Rings Related to Stable Range Conditions (Series in Algebra 11)

RINGS RELATED TO STABLE RANGE CONDITIONS

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SERIES IN ALGEBRA Series Editors: Derek J S Robinson (University of Illinois at Urbana-Champaign, USA) John M Howie (University of St. Andrews, UK) W Douglas Munn (University of Glasgow, UK)

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SERIES

RINGS RELATED TO STABLE RANGE CONDITIONS

IN

ALGEBRA VOLUME 11

Huanyin Chen Hangzhou Normal University, China

World Scientific NEW JERSEY

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RINGS RELATED TO STABLE RANGE CONDITIONS Series in Algebra — Vol. 11 Copyright © 2011 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

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ISBN-13 978-981-4329-71-2 ISBN-10 981-4329-71-1

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Preface

A ring R satisfies the n-stable range condition provided that, a1 R + · · · + an+1 R = R with a1 , · · · , an+1 ∈ R implies that there exists b1 , · · · , bn ∈ R such that (a1 + an+1 b1 )R + · · · + (an + an+1 bn )R = R. Over the past thirty years, this simple condition has been extensively studied from very different view points. It was first introduced so as to study stabilization in algebraic K-theory. Afterwards, it was studied to deal with the cancellation problem of modules. The earliest study along these lines is due to [394] in 1980. Meanwhile, this condition was also investigated in the theory of C ∗ -algebra. Problems utilizing the stable range condition are not merely fascinating but also enjoyable. A ring R is an exchange ring if, for every right R-module L A and any two decompositions A = M ⊕ N = i∈I Ai , where MR ∼ =R ′ and the index set I
is finite, there exist submodules Ai ⊆ Ai such that L ′ A = M ⊕ i∈I Ai . Such a ring was first introduced by Nicholson in 1977. The class of exchange rings is very large. It includes all regular rings, all π-regular rings, all strongly π-regular rings, all semiperfect rings, all left or right continuous rings, all clean rings, all unit C ∗ -algebras of real rank zero and all right semi-artinian rings. Many of the problems that are raised in exchange rings are the needs of, in particular, local rings, regular rings, C ∗ -algebras, etc. In [334], Nicholson proved that a ring R is an exchange ring if and only if for any x ∈ R, there exists an idempotent e ∈ xR such that 1−e ∈ (1−x)R. Using such element-wise characterization, [67] studied exchange rings with stable range one. Further, [175] and [402] developed the treatments of the n-stable range condition. This book will mainly restrict itself to studying exchange rings with various kinds of conditions related to stability. Here, we treat the stable range condition as it should be, as an element-wise condition on rings in its own right. Nevertheless, in the process we have tried to show its relationship to various topics: cancellation

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of modules, comparability of modules, cleanness, K-theory, monoid theory, matrix theory, topology, etc. In accordance with the logical development of the subject, the materials in this book fall roughly into seven parts. The first part, consisting of Chapter 1, is concerned with stable range one, i.e., the 1-stable range condition. Here the basic properties of stable range one are studied with a new perspective. In particular, we investigate this condition for exchange rings. Cancellation problems of modules are also discussed. The second part, consisting of Chapters 2 through 4, is concerned with conditions stronger than stable range one. These allow us to calculate K1 groups of such rings more accurately. In Chapters 5 through 12, we deal with conditions weaker than stable range one. From these, we consider analogues of stable range one for a large class of exchange rings. Also discussed in this third section are some recent results of other studies the author has done on the n-stable range condition. Part four which consists of Chapter 13 looks at the stable range for ideals. This allows us to discuss the stable range condition for a single element. In Chapter 14, we systematically investigate diagonal reduction of regular matrices over rings with stable range condition. We show that diagonal reduction over exchange rings behaves like a type of stable range. Chapters 15 and 16 are concerned with cleanness properties. We designed these chapters to cover some interesting challenges for such an elementary problem. The final part, consisting of Chapter 17, deals with the K0 group of abelian exchange rings. Certain particular aspects of the topological properties of such exchange rings are also discussed. This book has evolved from my lectures and research over the past several years. Its purpose is to give a general introduction to the theory of exchange rings with various conditions related to the stable range. It is also intended for use by instructors and graduate students to grasp the most recent progress in the subject. Thus, most of the results in this book are new. An effort has been made to keep the exposition as simple and as selfcontained as possible, and so some known results are involved. I thank the many authors whose works have been used but not cited. Though these elementary facts are included, it is hard to verify them by more simple proofs. It is presumed that the reader is at the least acquainted with the most elementary facts of the theory of rings (for example as given in Anderson and Fuller’s book).

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Finally, I wish to express my appreciation to my wife Xiaolin Zhang and my daughter Xue Chen for their devotion and unstinting support.

Huanyin Chen Department of Mathematics Hangzhou Normal University Hangzhou, China E-mail: [email protected] http://huanyinchens.blogbus.com June 6, 2010

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Notational Conventions

C H N R Z Zm ∅ AutR (A) AT A−1 A∼ =B A.B A .⊕ B A∝B A⊆B A ⊆e B A ⊆⊕ B L A B N A B

field of complex numbers Hamilton ring set of natural numbers field of real numbers ring of integers ring of integers modulo m empty set automorphism group of the R-module A transpose of the matrix A inverse of the matrix A A is isomorphic to B A is isomorphic to a submodule of B A is isomorphic to a direct summand of B A .⊕ nB for some n ∈ N A is a submodule of B A is an essential submodule of B A is a direct summand of B direct sum of modules A and B tensor product of modules A and B

R

B(R) Cl(R) coln (R)

set of all central idempotents in the ring R ideal class group of the ring R set of all completed n × 1 matrices over the ring R xi

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coker(f ) C(R) det(A) dimD (V ) D∗ EndR (A) F P (R) GLn (R) im(f ) I ER In J(R) ker(f ) K0 (R) K1 (R) ℓ(x) M ax(R) Mm×n (R) Mn (R) nA P ier(R) rad(M ) rown (R) Rp−1 Rq−1 r(x) R[x] R[[x]] Spec(R) |S| T Mn(R) tr(A) U (R) Ur (R) Uc,n (R) Ur,n (R) U + (R) U (R)′

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cokernel of the homomorphism f center of the ring R determinant of the matrix A dimension of the D-vector space V set of all nonzero elements in the division ring D ring of endomorphisms of the R-module A category of finitely generated projective R-modules group of all invertible n × n matrices over the ring R image of the function f I is an ideal of the ring R n × n identity matrix Jacobson radical of the ring R kernel of the homomorphism f K0 -group of the ring R K1 -group of the ring R left annihilator of x set of all maximal ideals of the ring R set of all m × n matrices with entries from the ring R ring of all n × n matrices with entries from the ring R direct sum of n copies of the module A Pierce space of the ring R radical of the module M set of all completed 1 × n matrices over the ring R set {u ∈ R | ∃ a, b ∈ R such that (1 − ua)♮(1 − bu)} set {u ∈ R | ∃ a, b ∈ R such that (1 − ua)⊥(1 − bu)} right annihilator of x ring of polynomials over R ring of formal power series over R set of all prime ideals of the ring R cardinal number of the set S ring of n × n lower triangular matrices with entries from R trace of the matrix A group of all units in the ring R set of all unimodular rows over the ring R set of all n-unimodular columns over the ring R set of all n-unimodular rows over the ring R set of all related units in the ring R subgroup of all commutators in the group U (R)

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Notational Conventions

V (R) W (R)

set {p(a, b)p(b, a)−1 | p(a, b) = 1 + ab ∈ U (R)} set {p(a, b, c)p(c, b, a)−1| p(a, b, c) = a + c + abc ∈ U (R)}

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Contents

Preface

vii

Notational Conventions

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1. Stable Range One

1

1.1 1.2 1.3 1.4

Substitutions of Modules . . . . . . . . Products of Three Triangular Matrices Exchange Rings, I . . . . . . . . . . . . Exchange Rings, II . . . . . . . . . . .

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2. Unit 1-Stable Range 2.1 2.2 2.3 2.4

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Elementary Properties . . . . Extensions and Substitutions . Examples . . . . . . . . . . . . Goodearl-Menal Condition . .

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3. On m-Fold Stable Rings 3.1 3.2 3.3 3.4

Symmetry . . . . . . . . . . . . . . . . . . . . . Morita Contexts . . . . . . . . . . . . . . . . . . Right Principal Ideals . . . . . . . . . . . . . . . Exchange Rings with Artinian Primitive Factors

Matrix Extensions . . Related Rings . . . . K1 -Groups . . . . . . Medium Stable Rings

41 48 53 58 69

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4. Strongly Stable Rings 4.1 4.2 4.3 4.4

2 10 18 28

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5. Weakly Stable Rings 5.1 5.2 5.3 5.4

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Comparability of Modules Exchange Rings . . . . . . Element Factorizations . . Matrix Factorizations . . .

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6. Related Comparability 6.1 6.2 6.3 6.4

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Related Units . . . . . . . . Equivalences . . . . . . . . . The Morita Invariance . . . Generalized s-Comparability

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7. Generalized Stable Rings 7.1 7.2 7.3 7.4

Criteria . . . . . . . . . Diagonal Reduction, I . Diagonal Reduction, II Invertible Matrices . .

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Matrices and Pseudo-Similarity Pseudo Substitutions . . . . . . Completions of Diagrams . . . . Quasi-Projective Modules . . .

Counterexamples . . Morita Equivalences . Exchange Conditions Subdirect Products .

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10.1 Power-Substitutions . . . . . . . 10.2 The Finite Exchange Properties 10.3 Unit π-Regularity . . . . . . . . 10.4 Stable Power-Substitutions . . .

11.1

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10. Power-Cancellation

11. Stably Euclidean Rings

194 208 216 223 235

9. P B-Rings 9.1 9.2 9.3 9.4

152 160 166 181 193

8. QB-Rings 8.1 8.2 8.3 8.4

119 125 138 142

265 268 277 282 291

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292 302 308 315 325

Divisible Chain Conditions . . . . . . . . . . . . . . . . . 326

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Contents

11.2 11.3 11.4

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2-Stable Range Condition . . . . . . . . . . . . . . . . . 332 Commutative Case . . . . . . . . . . . . . . . . . . . . . 336 Stably Free Modules . . . . . . . . . . . . . . . . . . . . 345

12. The n-Stable Range Condition 12.1 12.2 12.3 12.4

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Weak Substitutions . . . . . . Endomorphism Rings . . . . . Analogue of Stable Range One Rectangular Matrices . . . . .

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13. Stable Range for Ideals

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13.1 Strong π-Regularity . . . 13.2 Stable Ideals . . . . . . . 13.3 General Comparability . 13.4 Separative Regular Ideals

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14. Diagonal Reduction 14.1 14.2 14.3 14.4

Almost Hermitian Rings, I . Almost Hermitian Rings, II Strongly Separative Rings . Strongly Separative Ideals .

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15.1 Cleanness in Separative Ideals . . . . 15.2 Regular Rings in which 2 is Invertible 15.3 Cleanness in Exchange Rings . . . . . 15.4 Exchange P B-Ideals . . . . . . . . .

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16. Clean Properties, II Infinite-Dimensional Infinite-Dimensional Clean Matrices . . Strong Cleanness .

405 421 440 450 469

15. Clean Properties, I

16.1 16.2 16.3 16.4

361 376 383 393

511 519 527 535 541

Linear Transformations, I . Linear Transformations, II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17. Abelian Rings and Exchange

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17.1 Medium Spaces . . . . . . . . . . . . . . . . . . . . . . . 589 17.2 Abelian Exchange Rings . . . . . . . . . . . . . . . . . . 600 17.3 K0 -Groups . . . . . . . . . . . . . . . . . . . . . . . . . . 608

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17.4

Pierce Spaces . . . . . . . . . . . . . . . . . . . . . . . . 615

Bibliography

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Index

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Chapter 1

Stable Range One

Let R be an associative ring with an identity. We say that R has stable range one provided that aR + bR = R with a, b ∈ R implies that there exists some y ∈ R such that a + by ∈ U (R). In the literature, R is also said to be a B-ring. One of the most important features of stable range one is the cancellation of related modules from direct sums. We know that if EndR (A) has stable range one, then A ⊕ B ∼ = A ⊕ C implies that B ∼ =C for any right R-modules B and C. Stable range one plays an important rule in non-stable K-theory. It is well known that K1 (R) ∼ = U (R)/W (R) provided that R has stable range one (cf. [321, Theorem 1.2]). In [277], Keune has studied the K2 -groups of rings having stable range one. The symplectic group of a ring having stable range one was studied in [282]. Rings having stable range one include semilocal rings, unit-regular rings, strongly π-regular rings and exchange rings with artinian primitive factors (cf. [422, Theorem 1]). The main purpose of this chapter is to investigate necessary and sufficient conditions under which a ring R has stable range one. In Section 1.1, we characterize stable range one condition by means of the substitution property of modules. In Section 1.2, we study such condition via triangular factorization of invertible matrices. A right Rmodule A is said to have the finite exchange property if for every right L R-module M and any two decompositions M = A′ ⊕ N = i∈I Ai , where A′R ∼ set I
is finite, there exist submodules A′i ⊆ Ai = AR and the index L ′ ′ such that M = A ⊕ i∈I Ai . A ring R is said to be an exchange ring provided that R has the finite exchange property as a right R-module. In 1

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Section 1.3, we develop criteria for exchange rings to have stable range one. Finally, in Section 1.4, we continue our investigation by developing further equivalent conditions for exchange rings having stable range one.

1.1

Substitutions of Modules

The main purpose of this section is to study the stable range one condition via substitutions of modules. This implies that modules over rings having stable range one possess cancellation. Let     1∗ 10 B12 (∗) = and B21 (∗) = . 01 ∗1 We use [u, v] to denote the diagonal matrix diag(u, v) with u, v ∈ U (R). Lemma 1.1.1. Let R be a ring. Then the following are equivalent: (1) R has stable range one. (2) For any A ∈ GL2 (R), A = [∗, ∗]B21 (∗)B12 (∗)B21 (∗). Proof. (1) ⇒ (2) Given A = (aij ) ∈ GL2 (R), then a11 R + a12 R = R. Since R has stable range one, there exists some y ∈ R such that a11 + a12 y = u ∈ U (R). As a result, we get  
u b −1 B21 − (a21 + a22 y)u AB21 (y) = . 0 d − (c + dy)u−1 b It follows from A ∈ GL2 (R) that d − (c + dy)u−1 b ∈ U (R). Thus A = [∗, ∗]B21 (∗)B12 (∗)B21 (∗). (2) ⇒ (1) Given ax + b = 1 with a, x, b ∈ R, then    −1 a b x xa − 1 = ∈ GL2 (R). −1 x 1 a So we can find u, v ∈ R and y ∈ R such that   a b = [u, v]B21 (∗)B12 (∗)B21 (−y). −1 x This means that



a b −1 x



B21 (y) = [u, v]B21 (∗)B12 (∗),

and then a + by = u ∈ U (R). Therefore R has stable range one. Theorem 1.1.2. Let R be a ring. Then the following are equivalent:



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Stable Range One

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(1) R has stable range one. (2) Ra + Rb = R with a, b ∈ R implies that there exists a z ∈ R such that a + zb ∈ U (R). Proof. (1) ⇒ (2) Given xa + b = 1 with x, a, b ∈ R, then    −1 x b a ax − 1 = ∈ GL2 (R). −1 a 1 x

By virtue of Lemma 1.1.1, we have   x b = [∗, ∗]B21 (∗)B12 (∗)B21 (∗); −1 a

hence, we can find a z ∈ R such that   x b B21 (z) = [u, v]B12 (∗)B21 (∗). −1 a

This implies that a + zb = v ∈ U (R), as required. (2) ⇒ (1) Given any A = (aij ) ∈ GL2 (R), then Ra12 + Ra22 = R. By hypothesis, we can find some z ∈ R such that za12 + a22 = v ∈ U (R). This implies that   ∗∗ B21 (z)A = = [∗, ∗]B12 (∗)B21 (∗). ∗v

It follows that A = [∗, ∗]B21 (∗)B12 (∗)B21 (∗), and therefore we complete the proof by Lemma 1.1.1.  Let R be a ring. The notation Rop stands for the opposite ring of R. Theorem 1.1.2 means that stable range one condition is left-right symmetric. In other words, a ring R has stable range one if and only if the opposite ring Rop also has stable range one. Let A be an n × n matrix over a ring R. The notation AT stands for the transpose of the matrix A. Corollary 1.1.3. Let R be a ring. Then the following are equivalent: (1) R has stable range one. (2) For any A ∈ GL2 (R), A = [∗, ∗]B12 (∗)B21 (∗)B12 (∗).

Proof. (1) ⇒ (2) Since R has stable range one, so does Rop by Theorem 1.1.2. Given any (aij ) ∈ GL2 (R), then (aoij )T ∈ GL2 (Rop ). In view of Lemma 1.1.1, we have (aoij )T = [∗o , ∗o ]B21 (∗o )B12 (∗o )B21 (∗o ). As a result, we deduce that (aij ) = [∗, ∗]B12 (∗)B21 (∗)B12 (∗). (2) ⇒ (1) Given any (aoij ) ∈ GL2 (Rop ), then (aij )T ∈ GL2 (R). By assumption, we have (aij )T = [∗, ∗]B12 (∗)B21 (∗)B12 (∗). That is, (aoij ) =

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[∗o , ∗o ]B21 (∗o )B12 (∗o )B21 (∗o ). According to Lemma 1.1.1, Rop has stable range one, and therefore the result follows from Theorem 1.1.2.  Corollary 1.1.4. Let R be a ring. Then the following are equivalent: (1) R has stable range one. (2) aR + bR = dR with a, b, d ∈ R implies that there exist u ∈ U (R), v ∈ R such that au + bv = d. (3) Ra + Rb = Rd with a, b, d ∈ R implies that there exist u ∈ U (R), v ∈ R such that ua + vb = d. Proof. (1) ⇒ (2) Given aR + bR = dR with a, b, d ∈ R, then we have x, y, s, t ∈ R such that a = dx, b = dy and as + bt = d. Let c = 1 − xs − yt. Then xs + yt + c = 1. So we have some z ∈ R such that x + (yt + c)z = u ∈ U (R). Hence dx + d(yt + c)z = du, and then a + btz = du − dcz = du − (d − as − bt)z = du. Therefore au−1 + btzu−1 = d, as required. (2) ⇒ (1) is trivial. (1) ⇔ (3) follows by left-right symmetry.  Let A be a right R-module, and let E = EndR (A). Evidently, E has stable range one if and only if for any splitting epimorphism ϕ : A⊕H → A, there exists an R-morphism χ : A → H such that θ:A→ A,
a 7→ ϕ a, χ(a)

is an isomorphism. The following elementary result is due to Warfield [394, Theorem 2.1]. Theorem 1.1.5. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E has stable range one. (2) Given any right R-module decompositions M A1 ∼ =A∼ = A2 , there exists C ⊆ M such that (3) Given any right R-module decompositions M A1 ∼ =A∼ = A2 , there exists C ⊆ M such that

= A1 ⊕B1 = A2 ⊕B2 with M = C ⊕ B1 = C ⊕ B2 . = A1 ⊕B1 = A2 ⊕B2 with M = A1 ⊕ C = A2 ⊕ C.

Proof. (1)⇒(2) We are given M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ =A∼ = A2 . Using the decomposition M = A1 ⊕ B1 ∼ = A ⊕ B1 , we have projections p1 : M → A1 ∼ = A, p2 : M → B1 and injections q1 : A ∼ = A1 → M, q2 : B1 → M such that p1 q1 = 1A , q1 p1 + q2 p2 = 1M . Using the decomposition M = A2 ⊕ B2 ∼ = A ⊕ B2 , we have a projection f : M → A2 ∼ = A and an

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injection g : A ∼ = A2 → M such that f g = 1A . As (f q1 )(p1 g)+f q2 p2 g = 1A , there exists some y ∈ E such that f q1 + f q2 p2 gy ∈ U (E). This implies that M = ker(f ) ⊕ C, where C = im(q1 + q2 p2 gy). As p1 (q1 + q2 p2 gy) = 1A , we also get M = ker(p1 ) ⊕ C. Therefore M = C ⊕ B1 = C ⊕ B2 . (2) ⇒ (1) Suppose that ax + b = 1A with a, x, b ∈ E. Set M = 2A, and let pi : M → A, qi : A → M (for i = 1, 2) denote the projections and injections of this direct sum. Set A1 = q1 (A) and B1 = q2 (A), so that M = A1 ⊕ B1 with A1 ∼ = A. Define f = ap1 + bp2 from M to A and g = q1 x + q2 from A to M . Observing that f g = 1A , we get M = ker(f ) ⊕ g(A). Set A2 = g(A) and B2 = ker(f ), so that M = A2 ⊕ B2 and A2 ∼ = A. By assumption, M = C ⊕ B1 = C ⊕ B2 for some C ⊆ M . Let h : A ∼ = A1 ∼ = C → M be the natural injection. Then C = h(A). So M = ker(p1 ) ⊕ h(A), we infer that p1 h is an isomorphism. On the other hand, M = ker(f ) ⊕ h(A). Hence, f h is an isomorphism. Clearly, f h =
(ap1 + bp2 )h = a + bp2 h(p1 h)−1 p1 h ∈ U (E). Therefore a + bp2 h(p1 h)−1 ∈ U (E), as required. (1) ⇒ (3) As in (1) ⇒ (2), we have (f q1 )(p1 g) + f q2 p2 g = 1A . By virtue of Theorem 1.1.2, there exists some z ∈ E such that p1 g + zf q2p2 g ∈ U (E). This implies that M = im(g) ⊕ C, where C = ker(p1 + zf q2 p2 ). As (p1 + zf q2 p2 )q1 = 1A , we have M = im(q1 )⊕C. Therefore M = A1 ⊕C = A2 ⊕C. (3) ⇒ (1) As in (2) ⇒ (1), we have M = A1 ⊕ B1 with A1 ∼ = A. Define f = ap1 + p2 from M to A and g = q1 x + q2 b from A to M . Clearly, f g = 1A , and so M = ker(f ) ⊕ g(A). Set A2 = g(A) and B2 = ker(f ), so that M = A2 ⊕ B2 with A2 ∼ = A. By assumption, M = A1 ⊕ C = A2 ⊕ C for some C ⊆ M . Let h : M → A1 be the projection. Then C = ker(h), and so M = ker(h) ⊕ q1 (A). This implies that hq1 is an isomorphism. On the other hand, M = ker(h) ⊕ g(A). Thus, hg is
an isomorphism. Observe that hg = h(q1 x + q2 b) = (hq1 ) x + (hq1 )−1 (hq2 )b ∈ U (E). Thus x + (hq1 )−1 (hq2 )b ∈ U (E). According to Theorem 1.1.2, E has stable range one.  Corollary 1.1.6. Let A be a finitely generated projective right module over a ring R. If R has stable range one, then so does EndR (A). Proof. Let A be a finitely generated projective right R-module. Then we have an idempotent E ∈ Mn (R) (n ∈ N) such that A ∼ = E(nR). Hence EndR (A) ∼ = EMn (R)E. Given any right R-module decompositions M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = nR ∼ = A2 , then we have right R-module decompositions A1 = A11 ⊕ A12 ⊕ · · · ⊕ A1n and A2 = A21 ⊕ A22 ⊕ · · · ⊕ A2n

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with each A1i ∼ = R ∼ = A2i . In view of Theorem 1.1.5, we have a module D1 ⊆ M such that M = D1 ⊕A12 ⊕· · ·⊕A1n ⊕B1 ∼ = D1 ⊕A22 ⊕· · ·⊕A2n ⊕B2 . Analogously, we can find D2 , · · · , Dn ⊆ M such that M = D1 ⊕ D2 ⊕ · · · ⊕ Dn ⊕B1 = D1 ⊕D2 ⊕· · ·⊕Dn ⊕B2 . Using Theorem 1.1.5 again, EndR (nR) has stable range one, and so Mn (R) has stable range one. One easily checks that EMn (R)E has stable range one, and therefore so does EndR (A).  A ring S is said to be Morita equivalent to R in the case that there is an adjoint equivalence between the category of right R-modules and the category of right S-modules. By Corollary 1.1.6, we conclude that the stable range one condition is Morita invariant. Corollary 1.1.7. Let A be a finitely generated projective right module over a ring R. If R has stable range one, then for any right R-modules B and C, A ⊕ B ∼ = A ⊕ C implies that B ∼ = C.

Proof. Since ψ : A ⊕ B ∼ = A ⊕ C, we have A ⊕ C = ψ(A) ⊕ ψ(B). In view of Corollary 1.1.6, EndR (A) has stable range one. As A ∼ = ψ(A), it follows by Theorem 1.1.5 that there is a module D ⊆ A ⊕ C such that A ⊕ C = D ⊕ C = D ⊕ ψ(B). Thus B ∼  = ψ(B) ∼ = C, as asserted. We note that cancellation fails for general modules over rings having stable range one. Let F be a field and let   A   d   R = {  | A ∈ Mn (F ), n ≥ 1, d ∈ F }. d   .. .

Then R has stable range one. Choose its two ideals   A   0   I = {  | A ∈ Mn (F ), n ≥ 1}, J = HI 0   .. .   0   1  1 0     0   10  . Choose a =  where H =  . One easily checks    0 .   .   . 1 ..   . .. . that aR ⊕ I ∼ = J where I ∼ = J as right R-modules, while aR 6= 0.

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A right R-module M is said to be semisimple when M is the sum of its simple submodules. For example, every abelian group spanned by its elements of prime order is a semisimple Z-module and the socle of any right R-module is semisimple, where the socle of a right module is the unique largest semisimple submodule. Every vector space over a division ring D is a semisimple D-module. A right R-module M is semisimple if and only if every submodule of M is a direct summand. A ring R is semisimple if it is a semisimple right R-module. Let P be a right ideal of a ring R. We say that P is maximal provided that for any right ideal I, P ⊆ I implies that P = I. We use J(R) to denote the Jacobson radical of a ring R, i.e., T J(R) = {I | I is a maximal right ideal of R}. Then J(R) is an ideal of R and J(R) = {x ∈ R | 1 + xr ∈ U (R) for any r ∈ R}. We say that R is semilocal in the case when R/J(R) is semisimple. As is well known, a ring R is semilocal if and only if there exist some division rings D1 , · · · , Dn n L such that R/J(R) ∼ Mm (Di ). A commutative ring R is semilocal if = i

i=1

and only if it has only finite number of maximal ideals. A right R-module A is artinian provided that any descending chain of its submodules A ⊇ A1 ⊇ · · · An ⊇ · · · terminates. Corollary 1.1.8. Let A be an artinian right R-module. Then for any right R-modules B and C, A ⊕ B ∼ = A ⊕ C implies that B ∼ = C.

Proof. Let E = EndR (A). In light of the Camps-Dicks Theorem [68, Corollary 6], E is a semilocal ring; hence, E/J(E) is semisimple. Using the Wedderburn-Artinian Theorem, E/J(E) is a direct sum of a finite number of simple artinian ring Ri (1 ≤ i ≤ n). Clearly, we have division rings Di such that Ri = Mni (Di ). As Di has stable range one, so does Ri by Corollary 1.1.6. This implies that E/J(E) has stable range one, and then so does E. Similarly to Corollary 1.1.7, it follows by Theorem 1.1.5 that B∼  = C, as required. Lemma 1.1.9. If EndR (M1 ),
· · · , EndR (Mn ) have stable range one, then so does EndR M1 ⊕ · · · ⊕ Mn .

Proof. Given right R-module decompositions M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = M1 ⊕ · · · ⊕ Mn ∼ = A2 , then we have A1 = A11 ⊕ · · · ⊕ A1n and A2 = A21 ⊕ · · · ⊕ A2n with A1i ∼ = Mi ∼ = A2i (1 ≤
i ≤ n). So M = A11 ⊕
A12 ⊕· · ·⊕A1n ⊕B1 = A21 ⊕ A22 ⊕· · ·⊕A2n ⊕B2 with A11 ∼ = M1 ∼ = A12 . Since EndR (M1 ) has stable range one, by virtue of Theorem 1.1.5, we can

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find a submodule C1 ⊆ M
such that M = C1 ⊕ A12 ⊕ · · · ⊕ A1n ⊕ B1 = C1 ⊕ A22 ⊕· · ·⊕A2n ⊕B2 . Likewise,
we have submodules C2 , · · ·
, Cn ⊆ M such that M = C1 ⊕ C2 ⊕ · · · ⊕ Cn ⊕ B1 = C1 ⊕ C2 ⊕ · · · ⊕ Cn
⊕ B2 . By Theorem 1.1.5 again, we conclude that EndR M1 ⊕ · · · ⊕ Mn has stable range one.  A finite orthogonal set of idempotents e1 , · · · , en in a ring R is said to be complete in case e1 + · · · + en = 1 ∈ R. Theorem 1.1.10. Let {e1 , · · · , en } be a complete orthogonal set of idempotents in R. Then the following are equivalent: (1) R has stable range one. (2) ei Rei has stable range one for all i. Proof. (1) ⇒ (2) is clear. ∼ (2) ⇒ (1) Since ei Re
i = EndR (ei R), it follows from Lemma 1.1.9 that EndR e1 R ⊕ · · · ⊕ en R has stable range one. One easily checks that e1 R ⊕ · · · ⊕ en R = R. Hence R ∼  = EndR (R) has stable range one. Let e1 , · · · , en ∈ R be idempotents. It is easy to verify that (ei Rej ) := {(ei rij ej )| rij ∈ R (1 ≤ i, j ≤ n)} forms a ring with the identity being diag(e1 , · · · , en ). Corollary 1.1.11. Let e1 , · · · , en be idempotents of a ring R. If e1 Re1 , · · · , en Ren all have stable range one, then so does the ring (ei Rej ). Proof. Set T = (ei Rej ). Let f1 = diag(e1 , 0, · · · , 0), f2 = diag(0, e2 , · · · , 0), · · · , fn = diag(0, 0, · · · , en ). Then we have a complete orthogonal set of idempotents, {f1 , · · · , fn }, such that fi T fi ∼ = ei Rei has stable range one for each i. We obtain the result by Theorem 1.1.10.  Lemma 1.1.12. Let S have stable range one, and let K be an ideal of S. Let R be a subring of S which contains K. If R/K has stable range one, then so does R. Proof. Let ax+b = 1 in R. Then ax + b = 1 in R/K. Since R/K has stable range one, we can find some y ∈ R such that a + by is a unit in R/K. From ax + b = 1, we know that (a + by)x + b(1 − yx) = 1. Since a + by is a unit in R/K, there exist some w ∈ R, k ∈ K such that (a + by)w + k = 1. Thus
1 = (a+ by)w + k = (a+ by)w + 1 ·k = (a+ by)w + (a+ by)x+ b(1 − yx) k = (a+by)(w +xk)+b(1−yx)k. As S has stable range one, there is some z ∈ S

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such that (a+by)+b(1−yx)kz is a unit in S. So a+b(y +(1−yx)kz) v = 1 for a v ∈ S. On the other hand, it is easy to verify b(1 − yx)kz ∈ R. Clearly, (a + by)v = 1 in S/K. Since a + by is a unit in R/K, we can prove v ∈ R/K. Thus v ∈ R. Let t = y + (1 − yx)kz. Then t ∈ R since K an ideal of R. That is, we can find some t ∈ R such that a + bt is a unit in R, as required.  Recall that a ring R is a subdirect product of rings Ri (i ∈ I) provided that there exists an epimorphism ϕi : R ։ Ri for each i such that T ker(ϕi ) = 0.

i∈I

Theorem 1.1.13. The finite subdirect product of rings having stable range one has stable range one. Proof. Let R1 and R2 have stable range one, and let R be the subdirect product of R1 and R2 . Then there exist two epimorphisms ϕ1 : R ։ R1 T and ϕ2 : R ։ R2 with ker(ϕ1 ) ker(ϕ2 ) = 0. Thus, R/ker(ϕ1 ) ∼ = R1 and R/ker(ϕ2 ) ∼ = R2 . This implies that R/ker(ϕ1 )× R/ker(ϕ2 ) has stable range one. Let
φ : R → R/ker(ϕ1 ) × R/ker(ϕ2 ) given by φ(r) = r + ker(ϕ1 ), r + ker(ϕ2 )
for any r ∈ R. Then R ∼ = im(φ). Let K = { r + ker(ϕ1 ), r + ker(ϕ2 ) | r ∈ ker(ϕ2 )}. Then K is an ideal of R/ker(ϕ1 ) × R/ker(ϕ
2 ). Let ψ : im(φ) → R/ker(ϕ2 ) be given by ψ
r + ker(ϕ1 ), r + ker(ϕ2 ) = r + ker(ϕ2 ) for any r + ker(ϕ1 ), r + ker(ϕ2 ) ∈ im(φ). It is easy to verify that im(φ)/K ∼ = R/ker(ϕ2 ). By virtue of Lemma 1.1.12, we conclude that im(φ) has stable range one, as required.  Recall that an element a ∈ R is regular when there exists some x ∈ R such that a = axa. If one such x is a unit, then a is called unit-regular. A ring R is said to be regular if every element in R is regular. A ring R is said to be unit-regular if every element in R is unit-regular. It is well known that a regular ring R is unit-regular if and only if it has stable range one (cf. [217, Proposition 4.12]). As an immediate consequence, we derive that any finite subdirect product of unit-regular rings is unit-regular. Corollary 1.1.14. Let I be an ideal of a ring R. Then the following are equivalent : (1) R has stable range one. (2) R/I and R/r(I) have stable range one. (3) R/I and R/ℓ(I) have stable range one.

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Proof. (1)⇒(2) is trivial. (2)⇒(1) Clearly,



T T ∼ R/ I r(I)
/I/ I T r(I)
= R/I, T R/ I r(I) /r(I)/ I r(I) ∼ = R/r(I).
T In view of Theorem 1.1.13, we see that R/ I r(I) has stable range one.
2 T T As I r(I) = 0, we know that I r(I) ⊆ J(R). Therefore R has stable range one. (1) ⇔ (3) is proved in the same manner.  We end this section by noting that a commutative ring R has stable range one if and only if for any ideal I of R, the natural map U (R) → U (R/I) is surjective.

1.2

Products of Three Triangular Matrices

The m × n matrix A = (aij ), 1 ≤ i ≤ m, 1 ≤ j ≤ n, is said to be diagonal if aij = 0 for all i 6= j. The m× n matrix A admits diagonal reduction if there exist P ∈ GLm (R), Q ∈ GLn (R) such that P AQ is a diagonal matrix. We say that R is a right (left) hermitian ring if every 1 × 2 (respectively, 2 × 1) matrix admits a diagonal reduction. If R is both left and right hermitian, R is an hermitian ring. It is well known that every strongly separative regular ring is an hermitian ring (cf. [15, Proposition 3.2]). The set of all lower triangular matrices is denoted by L, i.e., L = {(aij ) | aij = 0 whenever i < j}. The set of all upper triangular matrices is denoted by U, i.e., U = {(aij ) | aij = 0 whenever i > j}. In 1999, Nagarajan et al. considered the problem of whether every square matrix over a field should be a product of three triangular matrices. They showed that any square matrix over a field is the product of at most three triangular matrices. After that, in 2002 they proved that a commutative ring R is a hermitian ring having stable range one if and only if every square matrix A can be written as A = LU M, L ∈ L, U ∈ U, M ∈ L and in U and M all the diagonal entries are equal to 1. [387, Theorem 1] showed that if R has stable range one, then every invertible matrix over R belongs to LUL, where LUL = {λρµ | λ, µ ∈ L, ρ ∈ U}. [331, Proposition 1.3] showed that the converse is true for 2 × 2 matrices over commutative rings. Linear equations have particularly transparent solutions when the matrix of coefficients is either lower triangular or upper triangular. The main

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purpose of this section is the study of products of three triangular matrices over associative rings having stable range one. Now we generalize [331, Proposition 1.3] as follows. Lemma 1.2.1. Let n ≥ 2 be a positive integer. Then the following are equivalent: (1) R has stable range one. (2) For any U ∈ GLn (R), we have    ∗ 1 ∗ ··· ∗ ∗   1 ···    U =. . .   ..  .. .. . .   . ∗ ∗ · · · ∗ n×n

 ∗ ∗  ..  .

1



 1 ∗ 1    . . . .   .. .. . .  ∗ ∗ · · · 1 n×n n×n

Proof. (1) ⇒ (2) Clearly, every right or left invertible element in R is invertible. Thus, it suffices to show that for any U ∈ GLn (R),       1 ∗ ∗ ··· ∗ 1 ∗ 1   ∗ ··· ∗ ∗ 1        U =. . . .    . . .  . . . . . . . . . . . . .   . . .  . . ∗ ∗ ··· 1

n×n

∗

n×n

∗ ∗ ··· 1

n×n

Proceeding by induction on n (the case n = 1 is trivial), we take an arbitrary matrix   η v V = ∈ GLn+1 (R), η ∈ Mn (R), v ∈ M1×n (R), u ∈ Mn×1 (R). uα Since R has stable range one, by Theorem 1.1.2, there exists u′ ∈ M1×n (R) such that u′ v + α = α′ ∈ U (R). Thus, we have a u′′ ∈ M1×n (R) such that      ′  In 0 η v In 0 η v . = u′ 1 uα u′′ 1 0 α′

By hypothesis, we can write η ′ = λρµ , where λ, µ ∈ Mn (R) are lower triangular matrices with diagonal entries 1 and ρ ∈ Mn (R) is an upper triangular matrix. Then     In 0 In 0 λρµ v V = −u′′ 1 −u′ 1 0 α′      λ 0 ρ λ−1 v µ 0 , = −u′′ 1 −u′ λ 1 0 α′ as required.

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(2) ⇒ (1) Suppose that ax + b = 1 with a, x, b ∈ R. Then    −1 a b x xa − 1  −1 x  1 a          1 1  =  ∈ GLn (R).     . . ..   ..   1

So we have  a b  −1 x   1   .. 

−1 . 1

     



1

c1 ∗   =∗  .  ..

c2 ∗ .. .

c3 .. . . . . ∗ ∗ ∗ · · · cn



1 ∗ ∗ ···  1 ∗ ···   1 ···   ..  .

  ∗ 1   ∗ ∗ 1  ∗ ∗ 1  ∗  . . . . .  ..  . . . .   . . . . .  1 ∗ ∗ ∗ ··· 1

Moreover, we get      a b 1 1 ∗ ∗ ··· ∗ c1  −1 x  ∗ 1  1 ∗ ··· ∗ ∗        ∗ ∗ 1   1 1 ··· ∗  =   ∗       . . .   .. .. .. . .   . . ..    . . .  .. . . . . 1 ∗ ∗ ∗ ··· 1 ∗ 1

Hence we have  a b  −1 x   1   ..  .

c2 ∗ .. .

c3 .. . . . . ∗ ∗ · · · cn

−1      

.

some y ∈ R such that      1 ∗ ∗ ··· ∗ 1 c1  1 ∗ ··· ∗  ∗ 1   y c2        ∗ ∗ 1   ∗ ∗ c3 1 ··· ∗ .  =    . . . .  . . . . . . ..    .. .. .. . .     .. .. .. . . . . 1 ∗ ∗ ∗ ··· 1 ∗ ∗ ∗ · · · cn 1

So ac1 +by = 1. Clearly, c1 d1 = 1 for some d1 ∈ R. Thus, a+byd1 = d1 ∈ R is left invertible. Since c1 d1 + 0 = 1, by the above consideration, we have some z ∈ R such that c1 + 0 · z ∈ R is left invertible. That is, c1 ∈ U (R), and so a + byd1 ∈ U (R). Therefore R has stable range one.  Theorem 1.2.2. Let n ≥ 2 be a positive integer. Then the following are equivalent: (1) If A ∈ Mn (R), then A can be written as A = LU M, L ∈ L, U ∈ U, M ∈ L and in U and M all the diagonal entries are equal to 1. (2) R is a right hermitian ring having stable range one.

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Proof. (1) ⇒ (2) In light of Lemma 1.2.1, R has stable range one. It suffices to show that R is a right hermitian ring. Letting a, b ∈ R, we have   a b 0 ··· 0 0 0 0 ··· 0   ∈ Mn (R). . . . . .  .. .. .. . . ..  0 0 0 ··· 0

n×n

Hence there exist c1 , c2 , · · · , cn ∈ R such that       a b 0 ··· 0 c1 1 ∗ ··· ∗ 1  0 0 0 · · · 0   ∗ c2  1 ··· ∗∗ 1        . . . . . .= . . .    . . . .. .   . . ..   .. .. .. . . ..   .. .. . .  . . . . .  0 0 0 ··· 0 ∗ ∗ · · · cn 1 ∗ ∗ ··· 1 This means that       a b 0 ··· 0 1 1 ∗ ··· ∗ c1 0 0 0 ··· 0∗ 1   1 · · · ∗   ∗ c2         . . . . .  . . .  = . . . . . . . . . . . . . . . . . . .  . . . . .  . . .   . .  . . . 0 0 0 ··· 0 ∗ ∗ ··· 1 1 ∗ ∗ · · · cn Thus we get       ab 10 1∗ c1 0 = . 00 ∗1 01 ∗ c2 Set      10 1∗ 1s = . ∗1 01 tz   1s Then ∈ GL2 (R). In addition, we see that a+bt = c1 and as+bz = 0. tz Therefore   1s (a, b) = (c1 , 0). tz Consequently, R is a right hermitian ring. (2) ⇒ (1) Let A = (aij ) ∈ Mn (R). Since R is a right hermitian ring, by induction, we can find a V1 ∈ GLn (R) such that (a11 , a12 , · · · , a1n )V1 = (∗, 0, · · · , 0). Further, there exist P ∈ L, V ∈ GLn (R) such that A = P V (cf. [331, Theorem 2.5]). By virtue of Lemma 1.2.1, we have       1 ∗ ··· ∗ 1 ∗ ∗ 1    1 ··· ∗ ∗ ∗       . V =. . .      . . . . ..    .. .. . . .   .. .. . .  . . 1 n×n ∗ ∗ · · · 1 n×n ∗ ∗ · · · ∗ n×n

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14

Therefore we get A = LU M , where    ∗ 1 ···  .. . .   .. L =  . .  ∈ L, U =  . as required.

rings

∗ ··· ∗

  ∗ 1 ..  ∈ U, M =  .. . . . . .

∗ ··· 1

1



  ∈ L,



In the preceding proof, it is shown that for any matrix A with elements in a right hermitian ring, there exists an invertible matrix U such that AU is a lower triangular matrix. It follows from this observation that for any right hermitian ring R, Mn (R) is right hermitian (cf. [275, Theorem 3.6]). Corollary 1.2.3. Let n ≥ 2 be a positive integer. Then the following are equivalent: (1) If A ∈ Mn (R), then A can be written as A = U LV, U ∈ U, L ∈ L, V ∈ U and in U and L all the diagonal entries are equal to 1. (2) R is a left hermitian ring having stable range one. Proof. (1) ⇒ (2) Let Ao ∈ Mn (Rop ). Then AT ∈ Mn (R), so AT can be written as AT = U LV, U ∈ U, L ∈ L, V ∈ U and in U and L all the diagonal entries are equal to 1. This means that Ao = (V o )T (Lo )T (U o )T . Obviously, (V o )T , (U o )T ∈ L and (Lo )T ∈ U. In addition, in (Lo )T and (U o )T all the diagonal entries are 1o . By Theorem 1.2.2, Rop is a right hermitian ring having stable range one, as required. (2) ⇒ (1) Clearly, Rop is a right hermitian ring having stable range one. Given any A ∈ Mn (R), we have (Ao )T ∈ Mn (Rop ). It follows from Theorem 1.2.2 that (Ao )T can be written as (Ao )T = LU M, L ∈ L, U ∈ U, M ∈ L and in U and M all the diagonal entries are equal to 1o . Hence A = (M o )T (U o )T (Lo )T . In addition, we have (M o )T , (Lo )T ∈ U, (U o )T ∈ L and in (M o )T and (U o )T all the diagonal entries are equal to 1.  A right hermitian ring is left hermitian provided that every finitely generated left ideal is principal. This yields, in particular, that any regular right hermitian ring is left hermitian ([320, Proposition 8]). 

 000 Example 1.2.4. Let F be a field, and let A =  0 1 0  . Assume A can be 100 expressed as LU M where L, M ∈ L, U ∈ U and in L and M all the diagonal entries are equal to 1. Then AM −1 = LU , while the (3, 1)-entry of AM −1 is

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Stable Range One

15

1, the (3, 1)-entry of LU is 0, a contradiction. This means that A cannot be expressed as LU M where L, M ∈ L, U ∈ U and in L and M all the diagonal entries with we have  are equal   to 1. But  in accordance    Theorem1.2.2,  100 1 0 −1 100 000 1 0 −1 100 A =  0 1 0   0 1 0   0 1 0  =  0 1 0   0 1 0   0 1 0 .  001 00 1 100 101 00 1 001 

0 0 ··· 0 0 ···   We set Γ =  ... ... . . .  0 1 ···

 01 1 0  .. ..  . .  0 0

1 0 ··· 0 0  0 0 ··· 0 0 0 ··· ∗   triangular matrices  ... ... . . . ...  0 ∗ ··· ∗

and use W to denote the set of all

n×n  ∗ ∗  ..  .  ∗

∗ ∗ ··· ∗ ∗

.

n×n

Lemma 1.2.5. Let n ≥ 2 be a positive integer. Then the following are equivalent: (1) R has stable range one. (2) For any U ∈ GLn (R), we have 

∗ ∗ ∗  U =. . .  .. .. . .

∗ ∗ ··· ∗

    



n×n

1 ∗  Γ .  ..



1 .. . . . .

∗ ∗ ··· 1

   



n×n

1 ∗ ···  1 ···   ..  .

 ∗ ∗  ..  .

1

. n×n

Proof. (1) ⇒ (2) For any U ∈ GLn (R), we have U Γ ∈ GLn (R). By virtue of Lemma 1.2.1, we have that 

∗ ∗   UΓ =  ∗ .  .. ∗

0 0 ··· ∗ 0 ··· ∗ ∗ ··· .. .. . . . . . ∗ ∗ ···

 1∗∗ 0 0 1 ∗ 0   0 0 0 1 . . . ..  .   .. .. .. 000 ∗

··· ··· ··· .. . ···

 0 0  ..  . .  ··· 1 0 ∗ ∗ ··· ∗ 1

 ∗ 10 ∗ 1 ∗  . . ∗   .. ..  ..  . ∗ ∗

1

··· ··· .. .

0 0 .. .

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16

Hence,



  0 1∗∗ 0 1 ∗ 0    0 Γ · Γ0 0 1 . . . ..   .. .. .. . 000 ∗ ∗ ∗ ··· ∗

∗00 ∗ ∗ 0   U = ∗ ∗ ∗ . . .  .. .. ..

··· ··· ··· .. .

  ∗ 1 0 ··· ∗ 1 ··· ∗   . . ∗  Γ · Γ  .. .. . . .  ..  ∗ ∗ ··· . ··· 1 ∗ ∗ ··· ··· ··· ··· .. .

as desired. (2) ⇒ (1) For any U ∈ GLn (R), we have that      ∗ 1 1 ∗ ··· ∗ ∗  ∗ 1   1 ···      UΓ =  . . . Γ . . .    ..  .. .. . .   .. .. . .   .

Hence,

∗ ∗ ··· ∗

∗ ∗ ··· 1

n×n

n×n



 00 0 0  .. ..  Γ, . .  1 0 ∗1

 ∗ ∗  ..  .

1

. n×n

     ∗ 1 1 ∗ ··· ∗ ∗ ∗  ∗ 1   1 ··· ∗       U =. . . Γ ·Γ Γ Γ . . .  . . ..   .. .. . .   .. .. . .   . . ∗ ∗ ··· ∗ ∗ ∗ · · · 1 n×n 1       1 1 ∗ ··· ∗ ∗  ∗ 1   1 ··· ∗ ∗ ∗       . =. . .   . . .   . . . . ..   .. .. . .    .. .. . . 

1 n×n ∗ ∗ · · · 1 ∗ ∗ · · · ∗ n×n According to Lemma 1.2.1, R has stable range one.

n×n



Theorem 1.2.6. Let n ≥ 2 be a positive integer. Then the following are equivalent: (1) If A ∈ Mn (R), then A can be written as A = W LU, W ∈ W, L ∈ L, U ∈ U and in L and U all the diagonal entries are equal to 1. (2) R is a right hermitian ring having stable range one. Proof. (1) ⇒ (2) By virtue of Lemma 1.2.5, R has stable range one. It will suffice to prove that R is a right hermitian ring. Let a, b ∈ R. Then there exist c1 , c2 , · · · , cn ∈ R such that        1 ∗ ∗ ··· ∗ c1 1 0 0 0 ··· 0  1 ∗ ··· ∗  ∗ 1  0 0 0 · · · 0   ∗ c2          ∗ ∗ 1  .. .. .. . . ..   .. .. . . 1 ··· ∗  . Γ . . . . .= . . .       . . ..   b a 0 · · · 0   ∗ ∗ · · · cn−1   ... ... ... . . .   . . 0 0 0 ··· 0

∗ ∗ ···

∗

cn

∗ ∗ ∗ ··· 1

1

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Stable Range One

This implies that    0 0 0 ··· 0 1 ∗ ∗ ··· ∗ 1 0 0 0 ··· 0 1 ∗ ··· ∗∗ 1      .. .. .. . . ..   1 ··· ∗ ∗ ∗  . . . . .     . . . . ..   b a 0 ··· 0 . .  .. .. 0 0 0 ··· 0 1 ∗∗



1 .. . . . . ∗ ··· 1

Thus we get

 Set

ba 00

 

1∗ 01



10 ∗1

17



=





c1  ∗    ∗ =   .   ..

c2 ∗ .. .



c3 .. . . . .

∗ ∗ ∗ · · · cn

0 cn−1 cn ∗



    Γ.  

.

   10 zs = . ∗1 t1       1t 01 zs 01 Then = ∈ GL2 (R). Furthermore, we get sz 10 t1 10 at + bz = 0 and a + bs = cn−1 . Therefore   1t (a, b) = (cn−1 , 0). sz 1∗ 01



That is, R is a right hermitian ring. (2) ⇒ (1) Let A ∈ Mn (R). As R is a right hermitian ring, we have some P ∈ L and some V ∈ GLn (R) such that A = P V . Using Lemma 1.2.5, we have       1 ∗ ··· ∗ 1 ∗   1 ··· ∗  ∗ 1 ∗ ∗       . Γ . . . V =. . .    . . ..    .. .. . .   .. .. . .  . . 1 n×n ∗ ∗ · · · 1 n×n ∗ ∗ · · · ∗ n×n

Thus, we see that  ∗  .. . . W =. .

A = W LU , where   1   .. . . Γ, L =  . . 

∗ ··· ∗

as required.

n×n

∗ ··· 1

  



n×n

1 ···  .. ,U =  .

 ∗ ..  .

1

, n×n



Let n ≥ 2 be a positive integer. Similarly, every A ∈ Mn (R) can be written as A = LU W, L ∈ L, U ∈ U, W ∈ W and in L and U all the diagonal

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entries are equal to 1 if and only if R is a left hermitian ring having stable range one.     01 10 Example 1.2.7. Let R be a field. Then 6∈ LU, 6∈ WU and 10 00   10 6∈ WL. Thus we need three matrices in Theorem 1.2.6. Let A = 01     000 001  0 1 0  . Suppose A can be expressed as W LU where W =  0 1 ∗  ∈ 001 1∗∗ W, L ∈ L, U ∈ U and in L all the diagonal entries are equal to 1. Then AU −1 = W L, while the (1, 3)-entry of AU −1 is 0 and the (1, 3)-entry of W L is 1. This gives a contradiction. So A cannot be expressed as the preceding  form.  Likewise, we see that A cannot be expressed as W LU where 001 W =  0 1 ∗  ∈ W, L ∈ L, U ∈ U and in U all the diagonal entries are 1∗∗ equal to 1.  A ring R is local if R/J(R) is a division ring. It is well known that a ring R is local if and only if R has a unique maximal right ideal if and only if x + y = 1 in R implies that either x or y is invertible. For instance, Z/pn Z is local for all prime p ∈ Z, and any n ≥ 1, e.g., Z4 . The ring m R = { | m, n ∈ Z, (m, n) = 1, n is odd } n is local. Let F be a field, the ring   ab R={ |a, b ∈ F } 0a is local. R is local if and only if for any A ∈ GL2 (R),  We  note that a ring   ∗0 0∗ A= B12 (∗)B21 (∗) or B21 (∗)B12 (∗). 0∗ ∗0 1.3

Exchange Rings, I

The class of exchange rings is very large. Examples of exchange rings include semiperfect rings, regular rings, π-regular rings, strongly π-regular rings, right continuous rings, clean rings and C ∗ -algebras of real rank zero.

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19

In [348], O’Meara showed that the ring B(R) is an exchange ring for all regular ring R. Here, B(R) is the ring of all ω × ω row-and-column-finite matrices over any ring R. More generally, the same is true of the multiplier ring of a non-unital regular ring with a countable unit. In this section, we characterize exchange rings having stable range one using cokernels of elements in the rings. Lemma 1.3.1. Let R be an exchange ring. Then the following are equivalent: (1) R has stable range one. (2) Every regular element in R is unit-regular. Proof. (1) ⇒ (2) Let a ∈ R be regular. Then there exists some x ∈ R such that a = axa. Since xa + (1 − xa) = 1, we have a y ∈ R such that u := x + (1 − xa)y ∈ U (R). Hence, a = axa = aua, as desired. (2) ⇒ (1) Assume that ax + b = 1 in R. In view of [382, Theorem 28.7], there exist s, t ∈ R such that e = bs and 1 − e = (1 − b)t for an idempotent e ∈ R. This implies that axt + e = 1, and then (1 − e)axt + e = 1. Hence (1−e)a ∈ R is regular. By hypothesis, there exists some u ∈ U (R) such that (1 − e)a = (1 − e)au(1 − e)a. Let f = u(1 − e)a. Then f xtu−1 + ueu−1 = 1, and so Thus,

f xtu−1 (1 − f ) + ueu−1 (1 − f ) = 1 − f.


u a + bs(u−1 (1 − f ) − a)
= u a + e u−1 (1 − f ) − a)
= u (1 − e)a + eu−1 (1 − f ) = f + ueu−1 (1 − f ) = 1 − f xtu−1 (1 − f ) ∈ U (R).
Therefore a + bs u−1 (1 − f ) − a ∈ U (R), as required.



As is well known, stable range one is Morita invariant. This fact should be contrasted to the following counterexample. Take W = F [x, y, δ0 ] where F is a field of characteristic 0, x, y are variables and δ0 the derivation given by xy − yx = 1, i.e., W is the Weyl-algebra. Then W is a noetherian simple domain. Thus, every regular element in W is unit-regular. Consider the matrix   x0 A= ∈ M2 (W). y0

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Then A ∈ M2 (W) is regular, but not unit-regular (cf. [267]). The following result is due to Yu [422, Theorem 10]. Theorem 1.3.2. Let A be a right R-module, and let E = EndR (A). If A has the finite exchange property, then the following are equivalent: (1) E has stable range one. (2) A = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 implies that B1 ∼ = B2 . (3) For any idempotents e, f ∈ E, eA ∼ = f A implies that (1 − e)A ∼ = (1 − f )A. Proof. (1) ⇒ (2) Given A = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 , then there exists an idempotent e ∈ E such that A1 = eA. Clearly, EndR (A1 ) ∼ = eEe has stable range one. According to Theorem 1.1.5, there is some C ⊆ A such that A = C ⊕ B1 = C ⊕ B2 , and therefore B1 ∼ = B2 . (2) ⇒ (3) For any idempotents e, f ∈ E with eA ∼ = f A, we see that ∼ A = eA ⊕ (1 − e)A = f A ⊕ (1 − f )A. Thus, (1 − e)A = (1 − f )A. (3) ⇒ (1) Given any regular a ∈ E, there exists some x ∈ E such that a = axa. Construct a map ϕ : aA = axA → xaA given by ϕ(ar) = xar for any r ∈ A. It is easy to check that ϕ : aA ∼ = xaA. Thus, ψ : (1 − ax)A ∼ = (1 − xa)A. Construct a map u : A = aA ⊕ (1 − ax)A → xaA ⊕ (1 − xa)A = A given by u(s + t) = ϕ(s) + ψ(t) for any s ∈ aA, t ∈ (1 − ax)A. We claim that a = aua with u ∈ U (E), as asserted.  The finite exchange property of the preceding theorem is necessary. Assume that Z = A1 ⊕B1 = A2 ⊕B2 with A1 ∼ = A2 . Then Z⊕B1 ∼ = Z⊕B2 . ∼ As Z is a principal ideal domain, we get B1 = B2 . That is, Z has the internal cancellation property. Clearly, 2 · 3 − 5 = 1. But 2 − 5y is never a unit in Z. Therefore, Z does not have stable range one. Corollary 1.3.3. Let A be a right R-module, and let E = EndR (A). If A has the finite exchange property, then the following are equivalent: (1) E has stable range one. (2) A = A1 ⊕B1 = A2 ⊕B2 with A1 for some C ⊆ A. (3) A = A1 ⊕B1 = A2 ⊕B2 with A1 for some C ⊆ A.

∼ = A2 implies that A = C ⊕B1 = C ⊕B2 ∼ = A2 implies that A = A1 ⊕C = A2 ⊕C

Proof. (1) ⇒ (2) Given A = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 , then there exists an idempotent e ∈ E such that A1 = eA. Hence, EndR (A1 ) ∼ = eEe

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21

has stable range one. In view of Theorem 1.1.5, A = C ⊕ B1 = C ⊕ B2 for some C ⊆ A. (2) ⇒ (1) Whenever A = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 , we have C ⊆ A such that A = C ⊕ B1 = C ⊕ B2 . Hence B1 ∼ B . = 2 According to Theorem 1.3.2, E has stable range one. (1) ⇔ (3) is proved in the same manner.  Theorem 1.3.4. Let R be an exchange ring. Then the following are equivalent: (1) R has stable range one. (2) For any a, b ∈ R, R/aR such that a = ubv. (3) For any a, b ∈ R, R/Ra such that a = ubv.

∼ = R/bR implies that there exist u, v ∈ U (R) ∼ = R/Rb implies that there exist u, v ∈ U (R)

Proof. (1) ⇒ (2) Since ϕ : R/aR ∼ = R/bR, there exists some c ∈ R such that ϕ(1 + aR) = c + bR. So R + bR = cR + bR; hence, R = cR + bR. Since R has stable range one, there exists some d ∈ R such that c + bd = u ∈ U (R). Clearly, bR = ϕ(aR) = ϕ(aR + aR) = caR + bR, and then caR ⊆ bR. Furthermore, uaR ⊆ bR. On the other hand, we have ϕ(1+aR) = (c+bd)+bR = u+bR. It follows that ϕ−1 (1+bR) = u−1 +aR. This implies that u−1 b + aR = (u−1 + aR)b = ϕ−1 (1 + bR)b = ϕ−1 (bR) = aR. Hence u−1 bR ⊆ aR, and then bR ⊆ uaR. Thus we can find x, y ∈ R such that ua = bx and b = uay. It follows from xy + (1 − xy) = 1 that there exists a z ∈ R such that we deduce that
x + (1 − xy)z = v ∈ U−1(R). Thus −1 bx = b x + (1 − xy)z = bv. As a result, a = u bx = u bv, as desired. (2) ⇒ (1) Given eR ∼ = f R with idempotents e, f ∈ R, then we get R/(1 − e)R ∼ = R/(1 − f )R. By assumption, there exist u, v ∈ U (R) such that 1 − e =
u(1 − f )v. Construct a map ϕ : (1 − e)R → (1 − f )R given by ϕ (1 − e)r = (1 − f )vr for any r ∈ R. Then ϕ : (1 − e)R ∼ = (1 − f )R. In light of Theorem 1.3.2, R has stable range one. (1) ⇔ (3) is obtained by symmetry.  In the proof of Theorem 1.3.4, we prove that an exchange ring R has stable range one if and only if for any regular a, b ∈ R, R/aR ∼ = R/bR implies that there exist u, v ∈ U (R) such that a = ubv if and only if for any regular a, b ∈ R, R/Ra ∼ = R/Rb implies that there exist u, v ∈ U (R) such that a = ubv. We note that Conditions (1) and (2) in Theorem 1.3.4 are

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not equivalent for some non-exchange rings. In [71, Example 6.7], Canfell supplied a principal ideal domain R which has elements a and b for which R/aR ∼ = R/bR but a 6= ubv for any u, v ∈ U (R). Corollary 1.3.5. Let R be an exchange ring. Then the following are equivalent: (1) R has stable range one. (2) For any regular a, b ∈ R, r(a) ∼ = r(b) implies that there exist u, v ∈ U (R) such that a = ubv. (3) For any regular a, b ∈ R, ℓ(a) ∼ = ℓ(b) implies that there exist u, v ∈ U (R) such that a = ubv. Proof. (1) ⇒ (2) Suppose that r(a) ∼ = r(b) with regular a, b ∈ R. Then there exist x, y ∈ R such that a = axa and b = byb. Hence (1 − xa)R = r(a) ∼ = r(b) = (1 − yb)R. As 1−xa, 1−ya ∈ R are idempotents, it follows that R(1−xa) ∼ = R(1−yb). ∼ ∼ Clearly, R(1 − xa) = R/Rxa and R(1 − yb) = R/Ryb. As a result, R/Ra ∼ = R/Rb. In view of Theorem 1.3.4, we can find u, v ∈ U (R) such that a = ubv. (2) ⇒ (1) Given eR ∼ = f R with idempotents e, f ∈ R, then r(1 − e) ∼ = r(1 − f ). By assumption, we can find some u, v ∈ U (R) such that 1 − e = u(1−f )v. Analogously to Theorem 1.3.4, we claim that (1−e)R ∼ = (1−f )R. Using Theorem 1.3.2, we prove that R has stable range one. (1) ⇔ (3) is symmetric.  It follows by Corollary 1.3.5 that a regular ring is unit-regular if and only if r(a) ∼ = r(b) implies that there exist u, v ∈ U (R) such that a = ubv if and only if ℓ(a) ∼ = ℓ(b) implies that there exist u, v ∈ U (R) such that a = ubv. Let a, b ∈ End(Z) be defined by left multiplication by 2 and 3, respectively. Then r(a) ∼ = r(b), while a 6= ubv for any u, v ∈ AutR (Z). Corollary 1.3.6. Let R be an exchange ring having stable range one, and let a, b ∈ R. Then the following are equivalent: (1) ϕ : aR ∼ = bR and ϕ(a) = ua for some u ∈ U (R). (2) There exist v, w ∈ U (R) such that a = vbw. Proof. (1) ⇒ (2) Suppose that ϕ : aR ∼ = bR and ϕ(a) = ua for some u ∈ U (R). Let ψ : R → R be given by ψ(r) = ur for any r ∈ R. Then ψ is an automorphism. So we have φ : R/aR → R/bR such that the following

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diagram 0 → aR ֒→ R → R/aR → 0 ϕ↓ ψ↓ φ↓ 0 → bR ֒→ R → R/bR → 0 commutates. Since ϕ and ψ are both isomorphisms, so is φ. That is, R/aR ∼ = R/bR. According to Theorem 1.3.4, we know that a = vbw for some v, w ∈ U (R). (2) ⇒ (1) Suppose that a = vbw with v, w ∈ U (R). Construct a map ϕ : aR → bR given by ϕ(ar) = v −1 (ar) for any r ∈ R. It is easy to verify that ϕ : aR ∼ = bR. In addition, ϕ(a) = v −1 a, and therefore the result follows.  Example 1.3.7. Let V be an infinite-dimensional vector space over a division ring D, and let R = EndD (V ). Let {x1 , x2 , · · · } be a basis of V. Define σ : V → V by σ(xi ) = xi+1 for i = 1, 2, 3, · · · . Define τ : V → V by τ (x1 ) = 0 and τ (xi ) = xi−1 (i = 2, 3, · · · ). Then τ σ = 1V and στ 6= 1V . ϕ Thus σ and τ are both regular and σR ∼ = τ R with ϕ(σ) = τ σ, while σ 6= uτ v

for any u, v ∈ AutD (V ). In this case, R is an exchange ring but it does not have stable range one. 

A right R-module A is quasi-projective provided that for any Repimorphism g : A → M and any R-morphism f : A → M , there exists h ∈ EndR (A) such that the diagram

A

A hւ↓f g

։M

commutates. Theorem 1.3.8. Let A be a quasi-projective right R-module, and let E = EndR (A). If A has the finite exchange property, then the following are equivalent: (1) E has stable range one. (2) For any U, V ⊆ A, A/U ∼ = A/V implies U ∼ =V. Proof. (1) ⇒ (2) Suppose that ϕ : A/U ∼ = A/V for some U, V ⊆ A. Let α : A → A/U and β : A → A/V be the canonical maps. Since A is

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quasi-projective, there exists λ ∈ E such that the diagram A λ ւ ↓ ϕα A

β

։ A/V

commutates, i.e., βλ
= ϕα. Set M = {(a1 , a2 ) ∈ A ⊕ A | ϕα(a1 ) = β(a2 )} and B = { a1 , λ(a1 ) | a1 ∈ A}. Clearly, B ⊆ M . Set A1 = {(a1 , 0) | a1 ∈ A} and A2 = {(0, a2 ) | a2 ∈ A}. Then M 2 = B ⊕ A2 with
:= A1 ⊕ AT T ∼ ∼ A1 = A = B. Thus, M = M B ⊕ A2 = B ⊕ M A2 ). Obviously, T M A2 = {(0, a2 ) | β(a2 ) = 0} ∼ = ker β; hence, M ∼ = A ⊕ ker β. Likewise, there exists µ ∈ E such that the diagram A µ ր ↓ ϕα A

β

։ A/V


commutates, i.e., ϕαµ = β. Set C = { µ(a2 ), a2 | a2 ∈ A}. Then C ⊆
M . T Clearly, M = C ⊕ A1 with C ∼ A∼ C ⊕ A1 = = A1 . Thus, M = M T T = C ⊕ M A1 ). Further, M A1 = {(a1 , 0) | ϕα(a1 ) = 0} ∼ = ker α; hence, M∼ = A ⊕ ker α. As a result, we get A⊕U ∼ = A⊕V . According to Theorem 1.1.5, U ∼ =V, as required. (2) ⇒ (1) Given A = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ =A∼ = A2 , then A/B1 ∼ = A/B2 .

By hypothesis, B1 ∼ = B2 . In view of Theorem 1.3.2, we obtain the result. Corollary 1.3.9. Let R be an exchange ring. Then the following are equivalent: (1) R has stable range one. (2) For any a, b ∈ R, R/aR ∼ = R/bR =⇒ aR ∼ = bR. (3) For any a, b ∈ R, R/Ra ∼ = R/Rb =⇒ Ra ∼ = Rb. Proof. (1) ⇒ (2) is trivial by Theorem 1.3.8. (2) ⇒ (1) Given eR ∼ = f R with idempotents e, f ∈ R, we see that ∼ R/(1 − e)R = R/(1 − f )R, whence (1 − e)R ∼ = (1 − f )R. In light of Theorem 1.3.2, R has stable range one. (1) ⇔ (3) is symmetric.  Lemma 1.3.10. Let R be a regular ring. Then the following are equivalent:

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(1) R is unit-regular. (2) Whenever ℓ(a) = ℓ(b), there exists u ∈ U (R) such that a = bu. (3) Whenever r(a) = r(b), there exists u ∈ U (R) such that a = ub. Proof. (1) ⇒ (2) Suppose that ℓ(a) = ℓ(b). Write a = axa and b = byb.

As 1 − ax ∈ ℓ(a), we get (1 − ax)b = 0. Thus b = axb. Likewise, a = bya. Hence, aR = bR. According to Corollary 1.1.4, there exists u ∈ U (R) such that a = bu, as desired. (2) ⇒ (1) For any a ∈ R, there exists a c ∈ R such that a = aca. It is easy to verify that ℓ(a) = ℓ(ac). By hypothesis, there exists a u ∈ U (R) such that a = acu. Hence, au−1 a = aca = a. Therefore R is unit-regular. (1) ⇔ (3) is obtained by symmetry.  Theorem 1.3.11. Let R be a regular ring. Then the following are equivalent: (1) R is unit-regular. (2) Whenever ℓ(a) = ℓ(b), there exists u ∈ U (R) such that a = bua. (3) Whenever r(a) = r(b), there exists u ∈ U (R) such that a = aub. Proof. (1) ⇒ (2) Suppose that ℓ(a) = ℓ(b). As R is regular, there exists x ∈ R such that a = axa. Obviously, ℓ(a) = ℓ(ax). Hence, ℓ(ax) = ℓ(b). By virtue of Lemma 1.3.10, we have a u ∈ U (R) such that b = axu; hence, ax = bu−1 . Therefore a = axa = bu−1 a, as desired. (2) ⇒ (1) For any a ∈ R, there exists some x ∈ R such that a = axa. It is easy to verify that ℓ(a) = ℓ(ax). By hypothesis, there exists a u ∈ U (R) such that ax = auax. Hence, a = axa = auaxa = aua. Therefore R is unit-regular. (1) ⇔ (3) follows by symmetry.  Corollary 1.3.12. Let R be a regular ring. Then the following are equivalent: (1) R is unit-regular. (2) Whenever ϕ : Ra ∼ = Rb, there exists u ∈ U (R) such that a = ϕ(a)ua.

Proof. (1) ⇒ (2) Suppose that ϕ : Ra ∼ = Rb. As R is regular, there exists some x ∈ R such that a = axa. Thus, we see that ϕ(a) = ϕ(axa) = aϕ(xa) ⊆ aR.

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On the other hand,
we have a y ∈ R such that ϕ(a) = ϕ(a)yϕ(a); hence, ϕ(a) = ϕ ϕ(a)ya . Since ϕ is an R-isomorphism, we get a = ϕ(a)y. Thus,
aR ⊆ ϕ(a)R. Therefore aR = ϕ(a)R. This implies that ℓ(a) = ℓ ϕ(a) . According to Theorem 1.3.11, there exists a u ∈ U (R) such that a = ϕ(a)ua, as required. (2) ⇒ (1) For any a ∈ R, there exists some x ∈ R such that a = axa and x = xax. It is easy to verify that ϕ : Rax ∼ = Rxa given by ϕ(rax) = (rax)a. By hypothesis, there exists a u ∈ U (R) such that ax = ϕ(ax)uax = axauax. Therefore a = axa = (axauax)a = aua, i.e., a ∈ R is unit-regular. As a result, R is unit-regular.  Let R be a unit-regular ring, and let a, c, d ∈R. Itis worth noting  that  ac a0 ax − yd = c has solutions x, y ∈ R if and only if is similar to 0d 0d (cf. [249]). The following result is due to Yu [422, Theorem 1]. Theorem 1.3.13. Every exchange ring with artinian primitive factors has stable range one. Proof. Assume that R does not have stable range one. Then there exist some a, b ∈ R such that aR + bR = R, while a + by 6∈ U (R) for any y ∈ R. Let Ω = {I E R | a + by 6∈ U (R/I) for any y ∈ R}. Then Ω 6= ∅. Suppose we are given an ascending chain A1 ⊆ A2 ⊆ · · · ⊆ ∞ S An ⊆ · · · in Ω. Set M = An . Then M is an ideal of R. If M is not n=1

in Ω, then there exists some s ∈ R such that a + bs ∈ U (R/M ). Hence (a + bs)u − 1, u(a + bs) − 1 ∈ M for some u ∈ R, and so we can find positive integers n1 , n2 such that (a + bs)u − 1 ∈ An1 and u(a + bs) − 1 ∈ An2 . Let p = max{n1 , n2 }. Then we see that a + bs ∈ U (R/Ap ). This contradicts the choice of Ap , so M ∈ Ω. Thus, Ω is inductive. By using Zorn’s Lemma, we have an ideal Q of R such that Q is maximal in Ω. Let S = R/Q. If R/Q is decomposable, then there exist two ideals A, B ⊆ R such that R/Q = A/Q ⊕ B/Q with A, B % Q. For any r ∈ R, write r + Q = (r0 , r∗ ), where r0 ∈ A/Q, r∗ ∈ B/Q. Construct two maps ()∗ : R/A r+A ()0 : R/B r+B

→ B/Q, 7 r∗ , ∀r ∈ R; → → A/Q, 7→ r0 , ∀r ∈ R.

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Clearly, B/Q and A/Q are rings with identities 1∗ and 10 , respectively. In addition, one easily checks that ()∗ and ()0 are both ring isomorphisms. By the maximality of Q ∈ Ω, we have some y, z ∈ R such that a + by ∈ U (R/A) and a + bz ∈ U (R/B). Applying ()∗ and ()0 , we get a∗ + b∗ y ∗ ∈ U (B/Q) and a0 + b0 z 0 ∈ U (A/Q).

Write (z 0 , y ∗ ) = d + Q for some d ∈ R. Moreover,

a + bd = (a0 , a∗ ) + (b0 , b∗ )(z 0 , y ∗ ) = (a0 + b0 z 0 , a∗ + b∗ y ∗ ) ∈ U (R/Q), contradicting the choice of Q. Therefore S is indecomposable as a ring. If J(S) 6= 0, we may assume that J(S) = N/Q with N % Q. By the maximality of Q, there exists some t ∈ R such that a + bt ∈ U (R/N ). Obviously, there exists a natural epimorphism ϕ : R/Q ։ R/N given by ϕ(r + Q) = r + N for any r ∈ R. Thus, S/ker(ϕ) ∼ = R/N , where
ker(ϕ) = J(S), and so we have (a + Q) + (b + Q)(t + Q) ∈ U S/J(S) . This implies that (a + Q) + (b + Q)(t + Q) ∈ U (S), a contradiction. Thus we see that J(S) = 0, so S is an indecomposable ring with J(S) = 0. Since R is an exchange ring with artinian primitive factors, by virtue of [424, Lemma 3.7], S is simple artinian. In view of [4, Theorem 13.4], R/Q ∼ = Mn (D) for a division ring D, whence, R/Q has stable range one. This contradicts the choice of Q and completes the proof.  Recall that a ring R is of bounded index provided that there exists n ∈ N such that xn = 0 for any nilpotent x ∈ R. Corollary 1.3.14. Every exchange ring of bounded index has stable range one. Proof. Let P be a primitive ideal of R, let S = R/P and let n be the bounded index of R. Then S is an exchange ring as well. Suppose that e1 , e2 , · · · , en+1 are orthogonal idempotents in S. In view of [334, Proposition 1.12], there exist orthogonal idempotents f1 , f2 , · · · , fn+1 in R such that each ei = fi + P . Without loss of generality we may assume that each fi 6∈ P . Since P is primitive, it is prime, so there exist x1 , x2 , · · · , xn ∈ R such that f1 x1 f2 , f1 x1 f2 x2 f3 , · · · , f1 x1 f2 x2 f3 · · · fn xn fn+1 6∈ P. Put x = f1 x1 f2 +f1 x1 f2 x2 f3 +· · ·+f1 x1 f2 x2 f3 · · · fn xn fn+1 . Then xn+1 = 0; hence, xn = 0. But xn = f1 x1 f2 x2 f3 · · · fn xn fn+1 6∈ P , a contradiction. Thus, S is an exchange ring which contains no infinite set of orthogonal idempotents. According to [66, Corollary 7], S is semiperfect; whence, S/J(S) is

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artinian. As P is primitive, J(S) = 0, and therefore S is artinian. In view of Theorem 1.3.13, R has stable range one.  Corollary 1.3.15. Every exchange ring with all idempotents central has stable range one. Proof. Since R is an exchange ring with all idempotents central, so is R/J(R). In view of [372, Lemma 4.10], R/J(R) is of bounded index 1. According to Corollary 1.3.14, R/J(R) has stable range one, and therefore the proof is true.  A ring R is semiregular in the case when R/J(R) is regular and every idempotent lifts modulo J(R). We end this section with the construction of an exchange ring having stable range one which is not semiregular. Example 1.3.16. Let Q be the field of rational numbers and L be the ring of all rational numbers with odd denominators. Define R(Q, L) = {(r1 , · · · , rn , s, s, · · · ) | n ≥ 1, r1 , · · · , rn ∈ Q, s ∈ L}. With componentwise operations, R(Q, L) is a ring. In view of [334, Example 1.7], R(Q, L) is a commutative exchange ring, while it is not semiregular.  1.4

Exchange Rings, II

By virtue of Theorem 1.3.2, an exchange ring R has stable range one if and only if R = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 implies that B ∼ = C. We now extend this fact as follows. Lemma 1.4.1. Let R be an exchange ring. Then the following are equivalent: (1) R has stable range one. (2) R = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 implies that B .⊕ C. Proof. (1) ⇒ (2) is trivial by Theorem 1.3.2. (2) ⇒ (1) Let a ∈ R be regular. Then we can find an x ∈ R such that a = axa and x = xax. So there are right R-module decompositions R = axR ⊕ (1 − ax)R = xaR ⊕ (1 − xa)R. Clearly, ϕ : aR = axR ∼ = xaR given by ϕ(ar) = xar for any r ∈ R. By assumption, we get (1 − ax)R .⊕ (1 − xa)R. Thus we have R-morphisms α : (1 − ax)R → (1 − xa)R and β : (1 − xa)R → (1 − ax)R such that βα = 1(1−ax)R . Construct a map θ :

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R → R given by θ(x1 , x2 ) = ϕ(x1 ) + α(x2 ) for any x1 ∈ aR, x2 ∈ (1 − ax)R. One easily checks that a = aua, and that u = θ(1) ∈ R is left invertible. Clearly, u ∈ R is regular. By the above consideration, we can find a left invertible v ∈ R such that u = uvu. As a result, vu = 1. Hence v ∈ U (R), and so u ∈ U (R). Thus, a ∈ R is unit-regular. In view of Lemma 1.3.1, R has stable range one.  Theorem 1.4.2. Let R be an exchange ring. Then the following are equivalent: (1) R has stable range one. (2) For any idempotent e ∈ R, aR + b(eR) = R implies that there exist u, v ∈ R such that au + b(ev) = 0 and (eR)u + (eR)(ev) = eR. Proof. (2) ⇒ (1) Given right R-module decompositions R = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 , then R ⊕ B ∼ = R ⊕ C and B = eR for an idempotent e ∈ R. Clearly, we have a splitting exact sequence σ

π

0 → C → R ⊕ B → R → 0. So there exists a map τ : R → R ⊕ B such that πτ = 1R . Let π1 = π |R and π2 = π |B . Let ϕ1 : R ⊕ B → R and ϕ2 : R ⊕ B → B be projections. Then one easily checks that π1 τ1 + π2 τ2 = 1R , where τ1 = ϕ1 τ and τ2 = ϕ2 τ. Thus, π1 τ1 + (π2 e)(eτ2 ) = 1R ; hence, π1 R + (π2 e)R = R. By assumption, we can find u, v ∈ R such that π1 u + (π2 e)(ev) = 0 and (eR)u
+ (eR)(ev) = eR. Let η : B → R ⊕ B be given by η(b) = ub, (ev)b for any b ∈ B. Clearly, πη = π1 u + (π2 e)(ev) = 0. Thus, there exists a right R-morphism θ : B → C such that η = σθ. Obviously, we can find some s, t ∈ R such that esu + etev = e. Let ζ : R ⊕ B → B be given by ζ(r, b) = esr + etb for any r ∈ R, b ∈ B. It is easy to verify that ζη = 1B , and so ςσθ = 1B . As a result, θ : B → C is an R-monomorphism. Therefore we see that B .⊕ C. According to Lemma 1.4.1, R has stable range one. (1) ⇒ (2) Suppose that aR + b(eR) = R for some idempotent e ∈ R. Then we can find some s, t ∈ R such that as + bet = 1. Choose B = eR. Let π : R ⊕ B → R be given by π(r, c) = ar + bec for any r ∈ R, c ∈ B, and let τ : R → R ⊕ B be given by τ (r) = (sr, etr) for any r ∈ R. Then πτ = 1R , and so we get a splitting exact sequence σ

π

0 → ker(π) ֒→ R ⊕ B → R → 0.

Thus, R ⊕ ker(π) ∼ = R ⊕ B. In view of Corollary 1.1.7, θ : B ∼ = ker(π). So we have a splitting R-morphism σθ : B → R ⊕ B, and then η(σθ) = 1B for some η : R ⊕ B → B.

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Let ϕ1 : R ⊕ B → R, ϕ2 : R ⊕ B → B be projections, and let φ1 : R → R⊕B, φ2 : B → R⊕B be injections. Then (ηφ1 )(ϕ1 σθ)+(ηφ2 )(ϕ2 σθ) = 1B . It follows from πσ = 0 that a(ϕ1 σθ) + be(ϕ2 σθ) = 0. Let u = ϕ1 σθ(e) and v = ϕ2 σθ(e). Then au + (be)(ev) = 0. Furthermore, we have (ηφ1 )(ϕ1 σθ)(e) + (ηφ2 )(ϕ2 σθ)(e) = e. Let s = (ηφ1 )(1) and t = (ηφ2 )(e). Then su + t(ev) = e and s, t ∈ eR. As a result, (eR)u + (eR)(ev) = eR, and therefore the result follows.  Corollary 1.4.3. Let R be an exchange ring. Then the following are equivalent: (1) R has stable range one. (2) For any idempotent e ∈ R, Ra + (Re)b = R implies that there exist u, v ∈ R such that ua + (ve)b = 0 and u(Re) + (ve)(Re) = Re. Proof. Clearly, R is an exchange ring if and only if so is the opposite ring Rop (cf. [335, Proposition]). Furthermore, R has stable range one if and only if Rop also has stable range one. Applying Theorem 1.4.2 to Rop , we obtain the result.  If R is an exchange ring having stable range one, it follows from Theorem 1.4.2 and Corollary 1.4.3 that the following hold: (1) aR + bR = R =⇒ there exist u, v ∈ R such that au + bv = 0 and Ru + Rv = R; (2) Ra + Rb = R =⇒ there exist u, v ∈ R such that ua + vb = 0 and uR + vR = R. But the converse is not true. Recall that a ring R is directly finite if xy = 1 implies yx = 1, for all x, y ∈ R. Of course any ring which is not directly finite is said to be directly infinite. Let R be the regular ring constructed in [22, Example 3.2]. Then R is directly finite and satisfies 2-comparability. By [22, Theorem 4.6], R has cancellation of small projectives, i.e., R ⊕ R ∼ = R ⊕ C =⇒ R ∼ C. Analogously, one easily checks that Conditions (1) and = (2) hold. By [22, Example 3.2] again, we see that R does not have stable range one. Theorem 1.4.4. Let R be an exchange ring. Then the following are equivalent: (1) R has stable range one. (2) For any idempotent e ∈ R, aR + b(eR) = R implies that there exist

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u, v ∈ R such that as + b(et) = 0 if and only if s = uz and t = vz for some z ∈ eR. Proof. (2) ⇒ (1) Given right R-module decompositions R = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 , then B = eR for an idempotent e ∈ R and we have a splitting exact sequence σ

π

0 → C → R ⊕ B → R → 0. So there is a map τ : R → R ⊕ B such that πτ = 1R . Let π1 = π |R and π2 = π |B . Analogously to Theorem 1.4.2, π1 R + (π2 e)(eR) = R. By assumption, we can find u, v ∈ R such that π1 s + (π2 e)(et) = 0 if and only if s = uz and t = vz for some z ∈ eR. We infer that
π1 ue + (π2 e)(eve) = 0. Let η : B → R ⊕ B be given
by η(b) = ub, (ev)b for any b ∈ B. Then πη(B) ⊆ π1 ue + (π2 e)(eve) (B) = 0, and so η(B) = im(η) ⊆ ker(π). If (s, t) ∈ ker(π), then π1 s + (π2 e)(et) = 0. By hypothesis, we can find
some z ∈ eR such that s = uz and t = vz. As a result, (s, t) = uz, (ev)z ∈ η(B); hence, C ∼ = ker(π) = η(B). Thus, there exists a right R-epimorphism θ : B → C. As C is projective, θ splits, and then C .⊕ B. In view of Lemma 1.4.1, R has stable range one. (1) ⇒ (2) Suppose that aR + b(eR) = R with an idempotent e ∈ R. Then ax + bey = 1 for some x, y ∈ R. Choose B = eR. Let π : R ⊕ B → R be given by π(r, c) = ar + bec for any r ∈ R, c ∈ B, and let τ : R → R ⊕ B be given by τ (r) = (xr, eyr) for any r ∈ R. Then πτ = 1R , and so we get a splitting exact sequence σ

π

0 → ker(π) ֒→ R ⊕ B → R → 0.

By virtue of Corollary 1.1.7, θ : B ∼ = ker(π). Let ϕ1 : R ⊕ B → R and ϕ2 : R ⊕ B → B be projections. Let u = ϕ1 σθ(e) and v = ϕ2 σθ(e). Then au + b(ev) = 0. If s = uz and t = vz for some z ∈ eR, then as + b(et) = 0. Conversely, assume that as + b(et) = 0 for some s, t ∈ R. Then (s, t) ∈ ker(π), and so (s, t) = θ(ez) for some z ∈ eR. Clearly, s = ϕ1 (s, t) = ϕ1 σθ(ez) = uz. Likewise, t = vz. This completes the proof.  Corollary 1.4.5. Let R be an exchange ring. Then the following are equivalent: (1) R has stable range one. (2) For any idempotent e ∈ R, Ra + (Re)b = R implies that there exist u, v ∈ R such that sa + (te)b = 0 if and only if s = zu and t = zv for some z ∈ eR.

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Proof. Obviously, R is an exchange ring having stable range one if and only if so is the opposite ring Rop . Applying Theorem 1.4.4 to Rop , we obtain the result.    Z/2Z Z/2Z Let R = . Then R is an exchange ring of bounded index, 0 Z/2Z   10 and so it has stable range one by Corollary 1.3.14. Choose a = and 00     11 10 b= . Then aR + bR = R. In this case, we can find u = and 01 01   10 v= such that as + bt = 0 if and only if s = uz and t = vz for some 00 z ∈ R. Lemma 1.4.6. Let R be an exchange ring. Then the following are equivalent: (1) R has stable range one. (2) For any idempotents e, f ∈ R, eR ∼ = f R implies that there exists some u ∈ U (R) such that e = uf u−1 . (3) For any idempotents e, f ∈ R, ϕ : eR ∼ = f R implies that there exists u ∈ U (R) such that e = euϕ(e). ∼ f R implies Proof. (1) ⇒ (2) For any idempotents e, f ∈ R, ϕ : eR = that Re = Rϕ(e) and ϕ(e)R = f R. Thus, we can find some x, y ∈ R such that e = xϕ(e) and ϕ(e) = ye. Since yx + (1 − yx) = 1, there is an element z ∈ R
such that v := x + z(1 − yx) ∈ U (R). Hence, e = xϕ(e) = x + z(1 − yx) ϕ(e) = vϕ(e). Likewise, we have an element w
∈ U (R) −1 such that ϕ(e) = f w. So v. Then
e = vf w. Let u = 1 − e + vf v −1 −1 −1 u =v 1 + e − vf v . Furthermore, eu = uf , and then e = uf u−1 . (2) ⇒ (1) For any idempotents e, f ∈ R, eR ∼ = f R implies that there exists an element u ∈ U (R) such that e = uf u−1. Hence, u−1 (1 − e) = (1 − f )u−1
. Construct an R-morphism ϕ : (1 − e)R → (1 − e)R given by ϕ (1 − e)r = u−1 (1 − e)r for any r ∈ R. It is easy to check that ϕ is an R-isomorphism, i.e., (1 − e)R ∼ = (1 − f )R. According to Theorem 1.3.2, R has stable range one. (1) ⇒ (3) Given ϕ : eR ∼ = f R with idempotents e, f ∈ R, we get Re = Rϕ(e) and ϕ(e)R = f R. We infer that ϕ(e) ∈ R is regular, whence
ϕ(e) = ϕ(e)uϕ(e) for a u ∈ U (R). Thus, ϕ(e) = ϕ euϕ(e) . As a result, e = euϕ(e).

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(3) ⇒ (1) For any regular a ∈ R, there exists some x ∈ R such that a = axa and x = xax. Thus, ϕ : xaR ∼ = axR given by ϕ(xar) = a(xar). Obviously, xa, ax ∈ R are both idempotents. By assumption, there exists a u ∈ U (R) such that xa = xauϕ(xa) = xauaxa. We infer that a = axa = aua. Therefore R has stable range one.  The following result is due to Nicholson [335, Theorem]. Lemma 1.4.7. A ring R is an exchange ring if and only if for any a ∈ R, there exist idempotents e, f ∈ R such that (1) e ∈ aR, 1 − e ∈ (1 − a)R; (2) f ∈ Ra, 1 − f ∈ R(1 − a). Proof. Let R be an exchange ring, and let a ∈ R. Then we have an idempotent e ∈ aR such that 1 − e ∈ (1 − a)R. Set g = 1 − e and b = 1 − a. Then e ∈ aR and g ∈ bR. Suppose that e = ar′ and g = bs′ . Set r = r′ e and s = s′ g. Then we have rar = r, sbs = s, rbs = 0 and sar = 0. Let r′′ = 1 − sb + rb and s′′ = 1 − ra + sa. Then we see that ar′′ = a(1 − sb) + arb = a(1 − sb) + (1 − bs)b = a(1 − sb) + b(1 − sb) = 1 − sb. It is easy to check that r′′ s = 0 and s′′ r = 0, and so r′′ ar′′ = r′′ . Likewise, bs′′ = 1 − ra, and then s′′ bs′′ = s′′ . As ab = ba, we get r′′ a + s′′ b = (1 − sb + rb)a + (1 − ra + sa)b = a + b = 1. Let f = r′′ a. Thus, we have f = f 2 ∈ Ra such that 1 − f = s′′ b ∈ R(1 − a), as required. The converse is obvious.  Proposition 1.4.8. Let R be an exchange ring. If R has stable range one, then for any a ∈ R, there exist idempotents e, f ∈ R such that (1) e ∈ aR, 1 − e ∈ (1 − a)R; (2) f ∈ Ra, 1 − f ∈ R(1 − a) and (3) e = uf u−1 for some u ∈ U (R). Proof. Using the notation of the proof of Lemma 1.4.7, there are idempotents e, f ∈ R such that (1) and (2) hold. As s′′ bR ∼ = bs′′ R = gR, ′′ in view of Lemma 1.4.6, we have gv = vs b for some v ∈ U (R). Hence ev = (1 − g)v = v(1 − s′′ b). On the other hand, 1 − s′′ b = ar′′ ; hence, we get (1 − s′′ b)R ∼ = r′′ aR = f R. By Lemma 1.4.6 again, we have (1 − s′′ b)w = wf for some w ∈ U (R). Therefore evw = v(1 − s′′ b)w = vwf . Set u = vw. Then e = uf u−1 , as required. 

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Recall that K and L ∈ L(A) are perspective in L(A) if there exists some right R-module D such that A = K ⊕ D = L ⊕ D, where L(A) denotes the set of direct summands of the right R-module A. Lemma 1.4.9. Let A be a finitely generated projective right module over a regular ring R, and let B, C ∈ L(A). Then the following are equivalent: (1) B andTC are perspective T in L(A). (2) B/(B C) ∼ = C/(B C). T
T
Proof. Clearly, B = B ′ ⊕ B C and C = C ′ ⊕ B C for some right R-modules B ′ and C ′ . (1) ⇒ (2) As B and C are perspective in L(A), we have a right
RT module D such that
A = B ⊕ D = C ⊕ D. Thus, A = B ′ ⊕ (B C)⊕ D = T C ′ ⊕ (B C) ⊕ D , and therefore B ′ ∼ = C ′ , as required. (2) ⇒ (1) By hypothesis, there exists an isomorphism f : B ′ → C ′ . Let D = {x + f (x) | x ∈ B ′ }. It is easy to verify that B + C = B ⊕ D = C ⊕ D. Since
R is regular, there exists
a right R-module
E such that A = B + C ⊕ E. Thus, A = B ⊕ D ⊕ E = C ⊕ D ⊕ E , and so B and C are perspective.  The following result gives a characterization of unit-regular ring by means of perspectivity, which was shown by Goodearl [217, Theorem 4.24]. Proposition 1.4.10. Let R be a regular ring. Then the following are equivalent: (1) R is unit-regular. (2) Perspectivity is transitive in L(2R). Proof. (1) ⇒ (2) Assume that 2R = B⊕E = C⊕E and 2R = C⊕F = D⊕F for right R-modules E and F . Then 2R = B ⊕ E = D ⊕ F with B ∼ = D. In view of Lemma 1.3.1, R has stable range one. Hence, EndR (B) has stable range one from Corollary 1.1.6. By virtue of Theorem 1.1.5, there exists G ⊆ 2R such that 2R = B ⊕ G = D ⊕ G, as desired. (2) ⇒ (1) Assume that eR ∼ = f R with idempotents e, f ∈ R. Set B = {(x, y) ∈ 2R | x ∈ (1 − e)R}, C = {(x, y) ∈ 2R | y ∈ (1 − e)R} and T D = {(x, y) ∈ 2R | x ∈ (1 − f )R}. Clearly, B C ∼ = (1 − e)R ⊕ (1 − e)R. ∼ ∼ Since B/(B ∩ C) = R/(1 − e)R = C/(B ∩ C), B and C are perspective in T L(2R) by Lemma 1.4.9. Obviously, C D = (1 − f )R ⊕ (1 − e)R, thus,
C/ C ∩ D) ∼ = R/(1 − f )R ∼ = fR ∼ = eR ∼ = R/(1 − e)R ∼ = D/ C ∩ D .

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By Lemma 1.4.9 again, C and D are perspective in L(2R). Thus, B and D are perspective in L(2R). Using Lemma 1.4.9, B/(B ∩ D) ∼ = D/(B ∩ D). That is,

∼ (1 − f )R/ (1 − e)R ∩ (1 − f )R . (1 − e)R/ (1 − e)R ∩ (1 − f )R =

Assume that (1 − e)R = (1 − e)R ∩ (1 − f )R ⊕ N1 and (1 − f )R = (1 − e)R ∩ (1 − f )R ⊕ N2 for right R-modules N1 and N2 . Then N1 ∼ = N2 , and therefore (1 − e)R ∼ (1 − f )R. According to Theorem 1.3.2, R has stable = range one.  Lemma 1.4.11. Let J be an ideal of an exchange ring R, and let A, B ∈ F P (R). Then the following hold: (1) If A/AJ .⊕ B/BJ, then there exist right R-module decompositions A = A1 ⊕ A2 and B = B1 ⊕ B2 with A1 ∼ = B1 and A2 = A2 J. (2) If A/AJ ∼ = B/BJ, then there exist right R-module decompositions A = A1 ⊕ A2 and B = B1 ⊕ B2 with A1 ∼ = B1 , A2 = A2 J and B2 = B2 J. Proof. (1) Since A and B are both finitely generated projective right Rmodules, we can find some idempotents e, f ∈ Mn (R) such that A ∼ = e(nR) and B ∼ f (nR). Because M (R) is also an exchange ring, it is enough to = n solve the n = 1 case. Since eR/eJ .⊕ f R/f J, we can find x ∈ eRf and y ∈ f Re such that xy ≡ e(mod J). Since eRe is an exchange ring with xy ∈ eRe, by [382, Theorem 28.7], there exists g = g 2 ∈ xyRe such that e − g ∈ (e − xy)Re. So e − g ∈ J and g = xyt for some t ∈ eRg. Set h = ytx. Then h = h2 ∈ f Rf . Since g = xyt and h = ytx, we see that gR ∼ = hR as right R-modules. On the other hand, we claim that (e−g)R = (e−g)RJ. Thus A = gR⊕(e−g)R and B = hR ⊕ (f − h)R such that gR ∼ = hR and (e − g)R = (e − g)RJ, as desired. (2) Construct e, f, x, y, t, g, h as in (1), then yx ≡ f (mod J). Hence, e ≡ g ≡ et = t(mod J). In addition, f ≡ yx = yex ≡ ytx = h(mod J). Therefore (f − h)R = (f − h)RJ, as required.  Lemma 1.4.12. Let R be an exchange ring having stable range one, let I, J and K be ideals of R such that K = IJ, and let A and B be finitely generated projective right R-modules. ∼ B/BI and A/AJ ∼ (1) If A/AI = = B/BJ, then A/AK ∼ = B/BK. (2) If A/AI .⊕ B/BI and A/AJ .⊕ B/BJ, then A/AK .⊕ B/BK.

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Proof. (1) Suppose that A/AI ∼ = B/BI and A/AJ ∼ = B/BJ. In view of Lemma 1.4.11, we can find right R-modules A1 , A2 , B1 , B2 such that A ∼ = A1 ⊕ B1 and B ∼ = A2 ⊕ B2 with A1 ∼ = A2 , B1 = B1 I, B2 = B2 I. Clearly, B1 J = B1 K and B2 J = B2 K. Thus, A1 /A1 J ⊕ B1 /B1 K ∼ = A/AJ ∼ = B/BJ ∼ = A2 /A2 J ⊕ B2 /B2 K

with A1 /A1 J ∼ = A2 /A2 J. Since R has stable range one, we get B1 /B1 K ∼ = B2 /B2 K by Corollary 1.1.7. Therefore A/AK ∼ = A1 /A1 K ⊕ B1 /B1 K ∼ = A2 /A2 K ⊕ B2 /B2 K ∼ = B/BK. (2) Suppose that A/AI .⊕ B/BI and A/AJ .⊕ B/BJ. In view of Lemma 1.4.11, we can find right R-modules A1 , A2 , B1 , B2 such that A ∼ = ∼ ∼ A1 ⊕ B1 and B = A2 ⊕ B2 with A1 = A2 , B1 = B1 I. Thus, A1 /A1 J ⊕ B1 /B1 J ∼ = A/AJ .⊕ B/BJ ∼ = A2 /A2 J ⊕ B2 /B2 J

with A1 /A1 J ∼ = A2 /A2 J. This gives that B1 /B1 J .⊕ B2 /B2 J. Hence, there exists a splitting epimorphism ϕ : B2 /B2 J ։ B1 /B1 J. Clearly, there exists an R-epimorphism ψ : B2 /B2 K ։ B2 /B2 J given by ψ(b + B2 K) = b+B2 J for any b+B2 K ∈ B2 /B2 K. As B1 J = B1 K, we can find a splitting R-epimorphism ϕψ : B2 /B2 K ։ B1 /B1 K. This implies that B1 /B1 K .⊕ B2 /B2 K. Thus, A/AK ∼ = A1 /A1 K ⊕ B1 /B1 K .⊕ A2 /A2 K ⊕ B2 /B2 K ∼ = A/AK.  Let P be an ideal of a ring R. We say that P is prime provided that for any ideals I, J, IJ ⊆ P implies that I ⊆ P or J ⊆ P . Proposition 1.4.13. Let R be an exchange ring having stable range one, and let A and B be finitely generated projective right R-modules. (1) If A/AP ∼ = B/BP for all prime ideals P of R, then A ∼ = B. (2) If A/AP .⊕ B/BP for all prime ideals P of R, then A .⊕ B. Proof. (1) Assume that A 6∼ = B. Set

Ω = {M | J is an ideal of R such that A/A ∼ 6 B/BJ}. =

Clearly, Ω 6= ∅. Given I1 ⊆ I2 · · · in Ω, then M =

∞ S

i=1

Ii is an ideal of R.

If M 6∈ Ω, then A/AM ∼ = B/BM . We may assume that A ∼ = e(nR) and ∼ B = f (nR) where e, f ∈ Mn (R) are idempotents. Thus, eMn (R/M ) ∼ = f Mn (R/M ). Hence, we have some a, b ∈ Mn (R) such that e = ab and f = ba. So we can find some i, j ∈ N such that e − ab ∈ Mn (Ii ) and

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f − ba ∈ Mn (Ij ). Choose p = max{i, j}. Then e = ab and f = ba in Mn (R/Ip ). This gives a contradiction. Therefore M ∈ Ω. Consequently, Ω is inductive. Thus we can find some Q ∈ Ω such that it is maximal in Ω. Obviously, Q is not prime. So we have ideals K and L such that Q $ K, L and KL ⊆ Q, whence, (A/AQ)/(A/AQ)(K/Q) ∼ = A/AK ∼ = B/BK ∼ = (B/BQ)/(B/BQ)(K/Q).

Likewise, we get (A/AQ)/(A/AQ)(L/Q) ∼ = (B/BQ)/(B/BQ)(L/Q). As KL ⊆ Q, we see that (K/Q)(L/Q) = 0. It follows from Lemma 1.4.12 that A/AQ ∼ = B/BQ, a contradiction. Thus A ∼ = B. (2) is similar to the above discussion.  Corollary 1.4.14. Let R be an exchange ring having stable range one, and let A be a finitely generated projective right R-module. If A/AP can be generated by n elements for all prime ideals P of R, then A can be generated by n elements. Proof. For all P ∈ Spec(R), A/AP can be generated by n elements; hence, there exists a right R/P -module Q such that A/AP ⊕ Q ∼ = n(R/P ). Thus, ⊕ ∼ A/AP . n(R/P ) = (nR)/(nR)P . In view of Proposition 1.4.13, A .⊕ nR, as desired.  We have seen that the cancellation of modules does not imply stable range one for non-exchange rings. We now give a large class of modules with cancellation. Let A and B be right R-modules, and let aA + bB = A with a ∈ EndR (A) and b ∈ HomR (B, A). Then we denote the submodule {r ∈ A | a(r) ∈ bB} of A by A(a,b) . Proposition 1.4.15. Let R be a ring, and let A, B be right R-modules. If for any a ∈ EndR (A), b ∈ HomR (B, A), aA + bB = A implies that there exists an R-morphism τ : A(a,b) → B such that a |A(a,b) +bτ : A(a,b) → bB is an R-isomorphism, then for any right R-modules C, A⊕B ∼ = A ⊕ C =⇒ B ∼ = C.

Proof. Assume that A ⊕ B ∼ = A ⊕ C. Then we have a splitting exact sequence i

ϕ

0 → C ֒→ A ⊕ B → A → 0,

whereϕ = (a, b), a ∈ EndR (A) and b ∈ HomR (B, A). Construct a map c θ = : A → A ⊕ B such that ϕθ = 1A , where c ∈ EndR (A), d ∈ d

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HomR (A, B). Hence, ac + bd = 1A , and so aA + bB = A. By assumption, there exists a map τ : A(a,b) → B such that u := a |A(a,b) +bτ : A(a,b) → bB is an R-isomorphism. Let M = A(a,b) ⊕ B. Construct an R-morphism  −1  u ψ = : bB → M. Clearly, a |A(a,b) u−1 + bτ u−1 = 1bB , whence, τ u−1 (ϕ |M )ψ = 1bB , and so M = ker(ϕ |M ) ⊕ im(ψ). Obviously, ker(ϕ |M ) ⊆ ker(ϕ). If (r1 , r2 ) ∈ ker(ϕ), then a(r1 ) + b(r2 ) = 0 with r1 ∈ A, r2 ∈ B. Then a(r1 ) ∈ bB; hence, r1 ∈ A(a,b) . As a result, (r1 , r2 ) ∈ M , and so (r1 , r2 ) ∈ ker(ϕ |M ). This implies that ker(ϕ) = ker(ϕ |M ). So M = ker(ϕ) ⊕ im(ψ). On the other hand, we have σψ = 1bB , where σ = (u, 0) : M = A(a,b) ⊕ B → bB. Consequently, we deduce that M = ker(σ) ⊕ im(ψ) = B ⊕ im(ψ), and so B ∼ = ker(ϕ) ∼ = C. Therefore we obtain the result.  Let aR + bB = R, where a ∈ R, b ∈ HomR (B, R), B is a right Rmodule. If R is a commutative ring, then R(a,b) = bB. Assume that R has stable range one. Then there exists an element y ∈ HomR (R, B) such that a + by ∈ U (R). Let τ = y|R(a,b) . Then a|R(a,b) + bτ : R(a,b) → bB is an isomorphism. Thus, the condition in Proposition 1.4.15 is an extension of stable range one for commutative rings. If R is a ring without zero divisors, T i.e., a domain, we note that R(a,b) is isomorphic to R or aR bB. Corollary 1.4.16. Let R be a commutative ring, and let B be a right R-module. If bB is projective for any b ∈ HomR (B, R), then for any right R-module C, R⊕B ∼ = R ⊕ C =⇒ B ∼ = C.

Proof. Let aR + bB = R with a ∈ R, b ∈ HomR (B, R). By assumption, R(a,b) = bB is a projective right R-module. Using
the Dual Basis Theorem, there exist {xi } ⊆ R(a,b) , fi ∈ HomR R(a,b) , R such that for any x ∈ P R(a,b) , x = xi fi (x), where only finite fi (x) are not zero. Let a∗ : B → B i

be given by a∗ (r) = ra for any r ∈ B. As R is commutative, a∗ is an R-morphism. It follows from aR + bB = R that 1 = ac + bd for some c ∈ R, d ∈ B. Hence, xi = axi c + bdxi ∈ bB. Thus, there exists pi
∈ B such P P that xi = b(pi ) for each i. Hence x = b(pi )fi (x) = b pi fi (x) . Define i i P a map h : R(a,b) → B given by h(p) = pi fi (p) for any p ∈ R(a,b) . Clearly, i

1R(a,b) = bh. One easily checks that a |R(a,b)
+b(1B − a∗ )h : R(a,b) → bB is an R-morphism. If a |R(a,b) +b(1B − a∗ )h (p) = 0 for some p ∈ R(a,b) ,

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then a(p) + b(1B − a∗ )h(p) = 0, and so p = 0. Thus, a |R(a,b) +b(1B − a∗ )h : R(a,b) → bB is an R-monomorphism. Given any bp ∈ bB,
we see that bp ∈ R(a,b) . Furthermore, we have a |R(a,b) +b(1B − a∗ )h (bp) = bp, i.e., a |R(a,b) +b(1B − a∗ )h : R(a,b) → bB is an R-epimorphism. It follows that a |R(a,b) +b(1B − a∗ )h : R(a,b) → bB is an R-isomorphism. Therefore we complete the proof by Proposition 1.4.15.  A ring R is a hereditary ring provided that every right ideal of R is projective. A commutative ring without zero divisors, i.e., a commutative domain, is said to be an integral domain. A hereditary integral domain √ is called a Dedekind domain. For example, R[x, y]/(x2 + y 2 − 1), Z[ 10] √ and Z[ −5] are Dedekind domains. In [299, Theorem 5.8], Hs¨ u showed that if R is a Dedekind domain, then for any right R-modules B and C, R⊕B ∼ = R ⊕ C =⇒ B ∼ = C. This also was proved in [436, Theorem 2.7]. In [435, Theorem 2.11], Zhang and Tong extended this result to commutative hereditary rings. In fact, we see that commutative hereditary rings satisfy the condition in Corollary 1.4.16. Let B be a finitely generated right Rmodule, then R(a,b) is a finitely generated right ideal of R. We say that a ring R is a commutative semihereditary ring if every finitely generated right ideal is projective. Thus, we see that for any finitely generated right R-modules B and C, R ⊕ B ∼ = R ⊕ C =⇒ B ∼ = C, provided that R is a commutative semihereditary ring. This also was proved in [435, Proposition 2.14] by a different route. Recall that a right R-module is hereditarily projective if all submodule of A is projective. Evidently, projective modules over a hereditary ring are hereditarily projective. We end this chapter with a result of Fuchs’ (cf. [209, Theorem 4]). Proposition 1.4.17. Let A be a hereditarily projective right R-module. Then the following are equivalent: (1) For any right R-modules B and C, A ⊕ B ∼ = A ⊕ C implies B ∼ = C. (2) For any U, V ⊆ A, A/U ∼ = A/V implies U ∼ =V. Proof. (1) ⇒ (2) is analogous to the proof of Theorem 1.3.8. (2) ⇒ (1) Suppose that A ⊕ B ∼ = A ⊕ C. Then M := A ⊕ B = D ⊕ E T with θ : A ∼ = D, E ∼ = C. Since B/(B E) ∼ = (B + E)/E is a submodule T ∼ ∼ of M/E = D = A, we see that B/(B E) is projective. Hence, B = T T T (B E) ⊕ U . Likewise, E = (B E) ⊕ V . Thus, M = A ⊕ (B E) ⊕ U = T D ⊕ (B E) ⊕ V . Let p1 : M → A and p2 : M → D be the projections

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onto A and D, respectively.
T Then p1 |V : V → A. If p1 (x) = 0Tand x ∈ V , T then x ∈ (B E) ⊕ U V . Write x = b + c where b ∈ B E, c ∈ U . T As U ⊆ B, x ∈ B. As V ⊆ E, x ∈ E. Thus, x ∈ B E, whence, x = 0. This implies that p1 |V is monomorphic. Likewise, p2 |U : U → D is monomorphic. Hence, p1 (V ) ∼ p (U ) ∼ easily checks that = V and = U . One T T T 2 p1 (V ) ⊕ (B E) ⊕ U = U + V + (B E) = p2 (U ) ⊕ (B E) ⊕ V . Thus,
T A/p1 (V ) ∼ = M/ p1 (V ) ⊕ (B E) ⊕ U
T ∼ = M/ p2 (U ) ⊕ (B E) ⊕ V ∼ = D/p2 (U ). Obviously, D/p2 (U ) ∼ = A/θ−1 p2 (U ). The hypothesis implies V ∼ = p1 (V ) ∼ = −1 ∼ ∼ ∼ θ p2 (U ) = p2 (U ) = U . Therefore U = V , as asserted. 

Notice that if R is a commutative ring, then for any ideals I1 and I2 , R/I1 ∼ = R/I2 implies I1 ∼ = I2 . Consequently, we can also show that a commutative hereditary ring enjoys the cancellation property as above.

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Chapter 2

Unit 1-Stable Range

An associative ring R is said to have unit 1-stable range if aR + bR = R implies there exists a u ∈ U (R) such that a + bu ∈ U (R). Many authors have studied this condition such as Chen ([85]), Goodearl and Menal ([224]), and Menal and Moncasi ([321]). As is well known, K1 (R) ∼ = U (R)/V (R) if R has unit 1-stable range [321, Theorem 1.2]. A ring R is said to satisfy the Goodearl-Menal condition provided that for any x, y ∈ R there exists some u ∈ U (R) such that x − u, y − u−1 ∈ U (R). We know that K1 (R) ∼ = U (R)ab if R satisfies the Goodearl-Menal condition, where Gab stands for G/G′ and G′ is the commutator subgroup of the group G. Rings satisfying the Goodearl-Menal condition provide a large class of rings having unit 1-stable range (cf. [224]). This chapter centers around the unit 1-stable range condition. Furthermore, we extend many of the results on unit 1stable range to the Goodearl-Menal condition.

2.1

Elementary Properties

We begin this section with a simple fact which will be frequently used in the sequel. Lemma 2.1.1 Let R be a ring. Then the following are equivalent: (1) R has unit 1-stable range. 41

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(2) For any x, y ∈ R, there exists some u ∈ U (R) such that 1 + x(y − u) ∈ U (R). (3) For any x, y ∈ R, there exists some a ∈ R such that 1+xa, y+a ∈ U (R). Proof. (1) ⇒ (2) Since (−x)y + (1 + xy) = 1, there exists some u ∈ U (R) such that 1 + xy − xu ∈ U (R), i.e., 1 + x(y − u) ∈ U (R). (2) ⇒ (3) For any x, y ∈ R, there exists some u ∈ U (R) such that 1 + x(−y − u) ∈ U (R). Let a = −y − u. Then 1 + xa, y + a ∈ U (R). (3) ⇒ (1) Given ax + b = 1 with a, x, b ∈ R, we then have an element c ∈ R such that 1 + ac, x + c ∈ U (R). Thus x + c = u ∈ U (R), hence c = −x
+ u. Consequently, a + bu−1 = (au + b)u−1 = (au + 1 − ax)u−1 = 1 + ac u−1 ∈ U (R), as desired. 

Lemma 2.1.1 should be contrasted to this fact: a regular ring R is unitregular if and only if whenever x = xyx, there exists a u ∈ U (R) such that 1 + x(y − u) ∈ U (R). It is well known that 1 + xa ∈ U (R) if and only if 1 + ax ∈ U (R). Thus, it follows from Lemma 2.1.1 that a ring R has unit 1-stable range if and only if the opposite ring Rop also does. Further, a ring R has unit 1-stable range if and only if ax + b = 1 in R implies that there exists some u ∈ U (R) such that x + ub ∈ U (R). Theorem 2.1.2. Let R be a ring. Then the following are equivalent: (1) R has unit 1-stable range. (2) Given ax + b = 1 in R, then there exists some y ∈ R such that a + by, 1 + xy ∈ U (R). (3) Given ax + b = 1 in R, then there exists some z ∈ R such that x + zb, 1 + xz ∈ U (R). Proof. (1) ⇒ (2) Given ax + b = 1 in R, then there exists some u ∈ U (R) such that x + ub = v ∈ U (R) from Lemma 2.1.1. Since    −1 a b x xa − 1 ∈ GL2 (R), = −1 x 1 a there exist α, β ∈ U (R), x, y ∈ R such that    10 a b = [α, β]B12 (x)B21 (y). u1 −1 x Thus we see that  
a b = [α, β]B21 − β −1 uα B12 (x)B21 (y). −1 x

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So a + b(−y) = α ∈ U (R) and 1 − x(−y) = uα ∈ U (R), as desired. (2) ⇒ (1) Given ax + b = 1 in R, then a + by = u ∈ U (R) and 1 − xy = v ∈ U (R) for some y ∈ R. It is easy to check that    
a b u b B21 (y) = = B21 vu−1 [∗, ∗]B12 (∗). −1 x −v x Hence

B21 (−vu−1 )



a b −1 x



= [∗, ∗]B12 (∗)B21 (∗).

Therefore we claim that x + (−vu−1 )b ∈ U (R), as desired. (1) ⇔ (3) Observe that 1 + xz ∈ U (R) if and only if 1 + zx ∈ U (R). By the symmetry of unit 1-stable range, the result follows.  Lemma 2.1.3. Let R have stable range one. For any right R-module M , if the two sets {a1 , · · · , an } and {b1 , · · · , bn } generate the same right R-submodule of M , then there exists some U ∈ GLn (R) such that (a1 , · · · , an ) = (b1 , · · · , bn )U . Proof. Let M be a right R-module with a1 , · · · , an , b1 , · · · , bn ∈ M . Since a1 R+· · ·+an R = b1 R+· · ·+bn R, we easily check that (a1 , · · · , an )Mn (R) = (b1 , · · · , bn )Mn (R) where (a1 , · · · , an ), (b1 , · · · , bn ) ∈ nM . Thus we have A, B ∈ Mn (R) such that (a1 , · · · , an )A = (b1 , · · · , bn ) and (a1 , · · · , an ) = (b1 , · · · , bn )B. Since R has stable range one, so does Mn (R) by Corollary 1.1.6. From BA + (In − BA) = In , we can find some C ∈ Mn (R) such that B + (In − BA)C =
U ∈ GLn (R). Therefore (b1 , · · · , bn )U = (b1 , · · · , bn ) B + (In − BA)C = (b1 , · · · , bn )B = (a1 , · · · , an ), as desired.  The following result is a slight generalization of a result due to Canfell [71, Theorem 2.9]. Theorem 2.1.4 Let R be a ring. Then the following are equivalent: (1) R has unit 1-stable range. (2) Whenever aR + bR = dR with a, b, d ∈ R, there exist u, v ∈ U (R) such that au + bv = d. (3) Whenever a1 R + · · · + am R = dR with a1 , · · · , am , d ∈ R(m ≥ 2), there exist u1 , · · · , um ∈ U (R) such that a1 u1 + · · · + am um = d. Proof. (1) ⇒ (2) Since R has unit 1-stable range, it has stable range one. Whenever aR + bR = dR with a, b, d ∈ R, the sets {a, b} and {d, 0}

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generate the same R-submodule of R. Therefore there exists a matrix U = (uij ) ∈ GL2 (R) such that (a, b) = (d, 0)U by Lemma 2.1.3. It is easy to verify that u11 R + u12 R = R. As R has unit 1-stable range, there exists some u ∈ U (R) such that u11 + u12 u = v ∈ U (R). Therefore a + bu = dv for some v ∈ U (R), as required. (2) ⇒ (3) Suppose a1 R + · · ·+ am R = dR where a1 , · · · , am , d ∈ R (m ≥ 2). If m = 2, then the result holds by hypothesis. Assume that the result holds for all m ≤ k, k ≥ 2. Let m = k + 1. Then there exist x1 , · · · , xk , xk+1 ∈ R such that a1 x1 + · · · + ak xk + ak+1 xk+1 = d. Obviously, a1 R + · · · + ak−1 R + (ak xk + ak+1 xk+1 )R = dR, and so there are v1 , · · · , vk−1 , vk ∈ U (R) such that a1 v1 + · · · + ak−1 vk−1 + (ak xk + ak+1 xk+1 )vk = d. Therefore we have a2 v2 + · · · + ak−1 vk−1 + ak xk vk + (a1 v1 + ak+1 xk+1 vk ) = d. We easily check that a2 R + · · · + ak−1 R + ak R + (a1 v1 + ak+1 xk+1 vk )R = dR. Thus we can find w1 , · · · , wk ∈ U (R) such that a2 w2 + · · · + ak−1 wk−2 + ak wk−1 + (a1 v1 + ak+1 xk+1 vk )wk = d. Consequently, we see that a3 R + · · · + ak R + ak+1 R + (a1 v1 wk + a2 w2 )R = dR. This implies that a3 u1 + · · · + ak+1 uk−1 + (a1 v1 wk + a2 w2 )uk ∈ U (R), as desired. (3) ⇒ (1) is obvious.  Corollary 2.1.5. Let R be a ring. Then the following are equivalent: (1) R has unit 1-stable range. (2) Whenever Ra + Rb = Rd with a, b, d ∈ R, there exist u, v ∈ U (R) such that ua + vb = d. (3) Whenever Ra1 + · · · + Ram = Rd with a1 , · · · , am , d ∈ R(m ≥ 2), there exist u1 , · · · , um ∈ U (R) such that u1 a1 + · · · + um am = d. Proof. In view of Lemma 2.1.1, R has unit 1-stable range if and only if the opposite ring Rop also does. Applying Theorem 2.1.4 to Rop , we obtain the result.  Corollary 2.1.6. Let A be a quasi-projective right R-module and let E = EndR (A). Then the following are equivalent: (1) E has unit 1-stable range.

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(2) Whenever f1 , f2 , f3 ∈ E satisfy f1 (A) + f2 (A) = f3 (A), there are automorphisms h1 , h2 ∈ E such that f1 h1 + f2 h2 = f3 . Proof. Assume that f1 E + f2 E = f3 E. Then there exist x, y, s, t ∈ E such that f1 = f3 x, f2 = f3 y and f1 s + f2 t = f3 . As a result, we deduce that f1 (A) + f2 (A) = f3 (A). Assume that f1 (A) + f2 (A) = f3 (A). Let i : f1 (A) ֒→ f3 (A) be the inclusion. Since A is quasi-projective, we have some s ∈ E such that the following diagram f1

A → f1 (A) s↓ ↓i f3

A ։ f3 (A) commutates. That is, f3 s = if1 = f1 ; hence, f1 E ⊆ f3 E. Likewise, f2 E ⊆ f3 E. This implies that f1 E+f2 E ⊆ f3 E. Let M = f3 (A)/f2 (A) and let g : f3 (A) → M be the canonical map. Since f1 (A) + ker(g) = f3 (A), gf1 and gf3 are both R-epimorphisms. By assumption, there exists an element h1 ∈ EndR (A) such that gf1 h1 = gf3 , and then im(f3 − f1 h1 ) ⊆ ker(g) = f2 (A). By assumption again, there exists an element h2 ∈ EndR (A) such that f3 − f1 h1 = f2 h2 . This implies that f3 E ⊆ f1 E + f2 E. Thus, we conclude that f1 (A) + f2 (A) = f3 (A) if and only if f1 E + f2 E = f3 E. Therefore the result follows from Theorem 2.1.4.  Theorem 2.1.7. Let R be a ring. Then the following are equivalent: (1) R has unit 1-stable range. (2) Whenever a, b ∈ R generate a principal right ideal of R, there exists some u ∈ U (R) such that aR + bR = (a + bu)R. (3) Whenever a, b ∈ R generate a principal left ideal of R, there exists some u ∈ U (R) such that Ra + Rb = R(a + ub). Proof. (1)⇒(2) is clear from Theorem 2.1.4. (2)⇒(1) Assume that aR + bR = R with a, b ∈ R. Then R = aR + bR = (a+ bu)R for some u ∈ U (R), so a+ bu = w is right invertible in R. Assume that wv = 1 for some v ∈ R. Since vR + (1 − vw)R = R, we can
find some t ∈ U (R) such
that R = vR + (1 − vw)R = v + (1 − vw)t R.
Thus v + (1 − vw)t s = 1 for some s ∈ R, and then w = w v + (1 − vw)t s = s. Thus a + bu ∈ U (R), as required. (1)⇔(3) By Lemma 2.1.1, we see that R has unit 1-stable range if and only if the opposite ring Rop also has unit 1-stable range. Applying (1) ⇔ (2) to Rop , we obtain the result. 

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Corollary 2.1.8. Let R be a regular ring. Then the following are equivalent: (1) R has unit 1-stable range. (2) For any a, b ∈ R and c ∈ aR + bR, there exist d ∈ R, u ∈ U (R) such that c = (a + bu)d. (3) For any a, b ∈ R, there exists u ∈ U (R) such that aR + bR = (a + bu)R. Proof. (1)⇒(2) Given any a, b ∈ R and c ∈ aR + bR, then since R is regular there exists some d ∈ R such that aR + bR = dR. From Theorem 2.1.7, we can find a u ∈ U (R) such that c ∈ aR + bR = (a + bu)R. Hence c = (a + bu)d for some d ∈ R. (2)⇒(3) Given any a, b ∈ R, there exists some c ∈ R such that c ∈ cR = aR + bR. Hence, there exist d ∈ R, u ∈ U (R) such that c = (a + bu)d. Thus aR + bR = cR ⊆ (a + bu)R. Clearly, we have (a + bu)R ⊆ aR + bR. So aR + bR = (a + bu)R, as desired. (3)⇒(1) Whenever a, b ∈ R generate a principal right ideal of R, there exists some u ∈ U (R) such that aR + bR = (a + bu)R. Thus the result follows from Theorem 2.1.7.  We note that a regular ring R has unit 1-stable range if and only if it is unit-regular and for any idempotents e, f ∈ R, eR + f R = R implies that there exist u, v ∈ U (R) such that eu + f v = 1. Now we investigate representations of general linear groups for the unit 1-stable range condition. Theorem 2.1.9. Let R be a ring. Then the following are equivalent: (1) R has unit 1-stable range. (2) For any A ∈ GL2 (R), there exists u ∈ U (R) such that A = [∗, ∗] B21 (∗)B12 (∗)B21 (u). (3) For any A ∈ GL2 (R), there exists u ∈ U (R) such that A = [∗, ∗] B21 (u)B12 (∗)B21 (∗). Proof. (1) ⇒ (2) Given any A = (aij ) ∈ GL2 (R), we have u1 , v1 ∈ U (R) −1 such that a11 u1 + a12 v1 = 1. Hence a11 + a12 v1 u−1 1 = u1 , and then  −1  u1 a12 −1 AB21 (v1 u1 ) = c a22 −1 −1 for some c ∈ R. Set u = u−1 1 , v = a22 − cu1 a12 and w = −v1 u1 . Then we get A = [u, v]B21 (∗)B12 (∗)B21 (w). (2) ⇒ (3) Given any A = (aij ) ∈ GL2 (R), we have that A−1 ∈ GL2 (R). By hypothesis, there are u, v, w ∈ U (R) such that A−1 =

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[u, v]B21 (∗)B12 (∗)B21 (w). Therefore A = B21 (−w)B12 (∗)B21 (∗)[u−1 , v −1 ] = [u−1 , v −1 ]B21 (−vwu−1 )B12 (∗)B21 (∗), as required.   a b (3) ⇒ (1) Given an equation ax + b = 1 in R, then ∈ GL2 (R). −1 x By hypothesis, there are some u, v, w ∈ U (R) such that   a b = [u, v]B21 (w)B12 (∗)B21 (∗). −1 x Thus,   a b B21 (−w)[u−1 , v −1 ] = B12 (∗)B21 (∗). −1 x This implies that x − vwu−1 b = v ∈ U (R). Therefore R has unit 1-stable range.  Corollary 2.1.10. Let R be a ring. Then the following are equivalent: (1) R has unit 1-stable range. (2) For any A ∈ GL2 (R), there exists u ∈ U (R) such that A = [∗, ∗] B12 (∗)B21 (∗)B12 (u). (3) For any A ∈ GL2 (R), there exists u ∈ U (R) such that A = [∗, ∗] B12 (u)B21 (∗)B12 (∗).
Proof. (1) ⇒ (2) Given any A ∈ GL2 (R), then (AT )o ∈ GL2 Rop . By virtue of Theorem 2.1.9, there are u, v, w ∈ U (R) such that (AT )o = [uo , v o ]B21 (uo )B12 (∗o )B21 (∗o ). As a result, A = [∗, ∗]B12 (∗)B21 (∗)B12 (u), as required. (2) ⇒ (3) Given any A = (aij ) ∈ GL2 (R), we have that A−1 ∈ GL2 (R). Hence, there are u, v, w ∈ U (R) such that A−1 = [u, v]B12 (∗)B21 (∗)B12 (w), and so A = B12 (−w)B21 (∗)B12 (∗)[u−1 , v −1 ] = [u−1 , v −1 ]B12 (−uwv −1 )B21 (∗)B12 (∗), as desired.   x −1 (3) ⇒ (1) Given ax + b = 1 in R, then ∈ GL2 (R). So we have b a   x −1 u, v, w ∈ U (R) such that = [u, v]B12 (∗)B21 (∗)B12 (w). Thus we b a get a − bw = v, and therefore R has unit 1-stable range. 

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Extensions and Substitutions

A Morita context (A, B, M, N, ψ, φ) consists of two rings A and B, two bimodules A NB and B MA , and a pair of bimodule homomorphisms ψ : N N N M → A and φ : M N → B which satisfy the following associativB AN
N
′ N
N
ity: ψ n m n = nφ m n′ and φ m n m′ = mψ n m′ for any m, m′ ∈ M, n,n′ ∈ N . These conditions ensure that the set T of generala n ized matrices ; a ∈ A, b ∈ B, m ∈ M, n ∈ N will form a ring, called mb the ring of the Morita context. Given any module R M ; the dual of M is denoted by M ∗ = HomR (M, R). Note that M ∗ becomes a right R-module. Write E = EndR (M ). Then M becomes a right E-module if, for m ∈ M and α ∈ E, then mα is the action of α. With this it is routine to see that N M =R ME is a bimodule. Moreover, we have a map ψ : M E M ∗ → R N given by function action, and a map φ : M ∗ R M → E. With these defini∗ tions, thequadruple  (R, E, M , M, ψ, φ) is a Morita context, and the formal R M matrices become an associative ring with the usual matrix opM∗ E erations. In addition, the class of rings of the Morita contexts includes all 2 × 2 matrix rings and all triangular matrix rings. Theorem 2.2.1. Let T be the ring of a Morita context (A, B, M, N, ψ, ϕ). If A and B have unit 1-stable range, then so does T .     a1 n1 a2 n2 Proof. Assume that , ∈ T. By virtue of Lemma 2.1.1, m1 b 1 m2 b 2 we can choose a ∈ A and b ∈ B such that
a1 − a = u1 ∈ U (A), 1A + N a2 a = v1 ∈ U (A), b1
− φ(m1 u−1 n1 ) − b = u2 ∈ U (B) and 1B + 1 N φ(m2 av1−1 n2 ) + b2 b = v2 ∈ U (B). Hence       a1 n1 a0 u 1 n1 − = m1 b 1 0b m1 b 1 − b and        1A 0 a2 n2 a0 v1 n2 b + = . 0 1B m2 b 2 0b m2 a 1 B + b 2 b Since −1 −1 N −1 −1 m1 u−1 m1 u−1 1 + m1 u1 ψ(n1 u2 1 ) − (b1 − b)u2 m1 u1 N
N −1 −1 = m1 u−1 ψ(n1 u−1 m1 u−1 u2 + φ(m1 u−1 n1 ) u−1 1 + m1 u 1 N 2 1 ) −N 1 2 m1 u 1 −1 −1 −1 −1 −1 −1 = m1 u1 ψ(n1 u2 m1 u1 ) − φ(m1 u1 n1 )u2 m1 u1 = 0,

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we check that    −1 −1 N −1 −1  u 1 n1 u1 + u−1 m1 u−1 1 ψ(n1 u2 1 ) −u1 n1 u2 −1 m1 b 1 − b −u−1 u−1 2 m1 u 1 2   1A 0 N = −1 0 φ(−m1 u−1 n1 u−1 1 2 ) + (b1 − b)u2   1A 0
N N = −1 0 φ(−m1 u−1 n1 u−1 n1 ) u−1 1 2 ) + u2 + φ(m1 u1 2   1A 0 = . 0 1B −1 N −1 −1 N Further, 1A + u−1 m1 u−1 m1 ) = 1A and 1 ψ(n1 u2 1 )u1 + ψ(−u1 n1 u2 N −1 −1 −1 −1 u−1 m1 u−1 1 n1 + u1 ψ(n1 u2 1 )n1 − u1 n1 u2 (b1 − b) −1 −1 −1 N −1 −1 −1 −1 N = u1 n1 + u1 ψ(n1 u2 m1 u1 )n1 − u1 n1 − u−1 n1 ) 1 n1 u2 ψ(m1 u1 = 0. Therefore  −1  −1 N −1 −1   u1 + u−1 m1 u−1 u 1 n1 1 ψ(n1 u2 1 ) −u1 n1 u2 −1 −u−1 u−1 m1 b 1 − b 2 m1 u 1 2   1A 0 N = 0 φ(−u−1 m1 u−1 n1 ) + u−1 2 1 2 (b1 − b)   1A 0 = . 0 1B

 −1 −1 −1 N −1 −1  u1 + u−1 m1 u−1 u 1 n1 1 ψ(n1 u2 1 ) −u1 n1 u2 = . Hence −1 −u−1 u−1 m1 b 1 − b 2 m1 u 1 2 Likewise, we check that  −1 v1 n2 b m2 a 1 B + b 2 b  −1  N v1 + ψ(−v1−1 n2 bv2−1 −m2 av1−1 ) −v1−1 n2 bv2−1 = . −v2−1 m2 av1−1 v2−1 

In view of Lemma 2.1.1, T has unit 1-stable range.



