom(Ai , Aj ), where i, j ∈ {1, 2}, μ21 f22 = 0,

μ12 f22 = 0,

f11 ∈ Reg(A1 , M1 ),

f22 α12 = 0, and

f11 α21 = 0, and

f22 ∈ Reg(A2 , M2 ).

We begin with a lemma that will considerably simplify and shorten future proofs. Lemma 2.12. Let A and M be abelian groups and let f ∈ Reg(A, M ). Then A = Ker(f ) ⊕ A1 and M = Im(f ) ⊕ M1 for subgroups A1 and M1 . Furthermore, 1) for any μ ∈ Hom(Im(f ), M1 ), necessarily Im(μ) ⊆⊕ M1 and Ker(μ) ⊆⊕ Im(f ), and

2. Hom(A, M ) and Regularity

111

2) for any α ∈ Hom(Ker(f ), A1 ), necessarily Ker(α) ⊆⊕ Ker(f ) and Im(α) ⊆⊕ A1 . Proof. We will again use the isomorphism f˜ : A1 x → f (x) ∈ Im(f ). Since f ∈ Reg(A, M ) we also have that, for every t ∈ End(A) and every s ∈ End(M ), the composites f t and sf are regular which means that the images and kernels of such maps are direct summands. 1) Let μ ∈ Hom(Im(f ), M1 ). The map μ extends to an endomorphism s of M by setting s Im(f ) = μ and s(M1 ) = 0. Then Im(sf ) = s(Im(f )) = Im(μ) and, sf being regular, this must be a direct summand of M and hence of M1 . Next, we claim that Ker(sf ) = Ker(f ) ⊕ f˜−1 (Ker(μ)). For the proof let x ∈ A and write x = y + z where y ∈ Ker(f ) and z ∈ A1 . Then x ∈ Ker(sf )

⇔ 0 = sf (x) = sf (z) ⇔ f (z) ∈ Ker(μ) ⇔ z ∈ f˜−1 (Ker(μ)).

The kernel Ker(sf ) must be a summand of A, and hence f˜−1 (Ker(μ)) is a direct summand of A1 , therefore Ker(μ) a summand of Im(f ). This establishes 1). 2) Let α ∈ Hom(Ker(f ), A1 ). The map α extends to a endomorphism t of A by setting t Ker(f ) = α and t(A1 ) = 0. Then f t must be regular. Therefore Im(f t) = f (Im(t)) ⊆⊕ Im(f ) and, applying f˜−1 , Im(α) = Im(t) ⊆⊕ A1 . It is easy to see that Ker(f t) = Ker(α) ⊕ A1 and Ker(f t) must be a summand of A, hence also Ker(α) is a summand of A, and this makes Ker(α) a summand of Ker(f ) as claimed. Remark. In connection with Lemma 2.11 we repeatedly use the following fact. Suppose that E is an elementary p-group, 0 = ξ ∈ End(E) and K a group with K/pK = 0. Then there exists γ ∈ Hom(K, E) such that ξγ = 0. This is true because γ is necessarily a composite of the natural epimorphism K K/pK with a map K/pK → E, and K/pK and E are just vector spaces over the prime ﬁeld Z/pZ so that maps K/pK → E are available whose image is not contained in Ker(ξ). We have established the machinery to settle the torsion-free and torsion cases. Theorem 2.13. Let A and M be torsion-free abelian groups. Then Reg(A, M ) = 0 unless A and M are both divisible. If both A and M are divisible, then Reg(A, M ) = Hom(A, M ).

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Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

Proof. Let 0 = f ∈ Reg(A, M ). Then Zf is regular and by Lemma 2.9 we have a decomposition Im(f ) = n Im(f ) ⊕ Im(f )[n]. As M is torsion-free, Im(f ) = n Im(f ) which says that Im(f ) is divisible and M = Im(f ) ⊕ D ⊕ M1 where D is divisible and M1 is reduced. If M1 = 0, then there is μ : M1 → Im(f ) and by Lemma 2.12 it follows that Im(μ) = 0 is a summand of Im(f ), so divisible and so M1 / Ker(μ) ∼ = Im(μ) is also divisible but also isomorphic to a nonzero direct summand of M1 . This contradiction shows that M is divisible. We also have that A = Ker(f ) ⊕ A1 where A1 ∼ = Im(f ) and we have seen that Im(f ) is divisible. There is a decomposition Ker(f ) = K1 ⊕ K2 such that K1 is divisible and K2 is reduced. If K2 = 0, then there is 0 = α ∈ Hom(K2 , Im(f )) and by Lemma 2.12 it follows that 0 = Im(α) ⊆⊕ Im(f ) and hence Im(α) is divisible. But Im(α) ∼ = K2 / Ker(α) is also isomorphic to a direct summand of the reduced group K2 which is a contradiction. Hence K2 = 0 and A is divisible. Suppose now that both A and M are divisible. Then both A and M are Qvector spaces, so Hom(A, M ) = HomQ (A, M ) and Reg(A, M ) = Hom(A, M ). By Lemma 2.3 there is no loss of generality in assuming that torsion groups are primary. Theorem 2.14. Let A and M be p-primary groups. 1) Suppose that M is not reduced. Then Reg(A, M ) = 0. 2) Suppose that M is reduced. There are decompositions A = A1 ⊕ A2 and M = M1 ⊕ M2 such that A1 and M1 are elementary, and A2 and M2 have no direct summands of order p. Then i) Reg(A, M ) = 0 if M2 = 0, ii) Reg(A, M ) = 0 if A2 is not divisible, iii) Reg(A, M ) = Hom(A, M ) if M2 = 0 and A2 is divisible, i.e., if A is the direct sum of an elementary group and a divisible group and M is elementary. Proof. 1) Write A = DA ⊕RA and M = DM ⊕RM such that DA , DM are divisible and RA , RM are reduced. Let f ∈ Hom(A, M ). Then fDD ∈ Hom(DA , DM ), fRD ∈ Hom(RA , DM ), fDD fRD , f= fDR fRR fDR ∈ Hom(DA , RM ), fRR ∈ Hom(RA , RM ). Suppose that f ∈ Reg(A, M ). Then fRD ∈ Reg(RA , DM ) = 0 by Lemma 2.5, fDR = 0 because epimorphic images of divisible groups are divisible, and fDD ∈ Reg(DA , DM ) = 0 (Proposition 2.4). Suppose that 0 = fRR ∈ Reg(RA , RD ). Then 0 = Im(fRR ) ⊆⊕ RM and DM = 0 by hypothesis. Hence there is 0 = μ ∈ Hom(Im(fRR ), DM ). Then 0 = μfRR ∈ Reg(RA , DM ) = 0, a contradiction. Hence f = fRR = 0.

2. Hom(A, M ) and Regularity

113

2) Let f ∈ Reg(A, M ). Then Im(f ) = p Im(f ) ⊕ Im(f )[p]. Hence p Im(f ) = p(p Im(f )) which means that p Im(f ) is divisible and thus p Im(f ) = 0 because M is reduced. So Im(f ) = Im(f )[p] is an elementary pgroup. Suppose there exists 0 = μ ∈ Hom(Im(f ), M2 ). Then Lemma 2.12 implies that Im(μ) ⊆⊕ M2 . But Im(μ) is elementary and M2 has no nonzero elementary summands. This is a contradiction and shows that f = 0 if M2 = 0. This proves i). Now suppose that M = M1 is elementary. Then pA ⊆ Ker(f ) and A = Ker(f ) ⊕ A3 where A3 ∼ = Im(f ) is elementary. There is a decomposition Ker(f ) = K1 ⊕ K2 such that K1 is elementary and K2 has no nonzero elementary direct summands. If K2 /pK2 = 0, we have a nonzero mapping proj

nat

α : Ker(f ) K2 K2 /pK2 → A3 . Lemma 2.12 says that Ker(α) ⊆⊕ Ker(f ) and pA ⊕ K1 ⊆ Ker(α) which makes the complementary direct summand of Ker(α) in Ker(f ) a nonzero elementary direct summand of K2 . But K2 has no such direct summands and this contradiction says that K2 = pK2 , i.e., K2 is divisible. We conclude that A = K1 ⊕ K2 ⊕ A3 is the direct sum of the divisible group K2 and the elementary group K1 ⊕ A3 . This proves ii). iii) Suppose that A2 is divisible. Then Hom(A, M ) = Hom(A1 , M ) and every element of Hom(A1 , M ) is regular because A1 and M are both elementary. We will next tackle the case of a splitting mixed group, reduced or not. Proposition 2.15. Let A and M be splitting mixed groups and consider decompositions A = A1 ⊕ A2 ⊕ A3 and M = M1 ⊕ M2 ⊕ M3 where A1 , M1 are elementary abelian groups, A2 , M2 are torsion groups without elementary direct summands, and A3 , M3 are torsion-free groups. Such decompositions exist. Let f ∈ Hom(A, M ) and represent f as ⎤ ⎡ f11 f21 f31 f = ⎣ f12 f22 f32 ⎦ where f13 f23 f33 f11 ∈ Hom(A1 , M1 ), f12 ∈ Hom(A1 , M2 ), f13 ∈ Hom(A1 , M3 ),

f21 ∈ Hom(A2 , M1 ), f22 ∈ Hom(A2 , M2 ), f23 ∈ Hom(A2 , M3 ),

f31 ∈ Hom(A3 , M1 ), f32 ∈ Hom(A3 , M2 ), f33 ∈ Hom(A3 , M3 ).

Then the following statements are true. 1) If f ∈ Reg(A, M ), then f21 = 0, f31 = 0, f12 = 0, f22 = 0, f32 = 0, f13 = 0, and f23 = 0.

114

Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

2) If f ∈ Reg(A, M ) and f11 = 0, then for every p for which (A1 )p = 0 necessarily A2 = pA2 , A3 = pA3 and M2 [p] = 0. 3) Assume that A2 = pA2 , A3 = pA3 and M2 [p] = 0 whenever (A1 )p = 0, then ⎡ ⎤ f11 0 0 f = ⎣ 0 0 0 ⎦ ∈ Reg(A, M ). 0 0 0 4) If f ∈ Reg(A, M ) and f33 = 0, then A3 and M3 are divisible. 5) Assume that A3 and M3 are divisible. Then ⎡ ⎤ 0 0 0 f = ⎣ 0 0 0 ⎦ ∈ Reg(A, M ). 0 0 f33 6) Assume that A2 = pA2 , A3 = pA3 and M2 [p] = 0 whenever (A1 )p = 0, and that A3 and M3 are divisible. Then ⎡ ⎤ f11 0 0 f = ⎣ 0 0 0 ⎦ ∈ Reg(A, M ). 0 0 f33 Proof. 1) Assume that f ∈ Reg(A, M ) and use Theorem II.6.14. Then f21 = 0, f31 = 0, f12 = 0, f22 = 0, and f32 = 0 by Corollary 1.5, f13 = 0 and f23 = 0 because there are no nonzero maps on a torsion group to a torsion-free group. 2) By Theorem II.6.14 we have μf11 ∈ Reg(A1 , M2 ) = 0, for every map μ ∈ Hom(M1 , M2 ). Suppose that f11 = 0. Then 0 = Im(f11 ) ⊆ M1 and if M2 [p] = 0, then there is μ ∈ Hom(M1 , M2 ) such that μf11 = 0, a contradiction. We conclude that M2 [p] = 0. Again by Theorem II.6.14 we have f11 α ∈ Reg(A2 , M1 ) = 0, for every map α ∈ Hom(A2 , A1 ). Suppose that f11 = 0. Then 0 = Im(f11 ) ⊆ M1 and, if A2 /pA2 = 0, there is α ∈ Hom(A2 , A1 ) such that f11 α = 0, a contradiction. We conclude that A2 = pA2 . Similarly, it follows that A3 = pA3 . 3) By Theorem II.6.14 we have to show that all possible maps of the form μf11 α where μ has domain M1 and α has codomain A1 lie in the appropriate Reg. The possibilities are depicted in the following diagram. A1 A2 A3

f11

−→ A1 −→ M1

M1

−→ M2 M3

3. Mixed Groups

115

By hypothesis we have Hom(A2 , A1 ) = 0, and Hom(M1 , M3 ) = 0 which settles afﬁrmatively ﬁve of the nine possible combinations μf11 α. But also Hom(A3 , A1 ) = 0 and Hom(M1 , M2 ) = 0 by our assumptions. This settles three more cases aﬃrmaf11

tively, and leaves the single composite μf11 α : A1 → A1 −→ M1 → M1 which is in Reg(A1 , M1 ) because Reg(A1 , M1 ) = Hom(A1 , M1 ). 4) Suppose that f ∈ Reg(A, M ) and f33 = 0. Then Reg(A3 , M3 ) = 0, and therefore A3 and M3 are divisible by Theorem 2.13. 5) By Theorem II.6.14 we have to show that all possible maps of the form μf33 α where μ has domain M3 and α has codomain A3 lie in the appropriate Reg. The possibilities are depicted in the following diagram. A1 A2

M1

f33

−→ A3 −→ M3

A3

−→ M2 M3

We have Hom(A1 , A3 ) = 0, Hom(A2 , A3 ) = 0, and Hom(M3 , M1 ) = 0 which settles aﬃrmatively seven of the nine possible combinations μf33 α. But also the f33 composite μf11 α : A3 → A3 −→ M3 → M2 is equal to 0 because M3 is divisible f33 by assumption. Finally, the composite μf11 α : A3 → A3 −→ M3 → M3 is regular because f33 ∈ Hom(A3 , M3 ) = Reg(A3 , M3 ). 6) is the combination of 3) and 5).

3

Mixed Groups

We now turn to reduced mixed groups and start with some simple examples. The ﬁrst example shows that A and M may have fairly general direct summands that are irrelevant for regularity. Example 3.1. Let A0 and B0 be elementary groups, D a divisible group and Z a reduced torsion-free group. Let A = A0 ⊕ D and let M := B ⊕ Z. Then Hom(A, M ) = Reg(A, M ) because Hom(A, M ) = Hom(A, B). The second example is a large example where regularity is due to large numbers of direct summands. Example 3.2. Let T1 , T2 be elementary groups, let P1 = p∈P (T1 )p and P2 = p∈P (T2 )p . For i = 1, 2 set P(Ti ) := {p ∈ P | (Ti )p = 0}. ∼ 1) Hom(P1 , P2 ) = p∈P Hom((T1 )p , (T2 )p ), so f ∈ Hom(P1 , P2 ) may be identiﬁed with f = p∈P fp where fp ∈ Hom((T1 )p , (T2 )p ) and the action is componentwise. 2) Hom(P1 , P2 ) ∼ Hom((T1 ) , (T2 ) ). = p∈P(T1 )∩P(T2 )

p

p

116

Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

3) Hom(P1 , P2 ) = Reg(P1 , P2 ).

Proof. 1) By homological algebra Hom(P1 , P2 ) ∼ = p∈P Hom(P1 , (T2 )p ). We have a short exact sequence T1 P1 P1 /T1 where P1 /T1 is torsion-free and divisible. Hence, for all p ∈ P, P1 P1 , (T2 )p ) Hom(P1 , (T2 )p ) → Hom(T1 , (T2 )p ) → Ext( , (T2 )p ) = 0 T1 T1 is exact, and Hom(T1 , (T2 )p ) ∼ = q∈P Hom((T1 )q , (T2 )p ) = Hom((T1 )p , (T2 )p ). The claim follows because all maps are natural. 0 = Hom(

2) follows immediately from 1). 3) Let = p∈P xp ∈ P1 we have f = p∈P fp ∈ Hom(P1 , P2 ). For any x f (x) = p∈P fp (xp ) which shows that Ker(f ) = p∈P Ker(fp ) and Im(f ) = p∈P Im(fp ). We also have decompositions (T1 )p = Ker(fp ) ⊕ (T1 )p ,

(T2 )p = Im(fp ) ⊕ (T2 )p

and we get P1 =

Ker(fp ) ⊕

p∈P

p∈P

(T1 )p , and P2 =

p∈P

Im(fp ) ⊕

Thus Ker(f ) and Im(f ) are direct summands and f is regular.

p∈P

(T2 )p .

The following “small” example is a modiﬁcation of the example [11, Vol.II, page 186, Ex.2]. Example p ∈ P let Tp := ap be cyclic of order p. Let 3.3. For each prime T = p∈P Tp and let P = p∈P Tp . Let a = Πp∈P ap ∈ P . Then for every p ∈ P and every positive integer n there exists a unique element a ˆp,n ∈ P such that a = pn a ˆp,n + ap . Let A = T +

p,n

Zˆ ap,n ⊆ P . Then the following hold.