Corollary 2.2.2. If R has unit 1-stable range, then so does Mn (R) for any n ≥ 1. Proof. The result holds for n = 2. Assume that the result holds for  R M12 n ≤ k (k ≥ 2). Let n = k + 1. Then Mn (R) ∼ , = M21 Mk (R) where M12 = {(a2 , · · · , ak+1 ) | a2 , · · · , ak+1 ∈ R} and M21 =

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      

  b2   ..  | b , · · · , b ∈ R . In view of Theorem 2.2.1, Mk+1 (R) has  2 k+1 .  

bk+1 unit 1-stable range. By induction, we obtain the result.



A natural question is to ask whether the unit 1-stable range condition is
Morita invariant. The answer is negative. We will see that M2 Z2 has unit
1-stable range, while Z2 does not have unit 1-stable range. Thus, M2 Z2 is a counterexample. Theorem 2.2.3. Let A and B be right R-modules. If End
R (A) and EndR (B) have unit 1-stable range, then so does EndR A ⊕ B .

Proof. Let e : A ⊕ B → A
⊕ B be given by e(a +
b) = a for any a ∈ A, b ∈ B. Then
eEndR A ⊕ B
e ∼ = End
R e(A ⊕ B) ∼ = EndR (A). Likewise, 1A⊕B − e EndR A ⊕ B 1A⊕B − e ∼ End (B). Furthermore, one easily = R checks that
 
eEndR A ⊕ B e ∗ ∼
EndR A ⊕ B = . ∗ (1 − e)EndR A ⊕ B (1 − e)
By hypothesis and Theorem 2.2.1, we conclude that EndR A ⊕ B has unit 1-stable range.  Corollary 2.2.4. Let R be a ring, and let m, n ∈ N. If Mm (R) and Mn (R) have unit 1-stable range, then so does Mm+n (R). Proof. Obviously, EndR (mR) ∼ = Mm (R), EndR (nR) ∼ = Mn (R) and
∼ EndR (m + n)R = Mm+n (R). Therefore we complete the proof by Theorem 2.2.3.  The following lemma is a known result in [4, Proposition 5.5]. Lemma 2.2.5. Let M = A ⊕ B and let p : M → A be the projection on A. If C ⊆ M , then the following are equivalent: (1) p|C is an isomorphism. (2) M = C ⊕ B. Proof. (1) ⇒ (2) Since p|C : C → A is an isomorphism, there exists an R-morphism ϕ : A → C such that ϕp|C = 1C and p|C · ϕ = 1A . Given T any m ∈ C B, then p|C (m) = 0; hence, m = ϕp|C (m) = ϕ(0) = 0. This T implies that C B = 0. For any m ∈ M , there are a ∈ A, b ∈ B such that

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m = a + b. Clearly, ϕ(a) ∈ C. Hence, p a − ϕ(a) = p(a) − p|C · ϕ(a) = 0, and so a − ϕ(a) ∈ ker(p) = B. Let d = b + a − ϕ(a). Then m = ϕ(a) + d with ϕ(a) ∈ C, d ∈ B. Therefore M = C + B. As a result, we conclude that M = C ⊕ B. (2) ⇒ (1) Clearly, p|C : C → A is an R-epimorphism. If p|C (m) = 0 for some m ∈ C, then we have some a ∈ A, b ∈ B such that m = a + b. Hence, T a = 0, and so m = b ∈ C B = 0. Therefore p|C is an isomorphism, as asserted.  Let A be a right R-module, and let E = EndR (A). Evidently, E has unit 1-stable range if and only if for any splitting epimorphism ϕ : A ⊕ A → A, there exists an R-isomorphism χ ∈ E such that θ:A→ A,
a 7→ ϕ a, χ(a)

is an isomorphism. Further, we can derive the following. Theorem 2.2.6. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E has unit 1-stable range. (2) Given any right R-module decompositions M = A1 ⊕ B1 = A2 ⊕ B2 = A1 ⊕ A2 with A1 ∼ =A∼ = A2 , then there exists some C ⊆ M such that M = C ⊕ B1 = C ⊕ B2 = C ⊕ A2 . (3) Given any right R-module decompositions M = A1 ⊕ B1 = A2 ⊕ B2 = B1 ⊕ B2 with A1 ∼ =A∼ = A2 , then there exists some C ⊆ M such that M = A1 ⊕ C = A2 ⊕ C = B1 ⊕ C. Proof. (1)⇒(2) We are given M = A1 ⊕ B1 = A2 ⊕ B2 = A1 ⊕ A2 with A1 ∼ = A ∼ = A2 . Using the decomposition M = A2 ⊕ B2 ∼ = A ⊕ B2 , we have projections p1 : M → A2 ∼ = A, p2 : M → B2 and injections q1 : A ∼ = A2 → M, q2 : B2 → M such that p1 q1 = 1A , q1 p1 + q2 p2 = 1M . Using the decomposition M = A1 ⊕ B1 ∼ = A ⊕ B1 , we have a projection ∼ f : M → A1 = A and an injection g : A ∼ = A1 → M such that f g = 1A . As (f q1 )(p1 g) + f q2 p2 g = 1A , there exists some u ∈ U (E) such that f q1 + f q2 p2 gu ∈ U (E). So we get M = ker(f ) ⊕ C, where C = im(q1 + q2 p2 gu). As p1 (q1 +q2 p2 gu) = 1A , M = ker(p1 )⊕C. Thus, M = C⊕B1 = C⊕B2 . As M = A1 ⊕ A2 , it follows by Lemma 2.2.5 that p2 |A1 is an isomorphism. We infer that p2 g is an isomorphism. It is easy to verify that p2 (q1 + q2 p2 gu) = (p2 g)u is an isomorphism. Thus, M = C ⊕ ker(p2 ) = C ⊕ A2 , as required.

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(2) ⇒ (1) Assume that ax + b = 1A in E. Set M = 2A, and let pi : M → A, qi : A → M (for i = 1, 2) denote the projections and injections of this direct sum. Define f = ap1 + bp2 and g = q1 x + q2 . Set A1 = g(A), B1 = ker(f ), A2 = q1 (A) and B2 = q2 (A). Along the lines of Theorem 1.1.5, M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A ∼ = A2 . Clearly, p2 g = 1; hence, p2 |A1 is an isomorphism. In view of Lemma 2.2.5, M = A1 ⊕ A2 . By assumption, M = C ⊕ B1 = C ⊕ B2 = C ⊕ A2 for some C ⊆ M . Let h : A ∼ = A1 ∼ = C → M be the injection. Then C = h(A). So M = ker(p1 ) ⊕ h(A) = ker(f ) ⊕ h(A); hence, p1 h and f h are isomorphisms. By Lemma 2.2.5 again, p2 |C is an isomorphism, i.e.,
p2 h is an isomorphism. Obviously, f h = (ap1 + bp2 )h = a + bp2 h(p1 h)−1 p1 h ∈ U (E). Therefore a + b(p2 h)(p1 h)−1 ∈ U (E), as desired. (1) ⇒ (3) We are given M = A1 ⊕ B1 = A2 ⊕ B2 = B1 ⊕ B2 with A1 ∼ =A∼ = A2 . Using the decomposition M = A1 ⊕ B1 ∼ = A ⊕ B1 , we have projections p1 : M → A1 ∼ = A, p2 : M → B1 and an injections q1 : A ∼ = A1 → M, q2 : B1 → M such that p1 q1 = 1A , q1 p1 + q2 p2 = 1M . Using the decomposition M = A2 ⊕B2 ∼ = A⊕B2 , we have projection f : M → A2 ∼ =A ∼ and injection g : A = A2 → M such that f g = 1A . As (f q1 )(p1 g)+f q2 p2 g = 1A , there exists some u ∈ U (E) such that p1 g + uf q2 p2 g ∈ U (E), and so (p1 + uf q2 p2 )g = 1A . This implies that M = im(g) ⊕ C = A2 ⊕ C, where C = ker(p1 + uf q2 p2 ). It follows from (p1 + uf q2 p2 )q1 = 1A that M = im(q1 ) ⊕ C = A1 ⊕ C. Let θ : A2 ∼ = A. In view of Lemma 2.2.5, θ−1 f |B1 : B1 → A2 is an isomorphism. Thus, f q2 is an isomorphism. We infer that (p1 + uf q2 p2 )q2 = u(f q2 ) is an isomorphism. Therefore M = im(q2 ) ⊕ ker(p1 + uf q2 p2 ) = B1 ⊕ C, as required. (3) ⇒ (1) As in (2) ⇒ (1), we have M = A1 ⊕ B1 with A1 ∼ = A. Define f = ap1 + p2 from M to A and g = q1 x + q2 b from A to M . Then M = ker(f ) ⊕ g(A). Set A2 = g(A) and B2 = ker(f ), so that M = A2 ⊕ B2 and A2 ∼ = A. Clearly, f q2 = 1, and so M = q2 (A) ⊕ ker(f ) = B1 ⊕ B2 . By assumption, M = A1 ⊕ C = A2 ⊕ C = B1 ⊕ C for some C ⊆ M . Let h : M → A1 be the projection. Then C = ker(h), and so M = ker(h)⊕ q1 (A). This implies that hq1 is an isomorphism. As M = A2 ⊕ C = g(A) ⊕ ker(h), hg is an isomorphism. On the other hand, M = ker(h) ⊕ q2 (A). Thus, hq2 is an
isomorphism. Observe that hg = h(q1 x + q2 b) = (hq1 ) x + (hq1 )−1 hq2 b ∈ U (E). Thus x + (hq1 )−1 hq2 b ∈ U (E). Therefore E has unit 1-stable range.  Example 2.2.7. Let R = Z/2Z. Choose A1 = B2 = (1, 0) · Z/2Z, B1 = (0, 1) · Z/2Z and A2 = (1, 1) · Z/2Z.

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Then 2R = A1 ⊕ B1 = A2 ⊕ B2 = A1 ⊕ A2 with A1 ∼ = Z/2Z ∼ = A2 , while B1 , B2 and A2 do not have a common complement. Proof. We directly verify that 2R = A1 ⊕ B1 = A2 ⊕ B2 = A1 ⊕ A2 with A1 ∼ = Z/2Z ∼ = A2 . Assume that 2R = C ⊕ B1 = C ⊕ B2 = C ⊕ A2 . Then ϕ : Z/2Z ∼ = A1 ∼ = C, and so ϕ(1) = (a, b) for some a, b ∈ Z/2Z. This implies C = im(ϕ) = ϕ(1) · Z/2Z = (a, b) · Z/2Z. Thus, C must be 0, B1 , B2 or A2 . This is absurd. Therefore B1 , B2 and A2 do not have a common complement. In fact, Z/2Z does not have unit 1-stable range though it does have stable range one.  2.3

Examples

Lemma 2.3.1. Let D be a division ring, and let n ∈ N (n ≥ 2). Then the following are equivalent: (1) Mn (D) has unit 1-stable range. (2) For any X = diag(Ir , 0) ∈ Mn (D) with r 6= 0, n, AX+B = In in Mn (D) ⇒ ∃ U ∈ GLn (D) such that A+BU ∈ GLn (D). Proof. (1) ⇒ (2) is clear. (2) ⇒ (1) Assume that AX + B = In in Mn (D). Since D is a division ring, we have U1 , V1 ∈ GLn (D) such that U1 XV1 = diag(Ir , 0). Likewise, we have U2 , V2 ∈ GLn (D) such that U2 AV2 = diag(Is , 0). Clearly,     0 −1 · · · 0 · · · 0 0 1 1 ··· 0 ··· 0 0 0 1 ··· 0 ··· 0 0  0 0 ··· 0 ··· 0 0      . . . . . . .  . . . . . . .   .. .. . . .. . . .. ..   .. .. . . .. . . .. ..          diag(Is , 0) =  0 0 · · · 1 · · · 0 0  +  0 0 · · · 0 · · · 0 0  .    . .  . . . . .. . . .. ..   .. .. . . .. . . .. ..  . . . . .  . . . . . . .  . .      0 0 · · · 0 · · · 0 1   0 0 · · · 0 · · · 0 −1  −1 0 · · · 0 · · · 0 0 1 0 ··· 0 ··· 0 0

Thus, there exist U, V ∈ GLn (D) such that A + U = V . If r = 0, then X = 0; hence, B = In . As a result, A + BU = V ∈ GLn (D). If r = n,
then A + U1 = V1 U1 ∈ GLn (D). If r 6= 0, n, then
BV1−1 −1 −1 V1 AU1 U1 XV1 + V1 BV1 = In . By hypothesis, there is some Y ∈ GLn (R) such that V1−1 AU1−1 + V1−1 BV1 Y ∈ GLn (D). This implies that A + BV1 Y U1 ∈ GLn (D). Therefore Mn (D) has unit 1-stable range.

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Lemma 2.3.2. Mn Z/2Z (n ≥ 2) has unit 1-stable range.

Proof. Step 1. Given AX + B = I2 with A = (aij ), X = diag(1, 0), B = (bij ) ∈ M2 (Z/2Z), then a11 + b11 = 1, b12 = 0, a21 + b21 = 0 and b22 = 1.   1 a12 (1) If a22 = 0, then A + B × I2 = . 0 1     11 1 a12 + b11 (2) If a22 = 1, a21 = 1, then A + B = . 01 0 1     10 11 (3) If a22 = 1, a21 = 0, a12 = 1, then A + B = . 11 10     01 10 (4) If a22 = 1, a21 = 0, a12 = 0, a11 = 1, then A + B = . 10 11     01 01 (5) If a22 = 1, a21 = 0, a12 = 0, a11 = 0, then A + B = . 10 11 In any case, there exists some U ∈ GL2 (Z/2Z)
such that A + BU ∈ GL2 (Z/2Z). According to Lemma 2.3.1, M2 Z/2Z has unit 1-stable range. Step 2. Given AX + B = I3 with A = (aij ), X = diag(1, 0, 0), B = (bij ) ∈ M3 (Z/2Z), then a11 + b11 = 1, b12 = 0, b13 = 0, a21 + b21 = 0, b22 = 1, b23 = 0, a31 + b31 = 0, b32 = 0, b33 = 1.

By Step 1, M2 (Z/2Z) has unit 1-stable range. Thus, we have U, V ∈ GL2 (Z/2Z) such that   a22 a23 + U = V. a32 a33 One easily checks that   1 a12 a13  ∈ GL3 (Z/2Z). A + Bdiag(1, U ) =  0 V 0

Given AX + B = I3 with A = (aij ), X = diag(1, 1, 0), B = (bij ) ∈ M3 (Z/2Z), then a11 + b11 = 1, a12 + b12 = 0, b13 = 0, a21 + b21 = 0, a22 + b22 = 1, b23 = 0, a31 + b31 = 0, a32 + b32 = 0, b33 = 1.

Obviously,



a22 a23 a32 a33



10 00



+



b22 b23 b32 b33



= I2 .

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Thus, we have V ∈ GL2 (Z/2Z) such that     a22 a23 b22 b23 + U = V. a32 a33 b32 b33 Therefore   1 ∗∗  ∈ GL3 (Z/2Z). A + Bdiag(1, U ) =  0 V 0

In view of Lemma 2.3.1, M3 (Z/2Z) has unit 1-stable range. Step 3. In view of Corollary 2.2.2 and Step 1, M2n (Z/2Z) (n ≥ 1) has unit 1-stable range. In view of Corollary 2.2.2, Step 1, Step 2 and Corollary 2.2.4, it follows from 2n + 1 = 2(n − 1) + 3 that M2n+1 (Z/2Z) (n ≥ 1) has unit 1-stable range. Therefore the proof is true.  Theorem 2.3.3. Let R be a semilocal ring. Then the following are equivalent : (1) R has unit 1-stable range. (2) There exist α, β ∈ U (R) such that α + β = 1. (3) R has no homomorphic image Z/2Z. Proof. (1) ⇒ (2) is obvious. (2) ⇒ (3) If R has Z/2Z as a homomorphic image, then there exists an ideal I of R such that R/I ∼ = Z/2Z. Clearly, α + β = 1 in R/I, a contradiction. Thus R does not have Z/2Z as a homomorphic image. (3) ⇒ (1) Since R is semilocal, there exist division rings D1 , · · · , Dm m L such that R/J(R) ∼ Mn (Di ). If | Di | ≥ 3, we claim that Di has unit = i

i=1

1-stable range. Let x, y ∈ Di . If y 6= 0, then 1 + x(y − u) = 1 ∈ U (Di ), where u = y ∈ U (Di ). If y = 0, x = 0, we choose u 6∈ {0, 1}, then 1 + x(y − u) ∈ U (Di ). If y = 0, x 6= 0, we choose u 6∈ {0, x−1 }, then 1 + x(y − u) ∈ U (Di ). In view of Lemma 2.1.1, Di has unit 1-stable range, and so does Mni (Di ) by Corollary 2.2.2. If ni ≥ 2 and | Di | = 2, then Mni (Di ) has unit 1-stable range by Lemma 2.3.2. If ni = 1 and | Di | = 2, then R/J(R) has Z/2Z as a homomorphic image. This gives a contradiction. Therefore R has unit 1-stable range. 

Corollary 2.3.4. Let A be an artinian right R-module. If EndR (A) has unit 1-stable range.

1 2

∈ R, then

Proof. Construct an R-morphism ϕ : A → A given by ϕ(a) = a · 12 for any a ∈ A. It is easy to verify that ϕ ∈ AutR (A). In addition, 1A =

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ϕ + ϕ. According to the Camps-Dicks Theorem [68, Corollary 6], EndR (A) is semilocal. Therefore the result follows from Theorem 2.3.3.  Theorem 2.3.5. Let R be an exchange ring with artinian primitive factors. Then the following are equivalent: (1) R has unit 1-stable range. (2) There exist α, β ∈ U (R) such that α + β = 1. (3) R has no homomorphic image Z/2Z. Proof. (1) ⇒ (2) and (2) ⇒ (3) are clear. (3) ⇒ (1) Assume that there are some x, y ∈ R such that 1 + x(y − u) 6∈ U (R) for any u ∈ U (R). Let Ω be the set of all ideals I of R such that 1 + x(y − u) is not a unit modulo A for any u + A ∈ U (R/A). Clearly, Ω 6= ∅. Choose an ascending chain A1 ⊆ A2 ⊆ · · · ⊆ An ⊆ · · · in Ω. Set ∞ S M = Ai . Then M is an ideal of R. Assume that M is not in Ω. We i=1
have u + M ∈ U (R/M ) such that 1 + x(y − u) + M ∈ U (R/M ). So there are positive integers ni (1 ≤ i ≤ 4) such that (1 + x(y − u))s − 1 ∈ An1 , s(1 + x(y − u)) − 1 ∈ An2 , ut − 1 ∈ An3 and tu − 1 ∈ An4 for some s, t ∈ R. Let n = max{n1 , n2 , n3 , n4 }. Then 1 + x(y − u) + An ∈ U (R/An ) for u + An ∈ U (R/An ), a contradiction. Hence M ∈ Ω. By using Zorn’s Lemma, there exists an ideal
Q of R which is maximal in Ω.
Set S = R/Q. If J R/Q 6= 0, then J R/Q = K/Q for some K % Q. Clearly, S/J(S) ∼ =
R/K. By the maximality of Q,
there is some (v + Q) +
J(S) ∈ U S/J(S) such
that (1 + x(y − v))
+ Q + J(S) ∈ U S/J(S) . Clearly, v + Q ∈ U R/Q . From 1 + x(y − v) + Q = (m + Q) + (r +
Q) for some m + Q ∈ U (S) and r + Q ∈ J(S), we claim that 1 + x(y − v) +Q ∈
U (S). This gives a contradiction, so J R/Q = 0. By the maximality of Q, one easily checks that R/Q is an indecomposable ring. It follows by [424, Lemma 3.7] that R/Q is a simple artinian ring. That is, R/Q ∼ = Mn (D) for a division ring D. Since R has no isomorphic image Z/2Z, neither does R/Q. It follows from Lemma
2.3.2 that R/Q has unit
1-stable range.
So we have w + Q ∈ U R/Q such that 1 + x(y − w) + Q ∈ U A/Q , a contradiction. Therefore, R has unit 1-stable range.  Corollary 2.3.6. Let R be an exchange ring of bounded index. Then the following are equivalent: (1) R has unit 1-stable range. (2) There exist α, β ∈ U (R) such that α + β = 1.

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(3) R has no homomorphic image Z/2Z. Proof. As in the proof of Corollary 1.3.14, R is an exchange ring with artinian primitive factors. Therefore we complete the proof by Theorem 2.3.5.  Corollary 2.3.7. Let R be an exchange ring with all idempotents central. Then the following are equivalent: (1) R has unit 1-stable range. (2) There exist α, β ∈ U (R) such that α + β = 1. (3) R has no homomorphic image Z/2Z. Proof. (1) ⇒ (2) and (2) ⇒ (3) are obvious. (3) ⇒ (1) As in the proof of Corollary 1.3.15, R/J(R) is an exchange ring of bounded index 1. According to Corollary 2.3.6, R/J(R) has unit 1-stable range, and therefore the result follows.  Let R be a π-regular ring with all idempotents central. If 2 ∈ R is nonnilpotent, then there exists a nonzero idempotent e ∈ R such that eRe has unit 1-stable range. Since R is an abelian π-regular ring with non-nilpotent 2 ∈ R, by virtue of [37, Theorem 7], we can find a nonzero idempotent e ∈ R such that every element in eRe is the sum of two units of R. Thus e is the sum of two units of eRe. Obviously, eRe is an exchange ring with all idempotents central. Using Corollary 2.3.7, we are done. Let R = Z6 . Then R is a π-regular ring with all idempotents central. Here, 2 ∈ R is a non-nilpotent element. In this case, R does not have unit 1-stable range, while the corner 4R4(∼ = Z3 ) has unit 1-stable range, where 4 ∈ R is an idempotent. Corollary 2.3.8. Let R be an exchange ring with artinian primitive factors, and let n ≥ 2. Then the ring of all n by n matrices over R has unit 1-stable range. Proof. Since R is an exchange ring with artinian primitive factors, so is Mn (R) by [9, Theorem 1.4]. In the ring Mn (R), we see that       0 −1 · · · 0 0 1 1 ··· 0 0 1 0 ··· 0 0 0 1 ··· 0 0 0 1 ··· 0 0  0 0 ··· 0 0        .. .. . . .. ..   .. .. . . .. ..   .. .. . . .. ..   . . . . . =  . . . . . +  . . . . ..        0 0 · · · 1 0   0 0 · · · 1 1   0 0 · · · 0 −1  −1 0 · · · 0 1 1 0 ··· 0 0 0 0 ··· 0 1

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By virtue of Theorem 2.3.5, we obtain the result.



As is well known, every n × n (n ≥ 2) matrix over an arbitrary ring is the sum of three invertible matrices (cf. [257, Theorem 3]). For an exchange ring with artinian primitive factors, we now derive the following. Corollary 2.3.9. Every n × n (n ≥ 2) matrix over an exchange ring with artinian primitive factors is the sum of two invertible matrices. Proof. Let R be an exchange ring with artinian primitive factors, and let A ∈ Mn (R). In light of Corollary 2.3.8, Mn (R) has unit 1-stable range. Since AMn (R) + In Mn (R) = Mn (R), we have some U ∈ GLn (R) such that A + In × U = V ∈ GLn (R). Thus A = (−U ) + V , and we are through.  A commutative ring R is said to be U -irreducible if the set of polynomials in R[x] which represents units in R is closed under multiplication ([272]). U -irreducible rings include many classes of rings such as commutative semilocal rings with infinite residue fields, commutative regular rings with infinite residue fields and algebraically closed integral domains having stable range one. Proposition 2.3.10. Every U -irreducible ring has unit 1-stable range. Proof. Let R be a U -irreducible ring, and let a, b, c ∈ R with ac + b = 1. Set f (x) = 1 + a(1 − c)x, g(x) = c + (1 − c)x ∈ R[x]. Clearly, f (0) = 1 = g(1). By the U -irreducibility of R,

we can find some r ∈ R such
that f (r)g(r) = a + a(1 − c)r c + (1 − c)r ∈ U (R). Thus a c + (1 − c)r + b =
−1 (ac+b)+a(1−c)r = 1+a(1−c)r ∈ U (R). Hence a+b c+(1−c)r ∈ U (R). So R has unit 1-stable range.  2.4

Goodearl-Menal Condition

A ring R is said to satisfy the Goodearl-Menal condition provided that for any x, y ∈ R, there exists some u ∈ U (R) such that x − u, y − u−1 ∈ U (R). In [224], Goodearl and Menal studied many classes of rings satisfying such condition. The main purpose of this section is to construct various classes of rings satisfying this condition. K1 groups of such rings are studied. Theorem 2.4.1. Let T be the ring of a Morita context (A, B, M, N, ψ, ϕ).

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If A and B satisfy the Goodearl-Menal condition, then so does T . Proof. Assume that 

a1 n1 m1 b 1

   a2 n2 , ∈ T. m2 b 2

Then we can choose a ∈ U (A) and b ∈ U (B) such
that a1 − a = u1 ∈ −1 N U (A), 1A + a2 a = v1 ∈ U (A), n1 ) − b = u2 ∈ U (B) and
b1 − φ(m1 u1 N 1B + φ(m2 av1−1 n2 ) + b2 b = v2 ∈ U (B). Hence       a1 n1 a0 u 1 n1 − = m1 b 1 0b m1 b 1 − b and



1A 0 0 1B



+



a2 n2 m2 b 2



a0 0b



=



v1 n2 b m2 a 1 B + b 2 b



.

As in the proof of Theorem 2.2.1, we see that     u 1 n1 v1 n2 b , ∈ U (T ). m1 b 1 − b m2 a 1 B + b 2 b   a0 Clearly, ∈ GL2 (T ). This proves that T satisfies the Goodearl-Menal 0b condition.  Corollary 2.4.2. Let R be a ring, and let m, n ∈ N. If Mm (R) and Mn (R) satisfy the Goodearl-Menal condition, then so does Mm+n (R). Proof. Clearly,
EndR (mR) ∼ = Mm (R), EndR (nR) ∼ = Mn (R) and ∼ EndR (m + n)R = Mm+n (R). Analogous to Theorem 2.2.3, we see that  
EndR (mR) ∗ ∼ EndR mR ⊕ nR = . ∗ EndR (nR)

According to Theorem 2.4.1, Mm+n (R) satisfies the Goodearl-Menal condition.  Lemma 2.4.3. Let D be a division ring, and let n ∈ N. Then the following are equivalent : (1) Mn (D) satisfies the Goodearl-Menal condition. (2) For any X = diag(Ir , 0, · · · , 0), Y ∈ Mn (D), there exists some U ∈ GLn (D) such that X − U, Y − U −1 ∈ GLn (D).

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Proof. (1) ⇒ (2) is clear. (2) ⇒ (1) For any X, Y ∈ Mn (D), there exist U, V ∈ GLn (D) such that U XV = diag(Ir , 0, · · · , 0) for some r. By hypothesis, we have some W ∈ GLn (D) such that diag(Ir , 0, · · · , 0)− W, V −1 Y U −1 − W −1 ∈ GLn (D). So,
−1 X − U −1 W V −1 , Y − U −1 W V −1 ∈ GLn (D), as required.  Clearly, Z/2Z and Z/3Z do not satisfy the Goodearl-Menal condition. Also we see that             01 01 10 10 11 11 GL2 (Z) = , , , , , . 10 11 01 11 01 10    
00 00 Choose A = ,B = . For any U ∈ GL2 Z/2Z , we have that 10 01


A − U 6∈ GL2 Z/2Z or B − U −1 6∈ GL2 Z/2Z . Thus, M2 Z/2Z does not satisfy the Goodearl-Menal condition.



Lemma 2.4.4. Mn Z/2Z (n ≥ 3, n 6= 5) and Mn Z/3Z (n ≥ 2) satisfy the Goodearl-Menal condition.
Proof. It
has been proved by a direct

computer search that M3 Z/2Z , M4 Z/2Z , M2 Z/3Z and M3 Z/3Z satisfy the Goodearl-Menal condition. Let n ≥ 2. Clearly, 2n
+ 1 = 2(n − 1) + 3. By virtue of Corollary 2.4.2, we deduce that Mn Z/3Z (n ≥ 2) satisfies
the Goodearl-Menal
condition. In light of Corollary 2.4.2, M3n Z/2Z and M4n Z/2Z satisfy the Goodearl-Menal condition. Clearly, we see that 3n + 1 = 3(n − 1) + 4(n ≥ 1), 3n + 2 = 3(n − 2) + 4 + 4(n ≥ 2).

By Corollary 2.4.2 again, M3n+1 Z/2Z (n ≥ 1) and M3n+2
Z/2Z (n ≥ 2) satisfy the Goodearl-Menal condition. Therefore Mn Z/2Z (n ≥ 3, n 6= 5) satisfies the Goodearl-Menal condition.  Lemma 2.4.5. Let D be a division ring. If | D | ≥ 4, then Mn (D) satisfies the Goodearl-Menal condition. Proof. Let x, y ∈ D. If y 6= 0, then we choose u 6∈ {0, x, y −1 }. If y = 0, then we choose u 6∈ {0, x}. In any case, x − u, y − u−1 ∈ D∗ . Thus, D satisfies the Goodearl-Menal condition. Therefore the result follows from Corollary 2.4.2. 

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Theorem 2.4.6. Let R be a
semilocal ring. If R has no homomorphic
images Z/2Z, Z/3Z, M2 Z/2Z , M5 Z/2Z , then R satisfies the GoodearlMenal condition. Proof. Since R is semilocal, there exist division rings D1 , · · · , Dm such that m L R/J(R) ∼ Mn (Di ). If | Di | ≥ 4, it follows from Lemma 2.4.5 that Di = i

i=1

satisfies the Goodearl-Menal condition, and so does Mni (Di ) by Corollary 2.2.2. If | Di | = 2, then Di ∼ = Z/2Z. If | Di | = 3, write Di = {0, 1, x}. If 1 + 1 = 0, then 1 + x 6∈ Di . Thus, 1 + 1 = x. Hence,
Di ∼ = Z/3Z.
Since R has no homomorphic images Z/2Z, Z/3Z, M2 Z/2Z , M5 Z/2Z , we see that ni ≥ 3, ni 6= 5 if | Di | = 2 and ni ≥ 2 if | Di | = 3. In any case, Mni (Di ) satisfies the Goodearl-Menal condition from Lemma 2.4.4. Therefore R/J(R) satisfies the Goodearl-Menal condition, as required.  Corollary 2.4.7. Let A be an artinian right R-module. If then EndR (A) satisfies the Goodearl-Menal condition.

1 1 1 2, 3, 5

∈ R,

Proof. Let S = EndR (A). Construct an R-morphism ϕ : A → A given by ϕ(a) = a · 21 for any a ∈ A. It is easy to verify that ϕ ∈ AutR (A). This means that 12 ∈ S. Likewise, 13 , 15 ∈ S. By virtue of [68, Corollary 6], S is semilocal. ideal I of S such that S/I ∼ = Z/2Z, Z/3Z,
If there exists an 1 1 1 M2 Z/2Z , or Z/5Z. Then 2 , 3 , 5 6∈ S/I. This gives a contradiction. In view of Theorem 2.4.6, EndR (A) satisfies the Goodearl-Menal condition. Recall that a ring A is an algebra over a commutative ring k provided that (A, +) is a left k-module, and that r(ab) = (ra)b = a(rb) for any a, b ∈ A, r ∈ k. Let k be a commutative noetherian semilocal ring, and R be a k-algebra which is finitely generated as a k-module. If A is a finitely generated right R-module, we note that EndR (A) is semilocal (cf. [297, Corollary 20.12]). Lemma 2.4.8. Let R be a ring. Then the following are equivalent: (1) R satisfies the Goodearl-Menal condition. (2) For any x, y ∈ R, there exists some u ∈ U (R) such that (x−u)(yu−1) ∈ U (R). Proof. (1) ⇒ (2) is clear. (2) ⇒ (1) Given ax + b = 1 in R, then x = u + v, a = u−1 + w, vs = 1, tw = 1 for some u, v, w, s, t ∈ R. Thus, av + b = a(x − u) + b = 1 − au = −wu. This implies that a + bs = −wus ∈ R. Since (vu−1 t)(−wus) = 1, we

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know that a + bs ∈ R is left invertible. By the above consideration, we also see that vu−1 t ∈ R is left invertible. Hence, a + bs ∈ R is invertible, and so every one-sided unit in R is invertible. Therefore x − u, y − u−1 ∈ U (R), as required.  Theorem 2.4.9. Let R be an exchange ring with artinian
primitive factors.
If R has no homomorphic images Z/2Z, Z/3Z, M2 Z/2Z , M5 Z/2Z , then R satisfies the Goodearl-Menal condition. Proof. Assume that there are some x, y ∈ R such that (x − u)(yu − 1) 6∈ U (R) for any u ∈ U (R). Let Ω be the set of all ideals I of R such that (x − u)(yu − 1) is not a unit modulo I for any u + I ∈ U (R/I). Clearly, Ω 6= ∅. Choose an ascending chain A1 ⊆ A2 ⊆ · · · ⊆ An ⊆ · · · in Ω. Set ∞ S Ai . Then M is an ideal of R. Assume that M is not in Ω. We M = i=1

have u + M ∈ U (R/M ) such that (x − u)(yu − 1) + M ∈ U (R/M ). So there are positive integers ni (1 ≤ i ≤ 4) such that (x − u)(yu − 1)s − 1 ∈ An1 , s(x − u)(yu − 1) − 1 ∈ An2 , ut − 1 ∈ An3 and tu − 1 ∈ An4 for some s, t ∈ R. Let n = max{n1 , n2 , n3 , n4 }. Then (x − u)(yu − 1) ∈ U (R/An ) for u + An ∈ U (R/An ), a contradiction. This implies that M ∈ Ω. By using Zorn’s Lemma, there exists an ideal Q of R such that it is maximal in Ω. Set S = R/Q. If J(R/Q) 6= 0, then J(R/Q) = K/Q for some K % Q. Clearly, S/J(S) ∼ is some (v + Q) + =
R/K. By the maximality of Q, there

J(S) ∈ U S/J(S) such that (x − v)(yv − 1) + Q + J(S) ∈ U S/J(S) . Clearly, v + Q ∈ U (R/Q). Furthermore, we claim that (x − v)(yv − 1) + Q ∈ U (S). This gives a contradiction, so J(R/Q) = 0. By the maximality of Q, R/Q is an indecomposable ring. It follows by [424, Lemma 3.7] that R/Q is a simple
artinian
ring. Since R has no isomorphic image Z/2Z, Z/3Z, M2 Z/2Z , M5 Z/2Z . In light of Theorem 2.4.6, R/Q satisfies the Goodearl-Menal condition. Thus, we have w + Q ∈ U (R/Q) such that (x − w)(yw − 1) ∈ U (R/Q), a contradiction. Therefore R satisfies the Goodearl-Menal condition from Lemma 2.4.8.  Corollary 2.4.10. Let R be an exchange ring of
bounded index. If R
has no homomorphic images Z/2Z, Z/3Z, M2 Z/2Z , M5 Z/2Z , then R satisfies the Goodearl-Menal condition. Proof. As in the proof of Corollary 1.3.14, R is an exchange ring with artinian primitive factors. Thus, we obtain the result by Theorem 2.4.9. 

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Corollary 2.4.11. Let R be an exchange ring with all idempotents central. If R has no homomorphic images Z/2Z, Z/3Z, then R satisfies the Goodearl-Menal condition. Proof. As in the proof of Corollary 1.3.15, R/J(R) is an exchange ring of bounded index 1. Since R has no homomorphic images Z/2Z, Z/3Z, neither has R/J(R). Let e ∈ R/J(R) be an idempotent. In view of [382, Theorem 29.2], there exists an idempotent f ∈ R such that e = f + J(R). Hence, every
idempotent
in R/J(R) is central. This implies that R/J(R) ≇ M2 Z/2Z , M5 Z/2Z . According to Corollary 2.4.10, R/J(R) satisfies the Goodearl-Menal condition. Since every unit lifts modulo J(R), R satisfies the Goodearl-Menal condition, and we are done.  Let A is an algebra over a field F . An element a of an algebra A over a field F is said to be algebraic over F if a is the root of some non-constant polynomial in F [x]. A is said to be an algebraic algebra over F if every element of A is algebraic over F . An algebra over a field F that is finite dimensional as a vector space over F is called a finite dimensional algebra over F . For instance, the 2 × 2 matrix ring over F [x]/(x2 ), where F is any field, is a finite dimensional algebra. If A is a finite dimensional algebra over a field, then A is an algebraic algebra. We now record a result of Goodearl and Menal (cf. [224, Theorem 3.1]). Proposition 2.4.12. Every algebraic algebra over an infinite field satisfies the Goodearl-Menal condition. Proof. Let R be an algebraic algebra over an infinite field F . For any a ∈ R, there exists a non-constant polynomial p(x) ∈ F [x] such that p(a) = 0. Assume that degp(x) = n, and let S be the set of all roots of p(x) in F [x]. Then | S | ≤ n. Let α ∈ F − S. Then p(α) 6= 0. Set q(x) = p(x + α) = n P qi xi with each qi ∈ F . Then 0 is not a root of q(x), and so q0 6= 0. Thus, i=0

n

i P 0 = q a− α·1R = q0 ·1R + qi a− α·1R . As a result, a− α·1R ∈ U (R). i=1

For any b, c ∈ R, by the proceeding discussion, there exist two non-constant polynomials s(x), t(x) ∈ F [x] such that s(b) = 0 and t(c) = 0. Let Sb and Sc be the sets of the roots of s(x) and t(x), respectively. Thus, we can find an S infinite number of elements α1 , · · · , αm , · · · such that each αi ∈ F −Sb Sc . S S Clearly, we have some αi0 ∈ F − Sb Sc such that α−1 Sc . This i0 ∈ F − Sb implies that b − αi0 · 1R , c − (αi0 · 1R )−1 ∈ U (R). Therefore R satisfies the

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A Banach space is a vector space V over the real or complex numbers with a norm k · k such that every Cauchy sequence (with respect to the metric d(x, y) = kx − y k) in V has a limit in V . A Banach algebra is an associative algebra A over the real or complex numbers which at the same time is also a Banach space. A C ∗ -algebra, A, is a Banach algebra over the field of complex numbers, together with a map, ∗ : A → A, called an involution. The image of an element x of A under involution is written x∗ . Involution has the following properties: (a) (b) (c) (d) (e)

(x∗ )∗ = x; (x + y)∗ = x∗ + y ∗ ; (xy)∗ = y ∗ x∗ ; (λx)∗ = λx∗ , λ ∈ C and k x∗ xk = kx k2 .

Lemma 2.4.13. If R satisfies the Goodearl-Menal condition, then R has unit 1-stable range. Proof. Given ax + b = 1 in R, then we have u, v, w ∈ U (R) such that a = u + v, x = u−1 + w. Hence vx + b = (a − u)x + b = 1 − ux = −uw, and then x + v −1 b ∈ U (R). Therefore R has unit 1-stable range from Corollary 2.1.5.  The following result is due to Goodearl and Menal [224, Theorem 4.1 and Corollary 4.2]. Theorem 2.4.14. Let R be a unital complex C ∗ -algebra. Then the following are equivalent : (1) R satisfies the Goodearl-Menal condition. (2) R has unit 1-stable range. (3) Every element in R is the sum of a unitary and a unit. Proof. (1) ⇒ (2) is clear by Lemma 2.4.13. (2) ⇒ (3) is obtained from [386, Theorem 4.2]. (3) ⇒ (1) For any x, y ∈ R, we have a unitary v ∈ R such that (1+ k y k) x − v is a unit. Let u = v/(1+ k y k). Then x − u ∈ U (R). As k v k= 1, we have k yu k< 1; hence, 1 − yu ∈ U (R), so y − u−1 ∈ U (R).  Recall that an element a in a unital C ∗ -algebra R satisfies unitary de-

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composition provided there exists a unitary u ∈ R such that au ≥ 0. Corollary 2.4.15. Let R be a unital complex C ∗ -algebra. If the set of elements in R which satisfy unitary decomposition is dense in R, then R satisfies the Goodearl-Menal condition. Proof. Given any a ∈ R, we have an element b ∈ R such that k a − bk < 1 and bu ≥ 0 for some unitary u ∈ R. Then 1 + bu ∈ U (R) and k (1 + bu)−1 k < 1. Since k (1 + au)(1 + bu)−1 − 1k ≤ k(1 + au) − (1 + bu) k k (1 + bu)−1 k ≤ ka − b kk uk < 1. Hence (1 + au)(1 + bu)−1 ∈ U (R), and then 1 + au ∈ U (R). We infer that a + u−1 ∈ U (R). In addition, −u−1 is a unitary as well. Therefore the result follows from Theorem 2.4.14.  In view of [243, Lemma 2], all elements in a finite AW ∗ algebra satisfy unitary decomposition. It follows from Corollary 2.4.15 that every finite AW ∗ algebra satisfies the Goodearl-Menal condition. Let R be an associative ring with an identity. Denote by GLn (R) the group of n × n matrices over R.  We embed GLn (R) in GLn+1 (R) by the a0 group homomorphism a 7→ . Let GL(R) denote the union of each 01 GLn (R) for n ≥ 1. We call an n × n matrix elementary if it has 1’s on the diagonal and at most one nonzero off-diagonal entry. The subgroup of GLn (R) generated by such matrices is denoted En (R). Via the usual embedding of En (R) in En+1 (R), the infinite union of the En (R) is denoted by E(R). According to the Whitehead Lemma, E(R) = E(R)′ , i.e., the subgroup of commutators, is the normal subgroup of GL(R). We define K1 (R) to be GL(R)/E(R). Many authors have studied K1 groups of rings with stable range conditions. We refer the reader to [363] for the general theory of K1 -groups. Let a, b, c ∈ R, p(a, b) = 1 + ab, p(a, b, c) = a + c + abc. If p(a, b, c) ∈ U (R), it is easy to verify that p(c, b, a)−1 = −p(b, c, d) where d = −p(a, b, c)−1 p(a, b). Thus, p(c, b, a) ∈ U (R). Obviously, p(a, b) = p(a, b − 1, 1) and p(b, a) = p(1, b − 1, a). Hence, p(a, b) ∈ U (R) implies that p(b, a) ∈ U (R). One easily verifies that p(a, b, c) = p(a, b)c + p(a), p(a, b, c)p(b, a) = p(a, b)p(c, b, a). Let V (R) and W (R) be the subgroups generated by {p(a, b)p(b, a)−1 | p(a, b) = 1 + ab ∈ U (R), a, b ∈ R} and {p(a, b, c)p(c, b, a)−1 | p(a, b, c) = a + c + abc ∈ U (R), a, b, c ∈ R}, respectively. Lemma 2.4.16. If R has stable range one, then K1 (R) ∼ = U (R)/W (R).

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Proof. See [321, Theorem 1.2].

Lemma 2.4.17. If R has unit 1-stable range, then K1 (R) ∼ = U (R)/V (R). Proof. Clearly, V (R) ⊆ W (R). For any a, b, c ∈ R with p(a, b, c) ∈ U (R), we see that p(c, b, a) ∈ U (R). If a ∈ U (R), then p(a, b, c) = = ≡ ≡ = =

a + c + abc a(1 + a−1 c + bc)
1 + (a−1 + b)c
a 1 + c(a−1 + b) a a + c + cba
p(c, b, a) mod V (R) .

For the general case, it follows from Lemma 2.1.1 that 1+b(c−u) ∈ U (R) for a u ∈ U (R). Let t = c − u. Then c = t + u and 1 + bt ∈ U (R). Observe that p(a, b, c) = (a + t + abt) + (1 + ab)u = (a + t + abt) + (1 + ab + tb + abtb)(1 + tb)−1 u = (a + t + abt) + (1 + tb)−1 u + (a + t + abt)b(1 +
tb)−1 u = (1 + tb)−1 p (1 + tb)(a + t + abt), b(1 + tb)−1 , u .
−1 By the preceding discussion, p (1+tb)(a+t+abt), b(1+tb) , u ≡ p u, b(1+

−1 tb) , (1 + tb)(a + t + abt) mod V (R) . Hence,
p(a, b, c) ≡ (1 + tb)−1 p(u, b(1 + tb)−1 )(1 + tb)(a + t + abt) + p(u)
= (1 + tb)−1 p(u, b(1 + tb)−1 )p(t, b)p(a, b, t) + p(u)

= (1 + tb)−1 p(u, b(1 + tb)−1 )p(t, b, a)p(b, t) + p(u) mod V (R) .

On the other hand,


p(c, b, a) = p u, (1 + bt)−1 b, (t + a + tba)(1 + bt) (1 + bt)
−1 −1 = p(u, (1 + bt)−1 b)(t + a + tba)(1 + bt) +
p(u) (1−1+ bt) −1 = p(u, (1 + bt) b)p(t, b, a)p(b, t) + p(u) (1 + bt) .

Since b(1 + tb)
−1 = (1 + bt)−1 b, we have (1 + tb)p(a, b, c) ≡ p(c, b, a)(1 + bt) mod V (R) . Therefore p(a, b, c)p(c, b, a)−1 ∈ V (R). According to Lemma 2.4.16, the result follows.  The following result is a slight extension of [321, Theorem 1.3]. Theorem 2.4.18. K1 (R) ∼ = U (R)ab .

If R satisfies the Goodearl-Menal condition, then

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Proof. For any u, v ∈ U (R), we have uv = 1 + av, vu = 1 + va where a = u−v −1 , so uvu−1 v −1 = p(a, v)p(v, a)−1 ∈ V (R). Thus, U (R)′ ⊆ V (R). Let a, b ∈ R such that p(a, b) = 1 + ab ∈ U (R). Since R satisfies the GoodearlMenal condition, there is a t ∈ U (R) such that v = 1 − at, w = 1 + t−1b are both units. Then p(a, b) = = ≡ = ≡ = =

1 + ab v + w − vw wv −1 (v + w − vw)w−1 v v + w − wv t(v + w − wv)t−1 1 + ba
p(b, a) mod U (R)′ .

Thus, p(a, b)p(b, a)−1 ∈ U (R)′ . Hence, V (R) ⊆ U (R)′ , and so V (R) = U (R)′ . In light of Lemma 2.4.13, R has unit 1-stable range, and we are done by Lemma 2.4.17. 

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Chapter 3

On m-Fold Stable Rings

A ring R is said to be m-fold stable if and only if whenever a1 R + b1 R = R, · · · , am R + bm R = R, there exists some y ∈ R such that a1 + b1 y, · · · , am + bm y are all units. The class of m-fold stable rings is very large. It includes the ring of all totally real algebraic integers in C and the ring of continuous complex valued functions on a 1-dimensional complex (cf. [272]). One easily observes the following facts: (1) R is 1-fold stable if and only if R has stable range one; (2) if R is 2-fold stable, it has unit 1-stable range; and (3) if R is 3-fold stable, it satisfies the Goodearl-Menal condition. Many authors have studied m-fold stable range condition such as [79], [84] and [314]. This condition was shown to be sufficient to describe the Whitehead group K1 (2, R) and the Steinberg group St(2, R). Further, it was shown that K2 (R) can be presented by Matsumoto’s relations for a 5-fold stable ring R (cf. [273]).

3.1

Symmetry

It is well known that a ring R has stable range one if and only if for any A ∈ GLn (R), there exists some y ∈ R such that A = [∗, ∗]B21 (∗)B12 (∗)B21 (y). Now we extend this result to m-fold stable rings. Furthermore, we observe that the m-fold stable condition is left-right symmetric. 69

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Lemma 3.1.1. Let R be a ring. Then the following are equivalent: (1) R is m-fold stable. (2) For any A1 , · · · , Am ∈ GL2 (R), there exists some y ∈ R such that A1 = [∗, ∗]B21 (∗)B12 (∗)B21 (y), · · · , Am = [∗, ∗]B21 (∗)B12 (∗)B21 (y). Proof. (1) ⇒ (2) For any A1 = (a1ij ), · · · , Am = (am ij ) ∈ GL2 (R), then a111 R + a112 R = R, · · · , a111 R + a112 R = R. By hypothesis, there exists some y ∈ R such that a111 + a112 y = u1 , · · · , a111 + a112 y = um ∈ U (R). Hence, we see that   u1 ∗ A1 B12 (y) = = [∗, ∗]B21 (∗)B12 (∗). ∗ ∗ Thus, A1 = [∗, ∗]B21 (∗)B12 (∗)B21 (−y). Similarly, we deduce that A2 = [∗, ∗]B21 (∗)B12 (∗)B21 (−y), · · · , Am = [∗, ∗]B21 (∗)B12 (∗)B21 (−y), as required. (2) ⇒ (1) Suppose that a1 R + b1 R = R, · · · , am R + bm R = R with a1 , · · · , am , b1 , · · · , bm ∈ R. Then we have x1 , · · · , xm , y1 , · · · , ym ∈ R such that a1 x1 + b1 y1 = 1, · · · , am xm + bm ym = 1. Hence,     a1 b 1 y 1 am b m y m ,··· , ∈ GL2 (R), −1 x1 −1 xm and so we have some y ∈ R such that   a1 b 1 y 1 = [∗, ∗]B21 (∗)B12 (∗)B12 (−y), −1 x1 .. .   am b m y m = [∗, ∗]B21 (∗)B12 (∗)B12 (−y). −1 xm This implies that a1 + b1 y1 y = u1 , · · · , am + bm ym y = um ∈ U (R). We easily check that       a1 b 1 0 u1 01 = [1, −1]B12 (−b1 ) B12 (−y1 y) . −y1 y 1 −1 0 10 Hence



a1 b 1 −y1 y 1





∈ GL2 (R). Analogously, we see that

a2 b 2 −y2 y 1



,··· ,



am b m −ym y 1



∈ GL2 (R).

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a1 b 1 −y1 y 1

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Therefore we have some z ∈ R such that = [∗, ∗]B21 (∗)B12 (∗)   am b m B21 (−z), · · · , = [∗, ∗]B21 (∗)B12 (∗)B21 (−z). Consequently, −ym y 1 a1 + b1 z, · · · , am + bm z ∈ U (R), as required.  Theorem 3.1.2. Let R be a ring. Then the following are equivalent: (1) R is m-fold stable. (2) Rop is m-fold stable. (3) Ra1 + Rb1 = R, · · · , Ram + Rbm = R implies that a1 + zb1 , · · · , am + zbm ∈ U (R) for some z ∈ R. Proof. (1) ⇒ (3) Suppose that Ra1 + Rb1 = R, · · · , Ram + Rbm = R. Since R is m-fold stable, it has stable range one. Thus we have some z1 ∈ R such that a1 + z1 b1 = u1 ∈ U (R). Hence,  −1  −1  1 −b1 1 − b1 u−1 1 z1 b1 u 1 = ∈ GL2 (R). z 1 a1 −u−1 u−1 1 z1 1 Likewise, we have  −1  −1  1 − bi u−1 1 −bi i zi bi u i = ∈ GL2 (R) −u−1 u−1 z i ai i zi i for i = 2, · · · , m. According to Lemma 3.1.1, there is a y ∈  −1 1 −bi = [∗, ∗]B21 (∗)B12 (∗)B21 (y) for all i. So R such that z i ai   1 −bi = B21 (−y)B12 (∗)B21 (∗)[∗, ∗] for all i. Therefore we have z i ai a1 + (−y)b1 , · · · , am + (−y)bm ∈ U (R), as required. (3) ⇒ (2) is trivial. (2) ⇒ (1) Applying (1) ⇒ (3) to the opposite ring Rop , we obtain the result.  Corollary 3.1.3. Let R be a ring. Then the following are equivalent: (1) R is m-fold stable. (2) For any A1 , · · · , Am ∈ GL2 (R), there exists some y ∈ R such that A1 = [∗, ∗]B12 (∗) B21 (∗)B12 (y), · · · , Am = [∗, ∗]B12 (∗)B21 (∗)B12 (y). Proof. (1) ⇒ (2) For any A1 , · · · , Am ∈ GL2 (R), we have that
T
T (Ao1 )−1 , · · · , (Aom )−1 ∈ GL2 (Rop ).

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By virtue of Theorem 3.1.2 and Lemma 3.1.1, there is an element y ∈ R such that
T (Ao1 )−1 = [∗o , ∗o ]B21 (∗o )B12 (∗o )B21 (y o ), .. .
o −1 T (Am ) = [∗o , ∗o ]B21 (∗o )B12 (∗o )B21 (y o ).

Thus, A1 = [∗, ∗]B12 (∗)B12 (∗)B12 (−y), · · · , Am = [∗, ∗]B12 (∗)B12 (∗) B12 (−y), as required. (2) ⇒ (1) For any Ao1 , · · · , Aom ∈ GL2 (Rop ), then
T
T A−1 , · · · , A−1 ∈ GL2 (R). m 1 By hypothesis, there exists an element y ∈ R such that
T A−1 = [∗, ∗]B12 (∗)B21 (∗)B12 (y), 1 .. .
−1 T Am = [∗, ∗]B12 (∗)B21 (∗)B12 (y).

Thus,

Ao1 = [∗o , ∗o ]B21 (∗o )B12 (∗o )B21 (−y o ), .. . Aom = [∗o , ∗o ]B21 (∗o )B12 (∗o )B21 (−y o ).

In view of Lemma 3.1.1, Rop is m-fold stable, therefore so is R by Theorem 3.1.2.  Lemma 3.1.4. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E has stable range one. (2) Given any right R-module decompositions M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 ∼ =A∼ = B2 ∼ = B1 , then there exists a C ⊆ M such that M = C ⊕ B1 = C ⊕ B2 . Proof. (1) ⇒ (2) is trivial by Theorem 1.1.5. (2) ⇒ (1) Assume that ax + b = 1A in E. Set M = 2A, and let pj : M → A, qj : A → M (for j = 1, 2) denote the projections and injections of this direct sum. Set A1 = q1 (A) and B1 = q2 (A). Define f = ap1 + bp2 and g = q1 x + q2 . Set A2 = g(A) and B2 = ker f . Analogously to Theorem 1.1.5, we get M = A1 ⊕ B1 = A2 ⊕ B2 with Ai ∼ = A(i = 1, 2). Furthermore, we see that B1 ∼ = A. As p2 g = 1A , M = ker p2 ⊕ g(A) = A1 ⊕ A2 = A2 ⊕ B2 . This implies that B2 ∼ = A1 ∼ = A. By hypothesis, M = C ⊕ B1 = C ⊕ B2

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for some C ⊆ M . Similarly to Theorem 1.1.5, we have an h such that a + b(p2 h)(p1 h)−1 ∈ AutR (A), as required.  Theorem 3.1.5. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E is m-fold stable. (2) If ϕi : M = A ⊕ A → A (i = 1, · · · , m) are splitting epimorphisms, then there exists a χ ∈ E such that each θi : A → A,
a 7→ ϕi a, χ(a)

is an isomorphism. (3) Given any right R-module decompositions M = A1 ⊕ B1 = · · · = Am+1 ⊕ Bm+1 with A1 ∼ = ··· ∼ = Am+1 ∼ = A ∼ = Bm+1 ∼ = ··· ∼ = B1 , then there exists a C ⊆ M such that M = C ⊕ B1 = · · · = C ⊕ Bm+1 . Proof. (1) ⇒ (2) Let ϕi : M = A ⊕ A → A (i = 1, · · · , m) be splitting epimorphisms. Then there exist R-morphisms  φi : A → M such that ci ϕi φi = 1A . Write ϕi = (ai , bi ) and φ = , where ai , bi , ci , di ∈ E. di Then ai E + bi E = E (i = 1, · · · , m). As E is m-fold stable, there exists a χ ∈ E such that each ai + bi χ ∈ U (E). Choose θi = ai + bi χ. Therefore we conclude that θi : A → A,
a 7→ ϕi a, χ(a)

is an isomorphism. (2) ⇒ (3) Given any right R-module decompositions M = A1 ⊕ B1 = · · · = Am+1 ⊕ Bm+1 with A1 ∼ = ··· ∼ = Am+1 ∼ = A ∼ = Bm+1 ∼ = ··· ∼ = B1 , then we have the projections ρi : M = Ai+1 ⊕ Bi+1 → Ai+1 (i = 1, · · · , m). Write µ : A ∼ = A1 , λ : A ∼ = B1 and π = (µ, λ) : A ⊕ A → M . Let σi = τi ρi π : A ⊕ A → A, where τi : Ai+1 ∼ = A. Then each σi is a splitting epimorphism. By hypothesis, we have an R-morphism χ : A → A such that each θi : A → A,
a 7→ σi a, χ(a)
is an isomorphism. Set C = {π a, χ(a) | a ∈ A}. Clearly, C + B1 ⊆ M . For any a1 ∈ A
1 , there is some a ∈ A such that a1 = µ(a), and so a1 = µ(a) + λχ(a) − λχ(a) ∈ C + B1 . This implies that M = C
+ B1 . T For any c ∈ C B1 , we have some a ∈ A such that c = π a, χ(a) ∈ B1 .

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T Hence, µ(a) + λχ(a) ∈ B1 . This implies that µ(a) ∈ A1 B1 = 0, and then a = 0. Moreover, c = 0; whence, M = C ⊕ B1 . Obviously, ρi | C : C → Ai+1 (i
= 1, · · · , m). For any c ∈ C, there exists a ∈
A such that c = π a, χ(a) . It is easy to verify that τi ρi (c) = σi a, χ(a) = θi (a). If ρi (c) = 0, then θi (a) = 0; hence, a = 0. Thus, c = 0. This implies that ρi | C is an R-monomorphism. For any ai+1 ∈ Ai+1 , τi (ai+1 ) ∈ A.
By hypothesis,
we can find some a ∈ A such
that τi (ai+1 ) =
σi a, χ(a) = τi ρi π a, χ(a) . Hence, ai+1 = ρi π a, χ(a) , where π a, χ(a) ∈ C. That is, ρi | C is an R-epimorphism. As a result, each ρi | C is an isomorphism. According to Lemma 2.2.5, M = C ⊕ Bi+1 . Therefore M = C ⊕ B1 = · · · = C ⊕ Bm+1 , as desired. (3) ⇒ (1) Given ai R + bi R = R (i = 1, · · · , m), then each ϕi := (ai , bi ) : M = A ⊕ A → A is a splitting epimorphism. Let A1 = {(a1 , 0) | a1 ∈ A}, A2 = {(0, a2 ) | a2 ∈ A}, and let N = A1 ⊕ A2 . Then φ : N ∼ = M , and so each ψi := ϕi φ is a splitting epimorphism. Let Ki = kerψi . Then there exist splitting exact sequences σi

ψi

0 → Ki ֒→ N = A1 ⊕ A2 → A → 0. Hence, N = A1 ⊕ A2 = Bi ⊕ Ki with A ∼ = Bi . In view of Lemma 3.1.4, we see that E has stable range one. Thus, we get each Ki ∼ = A from Theorem 1.1.5. By hypothesis, we have a C ⊆ M such that M = C ⊕ A2 = C ⊕ K1 = · · · = C ⊕ K2 . Write τi : A ∼ = Ai . For any a ∈ A, there exist c ∈ C, h ∈ A2 such that τ1 (a) = c − h. Let χ : A → A given by χ(a) = τ2−1 (h). One easily checks that χ is well defined. Construct an R-morphism θi : A → A,
a 7→ ψi τ1 (a) + τ2 χ(a) .

If θi (a) = 0, then τ1 (a) + τ2 χ(a) ∈ Ki . Write τ1 (a) = c − h, c ∈ C, h ∈ A2 . T Then τ1 (a) + τ2 χ(a) = τ1 (a) + h = c ∈ C Ki = 0; hence, τ1 (a) = T −τ2 χ(a) ∈ A1 A2 = 0. Thus, a = 0. That is, each θi is an Rmonomorphism. For any a ∈ A, we have a ci ∈ C such that a = ψi (ci ) as ψi is an R-epimorphism. Write ci = τ1 (di ) + hi ,
di ∈ A, hi ∈ A2 . Then χ(di ) = τ2−1 (hi ), and so a = ψi τ1 (di ) + τ2 χ(di ) = θi (di ). That is, θi is an Repimorphism. Therefore each θi is an isomorphism. Clearly, φ−1 = (τ1 , τ2 ), and thus, θi : A → A,
a 7→ ϕi a, χ(a)

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is an isomorphism. That is, ϕi bi χ ∈ U (E), as desired.



1A χ



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∈ AutR (A). Therefore, each ai + 

We note that a right R-module A has the preceding substitution property if and only if E = EndR (A) has the preceding substitution property as a right E-module. The following example was constructed in [299, Example 4.7]. Example 3.1.6. Let A1 = (1, 0) · Z, B1 = (0, 1) · Z, A2 = (7, 3) · Z and B2 = (5, 2) · Z be subgroups of the abelian group 2Z. Then 2Z = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 ∼ =Z∼ = B2 ∼ = B1 , while B1 and B2 do not have a common complement. Proof. One directly verifies that 2Z = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 ∼ = ∼ ∼ ∼ Z = B1 = B2 . Assume that 2Z = C ⊕ B1 = C ⊕ B2 . Then C = A1 ∼ = Z. Assume that ϕ : Z → C. Then ϕ(1) = (a, b) for some a, b ∈ Z. Thus, C = im(ϕ) = ϕ(1) · Z = (a, b) · Z. Let e1 = (1, 0), e2 = (0, 1); ξ = (a, b), η1 = (0, 1), η2 = (5, 2). Then (e1 , e2 ), (ξ, η1 ), (ξ, η2 ) are all bases of 2Z. Observing that     a0 a5 (ξ, η1 ) = (e1 , e2 ) and (ξ, η2 ) = (e1 , e2 ) , b1 b2 we see that 

a0 b1

   a5 , ∈ GL2 (Z). b2

This implies that a = ±1 and 2a − 5b = ±1, a contradiction. Therefore B1 and B2 do not have a common complement. In fact, the ring Z of integers does not have stable range one.  Lemma 3.1.7. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E has stable range one. (2) Given any right R-module decompositions M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 ∼ =A∼ = B2 ∼ = B1 , then there exists a C ⊆ M such that M = A1 ⊕ C = A2 ⊕ C.

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Proof. (1) ⇒ (2) is trivial by Theorem 1.1.5. (2) ⇒ (1) Assume that ax + b = 1A in E. Set M = 2A, and let pj : M → A, qj : A → M (for j = 1, 2) denote the projections and injections of this direct sum. Set A1 = q1 (A) and B1 = q2 (A). Define f = ap1 + p2 and g = q1 x + q2 b. Set A2 = g(A) and B2 = ker f . As f g = 1A , we have that M = A1 ⊕ B1 = A2 ⊕ B2 with Ai ∼ = A(i = 1, 2). Clearly, B1 ∼ = A. Obviously, f q2 = 1A , and so M = ker f ⊕ q2 (A) = B2 ⊕ B1 = A1 ⊕ B1 . This implies that B2 ∼ = A1 ∼ = A. By hypothesis, there exists a C ⊆ M such that M = A1 ⊕ C = A2 ⊕ C. Let h : M → A1 ∼ = A be the projection. Then C = ker h. Clearly, we have that M = q1 (A) ⊕ ker h = g(A) ⊕ ker h; hence, hq1 and hg are
isomorphisms. Observing that hg = h(q1 x + q2 b) = hq1 x + (hq1 )−1 hq2 b ∈ U (E), we get x + (hq1 )−1 hq2 b ∈ U (E). Therefore we obtain the result from Theorem 1.1.2.  Theorem 3.1.8. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E is m-fold stable. (2) Given any right R-module decompositions M = A1 ⊕ B1 = · · · = Am+1 ⊕ Bm+1 with A1 ∼ = ··· ∼ = Am+1 ∼ = A ∼ = Bm+1 ∼ = ··· ∼ = B1 , then there exists a C ⊆ M such that M = A1 ⊕ C = · · · = Am+1 ⊕ C. Proof. (1) ⇒ (2) Given any right R-module decompositions M = A1 ⊕B1 = · · · = Am+1 ⊕ Bm+1 with A1 ∼ = ··· ∼ = Am+1 ∼ =A∼ = Bm+1 ∼ = ··· ∼ = B1 , we have the projections ρi : M = Ai+1 ⊕ Bi+1 → Bi+1 (i = 1, · · · , m). Write µ:A∼ = A1 , λ : A ∼ = B1 and π = (λ, µ) : A ⊕ A → M . Let σi = τi ρi π : A ⊕ A → A, where τi : Bi+1 ∼ = A. Then each σi is a splitting epimorphism. According to Theorem 3.1.5, we have an R-morphism χ : A → A such that each θi : A → A,
a 7→ σi a, χ(a)
is an isomorphism. Set C = {π a, χ(a) | a ∈ A}. Obviously, A1 +C ⊆ M . For any b1 ∈ B1 , there
is a b ∈ A such that b1 = λ(b), and so b1 = −µχ(b) + λ(b) + µχ(b) ∈ A1 + C. This implies that M
= A1 + C. For T any r ∈ A1 C, we have a d ∈ A such that r = π d, χ(d) ∈ A1 . Hence, T λ(d) + µχ(d) ∈ A1 . This implies that λ(d) ∈ A1 B1 = 0, and then d = 0. Moreover, r = 0; whence, M = A1 ⊕ C. Obviously, ρi | C : C → Bi+1 (i
= 1, · · · , m). For any c ∈ C, there exists a ∈
A such that c = π a, χ(a)) . It is easy to verify that τi ρi (c) = σi a, χ(a) = θi (a). We infer that ρi | C is an R-monomorphism. For

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any bi+1 ∈ Bi+1 , τi (bi+1 ) ∈ A. As θi is
an R-isomorphism,
we have some a ∈ A such
that τi (bi+1 ) = σ a, χ(a) = τ ρ π a, χ(a) . Hence, bi+1 = i i i
ρi π a, χ(a) , where π a, χ(a) ∈ C. That is, ρi | C is an R-epimorphism. As a result, each ρi | C is an isomorphism. Hence, M = Ai+1 ⊕C. Therefore M = A1 ⊕ C = · · · = Am+1 ⊕ C, as required. (2) ⇒ (1) Assume that ai xi + bi yi = 1 in E(1 ≤ i ≤ m). Set M = 2A, and let pi : M → A, qi : A → M (for i = 1, · · · , m) denote the projections and injections of this direct sum. Set A1 = q1 (A) and B1 = q2 (A). Define fi = ai p1 + bi p2 and gi = q1 xi + q2 yi . Set Ai+1 = gi (A) and Bi+1 = kerfi . As fi gi = 1A , we get M = A1 ⊕ B1 = A2 ⊕ B2 = · · · = Am+1 ⊕ Bm+1 with each Ai ∼ = A. Clearly, B1 ∼ = A. In view of Lemma 3.1.7, E has stable range one. Thus, we get each Bi ∼ = B1 ∼ = A. By hypothesis, there is a C ⊆ M such that M = A1 ⊕ C = · · · = Am+1 ⊕ C. Let h : M → A1 ∼ = A be the projection. Then C = ker h. Clearly, we have that M = p1 (A) ⊕ ker h = gi (A) ⊕ ker h; hence, hp1 and hgi are isomorphisms. Thus, we see that
hgi = h(q1 xi + q2 yi ) = hqi xi + (hq1 )−1 hq2 yi ∈ U (E). As a result, we deduce that x1 +(hq1 )−1 hq2 y1 , · · · , xm +(hqm )−1 hq2 ym ∈ U (E). Therefore we complete the proof from Theorem 3.1.2. 

3.2

Morita Contexts

The aim of this section is to investigate the conditions under which the rings of Morita contexts are m-fold stable. As an application, we will prove that the m-fold stable property is Morita invariant. A right R-module M is cyclic provided that M = mR for some m ∈ M . We begin with the following simple fact. Lemma 3.2.1. Let R be m-fold stable, and let e ∈ R be an idempotent. Then eRe is m-fold stable. Proof. Suppose that a1 x1 + b1 y1 = e, · · · , am xm + bm ym = e with a1 , · · · , am , b1 , · · · , bm ∈ eRe. Then we have (a1 +1−e)(x1 +1−e)+b1y1 = 1, · · · , (am + 1 − e)(xm + 1 − e) + bm ym = 1. As R is m-fold stable, we can find some z ∈ R such that a1 + 1 − e + b1 z, · · · , am + 1 − e + bm z ∈ U (R). Assume now that (a1 + 1 − e + b1 z)u1 = 1 = u1 (a1 + 1 − e + b1 z), · · · , (am + 1 − e + bm z)um = 1 = um (am + 1 −
e + bm z). Clearly, (1 − e)u1 e = 0; hence, eu1 e = eu1 . So a
1 + b1 (eze) (eu1 e) = e. On the other hand, we have (eu1 e) a1 + b1 (eze) = e. Hence a1 + b1 (eze) ∈ U (eRe). Likewise,

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a2 + b2 (eze), · · · , am + bm (eze) ∈ U (eRe). Therefore eRe is m-fold stable.  Lemma 3.2.2. Let R be a ring. Then the following are equivalent: (1) R is m-fold stable. (2) Whenever a11 R + · · · + a1n R = R, · · · , am1 R + · · · + amn R = R(n ≥ 2), there exist y2 , · · · , yn ∈ R such that a11 + a12 y2 + · · ·+ a1n yn , · · · , am1 + am2 y2 + · · · + amn yn ∈ U (R). Proof. (2) ⇒ (1) is clear. (1) ⇒ (2) Obviously, the result holds for n = 2. Assume that the result holds for n ≤ k (k ≥ 2). Suppose that a11 R + · · · + a1(k+1) R = R, · · · , am1 R + · · · + am(k+1) R = R. Then we have a11 x11 + a12 x12 + · · · + a1k x1k + a1(k+1) x1(k+1) = 1, · · · , am1 x11 + am2 xm2 + · · · + amk xmk + am(k+1) xm(k+1) = 1; hence, there exists some y ∈ R such that a11 + a12 x12 y + · · · + a1k x1k y + a1(k+1) x1(k+1) y, · · · , am1 + am2 xm2 y + · · · +
amk xmk y + am(k+1) xm(k+1) y ∈ U (R). Thus we get a
11 + a1(k+1) x1(k+1) y R+a12 R+· · ·+a1k R = R, · · · , am1 +am(k+1) xm(k+1) y R+ am2 R + · · · + amk R = R, and so we have z2 , · · · , zk ∈ R such that a11 + a1(k+1) x1(k+1) y + a12 z2 + · · · + a1k zk , · · · , am1 + am(k+1) xm(k+1)
y + am2 z2 +· · ·+amk zk ∈ U (R). This implies that
a11 +a12 z2 +· · ·+a1k zk R+ a1(k+1) R = R, · · · , am1 +am2 z2 +· · ·+amk zk R+am(k+1) R = R. Therefore we have zk+1 ∈ R such that a11 +a12 z2 +· · ·+a1k zk +a1(k+1) zk+1 , · · · , am1 + am2 z2 +· · ·+amk zk +am(k+1) zk+1 ∈ U (R). By induction, the result follows.  Theorem 3.2.3. Let T be the ring of a Morita context (A, B, M, N, ψ, φ). If M and N are right cyclic modules, then T is m-fold stable if and only if so are A and B.   1A 0 Proof. Assume that T is m-fold stable. Set e = . Then A ∼ = eT e 0 0 is also m-fold stable by Lemma 3.2.1. Likewise, B is m-fold stable. Conversely, assume that A and B are both m-fold stable. For A1 = (A1ij ), · · · , Am = (Am GL (T ), we have some bkij
∈ A and ckij ∈ M ij ) ∈ N 1 2
N 1 1 1 1 m such that a11
b11 + ψ a12 c12 + a113
b113 + ψ a114 c14 = 1, · · · , am 11 b11 + N m N m m m m m ψ a12 c12 + a13 b13 + ψ a14 c14 = 1. Since M is a cyclic right Ak k module, we can find some y ∈ M such that all ckij = yrij for rij ∈ A. In
view N 1 of Lemma 3.2.2, we
can find y2 , y3 , y4 ∈ A such that a11 + ψ
a112 y y2 + N N m a113 y3 + ψ a114 y y4 = u1 ∈ U (R), · · · , am y y 2 + am 11 + ψ a12 13 y3 +

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ψ am 14

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N
y y4 = um ∈ U (R). So we see that           1A 0 00 1A 0 00  ∗ 1B   yy 1  00 00       A1   2 B      ∗0 1A 0 y3 0 1A 0  ∗0 0 1 yy4 0 0 1B    B u1 ∗ ∗ ∗ 1∗  0 a1∗  a1∗ 22 23 a24      =  , 0∗ ∗∗ 0∗ ∗∗ .. . 

         1A 0 00 1A 0 00  ∗ 1B   yy2 1B  00 00          Am     ∗0 1A 0 y3 0 1A 0  ∗0 0 1 yy4 0 0 1B    B um ∗ ∗ ∗  0 a1∗ a1∗ a1∗   22  23 24  . =   0∗ ∗∗ 0∗ ∗∗

1 Clearly, we
also have some dkij ∈ B and ekij ∈ N
such that a1∗ 22 d22 + N 1 N m 1∗ 1∗ 1 m∗ m m∗ m∗ m φ a23 e23 + a24 d24 = 1, · · · , a22 d22 + φ a23 e23 + a24 d24 = 1. Thus we can find a z ∈ N such that all ekij = ztkij for tkij ∈ B because N is a cyclic right B-module. From Lemma 3.2.2, there are z2 , z3 ∈ B and v1 , · · · , vm ∈ U (B) such that          1A 0 00 1A 0 00  0 1B  ∗ 1  00 00         B    A1    0∗ 1A 0 ∗0 1A 0  0∗ 0 1 ∗0 0 1   B     B   1A 0 00 1A 0 00  yy2 1B    00 0 1B 00             y3 0 1A 0 0 zz2 1A 0  yy4 0 0 1 0 z3 0 1B   B u1 ∗ ∗∗  0 v1 ∗∗   =  0 0 ∗ ∗, 00 ∗∗

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.. .  



Define

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      1A 0 00 1A 0 00  0 1B  ∗ 1  00 00         B    Am    0∗ 1A 0 ∗0 1A 0  0∗ 0 1 ∗0 0 1   B     B    1A 0 00 1A 0 00  yy2 1B   0 1B  0 0 00             y3 0 1A 0 0 zz2 1A 0  yy4 0 0 1 0 z3 0 1B    B um ∗ ∗∗  0 vm ∗∗   =  0 0 ∗ ∗. 00 ∗∗ 

     y3 zz2 1A 0 1A 0 δ= ,ε = and γ = . yy4 z3 yy2 1B 0 1B

Then A1 B21 δε−1 diag ε, γ = diag(∗, ∗)B21 (∗)B12 (∗). Hence
A1 = diag(∗, ∗)B21 (∗)B12 (∗)B21 − δε−1
, A2 = diag(∗, ∗)B21 (∗)B12 (∗)B21 − δε−1 , .. .
Am = diag(∗, ∗)B21 (∗)B12 (∗)B21 − δε−1 .

It follows by Lemma 3.1.1 that T is m-fold stable.



If M is a simple A-module and N is a simple B-module, it follows from Theorem 3.2.3 that T is m-fold stable if and only if so are A and B. Corollary 3.2.4. The m-fold stable condition is Morita-invariant. Proof. Let R be m-fold stable, and let S be Morita equivalent to R. In view of [4, Corollary 22.7], there exists a positive integer n and an idempotent matrix f ∈ Mn (R) such that S ∼ = f Mn (R)f . By Theorem 3.2.3, M2n (R) is m-fold stable. On the other hand, we can find an idempotent matrix e ∈ M2n (R) such that Mn (R) ∼ = eM2n (R)e. Using Lemma 3.2.1, Mn (R) is ∼ m-fold stable. Thus S = f Mn (R)f is m-fold stable, as asserted.  Let R be m-fold stable, and let P be a finitely generated projective right R-module. It follows from Corollary 3.2.4 that EndR (P ) is m-fold stable.

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Lemma 3.2.5. Let T be the ring of a Morita context (A, B, M, N, ψ, φ) with zero pairings. Then T /J(T ) ∼ = A/J(A) ⊕ B/J(B).       a1 n1 a n a2 n2 Proof. Given , , ∈ T with a1 , a2 ∈ A; b1 , b2 ∈ m1 b 1 mb m2 b 2 B; m1 , m, m2 ∈ M ; n1 , n, n2 ∈ N and a ∈ J(A), b ∈ J(B), then we have that       a1 n1 a n a2 n2 1A − a1 aa2 t 1T − = m1 b 1 mb m2 b 2 s 1B − b1 bb2 with s = −m1 a a2 − b1 ma2 − b1 bm2 ∈ M and t = −a1 an2 − a1 nb2 − n1 bb2 ∈ N . Set c = 1 − a1 aa2 and d = 1 − b1 bb2 . Since a ∈ J(A) and b ∈ J(B), c ∈ U (A) and d ∈ U (B). Furthermore,  −1   1A − a1 aa2 t c−1 −c−1 td−1 = . s 1B − b1 bb2 −d−1 sc−1 d−1       a n J(A) N a n Thus ∈ J(T ), hence ⊆ J(T ). Given any mb M J(B) mb ∈ J(T ) and a1 , a2 ∈ A, we have         1A 0 a1 0 a n a2 0 1A − a1 aa2 0 − = ∈ U (T ). 0 1B 0 0 mb 0 0 0 1B Thus, 1A − a1 aa2 ∈ U (A); hence, a ∈ J(A).   Likewise, b ∈ J(B). J(A) N J(A) N That is, J(T ) ⊆ . So J(T ) = . Construct M J(B) M J(B)       A/J(A) 0 a n a + J(A) 0 θ:T → given by 7→ 0 B/J(B) mb 0 b + J(B)   a n for any ∈ T . Then θ is a ring epimorphism, so we get T /J(T ) ∼ = mb   A/J(A) 0 ∼ T /ker(θ) ∼  = = A/J(A) ⊕ B/J(B). 0 B/J(B) Theorem 3.2.6. Let T be the ring of a Morita context (A, B, M, N, ψ, φ) with zero pairings. Then T is m-fold stable if and only if so are A and B.   1A 0 Proof. Assume that T is m-fold stable. Set e = . One easily 0 0 checks that A ∼ = eT e is m-fold stable by Lemma 3.2.1. Similarly, B is m-fold stable. Conversely, assume that A and B are both m-fold stable. Then A/J(A) and B/J(B) are also m-fold stable. It follows by Lemma 3.2.5 that T /J(T )

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is m-fold stable. Given a1 T + b1 T = T, · · · , am T + bm T = T , we have

a1 T /J(T ) + b1 T /J(T ) = T /J(T ), ..

. am T /J(T ) + bm T /J(T ) = T /J(T ).

So there is some y ∈ T such that a1 + b1 y, · · · , am + bm y are all units in T /J(T ). Since every unit lifts modulo J(T ), we can find r1 , · · · , rm ∈ J(T ) and u1 , · · · , um ∈ U (T ) such that a1 + b1 y = r1 + u1 , · · · , am + bm y = rm + um . Therefore a1 + b1 y, · · · , am + bm y ∈ U (T ), as required.  As an immediate consequence, the triangular matrix ring m-fold stable if and only if so are A and B.



A 0 M B



is

Corollary 3.2.7. The n × n lower triangular matrix ring over a ring R is m-fold stable if and only if so is R. Proof. Assume that R is m-fold stable. If n = 1, then the result follows. Suppose the result holds for n ≥ 1. Let T Mn(R) be the ring of all n×n lower   R  ..  triangular matrix over R and M =  .  . Then T Mn (R) is m-fold R

1×n 

 R 0 is M T Mn (R) (n+1)×(n+1) also m-fold stable. By induction, T Mn+1 (R) is m-fold stable from Theorem 3.2.6, as required. Conversely, we easily obtain the result by Lemma 3.2.1.  stable. It follows from Theorem 3.2.6 that

Analogously, we deduce that the upper triangular matrix ring over a ring R is m-fold stable if and only if so is R. Let A1 , A2 and A3 be associative rings with identities, let M21 , M31 , M32 be (A2 , A1 )-,(A3 , A1 )- and (A3 , A2 )-bimodules, respectively. Let φ : N M32 M21 → M31 be an (A3 , A1 )-homomorphism, and let T =   A2 A1 0 0  M21 A2 0  with usual matrix operations. Then T is an associative M31 M32 A3 ring with an identity. Theorem 3.2.8. The following are equivalent:

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(1) A1 , A2 , and A3 are m-fold stable.

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 A1 0 0 (2) The formal triangular matrix ring T =  M21 A2 0  is m-fold staM31 M32 A3 ble.     A2 0 M21 Proof. (1) ⇒ (2) Set B = ,M = . Since A2 and A3 are M32 A3 M31 m-fold stable, by virtue of Theorem 3.2.6, so is the ring  B. On  the other A1 0 hand, A1 is m-fold stable. By Theorem 3.2.6 again, is m-fold M B stable, as required. (2) ⇒ (1) In view of Lemma 3.2.1, we easily obtain the result.  3.3



Right Principal Ideals

In [71, Theorem 2.9], Canfell proved that R has stable range one if and only if aR + bR = dR implies that there exists some u ∈ U (R) such that a + by = du for some y ∈ R. Unfortunately, Canfell’s technique cannot be generalized to general m-fold stable rings. In this section, we investigate m-stable rings by means of right principal ideals and extend Canfell’s result to such rings. Theorem 3.3.1. Let R be a ring. Then the following are equivalent: (1) R is m-fold stable. (2) Whenever a1 R + b1 R = d1 R, · · · , am R + bm R = dm R, there exists some y ∈ R such that a1 + b1 y = d1 u1 , · · · , am + bm y = dm um for some u1 , · · · , um ∈ U (R). Proof. (2) ⇒ (1) Given a1 R + b1 R = R, · · · , am R + bm R = R, there exists a y ∈ R such that a1 + b1 y = u1 , · · · , am + bm y = um for some u1 , · · · , um ∈ U (R), as required. (1) ⇒ (2) Suppose that a1 R+b1 R = d1 R, · · · , am R+bm R = dm R. Then (a1 , b1 )M2 (R) = (d1 , 0)M2 (R), · · · , (am , bm )M2 (R) = (dm , 0)M2 (R). Obviously, we have A1 , · · · , Am , B1 , · · · , Bm ∈ M2 (R) such that (a1 , b1 )A1 = (d1 , 0), (a1 , b1 ) = (d1 , 0)B1 ; · · · , (am , bm )Am = (d1 , 0), (a1 , b1 ) = (d1 , 0)Bm . Clearly, M2 (R) has stable range one. From B1 A1 + (I2 − B1 A1 ) = I2 , we have W ∈ M2 (R) such that B1 + (I2 − B1 A
1 )W = U1 ∈ GL2 (R). Therefore (d1 , 0)U1 = (d1 , 0) B1 + (I2 − B1 A1 )W = (d1 , 0)B1 = (a1 , b1 ).

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That is, there exists U1 = (u1ij ) ∈ GL2 (R) such that (a1 , b1 ) = (d1 , 0)U1 . Likewise, we can find U2 , · · · , Um ∈ GL2 (R) such that (a2 , b2 ) = (d2 , 0)U2 , · · · , (am , bm ) = (dm , 0)Um . Since U1 , · · · , Um ∈ GL2 (R), we m have u111 R + u112 R = R, · · · , um 11 R + u12 R = R. Hence there is some 1 1 m y ∈ R such that u11 + u12 y = u1 , · · · , um 11 + u12 y = um ∈ U (R). Therefore (a1 , b1 )B21 (y) = (d1 , 0)U1 B21 (y) = (d1 u1 , ∗). Similarly, we have (a2 , b2 )B21 (y) = (d2 u2 , ∗), · · · , (am , bm )B21 (y) = (dm u2 , ∗). Consequently, we conclude that a1 + b1 y = d1 u1 , · · · , am + bm y = dm um , as asserted.  Corollary 3.3.2. Let R be a ring. Then the following are equivalent: (1) R is m-fold stable. (2) Whenever a1 R + b1 R = d1 R, · · · , am R + bm R = dm R, there exists some y ∈ R such that a1 R + b1 R = (a1 + b1 y)R, · · · , am R + bm R = (am + bm y)R. (3) Whenever Ra1 + Rb1 = Rd1 , · · · , Ram + Rbm = Rdm , there exists some z ∈ R such that Ra1 + Rb1 = R(a1 + zb1 ), · · · , Ram + Rbm = R(am + zbm ). Proof. (1) ⇒ (2) Assume that a1 R + b1 R = d1 R, · · · , am R + bm R = dm R. By virtue of Theorem 3.3.1, we can find a y ∈ R such that a1 + b1 y = d1 u1 , · · · , am + bm y = dm um for some u1 , · · · , um ∈ U (R). Thus a1 R + b1 R = d1 R = d1 u1 R = (a1 + b1 y)R. Similarly, a2 R + b2 R = (a2 + b2 y)R, · · · , am R + bm R = (am + bm y)R. (2) ⇒ (1) Given a1 R + b1 R = R, · · · , am R + bm R = R, there exists some y ∈ R such that a1 R + b1 R = R = (a1 + b1 y)R, · · · , am R + bm R = R = (am + bm y)R. So we know that a1 + b1 y, · · · , am + bm y ∈ R are all right invertible in R. Clearly, R has stable range one. Therefore a1 + b1 y, · · · , am + bm y ∈ U (R), as desired. (1) ⇔ (3) is proved by symmetry.  Recall that a right R-module M is quasi-injective provided that any homomorphism of a submodule of M into M extends to an endomorphism of M (cf. [214]). For example, all injective modules and all semisimple modules are quasi-injective. For quasi-injective modules, we can derive the following. Lemma 3.3.3. Let A be a right R-module. Then the following are equivalent: (1) A is quasi-injective.

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(2) For any f, g ∈ EndR (A), ker(g) ⊆ ker(f ) implies that there exists some h ∈ EndR (A) such that f = hg. Proof. (2) ⇒ (1) is trivial. (1) ⇒ (2) Let π1 : A ։ A/ker(f ) and π2 : A ։ A/ker(g) be the natural R-epimorphisms. Then there exist f ∗ : A/ker(f ) → A, g ∗ : A/ker(g) → A such that f = f ∗ π1 and g = g ∗ π2 . Let τ : A/ker(g) → A/ker(f ) be
given by τ a + ker(g) = a + ker(f ) for any a + ker(g) ∈ A/ker(g). As ker(g) ⊆ ker(f ), τ is well defined. Clearly, π1 = τ π2 . It is easy to verify that g ∗ is an R-monomorphism. By quasi-injectivity, there exists a map h : A → A such that the following diagram commutates, i.e., hg ∗ = f ∗ τ . f∗

A/ ker(f ) → A τ↑ ↑h g∗

A/ ker(g) ֌ A ∗ This implies that hg = hg π2 = f ∗ τ π2 = f ∗ π1 = f , as desired.



Lemma 3.3.4. Let A be a quasi-injective right R-module, let E = EndR (A), and let f1 , f2 , f3 ∈ E. Then the following are equivalent: (1) Ef1 + Ef T 2 = Ef3 . (2) ker(f1 ) ker(f2 ) = ker(f3 ).

Proof. (1) ⇒ (2) is obvious. (2) ⇒ (1) As ker(f3 ) ⊆ ker(f1 ), it follows from Lemma 3.3.3 that f1 = hf3 for some h ∈ E; hence, Ef1 ⊆ Ef3 . Likewise, Ef2 ⊆ Ef3 . Thus, Ef1 + Ef2 ⊆ Ef3 . Let i : ker(f1 ) → A be the inclusion. Then ker(f2 i) ⊆ ker(f3 i). In view of Lemma 3.3.3, we have some k ∈ E such that f3 i = kf2 i, and so im(i) ⊆ ker(f3 − kf2 ). That is, ker(f1 ) ⊆ ker(f3 − kf2 ). By Lemma 3.3.3 again, there exists some l ∈ E such that f3 − kf2 = lf1 . This shows that Ef3 ⊆ Ef1 + Ef2 , as required.  Theorem 3.3.5. Let R be a right self-injective ring. Then the following are equivalent: (1) R is m-fold stable. (2) Whenever ℓ(a1 ) ∩ ℓ(b1 ) = ℓ(d1 ), · · · , ℓ(am ) ∩ ℓ(bm ) = ℓ(dm ), there exists y ∈ R such that ℓ(a1 ) ∩ ℓ(b1 ) = ℓ(a1 + b1 y), · · · , ℓ(am ) ∩ ℓ(bm ) = ℓ(am + bm y). (3) Whenever r(a1 ) ∩ r(b1 ) = r(d1 ), · · · , r(am ) ∩ r(bm ) = r(dm ), there exists z ∈ R such that r(a1 ) ∩ r(b1 ) = r(a1 + zb1 ), · · · , r(am ) ∩ r(bm ) = r(am + zbm ).

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Proof. (1) ⇒ (3) Given r(a1 )∩r(b1 ) = r(d1 ), · · · , r(am )∩r(bm ) = r(dm ), we have Ra1 + Rb1 = Rd1 , · · · , Ram + Rbm = Rdm by Lemma 3.3.4. It follows from Corollary 3.3.2 that Ra1 + Rb1 = R(a1 + zb1 ), · · · , Ram + Rbm = R(am + zbm ) for some z ∈ R. By using Lemma 3.3.4 again, r(a1 ) ∩ r(b1 ) = r(a1 + zb1 ), · · · , r(am ) ∩ r(bm ) = r(am + zbm ), as required. (3) ⇒ (1) Assume that Ra1 + Rb1 = R, · · · , Ram + Rbm = R. It follows from Lemma 3.3.4 that \ \ r(a1 ) r(b1 ) = r(1), · · · , r(am ) r(bm ) = r(1). By hypothesis, we have some z ∈ R such that \ \ r(a1 ) r(b1 ) = r(a1 + zb1 ), · · · , r(am ) r(bm ) = r(am + zbm ).

Thus, r(a1 + zb1 ) ∩ r(0) = r(1). By Lemma 3.3.4 again, R(a1 + zb1 ) = R. This implies that a1 +zb1 ∈ U (R). Likewise, a2 +zb2 , · · · , am +zbm ∈ U (R), as asserted. (1) ⇔ (2) By Theorem 3.1.2, R is m-stable if and only if so is the opposite ring Rop . Applying (1) ⇔ (2) to Rop , the result follows. 

Proposition 3.3.6. Let R be a regular ring. Then the following are equivalent: (1) R is m-fold stable. (2) Whenever ℓ(a1 ) ∩ ℓ(b1 ) = ℓ(d1 ), · · · , ℓ(am ) ∩ ℓ(bm ) = ℓ(dm ), there exists y ∈ R such that ℓ(a1 ) ∩ ℓ(b1 ) = ℓ(a1 + b1 y), · · · , ℓ(am ) ∩ ℓ(bm ) = ℓ(am + bm y). (3) Whenever r(a1 ) ∩ r(b1 ) = r(d1 ), · · · , r(am ) ∩ r(bm ) = r(dm ), there exists z ∈ R such that r(a1 ) ∩ r(b1 ) = r(a1 + zb1 ), · · · , r(am ) ∩ r(bm ) = r(am + zbm ). Proof. Assume that aR + bR = dR. Then d = ax + by for some x, y ∈ R; hence, ℓ(a) ∩ ℓ(b) ⊆ ℓ(d). For any z ∈ ℓ(d), zd = 0, and so za = zb = 0. Thus, z ∈ ℓ(a) ∩ ℓ(b). Hence, we conclude that ℓ(a) ∩ ℓ(b) = ℓ(d). Assume that ℓ(a) ∩ ℓ(b) = ℓ(d). Since R is regular, there exists some c ∈ R such that aR + bR = cR. Write d = dxd and c = cyc. As 1 − dx ∈ ℓ(d), (1 − dx)a = (1 − dx)b = 0. Thus, a = dxa and b = dxb. Hence, aR + bR ⊆ dR. As (1 − cy)c = 0, we see that (1 − cy)a = (1 − cy)b = 0. That is, 1 − cy ∈ ℓ(a) ∩ ℓ(b). This implies that 1 − cy ∈ ℓ(d), and so d = cyd ∈ aR + bR. Hence, aR + bR = dR. Therefore aR + bR = dR if and only if ℓ(a) ∩ ℓ(b) = ℓ(d). Likewise, Ra + Rb = Rd if and only if T r(a) r(b) = r(d). Consequently, we complete the proof analogously to Theorem 3.3.5. 

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From the preceding proof, we see that a regular ring R is m-fold stable if and only if ℓ(a1 ) ∩ ℓ(b1 ) = 0, · · · , ℓ(am ) ∩ ℓ(bm ) = 0 implies that there exists y ∈ R such that ℓ(a1 + b1 y) = 0, · · · , ℓ(am + bm y) = 0 if and only if r(a1 ) ∩ r(b1 ) = 0, · · · , r(am ) ∩ r(bm ) = 0 implies that there exists z ∈ R such that r(a1 + zb1 ) = 0, · · · , r(am + zbm ) = 0. Thus, a regular ring R is T unit-regular if and only if ℓ(a) ℓ(b) = 0 implies that ℓ(a+by) = 0 for some T y ∈ R if and only if r(a) r(b) = 0 implies that r(a+zb) = 0 for some z ∈ R. There are many commutative rings R satisfying the property: aR + bR = R T if and only if ℓ(a) ℓ(b) = 0. For example, all zero-dimensional rings and all noetherian rings whose elements are either a unit or a zero divisor (cf. [186, Theorem 4 and Theorem 8]). Recall that a B´ezout ring is an integral domain in which every finitely generated ideal is principal. For example, every principal ideal domain is B´ezout, where an integral domain R is a principal ideal domain provided that every ideal is principal. Now we extend Theorem 3.3.1 to such rings. Let R be a commutative ring, and let a, b ∈ R. We use (a, b) to stand for the ideal generated by a and b. Theorem 3.3.7. Let R be an integral domain. Then the following are equivalent: (1) R is m-fold stable B´ezout. (2) For any a1 , b1 ; · · · ; am , bm ∈ R, d1 ∈ (a1 , b1 ), · · · , dm ∈ (am , bm ), there exist c1 , · · · , cm , y ∈ R such that d1 = c1 (a1 + b1 y), · · · , dm = cm (am + bm y). (3) For any a1 , b1 ; · · · ; am , bm ∈ R, there exists some y ∈ R such that (a1 , b1 ) = (a1 + b1 y), · · · , (am , bm ) = (am + bm y). Proof. (1) ⇒ (2) Since R is B´ezout, we have r1 , · · · , rm ∈ R such that (a1 , b1 ) = (r1 ), · · · , (am , bm ) = (rm ). By virtue of Theorem 3.3.1, there exist u1 , · · · , um ∈ U (R) such that a1 + b1 y = r1 u1 , · · · , am + bm y = rm um . On the other hand, we can find some s1 , · · · , sm ∈ R such that d1 = s1 r1 , · · · dm = sm rm . Therefore we claim that −1 d1 = s1 u−1 1 (a1 + b1 y), · · · , dm = sm um (am + bm y).

−1 Set c1 = s1 u−1 1 , · · · , cm = sm um . Then d1 = c1 (a1 + b1 y), · · · , dm = cm (am + bm y). (2) ⇒ (3) Clearly, (a1 , b1 ) ⊇ (a1 + b1 y), · · · , (am , bm ) ⊇ (am + bm y). Since b1 ∈ (a1 , b1 ), · · · , bm ∈ (am , bm ), we can find c1 , · · · , cm , y ∈ R

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such that b1 = c1 (a1 + b1 y), · · · , bm = cm (am + bm y). Hence a1 = (1 − c1 y)(a1 + b1 y), · · · , am = (1 − cm y)(am + bm y). Therefore (a1 , b1 ) ⊆ (a1 + b1 y), · · · , (am , bm ) ⊆ (am + bm y), as required. (3) ⇒ (1) Obviously, R is a B´ezout ring. Given a1 R + b1 R = R, · · · , am R + bm R = R, we have an element y ∈ R such that (a1 , b1 ) = (a1 +b1 y), · · · , (am , bm ) = (am +bm y). Thus a1 +b1 y, · · · , am +bm y ∈ U (R), as asserted.  Corollary 3.3.8. Let R be a B´ezout domain with quotient field K, and let R′ be an integral domain such that R ⊆ R′ ⊆ K. If R is m-fold stable, then so is R′ . Proof. Clearly, we may assume that R′ = RS for some multiplicative system S of R. For any a1 , b1 ; · · · ; am , bm ∈ R′ , d1 ∈ (a1 , b1 ), · · · , dm ∈ (am , bm ), we can find d′1 , a′1 , b′1 ; · · · ; d′m , a′m , b′m ∈ R and m1 ∈ S such that d1 = d′1 /m1 , a1 = a′1 /m1 , b1 = b′1 /m1 ; · · · ; dm = d′m /m1 , am = a′m /m1 , bm = b′m /m1 . Hence we have some n1 , · · · , nm ∈ S such that n1 d′1 ∈ (a′1 , b′1 ), · · · , nm d′m ∈ (a′m , b′m ). By virtue of Theorem 3.3.7, there exist y, c1 , · · · , cm ∈ R such that n1 d′1 = c1 (a′1 + b′1 y), · · · , nm d′m = cm (a′m + b′m y). Therefore we see that d1 = d′1 /m1 = (c1 /n1 )(a1 + b1 y). Likewise, d2 = (c2 /n2 )(a2 + b2 y), · · · , dm = (cm /nm )(am + bm y). By Theorem 3.3.7 again, the result follows.  Let R be a principal ideal domain with quotient field K, and let R′ be an integral domain such that R ⊆ R′ ⊆ K. If R is m-fold stable, then so is R′ from Corollary 3.3.8. Recall that a commutative ring R satisfies the primitive criterion if for each polynomial f (x) = a0 +a1 x+· · ·+an xn (n ≥ 0) with a0 R+· · ·+an R = R, i.e., f (x) ∈ R[x] is primitive, then there exists an α ∈ R such that f (α) ∈ U (R). The ring of continuous cross section of a sheaf of rings of many units over a Boolean space satisfies the primitive criterion. Hence, a zero-dimensional ring satisfies the primitive criterion if and only if its residue class fields are infinite. The ring of all algebraic integers in the field of complex numbers satisfies the primitive criterion. Let R be any commutative ring and S = {f | f ∈ R[x] is primitive}. Then the ring of fractions R(x) := S −1 R[x] satisfies the primitive criterion (cf. [272]). For other examples we refer to [199]. Proposition 3.3.9. Let R be a commutative ring. Then the following hold:

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(1) If R satisfies the primitive criterion, then R is U -irreducible. (2) If R is U -irreducible, then R is m-fold stable for all m ∈ N. Proof. (1) Suppose that R satisfies the primitive criterion. Let f1 (x), f2 (x) ∈ R[x] such that f1 (a), f2 (b) ∈ U (R). Then f1 (x), f2 (x) are primitive in m P R[x]. Let h(x) = f1 (x)f2 (x). Then h(x) = ci xi for some c0 , · · · , cm ∈ R. i=0

If h(x) ∈ R[x] is not primitive, then c0 R + · · · + cm R $ R. So we can find some maximal ideal M of R such that c0 R + · · · + cm R j M $ R. Write ni ni P P aij xj is primitive in (R/M )[x]. fi (x) = aij xj . Then each fi (x) = j=1

j=1 m P

As R/M is a field, we know that h(x) = This implies that

m P

i=0

ci xi is primitive in (R/M )[x].

i=0

ci (R/M ) = R/M , a contradiction. Therefore h(x) ∈

R[x] is primitive. By hypothesis, there exists an element y ∈ R such that h(y) ∈ U (R), i.e., f1 (y)f2 (y) ∈ U (R), as required. (2) Suppose that a1 R + b1 R = R, · · · , am R + bm R = R. Let fi (x) = ai + bi x (1 ≤ i ≤ m). In view of Proposition 2.3.10, there are some y1 , · · · , ym ∈ R such that each ai +bi yi ∈ U (R). Thus, f (y1 ), f (y2 ) ∈ U (R). By hypothesis, we can find a z1 ∈ R such that f1 (z1 )f2 (z1 ) ∈ U (R). As f3 (y3 ) ∈ U (R), we have a z2 ∈ R such that f1 (z2 )f2 (z2 )f3 (z2 ) ∈ U (R). By m Q iteration of this process, we can find a zm−1 such that fi (zm−1 ) ∈ U (R). i=1

Therefore f1 (zm−1 ), · · · , fm (zm−1 ) ∈ U (R), i.e., each ai + bi zm−1 ∈ U (R). Therefore R satisfies the m-fold stable range, as asserted. 

Clearly, the ring R[x] of all polynomials over R is not m-fold stable even if R is a field. But, we indeed have that the ring R[[x]] of all power series over R is m-fold stable if and only if so is R. We denote by H(R) the ring of Hurwitz series over R, defined as follows. The elements of H(R) are sequences of the form (an ) = (a0 , a1 , a2 , · · · ), where an ∈ R for each n ∈ N. Addition in H(R) is defined termwise, so that (an ) + (bn ) = (an + bn ). The product n P n of (an ) and (bn ) is given by (an )(bn ) = (cn ), where cn = ak bn−k . k k=0 The zero in H(R) is (0, 0, 0, · · · ) and the identity is (1, 0, 0, · · · ). The function ∂R : H(R) → H(R) given by ∂R (a0 , a1 , a2 , · · · ) = (a1 , a2 , a3 , · · · ) is a derivation of H(R). Rings of Hurwitz series have many interesting applications to differential algebra (cf. [276]). For m-fold stability, the following shows that ring of Hurwitz series closely mirrors that of the ground ring.

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Proposition 3.3.10. Let R be a ring. Then the following are equivalent: (1) R is m-fold stable. (2) H(R) is m-fold stable. (i)

(i)

(i)

Proof. (1) ⇒ (2) Given (an )(xn ) + (bn ) = (1, 0, 0, · · · ) (i = 1, 2, · · · , m) (i) (i) (i) in H(R), then a0 x0 + b0 = 1 (i = 1, 2, · · · , m) in R. Since R is m-fold (i) (i) stable, there exists some y0 ∈ R such that each a0 + b0 y0 = u(i) ∈ U (R). (i) (i) (i) Let y = (y0 , 0, 0, · · · ) ∈ H(R). Then each (an ) + (bn )y =
(u , ∗, ∗, · · · ). (i) (i) In view of [276, Proposition 2.5], (an ) + (bn )y ∈ U H(R) for all i. Thus, H(R) is m-fold stable. (2) ⇒ (1) Given a(i) x(i) + b(i) = 1 (i = 1, 2, · · · , m) in R, then each (i) (i) (i) (i) (i) (an )(xn ) + (bn ) = (1, 0, 0, · · · ), where (an ) = (a(i) , 0, 0, · · · ), (xn ) = (i) (i) (i) (x , 0, 0, · · · ) and (bn ) = (b , 0, 0, · · · ). As H(R) is m-fold stable, there
(i) (i) (i) exists some (yn ) ∈ H(R) such that (an ) + (bn )(yn ) = (un ) ∈ U H(R) . (i)

In view of [276, Proposition 2.5], each u0 ∈ U (R). Therefore all a(i) + (i) b(i) y0 = u0 ∈ U (R), as required. 

3.4

Exchange Rings with Artinian Primitive Factors

Lemma 3.4.1. Let R be a division ring with more than m elements, then R is m-fold stable. Proof. Assume that a1 R + b1 R = R, · · · , am R + bm R = R. If b1 · · · bm 6= 0, −1 then we choose y 6∈ {−b−1 1 a1 , · · · , −bm am }. If b1 · · · bm = 0, then we may assume that b1 = · · · = bs = 0, bi 6= 0 (s < i ≤ m). Then we choose −1 y 6∈ {−b−1 s+1 as+1 , · · · , −bm am }. In any case, a1 +b1 y, · · · , am +bm y ∈ U (R), and therefore R is m-fold stable.  The following results give more explicit descriptions for Yu’s results on exchange rings with artinian primitive factors (cf. [422, Theorem 1]). Theorem 3.4.2. Let R be an exchange ring with artinian primitive factors. 1 ∈ R, then R is m-fold stable. If m! Proof. Assume that there exist some ai , bi ∈ R (1 ≤ i ≤ m) such that a1 R + b1 R = R, · · · , am R + bm R = R, while one of ai + bi y is not a unit for any y ∈ R. Let Ω = {A | A is an ideal of R such that one of ai + bi y is not a unit for any y ∈ R modulo A}. Clearly, Ω 6= ∅. It is easy to verify that Ω is inductive. By using Zorn’s Lemma, we can find an ideal Q of R such that Q is maximal in Ω. Set S = R/Q.

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Then we easily show that S is an indecomposable ring. If J(S) 6= 0, by the maximality of Q, there is some z ∈ S such that all (ai + Q) + (bi + Q)z are units modulo J(S). Since units lift modulo J(S), this gives a contradiction to the choice of Q. Hence S is an indecomposable ring with J(S) = 0. In view of [424, Lemma 3.7], S is simple artinian. So we have a division ring D and a positive integer n such that S ∼ = Mn (D). Obviously, m! ∈ U (S). Thus m! ∈ D∗ . According to Lemma 3.4.1, D is m-fold stable. In view of Corollary 3.2.4, R/Q is m-fold stable, a contradiction. Therefore R is m-fold stable, as desired.  Corollary 3.4.3. Let R be an exchange ring of bounded index. If then R is m-fold stable.

1 m!

∈ R,

Proof. As in the proof of Corollary 1.3.14, R has artinian primitive factors. Therefore the proof is true from Theorem 3.4.2.  A ring R is a strongly π-regular ring in the case where every descending chain of right ideals xR ⊇ x2 R ⊇ · · · ⊇ xn R ⊇ · · · terminates, i.e, for any x ∈ R, there exists some n ∈ N such that xn ∈ xn+1 R. For instance, Z4 and Z8 are both strongly π-regular. Example 3.4.4. Let A = B = k[x]/(x2 ) = {a + bt | a, b ∈ k, t2 = 0} where k is a field of characteristic 5. Take M = N = k made into an A-module by α ∗ (a + bt) = αa with α, a, b ∈ k. Suppose a + bt ∈ A. If a 6= 0, then (a + bt)5 = (a + bt)6 (a − bt)a−2 . If a = 0, then (a + bt)2 = (a + bt)3 . Therefore A is a strongly π-regular ring. Assume that (a + bt)n = 0 in A.
5n 5n 5 n Then (a + bt) = 0, hence a = (a + bt) = 0. So a = 0. Therefore (a + bt)5 = a5 = 0. That is, A = B is a strongly π-regular ring of bounded 1 ∈ A, B. By virtue of Corollary 3.4.3, A and B are index 5. Clearly, 3! 3-fold stable. Clearly, M and N are cyclic right modules. It follows by Theorem 3.2.3 that T is 3-fold stable, whence T satisfies the GoodearlMenal condition. This shows that K1 (T ) ∼  = U (T )ab . Theorem 3.4.5. Let R be an exchange ring with all idempotents central. Then the following are equivalent: (1) R is m-fold stable. (2) | R/M | > m for all maximal ideals M of R. Proof. (1) ⇒ (2) Let M be a maximal ideal of R. If | R/M | ≤ m, we may assume that R/M = {x1 , · · · , xs } with 1 ≤ s ≤ m. Since xi R + (−1)R = R

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(1 ≤ i ≤ s), we can find some y ∈ R such that each xi − y ∈ U (R). This implies that y 6∈ {x1 , · · · , xs }, a contradiction. Therefore | R/M | > m, as required. (2) ⇒ (1) Assume that there exist some ai , bi ∈ R (1 ≤ i ≤ m) such that a1 R+b1 R = R, · · · , am R+bm R = R, while one of ai +bi y is not a unit for any y ∈ R. As in the proof of Theorem 3.4.2, we have a division ring D and a positive integer n such that S ∼ = Mn (D). Since R is an exchange ring with all idempotents central, so is S. This implies that n = 1. By hypothesis, | D| > m. In view of Lemma 3.4.1, D is m-fold stable, and so is R/Q. This gives a contradiction, and therefore R is m-fold stable.  1 Corollary 3.4.6. Let R be a ring with m! ∈ R (m ≥ 2). If every element in R is the sum of an idempotent and a central unit, then R is m-fold stable.

Proof. In view of [382, Theorem 30.2], R is an exchange ring. Clearly, 2 ∈ U (R). For any x = x2 ∈ R, there exists an idempotent e ∈ R and a central u ∈ U (R) such that x = e + u. Hence, (e + u)2 = e + eu + ue + u2 = e + u, and so e = 21 (1 − u) ∈ B(R). Thus, R is an exchange ring with all idempotents central. Obviously, |R/M | > m for any maximal ideal M of R. Therefore the result follows by Theorem 3.4.5.  Following Yu (cf. [419]), we call a ring R a left (right) quasi-duo ring if every maximal left (right) ideal of R is two-sided. For instance, every local ring is a left (right) quasi-duo ring. Clearly, all left (right) duo rings, all weakly left (right) duo rings and all exchange rings with all idempotents central are left (right) quasi-duo. It is claimed that every left (right) quasiduo ring is directly finite. Let R be a left quasi-duo ring, and let x, y ∈ R. If xy = 1, then Rx = R or there exists a maximal left ideal M of R such that Rx ⊆ M . Hence, x ∈ R is left invertible; otherwise, Rxy ⊆ M , and so R = M . This gives a contradiction, and we are done. Corollary 3.4.7. Let R be a left (right) quasi-duo exchange ring. Then the following are equivalent: (1) R is m-fold stable. (2) | R/M | > m for all maximal ideal M of R. Proof. (1) ⇒ (2) is similar to Theorem 3.4.5. (2) ⇒ (1) Set S = R/J(R). Given any e = e2 ∈ S and x ∈ S, there exists an idempotent f ∈ R such that e = f + J(R). Assume that x = y +J(R) with y ∈ R. One easily checks that (f y −f yf )2 = 0 = (yf −f yf )2 .

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Since R is a left (right) quasi-duo ring, by virtue of [419, Lemma 2.3], we know that f y − f yf ∈ J(R) and yf − f yf ∈ J(R). Therefore we
have f y −
yf = (f y − f yf
) − (yf −
f yf ) ∈ J(R), hence ex = f + J(R) y + J(R) = y + J(R) f + J(R) = xe. So S is an exchange ring with all idempotents central. It is easy to check that | S/N | > m for any maximal ideal N of S. Thus, by Theorem 3.4.5, S is m-fold stable. Furthermore, we conclude that R is m-fold stable, as asserted.  In [422, Theorem 1], Yu proved that every exchange ring R with artinian primitive factors is 1-fold stable. We extend this result to 2-fold rings and 3-fold stable rings. Theorem 3.4.8. Let R be an exchange ring with artinian primitive factors. Then the following are equivalent: (1) R is 2-fold stable.
(2) R has no homomorphic images Mn Z/2Z for all n ∈ N.

Proof. (1) ⇒ (2) Since R is 2-fold stable, so are its homomorphic images.
Clearly, Z/2Z is not 2-fold stable. In view of Corollary 3.2.4, Mn Z/2Z is not 2-fold stable, as desired. (2) ⇒ (1) Assume that there exist some ai , bi ∈ R (i = 1, 2) such that a1 R + b1 R = R, a2 R + b2 R = R, while one of ai + bi y is not a unit for any y ∈ R. Let Ω = {A | A is an ideal of R such that one of ai + bi y is not a unit for any y ∈ R modulo A}. As in the proof of Theorem 3.4.2, we have a division ring D and a positive integer n such that S ∼ = Mn (D). If |D| ≤ 2, then D ∼ = Z/2Z. Hence, |D| > 2. According to Lemma 3.4.1, D is 2-fold stable, hence so is R/Q. This gives a contradiction. Therefore, R is 2-fold stable.  Corollary 3.4.9. Let R be an exchange ring of bounded index. Then the following are equivalent: (1) R is 2-fold stable.
(2) R has no homomorphic images Mn Z/2Z for all n ∈ N.

Proof. As in the proof of Corollary 1.3.14, R has artinian primitive factors. Therefore we obtain the result from Theorem 3.4.8.  Corollary 3.4.10. Let R be a semilocal ring. Then the following are equivalent:

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(1) R is 2-fold stable.
(2) R has no homomorphic images Mn Z/2Z for all n ∈ N.

Proof. (1) ⇒ (2) is clear as in the proof of Theorem 3.4.8.
(2) ⇒ (1) Since R has no homomorphic images Mn Z/2Z , neither does R/J(R), where n = 1, 2, 3, · · · . As R is semilocal, R/J(R) is artinian. In view of Theorem 3.4.8, R/J(R) is 2-fold stable, and then so is R.  Theorem 3.4.11. Let R be an exchange ring with artinian primitive factors. Then the following are equivalent: (1) R is 3-fold stable.

(2) R has no homomorphic images Mn Z/2Z , Mn Z/3Z for all n ∈ N.

Proof. (1) ⇒ (2) Since R is 3-fold stable, so are its homomorphic images. It is easy to check that Z/2Z
and Z/3Z are
not 3-fold stable. By virtue of Corollary 3.2.4, Mn Z/2Z and Mn Z/3Z are not 3-fold stable, as required. (2) ⇒ (1) Assume that there exist some ai , bi ∈ R (i = 1, 2, 3) such that a1 R + b1 R = R, a2 R + b2 R = R and a3 R + b3 R = R, while one of ai + bi y is not a unit for any y ∈ R. Let Ω = {A | A is an ideal of R such that one of ai + bi y is not a unit for any y ∈ R modulo A}. As in the proof of Theorem 3.4.2, there is a division ring D and a positive integer n such that S∼ = Mn (D). If | D |≤ 3, then D ∼ = Z/2Z or Z/3Z. The assumption implies that | D |> 3. In view of Lemma 3.4.1, D is 3-fold stable, and then so is R/Q. This contradicts the choice of Q. Therefore R is 3-fold stable.  Corollary 3.4.12. Let R be an exchange ring of bounded index. Then the following are equivalent: (1) R is 3-fold stable.

(2) R has no homomorphic images Mn Z/2Z , Mn Z/3Z for all n ∈ N.

Proof. As in the proof of Corollary 1.3.14, R has artinian primitive factors. Thus we complete the proof by Theorem 3.4.11.  Corollary 3.4.13. Let R be an exchange ring with all idempotents central. Then the following hold: (1) R is 2-fold stable if and only if R has no homomorphic image Z/2Z. (2) R is 3-fold stable if and only if R has no homomorphic images Z/2Z, Z/3Z.

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Proof. (1) If R is 2-fold stable, then so are its homomorphic images; hence, R has no homomorphic image Z/2Z. Conversely, assume that R has no homomorphic image Z/2Z. Since R is an exchange ring with all idempotents central, so is every
homomorphic image of R. Then R has no homomorphic images Mn Z/2Z
(n ≥ 2). We infer that R/J(R) has no homomorphic images Mn Z/2Z (n ∈ N). In view of [372, Lemma 4.10], R/J(R) is an exchange ring of bounded index 1. According to Corollary 3.4.9, R/J(R) is 2-fold stable. Therefore R is 2-fold stable. (2) As in the above consideration, we obtain the result by Corollary 3.4.12.  The ring Z/2Z has stable range one, while it does not have unit 1-stable range. The ring Z/3Z has unit 1-stable range, while it does not satisfy the Goodearl-Menal condition. The ring
Z/3Z is 2-fold stable, while it is not 3-fold stable. The ring M2 Z/2Z
has unit 1-stable range, while it is not 2-fold stable. The ring M2 Z/3Z satisfies the Goodearl-Menal condition, while it is not 3-fold stable. Thus, we have the following relations among these conditions: 2-fold stable

⇒ :

unit 1-stable range

⇒ stable range one. :

6⇓ ⇑

6⇓ ⇑ ⇒ Goodearl-Menal condition 3-fold stable : Let R be an exchange ring with all idempotents central. It follows from Corollary 3.4.13 that (1) R is 2-fold stable if and only if it has unit 1-stable range; and (2) R is 3-fold stable if and only if R satisfies the Goodearl-Menal condition.

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Chapter 4

Strongly Stable Rings

This chapter is mainly concerned with the two classes of conditions that are stronger than stable range one. We are interested in the kernel of the natural map f : U (R) → K(R) especially in the case where the map is surjective. We will introduce strongly stable rings and medium stable rings, and focus our attention on K1 -groups of such new rings. It is shown that K1 (R) can be calculated by the normal subgroup L(R) of U (R) lying between V (R) and W (R) for the strongly stable ring R. Finally, we study one situation in which K1 (R) is determined by the normal subgroup Θ(R) of U (R), where we have the inclusions U ′ (R) ⊆ Θ(R) ⊆ V (R) ⊆ L(R) ⊆ W (R). For an exchange ring R with artinian primitive factors, we prove that K1 (R) ∼ = U (R)/V (R). If it has an additional condition of 12 ∈ R, then ∼ K1 (R) = U (R)ab . These extend the corresponding results on regular rings.

4.1

Matrix Extensions

Let us write Q(R) for the set {x ∈ R | ∃ e − e2 ∈ J(R), u ∈ U (R) such that x = eu}. One easily checks that Q(R) = {x ∈ R | ∃ e−e2 ∈ J(R), u ∈ U (R) such that x = ue}. Definition 4.1.1. A ring R is strongly stable provided that whenever aR + bR = R, there exists w ∈ Q(R) such that a + bw ∈ U (R). 97

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The following lemma is a slight generalization of a result due to Goodearl [219, Lemma 3.1], which will be repeatedly used in this book. Lemma 4.1.2. Given ax + b = 1, a, x, b ∈ R, then the following hold:

(1) If u(a+by) = 1, then
x+(1−xy)ub
a+y(1−xa) = 1. If (a+by)u = 1, then a + y(1 − xa) x + (1 − xy)ub
= 1.
(2) If (x+zb)v = 1, then
x+(1−xa)z a+bv(1−za) = 1. If v(x+zb) = 1,
then a + bv(1 − za) x + (1 − xa)z = 1.
Proof. (1)
Assume that u(a + by) = 1. Then we have x + (1 − xy)ub a + y(1 − xa) = xa + xy − xyxa + uba − xyuba + uby − xyuby − ubyxa + xyubyxa = xa + xy − xyxa + uba − xyuba + (1 − ua) − xy(1 − ua) − (1 − ua)xa + xy(1 − ua)xa = uba − xyuba + 1 − ua + xyua + uaxa − xyuaxa = u(1 − ax)a − xyu(1 − ax)a + 1
− ua + xyua + uaxa
− xyuaxa = 1. Similarly, we verify that a + y(1 − xa) x + (1 − xy)ub = 1. (2)
Assume that
(x + zb)v = 1. Then we can verify that x + (1 − xa)z a+bv(1−za) = xa+xbv −xbvza+za−xaza+zbv −zbvza−xazbv + xazbvza = xa+xbv−xbvza+za−xaza+(1−xv)−(1−xv)za−xa(1−xv)za = x(1 − ax)v − x(1 − ax)vza + za +
1 − xv − za + xvza + xaxv − xaxvza = 1.
Likewise, we have a + bv(1 − za) x + (1 − xa)z = 1, where v(x + zb) = 1.  Lemma 4.1.3. Let R be a ring. Then the following are equivalent: (1) R is strongly stable. (2) Whenever ax + b = 1, there exists w ∈ Q(R) such that aw + b ∈ U (R). (3) Whenever x, y ∈ R, there exists w ∈ Q(R) such that 1 + x(y − w) ∈ U (R). Proof. (1) ⇒ (2) is clear. (2) ⇒ (3) Given any x, y ∈ R, then (−x)y + (1 + xy) = 1. So we have w ∈ Q(R) such that 1 + x(y − w) = (−x)w + (1 + xy) ∈ U (R), as required. (3) ⇒ (1) Given aR + bR = R, then ax + by = 1 for some x, y ∈ R. For a, −x ∈ R, there exists w ∈ Q(R) such that 1+a(−x−w) = −aw+by = u ∈ U (R). So a(−wu−1 ) + byu−1 = 1. Since w ∈ Q(R), c := −wu−1 ∈ Q(R). By the definition of Q(R), we may assume that c = ve with e − e2 ∈ J(R) and v ∈ U (R); hence, ave + byu−1 = 1. Consequently, e + (1 − e)byu−1 = 1 − (1 − e)ave. From e(1 − e) ∈ J(R), we see that 1 − ave(1 − e) ∈ U (R). So 1 − (1 − e)ave ∈ U (R); hence, v −1 (−wu−1 ) + (1 − e)byu−1 ∈ U (R). That is, (−wu−1 ) + v(1 − e)byu−1 ∈ U (R). By Lemma 4.1.2, there exists

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z ∈ R such that a + byu−1 z = s ∈ U (R). For −b, yu−1 zs−1 ∈ R, there is an element t ∈ Q(R) such that 1 + (−b)(yu−1zs−1 + t) = as−1 − bt ∈ U (R). Thus, a + b(−t)s ∈ U (R) with −ts ∈ Q(R), as desired.  We note that if 1 + ab ∈ U (R), then also 1 + ba ∈ U (R). Because a ∈ Q(R) if and only if ao ∈ Q(Rop ), by Lemma 4.1.3, R is a strongly stable ring if and only if its opposite ring Rop is also a strongly stable ring. This actually shows that the notion of strongly stable rings is left-right symmetric. In other words, a ring R is a strongly stable ring if and only if whenever Ra + Rb = R, there exists w ∈ Q(R) such that a + wb ∈ U (R). Lemma 4.1.4. Let R be a ring. Then the following are equivalent: (1) R is strongly stable. (2) Whenever ax + b = 1, there exists w ∈ Q(R) such that a + bw ∈ U (R). (3) Whenever ax + b = 1, there exists y ∈ R such that a + by ∈ U (R) and 1 − xy ∈ Q(R). Proof. (1) ⇒ (2) is trivial. (2) ⇒ (3) From xa + (1 − xa) = 1, we have some z ∈ Q(R) such that x + (1 − xa)z = u ∈ U (R). Hence      −1 x 1 − xa u∗ a −b B21 (z) = B21 (z) = . −1 a ∗∗ 1 x Thus we have v ∈ U (R) such that       u∗ u0 1 0 B12 (∗), = B21 (∗) ∗∗ 01 0 v −1 and then 

u∗ ∗∗

−1

= B12 (∗)



10 0v



B21 (∗)



u−1 0 0 1



.

This implies that 

 ∗∗ = ∈ GL2 (R). ∗v     ∗∗ w∗ So we have w ∈ U (R) and y ∈ R such that = B21 (y). Hence ∗v 0 v       a −b w∗ w ∗ B21 (−y) = B21 (z) = . 1 x 0 v zw ∗ 1 0 −z 1



a −b 1 x





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Therefore a + by = w ∈ U (R) and 1 − xy = zw ∈ Q(R), as asserted. (3) ⇒ (1) Suppose that ax + b = 1 in R. Since xa + (1 − xa) = 1, by the hypothesis, we have some y ∈ R such that x + (1 − xa)y = u ∈ U (R) and 1 − ay = w ∈ Q(R). Hence  −1     a b x xa − 1 u∗ B21 (−y) = B21 (−y) = . −1 x 1 a w∗ So there is some v ∈ U (R) such that  −1     a b u0 10 = B21 (w) B12 (∗)B21 (y). −1 x 01 0v Thus we have      −1  a b 1 0 u 0 = B21 (−y)B12 (∗) B21 (−wu−1 ), −1 x 0 v −1 0 1 and then



a b −1 x



B21 (wu−1 ) =



u−1 ∗ ∗ ∗



.

Therefore we have some wu−1 ∈ Q(R) such that a + bwu−1 = u−1 ∈ U (R). Given aR + bR = R, then ax + by = 1 for some x, y ∈ R. By the above consideration, there exists w1 ∈ Q(R) such that b + axw1 = q ∈ U (R). Hence axw1 q −1 + bq −1 = 1. By the above consideration again, we can find w2 ∈ Q(R) such that a + bq −1 w2 ∈ U (R), as desired.  

 ab We note that if R is a ring and a ∈ U (R), then ∈ GL2 (R) if and cd only if d − ca−1 b ∈ U (R), which will be used in the following.

Theorem 4.1.5. Let e be an idempotent of R. If eRe and (1 − e)R(1 − e) are strongly stable, then so is R.   eRe eR(1 − e) Proof. Set T = . Clearly, we have a ring (1 − e)Re (1 − e)R(1 − e)   ere er(1 − e) ∼ isomorphism ϕ : R = T given by ϕ(r) = for (1 − e)re (1 − e)r(1 − e) any r ∈ R. It will suffice to show that T is a strongly  stable ring.  B D Given BC + D = diag(e, 1 − e) in T , then ∈ −diag(e, 1 − e) C GL2 (T ). Since eRe and (1 − e)R(1 − e) are strongly stable, they have stable range one. By virtue of Corollary 1.1.11, T has stable range one.

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Clearly, −diag(e, 1 − e)T + CT = T . Thus we have some E ∈ T such that −diag(e, 1 − e) + CE ∈ U (T ). Let   B D A = (Aij )2×2 = B21 (E) −diag(e, 1 − e) C

with all Aij = (aij st ) ∈ T (1 ≤ i, j ≤ 2, 1 ≤ s, t ≤ 2). Clearly, A21 ∈ U (T ). Moreover, we have x1 ∈ eRe, x2 ∈ (1 − e)Re, y1 ∈ eRe, y2 ∈ (1 − e)Re such that       x1 y1 e A11 + A12 = , x2 y2 0       x1 y1 0 A21 + A22 = . x2 y2 0

11 In view of Lemma 4.1.4, there is some z ∈ eRe such that a11 11 + a12 x2 z + 12 12 a11 y1 z + a12 y2 z = u ∈ U (eRe) and e − x1 z = w1 ∈ Q(eRe). So we have    e 0 0  x2 z 1 − e     A   y1 z 0 diag(e, 1 − e) y z0 2    u∗ A 12 . = ∗∗ A21 diag(w1 , 1) A22

Likewise, we can find v ∈ U (1 − e)R(1 − e) and w2 ∈ Q (1 − e)R(1 − e) such that   B D diag(∗, ∗) diag(∗, ∗)B21 (∗) −diag(e, 1 − e) C     u∗ A′12  0v = . A21 diag(w1 , w2 ) A22

Since the matrix obtained belongs to GL2 (T ), we have  −1 u∗ A′12 ∈ U (T ). G := A22 − A21 diag(w1 , w2 ) 0v Then

 B D diag(∗, ∗)B21 (∗) −diag(e, 1 − e) C = diag(∗, ∗)B21 (−W )B12 (∗)B21 (∗), diag(∗, ∗)



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where W = −G−1 A21 diag(w1 , w2 ) ∈ T . Consequently, we can find W1 , W2 ∈ U (T ) such that   B D B21 (W1 W W2 ) = diag(∗, ∗)B12 (∗)B21 (∗), −diag(e, 1 − e) C whence C +
W1 W W2 D ∈ U (T ). Since w1 ∈ Q(eRe) and w22 ∈ Q (1 − e)R(1 − e)
, we have f − f 2 ∈ J(eRe),
u1 ∈ U (eRe), g − g ∈ J (1 − e)R(1 − e) and u2 ∈ U (1 − e)R(1 − e) such that w1 = f u1 and w2 = gu2 . This shows that diag(w1 , w2 ) = diag(f, g)diag(u1 , u2 ). One easily checks that     J(eRe) 0 0
= eRe ∩ J(R) 0 J (1 −
e)R(1 − e) 0 (1 − e)R(1 − e) ∩ J(R) ⊆ T ∩ M2 J(R)
= T ∩ J M2 (R)
= J diag(e, 1 − e)M2 (R)diag(e, 1 − e) = J(T ). Hence diag(f, g) − diag(f, g)2 ∈ J(T ). Clearly, diag(u1 , u2 ) ∈ U (T ), and then W1 W W2 ∈ Q(T ). Therefore the result follows.  Corollary 4.1.6. Let e1 , · · · , en be idempotents of ,··· , a ring R. If e1 Re1 e1 Re1 · · · e1 Ren  ..  . .. en Ren are all strongly stable, then so is the ring  ... . .  



en Re1 · · · en Ren

e1 Re1 · · · e1 Ren .. ..  . Clearly, the result holds for n = 1. .. . . .  en Re1 · · · en Ren Assume the result holds for n = k ≥ 1. Now assume that n = k +1. Choose f = diag(e1 , 0, · · · , 0)(k+1)×(k+1) . Then f T f ∼ = e1 Re1 is strongly stable. Note that   e2 Re2 · · · e2 Ren

 . ..  , .. diag(e1 , · · · , en ) − f T diag(e1 , · · · , en ) − f ∼ =  .. . .   Proof. Set T = 

en Re2 · · · en Ren

which is strongly stable by the induction hypothesis. It follows from Theorem 4.1.5 that the result holds for n = k + 1. Therefore the proof is true. 

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A semiregular ring R is said to be unit semiregular provided that it has stable range one. It is easy to see that every unit semiregular ring is strongly stable. Corollary 4.1.7. Let e1 , · · · , en be idempotents of a unit semiregular ring  e1 Re1 · · · e1 Ren  ..  is strongly stable. .. R. Then the ring  ... . .  en Re1 · · · en Ren

Proof. One can easily prove that if R is unit semiregular and e is an idempotent of R, then eRe is also unit semiregular. So eRe is strongly stable. Therefore we complete the proof by Corollary 4.1.6.  Corollary 4.1.8. Let R be a ring. Then the following are equivalent: (1) R is strongly stable. (2) There exists a complete orthogonal set of idempotents, {e1 , · · · , en }, such that all ei Rei are strongly stable. Proof. (1) ⇒ (2) is trivial. (2) ⇒ (1) Suppose that {e1 , · · · , en } is a complete orthogonal set of idempotents such that all ei Rei are strongly stable. We note that R and the matrix ring (ei Rej )n×n are isomorphic, so the result follows from Corollary 4.1.6.  Corollary 4.1.9. Let M1 , · · · , Mn be right R-modules. If EndR (M1 ), · · · , EndR (Mn ) are strongly stable, then so is EndR (M1 ⊕ · · · ⊕ Mn ). Proof. Let M = M1 ⊕ · · · ⊕ Mn , and let ei be the element of EndR (M ) that projects onto the ith coordinate. Then EndR (M ) has a complete orthogonal set {e1 , · · · , en } of idempotents. Because ei EndR (M )ei ∼ = EndR (ei M ) ∼ = EndR (Mi ) for all i, we know that all ei EndR (M )ei are strongly stable. It follows from Corollary 4.1.8 that the result follows. 

Theorem 4.1.10. Let T be the ring of a Morita context (A, B, M, N, ψ, φ). If A and B are strongly stable, then so is T . Proof. Set e = diag(1A , 0). Since eT e ∼ = A and (1T − e)T (1T − e) ∼ = B, the result follows from Corollary 4.1.8.  If R is a strongly stable ring, then so is Mn (R) for any n ≥ 1. This is an immediate consequence of Theorem 4.1.10.

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Corollary 4.1.11. Let T be the ring of a Morita context (A, B, M, N, ψ, φ). If A and B are unit semiregular rings, then T is strongly stable. Proof. Clearly, A and B are strongly stable. In light of Theorem 4.1.10, the proof is true.  Theorem 4.1.12. The following are equivalent: (1) A1 , A2 and A3 are strongly stable.



 A1 0 0 (2) The formal triangular matrix ring T =  M21 A2 0  is strongly M31 M32 A3 stable.     A2 0 M21 Proof. (1) ⇒ (2) Set B = and M = . Since A2 and M32 A3 M31 A3 are strongly stable, by Theorem 4.1.10, so is the ring  B. On  the other A1 0 hand, A1 is strongly stable. By Theorem 4.1.10 again, is strongly M B stable, as desired. (2) ⇒ (1) Given ax + b = 1 in A2 , then diag(0, a, 0)diag(0, x, 0) +   w1 0 0 diag(1, b, 1) = 1T . Thus we can find some  ∗ w2 0  ∈ Q(T ) such that ∗ ∗ w3     u1 0 0 w1 0 0 diag(0, a, 0) + diag(1, b, 1)  ∗ w2 0  =  ∗ u2 0  ∈ U (T ). ∗ ∗ u3 ∗ ∗ w3       v1 0 0 e1 0 0 w1 0 0 Clearly, u2 ∈ U (R). Assume  ∗ w2 0  =  ∗ e2 0   ∗ v2 0 

∗ ∗ v3 ∗ ∗ e3  e1 0 0 e1 0 0 v1 0 0 with  ∗ e2 0  −  ∗ e2 0  ∈ J(T ),  ∗ v2 0  ∈ U (T ). One easily ∗ ∗ v3 ∗ ∗ e3 ∗ ∗ e3 checks that     J(A1 ) 0 0 U (A1 ) 0 0 J(T ) =  ∗ J(A2 ) 0  , U (T ) =  ∗ U (A2 ) 0  . ∗ ∗ U (A3 ) ∗ ∗ J(A3 ) 





2

∗ ∗ w3 

Hence e2 − e22 ∈ J(A2 ) and v2 ∈ U (A2 ). Therefore we have a + be2 v2 = u2 ∈ U (A2 ). This implies that A2 is strongly stable. Likewise, A1 and A3 are strongly stable, as asserted. 

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Corollary 4.1.13. A ring R is strongly stable if and only if so is the ring of all n × n lower triangular matrices over R. Proof. It is an immediate consequence of Theorem 4.1.12.



Analogously, we deduce that a ring R is strongly stable if and only if so is the ring of all n × n upper triangular matrices over R. 4.2

Related Rings

Lemma 4.2.1. Let I be an ideal of a ring R. If I ⊆ J(R), then the following are equivalent: (1) R is strongly stable. (2) R/I is strongly stable. Proof. (1) ⇒ (2) Given ax + b = 1 in R/I, then we have some r ∈ I such that ax + (b + r) = 1 in R. So there exist e − e2 ∈ J(R),
u ∈ U (R) a(eu) + b ∈ U R/I . Clearly, such that a(eu) + (b + r) ∈ U (R), whence
e − e2 ∈ J(R/I); hence, eu ∈ Q R/I . Therefore R/I is a strongly stable ring. (2) ⇒ (1) Given
ax + b = 1 in R, then ax +
b = 1 in R/I. Thus there is some w ∈ Q R/I
such that aw +
b ∈ U R/I . Assume that w = eu with e − e2 ∈ J R/I and u ∈ U R/I . Since units can be lifted modulo J(R), we assume that u ∈ U (R); hence, a(eu) + b = v + r with v ∈ U
(R) and r ∈ I. For any x, y ∈ R, we see that 1 − x(e − e2 )y + I ∈ U R/I ; hence, there exists some u ∈ R such that

1 − x(e − e2 )y + I (u + I) = 1 + I = (u + I) 1 − x(e − e2 )y + I . This implies that



1 − 1 − x(e − e2 )y u, 1 − u 1 − x(e − e2 )y ∈ I ⊆ J(R).

Therefore 1 − x(e − e2 )y ∈ U (R), and so e − e2 ∈ J(R). As a result, we deduce that a(eu)+b ∈ U (R) with e−e2 ∈ J(R) and u ∈ U (R). By Lemma 4.1.3, R is strongly stable.  Let R be a semilocal ring. Then R/J(R) is semisimple, so it is unitregular. It follows by Lemma 4.2.1 that R is strongly stable. Thus every semilocal ring is strongly stable.

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Let M be an R-R-bimodule. The trivial extension of R by M is the ring T (R, M ) of pairs (r, m), where r ∈ R and m ∈ M , and with the usual addition and multiplication given by (r1 , m1 )(r2 , m2 ) = (r1 r2 , r1 m2 +m1 r2 ) for r1 , r2 ∈ R and m1 , m2 ∈ M . So as to construct strongly stable rings, we observe the following fact. Theorem 4.2.2. Let R be a ring, and let M be a right R-module. Then the following are equivalent: (1) R is strongly stable. (2) T (R, M ) is strongly stable.
Proof. One directly verifies that J T (R, M ) = {(x, m) | x ∈
J(R), m ∈ M }. Next, construct a map ψ : R → T (R, M )/J T (R, M )
given by x 7→ (x, 0) for any x ∈ R. Let (x, m) ∈ T (R, M )/J T (R, M ) , we now have (x, m) = (x, 0) + (0, m) = ψ(x). Hence ψ is surjective. On the
other hand, ker(ψ) = {x ∈ R | (x, m) ∈ J T (R, M ) } = J(R). Thus,
∼ T (R, M )/J T (R, M ) = R/J(R). So we obtain the result by Lemma 4.2.1.  It follows from Theorem 4.2.2 that a ring R is strongly stable if and only if so is T (R, R). Proposition 4.2.3. Let R be a ring, then R[[x1 , · · · , xn ]] is strongly stable if and only if so is R. Proof. Construct a surjective ring homomorphism ϕ : R[[x1 , · · · , xn ]] → R
given by ϕ f (x1 , · · · , xn ) = f (0, · · · , 0). Then R[[x1 , · · · , xn ]]/ker(ϕ) ∼ = R. It is easy to check that ker(ϕ) = {f (x , · · · , x ) | f (0, · · · , 0) = 0} ⊆ 1 n
J R[[x1 , · · · , xn ]] . Therefore we complete the proof from Lemma 4.2.1.

Polynomial rings over a strongly stable ring are never strongly stable. Indeed, every polynomial
ring does not have stable range one. But, we claim that R[x]/
xn+1 is strongly stable if and only if so is R. Denote
n+1 n+1 ∼ n+1 x in R[x]/ x by u, so R[x]/ x = 0. As in the = R[u]
with u ∼ proof of [264, Proposition 2.3], R[u]/J R[u] = R/J(R), and the result immediately follows by Lemma 4.2.1. Theorem 4.2.4. If R/J(R) is an exchange ring with artinian primitive factors, then R is strongly stable.

Proof. Let S = R/J(R). Then S is an exchange ring with artinian primitive

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factors. Assume that there exist some a, b ∈ S such that aS + bS = S, while a + by 6∈ U (S) for any unit-regular y ∈ S. Let Ω = {I E S | a + by 6∈ U (S/I) for any unit-regular y ∈ S/I}. Then Ω 6= ∅. Given any ascending ∞ S chain A1 ⊆ A2 ⊆ · · · ⊆ An ⊆ · · · in Ω, set M = An . Then M is an n=1

ideal of S. If
M is not in Ω, then
there exist some e, u, v, w ∈ S such that 1 − a + b(eu) v, 1 − v a + b(eu) , e − e2, 1 − uw,
1 − wu ∈ M . Thus, we can
find n1 , n2 , n3 , n4 , n5 ∈ N such that 1− a+b(eu) v ∈ An1 , 1−v a+b(eu) ∈ An2 , e − e2 ∈ An3 , 1 − uw ∈ An4 , 1 − wu ∈ An5 . Let p = max {ni }. Then 1≤i≤5

a + b(eu) ∈ U (S/Ap ), where eu ∈ S/Ap is unit-regular. This contradicts the choice of Ap , so M ∈ Ω. That is, Ω is inductive. By using Zorn’s Lemma, we have an ideal Q of S such that Q is maximal in Ω. Let T = S/Q. As in the proof of Theorem 1.3.13, T is an indecomposable ring with J(T ) = 0. According to [424, Lemma 3.7], T is simple artinian. In view of [4, Theorem 13.4], S/Q ∼ = Mn (D) for a division ring D; whence, S/Q is unit-regular. This contradicts the choice of Q. Therefore aS + bS = S implies that there exists a unit-regular y ∈ R such that a + by ∈ U (S); hence, S is strongly stable. In view of Lemma 4.2.1, R is a strongly stable ring.  Corollary 4.2.5. If R/J(R) is an exchange ring of bounded index, then R is strongly stable. Proof. As in the proof of Corollary 1.3.14, R/J(R) is an exchange ring having artinian primitive factors. In view of Theorem 4.2.4, R is a strongly stable ring.  Corollary 4.2.6. If R is a right (left) quasi-duo exchange ring, then R is strongly stable. Proof. As in the proof of Corollary 3.4.7, R/J(R) is an exchange ring with all idempotents central. In view of [372, Lemma 4.10], R/J(R) is an exchange ring of bounded index 1. Therefore R is a strongly stable ring by Corollary 4.2.5. 

4.3

K1 -Groups

If H and K are subgroups of U (R), then the join of H and K is the subgroup W generated by H and K and is denoted by H K. Suppose that x ∈ J(R).

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Then 1 + x ∈ U (R). For
any y, z ∈ R, we have 1 − 1 − (1 +
x)−1 zy = (1 + x)−1 1 + x(1 + zy) ∈ U (R). So 1 − y 1 − (1 + x)−1 z ∈ U (R). This implies that (1 + x)−1 ∈ 1 + J(R). Therefore 1 + J(R) is a normal subgroup of U (R).
In fact, 1 + J(R) is the kernel of the natural
map W U (R) ։ U R/J(R) . Set L(R) = W (R) ∩ V (R) (1 + J(R)) . Then L(R) is a normal subgroup of U (R). Moreover, V (R) ⊆ L(R) ⊆ W (R). In this section, we study K1 -groups of strongly stable rings. Lemma 4.3.1. Let a, b, c ∈ R
with p(a, b, c) ∈ U (R). If a ∈ Q(R), then p(a, b, c) ≡ p(c, b, a) mod L(R) .

Proof. Since a ∈ Q(R), there exist e − e2 ∈ J(R), u ∈ U (R) such that a = eu. So p(a, b, c) = eu+c+eubc, and then p(a, b, c)u−1 = e+cu−1 +eubcu
−1.
Set w1 = p e, −ub(1−e) . From e(1−e) ∈ J(R), we have p −ub(1−e), e = 1 − ub(1 − e)e ∈ U (R). We see that if p(x, y) ∈ U (R), then p(y, x) ∈ U (R). Hence w1 = 1 − eub(1 − e) ∈ U (R). Let r = eub(1 − e)e + eub(1 − e)eubcu−1. Then

p(a, b, c)u−1 = w1−1 1 − eub(1 − e) e + cu−1
+ eubcu−1 = w1−1 e + cu−1 + eubecu−1 − r . As r ∈ J(R), we deduce that
k := e + cu−1 + eubecu−1 ∈ U (R). Therefore p(a, b, c)u−1 = w1−1 1 − rk −1 k. Set w2 = p(1 − e, −ube), t = e(1 − e)ube + cbe(1 − e)ube and l = e + cu−1 + cu−1 eube. We also have w2 , l ∈ U (R) and p(c, b, a)u−1 = e + cu−1 + cbe = (1 − tl−1 )lw2−1
−1
= (1 − tl−1 ) 1 − l−1 (1 − e)eube 1 + (cu−1 − 1 + e)(1 + eube) w2−1

−1 −1 ≡ (1 − tl−1 ) 1 − 1 + (1 + eube)(cu−1 − 1 + e)
l (1 − e)eube −1 w2 mod L(R)
−1
≡ (1 − tl−1 ) 1 − l−1 (1 − e)eube 1 − eube(1 − e)k −1 (1 − rk −1 )−1
w1 p(a, b, c)u−1 w2−1 mod L(R) .
−1
Set z = (1 − tl−1 ) 1 − l−1 (1 − e)eube 1 − eube(1 − e)k −1 (1 − rk −1 )−1 . Then z ∈ 1 + J(R). In addition, we have zw1 p(a, b, c)u−1 w2−1 up(c, b, a)−1 ∈ L(R). From [321, Lemma 1.1], we see that u−1 w2−1 uw2 ∈ U (R)′ ⊆ V (R) ⊆ L(R), so zw1 p(a, b, c)w2−1 p(c, b, a)−1 ∈ L(R). Set w = 1 − ub(e − e2 ). Then

−1 w ∈ U (R). Clearly, w1 w−1 = p − e, ub(1 − e) p ub(1 − e), −e ∈ V (R); hence, w1 w−1 ∈ L(R). Also we have w2−1 w = p(1 − e, −ube)−1 p(−ube, 1 − e) ∈ V (R) ⊆ L(R). Hence zp(a, b, c)p(c, b, a)−1 ≡ zwp(a, b, c)w−1 p(a, b, c)−1 p(a, b, c)p(c, b, a)−1 = zwp(a, b, c)w−1 p(c, b, a)−1 = zw1 w1−1 wp(a, b, c)w2−1 w2 w−1 p(c, b, a)−1 ≡ zw1 p(a, b, c)w2−1 p(c, b, a)−1 (mod L(R)).

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Since p(a, b, c) p(c, b, a)−1 ∈ W (R), we deduce that z ∈ W (R), whence, z ∈ L(R). Therefore p(a, b, c) ≡ p(c, b, a)(mod L(R)).  Theorem 4.3.2. If R is a strongly stable ring, then K1 (R) ∼ = U (R)/L(R). Proof. Let R be a strongly stable ring. Then it has stable range one. Using Lemma 2.4.16, we have K1 (R) ∼ = U (R)/W (R). Since L(R) ⊆ W (R), the only thing that remains to be proved is W (R) ⊆ L(R). For any a, b, c ∈ R with p(a, b, c) ∈ U (R), we see that p(c, b, a) ∈ U (R). Since R is a strongly stable ring, by using Condition (3) of Lemma 4.1.2, we have w ∈ Q(R) such that 1 + b(c − w) ∈ U (R). Let t = c − w. Then c = t + w and 1 + bt ∈ U (R). As in the proof of Lemma 2.4.17, we see that
−1 −1 p(a, b, c) = (1 + tb) p (1 + tb)(a + t + abt), b(1 + tb) , w . From Lemma 4.3.1, we have

p(a, b, c) ≡ (1 + tb)−1 p w, b(1 + tb)−1 , (1 + tb)(a + t + abt) mod L(R)
= (1 + tb)−1 p(w, b(1 + tb)−1 )(1 + tb)(a + t + abt) +
p(w) = (1 + tb)−1 p(w, b(1 + tb)−1 )p(t, b)p(a, b, t) + p(w)
= (1 + tb)−1 p(w, b(1 + tb)−1 )p(t, b, a)p(b, t) + p(w) .
On the other hand, p(c, b, a) = p(w, (1 + bt)−1 b)p(t, b, a)p(b, t) + p(w) (1 + −1 bt)−1 . Since b(1+tb) = (1+bt)−1 b, we have (1+tb)p(a, b, c) ≡ p(c, b, a)(1+
bt) mod L(R) . Therefore p(a, b, c)p(c, b, a)−1 ∈ L(R), as asserted. 

Corollary 4.3.3. If R/J(R) is an exchange ring with artinian primitive factors, then K1 (R) ∼ = U (R)/L(R).

Proof. By virtue of Theorem 4.2.4, R is a strongly stable ring, whence Theorem 4.3.2 applies.  Corollary 4.3.4. If R/J(R) is an exchange ring of bounded index, then K1 (R) ∼ = U (R)/L(R). Proof. As in the proof of Corollary 1.3.14, R/J(R) has artinian primitive factors. Hence the result follows from Corollary 4.3.3.  If R is an exchange ring with artinian primitive factors , then so is R/J(R). In general, the converse is not true. Let R = {m/n ∈ Q | 2 ∤ n and 3 ∤ n (m/n in lowest terms)}. Then R/J(R) is semisimple. Hence R/J(R) is an exchange ring with artinian primitive factors. But R is not an exchange ring because idempotents do not lift modulo J(R). We have more explicit results on the K1 -groups of exchange rings with artinian primitive factors.

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Lemma 4.3.5. Let a, b, c ∈ R with p(a, b,
c) ∈ U (R). If a ∈ R is unitregular, then p(a, b, c) ≡ p(c, b, a) mod V (R) .

Proof. Since a ∈ R is unit-regular, there exist e = e2 ∈ R, u ∈ U (R) such that a = eu. So
p(e, ub, cu−1) = 1 − eub(1 − e) p(e, ub, cu−1) ≡ e + cu−1 + eubecu−1
= 1 + (1 + eube) cu−1
− (1 − e) ≡ 1 + cu−1 − (1 − e) (1 + eube) = c + e + cu−1 eube
= p(cu−1 , ub, e) 1 − (1 − e)ube
≡ p(cu−1 , ub, e) mod V (R) . Thus, we see that p(a, b, c)u−1 = e + cu−1 + e(ub)(cu−1 ) = p(e, ub, cu−1) ≡ p(cu−1 , ub, e) = cu−1 + e + cube
= p(c, b, a)u−1 mod V (R) . Hence, p(a, b, c)p(c, b, a)−1 ∈ V (R), as desired. 

Lemma 4.3.6. Let R be an exchange ring with artinian primitive factors. Then for any x, y ∈ R, there exists a unit-regular w ∈ R such that 1 + x(y − w) ∈ U (R). Proof. As in the proof of Theorem 4.2.4, we show that aR+ bR = R implies that there exists a unit-regular s ∈ R such that a + bs ∈ U (R). For any x, y ∈ R, we have (−x)y + (1 − xy) = 1. Hence, there is a unit-regular w ∈ R such that (−x)w + (1 + xy) ∈ U (R). Thus, 1 + x(y − w) ∈ U (R), as asserted.  In [17, Theorem 2.8], Ara et al. proved that if R is a separative exchange ring, then the natural map U (R) → K1 (R) is surjective. For an exchange ring R with artinian primitive factors, the following result gives a more explicit description of the kernel of such a map. Theorem 4.3.7. Let R be an exchange ring with artinian primitive factors. Then K1 (R) ∼ = U (R)/V (R). Proof. In view of Theorem 1.3.13, R has stable range one. Using Lemma 2.4.16, we have K1 (R) ∼ = U (R)/W (R). Clearly, V (R) ⊆ W (R). It suffices to prove that W (R) ⊆ V (R).

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For any a, b, c ∈ R with p(a, b, c) ∈ U (R), we see that p(c, b, a) ∈ U (R). By Lemma 4.3.6, we have a unit-regular w ∈ R such that 1 + b(c − w) ∈ U (R). Let t = c − w. Then c = t + w and 1 + bt ∈ U (R). As in the proof −1 of Theorem
4.3.2, we get p(a, b, c) = (1 + tb) p (1 + tb)(a + t + abt), b(1 + −1 tb) , w . Furthermore, it follows by Lemma 4.3.5 that

p(a, b, c) ≡ (1 + tb)−1 p w, b(1 + tb)−1 , (1 + tb)(a + t + abt)
mod V (R) = (1 + tb)−1 p(w, b(1 + tb)−1 )p(t, b, a)p(b, t) + p(w) .

On the other hand,


p(c, b, a) = p w, (1 + bt)−1 b, (t + a + tba)(1 + bt) (1
+ bt)−1 = p(w, (1 + bt)−1 b)p(t, b, a)p(b, t) + p(w) (1 + bt)−1 .

−1 As b(1 + tb)−1 + tb)p(a, b, c) ≡ p(c, b, a)(1 +
= (1 + bt) b, we get (1 −1 bt) mod V (R) . Therefore p(a, b, c)p(c, b, a) ∈ V (R), and we are done. 

Corollary 4.3.8. Let R be an exchange ring of bounded index. Then K1 (R) ∼ = U (R)/V (R).

Proof. As in the proof of Corollary 1.3.14, R has artinian primitive factors. According to Theorem 4.3.7, we obtain the result.  4.4

Medium Stable Rings

In this section, we introduce a new class of stable rings: the medium stable ring. A ring R is said to be medium stable provided that ax + b = 1 implies that there exists some u ∈ U (R) such that a + bu ∈ R and 1 − xu ∈ R are unit-regular. We show that this new condition lies between the GoodearlMenal condition and the unit 1-stable range property. Furthermore, K1 (R) of a medium stable ring R is investigated. If R is medium stable, it is shown that K1 (R) ∼ = U (R)/Θ(R), where Θ(R) is an intermediate subgroup between V (R) and U (R)′ . If R is a regular ring having unit 1-stable range, then ax + b = 1 implies that there exists some u ∈ U (R) such that a + bu ∈ U (R). Since R is also unit-regular, 1 − xu = (1 − xu)v(1 − xu) for some v ∈ U (R). This proves that for regular rings, the notions of unit 1-stable range and medium stable condition coincide. Also we see that every ring satisfying the Goodearl-Menal condition is medium stable by Lemma 4.4.1. The ring Z/2Z is strongly stable, while it does not have unit 1-stable range. The ring Z/3Z is medium stable, while it does not satisfy the

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Goodearl-Menal condition. We may show the relations among these conditions: Goodearl-Menal condition 6⇑ ⇓ medium stable

strongly stable 6⇓ ⇑

⇒ unit 1-stable range. :

Lemma 4.4.1. Let R be a ring. Then the following are equivalent: (1) R is medium stable. (2) For any x, y ∈ R, there exists a unit-regular a ∈ R such that 1 + xa, y + a ∈ U (R). Proof. (1) ⇒ (2) For any x, y ∈ R, we see that xy + (1 − xy) = 1. Since R is medium stable, there exists some u ∈ U (R) such that x + (1 − xy)u ∈ U (R) and 1 − yu ∈ R is unit-regular. Let b = 1 − yu. Then y + bu−1 ∈ U (R). Furthermore, u + x(1 − yu) ∈ U (R); hence, 1 + x(1 − yu)u−1 ∈ U (R), i.e., 1 + xbu−1 ∈ U (R). Let a = bu−1 . Then 1 + xa, y + a ∈ U (R) and a ∈ R is unit-regular. (2) ⇒ (1) Given ax + b = 1 in R, we can find a unit-regular w ∈ R such that 1 + aw, x + w ∈ U (R). Let u = 1 + aw and v = x + w. Thus, 1 + a(v − x) = u, i.e., av + b = u. Hence, a + bv −1 ∈ U (R). In addition, 1 − xv −1 = wv −1 ∈ R is unit-regular, and we are done.  Theorem 4.4.2. Let e be an idempotent of R. If eRe and (1 − e)R(1 − e) are medium stable, then so is R.   eRe eR(1 − e) Proof. Let T = . It suffices to show that T (1 − e)Re (1 − e)R(1 − e) is medium stable.     a1 n1 a2 n2 Given any , ∈ T , we have unit-regular elements m1 b 1 m2 b 2 a ∈ eRe and b ∈ (1 − e)R(1 − e) such that a1 + a = u1 ∈ U (eRe),

1+ a2 a = v1 ∈ U (eRe),
b1 − m1 u−1 n + b = u ∈ U (1 − e)R(1 − e) and 1 2
1 1 + b2 − m2 av1−1 n2 b = v2 ∈ U (1 − e)R(1 − e) . So 

   a1 n1 u 1 n1 + diag(a, b) = , m1 b 1 m b +b    1 1  a2 n2 v1 n2 b 1T + diag(a, b) = . m2 b 2 m2 a 1 + b 2 b

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One easily checks that 

−1  −1 −1 −1 −1 −1  u1 + u−1 u 1 n1 1 n1 u2 m1 u1 −u1 n1 u2 = , −1 m1 b 1 + b −u−1 u−1 2 m1 u 1 2   −1  −1 v1 n2 b v1 + v1−1 n2 bv2−1 m2 av1−1 ) −v1−1 n2 bv2−1 . = −v2−1 m2 av1−1 v2−1 m2 a 1 + b 2 b Clearly, diag(a, b) ∈ T is unit-regular. Therefore T is medium stable, as needed.  Corollary 4.4.3. Let R be a ring. Then the following are equivalent: (1) R is medium stable. (2) There exists a complete orthogonal set of idempotents, {e1 , · · · , en }, such that all ei Rei are medium stable. Proof. (1) ⇒ (2) is trivial. (2) ⇒ (1) The result holds for n = 1, 2. Assume that the result holds for n ≤ k (k ≥ 2). Let n = k + 1. Since (e1 + e2 ) + e3 + · · · + ek+1 = 1, by induction, it suffices to show that (e1 + e2 )R(e1 + e2 ) is medium stable. Clearly, (e1 + e2 )e1 (e1 + e2 ) = e1 and (e1 + e2 )e2 (e1 + e2 ) = e2 . Thus, (e1 + e2 )e1 (e1 + e2 ) + (e1 + e2 )e2 (e1 + e2 ) = e1 + e2 . In view of Theorem 4.4.2, (e1 + e2 )R(e1 + e2 ) is medium stable, as required.  Corollary 4.4.4. Let T be the ring of a Morita context (A, B, M, N, ψ, φ). If A and B are medium stable, then so is T . Proof. Set e = diag(1A , 0). Then eT e ∼ = A and (1T − e)T (1T − e) ∼ = B. So the result follows from Theorem 4.4.2.  By induction, we see that if R is medium stable, then so is Mn (R) for each n ∈ N. Theorem 4.4.5. Let R be an exchange ring with artinian primitive factors. Then the following are equivalent: (1) R is medium stable. (2) R has unit 1-stable range. (3) 1R = α + β with α, β ∈ U (R). Proof. (1) ⇒ (2) is obvious. (2) ⇒ (3) is clear by Theorem 2.3.5.

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(3) ⇒ (1) Assume that there exist some x, y ∈ R such that (1 + xa)(y + a) 6∈ U (R) for any unit-regular a ∈ R. Let Ω = {I E R | (1 + xa)(y + a) 6∈ U (R/I) for any unit-regular a ∈ R/I}. Then Ω 6= ∅. Given any ascending chain A1 ⊆ A2 ⊆ · · · ⊆ An ⊆ · · · in Ω, ∞ S set M = An . Then M is an ideal of R. If M is not in Ω, then there n=1

exist some e, u, v ∈ R such that



1 + x(eu) y + (eu) v − 1, 1 − v 1 + x(eu) y + (eu) , e − e2 , 1 − uv, 1 − vu ∈ M.

Thus, we can find n1 , n2 , n3 , n4 , n5 ∈ N such that



1 + x(eu) y + (eu) v − 1 ∈ An1 , 1 − v 1 + x(eu) y + (eu) ∈ An2 , e − e2 ∈ An3 , 1 − uv ∈ An4 , 1 − vu ∈ An5 .

Let p = max {ni }. Then 1 + x(eu) y + (eu) ∈ U (R/Ap ), where eu ∈ 1≤i≤5

R/Ap is unit-regular. This contradicts the choice of Ap , so M ∈ Ω. Thus, Ω is inductive. By using Zorn’s Lemma, we have an ideal Q of R such that Q is maximal in Ω. Let S = R/Q. As in the proof of Theorem 1.3.13, we see that S is indecomposable as a ring. If J(S) 6= 0, we may assume that J(S) = N/Q with N % Q. By the maximality of Q, there exists a unit-regular c ∈ R/N such that (1 + xc)(y + c) ∈ U (R/N ). Obviously, there exists a natural epimorphism ϕ : R/Q ։ R/N given by ϕ(r + Q) = r + N for any r ∈ R. Thus, S/ker(ϕ) ∼ = R/N , where ker(ϕ) = J(S), and
so we have a unit-regular c ∈ S/J(S) such that (1 + xc)(y + c) ∈ U S/J(S) . Thus, we have a unitregular c ∈ R/Q such that (1 + xc)(y + c) ∈ U (R/Q), a contradiction. This implies that J(S) = 0, so S is an indecomposable ring with J(S) = 0. According to [424, Lemma 3.7], R/Q ∼ = Mn (D) for a division ring. By assumption, 1D = γ + δ with γ, δ ∈ D∗ . Clearly, |D| ≥ 3. Let s, t ∈ D. If s = 0, we choose a 6∈ {−t}. If s 6= 0, we choose a 6∈ {−s−1 , −t}. Then 1 + sa, t + a ∈ D∗ . In view of Corollary 4.4.4, Mn (D) is medium stable. This contradicts the choice of Q. Therefore for any x, y ∈ R, there exists a unit-regular a ∈ R such that (1 + xa)(y + a) ∈ U (R). Assume that xy = 1. Then we can find some a ∈ R such that (1+ya)(x+ a) ∈ U (R). Let u = 1 + ya and v = x + a. Then 1 + y(v − x) = u, and so 1 − yx + yv = u. Thus, v = x(1 − yx + yv) = xu. Clearly, there exists an element w ∈ U (R) such that uv = w. Hence, 1 = w−1 uv = w−1 uxu. This implies that u ∈ U (R). Thus, x = u−1 wu−1 , and so yx = 1. Thus, for any

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x, y ∈ R, there exists a unit-regular a ∈ R such that 1 + xa, y + a ∈ U (R). Consequently, R is medium stable.  Let R be an exchange ring with artinian primitive factors. Then Mn (R) (n ≥ 2) is medium stable. Since R is an exchange ring, so is Mn (R) by [9, Theorem 1.4]. As in the proof of Corollary 2.3.8, we see that 1Mn (R) = α + β with α, β ∈ GLn (R). By virtue of Theorem 4.4.5, we are done.  Corollary 4.4.6. Let R be an exchange ring of bounded index. Then the following are equivalent: (1) R is medium stable. (2) R has unit 1-stable range. (3) 1R = α + β with α, β ∈ U (R). Proof. As in the proof of Corollary 1.3.14, R is an exchange ring having artinian primitive factors. In view of Theorem 4.4.5, the result follows.  Corollary 4.4.7. Let R be a right (left) quasi-duo exchange ring. Then the following are equivalent: (1) R is medium stable. (2) R has unit 1-stable range. (3) 1R = α + β with α, β ∈ U (R). Proof. (1) ⇒ (2) and (2) ⇒ (3) are clear. (3) ⇒ (1) As in the proof of Corollary 3.4.7, R/J(R) is an exchange ring with all idempotents central. In view of [372, Lemma 4.10], R/J(R) is an exchange ring of bounded index 1. Thus, R/J(R) is a medium stable ring from Corollary 4.4.6. For any x, y ∈ R, there
exists a unit-regular w ∈ R/J(R) such that 1 + xw, y + w ∈ U R/J(R) . Clearly, every unit lifts modulo J(R). Since R is an exchange ring, it follows by [382, Theorem 29.2] that every idempotent lifts modulo J(R). Thus, we may assume that w ∈ R is unit-regular. Further, we have u, v ∈ U (R), s, t ∈ J(R) such that 1 + xw = u + s, y + w = v + t. Therefore 1 + xw, y + w ∈ U (R), and we are done. 

 Let G be a group, and let H ⊆ G. We use H to denote the subgroup  W generated by H. Let Θ(R) = U (R)′ 1 + er(1 − e) | e = e2 , r ∈ R , where U (R)′ denotes the commutator subgroup of U (R). For any

x ∈ U (R), e = e2 ∈ R and r ∈ R, x 1 + er(1 − e) x−1 = x 1 + er(1 − e) x−1 1 + er(1 −

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−1
e) 1 − er(1 − e) ∈ Θ(R). From this, we see that Θ(R) is a normal subgroup of U (R) such that U ′ (R) ⊆ Θ(R) ⊆ V (R). Now we determine the kernel of the natural map U (R) → K1 (R) for the medium stable ring R. Lemma 4.4.8. Let a, b ∈ R with
p(a, b) ∈ U (R). If a ∈ R is unit-regular, then p(a, b) ≡ p(b, a) mod Θ(R) .

Proof. Since a ∈ Q(R), there exist e = e2 ∈ R, u ∈ U (R) such that a = eu. Then p(a, b) = 1 + ueb −1 = u(1 + ebu)u−1 (1 + ebu) (1 + ebu)
′ ≡ 1 + ebu mod U
(R)
≡ 1 − ebu(1 − e) (1 + ebu) mod Θ(R) = 1 + ebue
= (1 + bue) 1 − (1 −
e))bue ≡ 1 + bue mod Θ(R) = p(b, a). Hence, p(a, b)p(b, a)−1 ∈ Θ(R), as required.



Theorem 4.4.9. If R is medium stable, then K1 (R) ∼ = U (R)/Θ(R). Proof. Let R be a medium stable ring. Then it has unit 1-stable range. In view of Lemma 2.4.17, K1 (R) ∼ = U (R)/V (R). Since Θ(R) ⊆ V (R), the only thing that remains to be proved is V (R) ⊆ Θ(R). For any a, b ∈ R with p(a, b) ∈ U (R), we see that p(b, a) ∈ U (R). Since R is medium stable, we have a unit-regular w ∈ R such that a − w, 1 + bw ∈ U (R). Let t = a − w. Then a = t + w and 1 + bw ∈ U (R). Observe that p(a, b) = 1 + ab = 1 + (w + t)b = 1 + wb + tb = 1 + wb + t(1 + bw)
−1 (1 + bw)b = 1 + t(1 + bw)−1
b (1 + wb) = p t, (1 + bw)−1 b (1 + wb).

Furthermore, we have

p(b, a) = 1 + ba = 1 + bw + bt −1 = 1 + bw + b(1 + wb)(1 + wb) t
−1 = (1 + bw)p b(1 + wb) , t .

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−1 −1 Since b(1 + bw)
+ wb) = (1−1
b, it follows
from Lemma 4.4.8 that p t, (1 + −1 bw) b ≡ p b(1 + wb) , t mod Θ(R)
. By Lemma 4.4.8 again, p(w, b) =
1 + wb ≡ 1 + bw = p(b, w) mod Θ(R) . Thus, p(a, b) ≡ p(b, a) mod Θ(R) . Therefore p(a, b)p(b, a)−1 ∈ Θ(R), as asserted. 

Corollary 4.4.10. Let R be an exchange ring with artinian primitive factors. If 1 = α + β with α, β ∈ U (R), then K1 (R) ∼ = U (R)/Θ(R). Proof. By Theorem 4.4.5, R is a medium stable ring, whence Theorem 4.4.9 applies.  Corollary 4.4.11. Let R be an exchange ring of bounded index. If 1 = α+β with α, β ∈ U (R), then K1 (R) ∼ = U (R)/Θ(R). Proof. As in the proof of Corollary 1.3.14, R has artinian primitive factors. Hence the result follows from Corollary 4.4.10.  Theorem 4.4.12. Let R be an exchange ring with artinian primitive factors. If 21 ∈ R, then K1 (R) ∼ = U (R)ab . Proof. Let e ∈ R be an idempotent, and let x ∈ R. Assume that e = u + v with some u, v ∈ U (eRe). Then we check that 1 + e(1 − u)x(1 − e)

= (u + 1 − e) 1 − ex(1 − e) (u−1 + 1 − e) 1 + ex(1 − e)

−1 = (u + 1 − e) 1 − ex(1 − e) (u + 1 − e)−1 1 − ex(1 − e) ∈ U (R)′ .

1 1 e = 2e + 2e with 2e ∈ U (eRe). This implies that

Clearly, 1R = 2 + 2 . Thus, 2 1 + er(1 − e) | e = e , r ∈ R ⊆ U (R)′ . Hence, Θ(R) = U (R)′ , and so K1 (R) ∼  = U (R)ab .

Corollary 4.4.13. Let R be an exchange ring of bounded index. If then K1 (R) ∼ = U (R)ab .

1 2

∈ R,

Proof. As in the proof of Corollary 1.3.14, R has artinian primitive factors, and therefore we complete the proof by Theorem 4.4.12.  Theorem 4.4.14. Let R be an exchange ring with all idempotents central. If 1 = α + β with α, β ∈ U (R), then K1 (R) ∼ = U (R)ab . Proof. In view of Corollary 4.4.7, R is medium stable. By virtue of Theorem 4.4.9, K1 (R) ∼ = U (R)/Θ(R). Obviously, < 1 + er(1 − e) | e = e2 , r ∈ R > = < 1 >⊆ U (R)′ . Thus, Θ(R) = U (R)′ , and so K1 (R) ∼  = U (R)ab .

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Corollary 4.4.15. Let R be an abelian regular ring. If 1 = α + β with α, β ∈ U (R), then K1 (R) ∼ = U (R)ab . Proof. Obviously, R is an exchange ring with all idempotents central. Therefore we complete the proof by Theorem 4.4.14.  If R is unit-regular, then K1 (R) ∼ = U (R)/V (R). This is done from Lemma 2.4.16 and Lemma 4.3.5. If R is unit-regular and 21 ∈ R, analogously to Theorem 4.4.12, K1 (R) ∼ = U (R)ab
. This should be contracted to the following fact. Let
R = M2 (Z/2Z . Then R is unit-regular. Clearly, K1 (R) ∼ = K1 Z/2Z ∼ = {1}. It is easy to verify that
U (R) = GL2 Z/2Z             01 01 10 10 11 11 = , , , , , , 10 11 01 11 01 10 U (R)′ = GL2 (Z/2Z)′       10 11 01 = , , . 01 10 11 Thus, we see that |U (R)ab | = 2, and so K1 (R) ≇ U (R)ab .

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Chapter 5

Weakly Stable Rings

This chapter is concerned with a condition which is slightly weaker than stable range one. A ring R is said to be weakly stable provided that aR + bR = R implies that there exists some y ∈ R such that a+by ∈ R is right or left invertible. In Section 5.1, we investigate comparability of modules over weakly stable rings. We prove that the weakly stable property is Morita invariant. In Section 5.2, we see that this condition on an exchange ring coincides with one-sided unit-regularity. It is shown that an exchange ring R is weakly stable if and only if it possesses a type of the comparability of modules. In Section 5.3, we discuss factorizations of regular elements in a weakly stable exchange ring. Finally, in Section 5.4, we develop a theory about the condition by means of factorizations of 2 × 2 invertible matrices. For papers on topics related to weakly stable rings, the reader is referred to [75], [197], [398] and [410].

5.1

Comparability of Modules

The main purpose of this section is to investigate the comparability of modules over a weakly stable ring. We begin with the following simple fact, which shows that the weakly stable property is right and left symmetric. Lemma 5.1.1. Let R be a ring. Then the following are equivalent: 119

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(1) R is weakly stable. (2) Ra + Rb = R implies that there exists some z ∈ R such that a + zb ∈ R is right or left invertible. Proof. This is clear by Lemma 4.1.2.



Theorem 5.1.2. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E is weakly stable. (2) Given any right R-module decompositions M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ =A∼ = A2 , there exist C, D ⊆ M such that M = C ⊕ D ⊕ B1 = C ⊕ B2 or M = C ⊕ B1 = C ⊕ D ⊕ B2 . (3) Given any right R-module decompositions M = A1 ⊕B1 = A2 ⊕B2 with A1 ∼ =A∼ = A2 , there exist C, D ⊆ M such that M = A1 ⊕ C = D ⊕ C, where D ⊆⊕ A2 or A2 ⊆⊕ D.

Proof. (1) ⇒ (2) Using the decomposition M = A1 ⊕ B1 ∼ = A ⊕ B1 , we have projections p1 : M → A, p2 : M → B1 and injections q1 : A → M, q2 : B1 → M such that p1 q1 = 1A , q1 p1 + q2 p2 = 1M and ker(p1 ) = B1 . Using the decomposition M = A2 ⊕B2 ∼ = A⊕B2 , we have a projection f : M → A and an injection g : A → M such that f g = 1A and ker(f ) = B2 . Now 1A = f g = f (q1 p1 + q2 p2 )g = (f q1 )(p1 g) + f q2 p2 g. There exists h ∈ E such that one of the following conditions holds: (a) f q1 + (f q2 p2 g)h is left invertible in E. (b) f q1 + (f q2 p2 g)h is right invertible in E. Assume that (a) is satisfied. Then there exists some k ∈ E such that kf (q1 + q2 p2 gh) = 1, and so M = (q1 + q2 p2 gh)(A)⊕ ker(kf ) = C ⊕ ker(kf ), where C = (q1 + q2 p2 gh)(A). Let θ : A2 ∼ = A. Then one easily checks that ker(kf ) = ker(kθ) ⊕ B2 ; hence, M = C ⊕ D ⊕ B2 , where D = ker(kθ). Since p1 (q1 + q2 p2 gh) = 1, we see that M = (q1 + q2 p2 gh)(A) ⊕ ker(p1 ) = C ⊕ B1 . Hence M = C ⊕ B1 = C ⊕ D ⊕ B2 . Assume that (b) is satisfied. Then there exists some k ∈ E such that (f q1 + f q2 p2 gh)k = 1. Hence, f (q1 + q2 p2 gh)k = 1. Thus, M = (q1 + q2 p2 gh)k(A) ⊕ ker(f ) = (q1 + q2 p2 gh)k(A) ⊕ B2 = C ⊕ B2 , where C = (q1 + q2 p2 gh)k(A). As p1 (q1 + q2 p2 gh) = 1, we get M = (q1 + q2 p2 gh)(A) ⊕ ker(p1 ) = (q1 + q2 p2 gh)(A) ⊕ B1 . Clearly, there exists an inclusion σ : (q1 + q2 p2 gh)k(A) → (q1 + q2 p2 gh)(A). Let τ = (q1 + q2 p2 gh)kf |(q1 +q2 p2 gh)(A) . Then it is easy to verify that τ σ = 1(q1 +q2 p2 gh)k(A) . That is, σ splits; hence, we can find a right R-module D such that (q1 + q2 p2 gh)(A) = (q1 + q2 p2 gh)k(A) ⊕ D = C ⊕ D. Therefore M = C ⊕ D ⊕ B1 = C ⊕ B2 , as

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asserted. (2) ⇒ (1) Suppose that ax+b = 1 with a, x, b ∈ E. Let M = 2A, and let pi : M → A, qi : A → M (for i = 1, 2) denote the projections and injections of this direct sum. Set A1 = q1 (A) and B1 = q2 (A), so that M = A1 ⊕ B1 with A1 ∼ = A. Define maps f = ap1 + bp2 from M → A and g = q1 x + q2 from A → M . Observing that f g = 1A , we have M = ker(f ) ⊕ g(A). Set A2 = g(A) and B2 = ker(f ), so that M = A2 ⊕ B2 with A2 ∼ = A. By hypothesis, we can find C, D ⊆ M such that M = C ⊕ B1 = C ⊕ D ⊕ B2 or M = C ⊕ D ⊕ B1 = C ⊕ B2 . Assume that M = C ⊕ B1 = C ⊕ D ⊕ B2 . Let h : A ∼ = A1 ∼ =C →M be the injection. Then C = h(A). Hence, M = h(A) ⊕ ker(p1 ) = h(A) ⊕ D ⊕ ker(f ). So, p1 h ∈ E is an isomorphism. If f h(a) = 0, then h(a) ∈ T h(A) ker(f ) = 0; hence, h(a) = 0. As h is monomorphic, a = 0. Thus, f h is an R-monomorphism. This implies that f h : A ∼ = im(f h). Observing that im(f h) = f (C) ⊆⊕ f (C) ⊕ f (D) = A, the inclusion im(f h) ֒→ A splits. Thus,
f h ∈ E is left invertible. Clearly, f h = (ap1 + bp2 )h = a + bp2 h(p1 h)−1 p1 h. This implies that a + bp2 h(p1 h)−1 ∈ E is left invertible. Assume that M = C ⊕ D ⊕ B1 = C ⊕ B2 . Let k : A ∼ = A2 ∼ = C → M be the natural injection. Then C = k(A). Thus, M = k(A) ⊕ D ⊕ ker(p1 ) = k(A) ⊕ ker(f ). This shows that f k ∈ E is invertible. If p1 k(a) = 0, then T k(a) ∈ k(A) ker(p1 ) = 0; hence, k(a) = 0. As k is monomorphic, a = 0. Thus, p1 k is an R-monomorphism. This implies that p1 k : A ∼ = im(p1 k). Observing that im(p1 k) = p1 (C) ⊆⊕ p1 (C) ⊕ p1 (D) = A, the inclusion im(p1 k) ֒→ A splits. Hence, p1 k ∈ E is left invertible. Assume that lp1 k = 1 for
some l ∈ E. Observe that f k = (ap1 + bp2 )k = ap1 k + bp2 k = a + bp2 kl p1 k, we see that a + bp2 kl ∈ E is right invertible. Therefore E is weakly stable. (1) ⇒ (3) As in (1) ⇒ (2), we have that 1A = (f q1 )(p1 g) + f q2 p2 g. In view of Lemma 5.1.1, there exists h ∈ E such that one of the following conditions holds: (a) p1 g + h(f q2 p2 g) is left invertible in E. (b) p1 g + h(f q2 p2 g) is right invertible in E. Assume that (a) is satisfied. Then there exists some k ∈ E such that k(p1 g + hf q2 p2 g) = 1, and so M = kerk(p1 + hf q2 p2 )(A) ⊕ g(A) = A2 ⊕ kerk(p1 + hf q2 p2 )(A). As (p1 + hf q2 p2 )q1 = 1, we see that M = ker(p1 + hf q2 p2 ) ⊕ q1 (A) = A1 ⊕ C, where C = ker(p1 + hf q2 p2 ). Let θ : C ֒→ ker(k(p1 + hf q2 p2 )) be the natural inclusion. Let α : M = A1 ⊕ C → C be the natural projection. It is easy to verify that α |ker(k(p1 +hf q2 p2 )) θ = 1. Thus, θ splits. So we have a right R-module E such that ker(k(p1 +

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hf q2 p2 )) = C ⊕ E. Let D = A2 ⊕ E. Then M = D ⊕ C with A2 ⊆⊕ D. Assume that (b) is satisfied. Then there exists some k ∈ E such that (p1 g + hf q2 p2 g)k = 1. Hence, (p1 + hf q2 p2 )gk = 1. Thus, M = ker(p1 + hf q2 p2 ) ⊕ gk(A). As (p1 + hf q2 p2 )q1 = 1, we see that M = ker(p1 + hf q2 p2 ) ⊕ q1 (A) = A1 ⊕ C, where C = ker(p1 + hf q2 p2 ). Let θ : gk(A) ֒→ g(A) be the inclusion. It is easy to verify that gk(p1 + hf q2 p2 ) |g(A) θ = 1; hence, θ splits. So we have a right R-module E such that gk(A) ⊕ E = A2 . Let D = gk(A). Then M = D ⊕ C with D ⊆⊕ A2 , as required. (3) ⇒ (1) Suppose that ax + b = 1 when a, x, b ∈ E. As in (2) ⇒ (1), we have that M = A1 ⊕ B1 with A1 ∼ = A. Define maps f = ap1 + p2 from M → A and g = q1 x + q2 b from A → M . Observing that f g = 1A , we have M = ker(f ) ⊕ g(A). Set A2 = g(A) and B2 = ker(f ), so that M = A2 ⊕ B2 with A2 ∼ = A. By hypothesis, we can find C, D ⊆ M such that M = A1 ⊕ C = D ⊕ C, where D ⊆⊕ A2 or A2 ⊆⊕ D. Let h : M = A1 ⊕ C → A1 ∼ = A be the projection. Then C = ker(h). Hence, M = q1 (A) ⊕ ker(h) = D ⊕ ker(h). So, hq1 ∈ E is an isomorphism. Assume that D ⊆⊕ A2 . Then there exists a right R-module E such that D ⊕ E = A2 . Define φ : A → A2 given by φ(x) = g(x) for any x ∈ A. φ−1

Then φ ∈ AutR (A). Let k : A ∼ = A1 ∼ = D ֒→ D ⊕ E = A2 ∼ = A. Then D = gk(A). Hence, M = gk(A) ⊕ ker(h), where gk ∈ E is monomorphic and h ∈ E is epimorphic. Thus, hgk ∈ AutD (A). So we may assume that hgkl = 1 for some l ∈ E, and so h(q1 x + q2 b)kl = 1. This implies that x + (hq1 )−1 hq2 b ∈ E is right invertible. Assume that A2 ⊆⊕ D. Then there exists a right R-module E such that A2 ⊕ E = D. Hence, M = g(A) ⊕ E ⊕ ker(h). Let k : A ∼ = A1 ∼ = ∼ D = A2 ⊕ E ։ A2 = A be the projection. Then E ⊕ ker(h) = ker(kh). This implies that M = ker(kh) ⊕ g(A), and so khg
∈ AutR (A). Observing that khg = kh(q1 x + q2 b) = khq1 x + (hq1 )−1 hq2 b ∈ AutR (A), we know that x + (hq1 )−1 hq2 b ∈ E is left invertible. Therefore E is weakly stable by Lemma 5.1.1.  Let A be a right R-module. If EndR (A) is weakly stable, then A ⊕ B ∼ = A ⊕ C implies that B .⊕ C or C .⊕ B for any right R-modules B and C. Given ϕ : A ⊕ B ∼ = A ⊕ C, then A ⊕ C = ϕ(A) ⊕ ϕ(B) with A∼ = ϕ(A). By virtue of Theorem 5.1.2, there exist D, E ⊆ A ⊕ C such that A ⊕ C = D ⊕ E ⊕ C = D ⊕ ϕ(B) or A ⊕ C = D ⊕ C = D ⊕ E ⊕ ϕ(B). Thus, E ⊕ C ∼ = ϕ(B) ∼ = B or C ∼ = E ⊕ ϕ(B) ∼ = E ⊕ B. Therefore B .⊕ C ⊕ or C . B, and we are done.

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Corollary 5.1.3. Let A be a right R-module such that EndR (A) is weakly stable. Then EndR (nA) is weakly stable for all n ∈ N.

Proof. Given M = A1 ⊕ B = A2 ⊕ C with A1 ∼ = nA ∼ = A2 , we have M = A11 ⊕ · · · ⊕ A1n ⊕ B = A21 ⊕ · · · ⊕ A2n ⊕ C with A1i ∼ =A∼ = A2i for all i. By virtue of Theorem 5.1.2, we
can find some D1 , E1 ⊆ M such that
M = D1 ⊕ E1 ⊕ A12 ⊕ · · ·
⊕ A1n ⊕ B = D1 ⊕ A22 ⊕ · · · ⊕ A2n ⊕ C or M =
D1 ⊕ A12 ⊕ · · ·
⊕ A1n ⊕ B = D1 ⊕ E1 ⊕
A22 ⊕ · · · ⊕ A2n ⊕ C . Thus we get
M = E1 ⊕A12 ⊕ A13 ⊕· · ·⊕A1n ⊕B⊕D1 =
A22 ⊕ A23 ⊕· ·
·⊕A2n ⊕C⊕D1 or M = A12 ⊕
A13 ⊕ · · · ⊕ A1n ⊕ B ⊕ D1 = E1 ⊕ A22 ⊕ A23 ⊕ · ·
· ⊕ A2n ⊕ C ⊕ D1 . As a result, M =
A′12 ⊕ A13 ⊕ · · · ⊕ A1n ⊕ B ⊕ D1 = A′22 ⊕ A23 ⊕ · · · ⊕ A2n ⊕ C ⊕ D1 , where A′12 = E1 ⊕ A12 , A′22 = A22 or A′12 = A12 , A′22 = E1 ⊕ A22 . Clearly, A′12 ∼ = A ∼ = A′22 . By Theorem 5.1.2 again, we can
find D2 ⊆ M such that M = A′13 ⊕ A14 ⊕
· · · ⊕ A1n ⊕ B ⊕ D1 ⊕ D2 = A′23 ⊕ A24 ⊕ · · · ⊕ A2n ⊕ C ⊕ D1 ⊕ D2 with A′13 ∼ =A∼ = A′23 . By iteration of this process, we get
D3 , · · · , Dn−1 ⊆ M such that M =
A′1n ⊕ D1 ⊕ D2 ⊕ · · · ⊕ Dn−1 ⊕ B = A′2n ⊕ D1 ⊕ D2 ⊕ · · · ⊕ Dn−1 ⊕ C with A′1n ∼ =A∼ =
A′2n . Thus we can find Dn , E ⊆ M
such that M = D1 ⊕ D2 ⊕ · · · ⊕
Dn ⊕ E ⊕ B = D1 ⊕ D2 ⊕
· · · ⊕ Dn ⊕ C or M = D1 ⊕ D2 ⊕ · · · ⊕ Dn ⊕ B = D1 ⊕ D2 ⊕ · · · ⊕ Dn ⊕ E ⊕ C. By Theorem 5.1.2 again, we prove that EndR (nA) is weakly stable.  Corollary 5.1.4. Every regular square matrix over a weakly stable ring is the product of an idempotent matrix and a right or left invertible matrix. Proof. Let R be weakly stable, and let A ∈ Mn (R) be regular. Then there exists some B ∈ Mn (R) such that A = ABA. In view of Corollary 5.1.3, Mn (R) is weakly stable. Since AB + (In − AB) = In , we have a Y ∈ Mn (R) such that U := A + (In − AB)Y ∈ GLn
(R). Let E = AB. Then E = E 2 ∈ Mn (R) and A = AB A + (In − AB)Y = EU , as asserted.  Lemma 5.1.5. Let e be an idempotent in R. If R is weakly stable, then so is eRe. Proof. Assume that a, x, b ∈ eRe with ax + b = e. Then (a + 1 − e)(x + 1 − e) + b = 1. So there exists some u ∈ R such that (a + 1 − e) + bu is right or
left invertible in R. If there exists a v
∈ R such that (a + 1 − e) + bu v = 1, then (1−e)v = (1−e) (a+1−e)+bu v = 1−e. Hence

(1−e)ve = 0, and so ve = eve. Thus a + b(eue) (eve) = (a + 1 − e) + bu ve = e. Let y = eue ∈

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eRe. Then a + by is right
invertible in eRe. If there exists some v ∈ R
such that v (a + 1 − e) +
bu = 1. Then ev(a + bue) = ev (a + 1 − e) + bu e = e. So (eve) a + b(eue) = e. Let y = eue ∈ eRe. Then a + by is left invertible in eRe, as required.  Theorem 5.1.6. The weakly stable property is Morita invariant. Proof. Let R be weakly stable and let S be Morita equivalent to R. Then there is a positive integer n and an idempotent matrix e ∈ Mn (R) such that S ∼ = eMn (R)e. Clearly, Mn (R) is weakly stable by Corollary 5.1.3. According to Lemma 5.1.5, S is weakly stable, as desired.  Corollary 5.1.7. Let A be a finitely generated projective right R-module over a weakly stable ring R. If B and C are any right R-modules such that A⊕B ∼ = A ⊕ C, then B .⊕ C or C .⊕ B.

Proof. Since ψ : A ⊕ B ∼ = A ⊕ C, we have A ⊕ C = ψ(A) ⊕ ψ(B) with A ∼ = ψ(A). By virtue of Theorem 5.1.6, EndR (A) is weakly stable. According to Theorem 5.1.2, there are some right R-modules D and E such that A ⊕ C = D ⊕ E ⊕ C = D ⊕ ψ(B) or A ⊕ C = D ⊕ C = D ⊕ E ⊕ ψ(B). Thus, E⊕C ∼ = ψ(B) ∼ = B or C ∼ = E ⊕ ψ(B) ∼ = E ⊕ B. Consequently, C .⊕ B or ⊕ B . C.  Proposition 5.1.8. Let R be a ring, and let M be an R-R-bimodule. Then T (R, M ) is weakly stable if and only if so is R.

Proof. Suppose that R is weakly stable. Given (a, m)(x, n) + (b, p) = (1, 0) in T (R, M ), then ax + b = 1 in R. So we can find an element y ∈ R such that a + by = u ∈ R is right or left invertible. Hence (a, m) + (b, p)(y, 0) = (u, m + py). If there
exists some v ∈ R such that uv = 1, then (u, m + py) v, −v(m + py)v =
(1, 0). If there exists some v ∈ R such that vu = 1, then v, −v(m + py)v (u, m + py) = (1, 0). Thus, (a, m) + (b, p)(y, 0) ∈ T (R, M ) is right or left invertible. Consequently, T (R, M ) is weakly stable. Conversely, assume that T (R, M ) is weakly stable. Given ax + b = 1 in R, then (a, 0)(x, 0) + (b, 0) = (1, 0) in T (R, M ). So we can find some (y, q) ∈ T (R, M ) such that (a, 0) + (b, 0)(y, q) = (a + by, bq) ∈ T (R, M ) is right or left invertible. Therefore a + by ∈ R is right or left invertible, as required.    ab Let R be a ring. Evidently, T (R, R) ∼ | a, b ∈ R}. Obviously, ={ 0a T (R, R) is weakly stable if and only if so is R.

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Proposition 5.1.9. Let R be a ring. Then R[[x1 , · · · , xn ]] is weakly stable if and only if so is R. Proof. It suffices to show that R[[x]] is weakly stable if and only if so is R. Suppose that R is weakly stable. If f (x)g(x) + h(x) = 1 in R[[x]], then f (0)g(0) + h(0) = 1 in R, and so we can find some y ∈ R such that u := f (0) + h(0)y ∈ R is right or left invertible. First, assume that uv = 1 for some v ∈ R. Choose b0 = v, b1 = −v(a1 b0 ), b2 = −v(a1 b1 + n
P a2 b0 ), · · · , bn = −v ai bn−i , · · · . Then it is easy to verify that f (x) +

g(x)y

∞
P

i=0

i

bi x




i=1

= 1. This implies that f (x) + g(x)y ∈ R[[x]] is right

invertible. A similar argument shows that if vu = 1 for some v ∈ R then f (x) + g(x)y ∈ R[[x]] is left invertible. This proves that R[[x]] is weakly stable. Conversely, if R[[x]] is weakly stable, then so is R because R is a factor ring of R[[x]].  5.2

Exchange Rings

An element a ∈ R is said to be one-sided unit-regular provided that there exists a right or left invertible u ∈ R such that a = aua. The purpose of this section is to investigate weakly stable rings by means of one-sided unit-regularity. Theorem 5.2.1. Let R be an exchange ring. Then the following are equivalent : (1) R is weakly stable. (2) Every regular element in R is one-sided unit-regular. Proof. (1) ⇒ (2) Given any regular x ∈ R, there exists a y ∈ R such that x = xyx. Since yx + (1 − yx) = 1, we have some z ∈ R such that y + (1 − yx)z =
u ∈ R is right or left invertible. Hence, x = xyx = x y + (1 − yx)z x = xux, as required. (2) ⇒ (1) Suppose that ax + b = 1 in R. In view of [382, Theorem 28.7], there exists an idempotent e ∈ bR such that 1 − e ∈ (1 − b)R. Assume that e = bs and 1 − e = axt for some s, t ∈ R. Then axt + e = 1, whence, (1 − e)a ∈ R is regular. By assumption, we can find a right or left invertible u ∈ R such that (1−e)a = (1−e)au(1−e)a. Since (1−e)axt+e = 1, we have

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that u(1 − e)axt + ue = u. Let f = u(1 − e)a. Then f = f 2 ∈ R. Clearly, f (xt + ue) + (1 − f )ue = u. So f (xt + ue) = f u and (1 − f )ue = (1 − f )u. Assume that uv = 1 or vu = 1 for some v ∈ R. Then
(1 − f )uv(1 − f )u = (1 − f ) uvu − uvu(1 − e)au = (1 − f )(u − f u) = (1 − f )u. Let g = (1 − f )uv(1 − f ). Then g = g 2 . As a result, we get (f + g)u = u with f g = gf = 0. One easily checks that

u (1 − e)a + ev(1 − f )(1 + f uev(1
− f ) 1 − f uev(1
− f) u = f + uev(1 − f )(1 + f uev(1 − f )
1 − f uev(1 − f ) u = f (1 − f uev(1 − f ) + uev(1 − f) u
= f + (1 − f )uev(1 − f ) u = f u + (1 − f )uv(1 − f )u = f u + gu = u. As u ∈ R is right or left invertible, it is easy to verify that
a + bs v(1 − f )(1 + f uev(1 − f )) −
a = a − ea + ev(1 − f ) 1 + f uev(1 − f )
= (1 − e)a + ev(1 − f ) 1 + f uev(1 − f ) ∈ R is right or left invertible, and therefore R is weakly stable.



Corollary 5.2.2. Let R be an exchange ring. Then the following are equivalent : (1) R is weakly stable. (2) For any regular x ∈ R, there exists a right or left invertible u ∈ R such that ux is an idempotent. (3) For any regular x ∈ R, there exists a right or left invertible u ∈ R such that xu is an idempotent. Proof. (1) ⇒ (2) For any regular x ∈ R, it follows by Theorem 5.2.1 that there exists a right or left invertible u ∈ R such that x = xux. So ux ∈ R is an idempotent. (2) ⇒ (1) For any regular x ∈ R, we have an element y ∈ R such that x = xyx and y = yxy. By hypothesis, there exists a right or left invertible u ∈ R such that uy is an idempotent. From yx + (1 − yx) = 1, we get uyx + u(1 − yx) = u. As in the proof of Theorem 5.2.1, we can find an

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element z ∈ R such that y +
(1 − yx)z = v ∈ R is right or left invertible. Hence, x = x y + (1 − yx)z x = xvx. According to Theorem 5.2.1, R is weakly stable. (1) ⇔ (3) Applying (1) ⇔ (2) to the opposite ring Rop , we obtain the result.  Theorem 5.2.3. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent: (1) E is weakly stable. (2) For any right R-modules B and C, A⊕B ∼ = A⊕C implies that B .⊕ C ⊕ or C . B. (3) A = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 implies that B .⊕ C or C .⊕ B. (4) For any idempotents e, f ∈ E, eA ∼ = f A implies that (1 − e)A .⊕ (1 − f )A or (1 − f )A .⊕ (1 − e)A. Proof. (1) ⇒ (2) is clear by Theorem 5.1.2. (2) ⇒ (3) Given A = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 , then A ⊕ B ∼ = ⊕ ⊕ A ⊕ C. By hypothesis, B . C or C . B. (3) ⇒ (4) For any idempotents e, f ∈ E, we see that A = eA⊕(1−e)A = f A ⊕ (1 − f )A. Thus, eA ∼ = f A implies that (1 − e)A .⊕ (1 − f )A or ⊕ (1 − f )A . (1 − e)A. (4) ⇒ (1) Given any regular x ∈ E, there exists a y ∈ E such that x = xyx and y = yxy. Clearly, ϕ : xyA ∼ = yxA given by ϕ(xya) = yxya for any a ∈ A. By hypothesis, we get (1 − xy)A .⊕ (1 − yx)A or (1 − yx)A .⊕ (1−xy)A. Thus, we have a splitting R-monomorphism ψ : (1−xy)A → (1− yx)A or a splitting R-epimorphism ψ : (1 − xy)A → (1 − yx)A. Construct an R-morphism φ : A = xyA
⊕ (1 − xy)A → yxA ⊕ (1 − yx)A = A given by φ(a) = ϕ(xya) + ψ (1 − xy)a for any a ∈ A. One easily checks that φ ∈ E is left or right invertible. Furthermore, x = xφx. In view of [382, Theorem 28.7], E is an exchange ring. By Theorem 5.2.1, we obtain the result.  Let A be a right R-module having the finite exchange property. It follows by Lemma 5.1.5, Theorem 5.1.2 and Theorem 5.2.3 that EndR (A) is weakly stable if and only if A = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 implies that there exist D, E ⊆ A such that A = D ⊕ E ⊕ B = D ⊕ C or A = D ⊕ B = D ⊕ E ⊕ C. Example 5.2.4. Let V be a vector space over a division ring D. Then EndD (V ) is weakly stable. Proof. Suppose that V = U1 ⊕ V1 = U2 ⊕ V2 with U1 ∼ = U2 . Clearly,

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dimD (V1 ) ≤ dimD (V2 ) if and only if V1 .⊕ V2 , and dimD (V2 ) ≤ dimD (V1 ) if and only if V2 .⊕ V1 . Therefore V1 .⊕ V2 or V2 .⊕ V1 , and we are done by Theorem 5.2.3.  In light of Corollary 5.1.4 and Example 5.2.4, we conclude that every row-finite matrix over a division ring is the product of an idempotent infinite matrix and a right or left invertible infinite matrix. Example 5.2.5. Let V be an infinite-dimensional  vector space over a EndD (V ) EndD (V ) division ring D, and let R = . Then R is not weakly 0 EndD (V ) stable. Proof. Let {x1 , x2 , · · · , xn , · · · } be a basis of V . Define σ : V → V given by σ(xi ) = xi+1 (i = 1, 2, · · · ) and τ : V → V given by τ (x1 ) = 0 and τ (xi ) = xi−1 (i = 2, 3, · · · ). Then τ σ = 1V and στ 6= 1V . Assume that R is weakly stable. Since      τ 0 σ0 0 0 + = diag(1V , 1V ), 0σ 0τ 0 1V − στ we have some



αγ 0β





∈ R such that

τ 0 0σ



+



0 0 0 1V − στ



αγ 0β



∈R

is right or left invertible. This implies that τ ∈ EndD (V ) is left invertible or σ
+ (1V − στ )β ∈ EndD (V ) is right invertible. Clearly, τ σ + (1V − στ )β = 1V . Thus, τ ∈ AutD (V ) or σ + (1V − στ )β ∈ AutD (V ). If σ + (1V − στ )β ∈ AutD (V ), then τ ∈ AutD (V ). In any case, τ ∈ AutD (V ), a contradiction. Therefore R is not weakly stable.  In the preceding example, we chose e = diag(1V , 0). Then eRe ∼ = EndD (V ) ∼ = (1R − e)R(1R − e). Thus, it is possible to have an exchange ring R, with an idempotent e, such that both eRe and (1 − e)R(1 − e)R are weakly stable, but R is not. Recall that a right R-module A is directly finite if A is not isomorphic to any proper direct summand of itself. Equivalently, A is directly finite if and only if B = 0 is the only module for which A ⊕ B ∼ = A. A module which is not directly finite is said to be directly infinite.

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Lemma 5.2.6. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Suppose that A is expressible as a direct sum of isomorphic indecomposable submodules. Then the following hold: (1) E is weakly stable. (2) E has stable range one if and only if A is a direct sum of many finite isomorphic indecomposable submodules. ∼ A2 , then Proof. (1) Assume that A = A1 ⊕ B = A2 ⊕ C with A1 = L A = A1 ⊕ B = Yi , where each Yi is isomorphic to an indecomposable i∈I

submodule Y of A. In view of [382, Lemma 28.1], A1 has the finite exchange property. Thus, we have some Yi′ ⊆ Yi such that M
A = A1 ⊕ Yi′ . i∈I

It is easy to verify that Yi′ ⊆⊕ Yi for all i ∈ I. As each Yi is indecomposable, we see that either Yi′ = 0 or Yi′ = Yi . Thus, there is a subset H1 of I such L L that B ∼ Yi . Likewise, there is a subset H2 of I such that C ∼ Yi . = = i∈H1

i∈H2

Clearly, |H1 | ≤ |H2 | or |H2 | ≤ |H1 |, whence either B .⊕ C or C .⊕ B. According to Theorem 5.2.3, E is weakly stable. (2) If E has stable range one, then A is directly finite. Hence, A is not isomorphic to a proper submodule of itself. But then the index set I is n L finite. Conversely, assume that A = Yi where each Yi is isomorphic to an i=1

indecomposable module Y . Since A has the finite exchange property, so has each Yi by [382, Lemma 28.1]. In view of [382, Theorem 29.5], EndR (Yi ) is local; hence, it has stable range one. According to Lemma 1.1.9, E has stable range one, as asserted.  A submodule B of a right R-module A is fully invariant provided that for any f ∈ EndR (A), f (B) ⊆ B. For instance, every direct summand of an abelian module M , i.e., all idempotents in EndR (M ) are central, is fully invariant (cf. [344, Lemma 1]). Theorem 5.2.7. Let A be a right R-module having the finite exchange L property, let E = EndR (A), and let A = Ai . Suppose that each Ai is a i∈I

fully invariant submodule, equal to a direct sum of isomorphic indecomposable submodules. Then the following hold : (1) E is weakly stable if and only if Ai is directly finite for all but (possibly) a single i ∈ I.

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(2) E has stable range one if and only if A is directly finite. Proof. (1) Suppose that E is weakly stable. If i1 , i2 ∈ I are two distinct indices such that Ai1 and Ai2 both fail to be directly finite, it follows from [217, Lemma 5.1] that EndR (Ai1 ) and EndR (Ai2 ) are both directly infinite. Thus, we can find some s1 , t1 ∈ EndR (Ai1 ) and s2 , t2 ∈ EndR (Ai2 ) such that s1 t1 = 1, t1 s1 6= 1, s2 t2 6= 1 and t2 s2 = 1. It is easy to check that (s1 , s2 ) = (s1 , s2 )(t1 , t2 )(s1 , s2 ), i.e., (s1 , s2 ) ∈ EndR (Ai1 ) ⊕ EndR (Ai2 ) is regular. If (s1 , s2 ) is one-sided unit-regular, there exists a right or left invertible (u1 , u2 ) such that (s1 , s2 ) = (s1 , s2 )(u1 , u2 )(s1 , s2 ); hence, s1 = s1 u1 s1 and s2 = s2 u2 s2 . As a result, s1 u1 = 1 and u2 s2 = 1. If (u1 , u2 ) is right invertible, u1 ∈ EndR (Ai1 ) is invertible. If (u1 , u2 ) is left invertible, u2 ∈ EndR (Ai2 ) is invertible. Thus, either s1 or s2 is invertible. This gives a contradiction. By Theorem 5.2.1, EndR (Ai1 ) ⊕ EndR (Ai2 ) is not weakly stable. Since each Ai is a fully invariant submodule, we see that Q HomR (Ai , Aj ) = 0 for i 6= j. In view of [43], E ∼ EndR (Ai ). According = i∈I

to Lemma 5.1.5, EndR (Ai1 ) ⊕ EndR (Ai2 ) is weakly stable, a contradiction. Therefore we conclude that Ai is directly finite for all but (possibly) a single i ∈ I. Conversely, assume that all but (possibly) a single one of the Ai is directly infinite. If all of the Ai are directly finite, then EndR (Ai ) has Q stable range one by Lemma 5.2.6. As E ∼ EndR (Ai ), we see that E has = i∈I

stable range one. If there exists some j ∈ I such that Aj is directly infinite while for all i 6= j(i ∈ I), Ai is directly finite, it follows by Lemma 5.2.6 that EndR (Aj ) is weakly stable, and for all i 6= j(i ∈ I), EndR (Ai ) has stable range one. From this, we show that E is weakly stable, as required. (2) If E has stable range one, it easily follows that E is directly finite. Conversely, suppose that E is directly finite. Then EndR (Ai ) is directly finite from [253, Corollary 1]. That is, Ai is not isomorphic to a proper submodule of itself. In view of Lemma 5.2.6, EndR (Ai ) has stable range Q one. As E ∼ EndR (Ai ), we see that E has stable range one.  = i∈I

Corollary 5.2.8. Let G be an abelian group such that End(G) is regular. If G is a nonreduced group, or a reduced torsion group, then End(G) has stable range one if and only if it is directly finite. Proof. A reduced abelian torsion group has a regular endomorphism ring if and only if it is a direct sum of cyclic groups of prime order; a nonreduced abelian group has a regular endomorphism ring if and only if it is a direct

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sum of cyclic groups of prime order and full rational groups (cf. [209]). Thus, the result follows by Theorem 5.2.7.  A ring R is separative provided that for all finitely generated projective right R-modules A, B, 2A ∼ =A⊕B ∼ = 2B =⇒ A ∼ = B (cf. [16]). Theorem 5.2.9. Every weakly stable exchange ring is separative. Proof. Let R be a weakly stable ring. Suppose that ϕ : 2A ⊕ B ∼ = 2A ⊕ C with A, B, C ∈ F P (R). Then M := A1 ⊕ A2 ⊕ C = A′1 ⊕ A′2 ⊕ ϕ(B)

∼ A ∼ with Ai = = A′i for each i. In view of Theorem 5.1.6, EndR (A1 ) is weakly stable. By virtue
of Theorem 5.1.2, there
exist D, E ⊆ M such that
M = D ⊕ E ⊕ A2
⊕ C = D ⊕ A′2 ⊕ ϕ(B) or M = D ⊕ A2 ⊕ C = D ⊕E ⊕ A′2 ⊕ϕ(B) . In any case, A⊕E ∼ = A. Thus, we get A⊕C ∼ = A⊕B. According to [16, Lemma 2.1], R is separative.  Corollary 5.2.10. Let R be a simple exchange ring. Then the following are equivalent : (1) R is weakly stable. (2) R is separative. Proof. (1) ⇒ (2) is obvious by Theorem 5.2.9. (2) ⇒ (1) Assume R is directly infinite. Then R ⊕ D ∼ = R for some nonzero right R-module D. Let e, f ∈ R be idempotents. If e = 0 or f = 0, then eR .⊕ f R or f R .⊕ eR. Now we assume that e 6= 0, f 6= 0. Clearly, there exists a nonzero idempotent g ∈ R such that D ∼ = gR. Since R is simple, we see that RgR = R. Thus, there are some si , ti ∈ R(1 ≤ n P i ≤ n) such that si gti = 1. Construct an R-morphism ϕ : n(gR) → R i=1

given by ϕ(gr1 , · · · , grn ) =

n P

i=1

si gri . It is easy to verify that ϕ is an R-

epimorphism. As R is projective, there exists a right R-module N such that R ⊕ N ∼ = n(gR). Hence, R .⊕ nD; whence, eR .⊕ nD. Thus ⊕ eR ⊕ R .
nD ⊕ R ∼ = R, and so eR ⊕ R .⊕ R .⊕ f R ⊕ R. Hence, ∼ R ⊕ eR ⊕ E = R ⊕ f R for a right R-module E. As
eR and f R are both nonzero, we also have R .⊕ s(eR) .⊕ s eR ⊕ E and R .⊕ t(f R) for some s, t ∈ N. Applying [16, Lemma 2.1], eR .⊕ eR ⊕ E ∼ = f R. In view of Theorem 5.2.3, R is weakly stable. Assume R is directly finite. Let R = A1 ⊕ B ∼ = A2 ⊕ C with A1 ∼ = A2 . If B = 0 or C = 0, then B ∼ = C.

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If B 6= 0 and C 6= 0, analogously to the preceding discussion, we have m ∈ N such that A .⊕ mB, mC. As R is separative, B ∼ = C. According to Theorem 1.3.2, R has stable range one, as required.  Theorem 5.2.11. Let R be an exchange ring. Then the following are equivalent : (1) R is weakly stable. (2) Whenever x = xyx, there exists a right or left invertible u ∈ R such that x = xyu. (3) Whenever x = xyx, there exists a right or left invertible u ∈ R such that x = uyx. Proof. (1) ⇒ (2) Suppose that x = xyx. Since xy + (1 − xy) = 1, we have that x + (1 − xy)z ∈
R is right or left invertible for some z ∈ R. Hence x = xy x + (1 − xy)z = xyu, where u := x + (1 − xy)z ∈ R is right or left invertible. (2) ⇒ (1) Suppose that x = xyx. Then there exists a right or left invertible u ∈ R such that x = xyu. Let e = xy. Then e ∈ R is an idempotent. Since xy +(1−xy) = 1, we have that euy +(1−xy) = 1, and so euy(1−e)+(1−xy)(1−e) = 1−e. Hence, e+(1−xy)(1−e) = 1−euy(1−e) ∈
U (R). Thus, we get x + (1 − xy)(1 − e)u = 1 − euy(1 − e) u ∈ R which is right or left invertible. By virtue of Lemma 4.1.2, we have an element z ∈ R such that w :=
y + z(1 − xy) ∈ R is right or left invertible. Therefore x = x y + z(1 − xy) x = xwx, and R is weakly stable by Theorem 5.2.1. (1) ⇔ (3) Applying (1) ⇔ (2) to the opposite ring Rop , we give the result at once.  Corollary 5.2.12. Let R be an exchange ring. Then the following are equivalent : (1) R is weakly stable. (2) For every regular x left invertible u ∈ R (3) For every regular x left invertible u ∈ R (4) For every regular x left invertible u ∈ R

∈ R, there exists an idempotent e and a right or such that x = eu. ∈ R, there exists an idempotent e and a right or such that x = ue. ∈ R, there exists an idempotent e and a right or such that x = eu or ue.

Proof. (1) ⇒ (2) is clear by Theorem 5.2.11. (2) ⇒ (4) is trivial. (4) ⇒ (1) Given any regular x ∈ R, we have an element y ∈ R such

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that x = xyx and y = yxy. By assumption, there is an idempotent e ∈ R and a right or left invertible u ∈ R such that y = eu or ue. Assume that y = eu. Since yx + (1 − yx) = 1, eux + (1 − yx) = 1. As in the
proof of Theorem 5.2.11, we have that y +(1−yx)(1−e)u = 1−eux(1−e) u ∈ R is right or left
invertible. This implies that x = xyx = xwx, where w := 1 − eux(1 − e) u ∈ R is right or left invertible. Assume that y = ue. Obviously, xy+(1−xy) = 1. Similarly, there exists a z ∈ R such that w := y+z(1−xy) is right or left invertible. It is easy to verify x = xyx = xwx. Theorem 5.2.1 applies. (1) ⇔ (3) Applying (1) ⇔ (2) to the opposite ring Rop , the proof is true.  Corollary 5.2.13. Let R be an exchange ring. Then the following are equivalent : (1) R is weakly stable. (2) For any a, b ∈ R, aR = invertible u ∈ R such that (3) For any a, b ∈ R, Ra = invertible u ∈ R such that

bR implies that there exists a right or left b = au. Rb implies that there exists a right or left b = ua.

Proof. (1) ⇒ (2) For any a, b ∈ R, aR = bR implies that ax = b and by = a for some x, y ∈ R. Since xy + (1 − xy) = 1, there exists some z ∈ R such that u := x + (1
− xy)z ∈ R is right or left invertible. Therefore b = ax = a x + (1 − xy)z = au, as asserted. (2) ⇒ (1) Given any regular x ∈ R, there exists a y ∈ R such that x = xyx. Hence, xyR = xR. By assumption, we have a right or left invertible u ∈ R such that x = xyu. According to Corollary 5.2.12, R is weakly stable. (1) ⇔ (3) is symmetric.  The following result was proved by Wei [400, Theorem 3.6]. Theorem 5.2.14. Let R be an exchange ring. Then the following are equivalent : (1) R is weakly stable. (2) Whenever x = xyx, there exists a right or left invertible u ∈ R such that x = xyu = uyx. (3) Whenever x = xyx, there exists a right or left invertible u ∈ R such that xyu = uyx.

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Proof. (1) ⇒ (2) Given any x = xyx, then we have x = xzx, z = zxz, where z = yxy. Since R is weakly stable, it follows by Theorem 5.2.1 that there exists a right or left invertible v ∈ R such that z = zvz. Let u = (1 − xz − vz)v(1 − zx − zv). One easily checks that (1 − xz − vz)2 = 1 = (1 − zx − zv)2 . Hence u ∈ R is right or left invertible. Clearly, xzu = = = =

−xzv(1 − zx − zv) −xzv + xzx + xzv xzx x

and uzx = = = = =

(1 − xz − vz)v(−zvzx) −(1 − xz − vz)vzx −vzx + xzx + vzx xzx x.

Thus, x = xzu = x(yxy)u = xyu and x = uzx = u(yxy)x = uyx. As a result, we see that x = xyu = uyx. (2) ⇒ (3) is trivial. (3) ⇒ (1) Given x = xyx, there exists a right or left invertible u ∈ R such that xyu = uyx. Thus, we can find some v ∈ R such that uv = 1 or vu = 1. Construct two maps ϕ : xR ⊕ (1 − xy)R → yxR ⊕ (1 − yx)R; ϕ xr + (1 − xy)s) = yxr + (1 − yx)v(1 − xy)s for any r, s ∈ R and φ : yR ⊕ (1 − yx)R → xR ⊕ (1 − xy)R, φ yr + (1 − yx)s) = xyr + u(1 − yx)s for any r, s ∈ R. One easily checks that xϕ(1)x = xϕ(x) = xyx = x. Furthermore, we see that
1 − φ(1)ϕ(1) = 1 − φ ϕ(1)
= 1 − φ yxy + (1 − yx)v(1 − xy)
= 1 − xyxy + u(1 − yx)v(1 − xy) = (1 − xy)(1 − uv)(1 − xy).

Likewise, we have that 1 − ϕ(1)φ(1) = (1 − yx)(1 − vu)(1 − yx). Thus, ϕ(1)φ(1) = 1 or φ(1)ϕ(1) = 1. Hence, ϕ(1) ∈ R is right or left invertible. According to Theorem 5.2.1, we obtain the result.  Corollary 5.2.15. Let R be an exchange ring. Then the following are equivalent :

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(1) R is weakly stable. (2) For any idempotents e, f ∈ R, eR ∼ = f R implies that there exists a right or left invertible u ∈ R such that eu = uf .

Proof. (1) ⇒ (2) For any idempotents e, f ∈ R, eR ∼ = f R implies that e = xy and f = yx, where x ∈ eRf, y ∈ f Re. Clearly, x = xyx. In view of Theorem 5.2.14, xyu = uyx, and so eu = uf . (2) ⇒ (1) Given x = xyx, then xyR ∼ = yxR. By hypothesis, there is a right or left invertible u ∈ R such that xyu = uyx. According to Theorem 5.2.14, R is weakly stable.  Lemma 5.2.16. Let R be an exchange ring. Then the following are equivalent : (1) R is weakly stable. (2) Whenever x = xyx, there exists some a ∈ R such that 1 + xa ∈ U (R) and y + a ∈ R is right or left invertible. Proof. (1) ⇒ (2) Assume that x = xyx. Since yx + (1 − yx) = 1, there exists some z ∈ R such that y + (1 − yx)z ∈ R is right or left invertible. Let a = (1 − yx)z. Then 1 + xa = 1 ∈ U (R), as required. (2) ⇒ (1) Given x = xyx, then there exists some a ∈ R such that u := 1 + xa ∈ U (R) and w := y + a ∈ R is right or left invertible. Thus, 1 + xw − xy = u, and so xwu−1 + (1 − xy)u−1 = 1. As wu−1 ∈ R is right or left invertible, it follows by Lemma 4.1.2 that t := x + (1 − xy)u−1 s ∈ R is right or
left invertible for some s ∈ R. Therefore x = xyx = xy x + (1 − xy)u−1 s = xyt, and we are done by Corollary 5.2.12. 

Proposition 5.2.17. Let R be an exchange ring. Then the following are equivalent : (1) R is weakly stable. (2) Whenever x = xyx, there exists a right or left invertible u ∈ R such that 1 − x(y + u) ∈ U (R).

Proof. (1) ⇒ (2) Whenever x = xyx, then −x = (−x)(−y)(−x). By virtue of Lemma 5.2.16, there exists some a ∈ R such that 1 + (−x)a ∈ U (R) and (−y) + a ∈ R is right or left invertible. Let u = −y + a. Then 1 − x(y + u) ∈ U (R). (2) ⇒ (1) Whenever x = xyx, there exists a right or left invertible u ∈ R such that 1 − x(y + u) ∈ U (R). Let a = −(y + u). Then 1 + xa ∈ U (R) and y + a = −u ∈ R are right or left invertible. According to Lemma 5.2.16,

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We use x+ to denote a reflexive inverse of an element x ∈ R, i.e., x = xx+ x and x+ = x+ xx+ . Corollary 5.2.18. Let R be an exchange ring. Then the following are equivalent : (1) R is weakly stable. (2) For any regular x ∈ R, there is an e ∈ ℓ(x+ ) and a right or left invertible u ∈ R such that x = e + u. (3) For any regular x ∈ R, there is an e ∈ ℓ(x) and a right or left invertible u ∈ R such that x+ = e + u. Proof. (1) ⇒ (2) For any regular x ∈ R, we have x+ ∈ R such that x = xx+ x and x+ = x+ xx+ . Since x+ x + (1 − x+ x) = 1, there exists a z ∈ R such that u := x + z(1 − x+ x) ∈ R is right or left invertible. Set e = z(x+ x − 1). Then x = e + u, where e ∈ ℓ(x+ ). (2) ⇒ (3) is trivial. (3) ⇒ (1) Given x = xyx, then x = xzx and z = zxz, where z = yxy. By hypothesis, there is an e ∈ ℓ(x) and a right or left invertible u ∈ R such that z = e + u. Let a = −e. Then u = z + a ∈ R is right or left invertible and 1 + ax = 1 ∈ U (R). It is well known that 1 + bc ∈ U (R) if and only if 1 + cb ∈ U (R). Hence, 1 + xa ∈ U (R). We infer that 1 − x(y − u) = 1 − x(z − u) = 1 + x(u − z) ∈ U (R). Therefore the proof is true from Proposition 5.2.17.  By symmetry, we deduce that an exchange ring R is weakly stable if and only if for any regular x ∈ R, we have an e ∈ r(x+ ) and a right or left invertible u ∈ R such that x = e + u if and only if for any regular x ∈ R, we have an e ∈ r(x) and a right or left invertible u ∈ R such that x+ = e + u. A ring R is one-sided unit-regular provided that for any x ∈ R, there exists a right or left invertible u ∈ R such that x = xux. Clearly, a regular ring is one-sided unit-regular if and only if it is weakly stable. We say that a ring R is a Boolean ring provided that every element in R is an idempotent. Proposition 5.2.19. Let R be regular. Then the following are equivalent: (1) Every nonzero element x ∈ R has a unique factorization x = eu for an idempotent e ∈ R, and a right or left invertible u ∈ R. (2) Every nonzero element x ∈ R has a unique factorization x = ue for an

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idempotent e ∈ R, and a right or left invertible u ∈ R. (3) R is a Boolean ring or R is a division ring. Proof. (1) ⇒ (3) By virtue of Corollary 5.2.12, R is one-sided unit-regular. Given 0 6= x ∈ R, there exists some y ∈ R such that x = xyx. Let e = xy. Then e = e2 ∈ R. Since xy + (1 − e) = 1, there exists some z ∈ R such that x + (1 − e)z = u is right or left invertible in R. So x = xyx = ex = e(x + (1 − e)z) = eu.
On the other hand,
it is easy to2 verify that xy 1 − x(1 − xy)
x = x. Let f = xy 1−x(1−xy) . Then f = f ∈ R and xy 1−x(1−xy) +(1−f ) = 1. So there exists an element w ∈ R such that x + (1 −
f )w = v is right or left invertible in R. Hence x = xy(1 − x(1 − xy)) x = f x = f x +
(1 − f )w = f v.
Since x = eu and x = f v, the uniqueness implies that xy 1−x(1−xy) = f = e = xy. This implies that x = xyx = xyx2 y = x2 y. Thus R is abelian regular, and so R is unit-regular. For any 0 6= x ∈ R, we assume that x = xux = xvx with u and v being units in R. Let e = xu, f = xv. Then e = e2 , f = f 2 ∈ R with x = eu−1 = f v −1 . By the uniqueness, this yields u−1 = v −1 , and so u = v. Thus every nonzero element of R has a unique unit inner inverse. Using [254, Theorem 4], R is a Boolean ring or R is a division ring. (3) ⇒ (1) is trivial. (2) ⇔ (3) is proved in the same manner.  Corollary 5.2.20. A ring R is a division ring if and only if every nonzero element in R is right or left invertible. Proof. One direction is obvious. Conversely, suppose that every nonzero element in R is right or left invertible. Then R is regular. Clearly, all idempotents in R are 0 and 1. Thus, we see that every nonzero element x ∈ R has a unique factorization x = ue for an idempotent e ∈ R, a right or left invertible u ∈ R. In view of Proposition 5.2.19, R is a Boolean ring or R is a division ring. If R is a Boolean ring, then R ∼ = Z/2Z. Therefore R is a division ring.  Proposition 5.2.21. Let R be regular. Then the following are equivalent: (1) For every nonzero x ∈ R, there exists a unique right or left invertible u ∈ R such that ux is an idempotent. (2) For every nonzero x ∈ R, there exists a unique right or left invertible u ∈ R such that xu is an idempotent. (3) R is a Boolean ring or R is a division ring.

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Proof. (1) ⇒ (3) By virtue of Corollary 5.2.2, R is one-sided unit-regular. Given 0 6= x ∈ R, there exists a right or left invertible u ∈ R such that xux = x, and so ux is
an idempotent. On the other hand, it is easy to
−1 verify xu 1 − x(1 − xu) x = x.
Since 1 − x(1 − xu) = 1 + x(1 − xu), we know that u 1 − x(1 −
xu) x is an idempotent with a right
or left invertible u 1 − x(1 − xu) ∈ R. Thus u = u 1 − x(1 − xu) . Hence x = xux = xux2 u = x2 u, and then R is abelian regular. Therefore R is unit-regular. Suppose that x = xux = xvx where u and v are units. Then ux and vx are both idempotents with invertible u, v ∈ R. So u = v. By [254, Theorem 4], R is a Boolean ring or R is a division ring. (3) ⇒ (1) is trivial. (2) ⇔ (3) is analogous to (1) ⇔ (3).  Similarly, we deduce that a ring R is a division ring if and only if for every nonzero x ∈ R, there exists a unique y ∈ R such that yx is an idempotent if and only if for every nonzero x ∈ R, there exists a unique y ∈ R such that xy is an idempotent. 5.3

Element Factorizations

An element a ∈ R is clean provided that it is the sum of an idempotent and a unit. A ring R is clean if every element in R is clean. In [63, Theorem 1], Camillo and Khurana proved every unit-regular ring is clean. In this section, we will extend this result to weakly stable exchange rings. Theorem 5.3.1. Let R be a weakly stable exchange ring. Then every regular element in R is the sum of an idempotent and a right or left invertible element. Proof. Let a ∈ R be regular. Then we have x ∈ R such that a = axa. So R = im(a) ⊕ (1 − ax)R = xaR ⊕ ker(a). Since R is an exchange ring, there exist right R-modules X1 , Y1 such that R = im(a) ⊕ X1 ⊕ Y1 with
X1 ⊆ ker(a) and Y1 ⊆ xaR. Clearly, ker(a) =
ker(a) ∩ X1 ⊕ im(a) ⊕ Y1 = X1 ⊕ X2 , where X2 = ker(a) ∩ im(a) ⊕ Y1 . Likewise, we have a right R-module Y2 such that xaR = Y1 ⊕ Y2 . Clearly, R = im(a) ⊕ X1 ⊕ Y1 = X1 ⊕ X2 ⊕ xaR. In addition, θ : im(a) = aR ∼ = xaR given by θ(ar) = xar for any r ∈ R.

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Thus, im(a) ⊕ X1 ∼ = X1 ⊕ xaR. In view of Theorem 5.2.3, we get X2 .⊕ Y1 ⊕ or Y1 . X2 . Thus, there exist right R-morphisms ψ : X2 → Y1 and ϕ : Y1 → X2 such that ϕψ = 1X2 or ψϕ = 1Y1 . Let

Let

k : X1 ⊕ X2 → X1 ⊕ Y1 ; x1 + x2 7→ x1 + ψ(x2 ), ∀x1 ∈ X1 , x2 ∈ X2 ; l : X1 ⊕ Y1 → X1 ⊕ X2 ; x1 + y1 7→ x1 + ϕ(y1 ), ∀x1 ∈ X1 , y1 ∈ Y1 . h : R = X1 ⊕ X2 ⊕ Y1 ⊕ Y2 → X1 ⊕ Y1 ⊕ X2 ⊕ Y2 = R; x1 + x2 + y1 + y2 7→ k(x1 + x2 ) + y1 , v : X1 ⊕ Y1 ⊕ X2 ⊕ Y2 → X1 ⊕ X2 ⊕ Y1 ⊕ Y2 ; x1 + y1 + x2 + y2 7→ l(x1 + y1 ) + ψ(x2 ), ∀x1 ∈ X1 , y1 ∈ Y1 , x2 ∈ X2 , y2 ∈ Y2 .

For any x1 ∈ X1 , y1 ∈ Y1 , x2 ∈ X2 , y2 ∈ Y2 , it is easy to verify that hvhv(x1 + y1 + x2 + y2 )
= hvh l(x1 + y1 ) + ψ(x2 )
= hv kl(x1 + y1 ) + ψ(x2 )
= h lkl(x1 + y1 ) + ϕψ(x2 ) = klkl(x1 + y1 ) + ψϕψ(x2 ) = kl(x1 + y1 ) + ψ(x2 ) = hv(x1 + y1 + x2 + y2 ).

Hence (hv)2 = hv. Let e = hv. Then e = e2 ∈ EndR (R). Assume that ϕψ = 1X2 . Let φ : R = im(a) ⊕ X1 ⊕ Y1 → R; z + x1 + y1 7→ xz + x1 + ϕ(y1 ), ∀z ∈ im(a), x1 ∈ X1 , y1 ∈ Y1 .

Then φ(a − hv)(x1 + y1 + x2 + y2 )
= φ a(y1 + y2 ) − kl(x1 + y1 ) − ψ(x2 )
= φ a(y1 + y2 ) − x1 − ψϕ(y1 ) − ψ(x2 ) = xa(y1 + y2 ) − x1 − ϕψϕ(y1 ) − ϕψ(x2 ) = y1 + y2 − x1 − ϕ(y1 ) − x2 , ∀x1 ∈ X1 , y1 ∈ Y1 , x2 ∈ X2 , y2 ∈ Y2 . Furthermore, we get

φ(a − hv)φ(a − hv)(x1 + y1 + x2 + y2 )
= φ(a − hv) y1 + y2 − x1 − ϕ(y1 ) − x2 = y1 + y2 + x1 − ϕ(y1 ) + ϕ(y1 ) + x2 = x1 + x2 + y1 + y2 , ∀x1 ∈ X1 , y1 ∈ Y1 , x2 ∈ X2 , y2 ∈ Y2 .

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This implies that φ(a − hv)φ(a − hv) = 1R , and so a − hv ∈ EndR (R) is left invertible. Assume that ψϕ
= 1Y1 . Given any t ∈ im(a), x1 ∈ X1 , y1 ∈ Y1 , we have t ∈ aR = a Y1 ⊕ Y2 . So we can find some y1′ ∈ Y1 and y2′ ∈ Y2 such that t = a(y1′ + y2′ ). Choose x′1 = −x1 ∈ X1 and x′2 = −ϕ(y1 + y1′ ) ∈ X2 . Then (a − hv)(x′1 + x′2 + y1′ + y2′ ) = a(y1′ + y2′ ) − x′1 − ψϕ(y1′ ) − ψ(x′2 ) = t + x1 − y1′ − ψ(x′2 ) = t + x1 + y1 .

This means that a − hv ∈ EndR (R) is a right R-epimorphism. Since R is a projective right R-module, a − hv ∈ EndR (R) is right invertible. Let u = a − hv. Thus, a = e + u, where u ∈ R is right or left invertible. Therefore the result follows.  Corollary 5.3.2. Let R be a weakly stable exchange ring. Then every regular matrix A ∈ Mn (R)(n ∈ N) is the sum of an idempotent matrix and a right or left invertible matrix. Proof. In view of [9, Theorem 1.4] and Theorem 5.1.6, Mn (R) is a weakly stable exchange ring. Therefore the proof is true by Theorem 5.3.1.  Corollary 5.3.3. Let R be an exchange ring. Then the following are equivalent: (1) R has stable range one. (2) For any regular a ∈ R, there exists an idempotent e ∈ R and a unit T u ∈ R such that a = e + u and aR eR = 0.

Proof. (1) ⇒ (2) Let a ∈ R be regular. By virtue of Theorem 5.3.1, there exists an idempotent e ∈ R and a right or left invertible u ∈ R such that a = e + u. Clearly, R is directly finite; hence, u ∈ R is a unit. Furthermore,
T T T we observe that aR eR = im(a) hvR ⊆ im(a) X1 ⊕ Y1 = 0. This T implies that aR eR = 0. (2) ⇒ (1) For any regular a ∈ R, there exists an idempotent e ∈ R and T a unit u ∈ R such that a = e+u and aR eR = 0. Hence, au−1 = eu−1 +1. T Thus, au−1 e = eu−1 e + e ∈ aR eR = 0. This yields au−1 (a − u) = 0, and so au−1 a = a. That is, a ∈ R is unit-regular. Thus, R has stable range one by Lemma 1.3.1.  Let R be a π-regular ring having stable range one. It follows from Corollary 5.3.3 that for any a ∈ R, there exists some n ∈ N such that

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an ∈ R is clean. Let R = M2 F [x]/(x2 ) , where F is a field. Then R is strongly π-regular, so it isa π-regular ring having stable range one by [382,    01 01 Theorem 30.9]. Let a = ∈ R, and let u = . Then a = aua 0x 10 with u ∈ U (R); hence, a is unit-regular. By Corollary 5.3.3, we have an idempotent e ∈ R and a unit u ∈ R such that a = e + u and aR ∩ eR = 0. As we know, a2 ∈ R is clean. But a2 cannot be written in the form above. This is because a2 is not regular. Corollary 5.3.4. Let R be a weakly stable exchange ring. If a ∈ R is regular, then 2a is the sum of two right or left invertible elements. Proof. Let a ∈ R. In view of Theorem 5.3.1, there exists an idempotent matrix e ∈ R such that a − e ∈ R is right or left invertible. As in the proof of Theorem 5.3.1, we also have that a + e ∈ R is right or left invertible. Therefore 2a = (a − e) + (a + e), as required.  Let R be a weakly stable exchange ring. If 21 ∈ R, it follows by Corollary 5.3.4 that every regular a ∈ R is the sum of two one-sided units. If R is an exchange ring having stable range one, analogously, we deduce that for any regular a ∈ R, 2a is the sum of two units. An element e ∈ R is infinite if there exist orthogonal idempotents f, g ∈ R such that e = f + g while eR ∼ = f R and g 6= 0. A simple ring is said to be purely infinite if every nonzero left ideal of R contains an infinite idempotent. It is well known that a ring R is a purely infinite simple ring if and only if it is not a division ring and for any nonzero a ∈ R there exist s, t ∈ R such that sat = 1. The class of purely infinite simple rings is rather large. All purely infinite simple C ∗ -algebras and all directly infinite regular rings over which every finitely generated projective right module is free are purely infinite simple rings (cf. [18, Theorem 1.6]). If R is any directly infinite simple ring, then the direct limit lim→ M2n (R) is a purely infinite simple ring (cf. [18, Example 1.3]). The exchange property of purely infinite rings is proved by Ara [11]. Theorem 5.3.5. Let R be a purely infinite simple ring. Then every regular element in R is the sum of an idempotent and a right or left invertible element. Proof. Let a ∈ R and set b = 1 − a. If a = a2 , then a ∈ Ra and T 1 − a ∈ R(1 − a). Assume that Ra Rb ⊇ R(ab) 6= 0. By definition

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of purely infinite simple, there is a nonzero idempotent g ∈ R such that T g ∈ Ra Rb. In addition, we have some s, t ∈ R such that sgt = 1. Let e = g + gta(1 − g). Then e ∈ Ra is an idempotent. Observing that se = sg+sgta(1−g) = sg+a−ag, we get a = se+(a−s)g. This implies that 1 = a + b = se + (a − s)g + b = se + kb for some k ∈ R. Let f = e + (1 − e)se. Then f ∈ Ra is an idempotent. Moreover, 1−f = (1−e)(1−se) ∈ R(1−a). Thus, R is an exchange ring. Let x, y ∈ R be regular. If x = 0, then xR .⊕ yR. If x 6= 0, then there exist s, t ∈ R such that sxt = 1. Construct a right R-morphism ϕ : xR → R given by ϕ(xr) = sxr for any r ∈ R. For any r ∈ R, we see that r = sxtr = ϕ(xtr), and so ϕ is an R-epimorphism. Since R is a projective right R-module, we get a splitting exact sequence: ϕ

0 → ker(ϕ) ֒→ xR → R → 0.

Thus, R .⊕ R ⊕ ker(ϕ) ∼ = xR. As y ∈ R is regular, yR .⊕ R. As a result, ⊕ yR . xR. According to Theorem 5.2.3, R is weakly stable, and therefore we complete the proof from Theorem 5.3.1.  Corollary 5.3.6. Let R be a purely infinite simple ring. If is generated by one-sided invertible elements.

1 2

∈ R, then R

Proof. As in the proof of Theorem 5.3.5, R is an exchange ring. Let a ∈ R. Then there exists an idempotent e ∈ aR such that 1 − e ∈ (1 − a)R. Thus, ea, (1 − e)(1 − a) ∈ R are both regular. According to Theorem 5.3.5, we have an idempotent f ∈ R and a right or left invertible element u ∈ R such that ea = f + u. Similarly to Corollary 5.3.4, we also have a right or left invertible element v ∈ R such that ea = −f + v. Thus, 2ea = (f + u) + (−f + v) = u + v. Likewise, we have two right or left invertible elements u′ , v ′ ∈ R such that 2(1 − e)(1 − a) = u′ + v ′ . Thus 2a = 2(1 − e) + 2(ea) − 2(1 − e)(1 − a) = 1 + (1 − 2e) + u + v − u′ − v ′ = 1 + (1 − 2e)−1 + u + v − u′ − v ′ .

Therefore a ∈ R is generated by six right or left invertible elements, as required.  5.4

Matrix Factorizations

It is instructive to represent an invertible matrix as a product of triangular matrices. Lemma 1.2.1 shows that a ring R has stable range one if and

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only if for any A ∈ GLn (R), there exist L ∈ L, U ∈ U, K ∈ L such that A = LU K, where U and K have diagonal entries 1. The aim of this section is to study triangular factorizations of 2×2 invertible matrices over a weakly stable ring. These extend the corresponding results on stable range one and give new characterizations of weakly stable rings. Theorem 5.4.1. Let R be a ring. Then the following are equivalent: (1) R is weakly stable. (2) For any A ∈ GL2 (R), there exist L ∈ L, U ∈ U, K ∈ L such that A = LU K, where two of L, U and K have diagonal entries 1. Proof. (1) ⇒ (2) Let A = (aij ) ∈ GL2 (R). Then a11 R + a12 R = R. Since R is a weakly stable ring, there exists y ∈ R such that a11 + a12 y = u ∈ R is right or left invertible. Assume that uv = 1 for some v ∈ R. Then      u a12 u0 1 va12 = a21 + a22 y a22 01 a +a y a    21   22  22 u0 10 1∗ = . 01 ∗∗ 01     ∗0 1∗ 1 0 Thus, we see that A = ∈ LUL. ∗∗ 01 −y 1 Assume that vu = 1 for some v ∈ R. Then      u a12 1 a12 u0 = a21 + a22 y a22 (a + a y)v a 01  21  22   22  10 1∗ u0 = . ∗1 0∗ 01     10 u∗ 1 0 Thus, we see that A = ∈ LUL, as required. ∗1 0∗ −y 1 (2) ⇒ (1) Given ax + b = 1 in R, then we have    −1 a b x xa − 1 = ∈ GL2 (R). −1 x 1 a   a b By assumption, we have a factorization = LU K ∈ LUL, where −1 x two of L, U and K have diagonal entries 1. If every diagonal entry of L and U is 1, then       a b 10 1∗ ∗0 = . −1 x ∗1 01 ∗∗

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Thus, we have z, u ∈ R such that       10 a b 1∗ ∗0 = . z1 −1 x 01 ∗u This implies that x + zb = u ∈ R is left invertible. In view of Lemma 4.1.2, there exists y ∈ R such that a + by ∈ R is right invertible. If every diagonal entry of U and K is 1, then       a b ∗0 1∗ 10 = . −1 x ∗∗ 01 ∗1 Thus, we have y, u ∈ R such that       a b 10 u∗ 1∗ = . −1 x y1 0∗ 01 This implies that a + by = u. Clearly, u ∈ R is left invertible. Hence, a + by ∈ R is left invertible. If every diagonal entry of L and K is 1, then       a b 10 ∗∗ 10 = . −1 x ∗1 0∗ ∗1 Thus, we have y, u ∈ R such that       a b 10 10 u0 = . −1 x y1 ∗1 ∗∗ This implies that a + by = u. In this case, u ∈ R is right invertible. Therefore we conclude that R is a weakly stable ring.  Corollary 5.4.2. Let R be a weakly stable ring. Then every 2×2 invertible matrix is similar to the product of two invertible triangular matrices. Proof. Let A ∈ GL2 (R). In view of Theorem 5.4.1, there exist L ∈ L, U ∈ U, K ∈ L such that A = LU K, where two of L, U and K have diagonal entries 1. Clearly, either L or K has diagonal entries 1. If L has diagonal entries 1, then L−1 AL = U (KL) is the product of an upper triangular matrix U and a lower triangular matrix KL. If K has diagonal entries 1, then KAK −1 = (KL)U is the product of a lower triangular matrix KL and an upper triangular matrix U , and therefore we obtain the result.  Recall that an exchange ring R satisfies the comparability axiom provided that for any idempotents e, f ∈ R, either eR .⊕ f R or f R .⊕ eR. Clearly, an exchange ring R satisfies the comparability axiom if and only

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if for any idempotents e, f ∈ R, there exist s ∈ eRf, t ∈ f Re such that st = e or ts = f . (cf. [217]). We note that every exchange ring satisfying the comparability axiom is weakly stable. Let R be the collection of ℵ0 × ℵ0 matrices with entries from a field F , the form   A  q    { | A ∈ Mn (F ) for some n ∈ N, q ∈ F }.  q   .. . ℵ0 ×ℵ0

Then R is a regular ring satisfying the comparability axiom. Hence, it is weakly stable. In view of Corollary 5.4.2, every 2 × 2 invertible matrix over R can be represented as the product of two invertible triangular matrices. Example 5.4.3. Let V be an infinite-dimensional vector space over a division ring D, and let R = EndD (V ). Then R is a weakly stable ring. Let {x1 , x2 , · · · , xn , · · · } be a basis of V . Define τ : V → V given by τ (xi ) = xi+1 (i = 1, 2, · · · ) and σ : V → V given by σ(x1 ) = 0 and σ(x 1V, while τ σ 6= 1V . Let A =  i ) = xi−1 (i = 2, 3, · · · ). Then  στ =  τ 1V − τ σ 1V 0 τ 1V − τ σ . Then A = ∈ GL2 (R). −2 · 1V σ −σ 1V −1V σ By virtue of Theorem 5.4.1, A can be expressed as the product of three triangular invertible matrices. In fact, we have     1V 0 τ 1V − τ σ 1V 0 A= . −σ 1V 0 σ −τ 1V In this case, A cannot be expressed as LU K, where L ∈ L, U ∈ U, K ∈ L and in L and U or in U and K all the diagonal entries are equal to 1V .  Theorem 5.4.4. Let R be a ring. Then the following are equivalent: (1) R is weakly stable. (2) For any A ∈ GL2 (R), there exist U ∈ U, L ∈ L, W ∈ U such that A = U LW , where two of U, L and W have diagonal entries 1.     01 01 Proof. (1) ⇒ (2) For any A ∈ GL2 (R), we have A ∈ GL2 (R). 10 10 In 5.4.1, we can find L ∈ L, U ∈ U, K ∈ L such that  view  of Theorem  01 01 A = LU K, where two of L, U and K have diagonal entries 10 10

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1. This implies that

Let U′ =



A=







01 10

01 10



LU K



01 10



.

     01 01 01 , L′ = U , 10 10 10     01 01 W′ = K . 10 10 L

Then A = U ′ L′ W ′ , where U ′ ∈ U, L′ ∈ L, W ′ ∈ U and two of L′ , U ′ and K ′ have diagonal entries 1.     01 01 (2) ⇒ (1) For any A ∈ GL2 (R), we have A ∈ GL2 (R). 10 10 By assumption, there exist U ∈ U, L ∈ L, W ∈ U such that     01 01 A = U LW, 10 10 where two of U, L and W have diagonal entries 1. We infer that     01 01 A= U LW . 10 10 Let ′

L =



01 10





01 U 10  0 K′ = 1



   01 01 ,U = L , 10 10    1 01 W . 0 10 ′



Then A = L′ U ′ K ′ , where L′ ∈ L, U ′ ∈ U, K ′ ∈ L and two of U ′ , L′ and W ′ have diagonal entries 1. Therefore R is a weakly stable ring by Theorem 5.4.1.  Theorem 5.4.4 implies that every 2 × 2 invertible matrix over a weakly stable ring admits a triangular matrix by two elementary transformations. Corollary 5.4.5. Let R be a ring. Then the following are equivalent: (1) R has stable range one. (2) R is directly finite and for any A ∈ GL2 (R), there exist L ∈ L, U ∈ U, K ∈ L such that A = LU K, where two of L, U and K have diagonal entries 1.

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(3) R is directly finite and for any A ∈ GL2 (R), there exist U ∈ U, L ∈ L, W ∈ U such that A = U LW , where two of U, L and W have diagonal entries 1. Proof. (1) ⇒ (2) Obviously, R is directly finite. For any A ∈ GL2 (R), it follows by Lemma 1.2.1 that there exist L ∈ L, U ∈ U, K ∈ L such that A = LU K, where U and K have diagonal entries 1, as required. (2) ⇒ (1) In view of Theorem 5.4.1, R is a weakly stable ring. Since R is directly finite, all right or left invertible elements in R are invertible; hence, R has stable range one. (1) ⇔ (3) As in the preceding consideration, we complete the proof by Theorem 5.4.4.  Example 5.4.6. Let V be an infinite-dimensional vector space over a division ring D,and let R = EndD (V ). Construct τ and σ as in Example σ −2 · 1V 5.4.3. Let A = . Then 1V − τ σ τ A=



0 1V 1V 0



τ 1V − τ σ −2 · 1V σ



0 1V 1V 0



∈ GL2 (R).

By virtue of Theorem 5.4.4, A can be expressed as the product of three triangular invertible matrices. In fact, we have     1V −σ σ 0 1V −τ A= . 0 1V 1V − τ σ τ 0 1V In this case, A cannot be expressed as U LW , where U ∈ U, L ∈ L, W ∈ U and in U and L or in L and W all the diagonal entries are equal to 1V .  Example 5.4.7. Let V be an infinite-dimensional vector space over a   0 1V division ring D, and let R = EndD (V ). Then 6∈ LU, UL, where 1V 0 all the diagonal entries of one of L and U are 1V . Thus, we need three matrices as shown in Theorem 5.4.1 and Theorem 5.4.4.  If



u0 ∗v



∈ GL2 (R), then 

u0 ∗v

−1

=



s 1 − su 0 u



1 0 ∗w



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 u∗ for some s ∈ R, w ∈ U (R). If ∈ GL2 (R), then 0v  −1    u∗ u 0 w∗ = 0v 1 − su s 0 1

for some s ∈ R, w ∈ U (R). Using these facts, it follows from Theorem 5.4.4 that for any A ∈ GL2 (R), there exist L, K ∈ L, U, W ∈ U such that LU AKW = I2 , where I2 denotes the diagonal matrix diag(1, 1). Now, we give explicit results by a new route. Theorem 5.4.8. Let R be a ring. Then the following are equivalent:

(1) R is weakly stable. (2) For any A ∈ GL2 (R), there exist L, K ∈ L, U ∈ U such that LAKU = I2 , where two of L, K and U have diagonal entries 1. Proof. (1) ⇒ (2) Let A = (aij ) ∈ GL2 (R). Then a11 R + a12 R = R. Since R is a weakly stable ring, we have y ∈ R such that a11 + a12 y = u ∈ R is right or left invertible. Assume that uv = 1 for some v ∈ R. Then        10 01 u 0 01 1∗ A = . y1 10 1 − vu v 10 ∗∗    −1 u 0 v 1 − vu ∈ GL2 (R). Hence It is easy to verify that = 1 − vu v 0 u we can find w ∈ U (R) such that           10 10 01 u 0 01 1∗ 1 0 A = . ∗1 y1 10 1 − vu v 10 01 0w Let

U=



01 10

   10 10 ,K = , ∗1 y1     u 0 01 1∗ 1 0 . 1 − vu v 10 01 0 w−1

L= 



Then L ∈ L, K ∈ L, U ∈ U and LAKU = I2 with every entry of L and K is 1. Assume that vu = 1 for some v ∈ R. Then       10 v 0 10 1∗ A = I2 . ∗1 1 − uv u y1 01

Let

L=



10 ∗1



v 0 1 − uv u



,K =



10 y1



,U =



1∗ 01



.

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Then L ∈ L, K ∈ L, U ∈ U and LAKU = I2 , where every entry of K and U is 1.   a b (2) ⇒ (1) Given ax + b = 1 in R, we have ∈ GL2 (R). By −1 x   a b assumption, we can find L ∈ L, K ∈ L, U ∈ U such that L KU = −1 x I2 , where two of L, K and U have diagonal entries 1. If every diagonal entry of L and K is 1, then we have w, y, z ∈ R such that      1 0 a b 10 z∗ = I2 . w1 −1 x y1 0∗ This implies that



a + by b −1 + xy x



z∗ 0∗



=



1 0 −w 1



.

Hence, (a + by)z = 1, and so a + by ∈ R is right invertible. If every diagonal entry of K and U is 1, then we can find w, y, z ∈ R such that      z0 a b 10 1w = I2 . ∗∗ −1 x y1 0 1 This implies that



z0 ∗∗



a + by b −1 + xy x



=



1 −w 0 1



.

Hence, z(a + by) = 1, and so a + by ∈ R is left invertible. If every diagonal entry of L and U is 1, then we can find w, y, z ∈ R such that      10 a b ∗0 1w = I2 . z1 −1 x ∗y 0 1

This implies that



a b za − 1 x + zb



∗0 ∗y



=



1 −w 0 1



.

Hence, (x + zb)y = 1, and so x + zb ∈ R is right invertible. In view of Lemma 4.1.2, there exists t ∈ R such that a + bt ∈ R is left invertible. Therefore R is weakly stable.  Corollary 5.4.9. Let R be an exchange ring satisfying the comparability axiom. Then for any A ∈ GL2 (R), there exist L, K ∈ L, U ∈ U such that LAKU = I2 , where two of L, K and U have diagonal entries 1.

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Proof. According to Theorem 5.2.3, R is a weakly stable ring. So the result follows from Theorem 5.4.8.  Theorem 5.4.10. Let R be a ring. Then the following are equivalent: (1) R is weakly stable. (2) For any A ∈ GL2 (R), there exist U, W ∈ U, K ∈ L such that U AW K = I2 , where two of U, W and K have diagonal entries 1.     01 01 Proof. (1) ⇒ (2) For any A ∈ GL2 (R), we have A ∈ GL2 (R). 10 10 In view of Theorem 5.4.8, we can find L, K ∈ L, U ∈ U such that LAKU = I2 , where two of L, K and U have diagonal entries 1. Thus,          2 01 01 01 01 01 L A KU = = I2 . 10 10 10 10 10 Let         01 01 01 01 U′ = L ,W′ = K , 10 10 10 10     01 01 K′ = U . 10 10 Then U ′ AW ′ K ′ = I2 and U ′ , W ′ ∈ U, K ′ ∈ L, where two of U ′ , W ′ and K ′ have diagonal entries 1.     01 01 (2) ⇒ (1) For any A ∈ GL2 (R), we have A ∈ GL2 (R). 10 10 By assumption, there exist U, W ∈ U, K ∈ L such that     01 01 U A W K = I2 , 10 10 where two of U, W and K have diagonal entries 1. Let         01 01 01 01 L′ = U , K′ = W , 10 10 10 10     01 01 U′ = K . 10 10 Then L′ AK ′ U ′ = I2 and L′ , K ′ ∈ L, U ′ ∈ U, where two of L′ , K ′ and U ′ have diagonal entries 1. According to Theorem 5.4.8, R is weakly stable. Corollary 5.4.11. Let R be an exchange ring satisfying the comparability axiom. Then for any A ∈ GL2 (R), there exist U, W ∈ U, K ∈ L such that U AW K = I2 , where two of U, W and K have diagonal entries 1. Proof. Obviously, R is a weakly stable ring, and therefore we complete the proof from Theorem 5.4.10. 

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Chapter 6

Related Comparability

Following Goodearl, a regular ring R satisfies general comparability, provided that, for any x, y ∈ R, there exists some u ∈ B(R) such that uxR . uyR and (1 − u)yR . (1 − u)(xR). This concept evolved from operator algebras and Baer rings, where it is one of the objectives of the axiomatic development (cf. [217]). Obviously, every regular ring satisfying the comparability axiom satisfies general comparability. In this chapter, we introduce related comparability over exchange rings as a generalization of the general comparability used by Goodearl. We say that a ring R satisfies related comparability provided that for any idempotents e, f ∈ R with e = 1 + ab and f = 1 + ba for some a, b ∈ R, there exists some u ∈ B(R) such that ueR .⊕ uf R and (1 − u)f R .⊕ (1 − u)eR. The class of rings satisfying related comparability is quite large. It includes regular rings satisfying general comparability, one-sided unit-regular rings and partially unit-regular rings (cf. [231] and [237]). First, we develop necessary and sufficient conditions for rings to satisfy related comparability. Then we observe that related comparability over exchange rings is Morita invariant. After that the general comparability over exchange rings is studied.

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6.1

rings

Related Units

The aim of this section is to introduce the notion of related units. The main emphasis is on developing the equivalent conditions for rings to satisfy related comparability by means of such elements. Definition 6.1.1. An element w ∈ R is called a related unit if there exists an e ∈ B(R) such that ew ∈ eRe is right invertible and (1 − e)w ∈ (1 − e)R(1 − e) is left invertible. We use U + (R) to denote the set of all related units in R. An element a ∈ R is called to be related unit-regular if there exists some w ∈ U + (R) such that a = awa. Obviously, every related unit is related unit-regular. The converse is not true. We claim that a ring R is a division ring if and only if it is a regular ring in which every nonzero related unit-regular element is a related unit. One direction is clear. Conversely, assume that R is a regular ring satisfying the preceding condition. For every 0 6= x ∈ R, there exists some y ∈ R such that xyx = x. Let e = xy. Then, 0 6= e = e2 is related unit-regular. Hence, e is a related unit. Thus, we have some f ∈ B(R) such that f e and (1 − f )e are right and left invertible in f Rf and (1 − f )R(1 − f ), respectively. Since (f e)2 = f ef e = f e2 = f e and f e is right invertible in f Rf , we deduce that f e = f . Likewise, (1 − f )e = 1 − f . Therefore, e = f e + (1 − f )e = 1. That is, xy = 1. Likewise, yx = 1. Thus, R is a division ring , and we are done. Lemma 6.1.2. Let A be a right R-module, let E = EndR (A), and let e, f ∈ E be idempotents. Then the following hold: (1) eA .⊕ f A if and only if there exist a ∈ eEf, b ∈ f Ee such that e = ab. (2) eA ∼ = f A if and only if there exist a, b ∈ E such that e = ab and f = ba. Proof. (1) Suppose that eA .⊕ f A. Then there exist R-morphisms α : eA → f A and β : f A → eA such that βα = 1eA . Let a : A = f A ⊕ (1 − f

β

e

α

f )A ։ f A → eA ֒→ A and b : A = eA ⊕ (1 − e)A ։ eA → f A ֒→ A. Then e = ab with a = eaf ∈ eEf and b = f be ∈ f Ee. Suppose that there exist a ∈ eEf, b ∈ f Ee such that e = ab. Construct two R-morphisms ϕ : eA → f A given by ϕ(er) = ber for any r ∈ A and φ : f A → eA given by φ(f r) = af r for any r ∈ A. It is easy to verify that φϕ = 1eA , i.e., ϕ is a splitting R-monomorphism. Thus, we have a right

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R-module D such that eA ⊕ D ∼ = f A. Therefore eA .⊕ f A. (2) Suppose that eA ∼ = f A. Then there exist R-morphisms α : eA → f A and β : f A → eA such that βα = 1eA and αβ = 1f A . Let a : A = f A ⊕ f

β

e

α

(1 − f )A ։ f A → eA ֒→ A and b : A = eA ⊕ (1 − e)A ։ eA → f A ֒→ A. Then e = ab and f = ba with a = eaf ∈ eEf and b = f be ∈ f Ee. Suppose that there exist a, b ∈ E such that e = ab and f = ba. Let c = eaf and d = f be. Then e = cd and f = dc with c ∈ eEf and d ∈ f Ee. Construct two R-morphisms ϕ : eA → f A given by ϕ(er) = der for any r ∈ A and φ : f A → eA given by φ(f r) = cf r for any r ∈ A. It is easy to verify that φϕ = 1eA and ϕφ = 1f A , i.e., ϕ is an isomorphism. Therefore eA ∼  = f A, as asserted. Theorem 6.1.3. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E satisfies related comparability. (2) Every regular element in E is related unit-regular. (3) A = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 implies that there exists some ⊕ u ∈ B(E) such that uB . uC and (1 − u)C .⊕ (1 − u)B.