1) t(A) = T , A/ t(A) ∼ = Q, and A is pure in P . 2) If f ∈ End(A) and Im(f ) ⊆ T , then Im(f ) is ﬁnite; furthermore, A = Im(f )⊕ Ker(f ) and f is regular. 3) The short sequence with natural maps Hom(A, T ) End(A) End(A/T ) = Q is exact. 4) Reg(A, A) = End(A).

3. Mixed Groups

117

Proof. 1) It is clear that t(A) = T because T ⊆ A ⊆ P and T = t(P ). It is also ˆp,n + T which says that the generators clear that A/T ∼ = Q because a + T = pn a of A/T are all dependent and A/T is divisible. 2) Let f = Πp∈P fp ∈ End(A) where we may and will assume that fp is (multiplication by) an element in Z/pZ because Tp is cyclic of order p. Suppose that Im(f ) ⊆ T . Then inparticular f (a) = Πp∈P fp ap ∈ T . Hence fp = 0 for almost all p and Im(f ) ⊆ fp =0 Tp which is ﬁnite. Moreover, it is easily seen that • actually Im(f ) = fp =0 Tp , • P = Im(f ) ⊕ fp =0 Tp , • A = Im(f ) ⊕ [A ∩ fp =0 Tp ], • and Ker(f ) = A ∩ fp =0 Tp , • hence A = Im(f ) ⊕ Ker(f ). 3) Let m ∈ Z and identify m with multiplication by m in A. Then Ker(m) = A[m] and Im(m) = mA. It is easily seen that A[m] = T [m] = P [m] = p|m Tp , pm Tp = mP , and P = T [m] ⊕ mP . Since T [m] ⊆ A and A is pure in P , it follows that A = A[m] ⊕ (A ∩ mP ) = A[m] ⊕ mA = Ker(m) ⊕ Im(m). This shows that multiplication by m on A is a regular endomorphism. Let g ∈ End(A) be such that mgm = m. Let g denote the endomorphism A/T induced by g and note that multiplication by m in A/T is an automorphism. Hence mgm = m implies that mg = 1A/T and shows that g induces multiplication by 1/m on End(A/T ) ∼ = Q. It is now clear that the natural map End(A) → End(A/T ) is surjective and the remaining exactness checks being routine, the claim is established. 4) This was shown in greater generality by Glaz-Wickless ([13, Theorem 4.1]) but we provide an independent proof. Let f ∈ End(A). Then the induced endomorphism f ∈ End(A/T ) ∼ = Q is regular. Hence there is g ∈ End(A) such that f gf = f and hence f gf − f ∈ Hom(A, T ) which by 2) is regular. By Lemma II.4.1 with M = A, the endomorphism f itself is regular. Proposition 3.5 generalizes an idea used in Example 3.3. We ﬁrst consider induced maps in general. Lemma 3.4. Let A and M be arbitrary groups with maximal torsion subgroups TA := t(A) and TM := t(M ). 1) There is a well-deﬁned homomorphism : Hom(A, M ) → Hom(A/TA , M/TM ) given by f (x + TA ) = f (x) + TM where f ∈ Hom(A, M ) and x ∈ A. Furthermore, Ker( ) = Hom(A, TM ) ⊆ Hom(A, M ).

118

Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

2) There is a well-deﬁned homomorphism : Hom(M, A) → Hom(M/TM , A/TA ) given by g(x + TM ) = g(x) + TA where g ∈ Hom(M, A) and x ∈ M . Furthermore, Ker( ) = Hom(M, TA ) ⊆ Hom(M, A). 3) Let φA : A → A/TA and φT : M → M/TM be the natural epimorphisms. Then the maps f and g deﬁned in 1) and 2) are the unique mappings satisfying f φA = φM f, gφM = φA g. Proof. 1) The map is well-deﬁned since for every f ∈ Hom(A, M ) it is true that is a homomorphism with kernel Hom(A, TM ) = {f ∈ f (TA ) ⊆ f (TM ). Clearly Hom(A, M ) | f (A) ⊆ TM }. 2) is identical with 1) except for the changed roles of A and M and 3) is just a rephrasing to the deﬁnition of . Proposition 3.5. Let A be a group such that pAp = 0, TA := t(A) = p∈P Ap ⊆ A ⊆ P := p∈P Ap , and A is pure in P . Let M be an arbitrary group with maximal torsion subgroups TM := t(M ). Let : Hom(A, M ) → Hom(A/TA , M/TM ) be the natural map. 1) The quotient A/TA is divisible, and Hom(A/TA , M/TM ) is a torsion-free divisible group and so a Q-vector space. 2) Multiplication on A by m ∈ N is a regular element of End(A) and the image : End(A) → End(A/TA ) is divisible as an abelian group. of 3) The image Hom(A, M ) in the Q-vector space Hom(A/TA , M/TM ) is a Qsubspace. 4) If f is regular in Hom(A, M ), then f is regular in Hom(A, M ). 5) If f is regular in Hom(A, M ) and Hom(A, TM ) is regular, then f is regular. Proof. 1) It is easy to see that P/TA is divisible and hence A/TA is divisible because A is pure in P . Since A/TA is divisible, therefore End(A/TA ) is divisible and so the right End(A/TA )-module Hom(A/TA , M/TM ) is also divisible and torsion-free because M/TM is torsion-free. ! ! and the 2) Let m ∈ N be given. Then A = p|m Tp ⊕ A ∩ pm Tp summand A ∩ pm Tp is uniquely m-divisible because A is pure in P and pm Tp is uniquely divisible in P . Deﬁne fm by stipulating that T 0 on p|m p . fm = 1 on A ∩ pm Tp m Then mfm m = m showing that m is regular and the quasi-inverse fm induces multiplication by 1/m in A/TA .

3. Mixed Groups

119

3) By 2) for every m ∈ N, the ring End(A) contains an element fm that induces multiplication by 1/m in End(A/TA ). In fact, having fm ∈ End(A) and 1 f ∈ Hom(A, M ) showing that Hom(A, M ) is f ∈ Hom(A, M ), we obtain f fm = m divisible. 4) Suppose that f gf = f for some g ∈ Hom(M, A). Then f gf = f

⇒

φM f gf = φM f ⇒ f φA gf = f φA ⇒ f gφM f = f φA ⇒ f gf φA = f φA ⇒ f gf = f ,

the last implication being valid because φA is surjective. 5) Suppose now that f gf = f for g ∈ Hom(M, A). Then (f gf − f )(A) ⊆ TM , i.e, f gf − f ∈ Hom(A, TM ) and so regular by hypothesis. By Lemma IV.4.1 with M = A, the homomorphism f itself is regular. If f ∈ Reg(A, M ), then pf must be regular for every prime p. This fact alone has far-reaching consequences. Theorem 3.6. Let A and M be reduced abelian groups and assume that Reg(A, M ) 0. Then P(A, M ) := {p ∈ P | Im(f )[p] = 0} is a non-void set of primes such = that ∀ p ∈ P(A, M ), A = pA ⊕ A[p], A[p] = 0, and

∀ p ∈ P(A, M ), M = M [p] ⊕ M ,

M [p] = 0.

Proof. Let 0 = f ∈ Reg(A, M ). Then there are decompositions (Proposition 2.9) A = Ker(f ) ⊕ A1 ,

M = Im(f ) ⊕ M1 ,

∀ p ∈ P : Im(f ) = p Im(f ) ⊕ Im(f )[p],

where A1 ∼ = Im(f ). Then P(A, M ) = ∅ as otherwise Im(f ) = p Im(f ) would be p-divisible for all primes p and hence divisible and so, A being reduced, we would have Im(f ) = 0. So far we only used that Zf is regular. In order to proﬁtably apply Lemma 2.12 we need to ﬁnd suitable maps α and μ. Let p ∈ P(A, M ) and write Ker(f ) = K1p ⊕ K2p such that K1p is an elementary p-group and K2p has no direct summand of order p. Such decompositions exist by Lemma I.3.1. We have Im(f ) = p Im(f ) ⊕ Im(f )[p] where p Im(f ) is p-divisible and hence (p Im(f ))p = 0. As A1 ∼ = Im(f ) we have correspondingly that A1 = pA1 ⊕ A1 [p] where pA1 is p-divisible and (pA1 )p = 0. We now have A = K1p ⊕ K2p ⊕ A1 [p] ⊕ pA1

(9)

where K1p and A1 [p] are elementary p-groups and pA1 contains no p-primary elements. We will show next that K2p = pK2p , i.e., K2p is p-divisible. In fact, consider the composite map α : K2p → K2p /pK2p → A1 [p]. By choice of p we have that A1 [p] = 0 and if K2p /pK2p were nonzero also, there must exist a nonzero map α whose kernel would have to be a summand of K2p with complementary summand a

120

Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

nonzero elementary p-group. This would contradict the choice of K2p , and we conclude that indeed K2p is p-divisible. This in turn implies that K2p [p] = 0 because p-divisible p-groups are divisible and A is reduced. Hence Ap = A[p] = K1p ⊕ A1 [p] and pA = pK2p ⊕p(pA1 ) = K2p ⊕pA1 . Substituting in (9) we get that A = pA⊕A[p]. p p p ⊕ M12 such that M11 is an eleSuppose p ∈ P(A, M ) and write M1 = M11 p mentary p-group and M12 has no direct summand of order p. Such decompositions exist by Lemma I.3.1. We now have p p ⊕ M12 , M = p Im(f ) ⊕ Im(f )[p] ⊕ M11

(10)

p p are elementary p-groups, (p Im(f ))[p] = 0 and M12 has no where Im(f )[p] and M11 p direct summand of order p. Suppose that M12 [p] = 0. Then there is a nonzero map proj

p μ : Im(f ) Im(f )[p] → M12 and it follows from Lemma 2.12 that Im(μ) ⊆⊕ p M12 . But Im(μ) is a nonzero p-elementary group and cannot be a summand by p p p ⊕ M12 . We conclude that (M12 )p = 0. choice of the decomposition M1 = M11 p Therefore M [p] = Im(f )[p] ⊕ M11 and substituting in (10) we obtain that M = p . M [p] ⊕ M where M = p Im(f ) ⊕ M12

Theorem 3.7. Let A and M be reduced abelian groups, S := End(M ), T := End(A), and let 0 = f ∈ Hom(A, M ). If Sf T is regular if and only if the following conditions 1) and 2) hold. 1) A = Ker(f ) ⊕ A1 , and M = Im(f ) ⊕ M1 for some groups A1 ⊆ A and M1 ⊆ M . 2) Using the decompositions in 1) we have the identiﬁcations 0 f21 μ11 μ21 α11 α21 f= ,μ = ∈ S, α = ∈ T, 0 0 μ12 μ22 α12 α22 where

and

(11)

μ11 : Im(f ) → Im(f ), μ21 : M1 → Im(f ), μ12 : Im(f ) → M1 , μ22 : M1 → M1 , α11 : Ker(f ) → Ker(f ), α21 : A1 → Ker(f ), α22 : A1 → A1 . α12 : Ker(f ) → A1 ,

With these notations, i) μ11 f21 α12 ∈ Reg(Ker(f ), Im(f )), ii) μ11 f21 α22 ∈ Reg(A1 , Im(f )), iii) μ12 f21 α12 ∈ Reg(Ker(f ), M1 ), iv) μ12 f21 α22 ∈ Reg(A1 , M1 ). Proof. 1) is simply Corollary II.1.3, and 2) is a special case of Theorem II.6.14.

3. Mixed Groups

121

Theorem 3.7 by far does not determine Reg(A, M ) for reduced mixed groups because Reg(Ker(f ), Im(f )), Reg(A1 , Im(f )), Reg(Ker(f ), M1 ), and Reg(A1 , M1 ) are not known. For one thing, it is quite unclear what Ker(f ) and M1 might be, however Im(f ) ∼ = A1 can be described quite well (Lemma 2.7, Proposition 3.10) which however does not mean that Reg(A1 , Im(f )) ∼ = Reg(End(Im(f ))) is known. Before we discuss further the determination of Reg(End(Im(f ))) we summarize what can be said at this point. Theorem 3.8. Let A and M be reduced abelian groups, S := End(M ), T := End(A), and let 0 = f ∈ Hom(A, M ). If Sf T is regular, then the following statements hold. 1) A = Ker(f ) ⊕ A1 , and M = Im(f ) ⊕ M1 for some groups A1 ⊆ A and M1 ⊆ M . 2) ∀ p ∈ P : Im(f ) = p Im(f ) ⊕ Im(f )[p]. 3) P(f ) := {p ∈ P : (Im(f )) [p] = 0} = ∅. 4) f21 : A1 x → f (x) ∈ Im(f ) is in Reg(A1 , Im(f )). 5) For every μ11 ∈ Hom(Im(f ), M1 ) it is true that Im(μ11 ) ⊆⊕ M1 and Ker(μ11 ) ⊆⊕ Im(f ). In particular, for p ∈ P(f ), it is true that (M1 )p = M1 [p]. 6) For every α12 ∈ Hom(Ker(f ), Im(f )) it is true that Im(α12 ) ⊆⊕ Im(f ) and Ker(α12 ) ⊆⊕ Ker(f ). In particular, decomposing Ker(f ) as Ker(f ) = K1p ⊕ K2p such that K1p is p-elementary and K2p has no direct summands of order p, it is true for p ∈ P(f ) that pK2 = K2 . 7) For every λ ∈ Hom(Ker(f ), M1 ) it is true that Im(λ) ⊆⊕ M1 and Ker(λ) ⊆⊕ Ker(f ). Proof. Suppose that Sf T is nonzero and regular. 1) and 2) are established in Corollary 2.9. 3) follows from 2) because P(f ) = ∅ would mean that Im(f ) = p Im(f ) for every prime p, i.e., Im(f ) would be a divisible subgroup of a reduced group and hence Im(f ) = 0, contrary to hypothesis. Using the decomposition in 1) we have the identiﬁcations μ11 μ21 α11 α21 0 f21 , μ= ∈ S, α = ∈ T, (12) f= 0 0 μ12 μ22 α12 α22 where

and

μ11 : Im(f ) → Im(f ), μ21 : M1 → Im(f ), μ22 : M1 → M1 , μ12 : Im(f ) → M1 , α11 : Ker(f ) → Ker(f ), α21 : A1 → Ker(f ), α22 : A1 → A1 . α12 : Ker(f ) → A1 ,

We now apply Theorem II.6.14 and Lemma 2.12 to get 4), 5), 6) and 7).

122

Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

Lemma 2.7 suggests investigating the consequences of the clearly very restrictive equations A = pA ⊕ A[p]. We will elucidate the structure of groups A satisfying A = pA ⊕ A[p]. Proposition 3.9. Let P be a set of primes and A be an abelian group such that, for every prime p ∈ P , there is a decomposition A = pA ⊕ Ap1 ,

Ap1 ⊆ A[p].

∼ Then A/D the maximal P -divisible = G where D is subgroup of A and G is a p group with p∈P A1 ⊆ G ⊆ p∈P Ap1 such that G/ p∈P Ap1 is P -divisible. Proof. Let p ∈ P . The decomposition A = pA ⊕ Ap1 with pAp1 = 0 implies pA = Let πp : A → Ap1 be the projection p(pA) ⊕ pAp1 = p(pA), i.e., pA is p-divisible. with kernel pA and let π : A → P := p∈P Ap1 be the canonical map induced by p the πp . Then π is the identity on p∈P A1 = t(P ). Secondly, for every p ∈ P , p ∼ A/A1 = pA is p-divisible and maps onto π(A)/ t(P ) which shows that the latter is P -divisible. It remains to show that Ker(π) = p∈P pA is P -divisible. Let x ∈ Ker(π). We will show that for any prime p ∈ P , there is y ∈ Ker(π) such that x = py. As x ∈ q∈P qA ⊆ pA there is y ∈ A such that x = py . Write y = x +y, where x ∈ Ap1 and y ∈ pA. Then x = py. Let q be a prime of P diﬀerent from p. We will see that y ∈ qA so that actually y ∈ q∈P qA = Ker(π). We have integers u, v such that 1 = up + vq and hence y = upy + vqy = ux + vqy ∈ qA where it is used that x ∈ qA. This establishes the embedding. The following corollaries are immediate. Corollary 3.10. Suppose that A is a reduced abelian group such that, for every prime p ∈ P, there is a decomposition A = pA ⊕ A[p]. ∼ Then p∈P A[p] ⊆ G ⊆ p∈P A[p] such that A = G where G is a group with G/ p∈P A[p] is divisible. Corollary 3.11. Suppose that f ∈ Reg(A, M ) and M is reduced. Then Im(f ) ∼ =G where G is a group with p∈P Im(f )[p] ⊆ G ⊆ p∈P Im(f )[p] such that G/ p∈P Im(f )[p] is divisible.

Proof. Lemma 2.7 and Corollary 3.10. Corollary 3.12. If in Proposition 3.9 the group A is reduced, then necessarily A[p] and hence A = A[p] ⊕ pA and D is torsion-free.

Ap1

=

Proof. We have pA = p(pA) and hence (pA)p = 0 because a p-divisible subgroup of p-group is divisible. For the same reason D must be torsion-free.

3. Mixed Groups

123

Theorem 3.6 and Proposition 3.9 suggest considering the following classes of groups. Deﬁnition3.13. For each p ∈ P, let Tp be a zero or nonzero elementary p-group, set P := p∈P Tp , T := p∈P Tp = t(P ), and let P(T ) := {p ∈ P | Tp = 0}. A group A is a G ∗ (T )-group if T ⊆ A ⊆ P such that A/T is (torsion-free) and P(T )divisible. Note that A/T = 0, i.e., A = T , is allowed. A G ∗ (T )-group A will be called slim if Tp is cyclic or zero for every p ∈ P. The symbol G ∗ (T ) also stands for the class of all G ∗ (T )-groups, so that A ∈ G ∗ (T ) if and only if A is a G ∗ (T )-group. A group is a G ∗ -group if it is a G ∗ (T )-group for some (elementary) group T . Let πp : P → Tp denote the projections. The restriction of πp to a subgroup of P will also be denoted by πp . The deﬁnition of G-group of Glaz-Wickless (see [13]) diﬀers from our deﬁnition of G ∗ -group in requiring that A/T be divisible and not just P(T )-divisible. We will ﬁrst get some insight into the homomorphisms between G ∗ -groups. Lemma 3.14. Let A ∈ G ∗ (T1 ) and M ∈ G ∗ (T2 ). Then there is a natural map σ : Hom(A, M ) → p∈P Hom((T1 )p , (T2 )p ) with Ker(σ) ∼ = Hom(A/T1 , M ). The mapping σ is an embedding if A/T1 = p (A/T1 ) whenever (T2 )p = 0. This is the case if A/T1 is divisible or if P(T2 ) ⊆ P(T1 ). Proof. The map σ is the composite of the restriction map Hom(A, M ) → Hom(T1 , M ) = Hom(T1 , T2 ) with the natural isomorphism Hom(T1 , T2 ) →

p∈P

Hom((T1 )p , (T2 )p ).

Let f ∈ Hom(A, M ) and let φ1 : A → A/T1 be the natural epimorphism. Assume that σ(f ) = 0. Then f (T1 ) = 0 and hence f = f1 φ1 for some f1 ∈ Hom(A/T1 , M ). Conversely, if f ∈ Hom(A, M ) and f = f1 φ1 for some f1 ∈ Hom(A/T1 , M ), then f (T1 ) = 0 and so σ(f ) = 0. If A/T1 is divisible, then Hom(A/T1 , M ) = 0 since M is reduced. More precisely, M contains no subgroup that is p-divisible for every p ∈ P(T2 ). On the other hand, by deﬁnition of G ∗ (T1 ), we have A/T1 = p(A/T1 ) for every prime p ∈ P(T1 ). Consequently, if P(T2 ) ⊆ P(T1 ), then Hom(A/T1 , M ) = 0. Recall that a group G is bounded if there exists a nonzero integer n such that nG = 0. Suppose that G is a torsion group with primary decomposition G = p∈P Gp . If G is bounded, then Gp = 0 for all but ﬁnitely p. Suppose that Gp is ﬁnite for every p. Then G is bounded if and only if it is ﬁnite. For example, a subgroup of a slim G ∗ -group is bounded if and only if it is ﬁnite. We begin with some basic observations.

124

Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

Lemma 3.15. Let A be a G ∗ (T )-group. Then the following statements hold. 1) A is P(T )-pure in P . More strongly, if p ∈ P(T ), x ∈ P , and px ∈ A, then x ∈ A. 2) If F ⊆ P(T ) is a ﬁnite set of primes, then " # A = p∈F Tp ⊕ A ∩ p∈F T p . / Further, ⎛ t ⎝A ∩

p∈F /

⎞ Tp ⎠ =

p∈F /

"

# A ∩ p∈F / Tp A ∼ , Tp and = t(A) p∈F / Tp

the latter group being torsion-free and P(T )-divisible. Furthermore, ∀p ∈ F, p(A ∩ p∈F / Tp ) = A ∩ p∈F / Tp . 3) If B is a bounded subgroup of A, then there is a ﬁnite set of primes F ⊆ P(T ), such that B ⊆ p∈F Tp and B ⊆⊕ A. 4) Every endomorphism f of A extends uniquely to an endomorphism fˆ of P and hence is of the form f = Πp∈P fp where fp ∈ End(Tp ). 5) Let m also stand for multiplication by m on A and assume that m is a P(T )number. Then m is regular. Proof. 1) Let x ∈ P , p ∈ P(T ) and px ∈ A. Then px + T ∈ A/T and as A/T is P(T )-divisible there is a ∈ A such that px + T = p(a + T ). Hence x − a ∈ T ⊆ A and so x ∈ A. " # 2) Clearly P = Tp ⊆ A, p∈F Tp ⊕ P where P := p∈F / Tp . Since " # p∈F the decomposition of A follows by intersecting P = p∈F Tp ⊕ P with A. Furthermore, t(A ∩ p∈F / Tp ) = p∈F / Tp and hence Tp A ∩ p∈F ( p∈F Tp ) ⊕ (A ∩ p∈F A / Tp ) ∼ / = = T ( T ) ⊕ ( T ) t(A) p p p∈F / p∈F p p∈F / is torsion-free and P(T )-divisible. Let p ∈ F . Then clearly P = pP and A ∩ P is p-pure in P because A = Tp ⊕ A ∩ P and A is p-pure in P , hence A ∩ P = (A ∩ P ) ∩ pP = p(A ∩ P ). 3) The bounded subgroup B has a primary decomposition B = p∈P Bp with 0 = Bp ⊆ Tp for p in some ﬁnite set of primes F . Since p∈F Tp is semisimple and contains B, it follows that B is a direct summand of p∈F Tp which in turn is a summand of A, so B is a summand of A.

3. Mixed Groups

125

4) The maximal torsion subgroup T of P is fully invariant in P . Therefore there is a well-deﬁned restriction map End(P ) → End(T ) ∼ = p∈P End(Tp ) that is an isomorphism. Hence elements of End(P ) ∼ = End(T ) are of the form f = Πp∈P fp . Let f ∈ End(A). First restrict f to t(A) = T = t(P ) and then extend to an endomorphism of P that restricts to f on A. 5) Let F := {p ∈ P | p divides m}. Then F ⊆ P(T ). Clearly Ker(m) = p∈F Tp and by 1) A=

" p∈F

# " # Tp ⊕ A ∩ p∈F / Tp .

= Ker(m) ⊕ Im(m) which shows We will show that mA = A ∩ p∈F / Tp . Then A that m is regular. It is clear that mA ⊆ A ∩ p∈F / Tp . Conversely, by 2), A ∩ T = m(A ∩ T ) ⊆ mA. p p p∈F / p∈F / The following theorem shows that it is quite common to have nonzero regular bimodules in Hom(A, M ) for G ∗ -groups A and M . Theorem 3.16. Suppose that A ∈ G ∗ (T1 ) and M ∈ G ∗ (T2 ). Then t(Hom(A, M )) is a regular bi-submodule of Hom(A, M ). Proof. Note that f ∈ t(Hom(A, M )) if and only if Im(f ) is bounded. Suppose that f ∈ t(Hom(A, M )), μ ∈ End(M ), and α ∈ End(A). Then (μf α)(A) ⊆ (μf )(A) = μ(f (A)) which is bounded. Hence t(Hom(A, M )) is a bi-submodule of Hom(A, M ). Let f ∈ t(Hom(A, M )), and let P1 := {p ∈ P | (Im(f ))p = 0}. Then P1 ⊕ is ﬁnite and Im(f ) ⊆⊕ (T ) . Then P1 = p∈P#1 (T2 )p ⊆ " M . Set P1 := p∈P " # " # " 1 p # ⊕ and A = ⊕ A ∩ (T ) (T ) (T ) (T ) 1 p 1 p 1 p 1 p . Note p∈P1 p∈P / 1 p∈P1 p∈P / 1 that P1 ⊆ P(T1 ). It now follows from the fact that f ( p∈P / 1 Tp ) = 0 and the ! fact that A ∩ p∈P / 1 (T1 )p / p∈P / 1 (T1 )p is P(T1 )-divisible, hence P1 -divisible, " # " # (T ) (T ) that A ∩ ⊆ Ker(f ). Hence Ker(f ) = Ker(f ) ∩ 1 1 p∈P / 1 p p∈P1 p ⊕ # " # " A∩ p∈P / 1 (T1 )p and since Ker(f ) ∩ p∈P1 (T1 )p is a direct summand of the elementary group p∈P1 (T1 )p , it follows that Ker(f ) ⊆⊕ A. This shows that f is regular. The following example shows that a G ∗ (T )-group may have epimorphic images of order p if p ∈ / P(T ). This means that the map σ in Lemma 3.14 need not be an embedding without proper assumptions. Example 3.17. Let P1 be a proper inﬁnite subset of P. For every p ∈ P1 let Tp be a group of order p, let T := T , and let P := p∈P1 p p∈P1 Tp . Let A ⊆ P ∈ Q | n is a P -number}. Then A/T is be the group given by A/T ∼ = Q := { m 1 n / P1 we obtain a mapping A A/T ∼ P1 -divisible, so A ∈ G ∗ (T ) but, for any p ∈ = Q Q/pQ ∼ = Z(p). Combining earlier results we obtain a characterization of G ∗ -groups.

126

Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

Theorem 3.18. Let A be an abelian group. Then A is a G ∗ (T )-group for T := p∈P Ap if and only if A is reduced and ∀ p ∈ P(T ), A = pA ⊕ A[p]. Proof. Suppose that A ∈ G ∗ . Then A is reduced and by Lemma 3.15.1with F = [A ∩ q =p Tq ] where {p} ⊆ P(T ) we have A = Tp ⊕ Tp = A[p] and A ∩ q =p Tq is p-divisible. Hence pA = p[A ∩ q =p Tq ] = [A ∩ q =p Tq ]. The converse is settled in Corollary 3.10. The following example shows that for A ∈ G ∗ (T ), the equality A = A[p] ⊕ pA need not hold for all p. Example 3.19. Let T = p∈P Tp be the elementary abelian group with T2 = 0, and, for primes p ≥ 3, let T = a be cyclic of order p. Let a = a2 = 0, p p p∈P ap ∈ P := p∈P Tp . Let A be the subgroup of P given by A/T = Z(2) (a + T ) where Z(2) is Z localized at the prime ideal (2) = 2Z. Then A ∈ G ∗ (T ) but A = A[2] ⊕ 2A. Proof. It is clear that A ∈ G ∗ (T ) and A2 = 0. Hence the equality A = A[2] ⊕ 2A is simply A = 2A. By way of contradiction assume that there is x ∈ A such that 2x = a. Since x ∈ A, there is m/n ∈ Z(2) such that x + T = m n (a + T ), or equivalently, nx − ma ∈ T where n is odd and gcd(m, n) = 1. It follows that (n − 2m)x ∈ T with n − 2m = 0, and this implies that x ∈ T , and so a = 2x ∈ T , a contradiction. The following example shows that there are nice groups in G ∗ (T ) yet there are non-regular homomorphisms between these groups. The example demonstrates that it will not be easy to determine those pairs of mixed groups A and M such that Hom(A, M ) is regular. Example p ∈ P, let Tp = ap be cyclic of order p, let T := p∈P Tp , 3.20. For each P := p∈P Tp , a = p∈P ap , and let A be the puriﬁcation of T + Za in P (see Lemma I.3.5). Split P into a disjoint union P = P1 ∪ P2 such that both parts P1 and P2 are inﬁnite. Let b := p∈P2 ap and let M be the puriﬁcation of T + Zb in P . Deﬁne 0 if p ∈ P1 . f = p∈P fp such that fp = 1 if p ∈ P2 Then f ∈ Hom(A, M ), M = p∈P1 Tp ⊕ Im(f ), Ker(f ) = p∈P1 Tp , and Ker(f ) is not a direct summand of A. Proof. To establish that f ∈ Hom(A, M ) requires only to check that f (a) ∈ M P because obviously f (T ) ⊆ T ⊆ M and f (T +Za) ⊆ M will imply that f ((T +Za)∗ ) ⊆ P (f (T ) + Zf (a))∗ ⊆ M . But clearly, f (a) = p∈P fp ap = p∈P2 ap = b ∈ M . It follows from thedeﬁnition of f that Ker(f ) = p∈P1 Tp and Im(f ) = M ∩ p∈P2 Tp so M = p∈P1 Tp ⊕ Im(f ). It remains to show that Ker(f ) is not a summand of A. By way of contradiction assume that for some A = Ker(f ) ⊕ B subgroup B of A. We have P = P1 ⊕ P2 where P1 = p∈P1 Tp and P2 = p∈P2 Tp

4. Regularity in Endomorphism Rings of Mixed Groups

127

subgroup of P and it is immediately seen that P1 is the maximal P2 -divisible subgroup of P . Now t(B) = and P2 is the maximal P1 -divisible p∈P2 Tp is P1 divisible and B/ t(B) ∼ = Q, = ( p∈P1 Tp ⊕ B)/( p∈P1 Tp ⊕ t(B)) = A/ t(A) ∼ hence by Lemma I.3.6, B is P1 -divisible and therefore B ⊆ P2 . It is now clear that a∈ / Ker(f ) ⊕ B and we have arrived at the desired contradiction.

4 Regularity in Endomorphism Rings of Mixed Groups It has been seen in Theorem 3.8 that Reg(A, A) = Reg(End(A)) comes into play when studying Hom(A, M ). Unfortunately, while there are several papers on the subject ([29], [12], [30], [13], [24]), it is not even known which mixed abelian groups have regular endomorphism rings. We include some of the more accessible results but the subject is wide open, and even more so the computation of Reg(A, M ) for mixed groups A, M . We begin with an immediate but attractive corollary of Proposition 2.9. Corollary 4.1. For any abelian group Z1A ⊆ End(A) is regular if and only if A = nA ⊕ A[n] for every n ∈ Z. The neatest result on regular endomorphism rings is due to Glaz and Wickless for which we provide a new proof. Theorem 4.2. ([13, Theorem 4.1]) Let A be a slim G ∗ (T )-group such that A/T has ﬁnite rank and is divisible. Then End(A) is regular. Proof. Let f ∈ End(A). Then f = p∈P fp where fp ∈ Z/pZ and we may choose fp = 0 if Tp = 0. Let P1 := {p ∈ P | fp = 0} and let P2 := {p ∈ P | fp = 0}. Then clearly Ker(f ) ⊆ p∈P1 Tp , and Im(f ) ⊆ p∈P2 Tp . Hence Ker(f ) ∩ Im(f ) = 0,

also

f (T ) =

Tp .

p∈P2

We also have a short exact sequence Ker(f ) + T A f (A) . T T f (T ) Observe next that f (T ) ⊆ T ∩ f (A)

=

[(⊕p∈P1 Tp ) ⊕ (⊕p∈P2 Tp )] ∩ f (A)

⊆

[Ker(f ) ⊕ f (T )] ∩ f (A) = f (T ),

and so T ∩ f (A) = f (T ). Therefore f (A) f (A) ∼ f (A) + T A = ⊆ . = f (T ) T ∩ f (A) T T

(13)

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Chapter VIII. Regularity in Homomorphism Groups of Abelian Groups

It follows that f (A)/f (T ) is torsion-free and therefore Ker(fT )+T is pure in A/T and therefore divisible. We conclude that the short exact sequence splits and dim

Ker(f ) + T f (A) A = dim + dim . T T f (T )

(14)

A Ker(f ) + T f (A) + T ⊇ ⊕ . T T T

(15)

We also have

That the sum is direct can be seen as follows. (Ker(f ) + T ) ∩ (f (T ) + T )

= [Ker(f ) ∩ (f (T ) + T )] + T = [Ker(f ) ∩ (f (T ) + ⊕p∈P2 Tp )] + T = (Ker(f ) ∩ f (T )) + ⊕p∈P2 Tp + T = T.