Proof. (1) ⇒ (3) Given A = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 , we have idempotents e, f ∈ E such that A1 = (1 − e)A, B = eA, A2 = (1 − f )A and C = f A. As (1 − e)A ∼ = (1 − f )A, it follows by Lemma 6.1.2 that there exist a, b ∈ E such that e = 1 + ab and f = 1 + ba. By hypothesis, there exists some u ∈ B(E) such that ueE .⊕ uf E and (1 − u)f E .⊕ (1 − u)eE. By Lemma 6.1.2 again, we have some s ∈ ueEuf and t ∈ uf Eue such that ue = st. According to Lemma 6.1.2, we get ueA .⊕ uf A. That is, uB .⊕ uC. Likewise, (1 − u)C .⊕ (1 − u)B, as required. (3) ⇒ (2) For any regular x ∈ E, there exists a y ∈ E such that x = xyx. Since xy and yx are both idempotents, A = yxA ⊕ (1 − yx)A = xyA ⊕ (1 − xy)A = xA ⊕ (1 − xy)A. Obviously, ϕ : xA → yxA, given by xr 7→ yxr, is an isomorphism. So, there exists f ∈ B(E) such that f (1−xy)A .⊕ f (1−yx)A and (1−f )(1−yx)A .⊕ (1−f )(1−xy)A. Thus, there exists a splitting R-monomorphism φ : f (1 − xy)A → f (1 − yx)A. Let α : f A → f A with α(b + c) = ϕ(b) + φ(c) for any b ∈ f xA, c ∈ f (1 − xy)A. It is easy to verify that α ∈ EndR (f A) is left invertible. Furthermore, we see that f x = xαf x. Moreover, there exists a splitting R-epimorphism ψ : (1 − f )(1 − xy)A → (1 − f )(1 − yx)A. Let β : (1 − f )A → (1 − f )A with β(b+c) = ϕ(b)+ψ(c) for any b ∈ (1−f )xA, c ∈ (1−f )(1−xy)A. One easily
checks that β ∈ EndR (1 − f )A is right invertible. In addition, we get

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(1−f )x = xβ(1−f )x. Define w : A = f A⊕(1−f )A → f A⊕(1−f )A given by w(s + t) = α(s) + β(t) for any s ∈ eA, t ∈ (1 − e)A. Then w ∈ U + (E). Furthermore, x = f x + (1 − f )x = xwx, as desired (2) ⇒ (1) For any idempotents e, f ∈ E with e = 1 + ab and f = 1 + ba for some a, b ∈ E, we see that 1−e = (−a)(1−f )b and 1−f = b(1−e)(−a). Let c = (1 − e)(−a)(1 − f ) and d = (1 − f )b(1 − e). Then 1 − e = cd and 1 − f = dc. In addition, dcd = (1 − f )d = d. By hypothesis, there exists some w ∈ U + (E) such that d = dwd. Set u = (e − wd)w(f − dw). Then (e − wd)2 = 1 = (f − dw)2 , whence, u ∈ U + (E). Furthermore, we see that eu = w − wdw = uf . As u ∈ U + (E), there is some g ∈ B(E) such that gus = g and t(1 − g)u = 1 − g. Thus, eg = uf sg = euf g · f seg. In view of Lemma 6.1.2, we get geE .⊕ gf E. Analogously, we deduce that (1 − g)f E .⊕ (1 − g)eE. Therefore E satisfies related comparability.  Theorem 6.1.3 shows that related comparability is right and left symmetric. That is, a ring R satisfies related comparability if and only if the opposite ring Rop also satisfies related comparability. Also we note that every commutative ring satisfies related comparability. In addition, an exchange ring R satisfies related comparability if and only if R = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 implies there exists some e ∈ B(R) such that Be .⊕ Ce and C(1 − e) .⊕ B(1 − e). Corollary 6.1.4. Let R be a ring. Then the following are equivalent: (1) R satisfies related comparability. (2) For any regular a ∈ R, aR + bR = R implies that there exists some y ∈ R such that a + by ∈ U + (R). (3) Whenever ax + b = 1 with ba = 0, then there exists some y ∈ R such that a + by ∈ U + (R). Proof. (1) ⇒ (2) For any regular a ∈ R, aR + bR = R implies that there exist x, y ∈ R such that ax + by = 1. In view of Theorem 6.1.3, a = awa for some w ∈ U + (R). Thus, we have an e ∈ B(R) such that ews = e and t(1 − e)w = 1 − e for some s, t ∈ R. Let f = wa. Then, f x + wc = w, where c = by. So f (x + wc) + (1 − f )wc = w. Clearly, (1 − f )wc
= (1 − f )w. It is easy to verify that (1 − f )w = (1 − f )w es + (1 − e)t (1 − f )w. Let
g = (1 − f )w es + (1 − e)t (1 − f ). Then g = g 2 and f g = gf = 0. This implies that f (x + wc) = f w and (1 − f )wc = gw. As a result,
we deduce that w a+c(es+(1−e)t)(1−f )(1+f wc(es+(1−e)t)(1−f )) 1−f wc(es+
(1 − e)t)(1 − f ) w = f + wc(es + (1 − e)t)(1 − f )(1 + f wc(es + (1 − e)t)(1 −

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f )) 1 − f wc(es + (1 − e)t)(1
− f ) w = f (1 − f wc(es + (1 − e)t)(1 −
f )) + wc(es + (1 − e)t)(1 − f ) w = f + (1 − f )w(es + (1 − e)t)(1 − f ) w = (f + g)w = w. As
w ∈ U + (R), we deduce that a + byz
∈ U + (R), where z = es + (1 − e)t (1 − f ) 1 + f wc(es + (1 − e)t)(1 − f ) . (2) ⇒ (3) Whenever ax + b = 1 with ba = 0, then axa = a, i.e., a ∈ R is regular. Thus, there exists some y ∈ R such that a + by ∈ U + (R). (3) ⇒ (1) Given any regular a ∈ R, there exists some x ∈ R such that a = axa and x = xax. Hence, xa + (1 − xa) = 1 with (1 − xa)x = 0. By hypothesis, we have an element y ∈ R such that u := x+(1−xa)y ∈ U + (R). Thus, a = axa = aua. According to Theorem 6.1.3, R satisfies related comparability.  Corollary 6.1.5. Let e ∈ R be an idempotent. If R satisfies related comparability, then so does eRe. Proof. Assume that ax + b = e and ba = 0, where a, x, b ∈ eRe. Then (a + 1 − e)(x + 1 − e) + b = 1 with b(a + 1 − e) = 0. Since R satisfies related comparability, by virtue of Corollary 6.1.4, there exists an element y ∈ R such that a + 1 − e + by ∈ U + (R). Thus, we have an f ∈ B(R) such that f (a + 1 − e + by) is right invertible in f Rf and (1 − f )(a + 1 − e + by) is left invertible in (1 − f )R(1 − f ). Assume that f (a + 1 − e + by)s = f for
some s ∈ R. Then f (1−e)se = 0; hence, f se = f ese. Thus, f a+b(eye) (ese) = f e. Assume that (1 −
f )t(a + 1 − e + by) = 1 − f for some t ∈ R. Then (1−f )(ete) a+b(eye) = (1−f )e. This implies that a+b(eye) ∈ U + (eRe). By Corollary 6.1.4 again, eRe satisfies related comparability.  If Mn (R) satisfies related comparability, then so does R from Corollary 6.1.5. It is well known that a ring R is unit-regular if and only if, for any a ∈ R, there exists an idempotent e and a unit u such that a = eu. We now generalize this result to related comparability. Theorem 6.1.6. Let R be a ring. Then the following are equivalent: (1) R satisfies related comparability. (2) For any regular a ∈ R, there exists an idempotent e and an element w ∈ U + (R) such that a = ew. (3) For any regular a ∈ R, there exists an idempotent e and an element w ∈ U + (R) such that a = we. Proof. (1) ⇒ (2) Let a ∈ R be regular. Then there exists some x ∈ R such that a = axa. Since aR + (1 − ax)R = R, it follows from Corollary

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6.1.4 that w := a
+ (1 − ax)y ∈ U + (R) for some y ∈ R. Hence, a = axa = ax a + (1 − ax)y = axw, as required. (2) ⇒ (1) Given ax + b = 1 with ba = 0 in R, then axa = a. By assumption, there exists an idempotent e ∈ R and a w ∈ U + (R) such that a = ew. Hence, ewx(1−e)+b(1−e) = 1−e. This implies that
e+b(1−e) = 1 − ewx(1 − e) ∈ U (R). Thus, a + b(1 − e)w = 1 − ewx(1 − e) w ∈ U + (R). According to Corollary 6.1.4, R satisfies related comparability. (1) ⇔ (3) is symmetric.  Corollary 6.1.7. Let R be a ring. Then the following are equivalent: (1) R satisfies related comparability. (2) Whenever a = axa, there exists some u ∈ U + (R) such that 1 − a(x + u) ∈ U (R). Proof. (1) ⇒ (2) Whenever a = axa, then ax + (1 − ax) = 1. So, we can find an element y ∈ R such that a + (1 − ax)y ∈ U + (R). In view
of Lemma 4.1.2, there exists some z ∈ R such that u := − x + z(1 − ax) ∈ U + (R). Let c = −z(1 − ax). Then 1 − ac = 1 ∈ U (R); hence, 1 − a(x + u) ∈ U (R), as required. (2) ⇒ (1) For any regular a ∈ R, there exists some x ∈ R such that a = axa. Hence, w := 1 − a(x + u) ∈ U (R) for a u ∈ U + (R). Thus, a(−uw−1 ) + (1 − ax)w−1 = 1. Obviously, −uw−1 ∈ U + (R). In view of Lemma 4.1.2, we have some y ∈ R such that
v := a+(1−ax)w−1 y ∈ U + (R). As a result, a = axa = ax a+(1−ax)w−1 y = axv. According to Theorem 6.1.6, R satisfies related comparability.  Lemma 6.1.8. Let R be a ring. Then the following hold: (1) a = aua, su = 1 in R implies that a = auw, wu = 1 for some w ∈ R. (2) a = aua, us = 1 in R implies that a = wua, uw = 1 for some w ∈ R. Proof. (1) Set w = a +
(1 − au)s(1 − ua), then wu = 1 and auw = au a + (1 − au)s(1 − ua) = aua = a. (2) Applying (1) to the opposite ring of R, we establish the result.  Let a, b ∈ R. We say that a is pseudo-similar to b if there exist x, y, z ∈ R such that xay = b, zbx = a and xyx = xzx = x. We denote it by a∼b. Now we observe a simple fact. Lemma 6.1.9. Let R be a ring, and let a, b ∈ R. Then the following are equivalent:

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(1) a∼b. (2) There exist some x, y ∈ R such that a = xby, b = yax, x = xyx and y = yxy. Proof. (2)⇒(1) is trivial. (1)⇒(2) As a∼b, we have x, y, z ∈ R such that b = xay, zbx = a and x = xyx = xzx. By replacing y with yxy and z with zxz, we can assume y = yxy and z = zxz. One easily checks that xazxy = xzbxzxy = xzbxy = xay = b, zxybx = zxyxayx = zxayx = zbx = a, zxy = zxyxzxy and x = xzxyx. This establishes the result.  Further, we see that a∼b if and only if there exist x, y ∈ R such that a = xby, b = yax, x = xyx if and only if there exist x, y ∈ R such that a = xby, b = yax and xy is an idempotent. Also we note that a∼b implies that an ∼bn for all n ∈ N (cf. [114]). In general, pseudo-similarity does not imply similarity as seen from the following example. Let V be a countably generated infinite-dimensional vector space over a division ring D, and let R = EndD (V ). Let {x1 , · · · , xn , · · · } be the basis of V . Define σ ∈ R such that α(xi ) = xi+1 for all i ≥ 1 and β(xi ) = xi−1 for all i ≥ 2 and β(x1 ) = 0. Set x = αβ. Then βxα = 1V , α = αβα and β = βαβ. Hence x and 1V are pseudo-similar, while they are not similar. If w ∈ U + (R), then there exists an e ∈ B(R) such that ewu = e and (1 − e)uw = 1 − e. We call u ∈ R a related inverse of w ∈ R, and denote it by w∗ . The following result is an extension of [231, Theorem 1]. Theorem 6.1.10. Let R be a ring. Then the following are equivalent: (1) R satisfies related comparability. (2) For any a, b ∈ R, a∼b if and only if there exists some w ∈ U + (R) such that a = wbw∗ , b = w∗ aw. (3) For any a, b ∈ R, a∼b implies that there exists some w ∈ U + (R) such that aw = wb. Proof. (1)⇒(2) Let a and b be pseudo-similar in R. By Lemma 6.1.9, there exist x, y ∈ R such that a = xby, b = yax, x = xyx and y = yxy. Using Theorem 6.1.3, we can find some u ∈ U + (R) such that y = yuy. Thus, we have an e ∈ B(R) such that eu is left invertible in eRe and (1 − e)u is right invertible in (1 − e)R(1 − e). Since ey = eyuy, from Lemma 6.1.8, we can find some c ∈ R such that ey = eyuc and ecu = e. Let k1 = (1 − xy − uy)u(1 − yx − yu), k2 = (1 − yx − yu)c(1 − xy − uy). Then ek1 bk2 = ea and ek2 k1 = e. Likewise, we have some d ∈ R, l1 :=

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(1 − xy − uy)u(1 − yx − yu) and l2 := (1 − yx − yu)d(1 − xy − uy) such that (1 − e)l1 bl2 = (1 − e)a and (1 − e)l1 l2 = 1 − e. Set w = ek1 + (1 − e)l1. Then w ∈ U + (R). In addition, ek2 + (1 − e)l2 = w∗ . Obviously, we have a = ek1 bk2 + (1 − e)l1 bl2 = wbw∗ . Analogously, we show that b = w∗ aw. Conversely, it is easy to verify that a = wbw∗ , b = w∗ aw implies that a∼b, as desired. (2) ⇒ (1) Given R = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 , then we have idempotents e, f ∈ R such that A1 = eR, B1 = (1 − e)R, A2 = f R and B2 = (1 − f )R. Since eR ∼ = f R, we have some a ∈ eRf and b ∈ f Re such that e = ab and f = ba. Hence, e = af b, f = bea, a = aba and b = bab. Thus, e∼f . By hypothesis, there is some w ∈ U + (R) such that f = w∗ ew. Assume that there exists g ∈ B(R) such that gw∗ w = g and (1 − g)ww∗ = 1 − g. Then we see that g(1 − f ) = gw∗ (1 − e)w = cd, where c = g(1 − f )w∗ g(1 − e) ∈ g(1 − f )Rg(1 − e) and d = g(1 − e)wg(1 − f ) ∈ g(1 − e)Rg(1 − f ), In view of Lemma 6.1.2, g(1 − f )R .⊕ g(1 − e)R. Likewise, (1 − g)(1 − e)R .⊕ (1 − g)(1 − f )R. That is, B2 g .⊕ B1 g and B1 (1 − g) .⊕ B2 (1 − g), as required.  (1) ⇒ (3) Given a∼b, it follows by Lemma 6.1.9 that we have x, y ∈ R such that a = xby, b = yax, x = xyx and y = yxy. In view of Theorem 6.1.3, there exists some v ∈ U + (R) such that y = yvy. Let u = (1 − xy − vy)v(1−yx−yv). It is easy to verify that (1−xy−vy)2 = 1 = (1−yx−yv)2 . Hence u ∈ U + (R), and au = ax = xb = ub. (3) ⇒ (1) Given any regular x ∈ R, there exists y ∈ R such that x = xyx. Since η : xyR = xR ∼ = yxR, we have xy∼yx. So there is + u ∈ U (R) such that yxu = uxy. Define α : (1−xy)R → (1−yx)R given by (1−xy)r → (1−yx)ur for any r ∈ R and β : (1−yx)R → (1−xy)R given by (1 − yx)r → (1 − xy)u∗ (1 − yx)r for any r ∈ R. Since (1 − yx)u = u(1 − xy), we easily check that α and β are right R-morphisms. Define φ : R = xR⊕(1−xy)R → yxR⊕(1−yx)R given by φ(x1 +x2 ) = η(x1 )+α(x2 ) for any x1 ∈ xR, x2 ∈ (1−xy)R and ψ : R = yxR⊕(1−yx)R → xR⊕(1−xy)R = R given by ψ(y1 + y2 ) = η −1 (y1 ) + β(y2 ) for any y1 ∈ yxR, y2 ∈ (1 − yx)R. Then (1 − ψφ)(x1 + x2 ) = (1 − xy)(1 − u∗ u)(1 − xy)x2 for any x1 ∈ xR and x2 ∈ (1 − xy)R and (1 − φψ)(y1 + y2 ) = (1 − yx)(1 − uu∗ )(1 − yx)y2 for any y1 ∈ yxR and y2 ∈ (1 − yx)R. As u∗ is a related inverse of u, we deduce that ψ is a related inverse of φ. This implies that φ(1) ∈ U + (R). Furthermore, we see that x = xφ(1)x. According to Theorem 6.1.3, we obtain the result.  The following result can be viewed as a generalization of [242, Theo-

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rem 2] though Handelman’s methods cannot be generalized to the related comparability. Corollary 6.1.11. Let R be a ring. Then the following are equivalent: (1) R satisfies related comparability. (2) For any idempotents e, f ∈ R, eR ∼ = f R implies that e = wf w∗ , f = ∗ + w ew for some w ∈ U (R). (3) For any idempotents e, f ∈ R, eR ∼ = f R implies that there exists some u ∈ U + (R) such that eu = uf .

Proof. (1) ⇒ (2) Given eR ∼ = f R with idempotents e, f ∈ R, then there exist a ∈ eRf, b ∈ f Re such that e = ab and f = ba; hence, e∼f . By virtue of Theorem 6.1.10, e = wf w∗ , f = w∗ ew for some w ∈ U + (R). (2) ⇒ (1) As in the proof of Theorem 6.1.10, R satisfies related comparability. (1) ⇒ (3) Whenever eR ∼ = f R with idempotents e, f ∈ R, then e∼f . It follows by Theorem 6.1.10 that there exists some u ∈ U + (R) such that eu = uf . (3) ⇒ (1) Suppose that R = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 . Then we have idempotents e, f ∈ R such that A1 = eR, B1 = (1 − e)R, A2 = f R and B2 = (1 − f )R. Since eR ∼ = f R, we have an element w ∈ U + (R) such that ew = wf . Assume that there exists g ∈ B(R) such that gw∗ w = g and (1 − g)ww∗ = 1 − g. Then g(1 − f ) = gw∗ (1 − e)w = cd, where c = g(1 − f )w∗ g(1 − e) ∈ g(1 − f )Rg(1 − e)

and d = g(1 − e)wg(1 − f ) ∈ g(1 − e)Rg(1 − f ).

In view of Lemma 6.1.2, we get g(1−f )R .⊕ g(1−e)R. Likewise, (1−g)(1− e)R .⊕ (1 − g)(1 − f )R. That is, B2 g .⊕ B1 g and B1 (1 − g) .⊕ B2 (1 − g), as required.  Example 6.1.12. If R is weakly stable, then

∞ Q

Ri , where Ri = R, satisfies

i=1

related comparability. If R is directly infinite, then

Ri is not weakly

i=1

stable. Proof. Given (ai )(xi ) + (bi ) = (1, 1, · · · , 1, · · · ) in

∞ Q

∞ Q

i=1

Ri , then ai xi + bi = 1

for all i ∈ N. For each i, since R is weakly stable, there exists some yi ∈ R such that ui := ai + bi yi ∈ R is right or left invertible. If ui ∈ R is

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right invertible, choose ei = 1. If ui ∈ R is left invertible, choose ei = 0. Then ei ui ∈ ei Rei is right invertible and (1 − ei )ui ∈ (1 − ei )R(1 − ∞
Q ei ) is left invertible. Let e = (ei ) and y = (yi ). Then e ∈ B Ri ,


and e (ai ) + (bi )y ∈ e





∞ Q

i=1

i=1


Ri e is right invertible and (1, 1, · · · , 1, · · · ) −

e (ai )+ (bi )y ∈ (1, 1, · · · , 1, · · · )− e invertible. Hence, (ai ) + (bi )y ∈ U +

∞
Q

∞ Q

i=1

i=1



Ri (1, 1, · · · , 1, · · · )− e is left

∞
Q Ri , and so Ri satisfies related i=1

comparability by Corollary 6.1.4. If R is directly infinite, as in the proof of Theorem 5.2.7, R is not weakly stable. 

6.2

Equivalences

The main purpose of this section is to investigate several necessary and sufficient conditions under which an exchange ring satisfies related comparability. Proposition 6.2.1. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies related comparability. (2) aR + bR = R implies that a + by ∈ U + (R) for some y ∈ R. (3) Ra + Rb = R implies that a + zb ∈ U + (R) for some z ∈ R. Proof. (1) ⇒ (2) Given aR + bR = R, then there exist some x, y ∈ R such that ax+by = 1. Since R is an exchange ring, by [382, Theorem 28.7], there exists an idempotent e ∈ byR such that 1 − e ∈ (1 − by)R. This implies that e = bys and 1 − e = axt for some s, t ∈ R. Thus, (1 − e)axt + e = 1, and so (1 − e)a ∈ R is regular. In view of Corollary 6.1.4, we have an element z ∈ R such that (1 − e)a + ez ∈ U + (R). That is, a + bys(z − a) ∈ U + (R), as required. (2) ⇒ (1) is obvious by Corollary 6.1.4. (1) ⇔ (3) is symmetric.  Corollary 6.2.2. Let R be an exchange ring. Then the following are equivalent: (1) R has stable range one.

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(2) R is directly finite and satisfies related comparability. Proof. (1) ⇒ (2) Obviously, R is directly finite. Suppose R = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 . From Theorem 1.3.2, we see that B1 ∼ = B2 . So we can find e = 1 ∈ B(R) such that B1 e .⊕ B2 e and B2 (1 − e) .⊕ B1 (1 − e). By virtue of Theorem 6.1.3, R satisfies related comparability. (2) ⇒ (1) As R is directly finite, U + (R) = U (R). Therefore we complete the proof by Proposition 6.2.1.  Using Theorem 6.1.3 and Corollary 6.2.2, we easily provide many classes of exchange rings satisfying related comparability, such as exchange rings with all idempotents central, exchange rings whose primitive factor rings are artinian, exchange rings of bounded index, exchange rings which satisfy the comparability axiom or general comparability, the regular self-injective rings, the continuous regular rings and ℵ0 -continuous regular rings. But, there indeed exist exchange rings which do not satisfy related comparability as the following shows. Example 6.2.3. Let F be a field and T = F [[t]] be the ring of formal power series over F in an indeterminate t, and let K denote the quotient field of T . Let S be the subring consisting of those x ∈ EndF (T ) such that (x − a)(tn T ) = 0 for some a ∈ K and some positive integer n. Since K is commutative, the map ϕ : S → K also defines a ring homomorphism ϕ : S op → K. Let R = {(x, y) ∈ S × S op | ϕx = ϕy} be a subring of S × S op . Then R is direct finite. From [217, Example 5.10], R is not unit-regular. By Corollary 6.2.2, R does not satisfy related comparability.  As usual, a group inverse of a ∈ R will be denoted by a# , which satisfies aa a = a, a# aa# = a# and aa# = a# a (cf. [254]). Let R be a ring. A group G in R is a set of R which is a group under multiplication of R. #

Lemma 6.2.4. Let R be a ring, and let a ∈ R. Then the following are equivalent: (1) a ∈ R has a group inverse. (2) There exists a group G in R such that a ∈ G. Proof. (1) ⇒ (2) Since a ∈ R has a group inverse a# , we see that a = aa# a, a# = a# aa# and aa# = a# a. Let G = {a, a# , aa# }. Then G is a group in R. (2) ⇒ (1) Assume that a ∈ G, where G is a group in R. If e ∈ G is

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the identity of G, then there exists some a# ∈ G such that aa# = e = a# a. Furthermore, we see that aa# a = ea = a and a# aa# = ea# = a# . Therefore a ∈ R has a group inverse a# ∈ R.  The following theorem generalizes [254, Theorem 2A] to related comparability. Theorem 6.2.5. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies related comparability. (2) For every regular x ∈ R, there exists a w ∈ U + (R) such that wx is an idempotent of R. (3) For every regular x ∈ R, there exists a w ∈ U + (R) and a group G in R such that wx ∈ G. Proof. (1)⇒(3) It is trivial from Theorem 6.1.3 and Lemma 6.2.4. (3) ⇒ (2) For any regular x ∈ R, there exists a w ∈ U + (R) and a group G in R such that wx ∈ G. Thus, the element wx ∈ R has a group
inverse (wx)# ∈ R. One easily checks that wx (wx)# + 1 − (wx)# wx wx = wx
−1 and (wx)# + 1 − (wx)#
wx = wx + 1 − (wx)# wx ∈ U (R).
Clearly, (wx)# + 1 − (wx)# wx w ∈ U + (R) and (wx)# + 1 − (wx)# wx wx is an idempotent of R. (2) ⇒ (1) Given any regular x ∈ R, there exists a w ∈ U + (R) such that wx = e is an idempotent of R. Assume x = xyx for some y ∈ R. From xy + (1 − xy) = 1, we have ey + w(1 − xy) = w. Assume f ww
∗ = f and (1 − f )w∗ w = 1 − f for f ∈ B(R). Then f eyw∗ + w(1 − xy)w∗ = f , and so

f e + w(1 − xy)w∗ (1 − e) = f 1 − eyw∗ (1 − e) . Hence,



f 1 + eyw∗ (1 − e) w x + (1 − xy)w∗ (1 − e) = f.


By Lemma 4.1.2, there exists some z1 ∈ R such that f y + z1 (1 −
xy) = u is right invertible in f Rf . In addition,
f x = f x y + z1 (1 − xy) x = xux. On the other hand, e y + w(1 − xy) + (1 − e)w(1 − xy) = w. Clearly, (1 − e)w(1 − xy) = (1 − e)w and (1 − f )(1 − e)w = (1 − f )(1 − e)ww∗(1 − e)w.
Set g = (1 − f )(1 − e)ww∗ (1 − e). Then (1 − f ) e(y + w(1 − xy)) + gw = (1 − f )w, e
= e2 , g = g 2 and eg = ge = 0. Consequently, (1 − f )e y + w(1 − xy) = (1 − f )ew. Therefore, (1 − f )(e + g)w
= (1 − f )w. This implies that x + (1 − xy)w∗ (1 − e) 1 + eww∗ (1 − e) is right invertible in

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(1 − f )R(1 − f ). In
view of Lemma 4.1.2, we can find z2 ∈ R such that (1−f ) y +z2 (1−xy) =
v is left invertible in (1−f )R(1−f ). So (1−f )x = (1 − f )x y + z2 (1 − xy) x = xvx. Thus x = f x + (1 − f )x = x(u + v)x, where u + v ∈ U + (R), as required.  Corollary 6.2.6. Let R be an exchange ring. Then the following are equivalent : (1) R satisfies related comparability. (2) For every regular x ∈ R, there exists a w ∈ U + (R) such that x ∈ wG for some group G in R. Proof. (1) ⇒ (2) is clear from Theorem 6.1.6 and Lemma 6.2.4. (2) ⇒ (1) For any regular x ∈ R, we have some y ∈ R such that x = xyx and y = yxy. So there is an element w ∈ U + (R) such that y ∈ wG for a group G in R. Hence y = wg, where g ∈ R has a group inverse g # . From xy + (1 − xy) = 1, we see that xwg + (1 − xy) = 1. Let u = g + 1 − gg #. Then u−1 = g # + 1 − gg #. Let e = gg # . Then g = ue with u ∈ U (R) and e = e2 ∈ R. Hence, xwue + (1 − xy) = 1, and so e + (1 − e)(1 − xy) = 1 − (1 − e)xwue. Set z = wu(1 − e).
−1 Then wg + z(1 − xy) = wu 1 + (1 − e)xwue ∈ U + (R). Therefore
−1
wu 1 + (1 − e)xwue x = wg + z(1 − xy) x = yx is an idempotent of R. By virtue of Theorem 6.2.5, R satisfies related comparability.  Corollary 6.2.7. Let R be an exchange ring. Then the following are equivalent : (1) R satisfies related comparability. (2) For every regular x ∈ R, there exists a w ∈ U + (R) such that x ∈ xwxR ∩ Rxwx. Proof. (1) ⇒ (2) is trivial by Theorem 6.1.3. T (2) ⇒ (1) For any regular x ∈ R, we have x ∈ xwxR Rxwx for some w ∈ U + (R). Thus, there exist s, t ∈ R such that x = xwxs and x = txwx. It is easy to verify that wx = (wx)2 s ∈ (wx)2 R. Furthermore, we see that wx = wtxwx = wt(txwx)wx = wt2 x(wx)2 ∈ R(wx)2 . Hence wx ∈ (wx)2 R ∩ R(wx)2 . It follows from [254, Theorem 1] that wx has a group inverse. This establishes the result from Theorem 6.2.5.  Theorem 6.2.8. Let R be an exchange ring. Then the following are equivalent:

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(1) R is a local ring or every regular element in R is an idempotent. (2) For every nonzero regular x ∈ R, there exists a unique related unit w ∈ R such that wx is an idempotent of R. Proof. (1) ⇒ (2) If R is a local ring, then each idempotent of R is trivial and each related unit of R is a unit. If every regular element in R is an idempotent, then each related unit is a unit. Thus, we claim that for every nonzero regular x ∈ R, there exists a unique related unit w ∈ R such that wx is an idempotent. (2)⇒(1) For any nonzero regular x ∈ R, there exists a unique related unit w ∈ R such that wx is an idempotent. Assume x = xyx, eww∗ = e, and (1 − e)w∗ w = 1 − e, where y ∈ R and e ∈ B(R). As wxwx = wx, we have (1 − e)xwx = (1 − e)x. Also we have wxy + w(1 − xy) = w. Thus, (ewx)yw∗ +ew(1−xy)w∗ = e, and then ewxyw∗ (1−wx)+ew(1−xy)w∗ (1− ∗ wx) = e − ewx. Therefore, we see that ewx + ew(1 − xy)w
(1 − wx) = ∗ ∗ e − ewxyw (1
− wx). Consequently, 1 + wxyw (1 − wx) ew x + (1 − xy)w∗ (1 − wx) = e. By virtue of Lemma 4.1.2, we can find z ∈ R such that ey + ez(1 −
xy) = ed is right invertible in eRe. Hence, ex = exyx
= x ey+ez(1−xy) x = x(ed)x, and then x = ex+(1−e)x = x ed+(1−e)w x.
Since ed + (1 − e)w ∈ U + (R) and ed + (1 − e)w x ∈ R is an idempotent, by uniqueness, we know that w = ed + (1 − e)w, this yields x = xwx.
−1 Since 1
− x(1 − xw) = 1 + x(1 − xw) ∈ U (R), we see that w 1 −
x(1 − xw) ∈ U + (R). By uniqueness, we get w = w 1 − x(1 − xw) . Hence, wx = wx2 w. On the other hand, one can verify that 1 − (1 −

2 wx)x wx = wx − x + wx2 and wx − x + wx2 = wx − x + wx2 . Clearly,

−1 1−(1−wx)x w ∈ U + (R) because 1−(1−wx)x = 1+(1−wx)x ∈ U (R).
Consequently, w = 1 − (1 − wx)x w. Therefore, xw = wx2 w. As wx = wx2 w and (1 − e)w∗ w = 1 − e, it follows that x(1 − e) = x2 (1 − e)w. Likewise, xe = ewx2 = x2 ew since wx = xw = wx2 w. Consequently, x = xe + x(1 − e) = x2 w is strongly π-regular. Since R is an exchange ring, by [8, Theorem 3], x is unit-regular. According to Lemma 1.3.1, R has stable range one, hence it is directly finite. So w ∈ U (R). Assume x 6= x2 . Set u = wxw. It is easy to verify that x(u + 1 − xu)x = x with u + 1 − xu = (x + 1 − xu)−1 ∈ U (R). By uniqueness, we have w = u + 1 − xu = wxw + 1 − xw, hence w(1 − xw) = 1 − xw is an idempotent. By uniqueness again, we know that w = 1 if 1 − xw 6= 0. Clearly, w = 1 implies x = x2 , a contradiction. So xw = 1, and then x ∈ U (R). Therefore, every nonzero regular element of R is an idempotent or a unit.

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Assume a ∈ R is a regular element but not an idempotent, and x, y ∈ R are nonzero regular elements such that xy = 0. By the above consideration, we see that a ∈ U (R), ax is regular, and x, y are idempotents. If ax = (ax)2 , then x = xax, which implies a = 1, a contradiction. If ax ∈ U (R), then x ∈ U (R). This implies y = 0, a contradiction. Therefore, we conclude that either all regular elements of R are idempotents, or for any regular x, y ∈ R, xy = 0 implies x = 0 or y = 0. Suppose there exist regular elements which are not idempotents. Given any a ∈ R, there is f = f 2 ∈ R such that f = am and 1 − f = (1 − a)n for some m, n ∈ R. As f (1 − f ) = 0, we see that f = 0 or f = 1. Thus, a or 1 − a is right invertible in R. Furthermore, a or 1 − a is invertible. In this case, R is a local ring.  We note that there exist exchange rings that satisfy the conditions in Theorem 6.2.8, for example, all perfect rings and all Boolean rings. Proposition 6.2.9. Let A be a right R-module having the finite exchange L property, let E = EndR (A), and let A = Ai . Suppose that each Ai is a i∈I

fully invariant submodule, equal to a direct sum of isomorphic indecomposable submodules. Then E satisfies related comparability. ∼ Q EndR (Ai ). According to Lemma 5.2.6, each Proof. In view of [43], E = i∈I

EndR (Ai ) is weakly stable. As in the proof of Example 6.1.12, E satisfies related comparability.  Example 6.2.10. Let G be an abelian group such that End(G) is regular. If G is a reduced torsion group, then End(G) satisfies related comparability.

Proof. A reduced abelian torsion group has a regular endomorphism ring if and only if it is a direct sum of cyclic groups of prime order (cf. [209]). So LL we may assume that G ∼ G(p,i) , where each G(p,i) is a cyclic group = p i L of prime order p. Clearly, each G(p,i) is fully invariant. In addition, i

G(p,i) ∼ = G(p,j) for any i, j. Thus, the result follows by Proposition 6.2.9. Take G = H ⊕ K, where H is the direct sum of infinitely many isomorphic copies of Zp , and K is the direct sum of infinitely many isomorphic copies of Zq (p, q distinct primes). In view of Example 6.2.10, End(G) is a regular ring satisfying related comparability. But End(G) is not one-sided unit-regular.

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The Morita Invariance

The main purpose of this section is to prove that the related comparability over exchange rings is Morita invariant. Let f : M → N be an R-morphism, and let e ∈ B(R). Define an R-morphism fe : M e → N e given by fe (me) = f (m)e for any m ∈ M . Now, we observe a kind of weak substitution of modules over such rings. Lemma 6.3.1. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies related comparability. (2) Given M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = R ∼ = A2 , there exist C, D, E ⊆ M such that M = C ⊕ D ⊕ B1 = C ⊕ E ⊕ B2 , De = 0 and E(1 − e) = 0 for some e ∈ B(R).

Proof. (1) ⇒ (2) Using the decomposition M = A1 ⊕ B1 ∼ = R ⊕ B1 , we obtain projections p1 : M → R, p2 : M → B1 and injections q1 : R → M, q2 : B1 → M such that p1 q1 = 1R , q1 p1 + q2 p2 = 1M and ker(p1 ) = B1 . Using the decomposition M = A2 ⊕ B2 ∼ = R ⊕ B2 , we obtain a projection f : M → R and an injection g : R → M such that f g = 1R and ker(f ) = B2 . Now 1R = f g = f (q1 p1 + q2 p2 )g = (f q1 )(p1 g) + (f q2 )(p2 g). Hence, we can + find some y ∈ R such that f q1 (1) + f q2 p2 g(1)y
∈ U (R). So, there exists an e ∈ B(R) such that
f q1 (1) + f q2 p2 g(1)y e is right invertible in eRe and f q1 (1) + f q2 p2 g(1)y (1 − e) is left invertible in (1 − e)R(1 − e). This implies that fe (q1 )e + (q2 )e (p2 )e ge ye )d = 1Re for some d ∈ End(Re), where ye : Re
→ Re given by ye (re) = yre for any r ∈ R. Let k = (q1 )e + (q2 )e (p2 )e ge ye d. Then fe k = 1Re . Thus M e = ker(f
e ) ⊕ k(Re) = C ⊕ B2 e with C = k(Re). Let ν = fe (q1 )e + (q2 )e (p2 )e ge ye . Then ν(p1 )e k = 1Re . So M e = ker(ν)(p1 )e ⊕ k(Re). Let α : B1 e ֒→ kerν(p1 )e be the inclusion. Then (p2 )e |kerν(p1 )e α = 1B1 e . That is, α splits. Thus, we can find a right Rmodule D such that kerν(p1 )e = D ⊕ B1 e. So M e = C ⊕ D ⊕ B1 e, whence, M e = C ⊕ D ⊕ B1 e = C ⊕ B2 e. Similarly, we obtain the corresponding result for 1 − e. That is, we have some right R-modules E, F ⊆ M such that M (1 − e) = E ⊕ B1 (1 − e) = E ⊕ F ⊕ B2 (1 − e). Therefore

M = M e ⊕ M (1 − e) = C ⊕ E ⊕ D ⊕ B1 = C ⊕ E ⊕ F ⊕ B2 . Obviously, we see that D(1 − e) = 0 and F e = 0, as required. (2) ⇒ (1) Suppose that ax + b = 1R with a, x, b ∈ R. Set M = 2R, and let pi : M → R, qi : R → M (for i = 1, 2) denote the projections

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and injections of this direct sum. Set A1 = q1 (R) and B1 = q2 (R), so that M = A1 ⊕ B1 with A1 ∼ = R. Define f = ap1 + bp2 from M to R and g = q1 x + q2 from R to M . Observing that f g = 1R , we get M = ker(f ) ⊕ g(R). Set A2 = g(R) and B2 = ker(f ), so that M = A2 ⊕ B2 and A2 ∼ = R. By assumption, there exist C, D, E ⊆ M such that M = C ⊕ D ⊕ B1 = C ⊕ E ⊕ B2 , where De = 0 and E(1 − e) = 0 for some e ∈ B(R). As De = 0, we get M e = Ce ⊕ B1 e = Ce ⊕ Ee ⊕ B2 e. As in the proof of Theorem 5.1.2, there exists an element y ∈ R such that e(a + by) ∈ eRe is left invertible. Analogously, there exists a z ∈ R such that (1 − e)(a + bz) ∈ (1−e)R(1−e) is right invertible. Let c = ey+(1−e)z. Then a+bc ∈ U + (R), and therefore R satisfies related comparability.  Theorem 6.3.2. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies related comparability. (2) For any finitely generated projective right R-module A and any right R-modules B, C, A ⊕ B ∼ = A ⊕ C implies that there exists an e ∈ B(R) ⊕ such that Be . Ce and C(1 − e) .⊕ B(1 − e). (3) For all finitely generated projective right R-modules A, B and C, A ⊕ B∼ = A ⊕ C implies that there exists an e ∈ B(R) such that Be .⊕ Ce and C(1 − e) .⊕ B(1 − e). Proof. (1) ⇒ (2) For any finitely generated projective right R-module A and any right R-modules B, C, A⊕B ∼ = A⊕C implies that nR⊕B ∼ = nR⊕C for some n ∈ N. Thus, we have decompositions

nR ⊕ C = A11 ⊕ A12 ⊕ · · · ⊕ A1n ⊕ B ′ = A21 ⊕ A22 ⊕ · · · ⊕ A2n ⊕ C ′ with each A1i ∼ = C. In view of Lemma 6.3.1, = B, C ′ ∼ =R∼ = A2i and B ′ ∼
there are D1 , E1 , F1 ⊆ M such that M = D1 ⊕E1 ⊕ A12 ⊕· · ·⊕A1n ⊕B ′ =
D1 ⊕ F1 ⊕ A22 ⊕ · · · ⊕ A2n ⊕ C ′ with E
1 e1 = 0 and F1 (1 − e1 ) = 0 for
an e1 ∈ B(R).
Thus we get M = E1 ⊕ A12 ⊕ A13 ⊕ · · · ⊕ A1n ⊕ B ′ ⊕ D1 =
F1 ⊕ A22 ⊕ A23 ⊕ · · · ⊕ A2n ⊕ C ′ ⊕ D1 . Clearly, D1 ⊕ E1 ∼ = A11 ∼ =R ∼ ∼ ∼ ∼ and D1 ⊕ F1 = A21 = R. Hence, D1 e = Re and D1 (1 − e) = R(1 − e), and so D1 ∼ = D1 e ⊕ D1 (1 − e) ∼ = R. This implies that E1 ⊕
R ∼ =R∼ = F1 ⊕ R. ′ ′ Therefore, M = A12
⊕ A13 ⊕ · · · ⊕ A1n ⊕ B ⊕ D1 = A′22 ⊕ A23 ⊕ · · · ⊕ A2n ⊕ C ′ ⊕ D1 , where A′12 ∼ =R∼ = A′22 . By Lemma 6.3.1 again, we can
find D2 ⊆ M such that M = A′13 ⊕ A14
⊕ · · · ⊕ A1n ⊕ B ′ ⊕ D1 ⊕ D2 = A′23 ⊕ A24 ⊕ · · · ⊕ A2n ⊕ C ′ ⊕ D1 ⊕ D2 with A′13 ∼ =R∼ = A′23 . By iteration of this process, there exist D3 , · · · , Dn−1 ⊆ M such that M =

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A′1n ⊕ D1 ⊕ D2 ⊕ · · · ⊕ Dn−1 ⊕ B ′ = A′2n ⊕ D1 ⊕ D2 ⊕ · · · ⊕ Dn−1 ⊕ C ′ with A′1n ∼ we have Dn , E, F ⊆ = R ∼ = A′2n . Furthermore,

M such′ that ′ M = D1 ⊕ D2 ⊕ · · · ⊕ Dn ⊕ E ⊕ B = D1 ⊕ D2 ⊕ · · · ⊕ Dn ⊕ F ⊕ C with E(1 − e) = 0 and F e = 0 for an e ∈ B(R). Consequently, we deduce that B ′ e .⊕ C ′ e. That is, Be .⊕ Ce. Analogously, we have that C(1 − e) .⊕ B(1 − e), as desired. (2) ⇒ (3) is trivial. (3) ⇒ (1) Given R = A1 ⊕B = A2 ⊕C with A1 ∼ = A2 , then R⊕B ∼ = R⊕C with B, C ∈ F P (R). By hypothesis, there exists an e ∈ B(R) such that Be .⊕ Ce and C(1 − e) .⊕ B(1 − e). According to Theorem 6.1.3, R satisfies related comparability.  Lemma 6.3.3. Let R be a ring. Suppose for any right R-module decompositions M = A1 ⊕B1 = A2 ⊕B2 with A1 ∼ = nR ∼ = A2 , there exist C, D, E ⊆ M such that M = C ⊕ D ⊕ B1 = C ⊕ E ⊕ B2 with De = 0 and E(1 − e) = 0 for some e ∈ B(R). Then Mn (R) satisfies related comparability. Proof. Take M = nR ⊕ nR. With respect to this decomposition, let pi : M → nR be the projections and qi : nR → M be the injections for i = 1, 2. Set A1 = q1 (nR) and B1 = ker(p1 ). Then M = A1 ⊕ B1 with A1 ∼ = nR. Suppose that ax + b = 1nR in EndR (nR). Set f = ap1 + bp2 and g = q1 x + q2 . We easily check that f g = ax + b = 1nR . Thus, M = g(nR) ⊕ ker(f ). Let A2 = g(nR) and B2 = ker(f ). Then M = A2 ⊕ B2 with A2 ∼ = nR. By hypothesis, we can find C, D, E ⊆ M such that M = C ⊕ D ⊕ B1 = C ⊕ E ⊕ B2 with De = 0 and E(1 − e) = 0 for some e ∈ B(R). Since De = 0, we have M e = Ce ⊕ B1 e = Ce ⊕ Ee ⊕ B2 e. Let h : nRe ∼ = A1 e ∼ = Ce ֒→ M e be the injection. Then Ce = h(nRe), and so M e = h(nRe) ⊕ (ker(p1 ))e = h(nRe) ⊕ ker(p1 )e . Therefore, (p1 )e h is an isomorphism. As in the proof of Theorem 5.1.2, we see that fe h ∈
EndR (nRe) is left invertible. Observing that ae (p1 )e + be (p2 )e h = ae +


−1 be (p2 )e h((p1 )e h)−1 (p1 )e h , ae + be (p2 )e h (p1 )e h is left invertible. Let
−1
k = (p2 )e h (p1 )e h . Then (a+bk)e∗ is left invertible in e∗ EndR (nR) e∗ , ∗ where e∗ : nR → nR
is given by e (x1 , · · · , xn ) = (ex1 , · · · , exn ). Clearly,
∗ e ∈ B EndR (nR) because
e ∈ B(R).
Likewise,
(a + bl) 1nR − e∗ is right invertible in 1nR
− e∗ EndR (nR) 1nR − e∗ . Thus, we show that a + b ke∗ + l(1nR − e∗ ) ∈ EndR (nR) is a related unit. By Corollary 6.1.4, Mn (R) ∼  = EndR (nR) satisfies related comparability. Theorem 6.3.4. Let R be an exchange ring. If R satisfies related compa-

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rability, then so does Mn (R) for any n ∈ N.

Proof. Suppose M = A1 ⊕B1 = A2 ⊕B2 with A1 ∼ = nR ∼ = A2 . Then we have decompositions A1 = A11 ⊕ · · · ⊕ A1n and A2 = A21 ⊕ · · · ⊕ A2n with all Aij isomorphic to R. By virtue of Lemma 6.3.1, we can
find C1 , D1 , E1 ⊆ M such that M =
C1 ⊕ D1 ⊕ A12 ⊕ · · · ⊕ A1n ⊕ B1 = C1 ⊕ E1 ⊕ A22 ⊕ · · · ⊕ A2n ⊕ B2 with D1 e1 = 0 and E1 (1 − e1 ) = 0 for some e1 ∈ B(R). As in the proof of Theorem 6.3.2, we get D1 ⊕ A12 ∼ = R ∼ = E1 ⊕ A22 . By iteration of this
process, we have C2 , · · · , Cn
, D, E ⊆ M such that M = C1 ⊕ · · · ⊕ Cn ⊕ D ⊕ B1 = C1 ⊕ · · · ⊕ Cn ⊕ E ⊕ B2 with De = 0 and E(1 − e) = 0 for some e ∈ B(R). In view of Lemma 6.3.3, we obtain the result.  Corollary 6.3.5. Let A be a finitely generated projective right module over an exchange ring R. If R satisfies related comparability, then so does EndR (A). Proof. Since A is a finitely generated projective right R-module, EndR (A) ∼ = eMn (R)e for a positive integer n and an idempotent matrix e ∈ Mn (R). In view of Theorem 6.3.4, Mn (R) satisfies related comparability. Therefore we complete the proof by Corollary 6.1.5.  As an immediate consequence, we show that the related comparability over exchange rings is Morita invariant. Corollary 6.3.6. Let R be an exchange ring and n a positive integer. Then the following are equivalent: (1) R satisfies related comparability. (2) For every regular A ∈ Mn (R),
there exists an idempotent matrix E ∈ Mn (R) and a W ∈ U + Mn (R) such that A = EW .
(3) For every regular A ∈ Mn (R), there exists a W ∈ U + Mn (R) such that W A is an idempotent matrix. Proof. (1) ⇔ (2) By virtue of Corollary 6.3.5, R satisfies related comparability if and only if Mn (R) also satisfies related comparability. Thus, the result follows from Theorem 6.1.6. (1) ⇔ (3) is similar by Theorem 6.2.5.  A monoid (M, +, 0) has refinement if, for all a, b, c and d in M , the equation a + b = c + d implies the existence of a1 , b1 , c1 , d1 ∈ M such that a = a1 + d1 , b = b1 + c1 , c = a1 + b1 and d = d1 + c1 . These equations are

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represented in the form of a refinement matrix c d



a a1 d1

b  b1 . c1

We use V(R) to denote the monoid of isomorphism classes of objects from F P (R). The following lemma was proved by Ara et al. [16, Lemma 2.7]. Lemma 6.3.7. Let A, B and C be finitely generated projective right modules over an exchange ring R. If A ⊕ C ∼ = B ⊕ C with C .⊕ A, B, then we have a refinement matrix A C in V(R) such that C1 .⊕ A1 , B1 .



BC  D1 A1 B1 C1

Proof. Suppose that ψ : A ⊕ C ∼ = B ⊕ C with C .⊕ A, B. Then we −1 −1 have M := C ⊕ A = ψ (C) ⊕ ψ (B). Clearly, C has the finite exchange property, and so we have A1 ⊆ ψ −1 (C) and D1 ⊆ ψ −1 (B) such that M = −1 ∼ C⊕ ⊆ D1 ⊕ C ⊕
D1 ⊕ A1 .−1Hence, A−1= D1T⊕ A1 . Since D1
⊆ ψ (B) −1 T A1
. Thus, ψ (B) = ψ (B) D1 ⊕ (C ⊕ A1 ) = D1 ⊕ ψ (B) C⊕ A1 . Hence, D1 is a direct summand of ψ −1 (B). Likewise, A1 is a direct summand of ψ −1 (C). Assume now that ψ −1 (B) = D1 ⊕ B1 and ψ −1 (C) = A1 ⊕ C1 . Thus we have B ∼ = D 1 ⊕ B1 , C ∼ = A1 ⊕ C1 . As D1 ⊕ B1 ⊕ A1 ⊕ C1 = ∼ C ⊕D1 ⊕A1 , we deduce that C = B1 ⊕C1 . Therefore we obtain a refinement matrix BC   A D1 A1 . C B1 C1

Since C .⊕ B, we have C1 .⊕ D1 ⊕ B1 . Similarly to the above consideration, we have C1 ∼ = C1′ ⊕ C1′′ with C1′ .⊕ B1 and C1′′ .⊕ D1 . Assume that ′′ ′ D1 ∼ = C1 ⊕ D1 for some right R-module D1′ . Therefore we get a refinement matrix BC   A D1′ A′1 , C B1′ C1′ where A′1 = A1 ⊕ C1′′ and B1′ = B1 ⊕ C1′′ . Clearly, C1′ .⊕ B1 .⊕ B1′ . Further, C1 .⊕ A1 implies that C1′ .⊕ C1 .⊕ A1 .⊕ A′1 . Since C .⊕ A, we need only repeat the procedure, and then get C1′ .⊕ A′1 . 

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In what follows, we observe that the separativity for exchange rings can be characterized by some kind of related comparability of modules. Theorem 6.3.8. Let R be an exchange ring. Then the following are equivalent: (1) R is separative. (2) For any A, B, C ∈ F P (R), A ⊕ C ∼ = B ⊕ C with C .⊕ A, B ⇒ Ae .⊕ ⊕ Be and B(1 − e) . A(1 − e) for some e ∈ B(R). (3) For any A, B ∈ F P (R), 2A ∼ = A⊕B ∼ = 2B ⇒ Ae .⊕ Be and B(1 − ⊕ e) . A(1 − e) for some e ∈ B(R). (4) For any A, B ∈ F P (R), 2A ∼ = 2B and 3A ∼ = 3B ⇒ Ae .⊕ Be and ⊕ B(1 − e) . A(1 − e) for some e ∈ B(R). Proof. (1) ⇒ (4) is trivial from [16, Lemma 2.1]. (4) ⇒ (3) Given any A, B ∈ F P (R) with 2A ∼ = A⊕B ∼ = 2B, we have ∼ ∼ 2A = 2B and 3A = 3B. By hypothesis, there exists an e ∈ B(R) such that Ae .⊕ Be and B(1 − e) .⊕ A(1 − e). (3) ⇒ (2) Suppose that A ⊕ C ∼ = B ⊕ C with C .⊕ A, B for A, B, C ∈ ∼ ∼ F P (R). Then A = C ⊕
D and B = C ⊕ E
for some right R-modules
D and E. Thus, 2A
∼ = A⊕C ⊕D ∼ = B⊕C ⊕D ∼ = A⊕B ∼ = A⊕C ⊕E ∼ = B⊕ C⊕E ∼ = 2B, as desired. (2) ⇒ (1) Suppose that A ⊕ C ∼ = B ⊕ C with C .⊕ A, B for A, B, C ∈ F P (R). Applying Lemma 6.3.7, we have a refinement matrix A C



BC  D1 A1 B1 C1

∼ C = ∼ B1 ⊕ C1 , we can find such that C1 .⊕ A1 , B1 . Since A1 ⊕ C1 = some e ∈ B(R) such that A1 e .⊕ B1 e and B1 (1 − e) .⊕ A1 (1 − e). As A1 e .⊕ B1 e, we have B1 e ∼ = A1 e ⊕ D for a right R-module D. We easily check that Ce ∼ = C1 e ⊕ B1 e ∼ = C1 e ⊕ A1 e ⊕ D ∼ = Ce ⊕ D. It follows that ⊕ Ae ∼ = Ae ⊕ D because C . A. Therefore Ae ∼ = Ae ⊕ D ∼ = D1 e ⊕ A1 e ⊕ D ∼ = D 1 e ⊕ B1 e ∼ Be. = On the other hand, B1 (1 − e) .⊕ A1 (1 − e), then A1 (1 − e) ∼ = B1 (1 − e) ⊕ E for a right R-module E. So C(1 − e) ∼ = C1 (1 − e) ⊕ A1 (1 − e) ∼ = C1 (1 − e) ⊕ B1 (1 − e) ⊕ E ∼ = C(1 − e) ⊕ E. It follows from C .⊕ B that B(1 − e) ∼ = B(1 − e) ⊕ E. Consequently, B(1 − e) ∼ = B(1 − e) ⊕ E ∼ = ∼ ∼ D1 (1 − e) ⊕ B1 (1 − e) ⊕ E = D1 (1 − e) ⊕ A1 (1 − e) = A(1 − e). Hence A∼ = Ae ⊕ A(1 − e) ∼ = Be ⊕ B(1 − e) ∼ = B. Therefore R is separative by [16, Corollary 2.9]. 

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Corollary 6.3.9. Every exchange ring satisfying related comparability is separative. Proof. Let R be an exchange ring satisfying related comparability. Suppose that 2A ∼ = A⊕B ∼ = 2B, where A, B ∈ F P (R). By virtue of Theorem 6.3.2, we have Ae .⊕ Be and B(1 − e) .⊕ A(1 − e) for some e ∈ B(R). Therefore the result follows from Theorem 6.3.8.  A monoid M is separative if, for all a, b ∈ M , 2a = a + b = 2b implies a = b. Separativity is a weak form of cancellativity for monoids. In general, we can extend Theorem 6.3.8 to refinement monoids as follows. Proposition 6.3.10. Let M be a refinement monoid. Then the following are equivalent: (1) M is separative. (2) For any a, b ∈ M , a = 2b and 3a = 3b implies that a ≤ b or b ≤ a. (3) For any a, b, c ∈ M , c + a = c + b with c ≤ a, b implies that a ≤ b or b ≤ a. Proof. (1) ⇒ (2) is clear. (2) ⇒ (3) Given c + a = c + b with c ≤ a, b in M , there exist e, f ∈ M such that a = c+e and b = c+f ; hence, 2a = a+(c+e) = (b+c)+e = b+a. Likewise, we have 2b = a + b. This implies that 2a = 2b. Furthermore, we get 3a = a + 2b = (a + b) + b = 3b. So either a ≤ b or b ≤ a. (3) ⇒ (1) Suppose that c + a = c + b with c ≤ a, b. It will suffice to show that a = b. As in the proof of Lemma 6.3.7, we have a refinement matrix b c   a d1 a1 over M : , where c1 ≤ a1 , b1 . From a1 + c1 = b1 + c1 with c b 1 c1 c1 ≤ a1 , b1 , we deduce that a1 ≤ b1 or b1 ≤ a1 . If a1 ≤ b1 , then b1 = a1 + e. Thus c = c1 + b1 = c1 + a1 + e = c + e. As c ≤ a, b, we have a = a + e and b = b + e; hence, a = a + e = a1 + d1 + e = b1 + d1 = b. The proof of the case b1 ≤ a1 is just symmetric to the case a1 ≤ b1 . So the result follows.  Corollary 6.3.11. Let M be a refinement monoid. Then the following are equivalent: (1) M is separative. (2) For any a, b ∈ M, n ∈ N, na = nb and (n + 1)a = (n + 1)b implies that a ≤ b or b ≤ a. (3) For any a, b ∈ M , 2a = a + b = 2b implies that a ≤ b or b ≤ a.

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Proof. (1) ⇒ (2) is obvious. (2) ⇒ (3) Suppose that 2a = a + b = 2b in M . Then 2a = 2b and 3a = a+(a+b) = 2a+b = 3b. Furthermore, we get 4a = 3b+a = 2b+(a+b) = 4b. Similarly, we deduce that na = nb and (n + 1)a = (n + 1)b (n ≥ 2). So a ≤ b or b ≤ a. (3) ⇒ (1) Given c + a = c + b with c ≤ a, b in M , we have e, f ∈ M such that a = c + e and b = c + f . It is easy to check that a + a = a + (c + e) = (b + c) + e = b + a. Similarly, b + b = a + b. Hence a ≤ b or b ≤ a. By virtue of Proposition 6.3.10, the result follows.  As a consequence, we observe that an exchange ring R is separative if and only if for all A, B ∈ F P (R), 2A ∼ = A⊕B ∼ = 2B implies that either ⊕ ⊕ A . B or B . A. We say that M is an ordered-separative monoid provided that (∀ a, b ∈ M )(a + b = 2b =⇒ a ≤ b). In [52, Proposition 9.5], Brookfield proved that every refinement order-separative monoid is separative. But the converse is not true. Let {0, ∞} be the monoid such that ∞ + ∞ = ∞, and let R++ be the subgroup of strictly positive real numbers. Let M be the monoid obtained from {0, ∞} × R++ by adding a zero element. Since {0, ∞} and R++ are separative refinement subgroups, we prove that M is a separative refinement monoid. Choose a = (0, 1) and b = (∞, 1). Then a + b = 2b, while a  b and b  a. Also we note that the refinement condition is necessary. Let M be the monoid generated by three elements a, b and c such that 2a = 0, a + b = c, a + c = b and b + c = 2b. Then M = {0, a, b, c, 2b, 3b, 4b, · · · } defined by the following addition: + 0 a b c

0 a b c 0a b c a0 c b b c 2b 2b c b 2b 2b

and a + mb = mb, c + mb = (m + 1)b(m ≥ 2). As 2b = b + c = 2c, M is not a separative monoid. But one checks M is an ordered-separative monoid. In [396, Theorem 1.4], Wehrung proved that if M is ”separative” then M is order-separative. The ”separativity” used in [396] differs from that in this book. Wehrung’s ”separativity” satisfies an additional condition: (∀ a, b, c ∈ M )(a + c ≤ b + c with c ∝ b ⇒ a ≤ b). Let M be a refinement monoid. We write M ∗ to denote the set of all nonzero elements of M . We say that M is conical if, for all x, y ∈ M, x+y =

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0 if and only if x = y = 0. The notation x ≤ y(x < y) means that there exists some z ∈ M (M ∗ ) such that x + z = y. A submonoid N of a conical monoid M is called an ideal provided that x ≤ y, y ∈ N implies that x ∈ N . Let a ∈ M , and let M (a) = {x ∈ M | x ≤ na for some n ∈ N}. Then M (a) is an ideal of M . M is called simple provided that M has only two ideals 0 and M . Let N be an ideal of a conical refinement monoid M . Set [a] = {b ∈ M | ∃x, y ∈ N such that a + x = b + y}. Let M/N = {[a] | a ∈ M }. Define [a] + [b] = [a + b] for any a, b ∈ M . Then M/N is a conical refinement monoid. Let s ∈ N. We say that M satisfies s-comparability if, for all x, y ∈ R, either x ≤ sy or y ≤ sx. Lemma 6.3.12. Let M be a conical refinement monoid, and let a, b, c ∈ M . k P If a + b = kc (k ∈ N), then there is a decomposition c = ci such that i=0

a=

k X

ici , b =

i=0

k X i=0

(k − i)ci .

Proof. The proof is true for k = 1. Assume that the result holds for k. Given a + b = (k + 1)c, we get a refinement matrix a b



c kc  u v . wx

By hypothesis, there is a decomposition c =

k P

di such that

i=0

v=

k X

idi , x =

i=0

k X i=0

(k − i)di .

Further, we have decompositions di = ei +fi (0 ≤ i ≤ k),

k P

i=0

ei = w,

k P

fi =

i=0

u. Let c0 = e0 , ck+1 = fk and ci = ei + fi−1 (1 ≤ i ≤ k). Then one checks that c=

k P

(ei + fi )

i=0

= e0 + =

k+1 P i=0

k P

(ei + fi−1 ) + fk

i=1

ci ,

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a= = = =

k P

fi +

i=0 k P

fi +

i=1 k+1 P

ici ,

i=0 k P

k P

i=0 k P

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175

idi i(ei + fi )

i=0

i(ei + fi−1 ) + (k + 1)fk

i=0

b = (k + 1)c − a k+1 k+1 P P = (k + 1) ci − ici =

k+1 P i=0

i=0

i=0

(k + 1 − i)ci .



By induction, the result follows.

An ideal N of a monoid M is proper provided that N 6= M . We say that a proper ideal N of M is maximal provided that N $ P ⊆ M implies that P = M . Lemma 6.3.13. Let M be a conical refinement monoid satisfying scomparability. Then L(M ), the set of all ideal of M , is totally ordered by inclusion. If, moreover, a 6= 0, then M (a) has a unique maximal ideal maxM (a). Proof. Let J, K be ideals of M . If J * K, then there exists some x ∈ J\K. For any y ∈ K, we see that sy ∈ K, and so x  sy. Hence, y ≤ sx. This implies that y ∈ J. Thus, K ⊆ J. Therefore L(M ) is totally ordered by inclusion. Let Ω = {J|J is a proper ideal of M (a)}. Then Ω 6= ∅. Suppose that ∞ S J1 ⊆ J2 ⊆ · · · in Ω. If Ji = M (a), then a ∈ Ji for some i ∈ N. Hence, i=1

M (a) = Ji , a contradiction. Thus,

∞ S

Ji is a proper ideal of M (a). By

i=1

Zorn’s Lemma, there exists an ideal Q of M (a) which is maximal in Ω. As L(M ) is totally ordered by inclusion, we see that Q is a unique maximal of M (a), as asserted.  Lemma 6.3.14. Let M be a conical refinement monoid satisfying scomparability, and let a, b ∈ M . Then the following hold:

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(1) If (s + 1)a ≤ 2b, then a ≤ b. (2) If M (a) ⊂ M (b), then na < b for any n ∈ N. (3) If N is an ideal of M , then [a] < [b] in M/N implies that a < b in M . Proof. (1) Given (s + 1)a ≤ 2b, then (s + 1)a = c + d, c, d ≤ b. By virtue of Lemma 6.3.12, there are decompositions a=

s+1 X

ai , c =

i=0

s+1 X

iai , d =

i=0

s+1 X (s + 1 − i)ai . i=0

By hypothesis, we get a0 ≤ sas+1 or as+1 ≤ sa0 . If a0 ≤ sas+1 , then s+1 s s P P P a= ai ≤ ai + (s + 1)as+1 ≤ iai + (s + 1)as+1 = c ≤ b. Thus, i=0

i=1

i=1

a ≤ b. If as+1 ≤ sa0 , then a = s P

i=1

s+1 P i=0

ai ≤ (s + 1)a0 +

(s + 1 − i)ai = d ≤ b. Thus, a ≤ b, as required.

s P

i=1

ai ≤ (s + 1)a0 +

(2) For any n ∈ N, M (na) = M (a) ⊂ M (b), and so b 6∈ M (na). Hence, b  s(s + 1)s na. By hypothesis, we get (s + 1)s na ≤ sb ≤ 2s b. By (1), we get na ≤ b. Therefore na < b, as required. (3) Given [a] < [b] in M/N , then there exist c ∈ M \N, d, e ∈ N such that a + c + d = b + e. Thus, we have a refinement matrix b e



acd  uvw . xy z

As d, e ∈ N and c 6∈ N , we see that x, y, z, w ∈ N and v 6∈ N ; hence, M (x) ⊆ N and M (v) * N . Thus, M (x) ⊆ N ⊂ M (v) from Lemma 6.3.13. By (2), we get x < v. Therefore a = u + x < u + v ≤ u + v + w = b, as asserted.  We say that a nonzero u ∈ M is an order-unit of M provided that for any x ∈ M , there exists some n ∈ N such that x ≤ nu. A conical monoid M is atomless provided that for any x ∈ M ∗ , there exists some y ∈ M ∗ such that y < x. Lemma 6.3.15. Let M be a conical simple, atomless, refinement monoid satisfying s-comparability, and let u ∈ M be an order-unit. Then the following hold: (1) For any a, b, c ∈ M ∗ , c + a < c + b =⇒ a < b.

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(2) If M is directly finite, then for any a, b, c ∈ M , 2c + a = c + b =⇒ ∃ e ∈ M such that 2c = c + e, b = a + e. (3) If M is directly finite, then for any a, b, c ∈ M ∗ , c + a = c + b =⇒ a = b. Proof. Consider any x ∈ M ∗ such that x ≤ u. As M is simple, we get M (x) = M (u), and so u ≤ 2k x for some k ∈ N. Let n = (s + 1)k . For any y ∈ M , ny ≤ u implies that (s + 1)k y ≤ 2k x. By virtue of Lemma 6.3.14, we have that y ≤ x. (1) Assume that c+a < c+b with a, b, c ∈ M ∗ . There exists m ∈ N such that c ≤ mu. Thus, mu+a < mu+b. Since M is conical, it suffices to prove the result for c = u. Assume that a + u < b + u. Write (a + u) + d = b + u for some d ∈ M ∗ . Thus, we get a refinement matrix b u

a+u d   zv . st

If v 6= 0, then there exist e, v ′ ∈ M ∗ such that v = e + v ′ . Set b′ = z + v ′ . Then b = b′ + e. Further, a + u = z + s < z + v ′ + s + t = b′ + u with 0 < b′ < b. If v = 0, it follows by [355, Lemma 5.1] that t = 2e+f, b = b′ +e for some e, f, b′ ∈ M ∗ . Therefore 0 < b′ < b, and a + u < a + u + e + f = b + s + e + f = s + 2e + f + b′ = b′ + u. As e ∈ M ∗ , then h < e for some h ∈ M ∗ . Clearly, h ≤ nu for some n ∈ N. n P Write h = hi , hi ≤ u. There exists some hi ∈ M ∗ such that hi ≤ h, u. i=1

By the preceding discussion, there is some k ∈ N such that, for q ∈ M , kq ≤ u implies that q ≤ hi < e. Write (a + r) + u = b′ + u for some r ∈ M . By [395, Lemma 1.11], there are decompositions a + r = a′ + t1 , b′ = a′ + t2 such that kt1 , kt2 ≤ u. Clearly, there exist some x1 ≤ a′ , x2 ≤ t1 such that a = x1 + x2 . Write x1 + x′1 = a′ . Then b′ = x1 + y1 , where y1 = x′1 + t2 . Further, kx2 ≤ kt1 ≤ u. This implies that x2 < e. Therefore a = x1 + x2 < x1 + y1 + e = b′ + e = b, as asserted. (2) Let a, b, c ∈ M such that 2c + a = c + b. Then we have a refinement matrix c b



cca  a1 a2 a3 . t1 t2 t3

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If a3 = 0, then 2c = a1 + a2 + t1 + t2 = c + (t1 + t2 ) and b = t1 + t2 + t3 = a + (t1 + t2 ). Thus, we may assume that a3 6= 0. Write a3 = x1 + x2 for some x1 , x2 ∈ M ∗ . For each i, ai + xi < a1 + a2 + x1 + x2 = c = ai + ti . If ai = 0, then xi < ti . As M is directly finite, ti 6= 0. Thus, we may assume that ai , xi , ti ∈ M ∗ . By (1), xi < ti . Write ti = xi + x′i for some x′i ∈ M ∗ . Let e = x′1 + x′2 . Then 2c = = = = b = = = =

a 1 + t1 + a 2 + t2 a1 + a2 + x1 + x′1 + x2 + x′2 (x′1 + x′2 ) + (a1 + a2 + a3 ) c + e; t1 + t2 + t3 x1 + x′1 + x2 + x′2 + t3 x′1 + x′2 + a3 + t3 e + a.

(3) Let a, b, c ∈ M ∗ such that c + a = c + b. By [355, Lemma 5.1], there exists some w ∈ M ∗ such that 2w ≤ a. As M (c) = M (w), we can find some k ∈ N such that c ≤ kw. Write c = c1 + · · · + cm with each 0 6= ci ≤ w(1 ≤ m ≤ k). In addition, 2ci ≤ 2w ≤ a. Thus, we can reduce the problem to the case in which 2c ≤ a. Write a = 2c + t for some t ∈ M . Then 2c + (c + t) = c + b. By (2), there exists e ∈ M such that 2c = c + e and b = e + (c + t). Therefore a = 2c + t = c + e + t = b, as asserted.  Lemma 6.3.16. Let M be a conical simple refinement monoid satisfying s-comparability, and let u ∈ M be an order-unit. Then for any a, b, c ∈ M ∗ , c + a = c + b =⇒ a = b. Proof. Let any a, b, c ∈ M ∗ such that c + a = c + b. If M is directly finite and atomless, then a = b by Lemma 6.3.15. If M is directly infinite and atomless, there exist x, y ∈ M ∗ such that x+ y = x. Clearly, M (c) = M (y); hence, there exists some p ∈ N such that c ≤ py. Thus, c + x ≤ py + x = x. As a result, c + x ≤ x < a + x. In view of Lemma 6.3.15, c < a. Likewise, c < b and 2c < c. Write a = c + a′ , b = c + b′ and c = 2c + c′ , where a′ , b′ , c′ ∈ M ∗ . Therefore a = c + a′ = 2c + c′ + a′ = c′ + c + a = c′ + c + b = c′ + 2c + b′ = c + b′ = b. Now assume that there exists an atom w of M . Then M = M (w) = {mw | m = 0 or m ∈ N}. If there exist some positive integers m, n such that mu = nu. If m < n, then m, n ≥ 0. Further, we n P have some aij ∈ M (1 ≤ i ≤ m, 1 ≤ j ≤ n) such that a = aij and j=1

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i=1

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aij . As a ∈ M ∗ is an atom, aij is 0 or a. Thus, a = ka for some

k ≥ 2. Hence, 0 < a < a, a contradiction. Whence n < m we deduce a similar condition. Thus, we get m = n. In this case, one verifies that a = b. Therefore the proof is true.  Lemma 6.3.17. Let M be a conical refinement monoid satisfying scomparability, and let w = 6 0. Then the following hold: (1) If M (w)/maxM (w) is directly infinite, then x < y, ∀x, y ∈ M (w)\maxM (w).

(2) If M (w)/maxM (w) is directly finite, then

c + x = c + y, 2c ≤ x =⇒ x = y, ∀c ∈ M (w)\maxM (w). Proof. (1) By hypothesis, there exist a, b ∈ M (w)\maxM (w) such that [a] + [b] = [a]. For any x, y ∈ M (w)\maxM (w), we see that M (x) = M (w) = M (b); hence, there exists some m ∈ N such that x ≤ mb. Thus, [x] + [a] ≤ m[b] + [a] = [a]. As a result, [x] + [a] ≤ [a] < [y] + [a]. Clearly, M (w)/maxM (w) is a conical refinement monoid satisfying s-comparability. In view of Lemma 6.3.16, [x] < [y]. Using Lemma 6.3.14, we get x < y, as asserted. (2) Let c ∈ M (w)\maxM (w). Then M (c) = M (w), and so M (c)/maxM (c) is directly finite. Given c + x = c + y with 2c ≤ x, then x = 2c + d for some d ∈ M . Hence, 2c + (c + d) = c + y. Thus, we get a refinement matrix 2c c + d   c zv . y st This implies that 2c + v = z + s + v = c + s. Thus, 2[c] + [v] = [c] + [s]. If [s] = 0, then [c] + [v] = 0 by the directly finiteness, a contradiction. Thus, [s] 6= 0. By Lemma 6.3.16, [c] + [v] = [s] in M (c)/maxM (c), and so [v] < [s]. In virtue of Lemma 6.3.14, v < s. Write s = v + e. Therefore x = 2c + d = z + s + d = z + v + e + d = c + e + d = e + v + t = s + t = y, as asserted.  Lemma 6.3.18. Let M be a conical refinement monoid satisfying scomparability. If 2a = a + b, then b ≤ a. Proof. As b ≤ 2a, we see that M (b) ⊆ M (a). If M (b) ⊂ M (a), it follows by Lemma 6.3.14 that b < a, and the result follows. Thus, we assume that

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M (b) = M (a). As in the proof of Lemma 6.3.7, we get a refinement matrix a a



ab  y z wx

with w ≤ y. If w = 0, then x = a = y; hence, b = z + x = z + y = a. Thus, we may assume that w 6= 0. Obviously, M (w) ⊆ M (a). (1) Suppose that M (w) ⊂ M (a). If M (x) ⊆ M (w), then M (a) ⊆ M (w), a contradiction. By virtue of Lemma 6.3.13, M (w) ⊂ M (x). Likewise, M (w) ⊂ M (y). Thus, 2w < x, y by Lemma 6.3.14. (i) Suppose that M (w)/maxM (w) is directly infinite. According to Lemma 6.3.17, 2w < w. Write w = 2w + w′ , x = 2w + x′ , y = 2w + y ′ . Then x = 3w + w′ + x′ = w + w′ + x = w′ + a and y = 3w + w′ + y ′ = w + w′ + y = w′ + a. Therefore b = z + x = z + y = a. (ii) Suppose that M (w)/maxM (w) is directly finite. Since w + x = a = w + y with 2w ≤ x, it follows by Lemma 6.3.17 that x = y. Therefore b = z + x = z + y = a. (2) Suppose that M (w) = M (a). Then M (w) = M (y) = M (b). (i) Suppose that M (w)/maxM (w) is directly infinite. Clearly, y ∈ M (w)\maxM (w). In view of Lemma 6.3.17, 2[w] < [w] < [y]. According to Lemma 6.3.14, 2w < w < y. As x ≤ b ∈ M (w), we see that x ∈ M (w). If x ∈ M (w)\maxM (w), then [x] < [y] by Lemma 6.3.5. It follows from Lemma 6.3.14 that x < y. Therefore b = x + z < y + z = a. If x ∈ maxM (w), then z ∈ M (w)\maxM (w). As w ∈ M (w)\maxM (w), it follows by Lemma 6.3.17 that [z] < [w]. This implies that z < w. Therefore b = x + z < x + w = a, and then b ≤ a. (ii) Suppose that M (w)/maxM (w) is directly finite. As M (w) = M (y), n P there exists some n ∈ N such that w ≤ ny. Write w = wi with each i=1

wi ≤ y. As w 6∈ maxM (w), we can find some wk 6∈ maxM (w). Since wk ≤ w, y, 2wk ≤ w + y = a. Clearly, M (wk ) = M (w) = M (a). Thus, a ≤ mwk . Write a+c = mwk . Then a+mwk = 2a+c = a+b+c = b+mwk . By virtue of Lemma 6.3.17, we obtain b = a. In any case, we deduce that b ≤ a, and therefore the proof is true. 

We have accumulated all the information necessary to prove the following result due to Pardo (cf. [23] and [354-355]). Theorem 6.3.19. Let M be a conical refinement monoid satisfying scomparability. Then

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(1) M is separative. (2) If M is directly finite, then M is strongly separative. Proof. (1) If 2a = a + b = 2b, then b ≤ a from Lemma 6.3.18. Therefore M is separative by Corollary 6.3.11. (2) If 2a = a + b, then b ≤ a from Lemma 6.3.18. If b < a, then there exists some x ∈ M ∗ such that a + x = b; hence, 2a + x = a + b = 2a. Thus, M is directly infinite. This gives a contradiction. Consequently, we deduce that a = b, and therefore M is strongly separative.  6.4

Generalized s-Comparability

Let s ∈ N. An exchange ring R is said to satisfy generalized s-comparability provided that for any idempotents x, y ∈ R, there exists some e ∈ B(R) such that xRe .⊕ s(yR)e and yR(1 − e) .⊕ s(xR)(1 − e). We say that an exchange ring R satisfies general comparability provided that it satisfies generalized 1-comparability. This concept is a generalization of the general comparability on regular rings. Many authors have studied generalized comparability. In [354], Pardo studied exchange rings that satisfying generalized s-comparability. In [405], Wu investigated general ℵ0 -comparability for exchange rings. Other extensions can be found in Chen [93, Theorem 7], [156, Theorem 3]   and Li [305, Theorem 2.4.4]. F F Let R = , where F is a field. Then R is an exchange ring satis0 F fying general comparability, while it is not regular (cf. [354, Example 3.7]). General comparability implies related comparability. But related comparability is not sufficient to imply general comparability, as the following example shows. Example 6.4.1. Choose a field F , and set Rn = M3n (F ) × M3n (F ) for n = 0, 1, 2, · · · . For each n, define a map ϕn : Rn → Rn+1 according to the rule     x00 x00 ϕn (x, y) =   0 x 0  ,  0 y 0   00y 00y and let R be the direct limit of the sequence ϕ0

ϕ1

R0 −→ R1 −→ · · · .

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By virtue of [217, Example 5.12], R does not satisfy general comparability. On the other hand, R is unit-regular. So it is a regular ring satisfying related comparability.  It is claimed that every exchange ring with all idempotents central satisfies general comparability. Let R be an exchange ring with all idempotents central, and let e, f ∈ R be idempotents. Choose u = f . Then ueR .⊕ uf R and (1−u)f R .⊕ (1−u)eR, and we are done. The main purpose of this section is to investigate exchange rings satisfying generalized s-comparability. Lemma 6.4.2. Let B be a finitely generated projective right module over an exchange ring R, and let s ∈ N. Then A .⊕ sB if and only if there is a decomposition A = A1 ⊕ A2 ⊕ · · · ⊕ As such that A1 .⊕ A2 .⊕ · · · .⊕ As .⊕ B. Proof. One direction is obvious. Conversely, assume that A .⊕ sB. It is trivial if s = 1. Assume that the result holds for k − 1(k ≥ 2). Suppose that A .⊕ kB. Then A ⊕ D ∼ = (k − 1)B ⊕ B for some right R-module D. By [16, Proposition 1.2], we can find right R-modules U, W, E and F such that A ∼ = U ⊕ W, D ∼ = E ⊕ F, U ⊕ E ∼ = (k − 1)B and W ⊕ F ∼ = B. ⊕ Thus U . (k − 1)B. By hypothesis, we have U = U1 ⊕ U2 ⊕ · · · ⊕ Uk−1 , where U1 .⊕ U2 .⊕ · · · .⊕ Uk−1 .⊕ B. Obviously, there exist right R-modules Dk , Dk−1 , · · · , D2 , D1 such that B = D1 ⊕ D2 ⊕ · · · ⊕ Dk , where Dk ⊕ Uk−1 ∼ = B, D1 = U1 and Di ⊕ Ui−1 ∼ = Ui (2 ≤ i ≤ k − 1). Therefore ∼ W ⊕ F = D1 ⊕ D2 ⊕ · · · ⊕ Dk . By [16, Proposition 1.2] again, we can find right R-modules X1 , X2 , · · · , Xk such that W = X1 ⊕ X2 ⊕ · · · ⊕ Xk with all Xi .⊕ Di . Consequently, we claim that A = U ⊕ W = (U1 ⊕ U2 ⊕ · · · ⊕ Uk−1 )⊕(X1 ⊕X2 ⊕· · ·⊕Xk ) = X1 ⊕(X2 ⊕U1 )⊕(X3 ⊕U2 )⊕· · ·⊕(Xk ⊕Uk−1 ). Obviously, we see that X1 .⊕ X2 ⊕ U1 .⊕ X3 ⊕ U2 .⊕ · · · .⊕ Xk ⊕ Uk−1 . Thus we complete the proof by induction.  Lemma 6.4.3. Let B be a finitely generated projective right module over an exchange ring R, and let s ∈ N. If A .⊕ sB, then there exists some right R-module V such that A .⊕ sV, V .⊕ A, B. Proof. By virtue of Lemma 6.4.2, we have a decomposition A = A1 ⊕ A2 ⊕ · · · ⊕ As with A1 .⊕ A2 .⊕ · · · .⊕ As .⊕ B. Set V = As . Then V .⊕ A, B. Moreover, we claim that A = A1 ⊕ A2 ⊕ · · · ⊕ As .⊕ sAs , as asserted. 

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Theorem 6.4.4. Let R be an exchange ring, and let s ∈ N. Then the following are equivalent: (1) R satisfies generalized s-comparability. (2) For any A, B ∈ F P (R), there exists e ∈ B(R) such that Ae .⊕ sBe and B(1 − e) .⊕ sA(1 − e). Proof. (2) ⇒ (1) For any idempotents e, f ∈ R, we have that eR, f R ∈ F P (R). So there is some u ∈ B(R) such that eRu .⊕ s(f R)u and f R(1 − u) .⊕ s(eR)(1 − u), as required. (1) ⇒ (2) Suppose that A, B ∈ F P (R). In view of [409, Theorem 2.1], there exist idempotents e′1 , · · · , e′n , e′′1 , · · · , e′′n ∈ R such that A = e′1 R ⊕ · · · ⊕ e′n R and B = e′′1 R ⊕ · · · ⊕ e′′n R. Assume that A, B .⊕ nR. It suffices to find some e ∈ B(R) such that Ae .⊕ sBe and B(1 − e) .⊕ sA(1 − e). If n = 1, the result follows. Assume that the result holds for n − 1 (n ≥ 2). Since we have right Rmodules E, F such that A ⊕ E ∼ = (n − 1)R ⊕ R and B ⊕ F ∼ = (n − 1)R ⊕ R, by [16, Proposition 1.2], we have decompositions A = A1 ⊕ A2 , B = B1 ⊕ B2 with A1 , B1 .⊕ (n − 1)R, A2 , B2 .⊕ R. Hence, there exist some f1 , f2 ∈ B(R) such that A1 f1 .⊕ sB1 f1 , B1 (1 − f1 ) .⊕ sA1 (1 − f1 ), A2 f2 .⊕ sB2 f2 , B2 (1 − f2 ) .⊕ sA2 (1 − f2 ). Set e1 = f1 f2 , e2 = (1 − f1 )(1 − f2 ). It is easy to verify that Ae1 .⊕ sBe1 , B2 e2 .⊕ sAe2 . Set g1 = f1 (1 − f2 ), g2 = f2 (1 − f1 ). We check that A1 g1 .⊕ B1 g1 , A2 g2 .⊕ B2 g2 , B1 g2 .⊕ A1 g2 , B2 g1 .⊕ A2 g1 . In view of Lemma 6.4.3, we can find right R-modules V1 , V2 , V3 and V4 such that A1 g1 A2 g2 B1 g 2 B2 g 1

.⊕ .⊕ .⊕ .⊕

sV1 , V1 sV2 , V2 sV3 , V3 sV4 , V4

.⊕ .⊕ .⊕ .⊕

A1 g1 , V1 A2 g2 , V2 B1 g2 , V3 B2 g1 , V4

.⊕ .⊕ .⊕ .⊕

B1 g 1 , B2 g 2 , A1 g2 , A2 g1 .

∼ V1 ⊕D1 , B2 g2 ∼ Thus, we may assume that B1 g1 = = V2 ⊕D2 , A1 g2 ∼ = V3 ⊕C1 ∼ and A2 g1 = V4 ⊕ C2 for some right R-modules C1 , C2 , D1 and D2 . Clearly, C1 ⊕ C2 , D1 ⊕ D2 .⊕ (n − 1)R. Hence there is some h ∈ B(R) such that (C1 ⊕ C2 )h .⊕ s(D1 ⊕ D2 )h, (D1 ⊕ D2 )(1 − h) .⊕ s(C1 ⊕ C2 )(1 − h). Set

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e3 = gh, e4 = g(1 − h). Then we claim that

Ae3 = A1 g1 h ⊕ A1 g2 h ⊕ A2 g1 h ⊕ A2 g2 h = A1 g1 h ⊕ V3 h ⊕ C1 h ⊕ V4 h ⊕ C2 h ⊕ A2 g2 h .⊕ sV1 h ⊕ B1 g2 h ⊕ B2 g1 h ⊕ sV2 h ⊕ s(D1 ⊕ D2 )h .⊕ s(V1 ⊕ B1 g2 ⊕ B2 g1 ⊕ V2 ⊕ D1 ⊕ D2 )h .⊕ s(B1 g1 ⊕ B2 g2 ⊕ B1 g2 ⊕ B2 g1 )h ∼ = sBe3 . Analogously, we show that Be4 = B1 g1 (1 − h) ⊕ B1 g2 (1 − h) ⊕ B2 g1 (1 − h) ⊕ B2 g2 (1 − h) = (V1 ⊕ D1 ⊕ V2 ⊕ D2 ⊕ B1 g2 ⊕ B2 g1 )(1 − h) .⊕ (V1 ⊕ B1 g2 ⊕ V2 ⊕ B2 g1 ⊕ sC1 ⊕ sC2 )(1 − h) .⊕ (sV3 ⊕ sV4 ⊕ sC1 ⊕ sC2 ⊕ sA1 g1 ⊕ sA2 g2 )(1 − h) ∼ = s(A1 g1 ⊕ A1 g2 ⊕ A2 g2 ⊕ A2 g1 )(1 − h) ∼ = sAe4 . Let e = e1 + e3 . Then e ∈ B(R) with 1 − e = e2 + e4 . Moreover, we show that Ae .⊕ sBe and B(1 − e) .⊕ sA(1 − e), as required.  Corollary 6.4.5. Let R be an exchange ring, and let A ∈ F P (R). If R satisfies generalized s-comparability, then so does EndR (A). Proof. Let E = EndR (A). Given any idempotents x, y ∈ E, we have u ∈ B(R) such that xAu .⊕ s(yA)u and yA(1 − u) .⊕ s(xA)(1 − u) by Theorem 6.4.4. Define an R-morphism g : A → A given by g(a) = au for any a ∈ A. Then g ∈ B(E) such that gxA .⊕ s(gyA) and (1 − g)yA .⊕
s (1 − g)xA . Let e = gx and f = gy. As eA .⊕ s(f A), there exist R-morphisms q : eA → s(f A) and p : s(f A) → eA such that pq = 1eA . For any r1 , · · · , rs ∈
E, let α(r1 ,··· ,rs ) : A → s(f A) given by α(r1 ,··· ,rs ) (a) = f r1 (a), · · · , f rs (a) for any a ∈ A. For any r ∈ E, let βr : A → eA given by βr (a) = er(a) for any a ∈ A. It is easy to verify that α(r1 ,··· ,rs ) and βr are well defined. Let r1 , · · · , rs ∈ E. As A is projective, there exists some γ ∈ E such that pα(r1 ,··· ,rs ) = eγ; hence, pα(r1 ,··· ,rs ) ∈ eE. Let pi : s(f A) → f A be the i′ th projection. Similarly, we can find some δi ∈ E such that pi qβr = f δi ∈ f E. Therefore qβr = (p1 qβr , · · · , ps qβr ) ∈ s(f E). Construct two maps ϕ:

ψ:

s(f E) → eE, (f r1 , · · · , f rs ) 7→ pα(r1 ,··· ,rs ) ; eE er

→ 7 →

s(f E), qβr .

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Thus, ϕ and φ are well defined. For any r ∈ E, a ∈ A, we verify that ϕψ(er)(a) = ϕ(qβr )(a). Write qβr = (f r1 , · · · , f rs ). Then
ϕψ(er)(a) = p(f r1 , · · · , f rs )(a) = pα(r1 ,··· ,rs ) (a) = p f r1 (a), · · · , f rs (a) = pqβr (a) = er(a). Hence, ϕψ = 1eE . As eE is
projective, we get eE .⊕ s(f E). Likewise, (1 − g)yE .⊕ s (1 − g)xE . Therefore E satisfies generalized scomparability.  As an immediate consequence, we observe that the generalized scomparability over exchange rings is Morita invariant. Corollary 6.4.6. Let R be an exchange ring, and let A, B ∈ F P (R). If R satisfies general comparability, then A ⊕ B is directly finite if and only if so are A and B. Proof. Assume that A ⊕ B is directly finite. It follows by [253, Corollary 1] that A and B are directly finite. Conversely, assume that A and B are both directly finite. According to Corollary 6.4.5, EndR (A) and EndR (B) satisfy general comparability; hence they have stable range one. Suppose that A ⊕ B ⊕ C ∼ = A ⊕ B. By Theorem 1.1.5, we have C = 0, as required. Following Lam (cf. [298]), a ring R is Mn -unique if, for any ring S, Mn (R) ∼ = Mn (S) implies that R ∼ = S. From [298, Theorem 9.2], we know that every commutative ring is Mn -unique for all n ≥ 1. Proposition 6.4.7. Let R be an exchange ring satisfying general comparability. If R is directly finite, then R is Mn -unique for all n ≥ 1.

Proof. Suppose that Mn (R) ∼ = Mn (S). Set T = Mn (R). Then there exist idempotents e, f ∈ T such that R ∼ = eT e and S ∼ = f T f . It will suffice to ∼ ∼ show that eT ∼ f T . Clearly, n(eT ) T n(f T ). By virtue of Corollary = = = 6.4.5, T is an exchange ring satisfying general comparability. So we can find a u ∈ B(T ) such that ueT .⊕ uf T and (1 − u)f T .⊕ (1 − u)eT . Assume that uf T ⊕ D ∼ = ueT for a right T -module D. Then n(uf T ) ⊕ nD ∼ = n(ueT ) ∼ = n(uf T ). Clearly, R has stable range one, and so has T . In view of Corollary 6.4.6, n(uf T ) is directly finite, and thus D = 0. This implies that ueT ∼ = uf T . Likewise, (1 − u)eT ∼ = (1 − u)eT . Consequently, ∼ eT = ueT ⊕ (1 − u)eT ∼  = uf T ⊕ (1 − u)f T ∼ = f T , as required. If R is a directly finite, right self-injective ring, it follows from Proposition 6.4.7 that R is Mn -unique for all n ≥ 1. Further, we derive the

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following. Corollary 6.4.8. Let R be an exchange ring with all idempotents central. Then R is Mn -unique for all n ≥ 1. Proof. In view of Corollary 1.3.15, R has stable range one; hence, it is directly finite. One easily checks that R satisfies general comparability. Therefore we complete the proof from Proposition 6.4.7.  Lemma 6.4.9. Let R be an exchange ring satisfying generalized scomparability, and let J be an ideal of R. Then the natural map B(R) →
B R/J is surjective.

Proof. Given any e ∈ B(R/J), there exists an idempotent x ∈ R such that x = e. By hypothesis, we can find f ∈ B(R) such that f (1 − x)R .⊕ sf xR and (1 − f )xR .⊕ s(1 − f )(1 − x)R. Hence, f (1 − e)(R/J) .⊕ sf e(R/J) and (1 − f )e(R/J) .⊕ s(1 − f )(1 − e)(R/J). From e ∈ B(R/J), we claim  that f (1 − e) = (1 − f )e = 0. Thus f = e, as required. Lemma 6.4.10. Let R be an exchange ring. Then R is separative if and only if every indecomposable factor R/P is separative.

Proof. One direction is obvious by [16, Proposition 2.2]. Suppose that every factor R/P with B(R/P ) = {0, 1} is separative. If R is not separative, then there exist some A, B ∈ F P (R) such that 2A ∼ = A⊕B ∼ = 2B, while A ≇ B. Let Ω = {I E R | A/AI ≇ B/BI}. Then Ω 6= ∅. ∞ S Given I1 ⊆ I2 ⊆ · · · in Ω, then J = Ii is a proper ideal of R. If J 6∈ Ω, i=1

then A/AJ ∼ = B/BJ. As in the proof of Lemma 1.4.11, we may assume that A ∼ = eR and B ∼ = f R, where e, f ∈ R are idempotents. Thus, we have some a, b ∈ R such that e ≡ ab, f ≡ ba(mod J). So, e − ab, f − ba ∈ Ik for some k ∈ N. This implies that Ik 6∈ Ω, a contradiction. Therefore, J ∈ Ω, and so Ω is inductive. By using Zorn’s Lemma, there exists an ideal Q which is maximal in Ω. By hypothesis, B(R/Q) 6= {0, 1}. In view of Lemma 6.4.9, we have an idempotent e ∈ B(R) such that e 6= 0, 1. This implies that e, 1 −
e 6∈ Q. By the maximality of Q, A/A(eR) ∼ = B/B(eR)
∼ and A/A (1 − e)R = B/B (1 − e)R . According to Lemma 1.4.11, there are decompositions A = A1 ⊕ A2 , B = B1 ⊕ B2 , A1 ∼ = A2 , A2 = A2 (eR), B2 = B2 (eR). This implies that A(1 − e) ∼ = B(1 − e). Likewise, we get Ae ∼ = Be. Hence, ∼ A = B, a contradiction. Therefore R is separative. 

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It is convenient at this stage to include the following result (cf. [354, Theorem 3.9]). Theorem 6.4.11. Every exchange ring satisfying generalized s- comparability is separative. Proof. Let R be an exchange ring satisfying generalized s-comparability. Let P E R such that B(R/P ) = {0, 1}. Then R/P is an exchange ring satisfying s-comparability. In view of Theorem 6.3.19, R/P is separative. Therefore we complete the proof by Lemma 6.4.10.  Lemma 6.4.12. Let A and B be finitely generated projective modules over an exchange ring R, and let J be an ideal of R. If A/AJ .⊕ s(B/BJ), then there are decompositions A = A1 ⊕ A2 and B = B1 ⊕ B2 with A1 .⊕ sB1 and A2 = A2 J. ∼ sB/(sB)J, by Lemma 1.4.11, we Proof. Since A/AJ .⊕ s(B/BJ) = have decompositions A = A1 ⊕ A2 and sB = C1 ⊕ C2 with A1 ∼ = C1 and A2 = A2 J. In view of Lemma 6.3.12, there are decompositions B = D0 ⊕ D1 ⊕· · ·⊕Ds , C1 ∼ = D1 ⊕2D2 ⊕· · ·⊕sDs , C2 ∼ = sD0 ⊕(s−1)D1 ⊕· · ·⊕Ds−1 . Set B1 = D1 ⊕ D2 ⊕ · · · ⊕ Ds . Obviously, A1 ∼ = C1 .⊕ sB1 . Set B2 = D0 . Then B = B1 ⊕ B2 with A1 .⊕ sB1 , as asserted.  Let J be an ideal of an exchange ring R. We say that J satisfies generalized s-comparability provided that for any idempotents x ∈ J, y ∈ R, there exists an e ∈ B(R) such that xRe .⊕ s(yR)e and yR(1−e) .⊕ s(xR)(1−e). Theorem 6.4.13. Let J be an ideal of an exchange ring R. Then R satisfies generalized s-comparability if and only if the following hold: (1) R/J satisfies generalized s-comparability. (2) J satisfies generalized s-comparability.
(3) The natural map B(R) → B R/J is surjective.

Proof. (1) and (2) are obvious. We get (3) by Lemma 6.4.9. Conversely, assume now that the above conditions hold. Given any idempotents x, y ∈ R, there exists f ∈ B(R/J) such that f x(R/J) .⊕ sf y(R/J) and (1 − f )y(R/J) .⊕ s(1 − f )x(R/J). By hypothesis, we may N assume that f = g with g ∈ B(R). Since gxR/(gxR)J ∼ = gxR R/J ∼ = N ⊕ ∼ ∼ f x(R/J) . sf y(R/J) = s(gyR R/J) = s(gyR/(gyR)J). In view of Lemma 6.4.12, we have decompositions gxR = x1 R ⊕ x2 R and gyR = y1 R ⊕ y2 R with x1 R .⊕ sy1 R and x2 R = x2 RJ ⊆ J. Thus we see that

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x2 ∈ J. By the above consideration, we can find h ∈ B(R) such that hx2 R .⊕ shy2 R and (1 − h)y2 R .⊕ s(1 − h)x2 R. For ring gR, we have (gx)(gR) = (gx1 )(gR)⊕(gx2 )(gR) and (gy)(gR) = (gy1 )(gR) ⊕(gy2 )(gR) with (gx1 )(gR) .⊕ s(gy1 )(gR), (gh)(gx2 )(gR) .⊕ s(gh) (gy2 )(gR) and (g(1−h))(gy2 )(gR).⊕ s(g(1−h))(gx2 )(gR). It is easy to show that (gx1 )(gR)(g(1 − h)) .⊕ s(gy1 )(gR)(g(1 − h)), (gy2 )(gR)(g(1 − h)) .⊕ s(gx2 )(gR)(g(1 − h)). In view of Lemma 6.4.3, we can find right gR-modules V1 , V2 such that (gx1 )(gR)(g(1 − h)) .⊕ sV1 , V1 .⊕ (gx1 )(gR)(g(1 − h)), V1 .⊕ (gy1 )(gR)(g(1 − h)), (gy2 )(gR)(g(1 − h)) .⊕ sV2 , V2 .⊕ (gy2 )(gR)(g(1 − h)), V2 .⊕ (gx2 )(gR)(g(1 − h)).

Thus, we may assume that (gy1 )(gR)(g(1 − h)) ∼ = V1 ⊕ C and (gx2 )(gR)(g(1−h)) ∼ = V2 ⊕D for some right R-modules C and D. Obviously, C ∼ = rR for some r ∈ R. On the other hand, we have D ∼ = tg(1 − h)x2 R with tg(1 − h)x2 ∈ J for some t ∈ R. By the above consideration, we can find k ∈ B(R) such that Dk .⊕ sCk and C(1 − k) .⊕ sD(1 − k). Set e1 = g(1 − h)k, e2 = g(1 − h)(1 − k). Then we check that gxRe1 = = .⊕ .⊕ .⊕ ∼ =

(gx1 )(gR)g(1 − h)k ⊕ (gx2 )(gR)g(1 − h)k (gx1 )(gR)g(1 − h)k ⊕ V2 k ⊕ Dk sV1 k ⊕ s(gy2 )(gR)g(1 − h)k ⊕ sCk s(V1 ⊕ (gy2 )(gR)g(1 − h) ⊕ C)k s((gy1 )(gR)g(1 − h) ⊕ (gy2 )(gR)g(1 − h))k sgyRe1.

Similarly, we have gyRe2 = ∼ = .⊕ .⊕ ∼ =

(gy1 )(gR)g(1 − h)(1 − k) ⊕ (gy2 )(gR)g(1 − h)(1 − k) V1 (1 − k) ⊕ C(1 − k) ⊕ (gy2 )(gR)g(1 − h)(1 − k) V1 (1 − k) ⊕ sD(1 − k) ⊕ sV2 s((gx1 )(gR) ⊕ (gx2 )(gR))g(1 − h)(1 − k) sgxRe2 .

Analogously, we claim that (1 − g)xRe3 .⊕ s(1 − g)yRe3 and (1 − g)yRe4 .⊕ s(1 − g)xRe4 with e3 = (1 − g)(1 − h)k, e4 = (1 − g)(1 − h)(1 − k). Consequently, we see that xR(e1 + e3 ) .⊕ syR(e1 + e3 ) and yR(e2 +e4 ) .⊕ sxR(e2 +e4 ). It is easy to verify that e1 +e3 , e2 +e4 ∈ B(R) with e1 + e3 + e2 + e4 = 1. Therefore R satisfies generalized s-comparability.

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An ideal J of a ring R is abelian provided that every idempotent in J is central. Corollary 6.4.14. Let J be an abelian ideal of an exchange ring R. Then R satisfies generalized s-comparability if and only if the following hold: (1) R/J satisfies generalized s-comparability.
(2) The natural map B(R) → B R/J is surjective.

Proof. Given any idempotents e ∈ J, f ∈ R, one easily checks that ef R .⊕ e · eR and (1 − e) · eR = 0 .⊕ (1 − e)f R. Hence, J satisfies generalized s-comparability. Therefore the result follows from Theorem 6.4.13.  Corollary 6.4.15. Let J be an ideal of an exchange ring R such that R .⊕ xR for any nonzero idempotent x ∈ J. Then R satisfies generalized s-comparability if and only if the following hold: (1) R/J satisfies generalized s-comparability.
(2) The natural map B(R) → B R/J is surjective.

Proof. For any idempotents x ∈ J, y ∈ R, we have yR .⊕ xR or xR = 0 .⊕ yR. Hence J satisfies generalized s-comparability. Therefore we complete the proof by Theorem 6.4.13. 

Let (aij ) ∈ Ms×t (R) and (bjk ) ∈ Mt×m (R). Define the product of (aij ) t P and (bjk ) by (aij )(bjk ) = (cik ), where cik = aij bjk . j=1

Lemma 6.4.16. Let e = e2 ∈ Ms (R) and f = f 2 ∈ Mt (R). Then we have: (1) e(sR) .⊕ f (tR) if and only if there exist a ∈ Ms×t (R), b ∈ Mt×s (R) such that e = ab, a = eaf and b = f be. (2) e(sR) ∼ = f (tR) if and only if there exist a ∈ Ms×t (R), b ∈ Mt×s (R) such that e = ab, f = ba, a = eaf and b = f be. Proof. (1) Suppose e(sR) .⊕ f (tR). Then we have a projection p : f (tR) → e(sR). Let ε1 = (1, 0, · · · , 0)T , ε2 = (0, 1, · · · , 0)T , · · · , εs = (0, 0, · · · , 1)T be a basis of sR and η1 = (1, 0, · · · , 0)T , η2 = (0, 1, · · · , 0)T , · · · , ηt = (0, 0, · · · , 1)T

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For 1 ≤ i ≤ s, we can find some xi ∈ tR t t

P P such that eεi = p(f xi ) = p f ηj ηjT f xi = p (f ηj )(ηjT f xi ) =

be a basis of tR.

j=1

t P

j=1

p(f ηj )(ηjT f xi ) =

a = e

t P

j=1

t P

j=1

j=1

t

P p(f ηj )ηjT f xi = e( p(f ηj )ηjT )f (f xi e). Let


p(f ηj )ηjT f, b = f (x1 , · · · , xs )e.

j=1

Then e = eaf and b =

f be. Furthermore, we have e = e(ε1 , · · · , εs ) = (eε1 , · · · , eεs ) = (af x1 e, · · · , af xs e) = af (x1 , · · · , xs )e = ab, as desired. Suppose e = ab with a ∈ Ms×t (R), b ∈ Mt×s (R), a = eaf and b = f be. We construct an R-morphism ψ : f (tR) → e(sR) given by ψ(f x) = af x for any x ∈ R. For any er ∈ e(sR) with r ∈ sR, there is x = br ∈ tR such that ψ(f x) = af x = af br = abr = er. So ψ is an epimorphism. Since e(sR) is a projective right R-module, we know that the exact sequence 0 −→ ker(ψ) −→ f (tR) −→ e(sR) −→ 0 splits. Hence e(sR) ⊕ ker(ψ) ∼ = f (tR). Therefore e(sR) .⊕ f (tR). (2) Suppose e(sR) ∼ = f (tR). Let ε1 = (1, · · · , 0)T , · · · , εs = (0, · · · , 1)T be a basis of sR and η1 = (1, · · · , 0)T , · · · , ηt = (0, · · · , 1)T be a basis of tR. Assume that eεi = ψ −1 (f xi ) for 1 ≤ i ≤ s and f ηj = ψ(eyj ) for 1 ≤ j ≤ t. Similarly to the preceding discussion, we have t P e = (e( ψ −1 (f ηj )ηjT )f )(f (x1 , · · · , xs )e) for some x1 , · · · , xs ∈ tR and j=1 s P



ψ(eεi )εTi )e e(y1 , · · · , yt )f for some y1 , · · · , yt ∈ sR. We easily i=1
check that f (x1 , · · · , xs )e = f (f x1 , · · · , f xs )e = f ψ(eε1 ), · · · , ψ(eεs ) e = s t

P P f ψ(eεi )εTi e. Likewise, we have e(y1 , · · · , yt )f = e ψ −1 (f ηj )ηjT f . f = f(

i=1

j=1

Set a = e(y1 , · · · , yt )f and b = f (x1 , · · · , xs )e. Then e = ab, f = ba, a = eaf and b = f be. Suppose e = ab, f = ba with a = eaf and b = f be. Then we can construct an R-morphism ψ : f (tR) → e(sR) given by ψ(f x) = af x for any x ∈ tR. By (1), we know that ψ is an R-epimorphism. Assume that ψ(f x) = 0 for some x ∈ tR. Then af x = 0. That is, ax = 0, and so f x = bax = 0. Therefore ψ is an R-monomorphism. Consequently, we conclude that e(sR) ∼  = f (tR). We say that x, y ∈ R are centrally orthogonal, in symbols x ⊥ y, if xRy = 0 = yRx. The following proposition is a slight generalization of a result of Goodearl [21, Proposition 8.11].

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Theorem 6.4.17. Let S be an exchange ring satisfying generalized scomparability, let J be an ideal of S, and let R be an exchange subring of S which contains J. Then R satisfies generalized s-comparability if and only if (1) R/J satisfies generalized s-comparability.
(2) The natural map B(R) → B R/J is surjective. (3) For any idempotents x, y ∈ R, x ⊥ y implies that there exists e ∈ B(R) such that ex = x and ey = 0. Proof. Assume that R satisfies generalized s-comparability. Then (1) and (2) hold by Theorem 6.4.13. Let x, y ∈ R be idempotents such that x ⊥ y. Since R satisfies generalized s-comparability, there exists
an e ∈ B(R) such that eyR .⊕ s(exR) and (1 − e)xR .⊕ s (1 − e)yR . Therefore eyR ⊆ eyReyR .⊕ s(exR)eyR ⊆ s(xRy)Re = 0; hence
ey = 0. Furthermore, (1 − e)xR ⊆ (1 − e)xR(1 − e)xR .⊕ s (1 − e)yR (1 − e)xR ⊆ s(yRx)R(1 − e) = 0, whence ex = x. Conversely, assume that the conditions above hold. Given any idempotents x ∈ J, y ∈ R, since S is an exchange ring satisfying generalized s-comparability, there is an
f ∈ B(S) such that f xS .⊕ s(f yS) and N (1 − f )yS .⊕ s (1 − f )xS . Clearly, (1 − f )yS (S/J) = 0; hence, S

(1 − f )y ∈ J. Clearly, f x, (1 − f )x ∈ R. In addition, f y = y − (1 − f )y ∈ R. In view of [17, Lemma 2.1], we can find some idempotents z, w ∈ R such that Rf xR+Rf yR = RzR and R(1−f )xR+R(1−f )yR = RwR. Furthermore, we check that zRw ⊆ f R(1 − f )R = 0; hence zRw = 0. Likewise, wRz = 0. That is, z ⊥ w. This implies that ez = z, ew = 0 for some e ∈ B(R). One easily shows that f xS = ef xS ⊆ exS ⊆ ef xS + e(1 − f )xS ⊆ ef xS = f xS, whence f xS = exS. Analogously, we have (1−f )xS = (1−e)xS, f yS = eyS and (1 − f )yS = (1 − e)yS. So exS
= f xS .⊕ s(f
yS) = s(eyS) and (1 − e)yS = (1 − f )yS .⊕ s (1 − f )xS = s (1 − e)xS . Let x′ = ex and y ′ = ey. Then x′ , y ′ ∈ S are idempotents such that x′ S .⊕ s(y ′ S). In view of Lemma 6.4.16, there are a ∈ M1×s (S) and b ∈ Ms×1 (S) such that x′ = ab, a = x′ ay ′ and b = y ′ bx′ . As x′ ∈ J, we see that a ∈ M1×s (J) and b ∈ Ms×1 (J). By using Lemma 6.4.16 again,
exR .⊕ s(eyR). Likewise, (1 − e)yR .⊕ s (1 − e)xR . Hence, J satisfies generalized s-comparability. According to Theorem 6.4.13, we establish the result.  An ideal J of a ring R is regular provided that every element in J is regular. Clearly, J is a regular ideal of R if and only if for any x ∈ J, there

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exists some y ∈ J such that x = xyx. Corollary 6.4.18. Let S be an exchange ring satisfying generalized scomparability, let J be an abelian regular ideal of S, and let R be an exchange subring of S which contains J. Then R satisfies generalized scomparability if and only if (1) R/J satisfies generalized s-comparability. (2) For any idempotents x, y ∈ R, x ⊥ y implies that there exists e ∈ B(R) such that ex = x and ey = 0. Proof. Let f ∈ B(R/J). Then there exists an idempotent e ∈ R such that f = e. For any x ∈ R, we see that ex(1 − e) = f x(1 − f )) = 0, and so ex(1 − e) ∈ J. Hence, there exists an element y ∈ R such that ex(1 − e) = ex(1 − e)yex(1 − e) = ex(1 − e)ex(1 − e)y = 0, and so ex = exe. Likewise, xe = exe. Thus, ex = xe, i.e., e ∈ B(R). This implies that the natural map B(R) → B(R/J) is surjective. According to Theorem 6.4.17, we are through. 

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Chapter 7

Generalized Stable Rings

So as to investigate more general rings such as regular, right self-injective rings, we generalize the stable range one to a generalized stable condition. We call a ring R a generalized stable ring provided that aR + bR = R implies that a + by ∈ K(R) for some y ∈ R, where K(R) = {x ∈ R | ∃s, t ∈ R such that sxt = 1}. Many of rings of interest are generalized stable rings. For instance: (1) All rings with stable range one. This includes all unit-regular rings and strongly π-regular rings, and hence all algebraic algebras over a field. (2) All exchange rings satisfying related comparability. This includes all exchange rings satisfying the comparability axiom and exchange rings satisfying general comparability, hence all one-sided unit-regular rings. (3) All right self-injective rings. (4) All right or left continuous regular rings. (5) All simple factors of unital complex AW ∗ -algebras and all simple exchange rings satisfying s-comparability. Moreover, all simple separative exchange rings. Thus we see that many known classes of exchange rings are generalized stable rings. This chapter will outline a few of the basic properties of generalized stable rings, and develop a number of equivalent characterizations of such rings. We will also study diagonal reduction, and derive the decompositions of invertible matrices over such rings.

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Criteria

The main purpose of this section is to investigate some equivalences of generalized stable rings. Lemma 7.1.1. Let R be a ring. Then the following are equivalent: (1) R is a generalized stable ring. (2) Whenever ax + b = 1, there exists some y ∈ R such that a + by ∈ K(R). Proof. (1)⇒(2) is clear. (2)⇒(1) Given aR + bR = R, then ax + by = 1 for some x, y ∈ R. Thus we can find a z ∈ R such that a + byz ∈ K(R), as desired.  Proposition 7.1.2. Let R be a ring. Then the following are equivalent: (1) R is a generalized stable ring. (2) R/J(R) is a generalized stable ring. Proof. (1)⇒(2) Given ax + b = 1 in R/J(R), we see that ax + (b + r) = 1 for some r ∈ J(R). Since R is a generalized stable ring, we can find
some y ∈ R such that a + (b + r)y ∈ K(R). Hence a + by ∈ K R/J(R) . Thus R/J(R) is a generalized stable ring from Lemma 7.1.1. (2)⇒(1) Given ax + b = 1 in R, then ax + b = 1 in R/J(R), so we s(a + by)t − 1 ∈ J(R). have s, t ∈ R such that s(a + by)t = 1. Hence
Therefore s(a + by)t = 1 + s(a + by)t − 1 = u ∈ U (R). Consequently, s(a + by)tu−1 = 1, and then a + by ∈ K(R), as required.  Corollary 7.1.3. Let M be a quasi-injective right R-module. EndR (M ) is a generalized stable ring.

Then

Proof. Since M is a quasi-injective right R-module, by virtue of [214,
Theorem 1], we see that EndR (M )/J EndR (M ) is
a regular ring satisfying general comparability. So EndR (M )/J EndR (M ) is a generalized stable ring. Thus the result follows from Proposition 7.1.2.  Let I be any nonzero ideal of a principal ideal domain R. We claim that EndR (R/I) is a generalized stable ring. In light of [299, Theorem 7.2], R/I is quasi-injective, and we are done by Corollary 7.1.3. Proposition 7.1.4. Let R be a ring. Then the following are equivalent:

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(1) R is a generalized stable ring. (2) Whenever aR+bR = dR, there exists some y ∈ R such that a+by = dw for some w ∈ K(R). Proof. (2)⇒(1) is trivial. (1)⇒(2) Assume that aR + bR = dR. Let g : dR → dR/bR be the canonical map, f1 : R → aR given by r 7→ ar for any r ∈ R, f2 : R → bR given by r 7→ br for any r ∈ R and f3 : R → dR given by r 7→ dr for any r ∈ R. Clearly, gf1 and gf3 are both epimorphisms. Since R is projective as a right R-module, we have some α ∈ EndR (R) such that gf1 = gf3 α. Since gf1 is epimorphic, there exists ψ ∈ EndR (R) such that gf3 αψ = gf3 . From αψ + (1 − αψ) = 1, we can find some y ∈ R such that α + (1 − αψ)y = w ∈ K(R). Thus we claim that gf1 = gf3 α = gf3 (α + (1 − αψ)y) = gf3 w, hence im(f1 − f3 w) ⊆ ker(g) = bR. Further, f2 β = f1 − f
3 w for some β ∈ EndR (R). Consequently, we conclude that a + b − β(1) = dw(1), as asserted.  Theorem 7.1.5. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E is a generalized stable ring. (2) Given any decompositions M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ =A∼ = A2 , there exist C, D, E ⊆ M such that M = C ⊕ D ⊕ B1 = C ⊕ E ⊕ B2 with C∼ = A. Proof. (1)⇒(2) Suppose M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A ∼ = A2 . ∼ Using the decomposition M = A1 ⊕ B1 = A ⊕ B1 , we obtain projections p1 : M → A, p2 : M → B1 and injections q1 : A → M, q2 : B1 → M such that p1 q1 = 1A , q1 p1 + q2 p2 = 1M and ker(p1 ) = B1 . Using the decomposition M = A2 ⊕ B2 ∼ = A ⊕ B2 , we obtain a projection f : M → A and an injection g : A → M such that f g = 1A and ker(f ) = B2 . From (f q1 )(p1 g) + f q2 p2 g = f (q1 p1 + q2 p2 )g = f g = 1A , we can find some s, t ∈ E such that s(f q1 + f q2 p2 gy)t = 1A for some y ∈ E. That is, sf (q1 + q2 p2 gy)t = 1. Thus M = ker(sf ) + (q1 + q2 p2 gy)t(A). Let C = (q1 + q2 p2 gy)t(A). It is easy to verify that C ∼ = A. Let ψ : A2 ∼ = A. Then we have ker(sf ) = ker(sψ) ⊕ ker(f ), so M = C ⊕ E ⊕ B2 with E = ker(sψ). Since sf (q1 + q2 p2 gy)p1 (q1 + q2 p2 gy)t = sf (q1 + q2 p2 gy)t = 1, we have
M = ker sf (q1 + q2 p2 gy)p1 ⊕ (q1 + q2 p2 gy)t(A).

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Let φ : A1 ∼ = A.
Then we can verify that ker sf (q1 + q2 p2 gy)p1 =
ker sf (q1 + q2 p2 gy)φ ⊕ ker(p1 ) = D ⊕ B1 with D = ker sf (q1 + q2 p2 gy)φ . Therefore M = C ⊕ D ⊕ B1 = C ⊕ E ⊕ B2 with C ∼ = A. (2) ⇒ (1) Given ax + b = 1A in E, set M = A ⊕ A. Let pi : M → A, qi : A → M (for i = 1, 2) denote the projections and injections of this direct sum. Set A1 = q1 (A) and B1 = q2 (A), so that M = A1 ⊕ B1 with A1 ∼ = A. Define f = ap1 + bp2 : M → A and g = q1 x + q2 : A → M . It is easy to verify that f g = ax + b = 1A . Set A2 = g(A), B2 = ker(f ). Then M = A2 ⊕ B2 with A2 ∼ = A. Thus we have C, D, E ⊆ M such that M = C ⊕ D ⊕ B1 = C ⊕ E ⊕ B2 with C ∼ = A. ∼ Let h : A = C → M be the injection and k : A ∼ = A2 ∼ = C ⊕E → C ∼ =A be the projection. Then we check that M = C ⊕ E ⊕ ker(f ) = h(A) ⊕ ker(kf ). Obviously, kf is epimorphic and h is monomorphic. Thus kf h is isomorphic. Let l : A ∼ = A1 ∼ = C⊕D → C ∼ = A be the projection. It is easy to verify that D ⊕ ker(p1 ) = ker(lp1 ). Hence we see that M = h(A) ⊕ ker(lp1 ). Observing that lp1 is epimorphic, we show that
lp1 h is isomorphic. From kf h = k(ap1 + bp2 )h = k a + bp2 h(lp1 h)−1 l p1 h is isomorphic. Therefore a + b(p2 h)(lp1 h)−1 l ∈ K(E), and we are done.  Corollary 7.1.6. Let P and Q be right R-modules. If End
R (P ) and EndR (Q) are generalized stable rings, then so is EndR P ⊕ Q .

Proof. Given any right R-module M and decompositions M = A1 ⊕ H = A2 ⊕ K with A1 ∼ = P ⊕Q ∼ = A2 , then we have A1 = A11 ⊕ A12 and A2 = A21 ⊕ A22 such that A11 ∼ =P ∼ = A21 and A12 ∼ =Q∼ = A22 . In light of Theorem 7.1.5, there exist right R-modules C1 , D1 , E1 and decompositions M = C1 ⊕ D1 ⊕ A12 ⊕ H = C1 ⊕ E1 ⊕ A22 ⊕ K such that C1 ∼ = P. Similarly, we have right R-modules C2 , D2 , E2 and decompositions M = C1 ⊕ D1 ⊕ C2 ⊕ D2 ⊕ H = C1 ⊕ E1 ⊕ C2 ⊕ E2 ⊕ K such that C2 ∼ = Q. Set C = C1 ⊕ C2 , D = D1 ⊕ D2 and E = E1 ⊕ E2 . Then M = C ⊕ D ⊕ H = C ⊕ E ⊕ K where C ∼ = P ⊕ Q. Using Theorem 7.1.5 again, we establish the result. 

A ring R is a right N F -ring provided that R and every right ideal of R are finite direct sums of quasi-injective right ideals. For instance, every artinian serial ring, e.g., the group ring Z3 S3 , is a right N F -ring (cf. [370]). Corollary 7.1.7. Every right N F -ring is a generalized stable ring. Proof. Since R is a right N F -ring, there exist quasi-injective right ideals

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I1 , · · · , In such that R =

n L

i=1

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Ii . Thus, we can find idempotents ei ∈ I such

that Ii ∼ = ei R. Since ei R is quasi-injective, it follows from Corollary 7.1.3 that EndR (ei R) is a generalized stable ring. According to Corollary 7.1.6, EndR e1 R ⊕ · · · ⊕ en R) is a generalized stable ring, and then so is R.  Lemma 7.1.8. Let R be a ring. Then the following are equivalent: (1) R is a generalized stable ring. (2) There exists a complete orthogonal set of idempotents, {e1 , · · · , en }, such that all ei Rei are generalized stable rings. Proof. (1) ⇒ (2) is obvious. (2) ⇒ (1) Clearly, the result holds for n = 1. We now assume the result holds for n ≤ m where m ≥ 1. Consider n = m + 1. Set f = e2 + e3 + · · · + em+1 . Clearly, f e2 f = e2 , · · · , f em+1 f = em+1 . Hence {f e2 f, · · · , f em+1 f } is a complete orthogonal set of idempotents in f Rf . By the hypothesis, f Rf is a generalized stable ring. In addition, {e1 , f } is a complete orthogonal set of idempotents in R. Furthermore, EndR (e1 R) and EndR (f R) are generalized stable rings. According to Corollary 7.1.6, R is a generalized stable ring. By induction, we obtain the result.  Let T be the ring of a Morita context (A, B, M, N, ψ, φ). If A and B are generalized stable rings, it follows from Lemma 7.1.8 that T is a generalized stable ring. By induction, we conclude that every matrix ring Mn (R) of a generalized stable ring R is a generalized stable ring. Proposition 7.1.9. Let T be the ring of a Morita context (A, B, M, N, ψ, φ) with zero pairings. Then the following are equivalent: (1) T is a generalized stable ring. (2) A and B are generalized stable rings. 

 a n
Proof. (1) ⇒ (2) Let θ : T → A given by θ = a. As T is the ring mb of (A, B, M, N, ψ, φ) with zero pairings, it is easy to verify that θ is a ring homomorphism. Clearly, θ is surjective; hence, A ∼ = T / ker θ. Therefore A is a generalized stable ring. Likewise, B is a generalized stable ring. (2) ⇒ (1) Set e = diag(1A , 0). Then e is an idempotent of T . Clearly, eT e ∼ = diag(A, 0). Because A is a generalized stable ring, so is diag(A, 0). Thus eT e
is a generalized stable ring. Similarly, diag(1A , 1B )−
e T diag(1A , 1B ) − e is also a generalized stable ring. Applying Lemma

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7.1.8, we are through.

Corollary 7.1.10. A ring R is a generalized stable ring if and only if so is the ring of all n × n lower triangular matrices over R. Proof. According to Proposition 7.1.9, we obtain the result by induction.  Similarly, we deduce that a ring R is a generalized stable ring if and only if so is the ring  of all n ×  n upper triangular  matrices over  R. Let  R be R 0 0 0 0 0 0 0 0 a ring, let A = B =  0 R 0  , M =  0 0 0  and N =  0 0 0 , and 0 0 R 0R0 N RR0 N let ψ : N M → A, ψ(n ⊗ m) = nm and φ : M N → B, φ(m, n) = mn. B

A

Then (A, B, M, N, ψ, φ) is a Morita context with zero pairings. Let T be the ring of (A, B, M, N, ψ, φ). In light of Proposition 7.1.9, T is a generalized stable ring if and only if so is R. But T is not a ring of all triangular matrices over R. For extensions of generalized stable rings, we refer the reader to [87], [150] and [438]. Though every exchange ring satisfying related comparability is a generalized stable ring, there indeed exist generalized stable rings which do not satisfy related comparability as the following shows.

Example 7.1.11. Let V be an infinite-dimensional vector space over a division ring D, and let R=



EndD (V ) EndD (V ) 0 EndD (V )



.

Then R is a generalized stable ring, while it does not satisfy related comparability. Proof. In view of [197, Corollary], EndD (V ) is one-sided unit-regular; hence, it is an exchange ring satisfying related comparability. By virtue of Proposition 7.1.9, R is a generalized stable ring. According to Example     00 1V 0 5.2.5, R is not weakly stable. Obviously, B(R) = { , }. 00 0 1V Therefore R does not satisfy related comparability.  Recall that an element a ∈ R is said to be full in the case of RaR = R. S One easily checks that a ∈ R is full if and only if a ∈ R − {I | I E R}.

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Proposition 7.1.12. Let R be an exchange ring. If 2(eR) .⊕ eR for all full idempotents e ∈ R, then R is a generalized stable ring. Proof. Given ax + b = 1 in R, then there exists an idempotent e ∈ R such that e = bs and 1 − e = (1 − b)t for some s, t ∈ R. Thus, (1 − e)axt + e = 1. This implies that

(1 − e) (1 − e)a + e xt + e (1 − e)a + e = 1.

Let c = a + bs(1 − a). Then x1 cR + x2 cR = R for some x1 , x2 ∈ R. Since R is an exchange ring, there are orthogonal idempotents g1 ∈ x1 cR and g2 ∈ x2 cR such that g1 + g2 = 1. Set gi = xi cyi with yi = yi gi . Let ei = cyi xi . Then e2i = ei and ei R ∼ = gi R. Since R = e1 R ⊕ (1 − e1 )R = e2 R ⊕ (1 − e2 )R, we can find some right R-modules A, A′ , B and B ′ such that R = e1 R ⊕ A ⊕ A′ , A ⊕ B = e2 R and A′ ⊕ B ′ = (1 − e2 )R.

Clearly, there exists an idempotent f ∈ R such that f R = e1 R ⊕ A, and so f ∈ e1 R + e2 R. This f ∈ cR. Obviously, e1 R .⊕ f R. Further,
implies that ′ ′ e1 R ∼ = R/ A ⊕ A ∼ = B ⊕ B . Hence, e2 R .⊕ A ⊕ e1 R ∼ = f R. So we see ⊕ that gi R . f R. In view of Lemma 6.1.2, there exist a ∈ gi Rf, b ∈ f Rgi such that gi = ab ∈ Rf R. As g1 + g2
= 1, we get Rf R = R. Thus, we have an idempotent f ∈ (a + bs(1 − a) R such that Rf R = R. So there are m P si , ti (1 ≤ i ≤ m) such that si f ti = 1R . Construct a map φ : m(f R) → R i=1

given by φ(f r1 , · · · , f rm ) =

m P

i=1

si f ri for any (f r1 , · · · , f rm ) ∈ m(f R).

Clearly, 1R = ϕ(f t1 , · · · , f tm ). As R is projective, φ splits, and so we can find a right R-module D such that R ⊕ D ∼ = m(f R). This implies that ⊕ ⊕ m R . m(f R) . 2 (f R). By hypothesis, we get 2m (f R) .⊕ 2m−1 (f R) .⊕ · · · .⊕ 2(f R) .⊕ f R. Therefore, R .⊕ f R. Hence, there exist right R-morphisms ψ : R → f R, ϕ : f R → R such that ϕψ = 1R . Hence, 1R = ϕψ(1R ) = ϕ(f )f ψ(1R ), and then f ∈ K(R). So, we have some s, t ∈ R
such that sf t = 1. Moreover, we have a d ∈ R such that s a + bs(1 − a) dt = 1. Consequently, R is a generalized stable ring by Lemma 7.1.1.  Let V be an infinite-dimensional vector space over a division D, let R = EndD (V ), and let e ∈ R be a full idempotent. Then dimD (eVD ) = ∞. Let pi : eV ⊕ eV → eV be the projections and qi : eV → eV ⊕ eV be the injections for i = 1, 2. Obviously, θ : eVD ∼ = eVD ⊕ eVD . Construct a map

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ϕ : eRR → eRR ⊕eRR given by ϕ(f ) = (p1 θf, p2 θf ) for any f ∈ eR. For any (g, h) ∈ eRR ⊕ eRR , we choose f = θ−1 q1 g + θ−1 q2 h. Then (g, h) = ϕ(f ). It is easy to verify that ϕ : eRR ∼ = eRR ⊕ eRR . Therefore R is an exchange ∼ ring such that eR = 2(eR) for any full idempotent e ∈ R. Recall that a ring R is a GEn -ring if every n × n invertible matrix over R can be reduced to diagonal form by elementary row (column) operations. A ring R is a GE-ring if it is a GEn -ring for all n. Theorem 7.1.13. Every generalized stable ring R is a GE-ring, and so the natural homomorphism U (R) → K1 (R) is surjective. Proof. It suffices to show that R is a GEn -ring for all n. Let A be an arbitrary invertible n × n matrix over R. Let (a1 , a2 , · · · , an ) be the first row of A. Obviously, we have a1 x1 + a2 x2 + · · · + an xn = 1 for some x1 , x2 , · · · , xn ∈ R. Since R is a generalized stable ring, by virtue of Lemma 7.1.1, there exists some z ∈ R such that a1 + a2 x2 z + · · · + an xn zn = w ∈ K(R). Assume that swt = 1 for s, t ∈ R. Then we verify that 

 0   sw 0 0 0  ..   1 − tsw t 0  . 0 0 In−2 0 xn z · · · 1

0 0 1  0 A .  ..



s 0  1 − wts wt 0 0 In−2





∗ ∗  =.  ..

1 x2 z .. .

··· ··· .. .

1 ∗ ··· ∗ ∗ ··· .. .. . . . . .

 ∗ ∗  .. , .

∗ ∗ ∗ ··· ∗

so we show that  1 0 ··· ∗ 1 ···  . . .  .. .. . . ∗ 0 ···

  0 0   s 0 0 1 0   ..   1 − wts wt 0  A  ..  . . 0 0 In−2 0 1 

1∗ sw 0 0 0 1  1 − tsw t 0   . .  .. .. 0 0 In−2 00 



··· ··· .. . ···

1 x2 z .. . xn z

  10 ∗ 0 ∗ 0   ..  =  .. .. . . . 0∗ 1

··· ··· .. . ···

··· ··· .. . ···

 0 0  ..  . 1

 0 ∗  ..  . . ∗

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One easily checks that   s 0 = B21 (−wt)B12 (s − 1)B21 (1)B12 (wt − 1); 1 − wts wt   sw 0 = B21 (−t)B12 (sw − 1)B21 (1)B12 (t − 1). 1 − tsw t Hence, 

s 0 1 − wts wt

  ,

sw 0 1 − tsw t



∈ E2 (R).

Therefore the result follows by induction on n.



Let ϕ : U (R) → K1 (R) be the proceeding surjective. Then K1 (R) ∼ = U (R)/kerϕ. Here, we list the kerϕ under the hypotheses of several conditions. R

ker ϕ

stable range one W (R)
T W strongly stable ring W (R) V (R) (1 + J(R)) unit 1-stable range V (R)

 W medium stable ring U (R)′ 1 + er(1 − e) | e = e2 , r ∈ R Goodearl-Menal condition U (R)′ Now we provide a connection between stable range one and the generalized stable property. Corollary 7.1.14. Let R be a ring. Then the following are equivalent: (1) R has stable range one. (2) R is a generalized stable ring and perspectivity is transitive in L(2R). Proof. (1)⇒(2) is clear. (2)⇒(1) Suppose 2R = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = B1 ∼ = R ∼ = A2 ∼ B . Assume that the bases of A , B , A and B are ξ , ξ , η = 2 1 1 2 2 1 2 1 and η2 respectively. Then there exists a matrix A ∈ GL2 (R) such that (ξ1 , ξ2 ) = (η1 , η2 )A. Since R is a generalized stable ring, by virtue of Theorem 7.1.13, it is a GE2 -ring. Thus we have α, β, x1 , x2 , · · · , xn , xn+1 ∈ R such that A = [α, β]B12 (x1 )B21 (x2 ) · · · B12 (xn )B21 (xn+1 ). Hence, (ξ1 , ξ2 )B21 (−xn+1 ) = (η1 , η2 )[α, β]B12 (x1 )B21 (x2 ) · · · B12 (xn ). Clearly, 2R = L(ξ1 , ξ2 ) = L(ξ1 , ξ2 )B21 (−xn+1 ). Let C1 = L(ξ1 − ξ2 xn+1 ). Then 2R = C1 ⊕ B1 . Likewise, we have C2 , · · · , Cn ⊆ 2R such that 2R =

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C1 ⊕ B1 = C1 ⊕ C2 = C3 ⊕ C2 = C3 ⊕ C4 = · · · = Cn ⊕ C(n−1) = Cn ⊕ B2 . Therefore 2R = D ⊕ B1 = D ⊕ B2 ; whence, R has stable range one by Lemma 3.1.4.  Proposition 7.1.15. Let R be an exchange ring. Then the following are equivalent: (1) R is a generalized stable ring. (2) For any regular x ∈ R, there exist e = e2 ∈ R, w ∈ K(R) such that x = ew. Proof. (1)⇒(2) For any regular x ∈ R, there exists some y ∈ R such that x = xyx. From xy + (1 − xy) = 1, we can find some z ∈ R such that x + (1 − xy)z = w ∈ K(R). Thus x = xyx = xy(x + (1 − xy)z) = ew, where e = e2 = xy ∈ R. (2)⇒(1) Suppose ax + b = 1 in R. Since R is an exchange ring, we have e = e2 ∈ R such that e = bs, 1 − e = (1 − b)t for some s, t ∈ R. Hence axt + e = (1 − b)t + e = 1, so (1 − e)axt + e = 1. Clearly, (1 − e)a is regular. Thus we see that (1 − e)a = f w with f = f 2 ∈ R, w ∈ K(R), and then f wxt(1 − f ) + e(1 − f ) = 1 − f . Consequently, a + bs(1 − f )w =

−1 f + e(1 − f ) w = 1 + f wxt(1 − f ) w ∈ K(R), as required.  Corollary 7.1.16. Let R be an exchange ring. Then the following are equivalent:

(1) R is a generalized stable ring. (2) Given aR = bR with a, b ∈ R, then there exists some w ∈ K(R) such that a = bw. Proof. (1) ⇒ (2) is obvious by Proposition 7.1.4. (2)⇒(1) Given any regular x ∈ R, then x = xyx for a y ∈ R. We easily check that xR = eR with e = e2 = xy ∈ R. By hypothesis, there exists some w ∈ K(R) such that x = ew. In view of Proposition 7.1.15, the result follows.  Lemma 7.1.17. Let A be a quasi-projective right R-module, and let E = EndR (A). Then the following are equivalent: (1) E is a generalized stable ring. (2) Whenever αA + K = A with α ∈ E and K ⊆ A, there exists β ∈ E such that βA ⊆ K and α + β ∈ K(E). Proof. (1) ⇒ (2) Suppose that αA + K = A with α ∈ E and K ⊆ A. Let

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M = A/K and let g : A → M be the canonical map. Then gα : A → M is an R-epimorphism. Since A is a quasi-projective right R-module, there exists some φ ∈ E such that gαφ = g; hence, g(1A −αφ) = 0. By hypothesis, we have that α + (1A − αφ)k ∈ K(E) for some k ∈ E. Let β = (1A − αφ)k. Then α + β ∈ K(E) and βA ⊆ ker(g) = K. (2) ⇒ (1) Given f E + gE = E with f, g ∈ E, then f A + gA = A as in the proof of Corollary 2.1.6. So there exists h ∈ E such that f + h ∈ K(E) and hA ⊆ gA. Since A is a quasi-projective right R-module, there exists some k ∈ E such that h = gk; hence f + gk ∈ K(E). Therefore E is a generalized stable ring.  Lemma 7.1.18. Let A be a quasi-projective right R-module, and let E = EndR (A). Then the following are equivalent: (1) E is a generalized stable ring. (2) For any R-epimorphisms f, g : A → M , there exists h ∈ K(E) such that f = gh. Proof. (1) ⇒ (2) Let f, g : A → M be R-epimorphisms. Since A is a quasiprojective right R-module, there exists some α ∈ E such that f = gα. As f is an R-epimorphism, αA + ker(g) = A. By virtue of Lemma 7.1.17, there exists β ∈ E such that α + β ∈ K(E) and βA ⊆ ker(g). Let h = α + β. Then gh = g(α + β) = gα = f with h ∈ K(E). (2) ⇒ (1) Given f E + gE = E with f, g ∈ E, then f A + gA = A as in the proof of Corollary 2.1.6. Let M = A/gA, and let h : A → M be the canonical map. Then f A + ker(h) = A; hence, hf and h are both R-epimorphisms. By assumption, we can find some α ∈ K(E) such that hf = hα, and so h(f − α) = 0. Thus, im(f − α) ⊆ ker(h) = gA. Since A is a quasi-projective right R-module, we have a k ∈ E such that the diagram A k ւ↓f −α g

A ։ gA

commutates. That is, f − α = −gk, and therefore f + gk = α ∈ K(E), as required.  Lemma 7.1.19. Let A be a quasi-projective right R-module, and let E = EndR (A). Then the following are equivalent: (1) E is a generalized stable ring.

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(2) For any R-epimorphisms f, g : A → M , there exist h ∈ K(E), e = e2 ∈ E such that f = g(eh). Proof. (1) ⇒ (2) is clear by Lemma 7.1.18. (2) ⇒ (1) Given f E + gE = E with f, g ∈ E, then f A + gA = A as in the proof of Corollary 2.1.6. Let M = A/gA, and let h : A → M be the canonical map. Then f A + ker(h) = A; hence, hf and h are both Repimorphisms. By assumption, we can find e = e2 ∈ E and α ∈ K(E) such that hf = h(eα), and so h(f − eα) = 0. Thus, im(f − eα) ⊆ ker(h) = gA. Since A is a quasi-projective right R-module, we can find some k ∈ E such that f − eα = −gk, and therefore f + gk = eα. Assume now that f x + gy = 1E with x, y ∈ E. Then (f + gk)x + g(y − kx) = 1; hence, eαx + g(y − kx) = 1. Thus, eαx(1 − e) + g(y − kx)(1 − e) = 1−e. This implies that e+g(y −kx)(1−e) = 1−eαx(1−e) ∈ E is invertible. So we deduce that
f + g k + (y − kx)(1 − e)α = (f + gk) + g(y − kx)(1 − e)α = eα + g(y − kx)(1
− e)α = 1 − eαx(1 − e) α ∈ K(E). Therefore E is a generalized stable ring.  Theorem 7.1.20. Let A be a quasi-projective right R-module, and let E = EndR (A). If A has the finite exchange property, then the following are equivalent: (1) E is a generalized stable ring. (2) For any R-morphism f : A → M with im(f ) ⊆⊕ M and any R- epimorphism g : A → M , there exist e = e2 ∈ E, h ∈ K(E) such that f = g(eh). Proof. (2) ⇒ (1) For any R-epimorphisms f, g : A → M , we see that im(f ) = M is a direct summand of M . By assumption, there exist e = e2 ∈ E, h ∈ K(E) such that f = g(eh). According to Lemma 7.1.19, E is a generalized stable ring. (1) ⇒ (2) Since im(f ) is a direct summand of M , we have R-morphisms σ : im(f ) ֒→ M and τ : M → im(f ) such that τ σ = 1im(f ) . In view of Lemma 7.1.19, there exist h ∈ K(E), g = g 2 such that f = (τ g)h. Clearly, (στ )2 = στ ∈ EndR (M ). Since A is a quasi-projective right R-module, we can find some α ∈ E such that στ g = gα. Hence gα2 = (στ g)α = (στ )2 g = (στ )g = gα,

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and then g α − α2 = 0. Obviously, E is an exchange ring. In light of [383, Theorem 29.2], we can find
an idempotent e ∈ R such that e − α ∈
2 α − α R. As a result, g e − α = 0, i.e., gα = ge. This means that the diagram g

A։M e↓ ↓ στ g

A։M

commutates. Therefore g(eh) = (ge)h = (στ )gh = σ(τ gh) = σf = f , as required.  Let A be a semisimple right R-module, and let E = EndR (A). Then for any R-morphism f : A → M and any R-epimorphism g : A → M , there exist e = e2 ∈ E, h ∈ K(E) such that f = g(eh). Lemma 7.1.21. Let A be a quasi-injective right R-module. If π : M → A is an R-monomorphism, then for any idempotent f ∈ EndR (M ), there exists an idempotent g ∈ EndR (A) such that gπ = πf . Proof. Let f ∈ EndR (M ) be an idempotent. As A is quasi-injective, we have a right R-endomorphism h of A such that πf = hπ. This implies that h2 π = h(πf ) = (hπ)f = πf 2 = πf = hπ, and so (h − h2 )π = 0. Clearly, E is an exchange ring. Using [382, Theorem 29.2], we have an idempotent g ∈ E such that h−g ∈ E(h−h2 ). As a result, we deduce that (h−g)π = 0, and then gπ = hπ = πf .  Theorem 7.1.22. Let A be a quasi-injective right R-module, and let E = EndR (A). Then for any R-morphism f : M → A with ker(f ) ⊆⊕ M and any R-monomorphism g : M → A, there exist h ∈ K(E), e = e2 ∈ E such that f = (he)g. Proof. As ker(f ) ⊆⊕ M , the exact sequence 0 → ker(f ) → M → im(f ) → 0 splits. Hence, we have an R-monomorphism σ : im(f ) → M such that f σ = 1im(f ) . Let i : im(f ) ֒→ A be the inclusion. Then i, gσ : im(f ) → A are R-monomorphisms. As A is a quasi-injective right R-module, we have some T α ∈ E such that i = αgσ. As i is an R-monomorphism, ker(α) im(g)σ = 0. Let h : im(g)σ → A be the inclusion map. Then αh : im(g)σ → A is an R-monomorphism. Thus, we have some φ ∈ E such that φαh = h;

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hence, (1A − φα)h = 0. Since E is a generalized stable exchange ring, it follows from φα + (1A − φα) = 1A that α + k(1A − φα) ∈ K(E) for some k ∈
E. Let β = k(1A −
φα). Then α + β ∈ K(E). Moreover, β im(g)σ = k(1A − φα) im(g)σ = 0. Let h = α + β. Then i = h(gσ). As (σf )2 = σf ∈ EndR (M ), it follows by Lemma 7.1.21 that there exists e = e2 ∈ E such that eg = g(σf ). As a result, we deduce that (he)g = h(eg) = h(gσf ) = (hgσ)f = if = f , as required.  Let A be a semisimple right R-module, and let E = EndR (A). Then for any R-morphism f : M → A and any R-monomorphism g : M → A, there exist h ∈ K(E), e = e2 ∈ E such that f = (he)g. Lemma 7.1.23. Let R be a generalized stable ring. Then ax + b = 1 in R implies that there exists some z ∈ R such that x + zb is the product of a right invertible element and a left invertible element. Proof. Assume that A = (aij ) ∈ GL2 (R) with a12 ∈ K(R). Then we have s, t ∈ R such that sa12 t = 1. So       s 0 sa12 0 ∗1 A = . 1 − a12 ts a12 t 1 − tsa12 t ∗∗

Clearly, we see that    −1 s 0 a12 t 1 − a12 ts ∈ GL2 (R), = 1 − a12 ts a12 t 0 s    −1 sa12 0 t 1 − tsa12 ∈ GL2 (R). = 1 − tsa12 t 0 sa12

Thus we show that  −1   −1 s 0 ∗1 sa12 0 . A= 1 − a12 ts a12 t ∗∗ 1 − txa12 t

Therefore

A−1 =



sa12 0 1 − tsa12 t



∗1 ∗∗

−1 

s 0 1 − a12 ts a12 t



Clearly, we can find some u ∈ U (R) such that        −1 10 ∗1 10 01 0 u−1 . = = ∗1 ∗∗ ∗1 u0 1 0

.

Consequently, we claim that   −1      10 ∗ u−1 ∗1 0 u−1 10 . = = ∗ ∗ ∗∗ ∗1 1 0 ∗1

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Hence, we have −1

A

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   sa12 0 ∗ u−1 s 0 = . 1 − tsa12 t ∗ ∗ 1 − a12 ts a12 t   ∗ sa12 u−1 a12 t = . ∗ ∗

Set w = sa12 u−1 a12 t. As sa12 t = 1, we see that sa12 is right invertible and u−1 a12 t is left invertible. That is, w is the product of a right invertible element and a left invertible element of R. Assume that B = (bij ) ∈ GL2 (R). Set B −1 = (cij ) ∈ GL2 (R). Then c12 R + c11 R = R. Since R is a generalized stable ring, we can find y ∈ R such that c12 + c11 y ∈ K(R). So     1y ∗ c12 + c11 y −1 B = . 01 ∗ ∗ By the above consideration, we can find the product w1 ∈ R of a right invertible element and a left invertible element of R such that      −1 1 −y 1 y
−1 ∗ c12 + c11 y B = B −1 = 0 1 01 ∗ ∗   ∗ w1 = . ∗ ∗ Given ax + b = 1 in R, then we easily check that    −1 −1 x xa − 1 x ∈ GL2 (R). = a b a 1 Therefore we can find z ∈ R such that      1z −1 x ∗ w2 = , 01 a b ∗ ∗ where w2 is the product of a right invertible element and a left invertible element of R. Consequently, we conclude that x + zb = w2 is the product of a right invertible element and a left invertible element of R.  Theorem 7.1.24. Let R be a generalized stable ring. Then every regular element in R is the product of a right invertible element, a left invertible element and an idempotent of R. Proof. Given any regular x ∈ R, there exists y ∈ R such that x = xyx. Since yx + (1 − yx) = 1, by virtue of Lemma 7.1.23, we can find a z ∈ R such that x + z(1 − yx) = uv with right invertible u ∈ R and left invertible

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v ∈ R. Thus x = xyx = x + z(1 − yx) yx = uve with e = e2 ∈ R, as asserted.  A right self-injective ring R is a right q-ring provided that every right ideal of R is quasi-injective. Clearly, every right q-ring is a N F -ring, but the converse is not true, e.g., M2 (Z4 ) (cf. [370, Example 3.5]). Corollary 7.1.25. Every q-ring is an exchange ring having stable range one. Proof. Let R be q-ring. Then R is a right self-injective ring. In view of [214, Theorem 1], R is an exchange ring. In light of Corollary 7.1.7, R is a generalized stable ring. It follows from [370, Theorem 3.7] that R is directly finite. Therefore R has stable range one by Theorem 7.1.24.  7.2

Diagonal Reduction, I

The m × n matrix A is said to admit diagonal reduction provided that there exist P ∈ GLm (R), Q ∈ GLn (R) such that P AQ is a diagonal matrix. Many authors have studied diagonal reduction such as [15], [349], [351] and [426]. In this section, we investigate diagonal reduction of regular matrices over generalized stable exchange rings. We say that two homomorphisms f1 , f2 : N → M are equivalent if f2 = uf1 v for some u ∈ Aut(M ) and v ∈ Aut(N ). A homomorphism f : N → M is regular if there exists some g : M → N such that f = f gf . Lemma 7.2.1. Let f1 , f2 : N → M be regular. Then f1 and f2 are equivalent if and only if ker(f1 ) ∼ = ker(f2 ), im(f1 ) ∼ = im(f2 ) and coker(f1 ) ∼ = coker(f2 ). Proof. Assume that f1 and f2 are equivalent. Then f2 = uf1 v for some u ∈ Aut(M ) and v ∈ Aut(N ). Construct R-morphisms ϕ : ker(f2 ) → ker(f1 ) given by ϕ(x) = v(x) for any x ∈ ker(f2 ) and φ : im(f1 ) → im(f2 ) given by φ(x) = u(x) for any x ∈ im(f1 ). One easily checks that ϕ and φ are R-isomorphisms. Further, there exists an R-morphism ψ : coker(f1 ) →

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coker(f2 ) such that the following diagram 0 → im(f1 ) ֒→ M → coker(f1 ) → 0 ↓φ ↓u ↓ψ 0 → im(f2 ) ֒→ M → coker(f2 ) → 0

commutates. As φ and u are both isomorphisms, so is ψ. That is, ψ : coker(f1 ) ∼ = coker(f2 ). Conversely, assume that ker(f1 ) ∼ = ker(f2 ), im(f1 ) ∼ = im(f2 ) and ∼ coker(f1 ) = coker(f2 ). Since f1 and f2 are both regular, there are g1 , g2 : M → N such that f1 = f1 g1 f1 and f2 = f2 g2 f2 . By hypothesis, we have ϕ : f1 g1 M = im(f1 ) ∼ = im(f2 ) = f2 g2 M and ψ : (1M − f1 g1 )M ∼ = ∼ coker(f1 ) = coker(f2 ) = (1M − f2 g2 )M . Let u : M → M given by u(x + y) = ϕ(x) + ψ(y) for any x ∈ f1 g1 M and y ∈ (1M − f1 g1 )M . Then u ∈ Aut(M ). Clearly, ϕi : gi fi N ∼ = fi gi M given by ϕi (gi fi x) = fi gi fi x for −1 any x ∈ N . Then, we get σ := ϕ−1 ϕ2 : g2 f2 N ∼ = g1 f1 N . By hypoth1 ϕ esis, τ : (1M − g2 f2 )N = ker(f2 ) ∼ ker(f ) = (1 − g1 f1 )N . Construct a = 1 M map v : N → N given by v(x + y) = σ(x) + τ (y) for any x ∈ g2 f2 N and y ∈ (1N − g2 f2 )N . Then v ∈ Aut(N ). One easily verifies that the diagram f1

N = g1 f1 N ⊕ (1N − g1 f1 )N −→ f1 g1 M ⊕ (1M − f1 g1 )M = M ↑v ↓u f2

N = g2 f2 N ⊕ (1N − g2 f2 )N −→ f2 g2 M ⊕ (1M − f2 g2 )M = M

commutates. Hence, f2 = uf1 v, and therefore f1 and f2 are equivalent, as asserted.  Now we identify the set Mm×n (R) of all m × n matrices over R with HomR (nR, mR) in the standard manner. The following elementary result is due to Ara et al. [15]. Proposition 7.2.2. Let R be an exchange ring, and let f ∈ Mm×n (R) be regular. (1) Suppose that n ≥ m. Then f admits a diagonal reduction if and only if there are decompositions ker(f ) = K1 ⊕· · ·⊕Kn , im(f ) = I1 ⊕· · ·⊕Im and coker(f ) = C1 ⊕· · ·⊕Cm such that Kj ⊕ Ij ∼ = Cj ⊕ Ij ∼ = R, j = 1, · · · , m and Kj ∼ = R, j =

m + 1, · · · , n. (2) Suppose that n ≤ m. Then f admits a diagonal reduction if and only if there are decompositions ker(f ) = K1 ⊕· · ·⊕Kn , im(f ) = I1 ⊕· · ·⊕In and coker(f ) = C1 ⊕· · ·⊕Cm

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such that Kj ⊕ Ij ∼ = Cj ⊕ Ij ∼ = R, j = 1, · · · , n and Cj ∼ = R, j = n + 1, · · · , m. Proof. (1) Assume that f admits a diagonal u ∈ AutR (mR), v ∈ AutR (nR) such that  r1 0 · · · 0 0 · · ·  0 r2 · · · 0 0 · · ·  uf v =  . . . . .  .. .. . . .. .. . . .

reduction. Then there exist  0 0  ..  .

0 0 · · · rm 0 · · · 0

. m×n

As f is regular, so is each ri . Thus, there are some si such that ri = ri si ri . We see that     x1 x1  .    .   ..  .  r1 0 · · · 0 0 · · · 0     .     0 r2 · · · 0 0 · · · 0     xm     xm  ker(uf v) = { | . . .   = 0}. . . . .  xm+1   .. .. . . .. .. . . ..   xm+1      .   ..  0 0 · · · rm 0 · · · 0   .   ..  xn xn m L

Thus, ker(uf v) =

(1−si ri )R

i=1

we have that 

  im(uf v) = {  Thus, im(uf v) =

r1 0 .. .

0 r2 .. .

··· ··· .. .

0 0 .. .


L

0 ··· 0 ··· .. . . . .

0 0 · · · rm 0 · · ·

m L

n L

i=m+1


Ri , where each Ri ∼ = R. Also

 x1  .  .  0   .    0   xm   |r1 , · · · , rm ∈ R}. ..    x .   m+1  .  0  ..   xn 

ri R. Furthermore, we get

i=1

coker(uf v) = mR/im(uf v) =

m M

R/ri R.

i=1

Obviously, each (1 − si ri )R ⊕ ri R ∼ = (1 − si ri )R ⊕ si ri R = R ∼ = ri R ⊕ R/ri R. According to Lemma 7.2.1, ker(f ) = K1 ⊕ · · · ⊕ Kn , im(f ) = I1 ⊕ · · · ⊕ Im and coker(f ) = C1 ⊕ · · · ⊕ Cm

such that Kj ⊕Ij ∼ = Cj ⊕Ij ∼ = R, j = 1, · · · , m and Kj ∼ = R, j = m+1, · · · , n.

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Conversely, assume that there are decompositions ker(f ) = K1 ⊕ · · · ⊕ Kn , im(f ) = I1 ⊕ · · · ⊕ Im and coker(f ) = C1 ⊕ · · · ⊕ Cm such that Kj ⊕Ij ∼ = Cj ⊕Ij ∼ = R, j = 1, · · · , m and Kj ∼ = R, j = m+1, · · · , n. Then we have decompositions Kj′ ⊕ Ij′ = Jj′ ⊕ Cj′ = R, Ij′ ∼ = Kj and Cj′ ∼ = Cj = Jj′ ∼ = Ij , Kj′ ∼ for j = 1, · · · , m. So, there are idempotents ej , fj ∈ R such that Kj′ = (1 − ej )R, Ij′ = ej R, Jj′ = fj R and Cj′ = (1 − fj )R. As ej R ∼ = fj R, there are some sj ∈ and fj = rj sj . Let  r1 0 · · ·  0 r2 · · ·  ϕ= . . .  .. .. . .

ej Rfj , rj ∈ fj Rej such that ej = sj rj 0 0 .. .

0 0 .. .

··· ··· .. .

 0 0  ..  .

0 0 · · · rm 0 · · · 0

Then we check that m
L L ker(ϕ) = (1 − si ri )R i=1

im(ϕ) =

m L

i=1

This implies that

n L

i=m+1

. m×n


Ri , where each Ri ∼ = R;

ri R and coker(ϕ) =

m L

R/ri R.

i=1

ker(ϕ) ∼ = ker(f ); im(ϕ) ∼ = im(f ) and coker(ϕ) ∼ = coker(f ). Therefore f and ϕ are equivalent by Lemma 7.2.1, as required. (2) is proved in the same manner.



We say that the decompositions nR ∼ = K ⊕ I and mR ∼ = I ⊕ C with n ≥ m have a diagonal refinement provided that there are decompositions K = K1 ⊕ · · · ⊕ Kn , I = I1 ⊕ · · · ⊕ Im and C = C1 ⊕ · · · ⊕ Cm such that Kj ⊕Ij ∼ = Cj ⊕Ij ∼ = R, j = 1, · · · , m and Kj ∼ = R, j = m+1, · · · , n. Lemma 7.2.3. Let R be an exchange ring. Consider decompositions nR ∼ = K ⊕ I and mR ∼ = I ⊕ C with n ≥ m. Suppose that K ∼ = K∗ ⊕ X ∗ ∗ ∼ ∼ ∼ and C = C ⊕ X. If nR = K ⊕ (X ⊕ I) and mR = (X ⊕ I) ⊕ C ∗ have a diagonal refinement, then so have nR = K ⊕ I and mR ∼ = I ⊕ C.

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Proof. By hypothesis, there are decompositions ∗ ∗ K ∗ = K1∗ ⊕ · · · ⊕ Kn∗ , X ⊕ I ∼ and C ∗ = C1∗ ⊕ · · · ⊕ Cm = I1∗ ⊕ · · · ⊕ Im

with Kj∗ ⊕ Ij∗ ∼ = Ij∗ ⊕ Cj∗ ∼ = R, j = 1, · · · , m and Kj∗ ∼ = R, j = m + 1, · · · , n. Further, we get I = I1 ⊕ · · · ⊕ Im and X = X1 ⊕ · · · ⊕ Xm

such that Ij ⊕ Xj ∼ = Ij∗ for all j ≤ m. Thus,

∗ ∗ K∼ ⊕ Xm ) ⊕ Km+1 ⊕ · · · ⊕ Kn∗ , = (K1∗ ⊕ X1 ) ⊕ · · · ⊕ (Km ∗ ∗ C∼ ⊕ Xm ) ⊕ Cm+1 ⊕ · · · ⊕ Cn∗ . = (C1∗ ⊕ X1 ) ⊕ · · · ⊕ (Cm

It is easy to verify that (Kj∗ ⊕Xj )⊕Ij ∼ = Kj∗ ⊕IJ∗ ∼ = Ij∗ ⊕Cj∗ ∼ = Ij ⊕(Cj∗ ⊕Xj ) for all j ≤ m. Therefore the proof is true.  Lemma 7.2.4. Let R be a generalized stable ring, and let n ∈ N. Then nR ⊕ M ∼ = nR ⊕ N implies that R ⊕ M ∼ = R ⊕ N for any right R-modules M and N . Proof. Suppose that ψ : nR⊕M ∼ = nR⊕N . Set K = A1 ⊕M = A2 ⊕ψ −1 (N ) ∼ ∼ with A1 = nR = A2 . Then we have decompositions A1 = A11 ⊕ A12 ⊕ · · · ⊕ A1n , A2 = A21 ⊕ A22 ⊕ · · · ⊕ A2n with all Aij ∼ = R. By virtue of Theorem 7.1.5, we can find C , D , E ⊆ K such that K = C1 ⊕D1 ⊕ A12 ⊕· · ·⊕A1n ⊕ 1 1 1

M = C1 ⊕ E1 ⊕ A22 ⊕ · · · ⊕ A2n ⊕ ψ −1 (N ) with C1 ∼ = R. Clearly, we have ∼ ∼ ∼ ∼ C1 ⊕ D1 = R ⊕ D1 = A11 = R. Thus, D1 ⊕ A12 = D1 ⊕ R ∼ = R. Similarly, we see that E1 ⊕ A22 ∼ = R. Likewise, we can find C2 , · · · , Cn , F, G ⊆ K such that K = C1 ⊕ C2 ⊕ · · · Cn ⊕ F ⊕ M = C1 ⊕ C2 ⊕ · · · Cn ⊕ G ⊕ ψ −1 (N ) with C1 ∼ = C2 ∼ = · · · Cn ∼ = R. In addition, R ⊕ F ∼ = R ∼ = R ⊕ G. Thus −1 ∼ R ⊕ M = R ⊕ ψ (N ), as desired.  Recall that (a1 , · · · , an ) is unimodular in R provided that a1 R + · · · + an R = R. Theorem 7.2.5. Let R be a generalized stable exchange ring, and let A ∈ Mn (R) be regular. If A has a unimodular row, then A admits a diagonal reduction. Proof. Let f ∈ EndR (nR) be a regular R-morphism corresponding to A, and let K = ker(f ), I = im(f ) and C = coker(f ). Then nR ∼ = K ⊕I ∼ = ∼ ∼ I ⊕ C. Since nR ⊕ C = (K ⊕ I) ⊕ C = nR ⊕ K, by Lemma 7.2.4, we derive that R ⊕ C ∼ = R ⊕ K. In view of [16, Proposition 1.2], we have R = R1 ⊕ R2 and C = C1 ⊕ C2 such that R1 ⊕ C1 ∼ = R and R2 ⊕ C2 ∼ = K. Therefore

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nR ∼ = R2 ⊕ (I ⊕ C2 ) ∼ = (I ⊕ C2 ) ⊕ C1 (∗). By virtue of Lemma 7.2.3 and Proposition 7.2.2, it suffices to show that the decompositions in (∗) have a diagonal refinement. Clearly, nR ⊕ R1 ∼ = K ⊕ I ⊕ R1 ∼ = R2 ⊕ C2 ⊕ I ⊕ R1 ∼ = R ⊕ I ⊕ C2 . By n P hypothesis, we may assume that aij R = R for some i. Since j=1



 r1   I = im(f ) = {A  ...  | r1 , · · · , rn ∈ R}, rn

we see that R .⊕ I. Hence, there exists a right R-module D such that I ⊕ C2 ∼ = R ⊕ D; hence, nR ⊕ R1 ∼ = 2R ⊕ D. In view of Lemma 7.2.4, (n − 1)R ⊕ R1 ∼ =R⊕D ∼ = I ⊕ C2 . Consequently, R2 ∼ = R2 ⊕ 0 ⊕ · · · ⊕ 0, I ⊕ C2 ∼ = R1 ⊕ R ⊕ · · · ⊕ R, C1 ∼ = C1 ⊕ 0 ⊕ · · · ⊕ 0 with R2 ⊕ R1 ∼ = C1 ⊕ R1 ∼ = R. Therefore the result follows.



Immediately, we may deduce that every square matrix over a division ring admits a diagonal reduction. Corollary 7.2.6. Let R be a generalized stable exchange ring, and let A ∈ Mn (R)(n ≥ 2) be regular. If A has an invertible entry, then A is the sum of two invertible matrices. Proof. As A has an invertible entry, it has a unimodular row. By virtue of Theorem 7.2.5, there exist some U, V ∈ GLn (R) such that U AV = diag(r1 , · · · , rn ) for r1 , · · · , rm ∈ R. Observing that   0 −1 · · · r1 1 · · · 00   0 0 ···  0 r2 · · · 0 0     .. ..  +  .. .. . . diag(r1 , · · · , rn ) =  ... ... . . .  . .   . . .   0 0 · · · rn−1 1   0 0 · · · −1 0 · · · 1 0 ··· 00 

we obtain the result.

 0 0  ..  , .  0 −1  0 0 .. .

0 rn



Recall that a right R-module G is a generator in the case where G generates every right R-module. A right R-module G is an R-progenerator in the case where it is a finitely generated projective generator.

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Proposition 7.2.7. Let R be a right self-injective ring, and let A be a finitely generated projective right R-module. If B and C are Rprogenerators such that A ⊕ B ∼ = A ⊕ C, then B ∼ = C. Proof. Since A is a finitely generated projective right R-module, there exists some n ∈ N such that nR ⊕ B ∼ = nR ⊕ C. As B and C are R-progenerators, we can find some k, l ∈ N such that R ⊕ D ∼ = kB, R ⊕ E ∼ = lC. Choose ′ ∼ ′ ∼ m = kl. Then R ⊕ D = mB, R ⊕ E = mC for some right R-modules D′ and E ′ . Thus, (n + 1)R ⊕ D′ ∼ = (n + 1)R ⊕ E ′ . By Corollary 7.1.3, R is a generalized stable ring. In view of Lemma 7.2.4, we get R ⊕ D′ ∼ = R ⊕ E′. That is, mB ∼ = mC. As B is a projective right R-module, we have some p ∈ N such that B ⊕ B ′ ∼ = pR. Thus, B is injective. Likewise, C is injective. According to [214, Theorem 3], B ∼  = C, as asserted. Lemma 7.2.8. Let R be an exchange ring, and let f ∈ Mn (R) be regular. Then diag(f, f ) admits diagonal reduction if and only if there exist decompositions 2ker(f ) ∼ = K1 ⊕ · · · ⊕ K2n , 2im(f ) ∼ = I1 ⊕ · · · ⊕ I2n and ∼ ∼ ∼ 2coker(f ) = C1 ⊕ · · · ⊕ C2n with all Ki ⊕ Ii = Ii = Ci ∼ = R.

Proof. It is easy to verify that ker(diag(f, f )) ∼ = 2ker(f ), im(diag(f, f )) ∼ = ∼ 2im(f ) and coker(diag(f, f )) ∼ 2nR/2im(f ) 2coker(f ). Therefore the = = result follows from Proposition 7.2.2. 

Theorem 7.2.9. Let R be a generalized stable exchange ring. Then for any regular A ∈ Mn (R), diag(A, A) admits a diagonal reduction.

Proof. Suppose nR ∼ = K ⊕I ∼ = I ⊕ C. By virtue of [16, Proposition 1.2], we have K = X1 ⊕ X2 , I = Y1 ⊕ Y2 such that X1 ⊕ Y1 ∼ = I and X2 ⊕ Y2 ∼ = C. ∼ ∼ Thus nR = X1 ⊕ (I ⊕ X2 ) = (I ⊕ X2 ) ⊕ Y2 . Obviously, X1 .⊕ I .⊕ I ⊕ X2 ; hence, nR .⊕ 2(I ⊕ X2 ). Thus there is no loss of generality in assuming that nR .⊕ 2I. Since nR ⊕ C ∼ = nR ⊕ K, by virtue of Lemma 7.2.4, R ⊕ C ∼ = R ⊕ K. In view of [16, Proposition 1.2], we have decompositions R = R1 ⊕ R2 , C = C1 ⊕C2 such that R1 ⊕C1 ∼ = R and R2 ⊕C2 ∼ = K. Thus nR ∼ = R2 ⊕(I⊕C2 ) ∼ = ∼ ∼ (I⊕C2 )⊕C1 . Obviously, R2 ⊕R1 = C1 ⊕R1 = R. Without loss of generality, we may assume that K ⊕ W ∼ =C ⊕W ∼ = R. We now have that nR ⊕ W ∼ = K ⊕I ⊕W ∼ = I ⊕ R, and so 2nR ⊕ 2W ∼ = 2R ⊕ 2I. Assume that nR ⊕ D ∼ = 2I for a right R-module D. Then 2nR ⊕ 2W ∼ = (n + 2)R ⊕ D. By Lemma 7.2.4 again, 2I ∼ = nR ⊕ D ∼ = 2(n − 1)R ⊕ 2W . Therefore, 2K ∼ = K ⊕K ⊕0⊕· · ·⊕0, 2I ∼ = W ⊕W ⊕R⊕· · ·⊕R, 2C ∼ = C ⊕C ⊕0⊕· · ·⊕0

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with K ⊕ W ∼ = W ⊕C ∼ = R. According to Proposition 7.2.2, diag(A, A) admits diagonal reduction.  Following Warfield [394], a ring R satisfies the n-stable range condition, whenever (a1 , · · · , an , an+1 ) is unimodular in R, there exist b1 , · · · , bn ∈ R such that (a1 + an+1 b1 , · · · , an + an+1 bn ) is unimodular in R. Clearly, R satisfies the 1-stable range condition if and only if it has stable range one. We note that Z satisfies the 2-stable range condition and EndD (V ) does not satisfy any finite stable range condition, where V is an infinite-dimensional vector space over a division ring D. Proposition 7.2.10. Let R be a generalized stable exchange ring. If R satisfies the m-stable range condition, then every regular A ∈ Mn (R)(n ≥ m + 1) admits diagonal reduction. Proof. Given any decompositions nR ∼ = K ⊕I ∼ = I⊕C. As in Theorem 7.2.9, ∼ we may assume that K ⊕ W = C ⊕ W ∼ = R. Furthermore, nR ⊕ W ∼ = I ⊕ R. As R satisfies the m-fold stable range condition, we get (n−1)R⊕W ∼ = I(cf. Corollary 12.1.7). Therefore, K∼ = K ⊕ 0 ⊕ · · · ⊕ 0, I ∼ = W ⊕ R ⊕ · · · ⊕ R, C ∼ = C ⊕ 0 ⊕ ···⊕ 0 with K ⊕ W ∼ =W ⊕C ∼ = R, as required.



So far one cannot find any exchange ring with finite stable range more than 2. Thus the following immediate consequence of Proposition 7.2.10 is interesting. Let R be a generalized stable exchange ring. If R satisfies the 2-stable range condition, then every A ∈ Mn (R)(n ≥ 3) admits a diagonal reduction. Proposition 7.2.11. Let R be an exchange ring, and let f ∈ Mm×n (R) be regular. (1) nR ⊕ coker(f ) ∼ = mR ⊕ ker(f ). (2) If n > m, then f admits a diagonal reduction if and only if ker(f ) ∼ = (n − m)R ⊕ coker(f ). (3) If n < m, then f admits a diagonal reduction if and only if coker(f ) ∼ = (m − n)R ⊕ ker(f ).

Proof. (1) Let K = ker(f ), I = im(f ) and C = coker(f ). Then nR ∼ = K ⊕I and mR ∼ = I ⊕ C. Therefore, nR ⊕ C ∼ = K ⊕I ⊕C ∼ = mR ⊕ K, as required. (2) Assume that n > m and f admits a diagonal reduction. By Propo-

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sition 7.2.2, we have that K = K1 ⊕ · · · ⊕ Kn , I = I1 ⊕ · · · ⊕ Im , C = C1 ⊕ · · · ⊕ Cm

such that Kj ⊕Ij ∼ = Cj ⊕Ij ∼ = R, j = 1, · · · , m and Kj ∼ = R, j = m+1, · · · , n. ∼ For j ≤ m, we have Kj ⊕ R ∼ K ⊕ I ⊕ C C ⊕ R; hence, = j j j = j m M j=1

Kj ⊕ R ∼ =

m M j=1

Cj ⊕ R.

Therefore, we get m L K∼ Kj ⊕ (n − m)R = j=1

m L ∼ Cj ⊕ (n − m)R =

∼ =

j=1

C ⊕ (n − m)R.

∼ (n − m)R ⊕ C. Then nR ∼ Conversely, assume that K = = (n − m)R ⊕ (I ⊕ C) and mR ∼ (I ⊕ C) ⊕ 0 (∗). By virtue of Lemma 7.2.3, it suffices to = show that the decompositions in (∗) have a diagonal refinement. Clearly, (n − m)R ∼ = 0 ⊕ · · · ⊕ 0 ⊕ R ⊕ R · · · ⊕ R, I ⊕C ∼ = R ⊕ · · · ⊕ R, 0 = 0 ⊕ · · · ⊕ 0. Therefore f admits a diagonal reduction. (3) is proved in the same manner.



Theorem 7.2.12. Let R be a generalized stable exchange ring. If R satisfies the 2-stable range condition, then every regular A ∈ Mm×n (R) (|m − n| 6= 0, 1) admits a diagonal reduction. Proof. Let f ∈ Mm×n (R) be regular. By Proposition 7.2.11, we have nR ⊕ coker(f ) ∼ = mR ⊕ ker(f ). If n > m, then R ⊕ ker(f ) ∼ = (n − m + 1)R ⊕ coker(f ) by Lemma 7.2.4. Since |n − m| ≥ 2, we see that ker(f ) ∼ = (n − m)R ⊕ coker(f ) (cf. Corollary 12.1.7). If n < m, we analogously get R ⊕ coker(f ) ∼ = (m − n + 1)R ⊕ ker(f ); hence, coker(f ) ∼ = (m − n)R ⊕ ker(f ). According to Proposition 7.2.11, f admits a diagonal reduction.  7.3

Diagonal Reduction, II

We say that eRe is a corner of a ring R where e ∈ R is an idempotent. The following result was proved by Ara et al. [15, Theorem 2.5].

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Lemma 7.3.1. Let R be a separative exchange ring. Then every square regular matrix over R admits a diagonal reduction. Proof. Suppose nR ∼ = K ⊕I ∼ = I ⊕ C. By virtue of [16, Proposition 1.2], we have K = X1 ⊕ X2 , I = Y1 ⊕ Y2 such that X1 ⊕ Y1 ∼ = I and X2 ⊕ Y2 ∼ = C. ∼ ∼ Thus nR = X1 ⊕ (I ⊕ X2 ) = (I ⊕ X2 ) ⊕ Y2 . Since R is separative, it follows from nR ⊕ Y2 ∼ = nR ⊕ X1 that R ⊕ Y2 ∼ = R ⊕ X1 . By [16, Proposition 1.2] again, R = R1 ⊕R2 , Y2 = C1 ⊕C2 such that R1 ⊕C1 ∼ = R and R2 ⊕C2 ∼ = X1 . ∼ ∼ Thus nR = R2 ⊕ (I ⊕ X2 ⊕ C2 ) = (I ⊕ X2 ⊕ C2 ) ⊕ C1 (∗). By Lemma 7.2.3 and Proposition 7.2.2, it suffice to show that the decompositions in (∗) have a diagonal refinement. One easily checks that R ⊕ (n − 1)R ⊕ R1 ∼ = R ⊕ I ⊕ X2 ⊕ C2 . As R .⊕ nR ∼ = I ⊕ X2 ⊕ Y2 .⊕ 2I ⊕ X2 .⊕ 2(I ⊕ X2 ⊕ C2 ), we get (n − 1)R ⊕ R1 ∼ = I ⊕ X2 ⊕ C2 . Consequently, we have that R2 ∼ = R2 ⊕ 0 ⊕ · · · ⊕ 0, I ⊕ X2 ⊕ C2 ∼ = R1 ⊕ R ⊕ · · · ⊕ R, C1 ∼ = C1 ⊕ 0 ⊕ · · · ⊕ 0

with R2 ⊕ R1 ∼ = C1 ⊕ R1 ∼ = R. Therefore the proof is true.



Theorem 7.3.2. Let R be an exchange ring whose corners are all generalized stable rings. Then every square regular matrix over R admits a diagonal reduction. Proof. Assume that A ⊕ C ∼ = B ⊕ C .⊕ R with C .⊕ A and C .⊕ B. ′ Then A ∼ = C ⊕ A and B ∼ = C ⊕ B ′ for some right R-modules A′ , B ′ . So we see that 2C ⊕ A′ ∼ = 2C ⊕ B ′ . Set M = C1 ⊕ C2 ⊕ A′ = C1′ ⊕ C2′ ⊕ D with ′ ∼ ∼ Ci = C = Ci and D ∼ = B ′ . Since C is isomorphic to a direct summand of R, we can find some idempotent e ∈ R such that EndR (C1 ) ∼ = eRe. Thus, EndR (C1 ) is a generalized stable ring. In view of Theorem 7.1.5, there exist E, F, G ⊆ M such that M = E ⊕ F ⊕ C2 ⊕ A′ = E ⊕ G ⊕ C2′ ⊕ D with E ∼ = C. Obviously, we check that E ⊕ F ∼ = C1 ∼ =C ∼ = C1′ ∼ = E ⊕ G, and ∼ ∼ ∼ ∼ ∼ ∼ then F ⊕ C2 = C ⊕ F = E ⊕ F = C = E ⊕ G = C ⊕ G = G ⊕ C2′ . Therefore C ⊕ A′ ∼ = C ⊕D ∼ = C ⊕ B ′ . Consequently, we derive that A ∼ = B, hence R is a separative ring from [16, Corollary 2.9]. By virtue of Lemma 7.3.1, we obtain the result.  Corollary 7.3.3. Let R be an exchange ring. If R .⊕ eR for any nonzero idempotent e ∈ R, then every square regular matrix over R admits a diagonal reduction. Proof. Let e ∈ R be an idempotent, and let ef e ∈ eRe be a nonzero idempotent. Then ef e ∈ R is an idempotent. By hypothesis, R .⊕ ef eR.

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In view of Lemma 6.1.2, there exist a ∈ Ref e, b ∈ ef eR such that 1 = ab. Thus, ea ∈ e(eRe)(ef e) and be ∈ (ef e)(eRe)e. Further, e = (ea)(be). By Lemma 6.1.2 again, eRe .⊕ (ef e)(eRe). As in the proof of Proposition 7.1.12, eRe is a generalized stable ring. According to Theorem 7.3.2, the result follows.  Theorem 7.3.2 shows that every exchange ring whose corners are generalized stable rings is separative. In the case where R is simple, we can derive the following element-wise characterization of separativity. Proposition 7.3.4. Let R be a simple exchange ring. Then the following are equivalent: (1) R is a separative ring. (2) R is weakly stable. (3) All corners of R are generalized stable rings. Proof. (1) ⇒ (2) is obvious by Corollary 5.2.10. (2) ⇒ (3) Since R is weakly stable, so is every corner of R by Lemma 5.1.5. Thus, all corners of R are generalized stable rings. (3) ⇒ (1) is obtained from Theorem 7.3.2.  Theorem 7.3.5. Let R be an exchange ring. Then the following are equivalent: (1) R is a generalized stable ring. (2) For any regular x ∈ R, there exists u ∈ K(R) such that x = xux. Proof. (1) ⇒ (2) Given any regular x ∈ R, there exists some y ∈ R such that x = xyx. Clearly, xR = (xy)R. Corollary 7.1.16 gives u ∈ K(R) such that xu = xy, and therefore x = xyx = xux. (2) ⇒ (1) Suppose that ax + b = 1 in R. Since R is an exchange ring, by virtue of Lemma 1.4.7, there exists an idempotent e ∈ R such that e = bc and 1 − e = (1 − b)d for some c, d ∈ R. Then axd = 1 − e. Hence, (1 − e)axd(1 − e)a = (1 − e)a, i.e., (1 − e)a ∈ R is regular. By assumption, there exists u ∈ K(R) such that (1 − e)a = (1 − e)au(1 − e)a. Assume that sut = 1 for some s, t ∈ R. Then
su (1 − e)a + (1 − (1 − e)au)ts ut
= su(1 − e)aut + su 1 − (1 − e)au tsut = sut = 1,

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and so w ∈ K(R), where w = (1 − e)a + 1 − (1 − e)au ts. In addition, (1 − e)a = (1 − e)au(1 − e)a = (1 − e)auw. Let f = (1 − e)au, so that f w = (1−e)a. It is easy to verify that (1−e)axd+e = 1; hence, f wxd+e = 1. It follows that f wxd(1 − f ) + e(1 − f ) = 1 − f . This implies that f + e(1 − f ) = 1 − f wxd(1 − f ) ∈ R is invertible, and so
a + bc (1 − f )w − a
= a + e (1 − f )w − a = (1 − e)a + e(1 − f )w = f w + e(1 − f )w
= 1 − f wxd(1 − f ) w ∈ K(R). Therefore R is a generalized stable ring.



Let R be an exchange ring. Then R is a generalized stable ring if and only if whenever ax + b = 1 in R, there exists some z ∈ R such that x + zb ∈ K(R). By Theorem 7.3.5, we see that generalized stable exchange rings are left-right symmetric. Thus we are done from Lemma 7.1.1. Corollary 7.3.6. Let R be an exchange ring. Then the following are equivalent: (1) R is a generalized stable ring. (2) For any idempotents e, f ∈ R, ϕ : eR ∼ = f R implies that there exists an element u ∈ K(R) such that ϕ(e) = ue = f u. (3) Whenever x = xyx, there exists u ∈ K(R) such that x = xyu = uyx.

Proof. (1) ⇒ (2) Let e, f ∈ R be idempotents such that ϕ : eR ∼ = f R. Then there exists r ∈ R such that f = ϕ(er) = ϕ(e)(erf ). One easily checks that (erf )ϕ(e) = ϕ−1 (f )ϕ(e) = e. Let a = erf and b = ϕ(e). Then e = ab and f = ba. Since ba + (1 − ba) = 1, we can find some y ∈
R such that v := b + (1 − ba)y ∈ K(R). Hence, b = ba b + (1 − ba)y = f v. Let u = (1 − f − va)v(1 − e − av). It is easy to verify that (1 − f − va)2 = 1 = (1 − e − av)2 . This implies that u ∈ K(R). Further, f u = −f vav(1 − e − av) = −b(1 − e − av) = bav = b. Also we have ue = −(1 − f − va)vave = −(1 − f − va)ve = −ve + b + vave = b. Therefore b = ue = f u, as required. (2) ⇒ (3) Whenever x = xyx, we have an R-isomorphism ϕ : yxR ∼ = xyR given by ϕ(yxr) = xyxr for any R ∈ R. By hypothesis, there exists some u ∈ K(R) such that ϕ(yx) = uyx = xyu. Therefore, x = xyu = uyx, as asserted.

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(3) ⇒ (1) Let x ∈ R be regular. Then x = xyx for a y ∈ R. Let z = yxy. Then z = zxz and x = xzx. So, we have a v ∈ K(R) such that z = zxv = vxz. This implies
that z + (1 − zx)v(1 − xz) = v. Thus, x = xzx = x z + (1 − zx)v(1 − xz) x = xvx. In view of Theorem 7.3.5, R is a generalized stable ring.  Theorem 7.3.7. Let R be a regular ring. Then the following are equivalent: (1) R is a generalized stable ring. (2) For any x ∈ R, there exists some w ∈ K(R) such that wx is an idempotent of R. (3) For any x ∈ R, there exists some w ∈ K(R) and a group G in R such that wx ∈ G. Proof. (1) ⇒ (3) Given any x ∈ R, it follows from Theorem 7.3.5 that there exists some w ∈ K(R) such that x = xwx. Hence, wx ∈ R is an idempotent, as required. (3) ⇒ (2) For any x ∈ R, there exists a w ∈ K(R) and a group G in R such that wx ∈ G. Thus, the element wx ∈ R has a group
inverse (wx)# ∈ R. One easily checks that wx (wx)# + 1 − (wx)# wx wx =
−1 # wx and (wx)# + 1 − (wx) wx = wx + 1 − (wx)# wx ∈
U (R). So,
(wx)# + 1 − (wx)# wx w ∈ K(R) and (wx)# + 1 − (wx)# wx wx ∈ R is an idempotent. (2) ⇒ (1) Given ax+ b = 1 in R, we have w ∈ K(R) such that wa = e = 2 e ∈ R. Thus, ex + wb = w. Since R is regular, there is an element c ∈ R such that (1−e)wb = (1−e)wbc(1−e)wb. Hence ex(1+wb)+gwb = w with g = (1 − e)wbc(1 − e). It is easy to verify that ex(1 + wb) = ew, gwb = gw. So
(e + g)w = e + (1 − e)wbc(1 − e) w
= e(1 − ewbc(1 − e)) + wbc(1 − e)
w
= w a + bc(1 − e)(1 + ewbc(1 − e) 1 − ewbc(1 − e) w = w. Assume that swt =
1 for some s, t ∈ R. Then sw a + bc(1 − e)(1 +
ewbc(1 −
e)) 1 − ewbc(1 − e) wt = 1, whence a + bc(1 − e) 1 + ewbc(1 − e) ∈ K(R), as asserted.  Corollary 7.3.8. Let R be a regular ring. Then the following are equivalent: (1) R is a generalized stable ring.

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(2) For any x ∈ R, there exists some w ∈ K(R) such that x ∈ wG for some group G in R. Proof. (1) ⇒ (2) Given any x ∈ R, there exists some y ∈ R such that x = xyx. Since yx+ (1 − yx) = 1, we have a z ∈ R such that x+ z(1 − yx) = w ∈ K(R). Hence
x = xyx = x + z(1 − yx) yx = w(yx) ∈ wG

for a group G in R. (2) ⇒ (1) For any x ∈ R, we have some y ∈ R such that x = xyx and y = yxy. So there is a w ∈ K(R) such that y ∈ wG for a group G in R. According to Lemma 6.2.4, there exists some g ∈ R such that y = wg, where g ∈ R has a group inverse. From xy + (1 − xy) = 1, we see that xwg + (1 − xy) = 1. As in the proof of Corollary 6.2.6, g ∈ R is unit-regular. Assume that g = ue with u ∈ U (R) and e = e2 ∈ R. Then xwue + (1 − xy) = 1, so e + (1 − e)(1 − xy) = 1 − (1 − e)xwue. Set z = wu(1−e). Then wg+z(1−xy) = wu(1+(1−e)xwue)−1 ∈ K(R). Therefore
−1
wu 1 + (1 − e)xwue x = wg + z(1 − xy) x = yx is an idempotent of R. By virtue of Theorem 7.3.7, R is a generalized stable ring.  Lemma 7.3.9. For any regular a, b ∈ R, if ϕ : aR ∼ = bR, then Ra = Rϕ(a) and ϕ(a)R = bR. Proof. Since ϕ : aR ∼ = bR, we have ϕ(a) ∈ bR. So ϕ(a)R ⊆ bR. On the other hand, there exists r ∈ R such that b = ϕ(ar) = ϕ(a)r ∈ ϕ(a)R; hence, bR ⊆ ϕ(a)R. Therefore ϕ(a)R = bR. Since b ∈ R is regular, we have an idempotent e ∈ R such that bR = eR. Hence ϕ(a)R = eR. This implies that ϕ(a) ∈ R is regular.
So we can find some c ∈ R such that ϕ(a) = ϕ(a)cϕ(a) = ϕ acϕ(a) . It follows that a = acϕ(a) ∈ Rϕ(a). Hence Ra ⊆ Rϕ(a). One the other hand, we have a = ada for some d ∈ R. This implies that ϕ(a) = ϕ(a)da ∈ Ra; hence, Rϕ(a) ⊆ Ra. Therefore we conclude that Ra = Rϕ(a), as asserted.  Theorem 7.3.10. Let R be a generalized stable exchange ring. Then for
any regular A ∈ Mn (R), there exist U, V ∈ K Mn (R) such that U AV is a diagonal matrix with idempotent entries. Proof. Given any regular A ∈ Mn (R), we have E = E 2 ∈ Mn (R) such that AMn (R) = EMn (R). Clearly, E(nR) is a finitely generated projective right R-module. Since R is an exchange ring, by [409, Theorem 2.1], there exist idempotents e1 , · · · , en ∈ R such that E(nR) ∼ =

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e1 R ⊕ · · · ⊕ en R ∼ so we have = diag(e1 , · · · , en )Rn as rightR-modules,  x1  .  ERn×1 ∼ = diag(e1 , · · · , en ) Rn×1 , where Rn×1 = { ..  | x1 , · · · , xn ∈ R}

xn is an Mn (R)-R-bimodule. Let R1×n = {(x1 , · · · , xn ) | x1 , · · · , xn N R}. Then R1×n is an R-Mn (R)-bimodule; hence, ERn×1 R1×n R N N diag(e1 , · · · , en )Rn×1 R1×n . It is easy to verify that Rn×1 R1×n R

∈ ∼ =

∼ =

Mn (R) as right Mn (R)-modules. Hence, we get ψ : AMn (R) ∼ = diag(e1 , · · · , en )Mn (R). In view of Lemma 7.3.9, Mn (R)A = Mn (R)ψ(A) and ψ(A)Mn (R) = diag(e1 , · · · , en )Mn (R). Since R is a generalized stable ring, so is Mn (R). In view of Theorem 7.3.5, Mn (R)op is a generalized
stable ring. According to Corollary 7.1.16, there exist U, V ∈ K Mn (R) such that U A = ψ(A) and ψ(A)V = diag(e1 , · · · , en ). Therefore U AV = diag(e1 , · · · , en ), as asserted.  Corollary 7.3.11. Let R be a purely infinite simple
ring. Then for any regular A ∈ Mn (R), there exist U, V ∈ K Mn (R) such that U AV is a diagonal matrix with idempotent entries. Proof. As in the proof of Theorem 5.3.5, R is a weakly stable exchange ring. Hence, it is a generalized stable ring. Therefore we complete the proof by Theorem 7.3.10.  Corollary 7.3.12. Let R be an exchange ring having stable range one. Then for any regular A ∈ Mn (R), there exist U, V ∈ GLn (R) such that U AV is a diagonal matrix with idempotent entries. Proof. Given any regular A
∈ Mn (R), it follows by Theorem 7.3.10 that there exist U, V ∈ K Mn (R) such that U AV = diag(e1 , · · · , en ) for some idempotents e1 , · · · , en ∈ R. Since R has stable range one, so is Mn (R) by Corollary 1.1.6. Hence, Mn (R) is directly finite. This implies that U, V ∈ GLn (R), as required.  Let R be a weakly stable exchange ring. By virtue of Theorem 5.2.9, R is a separative exchange ring. Using Lemma 7.3.1, we know that any regular square matrix over R admits a diagonal reduction. But we note that one can not reduce every square matrix over R to a diagonal matrix with idempotent entries. Let V be an infinite-dimensional vector space over a division ring D, and let R = EndD (V ). Let {x1 , x2 , · · · , xn , · · · }

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be a basis of V . Define σ : V → V given by σ(xi ) = xi+1 (i = 1, 2, · · · ) and τ : V → V given by τ (x1 ) =0 and  τ (xi ) = xi−1 (i = 2, 3, · · · ). Then τ 0 τ σ = 1V and στ 6= 1V . Let A = . Then A ∈ M2 (R) is regular. By 0τ Lemma 7.3.1, we can find U, V ∈ GL2 (R) such that U AV = diag(r1 , r2 ). Assume now that both ri ∈ R are idempotents. Set E = diag(r1 , r2 ). Then A = U −1 EV −1 , whence AV U A = A. That is, A ∈ GL2 (R); hence, τ ∈ AutD (V ), a contradiction. This shows that there exists some square matrix over R which does not admit a diagonal reduction with idempotent entries.

7.4

Invertible Matrices

An invertible matrix β ∈ Mn (R) is simple if there exists a column matrix   x1
 ..   .  and a row matrix y1 , · · · , yn such that xn

 x1
  β = In +  ...  y1 , · · · , yn . 

xn

Obviously, β ∈ GLn (R) if and only if 1 +

n P

i=1

yi xi ∈ U (R). For example,

reflections, involutions, transvections, axial affinities and hyperreflections are all simple. It is instructive to represent an invertible matrix as a product of simple matrices. Djokovi´c and Malzan studied the factorization of an invertible matrix by reflections and a simple matrix over division rings. Ellers and Ishibashi studied the factorization by simple matrices over commutative local rings and then over noncommutative local rings (cf. [198]). Lemma 7.4.1. Let R be a ring, and let n ≥ 2, x ∈ R, u ∈ U (R). Then 

10 x 1

In−2

are simple matrices.

 

1x ,0 1

In−2



,



In−1 u



n×n

∈ GLn (R)

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Proof. Since   0     10 x  x 1  = In +  0   .  ..  In−2 0



1, 0, 0, · · · , 0




,

n×1



10 x 1

1×n



1x  ∈ GLn (R) is a simple matrix. Likewise,  0 1

In−2 GLn (R) is a simple matrix. It is easy to verify that 

In−1 u



n×n

In−2



 ∈



 0  .. 
  = In +  .  0, · · · , 0 1 n×n ,  0  u − 1 n×n

and therefore the proof is true.



Lemma 7.4.2. Let R be a ring, and let a, b ∈ R, n ≥ 2. If ab = 1, then −1    a 0 a 0  are the products of  ∈ GLn (R) and  1 − ba b  1 − ba b In−2 In−2 simple matrices. Proof. One easily checks that       a 0 1 0 1a 0 −1 = . 1 − ba b −b 1 01 1 b Hence



a 0 1 − ba b





∈ GL2 (R). In addition, we have

−1 a 0 1 − ba b     b 1 1 −a 10 = −1 0 0 1 b1      1 1−b 1 0 1 1−a 10 = . 0 1 −1 1 0 1 b1

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Thus, 

a 0  1 − ba b

In−2

−1 

The result follows.



1 1−b = 0 1 

1 1−a 0 1

In−2

In−2



1 0   −1 1 

10 b 1

 

In−2 

In−2

.



Theorem 7.4.3. Let R be a generalized stable ring. Then every A ∈ GLn (R) (n ≥ 2) is the product of simple matrices. Proof. Let A = (aij )n×n ∈ GLn (R). Then we can find a B = (bij )n×n ∈ Mn (R) such that AB = In . Since R is a generalized stable ring, it follows from a11 b11 + a12 b21 + · · · + a1n bn1 = 1 that there exist some y2 , · · · , yn ∈ R such that a11 + a12 b21 y2 + · · · + a1n bn1 yn = u ∈ K(R). Thus, we get     1 u a12 a13 · · · a1n  b21 y2 1   a′21 a22 a23 · · · a2n      A . . . = . . . . ,  .. .. . .   .. .. .. . . . ..  a′n1 an2 an3 · · · ann bn1 yn 0 · · · 1 n P where each a′i1 = ai1 + aij bj1 yj (2 ≤ i ≤ n). As u ∈ K(R), we can find j=2

some s, t ∈ R such that sut = 1. So  u a12 a13   s 0  a′21 a22 a23   1 − uts ut  . . .  .. .. .. In−2 a′n1 an2 an3   1 c12 c13 · · · c1n  c21 c22 c23 · · · c2n      =  c31 c32 a33 · · · a3n  ,   . . . . .  .. .. .. . . ..  cn1 cn2 an3 · · · ann

 · · · a1n   t 1 − tsu · · · a2n    0 su . . ..  . .  In−2 · · · ann

where c12 = su(1 − tsu) + sa12 su,
c1i = sa1i (3 ≤ i ≤ n); ′
′ c21 = (1 − uts)u + uta
21 t, c22 = (1 − uts)u + uta21 (1 − tsu) + (1 − uts)a12 + uta22 su, c2i = (1 − uts)a1i + uta2i (3 ≤ i ≤ n); ci1 = a′i1 t(3 ≤ i ≤ n), ci2 = a′i1 (1 − tsu) + ai2 su(3 ≤ i ≤ n).

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Furthermore, we have     1 u a12 · · · a1n   s 0  −c21 1   a′21 a22 · · · a2n       . . .  1 − uts ut  . . . .   .. .. . .   .. .. . . ..  In−2 −cn1 0 · · · 1 a′ a · · · ann   n1 n2  1 0 ··· 0   1 −c12 · · · −c1n t 1 − tsu  0 1 · · · 0   0 d22 · · · d2n      0 su  . . . = . . . ...   ... ... . . . ...   .. .. In−2 0 0 ··· 1 0 dn2 · · · dnn

One easily checks that

d22 = c22 − c21 c12 , d2i = c2i − c21 c1i (3 ≤ i ≤ n); di2 = ci2 − ci1 c12 (3 ≤ i ≤ n); dij = aij − ci1 c1j (3 ≤ i, j ≤ n). −1  s 0  In view of Lemma 7.4.2,  1 − uts ut is the product of four In−2  −1   su 0 t 1 − tsu  is  =  1 − tsu t simple matrices. Likewise,  0 su In−2 In−2 the product of four simple matrices. Thus, we can find simple matrices ∆j1 , Λj1 ∈ GLn (R)(1 ≤ j ≤ 5) such that   u a12 a13 · · · a1n   Y 5 5  a′21 a22 a23 · · · a2n  Y
1
  ∆j1 Λj1 , =  . . . . . . . . . . D  . . . . .  j=1

j=1

a′n1 an2 an3 · · · ann

where D ∈ GLn−1 (R). As a result, we have some simple matrix Λ61 ∈ GLn (R) such that   Y 5 6 Y
1
A= ∆j1 Λj1 . D j=1

j=1

Similarly, we can find simple matrices ∆j2 , Λj2 , Λ62 ∈ GLn (R)(1 ≤ j ≤ 5) such that     Y 5 6 Y
I2
1 ∆j2 Λj2 , = D E j=1

j=1

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where E ∈ GLn−2 (R). By iteration of this process, we have some simple matrices ∆ji , Λji , Λ6i ∈ GLn (R)(3 ≤ i ≤ n − 1, 1 ≤ j ≤ 5) such that   n−1 n−1 5 6 YY YY
In−1
A= ∆ji Λji , u i=1 j=1

i=1 j=1

where u ∈ U (R). In view of Lemma 7.4.1, A is the product of simple matrices.  Let R be the collection of ℵ0 × ℵ0 matrices with entries from a field F , the form   A  q    { | A ∈ Mn (F ) for some n ∈ N, q ∈ F }.  q   .. . ℵ0 ×ℵ0

Then R is a regular ring satisfying the comparability axiom; hence, it is a generalized stable ring. In view of Theorem 7.4.3, every n × n(n ≥ 2) invertible matrix over R can be represented as the product of finitely many simple matrices. Corollary 7.4.4. Let R be a purely infinite simple ring. Then every A ∈ GLn (R)(n ≥ 2) is the product of simple matrices. Proof. As in the proof of Theorem 5.3.5, R is weakly stable; hence, it is a generalized stable ring. Therefore the result follows from Theorem 7.4.3. Example 7.4.5. Let V be an infinite-dimensional vector space over a division ring D, and let R = EndD (V ). Let {x1 , x2 , · · · , xn , · · · } be a basis of V . Define σ : V → V given by σ(xi ) = xi+1 (i = 1, 2, · · · ) andτ : V → V τ 0 given by τ (x1 ) = 0 and τ (xi ) = xi−1 (i = 2, 3, · · · ). Let A = . −IV σ   σ στ − IV Then A−1 = . Since R is a generalized stable ring, it follows IV τ by Theorem 7.4.3 that A is the finitely many product of simple matrices. In fact, we have the following factorization of simple matrices:      IV 0 IV 0 0 IV IV IV − σ A= σ IV 0 2στ − IV I 0 0 IV     V   IV 0 IV IV IV −τ IV 0 0 IV , −IV IV 0 IV 0 IV σ 1 IV 0

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where 2στ − IV = (2στ − IV )−1 ∈ AutD (V ).



A matrix A ∈ Mn (R) is cyclic if there exists a column α such that (α, Aα, · · · , An−1 α) ∈ GLn (R). A companion matrix is the n by n matrix over R with ones along the line just below and parallel to the main diagonal, the elements c1 , · · · , cn ∈ R as the entries of the last column, and zeros elsewhere. It is instructive to represent an invertible matrix as the product of some cyclic matrices. Now we investigate such a problem for generalized stable rings. Let ∆n = diag(1, · · · , 1, u)n×n for some u ∈ U (R). An upper or lower triangular matrix is called unit if all its diagonal entries are 1. The following lemmas are due to [387]. Lemma 7.4.6. Let R be a ring, and let A ∈ Mn (R). Then the following are equivalent: (1) A is cyclic. (2) A is similar to a companion matrix. Proof. (1) ⇒ (2) Let F = nR be a free right R-module, and let {e1 , · · · , en } be the standard basis of F . Define σ : F → F given by σ(e1 , · · · , en ) = (e1 , · · · , en )A. Since A is cyclic, there exists a column α such that (α, Aα, · · · , An−1 α) ∈ GLn (R). Thus, {α, Aα, · · · , An−1 α} is a basis of F . So there are some c1 , · · · , cn ∈ R such that An α = αc1 + Aαc2 + · · · + An−1 αcn . Clearly, (α, Aα, · · · , An−1 α) = (e1 , · · · , en )(α, Aα, · · · , An−1 α). Thus,
σ(α, Aα, · · · , An−1 α) = σ(e1 , · · · , en ) (α, Aα, · · · , An−1 α) = (e1 , · · · , en )A(α, Aα, · · · , An−1 α) = (Aα, A2 α, · · · , An−1 α, An α)   0 0 · · · 0 c1  1 0 · · · 0 c2      n−1 = (α, Aα, · · · , A α)  0 1 · · · 0 c3  . . . . . .   .. .. . . .. ..  n−1

Let γ = (α, Aα, · · · , A

α) ∈ GLn (R). Then   0 0 · · · 0 c1  1 0 · · · 0 c2      −1 γ Aγ =  0 1 · · · 0 c3  , . . . . .   .. .. . . .. ..  0 0 · · · 1 cn

0 0 · · · 1 cn

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as required. (2) ⇒ (1) Since A is similar to a companion matrix, there exists some γ ∈ GLn (R) such that   0 0 · · · 0 c1  1 0 · · · 0 c2      γ −1 Aγ =  0 1 · · · 0 c3  . . . . . .   .. .. . . .. ..  0 0 · · · 1 cn   1 0   Choose a column η =  . . Let α = γη. Then  .. 

0         1 0 0 0 0 1 0 0                 α = γ  0  , Aα = γ  0  , A2 α = γ  1  , · · · , An−1 α = γ  0  . . . . .  ..   ..   ..   ..  0

0

2

0

n−1

Therefore (α, Aα, A α, · · · , A

1

α) = γ ∈ GLn (R), as desired.



Lemma 7.4.7. Let K be a unit upper triangular matrix in GLn (R) and L be a unit lower triangular matrix in GLn (R). Then there exists a unit upper triangular matrix K ′ and a lower unit triangular matrix L′ such that KL∆n = ∆n K ′ L′ . Proof. Let



1 a12 0 1  K =. .  .. .. 0 0

  1 · · · a1n  a21 · · · a2n    ,L =  . . . ..    .. . . an1 ··· 1

0 1 .. . an2

··· ··· .. . ···

 0 0  ..  , . 1

and let ∆n = diag(1, · · · , 1, u) ∈ GLn (R). One easily checks that KL∆n = ∆n K ′ L′ , where     1 0 ··· 0 1 a12 · · · a1n u   0 1 · · · a2n u  1 ··· 0   ′  a21  . K′ =  . . . ,L =  . . . . . ..  .. ..   .. .. . . ..  . . u−1 an1 u−1 an2 · · · 1 0 0 ··· 1

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Therefore the result follows.

Lemma 7.4.8. Let K ∈ Mn (R) be an upper triangular matrix with diagonal entries g1 , · · · , gn ∈ U (R), and let Γ ∈ Mn (R) be a companion matrix. Then KΓ is cyclic. Proof. As KΓ is similar to ΓK. It will suffice to show that ΓK is similar to a companion matrix from Lemma 7.4.6. Choose δ = diag(gn gn−1 · · · g1 , gn gn−1 · · · g2 , · · · , gn ). Then

A := δΓKδ −1



 ∗∗ ∗ ∗  ∗ ∗ .  .. ..   . . 0 0 · · · 1 ∗ n×n

∗ 1   = 0 .  ..

∗ ··· ∗ ··· 1 ··· .. . . . .

  1 0   Choose α =  .  . Then (α, Aα, A2 α, · · · , An−1 α) ∈ GLn (R). Hence, A  .. 

0 is cyclic. By virtue of Lemma 7.4.6, A is similar to a companion matrix. Therefore ΓK is similar to a companion matrix, as required.  Lemma 7.4.9. Let K ∈ Mn (R) be a unit upper triangular matrix and L ∈ Mn (R) be a unit lower triangular matrix. Then ∆n KL is the product of two cyclic matrices. Proof. Let

Γn =





In−1

 01 1 0  .. ..  . . .   1 ··· 0 0 1 0 · · · 0 0 n×n

0   0 1  and Ξn =  ...  0

0 ··· 0 ··· .. . . . .

Then ∆n KL = (∆n KΓn )(Γ−1 n L). Since ∆n K is an upper triangular matrix with diagonal entries in U (R). In view of Lemma 7.4.8, ∆n KΓn is a cyclic matrix. It will suffice to prove that Γ−1 n L is a cyclic matrix.

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Clearly,

Γ−1 n

=



In−1 1

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231

 . Further, 

 ∗∗ ∗ ∗  .. ..  . . .  0 ··· ∗ ∗ 0 0 · · · 1 ∗ n×n

∗ 1  . Ξn (Γ−1 n L)Ξn =  ..  0

∗ ··· ∗ ··· .. . . . .

As in the proof of Lemma 7.4.8, there exists
a unit upper triangular matrix Θ ∈ GLn (R) such that Θ−1 Ξn (Γ−1 L)Ξ n Θ is a companion matrix. Since
−1 n −1 −1 −1 Θ Ξn = Θ Ξn = Ξn Θ , it follows by Lemma 7.4.6 that Γ−1 n L is a cyclic matrix, and therefore the result follows.  Theorem 7.4.10. Let R be a generalized stable ring, and let A ∈ GLn (R)(n ≥ 2). Then A is the product of cyclic matrices. Proof. Let A = (aij )n×n ∈ GLn (R). Then we can find some B = (bij ) ∈ Mn (R) such that AB = In . So there exist some y2 , · · · , yn ∈ R such that a11 + a12 b21 y2 + · · · + a1n bn1 yn = u ∈ K(R). Thus, we get     u a12 a13 · · · a1n 1   a′21 a22 a23 · · · a2n   b21 y2 1     A . . . = . . . . .  .. .. . .   .. .. .. . . . ..  a′n1 an2 an3 · · · ann bn1 yn 0 · · · 1 As in the proof of Theorem 7.4.3, we see that     u a12 · · · a1n 1   s 0  a′21 a22 · · · a2n    −c21 1       . . .  1 − uts ut  . . . .   .. .. . . ..   .. .. . .  In−2 a′ a · · · ann −cn1 0 · · · 1   n1 n2  1 0 ··· 0   1 −c12 · · · −c1n t 1 − tsu  0 1 · · · 0   0 d22 · · · d2n       0 su . = . . . . . ...   ... ... . . . ...   .. .. In−2 0 dn2 · · · dnn 0 0 ··· 1 By virtue of Lemma 7.4.2,  −1 s 0  1 − uts ut  = B12 (∗)B21 (∗)B12 (∗)B21 (∗), In−2

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where Bij (x) = In + xeij , eij is the matrix units (1 in the i, j position  −1 t 1 − tsu  and 0 elsewhere, i 6= j). Likewise, we have  0 su = In−2   su 0  1 − tsu t  = B21 (∗)B12 (∗)B21 (∗)B12 (∗). Thus, we see that In−2 



1 ∗  A = B12 (∗)B21 (∗)B12 (∗)B21 (∗)  .  .. 

1∗ 0 1  . .  .. .. 00

  1  1 .. . .  D . .  ∗ 0 ··· 1    ∗ 1 ∗ 1  0     B21 (∗)B12 (∗)B21 (∗)B12 (∗)  . . . , . . .   0 . . .  1 ∗ 0 ··· 1

··· ··· .. . ···

where D ∈ GLn−1 (R). As a result, we have unit upper triangular matrices K11 , K21 , U11 , U21 , U31 ∈ GLn (R) and unit lower triangular matrices L11 , L21 , V11 , V21 , V31 ∈ GLn (R) such that A=

2 Y

Kj1 Lj1

j=1






1 D

 Y 3

j=1


Uj1 Vj1 .

Likewise, we have a unit upper triangular matrices K12 , K22 , U12 , U22 , U32 ∈ GLn (R) and unit lower triangular matrices L12 , L22 , V12 , V22 , V32 ∈ GLn (R) such that 

1 D



=

2 Y

Kj2 Lj2

j=1






I2 E

 Y 3

j=1


Uj2 Vj2 ,

where E ∈ GLn−2 (R). By iteration of this process, we have some unit upper triangular matrices K1i , K2i , U1i , U2i , U3i ∈ GLn (R) and unit lower triangular matrices L1i , L2i , V1i , V2i , V3i ∈ GLn (R)(3 ≤ i ≤ n − 1) such that A=

n−1 2 YY

i=1 j=1

Kji Lji






In−1 u

 n−1 3 YY

i=1 j=1


Uji Vji ,

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where u ∈ U (R). Let ∆n = A = ∆n

n−1 2 Q Q



i=1 j=1

=

′ ∆n K11 L′11

In−1

′ Kji L′ji

233

 . In view of Lemma 7.4.7, we have

u 3
n−1 Q Q

2
n−1 Q Q

Uji Vji

i=1 j=1

i=2 j=1

′ Kji L′ji




3
n−1 Q Q

i=1 j=1


Uji Vji ,

′ where Kji and L′ji (1 ≤ i ≤ n − 1, 1 ≤ j ≤ 2) are unit upper and lower triangular matrices, respectively. According to Lemma 7.4.9, the result follows. 

Corollary 7.4.11. Let R be an exchange ring, and let A ∈ GLn (R)(n ≥ 2). If R satisfies related comparability, then A is the product of cyclic matrices. Proof. Clearly, R is a generalized stable ring. According to Theorem 7.4.10, we obtain the result.  Let R be a purely infinite simple ring, and let A ∈ GLn (R)(n ≥ 2). It follows from Theorem 7.4.10 that A is the product of cyclic matrices. For stable range one, we now derive a more explicit result. Proposition 7.4.12. Let R be a ring, and let A ∈ GLn (R)(n ≥ 2). If R has stable range one, then A is the product of two cyclic matrices. Proof. In view of Lemma 1.2.1, there exists some U ∈ GLn (R) such that     1 u1 ∗ · · · ∗  ∗ 1  u2 · · · ∗      , U AU −1 =    . . . . . . . ..   .. .. . .   un n×n ∗ ∗ · · · 1 n×n   1 where u1 , · · · , un ∈ U (R). Let Γn = . Then In−1     u1 ∗ · · · ∗ 1   u2 · · · ∗  ∗ 1    −1  Γ Γ . U AU −1 =    . . ..  n n  .. .. . .    . . . . . 

u1 ∗  u2  In view of Lemma 7.4.8,  

un

··· ··· .. .



∗ ∗ ··· 1

n×n

∗ ∗   ..  Γn is cyclic. As in the proof of .  un

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

 1 ∗ 1    Lemma 7.4.9, Γ−1 is cyclic. Therefore the result follows.  n  .. .. . . . . .  ∗ ∗ · · · 1 n×n 

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Chapter 8

QB-Rings

Ara et al. (2000) discovered a new class of rings, the QB-rings, an infinite analogue of rings having stable range one. A ring R is a QB-ring provided that aR + bR = R implies that there exists y ∈ R such that a + by ∈ Rq−1 . Here, Rq−1 is the set of all quasi-invertible elements in R, i.e., Rq−1 = {u ∈ R| ∃ a, b ∈ R such that (1 − ua)⊥(1 − bu)}. The class of QB-rings is very large. For example, all exchange rings satisfying related comparability are QB-rings. Let F be a field, and let B(F ) denote the algebra of all row- and column-finite matrices over F . Then B(F ) is a QB-ring (cf. [16, Example 8.8]). In this chapter, we investigate necessary and sufficient conditions on an exchange ring R under which it is a QB-ring. It is shown that an exchange ring R is a QB-ring if and only if for any regular x ∈ R, there exists some u ∈ Rq−1 such that ux ∈ R is an idempotent, if and only if for all idempotents e, f ∈ R, eR ∼ = f R implies that there exists some u ∈ Rq−1 such that eu = uf . As an application, we prove that every regular square matrix over exchange QB-rings admits a diagonal reduction by some quasi-inverse matrices.

8.1

Matrices and Pseudo-Similarity

Lemma 8.1.1. Let R be a ring, and let u ∈ R. Then the following are equivalent: 235

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(1) u ∈ Rq−1 . (2) There exists v ∈ R such that (1 − uv)⊥(1 − vu), u = uvu and v = vuv. Proof. (2) ⇒ (1) is trivial. (1) ⇒ (2) If u ∈ Rq−1 , then there are a, b ∈ R such that (1−ua)⊥(1−bu). Hence, (1 − ua)u(1 − bu) = 0. Let w = a + b − aub. Let v = wuw. Then u = uvu and v = vuv. Further, 1 − uv = (1 − ua)(1 − ub) and 1 − vu = (1 − au)(1 − bu). Thus, (1 − uv)⊥(1 − vu), as required.  Let p, q ∈ R be idempotents such that pRq 6= 0. Let (pRq)−1 = {u ∈ q pRq | ∃ v ∈ qRp such that (p − uv) ⊥ (q − vu)}. We say that pRq is a QB-corner in the case of (1) ax + b = p with a ∈ pRq, x ∈ qRp and b ∈ pRp implies that there exists y ∈ pRq such that a + by ∈ (pRq)−1 and (2) q ax + b = q with a ∈ qRp, x ∈ pRq and b ∈ qRq implies that there exists y ∈ qRp such that a + by ∈ (qRp)−1 q . Lemma 8.1.2. If p and q are idempotents of a QB-ring R such that (1 − p)R ∼ = (1 − q)R, then either pRq = 0 or pRq is a QB-corner.

Proof. Since (1 − p)R ∼ = (1 − q)R, there exist u ∈ (1 − p)R(1 − q), v ∈ (1 − q)R(1 − p) such that 1 − p = uv and 1 − q = vu. Assume that pRq 6= 0. Given ax + b = p with a ∈ pRq, x ∈ qRp and b ∈ pRp, then −1 (a + u)(x + v) + b = 1. So we can find
a y ∈ R such that a
+ u + by ∈ Rq . By Lemma 8.1.1, 1 −
(a + u + by)c ⊥ 1 − c(a
+ u + by) for some c ∈ R. Hence
p − (a + by)cp ⊥ q − (qcp)(a + b(pyq)) and (ucp) ⊥ q − (qcp)(a +


b(pyq)) . As p − a + b(pyq) (qcp) = p − (a + by)(cp) + (a + by)v (ucp),

we get p − (a + b(pyq))(qcp) ⊥ q − (qcp)(a + b(pyq)) . This implies that a + b(pyq) ∈ (pRq)−1  q , as required. As an immediate consequence of Lemma 8.1.2, we derive that every corner of a QB-ring is a QB-ring. Lemma 8.1.3. Let R be a QB-ring. If (1 − ux) ⊥ (1 − xu), (1 − vy) ⊥ (1 − yv), u = uxu and v = vyv with x, y, u, v ∈ R, then either (1 − ux)R(1 − yv) = 0 or (1 − ux)R(1 − yv) is a QB-corner.

Proof. Let I = R(1 − xu)R + R(1 − vy)R. Then (ux)(R/I) ∼ = (xu)(R/I) ∼ = R/I ∼ = (yv)(R/I). In view of Lemma 8.1.2, we have either (1) = (vy)(R/I) ∼ (1 − ux)(R/I)(1 − yv) = 0 or (2) (1 − ux)(R/I)(1 − yv) is a QB-corner. If (1) holds, then (1 − ux)R(1 − yv) ⊆ R(1 − xu)R + R(1 − vy)R. Thus, (1 − ux)R(1 − yv) ⊆ (1 − ux)R(1 − xu)R + R(1 − vy)R(1 − yv). This

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implies that (1 − ux)R(1 − yv) = 0. Assume now that (2) holds. Given ab + c = 1 − ux with a ∈ (1 − ux)R(1 − yv), b ∈ (1 − yv)R(1 − ux) and c ∈ (1 − ux)R(1 − ux), there exists y ∈ (1 − ux)R(1 − yv) such that
−1 a + cy ∈ (1 − ux)(R/I)(1 − yv) q . So we can find some w ∈ R such that


(1 − ux) − (a + cy)w R (1 − yv) − w(a + cy) , (1 − yv) − w(a + cy) R (1 −
ux) − (a + cy)w ⊆ I. This implies that

(1 − ux) − (a + cy)w R (1 − yv) − w(a + cy) = 0 and



(1 − yv) − w(a + cy) R (1 − ux) − (a + cy)w = 0.
−1 Consequently, we conclude that a+cy ∈ (1−ux)R((1−yv) q , as required.  We say that a unimodular row (a, b) is reducible if there exists an element y ∈ R such that a + by ∈ Rq−1 . Lemma 8.1.4. Let a, b, c ∈ R, u, v ∈ U (R). Then (a, b) is a reducible unimodular row if and only if so is (vau + vbc, vb). Proof. Assume that (a, b) is a reducible unimodular row. Then there exist some x, y ∈ R such that
ax + by = 1. This implies that (vau + vbc)u−1 xv −1 + (vb) y − cu−1 x v −1 = 1. Hence, (vau + vbc, vb) is unimodular. By hypothesis, there exists some y ∈ R such that a + by ∈ Rq−1 . Further, (vau + vbc) + vb(yu − c) = v(a + by)u ∈ Rq−1 , and so (vau + vbc, vb) is a reducible. Assume that (vau+vbc, vb) is a reducible unimodular row. Then we have x, y ∈ R such that (vau+vbc)x+(vb)y = 1. As a result, auxv +b(cx+y)v = 1, i.e., (a, b) is a unimodular row. By hypothesis, there exists some z ∈ R such that (vau + vbc) + vbz ∈ Rq−1 , and so a + b(c + z)u−1 ∈ Rq−1 , as required.  The following main result of this section was proved by Ara et al. ([16, Theorem 6.4]). Theorem 8.1.5. If R is a QB-ring, then so is Mn (R) for all n ∈ N. Proof. Given        a11 a12 x11 x12 b11 b12 10 + = a21 a22 x21 x22 b21 b22 01

(†)

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in M2 (R), then a11 x11 + (a12 x21 + b11 ) = 1 in R. Thus, we can find a y ∈ R such that a11 + (a12 x21 + b11 )y = u ∈ Rq−1 . By Lemma 8.1.4, (†) is reducible if and only if this is so for the row with elements         a11 a12 1 0 y0 b11 b12 b11 b12 + , . a21 a22 x21 y 1 b21 b22 00 b21 b22

So we assume that a11 ∈ Rq−1 . By Lemma 8.1.1, there exists x ∈ R such that (1 − a11 x) ⊥ (1 − xa11 ) and a11 = a11 xa11 , x = xa11 x. By Lemma 8.1.4 again, it suffices to prove the unimodular row with elements        1 0 a11 a12 1 −xa12 1 0 b11 b12 , −a21 x 1 a21 a22 0 1 −a21 x 1 b21 b22 is reducible. The first matrix has the form   a11 (1 − a11 x)a12 . a21 (1 − xa11 ) ∗

Thus, we may assume that a11 xa12 = 0 = a21 xa11 , a12 ⊥ a21 in (†). Computing the (2, 2)-element in (†), we obtain a21 x12 + (a22 x22 + b22 ) = 1 in R. Thus, we can find a z ∈ R such that a22 + (a21 x12 + b22 )z ∈ Rq−1 . By Lemma 8.1.4, (†) is reducible if and only if this is so for the row with elements         a11 a12 1 x12 z b11 b12 00 b b + , 11 12 . a21 a22 0 1 b21 b22 0z b21 b22

So we assume that a22 ∈ Rq−1 . By Lemma 8.1.1 again, there exists y ∈ R such that (1 − a22 y) ⊥ (1 − ya22 ) and a22 = a22 ya22 , y = ya22 y. For the (1, 2)-position, we compute

a11 x a12 + (a11 x12 + b12 )z = a11 x a11 x12 + a12 x22 + b12 z = 0.

Since a12 ⊥ a21 and
a11 x12 + a12 x22 + b12 = 0, one easily checks that a12 + (a11 x12 + b12 )z ⊥ a21 . By Lemma 8.1.4 again, it suffices to prove the unimodular row with elements        1 −a12 y a11 a12 1 0 1 −a12 y b11 b12 , 0 1 a21 a22 −ya21 1 0 1 b21 b22 is reducible. The first matrix has the form   a11 a12 (1 − ya22 ) . (1 − a22 y)a21 a22 Thus, we may assume that

a11 xa12 = 0 = a12 ya22 , a21 xa11 = 0 = a22 ya21 , a11 = a11 xa11 , x = xa11 x, a22 = a22 ya22 , y = ya22 y, (1 − a11 x) ⊥ (1 − xa11 ), (1 − a22 y) ⊥ (1 − ya22 ) in (†).

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Since a11 x11 + a12 x21 + b11 = 1, we obtain (1 − a11 x)a12 (1 − ya22 )x21 (1 − a11 x) + (1 − a11 x)b11 (1 − a11 x) = 1 − a11 x.

In view of Lemma 8.1.3, either (1) (1 − a11 x)R(1 − ya22 ) = 0 or (2) (1 − a11 x)R(1 − ya22 ) is a QB-corner. If (1) holds, then a12 = (1 − a11 x)a12 (1 − ya22 ) = 0. If (2) holds, then there exists an element q ∈ (1 − a11 x)R(1 −
−1 ya22 ) such that a12 +(1−a11 x)b11 (1−a11 x)q ∈ (1−a11 x)R(1−ya22 ) q . By Lemma 8.1.4 again, it suffices to prove the unimodular row with elements         a11 a12 1 −xb11 q b11 b12 0q b b + , 11 12 a21 a22 0 1 b21 b22 00 b21 b22 is reducible. The first matrix has the form   a11 a12 + (1 − a11 x)b11 q . a21 a22 + (b21 − a21 xb11 )q


−1 Thus, we may assume that a12 ∈ (1 − a11 x)R(1 − ya22 ) q . For the (2, 2)position, we compute

a22 + (b21 − a21 xb11 )q y a22
+ (b21 − a21 xb11 )q = a22 y a22 + (b21 − a21 xb11 )q = a22 + (b21 − a21 xb11 )q − (1 − a22 y)(b21 − a21 xb11 )q = a22 + (b21 − a21 xb11 )q;
1 − a22 + (b21 − a21 xb11 )q y = 1 − a22 y;
1 − y a22 + (b21 − a21 xb11 )q
= 1 − ya22 − y(b21 − a21 xb11 )q = 1 − y(b21 − a21 xb11 )q (1 − ya22 ).

Thus, we also have (1 − a22 y) ⊥ (1 − ya22 ) and a22 = a22 ya22 , y = ya22 y.
−1 Similarly, we assume that a21 = 0 or a21 ∈ (1 − a22 y)R(1 − xa11 ) q . If   a11 a12 a12 = 0 or a21 = 0, one easily checks that ∈ M2 (R)−1 q . Thus, by a21 a22 virtue of Lemma 8.1.1, we may assume that there exist s ∈ (1 − ya22 )R(1 − a11 x), t ∈ (1−xa11 )R(1−a22 y) such that (1−a11 x−a12 s) ⊥ (1−ya22 −sa12 ) and (1 − a22 y − a21 t) ⊥ (1 − xa11 − ta21 ). It is easy to verify that      a11 a12 xt a11 x + a12 s 0 = , a a sy 0 a21 t + a22 y  21 22     xt a11 a12 xa11 + ta21 0 = . sy a21 a22 0 sa12 + ya22   a11 a12 This implies that ∈ M2 (R)−1 q . It follows from Lemma 8.1.4 a21 a22 that M2 (R) is a QB-ring. By induction, M2n (R) is a QB-ring. According

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to Lemma 8.1.2, every corner of a QB-ring is a QB-ring. Therefore Mn (R) is a QB-ring, as asserted.  From Theorem 8.1.5 and Lemma 8.1.2, we see that the property of a ring to be a QB-ring is Morita invariant. Corollary 8.1.6. Every regular square matrix over a QB-ring is the product of an idempotent matrix and a quasi-invertible matrix. Proof. Let R be a QB-ring, and let A ∈ Mn (R) be regular. Then we have B ∈ Mn (R) such that A = ABA. In view of Theorem 8.1.5, Mn (R) is a QB-ring. As AB + (In − AB) = In , there exists Y ∈ Mn (R) such that A + (In − AB)Y = U ∈ Mn (R)−1 q . Let E = AB. Then A = EU with E = E 2 ∈ Mn (R), as required.  Corollary 8.1.7. Let e be a full idempotent of a ring R. Then the following are equivalent: (1) R is a QB-ring. (2) eRe is a QB-ring. Proof. (1) ⇒ (2) is clear by Lemma 8.1.2. (2) ⇒ (1) Since e ∈ R is a full idempotent, we have some si , ti (1 ≤ n P i ≤ n) such that 1 = si eti . Construct a map ϕ : n(eR) → R given by ϕ(er1 , · · · , ern ) =

n P

i=1

i=1

si eri for any (er1 , · · · , ern ) ∈ n(eR). Obviously, ϕ is θ

an R-epimorphism, and so R ⊕ kerϕ ∼ = n(eR). Let p : R ⊕ ker(ϕ) ։ R be the projection and i : R ֒→ R ⊕ ker(ϕ) be the inclusion, respectively. Let f = θipθ−1 : n(eR) ∼ = R ⊕ ker(ϕ) ։ R ֒→ R ⊕ ker(ϕ) ∼ = n(eR).

∼ Then f = f 2 ∈ EndR n(eR) . Further, R θ(R) = f n(eR) . It follows =

∼ ∼ that R = EndR f (n(eR)) = f EndR n(eR)
f . In view of Theorem 8.1.6, Mn (eRe) is a QB-ring, so is EndR n(eR) . Therefore we complete the proof by Lemma 8.1.2.  Let V be an infinite dimensional vector space over a field F , and set Q = EndF (V ) and J = {x ∈ Q | dimF (xV ) < ∞}. Set R = {(x, y) ∈ Q × Q | x − y ∈ J}. Choose an idempotent f ∈ Q which projects on a proper subspace of finite codimension, so that f 6= 1V and 1V − f ∈ J. Then dimF (V ) = dimF (f V ), and so V ∼ = f V . Hence, Q ∼ = f Q. This

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implies that f = xy, 1V = yx, where x ∈ f Q and y ∈ Qf . Set e = (f, 1). Then 1R = (1V , 1V ) = (y, 1V )(f, 1V )(x, 1V ), and so R = ReR. Thus, e ∈ R is a full idempotent. In view of [217, Example 4.21], eR ≇ R. Lemma 8.1.8. An exchange ring R is a QB-ring if and only if for every regular x ∈ R there exists some u ∈ Rq−1 such that x = xux. Proof. Let R be an exchange ring. Assume that R is a QB-ring. Given any regular x ∈ R, then there exists y ∈ R such that x = xyx. It follows from yx + (1 − yx) = 1 that there exists
some z ∈ R such that y + (1 − yx)z = u ∈ Rq−1 . Thus x = x y + (1 − yx)z x = xux. Conversely, assume that ax+b = 1 in R. By virtue of Lemma 1.4.7, there exists e = e2 ∈ R such that e = bs and 1 − e = (1 − b)t for s, t ∈ R. Thus, (1 − e)axt + e = (1 − e)(1 − b)t + e = 1. Obviously, (1 − e)a ∈ R is regular. So we have an element u ∈ Rq−1 such that (1 − e)au(1 − e)a = (1 − e)a. Let f = u(1 − e)a. Then f = f 2 ∈ R. Hence, f xt + ue = u, and then f (x + ue) + (1 − f )ue = u. As u ∈ Rq−1 , there exists v ∈ R such that (1 − uv) ⊥ (1 − vu), u = uvu and v = vuv. Let g = (1 − f )uev(1 − f ). 2 As (1 − f )ue = (1 − f )u, we
see that g = (1 − f )uv(1 − f )uv(1 − f ) = (1 − f ) uv − uvu(1 − e)auv) (1 − f ) = (1 − f )(uv − f uv)(1 − f ) = g. As a result, we deduce that

u a + bs(v(1 − f ) − a) 1
− f uev(1 − f ) u
= u(1 − e)a + uev(1 − f ) 1 − f uev(1
− f) u = f (1 − f uev(1 − f )) + uev(1 − f ) u
= f + (1 − f )uev(1 − f ) u = (f + g)u = u.

Let y = s v(1 − f ) − a and w = 1 − f uev(1 − f ) u. Then w(a + by)w = w with w ∈ Rq−1
. Suppose (1 − ws) ⊥ (1 − sw). Then 1 − w(a +
by) = 1 − w(a + by) (1 − ws) and
1 − (a + by)w = (1
− sw) 1 − (a + by)w . This implies that 1 − w(a + by) ⊥ 1 − (a + by)w . Therefore, a + by ∈ Rq−1 , the result follows.  Theorem 8.1.9. Let R be an exchange ring. Then the following are equivalent: (1) R is a QB-ring. (2) For any a, b ∈ R, a∼b implies that there exists u ∈ Rq−1 such that au = ub.

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(3) For any idempotents e, f ∈ R, eR ∼ = f R implies that there exists u ∈ Rq−1 such that eu = uf . Proof. (1) ⇒ (2) Let a and b be pseudo-similar in R. By Lemma 6.1.9, there exist x, y ∈ R such that a = xby, b = yax, x = xyx and y = yxy. In view of Lemma 8.1.8, we can find v ∈ Rq−1 such that x = xvx. Write (1−vw) ⊥ (1−wv). Set s = x+(1−xv)w(1−vx). Then x = xvs. Observing 1 − sv = (1 − xv)(1 − wv)(1 − xv) and 1 − vs = (1 − vx)(1 − vw)(1 − vx), we see that (1 − vs) ⊥ (1 − sv). Let u = (1 − xy − xv)s(1 − yx − vx). Then u ∈ Rq−1 and au = ax = xb = ub, as desired. (2) ⇒ (3) For any idempotents e, f ∈ R, eR ∼ = f R implies that e∼f . By hypothesis, there exists a u ∈ Rq−1 such that eu = uf . (3) ⇒ (1) Given any regular x ∈ R, there exists a y ∈ R such that x = xyx. Clearly, we have η : xyR = xR ∼ = yxR. By hypothesis, we can find some u ∈ Rq−1 such that yxu = uxy. In view of Lemma 8.1.1, there exists some v ∈ R such that (1 − uv) ⊥ (1 − vu), u = uvu and v = vuv. Define α : (1 − xy)R → (1 − yx)R given by (1 − xy)r → (1 − yx)ur for any r ∈ R and β : (1−yx)R → (1−xy)R given by (1−yx)r → (1−xy)v(1−yx)r for any r ∈ R. Since (1 − yx)u = u(1 − xy), we easily check that α and β are R-morphisms. Define φ : R = xR ⊕ (1 − xy)R → yxR ⊕ (1 − yx)R given by φ(x1 + x2 ) = η(x1 ) + α(x2 ) for any x1 ∈ xR, x2 ∈ (1 − xy)R and ψ : R = yxR ⊕ (1 − yx)R → xR ⊕ (1 − xy)R = R given by ψ(y1 + y2 ) = η −1 (y1 ) + β(y2 ) for any y1 ∈ yxR, y2 ∈ (1 − yx)R. Then (1 − ψφ)(x1 + x2 ) = x2 − (1 − xy)v(1 − yx)ux2 = (1 − xy)x2 − (1 − xy)vu(1 − xy)x2 = (1 − xy)(1 − vu)x2 for any x1 ∈ xR and x2 ∈ (1−xy)R. On the other hand, (1−φψ)(y1 +y2 ) = y2 − (1 − yx)uvy2 = (1 − yx)(1 − uv)y2 for any y1 ∈ yxR and y2 ∈ (1 − yx)R. Thus, we have φ(1) ∈ Rq−1 such that x = xφ(1)x. According to Lemma 8.1.8, we obtain the result.  It is claimed that every QB-ring is right and left symmetric. That is, a ring R is a QB-ring if and only if so is the opposite ring Rop . Let I and J be two ideals of a ring R. We say that I and J are orthogonal, I⊥J, provided that IJ = 0 = JI. Clearly, an element u ∈ Rq−1 if and only if there exist a pair of orthogonal ideals I and J such that uv = 1 in R/I and vu = 1 in R/J for some v ∈ R. By using Lemma 4.1.2, we are done. Recall that a unital C ∗ -algebra A is extremally rich if the set of quasiinvertible elements in A is dense in A. For such C ∗ -algebras, we now derive

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the following. Corollary 8.1.10. Let A be a unital C ∗ -algebra of real rank zero. Then the following are equivalent: (1) A is extremally rich. (2) For any idempotents e, f ∈ A, eA ∼ = f A implies that there exists an extreme partial isometry u ∈ A such that eu = uf . Proof. Using [16, Theorem 7.2], A is an exchange ring. By Theorem 8.1.9 and [25, Proposition 9.1], the proof is complete.  Lemma 8.1.11. Let R be an exchange ring. Then the following are equivalent: (1) R is a QB-ring. (2) For any a, b ∈ R, aR = bR implies that there exists u ∈ Rq−1 such that a = bu. (3) For any a, b ∈ R, Ra = Rb implies that there exists u ∈ Rq−1 such that a = ub. Proof. (1) ⇒ (2) Given aR = bR, then we have x, y ∈ R such that a = bx and ay = b. Since xy + (1 − xy) = 1, we can find a z ∈ R
such that x + (1 − xy)z = u ∈ Rq−1 . Thus, a = bx = b x + (1 − xy)z = bu, as required. (2) ⇒ (1) Given any regular x ∈ R, then x = xyx for some y ∈ R. From xR = (xy)R, we can find some u ∈ Rq−1 such that xy = xu; hence, x = xyx = xux. By virtue of Lemma 8.1.8, R is a QB-ring. (1) ⇔ (3) is symmetric.  We note that the exchange condition in Lemma 8.1.11 is necessary. For any a, b ∈ Z, aZ = bZ implies that a = ±b, while Z is not a QB-ring. In this case, a∼b implies that au = ub for a u ∈ U (Z). Theorem 8.1.12. Let R be an exchange QB-ring. For any regular A ∈ Mn (R), there exist U, V ∈ Mn (R)−1 such that U AV is a diagonal matrix. q Proof. Since R is an exchange QB-ring, so is Mn (R) by Theorem 8.1.5. Given any regular A ∈ Mn (R), there exists E = E 2 ∈ Mn (R) such that AMn (R) ∼ = EMn (R). As in the proof in Theorem 7.3.10, we can find idempotents e1 , · · · , en ∈ R such that ϕ : AMn (R) ∼ = diag(e1 , · · · , en )Mn (R). In view of Lemma 7.3.9, we show that Mn (R)A = Mn (R)ϕ(A) and

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ϕ(A)Mn (R) = diag(e1 , . . . , en )Mn (R). According to Lemma 8.1.11, we can find some U, V ∈ Mn (R)−1 such that U AV = ϕ(A)V = diag(e1 , · · · , en ), q as asserted.  8.2

Pseudo Substitutions

The main purpose of this section is to investigate a kind of pseudo substitution of modules over exchange QB-rings. Theorem 8.2.1. Let R be an exchange ring. Then the following are equivalent: (1) R is a QB-ring. (2) M = A1 ⊕ H = A2 ⊕ K with A1 ∼ =R∼ = A2 implies that there exist a pair of orthogonal ideals I1 and I2 , and M = E ⊕ B1 ⊕ H = E ⊕ B2 ⊕ K such that B1 I1 = B1 and B2 I2 = B2 . (3) M = A1 ⊕ H = A2 ⊕ K with A1 ∼ =R∼ = A2 implies that there exist a pair of orthogonal ideals I1 and I2 , and M = C1 ⊕ E = C2 ⊕ E, A1 = B1 ⊕ C1 , A2 = B2 ⊕ C2 such that B1 I1 = B1 and B2 I2 = B2 .

Proof. (1)⇒(2) Suppose M = A1 ⊕ H = A2 ⊕ K with A1 ∼ = R ∼ = A2 . ∼ Using the decomposition M = A1 ⊕ H = R ⊕ H, we have projections p1 : M → R, p2 : M → H and injections q1 : R → M, q2 : H → M such that p1 q1 = 1R , q1 p1 + q2 p2 = 1M and ker(p1 ) = H. Using the decomposition M = A2 ⊕ K ∼ = R ⊕ K, we have a projection f : M → R and an injection g : R → M such that f g = 1R and ker(f ) = K. It follows from (f q1 )(p1 g) + f q2 p2 g = 1R that there is an element u ∈ Rq−1 such that f q1 + f q2 p2 gy = u for some y ∈ R. That is, f (q1 + q2 p2 gy) = u. Thus, we get some v ∈ R such that (1 − uv) ⊥ (1 − vu), u = uvu and v = vuv from Lemma 8.1.1. Let α = vu, β = uv, ψ = q1 + q2 p2 gy. Clearly, α = α2 , β = β 2 . In addition, we get f ψ = u and p1 ψ = 1R . Let D1 = kerαp1 , D2 = kerβf and E = ψα(R). If m ∈ E ∩ D1 , then m = ψα(x) for some x ∈ R. Clearly, 0 = αp1 (m) = αp1 ψα(x) = α(x), and then m = 0. This means that
E ∩ D1 = 0. Given any m ∈ M , we have m = ψαp1 (m) + m − ψαp1 (m) ∈ E + D1 , and so M = E ⊕ D1 . Similarly, T E D2 = M = E ⊕ D1 → D1 be the projection onto D1 . Clearly,
0. Let T ker p|D2 = E D2 = 0; hence, p|D2 : D2 → D1 is an isomorphism. In view of Lemma 2.2.5, M = E ⊕ D2 . Let B1 = q1 (1 − α)(R) and B2 = g(1 − β)(R). One easily checks that D1 = B1 ⊕ H and D2 = B2 ⊕ K. Thus

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M = E ⊕ B1 ⊕ H = E ⊕ B2 ⊕ K. Let I1 = R(1 − α)R and I2 = R(1 − β)R. Then I1 and I2 are ideals of R. As (1 − α) ⊥ (1 − β), we deduce that I1 ⊥ I2 . Moreover, we have B1 I1 = B1 and B2 I2 = B2 , as required. (2)⇒(1) Suppose that a1 R + a2 R = R with a1 , a2 ∈ R. Choose M = A1 ⊕ H, where A1 = R = H. Then we have a splitting R-epimorphism ψ : M → R given by ψ(s, t) = a1 s + a2 t for any s ∈ A1 , t ∈ H; hence, M = A2 ⊕ K, where K = ker(ψ) and A2 ∼ = R. Therefore we get a pair of orthogonal ideals I1 and I2 and M = E ⊕ B1 ⊕ H = E ⊕ B2 ⊕ K such that B1 I1 = B1 and B2 I2 = B2 . Let ϕ : M = A1 ⊕ H → R be the projection onto the first factor. Write E1 = ϕ(E) and B1′ = ϕ(B1 ). Then A1 = E1 ⊕ B1′ . Let h : R = E1 ⊕ B1′ → E1 be the projection onto E1 . Then h ∈ R is an idempotent. In addition, 1 − h ∈ I1 . Let ψ : M = A2 ⊕ K → R be the projection onto the first factor. Write E2 = ψ(E) and B2′ = ψ(B2 ). We have R = E2 ⊕ B2′ . Let k : R = E2 ⊕ B2′ → E2 be the projection onto E2 . Then k ∈ R is an idempotent, and 1 − k ∈ I2 . Hence (1 − h) ⊥ (1 − k) because I1 ⊥ I2 . Clearly, ψ |E⊕B2 : E ⊕ B2 → R is an isomorphism. Let θ = (ψ |E⊕B2 )−1 , and let i : E ⊕ B2 → M be the injection. Since k ∈ R is an idempotent, we may assume that iθ(k) = (x1 , x2 ) with x1 ∈ A1 k and x2 ∈ Hk. Hence k = ψiθ(k) = ψ(x1 , x2 ) = a1 x1 + a2 x2 . As E1 = ϕ(E) and E2 = ψ(E), we have an isomorphism ϕθ : E2 → E1 . Evidently, E2 = k(R) and E1 = h(R). So we have r ∈ R such that h = ϕθ(kr) = hϕθ(k)krh. Clearly, x1 = ϕiθ(k) = ϕθ(k) = hϕθ(k)k. Let y1 = krh. Then h = x1 y1 . Moreover, y1 x1 = krhϕθ(k) = (ϕθ)−1 hϕθ(k) = (ϕθ)−1 (ϕθ)(k) = k. Hence (1 − y1 x1 ) ⊥ (1 − x1 y1 ). That is, x1 ∈ Rq−1 . In addition, (a1 + a2 x2 y1 )x1 = a1 x1 + a2 x2 k = a1 x1 k + a2 x2 k = k. So x1 (a1 + a2 x2 y1 )x1 = x1 k = x1 . As in the proof of Lemma 8.1.8, we show that a1 + a2 x2 y1 ∈ Rq−1 , as required. (1) ⇔ (3) is dual to (1) ⇔ (2).  Let I be an ideal of a ring R, and let ϕ : B → C be an R-morphism. Then there exists a unique R-morphism, denoted by ϕI , such that πC ϕ = ϕI πB , where πB : B → B/BI and πC : C → C/CI are both canonical maps. Corollary 8.2.2. Let R be an exchange ring. Then the following are equivalent: (1) R is a QB-ring. (2) Given any right R-module decompositions R = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 , there exist a pair of orthogonal ideals I, J and ϕ :

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B1 → B2 such that ϕI is a splitting epimorphism and ϕJ is a splitting monomorphism. Proof. (1) ⇒ (2) Given any right R-module decompositions R = A1 ⊕B1 = A2 ⊕ B2 with A1 ∼ = A2 , then we have ϕ : R ⊕ B1 ∼ = R ⊕ B2 . Hence R ⊕ B2 = ϕ(R)⊕ϕ(B1 ) with R ∼ ϕ(R). In view of Theorem 8.2.1, there exist a pair of = orthogonal ideals I, J such that R⊕B2 = C ⊕D ⊕B2 = C ⊕E ⊕ϕ(B1 ) with D = DJ and E = EI. This induces an isomorphism ψ : E ⊕ B1 ∼ = D ⊕ B2 . Let p : B1 → E ⊕ B1 be the injection and q : D ⊕ B2 → B2 be the projection. Set ϕ = qψp. Then ϕI = qI ψI pI . Clearly, pI and ψI are isomorphisms. Since qI is a splitting epimorphism, so is ϕI . Likewise, we have ϕJ = qJ ψJ pJ and qJ , ψJ are both isomorphisms. As pJ is a splitting monomorphism, so is ϕJ , as desired. (2) ⇒ (1) Let a ∈ R be regular. Then we have x ∈ R such that a = axa and x = xax. Since R = aR ⊕ (1 − ax)R = xaR ⊕ (1 − xa)R with φ : aR = axR ∼ = xaR, we get ψ : (1 − ax)R → (1 − xa)R such that ψI is a splitting epimorphism and ψJ is a splitting monomorphism. Construct a map ϕ : R → R given by ϕ(s + t) = φ(s) + ψ(t) for any s ∈ aR and t ∈ (1 − ax)R. Clearly, a = aϕ(1)a. Since ϕI : R/I → R/I is a splitting epimorphism, we get ϕI (1+I) = ϕ(1)+I ∈ R/I is right invertible. That is, ϕ(1) ∈ R is right invertible modulo I. Similarly, we prove that ϕ(1) ∈ R is left invertible modulo J. It follows that ϕ(1) ∈ Rq−1 . According to Lemma 8.1.8, R is a QB-ring.  As an immediate consequence, we derive that an exchange ring R is a QB-ring if and only if for any idempotents e, f ∈ R, eR ∼ = f R implies that there exist a pair of orthogonal ideals I, J and ϕ : (1 − e)R → (1 − f )R such that ϕI is a splitting epimorphism and ϕJ is a splitting monomorphism. Following Ara et al. (cf. [25]), we say that a regular a ∈ R extends to a regular b ∈ R, and write a ≤ b, if there exists x ∈ R such that a = axa = axb = bxa. Lemma 8.2.3. Let R be an exchange ring. Then the following are equivalent: (1) R is a QB-ring. (2) Every regular element in R extends to an element in Rq−1 . Proof. (1) ⇒ (2) Given any regular a ∈ R, then there exists x ∈ R such that a = axa and x = xax. Since R is a QB-ring, by Lemma 8.1.8, there exists

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some v ∈ Rq−1 such that x = xvx. Choose u = (1 − ax − vx)v(1 − xa − xv). Then u ∈ Rq−1 . One directly verifies that (ax)u = a = u(xa). Hence, a = axa = axu = uxa. Therefore a ≤ u, as required. (2) ⇒ (1) Given any regular a ∈ R, there exists some b ∈ R such that a = aba and b = bab. By hypothesis, there exists a u ∈ Rq−1 such that b ≤ u. This implies that b = bxb = bxu = uxb. Let e = bx. Then e = e2 ∈ R. It follows from ba + (1 − ba) = 1 that eua(1 − e) + (1 − ba)(1 − e) = 1 − e; hence, e + (1 − ba)(1 − e) = 1 − eua(1
− e) ∈ U (R). So we deduce that b + (1 − ba)(1 −
e)u = 1 − eua(1 − e) u
∈ Rq−1 . Thus, a = aba = a b + (1 − ba)(1 − e)u a = a (1 − eua(1 − e))u a, and then R is a QB-ring by Lemma 8.1.8.  Lemma 8.2.4. Let R be an exchange ring. Then the following are equivalent: (1) R is a QB-ring. (2) For any finitely generated projective right R-modules B1 and B2 , R ⊕ B1 ∼ = R ⊕ B2 implies that there exist a pair of orthogonal ideals I1 and I2 , and B1 ⊕ C1 ∼ = B2 ⊕ C2 such that C1 I1 = C1 and CI2 = C2 .

Proof. (1) ⇒ (2) Since ϕ : R ⊕ B1 ∼ = R ⊕ B2 , we have M := R ⊕ B2 = ∼ ϕ(R) ⊕ ϕ(B1 ) with R = ϕ(R). By virtue of Theorem 8.2.1, there exist a pair of orthogonal ideals I1 and I2 , and M = E ⊕ C1 ⊕ ϕ(B1 ) = E ⊕ C2 ⊕ B2 such that C1 I1 = C1 and C2 I2 = C2 . Thus B1 ⊕ C1 ∼ = B2 ⊕ C2 , as required. (2) ⇒ (1) Let a ∈ R be regular. Then there exists some b ∈ R such that a = aba and b = bab. Set p = ab and q = ba. It is easy to check that R ⊕ (1 − p)R ∼ = qR ⊕ (1 − q)R ⊕ (1 − p)R ∼ = pR ⊕ (1 − q)R ⊕ (1 − ∼ p)R = R ⊕ (1 − q)R. So we have right R-modules C1 and C2 such that (1 − p)R ⊕ C1 ∼ = (1 − q)R ⊕ C2 and a pair of orthogonal ideals I1 and I2 such that C1 I1 = C1 and C2 I2 = C2 . Clearly, there is a refinement matrix (1 − p)R C1  ′ ′ (1 − q)R A C1 . C2 C2′ C ′ N N As C1 = C1 I1 , we have C1 (R/I1 ) = 0, and so C ′ (R/I1 ) = 0. That R

R

is, C ′ I1 = C ′ . Likewise, C ′ = C ′ I2 . As a result, C ′ = C ′ I1 = (C ′ I2 )I1 = C ′ (I2 I1 ) = 0 because I1 ⊥ I2 . Hence we have idempotents e1 , f1 ∈ (1 − p)R(1 − p), e2 , f2 ∈ (1 − q)R(1 − q) such that 1 − p = e1 + f1 , 1 − q = e2 + f2 , e1 R ∼ = e2 R, and f1 R ∼ = C1 and f2 R ∼ = C2 . Since e1 R ∼ = e2 R, = A′ ∼ we can find c ∈ e1 Re2 and d ∈ e2 Re1 such that e1 = cd and e2 = dc.

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Clearly, a ∈ pRq and c ∈ (1 − p)R(1 − q) are both regular in R. It is easy to verify that a = aba = ab(a + c) = (a + c)ba, i.e., a ≤ a + c. Furthermore, it follows from b ∈ qRp, d ∈ (1 − q)R(1 − p) that 1 − (a + c)(b + d) = 1 − ab − cd = (1 − p) − e1 = f1 . Likewise, 1 − (b + d)(a + c) = f2 . Obviously, f1 RI1 = f1 R and
f2 RI2 = f2 R; hence, f
1 ∈ I1 and f2 ∈ I2 . As I1 ⊥ I2 , 1 − (a + c)(b + d) ⊥ 1 − (b + d)(a + c) . So a + c ∈ Rq−1 . Therefore we complete the proof from Lemma 8.2.3.  Theorem 8.2.5. Let R be an exchange ring. Then the following are equivalent: (1) R is a QB-ring. (2) For all finitely generated projective right R-modules A, B1 and B2 , A ⊕ B1 ∼ = A ⊕ B2 implies that there exist a pair of orthogonal ideals I1 and I2 , and B1 ⊕ C1 ∼ = B2 ⊕ C2 such that C1 I1 = C1 and C2 I2 = C2 . Proof. (2) ⇒ (1) is trivial by Lemma 8.2.4. (1) ⇒ (2) Let A, B1 and B2 be finitely generated projective right R-modules. Suppose that A ⊕ B1 ∼ = A ⊕ B2 . In light of [409, Theorem 2.1], there exist idempotents e1 , · · · , en ∈ R such that A ∼ = e1 R ⊕ n×1 ∼ · · · ⊕ en R; hence, diag(e1 , · · · , en )R ⊕ B = diag(e1 , · · · , en )Rn×1 ⊕ N 1×n1∼ B2 . So diag(e1 , · · · , en )Mn (R) ⊕ B1 R = diag(e1 , · · · , en )Mn (R) ⊕ R N 1×n B2 R . In view of Theorem 8.1.5, Mn (R) is a QB-ring. According to R

Lemma 8.2.4, there exist a pair of orthogonal ideals Mn (I1 ) and Mn (I2 ) of N N Mn (R), and B1 R1×n ⊕ C1′ ∼ = B2 R1×n ⊕ C2′ such that C1′ Mn (I1 ) = C1′ R R N and C2′ Mn (I2 ) = C2′ . Clearly, I1 ⊥ I2 . Let Ci = Ci′ Rn×1 (i = 1, 2). Mn (R) N Then B1 ⊕ C1 ∼ Rn×1 ∼ = R. As Rn×1 I ∼ = = B2 ⊕ C2 because R1×n Mn (R)

Mn (I)Rn×1 , we deduce that C1 I = C1 and C2 I = C2 , as required.



Corollary 8.2.6. Let A be a finitely generated projective right module over an exchange QB-ring R. If B1 and B2 are any right R-modules such that A ⊕ B1 ∼ = A ⊕ B2 , then there exist a pair of orthogonal ideals I1 and I2 , and B1 ⊕ C1 ∼ = B2 ⊕ C2 such that C1 I1 = C1 and CI2 = C2 . Proof. By virtue of [382, Proposition 28.6], A has the finite exchange property. If B1 and B2 are any right R-modules such that A ⊕ B1 ∼ = A ⊕ B2 , analogously to the proof of Lemma 6.3.7, we have right R-module decompositions A ∼ = A′ ⊕ B1′ ∼ = A′ ⊕ B2′ , B1 ∼ = B1′ ⊕ C ′ , B2 ∼ = B2′ ⊕ C ′ . Since

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A is a finitely generated projective right R-module, so are A′ , B1′ and B2′ . In view of Theorem 8.2.5, there exist a pair of orthogonal ideals I1 and I2 , and B1′ ⊕ C1 ∼ = B2′ ⊕ C2 such that C1 I1 = C1 and CI2 = C2 . Further, B1 ⊕ C1 ∼ = C ′ ⊕ B1′ ⊕ C1 ∼ = C ′ ⊕ B2′ ⊕ C2 ∼ = B2 ⊕ C2 , the result follows.



Theorem 8.2.7. Let R be an exchange ring. Then the following are equivalent: (1) R is a QB-ring. (2) For all finitely generated projective right R-modules A, B1 and B2 , A ⊕ B1 ∼ = A ⊕ B2 implies that there exist a pair of orthogonal ideals I and J, and ϕ : B1 → B2 such that ϕI : is a splitting epimorphism and ϕJ is a splitting monomorphism. Proof. (1) ⇒ (2) For all finitely generated projective right R-modules A, B1 and B2 , A⊕ B1 ∼ = A⊕ B2 implies that there exist a pair of orthogonal ideals I and J, and ψ : B1 ⊕ C1 ∼ = B2 ⊕ C2 such that C1 I = C1 and C2 J = C2 by Theorem 8.2.5. Let p : B1 → B1 ⊕C1 be the injection and q : B2 ⊕C2 → B2 be the projection. Set ϕ = qψp. Then ϕI = qI ψI pI and ϕI = qJ ψJ pJ . As in the proof of Corollary 8.2.2, pI and qJ are both isomorphisms. Therefore ϕI is a splitting epimorphism and ϕJ is a splitting monomorphism. (2) ⇒ (1) Given any right R-module decompositions R = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 , we deduce that R ⊕ B1 ∼ = R ⊕ B2 . By hypothesis, there exist a pair of orthogonal ideals I and J, and ϕ : B1 → B2 such that ϕI is a splitting epimorphism and ϕJ is a splitting monomorphism. This establishes the result from Corollary 8.2.2.  It is well known that an exchange ring R is separative if and only if for ∼ B⊕C all finitely generated projective right R-modules A, B and C, A⊕C = with C .⊕ A, B implies that A ∼ B. Now we give an analogue for exchange = QB-rings. Corollary 8.2.8. Let R be an exchange QB-ring. Then for all finitely generated projective right R-modules A, B and C, A ⊕ C ∼ = B ⊕ C with C .⊕ A, B implies that there exist a pair of orthogonal ideals I and J such that A/AI ∼ = B/BI and A/AJ ∼ = B/BJ. Proof. Let A, B, C ∈ F P (R) such that A ⊕ C ∼ = B ⊕ C with C .⊕ A, B.

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Applying Lemma 6.3.7, we have a refinement matrix A C such that C1 .⊕ A1 , B1 . Theorem 8.2.7 that there that A1 /A1 I .⊕ B1 /B1 I N A1 R/I ⊕ D for a right



BC  D1 A1 B1 C1

Since A1 ⊕ C1 ∼ = B1 ⊕ C1 , it follows by = C ∼ exist a pair of orthogonal ideals I and J such N and B1 /B1 J .⊕ A1 /A1 J. Thus, B1 R/I ∼ = R

R-module D. It is easy to check that

R

C

N R

N N R/I ∼ = C1 R/I ⊕ B1 R/I R R N N ∼ = C1 R/I ⊕ A1 R/I ⊕ D R NR ∼ R/I ⊕ D. =C R

So A

N R

R/I ∼ =A

N R

A

R/I ⊕ D because C .⊕ A. Therefore we get

N R

N R/I ∼ = A R/I ⊕ D R N N ∼ R/I ⊕ D = D1 R/I ⊕ A1 R R N N ∼ = D1 R/I ⊕ B1 R/I R NR ∼ = B R/I. R

This implies that A/AI ∼ = B/BI. Likewise, A/AJ ∼ = B/BJ because ⊕ B1 /B1 J . A1 /A1 J, and the result follows.  8.3

Completions of Diagrams

Let A be a quasi-projective right R-module. Canfell [71, Theorem 2.4] proved that EndR (A) has stable range one if and only if for any Repimorphisms f, g : A → M , there exists h ∈ AutR (A) such that the diagram A hւ↓f g

A ։ M

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commutates. Let A be a quasi-injective right R-module. He showed that EndR (A) has stable range one if and only if for any monomorphisms f, g : M → A, there exists h ∈ AutR (A) such that the diagram A hր↑f g

A ֋ M

commutates. These inspire us to investigate the completion of diagrams over QB endomorphism rings. Theorem 8.3.1. Let A be a quasi-projective right R-module, and let E = EndR (A). Then the following are equivalent: (1) E is a QB-ring. (2) For any R-epimorphisms f, g : A → M , there exists h ∈ Eq−1 such that f = gh. (3) Whenever αA + K = A with α ∈ E and K ⊆ A, there exists β ∈ E such that βA ⊆ K and α + β ∈ Eq−1 . (4) Whenever f A + gA = hA with f, g, h ∈ E, there exist k ∈ E, l ∈ Eq−1 such that f + gk = hl. Proof. (1) ⇒ (3) Suppose that αA + K = A with α ∈ E and K ⊆ A. Let M = A/K and g : A → M be the canonical map. Since αA + K = A, gα is an R-epimorphism. As A is quasi-projective, we have φ ∈ E such that gαφ = g; hence, g(1A − αφ) = 0. It follows from αφ + (1A − αφ) = 1A that α + (1A − αφ)k ∈ Eq−1 for some k ∈ E. Let β = (1A − αφ)k. Then α + β ∈ Eq−1 . Furthermore, gβ = g(1 − αφ)k = 0, and so β(A) ⊆ ker(g) = K, as required. (3) ⇒ (2) Let f, g : A → M be R-epimorphisms. Since A is quasiprojective, there exists α ∈ E such that gα = f . For any a ∈ A, there exists b ∈ A such that
g(a) = f (b) because f is an R-epimorphism. Thus, a = α(b) + a − α(b) . This implies that α(A) + ker(g) = A. By hypothesis, there exists β ∈ E such that α + β ∈ Eq−1 and β(A) ⊆ ker(g). Therefore, g(α + β) = gα = f , as desired. (2) ⇒ (4) Suppose that f A + gA = hA with f, g, h ∈ E. Let M = hA/gA and let ϕ : hA → M be the canonical map. Then ϕf and ϕh are both R-epimorphisms. So there exists l ∈ Eq−1 such that ϕf = ϕhl; hence, ϕ(f − hl) = 0. This implies that im(f − hl) ⊆ ker(ϕ) = gA. Since A is quasi-projective, there exists k ∈ E such that the diagram

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A k ւ ↓ f − hl −g

A ։ gA

commutates. That is, f + gk = hl, as required. (4) ⇒ (1) This is analogous to Corollary 2.1.6.



Corollary 8.3.2. Let R be a ring. Then the following are equivalent: (1) R is a QB-ring. (2) For any R-epimorphisms f, g : R → M , there exists u ∈ Rq−1 such that f (1) = g(1)u. (3) If M = m1 R = m2 R is a cyclic right R-module, then m1 = m2 u for some u ∈ Rq−1 . (4) For each finitely generated right R-module M and finitely generating sets {ui }1≤i≤n and {vi }1≤i≤n for M , there exists A ∈ Mn (R)−1 such q that (u1 , · · · , un ) = (v1 , · · · , vn )A. n n P P Proof. (1) ⇒ (4) Let M = ui R = vi R. Define i=1

f:

i=1

nR → M,     r1 r1  ..   ..  → 7 (u , · · · , u )  .  1 n  .  , ∀ r1 , · · · , rn ∈ R;

rn rn g : nR → M,     r1 r1  ..   ..   .  7→ (v1 , · · · , vn )  .  , ∀ r1 , · · · , rn ∈ R. rn

rn

Then f, g : nR → M are R-epimorphisms. In view of Theorem 8.3.1, there exists α ∈ EndR (nR)−1 such that f = gα. Let A ∈ Mn (R) be the q matrix corresponding to α. Then A ∈ Mn (R)−1 q . Further, (u1 , · · · , un ) = (v1 , · · · , vn )A, as required. (4) ⇒ (3) is trivial. (3) ⇒ (2) Suppose f, g : R → M are R-epimorphisms. Then M = f (1)R = g(1)R. By hypothesis, there exists some u ∈ Rq−1 such that f (1) = g(1)u. (2) ⇒ (1) is obvious by Theorem 8.3.1.  Corollary 8.3.3. Let R be a QB-ring, and let A ∈ Mm×n (R). Then the following hold:

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(1) If  A(m ≤ n) is right invertible, then there is a matrix of the form A ∈ Mn (R)−1 q . ∗ (2) If A (n ≤ m) is left invertible, then there is a matrix of the form (A, ∗) ∈ Mm (R)−1 q . Proof. (1) Assume that m ≤ n. Construct two maps ϕ:

nR  x1  ..  .  xn ψ : nR   x1  ..  .  

→ mR, 

   x1 x1  ..   ..  7→ A  .  , ∀  .  ∈ nR; xn xn → mR,     x1 x1  ..   ..  7→ (Im 0)m×n  .  , ∀  .  ∈ nR.

xn xn xn Then ϕ and ψ are both R-epimorphisms. Since R is a QB-ring, so is Mn (R) by Theorem 8.1.5. Clearly, nR is quasi-projective. In view of such that ϕ = ψφ. Let Theorem 8.3.1, we have some φ ∈ EndR (nR)−1 q U Mn (R) be the matrix corresponding to φ. Then A = (Im 0)U ; hence,  ∈ A = U ∈ Mn (R)−1 q , as required. ∗ (2) Assume that n ≤ m. Let A ∈ Mm×n (R) be left invertible. Then

o T A ∈ Mn×m Rop is right invertible. As Rop is a QB-ring, it follows  o T A ) ∈ Mm (Rop )−1 ∗) ∈ by (1) that q . Thus, we conclude that (A ∗o Mm (R)−1  q , the result follows. Lemma 8.3.4. Let A be a right R-module, and let n ∈ N. Then the following hold: (1) A is quasi-projective if and only if so is nA. (2) A is quasi-injective if and only if so is nA. Proof. (1) Suppose nA is quasi-projective. Let p : nA → A be the projection onto the first factor and i : A → nA be the injection. For any f : A → C and any R-epimorphism g : A → C, there exists some h ∈ EndR (nA) such that gph = f p. Hence, g(phi) = (gph)i = (f p)i = f . Therefore A is quasi-projective. Conversely, assume that A is quasi-projective. Let qi : A → nA be the injections (1 ≤ i ≤ n). For any f : nA → C and any R-epimorphism

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g : nA → C, it follows by [4, Proposition 16.12] that there exists some hi : A → nA such that ghi = f qi . Construct an R-morphism h : nA → nA n P given by h(x1 , · · · , xn ) = hi (xi ) for any (x1 , · · · , xn ) ∈ nA. Then gh = i=1

f , and therefore nA is quasi-projective. (2) Equivalence is established by [4, Proposition 16.13].



We note that Z/2Z and Q are both quasi-injective Z-modules whose direct sum is not quasi-injective. Proposition 8.3.5. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) For any R-epimorphisms f, g : 2A → M , there exists h ∈ EndR (2A)−1 q such that f = gh. (2) E is a QB-ring and A is quasi-projective. Proof. (1) ⇒ (2) Let h : A → M be an R-epimorphism   and k : A → M a1 an R-morphism. Define η : 2A → M given by η = h(a1 ) for any a2   a1 = k(a1 ) + (h − k)(a2 ) for a1 , a2 ∈ A and τ : 2A → M given by τ a2 any a1 , a2 ∈ A. Then η is an R-epimorphism.   Given any m ∈ M , we have a1 a1 ∈ A such that h(a1 ) = m. Hence τ = k(a1 ) + (h − k)(a1 ) = m, a1 and then τ is an R-epimorphism. By hypothesis, we have an R-morphism ρ : 2A → 2A such that τ = ηρ. Using matrix notation, there is some  ρ11 ρ12 such that ρ21 ρ22      a1 ρ11 ρ12 a1 ρ = , for any a1 , a2 ∈ A. a2 ρ21 ρ22 a2 Furthermore, we get the equation

(k, h − k) = (h, 0)



ρ11 ρ12 ρ21 ρ22



,

      a1 a1 a1 where η = (h 0) for any a1 , a2 ∈ A and τ = (h k − a2 a2 a2   a1 h) for any a1 , a2 ∈ A. Hence k = h̺11 . Thus, A is quasi-projective, a2 and so is 2A from Lemma 8.3.4. According to Theorem 8.3.1, EndR (2A) is a QB-ring. Therefore E is a QB-ring by Lemma 8.1.2.

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(2) ⇒ (1) By virtue of Lemma 8.3.4, 2A is quasi-projective. It follows from Theorem 8.1.5 that EndR (2A) is a QB-ring. Therefore we complete the proof by Theorem 8.3.1.  Lemma 8.3.6. Let A be a quasi-injective right R-module, and let E = EndR (A). Then the following are equivalent: (1) E is a QB-ring. T (2) Given α ∈ E and a submodule K ⊆ A with ker(α) K = 0, then there exists β ∈ E such that K ⊆ ker(β) and α + β ∈ Eq−1 . T Proof. (1) ⇒ (2) Suppose that ker(α) K = 0 with K ⊆ A. Let g : K → A be the inclusion map. Then αg : K → A is an R-monomorphism. Since A is a quasi-injective right R-module, we have an element φ ∈ E such that φαg = g; hence, (1A − φα)g = 0. Since E ia a QB-ring, it follows from φα + (1A − φα) = 1A that α + k(1A − φα) ∈ Eq−1 for some k ∈ E. Let β = k(1A − φα). Then α + β ∈ Eq−1 . Furthermore, we get β(K) = k(1A − φα)(K) = k(1A − φα)g(K) = 0, as required. T (2) ⇒ (1) Given Ef + Eg = E with f, g ∈ E, then ker(f ) ker(g) = 0 from Lemma 3.3.4. By assumption, we can find a β ∈ E such that f + β ∈ Eq−1 and ker(g) ⊆ ker(β). According to Lemma 3.3.3, we have that β ∈ Eg. This implies that f + hg ∈ Eq−1 for some h ∈ E. Therefore the proof is true.  Theorem 8.3.7. Let A be a quasi-injective right R-module, and let E = EndR (A). Then the following are equivalent: (1) E is a QB-ring. (2) For any R-monomorphisms f, g : M → A, there exists h ∈ Eq−1 such that f = hg. Proof. (1) ⇒ (2) Suppose that f, g : M → A are R-monomorphisms. As A is a quasi-injective right R-module, we have an element α ∈ E such T that f = αg. As f is an R-monomorphism, ker(α) im(g) = 0. In view of Lemma 8.3.6, there exists β ∈ E such that im(g) ⊆ ker(β) and α+β ∈ Eq−1 . Let h = α + β. Then hg = (α + β)g = αg = f , as desired. (2) ⇒ (1) Given Ef + Eg = E with f, g ∈ E, it follows by Lemma 3.3.4 T that ker(f ) ker(g) = 0. Let K = ker(g). Define ϕ : K → A given by ϕ(x) = f (x) for any x ∈ K. Let i : K ֒→ A be an inclusion. Then ϕ and i are both R-monomorphisms. By hypothesis, we can find h ∈ Eq−1 such that ϕ = hi. Hence (f − h)(K) = 0, and then ker(g) ⊆ ker(f − h). Since

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A is a quasi-injective right R-module, by Lemma 3.3.3, we can find some k ∈ E such that f − h = kg. This implies that f + (−k)g = h, and therefore we complete the proof by the symmetry of QB-ring.  Corollary 8.3.8. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) For any R-monomorphisms f, g : M → 2A, there exists h ∈ EndR (2A)−1 such that f = hg. q (2) E is a QB-ring and A is quasi-injective. Proof. (1) ⇒ (2) Let h : M → A be an R-monomorphism and k
: M → A an R-morphism. Define η : M → 2A given by η(a) =
h(a), 0 for any a ∈ M and τ : M → 2A given by τ (a) = k(a), (h − k)(a) for any a ∈ M . Clearly, η and τ are both R-monomorphisms. By assumption, we can find an R-morphismρ ∈ EndR (2A) such that  τ= ρη.  Using matrix   notation, ρ11 ρ12 a1 ρ11 ρ12 a1 we have some such that ρ = for any ρ21 ρ22 a2 ρ21 ρ22 a2 a1 , a2 ∈ A. Furthermore, we get the equation 

k h−k



=



ρ11 ρ12 ρ21 ρ22

  h , 0

    h k where η(a) = a for any a ∈ M and τ (a) = a for any a ∈ M. 0 h−k Hence k = ̺11 h, i.e., A is quasi-injective. According to Lemma 8.3.4, Theorem 8.3.7 and Lemma 8.1.2, E is a QB-ring. (2) ⇒ (1) The result follows from Proposition 8.3.5.  Corollary 8.3.9. Let A be a semisimple right R-module, and let E = EndR (A). Then the following hold: (1) For any R-epimorphisms f, g : A → M , there exists h ∈ Eq−1 such that f = gh. (2) For any R-monomorphisms f, g : M → A, there exists h ∈ Eq−1 such that f = hg. Proof. Since A is a semisimple right R-module, it is both quasi-projective and quasi-injective. In view of [217, Corollary 1.23], E is a QB-ring. Therefore we complete the proof by Theorem 8.3.1 and Theorem 8.3.7. 

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Quasi-Projective Modules

Let A be a quasi-projective right R-module. By Theorem 8.3.1, EndR (A) is a QB-ring if and only if for any R-epimorphisms π, σ : A → M and any R-isomorphism u ∈ EndR (M ), the diagram of π

A։M v↓ ↓u σ

A։M

form can be completed by a quasi-invertible v. The main purpose of this section is to extend this result to the general case. Let π : A → M be an R-morphism, and let S and T be subsets of EndR (A) and EndR (M ) respectively. Then πS := {πf | f ∈ S} and T π := {f π | f ∈ T }. The proof of the following lemma is obvious and therefore will be omitted. Lemma 8.4.1. Let πi : Ai →
Mi (i = 1, 2) be R-epimorphisms, and let σ : A1 → A2 . If σ ker(π1 ) ⊆ ker(π2 ), then there exists a unique σ ∗ : M1 → M2 such that the diagram π1

A1 ։ M1 σ↓ ↓ σ∗ π2

A2 ։ M2

commutates.