The dimension equation (14) together with (15) now implies that A = Ker(f ) + f (A) + T = Ker(f ) ⊕ Im(f ) and we have established that f is regular. Remark. The proof of Proposition 4.5 also shows that for a G ∗ -group A the ideal I = {f ∈ End(A) | f (A) is torsion} need not be regular. Inspection of the veriﬁcation of Example 3.3 and [13, Theorem 3.5]) suggest the following theorem. Theorem 4.3. Suppose that A is a G ∗ -group with the property that every torsion endomorphic image of A is bounded. Then if End(A) is regular, so is End(A). More generally, Reg(End(A)) = Reg(End(A)). Proof. Let f ∈ Reg(End(A)). Then f gf = f for some g ∈ End(A) and hence f gf = f , i.e., f is regular in End(A). Let g1 , g2 ∈ End(A)). Then g1 f g2 ∈ Reg(End(A)) and by the established result g1 f g2 is regular. So f ∈ Reg(End(A)). Conversely, let f ∈ End(A) and suppose that f ∈ Reg(End(A)). Then there is g ∈ End(A) such that f gf = f . It follows that f − f gf ∈ End(A) is such that (f − f gf )(A) ⊆ T . By hypothesis (f − f gf )(A) is bounded and by Theorem 3.16 it follows that f − f gf is regular in End(A). By Lemma II.4.1 f is regular. Suppose that f ∈ Reg(End(A)) and g1 , g2 ∈ End(A). Then g1 f g2 = g1 f g2 ∈ Reg(End(A)) and therefore g1 f g2 is regular by the previous result. Hence f ∈ Reg(End(A)). We conclude this section with two interesting results on regularity in endomorphism rings. Theorem 4.4. Let A be a reduced abelian group and assume that 0 = Reg(A, A). Then there exists a non-void set of primes P such that ∀ p ∈ P , A = pA ⊕ A[p],

A[p] = 0.

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129

Conversely, suppose that A is a group (reduced or not) and P a non-void set of primes such that ∀ p ∈ P , A = pA ⊕ A[p], and Ap = A[p] = 0. Then Reg(A, A) = 0. Proof. The ﬁrst part is the special case A = M of Theorem 3.6. Now suppose that A is a group and P a non-void set of primes such that ∀ p ∈ P , A = pA ⊕ A[p], and Ap = A[p] = 0. By induction we obtain for a ﬁnite set of primes p1 , . . . , pk ∈ P that A = A[p1 ] ⊕ · · · ⊕ A[pk ] ⊕ p1 · · · pk A.

(16)

Recall that n is a P -number if the prime factors of n all belong to P . Let I := {ξ ∈ End(A) | nξ = 0 for some nonzero P -number n}. We claim that I is a nonzero regular ideal of End(A). First of all I = 0 because P = ∅ and for p ∈ P the projector ξ of A onto A[p] along pA belongs to I. It is more or less evident that I is an ideal in End(A). It is left to show that every ξ ∈ I is regular. Let n = 0 be a P -number such that nξ = 0 and let p1 , . . . , pk be the prime factors of n. Then we have (16) and n(Im(ξ)) = 0. Hence Im(ξ) ⊆ A[p1 ]⊕· · ·⊕A[pk ] and the latter being elementary (= semisimple), Im(ξ) is a direct summand of it and in turn of A. The kernel Ker(ξ) must contain p1 · · · pk A and therefore it follows from (16) that Ker(ξ) = (Ker(ξ) ∩ (A[p1 ] ⊕ · · · ⊕ A[pk ])) ⊕ p1 · · · pk A. As Ker(ξ) ∩ (A[p1 ] ⊕ · · · ⊕ A[pk ]) is a direct summand of A[p1 ] ⊕ · · · ⊕ A[pk ] it follows from (16) that Ker(ξ) is a direct summand of A. Thus ξ is regular. The next proposition shows that there are also obstructions to regularity. Proposition 4.5. Let A be a G ∗ (T )-group with the property that every torsion direct summand of A is bounded and let B be an elementary group such that Bp = 0 whenever Tp = 0. Let G = A ⊕ B. Then End(G) is not regular and Reg(G, G) is a proper ideal of End(G). Proof. The hypotheses provide for the existence of an endomorphism f of G that is 0 on A and maps B onto an unbounded torsion summand of A. Then f (G) is not a direct summand of A and therefore not a summand of G. So f is not regular. Finally, by Theorem 4.4, Reg(G, G) = 0.

Chapter IX

Regularity in Categories 1 Regularity in Preadditive Categories The deﬁnition of a regular map makes perfectly good sense in any category. To wit, let C be a category. We write A ∈ C if A is an object of C and by C(A, B) we denote the set of morphisms in C for objects A and B in C. Then f ∈ C(A, B) is regular if and only if there exists g ∈ C(B, A) such that f gf = f . To do more we need more special categories. A preadditive category is a category C in which the morphism sets are abelian groups with the property that composition of morphisms distributes over addition. It is also postulated that the category contain a null object 0 such that C(0, B) = {0} and C(A, 0) = {0}. This assures that C(A) := C(A, A) is a ring with 1A ∈ C(A) for any object A in the preadditive category C and C(A, M ) is a C(M )-C(A)-bimodule. Surprisingly Reg(A, M ) exists in preadditive categories as the largest C(M )C(A)-submodule of C(A, M ) all of whose morphisms are regular. In fact, inspection of the proofs in the case of module categories reveal that they consist of computation with R-homomorphisms that are equally valid for morphisms in a preadditive category. In the following it is always assumed that C is a preadditive category. Lemma 1.1. Let f ∈ C(A, M ) and g ∈ C(M, A). If f − f gf is regular, then f is regular. Proof. The proof of Lemma II.4.1 applies literally. Deﬁnition 1.2. Let S := C(M ) and T := C(A). For f ∈ C(A, M ) let Reg(A, M ) := {f ∈ C(A, M ) | Sf T is regular}, where Sf T is the S-T -submodule of C(A, M ) generated by f . A subset X of C(A, M ) is regular if every element of X is regular.

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Chapter IX. Regularity in Categories

Theorem 1.3. Reg(A, M ) is the largest regular S-T -submodule of C(A, M ). Proof. The proof of Theorem II.4.3 applies.

2

Preadditive Categories

An object A in a preadditive category C is a biproduct of the objects A1 , . . . , An ∈ C if there exist morphisms, ιi : Ai → A,

and πi : A → Ai ,

called structural maps, such that πi ιi = 1Ai ,

πj ιi = 0 for i = j,

and ι1 π1 + · · · + ιn πn = 1A .

If this is the case we write A ∈C A1 ⊕ · · · ⊕ An , we say the A is a biproduct in C of the objects A1 , . . . , An , and think of A1 ⊕ · · · ⊕ An as the collection of all data sets (A, ιi , πi ). We write A ∈C A1 ⊕ · · · ⊕ An instead of A = A1 ⊕ · · · ⊕ An as is often done because there will later be situations (see Section 3, particularly Example 3.12) where we have to deal with true set-theoretic equality as well as with categorical notions. Let C be a preadditive category. If A ∈C A1 ⊕ · · · ⊕ An and B is isomorphic with A, then B ∈C A1 ⊕ · · · ⊕ An also. If A, B ∈C A1 ⊕ · · · ⊕ An , then A and B are isomorphic. If A ∈C A1 ⊕ · · · ⊕ An and Ai ∼ =C Ai , then A ∈C A1 ⊕ · · · ⊕ An . Further note that the apparent ordering of the summands is introduced by the choice of labels and is not intrinsic to the deﬁnition. A biproduct has the universal properties of both products and coproducts and is unique up to isomorphism in the category. We recall here the deﬁnitions of kernels, cokernels, images and coimages of morphisms in a category. Deﬁnition 2.1. Let C be a preadditive category and f ∈ C(A, M ). 1) A kernel of f is a morphism k : K → A such that f k = 0 and whenever χ : X → A is such that f χ = 0, then there is a unique morphism λ : X → K such that kλ = χ. We write k ∈ Ker(f ) if k is a kernel of f . 2) A cokernel of f is a morphism c : M → C such that cf = 0 and whenever χ : M → X is such that χf = 0, then there is a unique morphism λ : C → X such that λc = χ. We write c ∈ Coker(f ) if c is a cokernel of f . 3) An image of f is a kernel of a cokernel of f ; we write Im(f ) = Ker(Coker(f )). 4) A coimage of f is the cokernel of a kernel of f , Coim(f ) = Coker(Ker(f )). All these concepts are unique up to isomorphism. Recall that a morphism e is epic if it can be canceled on the right: f e = ge ⇒ f = g, and a morphism m is monic if it can be canceled on the left: mf = mg ⇒ f = g.

2. Preadditive Categories

133

Example 2.2. 1) Kernels and images are monic. 2) Cokernels and coimages are epic. Proof. Let k ∈ Ker(f : A → M ). Suppose that kk1 = kk2 for k1 , k2 ∈ C(B, A). Then f (kk1 ) = 0 and f (kk2 ) = 0. By deﬁnition of kernel the morphism kk1 = kk2 factors uniquely through k, so k1 = k2 . Images are monic because they are kernels. The epic property of cokernels and coimages is dual. Example 2.3. Let A ∈C A1 ⊕ A2 with structural maps ιi : Ai → A,

and πi : A → Ai .

Let i, j ∈ {1, 2}, i = j. Then 1) 0 : 0 → Ai is a kernel of ιi : Ai → A, 2) 0 : Ai → 0 is a cokernel of πi : A → Ai , 3) ιi : Ai → A is a kernel of πj : A → Aj , 4) πi : A → Ai is a cokernel of ιj : Ai → A, 5) ιi : Ai → A is an image of ιi : Ai → A, 6) 1Ai : Ai → Ai is an image of πi : A → Ai , 7) 1Ai : Ai → Ai is a coimage of ιi : Ai → A, 8) πi : A → Ai is a coimage of πi : Ai → A. We summarize the results in a table. Ker Coker Im Coim

ι1 0 π2 ι1 1A1

π1 ι2 0 1A1 π1

ι2 0 π1 ι2 1 A2

π2 ι1 0 1A2 π2

Proof. 1) and 2) follow from the fact that kernels are monic and cokernels are epic. 3) We have that πj ιi = 0. Suppose that χ : X → A is such that πj χ = 0. Then πi χ : X → Ai and ιi πi χ = (1 − ιj πj )χ = χ. So χ factors through Ai . Suppose that φ : X → Ai is such that ιi φ = χ. Then φ = πi ιi φ = πi χ, so φ is unique. 4) The proof is dual to 3). 5) Im(ιi ) = Ker(Coker(ιi ) = Ker(πj ) = ιi . 6) Im(πi ) = Ker(Coker(πi ) = Ker(0 : Ai → 0) = ιi . 7) Coim(ιi ) = Coker(Ker(ιi )) = Coker(0 : 0 → Ai ) = 1Ai . 8) Coim(πi ) = Coker(Ker(πi )) = Coker(ιj ) = πi .

134

Chapter IX. Regularity in Categories

Example 2.4. Let e ∈ C(A, A) be an idempotent and suppose that e and 1 − e have kernels and cokernels. Then Im(e) = Ker(1 − e),

and

Im(1 − e) = Ker(e).

Proof. Let γ : A → C be a cokernel of e. Since (1 − e)e = 0, there is ψ : C → A such that 1 − e = ψγ. (1) Let κ : K → A be a kernel of 1 − e. We will show that κ ∈ Ker(γ) = Ker(Coker(e)) = Im(e) thereby establishing in particular that Ker(γ) exists. We have γκ = γ(e + (1 − e))κ = γeκ + γ(1 − e)κ = 0. Suppose that χ : X → A is (1)

such that γχ = 0. Then (1 − e)χ = ψγχ = 0 and since κ ∈ Ker(1 − e), the map χ factors uniquely through κ showing that κ ∈ Ker(γ) = Ker(Coker(e)) = Im(e). We now assume that λ : L → A ∈ Im(e) = Ker(γ) and show that λ ∈ Ker(1−e). We have γλ = 0 and from γe = 0 it follows that there is ξ : A → L such that e = λ ξ. Suppose that χ : X → A is such that (1 − e)χ = 0. As κ ∈ Ker(1 − e) there is ζ : X → K such that χ = κζ. It follows that χ = eχ = λ(ξχ). So χ factors through λ and that this is uniquely so follows from the fact that λ is a kernel and hence monic. We have shown that λ ∈ Ker(1 − e). An additive category is a preadditive category that has biproducts for any ﬁnite set of objects. Given an idempotent e in a category, it is not automatic that it can be factored as e = ιπ, πι = 1 and that it produces a decomposition. Deﬁnition 2.5. Let A be an object in a category C and e = e2 ∈ C(A, A). Then the idempotent e splits in C if there exists an object M and mappings ι ∈ C(M, A), π ∈ C(A, M ) such that ιπ = e and πι = 1M . We say that idempotents split in C if all idempotents in C are splitting. We observe next that the splitting of idempotents means that every idempotent determines a direct decomposition. Lemma 2.6. Let C be a preadditive category. Suppose that A is an object of C, and e is an idempotent in C(A). Then 1A − e is another idempotent of C(A). Suppose further that there are objects and morphisms ι1 : A1 → A,

π1 : A → A1 ,

ι2 : A2 → A,

π2 : A → A2

such that ι1 π1 = e,

π1 ι1 = 1A1 ,

ι2 π2 = 1A − e,

π2 ι2 = 1A2 .

Then A ∈C A1 ⊕ A2 with structural maps ι1 , π1 , ι2 , π2 . In addition, ι1 ∈ Ker(1A − e) = Im(e), π1 ∈ Coker(1A − e),

ι2 ∈ Ker(e) = Im(1A − e), π2 ∈ Coker(e).

2. Preadditive Categories

135

Proof. We have ι1 π1 + ι2 π2 = e + 1A − e = 1A , next π1 ι2 = 1A1 π1 ι2 1A2 = π1 ι1 π1 ι2 π2 ι2 = π1 e(1A − e)ι2 = 0, and similarly π2 ι1 = 0. This shows that A is a biproduct of A1 and A2 . We show next that ι1 ∈ Ker(1−e). First (1−e)ι1 π1 = (1−e)e = 0 and since π1 is epic, (1 − e)ι1 = 0. Now suppose that χ : X → A is such that (1 − e)χ = 0. Then χ = 1A χ = ι1 π1 χ + ι2 π2 χ = ι1 (π1 χ) + (1 − e)χ = ι1 (π1 χ) and this factorization through ι1 is unique because ι1 is monic. By symmetry we have ι2 ∈ Ker(e). It was shown in Example 2.4 that Im(e) = Ker(1 − e), and Im(1 − e) = Ker(e). We show next that π1 ∈ Coker(1 − e). First ι1 π1 (1 − e) = e(1 − e) = 0 and so π1 (1 − e) = 0 because ι1 is monic. Suppose that χ : A → X is such that χ(1 − e) = 0. Then χ = χe = (χι1 )π1 so χ factors through π1 and uniquely so because π1 is epic. By symmetry π2 ∈ Coker(e). If idempotents split in a preadditive category C, then we have the familiar decompositions associated with regular maps. Theorem 2.7. Let C be a preadditive category in which idempotents split and suppose that f ∈ C(A, M ) is regular and f gf = f for g ∈ C(M, A). Then the following statements hold. 1) e := f g ∈ C(M, M ) is an idempotent. 2) d := gf ∈ C(A, A) is an idempotent. 3) There are structural maps ιAi : Ai → A, and πAi : A → Ai such that d = ιA1 πA1 , πA1 ιA1 = 1A1 , 1A − d = ιA2 πA2 , πA2 ιA2 = 1A2 , and A ∈C A1 ⊕ A2 . 4) There are structural maps ιMi : Mi → M , and πMi : M → Mi such that e = ιM1 πM1 , πM1 ιM1 = 1M1 , 1M − e = ιM2 πM2 , πM2 ιM2 = 1M2 , and M ∈C M1 ⊕ M2 . 5) πM1 f ιA1 : A1 → M1 is an isomorphism with inverse πA1 gιM1 . 6) ιA2 ∈ Ker(f ) and ιM2 ∈ Ker(f g). 7) ιM1 ∈ Im(f ) and ιA1 ∈ Im(gf ). Proof. 1) and 2). It follows from f gf = f that e = f g and d = gf are idempotents. 3) and 4) As idempotents split in C we have the indicated structural maps and by Lemma 2.6 we have the claimed biproducts. 5) We compute: πA1 gιM1 πM1 f ιA1 = πA1 gef ιA1 = πA1 gf gf ιA1 = πA1 gf ιA1 = πA1 ιA1 πA1 ιA1 = 1A1 1A1 = 1A1 , and πM1 f ιA1 πA1 gιM1 = πM1 f dgιM1 = πM1 f gf gιM1 = πM1 f gιM1 = πM1 ιM1 πM1 ιM1 = 1M1 1M1 = 1M1 . 6) To show that ιA2 ∈ Ker(f ) we ﬁrst note that f ιA2 πA2 = f (1A − d) = f (1A − gf ) = 0 which implies that f ιA2 = 0 because πA2 is epic. Suppose that χ : X → A is such that f χ = 0. Then ιA2 (πA2 χ) = (1A − d)) χ = χ − gf χ = χ. This shows that χ factors through ιA2 . To show that this is uniquely so, suppose that ιA2 φ = χ for some φ. Then πA2 ιA2 φ = πA2 χ, therefore φ = πA2 χ.