Let π : A → M be an R-epimorphism, and let Kerπ = {f ∈ EndR (A) | πf = 0}. Then Kerπ is a right ideal of EndR (A). Especially, Kerπ is an ideal if ker(π) is fully invariant. Lemma 8.4.2. Let A be a quasi-projective right R-module, and let π : A → M be an R-epimorphism. If ker(π) is fully invariant, then EndR (A)/Kerπ ∼ = EndR (M ). Proof. Since ker(π) is fully invariant, by Lemma 8.4.1, we can construct a map ϕ : EndR (A) → EndR (M ) given by ϕ(f ) = f ∗ for any f ∈ EndR (A). By the uniqueness, ϕ is a ring homomorphism. Let g ∈ EndR (M ). As A is quasi-projective, there exists some h ∈ EndR (A) such that gπ = πh. By Lemma 8.4.1 again, we see that h∗ = f . Therefore ϕ(h) = f , i.e., ϕ is surjective. In addition, we see that ker(ϕ) = {f ∈ EndR (M ) | f ∗ = 0} =

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{f ∈ EndR (M ) | f ∗ π = 0} = {f ∈ EndR (M ) | πf = 0} = Kerπ. Therefore EndR (A)/Kerπ ∼  = EndR (M ), as asserted. Lemma 8.4.3. Let A be a quasi-projective right R-module, let π : A → M be an R-epimorphism and let ker(π) be fully invariant. If u ∈ EndR (A)−1 q , ∗ −1 then u ∈ EndR (M )q . Proof. Let E = EndR (A), F = EndR (M ) and u ∈ Eq−1 . It follows from Lemma 8.1.1 that (1A − uv) ⊥ (1A − vu) for a v ∈ E. Since A is quasiprojective, we see that F π ⊆ πE. In view of Lemma 8.4.1, one easily checks that (1M − u∗ v ∗ )F (1M − v ∗ u∗ )(M ) ⊆ (1M − u∗ v ∗ )F (1M − v ∗ u∗ )F π(A) = (1M − u∗ v ∗ )F (1M − v ∗ u∗ )πE(A) = π(1A − uv)E(1A − vu)E(A) = 0. Hence, (1M − u∗ v ∗ ) ⊥ (1M − v ∗ u∗ ). Likewise, (1M − v ∗ u∗ ) ⊥ (1M − u∗ v ∗ ). Then u∗ ∈ Fq−1 , the result follows.  Lemma 8.4.4. Let A be a quasi-projective right R-module, let π : A → M be an R-epimorphism and let ker(π) be fully invariant. If EndR (A) is a −1 QB-ring, then for any u ∈ EndR (M )−1 q , there exists some u∗ ∈ EndR (A)q such that uπ = πu∗ , i.e., the diagram π

A։M u∗ ↓ ↓u π

A։M

commutates. Proof. Let E = EndR (A), and let u ∈ EndR (M )−1 q . By Lemma 8.1.1, we have α ∈ EndR (M ) such that (1M − uα) ⊥ (1M − αu), u = uαu, α = αuα. Since A is quasi-projective, there are w, β ∈ E such that uπ = πw and απ = πβ, i.e., the following diagrams commutate: π

π

A։M A։M w↓ ↓ u and β ↓ ↓α π

A։M

π

A ։ M.

Hence uαπ = uπβ = πwβ and αuπ = απw = πβw. This implies that (1 − uα)π = π(1 − wβ) and (1 − αu)π = π(1 − βw). As E is a QB-ring, it follows from wβ + (1A − wβ) = 1A that there exists some z ∈ E such that p = β + z(1A − wβ) ∈ Eq−1 . According to Lemma 8.1.1, there exists an element q ∈ E such that (1 − pq) ⊥ (1 − qp) and p = pqp, q = qpq. Let u∗ = q + w(1A − pq) + (1A − qp)w. Then u∗ ∈ Eq−1 . Since u = uαu, we get πw = uπ = uαuπ = uαπw = uπβw = πwβw. Likewise, πβ = πβwβ. Let z ∈ E. In view of Lemma 8.4.1, there exists

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some z ∗ ∈ EndR (M ) such that z ∗ π = πz. In addition, we have π(1A − βw)z(1A − wβ) = (1 − αu)πz(1A − wβ) = (1 − αu)z ∗ π(1A − wβ) = (1 − αu)z ∗ (1 − uα)π = 0. Hence,



πpwp = π β + z(1A − wβ) w
β + z(1A − wβ)
= πβ + z ∗ π(1A − wβ)
w β + z(1A − wβ) = πβw β + z(1A − wβ) = πβ + πβwz(1A − wβ) = πβ + πz(1A − wβ) − π(1 − βw)z(1 − wβ) = πβ + πz(1A − wβ) = πp.

Since (1A − qp)w(1A − pq) = 0, we get w = q + w(1A − pq) + (1A − qp)w + (qpwpq − q). Clearly, we have some q ∗ ∈ EndR (M ) such that q ∗ π = πq. It follows that π(qpwpq − q) = = = = =

q ∗ (πpwpq − π) q ∗ (πpq − π) q ∗ πpq − q ∗ π πqpq − πq 0.

Thus, πw = πq + πw(1A − pq) + π(1A − qp)w = πu∗ and the result follows.  Let I be an ideal of a QB-ring R. For any u ∈ EndR (R/I)−1 q , there exists some u∗ ∈ Rq−1 such that u(1) = u∗ . As every ideal is fully invariant, we are done by Lemma 8.4.4. Let cr(Rq−1 ) = {a ∈ R | aR + bR = R ⇒ ∃ y ∈ R such that a + by ∈ Rq−1 }. Further, we can generalize [25, Theorem 7.2] as follows. Theorem 8.4.5. Let A be a quasi-projective right R-module, and let π : A → M be an R-epimorphism. If ker(π) is fully invariant, then EndR (A) is a QB-ring if and only if the following hold: (1) EndR (M ) is a QB-ring. −1 (2) EndR (M )−1 q π = πEndR (A)q .

−1 (3) EndR (A)q + HomR A, ker(π) ⊆ cr EndR (A)−1 . q

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Proof. Suppose that EndR (A) is a QB-ring. By virtue of Lemma 8.4.2, EndR (M ) ∼ = EndR (A)/Kerπ. Thus, EndR (M ) is a QB-ring. In view of −1 Lemma 8.4.4, EndR (M )−1 q π ⊆ πEndR (A)q . It follows from Lemma 8.4.3 −1 that πEndR (A)−1 q ⊆ EndR (M )q π, and so the conditions above hold. Suppose that (1), (2) and (3) hold. Given ax + b = 1A in EndR (A), then πax + πb = π, and so a∗ x∗ π + b∗ π = π. As π is surjective, we get a∗ x∗ + b∗ = 1M . By (1), there exists some y ∈ EndR (M ) such that −1 a∗ + b∗ y ∈ EndR (M )−1 such q . By (2), there exists some u ∈ EndR (A)q ∗ ∗ that a π + b yπ = πu. As A is quasi-projective, we have an element y∗ ∈ EndR (A) such that yπ = πy∗ , thus, π(a
+ by∗ ) = πu. Let h = a + by∗ − u. Then h ∈ Kerπ = HomR A, ker(π) . Since ax + b = 1A , (a + by∗ )x + b(1A − y∗ x) = 1A . As a + by∗ = u + h ∈ EndR (A)−1 q + Kerπ, it follows by −1 (3) that a + by∗ + b(1A − y∗ x)z ∈ EndR (A)q for some z ∈ EndR (A). Let t = y∗ + (1A − y∗ x)z. Then a + bt ∈ EndR (A)−1 q , and therefore EndR (A) is a QB-ring.  For any right R-module A, we define rad(A) to be the intersection of all the maximal submodules of A. If there are no submodules in A, we define rad(A) to be A. Corollary 8.4.6. Let A be a quasi-projective right R-module, and let π : A → A/rad(A) be the canonical map. Then EndR (A) is a QB-ring if and only if the following hold:
(1) EndR A/rad(A) is a QB-ring.
−1 (2) EndR A/rad(A) q π = πEndR (A)−1 .
q
(3) EndR (A)−1 + Hom A, rad(A) ⊆ cr EndR (A)−1 . R q q
Proof. If f : A → A, then f rad(A) ⊆ rad(A); hence, rad(A) is fully invariant. The result follows from Theorem 8.4.5.  The following facts are dualities of QB properties on quasi-projective modules. Lemma 8.4.7. Let πi : M
i → Ai (i = 1, 2) be R-monomorphisms, and let σ : A1 → A2 . If σ im(π1 ) ⊆ im(π2 ), then there exists a unique σ∗ : M1 → M2 such that the diagram π2

M2 ֌ A2 σ∗ ↑ ↑σ π1

M1 ֌ A1

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Let π : M → A be a right R-morphism, and let Cokerπ = {f ∈ EndR (A)| f π = 0}. Then Cokerπ is a left ideal of EndR (A). Especially, Cokerπ is an ideal of EndR (A) if im(π) is full invariant. Lemma 8.4.8. Let A be a quasi-injective right R-module, and let π : M → A be an R-monomorphism. If im(π) is fully invariant, then EndR (A)/Cokerπ ∼  = EndR (M ). Lemma 8.4.9. Let A be a quasi-injective right R-module, let π : M → A be an R-monomorphism and let im(π) be fully invariant. If u ∈ EndR (A)−1 q , then u∗ ∈ EndR (M )−1 .  q Lemma 8.4.10. Let A be a quasi-injective right R-module, let π : M → A be an R-monomorphism and let im(π) be fully invariant. If EndR (A) is a ∗ −1 QB-ring, then for any u ∈ EndR (M )−1 q , there exists some u ∈ EndR (A)q ∗ such that πu = u π.  Proposition 8.4.11. Let A be a quasi-injective right R-module, and let π : M → A be an R-monomorphism. If im(π) be fully invariant, then EndR (A) is a QB-ring if and only if the following hold: (1) EndR (M ) is a QB-ring. −1 (2) EndR (A)−1 q π = πEndR (M )q .
−1 (3) EndR (A)q + Cokerπ ⊆ cr EndR (A)−1 . q



Corollary 8.4.12. Let A be a quasi-injective right R-module, and let π : rad(A) → A be the inclusion. Then EndR (A) is a QB-ring if and only if the following hold:
(1) EndR rad(A) is a QB-ring.
−1 (2) EndR (A)−1 q π = πEndR rad(A) q .
−1 (3) EndR (A)−1 . q + Cokerπ ⊆ cr EndR (A)q

Proof. Obviously, rad(A) is fully invariant, and therefore we complete the proof by Proposition 8.4.11. 

Let M be a right R-module. We say that E(M ) is an envelope of M in the case where E(M ) is an injective right R-module and there exists an inclusion π : M ֒→ E(M ) such that π is an essential R-monomorphism,

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i.e., im(π) is a essential submodule of E(M ). For example, E(Z) = Q. It is well known that every right R-module has an envelope. This also shows that every right R-module can be seen as a submodule of a quasi-injective module. Corollary 8.4.13. Let M be a quasi-injective right
R-module, and let π : M → E(M ) be the inclusion. Then EndR E(M ) is a QB-ring if and only if the following hold: (1) EndR (M ) is
a QB-ring. −1 (2) EndR E(M ) q π = πEndR (M )−1 q .
−1
−1
(3) EndR E(M ) q + Cokerπ ⊆ cr EndR E(M ) q .
Proof. Let f ∈ EndR E(M ) , and let N = {m ∈ M | f (m) ∈ M }. Then N ⊆ M . As M is quasi-injective, there exists some g ∈ EndR (M ) such that g|N = f |N . This implies that \ im(f |M − g) M = 0.

As M is a essential submodule of E(M ), im(f |M − g) = 0, i.e., g = f |M . Hence, f (M ) = g(M ) ⊆ M . Therefore, M is a fully invariant submodule of E(M ). Clearly, E(M ) is quasi-injective, and so the result follows by Proposition 8.4.11. 


Let M be a quasi-injective right R-module. If EndR E(M ) is a QB-ring, it
follows by Corollary 8.4.13 that there is a bijection from −1 EndR E(M ) q to EndR (M )−1 q .

Lemma 8.4.14. Let A be a quasi-projective right R-module, and let E = EndR (A). Suppose that A = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 implies that there exist orthogonal ideals I, J and ψ : B1 → B2 such that ψI is a splitting epimorphism and ψJ is a splitting monomorphism. If E is an exchange ring, then E is a QB-ring. Proof. Let a ∈ E be regular. Then we have x ∈ E such that a = axa and x = xax. Since A = aA ⊕ (1 − ax)A = xaA ⊕ (1 − xa)A with φ : aA = axA ∼ = xaA, we get ψ : (1 − ax)A → (1 − xa)A such that ψI is a splitting epimorphism and ψJ is a splitting monomorphism, where I and J are orthogonal ideals of R. Construct a map ϕ : A → A given by ϕ(s + t) = φ(s) + ψ(t) for any s ∈ aA and t ∈ (1 − ax)A. Then a = aϕa. Let π : A → A/AI. By virtue of Lemma 8.4.2, we have an isomorphism α∗ : EndR (A)/K ∼ = EndR (A/AI), where K := Kerπ = {σ ∈

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E | σ(A) ⊆ AI}. Likewise, we have L = {τ ∈ E | τ (A) ⊆ AJ} such that β ∗ : EndR (A)/L ∼ = EndR (A/AJ). It is easy to check that K and L are orthogonal ideals of E. In addition, α∗ (ϕ + K) = ϕI and β ∗ (ϕ + L) = ϕJ . As α∗ and β ∗ are R-isomorphisms, we deduce that ϕ + K is right invertible in E/K and ϕ+L is left invertible in E/L. Thus, we can find some λ, µ ∈ E such that 1−ϕλ ∈ K and 1−µϕ ∈ L. This implies that (1−ϕλ) ⊥ (1−µϕ), and so ϕ ∈ Eq−1 . According to Lemma 8.1.8, E is a QB-ring.  We end this chapter by giving an extension of Theorem 8.2.7. Proposition 8.4.15. Let A be an R-progenerator of an exchange ring R. Then the following are equivalent: (1) R is a QB-ring. (2) For any right R-modules B1 and B2 , A ⊕ B1 ∼ = A ⊕ B2 implies that there exist orthogonal ideals I, J and ψ : B1 → B2 such that ψI is a splitting epimorphism and ψJ is a splitting monomorphism. Proof. (1) ⇒ (2) is obvious from Theorem 8.2.7. (2) ⇒ (1) Suppose that A = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 . Then A ⊕ B1 ∼ = A2 ⊕ B2 ⊕ B1 ∼ = A1 ⊕ B1 ⊕ B2 ∼ = A ⊕ B2 ; hence, there exist orthogonal ideals I, J and ψ : B1 → B2 such that ψI is a splitting epimorphism and ψJ is a splitting monomorphism. According to Lemma 8.4.14, EndR (A) is a QB-ring. Since A is an R-progenerator, it is a finitely generated projective right R-module; hence, there exist idempotents e1 , · · · , en ∈ R such that A∼ = e1 R ⊕ · · · ⊕ en R. In addition, we have a right R-module D and m ∈ N such that R⊕D ∼ = mA. Thus, we have

R⊕D ∼ = e1 R ⊕ · · · ⊕ en R ⊕ · · · ⊕ e1 R ⊕ · · · ⊕ en R ∼ = diag(E, · · · , E)(mnR), where E = diag(e1 , · · · , en ). It follows that N N R1×mn ⊕ D R1×mn ∼ = diag(E, · · · , E)(mnR) R1×mn R

R

∼ = diag(E, · · · , E)Mmn (R). Let ϕ : diag(E, · · · , E)Mmn (R) → R1×mn be the projection onto R1×mn . Then we can find some A1 , · · · , Amn ∈ Mmn (R) such
that (1, 0, · · · , 0) = ϕ diag(E, · · · , E)A1
, (0, 1, · · · , 0) = ϕ diag(E, · · · , E)A2 , .. .
(0, 0, · · · , 1) = ϕ diag(E, · · · , E)Amn .

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Thus we get  In 0 · · ·  0 In · · ·   . . .  .. .. . .

0 0 .. .

    


 ϕ diag(E, · · · , E)   0   =  ..   . 

0 0 · · · In mn   0
 ϕ diag(E, · · · , E)    +  ..   . 0

    



0 0 .. .


ϕ diag(E, · · · , E)

0

diag(E, · · · , E)A1

mn

diag(E, · · · , E)A2 + · · · +

mn

   diag(E, · · · , E)Amn , 

where In = diag(1, 1, · · · , 1)n×n .    In 0 · · · In 0 · · · 0  0 0n · · · 0   0 In · · ·     . . . .   . .  .. .. . . ..   .. .. . . .

Furthermore, we see that    In 0 · · · 0 0   0   0 0n · · · 0    .  . . . .. .   .. .. . . .. 

0 0 · · · 0n mn 0 0 · · · In mn 0 0 · · · 0n mn     L11 0n · · · 0n K11 K12 · · · K1m  L21 0n · · · 0n   0n 0n · · · 0n      diag(E, · · · , E) = .   . . . .  . . .. . . . ..   .. .. . . ..   ..

0n 0n · · · 0n  Kn1 Kn2 · · · Knm  0n 0n · · · 0n     . .. . . ..   .. . .  . 

0n

0n · · · 0n

mn

mn



  diag(E, · · · , E)  

Lm1 0n · · · 0n  L1n 0n · · · 0n L2n 0n · · · 0n   .. . . ..  .. . .  . .

Lmn 0n · · · 0n

+ ···+ mn

. mn

As a result, we can find some Kij , Lji ∈ Mn (R)(1 ≤ i ≤ n, 1 ≤ j ≤ m) such P that In = Kij ELji . Hence E ∈ Mn (R) is a full idempotent 1≤i≤n,1≤j≤m

matrix. As A ∼ = E(nR), we get EndR (A) ∼ = EMn (R)E, and so EMn (R)E is a QB-ring. According to Corollary 8.1.7, Mn (R) is a QB-ring. Clearly, R ∼ = diag(1, 0, · · · 0)n×n Mn (R)diag(1, 0, · · · 0)n×n . Therefore R is a QBring by Lemma 8.1.2. 

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Chapter 9

P B-Rings

In this chapter, we introduce a new class of rings. A ring belonging to this class is called a P B-ring if some unimodular row can be reduced in a similar way as it can be done for QB-rings. We say that x, y ∈ R are pseudoorthogonal, in symbols x♮y, if there exists n ∈ N such that (RxRyR)n = 0. Clearly, x⊥y implies that x♮y. But the converse is not true. Let F be a field and R = F [x]/(x2 ). Choose u = 1+(x2 ) and v = x+(x2 ). Then u♮v, while u and v are not centrally orthogonal. Now we replace centrally orthogonality with pseudo-orthogonality in QB-rings. Let Rp−1 = {u ∈ R | ∃ a, b ∈ R such that (1 − ua)♮(1 − bu)}. The condition used to define a P B-ring R is the following: aR + bR = R =⇒ a + by ∈ Rp−1 for some y ∈ R. The main purpose of this chapter is to extend as much of the theory of QB-rings as possible to the much larger class of P B-rings.

9.1

Counterexamples

A ring R is prime in the case where xRy = 0 implies that x = 0 or y = 0. A ring R is semiprime in the case where xRx = 0 implies that x = 0. For semiprime rings, the notions of P B-rings and QB-rings coincide. In this section, we find examples of P B-rings that are not QB-rings. Lemma 9.1.1. A ring R is a P B-ring if and only if so is T Mn (R) 265

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Proof. It is straightforward.



Lemma 9.1.2  Let  R be a prime ring and  u,v, z ∈ R. If uv = 1 and u0 u0 vu 6= 1, then 6∈ T M2 (R)−1 ∈ T M2 (R)−1 q , while p . zv zv     u0 u0 −1 Proof. Clearly, ∈ T M2 (R)p . If ∈ T M2 (R)−1 q , then we can zv zv   x0 find ∈ T M2 (R) such that t y       u0 u0 x0 u0 = zv zv t y zv and           10 u0 x0
10 x0 u0
− ⊥ − . 01 zv t y 01 t y zv We infer that u = uxu and v = vyv; hence, ux = 1 and yv = 1. Furthermore, we have       1 − ux 0 0 0 1 − xu 0 00 = . −(zx + vt) 1 − vy R0 −(tu + yz) 1 − yv 00 This implies that (1−vy)R(1−xu) = 0. As a result, xu= 1 or  vy = 1. Thus, u0 either u or v is invertible, a contradiction. Therefore 6∈ T M2 (R)−1 q . zv  Example 9.1.3. Let R be a purely infinite simple ring. Then T M2 (R) is a P B-ring, while it is not a QB-ring. Proof. By the proof of Theorem 5.3.5, R is weakly stable; hence, it is a QBring. Clearly, R is semiprime. It follows from Lemma 9.1.1 that T M2 (R) is a P B-ring. On the other hand, R is not a division ring; hence, there exists some nonzero element u ∈ R, but it is not right invertible.  Inaddition, we y 0 have s, t ∈ R such that sut = 1 and uts 6= 1. For any ∈ T M2 (R), wz we see  that      s 0 0 0 y 0 s 0 + = . 0 ut 0 1 − uts wz (1 − uts)w ut + (1 − uts)z

One easily checks that s ut + (1 − uts)z = 1 and ut + (1 − uts)z s 6= 1. As R is simple, it is prime. In view of Lemma 9.1.2,      s 0 0 0 y 0 + 6∈ T M2 (R)−1 q . 0 ut 0 1 − uts wz

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Therefore T M2 (R) is not a QB-ring.

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Let V be an infinite-dimensional vector space over a division ring D, and let M = {f ∈ EndD (V ) | dimD (f V ) < ∞}. Then M is the unique maximal of EndD (V ). By [18, Example 1.3], EndD (V ) modulo M is purely
infinite. According to Example 9.1.3, T M2 EndD (V )/M is a P B-ring but it is not a QB-ring. Example 9.1.4. Let V be an infinite-dimensional vector space over a
division ring D. Then T M2 EndD (V ) is a P B-ring, while it is not a QB-ring.
Proof. In view of Lemma 9.1.1, T M2 EndD (V ) is a P B-ring. Let {x1 , x2 , · · · , xn , · · · } be a basis of V . Define σ : V → V given by σ(xi ) = xi+1 (i = 1, 2, · · · ) and τ : V → V given by τ (x1 )= 0  and τ (xi ) = xi−1 (i = 2, 3, · · · ).
x0 Then τ σ = 1V and στ 6= 1V . For any ∈ T M2 EndR (V ) , we have zy        τ 0 0 0 x0 τ 0 + = . 0σ 0 1 − στ zy (1 − στ )z σ + (1 − στ )y Clearly, EndR (V ) is a prime ring. Analogously to Example 9.1.3, we prove that  
−1 τ 0 6∈ T M2 EndR (V ) q , (1 − στ )z σ + (1 − στ )y as required.




In Example 9.1.4, we
set e = diag(1V , 0). Then eT M2 EndR (V ) e and
(1 − e)T M2 EndR (V ) (1 − e) are both QB-rings, but T M2 EndR (V ) is
not. This means that EndR A ⊕ B being a QB-ring is not equivalent to EndR (A) and EndR (B) being QB-rings. Example 9.1.5. Let R be a regular QB-ring. Then R[x]/(xn+1 ) is an exchange P B-ring. Proof. Denote x in R[x]/(xn+1 ) by u, so R[x]/(xn+1 ) ∼ = R[u], where un+1 = 0 and u commutates with any elements of R. As in the proof
of [264, Proposition 2.3], J R[u] = J(R) + (u) = (u) and R[u]/J(R[u]) ∼ = R/J(R) ∼ = R. It follows by [382, Theorem 29.2] that R[u] is an exchange
n+1 ring. As un+1 = 0, we get J R[u] = 0, and therefore R[u] is a P B-ring. 

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Morita Equivalences

In several respects, P B-rings behave like QB-rings. We now develop the analogues of the results proved in the preceding chapter for QB-rings. Following the line of proof of Theorem 8.1.5, we shall prove that the notation of P B-rings has the Morita invariant property. Let I be an ideal of a ring R, and let Q(I) = {x ∈ R | ∃ m ∈ N such that (RxR)m ⊆ I}. Then Q(I) is an ideal of R. Let N (R) = {x ∈ R | RxR is nilpotent}. Then N (R) = Q(0) is an ideal of R. We begin this section with a simple fact. Lemma 9.2.1. Let R be a ring, and let u ∈ R. Then the following are equivalent: (1) u ∈ Rp−1 . (2) There exists v
∈ R such that (1 − uv)♮(1 − vu) and u ≡ uvu, v ≡ vuv mod N (R) .

Proof. (2) ⇒ (1) is trivial. (1) ⇒ (2) As u ∈ Rp−1 , we have a, b ∈ R such that (1 − ua)♮(1 − bu). Let w = a + b − aub. Then 1 − uw = (1 − ua)(1 − ub) and 1 − wu = (1 − au)(1 − bu). This implies that (1 − uw)♮(1 − wu), and so R(1 − uw)R(1 − wu)R is nilpotent. Hence, (1 − uw)u(1 − wu) ∈ N (R), and then
u ≡ 2uwu − uwuwu mod N (R) . Let q = 2w − wuw. Then 1 − uq = (1 − uw)2 and 1 − qu = (1 − wu)2 . Thus, (1 − uq)♮(1 − qu). In addition,
we have uqu = 2uwu − uwuwu ≡ u mod N (R) . Furthermore, we get
u−u(quq)u = (u−uqu)+(u−uqu)qu ∈ N (R), and so u ≡ uvu mod N (R)
, where v = quq. On the other hand,
v −vuv = q (u−uqu)+(u−uqu)qu q ∈ N (R). That is, v ≡ vuv mod N (R) . Since 1 − uv = (1 − uq)(1 + uq) and 1 − vu = (1 − qu)(1 + qu), we deduce that (1 − uv)♮(1 − vu), as asserted. We say that an ideal I of a ring R is locally nilpotent provided that for any x ∈ I, RxR is nilpotent. Thus, N (R) is the largest locally nilpotent ideal of R. In addition, Q(I) = {x ∈ R | x ∈ N (R/I)} where I is an ideal of R. Let p, q ∈ R be idempotents such that p 6 ♮q. We say that x ∈ pRp and y ∈ qRq are pseudo-orthogonal in R, in symbols x♮R y, if there exists n ∈ N such that (RxRyR)n = 0. Let [pRq]−1 p = {u ∈ pRq | ∃ v ∈ qRp such that (p − uv)♮R (q − vu)}.

We say that pRq is a P B-corner in the case where (1) ax + b = p with a ∈ pRq, x ∈ qRp and b ∈ pRp implies that there exists y ∈ pRq such that

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a + by ∈ [pRq]−1 p and (2) ax + b = q with a ∈ qRp, x ∈ pRq and b ∈ qRq implies that there exists y ∈ qRp such that a + by ∈ [qRp]−1 p . The notion of P B-corner parallels the one introduced in Chapter 8. Lemma 9.2.2. If p and q are idempotents of a P B-ring R such that (1 − p)R ∼ = (1 − q)R, then either p♮q or pRq is a P B-corner.

Proof. Since (1 − p)R ∼ = (1 − q)R, there exist u ∈ (1 − p)R(1 − q), v ∈ (1 − q)R(1 − p) such that 1 − p = uv and 1 − q = vu. Given ax + b = p with a ∈ pRq, x ∈ qRp and b ∈ pRp, then (a + u)(x + v) + b = 1. So we can
find −1 y ∈ R such that a + u + by ∈ R . By Lemma 9.2.1, 1 − (a + u + by)c p


♮ 1 − c(a+ u + by) for a c ∈ R. Hence p− (a+ by)cp ♮ q − (qcp)(a+ b(pyq)) and


(ucp)♮ q −
(qcp)(a+ b(pyq)) . As p− a+ b(pyq)
(qcp) = p− (a+ by)(cp) +
(a + by)v (ucp), we get p − (a + b(pyq))(qcp) ♮R q − (qcp)(a + b(pyq)) . This implies that a + b(pyq) ∈ [pRq]−1 p . The symmetry is true as well.  Corollary 9.2.3. Every corner of a P B-ring is also a P B-ring. Proof. Let R be a P B-ring, and let p ∈ R be a nonzero idempotent. Then p 6 ♮p. In view of Lemma 9.2.2, pRp is a P B-corner. Given ax + b = p with a, x, b ∈ pRp, then there exists an element y ∈ pRp
such that a + by
∈ [pRp]−1 p , i.e., we have a u ∈ pRp such that p −
(a + by)u ♮R p − u(a + by) . n That is, R(p − (a + by)u)R(p − u(a + by))R = 0 for some n ∈
N. As n a result, we get (pRp)(p − (a + by)u)(eRe)(p − u(a + by))(pRp) = 0. −1 Therefore a + by ∈ (pRp)p , as desired.  Lemma 9.2.4. Let R be a P B-ring. If (1 − ux)♮(1 − xu), (1 − vy)♮(1 − yv), u = uxu, v = vyv with x, y, u, v ∈ R, then either (1 − ux)♮(1 − yv) or (1 − ux)R(1 − yv) is a P B-corner. Proof. Let I = R(1 − xu)R + R(1 − vy)R. If I = R, then we have R(1 − ux)R(1 − yv)R ⊆ R(1 − ux)R(1 − xu)R + R(1 − vy)R(1 − yv);

hence, (1 − ux)♮(1 − yv). Otherwise, (ux)(R/I) ∼ = (yv)(R/I). In view of Lemma 9.2.2, we have either (1) (1 − ux)♮(1 − yv) or (2) (1 − ux)(R/I)(1 − yv) is a P B-corner.
If (1) holds, then there exists some p p ∈ N such that R(1 − ux)R(1 − yv)R ⊆ I. Thus,
p+1 R(1 − ux)R(1 − yv)R ⊆ R(1 − xu)R(1 − ux)R + R(1 − vy)R(1 − yv)R.

This implies that R(1 − ux)R(1 − yv)R is a nilpotent ideal. That is, (1 − ux)♮(1 − yv). Assume that (2) holds. Given ab + c = 1 − ux with a ∈

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(1 − ux)R(1 − yv), b ∈ (1 − yv)R(1 − ux) and c ∈ (1 − ux)R(1 − yv), there  −1 exists y ∈ (1 − yv)R(1 − ux) such that a + cy ∈ (1 − ux)(R/I)(1 − yv) p . So we have n ∈ N and w ∈ (1 −
nyv)R(1 − ux) such that R((1 − ux) − (a + cy)w)R((1 − yv)− w(a+ cy))R ⊆ I. Thus, R((1 − ux)− (a+ cy)w)R((1 −
n+1 yv) − w(a + cy))R ⊆ R(1 − xu)R(1 − ux)R + R(1 − vy)R(1 − yv)R. This
m(n+1) implies that R((1 − ux) − (a + cy)w)R((1 − yv) − w(a + cy))R =0  −1 for some m ∈ N. Consequently, a + cy ∈ (1 − ux)R((1 − yv) p . The symmetry is similar, and therefore the proof is true.  We say that a unimodular row (a, b) is pseudo reducible if there exists some y ∈ R such that a + by ∈ Rp−1 . As in the proof of Lemma 8.1.4, a unimodular (a, b) is pseudo reducible if and only if so is (vau + vbc, vb), where u, v ∈ U (R) and c ∈ R. Let I be a nilpotent ideal of a ring R. We −1 note that x ∈ (R/I)−1 p if and only if x ∈ Rp . This will be repeatedly used in the sequel. Theorem 9.2.5. If R is a P B-ring, then so is Mn (R) for all n ∈ N. Proof. Given 

a11 a12 a21 a22



x11 x12 x21 x22



+



b11 b12 b21 b22



=



10 01



(‡)

in M2 (R), then a11 x11 + (a12 x21 + b11 ) = 1 in R. Thus, we can find a y ∈ R such that a11 + (a12 x21 + b11 )y = u ∈ Rp−1 . (‡) is pseudo reducible if and only if this is so for the row with elements         a11 a12 1 0 b11 b12 y0 b b + , 11 12 . a21 a22 x21 y 1 b21 b22 00 b21 b22 So we assume that a11 ∈ Rp−1 . By Lemma 9.2.1, there exists x ∈
R such that (1 − a11 x)♮(1 − xa11 ) and a11 ≡ a11 xa11 , x ≡ xa11 x mod N (R) . It suffices to prove
that (‡) is pseudo reducible modulo M2 R(a11 −a11 xa11 )R+R(x− xa11 x)R . So we may assume that a11 = a11 xa11 and x = xa11 x. Thus, it suffices to prove the unimodular row with elements        1 0 a11 a12 1 −xa12 1 0 b11 b12 , −a21 x 1 a21 a22 0 1 −a21 x 1 b21 b22 is pseudo reducible. Assume that a11 xa12 = 0 = a21 xa11 in (‡). Likewise, we assume that a11 xa12 = 0 = a12 ya22 , a21 xa11 = 0 = a22 ya21 , a11 = a11 xa11 , x = xa11 x, a22 = a22 ya22 , y = ya22 y, (1 − a11 x)♮(1 − xa11 ) and (1 − a22 y)♮(1 − ya22 ) in (‡).

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Since a11 x11 + a12 x21 + b11 = 1, we have (1 − a11 x)a12 (1 − ya22 )x21 (1 − a11 x) + (1 − a11 x)b11 (1 − a11 x) = 1 − a11 x. In view of Lemma 9.2.4, either (1) (1 − a11 x)♮(1 − ya22 ) or (2) (1 − a11 x)R(1 − ya22 ) is a P B-corner. If (2) holds, then there exists some q ∈ (1 − a11 x)R(1 − ya22 ) such that  −1 a12 + (1 − a11 x)b11 (1 − a11 x)q ∈ (1 − a11 x)R(1 − ya22 ) p . Thus, it suffices to prove the unimodular row with elements         a11 a12 1 −xb11 q b11 b12 0q b11 b12 + , a21 a22 0 1 b21 b22 00 b21 b22

is pseudo reducible. Modulo the nilpotent ideal
R (a22 + b21 (1 − a11 x)q) − (a22 + b21 (1 − a11 x)q)y(a22 + b21 (1 − a11 x)q) R, we may assume that a11 = a11 xa11 , x = xa11 x, a22 = a22 ya22 , y = ya22 y, (1 − a11 x)♮(1 − xa11 ), (1 − a22 y)♮(1 − ya22 ), a11 xa21 = 0 = a21 ya22 , a21 xa11 = 0 = a22 ya12 , and that  −1 either (1 − a11 x)♮(1 − ya22 ) or a12 ∈ (1 − a11 x)R(1 − ya22 ) p . Similarly,  −1 we assume that (1 − a22 y)♮(1 − xa11 ) or a21 ∈ (1 − a22 y)R(1 − xa11 ) p . In view of Lemma 9.2.1, there exist s ∈ (1 − ya22 )R(1 − a11 x), t ∈ (1 − xa11 )R(1 − a22 y) such that (1 − a11 x − a12 s)♮(1 − ya22 − sa12 ) and (1 − a22 y − a21 t)♮(1 − xa11 − ta21 ). One easily checks that      a11 a12 xt a11 x + a12 s 0 = , a a sy 0 a21 t + a22 y  21 22     xt a11 a12 xa11 + ta21 0 = . sy a21 a22 0 sa12 + ya22   a11 a12 This implies that ∈ M2 (R)−1 p . Thus, M2 (R) is a P B-ring. As a21 a22 in the proof of Theorem 8.1.5, we obtain the result. 

Let R be a P B-ring. Then every n × n regular matrix over R is the product of an idempotent matrix E ∈ Mn (R) and a matrix U ∈ Mn (R)−1 p . In view of Theorem 9.2.5, Mn (R) is a P B-ring. Let A ∈ Mn (R) be regular. Then we have B ∈ Mn (R) such that A = ABA. As AB + (In − AB) = In , there exists Y ∈ Mn (R) such that A + (In − AB)Y = U ∈ Mn (R)−1 p . Let E = AB. Then A = EU , and we are done. Let I1 and I2 be ideals of a ring R. I1 and I2 are pseudo-orthogonal provided that there exists n ∈ N such that (I1 I2 )n = 0. One easily checks T that I1 and I2 are pseudo-orthogonal if and only if I1 I2 is a nilpotent ideal of R. Lemma 9.2.6. Let R be a ring, and let a, x, b ∈ R. If ax + b = 1, then the following are equivalent :

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(1) There exists some y ∈ R such that a + by ∈ Rp−1 . (2) There exists some z ∈ R such that x + zb ∈ Rp−1 . Proof. (1) ⇒ (2) By
virtue of Lemma
9.2.1, there exists some u ∈
R such that 1 − (a + by)u ♮
1 − u(a + by) . Let I = R 1 − (a + by)u R and J = R 1 − u(a + by) R. Then I♮J. Thus, (a + by)u = 1 in R/I and u(a + by) = 1 in R/J. In view of Lemma 4.1.2, we get

a + y(1 − xa) x + (1 − xy)ub = 1 in R/I;

x + (1 − xy)ub a + y(1 − xa) = 1 in R/J.
Therefore R 1 − (a + y(1 − xa))(x + (1 − xy)ub) R 1 − (x + (1 − xy)ub)(a +
y(1 − xa)) R ⊆ IJ is nilpotent. This implies that x + (1 − xy)ub ∈ Rp−1 , as required. (2) ⇒ (1) is symmetric.  Lemma 9.2.7. Let R be an exchange P B-ring. Then for any regular a, b ∈ R, aR ∼ = bR implies that there exist u, v ∈ Rp−1 such that a = ubv.

Proof. Let a, b ∈ R be regular elements such that ψ : aR ∼ = bR. Then ψ(a)R = bR and Ra = Rψ(a) by Lemma 7.3.9. Thus, we have some x, y ∈ R such that a = xψ(a) and ψ(a) = ya. Since yx+ (1 − yx) = 1, it follows by Lemma 9.2.6 that there exists some
z ∈ R such that x+z(1−yx) = u ∈ Rp−1 ; hence, a = xψ(a) = x + z(1 − yx) ψ(a) = uψ(a). Analogously, there exists v ∈ Rp−1 such that ψ(a) = bv. Therefore we conclude that a = ubv.  Theorem 9.2.8. Let R be an exchange P B-ring. Then for any regular A ∈ Mn (R) there exist U, V ∈ Mn (R)−1 such that U AV is a diagonal p matrix over R. Proof. Given any regular A ∈ Mn (R), there exists E = E 2 ∈ Mn (R) such that AMn (R) = EMn (R). As in the proof of Theorem 7.3.10, we have idempotents e1 , · · · , en ∈ R such that ARn×1 ∼ = diag(e1 , · · · , en )Rn×1 . Hence, AMn (R) ∼ = diag(e1 , · · · , en )Mn (R). According to Theorem 9.2.5, Mn (R) is a P B-ring. In view of Lemma 9.2.7, there are U, V ∈ Mn (R)−1 p such that U AV = diag(e1 , · · · , en ). 


Lemma 9.2.9. Let I be an ideal of a P B-ring R. Then Q(I) + Rp−1 /I = (R/I)−1 p .

Proof. Given any u ∈ Rp−1 ,
r ∈ Q(I), we can find an element v ∈ R such m that
R(1 − uv)R(1 − vu)R = 0; hence, R(1 − (u + r)v)R(1 − v(u + m r))R ⊆ R(1−uv)R(1−vu)R)m +RrR. As r ∈ Q(I), we have some n ∈ N

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mn such that (RrR)n ⊆ I; hence, R(1 − (u + r)v)R(1
− v(u + r))R ⊆ I. mn = 0, and so u + r ∈ Thus (R/I)(1−(u + r)v)(R/I)(1−v(u +
r))(R/I) −1 (R/I)−1 /I ⊆ (R/I)−1 p . This implies that Q(I) + Rp p . Let π : R → R/I be the quotient morphism. Given any π(a) ∈
(R/I)−1 p , by Lemma 9.2.1, there exists π(b) ∈ R/I such that π(1)−π(a)π(b) ♮ π(1)−

π(b)π(a) and π(a) ≡ π(a)π(b)π(a), π(b) ≡ π(b)π(a)π(b) mod N (R/I) .
k Hence R(1 − ab)R(1 − ba)R ⊆ I for some k ∈ N. Since ab + (1 − ab) = 1, we have v := b + y(1 − ab) ∈ Rp−1 for some y ∈ R. By Lemma 9.2.1 again, we can find an element
u ∈ R such that (1 − vu)♮(1 − uv) and u ≡ uvu, v ≡ vuv mod N (R) . Let w = u + a(1 − vu) + (1 − uv)a.
Then 1 − vw ≡ (1 − va)(1 − vu), 1 − wv ≡ (1 − uv)(1 − av) mod N (R) . Hence −1 (1 − vw)♮(1
− wv), and so w ∈
Rp . Clearly, π(w) = π(u) + π(a) π(1) − π(v)π(u) + π(1) − π(u)π(v) π(a). Let r = (1 − ba)y(1 − ab). Then r ∈ Q(I), and so we deduce that π(v)π(a)π(v)
= π (b + y(1 − ab))a(b
+ y(1 − ab)) ≡ π ba(b + y(1 − ab))
≡ π b + y(1 − ab) − r)
= π(v − r) mod N (R/I) .

As (1
− vu)♮(1 − uv), we can find some p ∈ N such that R(1 − uv)R(1 − p vu)R = 0. Let t = (1 − uv)a(1 − vu). Then t ∈ N (R). As a result, we get π(a) = π(u)π(v)π(a) + π(a)π(v)π(u) − π(u)π(v)π(a)π(v)π(u) + π(t)
≡ π(u)π(v)π(a) + π(a)π(v)π(u) − π(u) + π(uru + t) mod N (R/I) .

Hence, π(a) ≡ π(u) + π(a) π(1) − π(v)π(u) + π(1) − π(u)π(v) π(a) −


π(uru + t) mod N (R/I) , and so π(a) ≡ π w − (uru + t) mod N (R/I) . Thus a − w + (uru + t) ∈ Q(I).
Clearly, −(uru + t) ∈ Q(I). Let q = a − w. Then q = a − w + (uru + t) − (uru + t) ∈ Q(I). In addition, we have a = w + q, and then π(a) = π(w + q). Therefore the result follows.  Let cr(Rp−1 ) = {a ∈ R | aR + bR = R implies that there exists an element y ∈ R such that a + by ∈ Rp−1 }. Theorem 9.2.10. Let I be an ideal of a ring R. Then R is a P B-ring if and only if the following hold: (1) R/I is a P B-ring.
(2) Q(I) + Rp−1 /I = (R/I)−1 p .

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(3) Q(I) + Rp−1 ⊆ cr(Rp−1 ). Proof. One direction is obvious by Lemma 9.2.9. Conversely, assume that Conditions (1), (2) and (3) hold. Let π : R → R/I be the quotient morphism. Suppose that ax + b = 1 in R. Then π(a)π(x) + π(b) = π(1) in R/I. Since R/I is a P B-ring, we have an element y ∈ R such that π(a) + π(b)π(y) ∈ (R/I)−1 p .

By Condition (2), there exist w ∈ Rp−1 , r ∈ Q(I) such that π(a)+π(b)π(y) = π(w + r). Hence a + by − w − r ∈ I, and then a + by ∈ Q(I) + Rp−1 . From
ax+b = 1, we get (a+by)x+b(1−yx) = 1. Therefore a+b y +(1−yx)z = a + by + b(1 − yx)z ∈ Rp−1 , as required.  Recall that an ideal I of a ring R is a B-ideal provided that aR+bR = R with a ∈ 1 + I, b ∈ R implies that there exists some y ∈ R such that a + by ∈ U (R). As is well known, I is a B-ideal of a ring R if and only if Ra + Rb = R with a ∈ 1 + I, b ∈ R implies that there exists a z ∈ R such that a + zb ∈ U (R). Corollary 9.2.11. Let I be a B-ideal of a ring R. Then R is a P B-ring if and only if the following hold: (1) R/I is a P B-ring.
(2) Q(I) + Rp−1 /I = (R/I)−1 p .

Proof. One direction is obvious. Conversely, assume that Conditions (1) and (2) hold. Take u ∈ Rp−1 and t ∈ Q(I) and assume that (u − t)x + b = 1 with x, b ∈ R. In view of Lemma 9.2.1, there exists
some v ∈ R such that (1 − uv)♯(1 − vu) and u ≡ uvu, v ≡ vuv mod N (R) . It is easy to see that 1 = (1 − tv)ux − t(1 − vu)x + b.

As t ∈ Q(I), tv ∈ N (R/I); hence, 1 − tv ∈ U (R/I). Since I is a B-ideal, it is easy to find an element w ∈ U (R) such that 1 − tv ≡ w(mod I). That is, 1 − tv = w + r for some r ∈ I. As a result, we deduce that (w + r)ux + b − t(1 − vu)x = 1. This implies that (1 + rw−1 )wux + b − t(1 − vu)x = 1.

Since 1 + rw−1 ∈ 1 + I, we have some z ∈ R such that
1 + rw−1 + b − t(1 − vu)x z ∈ U (R);

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and then w + r + b − t(1 − vu)x zw ∈ U (R). That is,
w1 := 1 − tv + b − t(1 − vu)x zw ∈ U (R). Clearly,


w1 u = u − tvu + b − t(1 − vu)x zwu
= u − t + bzwu + t(1 − vu) 1 − (1 − vu)xzwu .
2 As (1 − vu)xzwu ∈ N (R), we see that (1 − vu)xzwu ∈ R is nilpotent,
−1 and so 1 − (1 − vu)xzwu ∈ U (R). Let
w2 = 1 − (1 − vu)xzwu
. Then w2 ≡ 1 + (1 − vu)xzwu mod N (R) , and so uw2 ≡ u mod N (R) . This implies that w1−1 (u − t + bzwu)w2 = uw2 − w1−1 t(1 − vu)
≡ u − w1−1 t(1 − vu) mod N (R) .

Thus, we have some s ∈ N (R) such that w1−1 (u−t+bzwu)w2 = u−w1−1
t(1− ′ vu)+s. Let u′ = u−w1−1 t(1−vu)+s. Then
1−u v ≡ 1−uv mod N (R) and ′ ′ 1 − vu ≡ (1 + vw )(1 − vu) mod N (R) . So we can find some c, d ∈ N (R) such that 1 − u′ v = (1 − uv) + c and 1 − vu′ = (1 + vw′ )(1 − vu) + d. Thus, R(1 − u′ v)R(1 − vu′ )R ⊆ R(1 − uv)R(1 − vu)R + RcR + RdR.

Clearly, RcR and RdR are both nilpotent. As (1−uv) ⊥ (1−vu), we deduce that (1 − u′ v) ⊥ (1 − vu′ ), whence, u′ ∈ Rp−1 . Therefore u − t+ bzwu ∈ Rp−1 . According to Theorem 9.2.10, R is a P B-ring.  Corollary 9.2.12. Let I be a B-ideal of a ring R, and S be a P B-subring of R containing 1. If R = I + S and Sp−1 ⊆ Rp−1 , then R is a P B-ring. T Proof. Clearly, R/I = (I + S)/I ∼ S/(I S). As S is a P B-ring, so is = T S/(I S) by Theorem 9.2.10. Thus, R/I is a P B-ring. Obviously,
Q(I) + Rp−1 /I ⊆ (R/I)−1 p .

Given any u ∈ (R/I)−1 p , then there exist a ∈ I, b ∈ S such that u = a + b. T
−1 Thus, u + I = b + I. It is easy to verify that b ∈ S/(I S) p . In view of Theorem 9.2.10, \ \

b ∈ Q(I S) + Sp−1 / I S . T T
Thus, we have a c ∈ Sp−1 such that (b − c) + I S ⊆ N S/(I S) . This −1 −1 −1 implies that (b − c) + I ∈ N (R/I).
As Sp ⊆ Rp , we see that c ∈ Rp , −1 −1 and then (R/I)p ⊆ Q(I) + Rp /I. By Corollary 9.2.11, R is a P B-ring. 

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Further, we note that {QB-Rings} ⊆ {P B-Rings} ⊆ {R | R/J(R) is a QB-ring}. In fact, we have the following corollaries. Corollary 9.2.13. A ring R is a QB-ring if and only if the following hold: (1) R is a P B-ring.

−1 (2) J(R) + Rq−1 /J(R) = R/J(R) q .



Corollary 9.2.14. A ring R is a P B-ring if and only if the following hold: (1) R/J(R) is a
QB-ring.
−1 (2) J(R) + Rp−1 /J(R) = R/J(R) p .



Example 9.2.15. Let R be an exchange QB-ring. Then the ring   a0b T = { 0 a 0  | a, b ∈ R} 00a

is an exchange P B-ring. Proof. Clearly,

 a0b J(T ) = { 0 a 0  | a ∈ J(R), b ∈ R}. 00a 

Then T /J(T ) ∼ = R/J(R), and so T /J(T ) is a QB-ring. One easily checks that idempotents lift modulo J(T ). Therefore T is an exchange ring by [382,     a00 a0b
−1 Theorem 29.2]. For any  0 a 0  + J(T ) ∈ T /J(T ) p , then  0 a 0  + 00a 00a
−1 J(T ) ∈ T /J(T ) p . Thus, we can find some c, d ∈ R and m ∈ N such that

    a00 d c0d
m ⊆ J(T ). 0 )T (1 −  0 c 0   0 a 0 )T 00a c 00c
−1 Hence, (1 − ac)♮(1 − ca) in R/J(R), i.e., a ∈ R/J(R) p . In view of Corollary 9.2.14, we have some z ∈ Rp−1 such that a − z ∈ J(R). Write  a00 c0    T (1 − 0 a 0 0c 00a 00 

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(1 − zu)♮(1 − uz) for some
m u ∈ R. Then there exists some m ∈ N such that R(1 − zu)R(1 − uz)R = 0. Hence, 

     z00 u00 u00 z00
2m T (1T −  0 z 0   0 u 0 )T (1T −  0 u 0   0 z 0 )T = 0. 00z 00u 00u 00z 

 z00 This implies that  0 z 0  ∈ Tp−1 . Obviously, 00z 

   a0b z00  0 a 0  + J(T ) =  0 z 0  + J(T ). 00a 00z


Therefore, J(T ) + Tp−1 /J(T ) = 9.2.14, T is a P B-ring, as asserted. 9.3


−1 T /J(T ) p . According to Corollary 

Exchange Conditions

The aim of this section is to give some equivalent characterizations of P Brings under exchange conditions. Proposition 9.3.1. An exchange ring R is a P B-ring if and only if for every regular x ∈ R there exists some u ∈ Rp−1 such that x = xux. Proof. One direction is obvious. Conversely, assume that ax + b = 1 in R. By Lemma 1.4.7, there exists e = e2 ∈ R such that e = bs and 1−e = (1−b)t for s, t ∈ R. Thus, (1 − e)axt + e = 1, and so (1 − e)a ∈ R is regular. Hence, we have some u ∈ Rp−1 such that u(1 − e)a = f is an idempotent of R. Thus, f xt + ue = u, and then f (x + ue) + (1 − f )ue = u. As u ∈ Rp−1 , there
exists q ∈ R such that (1 − uq)♮(1 − qu) and u ≡ uqu, q ≡ quq mod N (R) . Let g = (1 − f )ueq(1 − f ). Then (1 − f )uq(1 − f )u = (1 − f )uqu − (1 − f )uqu(1 − e)au = (1 − f )uqu + (1 − f )(u − uqu)(1
− e)au − (1 − f )u(1 − e)au = (1 − f )u − (1 − f )(u − uqu) + (1 − f )(u − uqu)(1 − e)au.

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As a result, we deduce that

u a + bs(q(1 − f ) − a) 1
− f ueq(1 − f ) u
= u(1 − e)a + ueq(1 − f ) 1 − f ueq(1
− f) u = f (1 − f ueq(1 − f )) +
ueq(1 − f ) u = f + (1 − f )ueq(1 − f) u
≡ u mod N (R) .

Let y = s q(1
− f ) − a and w = 1 − f ueq(1 − f ) u. Then w(a + by)w ≡ w mod N (R) , where w ∈ Rp−1 .
Suppose (1 − ws)♮(1 − sw). Then 1 − w(a
+ by) ≡
1 − w(a + by) (1 − ws), 1 − (a + by)w ≡
(1 − sw) 1 − (a + by)w mod N (R) . Let r = 1 − w(a +
by)− 1 − w(a+ by) (1 − ws) and t = 1 − (a+ by)w − (1 − sw) 1 − (a+ by)w . Then
RrR and RtR
are both nilpotent. This implies
that R 1 − (a +
by)w R 1 − (a + by)w R is nilpotent, i.e., 1 − w(a + by) ♮ 1 − (a + by)w . As a result, a + by ∈ Rp−1 , as required.  Corollary 9.3.2. Let R be an exchange ring. Then the following are equivalent: (1) R is a P B-ring. (2) Every regular element in R is the product of an idempotent in R and an element in Rp−1 . Proof. (1) ⇒ (2) is clear. (2) ⇒ (1) Given any regular x ∈ R, we have some y ∈ R such that x = xyx and y = yxy. By hypothesis, there exist e = e2 ∈ R, u ∈ Rp−1 such that y = eu. Hence eux+(1−yx) = 1, and so e+(1−yx)(1−e) = 1−eux(1−
e) ∈ U (R). As a result, y + (1 − yx)(1 − e)u = 1 − eux(1 − e) u ∈ Rp−1 . −1 Thus, there exists z ∈ R such that
y + (1 − yx)z = v ∈ Rp . So we deduce that x = x y + (1 − yx)z x = xvx. Therefore the result follows from Proposition 9.3.1.  Theorem 9.3.3. Let R be an exchange ring. Then the following are equivalent: (1) R is a P B-ring. (2) Whenever a∼b, there exists u ∈ Rp−1 such that au = ub. (3) For all idempotents e, f ∈ R, eR ∼ = f R implies that there exists u ∈ Rp−1 such that eu = uf . Proof. (1) ⇒ (2) and (2) ⇒ (3) are analogous to Theorem 8.1.9.

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(3) ⇒ (1) Given any regular x ∈ R, there exists y ∈ R such that x = xyx. Clearly, we have η : xyR = xR ∼ = yxR given by η(xr) = yxr for any r ∈ R. So we can find u ∈ Rp−1 such that yxu = uxy. In view of Lemma 9.1.1, there exists
some v ∈ R such that (1 − uv)♮(1 − vu), u ≡ uvu, v ≡ vuv mod N (R) . Define α : (1 − xy)R → (1 − yx)R given by (1 − xy)r → (1 − yx)ur for any r ∈ R and β : (1 − yx)R → (1 − xy)R given by (1 − yx)r → (1 − xy)v(1 − yx)r for any r ∈ R. Then α and β are R-morphisms. Define φ : R = xR ⊕ (1 − xy)R → yxR ⊕ (1 − yx)R given by φ(x1 + x2 ) = η(x1 ) + α(x2 ) for any x1 ∈ xR, x2 ∈ (1 − xy)R and ψ : R = yxR ⊕ (1 − yx)R → xR ⊕ (1 − xy)R = R given by ψ(y1 + y2 ) = η −1 (y1 ) + β(y2 ) for any y1 ∈ yxR, y2 ∈ (1 − yx)R. As in the proof of Theorem 8.1.9, (1 − ψφ)(x1 + x2 ) = (1 − xy)(1 − vu)x2 for any x1 ∈ xR and x2 ∈ (1 − xy)R and (1 − φψ)(y1 + y2 ) = (1 − yx)(1 − uv)y2 for any y1 ∈ yxR and y2 ∈ (1 − yx)R. As (1 − uv)♮(1 − vu), we get (1 − φψ)♮(1 − ψφ). Thus, x = xφ(1)x, where φ(1) ∈ Rp−1 . Therefore the proof is true by Proposition 9.3.1.  Let R be an exchange P B-ring. Analogously to the preceding theorem, we derive that a∼b with a, b ∈ R implies that there exist u, v ∈ Rp−1 such that a = ubv, where (1 − uv)♮(1 − vu). Corollary 9.3.4. Let R be an exchange ring. Then the following are equivalent : (1) R is a P B-ring. (2) Whenever x = xyx, there exists u ∈ Rp−1 such that x = xyu = uyx. (3) Whenever x = xyx, there exists u ∈ Rp−1 such that xyu = uyx. Proof. (1) ⇒ (2) Given any x = xyx, then we have x = xzx, z = zxz, where z = yxy. Since R is a P B-ring, it follows by Proposition 9.3.1 that there exists some v ∈ Rp−1 such that z = zvz. Let u = (1 − xz − vz)v(1 − zx− zv). One easily checks that (1 − xz − vz)2 = 1 = (1 − zx − zv)2 .

Hence u ∈ Rp−1 . As in the proof of Theorem 5.2.14, x = xzu = x(yxy)u = xyu and x = uzx = u(yxy)x = uyx, as desired. (2) ⇒ (3) is trivial. (3) ⇒ (1) For any idempotents e, f ∈ R, eR ∼ = f R implies that there exist some x ∈ eRf, y ∈ f Re such that e = xy and f = yx. As x = xyx, there exists some u ∈ Rp−1 such that xyu = uyx. That is, eu = uf . According to Theorem 9.3.3, we obtain the result. 

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Proposition 9.3.5. Let R be a ring. Then the following are equivalent: (1) R is a P B-ring. (2) For any R-epimorphism f, g : R → M , there exists h ∈ EndR (R) such that f = gh and h(1) ∈ Rp−1 . (3) For any a, b, d ∈ R, aR + bR = dR implies that there exist u ∈ Rp−1 , v ∈ R such that au + bv = d. Proof. (1) ⇒ (2) Let f, g : R → M be R-epimorphisms. Since R is a projective right R-module, there exists some α : R → R such that gα = f . Similarly, there exists some β : R → R such that f β = g. As R is a P Bring, it follows from αβ + (1R − αβ) = 1R that there exists γ : R → R such that h := α
+ (1R − αβ)γ ∈ EndR (R)−1 p . Therefore f = gα = g α + (1R − αβ)γ = gh and h(1) ∈ Rp−1 , as required. (2) ⇒ (3) Suppose that aR + bR = dR with a, b, d ∈ R. Let f : R → aR given by r → ar for any r ∈ R, g : R → bR given by r → br for any r ∈ R and h : R → dR given by r → dr for any r ∈ R. Then f R + gR = hR with f, g, h ∈ EndR (R). Let M = hR/gR and let ϕ : hR → M be the canonical map. Then ϕf, ϕg : R → M . Since f R + ker(ϕ) = hR, one easily checks that ϕf and ϕh are both R-epimorphisms. So there exists k ∈ EndR (R)p−1 such that ϕf k = ϕh; hence, ϕ(f k − h) = 0. This implies that im(f k − h) ⊆ ker(ϕ) = gR. As R is a projective right R-module, there exists l : R → R such that f k − h = gl, and then f k + g(−l) = h. Thus ak(1) + b(−l)(1) = d. Choose u = k(1) and v = (−l)(1). Then u ∈ Rp−1 and au + bv = 1, as desired. (3) ⇒ (1) Given aR + bR = R with a, b ∈ R, then there exist u ∈ Rp−1 , v ∈ R such that au + bv = 1. By Lemma 9.2.1, there exists
some w ∈ R such that (1 − uw)♮(1 − wu), u ≡ uwu, w ≡ wuw mod N (R) . One
easily checks that 1−(a+bvw)u ≡ 1−wu mod N (R) and 1−u(a+bvw) = (1 − ua)(1 − uw). Hence a + bvw ∈ Rp−1 , and therefore R is a P B-ring.  Corollary 9.3.6. An exchange ring R is a P B-ring if and only if ψ : aR ∼ = bR implies that there exists some u ∈ Rp−1 such that b = ψ(au).

Proof. Let R be an exchange P B-ring. Suppose that ψ : aR ∼ = bR. Let ϕ : R → aR given by φ(r) = ar for any r ∈ R and ϕ : R → bR given by ψ(r) = br for any r ∈ R. In view of Proposition 9.3.5, there exists −1 h : R → R that
such that ψφh = ϕ and h(1) ∈ Rp . Then we have −1 ψ φ(1)h(1) = ϕ(1), and therefore b = ψ(au), where u = h(1) ∈ Rp . Conversely, let x ∈ R be regular, then there exists some y ∈ R such that x = xyx. As xyR = xR, we can find an element u ∈ Rp−1 such that

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x = xyu. Therefore R is a P B-ring from Corollary 9.3.2.

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Proposition 9.3.7. Let R be an exchange ring. Then the following are equivalent : (1) R is a P B-ring. (2) Whenever x = xyx, there exists some a ∈ R such that 1 + xa ∈ U (R) and y + a ∈ Rp−1 . Proof. (1) ⇒ (2) Since x = xyx, it follows from yx+ (1 − yx) = 1 that there exists some z ∈ R such that y + (1 − yx)z ∈ Rp−1 . Let a = (1 − yx)z. Then y + a ∈ Rp−1 . In addition, we have 1 + xa = 1 + x(1 − yx)z = 1 ∈ U (R). (2) ⇒ (1) Given x = xyx, then x = xzx and z = zxz, where z = yxy. By assumption, we have some c ∈ R such that x + c ∈ Rp−1 and 1 + zc ∈ U (R). Thus, 1 + z(u − x) ∈ U (R) for some u ∈ Rp−1 . Let w = 1 + z(u − x). Then zuw−1 + (1 − zx)w−1 = 1. As uw−1 ∈ Rp−1 , it follows from Lemma −1 9.2.6 that v := z + (1 − zx)w−1
t ∈ Rp for some t ∈ R. As a result, −1 x = xzx = x z + (1 − zx)w t x = xvx. According to Proposition 9.3.1, R is a P B-ring.  Corollary 9.3.8. Let R be an exchange ring. Then the following are equivalent : (1) R is a P B-ring. (2) Whenever x = xyx, there exist e ∈ r(x), u ∈ Rp−1 such that y = e + u. Proof. (1) ⇒ (2) Write x = xyx. Since yx + (1 − yx) = 1, we can find some z ∈ R such that u := y + (1 − yx)z ∈ Rp−1 . Thus, y = (yx − 1)z + u. Let e = (yx − 1)z. Then y = e + u, where e ∈ r(x) and u ∈ Rp−1 . (2) ⇒ (1) Given any regular x ∈ R, we have a y ∈ R such that x = xyx. Hence, there exists an element e ∈ r(x) and a u ∈ Rp−1 such that y = e + u. Let a = −e. Then y + a ∈ Rp−1 and 1 + xa = 1 ∈ U (R), as required.  Corollary 9.3.9. Let R be an exchange ring. Then the following are equivalent : (1) R is a P B-ring. (2) Whenever x = xyx, there exists u ∈ Rp−1 such that 1−x(y +u) ∈ U (R). Proof. (1) ⇒ (2) Whenever x = xyx, then −x = (−x)(−y)(−x). By Proposition 9.3.7, there exists a ∈ R such that 1 − xa ∈ U (R) and −y + a ∈ Rp−1 . Let u = −y + a. Then 1 − x(y + u) ∈ U (R).

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(2) ⇒ (1) Whenever x = xyx, there exists u ∈ Rp−1 such that 1 − x(y + u) ∈ U (R). Let a = −(y + u). then 1 + xa ∈ U (R) and y + a = −u ∈ Rp−1 . According to Proposition 9.3.7, we obtain the result.  9.4

Subdirect Products

Let I be an ideal of a ring R, and let e ∈ R be an idempotent. We say that I is a P B-ideal provided that aR + bR = R with a ∈ 1 + Q(I), b ∈ R =⇒ ∃ y ∈ R such that a + by ∈ Rp−1 . Let Z[[x]] be the ring of formal power series in one variable over Z. Then the ideal xZ[[x]] is a P B-ideal. Assume that xZ[[x]] is an ideal of a P B-ring R. Then 3 · 2 − 5 = 1R implies that 2 − 5y ∈ Rp−1 for some y ∈ R. Hence, we can find a v ∈ R such that
n R(1R − (2 − 5y)v)R(1R − v(2 − 5y))R = 0 for some n ∈ N. This implies that


n (x2 − (2x − 5xy)vx)(x2 − xv(2x − 5yx)) = 0.

As Z[[x]] is prime, we get

x2 − (2x − 5xy)vx = 0 or x2 − xv(2x − 5yx) = 0. As in the proof of [25, Example 4.11], we obtain 1R = (2 − 5α)β in Z, which is impossible, and so R is not a P B-ring. Therefore xZ[[x]] is a P B-ideal of Z[[x]], while it can not be seen as an ideal of a P B-ring. This shows that the concept of the P B-ideal is a nontrivial generalization of that of the P B-ring. We say that eIe is a P B-corner provided that ax + b = e with a ∈ e + eQ(I)e, x, b ∈ eRe implies that there exists y ∈ eRe such that a + by ∈ [eRe]−1 p . Lemma 9.4.1. Let I be a P B-ideal of a ring R, and let e ∈ R be an idempotent. Then eIe is a P B-corner. Proof. Given ax + b = e with a ∈ e + eQ(I)e, x, b ∈ eRe, then (a + 1 − e)(x + 1 − e) + b = 1, where a + 1 − e ∈ 1 + Q(I). Thus, we can find some y ∈ R such that a + 1 − e + by ∈
Rp−1 . Hence, we have an
element u ∈ R such that R 1 − (a + 1 − e + by)u R 1
− u(a + 1 − e + by) R is a nilpotent
ideal. This implies that R (1 − e)ue (eRe) e − (eue)(a + b(eye) R and

R e − (a + by)ue (eRe) e − (eue)(a
+ b(eye)) R are nilpotent
ideals, and then so is R e − (a + b(eye))(eue) (eRe) e − (eue)(a + b(eye) R. Therefore a + b(eye) ∈ [eRe]−1  p , as desired.

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Lemma 9.4.2. Let I be a P B-ideal of a ring R, and let e ∈ R be an idempotent. If ax + b = e with a ∈ e + eQ(I)e, x, b ∈ eRe, then there exists y ∈ eQ(I)e such that a + by ∈ [eRe]−1 p . Proof. Given ax + b = e with a ∈ e + eQ(I)e, x, b ∈ eRe, then a(x + b) + (e − a)b = e. By Lemma 9.4.1, there exists s ∈ eRe such that a + (e − a)bs ∈ [eRe]−1 p . According to Lemma 9.2.6, we can find t ∈ eRe such that −1 x + b + t(e − a)b ∈ [eRe]−1 p . Let z = e + t(e − a)b. Then x + zb ∈ [eRe] p
with z ∈ e+eQ(I)e, and so we have v ∈ eRe such that e−(x+zb)v ♮ e− R
v(x + zb) . Let y = v(e − za) and d = x + (e − xa)z. Clearly, y ∈ eQ(I)e.

Similarly to Lemma 9.2.6, we prove that e − (a + by)d ♮R e − d(a + by) , as required. 
n Suppose that f − f 2 ∈ N (R). Then f − f 2 = 0 for some n ∈ N. That is, f n (1 − f )n = 0. Working in the polynomial ring Z[x], we have  2n 
2n X 2n 1 = x + (1 − x) = x2n−k (1 − x)k . k k=0   n
P 2n Let g(x) = x2n−k (1 − x)k . Then g(x) ≡ 0 mod xn and g(x) ≡ k=0
k
2 1 mod (1−x)n . Hence g(x) ≡ g(x) mod xn (1−x)n . Put p = g(f ). Then
p2 ≡ p mod f n (1 −
f )n , i.e., p = p2 . Since f (1 − f ) ∈ N (R), we have k k+1 k p ≡ f 2n mod N (R)
. Furthermore, we see that f = f
+ (1 − f )f = k+1 f mod N (R) for all k ∈ N. Therefore p ≡ f mod N (R) . We say that p is an idempotent corresponding to f . In addition, we see that p ∈ f Rf , this will be used in the sequel. Let I be a P B-ideal of a ring R, and let u ∈ Rp−1 . Then there
exists v ∈ R such that (1 − uv)♮(1 − vu) and u ≡ uvu, v ≡ vuv mod N (R) . Let p′ = uv, q ′ = vu. Let p and q be idempotents corresponding to p′ and q ′ , ′ respectively. Let w ∈ [pRp]−1 p . Then, we can find some w ∈ pRp
such that ′ ′ ′ ′ ′ ′ (p − ww )♮R (p − w w) and w ≡ ww w, w ≡ w ww mod N (R) . Let p′2 = p−ww′ and q1′ = q−vw′ wu. Let p2 and q1 be the idempotents corresponding to p′2 and q1′ . Let p1 = 1 − p and q2 = 1 − q. Then p1 , p2 , q1 , q2 ∈ R are all idempotents. Conveniently, we say that (p1 , q1 ) and (p2 , q2 ) are two pairs of perfect idempotents.
Let p 6 ♮q, and let cr [pRq]−1 = {a ∈ pRq | ax+ b = p with x ∈ qRp, b ∈ p pRp implies that there exists some y ∈ pRq such that a + by ∈ [pRq]−1 p }. Using the previous notations, we can derive the following lemmas. Lemma 9.4.3. Let a = wu + t1 + t2 with t1 ∈ p1 Rq1 , t2 ∈ p2 Rq2 . Then

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the following hold: −1 (1) If t1 ∈ [p1 Rq1 ]−1 Rq2 ]−1 p and t2 ∈ [p2
p , then a ∈ Rp .
−1 (2) If p1 ♮q1 or t1 ∈ cr [p1 Rq1 ]p , and p1 ♮q2 or t2 ∈ cr [p2 Rq2 ]−1 , then p a ∈ cr(Rp−1 ).

Proof. (1) Let t1 ∈ [p1 Rq1 ]−1 and t2 ∈ [p2 Rq2 ]−1 Analogously to p p . Lemma 9.2.1, we have t′1 ∈
q1 Rp1 and t′2 ∈ q2 Rp2 such that t1 ≡ t1 t′1 t1 , t′1 ≡ t′1 t1 t′1
mod N (R) , (p1 − t1 t′1 )♮R (q1 − t′1 t1 ), t2 ≡ t2 t′2 t2 , t′2 ≡ t′2 t2 t′2 mod N (R) and (p2 − t2 t′2 )♮R (q2 − t′2 t2 ). Choose a
′ = vw′ + t′1 + t′2 . Then 1 − aa′ ≡ (p1 − t1 t′1 ) + (p2
− t2 t′2 ) mod N (R) and 1 − a′ a ≡ (q1 − t′1 t1 ) + (q2 − t′2 t2 ) mod N (R) . Clearly, p1 ♮q2 and p2 ♮q1 . This implies that (1 − aa′ )♮(1 − a′ a), as required.

(2) Assume that t1 ∈ cr [p1 Rq1 ]−1 , and t2 ∈ cr [p2 Rq2 ]−1 . Given p p ax + b = 1 with a, x, b ∈ R, then t1 (q1 xp1 ) + p1 bp1 = p1 , and so we have c1 ∈ p1 Rq1 such that t1 + p1 bc1 ∈ [p1 Rq1 ]−1 p . On the other hand, there exists some r ∈ N (R) such that t2 q2 xp2 + p2 (b + r)p2 = p2 . This implies that t2 + p2 bc2 ∈ [p2 Rq2 ]−1 p for some c2 ∈ p2 Rq2 . Clearly,
bc1 ≡ p2 bc1 + ww′ bc1 + p1 bc1 , bc2 ≡ p2 bc2 + ww′ bc2 + p1 bc2 mod N (R) . Thus,



a+b(c1 +c2 ) ≡ wu 1+vw′ b(c1 +c2 ) +(t1 +p1 bc1 )+(t2 +p2 bc2 ) mod N (R) .
2k+1 Obviously, we have some k ∈ N such that vw′ b(c1 + c2 ) = 0; hence, 2k
−1 X
i 1 + vw′ b(c1 + c2 ) = (−1)i vw′ b(c1 + c2 ) ∈ U (R). i=0

Since



−1 a + b(c1 + c2 ) 1 + vw′ b(c1 + c2 )
≡ wu + (t1 + p1 bc1 ) + (t2 + p2 bc2 ) mod N (R) ,

it follows by (1) that wu + (t1 + p1 bc1 ) + (t2 + p2 bc2 ) ∈ Rp−1 , and therefore a + b(c1 + c2 ) ∈ Rp−1 . In other cases, we easily find some c ∈ R such that a + bc ∈ Rp−1 . Therefore the proof is true.  Lemma 9.4.4. Let I be a P B-ideal of a ring R, and assume that for every pair of defect idempotents (p, q) with p or q in Q(I), p♮q or pRq is a P B-corner. Then Q(I) + Rp−1 ⊆ cr(Rp−1 ). Proof. Given (u − t)x + b = 1 with u ∈ Rp−1 and t ∈ Q(I), then we
have v ∈ R such that (1 − uv)♮(1 − vu) and u ≡ uvu, v ≡ vuv mod N (R) . Let

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f = uv and h = vu. Then we have idempotents p, q ∈ R which correspond to f and h, respectively.
Clearly, p(1 − tv)p (puxp) + p(u − t)(1 − q)xp + pbp + prp = p for some r ∈ N (R). By Lemma 9.4.2, pIp is a P B-corner, and so we can
find some z ∈ pQ(I)p such that w := p(1 − tv)p + p(u − t)(1 − q)xp + pbp z ∈ [pRp]−1 p . Thus, wu
≡ u − t + (1 − p)(tq − bpzu) + t(1 − q) − pt(1 − q)xpzu + bpzu mod N (R) . Let t′ = tq − bpzu
+ t(1 − q)xpzu. Then
wu ≡ u − t + (1 − p)t′ + t(1 − q) 1 − (1 − q)xpzu + bpzu mod N (R) . Choose w1 =
1 − (1 − q)xpzu. Then
w1 ∈ U (R) and w1−1 ≡ 1 + (1 − q)xpzu mod N
(R) ; hence, u − t + bpzu w1−1 ≡ wu − (1 − p)t′ − t(1 − q) mod N (R) . As ′ ′ ′ w ∈ [pRp]−1 p , we have w ∈ pRp
such that (p − ww )♮R (p − w w) and ′ ′ ′ ′ ′ ′ w ≡ ww w, w ≡ w ww mod N (R) . Let w2 = 1 − (1 − p)t vw and w3 = 1 − vw′ t(1 − q). Then w2 , w and w2−1 ≡ 1 + (1 − p)t′ vw′ , w3−1 ≡
3 ∈ U (R) ′ ′ 1 + vw t(1 − q) mod N (R) . Let p2 = p − ww′ , q1′ = q − vw′ wu. Then p′2 − (p′2 )2 , q1′ − (q1′ )2 ∈ N (R). Let p2 , q1 ∈ R be idempotents corresponding to p′2 and q1′ , respectively. Let p1 = 1−p and q2 = 1−q. Then p2 , q1 , q2 ∈
−1p1 ,−1 −1 R are all idempotents.
Furthermore, w2 u − t + bpuz w1 w3 ≡ wu − p1 t′ q1 −p2 tq2 mod N (R) . As z ∈ pQ(I)p, w ∈ p+pQ(I)p. This implies that w′ ∈ p + pQ(I)p, and so p2 ∈ Q(I). Likewise, q1 ∈ Q(I). By hypothesis, pi ♮qi or pi Rqi (i = 1, 2) is a P B-corner. So the result follows by Lemma 9.4.3.  Let R be the subdirect product of two P B-rings A and B. Then we have T two epimorphisms δ : R → A and γ : R → B such that ker(δ) ker(γ) = 0. Let I = ker(γ). Clearly, δ|I : I → A is monomorphic. Let C = A/δ(I) and α : A → C be the natural epimorphism. Define β : B → C given by β(x) = δ(y) + δ(I), where γ(y) = x. So we get commutative diagrams with exact rows γ 0 → I ֒→ R → B → 0 k δ↓ ↓β δ|I

α

0 → I → A → C → 0.

Theorem 9.4.5. Every finite subdirect product of P B-rings is a P B-ring. Proof. Using the previous diagram, it will suffice to prove that R is a P B-ring. Given any u ∈ Bp−1 , we have v ∈ B such that
(1B − uv)♮(1B − vu) and u ≡ uvu, v ≡ vuv mod N (B) .

Assume that γ(u′ ) = u and γ(v ′ ) = v with u′ , v ′ ∈ R. Let δ(u′ ) = a and δ(v ′ ) = b. It follows from ab+(1A −ab) = 1A that v1 = b+y(1A −ab) ∈ A−1 p

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for some y ∈ A. Hence, a ≡ aba ≡ av1 a mod δQ(I) . Assume that


(1A − u1 v1 )♮(1A − v1 u1 ) and u1 ≡ u1 v1 u1 , v1 ≡ v1 u1 v1 mod N (A) .
Let r = (1A − ba)y(1A − ab). Then r ∈ δ Q(I) . In addition,
α(v1 )α(a)α(v1 ) ≡ α(v1 − r) mod αδQ(I) .

Let q1 = (1A −u1 v1 )a(1A −v1 u1 ) and q2 = u1 −u1 v1 u1 . Then q1 , q2 ∈ N (A). Let x = u1 + a(1A − v1 u1 ) + (1A − u1 v1 )a. Then
α(a) ≡ α(x − q1 − q2 − u1 ru1 ) mod αδQ(I) ,

whence α(a − x + q1 + q2 + u1 ru1 ) = αδ(s) for some s ∈ Q(I). Let r1 = δ(s) − u1 ru1 and
x′ = x − q1 − q2 . Then α(a) = α(x′ + r1 ) with x′ ∈ A−1 p and r1 ∈ δ Q(I) . Clearly, there exists w ∈ A such that
(1 − x′ w)♮(1 − wx′ ) and x′ ≡ x′ wx′ , w ≡ wx′ w mod N (A) .

Let p1 = (1A −wx′ )b(1A −x′ w), p2 = w−wx′ w, y0 = w+b(1A −x′ w)+(1
A− wx′ )b and let y ′ = y0 − p1 − p2 . Similarly, we can find r2 ∈ δ Q(I) such that α(b) = α(y ′ +r2 ) with (1−x′ y ′ )♮(1−y ′ x′ ). Assume that δ(z1 ) = x′ +r1 and δ(z2 ) = y ′ + r2 for some z1 , z2 ∈ R. Then z1 − u′ ∈ kerβγ = kerαδ ⊆ ker(δ) + ker(γ).

So we assume that z1 − u′ = e + f with e ∈ ker(δ) and f ∈ ker(γ). Let u′′ = u′ + f . Then γ(u′′ ) = u and δ(u′′ ) = x′ + r1 . Likewise, we have v ′′ ∈ R such that γ(v ′′ ) = v and δ(v ′′ ) = y ′ + r2 . As (1B − uv)♮(1
m B − vu), we can find m ∈ N such that γ R(1R − u′′ v ′′ )R(1R − v ′′ u′′ )R = 0. Since (1A − x′ y ′ )♮(1A − y ′ x′ ), we have n ∈ N such that
n A(1A − (δ(u′′ ) − r1 )(δ(v ′′ ) − r2 ))A(1A − (δ(v ′′ ) − r2 )(δ(u′′ ) − r1 ))A = 0.

Assume that r1 = δ(s1 ), r2 = δ(s2 ) with s1 , s2 ∈ Q(I).
Let u0 = u′′ − s1 n and v0 = v ′′ − s2 . Then R(1
R − u0 v0 )R(1R − v0 u0 )R ⊆ ker(δ), and so T mn R(1R − u0 v0 )R(1R − v0 u0 )R ⊆ ker(γ) ker(δ) = 0. Thus, we see that u0 ∈ Rp−1 . Since u = γ(u′′ − f ) = γ(u0 + s1 − f ) with s1 − f ∈ Q(I),

we −1 −1 −1 −1 get (R/I)p ⊆ Q(I) + Rp /I. Therefore, (R/I)p = Q(I) + Rp /I. Let (p, q) be a pair of defect idempotents with p or q in Q(I), say (p1 , q1 ). As in the proof of Lemma 9.2.4, Aδ(p)Aδ(q)A is nilpotent or δ(p)Aδ(q) is a P B-corner. If Aδ(p)Aδ(q)A is nilpotent, then there exists some t ∈ N T such that (RpRqR)t ⊆ I ker(δ) = 0, and so RpRqR is nilpotent. That is, p♮q. Now assume that δ(p)Aδ(q) is a P B-corner. Given ax + b = p with a ∈ pRq, x ∈ qRp and b ∈ pRp, then δ(a)δ(x) + δ(b) = δ(p). So, we have

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 −1 some d ∈ R such that δ(a) + δ(b)δ(pdq) ∈ δ(pRq) p , whence, we have k ∈ N and w′ ∈ qRp such that
k R(p − (a + bpdq)w′ )R(q − w′ (a + bpdq))R ⊆ ker(δ).

Clearly, R p − (a + bpdq)w′ R q − w′ (a + bpdq) R ⊆ RpRqR, so we can find l ∈ N such that
l R(p − (a + bpdq)w′ )R(q − w′ (a + bpdq))R ⊆ I. As a result, we get

R(p − (a + bpdq)w′ )R(q − w′ (a + bpdq))R


kl

⊆ ker(δ)

\

I = 0;

hence a + b(pdq) ∈ [pRq]−1 . This means that pRq is a P B-corner. Given ax + b = 1 with a ∈ 1 + Q(I), x, b ∈ R, then δ(a)δ(x) + δ(b) = 1A . As in the proof of Lemma 9.4.2, there exists some y ∈ Q(I) such that δ(a) + δ(b)δ(y) ∈ A−1 p . Thus, we can find some c ∈ R such that

δ(a) + δ(b)δ(y) = δ(a) + δ(b)δ(y) δ(c) δ(a) + δ(b)δ(y)
;
1A − (δ(a) + δ(b)δ(y))δ(c) ♮ 1A − δ(c)(δ(a) + δ(b)δ(y)) .

This implies that a + by − (a + by)c(a + by) ∈ ker(δ). Hence, c ∈ 1 + ker(δ) + Q(I). Say c = c1 + c2 , where c1 ∈ ker(δ), c2 ∈ 1 + Q(I). \
m R(1 − (a + by)c2 )R(1 − c2 (a + by))R ∈ kerδ I for some m ∈ N.

This implies that a + by ∈ Rp−1 , and so I is a P B-ideal of R. It follows from Lemma 9.4.4 that Q(I) + Rp−1 ⊆ cr(Rp−1 ). According to Theorem 9.2.10, R is a P B-ring, as required.  Corollary 9.4.6. Let I and J be ideals of a ring R. Then the following are equivalent: (1) R/I and R/J are P B-rings. (2) R/(IJ) T is a P B-ring. (3) R/(I J) is a P B-ring. T T
T
Proof. (1) ⇒ (3) Let ϕ
: R/(I J) ։ R/(I J) / I/(I J) and ψ :
T T T R/(I J) ։ R/(I J) / J/(I J) be the quotient morphisms. Then T we get ker(ϕ) ker(ψ) = 0. One easily checks that \
\
\
\
R/(I J) / I/(I J) ∼ J) / J/(I J) ∼ = R/I and R/(I = R/J. T Hence R/(I J) is the subdirect product of two P B-rings, and therefore T R/(I J) is a P B-ring by Theorem 9.4.5.

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T T (3) ⇒ (2) Let K = (I J)/(I J)2 . One easily checks that K T T is an ideal of R/(I J) and K 2 = 0. It follows from R/(I J) T T ∼ Given ax + = R/(I J)2 /K that R/(I J)2 /K is a P B-ring. T 2 T b = 1 in R/(I J) , then ax + b = 1 in R/(I J)2 /K. So we T T 2 can find d, u ∈ R/(I J)2 such that (R/(I
m J) /K)(1 − (a + by)d) T 2 T 2 (R/(I J) /K)(1 − d(a + by))(R/(I J) /K) = 0 for some m ∈ N. Hence, \ \ \
m (R/(I J)2 )(1−(a+by)d)(R/(I J)2 )(1−d(a+by))(R/(I J)2 ) ⊆ K.

As a result, we obtain \ \ \



2m R/(I J)2 1−(a+by)d R/(I J)2 1−d(a+by)R/(I J)2 = 0,


−1 T T whence, a + by ∈ R/(I J)2 p . This means that R/(I J)2 is a P B
T 2 T T ring. As (I J) ⊆ IJ, we have R/(IJ) ∼ = R/(I J)2 / (IJ)/(I J)2 . Therefore R/(IJ) is a P B-ring.

(2) ⇒ (1) Clearly, R/I ∼ = R/(IJ) / I/(IJ) . So R/I is a P B-ring. Likewise, we prove that R/J is a P B-ring.  Let I be an ideal of a ring R. In Corollary 9.4.6, we choose J = I. Then R/I is a P B-ring if and only if so is R/I 2 . Let V bean infinite-dimensional  EndD (V ) EndD (V ) vector space over a division ring D, and let R = 0 EndD (V )   0 EndD (V ) and I = . Then R/I is a QB-ring, while R/I 2 is not a 0 0 QB-ring by Example 9.1.3. Example 9.4.7. Let V be an infinite-dimensional vector space over a field F , let Q = EndF (V ), and let J = {x ∈ Q | dimF (xV ) < ∞}. Set R = {(x, y) ∈ Q × Q | x − y ∈ J}. Then R is a regular QB-ring. Proof. Clearly, R is a subring of Q × Q. Since J × J is an ideal of R and R/(J × J) ∼ = Q/J is regular. Thus, R is regular by [217, Lemma 1.3]. Set P1 = J × 0 and P2 = 0 × J. Since R/P1 ∼ = R/P2 ∼ = Q, R/P1 and R/P2 are P B-rings. In addition, P1 P2 = 0. According to Corollary 9.4.6, R is a P B-ring. Therefore, R is a QB-ring.  Proposition 9.4.8. Let A and B be fully invariant submodules of a quasiT projective R-module M . If A B = 0, then the following are equivalent: (1) EndR (M ) is a P B-ring.

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(2) EndR (M/A) and EndR (M/B) are P B-rings. Proof. (1) ⇒ (2) is trivial by Lemma 8.4.2 and Theorem 9.2.10. (2) ⇒ (1) Let π : M → M/A and σ : M → M/B be the canonical maps. By assumption, ker(π) = A and ker(σ) = B are both fully invariant. According to Lemma 8.4.2, we get EndR (M/A) ∼ = EndR (M )/Kerπ, EndR (M/B) ∼ = EndR (M )/Kerσ. T For any f ∈ Kerπ Kerσ, the πf = 0 = σf ; hence, \ \ im(f ) ⊆ ker(π) ker(σ) = A B = 0. T This implies that f = 0. As a result, Kerπ Kerσ = 0. Consequently, EndR (M ) is the subdirect product of EndR (M/A) and EndR (M/B). According to Theorem 9.4.5, EndR (M ) is a P B-ring.  Corollary 9.4.9. Let M be a quasi-projective right R-module. If M = A ⊕ B and A, B are fully invariant, then EndR (M ) is a P B-ring if and only if so are EndR (A) and EndR (B). Proof. Since M = A ⊕ B, we see that A ∼ = M/B and B ∼ = M/A. Therefore the proof is true from Proposition 9.4.8.  Proposition 9.4.10. Let M be a quasi-injective right R-module. If M = A + B and A, B are fully invariant, then EndR (M ) is a P B-ring if and only if so are EndR (A) and EndR (B). Proof. One direction is obvious by Lemma 8.4.8. Assume now that EndR (A) and EndR (B) are P B-rings. Let π : A ֒→ M and σ : B ֒→ M be inclusions. Then im(π) = A and im(σ) = B are both fully invariant. By virtue of Lemma 8.4.8, we have that EndR (A) ∼ = EndR (M )/Cokerπ, EndR (B) ∼ = EndR (M )/Cokerσ. T For any f ∈ Cokerπ Cokerσ, the f π = 0 = f σ; For any m ∈ M , by assumption, there exist a ∈ A, b ∈ B such that m = a + b. Thus, f (m) = T f (a) + f (b) = f π(a) + f σ(b) = 0; hence, f = 0. So Cokerπ Cokerσ = 0. Therefore EndR (M ) is isomorphic to the subdirect product of EndR (A) and EndR (B). Using Theorem 9.4.5, EndR (M ) is a P B-ring.  Let R be a right self-injective ring. If there exist two right ideals I and J such that R = I + J, then R is a P B-ring if and only if so are EndR (I) and EndR (J).

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Proposition 9.4.11. Let M be a quasi-projective, quasi-injective right σ π R-module, and let 0 → A → M → B → 0 be an exact sequence of right R-modules. If ker(π) is full invariant, then EndR (M ) is a P B-ring if and only if so are EndR (A) and EndR (B). Proof. One direction is obvious. Conversely, we get EndR (M )/Kerπ ∼ = EndR (B), EndR (M )/Cokerσ ∼ = EndR (A). One easily checks that  T T ∼ EndR (M )/Kerπ Cokerσ  Kerπ/Kerπ T Cokerσ = EndR (M )/Kerπ, T EndR (M )/Kerπ Cokerσ Cokerσ/Kerπ Cokerσ ∼ = EndR (M )/Cokerσ.
T
T T As Kerπ/Kerπ Cokerσ Cokerπ/Kerπ Cokerσ = 0, it follows by T Theorem 9.4.5 that EndR (M )/Kerπ Cokerσ is a P B-ring. T Let f ∈ Kerπ Coker. Then πf = 0 and f σ = 0. For any m ∈ M , πf (m) = 0; hence, f (m) ∈ ker(π). Clearly, ker(π) = im(σ). Thus, f (m) = σ(a) for some a ∈ A. It follows that f 2 (m) = f σ(a) = 0. That is, f 2 = 0.
2 T As a result, we see that Kerπ Cokerσ = 0. Therefore we conclude that EndR (M ) is a P B-ring by Corollary 9.4.6.  Let M be a quasi-projective, quasi-injective right R-module.
Then EndR (M ) is a P
B-ring if and only if so are EndR rad(M ) and EndR M/rad(M ) . Since there is an exact sequence 0 → rad(M ) ֒→ M → M/rad(M ) → 0, we are done by Proposition 9.4.11.

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Chapter 10

Power-Cancellation

Let R be an associative ring with identity, and let A be a right R-module. It is well known that cancellation fails even in the case where A is a finitely generated projective module over a commutative ring. But Goodearl showed that for all finitely generated projective modules A, B and C over a commutative ring, A ⊕ B ∼ = A ⊕ C implies that nB ∼ = nC for some n ∈ N (cf. [298, Theorem 5.1]). This inspires us to introduce a new kind of weakly stable range condition. We say that A satisfies power-cancellation provided that for any right R-modules B and C, A⊕B ∼ = A⊕C implies that nB ∼ = nC for some n ∈ N. Many authors have studied power-substitution such as [69], [215], [230] and [298]. The main purpose of this chapter is to investigate power-cancellation of modules over an exchange ring. Let f : M → N be a right R-morphism, and letn ∈N.  We use  f ∗ to der1 f (r1 )  ..  ∗  ..  note the R-morphism nM → nN given by f  .  =  .  for any 

 r1  ..   .  ∈ nM . That is, f ∗ = diag(f, · · · , f ). rn

291

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Power-Substitutions

We say that R satisfies power-substitution in the case where aR + bR = R with a, b ∈ R implies that there exist n ∈ N, Y ∈ Mn (R) such that aIn + bY ∈ GLn (R). Theorem 10.1.1. Let R be a ring. Then the following are equivalent: (1) R satisfies power-substitution. (2) Whenever aR + bR = dR, there exist n ∈ N, Q ∈ Mn (R) such that aIn + bQ = dU for some U ∈ GLn (R). Proof. (2)⇒(1) is trivial. (1)⇒(2) Assume that a, b, d ∈ R such that aR + bR = dR. Let g : dR → dR/bR be the canonical map, f1 : R → aR given by r 7→ ar for any r ∈ R, f2 : R → bR given by r 7→ br for any r ∈ R, f3 : R → dR given by r 7→ dr for any r ∈ R. It follows from aR + bR = dR that gf1 , gf3 are both R-epimorphisms. By the projectivity of R, we can find α : R → R such that gf1 = gf3 α. As gf1 is an R-epimorphism, there exists some ψ : R → R such that gf3 αψ = gf1 ψ = gf3 . Since αψ + (1 − αψ) = 1R , by hypothesis, there exists a positive integer n and a matrix W ∈ Mn (R) such that U := αIn + (1R − αψ)W ∈ GLn (R). Thus we have g ∗ f1∗ = g ∗ f3∗ α∗ =
∗ ∗ ∗ ∗ g f3 αIn + (1R − αψ)W = g f3 U , so g ∗ (f1∗ − f3∗ U ) = 0. Therefore we see that im(f1∗ − f3∗ U ) ⊆ ker(g ∗ ) = n(bR). As nR is projective, we have some Q ∈ Mn (R) such that −f2∗ Q = f1∗ − f3∗ U . Consequently, we conclude that aIn + bQ = dU .  Corollary 10.1.2. Let R be a commutative ring. Then the following are equivalent: (1) R satisfies power-substitution. (2) Whenever aR + bR = R, there exist n ∈ N, y ∈ R such that an + by ∈ U (R). (3) Whenever aR + bR = dR, there exist n ∈ N, y ∈ R and u ∈ U (R) such that an + by = dn u. Proof. (1) ⇒ (3) Assume that a, b, d ∈ R such that aR + bR = dR. In view of Theorem 10.1.1, we can find a positive integer n and a matrix Q ∈ Mn (R) such that aIn + bQ = dU for some U ∈ GLn (R). Observing that there is some y ∈ R such that det(aIn + bQ) = an + by, we have that an + by = det(dU ) = dn det(U ). Let u = det(U ). Then an + by = du, as

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required. (3) ⇒ (2) is trivial. (2) ⇒ (1) Given ax + b = 1 in R, then there exists a positive integer n and a y ∈ R such that an + by ∈ U (R). Now an + by = an + by(ax + b)n−1 = an + a1 an−1 b + a2 an−2 b2 + · · · + an bn for suitable a1 , · · · , an ∈ R. Set a1 −a2 a3 · · · (−1)n an−1 (−1)n+1 an 1 0 0 ··· 0 0 0 1 0 ··· 0 0 Q= . . . . . . . .. .. .. .. .. . . 0 0 0 ··· 1 0 Then

a + a1 b −a2 b b a 0 b det(aIn + bQ) = .. .. . . 0 0

a3 b · · · (−1)n an−1 b (−1)n+1 an b 0 ··· 0 0 a ··· 0 0 . .. . . .. .. . . . . 0 ··· b a

Hence, det(aIn + bQ) = an + a1 an−1 b + a2 an−2 b2 + · · · + an bn = an + by ∈ U (R), and so aIn +bQ ∈ GLn (R). Therefore R satisfies power-substitution.  Let R be a commutative ring. For each x ∈ R, there exists a natural group homomorphism θx : U (R) → U (R/xR). Set Hx = U (R/xR) Imθx .

Corollary 10.1.3. Let R be a commutative ring. Then the following are equivalent : (1) R satisfies power-substitution. (2) For each x ∈ R, Hx is torsion.

Proof. (1) ⇒ (2) For any a ∈ U (R/xR), we have some b ∈ R such that ab = 1. Hence ab + xy = 1 for a y ∈ R. It follows from Corollary 10.1.2 that there exists some n ∈ N such that u := an + xz ∈ U (R) for a z ∈ R. This implies that an ≡ u(mod xR). As u ∈ Imθx , we see that Hx is torsion. (2) ⇒ (1) Given ax + b = 1 in R, then ax ≡ 1(mod bR); hence, a ∈
n U (R/bR). As Hb is torsion, we can find some n ∈ N such that aImθb = Imθb . As a result, an = θb (u) = u for a u ∈ U (R). Consequently, an − u ∈

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bR, and so an + by ∈ U (R) for a y ∈ R. Therefore the assertion is true from Corollary 10.1.2.  Analogously, we deduce that a commutative ring R has stable range one if and only if for each x ∈ R, Hx = {1}. Corollary 10.1.4. The ring Z of all integers satisfies power-substitution. Proof. Let 0 6= n ∈ Z. For any m ∈ N, there exist some q, r ∈ Z such that m = nq + r, where 0 ≤ r < n. Hence, | Z/nZ | ≤ n. This implies that the group of units of Z/nZ is torsion. Further, Hn is torsion. Clearly, H0 = {1} is torsion. Hence, Z satisfies power-substitution by Corollary 10.1.3.  Let R be a commutative ring such that for any nonzero b in R, the group of units of R/bR is torsion. As in the preceding proof, R satisfies power-substitution. An a consequence, we deduce that the ring Z[i] = {a + bi | a, b ∈ Z, i2 = −1} satisfies power-substitution. Proposition 10.1.5. Let e be an idempotent of a ring R. If R satisfies power-substitution, then so does eRe. Proof. Given ax+b = e with a, x, b ∈ eRe, then (a+1−e)(x+1−e)+b = 1. Since R satisfies power-substitution, there exist n ∈ N, Y ∈ Mn (R) such that (a + 1 − e)In + bY ∈ GLn (R). This implies that

U (a + 1 − e)In + bY = (a + 1 − e)In + bY U = In .

for some U ∈ GLn (R). Clearly, (1 − e)U
= (1 − e)In .
Hence eU e = U e. Thus we have that (eU e) aIn + b(eY e) = aIn + b(eY e) (eU e) = eIn , and so aIn + b(eY e) ∈ GLn (eRe). Therefore eRe satisfies power-substitution.  Example 10.1.6. Let X be a topological space, and let CR (X) be the ring of all real-valued functions on X. Then CR (X) satisfies power-substitution. Proof. Given ax+b = 1 in CR (X), then a2 +b2 > 0 everywhere. As a result,  
0 1 aI2 + b , which has determinant a2 + b2 , is a unit in M2 CR (X) . −1 0 Consequently, CR (X) satisfies power-substitution.  The following elementary fact is due to Goodearl (cf. [215, Theorem 2.1]).

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Theorem 10.1.7. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E satisfies power-substitution. (2) M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ =A∼ = A2 implies that there exist n ∈ N, C ⊆ nM such that nM = C ⊕ nB1 = C ⊕ nB2 . (3) M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ =A∼ = A2 implies that there exist n ∈ N, C ⊆ nM such that nM = nA1 ⊕ C = nA2 ⊕ C. Proof. (1)⇒(2) Suppose M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A ∼ = A2 . Using the decomposition M = A1 ⊕ B1 ∼ = A ⊕ B1 , we have projections p1 : M → A1 ∼ = A, p2 : M → B1 and injections q1 : A ∼ = A1 → M, q2 : B1 → M such that p1 q1 = 1A , q1 p1 + q2 p2 = 1M . Using the decomposition M = A2 ⊕ B2 ∼ = A ⊕ B2 , we have projection f : M → A2 ∼ = A and injection ∼ g : A = A2 → M such that f g = 1A . As (f q1 )(p1 g) + f q2 p2 g = 1A , there exists some Y ∈ Mn (E) such that f q1 In + f q2 p2 gY ∈ GLn (E). That is, f ∗ q1∗ + f ∗ q2∗ p∗2 g ∗ y ∈ AutR (nA), where y ∈ EndR (nA) is an Rmorphism corresponding to Y . This implies that nM = ker(f ∗ ) ⊕ C, where C = im(q1∗ + q2∗ p∗2 g ∗ y). As p∗1 (q1∗ + q2∗ p∗2 g ∗ y) = 1nA , nM = ker(p∗1 ) ⊕ C. Therefore nM = C ⊕ nB1 = C ⊕ nB2 . (2) ⇒ (1) Suppose that ax + b = 1A with a, x, b ∈ E. Set M = 2A, and let pi : M → A, qi : A → M (for i = 1, 2) denote the projections and injections of this direct sum. Set A1 = q1 (A) and B1 = q2 (A), so that M = A1 ⊕ B1 with A1 ∼ = A. Define f = ap1 + bp2 from M to A and g = q1 x + q2 from A to M . Set A2 = g(A) and B2 = ker(f ). Then M = A2 ⊕ B2 with A2 ∼ = A. By assumption, there exists n ∈ N such that nM = C ⊕ nB1 = C ⊕ nB2 for some C ⊆ nM . Let h : nA ∼ = nA1 ∼ = C → nM be the injection. Then C = h(nA). So nM = ker(p∗1 ) ⊕ h(nA). This implies that p∗1 h is an isomorphism. In addition, nM = ker(f ∗ ) ⊕ h(nA); hence, f ∗ h is an isomorphism. Observing
that f ∗ h = (a∗ p∗1 + b∗ p∗2 )h = a∗ + b∗ p∗2 h(p∗1 h)−1 p∗1 h ∈ AutR (nA), we get aIn + bQ ∈ GLn (E), where Q is the matrix corresponding to the Rmorphism p∗2 h(p∗1 h)−1 . Therefore E satisfies power-substitution. (1) ⇒ (3) As in (1) ⇒ (2), we have that (f q1 )(p1 g) + f q2 p2 g = 1A . By virtue of Lemma 4.1.2, there exists some z ∈ EndR (nA) such that p∗1 g ∗ + zf ∗ q2∗ p∗2 g ∗ ∈ AutR (nA). This implies that nM = im(g ∗ ) ⊕ C, where C = ker(p∗1 + zf ∗ q2∗ p∗2 ). As (p∗1 + zf ∗ q2∗ p∗2 )q1∗ = 1nA , we have nM = im(q1∗ ) ⊕ C. Therefore nM = nA1 ⊕ C = nA2 ⊕ C. (3) ⇒ (1) As in (2) ⇒ (1), we have M = A1 ⊕ B1 with A1 ∼ = A. Define f = ap1 + p2 from M to A and g = q1 x + q2 b from A to M . Set A2 = g(A)

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and B2 = ker(f ). Then M = A2 ⊕ B2 with A2 ∼ = A. By assumption, there exists n ∈ N such that nM = nA1 ⊕ C = nA2 ⊕ C for some C ⊆ nM . Let h : nM → nA1 be the projection. Then C = ker(h), and so nM = ker(h) ⊕ q1∗ (nA). This implies that hq1∗ is an isomorphism. On the other hand, nM = ker(h) ⊕ g2∗ (nA). Thus, hg2∗ is an isomorphism. Observing
that hg2∗ = h(q1∗ x∗ + q2∗ b∗ ) = (hq1∗ ) x∗ + (hq1∗ )−1 hq2∗ b∗ ∈ AutR (nA). Thus xIn + Qb ∈ GLn (E), where Q is the matrix corresponding to the right R-morphism (hq1∗ )−1 hq2∗ . According to Lemma 4.1.2, E satisfies powersubstitution.  Corollary 10.1.8. Let A be a right R-module such that EndR (A) satisfies power-substitution. If B and C are any right R-modules such that A ⊕ B ∼ = A ⊕ C, then there exists some n ∈ N such that nB ∼ = nC.

Proof. Assume that ϕ : A ⊕ B ∼ = A ⊕ C. Then A ⊕ C = ϕ(A) ⊕ ϕ(B) with A∼ = ϕ(A). In view of Theorem 10.1.7, there exist n ∈ N, D ⊆ n(A ⊕ C) such that n(A⊕ C) = D ⊕ nC = D ⊕ nϕ(B). Therefore nB ∼ = nϕ(B) ∼ = nC, as asserted. 

Let S be a commutative ring with identity and R a subring of S containing 1S . Then S is said to be an extension ring of R. A polynomial f (x) ∈ R[x] is monic provided that f (x) = xn + cn−1 xn−1 + · · · + c1 x + c0 for some cn−1 , · · · , c0 ∈ R. Let S be an extension ring of R. We say that s ∈ S is integral over R provided that there exists a monic polynomial f (x) ∈ R[x] such that f (s) = 0. If every element of S is integral over R, S ˆ be the set of all elements of is said to be an integral extension of R. Let R ˆ is an integral extension ring of R which S that are integral over R. Then R ˆ = R, then R is contains every subring of S that is integral over R. If R said to be integrally closed in S. An integral domain is said to be integrally closed provided it is integrally closed in its quotient field. As is well known, every Dedekind domain is integrally closed. A field K is said to be algebraically closed provided that every nonconstant polynomial f (x) ∈ K[x] has a root in K. As is well known, K is an algebraically closed field if and only if every irreducible polynomial in K[x] has degree one. An algebraically closed field K is said to be an algebraic closure of a field F provided that K is an integral extension of F . We note that every field F has an algebraic closure which is unique under F -isomorphism. We say that an integrally closed domain D is algebraically closed provided that its quotient field is an algebraically closed field. Let D

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be an integral domain, let F be its quotient field, and let F be the algebraic closure of F . Set D = {α ∈ F | α is integral in D}. In the remainder of this section, we investigate several conditions related to power substitution of integral closed domains, which are due to Estes and Guralnick ([199]). Lemma 10.1.9. Let D be an integral domain, let a, b ∈ D, and let f (x) = xn + z1 xn−1 + · · · + zn ∈ D[x]. Then the following are equivalent: (1) For each α ∈ D, f (α) = 0 implies that a − bα ∈ U (D). (2) an + z1 an−1 b + · · · + zn bn ∈ U (D). Proof. Let F be the quotient field of D, and let F be the algebraic closure of F . Then f (x) ∈ F [x]. As F is an algebraically closed field, there are roots α1 , · · · , αn ∈ F . Since f (αi ) = 0, αi is integral over D. Hence, each n Q αi ∈ D. Write f (x) = (x − αi ). If (1) holds, then each a − bαi ∈ U (D). i=1

n

Hence, v := a +z1 a

n−1

b+· · ·+zn bn =

n Q

i=1

(a−bαi ) ∈ U (D). Clearly, v ∈ D.

As v −1 ∈ D is integral over D, we deduce that v ∈ U (D), as required. If (2) holds, then each a − bαi ∈ U (D). If α ∈ D, f (α) = 0 implies that α ∈ F is a root of f (x) ∈ F [x]; hence, α = αi for some i. Therefore the assertion is true.  Theorem 10.1.10. Let D be an integrally closed domain. Then the following are equivalent: (1) D satisfies power-substitution. (2) Whenever aD+bD = D with a, b ∈ D, there exists a monic f (x) ∈ D[x] such that a + bα ∈ U (D) for all roots α of f (x) ∈ F [x]. Proof. (1) ⇒ (2) Suppose that ac + b = 1 with a, c, b ∈ D. As D satisfies power-substitution, it follows from Corollary 10.1.2 that there exists a positive integer n and a z ∈ D such that u := an + bz ∈ U (D). Now u = an + bz(ac + b)n−1 = an + c1 an−1 (−b) + c2 an−2 (−b)2 + · · · + cn (−b)n for suitable c1 , · · · , cn ∈ R. Choose f (x) = xn + c1 xn−1 + · · · + cn ∈ D[x]. Let α ∈ F be a root of f (x) ∈ F [x]. Then α ∈ D. According to Lemma 10.1.9, a + bα ∈ U (D), as required. (2) ⇒ (1) Suppose that ac + b = 1 with a, c, b ∈ D. By hypothesis, there exists a monic f (x) = xn + c1 xn−1 + · · · + cn ∈ D[x] such that a + by ∈ U (D) for all roots α of f (x) ∈ F [x]. According to Lemma 10.1.9,

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an + c1 an−1 b + · · · + cn bn ∈ U (D). Thus, an + bz ∈ U (D) for some z ∈ D. In view of Corollary 10.1.2, D satisfies power-substitution.  Corollary 10.1.11. Let D be a Dedekind domain such that every residue field is finite. Then D has stable range one. Proof. Suppose that ax + b = 1 with a, x, b ∈ D. Let S = D[a, x]. Let E be the integral closure of S in its quotient field. Then E is a Dedekind domain. Further, every residue field of E is finite. Thus, E satisfies powersubstitution by Corollary 10.1.3. In view of Corollary 10.1.2, there exists some n ∈ N such that an + by ∈ U (E) for a y ∈ E. Thus, an + by(ax + b)n−1 ∈ U (E). Therefore an + z1 an−1 b + · · · + zn bn ∈ U (E). Clearly, f (x) = xn + z1 xn−1 + · · · + zn has solutions in E. Hence, a + bz ∈ U (E) for a z ∈ E. As E ⊆ D, we see that a + bz ∈ U (D) for a z ∈ D. Consequently, D has stable range one.  Let D = Z or k[x], k is finite field. It follows from Corollary 10.1.11 that D has stable range one. A complex number is said to be an algebraic integer if it is the root of a monic polynomial in Z[x]. Thus, the ring Z of n P all algebraic integers has stable range one. Let f (x) = ai xi ∈ D[x]. We use C(f ) to denote the ideal

n P

i=0

ai D.

i=0

Lemma 10.1.12. Let D be an integral closed domain. If f (x) ∈ D[x] represents a unit in D, then f = f1 · · · fm with fi (x) ∈ D[x] irreducible over the quotient field F of D. T Proof. Since D is an integral closed domain, D = Rv , where each Rv is a valuation ring. In particular, Rv is a B´ ezout domain. It suffices to consider the case where f is irreducible since any factor of f also represents a unit. Suppose that f is not irreducible over F [x]. Then f is the product of two polynomials of lower degree than f over F [x]. Thus, df (x) = g(x)h(x) for some d ∈ D and g, h ∈ D[x] of lower degree than f . Choose b ∈ D so that a = f (b) ∈ U (D). Since Rv is a B´ ezout domain, g(b)Rv = C(g)Rv T and h(b)Rv = C(h)Rv . Hence, C(g) ⊆ g(b)Rv = g(b)D ⊆ C(g), and so C(g) = g(b)D. Likewise, C(h) = h(b)D. This implies that g(b) and h(b) divide any coefficients of g(x) and h(x), respectively. Thus, d ∈ D divides any coefficients of g(x)h(x) in D. Therefore f (x) is not irreducible in D[x], a contradiction. Consequently, f is irreducible in F [x], as required. 

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Lemma 10.1.13. If D is an algebraically closed domain, then D is m-fold stable for all m ∈ N if and only if D is U -irreducible. Proof. In view of Proposition 3.3.9, U -irreducibility implies that D is m-fold stable for all m ∈ N. Conversely, suppose f (x), g(x) ∈ D[x] each represent a unit. By virtue of Lemma 10.1.12, f = f1 · · · fm and g = g1 · · · gn , where each fi (x) ∈ D[x] and gj (x) ∈ D[x] is irreducible over the quotient field F of D. As D is an algebraically closed domain, F is an algebraically closed field. Thus, we may assume that fi (x) = ai + bi x ∈ D[x] and gj (x) = cj + dj x ∈ D[x]. Obviously, fi (x) and gj (x) each represent a unit. Hence, ai D + bi D = D and cj D + dj D = D. Since D is m-fold stable for all m ∈ N, we can find a c ∈ D such that fi (c), gj (c) ∈ U (D). Therefore m n Q Q f (c)g(c) = fi (c) gj (c) ∈ U (D), as required.  i=1

j=1

Lemma 10.1.14. Let D be an integral domain, a, b ∈ D, and let A = s−j j
ai bi ∈ Ms (D). Set dij = ai bj − aj bi . Then Q (1) detA = ±(b1 · · · bs ) i<j dij .   0  . s−1 Q . (2) If detA 6= 0 and bs dis y, then Az =  .  has a solution z ∈ s D. 0 i=1 y

Proof. (1) follows as A is essentially a Vand der Monde matrix. (2) is a consequence of Cramer’s result.



Lemma 10.1.15. Let D be an algebraically closed domain. If D has stable range one, then D is m-fold stable for all m ∈ N. Proof. If m = 1, then the result follows. Assume that the result holds for m = s−1(s ≥ 2). It suffices to show that D is s-stable. Suppose ai D+bi D = D for 1 ≤ i ≤ s. If dij = ai bj − aj bi = 0, then aj = uai , bj = ubi for a u ∈ U (D). Thus, without loss of generality, we can assume that each bi 6= 0 and dij 6= 0 if i 6= j. Then there exists an e ∈ D such that bei − ai ∈ U (D) for each 1 ≤ i ≤ s − 1. Set g(x) = (x − e)s = xs + z1 xs−1 + · · · + zs . In view of Lemma 10.1.9, vi := asi + z1 as−1 bi + · · · + zs bsi ∈ U (D) for all i 1 ≤ i ≤ s − 1. If vs D + dis D $ D, then there exists a maximal ideal P such that vs D + dis D ⊆ P $ D. In the field D/P , dis = 0, bi e − ai ∈ U (D/P ) for 1 ≤ i ≤ s − 1. Hence, we have a u ∈ U (D/P ) such that bs = ubi and as = uai . Therefore we get bs e − as ∈ U (D/P ). Applying Lemma 10.1.9,

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we conclude that vs ∈ U (D/P ); and so vs 6∈ P . Hence, vs D + dis D = D s−1
Q for 1 ≤ i ≤ s − 1, and then vs D + dis D = D. On the other hand, i=1

ass D + bs D = D and vs ≡ ass (mod bs D). Therefore vs D + uD = D, where s−1 Q u = bs dis . As D has stable range one, there exists an r ∈ D such that i=1
vs + ru ∈ U (D). Let A = as−j bji ∈ Ms (D), and let i       w1 z1 v1 − as1   ..   .   .. .  .  := A  ..  =  . ws

zs

vs − ass

In view of Lemma 10.1.14, there exist x1 , · · · , xs ∈ D such that     0 x1  ..      A  ...  =  .  .  0  xs ru

Choose yi = xi + zi for 1 ≤ i ≤ s. Then     w1 y1   ..     . A  ...  =  .  ws−1  ys ws + ru

Let vi′ = asi + y1 as−1 bi + · · · + ys bsi . For 1 ≤ i ≤ s − 1, vi′ = asi + wi = vi ∈ i U (D). In addition, ws + ru = vs′ − ass . Hence, vs′ = ass + ws + ru = vs + ru ∈ U (D). According to Lemma 10.1.9, each ai − bi θ ∈ U (D)(i = 1, · · · , s) for a root θ of xs + y1 xs−1 + · · · + ys . Obviously, θ ∈ D. For any c ∈ D, c ∈ F is integral over D; hence, c ∈ D. This implies that D = D. Therefore ai − bi θ ∈ U (D)(i = 1, · · · , s) for a θ ∈ D, and so the result follows by induction.  An R-algebra over a commutative ring R can be seen as an extension ring S of R with R central in S. If, in addition, S is a finitely generated R-module, then S is termed a module finite R-algebra. Theorem 10.1.16. Let D be an integrally closed domain. Then the following are equivalent: (1) Every module finite D-algebra satisfies power-substitution.

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(2) D has stable range one. (3) D is m-fold stable for all m ∈ N. (4) D is U -irreducible. Proof. (1) ⇒ (2) Assume that ax + b = 1 in D. Set S = D[a, x]. As a, x ∈ D, S is a module finite D-algebra. Hence, S satisfies power-substitution. In view of Corollary 10.1.2, there exists a y ∈ S such that an + by ∈ U (S) for some n ∈ N. Thus, an + by(ax + b)n−1 ∈ U (S), and so an + z1 an−1 b + · · · + zn bn ∈ U (S). According to Lemma 10.1.9, there exists a z ∈ S such that a + bz ∈ U (S). As S ⊆ D, we see that a + bz ∈ U (D) for a z ∈ D. Consequently, D has stable range one. (2) ⇒ (3) is obvious by Lemma 10.1.15. (3) ⇒ (4) is proved by Lemma 10.1.13. (4) ⇒ (1) Let S be a module finite D-algebra. Then S is the homomorphic image of a module finite R-subalgebra T of Mn (D) (cf. [199, Lemma 2.1]). As power-substitution property is preserved by homomorphic images, we assume that
S = T . Let ac + b
= 1 in T . Set f (x) = det ac + ax(1 − c) + b , g(x) = det c + x(1 − c) ∈ D[x]. Clearly, f (0) = 1 = g(1). As D[x] ⊆ D[x], it follows by hypothesis that there exists r ∈ D such that f (r)g(r) ∈ U (D).
Hence, g(r) ∈ U (D). Thus,
c + r(1 − c) ∈ U T [r] , and so a + bd ∈ U T [r] for a d ∈ T [r]. However T [r] embeds in Mn (T ) where n is the degree of m(x), the minimal polynomial of r over D via the map ϕ which sends α ∈ T to αIn and r to the companion matrix of m(x). Let Q = ϕ(d) ∈ Mn (T ). Then aIn + bQ ∈ GLn (T ). Consequently, T satisfies power-substitution, as required.  Corollary 10.1.17. Let D be an algebraically closed domain. Then the following are equivalent: (1) D satisfies the primitive criterion. (2) D has stable range one and every primitive polynomial is the product of linear polynomials. Proof. (1) ⇒ (2) In view of Proposition 3.3.9, D has stable range one. Let f (x) ∈ D[x] be a primitive polynomial. Then f (x) represents a unit in D. By virtue of Lemma 10.1.12, f = f1 · · · fn , where each fi ∈ D[x] is irreducible over the quotient field F of D. As D is algebraically closed, we see that F is an algebraically closed field. If fi (x) = ai xi + · · · + a2 x2 + a1 x + a0 ∈ D[x], then fi (x) ∈ F [x]. As fi (x) is irreducible over F , we deduce that ai = · · · = a2 = 0 and a1 6= 0. Therefore each fi ∈ D[x] is a linear polynomial, as required.

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n Q

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(2) ⇒ (1) Let f (x) ∈ D[x] be a primitive polynomial. Then f (x) =

(ai x + bi ), where each ai , bi ∈ D. Clearly, ai D + bi D = D for 1 ≤

i ≤ n. As D has stable range one, there exists some ci ∈ D such that fi (ci ) ∈ U (D), where each fi (x) = ai x + bi . Since D is an algebraically closed domain, we deduce that D = D. According to Theorem 10.1.16, D n Q is U -irreducible. Thus, we have a d ∈ D such that f (d) = fi (d) ∈ U (D). Therefore D satisfies the primitive criterion. 10.2

i=1



The Finite Exchange Properties

In this section, we establish several equivalent characterizations of powersubstitution over exchange rings. Theorem 10.1.7 gives the external powersubstitution property. Now we observe that this property is equivalent to the internal power-substitution for modules with the finite exchange property. Lemma 10.2.1. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies power-substitution. (2) For any regular x ∈ R, there exists some n ∈ N such that xIn ∈ Mn (R) is unit-regular. Proof. (1) ⇒ (2) Let x ∈ R be regular. Then there exists some y ∈ R such that x = xyx. In view of Lemma 4.1.2, we have n ∈ N and Q ∈ Mn (R) such that U := yIn + (1 − yx)Q ∈ GLn (R). Hence, xIn = xyxIn = xU x, as required. (2) ⇒ (1) Assume that ax+b = 1 in R. By virtue of Lemma 1.4.7, there exist s, t ∈ R such that e = bs and 1 − e = (1 − b)t for an idempotent e ∈ R. Hence, axt + e = 1, and then (1 − e)axt + e = 1 with regular (1 − e)a ∈ R. By hypothesis, there exist n ∈ N, U ∈ GLn (R) such that (1 − e)aIn = (1 − e)aU (1 − e)a. Let F = U (1 − e)a. Then F xtU −1 + U eU −1 = In , whence F xtU −1 (In − F ) + U eU −1 (In − F ) = In − F.

Thus, U aIn + bs(U −1 (In −
F ) − aIn ) = U aIn + e U −1 (In − F ) − aIn ) = U (1−e)aIn +eU −1 (In −F ) = F +U eU −1 (In −F
) = In −F xtU −1 (In −F ) ∈ GLn (R). Therefore aIn + bs U −1 (In − F ) − aIn ∈ GLn (R), as desired. 

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Theorem 10.2.2. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent: (1) E satisfies power-substitution. (2) A = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 implies that there exist n ∈ N, C ⊆ nA such that nA = C ⊕ nB1 = C ⊕ nB2 . (3) A = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 implies that there exist n ∈ N, C ⊆ nA such that nA = nA1 ⊕ C = nA2 ⊕ C. Proof. (1)⇒(2) Given right R-module decompositions A = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 , then we can find an idempotent e ∈ E such that A1 ∼ eA. Since E satisfies power-substitution, it follows from Proposition = 10.1.5 that EndR (A1 ) ∼ = eEe satisfies power-substitution. According to Theorem 10.1.7, we can find a positive integer n and some C ⊆ nA such that nA = C ⊕ nB1 = C ⊕ nB2 . (2) ⇒ (1) Given any regular x ∈ E, we have some y ∈ E such that x = xyx. Clearly, A = yxA ⊕ ker(x) = xA ⊕ (1 − xy)A with yxA ∼ = xA. So there exists a positive integer n and a C ⊆ nA such that nA = C⊕nker(x) = C ⊕ n(1 − xy)A, hence nker(x) ∼ = n(1 − xy)A. Also, x∗ restricts to an isomorphism of n(yxA) onto n(xA). Define ϕ ∈ EndR (nA) so that ϕ restricts to (x∗ )−1 : n(xA) → n(yxA) and ϕ restricts to an isomorphism of n(1 − xy)A onto nker(x). It is easy to check that x∗ = x∗ ϕx∗ with ϕ ∈ AutR (nA). Let U be the matrix corresponding to ϕ. Then xIn = xU x with U ∈ GLn (E). According to Lemma 10.2.1, E satisfies power-substitution. (1) ⇒ (3) Given right R-module decompositions A = A1 ⊕B1 = A2 ⊕B2 with A1 ∼ = A2 , as in (1) ⇒ (2), we see that EndR (A1 ) satisfies powersubstitution. By virtue of Theorem 10.1.7, we can find a positive integer n and some C ⊆ nA such that nA = nA1 ⊕ C = nA2 ⊕ C. (3) ⇒ (1) Given any regular x ∈ E, we have some y ∈ E such that x = xyx. From the right R-module decompositions A = yxA ⊕ ker(x) = xA ⊕ (1 − xy)A with yxA ∼ = xA, we can find a positive integer n and some C ⊆ nA such that nA = nyxA ⊕ C = nxA ⊕ C. Thus we see that nker(x) ∼ =C ∼ = n(1 − xy)A. Analogously to the consideration in (2) ⇒ (1), we have some U ∈ GLn (E) such that xIn = xU x, as desired.  Corollary 10.2.3. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent: (1) E satisfies power-substitution. (2) A = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 implies that there exists some n ∈ N such that nB1 ∼ = nB2 .

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(3) For any idempotents e, f ∈ E, eA ∼ = f A implies that there exists some n ∈ N such that n(1 − e)A ∼ = n(1 − f )A.

Proof. (1) ⇒ (2) Given e = e2 , f = f 2 ∈ E such that eA ∼ = f A, then we see that A = eA ⊕ (1 − e)A = f A ⊕ (1 − f )A with eA ∼ = f A. By Theorem 10.2.2, we can find a positive integer n and a right R-module C such that nA = C ⊕ n(1 − e)A = C ⊕ n(1 − f )A. Hence n(1 − e)A ∼ = n(1 − f )A. (3) ⇒ (2) is obvious. (3) ⇒ (1) Given any regular x ∈ E, we have some y ∈ E such that x = xyx. Clearly, A = yxA ⊕ ker(x) = xA ⊕ (1 − xy)A with yxA ∼ = xA. So ∼ there is a positive integer n such that nker(x) = n(1−xy)A. Analogously to the discussion in Theorem 10.2.2, it is easy to check that xIn = xU x, where U ∈ GLn (E). According to Lemma 10.2.1, E satisfies power-substitution, hence completing the proof.  Corollary 10.2.4. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent: (1) E satisfies power-substitution. (2) For any R-modules B and C, A ⊕ B ∼ = A ⊕ C implies that there exists ∼ some n ∈ N such that nB = nC. Proof. (1) ⇒ (2) is clear by Corollary 10.1.8. (2) ⇒ (1) Given A = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = A2 , then A ⊕ B1 ∼ = A ⊕ B2 ; hence, there exists some n ∈ N such that nB1 ∼ = nB2 . According to Corollary 10.2.3, E satisfies power-substitution.  Theorem 10.2.5. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies power-substitution. (2) Whenever a∼b with a, b ∈ R, there exist n ∈ N, U ∈ GLn (R) such that aU = U b. Proof. (1) ⇒ (2) Let a and b be pseudo-similar in E. In view of Lemma 6.1.9, there exist x, y ∈ E such that a = xby, b = yax, x = xyx and y = yxy. By virtue of Lemma 10.2.1, we can find a positive integer n and a V ∈ GLn (E) such that yIn = yV y. Let U = (In −xyIn −V y)V (In −yxIn −yV ). Then aU = U b, where U −1 = (In − yxIn − yV )V −1 (In − xyIn − V y) ∈ GLn (R), as required. (2)⇒(1) Suppose that A = A1 ⊕ B1 = A2 ⊕ B2 with ϕ : A1 ∼ = A2 . Let p1 : A → A1 , p2 : A → A2 be the projections. Let x : A → A

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given by x(a + b) = ϕ(a) for any a ∈ A1 , b ∈ B1 and y : A → A given by y(c + d) = ϕ−1 (c) for any c ∈ A2 , d ∈ B2 . It is easy to verify that p1 = yp2 x, p2 = xp1 y, x = xyx and y = yxy. Thus p1 ∼p2 , and then p1 U = U p2 for some U ∈ GLn (E). We check that nB1 = nker(p1 ) = ker(p∗1 ) ∼ = U −1 ker(p∗1 ) = ker(p1 U ) = ker(U −1 p1 U ) = ker(p∗2 ) = nker(p2 ) = nB2 . According to Corollary 10.2.3, E satisfies power-substitution.  Corollary 10.2.6. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent: (1) E satisfies power-substitution. (2) For any idempotents e, f ∈ E, eA ∼ = f A implies that there exist n ∈ N, u ∈ AutR (nA) such that e∗ u = uf ∗ . (3) For any idempotents e, f ∈ E, eA ∼ = f A implies that there there exist n ∈ N, u, v ∈ AutR (nA) such that e∗ u = vf ∗ . Proof. (1) ⇒ (2) Let e, f ∈ E be idempotents such that eA ∼ = f A. In view of Lemma 6.1.2, there are a, b ∈ E such that e = ab and f = ba. Hence, e = af b and f = bea. Let c = eaf and d = f be. Then a = cbd, b = dac, c = cdc and d = dcd. That is, e∼f in E. Thus, there exists a positive integer n and a U ∈ GLn (E) such that eU = U f by Theorem 10.2.5. Let u ∈ AutR (nA) be the R-morphism corresponding to the matrix U . Then e∗ u = uf ∗ , as required. (2) ⇒ (3) is trivial. (3) ⇒ (1) Given any regular x ∈ E, there is some y ∈ E such that x = xyx and y = yxy. Clearly, xyA ∼ = yxA. Let e = xy and f = yx. Then we can find n ∈ N and u, v ∈ AutR (nA) such that e∗ u = vf ∗ . Set z = vf ∗ v −1 , and w = 1nA − e∗ + z. Then ze∗ = vf ∗ v −1 e∗ = vf ∗ v −1 vf ∗ u−1 = vf u−1 = e∗ and z = vf ∗ v −1 = vf ∗ u−1 uv −1 = e∗ uv −1 . So e∗ z = z. It is easy to check that w−1 = 1nA + e∗ − z. Since e∗ vf ∗ = vf ∗ , we deduce that wvf ∗ = (1nA − e∗ + z)vf ∗ = vf ∗ − e∗ vf ∗ + zvf ∗ = vf ∗ . Also we have e∗ wv = e∗ (1nA − e∗ + z)v = e∗ zv = e∗ vf ∗ = vf ∗ . Thus, e∗ wv = wvf ∗ . Let U ∈ GLn (R) be the matrix corresponding to wv. Then eU = U f , and so (1 − e)U = U (1 − f ). Let c = (1 − e)U (1 − f ) and d = (1 − f )U −1 (1 − e). Then (1 − e)In = cd and (1 − f )In = dc. In view of Lemma 6.1.2, (1−e)(nA) ∼ = (1−f )(nA). As in the proof of Theorem 10.2.2, xIn = xV x for some V ∈ GLn (R). Therefore E satisfies power-substitution from Lemma 10.2.1.  Recall that an element a ∈ R is strongly π-regular if there exist n ∈

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N, x ∈ R such that an = an+1 x, ax = xa and x = xax. We say that the solution x ∈ R is the Drazin inverse of a. Now we investigate powersubstitution by Drazin inverses for exchange rings. Lemma 10.2.7. Let R be a ring, and let a ∈ R. If a is strongly π-regular, then the Drazin inverse of a is unique, and we denote it by ad . Proof. Let x and y be the Drazin inverses of a. Then there exist some m, n ∈ N such that am = am+1 x, ax = xa, x = xax; an = an+1 y, ay = ya, y = yay. Thus, am+n = am+n+1 x = am+n+1 y. One easily checks that x = ax2 = · · · = am+n xm+n+1 = yam+n+1 xm+n+1 = ya · yam+n+1 · xm+n+1 = · · · = y m+n+1 a2m+2n+1 xm+n+1 = y m+n+1 am+n (am+n+1 x)xm+n = y m+n+1 a2m+2n xm+n = · · · = y m+n+1 am+n+1 x = y m+n+1 am+n = y, and therefore the uniqueness is proved.  Proposition 10.2.8. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies power-substitution. (2) If ab, ba ∈ R are strongly π-regular, then there exist n ∈ N, U ∈ GLn (R) such that (ab)d U = U (ba)d . ∼ f R with idempotents e, f ∈ R, then there exist Proof. (2) ⇒ (1) If eR = a, b ∈ R such that e = ab and f = ba. Clearly, ed = e and f d = f . Hence eU = U f for some U ∈ GLn (R). According to Corollary 10.2.6, R satisfies power-substitution. (1) ⇒ (2) Suppose that ab and ba are strongly π-regular. Then we have k ∈ N such that (ba)k = (ba)k+1 (ba)d , (ba)(ba)d = (ba)d (ba) and (ba)d = (ba)d (ba)(ba)d .