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To show that ιM2 ∈ Ker(f g) we ﬁrst note that f gιM2 πM2 = f g(1M − e) = f g − f gf g = 0. Suppose that χ : X → M is such that f gχ = 0. Then ιM2 πM2 χ = (1 − e)χ = χ − f gχ = χ, so χ factors through ιM2 . To show that this is uniquely so, suppose that ιM2 φ = χ. Then φ = πM2 ιM2 φ = πM2 χ. 7) We show ﬁrst that πM2 ∈ Coker(f ). Indeed, ιM2 πM2 f = (1 − f g)f = 0, and, ιM2 being monic, πM2 f = 0. Suppose that χ : M → X is such that χf = 0. Then χ = χ1M = χιM1 πM1 + χιM2 πM2 = χf g + χιM2 πM2 = (χιM2 )πM2 , thus χ factors through πM2 and because πM2 is epic, uniquely so. We now have ιM1 ∈ Ker(πM2 ) = Ker(Coker(f )) = Im(f ). It follows from Lemma 2.6 that ιA1 ∈ Im(gf ) because gf = d is an idempotent. Let f ∈ C(A, M ). We wish to generalize to categories the result “f is regular if Ker(f ) ⊆⊕ A and Im(f ) ⊆⊕ M ”. Theorem 2.8. Let C be a preadditive category and let f ∈ C(A, M ). Assume that there exists a biproduct ιK πL K A L πK ιL with ιK ∈ Ker(f ) and assume that ιI πJ I M J πI ιJ is a biproduct with πJ ∈ Coker(f ), and ιI ∈ Ker(πJ ) = Im(f ). Then πI f ιL : L → I is both epic and monic. Assume that there is a morphism h : I → L such that h (πI f ιL ) = 1L ,

and

(πI f ιL ) h = 1I .

Then g : M → A : g := ιL hπI is such that f gf = f . Proof. We ﬁrst record two immediate consequences of our assumptions that we will have to use: (2) f ιK = 0, πJ f = 0. Suppose that χ : X → L is such that πI f ιL χ = 0. Then 0 = ιI πI f ιL χ = (1M −ιJ πJ )f ιL χ = f ιL χ−ιJ (πJ f )ιL χ = f ιL χ. Since f (ιL χ) = 0 and ιK ∈ Ker(f ), there is φ : X → K such that (3) ιL χ = ιK φ. (3)

We conclude that χ = 1L χ = πL ιL χ = πL ιK φ = 0 showing that πI f ιL is monic. Now suppose that χ : I → X is such that χπI f ιL = 0. Then 0 = χπI f ιL πL = χπI f (1A − ιK πK ) = χπI f using (2). Since (χπI )f = 0 and πJ ∈ Coker(f ), there is ψ : J → X such that (4) χπI = ψπJ .

3. Quasi-Isomorphism Category

137 (4)

We conclude that χ = χ1I = χπI ιI = ψπJ ιI = 0 showing that πI f ιL is epic. By assumption there is h : I → L such that h (πI f ιL ) = 1L and (πI f ιL ) h = 1I . Let g : M → A : g = ιL hπI . Note that πI f g = πI f ιL hπI = 1I πI = πI .

(5)

Note further that (2)

f = 1M f = (ιI πI + ιJ πJ )f = ιI πI f. (6)

(5)

(6)

Then f gf = ιI πI f gf = ιI πI f = f .

(6)

Question 2.9. In a module category the isomorphism HomR (R, M ) ∼ = M leads to regularity in modules. Is there an analogue in categories? Suppose there is an object A ∈ C that is free on one generator. Then C(A, M ) ∼ = M.

3

The Quasi-Isomorphism Category of Torsion-free Abelian Groups

In group and module theory there is a special interest in decompositions into direct sums of indecomposable subobjects. For convenience we call a direct decomposition an indecomposable decomposition if the direct summands are all indecomposable. In general, torsion-free abelian groups are notorious for their essentially diﬀerent indecomposable decompositions. In this section we demonstrate that the lack of uniqueness can be salvaged at the expense of introducing an equivalence of groups that is weaker than isomorphism. This leads to the so-called “quasi-isomorphism category”. We begin with an example, that may be considered folklore, of a group X that has two essentially diﬀerent indecomposable decompositions. The group X is a ﬁnite essential extension of a completely decomposable group A, meaning that A is large in X and X/A is ﬁnite. The fact that A is essential in X is equivalent to saying that X is also torsion-free. The group A has a homogeneous direct summand of rank 2 that can be decomposed in diﬀerent ways and as a consequence the elements of X (“clamps”) that tie the summands of A together are replaced by clamps that tie together diﬀerent summands of A. The group may be depicted as follows. The homogeneous block of rank 2 pictured on the left is decomposed in two diﬀerent ways which replaces a single 3-pronged clamp by two 2-pronged clamps.

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r r

r

r

r

= r

r

Example 3.1. Let V be a four-dimensional Q-vector space with basis {v1 , v2 , v3 , v4 }. Then V = Qv1 ⊕ Qv2 ⊕ Qv3 ⊕ Qv4 . The example will be obtained as an additive subgroup of V . Let A0 := Z[5−1 ], A1 := Z[7−1 ], A2 := Z[11−1 ], A = A0 v1 ⊕ A0 v2 ⊕ A1 v3 ⊕ A2 v4 , and X = A + Z

1 (v2 + 3v3 + 2v4 ). 2·3

Then X

1 = A0 v1 ⊕ (Av2 ⊕ A1 v3 ⊕ A2 v4 ) + Z (v2 + 3v3 + 2v4 ) 2·3 1 = (A0 (2v1 + v2 ) ⊕ A1 v3 ) + Z (2v1 + v2 + 3v3 ) 2 1 ⊕ (A(3v1 + v2 ) ⊕ A2 v4 ) + Z (3v1 + v2 + 2v4 ) 3

are indecomposable decompositions with summands of rank 1 and 3 in the ﬁrst decomposition and two summands of rank 2 in the second decomposition. Proof. Details can be found in [20, Chapter V, Example 3.1].

A very striking and easily stated result on “pathological decomposition” is as follows. Theorem 3.2 (Corner [6], or [23, page 281]). Given integers n ≥ k ≥ 1, there exists a torsion-free group X of rank n such that for any partition n = r1 + · · · + rk , there is a decomposition of X into a direct sum of k indecomposable subgroups of ranks r1 , . . . , rk respectively. Let A denote the category of torsion-free abelian groups of ﬁnite rank with the usual homomorphisms A(A, M ) = Hom(A, M ) as morphisms. The torsionfree groups are exactly the additive subgroups of Q-vector spaces. Formally the torsion-free group A is naturally embedded in Q ⊗Z A, a Q-vector space. This can

3. Quasi-Isomorphism Category

139

be ignored if we simply assume that A is an additive subgroup of some Q-vector space V . If so, the subspace of V spanned by A is denoted by QA and QA = {ra | r ∈ Q, a ∈ A}(∼ = Q ⊗Z A). The rank of A is rk(A) = dim(QA). We insert here a fact that is very helpful and will be used in the following without explicit reference. Lemma 3.3. Let f ∈ Hom(A, M ). Then f has a unique extension F : QA → QM that is a linear mapping. Hence any homomorphism f : A → M may be considered to be the restriction of a linear mapping F : QA → QM . Proof. Let f ∈ Hom(A, M ) be given. We deﬁne F : QA → QM as follows. For x = n1 y ∈ QA, y ∈ A, let F (x) = n1 f (y) ∈ QM . It is an easy exercise to check that F is well-deﬁned (independent of the choice of n and y), linear, unique and restricts to f on A. Notational Convention. We will use A, B, C, M, . . . to denote torsion-free abelian groups in A, letters f, g, h, . . . for ordinary homomorphisms in A and capitals F, G, H, . . . for the linear transformations that restrict to f, g, h, . . . respectively. The problem with “pathological decompositions” can be remedied by changing the category as follows. This idea goes back to Bjarni Jonsson ([16], [17]). Let A and M be groups in A. Recall Lemma 3.3 which can be interpreted to mean that Hom(A, M ) ⊆ HomQ (QA, QM ) = Hom(QA, QM ), where HomQ (QA, QM ) is the vector space of linear transformations on QA to QM . Let Q Hom(A, M ) = {rf | r ∈ Q, f ∈ Hom(A, M )} ⊆ Hom(QA, QM ) be the subspace of Hom(QA, QM ) spanned by Hom(A, M ). If A, M , K are three groups in A, F ∈ Q Hom(A, M ) ⊆ Hom(QA, QM ), and G ∈ Q Hom(M, K) ⊆ Hom(QM, QK), then the composite GF of the linear mappings G, F is not only in Hom(QA, QM ) but in Q Hom(A, M ). Also, if F, G ∈ Q Hom(A, M ), then F + G ∈ Q Hom(A, M ). This says that the “category” deﬁned next is indeed a category. Deﬁnition 3.4. Let QA denote the category whose objects are those of A, and whose morphism sets are the abelian groups Q Hom(A, M ) with addition and composition inherited from Hom(QA, QM ). This is the quasi-isomorphism category of torsion-free abelian groups of ﬁnite rank. Proposition 3.5. The quasi-isomorphism category QA is an additive category. Proof. The algebraic laws of the morphisms are automatic because the operations are inherited from vector space laws, each morphism set Q Hom(A, M ) is an abelian group, and the group {0} is the zero object of the category. Let A1 , . . . An be in QA. Then the ordinary biproduct A1 ⊕ · · · ⊕ An in A is a biproduct in QA.

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Chapter IX. Regularity in Categories

The following lemma contains the basic properties of the quasi-isomorphism category that are essential for working purposes. The routine veriﬁcations are left to the reader. Lemma 3.6. 1) Q Hom(A, M ) = {F ∈ Hom(QA, QM ) | ∃ 0 = n ∈ N, nF (A) ⊆ M } = { n1 f | 0 = n ∈ N, f ∈ Hom(A, M )}. 2) Let F, G ∈ Q Hom(A, M ). Choose 0 = n ∈ N such that nF ∈ Hom(A, M ) and nG ∈ Hom(A, M ). Then F = G in QA if and only if nF = nG in Hom(A, M ). 3) Let F, G ∈ Q Hom(A, M ). Choose 0 = n ∈ N such that nF ∈ Hom(A, M ) 1 and 0 = m ∈ N such that mG ∈ Hom(A, M ). Then F + G = nm (nF + mG). 4) Let F ∈ Q Hom(A, M ) and G ∈ Q Hom(M, K). Choose 0 = n ∈ N such that f := nF ∈ Hom(A, M ) and 0 = m ∈ N such that g := mG ∈ Hom(M, K). 1 Then GF = nm gf . Hence GF = 0 if and only if gf = 0. 5) Let F ∈ Q Hom(A, M ) and G ∈ Q Hom(M, K). Choose 0 = n ∈ N such that f := nF ∈ Hom(A, M ) and 0 = m ∈ N such that g := mG ∈ Hom(M, K). If GF = H ∈ Q Hom(A, K), then gf = mnH ∈ Hom(A, K). Two groups A, M that are isomorphic in the category QA are quasi-isomorphic and we write A ∼ =qu M . All QA notions can be expressed in A. It is immediate that isomorphic groups are quasi-isomorphic. The interpretation of isomorphism in QA is as follows. Proposition 3.7. Let A, M ∈ QA. Then the following statements are equivalent. 1) A ∼ =qu M . 2) There exist 0 = n ∈ N, f ∈ Hom(A, M ), g ∈ Hom(M, A) such that f g = n · 1M and gf = n · 1A . 3) There exists a monomorphism f : A → M and 0 = n ∈ N such that nM ⊆ f (A) ⊆ M . 1 Proof. 1) ⇒ 2) Suppose A ∼ g ∈ =qu M . Then there are k1 f ∈ Q Hom(A, M ), m 1 1 1 1 Q Hom(M, A) such that k f · m g = 1QM and m g · k f = 1QA . Hence n = km will do. 2) ⇒ 3) Suppose that f g = n · 1M and gf = n · 1A , then f is injective and nM = n · 1M M = f g(M ) ⊆ f (A) ⊆ M . 3) ⇒ 1) Suppose that f : A → M is a monomorphism and 10 = n ∈ N such −1 that nM ⊆ f (A) ⊆ M . Deﬁne g : M → A : g := f n. Then f g = 1M and n g n1 f = 1A hence A ∼ M . =qu

Example 3.8. The group X in Example 3.1 is quasi-isomorphic to its completely decomposable subgroup A: X ∼ =qu A.

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141

Proof. The inclusion of A in X is a monomorphism and the index [X : A] is ﬁnite, in fact, a short computation shows that [X : A] = 6. Our next example deals with rational groups that are deﬁned to be additive subgroups of Q that contain Z. Every additive subgroup of Q is isomorphic with a rational group and it is convenient to have 1 as an element of the group because of Lemma 3.10. We need a standard result from number theory. Lemma 3.9 (Partial Fraction Decomposition). Let n be a positive integer and let n = p pnp be its prime factorization. Then there exist integers up such that 1 n

=

up p pnp

.

Rational groups can be( obtained by choosing some set of generators. For ) 1 example, Z = 1 and Q = pn : p ∈ P, n ∈ N0 . The goal is to gain control of the generating sets of rational groups. Lemma 3.10. Let A be a rational group. 1) If m n ∈ A with gcd(m, n) = 1, then of the form p1k where p ∈ P. 2) If

1 1 m, n

∈ A, then

1 lcm(m,n)

1 n

∈ A and A is generated by its elements

∈ A.

m n

∈ A with gcd(m, n) = 1. Choose integers u, v such that Proof. 1. Assume that 1 um + vn = 1. Then n1 = u m n + v ∈ A. Hence A is generated by its fractions np n . 1 p . By Let n be such a generator and consider its prime factorization n = Lemma 3.9,

up 1 = ∈ A. n p np p This shows that the special fractions 1/pnp ∈ A generate A. 2. Write gcd(m, n) = um + vn. Then 1 gcd(m, n) 1 1 = = u + v ∈ A. lcm(m, n) mn n m

We can now clarify quasi-isomorphism for groups of rank 1. Example 3.11. Two rank-1 groups are quasi-isomorphic if and only if they are isomorphic. Proof. We can assume without loss of generality that the two quasi-isomorphic groups are rational groups because every rank-1 group is isomorphic to a rational group. Let A and B be rational groups that are quasi-isomorphic. Any homomorphism A → B is multiplication by a rational number and hence injective. By assumption, there is r ∈ Q such that mB ⊆ rA ⊆ B (Proposition 3.7.3). Since A ∼ = rA we assume without loss of generality that A = rA, so that we have

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Chapter IX. Regularity in Categories

mB ⊆ A ⊆ B. We now look at generators of the form 1/ps , p a prime. All such generators of A are contained in B, and if 1/ps ∈ B, then m/ps ∈ A, and if gcd(m, p) = 1, then in fact 1/ps ∈ A. Hence it is only generators 1/ps with p dividing m, that may prevent A from being equal to B. One easily convinces oneself that 1/ps ∈ B for all s implies that 1/ps ∈ A for all s. This leaves the case that p divides m and not all 1/ps are in B. In this case there is a largest S such that 1/pS ∈ B. Because A ⊆ B there is a largest T such that 1/pT ∈ A and necessarily T ≤ S. Now pS−T B ∼ = B and the largest generator 1/pr ∈ pS−T B is 1/pT , the same as A, while the other generators have not changed because of Lemma 3.10.1. Hence mpS−T B ⊆ mB ⊆ A ⊆ pS−T B. This process can be repeated for other prime factors of m and we end up with A = m B for some factor m of m. We arrived at A = m B ∼ = B as desired. The following example may serve as motivation and explanation for writing A ∈C A1 ⊕ · · · ⊕ An instead of A = A1 ⊕ · · · ⊕ An as is often done. Example 3.12. Let A = Zv1 ⊕ Zv2 in A which is true set-theoretic equality. Let M = K = Z(v1 + v2 ). Then A ∈QA M ⊕ K but A = M ⊕ K would not make sense strictly speaking. Proof. Structural maps for A ∈QA M ⊕ K are ιM : M → A : ιM (v1 + v2 ) = v1 , πM : A → M : πM (v1 ) = v1 + v2 , πM (v2 ) = 0; ιK : K → A : ιK (v1 + v2 ) = v2 , πK : A → M : πM (v1 ) = 0, πK (v2 ) = v1 + v2 . Of course, it also follows from A ∈QA Zv1 ⊕ Zv2 and Zv1 ∼ =qu Zv2 =qu Z(v1 + v2 ) ∼ that A ∈QA M ⊕ K. We now interpret the meaning of biproduct in QA in the category A. While M ⊕ K as such is meaningful in A, we write A ∈QA M ⊕ K for the biproduct of M and K in QA. Proposition 3.13. Let A, M , and K be torsion-free abelian groups. 1) Let nA ⊆ M ⊕ K ⊆ A in A for some positive integer n. Then A ∈QA M ⊕ K. In particular, if A = M ⊕ K in A, then A ∈QA M ⊕ K. 2) Suppose that A ∈QA M ⊕K. Then there are a positive integer n and subgroups M , K of A such that M ∼ = M , K ∼ = K and nA ⊆ M ⊕ K ⊆ A. Proof. 1) By hypothesis and Proposition 3.7 we have A ∼ =qu M ⊕ K, and hence A ∈QA M ⊕ K.

3. Quasi-Isomorphism Category

143

2) Let ιM ∈ Q Hom(M, A), πM ∈ Q Hom(A, M ), ιM ∈ Q Hom(M, A), and πM ∈ Q Hom(A, M ) be a set of structural maps belonging to the biproduct A ∈QA M ⊕ K. Then there is a positive integer m such that mιM ∈ Hom(M, A), mπM ∈ Hom(A, M ), mιM ∈ Hom(M, A), and mπM ∈ Hom(A, M ). Let n = m2 , M = (mιM )(M ) and K = (mιK )(K). Then it follows from ιM πM + ιK πK = 1QA that (mιM )(mπM ) + (mιK )(mπK ) = m2 · 1A , and therefore for every x ∈ A, m2 x = (mιM )(mπM )(x) + (mιK )(mπK )(x) ∈ M ⊕ K . Hence m2 A ⊆ M + K ⊆ A. Suppose that x ∈ M ∩ K . Then x = (mιM )(h) = (mιK )(k) for some h ∈ M and some k ∈ K. Using that πK ιM = 0 and πM ιK = 0 we obtain m2 x = (mιM )(mπM )((mιK )(k)) + (mιK )(mπK )((mιM )(h)) = 0. Hence M ∩ K = 0. Finally, the maps M x → (mιM )(x) ∈ M and K x → (mιK )(x) ∈ K are isomorphisms because they are surjective by deﬁnition and also injective because, e.g., if h ∈ M and mιM (h) = 0, then 0 = (mπM )(mιM )(h) = m2 · 1M (h) = m2 h, so h = 0. Deﬁnition 3.14. We call A ∈QA M ⊕ K a quasi-decomposition of A. A group M is called a quasi-summand of A if and only if there exists a quasi-decomposition A ∈QA M ⊕K. An indecomposable group of QA is called strongly indecomposable. Strongly indecomposable groups are plentiful. Simple examples are the groups of rank 1. Example 3.15. Let A be an additive subgroup of Q, so rk(A) = 1. Then A is strongly indecomposable because of rank. It is well-known that End(A) = { m n : m nA = A} where the fraction m n is assumed to be reduced and n acts by multiplication on elements of A. Hence Q End(A) = Q. We will now give an example showing that indecomposable groups need not be strongly indecomposable. Example 3.16. The group X = Z[2−1 ]v1 ⊕ Z[3−1 ]v2 + Z 15 (v1 + v2 ) is indecomposable but not strongly indecomposable as X ∈QA Z[2−1 ]v1 ⊕ Z[3−1 ]v2 . Proof. Clearly 5X ⊆ Z[2−1 ]v1 ⊕ Z[3−1 ]v2 ⊆ X, hence X ∈QA Z[2−1 ]v1 ⊕ Z[3−1 ]v2 by Proposition 3.13. We show next that X is indecomposable. Note that every element x ∈ X is of the form b m 1 x = 2aa12 + m 5 v1 + 3b2 + 5 v2 where m ∈ Z, a1 is an odd integer, a2 is a non-negative integer, b1 is an integer not divisible by 3, and b2 is a non-negative integer. Using this representation it is not hard to show that Z[2−1 ]v1 is the unique maximal 2-divisible subgroup of X and Z[3−1 ]v2 is the unique maximal 3-divisible subgroup of X. Both these groups are pure fully invariant subgroups of X and indecomposable (having rank 1).

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Chapter IX. Regularity in Categories

Now suppose that X = Y ⊕ Z for subgroups Y and Z of X. By Lemma 2.5 we conclude that Z[2−1 ]v1 = Y ∩ Z[2−1 ]v1 ⊕ Z ∩ Z[2−1 ]v1 . As Z[2−1 ]v1 is indecomposable we assume without loss of generality that Y ∩ Z[2−1 ]v1 = Z[2−1 ]v1 , i.e., Z[2−1 ]v1 ⊆ Y . Similarly, either Z[3−1 ] ⊆ Y or Z[3−1 ] ⊆ Z. If Y contains both Z[2−1 ]v1 and Z[3−1 ]v2 , then 5X ⊆ Y and Y = Y∗X = X. In this case we are done. The other possibility is that Z[2−1 ]v1 ⊆ Y and Z[3−1 ]v2 ⊆ Z. In this case 5Z[2−1 ]v1 ⊕ 5Z[3−1 ]v2 = 5X ⊆ Z[2−1 ]v1 ⊕ Z[3−1 ]v2 ⊆ X = Y ⊕ Z and it follows that 5Y ⊆ Z[2−1 ]v1 ⊆ Y and 5Z ⊆ Z[3−1 ]v2 ⊆ Z and so Y = Z[2−1 ]v1 and / Y ⊕ Z = X. Z = Z[3−1 ]v2 . This is a contradiction because then 12 (v1 + v2 ) ∈ Remark. The groups in Example 3.1 and Example 3.16 are examples of almost completely decomposable groups. A group is completely decomposable if it is a direct sum of rank-1 groups. A torsion-free group of ﬁnite rank is almost completely decomposable if it contains a completely decomposable subgroup of ﬁnite index. As the examples show such groups are fairly concrete and can be approached computationally, but the computations become very nasty very soon. Luckily there is by now an extensive theory of almost completely decomposable groups that goes a long way although it also shows that these groups are very complicated. See the survey article [22] or the comprehensive book [23]. Lemma 3.17. Idempotents split in QA. Moreover, if A ∈ QA, E 2 = E ∈ Q End(A) and e = nE ∈ End(A), then nA ⊆ Ker(e)⊕Im(e) ⊆ A, i.e., A ∈QA Ker(e)⊕Im(e). Proof. Let A ∈ QA, and E 2 = E ∈ Q End(A). Then we take E to be an idempotent linear transformation E : QA → QA and we get QA = E(QA) ⊕ (1 − E)(QA) with structural maps ι : E(QA) x → x ∈ QA and π : QA x → E(x) ∈ E(QA). We have (7) πι = 1E(QA) and ιπ = E. As E is a quasi-endomorphism of A, there exists a positive integer n such that e := nE ∈ End(A). Then (ι eA ) : eA → A is a well-deﬁned homomorphism because ι(x) = x, and n(π A ) : A → eA is a well-deﬁned homomorphism because nπ(A) = nE(A) = e(A) ⊆ A, so (π A ) ∈ Q Hom(A, eA). By (7) we have that (π A )(ι eA ) = 1eA and (ι eA )(π A ) = E and this shows that E splits in QA. Observe that ne = e2 and so (n − e)e = 0. We claim that nA ⊆ Ker(e) ⊕ e(A) ⊆ A. In fact, let x ∈ A be given. Then nx = (n − e)(x) + e(x), hence nA ⊆ (n − e)(A) + e(A) ⊆ A. Also (n − e)(A) ⊆ Ker(e) because e(n − e) = 0. So nA ⊆ Ker(e) + Im(e) ⊆ A. The sum is direct because eA = nEA ⊆ E(QA) and Ker(e) = Ker(nE) ⊆ Ker(E) = (1 − E)QA. The ring Q End(A) for A ∈ QA is a ﬁnite-dimensional Q-algebra, hence has ﬁnite length. Proposition 3.18. Let A ∈ QA. Then A is indecomposable in QA if and only if Q End(A) is local.

3. Quasi-Isomorphism Category

145

Proof. (Arnold [3]) Let A be an indecomposable object in QA. Then Q End(A) contains no idempotents other than 0 and 1 since idempotents split in QA. Also Q End(A) is Artinian and therefore its radical Rad(Q End(A)) is nilpotent. It follows that Q End(A)/ Rad(Q End(A)) contains no idempotents other than 0 and 1, since idempotents lift modulo the nilpotent radical Rad(Q End(A)). Thus the quotient ring Q End(A)/ Rad(Q End(A)) is a semisimple Artinian ring without proper idempotents and therefore a division ring. Consequently, Q End(A) is local. Conversely, a local ring contains no proper idempotents, hence A is indecomposable. The existence of indecomposable decompositions in QA is guaranteed by rank arguments. The uniqueness of decompositions is a consequence of the fact that strongly indecomposable groups have local quasi-endomorphism rings. The proper tool is provided by the so-called “Azumaya Unique Decomposition Theorem”. A detailed proof is given in [23] for a preadditive category because this version was needed in the context of almost completely decomposable groups. The Azumaya Theorem is usually proved for additive categories. The diﬀerence between additive and preadditive categories is the lack of a biproduct for every given ﬁnite set of objects in the latter, but it is intuitively clear that the uniqueness question that deals with two given biproducts should not need the existence of arbitrary biproducts, only the existence of subsums. Theorem 3.19 (Azumaya Unique Decomposition Theorem). Let C be a preadditive category in which idempotents split. Suppose that A ∈ A1 ⊕ · · · ⊕ Am and also A ∈ B1 ⊕ · · · ⊕ Bn . 1) If each HomC (Ai , Ai ) is a local ring, then, for, for every j there exist objects Bj1 , . . . , Bjs such that Bj ∈ Bj1 ⊕ · · · ⊕ Bjs and each Bjk is isomorphic with one of the Ai . 2) If, in addition, each of the Bj is indecomposable, then n = m and, after relabeling if necessary, Bi ∼ = Ai for i ∈ {1, . . . , n}. Corollary 3.20. If A ∈QA A1 ⊕ · · · ⊕ Am and also A ∈QA B1 ⊕ · · · ⊕ Bn are indecomposable decompositions in QA, then m = n and, after relabeling if necessary, Ai ∼ =qu Bi for every i. Before getting into regularity we look at kernels and images in the category QA. We are confronted with a notational conﬂict. For a morphism F : A → B in QA we have the category theoretical Ker(F ) and Im(F ) and we also have the usual kernel and image of the linear transformation F : QA → QB. We will denote the latter by KerQ (F ) and ImQ (F ). For a map f ∈ Hom(A, M ) ⊆ Q Hom(A, M ), the symbols KerA (f ) and ImA (f ) will be the usual kernel and image. The connections between the usual kernels and images are as follows. Lemma 3.21. Let F ∈ Q Hom(A, M ) and let 0 = n ∈ N be such that f := nF ∈ Hom(A, M ). Let KerQ (F ) := {v ∈ QA | F (v) = 0},

and

ImQ (F ) := F (QA),

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Chapter IX. Regularity in Categories

i.e., KerQ (F ) and ImQ (F ) are the kernel and image of the linear transformation F : QA → QM . Let KerA (f ) := {x ∈ A | f (x) = 0},

and

ImA (f ) := f (A),

i.e., KerA (f ) and ImA (f ) are the kernel and image of the homomorphism f : A → M . Then KerA (f ) = A ∩ KerQ (F ), Q ImA (f ) = ImQ (F ),

and

and

Q KerA (f ) = KerQ (F ), M

ImA (f ) ⊆ A ∩ ImQ (F ) = (Im(f ))∗ .

Proof. Let x ∈ KerA (f ). Then nF (x) = f (x) = 0, hence F (x) = 0. Thus KerA (f ) ⊆ A ∩ KerQ (F ). Let x ∈ A ∩ KerQ (F ). Then 0 = nF (x) = f (x) so x ∈ KerA (f ). Next, let x ∈ QA and assume that F (x) = 0. There is 0 = m ∈ N such that mx ∈ A. Then mx ∈ A ∩ KerQ (F ) = KerA (f ). This shows that KerQ (F ) ⊆ Q KerA (f ). The containment Q KerA (f ) ⊆ KerQ (F ) follows from the fact that KerA (f ) ⊆ KerQ (F ) and KerQ (F ) is a Q-vector space. Turning to images, let x ∈ ImQ (F ). Then there is y ∈ QA such that x = F (y). Choose 0 = m ∈ N such that my ∈ A. Then nmx = nmF (y) = (nF )(my) = f (my) ∈ ImA (f ). Hence Qf (A) ⊆ F (QA) ⊆ Qf (A), and consequently, f (A) ⊆ A A ∩ F (QA) = A ∩ Qf (A) = (f (A))∗ . We will use Deﬁnition 2.1. Lemma 3.22. Morphisms in QA have kernels and cokernels. More precisely, let F ∈ Q Hom(A, B) and let f := mF ∈ Hom(A, B) for 0 = m ∈ N. 1) Let ι : KerA (f ) → A be the insertion in A. Then ι ∈ Ker(F ). 2) Let ν : B → B/(Im(f ))B ∗ be the natural epimorphism in A. Then ν ∈ Coker(F ). Proof. We will use Lemma 3.6 without explicit reference. 1) Note that ι is monic. From f ι = 0, it follows that F ι = 0. Suppose that Ξ ∈ Q Hom(X, A) such that F Ξ = 0. Let 0 = n ∈ N be such that ξ := nΞ ∈ Hom(X, A). Then f ξ = (mF )(nΞ) = mnF Ξ = 0, and hence ImA ξ ⊆ KerA (f ) ˜ Then ˜ = 1 ξ. and ξ˜ : X x → ξ(x) ∈ KerA (f ) is in Hom(X, KerA (f )). Let Ξ n ˜ ˜ ιξ = ξ, so also ιΞ = Ξ. Uniqueness: Suppose that ιΞ1 = G and ιΞ2 = G. Then ιΞ1 = ιΞ2 and ι being monic it follows that Ξ1 = Ξ2 . 2) Note that ν is epic. From νf = 0, it follows that νF = 0. Let Ξ ∈ Q Hom(B, X) be such that ΞF = 0. Then Ξ(F (QA)) = 0. Deﬁne Φ : B/(Im(f ))B ∗ → X : Φ(ν(b)) = Ξ(b), where b ∈ B. We will show that Φ is a well-deﬁned map in Q Hom(B/(Im f )B ∗ , X) and then by deﬁnition, Φν = Ξ.