So we check that (ab)k+2 a(ba)d (ba)d b = (ab)a(ba)k+1 (ba)d (ba)d b = (ab)a(ba)k (ba)d b = a(ba)k+1 (ba)d b = a(ba)k b = (ab)k+1 ;
(ab) a(ba)d (ba)d b = a(ba)d (ba)d (ba)b = a(ba)d b
= a(ba)d (ba)d b (ab);

a(ba)d (ba)d b (ab) a(ba)d (ba)d b = a(ba)d (ba)d b.

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By the uniqueness of the Drazin inverse, (ab)d = a(ba)d (ba)d b. In addition, one checks that (ba)d = (ba)d (ba)(ba)d = (ba)d ba(ba)d (ba)d ba = (ba)d b(ab)d a. Clearly, (ba)d ba(ba)d b = (ba)d b. Thus, (ba)d ∼(ab)d , and therefore we complete the proof by Theorem 10.2.5.  Proposition 10.2.9. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies power-substitution. (2) Whenever ψ : aR ∼ = bR, there exist n ∈ N, U ∈ GLn (R) such that ψ(a)In = bU . ∼ bR with a, b ∈ R, then b = ψ(as), a = Proof. (1) ⇒ (2) Given ψ : aR = ψ −1 (bt) for some s, t ∈ R. Since R satisfies power-substitution, we can find a positive integer n and a matrix Q ∈ M
n (R) such that tIn + (1 − ts)Q = U ∈ GLn (R). Clearly, b = ψ ψ −1 (bt)s = bts. Hence, ψ(a)In = btIn =
b tIn + (1 − ts)Q = bU . (2) ⇒ (1) Given any regular x ∈ R, there is some y ∈ R such that x = xyx. Obviously, xR = xyR. Thus we can find a positive integer n and a matrix U ∈ GLn (R) such that xIn = xyU ; hence xIn = xyxIn = xU −1 x. In view of Lemma 10.2.1, the result follows.  We note that an exchange ring R satisfies power-substitution if and only if for any regular a, b ∈ R, aR ∼ = bR implies that there exist n ∈ N, U, V ∈ GLn (R) such that aU = V b. The following power-cancellation result is due to Blackadar (cf. [48] and [298]). Theorem 10.2.10. Let A, B and C be right R-modules. If A .⊕ nB, nC for some n ∈ N, then A ⊕ B ∼ = A ⊕ C implies that mB ∼ = mC for all m ≥ 2n.

Proof. Write nB ∼ = A⊕D and nC ∼ = A⊕E. Then (n+1)B ∼ = B ⊕(A⊕D) ∼ = C ⊕(A⊕D) ∼ = C ⊕nB. Likewise, (n+1)C ∼ = B ⊕nC. For any m ≥ 2n, write m = 2n + p, where p ≥ 0. Then mB ∼ = nB ⊕ (n + p)B ∼ = nB ⊕ (n + p)C ∼ = ∼ ∼ nB ⊕ nC ⊕ pC = (n + n)C ⊕ pC = mC, as asserted.  Corollary 10.2.11. Every directly finite simple exchange ring satisfies power-substitution.

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Proof. Let R be a directly finite simple exchange ring. Given R = A1 ⊕B1 = A2 ⊕ B2 with A1 ∼ = A2 , then there exist idempotents e, f ∈ R such that B1 = eR and B2 = f R. If B1 = 0, then R ∼ = R ⊕ B2 ; hence, B2 = 0. Likewise, B2 = 0 implies that B1 = 0. Thus, we may assume that B1 6= 0 and B2 6= 0. Since R is simple, ReR = Rf R = R. As in the proof of Corollary 8.1.7, R .⊕ n(eR), n(f R). According to Theorem 10.2.10, there exists m ∈ N such that mB1 ∼ = mB2 . Therefore R satisfies powersubstitution by Corollary 10.2.3. 

10.3

Unit π-Regularity

Since there exist exchange rings which do not satisfy power-substitution such as the endomorphism ring of an infinite-dimensional vector space over a field, it is interesting to investigate the conditions under which an exchange ring satisfies power-substitution. We say that x ∈ R is unit π-regular provided that there exists some n ∈ N such that xn is unit-regular. A ring R is unit π-regular in the case where every element in R is unit π-regular.  Z2 Z2 For instance, is unit π-regular, while it is not regular. The main 0 Z2 purpose of this section is to investigate power-substitution of exchange rings by virtue of unit π-regularity. Lemma 10.3.1. Let R be an exchange ring. If for any regular x, y ∈ R, there exist n ∈ N, A ∈ Mn (R) such that xIn − A is unit-regular and In − yA ∈ GLn (R), then R satisfies power-substitution. Proof. Suppose that ax + b = 1 in R. Since R is an exchange ring, there exists an idempotent e of R such that e = bs and 1 − e = (1 − b)t for some s, t ∈ R from Lemma 1.4.7. Thus axt + e = (1 − b)t + e = 1; hence (1 − e)axt(1 − e) + e = 1. This implies that (1 − e)a and xt(1 − e) are both regular. So there exists a positive integer n and a matrix A ∈ Mn (R) such that xt(1 − e)In − A = W is unit-regular and In − (1 − e)aA = U ∈ GLn (R). Therefore, (1 − e)aW + eIn = U ∈ GLn (R). We assume W U −1 = V F for some V ∈ GLn (R), F = F 2 ∈ Mn (R). Then (In − F )(1 − e)aV F + (In − F )eU −1 = In
− F , whence W U −1 + V (In − F )eU −1 = V In − (In − F )(1 − e)aV F ∈ GLn (R). By virtue of Lemma 4.1.2, we can find some

S ∈ Mn (R) such that aIn + bs U −1 S − aIn = aIn + e U −1 S − aIn = (1 − e)aIn + eU −1 S ∈ GLn (R). Therefore R satisfies power-substitution.

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Theorem 10.3.2. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies power-substitution. (2) For any regular x ∈ R, there exists a positive integer n such that xIn is unit π-regular in Mn (R). Proof. (1) ⇒ (2) is trivial from Lemma 10.2.1. (2) ⇒ (1) Given any regular x, y ∈ R, there are positive integers m, n and a matrix U ∈ GLn (R) such that xm In = xm U xm . Set 

0n  In  A= .  ..

··· ··· .. .

0n 0n .. .

  0n In   0n   0n ..  , B =  ..  . . 

0n · · · In 0n 

0n 0n

y m−1 In  In  C= ..  . 0n

xIn In .. .

 · · · xm−1 In ··· 0n   , .. ..  . . ··· In

 · · · yIn In · · · 0n 0n   ∈ Mmn (R). . . .. ..  . . .  · · · In 0n

One easily checks that  0n · · · 0n xm In xIn · · · 0n ∗   ..  , .. . . .. . . .  .  0n · · · xIn ∗  0n 0n · · · −In ∗   0n · · · 0n 0n In  In · · · 0n 0n ∗      C(Imn − yA) =  −yIn · · · 0n 0n ∗  .   . . . . . . . .. .. ..   .. 

0n  −In   B(xImn − A) =  ...   0n

0n · · · −yIn In ∗

Clearly, B, C ∈ GLmn (R), and so Imn − yA ∈ GLmn (R). As xm In is unitregular, we may assume that xm In = eu for some e = e2 ∈ Mn (R), u ∈

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GLn (R). So we have 

 0n 0n · · · 0n eu  −In xIn · · · 0n ∗    .. . . .. ..  −1  .. xImn − A = B  . . . .  .    0n 0n · · · xIn ∗  0n 0n · · · −In ∗    0 0 e 0n · · · 0n  n n −I  0n In · · · 0n   n xIn  . .. −1  =B  . . . .  . .  .. .. . . ..   .  0n 0n 0n 0n · · · In 0n 0n

··· ··· .. . ··· ···

0n 0n .. . xIn −In

 u ∗  ..  . .  ∗ ∗

Consequently, xImn −A = (xImn −A)U −1 B(xImn −A), because xImn −A = B −1 EU and E = E 2 , where     0n 0n · · · 0 u e 0n · · · 0n  −In xIn · · · 0n ∗    0n In · · · 0n     ..  .. . . .. ..  . , U = E= . . .   . . . . .  .   .. .. . . ..   0n 0n · · · xIn ∗  0n 0n · · · In 0n 0n · · · −In ∗

This shows that xImn − A ∈ Mmn (R) is unit-regular. As Imn − yA ∈ GLmn (R), it follows from Lemma 10.3.1 that R satisfies power-substitution. 

Corollary 10.1.2 shows that a commutative ring R satisfies powersubstitution if and only if ax + b = 1 in R implies that there exists n ∈ N such that an + by ∈ U (R) for some y ∈ R. Now we generalize this result to exchange rings. Corollary 10.3.3. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies power-substitution. (2) If ax + b = 1 in R, then there exist m, n ∈ N and Q ∈ Mn (R) such that am In + bQ ∈ GLn (R). Proof. (1)⇒(2) is obvious. (2)⇒(1) Given any regular x ∈ R, there is some y ∈ R such that x = xyx. Obviously, x = xzx and z = zxz with z = yxy. From xz+(1−xz) = 1, we can find positive integers m, n such that xm In + (1 − xz)Q = U ∈

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GLn (R) for easily check that zU = z xm In +
somemQ ∈ Mn (R). We m −1 (1 − xz)Q = zx In , so zIn = zx U . Hence xm In = xzxxm−1 In = xzxm U −1 xxm−1 = xm U −1 xm , i.e., xm In ∈ Mn (R) is unit-regular. By Theorem 10.3.2, we conclude that R satisfies power-substitution.  As in the proof of Corollary 10.3.3, we derive that an exchange ring R satisfies power-substitution if and only if aR+ bR = R with a, b ∈ R implies that there exist m, n ∈ N and Q ∈ Mn (R) such that am In +bm Q ∈ GLn (R). Theorem 10.3.4. Let A be a right R-module having the finite exchange property, and let E = EndR (A). If every regular element of E is unit π-regular, then A satisfies power-cancellation. Proof. By virtue of [382, Proposition 28.6], E is an exchange ring. According to Theorem 10.3.2, E satisfies power-substitution. Therefore A satisfies power-cancellation from Corollary 10.2.4.  In [69, Theorem 2.3], Camps and Menal showed that every strongly πregular ring satisfies power-substitution. Ara [8] generalized this result and showed that every strongly π-regular ring has stable range one. Clearly, every strongly π-regular ring is unit π-regular. But the converse is not ∞ Q true. Let R = Mn (F ) where F is a field. Then R is unit-regular; n=1

hence, it is unit π-regular. Let a1 = 0 and an = e12 + e23 + · · · + e(n−1)n , where eij is an n × n matrix with 1 in the (i, j) position and 0′ s elsewhere (n ≥ 2). One easily checks that eij ekl = eil (j = k) and eij ekl = 0(j 6= k). Thus, ann = 0 and an−1 6= 0(n ∈ N). Choose a = (a1 , a2 , · · · ). Then n aR % a2 R % · · · . Therefore R is not strongly π-regular. Also we notice that the center, the set of all elements commutative with all elements, of any unit π-regular ring is strongly π-regular. Let C(R) be the center of a unit π-regular ring R. For any x ∈ C(R), there exists a y ∈ R such that xn = xn yxn for some n ∈ N. Set z = yxn y. Then xn = xn zxn . For any r ∈ R, zr = yxn yr = y 2 xn r = y 2 rxn yxn = yxn yxn ry = yxn ry. Likewise, rz = yrxn y. Hence, zr = rz, and so z ∈ C(R). In addition, xn = x2n z, and we are done. Corollary 10.3.5. Every unit π-regular ring satisfies power-substitution. Proof. Let R be a unit π-regular ring. Then R is an exchange ring. By Theorem 10.3.4, we see that R satisfies power-substitution. 

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Corollary 10.3.6. Let R be a separative exchange ring, and let I be an ideal of R. If every regular element of R is unit π-regular, then the cokernel of the map α∗ : K1 (R) → K1 (R/I) induced by the canonical projection α : R → R/I is a torsion group. Proof. In view of Theorem 10.3.4, R satisfies power-substitution. Given nR ⊕ B ∼ = nR ⊕ C(n ∈ N, then So R ⊕ ((n − 1)R ⊕ B) ∼ = R ⊕ ((n − 1)R ⊕ C) ⊕ ⊕ with R . n − 1R⊕B and R . n − 1R⊕C(n ≥ 2). Since R is separative, we claim that (n − 1)R ⊕ B ∼ = (n − 1)R ⊕ C. By iteration of this process, R⊕B ∼ = R ⊕ C. In view of Theorem 10.3.4, R satisfies power-substitution. Hence, it satisfies power-cancellation. So, we can find a positive integer m such that mB ∼ = mC. Using Corollary 10.2.4, Mn (R) satisfies powersubstitution. Let K be an ideal of R. Then Mn (R/K) ∼ = Mn (R)/Mn (K) satisfies power-substitution. This implies that Mn (R/K) is directly finite. That is, every factor ring of R is stably finite. On the other hand, we easily check that R has stable weak power-substitution. By [70, Lemma 3.1], the result follows.  Using the technique above, we can relate unit-regularity and separativity over regular rings. We claim that a regular ring R is unit-regular if and only if R is separative and for any x ∈ R, there exists n ∈ N such that xn , xn+1 are both unit-regular. Suppose that for any x ∈ R, there exists n ∈ N such that xn , xn+1 are both unit-regular. If A, B and C are finitely generated projective right R-modules such that A ⊕ B ∼ = A ⊕ C, analogously to the above consideration, we have nB ∼ nC and (n + 1)B ∼ = = (n + 1)C. By the ∼ separativity of R, we can derive A = B, and we are done. Lemma 10.3.7. Let R be an exchange ring. If for any regular x, y ∈ R, there exist n ∈ N, A ∈ Mn (R) such that xIn − A ∈ GLn (R) and In − yA is unit-regular, then R satisfies power-substitution. Proof. Suppose that ax + b = 1 in R. Then there exists an idempotent e ∈ R such that e = bs and 1 − e = (1 − b)t for some s, t ∈ R. Hence, (1 − e)axt(1 − e) + e = 1. In addition, (1 − e)a, xt(1 − e) ∈ R are both regular. So we can find a positive integer n and a matrix A ∈ Mn (R) such that xt(1 − e)In − A = U ∈ GLn (R) and In − (1 − e)aA = W is unitregular. Thus (1 − e)aIn + eU −1 = W U −1 is unit-regular. Assume that

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W U −1 = F V for some F = F 2 ∈ Mn (R), V ∈ GLn (R). We have that
F V xt(1 − e) + e In −
eU −1 xt(1 − e)
= (1 − e)aIn + eU −1 xt(1 − e) + e In − eU −1 xt(1 − e) = In ;
hence, F V xt(1 − e)(In − F ) + e In − eU −1 xt(1 − e) (In − F ) = In − F . Thus

(1 − e)aIn + eU −1 + e In −
eU −1 xt(1 − e) (In − F )V = F V + e In − eU −1 xt(1 − e)
(In − F )V = In − F V xt(1 − e)(In − F ) V ∈ GLn (R). Consequently, we get
aIn + bs U −1 + (In − eU −1 xt(1 − e))(In − F )V − aIn
= (1 − e)aIn + e U −1 + (In − eU −1 xt(1 − e))(In − F )V ∈ GLn (R), as required.  Theorem 10.3.8. Let R be an exchange ring. If for any regular x, y ∈ R, there exist m, n ∈ N and U ∈ GLn (R) such that In + xm (y m In − U ) is unit-regular, then R satisfies power-substitution. Proof. Given any regular x, y ∈ R, there exist positive integers m, n and a matrix U ∈ U (R) such that In + y m (xm In − U ) ∈ Mn (R) is unit-regular. Thus xm In − Z is invertible and In + y m Z is unit-regular, where Z = xm In − U . Set     In xIn · · · xm−1 In 0n · · · 0n −Z  0n In · · ·  In · · · 0n 0n  0n      , B = A= . . ,   .. . . . .   .. .. . . .  .. . . .. ..  . 0n · · · In 0n  m−1 y In  In  C= ..  .

0n Then, it is easy to check that 

0n 0n · · · 

In

· · · yIn In · · · 0n 0n   ∈ Mmn (R). . . .. ..  . . .  · · · In 0n

0n  −In   B(xImn − A) =  ...   0n

0n xIn .. .

0n 0n 0n

 · · · 0n xm In − Z  · · · 0n ∗   .. .. .. , . . .   · · · xIn ∗ · · · −In ∗

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

0n  I  n   −yIn C(Imn − yA) =   ..  .   0n 0n

 0n 0n In + y m Z  0n 0n ∗   0n 0n ∗  . .. .. ..   . . .   0n · · · In 0n ∗ 0n · · · −yIn In ∗

0n 0n In .. .

··· ··· ··· .. .

As B, C ∈ GLmn (R), we see that xImn − A ∈ GLmn (R) and Imn − yA is unit-regular. According to Lemma 10.3.7, we obtain the result.  Lemma 1.2.1 shows that a ring has unit 1-stable range if and only if for any x, y ∈ R, there exists some u ∈ U (R) such that 1 + x(y − u) ∈ U (R). As a consequence of Theorem 10.3.8, we derive the following. Corollary 10.3.9. Let R be an exchange ring. If for any regular x, y ∈ R, there exists a positive integer m and an element u ∈ U (R) such that 1 + xm (y m − u) ∈ U (R), then R satisfies power-substitution. Now we extend Theorem 10.3.8 to a general ring as follows. Theorem 10.3.10. Let R be a ring. If for any x, y ∈ R, there exist m, n ∈ N and U ∈ GLn (R) such that In + xm (y m In − U ) is unit-regular, then R satisfies power-substitution. Proof. Given any c, d ∈ R, there exist positive integers m, n and a matrix U ∈ U (R) such that In + cm (dm In − U ) ∈ Mn (R) is unit-regular. Thus dm In − Z is invertible and In + cm Z is unit-regular, where Z = dm In − U . As in the proof of Theorem 10.3.8, there is some B ∈ Mmn (R) such that cImn − B ∈ GLmn (R) and Imn − dA is unit-regular. Suppose that ax + b = 1 in R. By the preceding discussion, we have n ∈ N and A ∈ Mn (R) such that xIn − A = U ∈ GLn (R) and In − aA = W is unit-regular. Thus aIn + bU −1 = W U −1 is unit-regular. Assume that W U −1 = F V for some F = F 2 ∈ Mn (R), V ∈ GLn (R). Thus, (aIn +
bU −1 )x+b(In −U −1 x) = In ; hence, F V x(In
−F )+b In −U −1 x (In −F ) = In − F . This implies that F + b In − U −1 x (In − F ) = In − F V x(In −
F ). −1 −1 As a result, we deduce that x)(I
n − F )V =
aIn + b U + (In − U −1 −1 −1 aIn + bU + b In − U x (I − F )V = F V + b I − U x (In − F )V = n n
In − F V x(In − F ) V ∈ GLn (R). Therefore the result follows.  Corollary 10.3.11. Let R be a ring. If for any x, y ∈ R, there exist

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m, n ∈ N and U ∈ GLn (R) such that xm In − U, y m In − U −1 ∈ GLn (R), then R satisfies power-substitution. Proof. For any x, y ∈ R, there exist m, n ∈ N and V ∈ GLn (R) such that y m In − V, xm In − V −1 ∈ GLn (R). Let U = y m In − V . Then xm V − In ∈ GLn (R); hence, In − xm (y m In − U ) ∈ GLn (R). Therefore we complete the proof by Theorem 10.3.10.  Corollary 10.3.12. Let R be a unital complex C ∗ -algebra. Suppose that for any x ∈ R, there exists m ∈ N such that xm is the sum of a unitary and a unit. Then R satisfies power-substitution. Proof. For any x, y ∈ R, we have a unitary v ∈ R such that (1+ k y k)m m xm − v ∈ U (R) for some m ∈ N. Let u = v/(1+ k y k)m . Then
m x − u ∈ m U (R). As k v k= 1, we see that k y u k= k y k /(1+ k y k) k v k< 1; m m −1 hence, 1 − y u ∈ U (R). This implies that y − u ∈ U (R). Therefore xm − u, y m − u−1 ∈ U (R). According to Corollary 10.3.11, R satisfies power-substitution. 

10.4

Stable Power-Substitutions

A ring R is said to satisfy stable power-substitution provided that Mn (R) satisfy power-substitution for all n ∈ N. Example 10.4.1. Let X = [−1, 1]4 . Then CR (X) satisfies powersubstitution, while it does not satisfy stable power-substitution. Proof. In view of Example 10.1.6, CR (X) satisfies power-substitution.
According to [70, Theorem 2.7], M3 CR (R) does not satisfy powersubstitution. Therefore CR (X) does not satisfy stable power-substitution.  The preceding example of the ring of real continuous functions shows that the power-substitution property is not Morita invariant. Following Goodearl [219], a ring homomorphism ϕ : R → S has stable range one provided that aR + bR = R with a, b ∈ R implies that there exists some y ∈ S such that ϕ(a) + ϕ(b)y ∈ U (S). The following
result is due to Goodearl, Camps and Menal, which shows that M2 CR (X) satisfies powersubstitution.

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Example 10.4.2. Let X be a topological space. Then M2 CR (X) satisfies power-substitution.

Proof. Let α : M2 CR (X) → M2 CH (X) be the inclusion map. We shall prove that α has stable range one. Suppose that ad + b = 1 for some
a = (aij ), d = (dij ), b = (bij ) ∈ M2 CR (X) . Then a11 d11 +a12 d21 +b11 = 1. Thus, a11 + (a12 d21 + b11 )i is a unit in CC (X). Set       1 0 i0 1 0 A = (Aij ) = a +b , D = (Dij ) = d, d21 i 1 00 −d21 i 1  
1 − id11 −id12 B = (Bij ) = b ∈ M2 CC (X) . 0 1 Then AD +
B = I2 and A11 is a unit in CC (X). It suffices to find Y ∈ M2 CH (X) such that A + BY is a unit. By replacing A, D and B by         1 0 1 0 1 0 1 0 A, D and B , −A21 A−1 A21 A−1 −A21 A−1 A21 A−1 11 1 11 1 11 1 11 1

we may assume that A21 = 0. Thus, A22 D22 + B22 = 1 in CC (X). This implies that A22 + B22 j ∈ CH (X) is a unit. Therefore     00 A11 A12 + B22 j A+B = 0j 0 A22 + B22 j
is a unit in M2 CH (X) . Thus, α has the stable range one. Let us fix an R-algebra embedding of H into
M4 (R). This induces
a CR (X)-algebra monomorphism β : M2 CH (X) → M2 M4 (CR (X))
. Thus, we get a
CR (X)-algebra monomorphism γ := βα : M2 CR (X) → M2 M4 (CR (X)) . Further, γ has stable range one. Since γ is a CR (X)algebra monomorphism, it follows from the Skolem-Noether
Theorem that for some g ∈ GL8 (R), gγ(c)g −1 = cI4 for all c ∈ M2 CR (X) . Therefore we can easily obtain the result.  Lemma 10.4.3. Let R satisfy stable power-substitution, and let A be a finitely generated projective right R-module. If B and C are any right Rmodules such that A ⊕ B ∼ = A ⊕ C, then there exists n ∈ N such that nB ∼ nC. = Proof. Let E = EndR (A). Then we have m ∈ N and F = F 2 ∈ Mm (R) such that E ∼ = F Mm (R)F . Since R satisfies stable power-substitution, Mm (R) satisfies power-substitution. In view of Proposition 10.1.5, E satisfies power-substitution. Therefore the result follows by Corollary 10.1.8. 

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Theorem 10.4.4. Let R be an exchange ring. Then R satisfies stable power-substitution if and only if for all finitely generated projective right R-modules A, B and C, A ⊕ B ∼ = A ⊕ C implies that there exists n ∈ N such that nB ∼ nC. = Proof. Suppose that R satisfies stable power-substitution. If A, B and C are finitely generated projective right R-modules such that A ⊕ C ∼ = B ⊕ C, ∼ then there exists n ∈ N such that nB = nC from Lemma 10.4.3. Conversely, assume that nR = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 . Then nR ⊕ B ∼ = A2 ⊕ C. In addition, B and C are finitely generated projective right R-modules. By hypothesis, there exists some m ∈ N such that mB ∼ = mC. According to Corollary 10.2.3, Mn (R) satisfies power-substitution, as required.  Corollary 10.4.5. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies stable power-substitution. (2) For any A ∈ F P (R), A = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 implies nB ∼ = nC for some positive integer n. Proof. (1) ⇒ (2) Let A ∈ F P (R). Suppose that A = A1 ⊕ B = A2 ⊕ C with A1 ∼ = A2 . Then A ⊕ B ∼ = A ⊕ C with A, B, C ∈ F P (R). It follows by Theorem 10.4.4 that nB ∼ = nC for some positive integer m. (2) ⇒ (1) By Corollary 10.2.3, EndR (nR) satisfies power-substitution, and therefore so does Mn (R), as required.  Immediately, we deduce that a regular ring R satisfies stable powersubstitution if and only if for any A ∈ F P (R), A = A1 ⊕ B = A2 ⊕ C with T A1 ∼ = A2 , B C = 0 implies nB ∼ = nC for some positive integer n. A right R-module P is a stably free module of rank n provided that there exist some s, t ∈ N such that P ⊕ sR ∼ = tR, where t − s = n. We say that a stably free right R-module of rank n is free when P ∼ = nR. We say that a stably free right R-module of rank n is power-free in the case where sP ∼ = snR for a positive integer s. An interesting problem is when a stably free module is power-free. If R is a right noetherian ring or a commutative ring, then every stably free module is power-free (cf. [298, Theorem 5.10 and Theorem 5.11]). Corollary 10.4.6. Let R be an exchange ring. If R satisfies stable powersubstitution, then every stably free right R-module of positive rank is power-

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free. Proof. Let P be a right R-module with Then nR ∼ = P ⊕ mR ∼ = (n − m)R ⊕ mR. satisfies power-substitution for all n ∈ N. that sP ∼ = s(n − m)R is free for some s follows.

P ⊕ mR ∼ = nR with m < n. Clearly, EndR (nR) ∼ = Mn (R) It follows by Corollary 10.4.5 ∈ N, and therefore the result 

Lemma 10.4.7. Let I be an ideal of a ring S, and let R be a subring of S containing I. If R/I and S satisfies stable power-substitution, then so does R. Proof. Given AX + B = In in Mn (R), then AX + B = In in Mn (R/I). By hypothesis, there exist m ∈ N, Y ∈ Mmn (R) such that AImn + BY ∈ GLmn (R/I), where AImn = diag(A, · · · , A)m×m , Y = (Yij )m×m , Yij ∈ Mn (R) and BY = (BYij )m×m . That is, U (AImn + BY ) = (AImn + BY )U = Imn for some U ∈ Mmn (R). This implies that C := Imn − (AImn + BY )U ∈ Mmn (I). Hence, (AImn + BY )U + C = Imn . Clearly, (AImn + BY )X + B(Imn − Y X) = Imn , where Y X = (Yij X)m×m . Thus, (AImn + BY )U + (AImn + BY )(XC) + B(Imn − Y X)C = Imn , where C = (Cij )m×m , Cij ∈ Mn (R), XC = (XCij )m×m . It follows that (AImn + BY )(U + XC) + B(Imn − Y X)C = Imn . Since S satisfies stable power-substitution, Mmn (S) satisfies powersubstitution. Thus, there exist p ∈ N, Z ∈ Mpmn (S) such that AIpmn + BY Ipmn + B(Imn − Y X)CZ ∈ GLpmn (S). That is, we have some V ∈ Mpmn (S) such that
AIpmn + BY Ipmn + B(Imn − Y X)CZ V
= V AIpmn + BY Ipmn + B(Imn − Y X)CZ = Ipmn . As a result,
AIpmn + BY Ipmn V
= U AIpmn + BY Ipmn = Ipmn in Mpmn (S/I). Therefore, U Ipmn = U · Ipmn
= U · AIpmn + BY Ipmn V
= U AIpmn + BY Ipmn · V = Ipmn · V = V,

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and then U Ipmn − V ∈ Mpmn (I). This implies that V ∈ Mpmn (R). Consequently,
AIpmn + B Y Ipmn + (Imn − Y X)CZ ∈ GLpmn (R).

That is, Mn (R) satisfies power-substitution, and therefore R satisfies stable power-substitution. 

Theorem 10.4.8. Every finite subdirect product of rings satisfying stable power-substitution satisfies stable power-substitution. Proof. Let R1 and R2 satisfy stable power-substitution, and let R be the subdirect product of R1 and R2 . It suffices to prove that R satisfies stable power-substitution. As in the proof of Theorem 1.1.13, we may assume that R is a subring of R1 ⊕ R2 . Let K = {(r1 , r2 ) | r2 = 0}. Then K is an ideal of R1 ⊕ R2 which is contained in R. Further, R/K ∼ = R1 . One easily checks that R1 ⊕ R2 satisfies stable power-substitution. According to Lemma 10.4.7, we obtain the result.  Corollary 10.4.9. Let I be an ideal of a ring R. Then the following are equivalent : (1) R satisfies stable power-substitution. (2) R/I and R/r(I) satisfy stable power-substitution. (3) R/I and R/ℓ(I) satisfy stable power-substitution. Proof. (1)⇒(2) is trivial.
T (2) ⇒ (1) As in the proof of Corollary 1.1.14, R/ I r(I) is isomorphic to the subdirect product of R/I and R/r(I). By virtue of Theorem 10.4.8,

2 T T R/ I r(I) satisfies stable power-substitution. Since I r(I) = 0, we T see that I r(I) ⊆ J(R). Therefore R satisfies stable power-substitution. (1) ⇔ (3) is proved in the same manner.  Lemma 10.4.10. Let e ∈ R be an idempotent, and let C be a right Rmodule which is isomorphic to a direct summand of n(eR)(n ∈ N). Then N
N C Re eR ∼ = C. R

eRe

Proof. Since C is isomorphic to a direct summand of n(eR), we
have an idempotent matrix E ∈ Mn (eRe) such that C ∼ = E n(eR)
N . So N
N N
N N ∼ C Re eR = E(n(eR)) Re eR ∼ eR ∼ = E(n(eR) Re) = R eRe R eRe R eRe


N N N eR ∼ E (n(eR) Re) eR ∼ = E n(eR) ∼ = C, as as= E n(eRe) serted.

R

eRe

eRe



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Theorem 10.4.11. Let R be an exchange ring, and let e ∈ R be an idempotent. Then R satisfies stable power-substitution if and only if so eRe and (1 − e)R(1 − e) also satisfy stable power-substitution. Proof. Assume that R satisfies stable power-substitution. Then Mn (R) satisfies power-substitution. Let E = diag(e, · · · , e)n×n . Then Mn (eRe) ∼ = EMn (R)E. In light of Proposition 10.1.5, Mn (eRe) satisfies powersubstitution. Therefore eRe satisfies stable power-substitution. Likewise, (1 − e)R(1 − e) satisfies stable power-substitution. Conversely, assume that eRe and (1 − e)R(1 − e) satisfy stable powersubstitution. For all finitely generated projective right R-modules, A⊕B
∼ = ∼ A ⊕ C implies that nR ⊕
B = nR ⊕ C; hence, n(eR) ⊕
n (1 − e)R ⊕ B ∼ = n(eR)
⊕ n (1 − e)R ⊕ C. Let B ′ = n (1 − e)R ⊕ B and C ′ = n (1 − e)R ⊕ C. Then n(eR) ⊕ B ′ ∼ = n(eR) ⊕ C ′ . Since R is an exchange ring, by [16, Proposition 1.2], there exist decompositions n(eR) = C11 ⊕ C12 , B ′ = C21 ⊕ C22 , n(eR) ∼ = C11 ⊕ C21 and C ′ = C12 ⊕ C22 . As a N ∼ result, n(eR) = C11 ⊕ C12 = C11 ⊕ C21 . This implies that n(eR) Re ∼ = R N N N N N ∼ C11 ∼ C11 Re ⊕ C12 Re = Re ⊕ C21 Re. Clearly, n(eR) Re = R

R

R

R

R

n(eRe) as right eRe-modules. Since eRe satisfies stable power-substitution, N
N
∼ Re . Furthermore we can find m ∈ N such that m C12 Re = m C21 R R N
N N
N we get m C12 Re eR ∼ Re eR. Since C12 and C21 = m C21 R

eRe

R

eRe

are both isomorphic to direct
summands of n(eR), by Lemma
10.4.10, we N N N N Re eR ∼ check that C12 Re eR ∼ = C21 ; hence, = C12 and C21 R eRe
R eRe ′ ′ ∼ ∼ ∼ mC12
= mC21 . Thus mB = mC . That is, mn (1−e)R ⊕mB = mn (1− e)R ⊕mC. Since (1−e)R(1−e) has stable power-substitution, analogously to the above consideration, we can find p ∈ N such that mpB ∼ = mpC. According to Theorem 10.4.4, R satisfies stable power-substitution.  Corollary 10.4.12. Let A and B be
right R-modules having the finite exchange property. Then EndR A ⊕ B satisfies stable power-substitution if and only if EndR (A) and EndR (B) also satisfy stable power-substitution. Proof. Let e : A ⊕ B
→ A ⊕ B given by e(a, b) = a
for any (a, b)
∈ A ⊕ B. Then eEndR A ⊕ B e ∼ = EndR (A) and 1A⊕B − e EndR A ⊕ B 1A⊕B −
∼ e = EndR (B). In view of [382, Theorem 28.7], A ⊕ B has the finite
exchange property. Hence, EndR A ⊕ B is an exchange ring from [382, Proposition 28.6]. Therefore we complete the proof by Theorem 10.4.11.

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If {e1 , · · · , en } is a complete set of idempotents in R, it follows by induction and Theorem 10.4.11 that R satisfies stable power-substitution if and only if all ei Rei also satisfy stable power-substitution. Corollary 10.4.13. Let T be the ring of a Morita context (A, B, M, N, ψ, φ). If A and B are exchange rings, then T satisfies stable powersubstitution if and only if A and B also satisfy stable power-substitution. Proof. Let e = diag(1, 0). Then eT e ∼ = A and (1T − e)T (1T − e) ∼ = B. In view of [382, Theorem 28.7], T is an exchange ring. So the result follows by Theorem 10.4.11.  By induction and Corollary 10.4.13, we prove that the ring of all lower (upper) triangular matrices over an exchange ring satisfying stable powersubstitution satisfies stable power-substitution. Let A ∈ Mm×n (R). We say that A admits a diagonal power-reduction in the case where there exists some s ∈ N such that (aij Is ) ∈ Mms×ns (R) admits a diagonal reduction. Theorem 10.4.14. Let R be an exchange ring satisfying stable powersubstitution. Then every regular matrix over R admits a diagonal powerreduction. Proof. Let A = (aij ) ∈ Mm×n (R) be regular. Then we have B ∈ Mn×m (R) such that A = ABA. Let E = (eij ) = AB. Then A(nR) = E(mR). If m = n, it follows from [409, Theorem 2.1] that E(mR) ∼ = e1 R ⊕ · · · ⊕ en R ∼ = diag(e1 , · · · , en )(nR) for some idempotents ei ∈ R. Since R satisfies stable power-substitution, Mn (R) satisfies power-substitution. In view of Corollary 10.2.6, diag(E, · · · , E)s×s is similar to diag(F, · · · , F )s×s , where F = diag(e1 , · · · , en )n×n . That is, there exists some U ∈ GLns (R) such that diag(E, · · · , E)s×s = U diag(F, · · · , F )s×s U −1 . As A(nR) = E(nR), we deduce that AMn (R) = EMn (R). So, there exist X, Y ∈ Mn (R) such that AX = E and A = EY . Since XY + (In − XY ) = In , there exist p ∈ N, Z ∈ Mnp (R) such that diag(X, · · · , X)p×p + diag(In − XY, · · · , In − XY )p×p Z = V ∈ GLnp (R). Hence diag(E, · · · , E)p×p = diag(A, · · · , A)p×p diag(X, · · · , X)p×p = diag(A, · · · , A)p×p V . Consequently, diag(A, · · · , A)ps×ps = diag(E, · · · , E)ps×ps diag(V −1 , · · · , V −1 )s×s = diag(U, · · · , U )p×p diag(F, · · · , F )ps×ps diag(U −1 , · · · , U −1 )p×p

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diag(V −1 , · · · , V −1 )s×s . Therefore (aij Ips ) ∈ Mpsn (R) can be reduced to a diagonal matrix, as desired. Assume that the corresponding map of A is f . If m < n, by virtue of Proposition 7.2.11, we have nR ⊕ coker(f ) ∼ = mR ⊕ ker(f ). Hence,


mR ⊕ (n − m)R ⊕ coker(f ) ∼ = mR ⊕ ker(f ). ∼ As EndR (mR) = Mm (R) has power-substitution, by Corollary 10.2.4, there exists s ∈ N such that
s (n − m)R ⊕ coker(f ) ∼ = s(ker(f )).

Since s(coker(f )) ∼ = coker(diag(f, · · · , f )s×s ) and s(ker(f )) ∼ = ker(diag (f, · · · , f )s×s ), it follows from Proposition 7.2.11 that diag(A, · · · , A)s×s admits a diagonal reduction. Furthermore, we prove that (aij Is ) ∈ Msn (R) admits a diagonal reduction. If n < m, by a similar route, we prove that there exists t ∈ N such that (aij It ) ∈ Mtn (R) admits a diagonal reduction.  A space X is said to be a P -space if every prime ideal of CR (X) is a maximal ideal. Let X be a topological P -space with d(X) ≤ 3. By virtue of [384, Remark 3.1], CR (X) is regular. In view of [70, Proposition 2.8],
Mn CR (X) satisfies power-substitution. As an immediate consequence, we show that every square matrix over CR (X) admits a diagonal powerreduction. Corollary 10.4.15. Let R be an exchange ring satisfying stable powersubstitution. Then for any regular (aij ) ∈ Mm×n (R) there exists some s ∈ N such that (aij Is ) can be completed to a sum of two invertible matrices. Proof. Let A = (aij ) ∈ Mm×n (R) be regular. By virtue of Theorem 10.4.14, there exist s ≥ 2, U ∈ GLms (R) and V ∈ GLns (R) such that U (aij Is )m×n V is a diagonal matrix. If m = n, then we have some ri ∈ R such that U (aij Is )m×n V = diag(r1 , · · · , rms ). As ms ≥ 2, we see that     0 −1 · · · 0 0 r1 1 · · · 0 0    0 r2 · · · 0 0   0 0 ··· 0 0      .. .. . . . . . . . . . .. ..  +  .. .. . . .. ..  diag(r1 , · · · , rms ) =  . . . .      0 0 · · · rms−1 1   0 0 · · · 0 −1  1 0 ···

0

0

−1 0 · · · 0 rms

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Therefore (aij Is )m×n  r1 0   = U −1  ...  0 1

1 r2 .. . 0 0

··· 0 ··· 0 .. .. . . · · · rms−1 ··· 0

  0 0 −1 · · ·  0 0 ··· 0   ..  V −1 + U −1  .. .. . .   . . . .    0 0 ··· 1 0 −1 0 · · ·

is the sum of two invertible matrices. If m < n, then we have some ri ∈ R such that

Let V −1

 0 0 0 0   .. ..  V −1 . .   0 −1  0 rms


U (aij Is )m×n V = diag(r1 , · · · , rms ) 0ms×(n−m)s . ! ′ Vms×ns = . Then ′ W(n−m)s×ns ! aij Is diag(U, I(n−m)s×(n−m)s ) V ′ W(n−m)s×ns   diag(r1 , · · · , rms ) 0ms×(n−m)s = , 0(n−m)s×ms I(n−m)s×(n−m)s

which is the sum of two invertible matrices. Therefore (aij Is ) can be completed to a sum of two invertible matrices. If n < m, we have s ∈ N, U ∈ GLms (R), V ∈ GLns (R) and some ri ∈ R such that   diag(r1 , · · · , rns ) U (aij Is )m×n V = . 0(m−n)s×ns Analogously to the above consideration, (aij Is ) can be completed to a sum of two invertible matrices, and we are done.  Let R be an exchange ring satisfying stable power-substitution. It follows from Corollary 10.4.15 that for any regular square (aij ) there exists some s ∈ N such that (aij Is ) is the sum of two invertible matrices.

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Chapter 11

Stably Euclidean Rings

A norm over a ring R is defined as a function ϕ : R → Z satisfying ϕ(0) = 0, ϕ(a) > 0 for a 6= 0. A commutative ring R is a euclidean ring if there exists a norm ϕ and if for any a, b ∈ R, b 6= 0, there exists an equation: a = bq + r with ϕ(r) < ϕ(b). Following Cooke, a commutative ring R is an ω-stage euclidean ring provided that there exists a norm ϕ and that for any a, b ∈ R, b 6= 0, there exists a sequence of equations: a = bq1 + r1 , b = r1 q2 +r2 , r1 = r2 q3 +r3 , · · · , rn−2 = rn−1 qn +rn (⋆) with ϕ(rn ) < ϕ(b). This concept is a generalization of 2-stage euclidean rings and k-stage euclidean rings (cf. [189], [362] and [430]). The class of ω-stage euclidean rings is very large. In [189], Cooke provided many examples √ domains √ of integral 1+ 53 which satisfy ω-stage euclidean condition, e.g., Z[ 22] and Z[ 2 ]. The aim of this chapter is to introduce a new concept, stably euclidean rings as a natural extension of ω-stage euclidean rings and stable range one. A ring R is a stably euclidean ring if, there exists a norm ϕ on R such that aR + bR = R with a, b ∈ R, b 6= 0 implies that there exists a sequence of equations (⋆) with ϕ(rn ) < ϕ(b). We will establish various properties of stably euclidean rings. Diagonal reduction of matrices over such rings is studied as well.

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Divisible Chain Conditions

The main purpose of this section is to investigate equivalent conditions on a stably euclidean ring. We see that such rings can be characterized by some kind of divisible chain condition. Theorem 11.1.1. A ring R is a stably euclidean ring if and only if aR + bR = R with a, b ∈ R implies that there exists a sequence of equations: a = bq1 +r1 , b = r1 q2 +r2 , · · · , rn−3 = rn−2 qn−1 +rn−1 , rn−2 = rn−1 qn .

(∗)

Proof. Suppose that aR + bR = R with a, b ∈ R, b 6= 0. Then there exists a sequence of equations (∗). Let rn = 0. Define  0 x = 0; ϕ(x) = 1 x 6= 0. Then ϕ is a norm on R. Clearly, ϕ(rn ) = 0 < ϕ(b). This implies that R is a stably euclidean ring. Assume now that R is a stably euclidean ring. Suppose that aR+bR = R with a, b ∈ R. If b = 0, then a = b × 1 + a, b = a × 0. If b 6= 0, then there exists a sequence of equations: a = bq1 + r1 , b = r1 q2 + r2 , r1 = r2 q3 + r3 , · · · , rn1 −2 = rn1 −1 qn1 + rn1 with ϕ(rn1 ) < ϕ(b). Since aR + bR = R, one easily checks that bR + r1 R = R, r1 R + r2 R = R, · · · , rn1 −1 R + rn1 R = R. If rn1 = 0, then the result follows. If rn1 6= 0, by hypothesis, there exists a sequence of equations: rn1 −1 = rn1 p1 +s1 , rn1 = s1 p2 +s2 , s1 = s2 p3 +s3 , · · · , sn2 −2 = sn2 −1 pn2 +sn2 with ϕ(sn2 ) < ϕ(rn1 ). If sn2 = 0, then the result follows. If sn2 6= 0, there exists a new finite divisible chain. By iteration of this process, either the result holds or there exist n1 , n2 , · · · , nm , · · · ∈ N such that · · · < ϕ(tnm ) < · · · < ϕ(sn2 ) < ϕ(rn1 ) < ϕ(b) is an infinite inequality. This gives a contradiction, and therefore the result follows.  Note that the stably euclidean condition is independent of the norm ϕ by Theorem 11.1.1. Analogously, we see that a commutative ring R is a

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ω-stage euclidean ring if and only if for any pair (a, b) with b 6= 0, there exists a sequence of equations: a = bq1 + r1 , b = r1 q2 + r2 , · · · , rn−3 = rn−2 qn−1 + rn−1 , rn−2 = rn−1 qn . Corollary 11.1.2. A ring R has stable range one if and only if aR + bR = R with a, b ∈ R implies that there exists a sequence of equations: a = bq1 + r1 , b = r1 q2 ; hence, it is a stably euclidean ring. Proof. Let R have stable range one. Assume that aR + bR = R with a, b ∈ R. Then there exists some y ∈ R such that a + by = u ∈ U (R). Thus, a = b(−y) + u and b = u(u−1 b). Conversely, aR + bR = R implies that a = bq1 + r1 , b = r1 q2 . Hence, a = r1 (q2 q1 + 1), and so r1 R = R. This implies that a + b(−q1 ) ∈ R is right invertible. One easily checks that R has stable range one. In this case, it follows from Theorem 11.1.1 that R is a stably euclidean ring.  √ Let R = Z⊕Z[ −19]. Then R is a stably euclidean ring which is neither a euclidean ring nor a ring having stable range one. We say that a ring R is a U Cn -ring provided that a1 R + a2 R + · · · + an+1 R = R implies that there exists an invertible matrix whose first row is (a1 , a2 , · · · , an+1 ). Now we give several elementary characterizations of stably euclidean rings. Theorem 11.1.3. Let R be a ring. Then the following are equivalent: (1) R is a stably euclidean ring. (2) aR + bR = R with a, b ∈ R implies that there exists a sequence of equations: a = bq1 + r1 , b = r1 q2 + r2 , r1 = r2 q3 + r3 , · · · , rn−2 = rn−1 qn + rn ,

where rn ∈ R is invertible. (3) aR + bR = R with a, b ∈ R implies that there exists some U ∈ E2 (R) such that (a, b)U = (1, 0). (4) R is both a GE2 -ring and a U C1 -ring. Proof. (1) ⇒ (2) Let R be a stably stable ring. Given aR + bR = R with a, b ∈ R, it follows from Theorem 11.1.1 that there exists a sequence of equations (∗). Clearly, rn−2 ∈ rn−1 R and rn−3 ∈ rn−2 R + rn−1 R ⊆ rn−1 R. Repeating this process, we see that rn−4 , · · · , a, b ⊆ rn−1 R. Hence, R = aR + bR ⊆ rn−1 R ⊆ R. This implies that rn−1 R = R, and thus rn−1 pn = 1 for some pn ∈ R. As a result, we have rn−2 = rn−1 pn (rn−2 − 1) + 1. Let qn′ = pn (rn−2 − 1) and rn = 1. Then rn−2 = rn−1 qn′ + rn , where rn ∈ R is invertible.

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(2) ⇒ (3) Suppose aR + bR = R with a, b ∈ R. By hypothesis, there exists a finite divisible chain: a = bq1 + r1 , b = r1 q2 + r2 , · · · , rn−3 = rn−2 qn−1 + rn−1 , where rn−1 ∈ R is invertible. Assume that rn−1 tn−1 = 1 for some tn−1 ∈ R. Then we get a sequence of equations: a = bq1 +r1 , b = r1 q2 +r2 , · · · , rn−3 = rn−2 qn−1 +rn−1 , rn−2 = rn−1 qn +rn ,
where qn = tn−1 rn−2 − (−1)n(n+1)/2 , rn = (−1)n(n+1)/2 . Let qn+1 =   0 −1 rn rn−1 . Then rn−1 = rn qn+1 . Hence, (a, b) = (b, −r1 ), (b, −r1 ) 1 q     1 0 −1 0 −1 = (−r1 , −r2 ), (−r1 , −r2 ) = (−r2 , r3 ), (−r2 , r3 ) 1 −q2 1 q3     0 −1 0 −1 = (r3 , r4 ), (r3 , r4 ) = (r4 , −r5 ), · · · , (−1)(n−2)(n−1)/2 · 1 −q4 1 q5  
0 −1 rn−2 , (−1)(n−1)n)/2 · rn−1 = (−1)(n−1)n/2 · rn−1 , 1 (−1)n−1 qn

(−1)n(n+1)/2 · rn , and (−1)(n−1)n/2 · rn−1 , (−1)n(n+1)/2 · rn  
0 −1 = (−1)n(n+1)/2 · rn , 0 . Let F (a) denote the matrix n 1 (−1) qn+1   0 −1 . Then 1 a      1 −1 10 1 −1 1a F (a) = ∈ E2 (R). 0 1 11 0 1 01 One easily checks that

(a, b)F (q1 )F (−q2 ) · · · F
(−1)i−1 qi · · · F (−1)n qn+1 = (−1)n(n+1)/2 · rn , 0 = (1, 0).

Let U = F (q1 )F (−q2 ) · · · F (−1)i−1 qi · · · F (−1)n qn+1 . Then (a, b)U = (1, 0) for some U ∈ E2 (R),  asasserted. ab (3) ⇒ (4) Let A = ∈ GL2 (R). Then aR + bR = R, and so we cd have some U ∈ E 0). Hence, are c′ ,d′ ∈ R 2 (R) such  that (a, b)U = (1,   there  1 0 1 0 1 0 , and such that AU = . This implies that AU = c′ d′ −c′ 0 0 d′ thus R is a GE2 -ring. Let U −1 = (uij )2×2 . Then (a, b) = (1, 0)U −1 ; hence,

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(a, b) = (u11 , u12 ). This implies that



a b u21 u22

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∈ GL2 (R), and therefore

R is a U C1 -ring. (4) ⇒ (1) Suppose aR + bR = R with a, b ∈ R. By hypothesis, there exists some V ∈ GL2 (R) such that (a, b)V = (1, 0). Since R is a GE 2 -ring,  10 there is some U ∈ E2 (R) and some d ∈ U (R) such that V = U . 0d Hence,  −1 10 (a, b)U = (1, 0) = (1, 0), 0d and so we can find some         1 0 1 x2 1 0 1 xn , ,··· , , ∈ E2 (R) x1 1 0 1 xn−1 1 0 1 such that



 1 xn A ··· = (1, 0). xn−1 0 1   1 0 We may assume that n ∈ N is even. Let (a, b) = (y1 , z1 ). Then x1 1   1 x2 a + bx1 = y1 , z1 = b. Let (y1 , z1 ) = (y2 , z2 ). Then z1 + y1 x2 = 0 1 z2 , y1 = y2 . By iteration of this process, we let   1 0 (yn−2 , zn−2 ) = (yn−1 , zn−1 ), x 1  n−1 1 xn (yn−1 , zn−1 ) = (1, 0). 0 1 1 0 x1 1



1 x2 0 1





1

0 1



Then yn−2 +zn−2 xn−1 = yn−1 , zn−2 = zn−1 , yn−1 = 1 and zn−1 +yn−1 xn = 0. Let qi = −xi , ri = yi (i is odd) and ri = zi (i is even). Then we get a sequence of equations: a = bq1 + r1 , b = r1 q2 + r2 , · · · , rn−3 = rn−2 qn−1 + rn−1 , rn−2 = rn−1 qn . According to Theorem 11.1.1, R is a stably euclidean ring.



Corollary 11.1.4. Let R be a stably euclidean   ring. If aR + bR = R ab with a, b ∈ R, then the matrix equation X X = X(c, d ∈ R) has a cd nonzero solution.

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Proof. Given aR + bR = R with a, b ∈ R, it follows from Theorem 11.1.3 that there exists some Q ∈ E2 (R) such that b)Q =(1, 0).  (a,   For any ab 10 c, d ∈ R, we can find some e, f ∈ R such that Q= ; hence, cd ef      1 0 ab 10 Q= . −e 1 cd 0f        ab 10 1 0 10 Let A = and B = Q . Then AB = ∈ M2 (R) cd 00 −e 1 e0 is an idempotent. Thus, BABABAB = BAB. Let X =  BAB.  Then 10 XAX = X. If X = 0, then AB = AX = 0; hence, = 0, a e0 contradiction. Consequently, we conclude that XAX = X has a nonzero solution.  Corollary 11.1.5. Let R be a stably euclidean ring. If aR + bR = R with T a, b ∈ R, then aR bR is principal.

Proof. Given aR + bR = R with  a,b ∈ R, it follows from Theorem 11.1.3 pr ∈ E2 (R) such that (a, b)U = (1, 0). that there exists some U = qs T Let m = ar. Then ar + bs = 0, and so m ∈ aR bR. For any z ∈ T aR bR, there are x, y ∈ R such that z = ax = −by. This implies that    −1   px pr px (a, b) = (1, 0). Hence, (1, 0) = (1, 0). Let qy qs qy  −1     pr px ef = . qs qy gh   ef Then (1, 0) = (1, 0); whence, e = 1 and f = 0. Observing that gh        px pr 10 ∗ rh = = , qy qs gh ∗ ∗ we get x = rh. Furthermore, z = ax = arh = mh ∈ mR. Therefore T aR bR = mR, as desired. 

 α β | α, β ∈ C , where −β α α is the complex conjugate of α ∈ C. Then R := H[x1 , x2 ] is not a stably euclidean ring. Choose elements a, b ∈ H such that ab 6= ba. Note that Let H be the Hamilton ring, i.e, H =





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ab−ba = (x2 +a)(x1 +b)−(x1 +b)(x2 +a) and so (x2 +a)R+(x1 +b)R = R. T In light of [313, Theorem 11.2.7], (x2 + a)R (x1 + b)R is nonfree. By using T [313, Theorem 11.2.7] again, it is not cyclic. Hence, (x2 + a)R (x1 + b)R is not a principal ideal. According to Corollary 11.1.5, we conclude that R is not a stably euclidean ring. We note that a stably euclidean ring may be not a right principal ideal ring. Let F be a filed, and let   f (0) g(x) R={ | f (x), g(x) ∈ F [x]}. 0 f (x)   0 g(x) Then R is a stably euclidean ring. Let I = { | f (x), g(x) ∈ 0 xf (x) F [x]}. Then I is a right ideal of R. One directly verifies that I is not a principal right ideal, and we are through. Corollary 11.1.6. A commutative ring R is a stably euclidean ring if and only if it is a GE2 -ring. Proof. Let R be a commutative ring. Given aR + bR  = R with a, b ∈ R, a b then ax + by = 1 for some x, y ∈ R. Let U = . Then U ∈ GL2 (R) −y x and (a, b)U −1 = (1, 0). This implies that R is a U C1 -ring. Therefore we complete the proof by Theorem 11.1.3.  Now let us take a particular example. Let R =F [x, y], where  F is a 1 + xy x2 field, with the usual degree function, and let A = . By −y 2 1 − xy an argument due to Cohn, A ∈ GL2 (R), while A 6∈ GE2 (R). According to Corollary 11.1.6, R is not a stably euclidean ring. Let J(R) be the Jacobson radical of a ring R. Note that R is a stably euclidean ring if and only if so is R/J(R). The definition of euclidean ring applies, of course, to non commutative rings as well. It is obvious by considering examples that a left euclidean ring is not necessarily right euclidean. But we observe that the stably euclidean property is right and left symmetric. Corollary 11.1.7. A ring R is a stably euclidean ring if and only if so is the opposite ring Rop . Proof. Suppose R is a stably euclidean ring. By Theorem 11.1.3, R is a GE2 -ring, and so is Rop . Given Ra + Rb = R, then xa + yb = 1 for some

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x, y ∈ R. By Theorem 11.1.3 again, R is a U C1 -ring; hence, we have some    −1 xy ef c, d ∈ R such that = ∈ GL2 (R). It is easy to verify that cd gh     1 0 xy af = diag(1, 1). −ca − db 1 cd bh        1 0 xy a 1 Let Q := ∈ GL2 (R). Then Q = . This −ca − db 1 cd b 0 op op means that R is a U C1 -ring. Applying Theorem 11.1.3 to R , we see that Rop is a stably euclidean ring. The converse is symmetric.  Example 11.1.8. Here are some examples. (1) Any finite direct product of stably euclidean rings is a stably euclidean ring. (2) R[[x1 , · · · , xn ]] is a stably euclidean ring if and only if so is R. (3) T Mn(R) (n ∈ N) is a stably euclidean ring if and only is so is R. (4) R[x]/(xn ) is a stably  euclidean ring if and only so is R. ab (5) The ring T = { | a, b ∈ R} is a stably euclidean ring if and only 0a if so is R.  11.2

2-Stable Range Condition

We say that a ring R satisfies the 2-stable range condition provided that a1 R + a2 R + a3 R = R implies that there exist some b1 , b2 ∈ R such that (a1 +a3 b1 )R+(a2 +a3 b2 )R = R. This is a generalization of stable range one. In fact, a ring R has stable range one if and only if a1 R + a2 R + a3 R = R implies that there exists some b2 ∈ R such that a1 R + (a2 + a3 b2 )R = R. The purpose of this section is to investigate stably euclidean rings under the 2-stable range condition. Theorem 11.2.1. Let R be a stably euclidean ring satisfying the 2-stable range condition. If a1 R + a2 R + · · · + an R = R (n ≥ 2), then there exists some U ∈ En (R) such that (a1 , a2 , · · · , an )U = (1, 0, · · · , 0). Proof. By virtue of Theorem 11.1.3, the result holds for n = 2. Assume that the result holds for n = k(k ≥ 2). Let n = k + 1. As R satisfies the 2-stable range condition, we can find some c1 , · · · , ck ∈ R such that

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(a1 + ak+1 c1 )R + · · · + (ak + ak+1 ck )R = R. This implies that   1 0 ··· 0 0  0 1 ··· 0 0     (a1 , a2 , · · · , an )  0 0 · · · 0 0  = (a1 + ak+1 c1 , · · · , ak + ak+1 ck , an ).  . . .  . .  .. .. . . .. ..  c1 c2 · · · ck 1

By hypothesis, there exists some Q ∈ Ek (R) such that (a1 + ak+1 c1 , · · · , ak + ak+1 ck )Q = (1, 0, · · · , 0). Therefore   1 0 ··· 0 0  0 1 ··· 0 0     0 0 ··· 0 0 Q (a1 , a2 , · · · , an )    . . .  1  .. .. . . ... ...  c1 c2 · · · ck 1

= (a1 + ak+1 c1 , · · · , ak + ak+1 ck , an )

Let

= (1, 0, · · · , 0, an ).



Q

1



   1 0 · · · 0 −an 0   0   0 1 ··· 0 0    Q 0 0 ··· 0 0   0 U = .    1 . . . . .  ..  . . . . .    . . . . .  . 0 0 ··· 0 1 c1 c2 · · · ck 1 

1 0 0 .. .

0 1 0 .. .

··· ··· ··· .. .

0 0 0 .. .

Then (a1 , a2 , · · · , an )U = (1, 0, · · · , 0). By induction, we obtain the result.  Let P ⊕ R ∼ = nR(n ≥ 2). Then there is an exact sequence 0 → P → ϕ nR → R → 0. Let (a1 , · · · , an ) be the matrix corresponding to ϕ. If there n−1 P exist some z, · · · , zn−1 ∈ R such that (ai +an+1 zi )R = R, then P is free. i=1

As in the proof of Theorem 11.2.1, we can find some Q ∈ GEn (R) such that (a1 , · · · , an )Q = (1, 0, · · · , 0). That is, (a1 , · · · , an ) can be transformed to (1, 0, · · · , 0) via elementary transformations. Let ψ be the R-morphism corresponding to (1, 0, · · · , 0). Therefore P ∼ = ker(ϕ) ∼ = ker(ψ) ∼ = (n − 1)R, and we are done. Corollary 11.2.2. Every stably euclidean ring satisfying the 2-stable range condition is a GE-ring.

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Proof. Let A = (aij ) ∈ GLn (R). Then a11 R + · · · + a1n R = R. In view of Theorem 11.2.1, there exists some U ∈ En (R) such that (a11 , · · · , a1n )U = (1, 0, · · · , 0). Thus we deduce that   1 0 ··· 0 ∗ ∗ ··· ∗   AU =  . . . .  . . . . . . . . . ∗ ∗ ··· ∗

This implies that



1 ∗  .  ..

0 ··· 0 ··· .. . . . .

 0   0 1  AU = ..  B .

∗ 0 ··· 0

for some B ∈ GLn−1 (R). By iteration of this process, we can find V, W ∈ En (R) such that V AW = diag(1, 1, · · · , 1, u) for an invertible u ∈ R, and therefore R is a GE-ring.  Let R be a GE-ring, and let A ∈ GLn (R)(n ∈ N). We note that there exist c1 , c2 ∈ U (R) and E1 , E2 ∈ En (R) such that A = diag(1, · · · , 1, c1 )E1 = E2 diag(1, · · · , 1, c2 ). Let R be a stably euclidean ring satisfying the 2-stable range condition. By virtue of Corollary 11.2.2, R is a GE-ring. Hence, the canonical map K1 (R) → U (R) is surjective. Corollary 11.2.3. Let R be a stably euclidean ring satisfying the 2-stable range condition, and let n ∈ N. Then the following are equivalent: (1) R is directly finite. (2) Mn (R) is directly finite. Proof. (1) ⇒ (2) Given A = (aij ), B = (bij ) ∈ Mn (R) with AB = In (n ≥ 2), then a11 R + · · · + a1n R = R. In view of Theorem 11.2.1, there exists some U ∈ E n (R) such that (a11 , · · · , a1n )U = (1, 0, · · · , 0). This implies 1 0 that AU = , where D is some (n−1)×1 matrix and C ∈ Mn−1 (R). DC   p Q Let U −1 B = . As AB = In , we see that EF    1 0 p Q = (AU )(U −1 B) = In , DC EF

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where E is an (n − 1) × 1 matrix, Q is a 1 × (n − 1) matrix and F ∈ Mn−1 (R). Clearly, p = 1 and Q = 01×(n−1) . Furthermore, CF = In−1 . By induction on C, we see that C ∈ Mn−1 (R) is invertible. Thus, A ∈ Mn (R) is invertible. So Mn (R) is directly finite for any n ∈ N. (2) ⇒ (1) Let S be a directly finite ring, and let e ∈ S be an idempotent. Assume that ab = e with a, b ∈ eSe. Then (a+1−e)(b+1−e) = 1 in R. As S is directly finite, we get (b + 1 − e)(a + 1 − e) = 1; and so ba = e. Thus, we conclude that every corner of a directly finite ring is directly finite. Choose S = Mn (R) and e = diag(1, 0, · · · , 0) ∈ S. Then R ∼ = eSe. Consequently, R is directly finite by the above discussion.  It is worth noting that every subring of a directly finite ring is directly finite. Let S be directly finite, and let S ⊆ R be a subring with an identity e. Then e = e2 ∈ R. Given ab = e in S, then (eae)(ebe) = ab = e in eSe. As in the preceding discussion, eSe is directly finite; hence, (ebe)(eae) = e, i.e., ba = e. Therefore R is directly finite. Suppose a1 R + · · · + an R = dR. Then d = a1 x1 + · · · + an xn , ai = dyi , i = 1, 2, · · · , n, for some x1 , · · · , xn , y1 , · · · , yn ∈ R. Write   x1  ..  X =  .  , Y = (y1 , · · · , yn ). Then we have U :=



xn

X In − XY 1 −Y



∈ GLn+1 (R) such that (a1 , · · · , an , 0)U

= (d, 0, · · · , 0). Analogously to the proof of Corollary 11.2.3, we deduce that every stably euclidean ring R satisfying the 2-stable range condition has the invariant basis property, i.e., mR ∼ = nR implies that m = n. For stably euclidean rings, we now derive the following result. Theorem 11.2.4. Let R be a stably euclidean ring satisfying the 2-stable range condition. If a1 R + a2 R + · · · + an R = dR (n ≥ 2), then there exists a U ∈ En (R) such that (a1 , a2 , · · · , an )U = (d, 0, · · · , 0). Proof. Given a1 R + a2 R + · · ·+ an R = dR (n ≥ 2), there exist d, x1 , · · · , xn , y1 , · · · , yn ∈ R such that a1 = dx1 , a2 = dx2 , · · · , an = dxn , a1 y1 + · · · + an yn = d. Let e = x1 y1 + · · · + xn yn − 1. Then x1 R + · · · + xn R + eR = R with de = 0. Since R satisfies the 2-stable range condition, it satisfies the n-stable range condition. So we have some c1 , · · · , cn ∈ R such that (x1 + ec1 )R + · · · + (xn + ecn )R = R. In view of Theorem 11.1.3, there

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exists some U ∈ En (R) such that (x1 + ec1 , · · · , xn + ecn )U = (1, 0, · · · , 0). Multiplying the equality from the left side by d, we get d(x1 + ec1 , · · · , xn + ecn )U = (a1 , · · · , an )U = (d, 0, · · · , 0), and the proof is true.  Corollary 11.2.5. Let R be a B´ezout, stably euclidean ring. If R satisfies the 2-stable range condition, then every square matrix over R admits a triangular reduction by elementary transformations. Proof. Let A = (aij ) ∈ Mn (R). Since R is a B´ezout ring, there exists some d1 ∈ R such that a11 R + a12 R + · · · + a1n R = d1 R. According to Theorem 11.2.4, we have an element U1 ∈ En (R)  such that  d1 (a11 , a12 , · · · , a1n )U1 = (d1 , 0, · · · , 0). This implies that AU1 = , ∗ B where B = (bij ) ∈ Mn−1 (R). Similarly, we have a U2 ∈ En−1 (R) such that   d2 BU2 = for some d2 ∈ R. Hence, ∗ C     d1 1 AU1 =  ∗ d2  . U2 ∗ ∗ ∗ By iteration of this process, we can admits a triangular matrix  d1  ∗ d2   . .  .. .. ∗ ∗

and therefore the result follows.

11.3

find some d3 , · · · , dn ∈ R such that A

..

. · · · dn



  , 



Commutative Case

The main purpose of this section is to study commutative stably euclidean rings. These results reveal new properties of GE2 -rings as well. Let R be a commutative ring with an identity 1. We denote by SL2 (R) the 2dimensional special linear group of R, i.e., the group of 2 × 2 matrices over R whose determinants are equal to 1. Lemma 11.3.1. Let I be an ideal of a stably euclidean ring R. If R satisfies the 2-stable range condition, then R/I is a stably euclidean ring.

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Proof. Given a(R/I)+b(R/I) = R/I with a, b ∈ R, there exist two elements x, y ∈ R such that ax + by − 1 ∈ I, and so ax + by + z = 1 for some z ∈ I. Since R satisfies the 2-stable range condition, we can find some s, t ∈ R such that (a + zs)R + (b + zt)R = R. In view of Theorem 11.1.1, there exists a sequence of equations (∗). This implies that a = bq1 + r1 , b = r1 q2 + r2 , r1 = r2 q3 + r3 , · · · , rn−2 = rn−1 qn in R/I. By using Theorem 11.1.1 again, R/I is a stably euclidean ring.  Theorem 11.3.2. Let R be a commutative ring. Then the following are equivalent: (1) R is a stably euclidean ring satisfying the 2-stable range condition. (2) Every factor of R is a stably euclidean ring. Proof. (1) ⇒ (2) is obvious from Lemma 11.3.1. (2) ⇒ (1) It will suffice to prove that R satisfies the 2-stable range condition. Suppose that aR + bR + cR = R with
a, b, c ∈ R. Let
I = cR. If I = R, then c ∈ U (R), and so a + c(1 − c−1 a) R + b + c × 0 R = R. If I 6= R, then we have a(R/I) + b(R/I) = R/I. By hypothesis, there are some x, y ∈ R such that ax + by = 1. Thus, x y −b a = 1. That is,



x y −b a



∈ SL2 (R/I).

By hypothesis, R/I is a stably euclidean ring; hence, it is a GE2 -ring. As a result, SL2 (R/I) = E2 (R/I). This implies that   x y ∈ E2 (R/I). −b a Clearly, there exists an epimorphism E2 (R) → E2 (R/I). We can find some x′ , y ′ , a′ , b′ ∈ R such that  ′ ′ x y ∈ E2 (R) ⊆ SL2 (R), −b′ a′ where x − x′ , y − y ′ , a − a′ , b − b′ ∈ I. Hence, a′ x′ + b′ y ′ = 1, and thus ax′ + by ′ − 1 ∈ I = cR. So ax′ + by ′ + cz = 1 for some z ∈ R. As c = 1 × c = (a′ x′ + b′ y ′ )c, we deduce that ax′ + by ′ + c(a′ x′ + b′ y ′ )z = 1.

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Thus, (a + ca′ z)x′ + (b + cb′ z)y ′ = 1, i.e., (a + ca′ z)R + (b + cb′ z)R = R. Therefore R satisfies the 2-stable range condition.  Clearly, Z[x] is a euclidean integral domain; hence, it is a Dedekind domain. Thus, Z[x] is a stably euclidean ring satisfying the 2-stable range 2 condition (cf. Example 12.1.14). In light of Theorem
11.3.2, Z[x]/(x ) 2 2 ∼ is a stably euclidean ring. As Z[x]/(x )/J Z[x]/(x ) = Z, we see that Z[x]/(x2 ) does not have stable range one. Further, we claim that it is not a euclidean ring. Otherwise, I = {a+bx | a ∈ 2Z, b ∈ Z, x2 = 0} is a principal ideal of R = {a + bx | a, b ∈ Z, x2 = 0}. This gives a contradiction, and we are done. Corollary 11.3.3. Let R be a commutative ring. Then the following are equivalent: (1) R is an ω-stage euclidean ring. (2) R is a B´ezout, stably euclidean ring satisfying the 2-stable range condition. Proof. (1) ⇒ (2) Let R be a factor of R. Suppose that aR + bR = R. If b = 0, then a = b × 1 + a, b = a × 0. If b 6= 0, then b 6= 0. Since R is an ω-stage euclidean ring, analogously to Theorem 11.1.1, there exists a sequence of equations (∗). Furthermore, we have a = bq1 + r1 , b = r1 q2 + r2 , r1 = r2 q3 + r3 , · · · , rn−2 = rn−1 qn in R. By Theorem 11.1.1 again, every factor of R is a stably euclidean ring. According to Theorem 11.3.2, R is a stably euclidean ring satisfying the 2-stable range condition. For any a, b ∈ R, b 6= 0, there is a sequence of equations (∗). Clearly, rn−2 ∈ rn−1 R and rn−3 ∈ rn−2 R + rn−1 R ⊆ rn−1 R. Hence, rn−3 ∈ rn−1 R. By iteration of this process, we see that rn−4 , · · · , a, b ⊆ rn−1 R. Hence, aR + bR ⊆ rn−1 R ⊆ aR + bR. This implies that aR + bR = rn−1 R, and therefore R is a B´ ezout ring, as required. (2) ⇒ (1) For any a, b ∈ R, b 6= 0, we can find a d ∈ R such that aR + bR = dR. So there exist some x, y, s, t ∈ R such that a = dx, b = dy, as + bt = d. Let e = xs + yt − 1. Then xs + yt + e = 1 with de = 0. Since R satisfies the 2-stable range condition, there are z1 , z2 ∈ R such that (x + ez1 )R + (y + ez2 )R = R. In view of Theorem 11.1.1, we get a finite divisible chain: x+ez1 = (y+ez2 )q1 +r1 , y+ez2 = r1 q2 +r2 , r1 = r2 q3 +r3 , · · · , rn−2 = rn−1 qn . Multiplying these equalities from the left side by d, we see that a = bq1 +dr1 , b = (dr1 )q2 +dr2 , dr1 = (dr2 )q3 +dr3 , · · · , drn−2 = (drn−1 )qn .

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Therefore R is an ω-stage euclidean ring.

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Analogously, we deduce that an integral domain R is an ω-stage euclidean if and only if R is a B´ ezout, stably euclidean ring. According to Corollary 11.3.3 and Corollary 11.2.2, we conclude that every ω-stage euclidean ring is a GE-ring. Furthermore, we can derive the following. Corollary 11.3.4. Every square matrix over an ω-stage euclidean ring admits a triangular reduction by elementary transformations. Proof. By virtue of Corollary 11.3.3, R is a B´ezout, stably euclidean ring satisfying the 2-stable range condition. Therefore we obtain the result from Corollary 11.2.5.  Now we observe a new equivalent characterization of ω-stage euclidean rings. Theorem 11.3.5. A commutative ring R is an ω-stage euclidean ring if and only if for any a, b ∈ R, there exists some Q ∈ E2 (R) such that (a, b)Q = (d, 0). Proof. Suppose that R is an ω-stage euclidean ring. In view of Corollary 11.3.3, R is a B´ezout, stably euclidean ring satisfying the 2-stable range condition. For any a, b ∈ R, there is some d ∈ R such that aR + bR = dR. In view of Theorem 11.2.4, we have some element Q ∈ E2 (R) such that (a, b)Q = (d, 0). Now we prove the converse. Let I be an ideal of R. Given a(R/I) + b(R/I) = R/I, then we can find a Q ∈ E2 (R) such that (a, b)Q = (d, 0) for some d ∈ R. Hence, (a, b)Q = (d, 0). So a(R/I) + b)(R/I) = d(R/I). Thus, d ∈ U (R/I). As a result, we obtain ! −1 d (a, b)Q = (1, 0). d One easily checks that −1

d

!

= d   1 −1 1 0 1 1

! !  −1 −1 1 0 1d 1d −d 1 0 1 0 1   0 1 −1 ∈ E2 (R/I). 1 0 1

By Theorem 11.1.3, R/I is a stably euclidean ring. According to Theorem 11.3.2, R is a stably euclidean ring satisfying the 2-stable range condition.

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By hypothesis, one easily checks that R is a B´ezout ring. Consequently, R is an ω-stage euclidean ring from Corollary 11.3.3.  Corollary 11.3.6. A commutative ring R is an ω-stage euclideanring  if a and only if for any a, b ∈ R, there exists a Q ∈ E2 (R) such that Q = b   d . 0

Proof. As R is commutative, R = Rop . Applying Theorem 11.3.5 to Rop , we obtain the result. 

Theorem 11.3.7. Let R be a commutative, stably euclidean ring. Then the following are equivalent:   ab (1) Every ∈ M2 (R) with aR + bR + cR = R admits a diagonal 0c reduction by elementary transformations. (2) aR + bR + cR = R with a, b, c ∈ R implies that there exist p, q ∈ R such that (pa)R + (pb + qc)R = R. Proof. (1)⇒ (2) Given aR + bR + cR = R with a, b, c ∈ R, then we ab have ∈ M2 (R). By assumption, there are P ′ , Q′ ∈ E2 (R) such that 0c     ab d0 P′ Q′ = for some d, e ∈ R. As aR + bR + cR = R, we see 0c 0e that dR + eR = R; and so dx + ey = 1 for some x, y ∈ R. We observe that      0 1 10 d0 10 −1 0 x1 0e y1    0 1 d0 = −1 0 1e   1 e = . −d 0 

ab 0c





1 e −d 0



Thus, we can find some P, Q ∈ E2 (R) such that P Q= .   x Let (p, q) be the first row of P and the first column of Q. Then y pax + pby + qcy = 1, and thus (pa)R + (pb + qc)R = R.

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 ab (2) ⇒ (1) Given ∈ M2 (R) with aR + bR + cR = R, there are 0c p, q ∈ R such that (pa)R + (pb + qc)R = R. Obviously, pR + qR = R. In view of Theorem 11.1.3, there exists an element Q ∈ E2 (R) such that (p, q)Q = (1, 0), i.e., (p, q) is the first row of Q−1 . Thus, we see that     ab pa pb + qc Q−1 = . 0c ∗ ∗ As (pa)R+ (pb + qc)R = R, by Theorem 11.1.3 again, there ∈  exists  some  P a b 1 0 E2 (R) such that (pa, pb + qc)P = (1, 0). Hence, Q−1 P = 0c d∗ for some d ∈ R. Furthermore, we get     1 0 ab Q−1 P −d 1 0c    1 0 10 = −d 1 d∗   10 = , 0∗ and therefore the result follows.  Corollary 11.3.8. Let R be an ω-stage euclidean ring. Then the following are equivalent: (1) Every 2 × 2 matrix over R admits a diagonal reduction by elementary transformations.   ab (2) Every ∈ M2 (R) with aR + bR + cR = R admits a diagonal 0c reduction by elementary transformations. (3) aR + bR + cR = R with a, b, c ∈ R implies that there exist p, q ∈ R such that (pa)R + (pb + qc)R = R. Proof. (1) ⇒ (2) is obvious. (2) ⇒ (3) is clear by Theorem 11.3.7. (3) ⇒ (1) Let A = (aij ) ∈ M2 (R). In view of Corollary 11.3.4, there   a0 exists some H ∈ E2 (R) such that AH = . By virtue of Corollary bc 11.3.3, R is a B´ezout, stably euclidean ring satisfying the 2-stable range condition. Thus, we can find some d ∈ R such that aR + bR + cR = dR. So a = da1 , b = db1 , c = dc1 , as + bt + ck = d for some a1 , b1 , c1 , s, t, k ∈ R.

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Hence, d(a1 s + b1 t + c1 k − 1) = 0. Let l = a1 s + b1 t + c1 k − 1. Then a1 R+b1 R+c1 R+lR = R, and so (a1 +lz1)R+(b1 +lz2)R+(c1 +lz3 )R = R for some z1 , z2 , z3 ∈ R. In view of Theorem 11.3.7, every   a1 + lz1 b1 + lz2 ∈ M2 (R) 0 c1 + lz3 admits a diagonal reduction by elementary transformations. That is, there exist P, Q ∈ GE2 (R) such that   a1 + lz1 b1 + lz2 P Q = diag(d1 , d2 ). 0 c1 + lz3 Furthermore,       d0 a1 + lz1 b1 + lz2 d0 P Q= diag(d1 , d2 ). 0d 0 c1 + lz3 0d This implies that P



da1 db1 0 dc1



P



Q = diag(dd1 , dd2 ).

Q = diag(dd1 , dd2 ),

i.e., ab 0c



Hence, there exist P ′ , Q′ ∈ GL2 (R) such that P ′ AHQ′ = diag(dd1 , dd2 ). Clearly, R is a GE-ring. Thus, we have c1 , c2 ∈ U (R) and E1 , E2 ∈ E2 (R) such that     1 1 E1 AHE2 = diag(dd1 , dd2 ). c1 c2 −1 This implies that E1 AHE2 = diag(dd1 , c−1 1 dd2 c2 ), and then the proof is completed. 

Corollary 11.3.9. Let R be an ω-stage euclidean integral domain. Then the following are equivalent: (1) Every 2 × 2 matrix over R admits a diagonal reduction by elementary transformations.   ab (2) Every principal right ideal generated by ∈ M2 (R) with aR + 0c bR + cR = R contains a nontrivial idempotent of M2 (R).

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 ab Proof. (1) ⇒ (2) Consider the matrix A = ∈ M2 (R) with aR+bR+ 0c cR = R. Then there exist P1 , Q1 ∈ E2 (R) such that P1 AQ1 = diag(d1 , d2 ). It is easy to check that d1 R + d2 R = R. Assume that d1 r + d2 s = 1. Then   01 B21 (−d2 )B12 (s)P1 AQ1 B12 (r) B12 (−d1 ) = diag(1, −d1 d2 ). 10 Thus, there are some P2 , Q2 ∈ GL2 (R) such that P2 AQ2 = diag(1, −d1 d2 ).
2 One easily verifies that AQ2 diag(1, 0)P2 = AQ2 diag(1, 0)diag(1, −d1d2 ) diag(1, 0)P2 = AQ2 diag(1, 0)P2 ∈ M2 (R). Further, AQ2 diag(1, 0)P2 6= 0, I2 . Therefore AQ2 diag(1,  0)P  2 ∈ AM2 (R) is a required idempotent. ab (2) ⇒ (1) Let A = ∈ M2 (R) with aR + bR + cR = R. Then 0c   rk there exists ∈ M2 (R) such that st      ab rk ar + bs ak + bt F = = ∈ M2 (R) 0c s t cs ct is a nontrivial idempotent. As R is an integral domain, detF = 0 and trF = 1. If ac = 0, then either a = 0 or c = 0. In these cases, it follows from Theorem 11.3.5 and Corollary 11.3.6 that A admits a diagonal reduction by elementary transformations. Now we assume that ac 6= 0. Thus, we get ar + bs + ct = 1 and sk = rt. According to Corollary 11.3.3, R is a B´ezout ring. Thus, rR + sR = dR for some d ∈ R. If d = 0, then c ∈ U (R), whence, A admits a diagonal reduction by elementary transformations. Now assume that d 6= 0. Hence, we have some p, q ∈ R such that r = dp, s = dq and pR + qR = R. As sk = rt, we see that pt = qk. Clearly, there exist some p′ , q ′ ∈ R such that pp′ + qq ′ = 1, and then t = qm, where m = p′ k + q ′ t. Let     d m p −bd − cm P = ,Q = . −cq ap + bq q ad Clearly, P, Q ∈ SL2 (R). As R is a GE2 -ring, SL2 (R) = E2 (R). Therefore, P and Q are invertible matrices that are finite products of elementary matrices over R. It is easy to verify that P AQ = diag(1, ac), and we are done.  Let a ∈ R. We use mspec(a) to denote the set of all maximal ideals which contain a.

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Corollary 11.3.10. Let R be an ω-stage euclidean ring of which the set of maximal ideals is at most countable. Then every 2 × 2 matrix over R admits a diagonal reduction by elementary transformations. Proof. According   to Corollary 11.3.8, it suffices to consider matrices of the a0 form A = , where aR + bR + cR = R. If a ∈ U (R), then bc   1 0 A = diag(a, c) −ba−1 1 is a diagonal matrix. Assume that a 6∈ U (R). Then aR 6= R; hence, mspec(a) 6= ∅. Set mspec(a) = {M1 , M2 , M3 , · · · }. If b ∈ M1 , then c 6∈ M1 ; hence, b + c 6∈ M1 . By an elementary transformation of column, b can be replaced by the element b + c. Thus, we may assume that b 6∈ M1 .  According  to Corollary a1 b 1 11.3.6, there exists P1 ∈ E2 (R) such that P1 A = . This implies 0 c1 that aR + bR = a1 R. If a1 ∈ U (R), then   1 −a−1 1 b1 P1 A = diag(a1 , c1 ) 0 1 is a diagonal matrix. Assume that a1 6∈ U (R). Then mspec(a1 ) 6= ∅. As mspec(a1 ) ⊆ mspec(a), we may assume that a1 ∈ M2 . If b1 ∈ M2 , then c1 6∈ M2 since a1 R + b1 R + c1 R = R. Hence, b1 + c1 6∈ M2 . By an elementary transformation of row, b1 can be replaced by the element b1 + c1 . Thus, we may assume that b1 6∈ M2 . According   to Theorem 11.3.5, there exists a2 0 Q1 ∈ E2 (R) such that P1 AQ1 = . In addition, a1 R + b1 R = a2 R. b 2 c2
S Further, we observe that a ∈ M1 , a1 ∈ M2 \M1 , a2 ∈ M3 \ M1 M2 . By iteration of this process, there is a collection of matrices of the form   ai ∗ Pk AQk = . ∗ ∗ If there exists some ai ∈ U (R), then A admits a diagonal reduction by elementary transformations. Otherwise, we get an infinite chain of ideals aR $ a1 R $ a2 R $ · · · . Further, each an 6∈ Mn (n ≥ 1). Let I =

∞ S

i=1

ai R. Then I 6= R; hence, there

exists a maximal ideal J of R such that I ⊆ J $ R. It follows that a ∈ J,

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and so J ∈ mspec(a). Thus, J = Mk for some k ∈ N. As ak 6∈ Mk , we deduce that ak 6∈ J, a contradiction. Therefore the proof is true.  Let R be an integral domain, and let F be the fraction field of R. Let q1 , q2 , · · · , qn ∈ R. We use [q1 , q2 , · · · , qn ] to denote the continued fraction q1 + q2 + 1 1 . .. . + q1n Proposition 11.3.11. An integral domain R is a stably euclidean ring if and only if every element ab , (a, b) = 1, of the fraction field can be expressed as a continued fraction ab = [q1 , · · · , qn ] where q1 , · · · , qn ∈ R. Proof. Let R be an integral domain. Suppose that R is a stably euclidean ring with the fraction field F . Let ab ∈ F with (a, b) = 1. Then aR + bR = R. By virtue of Theorem 11.1.1, we get a sequence of equations: a = bq1 + r1 , b = r1 q2 + r2 , · · · , rn−3 = rn−2 qn−1 + rn−1 , rn−2 = rn−1 qn , where r1 , · · · , rn−1 6= 0. This implies that a r1 b r2 rn−3 rn−1 rn−2 = q1 + , = q2 + , · · · , = qn−1 + , = qn , b b r1 r1 rn−2 rn−2 rn−1 and therefore ab = [q1 , · · · , qn ]. Conversely, assume that aR+bR = R and ab = [q1 , · · · , qn ]. Choose r1 = rn−2 r1 b [q2 ,··· ,qn ] , r2 = [q3 ,··· ,qn ] , · · · , rn−1 = [qn ] . Then a = bq1 + r1 , b = r1 q2 + r2 , · · · , rn−3 = rn−2 qn−1 + rn−1 , rn−2 = rn−1 qn . According to Theorem 11.1.1, R is a stably euclidean ring. 

11.4

Stably Free Modules


Let R = R[x1 , x2 , x3 ]/ x21 + x22 + x23 − 1 . Then there exists a stably free right R-module of positive rank which is not free (cf. [298, Example 3.2]). A natural question to ask is over what kind of rings a stably free module is free. The main purpose of this section is to study stably free modules by means of completeness of unimodular rows. A ring R is a U C-ring provided that it is a U Cn -ring for all n ∈ N. In view of Theorem 11.1.3, a ring R is stably euclidean if and only if it is both a GE2 -ring and a U C1 -ring. If R satisfies the n-stable range condition, then it is a U Cn -ring. If R has stable range one, then R is a U C-ring. Also we see that every commutative hermitian ring is a U C-ring.

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Theorem 11.4.1. Let R be a ring. Then the following are equivalent: (1) For any s ∈ N, every stably free right R-module of rank s is a free module of rank s. (2) R is a U C-ring. Proof. (2) ⇒ (1) Let P be a right R-module such that P ⊕ R ∼ = mR(m ≥ 2)). Then we get a short exact sequence: α

0 −→ P −→ mR −→ R −→ 0. As R is projective, the above sequence splits; hence, there exists some β : R → mR such that αβ = 1R . Let ε and η1 , · · · , ηm be bases of R and mR, respectively. Assume that α(η1 ) = εa1 , · · · , α(ηm ) = εam ; β(ε) = η1 b1 + · · · + ηm bm . Then α(η1 , · · · , ηm ) = ε(a1 , · · · , am ), 

 b1   β(ε) = (η1 , · · · , ηm )  ...  . bm

Further,

  b1    αβ(ε) = α(η1 , · · · , ηm )  ...  = ε(a1 , · · · , am )  

bm





 b1 ..  . . 

bm

b1  ..  Observing that αβ = 1R , we get (a1 , · · · , am )  .  = 1. As R is bm   a1 · · · am   a U C-ring, there is an invertible matrix  ... . . . ... , and so we have ∗ ··· ∗   c1 · · · ∗  .. . . ..   . . .  such that cm · · · ∗



a1  ..  . ∗

··· .. . ···

 am c1 ..   .. .  . ∗ cm

··· .. . ···

 ∗ ..  = I . m . ∗

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Thus,



c1 · · ·  .. . . α(η1 , · · · , ηm )  . . cm · · · Let

  ∗ ..  = ε(a , · · · , a )  1 m  . ∗

= ε(1, 0, · · · , 0).

347

c1 · · · .. . . . . cm · · ·

 ∗ ..  . ∗



 c1 · · · ∗   (ξ1 , · · · , ξm ) = (η1 , · · · , ηm )  ... . . . ...  . cm · · · ∗   c1 · · · ∗  .. . . ..  Since  . . .  is invertible, (ξ1 , · · · , ξm ) is a basis of mR. In addition, cm · · · ∗ α(ξ1 , · · · , ξm ) = ε(1, 0, · · · , 0). It follows that P ∼ = ker(α)     r1 r1     = {(ξ1 , · · · , ξm )  ...  | α(ξ1 , · · · , ξm )  ...  = 0, r1 , · · · , rm ∈ R} rm r   m r1 r1     = {(ξ1 , · · · , ξm )  ...  | ε(1, 0, · · · , 0)  ...  = 0, r1 , · · · , rm ∈ R} rm r  m r1   = {(ξ1 , · · · , ξm )  ...  | r1 = 0, r2 , · · · , rm ∈ R} 

rm = {ξ2 r2 + · · · ξm rm | r2 , · · · , rm ∈ R} = (m − 1)R. Let n ∈ N. Assume that Q ⊕ sR ∼ = tR with t − s ≥ n. If
s = 1, by the above consideration, Q is free. If s ≥ 2, then Q ⊕ (s − 1)R ⊕ R ∼ = tR. By the ∼ preceding discussion again, we get Q ⊕ (s − 1)R = (t − 1)R. By iteration of this process, we see that Q ∼ = (t − s)R is free, as desired.   b1  ..  (1) ⇒ (2) Let m ≥ 2. Suppose that α = (a1 , · · · , am ) and β =  .  bm such that αβ = 1R . Let ε and η1 , · · · , ηm be bases of R and mR, respectively. Construct two R-morphisms ϕ : R → mR, ϕ(ε) = (η1 , · · · , ηm )β;

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φ : mR → R, φ(η1 , · · · , ηm ) = εα. Obviously, φϕ(ε) = φ(η1 , · · · , ηm )β = εαβ = ε. Hence, φϕ = 1R , i.e., the exact sequence φ

0 −→ ker(φ) −→ mR −→ R −→ 0

∼ R such that mR = ker(φ) ⊕ C. Hence, splits. Thus, we have some C = ∼ ker(φ) ⊕ R = mR, i.e., ker(φ) is stably free of rank m − 1. By hypothesis, ker(φ) ∼ = (m − 1)R. Let δ1 , · · · , δm−1 and δm be bases of ker(φ) and C, respectively. Then (δ1 , · · · , δm ) and (η1 , · · · , ηm ) are both bases of mR.   c11 · · · c1m   Thus, we have an invertible matrix  ... . . . ...  such that cm1 · · · cmm   c11 · · · c1m   (η1 , · · · , ηm ) = (δ1 , · · · , δm )  ... . . . ...  . cm1 · · · cmm

This implies that

   r1 r1     ker(φ) = {(η1 , · · · , ηm )  ...  | εα  ...  = 0, eachri ∈ R} rm r      m r1 r1 c11 · · · c1m      = {(δ1 , · · · , δm )  ... ... . . .   ...  | εα  ...  = 0}. 

cm1 · · · cmm

Let

Then

rm

rm

    s1 r1 c11 · · · c1m  .. .. . .   ..   ..   . . .  .  =  . . rm sm cm1 · · · cmm 

  s1    ker(φ) = {(δ1 , · · · , δm )  ...  | εα 

−1   c11 · · · c1m s1 .. .. . .   ..  = 0} .   .  . . sm cm1 · · · cmm sm    −1   s1 c11 · · · c1m s1       = {(δ1 , · · · , δm )  ...  | α  ... ... . . .   ...  = 0}. sm cm1 · · · cmm sm 

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Assume that 

−1 c11 · · · c1m   α  ... ... . . .  = (t1 , · · · , tm ). cm1 · · · cmm

Then



 s1   ker(φ) = {(δ1 , · · · , δm )  ...  | t1 s1 + · · · + tm sm = 0, each si ∈ R} sm = δ1 R ⊕ · · · ⊕ δm−1 R.

Clearly, δ1 ∈ ker(φ) whence

  1 0   (δ1 , · · · , δm )  .  ∈ ker(φ).  ..  0

This implies that t1 = 0. Similarly, t2 = · · · = tm−1 = 0. Therefore,   c11 · · · c1m   αβ = (0, · · · , 0, tm )  ... . . . ...  β cm1 · · · cmm = tm (cm1 , · · · , cmm )β.

As αβ = 1, tm is right invertible. So we have some k such that tm k = 1. From this, we see that tm (1 − ktm ) = 0, and thus, δm (1 − ktm ) ∈ ker(φ). Moreover, 1 − ktm = 0; hence, tm is invertible. This implies that α = (tm cm1 , · · · , tm cmm ) is the last row of an invertible matrix    c11 c12 · · · c1(m−1) c1m 1 0 ··· ··· 0  .. .. .. ..  .. .. . . .. ..   ..   . . . . .  . . . . . ,    0 0 · · · 1 0   c(m−1)1 c(m−1)2 · · · c(m−1)(m−1) c(m−1)m  cm1 cm2 · · · cm(m−1) cmm 0 0 · · · 0 tm 

as required.





a b −y x

−1

Let R be a commutative ring. If ax + by = 1, then =  x −b ; hence, R is a U C1 -ring. As in the proof of Theorem 11.4.1, we y a

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see that P ⊕ R ∼ = 2R if and only if P is free of rank 1. Let R be a stably euclidean ring. In view of Theorem 11.1.3, we also see that P ⊕ R ∼ = 2R if and only if P is free of rank 1. We say that (x1 , · · · , xn ) is a n-unimodular row over R provided that x1 R + · · · + xn R = R. The n-unimodular column is defined in the same way. We use Ur,n (R) and Uc,n (R) to denote the sets of all n-unimodular rows over R and all n-unimodular columns over R, respectively. As in the preceding proof, we deduce that a ring R is a U Cn -ring if and only if P ⊕ R ∼ = (n+ 1)R implies that P ∼ = nR if and only if for any α, β ∈ Ur,n+1 (R), there exists a U ∈ GLn+1 (R) such that α = βU . Let D be a noncommutative division, and let R = D[x, y]. Let a, b be a pair of noncommutative elements in D, and let ϕ : 2R → R given by ϕ(1, 0) = x + a and ϕ(0, 1) = y + b. Choose P = kerϕ. Then P ⊕ R ∼ = 2R, ∼ while P = 6 R (cf. [298, Example 2.3]). Corollary 11.4.2. If R is a U C-ring, then so is Mn (R) for all n ∈ N.

Proof. Given P ⊕ sMn (R) ∼ tMn (R) with t > s, then O= P Rn×1 ⊕ sRn×1 ∼ = tRn×1 . Mn (R)

N

Rn×1 ⊕ snR ∼ = tnR. Since R is a U C-ring, by virtue of Mn (R) N Theorem 11.4.1, P Rn×1 ∼ = (t − s)nR ∼ = (t − s)Rn×1 . This implies That is, P

Mn (R)

that

P

O

Rn×1

P ∼ =P

O

R

Mn (R)

Therefore

O

Mn (R)

R1×n ∼ = (t − s)Rn×1

Rn×1

O R

O

R1×n .

R

R1×n ∼ = (t − s)Mn (R).

By using Theorem 11.4.1 again, Mn (R) is a U C-ring, as desired.



Lemma 11.4.3. A ring R is a U Cn -ring if and only if so is the opposite ring Rop . Proof. Let R be a U Cn -ring. Given Ra1 + · · · + Ran+1 = R, then x1 a1 + · · · + xn+1 an+1 = 1 for some xi ∈ R. Hence, we have some bij , cij ∈ R such that   −1  x1 · · · xn+1 c11 · · · c1(n+1)  b21 · · · b2(n+1)   c21 · · · c2(n+1)      =  ∈ GLn+1 (R).  .. .. .. .. .. ..     . . . . . . b(n+1)1 · · · b(n+1)(n+1) c(n+1)1 · · · c(n+1)(n+1)

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This implies that  x1 · · · xn+1  b21 · · · b2(n+1)   .. .. ..  . . .

b(n+1)1 · · · b(n+1)(n+1)

Therefore,  Let



 1 0   ∗ In 

x1 b21 .. .

   

a1 a2 .. .



xn+1 b2(n+1) .. .



1 0 ∗ In



   

c12 c22 .. .

··· ··· .. .

c1(n+1) c2(n+1) .. .

an+1 a(n+1)2 · · · c(n+1)(n+1)

b(n+1)1 · · · b(n+1)(n+1)

Q=

Then

··· ··· .. .



351

x1 b21 .. . b(n+1)1

   

a1 a2 .. .

··· ··· .. .



   1 0  = .  ∗ In 

c1(n+1) c2(n+1) .. .

a(n+1) · · · c(n+1)(n+1)



   = In+1 . 

 ··· xn+1 · · · b2(n+1)    ∈ GLn+1 (R). .. ..  . . · · · b(n+1)(n+1)

  1  0      Q  =  . .   ..   0 an+1 

a1 a2 .. .



This implies that Rop is a U Cn -ring. The converse is symmetric.



We say that A ∈ Mm×n (R) is rectangular provided that m 6= n. A rectangular matrix A ∈ Mm×n (R) iscalled  completable if either (1) m < n A and for some B ∈ M(n−m)×n (R), is invertible or (2) m > n and B for some B ∈ Mm×(m−n) (R), (A, B) is invertible. Now we characterize U C-rings by means of rectangular matrices. Theorem 11.4.4. Let R be a ring. Then the following are equivalent: (1) R is a U C-ring. (2) Every right invertible rectangular matrix is completable. (3) Every left invertible rectangular matrix is completable.

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Proof. (1) ⇒ (2) Let A = (aij ) ∈ Mm×n (R)(m 6= n) be right invertible. Assume that m < n. Let ϕ be the R-morphism corresponding to A, and let K = ker(ϕ). Then ϕ is an R-epimorphism. Thus, we can find a right R-module D such that nR = K ⊕ D with σ = ϕ|D : D ∼ = mR. This means that K is a stably free right R-module of positive rank. In view of Theorem 11.4.1, φ : K ∼ = (n−m)R. Let θ be the R-morphism corresponding to (Im , 0(n−m)×n ). Construct an R-isomorphism ψ : nR = K ⊕ D → (n − m)R ⊕ mR = nR given by ψ(x, y) = φ(x) + σ(y) = nR for any x ∈ (n − m)R, y ∈ mR. Then we get the following commutative diagram: 0→

K φ↓

ϕ

֒→ nR → mR → 0 ψ↓ k θ

0 → (n − m)R ֒→ nR ։ mR → 0. Let B be the matrix corresponding to ψ. Then A = (Im , 0m×(n−m) )B. Therefore A can be completed to an invertible matrix B. Assume that m > n. Clearly, we have some B ∈ Mn×m (R) such that AB = Im . Let ϕ be the R-morphism corresponding to B, and let K = coker(ϕ). Then ϕ is a splitting R-monomorphism. Thus, we can find a right R-module D such that mR = im(ϕ) ⊕ D with τ |D : D ∼ = K, where τ : mR → K is a canonical map. This means that D is a stably free right ∼ R-module of positive rank. In view of Theorem  11.4.1, φ : (n − m)R = D. In Let θ be the R-morphism corresponding to . Construct an 0(m−n)×n R-morphism ψ : mR = nR ⊕ (m − n)R → im(ϕ) ⊕ D = mR given by ψ(x, y) = ϕ(x) + φ(y) = mR for any x ∈ nR, y ∈ (m − n)R. Then we get the following commutative diagram: ϕ

τ

θ

π

0 → nR → mR ։ K → 0 k ψ↑ τ |D φ ↑

0 → nR ֒→ mR ։ (m − n)R → 0.

One easily checks that ψ is anR-isomorphism. Let B be the matrix corre In sponding to ψ. Then A = B . Then A can be completed to 0(m−n)×n an invertible matrix B. (2) ⇒ (1) Given a1 R + · · · + an+1 R = R (n ∈ N), then (a1 , · · · , an+1 ) ∈ M1×(n+1) (R) is right invertible. By hypothesis, (a1 , · · · , an+1 ) is the first row of an invertible matrix. Therefore R is a U Cn -ring, as required. (1) ⇔ (3) Applying (1) ⇔ (2) to the opposite ring Rop , we complete the proof by Lemma 11.4.3. 

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By the Quillen-Suslin Theorem, every polynomial ring over a principal ideal domain is a U C-ring (cf. [364, Theorem 4.63]). σ

Lemma 11.4.5. Suppose that 0 → P → mR → nR → 0 is an exact sequence. If {ε1 , · · · , εm } is a basis of mR such that nR is generated by {σ(ε1 ), · · · , σ(εt )} (t ≤ m − n), then P is free. Proof. Let {η1 , · · · , ηn } be a basis of nR. Then there exists some A ∈ Mn×m (R) such that σ(ε1 , · · · , εm ) = (η1 , · · · , ηn )A. As nR is generated by {σ(ε1 ), · · · , σ(εt )}(t ≤ m−n), there exists some C ∈ Mt×n (R) such that (η1 , · · · , ηn ) = σ(ε1 , · · · , εt )C. Set A = [B, D], where B ∈ Mn×t (R). Then σ(ε1 , · · · , εm ) = σ(ε1 , · · · , εt )CA = σ(ε1 , · · · , εt )C[B, D]; hence, σ(ε1 , · · · , εt ) = σ(ε1 , · · · , εt )CB. Further, we see that (η1 , · · · , ηn ) = σ(ε1 , · · · , εt )C = σ(ε1 , · · · , εt )CBC = (η1 , · · · , ηn )BC.

Thus, BC = In . As m − t ≥ n, we set D = [E, F ], where E ∈ Mn (R). By elementary transformations, we see that A = [B, E, F ] → [B, E − BC(E − In ), F ] → [B, In , F ] → [In , 0].

Therefore, we have some U ∈ GLn (R) and V ∈ GLm (R) such that U AV = [In , 0]. It follows that ∼ ker(σ) P =   r1 m P   = { εi ri | (η1 , · · · , ηn )A  ...  = 0, r1 , · · · , rm ∈ R} i=1

={ 



m P

i=1





rm

r1  ..  εi ri | A  .  = 0, r1 , · · · , rm ∈ R}. 



rm

s1 r1  ..   −1  .. Let  .  = V  .  and (ε′1 , · · · , ε′m ) = (ε1 , · · · , εm )V . Then sm

rm

   s1 s1 .  .  P ∼ = {(ε1 , · · · , εm )V  ..  | (In , 0)  ..  = 0, s1 , · · · , sm ∈ R} 

s sm m s1   = {(ε′1 , · · · , ε′m )  ...  | s1 = · · · = sn = 0, sn+1 , · · · , sm ∈ R}. sm 

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Obviously, {ε′1 , · · · , ε′m } is a basis of mR. Therefore, we conclude that P ∼  = ε′n+1 R ⊕ · · · ⊕ εm R is free. Theorem 11.4.6. Let R be a commutative ring. If P ⊕ nR ∼ = mR(m > n), n . then sP ∼ s(m − n)R for all s ≥ n + = m−n Proof. Given P ⊕ nR ∼ = mR(m > n), there is an exact sequence σ

0 −→ P −→ mR −→ nR −→ 0. Let ε1 , · · · , εm and η1 , · · · , ηn be the bases of mR and  nR, respectively.  a11 · · · am1   Then σ(ε1 , · · · , εm ) = (η1 , · · · , ηn )A, where A =  ... . . . ...  ∈ a1n · · · amn Mn×m (R). Clearly, there is some C ∈ Mm×n (R) such that AC = In . Let 

a11  .. M = . a1n

  · · · an1 a(n+1)1  .. . . ..  . . , N =  . · · · ann a(n+1)n

 · · · am1 .  .. . ..  . · · · amn

Then A = [M, N ]. Obviously, P ∼ = ker(σ)

 r1   = {ε1 r1 + · · · + εm rm | σ(ε1 , · · · , εm )  ...  = 0} 

r  m  r1 r1  ..   ..  = {(ε1 , · · · , εm )  .  | (η1 , · · · , ηn )A  .  = 0} 

rm   r1    = {(ε1 , · · · , εm )  ...  | A  

rm



rm

r1 ..  = 0}. . 

rm

Further, we get an exact sequence ϕ

0 −→ P ⊕ P −→ mR ⊕ mR −→ nR ⊕ nR −→ 0,

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where the corresponding matrix of ϕ is diag(A, A). Hence, 2P ∼ = ker(ϕ) 

  = {(δ1 , δ2 , · · · , δ2m )  

r1 r2 .. . r2m



  | 



 r1  .   ..       A 0  rm    = 0}, 0 A  rm+1     ..   .  r2m

where {δ1 , δ2 , · · · , δ2m } is a basis of 2mR. Observing the elementary transformations,         A 0 A 0 0 A −AC 0 A −In 0 = → = 0 A 0 M N 0 M N 0 M N →



A −In 0 MA 0 N



→



0 −In 0 MA 0 N



→





In 0 0 0 MA N

,

we get 

 2P ∼ = {(τ1 , · · · , τ2m−n ) 

r1 .. . r2m−n





  2  | [M , M N, N ] 



r1 .. . r2m−n

  = 0},

where {τ1 , τ2 , · · · , τ2m−n } is a basis of (2m − n)R. By iteration of this process,     r1 r1  ..   ..  t t−1 ∼ tP = {(α1 , · · · , αk )  .  | [M , M N, · · · , M N, N ]  .  = 0}, rk

rk

where k = n + t(m − n) and {α1 , · · · , αk } is a basis of kR. As a result, we get an exact sequence
φ 0 −→ tP −→ n + t(m − n) R −→ nR −→ 0,

where the corresponding matrix of φ is [M t , M t−1 N, · · · , M N, N ]. Since R is a commutative ring, there are some ai ∈ R such that det(M − λE) = λn −a1 λn−1 −· · ·−an ; hence, M n = a1 M n−1 +· · ·+an In . Moreover, M n N = a1 M n−1 N +· · ·+an N. It follows that whenever i ≥ n, M i N can be obtained by elementary column transformations from [M n−1 N, · · · , N ]. Thus, nR
is generated by n + t(m − n) image elements of a basis of n + t(m − n) R.

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According to Lemma 11.4.5, whenever n + n(m − n) ≤ n + t(m − n) − n, n , tP is free, as asserted. tP is free. In other words, whenever t ≥ n + m−n Let R be a commutative ring. According to Theorem 11.4.6, every stably free right R-module of positive rank is power-free. It is worth noting that a stably Ofree right O R-module of positive rank is power-free if and only if P ⊗m := P ··· P is free for some m ∈ N (cf. [41, Theorem]). Further, |

R

{z

R

m

}

we can derive the following. Corollary 11.4.7. Let R be a commutative ring. If P ⊕nR ∼ = mR(m > n), then sP ∼ = s(m − n)R for all s ≥ 3n/2.

Proof. Let P ⊕ nR ∼ = mR(m > n). If m − n = 1, then we get a short exact sequence α

0 → P → (n + 1)R → nR → 0,

where P = ker(α). Let A be the n ×
(n + 1) matrix corresponding to α, and let di = det A − [i − th column] (1 ≤ i ≤ n + 1). Clearly, A is right invertible, that is, there exists
some B ∈ M(n+1)×n (R) such that AB = In . Let yi = det B −[i−th row] (1 ≤ i ≤ n+1). By the Binet-Cauchy formula, n+1 P 1 = det(AB) = di yi . Consider the matrix i=1

A=



A (−1)n y1 · · · (−1)n+n yn+1

By expansion along the last row, we get det(A) =



.

n+1 X

(−1)(n+1)+i (−1)n+i−1 yi di = 1.

i=1

Thus, A ∈ GLn+1 (R). Let α : (n+1)R → (n+1)R be the R-morphism corresponding to A. Let π : (n + 1)R → nR be the projection onto nR. Then the matrix corresponding to π is (In , 0) ∈ Mn+1 (R). Clearly, (In , 0)A = A. Thus, πα = α. Therefore P = ker(α) = ker(π)α ∼ = ker(π) = R is free. If n m−n ≥ 2, then s ≥ 3n/2 implies that s ≥ n+ m−n . Therefore we complete the proof by Theorem 11.4.6.  Corollary 11.4.8. Let R be a commutative ring. Then for any right (left) rectangular invertible matrix (aij ) ∈ Mm×n (R) there exists some s ∈ N such that (aij Is ) can be completed to an invertible matrix.

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Proof. Let A = (aij ) ∈ Mm×n (R)(m 6= n) be right invertible. Assume that m < n. Let ϕ be the R-morphism corresponding to A, and let K = ker(ϕ). Then there exists a right R-module D such that nR = K ⊕ D with σ = ϕ|D : D ∼ = mR. That is, K is a stably free right R-module of positive rank. In view of Theorem 11.4.6, φ : sK ∼ = s(n − m)R for some s ∈ N. Let θ be the R-morphism corresponding to (Ism , 0s(n−m)×sn ). Construct an R-isomorphism ψ : snR = sK ⊕ sD → s(n − m)R ⊕ smR = snR given by ψ(x, y) = φ∗ (x) + σ ∗ (y) = snR for any x ∈ s(n − m)R, y ∈ smR. As in the proof of Theorem 11.4.4, the following diagram 0→

sK φ↓

ϕ∗

֒→ snR → smR → 0 ψ↓ k θ

0 → s(n − m)R ֒→ snR ։ smR → 0

commutates. Let B be the matrix corresponding to ψ. Then diag(A, · · · , A)s×s = (Ism , 0sm×s(n−m) )B. Thus, diag(A, · · · , A)s×s can be completed to an invertible matrix B. Therefore (aij Is ) can be completed to some invertible matrix. Assume that m > n. As A ∈ Mm×n (R) is right invertible, there is some B ∈ Mn×m (R) such that AB = Im . Let ϕ be the R-morphism corresponding to B, and let K = coker(ϕ). Then ϕ is a splitting R-monomorphism. So, we have a right R-module D such that mR = im(ϕ) ⊕ D with τ |D : D ∼ = K, where τ : mR → K is a canonical map. According to Theorem 11.4.6, φ : s(n − m)R ∼ = sD for some s ∈ N. Let θ Isn be the R-morphism corresponding to . Construct an R0s(m−n)×n morphism ψ : smR = snR ⊕ s(m − n)R → sim(ϕ) ⊕ sD = smR given by ψ(x, y) = ϕ∗ (x) + φ∗ (y) = smR for any x ∈ snR, y ∈ s(m − n)R. As in the proof of Theorem 11.4.4, the following diagram ϕ∗

τ∗

θ∗

π∗

0 → snR → smR ։ sK →0 k ψ↑ τ ∗ |sD φ ↑ 0 → snR ֒→ smR ։ s(m − n)R → 0 commutates.

Further, ψ is an R-isomorphism.

Let  B be the matrix Isn corresponding to ψ. Then diag(A, · · · , A)s×s = B . Thus 0s(m−n)×sn diag(A, · · · , A)s×s can be completed to an invertible matrix B. Therefore (aij Is ) can be completed to some invertible matrix. Other cases can be proved in the same manner. So the proof is true. 

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Let R be a commutative ring. Then for all finitely generated projective right R-modules A, B and C, A ⊕ B ∼ = A ⊕ C implies that nB ∼ = nC for all sufficiently large n ([298, Theorem 5.1]). But Goodearl’s proof is based on an application of the standard cancellation theorem, plus a reduction to the case of noetherian rings. It provides no information on the number n. Proposition 11.4.9. Let R be a commutative ring. Then the following are equivalent: (1) Every nonzero finitely generated projective R-module is free. (2) R is a U C-ring and every finitely generated projective R-module is stably free. (3) For any regular  A ∈ Mn (R), there exist U, V ∈ GLn (R) such that Ir 0 U AV = for some r. 0 0 (4) For any idempotent   E ∈ Mn (R), there exists W ∈ GLn (R) such that I 0 r W EW −1 = for some r. 0 0 Proof. (1) ⇒ (2) Clearly, every finitely generated projective R-module is stably free. Let P be a stably free R-module of rank s. Write P ⊕ mR ∼ = (m + s)R with s ∈ N. By hypothesis, P ∼ = tR for some t ∈ N. Thus, (m + t)R ∼ = (m + s)R. As R is a commutative ring, t = s. Hence, P is a free R-module of rank s. Thus, R is a U C-ring by Theorem 11.4.1. (2) ⇒ (3) Let P ∈ F P (R). Then P ⊕ mR ∼ = nR. If m = n, then O M P RQ mRQ ∼ = mRQ

for all Q ∈ Spec(R). As RQ is local, it has stable range one; hence, PQ = 0. If P 6= 0, then annR (P ) := {x ∈ R | P x = 0} $ R.

Thus, we can find a maximal ideal M of R such that annR (P ) ⊆ M $ R. Assume that P is generated by the set {x1 , · · · , xn }. Then x11 , · · · , x1n ∈ PM = 0. Thus, there are some s1 , · · · , sn 6∈ M such that each xi si = 0. Set s = s1 · · · sn . Then P s = 0, and so s ∈ annR (P ). We infer that s ∈ M , a contradiction. Accordingly, P = 0. Further, we see that m  n. Hence, m < n. By hypothesis, P is a free R-module of rank n − m, which implies that P is free or a zero. Let A ∈ Mn (R) be regular. Then there exists some B ∈ Mn (R) such that A = ABA. Write nR = A(nR) ⊕ (In − AB)(nR) = B(nR) ⊕ (In − BA)(nR), where A(nR), (In −

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AB)(nR),   B(nR), (In − BA)(nR) ∈ F P (R). Thus, A(nR) Ir 0 (nR) for some r. Further, 0 0    0 0 Ir ∼ ∼ ∼ coker A = (In − AB)(nR) = = coker 0 In−r 0

∼ = B(nR) ∼ =

0 0





.

.

In addition, ker A ∼ = (In − BA)(nR) ∼ =



0 0 0 In−r



∼ = ker 



Ir 0 0 0

 Ir 0 According to Theorem 7.2.1, A is equivalent to , as required. 0 0 (3) ⇒ (4) For anyidempotent E ∈ Mn (R), there exist U, V ∈ GLn (R)  Ir 0 such that U EV = for some r. Let Y = U EU −1 and F = 0 0   Ir 0 . Then Y F = (U EU −1 )(U EV ) = F and F Y = F (U EU −1 ) = 0 0 F U E(V V −1 )U −1 = F (U EV )V −1 U −1 = (U EV )V −1 U −1 = U EV = Y . Let W = In − F + Y . Then one easily checks that W −1 = In + F − Y . Further, we see that W U E = (In −F +Y )U E = U E −F U E +Y U E = Y U E = (U EU −1 )U E = U E and F W U = F (In − F + Y )U = F Y U = Y U = U E. Therefore (W U )E(W U )−1 = F , as required. (4) ⇒ (1) Let 0 6= P ∈ F P (R). Then we have an idempotent E ∈ Mn (R) such that P ∼ By hypothesis, there exist U, V ∈ GLn (R) = E(nR).  Ir 0 such that U EV = . Thus, P ∼ = U EV (nR) ∼ = rR, where r ∈ N. 0 0 Therefore P is free, as asserted.  Corollary 11.4.10. Let R be a principal ideal domain. Then  for any  Ir 0 regular A ∈ Mn (R), there exist U, V ∈ GLn (R) such that U AV = 0 0 for some r. Proof. In view of Proposition 11.4.9, it suffices to show that every nonzero finitely generated projective R-module is free. Let M = nR. We will prove that every submodule of M is free or a zero module. If n = 1 and 0 6= N ⊆ M , then N is a right ideal of R. As R is a principal ideal domain, N = xR for some 0 6= x ∈ R. Hence, N ∼ = R is free. Assume that every submodule of kR (k ≥ 1) is free or a zero module. Let N ⊆ (k + 1)R. Choose N1 = {(x1 , x2 , · · · , xk , 0) ∈ N | x1 , · · · , xk ∈ R}. Then N1 is

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isomorphic to a submodule of kR. By hypothesis, N1 is free or a zero module. Construct a map ϕ : N → R given by (x1 , x2 , · · · , xk , xk+1 ) ֌ xk+1 . Then N/N1 ∼ = Imϕ ⊆ R. Hence, N/N1 is free or a zero module. We infer that N = N1 or N ∼ = N/N1 ⊕ N1 is free. By induction, the result follows. 

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Chapter 12

The n-Stable Range Condition

The basic theme of this chapter is to examine ”high” stable range condition. The class of rings satisfying the n-stable range condition is very large. For example, any commutative ring of Krull dimension d satisfies the (d + 2)stable range condition (cf. [393]), and R[x1 , · · · , xn ] satisfies the (n + 1)stable range condition (cf. [313, Corollary 5.10]). Warfield [394, Theorem 1.2] proved that if EndR (A) satisfies the n-stable range condition, then we have the cancellation of modules: A⊕X ∼ = A⊕Y with nA .⊕ X =⇒ X ∼ =Y for any right R-modules X and Y . This condition is a useful one in algebraic K-theory. For example, if R satisfies the n-stable range condition then the central extension 1 → K2 (m, R) → St(m, R) → E(m, R) → 1 is a universal central extension for all m ≥ max(n + 3, 5) and K2 (m, R) → K2 (m + 1, R) is surjective for all m ≥ n + 2. Further, one sees that K2,m (R) ∼ = K2 (R) provided that m ≥ n + 2. The results for the n-stable range condition will make the behavior of stable range one and the comparability of modules clear.

12.1

Weak Substitutions

We use Rn to denote the set M1×n (R) of all 1 × n matrices and n R to denote the set Mn×1 (R) of all n × 1 matrices. Call x ∈ Rn (resp. x ∈ n R) is unimodular provided that xy = 1 (resp. yx = 1) for some y ∈ n R (resp. 361

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y ∈ Rn ). We know that a ring R satisfies the n-stable range condition if and only if ax + b = 1 with a ∈ Rn , x ∈ n R, b ∈ R implies that there exists some y ∈ Rn such that a+by ∈ Rn is unimodular. The main purpose of this section is to investigate the weak substitution property of such a condition. We will deduce a type of cancellation of modules over such rings. Lemma 12.1.1. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E satisfies the n-stable range condition. (2) For any f ∈ HomR (nA, A), g ∈ HomR (A, nA) and h ∈ EndR (A), f g + h = 1A implies that there exist k ∈ HomR (nA, A), l ∈ HomR (A, nA) such that (f + hk)l = 1A . (3) For any f ∈ HomR (nA, A), g ∈ HomR (A, nA) and h ∈ EndR (A), f g + h = 1A implies that there exist k ∈ HomR (A, nA), l ∈ HomR (nA, A) such that l(g + kh) = 1A . Proof. (1) ⇒ (2) Let f ∈ HomR (nA, A), g ∈ HomR (A, nA) and h ∈ EndR (A) such that f g +h = 1A . Let pi : nA → A be the projection and qi : n n P P A → nA be the injection. Then qi pi = 1nA . Hence, (f qi )(pi g)+h = 1. i=1

i=1

By hypothesis, there exist some y1 , · · · , yn ∈ E such that n X

(f qi + hyi )E = E.

i=1

That is, n X (f qi + hyi )zi = 1A i=1

for some zi ∈ E. Construct two maps k:

l:

nA → A;     r1 r1  ..   ..   .  7→ (y1 , · · · , yn )  .  ; rn rn A

→

r

7→

nA;  z1  ..   .  r. zn 

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For any r ∈ R, it is easy to check that   z1  ..  (f + hk)l(r) = (f + hk)  .  r zn

   z1 z1  ..   ..  = (f q1 , · · · , f qn )  .  r + h(y1 , · · · , yn )  .  r =

n P



zn

zn

(f qi + hyi )zi (r)

i=1

= r.

Therefore (f + hk)l = 1A , as required. (2) ⇒ (3) Suppose that there exist k ∈ HomR (nA, A), l HomR (A, nA) such that (f + hk)l = 1A . Then we check that

f + k(1nA − gf ) g + (1nA − gk)lh = 1A ,

∈

as asserted. (3) ⇒ (1) Let a ∈ Rn , x ∈ n R, b ∈ R such that ax + b = 1. Construct two maps f:

nA →   r1  ..   .  7→ rn

A;  r1   a  ...  ; 

rn

g : A → nA; h : A → A; r 7→ xr; r → 7 br. Then f g + h = 1A . By hypothesis, there exist k ∈ HomR (A, nA), l ∈ HomR (nA, A) such that l(g + kh) = 1A . It is easy to verify that

f + hl(1nA − kf ) g + (1nA − gf )k = 1A .

Let α = l(1nA − kf ) : nA → A. Then (f + hα)β = 1A for a β ∈ HomR (A, nA). Let pi : nA → A be the projection and qi : A → nA be   p1 β   the injection. Let y = (αq1 , · · · , αqn ) ∈ E n and u =  ...  ∈ n E. For pn β

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any r ∈ A, we see that (f + hα)β(r) = r; hence, aβ(r) + bαβ(r) = r. This implies that     p1 β p1 β     (a + by)u(r) = a  ...  (r) + b(αq1 , · · · , αqn )  ...  (r) pn β pn β n
P = aβ(r) + bα qi pi β(r) i=1

= r.

Hence, (a + by)u = 1A , i.e., a + by ∈ E n is unimodular. Therefore the result follows.  As an immediate consequence, we deduce that a ring R satisfies the nstable range condition if and only if ax + b = 1 with a ∈ Rn , x ∈ n R, b ∈ R implies that there exists an element z ∈ n R such that x + zb ∈ n R is unimodular. This means that the n-stable range condition is right-left symmetric. That is, a ring R satisfies the n-stable range condition if and only if the opposite ring Rop also satisfies the n-stable range condition. Proposition 12.1.2. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent: (1) E satisfies the n-stable range condition. (2) For any regular f ∈ HomR (nA, A), there exist g ∈ HomR (A, nA), h ∈ HomR (nA, A) such that f = f gf and hg = 1A . (3) For any regular f ∈ HomR (A, nA), there exist g ∈ HomR (nA, A), h ∈ HomR (A, nA) such that f = f gf and gh = 1A . Proof. (1) ⇒ (2) Given any regular f ∈ HomR (nA, A), there exists k ∈ HomR (A, nA) such that f = f kf . Since f k + (1A − f k) = 1A , by Lemma 12.1.1, we can find some l ∈ HomR (nA, A) such that k + l(1A − f k) = g with hg =
1A for some h ∈ HomR (nA, A). Therefore f = f kf = f k + l(1 − f k) f = f gf , as required. (2) ⇒ (1) Suppose f g + h = 1A with f ∈ HomR (nA, A), g ∈ HomR (A, nA) and h ∈ E. Since A has the finite exchange property, E is an exchange ring by [382, Theorem 28.7]. Thus we have an idempotent e ∈ E such that e = hs and 1A − e = (1A − h)t for some s, t ∈ E from Lemma 1.4.7. So f gt + e = 1A , and then (1A − e)f gt + e = 1A . Clearly, (1A − e)f ∈ HomR (nA, A) is regular. Therefore we have k ∈ HomR (A, nA), l ∈ HomR (nA, A) such that (1A −e)f = (1A −e)f k(1A −e)f and lk = 1A . Set p = k(1A − e)f . Then pgt + ke = k, whence

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p(gt + ke) + (1A − p)ke = k. Clearly, (1A − p)ke = (1A − p)kel(1A − p)ke. Set q = (1A − p)kel(1A − p). Then p(gt + ke) + qke = k. It is easy to verify that p = p2 , q = q 2 and pq = qp = 0. Thus, we see

k (1A − e)f + el(1A − p)(1A + pkel(1
A − p)) 1A − pkel(1
A − p) k = p + kel(1A − p)(1A + pkel(1A − p))
1A − pkel(1A − p) k = p(1A − pkel(1A − p)) + kel(1 A − p) k
= p + (1A − p)kel(1A − p) k = pk + qk = k. As lk = 1A , we know that



(1A − e)f + el(1A − p)(1A + pkel(1A − p)) 1A − pkel(1A − p) k = 1A .

Consequently, we deduce that



f + hs(l(1A − p)(1A + pkel(1A − p)) − f ) 1A − pkel(1A − p) k = 1A .

In light of Lemma 12.1.1, E satisfies the n-stable range condition. (1) ⇔ (3) is analogous to the above consideration.



Theorem 12.1.3. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E satisfies the n-stable range condition. (2) Given any right R-module decompositions M = A1 ⊕B1 = A2 ⊕B2 with A1 ∼ = nA, A ∼ = A2 , there exist C, D ⊆ M such that M = C ⊕ D ⊕ B1 = C ⊕ B2 . (3) Given any right R-module decompositions M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = nA, A ∼ = A2 , there exist C, D ⊆ M such that M = A1 ⊕ C = A2 ⊕ C ⊕ D.

Proof. (1) ⇒ (2) Suppose M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = nA, A ∼ = A2 . ∼ From M = A1 ⊕B1 = nA⊕B1 , we have projections p1 : M → nA, p2 : M → B1 and injections q1 : nA → M, q2 : B1 → M . From M = A2 ⊕B2 ∼ = A⊕B2 , we have a projection f : M → A and an injection g : A → M such that f g = 1A and ker(f ) = B2 . Since 1A = (f q1 )(p1 g) + f q2 p2 g, by Lemma 12.1.1, we have two Rmorphisms h : nA → A and k : A → nA such that f (q1 + q2 p2 gh)k = 1A . Hence M = ker(f ) ⊕ (q1 + q2 p2 gh)k(A) = C ⊕ B2 , where C = (q1 + q2 p2 gh)k(A). As p1 (q1 + q2 p2 gh) = 1nA , M = ker(p1 ) ⊕ (q1 + q2 p2 gh)(nA). Let σ : (q1 + q2 p2 gh)k(A) → (q1 + q2 p2 gh)(nA) be the inclusion and τ = (q1 + q2 p2 gh)kf |(q1 +q2 p2 gh)(nA) . Then τ σ = 1(q1 +q2 p2 gh)k(A) . Hence,

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there exists a right R-module D such that (q1 + q2 p2 gh)(nA) = (q1 + q2 p2 gh)k(A) ⊕ D. Therefore M = C ⊕ D ⊕ B1 = C ⊕ B2 . (2) ⇒ (1) Let ax + b = 1A with a : nA → A, x : A → nA and b : A → A. Set M = nA ⊕ A. Let p1 : M → nA, p2 : M → A, q1 : nA → M and q2 : A → M denote the projections and injections of this direct sum. Set A1 = q1 (nA) and B1 = q2 (A). Then M = A1 ⊕ B1 with A1 ∼ = nA. Define f = ap1 + bp2 and g = q1 x + q2 . Since f g = ax + b = 1A , M = A2 ⊕ B2 with A2 ∼ = A, where A2 = g(A) and B2 = ker(f ). Therefore we can find C, D ⊆ M such that M = C ⊕ D ⊕ B1 = C ⊕ B2 . Let h:A∼ = A2 ∼ = C → M be the injection. Then C = h(A). Consequently, M = h(A) ⊕ D ⊕ ker(p1 ) = h(A) ⊕ ker(f ). Hence, f h ∈ E is invertible. T If p1 h(a) = 0, then h(a) ∈ h(A) ker(p1 ) = 0, whence, h(a) = 0. As h is monomorphic, a = 0. Thus, p1 h : A → nA is an R-monomorphism. This implies that p1 h : A ∼ = im(p1 h). Observing that im(p1 h) = p1 (C) ⊆⊕ p1 (C) ⊕ p1 (D) = nA, the inclusion im(p1 h) ֒→ nA splits. It therefore follows that p1 h : A → nA is left invertible, and so qp1 h is isomorphic for some q : nA → A.
Observing that f h = ap1 h
+ bp2 h = a + bp2 h(qp1 h)−1 q p1 h, we conclude that a + bp2 h(qp1 h)−1 q p1 h(f h)−1 = 1A . Therefore E satisfies the n-stable range condition by Lemma 12.1.1. (1) ⇒ (3) Let M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = nA, A ∼ = A2 . As in (1) ⇒ (2), we get 1A = (f q1 )(p1 g) + f q2 p2 g. By virtue of Lemma 12.1.1, there are R-morphisms h : A → nA and k : nA → A such that k(p1 + hf q2 p2 )g = 1A . Hence M = ker(k(p1 + hf q2 p2 )) ⊕ g(A). In addition, (p1 + hf q2 p2 )q1 = 1nA . Thus, M = ker(p1 + hf q2 p2 ) ⊕ q1 (nA) = A1 ⊕ C, where C = ker(p1 + hf q2 p2 ). Let ψ = (p1 + hf q2 p2 ) |A1 : A1 → nA be the restriction of p1 + hf q2 p2 . For any m ∈ ker(k(p1 + hf q2 p2 )), there exist a ∈ A1 , c ∈ C such that m = a + c. Hence, c ∈ ker(p1 + hf q2 p2 ), and so ker(kψ(a)) = 0. This implies that ker(k(p1 + hf q2 p2 )) = ker(p1 + hf q2 p2 )⊕ ker(kψ). Set D = ker(kψ). Then M = A2 ⊕ C ⊕ D. Consequently, we can find some C, D ⊆ M such that M = A1 ⊕ C = A2 ⊕ C ⊕ D. (3) ⇒ (1) Suppose ax + b = 1A with a : nA → A, x : A → nA and b : A → A. Set M = nA⊕A. Let p1 : M → nA, p2 : M → A, q1 : nA → M and q2 : A → M denote the projections and injections of this direct sum. Define f = ap1 + p2 and g = q1 x + q2 b. Set A1 = q1 (nA), B1 = q2 (A), A2 = g(A) and B2 = ker(f ). Then M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = nA, A ∼ = A2 . So we have C, D ⊆ M such that M = A1 ⊕ C = A2 ⊕ C ⊕ D. Let h : M = A1 ⊕ C → A1 ∼ = nA be the projection. Then C = ker(h). Hence, M = q1 (nA) ⊕ ker(h). This implies that hq1 ∈ AutR (nA). If hg(a) = 0,

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T then g(a) ∈ A2 C = 0, whence, a = 0. This implies that hg : A → nA π is an R-monomorphism. Thus, A ∼ = imhg ֒→ nA. It is easy to see that nA = h(M ) = im(hg) ⊕ h(D), and so π splits. Therefore hg is right invertible, i.e., khg = 1A for some k ∈ Hom
R (nA, A). Observing that khg = kh(q1 x + q2 b) = khq1 x + (hq1 )−1 hq2 b = 1A , we conclude that E satisfies the n-stable range condition from Lemma 12.1.1.  Corollary 12.1.4. Let A be a right R-module, and let E = EndR (A). If E satisfies the n-stable range condition, then nA ⊕ B ∼ = A ⊕ C implies that B .⊕ C. Proof. As ϕ : A ⊕ C ∼ = nA ⊕ B, we have that nA ⊕ B = ϕ(A) ⊕ ϕ(C) with ϕ(A) ∼ = A. By virtue of Theorem 12.1.3, there exist D, E ⊆ nA ⊕ B such that nA ⊕ B = nA ⊕ D = ϕ(A) ⊕ D ⊕ E. Therefore B ∼ = D .⊕ D ⊕ E ∼ = ϕ(C) ∼  = C, as asserted. Lemma 12.1.5. A ring R satisfies the n-stable range condition if and only if for any unimodular (a1 , · · · , an+1 ), there exist x1 , · · · , xn+1 ∈ R n+1 n P P such that ai xi = 1 and Rxi = R. i=1

i=1

Proof. Suppose that R satisfies the n-stable range condition. For any unimodular (a1 , · · · , an+1 ), there exist b1 , · · · , bn ∈ R such that (a1 + an+1 b1 , · · · , an + an+1 b1 ) is unimodular. Hence, we can find some n P x1 , · · · , xn ∈ R such that (ai + an+1 bi )xi = 1. This implies that i=1

n P

Rxi = R. Further,

i=1

n+1 P

ai xi = 1, where xn+1 = bn xn . Therefore one

i=1

direction is proved. For any unimodular (a1 , · · · , an+1 ), there exist x1 , · · · , xn+1 ∈ R such n+1 n P P that ai xi = 1 and Rxi = R. So we have some r1 , · · · , rn ∈ R i=1

such that

Thus,

n P

n P

i=1

i=1

ri xi = 1. This implies that

n P

i=1

ai xi + an+1 xn+1

n P

ri xi = 1.

i=1

(ai + an+1 xn+1 ri )xi = 1, i.e., (a1 + an+1 b1 , · · · , an + an+1 bn ) is

i=1

unimodular, where each bi = xn+1 ri . Therefore R satisfies the n-stable range condition.  We now apply this lemma to obtain the following useful result due to Warfield [394, Theorem 1.6].