3. Quasi-Isomorphism Category

147

Suppose that ν(b) = ν(b ). Then b − b ∈ (ImA f )B ∗ = A ∩ F (QA) and as Ξ(F (QA)) = 0 it follows that Ξ(b) − Ξ(b ) = 0. Hence Φ is well-deﬁned as a linear transformation. To show that Φ is a quasi-homomorphism, choose 0 = n ∈ N such that nΞ ∈ Hom(B, X). Then (nΦ)(ν(B)) = nΞ(B) ⊆ X. Uniqueness follows from the fact that ν is epic. A category is preabelian if it is additive and has kernels and cokernels. Proposition 3.23. The quasi-isomorphism category QA is a preabelian category. We will give an example showing that QA is not an abelian category. To do so we need a lemma from category theory. Lemma 3.24. Let f : A → M be a morphism, assume that kernels and cokernels exist as needed and let (k : K → A) ∈ Ker(f ),

(c : M → C) ∈ Coker(f ),

(c : A → C ) ∈ Coker(k) = Coim(f ), (k : K → M ) ∈ Ker(c) = Im(f ). Then there is a unique morphism f : C → K such that f = k f c . K

k

−→

f

A c ↓

−→

C

−→

f

M ↑ k

c

−→ C

K

Proof. The uniqueness follows because k is monic and c is epic and therefore these maps can be canceled in k f c = k f c . For the existence we use that cf = 0 and k ∈ Ker(c) to get a morphism f1 : A → K such that f = k f1 . We now ﬁnd that k f1 k = f k = 0 and since k is monic, already f1 k = 0. But c ∈ Coker(k) hence there is f2 : C → K such that f1 = f2 c . The morphism f := f2 is the desired map. For a preabelian category to be abelian it is required that for every morphism f ∈ C(A, M ) the induced map f : C → K is an isomorphism. Let us clear up what the induced map is in our case. By deﬁnition Im(F ) = Ker(Coker(F )) and Coim(F ) = Coker(Ker(F )). Let f := mF ∈ Hom(A, B). Then (Lemma 3.22) (ι : KerA (f ) → A) ∈ Ker(F ),

and (ν : B → B/(ImA f )B ∗ ) ∈ Coker(F ).

We now see that for the insertion ins and the natural epimorphism nat (ins : (ImA f )B ∗ → B) ∈ Im(F ) and

(nat : A → A/ KerA (f )) ∈ Coim(F )

and we have the induced map F : A/ KerA (f ) a + KerA (f ) → f (a) ∈ (ImA f )B ∗.

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Chapter IX. Regularity in Categories

As A/ KerA (f ) ∼ = ImA f it is evident that F need not be surjective although it is injective and there may not be an inverse of F in QA. We give a concrete example. Example 3.25. Let F : Z ⊕ Z (a, b) → a − b ∈ Q. Then F : 0 → Q which is not invertible. Proof. In our example the diagram in Lemma 3.24 becomes ins

F

Q ↑ 1Q

F

Q

KerA (f ) −→ Z ⊕ Z −→ 0↓ 0

−→

0

−→ 0

and F : 0 → Q does not have an inverse.

We now steer toward exact sequences in QA. We settle ﬁrst the question which quasi-homomorphisms are monic and which are epic. Lemma 3.26. Let F ∈ Q Hom(A, B), and n = 0 such that f = nF ∈ Hom(A, B). 1) F is monic if and only if f is injective if and only if F ∈ Hom(QA, QB) is injective as a linear transformation. 2) F is epic if and only if B/f (A) is torsion if and only if F (QA) = QB, i.e., if and only if F is surjective as a linear transformation. Proof. 1) Suppose that U, V ∈ Q Hom(C, A) and F U = F V . There is n = 0 such that f := nF ∈ Hom(A, B), u := nU, v := nV ∈ Hom(C, A). Then f u = f v. If f is injective, then u = v and hence also U = V , so F is monic. Now suppose that F is monic and choose n = 0 such that f := nF ∈ Hom(A, B). Assume by way of contradiction that f is not injective. Let u : KerA (f ) → A be the insertion and let v : KerA (f ) → A be the zero map. Then f u = 0 = f v and also F u = F v. By cancellation we obtain the contradiction u = v. Finally, if f = nF for some nonzero integer n, then nF considered as the unique linear transformation that extends f to QA is injective if and only if f is injective, and nF is injective if and only if F is injective. B

B

2) Suppose that F is epic. Then B/ (f (A))∗ is either 0 or B/ (f (A))∗ is a nonzero torsion-free group. Assume by way of contradiction that the latter is B the case. Then we have map u := 0 : B → B/ (f (A))∗ and 0 = v : B b → B B b + (f (A))∗ ∈ B/ (f (A))∗ . It follows that uF = vF but F cannot be canceled, a contradiction. Now suppose that F is surjective as a linear transformation. Then it can be canceled on the right. We can now deal with exact sequences. We ﬁrst recall the deﬁnition.

3. Quasi-Isomorphism Category f

149 g

Deﬁnition 3.27. Let E : A → B → C be a sequence of morphisms in a category with kernels and cokernels. The sequence is exact at B if Im(f ) = Ker(g). i.e., if (c : B → C) ∈ Coker(f ) and (k : K → B) ∈ Ker(c), so k ∈ Im(f ), then E is exact at B if and only if k ∈ Ker(g).

A

f

K k↓

−→

F

B c↓ C

g

−→ C

G

Proposition 3.28. Let E : A → B → C be a sequence of morphisms in QA. Choose nonzero integers m, n such that f := mF ∈ A and g := nG ∈ A. Then the following statements are equivalent. 1) E is exact at B. 2) F (A) ∩ B ⊆ KerQ (G) and (KerQ (G) ∩ B)/(F (A) ∩ B) is bounded (so ﬁnite). 3) There exist integers e1 and e2 such that e1 Im(f ) ⊆ Ker(g) and e2 Ker(g) ⊆ Im(f ), i.e., Im(f ) and Ker(g) are “quasi-equal”. Proof. 1) ⇔ 2). By Lemma 3.22 we know kernels and cokernels in the quasiisomorphism category. The relevant diagram is as follows.

F

A −→

F (A) ∩ B ins ↓

G

B −→ C nat ↓ B/(F (A) ∩ B)

By deﬁnition E is exact at B if and only if ins ∈ Ker(G). Suppose that E is exact. Then G ◦ ins = 0 which is equivalent to F (A) ∩ B ⊆ KerQ (G). We also have the insertion ι : KerQ (G) ∩ B → B with the property that Gι = 0. Using that ins ∈ Ker(G) we obtain a morphism Φ : KerQ (G) ∩ B → F (A) ∩ B such that ι = ins ◦ Φ. There exists 0 = m ∈ N such that φ := mΦ ∈ Hom(KerQ (G) ∩ B, F (A) ∩ B). Then m(KerQ (G) ∩ B) = mι(KerQ (G) ∩ B) = m ins Φ(KerQ (G) ∩ B) = φ((KerQ (G) ∩ B) ⊆ F (A) ∩ B. Conversely, assume that F (A) ∩ B ⊆ KerQ (G) and for some positive integer m(KerQ (G)∩B) ⊆ F (A)∩B. We must show that ins ∈ Ker(G). The ﬁrst condition implies at once that G ◦ ins = 0. Now suppose that Φ : X → B is a morphism such that GΦ = 0. We must show that Φ factors through ins. Let n be a positive integer such that φ = nΦ ∈ Hom(X, B). Then Gφ = 0 hence φ(X) ⊆ KerQ (G) ∩ B and nmΦ(X) = mφ(X) ⊆ m(KerQ (G) ∩ B) ⊆ F (A) ∩ B. This means that Ψ : QX x → Φ(x) ∈ Q(F (A) ∩ B) is a morphism Ψ ∈ Q Hom(X, F (A) ∩ B). Clearly, Φ = ins ◦ Ψ which was needed.

150

Chapter IX. Regularity in Categories 2) ⇔ 3). F (A) ∩ B ⊆ KerQ (G) and KerQ (G) ∩ B/F (A) ∩ B is bounded is equivalent with F (A) ∩ B ⊆ KerA (g) and e KerQ (G) ∩ B ⊆ F (A) for some 0 = e ∈ N, which in turn is equivalent with

f (A) = mF (A)∩B ⊆ F (A)∩B ⊆ KerA (g) and me KerA (g) ⊆ m(F (A)∩B) ⊆ mF (A) ∩ B = f (A). We record two special cases. Corollary 3.29. F

G

F

G

1) Let E : 0 → B → C be a sequence of morphisms in QA. Then E is exact at B if and only if G is monic, or, equivalently, if and only if G : QB → QC is injective as a linear transformation. 2) Let E : A → B → 0 be a sequence of morphisms in QA. Then E is exact at B B if and only if F is epic, or, equivalently, if and only if B = (ImQ (F ) ∩ B)∗ . Proof. 1) Suppose that E is exact at B. As F (A) = 0 in this case it follows that KerQ (G) ∩ B is bounded, hence KerQ (G) ∩ B = 0 and by Lemma 3.21 it follows that KerQ (G) = 0 and so G is monic. The converse is immediately seen by looking at Lemma 3.28. 2) Suppose that E is exact at B. As KerQ (G) = B in this case, B/(F (A)∩B) = B/(ImQ (F ) ∩ B) is bounded. But ImQ (F ) ∩ B is pure in B and therefore B = ImQ (F ) ∩ B. Thus QB ⊆ ImQ (F ) which means that F : QA → QB is surjective and so epic in QA. Again the converse is easily seen.

4

Regularity in QA

Let F ∈ Q Hom(A, M ) be regular and let G ∈ Q Hom(M, A) be such that F GF = F . Then E := GF is an idempotent in Q End(A) and D := F G is an idempotent in Q End(M ). Choose 0 = n ∈ N such that f := nF ∈ Hom(A, M ) and g := nG ∈ Hom(M, A). Then e := n2 E ∈ End(A) and d := n2 D ∈ End(M ). By Lemma 3.17, n2 A ⊆ Ker(e) ⊕ Im(e) ⊆ A,

and n2 M ⊆ Ker(d) ⊕ Im(d) ⊆ M.

In other words, A ∈QA Ker(e) ⊕ Im(e),

and M ∈QA Ker(d) ⊕ Im(d).

Theorem 4.1. 1) Suppose that 0 = F ∈ Q Hom(A, M ) is regular and μF is regular for all μ ∈ Q End(M ). If M is strongly indecomposable, then Q End(M ) is a division ring.

4. Regularity in QA

151

2) Suppose that there is 0 = F ∈ Hom(A, M ) that is regular and F α is regular for all α ∈ Q End(A). If A is strongly indecomposable, then Q End(A) is a division ring. Proof. 1) Let M be strongly indecomposable, assume that F ∈ Q Hom(A, M ) is regular, F = 0, and that μF is regular for every μ ∈ Q End(M ). By hypothesis, we have F G1 F = F = 0 for some G1 , hence F G1 = 0 is a nonzero idempotent of Q End(M ) and since M is strongly indecomposable and idempotents split in QA, it follows that F G1 = 1QM . Let 0 = μ ∈ Q End(M ). Then μF G1 = μ = 0, and hence also μF = 0. By hypothesis there is G ∈ Q Hom(M, A) such that μF GμF = μF = 0. Then μF G is a nonzero idempotent of Q End(M ) and as M is strongly indecomposable μF G = 1. Hence every nonzero element in Q End M has a right inverse and Q End(M ) is a division ring by Lemma III.1.2. 2) Let A be strongly indecomposable, and assume that 0 = F ∈ Q Hom(A, M ) is regular. Then F G1 F = F = 0 for some G1 ∈ Q Hom(M, A). Then G1 F is a nonzero idempotent in Q End(A). Since A is strongly indecomposable and idempotents split in QA, it follows that G1 F = 1. Let 0 = α ∈ Q End(A). Then G1 F α = α and so F α = 0. By assumption F α is regular, hence F αGF α = F α for some G ∈ Q Hom(M, A). Hence GF α is a nonzero idempotent of Q End(A) and we get GF α = 1 and GF ∈ Q End(A) is a left inverse of α. By Lemma III.1.2 the quasi-endomorphism ring Q End(A) is a division ring. While Q is up to isomorphism the only group in the category of torsion-free abelian groups whose endomorphism ring is a division ring, there is an abundance of strongly indecomposable groups whose quasi-endomorphism ring is a division ring. There also exist many strongly indecomposable groups whose quasiendomorphism ring is not a division ring. There are many results in abelian group theory that state that, given a ring subject to certain conditions, that ring is isomorphic to an endomorphism ring of an abelian group. These results are known as realization theorems. The most celebrated realization theorem is due to A.L.S. Corner. Theorem 4.2 ([7], [11, Theorem 110.1]). If R is a reduced torsion-free ring with rk(R) = n < ∞, then R is isomorphic to the endomorphism ring of a torsion-free abelian group of rank 2n. R.S. Pierce and C. Vinsonhaler ([28]) observed the following corollary. Corollary 4.3. Let E be a ﬁnite-dimensional Q-algebra with 1E . Then there is a group A in QA such that Q End(A) ∼ = E. Proof. In order to apply Corner’s Theorem we need to construct a reduced torsion-free ring contained in E. Let (v1 , . . . , vn ) be a Q-basis of E. Then vi vj = n k=1 rijk vk is a rational linear combination of the basis elements vk and there exists a positive integer m such that mrijk ∈ Z for all i, j, k. Let wi := mvi and

152

Chapter IX. Regularity in Categories

let R := Z1E + Zw1 + · · · + Zwn . The R is free as a ﬁnitely generated torsion2 free nclosed under multiplication because wi wj = m vi vj = n abelian group and k=1 (mrijk )mvk = k=1 (mrijk )wk ∈ R. By Theorem 4.2 there is a group G such that End(G) ∼ = R and hence Q End(G) ∼ = QR = E. In particular, if F is a ﬁnite ﬁeld extension of Q, then there are groups G in QA with Q End(G) ∼ = F . Thus the variety of strongly indecomposable groups is huge and little is known about them. On the other hand, QA is a Krull-Schmidt category (Corollary 3.20) and the computation of Reg(A, M ) reduces to determination of Reg(A, M ) for strongly indecomposable groups A and M (Theorem II.6.17) and questions about maps between strongly indecomposable groups. The ﬁrst question is easy. Proposition 4.4. Let A and M be strongly indecomposable groups. Then Reg(A, M ) = 0 except when A ∼ = M and Q End(A) is a division ring in which case Reg(A, M ) ∼ = Q End(A). Proof. Suppose that f ∈ Q End(A, M ) is regular. Then A and M are quasiisomorphic by Theorem 2.7 and the strong indecomposability. Finally, Q End(A) is a division ring by Theorem 4.1. An obvious simple case is the following. Proposition 4.5. Let A ∈QA A1 ⊕ · · · ⊕ Am and M ∈QA M1 ⊕ · · · ⊕ Mn be such that Q Hom(Ai , Mj ) = 0 unless Ai ∼ =qu Mj and Q End(A) is a division ring. Then Reg(A, M ) =

{Q Hom(Ai , Mj ) | Ai ∼ =qu Mj }.

Unfortunately there may well be non-trivial quasi-homomorphisms between groups whose quasi-endomorphism rings are division rings. Recall that Q is a strongly indecomposable group whose quasi-endomorphism ring is Q, so even a ﬁeld. Example 4.6. Let A be any group in QA. Then Q Hom(A, Q) ∼ = Qrk(A) . Proof. Here Q Hom(A, Q) = Hom(A, Q). Choose a free subgroup F of A such that A/F is a torsion group and consequently rk(F ) = rk(A). Then Hom(A/F, Q) = 0 and since Q is injective, we have Hom(A, Q) ∼ = Hom(F, Q) ∼ = Qrk(F ) by standard homological algebra. Another easy case is that of completely decomposable groups, i.e., direct sums of rank-1 groups. Note that for any rank-1 group A we have Q End(A) = Q. Proposition 4.7. Let A and M be completely decomposable groups in QA. Then Reg(A, M ) = Hom(A, M ).

4. Regularity in QA

153

In general, little is known about strongly indecomposable groups and homomorphisms between them. Existence theorems for strongly indecomposable groups whose quasi-endomorphism rings are division rings are rarely constructive. We give a construction to illustrate the situation.

4.1

√ Realizing Q( p)

Let M2 (Q) be the 2 × 2-matrix ring with rational entries. Fix a prime p ≥ 3. Note that 2 1 1 1 0 =p . p − 1 −1 0 1 Therefore

Q

1 0

0 1

⊕Q

1 p−1

1 −1

is a subalgebra of M2 (Q) that is isomorphic in the obvious way with the ﬁeld √ extension Q( p).

4.2

Constructing the Group

Let V = Qv1 ⊕ Qv2 be a Q-vector space, let Pp be a set of primes and for each q ∈ Pp let cq ∈ Z, all of these to be speciﬁed later. Deﬁne G := (Zv1 ⊕ Zv2 ) +

q∈Pp

Z 1q (v1 + cq v2 ).

Set B := Zv1 ⊕ Zv2 . Then G B

∼ = q∈Pp

Z q1 (v1 +cq v2 )+B B

and

Z q1 (v1 +cq v2 )+B B

∼ =

Z Zq .

We now choose Pp and the cq . Let Pp := {q ∈ P | q = p, q divides p − n2 for some n ∈ N}. Lemma 4.8. Pp is inﬁnite. Proof. Suppose that q1 , . . . , qk ∈ Pp . Then any prime factor of p − (q1 q2 · · · qk )2 is diﬀerent from the qi and from p, hence belongs to Pp . Hence Pp cannot be a ﬁnite set. By deﬁnition of Pp , for any q ∈ Pp , the congruence p ≡ (1 + x)2 mod q has a solution. For each q ∈ Pp choose a solution cq of p ≡ (1 + x)2 mod q. We then have ∀ q ∈ Pp : p ≡ (1 + cq )2 mod q. (8)

154

4.3

Chapter IX. Regularity in Categories

Computing the Quasi-Endomorphism Ring

Proposition 4.9. Q End(G) = Q

1 0

0 1

⊕Q

1 p−1

1 −1

√ ∼ = Q( p).

Proof. We ﬁrst show that G is invariant under the ring 1 0 1 1 Z ⊕Z 0 1 p − 1 −1 of lineartransformations on V that acts by left multiplication on the coordinate x vectors with respect to the basis {v1 , v2 }. Clearly, we only need to check y 1 1 that the linear transformation maps each element 1q (v1 + cq v2 ) = p − 1 −1 1 1 into G. By a straightforward computation q cq 0 1 1 1 1 1+cq 1 = + 1 2 q cq p − 1 −1 q cq q (p − (1 + cq ) ) where the element 1q (p − (1 + cq )2 ) is integral by the choice of q and cq (see (8)). This shows that 1 0 1 1 Q ⊕Q ⊆ Q End(G). 0 1 p − 1 −1 r11 r12 ∈ Q End(G) ⊆ M2 (Q). There To show the converse containment let r r22 21 r11 r12 ∈ End(G) and all aij := mrij exists a positive integer m such that m r21 r22 are integers. We will show that a11 a12 1 0 1 1 α := + a12 = (a11 − a22 ) , (9) a21 a22 0 1 p − 1 −1 r11 r12 ∈ Q End(G) as needed. The assumption that and this shows that r21 r22 α ∈ End(G) implies that for every q ∈ Pp , there exist integers xq , yq , zq such that 1 1 yq a11 a12 1 xq = + . q a21 a22 q cq zq cq This is equivalent to a11 + a12 cq = xq + qyq ,

a21 + a22 cq = xq cq + qzq ,

5. Regularity in the Category of Groups

155

and we get the equivalent system xq = a11 + a12 cq − qyq ,

q(zq − cq yq ) = a21 + cq (a22 − a11 ) − a12 c2q .

There also is an integer uq such that 1+2cq +c2q = (1+cq )2 = p+quq . Substituting for c2q in the second equation we obtain q(zq − cq yq + a12 uq ) = a21 − a12 (p − 1) + cq (a22 − a11 + 2a12 ). Setting a := a21 − a12 (p − 1), b := a22 − a11 + 2a12 and mq := zq − cq yq + a12 uq , then we have the equations qmq = a + cq b,

p − (1 + cq )2 = quq .

Multiplying the second equation by b2 and substituting cq b from the ﬁrst equation we get (a + b)2 + (p − 2)b2 = q(qm2q − 2amq + 2bmq + uq b2 ) which must hold for all q ∈ Pp . As Pp is inﬁnite and p ≥ 3, it follows that (a + b)2 + (p − 2)b2 = 0, b = a22 − a11 + 2a12 = 0 and a = a21 − a12 (p − 1) = 0. These are the relations needed to validate (9).

5

Regularity in the Category of Groups

As another example we look at the category of general groups. We begin with the observation that idempotents split in the category of groups in the sense that the existence of an idempotent endomorphism of G implies that G is a semidirect product. Lemma 5.1. Let G be a group and let e = e2 ∈ End(G). Then G = Ker(e) Im(e). Proof. Let x ∈ G. Then x = (xe(x)−1 )e(x) and it is immediate that xe(x)−1 ∈ Ker(e). Suppose that e(x) ∈ Ker(e). Then 1 = e(e(x)) = e(x), hence Ker(e) ∩ Im(e) = {1}. Theorem 5.2. Let G and H be groups and f ∈ Hom(G, H). 1) Assume that f is regular. Let g ∈ Hom(H, G) such that f gf = f . Then e := gf ∈ End(G) is an idempotent, d := f g ∈ End(H) is an idempotent and G = Ker(f ) Im(e),

Im(e) ∼ = Im(f ),

and

H = Ker(d) Im(f ).

2) Suppose that G = Ker(f ) K and H = N Im(f ). Then f is regular. Proof. 1) That e and d are idempotents follows immediately from the equality f gf = f . By Lemma 5.1 we have that G = Ker(e) Im(e). Also Ker(e) = Ker(f ) because e(x) = 1 if and only if gf (x) = 1 if and only if f gf (x) = f (1) = 1. We

156

Chapter IX. Regularity in Categories

have shown that G = Ker(f ) Im(e) and Im(e) ∼ = G/ Ker(e) ∼ = Im(f ) by the ﬁrst isomorphism theorem. Again by Lemma 5.1 we have that G = Ker(d) Im(d). Clearly Im(d) ⊆ Im(f ). Suppose that u ∈ Ker(d) ∩ Im(f ). Then u = f (x) for some x ∈ G, and u = f (x) = f gf (x) = d(u) = 1. This shows that H = Ker(d) Im(f ). 2) Suppose that G = Ker(f ) K and H = N Im(f ). Then f K : K → Im(f ) is invertible. Let π : H → Im(f ) be the projection along N which is a homomorphism. Then g := (f K )−1 π ∈ Hom(H, G) and f gf = f .

Bibliography [1] F. W. Anderson and K. R. Fuller. Rings and Categories of Modules, Volume 13 of Graduate Texts in Mathematics. Springer Verlag, 1992. [2] D. M. Arnold. Finite Rank Torsion Free Abelian Groups and Rings, volume 931 of Lecture Notes in Mathematics. Springer Verlag, 1982. [3] D. M. Arnold. Abelian Groups and Representations of Finite Partially Ordered Sets, volume 2 of CMS Books in Mathematics. Springer Verlag, 2000. [4] B. Brown and N.H. McCoy. The maximal regular ideal of a ring. Proc. Am. Math. Soc., 1:165–171, 1950. [5] S.U. Chase. Direct products of modules. Trans. Am. Math. Soc., 97:457–473, 1960. [6] A. L. S. Corner. A note on rank and direct decompositions of torsion-free abelian groups. Proc. Cambridge Philos. Soc., 57:230–33, 1961. [7] A.L.S. Corner. Every countable reduced torsion-free ring is an endomorphism ring. Proc. London Math. Soc., 13:687–710, 1963. [8] P. Crawley and R.P. Dilworth. Algebraic Theory of Lattices. Prentice Hall, Inc., 1973. [9] Carl Faith. Rings and Things and a Fine Array of Twentieth Century Associative Algebra Mathematical Surveys and Monographs, Volume 65 American Mathematical Society, 1999. [10] D.J. Fieldhouse. Pure theories. Math. Ann., 184:1–18, 1969. [11] L. Fuchs. Inﬁnite Abelian Groups, Vol. I, II. Academic Press, 1970 and 1973. [12] L. Fuchs and K.M. Rangaswamy. On generalized regular rings. Math. Z., 107:71–81, 1968. [13] S. Glaz and W. Wickless. Regular and principal projective endomorphism rings of mixed abelian groups. Communications in Algebra, 22:1161–1176, 1994.

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[14] K.R. Goodearl. Von Neumann regular Rings. Pitman Publishing Limited, 1979. [15] K.R. Goodearl. Von Neumann regular Rings and Direct Sum Decomposition Problems. In Abelian Groups and Modules, Proceedings of the 1994 Padova Conference, Kluwer Academic Publishers, 249–255, 1995. [16] B. Jonsson. On direct decompositions of torsion-free abelian groups. Math. Scand., 5:230–235, 1957. [17] B. Jonsson. On direct decompositions of torsion-free abelian groups. Math. Scand., 7:361–371, 1959. [18] F. Kasch. Modules and Rings. A translation from the German “Moduln und Ringe”. Academic Press, 1982. [19] F. Kasch. Regularity in hom. Algebra Berichte, 75:1–11, 1996. [20] F. Kasch and A. Mader. Rings, Modules, and the Total. Birkh¨ auser Verlag, 2004. [21] F. Kasch and A. Mader. Regularity and substructures of hom. Communications in Algebra, 34:1459–1478, 2006. [22] A. Mader. Almost completely decomposable torsion-free abelian groups. In Abelian Groups and Modules, Proceedings of the 1994 Padova Conference, Kluwer Academic Publishers, 343–366, 1995. [23] A. Mader. Almost Completely Decomposable Groups, volume 13 of Algebra, Logic and Applications. Gordon and Breach Science Publishers, 2000. [24] A. Mader. Regularity in endomorphism rings. Communications in Algebra, to appear. [25] S.H. Mohamed and B. M¨ uller. Continuous and discrete modules. Lecture Notes, 147, 1990. [26] W.K. Nicholson. Semiregular modules and rings. Canadian J. Math., 28:1105– 1120, 1976. [27] R.S. Pierce. Modules over Commutative Regular Rings. Memoirs of the American Math. Soc., 70:1–112, 1967. [28] R.S. Pierce and C. Vinsonhaler. Realizing central division algebras. Pac. J. Math., 109:165–177, 1983. [29] K.M. Rangaswamy. Abelian groups with special endomorphism rings. J. Algebra, 6:271–280, 1967. [30] K.M. Rangaswamy. Regular and Baer rings. Proc. Amer. Math. Soc., 42:354– 358, 1974.

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Index ⊆∗ , 81 ⊆◦ , 81 A ∈ A1 ⊕ · · · ⊕ An , 134 abelian group A[n], 6 A[p], 6 divisible, 6 divisible hull, 7 elementary, 6 Hom(QA, QB), 7 maximal torsion subgroup, 6 mixed, 6 n-bounded, 6 P -divisible, 6 P0 -pure, 7 p-component, 6 p-divisible, 6 p-elementary, 6 p-group, 6 p-primary, 6 p-pure, 7 p-socle, 6 primary decomposition, 6 primary group, 6 pure, 6 pure hull, 8 puriﬁcation, 8 reduced, 6 Tp , 6 torsion, 6 torsion-free, 6 additive category, 134 A[n], 6 Anderson, 4, 59

A[p], 6 Arnold, 144, 145 Azumaya, 145 Unique Decomposition Theorem, 145 biproduct, 132 C, 131, 145 category C, 145 preadditive, 145 characterization of a regular element, 51 of regularity, 64 Chase, 63 complement, 2 Corner, 138, 151 cosingular submodule, 81 Δ(A, M ), 81 decomposition unique, 145 direct complements, 2 directly indecomposable, 41 divisible, 6 divisible hull, 7 element partially invertible, 54 regular, 49, 51 semiregular, 73 U -regular, 72 elementary, 6

162 example 0 = Reg(A, M ) = Hom(A, M ), 84, 85 Hom(A, M ) = Reg(A, M ), 115, 116 Reg(A, A) = End(A), 116 f regular in H does not follow from f regular in S H, 63 abelian groups whose endomorphism ring is a division ring, 42 diﬀerent indecomposable decompositions, 138 indecomposable weaker than strongly indecomposable, 143 non-regular function between nice groups, 126 partially invertible but not regular, 17 quasi-isomorphic groups, 140 restriction of regular map to direct summand not regular, 33 strongly indecomposable group, 143 substructure when HomR (M, A) = 0, 83 transfer dependent on quasi-inverse, 26 existence of regular elements, 51 Faith, 25 Fieldhouse, 49 Fuchs, 6, 103, 104, 116, 127, 151 Fuller, 4, 59 Glaz, 103, 117, 123, 127, 128 Goodearl, 33 Gp , 6 groups of matrices, 13 H := HomR (A, M ), 11 Hom(A, M ), 139 Hom(QA, QB), 7

Index homogeneous decomposition, 46 homomorphism S-semiregular, 76 T -semiregular, 76 semiregular, 76 Idl, 97, 100 Idl(K), 95 IdlR (K), 100 IdlS (K), 100 idempotents associated with regular maps, 12 idempotents split, 134 indecomposable, 41 injective, 13 Jonsson, 139 Kasch, 2, 4, 17, 69, 77, 83, 85, 86, 138 large, 81 large restricted, 85 largest regular S-T -submodule, 21 largest regular submodule, 52, 65 lattice of all bi-submodules, 97 lattice of all two-sided ideals, 97 left perfect, 25 L(H), 97 locally injective, 85 locally projective, 85 L(S), 97 lying over, 74, 77 M ∗ , 52 Mader, 2, 17, 77, 83, 85, 86, 103, 127, 138, 144, 145 map structural, 132 matrix groups, 13 matrix rings, 13 maximal torsion subgroup, 6 Mdl, 97, 100 Mdl(I), 95, 100 Mdl(J), 100 mixed, 6

Index Mod-R, 11 module large restricted, 85 locally injective, 85 locally projective, 85 regular, 49 small restricted, 86 ∇(A, M ), 81 n-bounded, 6 Nicholson, 73–76 p-component, 6 p-divisible, 6 p-elementary, 6 p-group, 6 p-primary group, 6 p-pure, 7 p-socle, 6 P -divisible, 6 P0 -pure, 7 Partial Fraction Decomposition, 141 partially invertible, 17, 54, 77 partially invertible element, 54 Pierce, 151 preadditive category, 131, 145 primary decomposition, 6 primary group, 6 product semidirect, 155 projective, 13 projective direct summand, 55 projective direct summands, 55 pure, 6 pure hull, 8 puriﬁcation, 8 QA, 139, 147 biproduct, 142 direct sum, 142 Q End(A), 144 Q Hom(A, M ), 139, 140 quasi-concepts, 140 quasi-decomposition, 142

163 quasi-endomorphism ring, 153 quasi-inverse, 11, 14, 16, 49 unique, 18, 19 quasi-inverse of an element, 52 quasi-inverses, 65 quasi-isomorphic, 140 quasi-isomorphism, 140 quasi-isomorphism category, 139 r-inverse, 11 Rad(A, M ), 82 Rad(A, M )-regular, 77 Rad(H), 82 Rad(HT ), 82 Rad(M ), 2 Rad(R), 3 Rad(S H), 82 radical, 2, 3, 82 Rangaswamy, 103, 127 reduced, 6 Reg(A, M ), 21, 23–25 structure, 23, 24 Reg(A, M )T , 64 Reg(HT ), 65 Reg(HT , 65 Reg(HT ), 65 Reg(MR ), 52, 55 Structure, 55 structure, 55 Reg(S H), 65 Reg(S H), 65 Reg(S H), 65 regular, 17 regular element, 49, 51 characterization, 51 existence, 51 regular homomorphism, 11 regular map, 11 characterization, 12, 13 regular module, 49 regular pair, 25, 26, 57 regular ring element, 11 regular subset, 11

164 regularity characterization, 64 ρS , 4 ρT , 4 right perfect, 25 ring left perfect, 25 right perfect, 25 S := End(MR ), 11 S-ﬂat, 64 S-projective, 25 S-projective direct summand, 19 S-regular, 61 S -regular, 61 S -regular, 61 S-semiregular homomorphism, 76 semidirect product, 155 semiregular, 73, 76 semiregular element, 73 semiregular homomorphism, 76 semisimple, 13 singular submodule, 81 small, 81 small restricted, 86 split, 134 splitting idempotents, 155 S Reg(A, M ), 64 strongly indecomposable, 144 structural map, 132 structure of Reg(MR ), 55 structure of Reg(A, M ), 23, 24 structure of Reg(MR ), 55 submodule largest regular, 21 supplement, 2 T := End(AR ), 11 T, 46 t(A), 6 T -ﬂat, 64 T -projective, 25 T -projective direct summand, 19 T -regular, 61

Index T -semiregular homomorphism, 76 T -regular, 61 T -regular, 61 τ -homogeneous component, 46 Tcr (A), 46 Tp , 6 torsion, 6 torsion-free, 6 Tot(A, M ), 17 Tot(M, R), 54 Tot(MR ), 54 total, 17, 54 tp A, 46 transfer, 25, 57 transfer rule, 25, 57 trf, 25 type, 46 (U-Reg(A, M ))- Reg(A, M ), 71 U-Reg(A, M ), 69, 70 U-Reg(MR ), 72 U -regular, 69 U -regular element, 72 unique decomposition, 145 unique quasi-inverse, 18, 19 Vinsonhaler, 151 Ware, 49, 57, 63 Wickless, 103, 117, 123, 127, 128 Zelmanowitz, 25, 49, 52, 56, 57, 63 Z(p∞ ), 6