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Theorem 12.1.6. Let A be a right R-module, and let E = EndR (A). Then the following are equivalent: (1) E satisfies the n-stable range condition. n
L (2) If ϕ : M = Ai ⊕ H → A with each Ai ∼ = A is a splitting epii=1

morphism, then there exists a map φ : A → M such that ϕφ = 1A and n L M = φ(A) ⊕ L ⊕ H, where L ⊆ Ai . i=1

Proof. (1) ⇒ (2) Let ϕ = (ϕ1 , · · · , ϕn , f ) : M = 

n L


Ai ⊕ H → A with

i=1  ψ1  ..    each αi : A ∼ = Ai . Then there exists some ψ =  .  : A → M such that  ψn 

ϕψ = 1A . Hence,

n P

ϕi ψi + f g = 1A , and then

i=1

g n P

i=1



ϕi αi α−1 i ψi + f g =

1A . By Lemma 12.1.5, there exist x1 , · · · , xn+1 ∈ E such that n X i=1

n X
ϕi αi xi + f gxn+1 = 1A and θi xi = 1A i=1

for some θi ∈ E.

Define two maps θ : M → A given  α1 x1  ..  . −1 (θ1 α−1 1 , · · · , θn αn , 0) and φ : A → M given by φ =   αn xn

by θ = 

  . It is 

gxn+1 easy to verify that θφ = 1A , and thus, M = φ(A) ⊕ ker(θ) = φ(A) ⊕ L ⊕ H, n L where L ⊆ Ai . Moreover, we check that ϕφ = 1A , as required. i=1

(2) ⇒ (1) Let

n+1 P

ai xi = 1A in E.

Then we have a splitting R-

i=1

epimorphism ϕ = (a1 , · · · , an+1 ) : nA ⊕ A → A. By hypothesis, there   y1   exists some φ =  ...  : A → nA ⊕ A such that ϕφ = 1A and yn+1

nA ⊕ A = φ(A) ⊕ L ⊕ A, where L ⊆ nA.

Let β : nA ⊕ A = φ(A) ⊕ L ⊕ A → φ(A) ∼ = A be the projection. Then β = (b1 , · · · , bn , 0) for some bi ∈ E. In addition, nA ⊕ A = φ(A) ⊕ ker(β).

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This implies that τ := βφ ∈ AutR (A), and so Moreover,

n+1 P

n P

369

(τ −1 bi )yi = τ −1 βφ = 1A .

i=1

ai yi = 1. According to Lemma 12.1.5, E satisfies the n-stable

i=1



range condition.

Corollary 12.1.7. Let A be a right R-module, and let E = EndR (A). If E satisfies the n-stable range condition, then A⊕ B ∼ = A⊕ C with nA .⊕ B ∼ implies that B = C. Let A ⊕ B ∼ = A ⊕ C with nA .⊕ B. Then there exist some n
L right R-modules Ai ∼ Ai ⊕ B ′ . Hence, = A and B ′ such that B ∼ =

Proof.

n L

ϕ :

Ai

i=1




i=1

⊕B ⊕A ∼ = A ⊕ C. Let p : A ⊕ C → A be the projection. ′

Then there is a splitting R-epimorphism pϕ :

n L

i=1


Ai ⊕ B ′ ⊕ A → A,

where ker(pϕ) ∼ = C. By virtue of Theorem 12.1.6, there is some φ : A → n
L Ai ⊕ B ′ ⊕ A such that pϕφ = 1A and i=1

n M i=1

n M
Ai ⊕ B ′ ⊕ A = φ(A) ⊕ L ⊕ B ′ ⊕ A, where L ⊆ Ai . i=1

n L This implies that L ⊕ A ∼ Ai , and so B ∼ = L ⊕ φ(A) ∼ = = L ⊕ A ⊕ B′.

Further, we get

n L

i=1




i=1

′

Ai ⊕ B ⊕ A = φ(A) ⊕ ker(pϕ), and then

n M
C∼ Ai ) ⊕ B ′ ⊕ A /φ(A) ∼ = ker(pϕ) ∼ = ( = L ⊕ B′ ⊕ A ∼ = B. i=1

Therefore the proof is true.



In [411, Theorem 13], Wu and Xu proved the following. Theorem 12.1.8. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent: (1) E satisfies the n-stable range condition. (2) Given any right R-module decompositions A = A1 ⊕ B1 , nA = A2 ⊕ B2 with A1 ∼ = A2 , B1 .⊕ B2 . Proof. (1) ⇒ (2) Given any right R-module decompositions A = A1 ⊕ B1 , nA = A2 ⊕ B2 with A1 ∼ = A2 , then ϕ : A ⊕ B2 ∼ = nA ⊕ B1 . Thus,

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nA ⊕ B1 = ϕ(A) ⊕ ϕ(B2 ). By virtue of Proposition 12.1.2, there exist C, D ⊆ nA ⊕ B1 such that nA ⊕ B1 = C ⊕ D ⊕ B1 = C ⊕ ϕ(B2 ). Therefore B1 . ⊕ D ⊕ B1 ∼ = ϕ(B2 ) ∼ = B2 , as required. (2) ⇒ (1) For any regular f ∈ HomR (A, nA), there exists g ∈ HomR (nA, A) such that f = f gf . Hence, A = gf (A) ⊕ (1A − gf )(A), nA = f g(nA)
⊕ (1nA − f g)(nA). Clearly, ϕ : gf (A) ∼ = f g(nA) given by ϕ gf (r) = f gf (r) for any r ∈ A. By hypothesis, we get (1A − gf )(A) .⊕ (1nA − f g)(nA). Thus, we have φ : (1A − gf )(A) → (1nA − f g)(nA) and ψ : (1nA − f g)(nA) → (1A − gf )(A) such that ψφ = 1(1A −gf )(A) . Construct two maps h:

A = gf (A) ⊕ (1A − gf )(A) x+y

→ f g(nA) ⊕ (1nA − f g)(nA) = nA, 7 → ϕ(x) + φ(y);

k : nA = f g(nA) ⊕ (1nA − f g)(nA) → x+y 7→

gf (A) ⊕ (1A − gf )(A) = A, ϕ−1 (x) + ψ(y).

It is easy to verify that f = f kf and kh = 1A . Therefore we complete the proof by Proposition 12.1.2.  Corollary 12.1.9. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent: (1) E satisfies the n-stable range condition. (2) nA ⊕ B ∼ = A ⊕ C implies that B .⊕ C. Proof. (1) ⇒ (2) is clear by Corollary 12.1.4. (2) ⇒ (1) Given any right R-module decompositions A = A1 ⊕B1 , nA = A2 ⊕ B2 with A1 ∼ = A2 , then nA ⊕ B1 ∼ = A2 ⊕ B2 ⊕ B1 ∼ = A1 ⊕ B1 ⊕ B2 ∼ = ⊕ A ⊕ B2 . By hypothesis, B1 . B2 . Therefore E satisfies the n-stable range condition from Theorem 12.1.8.  Lemma 12.1.10. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies the n-stable range condition. (2) For all finitely generated projective right R-modules B and C, nR⊕B ∼ = ⊕ R ⊕ C implies that B . C. Proof. (1) ⇒ (2) is obvious from Corollary 12.1.9. (2) ⇒ (1) Given any right R-module decompositions R = A1 ⊕B1 , nR = A2 ⊕ B2 with A1 ∼ = A2 , then nR ⊕ B1 ∼ = R ⊕ B2 . Clearly, B1 and B2

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are finitely generated projective right R-modules, and then B1 .⊕ B2 . Therefore E satisfies the n-stable range condition by Theorem 12.1.8.  Lemma 12.1.11. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent: (1) E satisfies the n-stable range condition. (2) nA ⊕ B ∼ = A ⊕ C implies that there exists a right R-module D such that nA ∼ = A ⊕ D and C ∼ = B ⊕ D. Proof. (1) ⇒ (2) Given ϕ : A ⊕ C ∼ = nA ⊕ B, then nA ⊕ B = ϕ(A) ⊕ ϕ(C) with ϕ(A) ∼ = A. In view of Theorem 12.1.3, there exist D, E ⊆ nA ⊕ B such that nA ⊕ B = E ⊕ D ⊕ B = E ⊕ ϕ(C). Thus, C ∼ = ϕ(C) ∼ = B⊕D ∼D⊕E = ∼ A ⊕ D, as required. and nA = (2) ⇒ (1) nA ⊕ B ∼ = A ⊕ C implies that there exists a right R-module D such that nA ∼ = A ⊕ D and C ∼ = B ⊕ D. This implies that B .⊕ C. According to Corollary 12.1.9, we establish the result.



Theorem 12.1.12. Let R be a separative exchange ring. Then the following are equivalent: (1) R satisfies the finite stable range condition. (2) R satisfies the 2-stable range condition. (3) For any A, B ∈ F P (R), 2R ⊕ A ∼ = R ⊕ B implies that R ⊕ A ∼ = B. Proof. (1) ⇒ (3) Assume R satisfies the n-stable range condition. Given 2R ⊕ A ∼ = R ⊕ B with A, B ∈ F P (R), we have that R ⊕ nB ∼ = 2R ⊕ A ⊕ (n − 1)B ∼ = ··· ∼ = nR ⊕ (R ⊕ nA). ∼ ∼ By Lemma 12.1.11, nR = R ⊕ D and nB = (R ⊕ nA) ⊕ D for some right R-module D. Therefore, R .⊕ nB. Since R is a separative ring, by [382, Lemma 34.5], R ⊕ A ∼ = B. (3) ⇒ (2) is obvious from Lemma 12.1.10. (2) ⇒ (1) is trivial.  Let V be an infinite-dimensional vector space over a division ring D, and let R = EndD (V ). Then R is a separative exchange ring which does not satisfy the finite stable range condition. Analogously to [16, Theorem 3.2], we now extend Lemma 12.1.11 as follows. Proposition 12.1.13. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent:

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(1) E satisfies the n-stable range condition. (2) nA ⊕ B ∼ = A ⊕ C with B .⊕ A, C .⊕ nA implies that there exists a right R-module D such that nA ∼ = A ⊕ D and C ∼ = B ⊕ D. Proof. (1) ⇒ (2) is trivial by Lemma 12.1.11. (2) ⇒ (1) Let E = a1 E + · · · + an+1 E. Since A has the finite exchange property, E is an exchange ring. In view of [382, Theorem 28.7], there are orthogonal idempotents e1 , · · · , en+1 ∈ E such that e1 + · · · + en+1 = 1A and ei E ⊆ ai E. Assume ei = ai xi where xi ∈ Eei . Let fi = xi ai . Then fi2 = xi ai xi ai = xi ei ai = xi ai = fi , whence ei A ∼ = fi A for all i. Since nA ⊕ en+1 A ∼ = A ⊕ (1 − f1 )A ⊕ · · · ⊕ (1 − fn )A, we can find a right Rmodule D such that nA ∼ = A ⊕ D and (1 − f1 )A ⊕ · · · ⊕ (1 − fn )A ∼ = n L en+1 A ⊕ D. Thus there exist R-morphisms α : en+1 A → (1 − fi )A and β:

n L

i=1

(1 − fi )A → en+1 A such that βα = 1en+1 A . Let

i=1

1−fi

ti : A = fi A ⊕ (1 − fi )A →

n L

i=1 n αen+1 L

si : A = en+1 A ⊕ (1 − en+1 )A →

β

(1 − fi )A → en+1 A ֒→ A; (1 − fi )A ։ (1 − fi )A ֒→ A.

i=1

Then we have ti ∈ en+1 E(1−fi ) and si ∈ (1−fi )Een+1 such that

n P

ti s i =

i=1

en+1 . For i = 1, · · · , n, set zi = en+1 ai (1 − fi ), ci = xn+1 (ti − zi ), and n P di = si + xi − xi aj sj . Then we verify that ai + an+1 ci = ai + en+1 (ti − j=1

j6=i

zi ) = ai + ti − zi ; ai fi = ai xi ai = ei ai ; ai si = ai (1 − fi )si = (1 − ei )ai si ; (1 − ei − en+1 )ai si = ai si − zi si ; xi = fi xi and ti fi = zi fi = 0. Therefore n n n
P P P (ai + an+1 ci )di = (ai + ti − zi )si + ei − ei aj sj i=1

i=1

=

n P

j=1

ti s i +

i=1

= 1+

i=1 n P

i=1

= 1.

n P

ei +

n P

j6=i

(ai − zi )si −

i=1

(ai − zi )si −

n P

n P

e j ai si

i,j=1

i6=j

(1 − ei − en+1 )ai si

i=1



A commutative ring R satisfies the n-stable range condition if and only
if for every ideal I of R and every unimodular a1 , · · · , an over R/I, there exists a unimodular (x1 , · · · , xn ) over R such that each ai −xi ∈ I (cf. [190, Proposition]).

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It is well known that an integral domain R is a Dedekind domain if and only if every ideal in R is uniquely a product of a finite number of prime ideals. Also we know that every nonzero prime ideal of a Dedekind domain is maximal. The following example provides a class of commutative rings which satisfy the 2-stable range condition. Example 12.1.14. Every Dedekind domain satisfies the 2-stable range condition. Proof. Suppose that aR + bR + cR = R with a, b, c ∈ R. Let I = cR. If I = 0, then c = 0; hence, (a +
c)R + (b + c)R
= R. If I = R, then c ∈ U (R); hence, a + c(1 − c−1 a) R + b + c × 0 R = R. Now we assume that 0 6= I 6= R. Then there exist some maximal ideals M1 , · · · , Mk of R such that I = M1n1 · · · Mknk . By the Chinese Remainder Theorem, R/I ∼ = R/M1n1 ⊕ · · · ⊕ R/Mknk .

Clearly, Mi /Mini is a maximal ideal of R/Mini . If N/Mini is a maximal ideal of R/Mini other than Mi /Mini , then N is a maximal ideal of R other than Mi . Hence, Mi + N = 1. Write 1 = x + y for some x ∈ Mi , y ∈ N . Then xni = (1 − y)ni ∈ 1 + N , and so N = R, a contradiction. Thus, Mi /Mini has a unique maximal ideal. Consequently, each R/Mini is lo

ni ni cal. Therefore SL2 R/Mi = E2 R/Mi . Thus, SLn (R/I) = En (R/I). Clearly, a(R/I) + b(R/I) = R/I. Hence, there are some x, y ∈ R such that ax + by = 1. As in the proof of Theorem 11.3.2, we have some x′ , y ′ , a′ , b′ ∈ R such that  ′ ′ x y ∈ E2 (R) ⊆ SL2 (R), −b′ a′ where x − x′ , y − y ′ , a − a′ , b − b′ ∈ I. Hence, a′ x′ + b′ y ′ = 1, and thus ax′ + by ′ − 1 ∈ I = cR. So ax′ + by ′ + cz = 1 for some z ∈ R. Thus, (a + ca′ z)x′ + (b + cb′z)y ′ = 1, i.e., (a + ca′ z)R + (b + cb′z)R = R. Therefore R satisfies the 2-stable range condition.  Analogously, we claim that every commutative ring whose proper homomorphic images have stable range one satisfies the 2-stable range condition. This is clear from the fact that SLn (R) = En (R) if R is a commutative ring having stable range one. Example 12.1.14 also proves that every proper homomorphic image of a Dedekind domain has stable range one (cf. [314]). It is worth noting that every right or left hermitian ring satisfies the 2-stable range condition. Let R be a right hermitian ring. Suppose x1 R +

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x2 R + x3 R = R  with x1 , x2 , x3 ∈ R. Then there exists a U ∈ GL2 (R) such U 0 that (x1 , x2 , x3 ) = (0, d, x3 ) for a d ∈ R. Clearly, there are some 0 1     1 00 U 0  y1 , y2 ∈ R such that dy1 +x3 y2 = 1. Thus, (x1 , x2 , x3 ) y1 0 0  = 0 1 y2 0 0     1 00  −1 U 0  U 0 (1, 0, 0). Set V = . Then (x1 , x2 , x3 )V = y1 0 0  0 1 0 1 y2 0 0    −1  1 00 U 0 (1, 0, 0) , where V =  z1 0 0  for some z1 , z2 ∈ R. Hence, 0 1 z2 0 0 (x1 + x2 z1 + x3 z2 , 0) is the first row of the invertible U −1 . Thus, we can find some t ∈ R such that (x1 + x2 z1 + x3 z2 )t = 1, and so (x1 + x2 z1 )t+ (x3 + x2 · 0)(z2 t) = 1. Therefore R satisfies the 2-stable range condition, and we are through. It is an open question whether regular rings satisfying the 2-stable range condition are right hermitian. But we record that every right ℵ0 continuous regular ring satisfying the 2-stable range condition is hermitian ([5, Theorem 6]). Rings satisfying the higher stable range condition are abundant. Let R = R/(x2 + y 2 + z 2 − 1). Then R does not satisfy the 2-stable range condition, but it satisfies the 3-stable range condition (cf. [313, Proposition 5.4]. Proposition 12.1.15. Let A be a right R-module. If EndR (A) satisfies the n-stable range condition, then M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = mA ∼ = A2 (m ≥ n) implies that M = C ⊕ D ⊕ H ⊕ B1 = C ⊕ D ⊕ K ⊕ B2 ,

where D ∼ = A, D ⊕ H ∼ = nA ∼ = D ⊕ K.

Proof. Suppose M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = mA ∼ = A2 (m ≥ n). If m = n, then M = A1 ⊕ B1 = A21 ⊕ (A22 ⊕ B2 ), where A21 ∼ = A, A22 ∼ = (n − 1)A. By virtue of Theorem 12.1.3, there are some D, H ⊆ M such that M = D ⊕ H ⊕ B1 = D ⊕ (A22 ⊕ B2 ). Choose C = 0, K = A22 . Then M = C ⊕ D ⊕ H ⊕ B1 = C ⊕ D ⊕ K ⊕ B2 ,

where D ∼ = A21 ∼ = A, D ⊕ H ∼ = A1 ∼ = nA ∼ = A ⊕ (n − 1)A ∼ = D ⊕ K.

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Assume that the result holds for m(m ≥ n). Given M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = (m + 1)A ∼ = A2 , then M = A11 ⊕ (A12 ⊕ B1 ) = A21 ⊕ (A22 ⊕ B2 ) with A11 ∼ = mA ∼ = A21 and A12 ∼ =A∼ = A22 . By hypothesis, M = C ′ ⊕ D′ ⊕ H ′ ⊕ (A12 ⊕ B1 ) = C ′ ⊕ D′ ⊕ K ′ ⊕ (A22 ⊕ B2 ), where D′ ∼ = A, D′ ⊕ H ′ ∼ = nA ∼ = D′ ⊕ K ′ . This means that M = (H ′ ⊕ A12 ) ⊕ (C ′ ⊕ D′ ⊕ B1 ) = (K ′ ⊕ A22 ) ⊕ (C ′ ⊕ D′ ⊕ B2 ), where H ′ ⊕ A12 ∼ = nA ∼ = K ′ ⊕ A22 . By the discussion above, M = C ′′ ⊕ D′′ ⊕ H ′′ ⊕ (C ′ ⊕ D′ ⊕ B1 ) = C ′′ ⊕ D′′ ⊕ K ′′ ⊕ (C ′ ⊕ D′ ⊕ B2 ), where D′′ ∼ = A, D′′ ⊕ H ′′ ∼ = nA ∼ = D′′ ⊕ K ′′ . Let C = C ′′ ⊕ C ′ ⊕ D′ . Then M = C ⊕ D′′ ⊕ H ′′ ⊕ B1 = C ⊕ D′′ ⊕ K ′′ ⊕ B2 ; thus the result holds for m + 1. By induction, we obtain the result.



Corollary 12.1.16. Let R satisfy the n-stable range condition. If Mn (R) is a generalized stable ring, then so is Mm (R) for all m ≥ n. Proof. Suppose Mn (R) is a generalized stable ring. Let m ≥ n. Given M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = mR ∼ = A2 , it follows from Proposition 12.1.15 that M = C ⊕ D ⊕ H ⊕ B1 = C ⊕ D ⊕ K ⊕ B2 , D∼ = R and D ⊕ H ∼ = nR ∼ = D ⊕ K. In view of Theorem 7.1.5, we derive M = E ⊕ F ⊕ (C ⊕ B1 ) = E ⊕ G ⊕ (C ⊕ B2 ), where E ∼ = nR. This means that M = (C ⊕ E) ⊕ F ⊕ B1 = (C ⊕ E) ⊕ G ⊕ B2 . Clearly, C ⊕ E ∼ = C ⊕ nR ∼ = C⊕D⊕H ∼ = mR. According to Theorem 7.1.5, Mm (R) is a generalized stable ring.  Corollary 12.1.17. If R satisfies the n-stable range condition, then the following are equivalent: (1) R satisfies stable power-substitution. (2) Mn (R) satisfies power-substitution.

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Proof. (1) ⇒ (2) is trivial. (2) ⇒ (1) Suppose Mn (R) satisfies power-substitution. Then we see that Mm (R)(1 ≤ m ≤ n) satisfies power-substitution by Proposition 10.1.5. So we may assume that m ≥ n. Given M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = mR ∼ = A2 , it follows from Proposition 12.1.15 that M = C ⊕ D ⊕ H ⊕ B1 = C ⊕ D ⊕ K ⊕ B2 , D∼ = R and D ⊕ H ∼ = nR ∼ = D ⊕ K. As Mn (R) satisfies power-substitution, by Theorem 10.1.7, there exists some s ∈ N such that sM = L ⊕ s(C ⊕ B1 ) = L ⊕ s(C ⊕ B2 ), where L ⊆ sM . Let X = L ⊕ sC. Then sM = X ⊕ sB1 = X ⊕ sB2 . By Theorem 10.1.7 again, Mm (R) satisfies power-substitution, and the result follows.  12.2

Endomorphism Rings

The main purpose of this section is to show that the n-stable range condition can be inherited by general Morita contexts. This means that the n-stable range condition over Morita contexts are not related to their map pairings. Extensions of strongly separative exchange rings are studied as well. Theorem 12.2.1. Let P and Q be right R-modules. If EndR (P ) and EndR (Q) satisfy the n-stable range condition, then so does EndR (P ⊕ Q). Proof. Given any R-module decompositions M = A1 ⊕B1 = A2 ⊕B2 , where A1 ∼ = n(P ⊕ Q) and A2 ∼ = P ⊕ Q, there exist A11 ∼ = nP, A12 ∼ = nQ, A21 ∼ =P ∼ and A22 = Q such that A1 = A11 ⊕ A12 and A2 = A21 ⊕ A22 . Thus M = A11 ⊕ (A12 ⊕ B1 ) = A21 ⊕ (A22 ⊕ B2 ) with A11 ∼ = nP and A21 ∼ = P. By Theorem 12.1.3, we can find C, D ⊆ M such that M = C ⊕ D ⊕ (A12 ⊕ B1 ) = C ⊕ (A22 ⊕ B2 ). So M = A12 ⊕ (C ⊕ D ⊕ B1 ) = A22 ⊕ (C ⊕ B2 ) with A12 ∼ = nQ and A22 ∼ = Q. By Theorem 12.1.3 again, we can find E, F ⊆ M such that M = E ⊕ F ⊕ (C ⊕ D ⊕ B1 ) = E ⊕ (C ⊕ B2 ). Therefore M = (C ⊕ E) ⊕ (D ⊕ F ) ⊕ B1 = (C ⊕ E) ⊕ B2 . According to Theorem 12.1.3, EndR (P ⊕ Q) satisfies the n-stable range condition.  Corollary 12.2.2. Let e ∈ R be an idempotent. If eRe and (1 − e)R(1 − e) satisfy the n-stable range condition, then so does R.

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Proof. Since R ∼ = eR ⊕ (1 − e)R as right R-modules, by Theorem 12.2.1, we see that R satisfies the n-stable range condition because EndR (eR) ∼ = eRe and EndR ((1 − e)R) ∼  = (1 − e)R(1 − e). Corollary 12.2.3. Let T be the ring of a Morita context (A, B, M, N, ψ, φ). If A and B satisfy the n-stable range condition, then so does T . ∼ eT e and B ∼ Proof. Set e = diag(1A , 0). Then A = = (1T − e)T (1T − e). So eT e and (1T − e)T (1T − e) both satisfy the n-stable range condition. Thus the result follows by Corollary 12.2.2.  If R satisfies the n-stable range condition, by induction, then so does Mn (R) for all n ∈ N. Proposition 12.2.4. The following are equivalent: (1) A1 , A2 and A3 satisfy the n-stable range  condition.  A1 0 0 (2) The formal triangular matrix ring T =  M21 A2 0  satisfies the M31 M32 A3 n-stable range condition.     A2 0 M21 Proof. (1) ⇒ (2) Set B = and M = . Since A2 and A3 M32 A3 M31 satisfy the n-stable range condition, by Corollary 12.2.3, so does the ring B. On the other hand,A1 satisfies the n-stable range condition. Using  A1 0 Corollary 12.2.3 again, satisfies the n-stable range condition, as M B required. the n-stable  (2) ⇒ (1) SupposeT satisfies   range  condition. Set B = A2 0 M21 A1 0 and M = . Then T = . For any unimodular M32 A3 M31 M B
(a1 , · · · , an+1 ) ∈ An+1 , we see that diag(a1 , 1B ), · · · , diag(an+1 , 1B ) ∈ 1  ′  ai 0 T n+1 is unimodular. By hypothesis, there are some c′i b′i  ′  a1 0 , · · · , diag(an , 1B ) + such that diag(a1 , 1B ) + diag(an+1 , 1B ) c′1 b′1  ′  an 0
diag(an+1 , 1B ) ∈ T n is unimodular. Thus we have c′n b′n     x1 0 xn 0 ,··· , ∈T s1 y 1 sn y n

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such  that      a1 + an+1 a′1 0 x1 0 an + an+1 a′n 0 xn 0 + ··· + c′1 1 + b′1 s1 y 1 c′n 1 + b′n sn y n = diag(1A , 1B ). It follows that (a1 + an+1 a′1 )x1 + · · · + (an + an+1 a′n )xn = 1A ; hence (a1 + an+1 a′1 , · · · , an + an+1 a′n ) ∈ An1 is unimodular. Therefore A1 satisfies the n-stable range condition. Likewise, A2 and A3 satisfy the n-stable range condition. So the proposition is true.  Corollary 12.2.5. A ring R satisfies the n-stable range condition if and only if the ring of all n × n lower triangular matrices over R also satisfies the n-stable range condition. Proof. By induction and Proposition 12.2.4, we easily obtain the result.  Analogously, we deduce that a ring R satisfies the n-stable range condition if and only if the ring of all n × n upper triangular matrices over R also satisfies the n-stable range condition. Now we observe the following simple fact. Example 12.2.6. A ring R satisfies the n-stable range condition if and only if the ring R[x]/(xn+1 ) also satisfies the n-stable range condition. Proof. Clearly, R[x]/(xn+1 ) = R[u] = {a0 + a1 u + · · · + an un | n+1 each ai ∈ R, = 0 and u commutates with elements of R}. Further,
u ∼  R[u]/J R[u] = R/J(R); so the result follows.

Ara et al. [16] extended separative cancellation of modules to strong separativity. A ring R is a strongly separative ring provided that A ⊕ A ∼ = A⊕B implies A ∼ = B for any finitely generated projective right R-modules A and B. Many exchange rings are strongly separative, for instance, directly finite exchange rings satisfying the s-comparability (cf. [354, Theorem 2.3]). For future use, we now record the following. Lemma 12.2.7. Let R be an exchange ring. Then the following are equivalent: (1) (2) (3) (4)

R is strongly separative. For all A, B ∈ F P (R), 2A ∼ = A ⊕ B ⇒ A .⊕ B. For all A, B, C ∈ F P (R), A ⊕ C ∼ = B ⊕ C with C .⊕ A ⇒ A .⊕ B. For all A, B, C ∈ F P (R), A ⊕ C ∼ = B ⊕ C with C .⊕ A ⇒ ∃X such ∼ ∼ that C = C ⊕ X and A ⊕ X = B.

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Proof. (1) ⇒ (2) is trivial. (2) ⇒ (3) Given A ⊕ C ∼ = B ⊕ C with A, B, C ∈ F P (R), C .⊕ A, then A∼ = C ⊕ D for a right R-module D. Thus, 2A ∼ = A⊕C ⊕D ∼ = B ⊕C ⊕D ∼ = ⊕ A ⊕ B. It follows that A . B, as required. (3) ⇒ (4) Suppose A, B, C ∈ F P (R) such that A ⊕ C ∼ = B ⊕ C with C .⊕ A. As in the proof of Lemma 6.3.7, we have a refinement matrix

A C



BC  D1 A1 B1 C1

such that C1 .⊕ A1 . Assume that A1 ∼ = C1 ⊕ E for a right R-module E. Then 2A1 ∼ = A1 ⊕ C1 ⊕ E ∼ = C⊕E ∼ = B1 ⊕ C1 ⊕ E ∼ = B1 ⊕ A1 ; hence ⊕ A1 . B1 . Thus we can find a right R-module X such that B1 ∼ = A1 ⊕ X. It follows from A1 ⊕ C1 ∼ =C ∼ = B1 ⊕ C1 that C ∼ = C ⊕ X. On the other hand, we see that B ∼ = D 1 ⊕ B1 ∼ = D1 ⊕ A1 ⊕ X ∼ = A ⊕ X, as required. (4) ⇒ (1) Suppose A, B ∈ F P (R) such that A ⊕ A ∼ = A ⊕ B. As in the proof of Lemma 6.3.7, we have a refinement matrix

A A



BA  D1 A1 B1 C1

such that C1 .⊕ A1 . Thus, A1 ⊕ C1 ∼ =A∼ = B1 ⊕ C1 . By hypothesis, there exists some right R-module X such that C1 ∼ = C1 ⊕ X and A1 ⊕ X1 ∼ = B1 ; ⊕ ∼ whence, A1 . B1 . Assume that B1 = A1 ⊕ E for a right R-module E. Then A ∼ = C1 ⊕ B1 ∼ = C1 ⊕ A1 ⊕ E ∼ = A ⊕ E. Therefore we conclude that ∼ ∼ ∼ A∼ A ⊕ E D ⊕ A ⊕ E D ⊕ B  = = 1 = 1 1 1 = B, the result follows. Theorem 12.2.8. Let R be an exchange ring. Then the following are equivalent: (1) R is strongly separative. (2) All corners of R satisfy the 2-stable range condition. (3) All corners of R are strongly separative. Proof. (1) ⇒ (2) Let e ∈ R be an idempotent. Given 2(eR) ⊕ B ∼ = eR ⊕ C with B, C ∈ F P (R), then eR ⊕ B ∼ = C. This implies that B .⊕ C. In view of Corollary 12.1.9, eRe satisfies the 2-stable range condition, as desired. (2) ⇒ (1) Suppose A ⊕ C ∼ = B ⊕ C with C .⊕ A and A, B, C ∈ F P (R). As in the proof of Lemma 6.3.7, we have a refinement matrix

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A C



BC  D1 A1 B1 C1

such that C1 .⊕ A1 . Thus 2C1 .⊕ C1 ⊕ A1 ∼ = C .⊕ A, so A ∼ = 2C1 ⊕ E for a right R-module E. It is easy to check that A ⊕ C1 ∼ = D1 ⊕ A1 ⊕ C1 ∼ = D1 ⊕ C ∼ = D1 ⊕ B1 ⊕ C1 ∼ = B ⊕ C1 . Therefore 3C1 ⊕ E ∼ = C1 ⊕ B. Clearly, C1 is a finitely generated projective right R-module. Since all corners of R satisfy the 2-stable range condition, so does EndR (C1 ). Hence 2C1 ⊕E ∼ =B by Corollary 12.1.7. That is, A ∼ = B. According to Lemma 12.2.7, R is strongly separative. (2) ⇒ (3) Let e ∈ R be an idempotent. For any ef e = (ef e)2 ∈ eRe, ef e(eRe)ef e = (ef e)R(ef e) satisfies the 2-stable range condition. By (1) ⇔ (2), eRe is strongly separative, as required. (3) ⇒ (1) is trivial.  Corollary 12.2.9. Let R be an exchange ring. Then the following are equivalent: (1) R is strongly separative. (2) T Mn(R) is strongly separative. (3) The ring of eventually constant sequence {(a1 , · · · , an , b, b, · · · ) | ai , b ∈ R}, where n ∈ N depends on the element (a1 , · · · , an , b, b, · · · ), is strongly separative. Proof. (1) ⇒ (2) We will prove by induction. If n = 1, the result is obvious. Assume that the result holds for n. For any e = e2 ∈T Mn+1  (R), f ∗ there exist f = f 2 ∈ T Mn(R), g = g 2 ∈ R such that e = . By 0g hypothesis, T Mn (R) is strongly separative; hence, f T Mn(R)f satisfies the 2-stable range condition from Theorem 12.2.8. In addition, gRg satisfies ∼ the 2-stable range   condition. According to Corollary 12.2.3, eT Mn+1(R)e = f T Mn f ∗ satisfies the 2-stable range condition. Therefore T Mn (R) 0 gRg is strongly separative by Theorem 12.2.8. (2) ⇒ (1) Let e = diag(1, 0, · · · , 0) ∈ T Mn (R). Then R ∼ = eT Mn(R)e. According to Theorem 12.2.8, R is strongly separative. (1) ⇒ (2) Let S = {(a1 , · · · , an , b, b, · · · ) | ai , b ∈ R}, and let f = (e1 , · · · , em , em+1 , em+2 , · · · ) ∈ S be an idempotent, where em+1 = em+2 = · · · . It will suffice to show that f Sf satisfies the 2-stable range condition by Theorem 12.2.8. Given (ai )(xi ) + (bi )(yi ) + (ci )(zi ) = f in f Sf ,

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then ai xi + bi yi + ci zi = ei in ei Rei . Choose some p ∈ N such that ai+1 = ai+2 = · · · , xi+1 = xi+2 = · · · , bi+1 = bi+2 = · · · , yi+1 = yi+2 = · · · , ci+1 = ci+2 = · · · , zi+1 = zi+2 = · · · for all i ≥ p. By Theorem 12.2.8, all ei Rei satisfies the 2-stable range condition. Thus, we can find some r1i , r2i ∈ R such that (ai + ci r1i , bi + ci r2i ) ∈ (ei Rei )2 is unimodular, where r1(i+1) = r1(i+2) = · · · , r2(i+1) = r2(i+2) = · · · for all
i ≥ p. Choose r1 = (r1i ), r2 = (r2i ) ∈ f Sf . Then (ai )+(ci )r1 , (bi )+ (ci )r2 ∈ (f Sf )2 is unimodular. Therefore f Sf satisfies the 2-stable range condition, as desired. (3) ⇒ (1) Choose e = (1, 0, 0, · · · ). Then R ∼ = eSe. In view of Theorem 12.2.8, R is strongly separative.  Theorem 12.2.10. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent: (1) E is strongly separative. (2) For all e = e2 ∈ E, eA = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = 2A2 , then A2 ⊕ B1 ∼ = B2 . Proof. (1) ⇒ (2) Let e ∈ E be an idempotent. Given any right R-module decompositions eA = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = 2A2 , then there exists ∼ an idempotent f ∈ E such that A2 = f A. Hence, eA ∼ = 2(f A) ⊕ B1 ∼ = f A⊕B2 . In view of [382, Lemma 28.1], f A has the finite exchange property. As in the proof of Lemma 6.3.7, we have a refinement matrix

f A ⊕ B1 fA



B2 f A  D1 A1 B1 C1

such that C1 .⊕ A1 . Thus 2C1 .⊕ C1 ⊕ A1 ∼ = f A .⊕ f A ⊕ B1 , so f A ⊕ B1 ∼ = 2C1 ⊕ E for a right R-module E. It is easy to check that f A ⊕ B1 ⊕ C1 ∼ = D1 ⊕ A1 ⊕ C1 ∼ = D1 ⊕ f A ∼ = D1 ⊕ B1 ⊕ C1 ∼ = B2 ⊕ C1 . ∼ Therefore 3C1 ⊕ E = C1 ⊕ B2 . Clearly, C1 = gA for an idempotent g ∈ E. Thus, EndR (C1 ) ∼ = gEg. By virtue of Theorem 12.2.8, EndR (C1 ) satisfies the 2-stable range condition. Hence 2C1 ⊕ E ∼ = B2 by Corollary 12.1.7. That is, A2 ⊕ B1 ∼ = f A ⊕ B1 ∼ = 2C1 ⊕ E ∼ = B2 , as required. (2) ⇒ (1) Let e ∈ E be an idempotent. Given eA = A1 ⊕ B1 , 2(eA) = A1 ⊕ B2 with A1 ∼ = A2 , then eA ⊕ (eA ⊕ B1 ) ∼ = A2 ⊕ B2 ⊕ B1 ∼ = eA ⊕ B2 . As A has the finite exchange property, so has eA. As in the proof of Lemma

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6.3.7, we have a refinement matrix eA ⊕ B1 eA



B2 eA  D1 A1 B1 C1

such that C1 .⊕ A1 . Assume that A1 ∼ = C1 ⊕ D. Hence, eA ∼ = 2C1 ⊕ D ∼ = ∼ C1 ⊕ B1 . By hypothesis, C1 ⊕ D = B1 . This implies that A1 ∼ = B1 . It follows that B1 .⊕ eA ⊕ B1 ∼ = D1 ⊕ A1 ∼ = D 1 ⊕ B1 ∼ = B2 . According to Theorem 12.1.8, eEe satisfies the 2-stable range condition. Therefore we complete the proof by Theorem 12.2.8.  Corollary 12.2.11. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent: (1) E is strongly separative. (2) For all idempotents e ∈ E, eA ⊕ B ∼ = eA ⊕ C with eA .⊕ B implies ∼ that B = C. (3) For all idempotents e, f ∈ E, eA⊕eA ∼ = eA⊕f A implies that eA ∼ = f A. Proof. (1) ⇒ (2) Let e ∈ E be an idempotent. Given any right R-module decompositions eA⊕ B ∼ = eA⊕ C with eA .⊕ B, then we have a refinement matrix B eA



C eA  E1 B1 C1 D1

such that D1 .⊕ B1 . Assume that B1 ∼ = D1 ⊕ F . Then eA ∼ = 2D1 ⊕ F ∼ = ∼ D1 ⊕ C1 . By virtue of Theorem 12.2.10, B1 ∼ D ⊕ F C . Therefore = 1 = 1 B∼ = E1 ⊕ B1 ∼ = E1 ⊕ C1 ∼ = C, as desired. (2) ⇒ (3) is trivial. (3) ⇒ (1) Let e ∈ E be an idempotent, and let eA = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = 2A2 . It is easy to see that A2 ⊕ B1 . A1 ⊕ B1 ∼ = eA, whence, 2 ∼ there is an idempotent g = g ∈ E such that A2 ⊕ B1 = gA. In addition, (A2 ⊕B1 )⊕(A2 ⊕B1 ) ∼ = (A1 ⊕B1 )⊕B1 ∼ = (A2 ⊕B1 )⊕B2 . Clearly, B2 = f A for an idempotent f ∈ E. By hypothesis, A2 ⊕ B1 ∼ = B2 . Therefore the result follows from Theorem 12.2.10.  Chen observed that a regular ring R is separative if and only if for all idempotents e, f ∈ R, eR ⊕ eR ∼ = eR ⊕ f R ∼ = f R ⊕ f R =⇒ eR ∼ = f R (cf. [139, Corollary 4.6]). But that method depends on regularity. It follows by Corollary 12.2.11 that an exchange ring R is strongly separative if and only

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if for all idempotents e, f ∈ R, eR ⊕ eR ∼ = eR ⊕ f R implies that eR ∼ = f R. Further, we claim that strong separativity over exchange rings is Morita invariant. Let R be a strongly separative exchange ring, let e = e2 ∈ R, and let n ∈ N. In view of Theorem 12.2.8 and [382, Theorem 29.2], eRe is a strongly separative exchange ring. For all idempotents f, g ∈ Mn (R), f (nR)⊕f (nR) ∼ = f (nR)⊕g(nR) implies that f (nR) ∼ = g(nR). According to Corollary 12.2.11 and [382, Theorem 28.7], Mn (R) is a strongly separative exchange ring, and we are done. Let A and B be right R-modules having the finite exchange property. If EndR (A) and EndR (B) are strongly separative, we know, from Corollary 12.2.11, that for any right R-module C, 2(A ⊕ B) ∼ = (A ⊕ B) ⊕ C implies that A ⊕ B ∼ = C. Corollary 12.2.12. Let A be a right R-module having the finite exchange property, and let E = EndR (A). Then the following are equivalent: (1) E is strongly separative. (2) For all idempotents e ∈ E, 2(eA) = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = 2A2 , ⊕ we have B1 . B2 . ∼ Proof. (1) ⇒ (2) Let 2(eA) = A1 ⊕ B1 = A2 ⊕ B2 , where e = e2 ∈ E, A1 = 2A2 . Then EndR (A2 ) is strongly separative from the preceding remark. In view of Corollary 12.2.11, A2 ⊕ B1 ∼ = B2 . Therefore B1 .⊕ B1 . (2) ⇒ (1) Let eA = A1 ⊕ B1 , 2(eA) = A2 ⊕ B2 , where e = e2 ∈ E. Then 2(eA) = 2A1 ⊕ 2B1 = A2 ⊕ B2 with 2A1 ∼ = 2A2 . By hypothesis, 2B1 .⊕ B2 . We infer that B1 .⊕ B2 . According to Theorem 12.1.8, eEe ∼ = EndR (eA) satisfies the 2-stable range condition. So the proof is true from Theorem 12.2.8.  12.3

Analogue of Stable Range One

The aim of this section is to investigate the necessary and sufficient conditions on an exchange ring R under which it satisfies the n-stable range condition. As an application, we observe that every separative simple exchange ring satisfying the finite stable range has stable range one. Lemma 12.3.1. Let R be an exchange ring. Then the following are equivalent:

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(1) R satisfies the n-stable range condition. (2) For all idempotents e ∈ R, f ∈ Mn (R), eR ∼ = f (nR) implies that (1 − e)R .⊕ (In − f )(nR). (3) For any maximal right ideal J of R and any idempotents e ∈ 1 + J, f ∈ Mn (R), eR ∼ = f (nR) implies that (1 − e)R .⊕ (In − f )(nR).

Proof. (1) ⇒ (2) Let eR ∼ = f (nR) with idempotents e ∈ R, f ∈ Mn (R). Then R = eR ⊕ (1 − e)R and nR = f (nR) ⊕ (In − f )(nR). By Corollary 12.1.9, (1 − e)R .⊕ (In − f )(nR). (2) ⇒ (3) is trivial. (3) ⇒ (1) Suppose that ab + c = 1, where a ∈ Rn , b ∈ n R, c ∈ R. Since R is an exchange ring, we can find an idempotent e ∈ R such that e = cs and 1 − e = (1 − c)t for some s, t ∈ R. Hence, (1 − e)abt + e = 1. Set x = (1 − e)a and y = bt. Then xyx = x and xy + e = 1.
If eR = R, then er = 1 for some r ∈ R. Hence, a + csr (1, 0, · · · , 0) − a = (1, 0, · · · , 0) ∈ Rn is unimodular. If eR 6= R, then there exists a maximal right ideal J of R such that eR ⊆ J $ R. Hence, xy ∈ 1 + J. Let z = yxy. Then x = xzx and z = zxz. In addition, xz = xy ∈ 1 + J. Obviously, xz ∈ R, zx ∈ Mn (R) are idempotents, and xzR ∼ = zx(nR). By hypothesis, we have (1 − xz)R .⊕ (In − zx)(nR). In view of Lemma 6.4.16, we can find v ∈ (1 − xz)Rn (In − zx) and w ∈ (In − zx) n R(1 − xz) such that 1 − xz = vw. Clearly, xw = vz = 0; hence, (x + v)(z + w) = xz + vw = 1. That is, x+ v ∈ Rn is unimodular. Write v = (1 − xz)d, where d ∈ R. Then a+cs(d−a) = a+e(d−a) = (1−e)a+ed = x+ed = x+(1−xy)d = x+v ∈ Rn is unimodular, as required.  Proposition 12.3.2. Let J be an ideal of an exchange ring R. Then the following are equivalent: (1) R satisfies the n-stable range condition. (2) R/J satisfies the n-stable range condition and (1 − e)R ∼ = (In − f )(nR) ⊕ with idempotents e ∈ J, f ∈ Mn (R) implies that eR . f (nR). Proof. (1) ⇒ (2) Clearly, R/J satisfies the n-stable range condition. Thus the result follows by Lemma 12.3.1. 2 ∼ (2) ⇒ (1) Let g = g 2 ∈ R and
h = h ∈ Mn (R) such that gR = ∼ h(nR). Then g(R/J) = h n(R/J) . Since R/J satisfies the n-stable range
condition, by Lemma 12.3.1, (1 − g)(R/J) .⊕ (In − h) n(R/J) . In view of Lemma 1.4.11, we have right R-module decompositions (1 − g)R = A1 ⊕ A2 and (In − h)(nR) = B1 ⊕ B2 with A1 ∼ = B1 and A2 = A2 J. Clearly, there exists an idempotent e ∈ R such that eR = A2 , and then e ∈ J. Since

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R = (1 − e)R ⊕ eR = gR ⊕ A1 ⊕ eR, we see that (1 − e)R ∼ = gR ⊕ A1 . Similarly, we have some f = f 2 ∈ Mn (R) such that f (nR) = B2 . Thus (In − f )(nR) = h(nR) ⊕ B1 . Hence, we obtain (1 − e)R ∼ = (In − f )(nR) 2 2 with e = e ∈ J and f = f ∈ Mn (R). This yields that eR .⊕ f (nR), and so (1 − g)R = A1 ⊕ A2 .⊕ B1 ⊕ B2 = (In − h)(nR). According to Lemma 12.3.1, R satisfies the n-stable range condition.  Corollary 12.3.3. Let J be an ideal of an exchange ring R. Then the following are equivalent: (1) R has stable range one. (2) R/J has stable range one and (1 − e)R ∼ = (1 − f )R with idempotents ∼ e ∈ J, f R implies that eR = f R. Proof. (1) ⇒ (2) is trivial from Theorem 1.3.2. (2) ⇒ (1) is proved by Proposition 12.3.2.



Following Ara (cf. [9]), an ideal I of a ring R is said to be an exchange ideal provided that for every x ∈ I there exists an idempotent e ∈ I and elements r, s ∈ I such that e = xr = x + s − xs. Lemma 12.3.4. Let J be an ideal of a ring S, and let R be a subring of S which contains J. If R/J and S are exchange rings satisfying the n-stable range condition, then so is R. Proof. Since S is an exchange ring, so is J by [2, Theorem 2.2]. Assume that e + J = (e + J)2 ∈ R/J with e ∈ R. Since J ⊆ R ⊆ S, we know that e + J = (e + J)2 ∈ S/J. So we can find some f = f 2 ∈ S such that e − f ∈ J. Therefore e + J = f + J and f = f 2 ∈ e + J ⊆ R. That is, idempotents can be lifted modulo J. In view of [2, Theorem 2.2], we see that R is an exchange ring because R/J and J are both exchange rings. If e = e2 ∈ J and f = f 2 ∈ Mn (R) such that (1 − e)R ∼ = (In − f )(nR), by Lemma 6.4.16, we can find a ∈ M1×n (R) and b ∈ Mn×1 (R) such that 1 − e = ab, In − f = ba, a = (1 − e)a(In − f ) and b = (In − f )b(1 − e). Clearly, a ∈ M1×n (S) and b ∈ Mn×1 (S). By Lemma 6.4.16 again, we see that (1 − e)S ∼ = (In − f )(nS). Since S is an exchange ring satisfying the n-stable range condition, by Lemma 12.3.1, we have eS .⊕ f (nS). From Lemma 6.4.16, there are c ∈ M1×n (S) and d ∈ Mn×1 (S) such that e = cd, c = ecf and d = f de. Since e ∈ J, we see that c ∈ M1×n (R) and d ∈ Mn×1 (R). Hence eR .⊕ f (nR) by Lemma 6.4.16. It follows from Proposition 12.3.2 that R satisfies the n-stable range condition. 

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Theorem 12.3.5. Every finite subdirect product of exchange rings satisfying the n-stable range condition is also an exchange ring satisfying the n-stable range condition. Proof. Let R be the subdirect product of two exchange rings R1 and R2 , where R1 and R2 satisfy the n-stable range condition. It suffices to show that R is an exchange ring satisfying the n-stable range condition. Set J = {(r, r2 ) ∈ R | r2 = 0}. Then J is an ideal of R1 × R2 . Clearly, R1 × R2 is an exchange ring satisfying the n-stable range condition and J ⊆ R. Since R/J ∼ = R1 is an exchange ring satisfying the n-stable range condition, by Lemma 12.3.4, R is an exchange ring satisfying the n-stable range condition, as asserted.  Corollary 12.3.6. Let J be an ideal of an exchange ring R. Then the following are equivalent: (1) R satisfies the n-stable range condition. (2) R/J and R/r(J) both satisfy the n-stable range condition. (3) R/J and R/ℓ(J) both satisfy the n-stable range condition. Proof. (1)⇒(2) is trivial. (2)⇒(1) As in the proof of Corollary 1.1.4, R/(J ∩r(J)) is isomorphic to the subdirect product of R/J and R/r(J). Thus it follows from Theorem 12.3.5 that R/(J ∩ r(J)) is an exchange ring satisfying the n-stable range condition. On the other hand,
2 J ∩ r(J) ⊆ Jr(J) = 0 ⇒ J ∩ r(J) ⊆ J(R), where J(R) denotes the Jacobson radical of R. Therefore R satisfies the n-stable range condition. (1)⇔(3) is proved by the symmetry of the n-stable range condition for exchange rings.  Proposition 12.3.7. Suppose that R is a subring of S with nS ⊆ R ⊆ S for some 0 6= n ∈ Z and that S/nS is finite. If S is an exchange ring satisfying the n-stable range condition, then so is R. Proof. Since R/nS is a subring of S/nS, we see that R/nS is a finite ring, hence it is artinian. So R/nS is an exchange ring having stable range one. Since S is an exchange ring satisfying the n-stable range condition, so is R by Lemma 12.3.4.  Recall that an abelian group A is said to have finite rank provided there exists a positive integer n such that every finitely generated subgroup of A

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can be generated by n elements. Note that if A has finite rank, then each of the primary components of T (A) has DCC on subgroups, where T (A) is the torsion subgroup of A. We say a ring R is a finite rank Z-algebra if its additive group has finite rank. It is well known that if A is a torsionfree abelian group of finite rank, then End(A) satisfies the 2-stable range condition (cf. [394, Theorem 5.6]). Now we extend this result to exchange finite rank Z-algebras as follows. Proposition 12.3.8. Every exchange finite rank Z-algebra satisfies the 2-stable range condition. Proof. Let R be an exchange finite rank Z-algebra. It suffices to show that R/J(R) satisfies the 2-stable range condition. So we may assume that R is semiprime. Let T be the torsion subgroup of R and Tp be the p-primary component of R for all prime integer p. Then T and Tp are ideals of R. Since R is semiprime, analogously to [215, Theorem 4.12], we claim that Tp = ep R for L some central idempotent ep . Furthermore, we have T = ep R.
Set S = (R/T ) × Πep R . Since R has finite rank over Z, Tp must have DDC on subgroups, hence Tp is an artinian ring. So we see that Πep R has stable range one. On the other hand, R/T is an exchange ring which is a torsion-free finite rank Z-algebra. In the proof of [394, Theorem 5.6], Warfield showed that R/T satisfies the 2-stable range condition. Therefore S is an exchange ring satisfying the 2-stable range condition.
T T Since T (1 − ep )R = 0, we
have an injective ring map ψ : R → S. L Clearly, ψ(T ) = {0} × ep R is an ideal of S. As ψ(R)/ψ(T ) ∼ = R/T is an exchange ring satisfies the 2-stable range condition, by Lemma 12.3.4, R∼ = ψ(R) is an exchange ring satisfying the 2-stable range condition, as asserted.  Theorem 12.3.9. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies the n-stable range condition. (2) For all idempotents e ∈ R and f ∈ Mn (R), eR ∼ = f (nR) implies that n there exists a unimodular u ∈ R such that eu = uf . (3) For any maximal right ideal J of R and any idempotents e ∈ 1 + J and f ∈ Mn (R), eR ∼ = f (nR) implies that there exists a unimodular u ∈ Rn such that eu = uf . Proof. (1) ⇒ (2) Let ψ : eR ∼ = f (nR) with idempotents e ∈ R, f ∈ Mn (R).

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Then R = eR ⊕ (1 − e)R and nR = f (nR) ⊕ (In − f )(nR). Let p1 : R → eR ֒→ R, p2 : nR → f (nR) ֒→ nR be the projections. Let x : R → nR given by x(a+b) = ψ(a) for any a ∈ eR and b ∈ (1−e)R. Let y : nR → R given by y(c + d) = ψ −1 (c) for any c ∈ f (nR) and d ∈ (In − f )(nR). We easily check that p1 = yp2 x, p2 = xp1 y, x = xyx and y = yxy. In view of Proposition 12.1.2, there exist s ∈ HomR (R, nR), t ∈ HomR (nR, R) such that y = ysy and ts = 1R . Let w = y + (1 − ys)t(1nR − sy). Then y = ysw and ws = 1. Let α = (1−yx−ys)w(1nR −xy −sy) and β = (1nR −xy −sy)s(1−yx−ys). One easily checks that p2 = βp1 α with αβ = 1. Let u ∈ Rn and v ∈ n R be the matrices corresponding to α and β, respectively. Clearly, e ∈ R and f ∈ Mn (R) are the matrices corresponding to p1 and p2 , respectively. Therefore f = veu with uv = 1. Consequently, eu = uf for a unimodular u ∈ Rn . (2) ⇒ (3) is trivial. (3) ⇒ (1) For any maximal right ideal J of R and any idempotents e ∈ 1 + J and f ∈ Mn (R), eR ∼ = f (nR) implies that there exists a unimodular n u ∈ R such that eu = uf . Write uv = 1 for a v ∈ n R. Construct two maps ϕ : (1 − e)R → (In − f )(nR) given by (1 − e)r → (In − f )ver for any r ∈ R and φ : (In − f )(nR) → (1 − e)R given by (In − f )s → (1 − e)u(In − f )s for any s ∈ nR. It is easy to verify that φϕ = 1(1−e)R . Therefore (1 − e)R .⊕ (In − f )(nR). Lemma 12.3.1 establishes the result.  Analogously, we deduce that an exchange ring R is weakly stable if and only if for any maximal right ideal J and any idempotents e ∈ 1 + J and f ∈ R, eR ∼ = f R implies that (1 − e)R .⊕ (1 − f )R or (1 − f )R .⊕ (1 − e)R if and only if for any maximal right ideal J and any idempotents e ∈ 1 + J and f ∈ R, eR ∼ = f R implies that there exists a right or left invertible u ∈ R such that eu = uf . Lemma 12.3.10. Let R be a directly infinite simple exchange ring. Then 2R .⊕ R. Proof. Since R is directly infinite, there exist x, y ∈ R such that xy = 1 and yx 6= 1. Set e1 = 1 − yx. Then 0 6= e1 = e21 ∈ R. Since R is a simple ring, Re1 R = R. So we have y1 , · · · , yn ∈ R such that 1 ∈ y1 e1 R + · · · + yn e1 R. Thus we can construct an R-epimorphism ψ : n(e1 R) → y1 (e1 R) ⊕ · · · ⊕ yn (e1 R) = R. Since R is projective, we have a splitting exact sequence ψ

0 −→ ker(ψ) −→ n(e1 R) −→ R −→ 0.

Therefore R .⊕ n(e1 R) for some n ∈ N; whence 2R .⊕ (2n)(e1 R).

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Set e2 = yx−y 2 x2 , e3 = y 2 x2 −y 3 x3 , · · · , e2n = y 2n−1 x2n−1 −y 2n x2n . It is easy to check that all ei are idempotents of R, and that ei R ∼ = ej R, ei ej = 0(i 6= j). Further, R = e1 R ⊕ e2 R ⊕ · · · ⊕ e2n R ⊕ (1 − e1 − e2 − · · · − e2n )R. Hence (2n)(e1 R) .⊕ R. Therefore 2R .⊕ R, the result follows.  Theorem 12.3.11. Let R be a simple separative exchange ring. Then R has stable range one if and only if it satisfies the finite stable range condition. Proof. One direction is obvious. Conversely, assume that R satisfies the nstable range condition. In view of Theorem 12.1.12, R satisfies the 2-stable range condition. Suppose R is directly infinite. We see from Lemma 12.3.10 that 2R .⊕ R. So there is an idempotent e ∈ R such that 2R ∼ = eR, i.e., eR ∼ = diag(1, 1)(2R). According to Theorem 12.3.9, we can find u ∈ R2 and v ∈ 2 R such that e = udiag(1, 1)v and uv = 1. That is, e = 1, and so 2R ∼ = R. This implies that 3R ∼ = R. Hence, there exist pairwise orthogonal idempotents e1 , e2 , e3 ∈ R such that e1 + e2 + e3 = 1 and each ei R ∼ = R. In view of Lemma 6.4.16, there exist ai ∈ ei R, ci ∈ Rei such that ei = ai ci and ci ai = 1. This yields that a1 R + a2 R + a3 R = R, and so there exist b1 , b2 , d1 , d2 ∈ R such that (a1 + a3 b1 )d1 + (a2 + a3 b2 )d2 = 1. It follows that each ci = di . Thus d1 e3 = d2 e3 = 0; whence, e3 = 0. This gives a contradiction. Therefore R is directly finite, and the result follows by Corollary 5.2.10.  Similarly, it follows from Corollary 10.2.11 that every simple exchange ring satisfying the finite stable range condition satisfies power-substitution. Lemma 12.3.12. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies the n-stable range condition. (2) For any regular x ∈ Rn , there exists unimodular u ∈n R such that x = xux. (3) For any regular x ∈ n R, there exists unimodular u ∈ Rn such that x = xux. Proof. (1) ⇒ (2) For any regular x ∈ Rn , there exists some y ∈ n R such that x = xyx. Since xy + (1 − xy) = 1, there exists some z ∈ n R such
that y + z(1 − xy) = u ∈ n R is unimodular. Therefore x = x y + z(1 − xy) x = xux. (2) ⇒ (1) Suppose that ax + b = 1 with a ∈ Rn , x ∈ n R and b ∈ R.

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In view of [382, Theorem 28.7], there exist s, t ∈ R such that e = bs and 1 − e = (1 − b)t for an idempotent e ∈ R. Thus, axt + e = 1, and then (1 − e)a ∈ Rn is regular. By hypothesis, there exist u ∈ n R, v ∈ Rn such that (1 − e)a = (1 − e)au(1 − e)a and vu = 1. Let f = u(1 − e)a ∈ Mn (R). Then f xt + ue = u, and so f (xt + ue) + (In − f )ue = u. This implies that f (xt + ue) = f u and (In − f )ue = (In − f )u. Clearly, (In − f )u = (In − f )uv(In − f )u. Let g = (In − f )uev(In − f ). Then (f + g)u = u. We check that
u = f + (In − f )uev(In − f ) u
= f (In − f uev(In − f )) + uev(In − f )
u
= f + uev(In − f )(In + f uev(In − f )) In −
f uev(In − f ) u
= u (1 − e)a + ev(In − f )(In + f uev(In − f )) In − f uev(In − f ) u.
As vu = 1, we deduce that a + bs v(In − f )(In + f uev(In − f )) − a ∈ Rn is unimodular. Therefore R satisfies the n-stable range condition. (1) ⇔ (3) is symmetric.  Theorem 12.3.13. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies the n-stable range condition. (2) For any regular x ∈ Rn , there exists a unimodular u ∈ Rn and an idempotent e ∈ R such that x = eu. (3) For any regular x ∈ n R, there exists a unimodular u ∈n R and an idempotent e ∈ R such that x = ue.

Proof. (1)⇒(2) Given any regular x ∈ Rn , there exists some y ∈ n R such that x = xyx. Since xy + (1 − xy) = 1, we have z ∈ Rn such that
x+(1−xy)z = u ∈ Rn is unimodular. Thus we have x = xy x+(1−xy)z = eu, where e = e2 = xy ∈ R. (2)⇒(1) Given any regular x ∈ Rn , there exists some y ∈ n R such that x = xyx and there exists a unimodular u ∈ Rn and an idempotent e ∈ R such that x = eu. Since xy+(1−xy) = 1, euxy(1−e)+(1−xy)(1−e) = 1−e.
−1 Hence x + (1 − xy)(1 − e)u = 1 + euxy(1 − e) u ∈ Rn is unimodular. As in the proof of Proposition 12.1.2, we can find some z ∈ n R such that y+z(1−xy) = v ∈ n R is unimodular. Hence x = xyx = x(y+z(1−xy))x = xvx. Consequently, R satisfies the n-stable range condition from Lemma 12.3.12. (1) ⇔ (3) is symmetric.  Proposition 12.3.14. Let R be an exchange ring. Then the following are

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equivalent: (1) R satisfies the n-stable range condition. (2) For any regular x ∈ Rn , there exists a unimodular u ∈ n R such that ux ∈ Mn (R) is an idempotent. (3) For any regular x ∈ n R, there exists a unimodular u ∈ Rn such that xu ∈ Mn (R) is an idempotent. Proof. (1)⇒(2) is trivial by Lemma 12.3.12. (2)⇒(1) Given any regular x ∈ Rn , there exists a unimodular u ∈ n R such that ux = e ∈ Mn (R) is an idempotent. Let x = xyx for some y ∈ n R. From xy + (1 − xy) = 1, we see that ey + u(1 − xy) = u. So e(y + u(1 − xy)) + (In − e)u(1 − xy) = u. Assume that vu = 1 for some v ∈ Rn . As in of Lemma 12.3.12, x + (1 − xy)v(In − e) 1 + eu(1 −
the proof n xy)v(In − e) ∈ R is unimodular. As in the proof of Lemma 4.1.2, there is some z ∈ n R such
that y + z(1 − xy) = u ∈ n R is unimodular. So x = x y + z(1 − xy) x = xux. Therefore R satisfies the n-stable range condition. (1)⇔(3) Applying (1)⇔(2) to the opposite ring Rop , we obtain the result.  Lemma 12.3.15. Suppose that R satisfies the n-stable range condition. Then AX + B = In with A, X, B ∈ Mn (R) implies that there exists some Y ∈ Mn (R) such that the final row of A + BY is unimodular. Proof. Let A = (aij ), X = (xij ) and B = (bij ) such that AX + B = In . Then (an1 , · · · , ann ) (xn1 , · · · , xnn )T + bnn = 1. Since R satisfies the n-stable range condition, we can find some (y1 , · · · , yn ) ∈ Rn such that (an1 , · · · , ann ) + bnn (y1 , · · · , yn ) ∈ Rn is unimodular. Set   0 ··· 0  .. . . ..   . .  Y = . .  0 ··· 0  y1 · · · yn

Then we check that the final row of A + BY is unimodular, as desired. 

Now we generalize Corollary 1.1.4 to rings satisfying the n-stable range condition as follows. Theorem 12.3.16. Let R be an exchange ring. Then the following are equivalent:

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(1) R satisfies the n-stable range condition. (2) Whenever a(n R) + bR = dR with a ∈ Rn and b, d ∈ R implies that there exists a unimodular u ∈ n R and a y ∈ R such that au + by = d. (3) Whenever (Rn )a + Rb = Rd with a ∈ n R and b, d ∈ R implies that there exists a unimodular u ∈ Rn and a z ∈ R such that ua + zb = d. Proof. (1) ⇒ (2) Given a1 R + · · · + an+1 R = dR with a1 , · · · , an+1 , d ∈ R, then (a1 , · · · , an , an+1 )Mn+1 (R) = (0, · · · , 0, d) Mn+1 (R). Assume (0, · · · , 0, d)A = (a1 , · · · , an+1 ) and (a1 , · · · , an+1 )B = (0, · · · , 0, d) for some A, B ∈ Mn+1 (R). Clearly, R satisfies the (n + 1)-stable range condition. As AB + (In+1 − AB) = In+1 , it follows by Lemma 12.3.15 that there exists some Y ∈ Mn+1 (R) such that the final row of A + (In+1 − AB)Y = (uij ) is unimodular. Thus we have
(a1 , · · · , an+1 ) = (0, · · · , 0, d) A + (In+1 − AB)Y = (0, · · · , 0, d)(uij ). Thus, each ai = du(n+1)i . Since (u(n+1)1 , · · · , u(n+1)n )(n R) + u(n+1)(n+1) R = R, there are some y1 , · · ·
, yn ∈ R such that u(n+1)1 + u(n+1)(n+1) y1 , · · · , u(n+1)n + u(n+1)(n+1) yn is unimodular. Assume that n X i=1

Then


u(n+1)i + u(n+1)(n+1) yi vi = 1 for some vi ∈ R.  v1 n P   yi vi (a1 , · · · , an )  ...  + an+1 i=1 vn n P = (ai vi + an+1 yi vi ) 

= =

i=1 n P i=1 n P

(ai + an+1 yi )vi

(du(n+1)i + du(n+1)(n+1) yi )vi

i=1 n P

=d

i=1

(u(n+1)i + u(n+1)(n+1) yi )vi

= d,

as required. (2)⇒(1) Let a(n R) + bR = R with a ∈ Rn , b ∈ R. Then we can find a unimodular u ∈ n R such that au + by = 1 for some y ∈ R. Assume that vu = 1 for some v ∈ Rn . Then (a + byv)u = 1. Therefore a + byv ∈ Rn is unimodular, as desired.

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(1) ⇔ (3) Applying (1) ⇔ (2) to the opposite ring R op , we obtain the result.  Proposition 12.3.17. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies the n-stable range condition. (2) Whenever a(n R) + bR = dR with a ∈ Rn , b, d ∈ R, there exists y ∈ Rn such that a + by = du for some unimodular u ∈ R n . (3) Whenever (Rn )a+Rb = Rd with a ∈ n R, b, d ∈ R, there exists z ∈ n R such that a + zb = ud for some unimodular u ∈n R. Proof. (1) ⇒ (2) Suppose a(n R)+bR = dR with a ∈ Rn and b, d ∈ R. Similarly to the discussion in Theorem 12.3.16, we can find some U = (uij ) ∈ Mn+1 (R) such that the final row of U is unimodular, and that (a, b) = (0, · · · , 0, d)U . This implies that a = d(u(n+1)1 , · · · , u(n+1)n ) and b = du(n+1)(n+1) . Since (u(n+1)1 , · · · , u(n+1)n )(n R)+u(n+1)(n+1) R = R, we can find an element y ∈ Rn such that (u(n+1)1 , · · · , u(n+1)n ) + u(n+1)(n+1) y = u ∈ Rn is unimodular. Therefore, a + by = du, as desired. (2) ⇒ (1) Let ax + b = 1 with a ∈ Rn , x ∈ n R, b ∈ R. Then there exists some y ∈ Rn such that a + by = u ∈ Rn is unimodular, as required. (1) ⇔ (3) Applying (1) ⇔ (2) to the opposite ring R op , we easily obtain the result. 

12.4

Rectangular Matrices

Lemma 12.4.1. Let (a1 , · · · , an−1 , an ) ∈ Rn . If there exist some y1 , · · · , yn−1 ∈ R such that (a1 + an y1 , · · · , an−1 + an yn−1 ) is unimodular, then there exists some Q ∈ En (R) such that (a1 , · · · , an−1 , an )Q = (1, 0, · · · , 0). Proof. As (a1 + an y1 , · · · , an−1 + an yn−1 ) is n−1 P x1 , · · · , xn ∈ R such that (ai + an yi )xi i=1   1 0 ··· 0 0 1 ···  0 1 · · · 0 0   .. . .   . . (a1 , · · · , an−1 , an )  . . . . .   .. .. . . .. ..   0 · · ·

unimodular, we have some = 1.

Thus, we see that  0 x1 (1 − an )  .. ..  . .  = (a1 + 1 xn−1 (1 − an ) 

y1 y2 · · · yn−1 1 0 ··· 0 1 an y1 , · · · , an−1 + an yn , 1). Therefore we can find some Q ∈ En (R) such

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that (a1 , · · · , an−1 , an )Q = (1, 0, · · · , 0).



Lemma 12.4.2. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies the n-stable range condition. (2) For any right invertible (α1 , · · · , αs+n ) ∈ Ms×(s+n) (R), there exist
y1 , · · · , ys+n−1 ∈ R such that α1 + αs+n y1 , · · · , αs+n−1 + αs+n ys+n−1 is right invertible. Proof. (2) ⇒ (1) Choose s = 1. Then the result is obvious. (1) ⇒ (2) We will prove this by induction on s. If s = 1, the result holds. Assume the result holds for s(s ≥ 1). Let α = (α1 , · · · , αs+1+n ) ∈   a1i   M(s+1)×(s+1+n) (R) be right invertible, where αi =  ... . Then a(s+1)i

s+1+n P

a1i R = R. As R satisfies the n-stable range condition, there exist

i=1

some y1 , · · · , ys+n ∈ R such that s+n X i=1


a1i + a1(s+1+n) yi R = R.

By virtue of Lemma 12.4.1, there exists some
C ∈ Es+n (R) such that a11 +a1(s+1+n) y1 , · · · , a1(n+s) +a1(s+1+n) ys+n C = (1, 0, · · · , 0). Therefore 

1   0 1  α . ∗ Is+n  .. y1

0 1 .. . y2

··· ··· .. . ···

   1 (0, · · · , 0, ∗) 0      0  0  C   =   ,   .. .. B 1      . . 1 0

where B = (bij ) ∈ Ms×(s+n) (R). Clearly, B is right invertible. By hypothesis, there exist some z1 , · · · , zs−1+n ∈ R such that 

1 0  B .  .. z1

0 1 .. . z2

··· 0 ··· 0 .. .. . . · · · zs−1+n

 0 0
 ..  = Ds×(s−1+n) , βs×1 , . 1

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where D ∈ Ms×(s−1+n) (R) is right invertible. This implies that     1   1 0 ··· 0 1 0 ··· 0 0    0 1 ··· 0     C    0 1 ··· 0 0 α . . . .      1   . . .  .. .. . . ..  .. ..     .. .. . .  . .  y1 y2 · · · 1 z1 z2 · · · zs−1+n 1   1 (0, · · · , 0, ∗)  0  1 
 =  .  ; ∗ Is+n   ..  Ds×(s−1+n) , βs×1  0

hence,



  0 ··· 0 1 0 ··· 0 1 ···   1 · · · 0   C  α . . . .. . . ..  1    .. .. . . . . . y y ··· 1 0 z1 · · ·   1 2   1 1 ∗ . = C −1 ∗ ∗ Is+n D 1 0 .. .

As a result, we deduce that    α 

Is+n

 0   0  C −1 ..  1 . 1

  0  ..    .   0 


w1 , w2 , · · · , ws+n 1     1 1 ∗ = C −1 , ∗ Is+n D ∗


−1 where w1 , w2 , · · · , ws+n = y1 , y2 , · · · , ys+n + 0,
z1 , · · · , zs−1+n C . Therefore α1 + αs+n+1 w1 , · · · , αs+n + αs+n+1 ws+n is right invertible, as required.  Lemma 12.4.3. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies the n-stable range condition. (2) For any regular A ∈ Ms×t (R)(t−s ≥ n−1), there exists a left invertible U ∈ Mt×s (R) such that A = AU A. (3) For any regular A ∈ Ms×t (R)(s − t ≥ n − 1), there exists a right invertible U ∈ Mt×s (R) such that A = AU A.

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Proof. (1) ⇒ (2) Given any regular A ∈ Ms×t (R) with t − s ≥ n − 1, then A = ABA for some B ∈ Mt×s (R). Since AB + (Is − AB) = Is , (A, Is −AB) is right invertible. Set A = (α1 , · · · , αt ) with all αi ∈ Ms×1 (R) and Is − AB = (β1 , · · · , βs ) with all βi ∈ Ms×1 (R). Since t ≥ s + n − 1 ≥ n, R satisfies the t-stable range condition. By virtue of Lemma 12.4.2, we can find y1 , · · · , yt ∈ R and z1 , · · · , zs−1 ∈ R such that
α1 + βs y1 , · · · , αt + βs yt , β1 + βs z1 , · · · , βs−1 + βs zs−1

is right invertible. If s ≥ 2, then t − 1 ≥ s + n − 2 ≥ n. So R satisfies the (t − 1)-stable range condition. Using Lemma 12.4.2 again, we have v1 , · · · , vt ∈ R and w1 , · · · , ws−2 ∈ R such that α1 + βs y1 + (βs−1 + βs zs−1 )v1 , · · · , αt + βs yt + (βs−1 + βs zs−1 )vt ,
β1 + βs z1 + (βs−1 + βs zs−1 )w1 , · · · , βs−2 + βs zs−2 + (βs−1 + βs zs−1 )ws−2 is right invertible. That is, α1 + βs q11 + βs−1 q12 , · · · , αt + β
s qt1 + βs−1 qt2 , β1 + βs c11 + βs−1 c12 , · · · , βs−2 + βs c(s−2)1 + βs−1 c(s−2)2 is right invertible for some qij , cij ∈ R. Clearly, R also satisfies the (t − s + 1)-stable range condition. Following this route, we can find δij (1 ≤ i ≤ t, 1 ≤ j ≤ s) such that
α1 + βs δ11 + βs−1 δ12 + · · · + β1 δ1s , · · · , αt + βs δt1 + βs−1 δt2 + · · · + β1 δts is right invertible. That is,

(α1 , · · · , αt ) + (β1 , · · · , βs )(δij )s×t is right invertible, where 

δ1s

δ2s

 δ1(s−1) δ2(s−1)  (δij )s×t =  . ..  .. . δ11 δ21

 · · · δts · · · δt(s−1)   ..  . .. . .  · · · δt1

Thus there exists some U ∈ Mt×s (R) such that
A + (Is − AB)(δij ) U = Is . Therefore


AB A + (Is − AB)(δij ) U = AB,

and then AU A = ABAU A = ABA = A, as required. (2) ⇒ (1) Let s = 1 and t = n. Then we see that for any regular x ∈ n R, there exists unimodular u ∈ n R such that x = xux. By Lemma 12.3.12, R satisfies the n-stable range condition. (1) ⇔ (3) Applying (1) ⇔ (2) to the opposite ring Rop , we obtain the result. 

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We now come to the main result of this section. Theorem 12.4.4. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies the n-stable range condition. (2) For all idempotents e ∈ Ms (R) and f ∈ Mt (R), e(sR) ∼ = f (tR)(t − s ≥ n − 1) implies that there exists a right invertible u ∈ Ms×t (R) such that eu = uf . (3) For all idempotents e ∈ Ms (R) and f ∈ Mt (R), e(sR) ∼ = f (tR)(t − s ≥ n − 1) implies that (Is − e)(sR) .⊕ (It − f )(tR). Proof. (1) ⇒ (2) Given e(sR) ∼ = f (tR) with e = e2 ∈ Ms (R), f = f 2 ∈ Mt (R) and t − s ≥ n − 1, then there exist a ∈ Ms×t (R), b ∈ Mt×s (R) such that e = ab, f = ba, a = eaf and b = f be from Lemma 6.4.16. Clearly, we have e = af b, f = bea, a = aba and b = bab. Since a ∈ Ms×t (R) is regular, by Lemma 12.4.3, we can find v ∈ Mt×s (R) such that a = ava with cv = Is for c ∈ Ms×t (R). Let w = a + (Is − av)c(It − va). Then a = avw with wv = Is . Set u = (Is − e − av)w(It − f − va) ∈ Ms×t (R) and v = (It − f − va)v(Is − e − av) ∈ Mt×s (R). It is easy to check that (Is − e − av)2 = Is and (It − f − va)2 = It , and thus uv = Is . Furthermore, we check that eu = a = uf , as required. (2) ⇒ (3) Given e(sR) ∼ = f (tR) with e = e2 ∈ Ms (R), f = f 2 ∈ Mt (R) and t − s ≥ n − 1, we can find u ∈ Ms×t (R) and v ∈ Mt×s (R) such that e = uf v with
uv = Is . It is easy
to check that Is − e = u(It − f )v = (Is − e)u(It − f ) (It − f )v(Is − e) . In view of Lemma 6.4.16, we conclude that (Is − e)(sR) .⊕ (It − f )(tR). (3) ⇒ (1) Choose s = 1 and t = n. Then eR ∼ = f (nR) with e = e2 ∈ R 2 ⊕ and f = f ∈ Mn (R) implies that (1 − e)R . (In − f )(nR). Thus the result follows by Lemma 12.3.1.  Corollary 12.4.5. Let R
be a strongly separative exchange ring and n ∈ N. Then e(nR) ∼ = f (n + 1)R with idempotents e ∈ Mn (R) and f ∈ Mn+1 (R) implies that there exists a right invertible u ∈ Mn×(n+1) (R) such that eu = uf . Proof. Since R is a strongly separative exchange ring, it follows from Theorem 12.2.8 that R satisfies the 2-stable range condition. Thus we conclude the result by Theorem 12.4.4. 

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Theorem 12.4.6. Let R be an exchange ring. Then the following are equivalent: (1) R satisfies the n-stable range condition. (2) For any regular A ∈ Ms×t (R)(t − s ≥ n − 1), there exists an idempotent E ∈ Ms (R) and a right invertible U ∈ Ms×t (R) such that A = EU . (3) For any regular A ∈ Ms×t (R)(s − t ≥ n − 1), there exists an idempotent E ∈ Ms (R) and a left invertible U ∈ Ms×t (R) such that A = EU . Proof. (1) ⇒ (2) For any regular A ∈ Ms×t (R) with t − s ≥ n − 1, it follows by Lemma 12.4.3 that there exists a left invertible W ∈ Mt×s (R) such that A = AW
A. Assume that V W = Is for some V ∈ Ms×t (R). Then A + (Is − AW )V W = Is . Let U = A + (Is − AW )V and E = AW . Then A = EU and E = E 2 , as desired. (2) ⇒ (1) Choose s = 1 and t = n. Then R satisfies the n-stable range condition by Theorem 12.3.13. (1) ⇒ (3) For any regular A ∈ Ms×t (R) with s − t ≥ n − 1, there are W ∈ Mt×s (R) and U ∈ Ms×t (R) such that A = AW A with W U = It . Let E = AW . Then A = EU , as required. (3) ⇒ (1) For any x ∈ n R, there are e = e2 ∈ Mn (R), u ∈ n R and v ∈ Rn such that x = eu and vu = 1. Since xy + (In − xy) = In , we get euy + (In − xy) = In ; hence, euy(In − e) + (In − xy)(In − e) = In − e. This implies that
x + (In − xy)(In − e)u = eu + (In − xy)(In − e)u = In − euy(In − e) u ∈n R is unimodular. As in the proof of Lemma 12.1.1, we have a z ∈ Rn such that w := y +
z(In − xy) ∈ Rn is unimodular. Therefore x = xyx = x y + z(In − xy) x = xwx. The result now follows by Lemma 12.3.12.  Corollary 12.4.7. Let R be a strongly separative regular ring. Then every rectangular matrix over R is the product of an idempotent and a right or left invertible matrix.   A Proof. Let A ∈ Ms×t (R)(t − s ≥ 1). Then ∈ Mt (R). 0(t−s)×t According to [217, Theorem 1.7], Mt (R) is regular, thus, we can find a
Ct×s , Dt×(t−s) ∈ Mt (R) such that      
A A A = C, D . 0(t−s)×t 0(t−s)×t 0(t−s)×t

Therefore, A = ACA, i.e., A ∈ Ms×t (R) is regular. Analogously, we show that every s×t matrix with s−t ≥ 1 is regular as well. By virtue of Theorem

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12.2.8, R satisfies the 2-stable range condition. Therefore we complete the proof from Theorem 12.4.6.  Let R be a strongly separative regular ring. If 12 ∈ R, then every rectangular matrix over R is generated by two right or left invertible matrices from Corollary 12.4.7. Let p, q ∈ R be idempotents. We say that pRq has stable range one provided that ax + b = p with a ∈ pRq, x ∈ qRp and b ∈ pRp implies that there exists a y ∈ pRq and a z ∈ qRp such that (a + by)z = p. We conclude this chapter by investigating the n-stable range condition on corners of full idempotents. The key is to reduce stable range calculations to a question of stable range one for a kind of matrix ring. Lemma 12.4.8. Let p, p′ , q, q ′ ∈ R be idempotents such that pR ∼ = p′ R ′ ′ ′ ∼ and qR = q R. Then pRq has stable range one if and only if p Rq also has stable range one. Proof. Assume that pRq has stable range one. As pR ∼ = p′ R and qR ∼ = q ′ R, ′ ′ ′ ′ there exist c ∈ pRp , d ∈ p Rp, u ∈ qRq and v ∈ q Rq such that p = cd, p′ = dc, q = uv and q ′ = vu. Given ax + b = p′ with a ∈ p′ Rq ′ , x ∈ q ′ Rp′ and b ∈ p′ Rp′ , then (cav)(uxd) + cbd = cp′ d = cdcd = p with cav ∈ pRq, uxd ∈ qRp and cbd ∈ pRp. Thus, we can find y ∈ pRq and z ∈ qRp such that (cav = p. Therefore, d(cav + cbdy)zc = dpc = p′ , i.e.,
+ cbdy)z a + b(dyu) (vzc) = p′ . This implies that p′ Rq ′ has stable range one. The conversely is symmetric.  Lemma 12.4.9. Let p, q ∈ R be idempotents such that pRq has stable range one. Then the following hold: (1) If qr = rq = 0, then pR(q + r) has stable range one. (2) If pr = rp = qr = rq = 0, then (p + r)R(q + r) has stable range one. Proof. (1) Given ax + b = p with a ∈ pR(q + r), x ∈ (q + r)Rp and b ∈ pRp, then (aq)(qx) + (b + arx) = p. Since pRq
has stable range one, we can find y ∈ pRq such that aq + (b + arx)y
z = p for some z ∈ qRp. As rq = 0, we see that a(q + r + rxy) + by z = p. It is easy to verify that (q + r − rxy)(q + r + rxy) = q + r. Thus
a + by(q + r − rxy) (q + r + rxy)z = p, and pR(q + r) has stable range one. (2) Applying (1) to the opposite ring Rop , we obtain the result.



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Lemma 12.4.10. Let p, q, r ∈ R be idempotents such that pr = qr = 0. If (p + r)R(q + r) has stable range one, then so has pRq. Proof. Given ax + b = p with a ∈ pRq, x ∈ qRp and b ∈ pRp, then (a+ r)(x+ r)+ b = p+ r with a+ r ∈ (p+ r)R(q + r), x ∈ (q + r)R(p+ r) and b ∈ (p + r)R(p + r). Since (p + r)R(p + r) has stable range one, , we can find y ∈ (p+r)R(q+r) such that (a+r+by)z = p+r for some z ∈ (q+r)R(p+r). Clearly, (a + by)z = p(a + r + by)z = p(p + r) = p. On the other hand, z = (q + r)z = qz + rz = qz + r(a + r + by)z =
qz + r(p + r) = qz + r. This implies that zp = qzp. Hence a + b(pyq) (qzp) = (a + by)(qzp) = (a + by)zp = p. This shows that pRq has stable range one.  Lemma 12.4.11. Let p = diag(1, 0, · · · , 0), q = diag(1, 1, · · · , 1) ∈ Mn (R). Then pMn (R)q has stable range one if and only if R satisfies the n-stable range condition. Proof. Assume that pMn (R)q has stable range one. Given a1 x1 + · · · + an xn + an+1 xn+1 = 1 in R, then αβ + γ = p, where       an+1 xn+1 0 · · · 0 a1 · · · an x1 0 · · · 0   0 ··· 0   x2 0 · · · 0  0 0 ··· 0       α= .  and γ =  ,β =  . . . . . .. . . ..   .. · · · ..   .. .. · · · ..  . · · · ..  0 ··· 0

xn 0 · · · 0

0

0 ··· 0

Obviously, α ∈ pMn (R)q, β ∈ qMn (R)p, γ ∈ pMn (R)p. Thus, there exist     u1 0 · · · 0 y1 · · · yn  u2 0 · · · 0   0 ··· 0      δ= .  ∈ pMn (R)q, ε =  . . .  ∈ qMn (R)p (∗) .  .. .. · · · ..   .. · · · ..  un 0 · · · 0 0 ··· 0 such that (α + γδ)ε = p. This implies that

n P

(ai + an+1 xn+1 yi )ui = 1.

i=1

Therefore R satisfies the n-stable range condition. Conversely, assume that R satisfies the n-stable range condition. Given αβ + γ = p with α ∈ pMn (R)q, β ∈ qMn (R)p and γ ∈ pMn (R)p, we have the forms       b 0 ··· 0 x1 0 · · · 0 a1 · · · an 0 0 ··· 0  x2 0 · · · 0   0 ··· 0        α= .  and γ =  . . ,β =  . . . . . .  .. .. · · · ..   .. .. · · · ..   .. · · · ..  0 0 ··· 0 xn 0 · · · 0 0 ··· 0

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Further, n P

n P

i=1

i=1

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ai xi + b = 1. Thus, we can find some y1 , · · · , yn ∈ R such that

(ai + byi )ui = 1 for some ui ∈ R. Let δ ∈ pMn (R)q and ε ∈ qMn (R)p

as in (∗) above. Then (α + γδ)ε = p. Therefore pMn (R)q has stable range one, as asserted.  We have gathered all the information necessary to prove the following result due to Ara and Goodearl (cf. [14, Theorem 7]). Theorem 12.4.12. A ring R satisfies the n-stable range condition if and only if eRe also satisfies the n-stable range condition for some full idempotent e ∈ R. Proof. One direction is obvious. Conversely, assume now that there exists a full idempotent e ∈ R such that eRe satisfies the n-stable range condition. Since ReR = R, analogously to Corollary 8.1.7, there exists a right R-module D such that t(eR) ∼ = D ⊕ R. Let p = diag(1, 0, · · · , 0), q = N diag(e, e, · · · , e) ∈ Mt (R). Then p(tR)⊕D ∼ = q(tR); hence, p(tR) R1×t ⊕ R N N N D R1×t ∼ = Mn (R), we obtain = q(tR) R1×t . As Rn×1 R1×n ∼ R

R

R

pMt (R) ⊕ D

O R

R1×t ∼ = qMt (R).

So, there is a projection φ : qM
t (R) → pMt (R). Further, we have some A ∈ Mt (R) such that p = φ qA = BC, where B = pφ(q)q and C = qAp. Let f = CB and g = q − f . Then q = f + g with orthogonal idempotents f, g ∈ Mt (R). According to Lemma 12.4.10,     e e  0   e      M (eRe)    n ..  ..    . .  0

n×n

e

n×n

has stable range one. This implies that     e e  0   ..      . M (R)     n+t−1 ..     . e 0t−1 (n+t−1)×(n+t−1) 0t−1 (n+t−1)×(n+t−1)

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has stable range one. According to Lemma 12.4.9 and Lemma 12.4.8, we see that     q q  0   e      M (R)    n+t−1 ..  ..    . .  0

(n+t−1)×(n+t−1)

e

(n+t−1)×(n+t−1)

has stable range one. As f and g are orthogonal idempotents, it follows by Lemma 12.4.10,     f f  0   e      M (R)    n+t−1 ..  ..    . .  0 (n+t−1)×(n+t−1) e (n+t−1)×(n+t−1) has stable range one. By Lemma 12.4.9 again,    f f  0   1    Mn+t−1 (R)   ..  ..   .  . 0 (n+t−1)×(n+t−1)

 1

   

(n+t−1)×(n+t−1)

∼ pMn (R), it follows by Lemma 12.4.8 has stable range one. Since f Mn (R) = that     p p    1  0     Mn+t−1 (R)     . . ..  ..    1 (n+t−1)×(n+t−1) 0 (n+t−1)×(n+t−1) has stable range one. By using Lemma 12.4.8 again,    1 1   ..  0    . Mn+t−1 (R)    ..    . 1 0t−1

(n+t−1)×(n+t−1)

 0t−1

has stable range one. Therefore     1 1  0   ..      . M (R)     n ..    . 1  0 n×n 1 n×n

   

(n+t−1)×(n+t−1)

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has stable range one. Consequently, R satisfies the n-stable range condition from Lemma 12.4.11.  Simple counterexamples show that the hypothesis that e ∈ R be full is necessary: for example, R = B(H) for infinite-dimensional H, e is a finite-rank projection (cf. [46]).

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Chapter 13

Stable Range for Ideals

In this chapter we establish some elementary properties of stable range condition for exchange ideals. Usually, we consider a non-unital exchange ring which is an ideal of a unital ring R∗ , where R∗ = R ⊕ Z with addition and multiplication defined by (r, m) + (s, n) = (r + s, m + n), (r, m)(s, n) = (rs + ms + nr, mn), for all r, s ∈ R and m, n ∈ Z. Thus, these results can be applied to non-unital exchange rings. It is well known that a regular ideal of a ring R is a B-ideal if and only if eRe is unit-regular for any idempotent e ∈ I. The B-ideal is an extension of stable range one for a unital ring. In fact, a ring R has stable range one if and only if (1) I is a B-ideal; (2) R/I has stable range one and (3) the natural map U (R) → U (R/I) is surjective. We will extend B-ideals to various kinds of other conditions.

13.1

Strong π-Regularity

As is well known, a ring R is a strongly π-regular ring if and only if every element in R is strongly π-regular. In [8, Theorem 4], Ara solved a known conjecture and proved that every strongly π-regular ring has stable range one. In this section, we introduce two classes of ideals, as extensions of strongly π-regular rings. As an application, we investigate diagonal reduction for regular matrices over such ideals. 405

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Definition 13.1.1. Let I be an ideal of a ring R. We say that I is a strongly π ∼regular ideal of R when every element in 1 + I is strongly π-regular. Let D be a division ring, let V1 be a finite-dimensional vector space over D, and let V2 be an infinite-dimensional vector space over D. Set R = EndD (V1 ) ⊕ EndD (V2 ) and I = EndD (V1 ) ⊕ 0. Then I is a strongly π ∼regular ideal of R, while R is not a strongly π-regular ring. One easily checks that every nil ideal of a ring is strongly π ∼regular. Proposition 13.1.2. Let I be an ideal of a ring R. Then the following are equivalent: (1) I is strongly π ∼regular. T (2) For any x ∈ 1 + I, there exists n ∈ N such that xn ∈ xn+1 R Rxn+1 .

Proof. (1) ⇒ (2) is obvious. (2) ⇒ (1) For any x ∈ 1 + I, there exists n ∈ N such that xn ∈ T n+1 x R Rxn+1 . Write xn = xn+1 a = bxn+1 . Thus, xn a = bxn+1 a = bxn . By induction, we get xn ak = bk xn for any k ∈ N. Let c = xn an+1 . Then c = bn+1 xn . It is easy to verify that xc = xn+1 an+1 = (xn+1 a)an = xn an = bn xn = bn (bxn+1 ) = (bn+1 xn )x = cx; xn+1 c = x2n+1 an+1 = xn (xn+1 a)an = x2n an = · · · = xn+1 a = xn ; c2 x = c(xc) = c(xn+1 an+1 ) = xn+1 can+1 = xn an+1 = c.

Therefore x ∈ 1 + I is strongly π-regular, as asserted.



Lemma 13.1.3. Every strongly π ∼regular ideal of a ring is an exchange ideal. Proof. Let I be a strongly π ∼regular ideal of R. For any x ∈ I, we have n ∈ N and y ∈ R such that (1 − x)n = (1 − x)n+1 y and xy = yx. Hence (1 − x)n = (1 − x)n z(1 − x)n , where z = y n . Let g = z(1 − x)n and e = g + (1 − g)(1 − x)n g. Then e ∈ R(1
− x) is an idempotent.
In addition, we have 1 − e = (1 − g) 1 − (1 − x)n g = (1 − g) 1 − (1 − x)n ∈ Rx. Set f = 1 − e. Then there exists an idempotent f ∈ I such that f ∈ Rx and 1 − f ∈ R(1 − x). It follows from [9, Lemma 1.1] that I is an exchange ideal of R.  Lemma 13.1.4. Let I be an exchange ideal of a ring R. Then eRe is an exchange ring for all idempotents e ∈ I.

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Proof. Let e ∈ I be an idempotent. Given any x ∈ eRe, then we have x ∈ I. As I is an exchange ideal of R, we have an idempotent f ∈ I such that f ∈ Rx and 1 − f ∈ R(1 − x). Thus f e = f , and then (ef e)2 = ef e ∈ (eRe)x. In addition, we have e − ef e = e − ef = e(1 − f ) ∈ (eRe)(e − x). According to [382, Theorem 29.2], eRe is an exchange ring.  Lemma 13.1.5. Let a ∈ R be regular, let K = r(a) and R = aR ⊕ E. If K has the finite exchange property, then there exist right ideals Ai , A′i , Bi , Bi′ , Ci and Ci′ for i ≥ 1 such that the following conditions are satisfied: i
L (1) R = K ⊕ (Aj ⊕ Bj ) ⊕ Ci for all i ≥ 1. j=1

(2) (3) (4) (5)

E∼ = (Ai ⊕ Bi ) ⊕ (A′i ⊕ Bi′ ) for all i ≥ 1. K∼ = A′i ⊕ Bi′ ⊕ Ci′ for all i ≥ 1. ′ Ai ⊕ Bi′ ∼ = Ai+1 ⊕ A′i+1 for all i ≥ 1. ′ ′ aR = C1 ⊕ C1′ and aAi ⊕ aBi = Bi+1 ⊕ Bi+1 and aCi = Ci+1 ⊕ Ci+1 for all i ≥ 1.

Proof. Write R = K ⊕ L = E ⊕ aR. Since K has the finite exchange property, we have A1 , A′1 , C1 , C1′ such that R = K ⊕ A1 ⊕ C1 , E = A1 ⊕ A′1 and aR = C1 ⊕ C1′ . Set B1 = B1′ = 0. Clearly, we have R= = = =

K ⊕ (A1 ⊕ B1 ) ⊕ C1 E ⊕ aR (A1 ⊕ A′1 ) ⊕ (aA1 ⊕ aB1 ) ⊕ aC1 (A1 ⊕ B1 ) ⊕ (A′1 ⊕ B1′ ) ⊕ (aA1 ⊕ aB1 ) ⊕ aC1 .

As R = K ⊕ A1 ⊕ C1 = A1 ⊕ A′1 ⊕ C ⊕ C1′ , we deduce that K∼ = A′1 ⊕ B1′ ⊕ C1′ . By the finite exchange property of K and [442, Lemma 2], we can find right R-modules A2 , A′2 , B2 , B2 , C2 and C2′ such that R = K ⊕(A1 ⊕B1 )⊕(A2 ⊕B2 )⊕C2 , A2 ⊕A′2 = A′1 ⊕B1′ , B2 ⊕B2′ = aA1 ⊕aB1 and C2 ⊕ C2′ = aC1 . Thus this lemma holds for i = 1.

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Now assume that, for some n ∈ N, we have constructed Ai , A′i , Bi , Bi′ , Ci and Ci′ , with 1 ≤ i ≤ n, satisfying the desired conditions. Clearly, we have n
L R=K⊕ (Ai ⊕ Bi ) ⊕ Cn i=1

= E ⊕ aR

= (A1 ⊕ A′1 ) ⊕ = (A1 ⊕ =

n L

A′1 )

⊕


(aAi ⊕ aBi ) ⊕ aAn ⊕ aBn ⊕ aCn

n−1 L

i=1 n−1 L i=1


′ (Bi+1 ⊕ Bi+1 ) ⊕ aAn ⊕ aBn ⊕ aCn


(Ai ⊕ Bi ) ⊕ (A′n ⊕ Bn′ ) ⊕ aAn ⊕ aBn ⊕ aCn .

i=1

By the finite exchange property, we obtain the decomposition n M
R=K⊕ (Ai ⊕ Bi ) ⊕ An+1 ⊕ Bn+1 ⊕ Cn+1 i=1

such that

An+1 ⊕ A′n+1 = A′n ⊕ Bn′ , ′ Bn+1 ⊕ Bn+1 = aAn ⊕ aBn , ′ Cn+1 ⊕ Cn+1 = aCn .

Furthermore, we check that

E = (An ⊕ Bn ) ⊕ (A′n ⊕ Bn′ ) ∼ = aAn ⊕ aBn ⊕ An+1 ⊕ A′n+1 ′ = Bn+1 ⊕ Bn+1 ⊕ An+1 ⊕ A′n+1 ′ = (An+1 ⊕ Bn+1 ) ⊕ (A′n+1 ⊕ Bn+1 ). ∼ ∼ Therefore we get K = A2 ⊕ B2 ⊕ C2 . Obviously, A1 = aA1 , and so E ∼ = (A2 ⊕ B2 ) ⊕ (A′2 ⊕ B2′ ). By induction, the result follows.  Lemma 13.1.6. Let a ∈ R be a nilpotent, regular element. If K = r(a) has the finite exchange property, then a is unit-regular. Proof. Assume that an+2 = 0 for some n ∈ N. Since a ∈ R is regular, by Lemma 13.1.5, we have Ai , A′i , Bi , Bi′ , Ci and Ci′ for i ≥ 1 such that the conditions (1)–(5) in Lemma 13.1.5 hold. By Condition (5), we have Cn+1 ⊆ an+1 R; hence, Cn+1 ⊆ K. As K ∩ Cn+1 = 0, we deduce that ′ Cn+1 = 0, so Cn+1 = aCn . Let ψ : Cn → aCn given by ψ(r) = ar for any r ∈ Cn . Clearly, ψ is an R-epimorphism. On the other hand, ker(ψ) = {r ∈ Cn | ar = 0} ⊆ Cn ∩ K = 0; hence, ψ : aCn ∼ = Cn . As n n

L L R = K⊕ (Ai ⊕ Bi ) ⊕ Cn = K ⊕ (Ai ⊕ Bi ) ⊕ An+1 ⊕ Bn+1 ⊕ Cn+1 , i=1

i=1

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′ ∼ we see that Cn+1 = Cn ∼ = An+1 ⊕ Bn+1 ⊕ Cn+1 = An+1 ⊕ Bn+1 . Therefore, ′ ′ ′ ′ ∼ ∼ we have K = An+1 ⊕ Bn+1 ⊕ Cn+1 ⊕ An+1 ⊕ Bn+1 ∼ = A′n+1 ⊕ Bn+1 = E. This implies that a ∈ R is unit-regular. 

Theorem 13.1.7. Every strongly π ∼regular ideal of a ring is a B-ideal. Proof. Let I be a strongly π ∼regular ideal of R. Suppose that aR+bR = R with a ∈ 1 + I, b ∈ I. Then we have x, y ∈ R such that ax + by = 1. By Lemma 13.1.3, I is an exchange ideal of R. As b ∈ I, we have e ∈ I such that e = bys and 1 − e = (1 − by)t for some s, t ∈ R. So axt + e = 1; hence, (1 − e)axt + e = 1 and (1 − e)axt(1 − e)a = (1 − e)a. This implies that (1 − e)a ∈ 1 + I is regular. Thus, we can find n ∈ N and c ∈ R
n
n+1 such that (1 − e)a = (1 − e)a c and (1 − e)ac = c(1 − e)a. Set
n n
−1
n f = (1 − e)a c . Then f (1 − e)a = (1 − e)a cn+1 in f Rf . In addition, (1 − f )(1 − e)a = (1 − e)a(1 − f ) ∈ (1 − f )R(1 − f ). Furthermore, we see that (1 − f )(1 − e)a
is a regular nilpotent in (1 − f )R(1
− f ). Let K = r (1 − f )(1 − e)a(1 − f ) . Then K = (1 − f ) 1 − xt(1 − e)a (1 − f ) (1 −
f )R(1−f ) . Since (1−e)axt(1−e)a = (1−e)a and (1−e)a ∈ 1+I, we
deduce that xt ∈ 1+I. Hence 1−xt(1−e)a ∈ I. Set g = (1−f ) 1−xt(1−e)a (1−f ). Then g ∈ (1 − f )R(1 − f ). One easily checks that

g 2 = (1 − f ) 1 − xt(1 − e)a (1 − f ) 1 −
xt(1 − e)a (1 − f
) = (1 − f ) (1 − f ) − xt(1 − f )(1
− e)a 1 − xt(1
− e)a (1 − f ) = (1 − f ) 1 − xt(1 − f )(1 − e)a
1 − xt(1 − e)a (1 − f ) = (1 − f ) 1 − xt(1 − f )(1 − e)a (1 − f ) = g ∈ I.
∼ g(1 − f )R(1 − f )g = gRg. Clearly, End(1−f )R(1−f ) g(1 − f )R(1 − f ) = In light of Lemma 13.1.4, gRg is an exchange ring. Therefore K ∼ = g(1 − f )R(1 − f ) has the finite exchange property as a right (1 − f )R(1 − f )module. It follows by Lemma 13.1.6 that (1 − f )(1 − e)a is unit-regular in (1 − f )R(1 − f ). Therefore (1 − e)a = f (1 − e)a + (1 − f )(1 − e)a ∈ R is unit-regular. So we have an idempotent h ∈ R and a unit v ∈ R such that (1 − e)a = hv, and then hvxt(1 − h) + e(1 − h) = 1 − h. As a result, we get

a + bys (1 − h)v − a = a + e (1 − h)v − a = (1 − e)a + e(1
− h)v = h + e(1 − h) v
−1 = 1 + hvxt(1 − h) v ∈ U (R). Suppose now that aR+bR = R with a ∈ 1+I, b ∈ R. Then we have x, y ∈ R such that ax + by = 1; hence, a(x + by) + (1 − a)by = 1. As a ∈ 1 + I,

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(1 − a)by ∈ I. By the above consideration, we can find an element z ∈ R such that a + (1 − a)byz ∈
U (R). Using Lemma 4.1.2, we have some w ∈ R such that x + 1 + w(1 − a) by = x + by + w(1 − a)by ∈ U (R). Using Lemma 4.1.2 again, we have some t ∈ R such that a + byt ∈ U (R), as required.  Immediately, every strongly π-regular ring has stable range one. Let A be an algebraic algebra over a field F , and let a ∈ A. Then a is the root of some non-constant polynomial in F [x]. Thus, there exist some am , · · · , an ∈ F such that an an + an−1 an−1 + · · · + am am = 0, where am 6= 0. Thus, n m+1 n−m−1 xm = −a−1 ) = −a−1 +· · ·+am+1 )am+1 . m (an a +· · ·+am+1 a m (an a −1 n−m−1 m+1 m Set b = −am (an a + · · · + am+1 )a . Then a = bam+1 , and so A is strongly π-regular. Thus, any algebraic algebra over a field has stable range one. Let I be an ideal of a ring R. We use F P (I) to denote the set of all P ∈ F P (R) such that P = P I. Lemma 13.1.8. Let I be an exchange ideal of a ring R, and let P ∈ F P (I). Then there exist idempotents e1 , . . . , en ∈ I such that P ∼ = e1 R ⊕ · · · ⊕ en R.

Proof. Since P ∈ F P (I), we have a right R-module Q such that P ⊕ Q ∼ = nR for some n ∈ N. Let e : nR → P be the projection onto P . Then P ∼ = e(nR), whence EndR (P ) ∼ = eMn (R)e. Inasmuch as P = P I, we have e(nR) = e(nR)I ⊆ nI. Set e = (α1 , · · · , α1 ) ∈ Mn (R). We have e(1, 0, · · · , 0)T ∈ nI. Hence α1 ∈ nI. Likewise, we have α2 , · · · , αn ∈ nI. Therefore e ∈ Mn (I). As I is an exchange ideal of R, then so is Mn (I) as an ideal of Mn (R) from [9, Theorem 1.4]. According to Lemma 13.1.4, EndR (P ) is an exchange ring. That is, P has the finite exchange property. n L Set M = P ⊕ Q. Then M = P ⊕ Q = Ri with all Ri ∼ = R. By the i=1

finite exchange property of P , we have Qi (1 ≤ i ≤ n) such that M = n
L P⊕ Qi , where all Qi are direct summands of Ri respectively. Assume i=1

that Qi ⊕ Pi = Ri for all i. Then P ⊕

n L

i=1


Qi =

n L

i=1


Pi ⊕

n L

i=1


Qi .

Hence P ∼ = P1 ⊕ · · · ⊕ Pn , where Pi is isomorphic to a direct summand of R as a right R-module for all i. So we have idempotents ei such that Pi ∼ = ei R. Clearly, ei R is a finitely generated projective right R-module. It N N follows from P = P I that P (R/I) = 0; hence, Pi (R/I) = 0. That is, R R N (ei R) (R/I) = 0, and so ei R = ei RI ⊆ I. Furthermore, each ei ∈ I for R

all i. Therefore P ∼ = e1 R ⊕ · · · ⊕ en R with all ei ∈ I.



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Lemma 13.1.9. Let I be an exchange ideal of a ring R, and let A ∈ F P (I). If I is a B-ideal, then for any right R-modules B and C, A ⊕ B ∼ = A⊕C implies that B ∼ = C. Proof. As A ∈ F P (I), it follows by Lemma 13.1.8 that there exist idempotents e1 , · · · , en ∈ I such that A ∼ = e1 R ⊕ · · · ⊕ en R. Given ax + b = e1 in e1 Re1 , then (a + 1 − e1 )(x + 1 − e1 ) + b = 1 in R. Since a + 1 − e1 ∈ 1 + I, we have y ∈ R such that a + 1 − e1 + by ∈ U (R). Assume that u(a + 1 − e1 + by)
= (a + 1 − e1
+ by)u = 1. One easily checks that (e1 ue1 ) a + b(e1 ye1 ) = a + b(e1 ye1 ) (e1 ue1 ) = e1 . Thus a + b(e1 ye1 ) ∈ U (e1 Re1 ), and then EndR (e1 R) ∼ = e1 Re1 has stable range one. In view of Corollary 1.1.7, we have e2 R ⊕ · · · ⊕ en R ⊕ B ∼ = e2 R ⊕ · · · ⊕ en R ⊕ C. Likewise, we have e3 R ⊕ · · · ⊕ en R ⊕ B ∼ = e3 R ⊕ · · · ⊕ en R ⊕ C. Furthermore, we get B ∼  = C, as asserted. Proposition 13.1.10. Let I be a strongly π ∼regular ideal of a ring R, and let A ∈ F P (I). Then for any right R-modules B and C, A⊕B ∼ = A⊕C ∼ implies that B = C. Proof. In view of Theorem 13.1.7, I is a B-ideal. Therefore we complete the proof by Lemma 13.1.9.  Let I be an ideal of a ring R, and let π : R → R/I be the natural quotient map. Define
the Fredholm elements relative to I as the set F (I, R) = π −1 U (R/I) . Denote δ : K1 (R/I) → K0
(I) as the connecting map in algebraic K-Theory. Set index(x) = δ [π(x)] . Let I be a strongly π ∼regular ideal of a ring R, and let x ∈ R be a Fredholm element relative to I. Then there exists a y ∈ U (R) such that x − y ∈ I if and only if index(x) = 0. According to Theorem 13.1.7, I is a separative exchange ideal of R, and so we are done by [356, Theorem 2.4]. Lemma 13.1.11. If for any α ∈ R, there exists an n ∈ N and an idempotent e ∈ R such that Rαn + J(R) = Re + J(R), then R is an exchange ring. ∼ RR . Let the Proof. Suppose that G = M ⊕ N = A ⊕ B with M = projections of G onto M, A and B be π, ξ and ζ, respectively. Let α = πξ|M and β = πζ|M . Then α+β = 1M . By hypothesis, there exists an n ∈ N and an idempotent e ∈ R such that Rαn +J(R) = Re+J(R). Thus, we can find

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some ρ ∈ R and r ∈ J(R) such that e = ρα+r. Let ϕ : R → R/J(R) be the n n natural homomorphism. Then ϕ(e) = ϕ(ρ)ϕ(α) and ϕ(α

)ϕ(e) = ϕ(α
). n n Thus, ϕ(α ) ϕ(1) − ϕ(e) = 0, and so ϕ(1) − ϕ(α ) ϕ(1) − ϕ(e) = ϕ(1) − ϕ(e). Let h = 1 + α + · · · + αn−1 . then ϕ(h)ϕ(β)ϕ(1 − e)
= ϕ(1 − e). One easily checks that ϕ(eραe) = ϕ(e) and ϕ (1 − e)hβ(1 − e) = ϕ(1 − e). Hence, there exist some u ∈ eRe, v ∈ (1 − e)R(1 − e) such that uραe = e and vhβ(1 − e) = 1 − e. Let σ = ξuρπξ and τ = ζvhπζ. Then σ = σ 2 , τ = T T τ 2 ∈ EndR (G). Let A′ = A ker(σ), B ′ = B kerτ ), A′′ = im(σ) and B ′′ = im(τ ). Then G = im(σ) ⊕ ker(σ), and so \ \ \
A=A G=A im(σ) ⊕ ker(σ) = im(σ) ⊕ A ker(σ) = A′ ⊕ A′′ .

Likewise, B = B ′ ⊕ B ′′ . Thus, G = A′ ⊕ B ′ ⊕ A′′ ⊕ B ′′ . Obviously, σ + τ : G → A′′ ⊕ B ′′ is the projection onto A′′ ⊕ B ′′ . It will suffice to show that θ := (σ + τ )|M is an isomorphism. Define f : A′′ ⊕ B ′′ → M by f (a + b) = uρπ(a) + vhπ(b) for any a ∈ A′′ , b ∈ B ′′ . Then f θ = f (σ + τ )|M = uρπξuρπξ|M + vhπζvhπζ|M = uρα + vhβ.

As ϕ(ρα) = ϕ(u) = ϕ(e), we get ϕ(f θ) = ϕ(e + vhβ). If (e + vhβ)(x)
= 0, then ex = 0, and so x = (1 − e)x. This implies that x = vhβ(1 − e) (x) = (e + vhβ)(x) = 0, i.e., e + vhβ is injective. As in the
proof of [392, Theorem 3], we see that x = (e + vhβ) x + (1 − e)x − vhβx ; hence, e + vhβ is surjective. Therefore e + vhβ ∈ R is an automorphism. This implies that f θ is a unit in R, whence, f is a splitting epimorphism. If f (a + b) = 0 for some a ∈ A′′ and b ∈ B ′′ , then there exist x, y ∈ G such that a = σ(x) and b = τ (y). Thus, uρπ(a) + vhπ(b) = 0. This implies that uρπ(a) = vhπ(b) = 0. Hence,
0 = uρπ(a) = uρπ σ(x) = uρπξuρπξ(x) = uρπξ(x),
and so a = ξ uρπξ(x) = 0. Likewise, b = 0. As a result, f is an automorphism. Therefore θ is an automorphism, and then G = M ⊕ A′ ⊕ B ′ with A′ ⊆ A, B ′ ⊆ B, as required.  Let R be a ring, and suppose that for any α ∈ R, there exists an n ∈ N and an idempotent e ∈ R such that αn R + J(R) = eR + J(R). Then R is an exchange ring. This is proved by symmetry. Theorem 13.1.12. If for any α ∈ R, there exists an n ∈ N and an idempotent e ∈ R such that Rαn + J(R) = Re + J(R), αn R + J(R) = eR + J(R),

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then R is an exchange ring having stable range one. Proof. According to Lemma 13.1.11, R is an exchange ring. Let S = R/J(R). For any x ∈ S, by hypothesis, there exists an idempotent e ∈ R such that Sxn = Se and xn S = eS. Let f = e. Then we have some y, z ∈ S such that xn = yf and xn z = f . This implies that xn = xn f = xn (xn z) = xn+1 (xn−1 z). Hence, S is a strongly π-regular ring. In view of Theorem 13.1.7, S has stable range one. Therefore R has stable range one.  Let R = { m n | m, n ∈ Z, n is odd }. Then J(R) = 2R. Further, for any α ∈ R, there exists an idempotent e ∈ R such that Rα + J(R) = Re + J(R), αR + J(R) = eR + J(R). In this case, e = 0 if α ∈ J(R); e = 1 if α 6∈ J(R). This means that R satisfies the condition in Theorem 13.1.12. But R is not a strongly π-regular ring as J(R) is not nil.  We say that M is a superfluous submodule of a right module A, i.e., M ⊆s A, provided that M + N = A implies that N = A. As is well known, J(A) is also equal to the sum of all superfluous submodules of A. Let A be a quasi-projective right R-module in which every submodule is contained in a maximal submodule (e.g., finitely generated quasi-projective right modules), then J(A) is a superfluous submodule of A. Proposition 13.1.13. Let A be a quasi-projective right R-module. If for any α ∈ EndR (A), there exist n ∈ N, e = e2 ∈ EndR (A) and some M ⊆s A such that αn (A) + M = e(A) + M, then A has the finite exchange property. Proof. Let α ∈ EndR (A). By hypothesis, there exist n ∈ N, e = e2 ∈ EndR (A) and some M ⊆s
A such that αn (A) + M = e(A)
+ M. Let π1 : αn (A) → αn (A) + M /M and π2 : e(A) → e(A) + M /M be the natural epimorphisms, respectively. Since A is quasi-projective, there exists some ρ ∈ EndR (A) such that the following diagram
π e A → e(A) →2 e(A) + M /M ρ↓ k
π1 αn n A → α (A) ։ e(A) + M /M

commutates, i.e., π1 αn ̺ = π2 e. This implies that im(e − αn ρ) ⊆ M . Let γ = e − αn ρ. Then e = αn ρ + γ with im(γ) ⊆ M . Since M is a superfluous submodule of
A, so is im(γ). As in the proof [4, Proposition 17.1], γ ∈
J EndR (A) . Similarly, we can find some ρ′ ∈ EndR (A), γ ′ ∈ J EndR (A)

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such that αn = eρ′ + γ ′ . This implies that αn E + J(E) = eE + J(E), where E = EndR (A). According to Lemma 13.1.11, E is an exchange ring. By virtue of [382, Theorem 28.7], A has the finite exchange property.  Let A be a quasi-projective right R-module. If for any α ∈ EndR (A), there exists an n ∈ N and a unit-regular x ∈ EndR (A) such that (αn − x)(A) ⊆s A, then A has the finite exchange property. For any α ∈ EndR (A), there exists an n ∈ N and a unit-regular x ∈ EndR (A) such that (αn − x)(A) ⊆s A. Let M = (αn − x)(A). Then M ⊆s A. As x ∈ R is unit-regular, there exists an idempotent e ∈ R and an invertible u ∈ R such that x = eu. Hence, an (A) ⊆ (an − x)(A) + x(A) = e(A) + M , and so an (A) + M ⊆ e(A) + M . Likewise, e(A) + M ⊆ an (A) + M . Therefore we have an (A) + M = e(A) + M . According to Proposition 13.1.13, we are done. Definition 13.1.14. Let I be an ideal of a ring R. We say that I is a strongly π-regular ideal of R when every element in I is strongly π-regular. Proposition 13.1.15. Let I be an ideal of a ring R. Then the following are equivalent: (1) I is strongly π-regular. (2) For any x ∈ I, there exist n ∈ N, y ∈ I such that xn = xn+1 y. Proof. (1) ⇒ (2) is trivial. (2) ⇒ (1) Given any x ∈ I, then we have n ∈ N and y ∈ I such that xn = xn+1 y, and then xn = xn+i y i for any i ∈ N. Also we have some m ≥ n and z ∈ I such that y m = y m+1 z; hence, y m = y m+j z j for any j ∈ N. Furthermore, we get xm = xm−n xn = xm−n xn+m y m = x2m y m . Set a = xm , b = y m and c = z m . Then a = a2 b and b = b2 c. As c, a ∈ I, we have c − a ∈ I. By hypothesis, there exist k ∈ N, d ∈ I such that (c − a)k = (c − a)k+1 d. One easily checks that ac = (a2 b)c = a(a2 b)bc = a3 (b2 c) = a3 b = a2 and abc = a2 (b2 c) = a2 b = a. Hence (c − a)2 = c2 − ca − ac + a2 = c(c − a). We infer that ab(c − a)2 = abc(c − a) = a(c − a) = 0. It follows from b2 (c − a)2
= b2 c(c − a) = b(c − a) that b(c − a) =
b b(c−a) (c−a) = b b2 (c−a)2 (c−a) = b3 (c−a)3 . Likewise, we deduce that bk (c−a)k = b(c−a). Thus b(c−a)2 d = bk (c−a)k+1 d = bk (c−a)k = b(c−a). So 0 = ab(c − a)2 d = ab(c − a), whence,
a = abc = aba. In addition, 0 = ab(c−a)d = ab2 (c−a)2 d = ab b(c−a) = ab2 (c−a); hence, ab2 c = ab2 a. T This implies that a = aba = a(b2 c)a = ab2 a2 . Therefore a ∈ a2 R Ra2 ,

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T and so xm ∈ xm+1 R Rxm+1 . As in the proof of Proposition 13.1.2, x ∈ I is strongly π-regular. Consequently, I is strongly π-regular.  

     QQ QQ 00 Let R = , and let I = . Write r = . Then 0 Z 0 0 02 rn 6∈ rn+1 R for all n ∈ N. This implies that R is not a strongly π-regular    2  3 xy xy xy ring. Let ∈ I. If x = 0, then = . If x 6= 0, then 00 00 00    2  −1 −1  xy xy x x = . Therefore, I is a strongly π-regular ideal 00 00 0 0 of R. This means that the strongly π-regular ideal is the other kind of extension of strongly π-regular rings. In view of Proposition 13.1.15, an ideal I of a ring R is strongly πregular if and only if for any a ∈ I the chain aR ⊇ a2 R ⊇ · · · terminates. T If {Ii }(i ∈ K) is a set of strongly π-regular ideals of a ring R, then Ii is i∈K

a strongly π-regular ideal of R. Also we note that every nil ideal of a ring is a strongly π-regular ideal. Let p ∈ Z be a prime and set Z(p) = {a/b | b 6∈ Zp (a/b in lowest terms)}.

Then Z(p) is a local ring with maximal pZ(p) . Thus, the Jacobson radical pZ(p) is
strongly π ∼regular. Choose p/(p + 1) ∈ pZ(p) . Then p/(p + 1) ∈ J Z(p) is not nilpotent. This implies that pZ(p) is not strongly π-regular. Proposition 13.1.16. Let I be an ideal of a ring R. Then the following hold:

(1) I is strongly π ∼regular if and only if for any x ∈ 1 + I, there exists n ∈ N such that xn ∈ R has a group inverse. (2) I is strongly π-regular if and only if for any x ∈ I, there exists n ∈ N such that xn ∈ R has a group inverse.

Proof. Write xn = xn+1 z, xz = zx and z = z 2 x. Let y = z n . Then xn = x2n y, xy = yx and y = y 2 xn . Set u = y + 1 − xn y. Then xn u = xn y +xn (1−xn y) = xn y is an idempotent. In addition, u−1 = xn +1−xn y. Let e = xn y. Then {xn , ue, e} is a group in R. Therefore xn ∈ R has a group inverse by Lemma 6.2.4. Conversely, assume that xn ∈ R has a group inverse. Then xn = xn yxn , y = yxn y and xn y = yxn for some y ∈ R. T Thus, xn ∈ xn+1 R Rxn+1 , and the result follows by Proposition 13.1.2 and Proposition 13.1.15. 

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Proposition 13.1.17. Every square matrix over a strongly π-regular ideal of a ring is the sum of an idempotent matrix and an invertible matrix. Proof. Let I be a strongly π-regular ideal of a ring R, and let x ∈ I. As in the proof of Proposition 13.1.16, there exists n ∈ N such that xn = f w = wf , where f = f 2 ∈ R, w ∈ U (R) and f, w and x all commutate. Let e = 1 − f and u = x − e. Choose v = xn−1 w−1 f − (1 + x + · · · + xn−1 )e. Then
uv = (x − e) xn−1 w−1 f − (1 + x + · · · + xn−1 )e = xn w−1 f + (1 − x)(1 + x + · · · + xn−1 )e = f + (1 − xn )e = f +e = 1. As uv = vu, we get vu = 1. Therefore x is the sum of an idempotent e ∈ R and a unit u ∈ R.   a11 a12 Assume that the result holds for n × n matrix. Let A = ∈ a21 a22 Mn+1 (I) with a11 ∈ I, a12 ∈ M1×n (I), a21 ∈ Mn×1 (I) and a22 ∈ Mn (I). Clearly, there exists an idempotent e ∈ R and a unit u ∈ R such that a11 = e + u. By hypothesis, there exists an idempotent matrix E and an invertible U such that a22 − a21 u−1 a12 = E + U . Hence   u a12 A = diag(e, E) + . a11 U + a21 u−1 a12   u a12 Clearly, diag(e, E) is an idempotent matrix and ∈ a11 U + a21 u−1 a12 GLn+1 (R). Therefore the proof is true by induction.  Proposition 13.1.18. Let I be an ideal of a ring R. Then the following are equivalent: (1) I is strongly π-regular. (2) For any element a ∈ I, there exist e = e2 ∈ R, u ∈ U (R) and a nilpotent w ∈ R such that a = eu + w and e, u, w commutate. Proof. (1) ⇒ (2) Given any a ∈ I, we have n ∈ N and x ∈ R such that an = an+1 x, ax = xa. Set w = (a − a2 x) + (a − a2 x)ax + · · · + (a − a2 x)an−2 xn−2 and z = an xn−1 . Then (a−a2 x)n = an (1−ax)n = an (1−ax)(1−ax)n−1 = 0. That is, w ∈ I is nilpotent. Since z = zxz and xz = zx, we deduce that z = zvz, vz = zv and v = v 2 z, where v = xzx. Set u = z + 1 − zv.

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Then u−1 = v + 1 − zv. Moreover, we have z = eu, where e = zv ∈ I is an idempotent. Obviously, a = eu + w and e, u, w commutate. (2) ⇒ (1) For any element a ∈ I, there exist e = e2 ∈ R, u ∈ U (R) and a nilpotent w ∈ R such that a = eu + w and e, u, w commutate. Let x = eu. Then x = x2 u−1 , xu−1 = u−1 x and wu−1 = u−1 w. One easily checks that a − a2 u−1 = (w + eu) − (w + eu)2 u−1 = w − w(2e + wu−1 ) is nilpotent. Clearly, au−1 = u−1 a. Consequently, there is some n ∈ N such that an ∈ an+1 R. According to Proposition 13.1.15, I is strongly π-regular.  Let I be an ideal of a ring R. One easily checks that I is strongly πregular if and only if for any element a ∈ I, there exists n ∈ N such that R = an R + r(an ) if and only if for any element a ∈ I, there exists n ∈ N such that R = Ran + ℓ(an ). Let e be an idempotent in a strongly π-regular ideal I, and let f ∈ EndR (eR). We claim that if f is an R-monomorphism then f is an R-isomorphism. Clearly, f (e) = ex for some x ∈ R. Let y = ex. Then y ∈ I. Further, f k (e) = ey k for all k ∈ N. As I is strongly π-regular, there exists some t ∈ N such that y t = y t+1 z for some z ∈ I.
Thus, f t (e) = ey t = ey t+1 z = f t+1 (ez), and so f t e − f (ez) = 0. As f t is an R-monomorphism, we get e = f (ez). This implies that f is an isomorphism, and we are done. Lemma 13.1.19. Let I be a regular ideal of a ring R and x1 , x2 , · · · , xm ∈ I. Then there exists an idempotent e ∈ I such that xi ∈ eRe for all i = 1, 2, · · · , m. Pm Proof. Clearly there exist idempotents u, v ∈ I such that uR = i=1 xi R Pm and Rv = i=1 Rxi . It is enough to show that there exists e = e2 ∈ I with eu = u = ue and ev = v = ve. Next, f R = uR + vR for some f = f 2 ∈ I. Clearly, f u = u and f v = v. Set g = f + u(1 − f ). Obviously, g 2 = g ∈ I, ug = u = gu and gv = v. It is enough to show that there exists e = e2 ∈ I with eg = g = ge and ev = v = ve. Pick an idempotent h ∈ I with Rg + Rv = Rh. Clearly, gh = g and vh = v. Set e = h + (1 − h)g. Obviously, e = e2 ∈ I, eg = g = ge and ev = v = ve because gv = v.  The index of a nilpotent element in a ring is the least positive integer n such that xn = 0. The index i(I) of an ideal I of a ring R is the supremum of the indices of all nilpotent elements of I. An ideal I of a ring R is of bounded index if i(I) < ∞. It is well known that i(I) ≤ n if and only if I

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contains no direct sums of n + 1 nonzero pairwise isomorphic right ideals (cf. [217, Theorem 7.2]). We claim that every regular ideal of bounded index is a strongly π-regular ideal. Let I E R be a regular ideal of bounded index. Given any x ∈ I, by Lemma 13.1.19, we have an idempotent e ∈ I such that x ∈ eRe. Since I is regular, eRe is a regular ring of bounded index. It follows by [217, Theorem 7.15] that eRe is a strongly π-regular ring. So there exist n ∈ N, y ∈ eRe ⊆ I such that xn = xn+1 y. Therefore I is a strongly π-regular ideal, and we are through. Example 13.1.20. Let R be a regular ring, and let I = {a ∈ R | i(RaR) < ∞}.

Then I is a strongly π-regular ideal of R.

Proof. Let x, y ∈ I and z ∈ R. Then RxzR, RzxR ⊆ RxR. This implies that i(RxzR), i(RzxR) < ∞. That is, xz, zx ∈ I. Obviously, R(x + y)R ⊆ RxR + RyR. Assume that i(RxR), i(RyR) ≤ m. Then i(RxR + RyR) ≤ m by [217, Lemma 7.6 and Proposition 7.7]. Therefore, x + y ∈ I. Consequently, I is an ideal of R. For any a ∈ I, there exists an idempotent e ∈ I such that x ∈ eRe from Lemma 13.1.19. By hypothesis, i(ReR) ≤ n for some n ∈ N. If ebe ∈ eRe is nilpotent, then so is ebe ∈ ReR. Thus, (ebe)n = 0. This implies that eRe is of bounded index. Therefore, EndR (eR) is a regular ring of bounded index. It follows from [217, Theorem 7.15] that eRe is a strongly π-regular ring. Hence xn = xn+1 (eye) for some n ∈ N, y ∈ R. Let z = eye. Then xn = xn+1 z for some z ∈ I. According to Proposition 13.1.15, I is strongly π-regular, as asserted.  Theorem 13.1.21. Let T = (A, B, M, N, ψ, φ) with zero pairings, and let I = (K, L, M, N, ψ, φ), where K and L are ideals of A and B respectively. Then I is a strongly π-regular ideal of bounded index if and only if so are K and L. Proof. Let I be a strongly π-regular ideal of bounded index. Set e =  1A 0 . Then e = e2 ∈ T . It is easy to check that K ∼ = eIe is also a 0 0 strongly π-regular ideal of A. Further, it is of bounded index. Likewise, B is a strongly π-regular ideal of bounded index. Conversely, let K and L be strongly π-regular ideals of bounded inT T T dex. Set J(K) = K J(A), J(L) = L J(B) and J(I) = I J(T ), i.e., they are Jacobson radicals of the corresponding ideals. Then K/J(K)

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and L/J(L) are strongly π-regular ideals of A/J(K)  and B/J(L),  respecJ(A) N tively. As in the proof of Lemma 3.2.5, J(T ) = ; hence, M J(B)   J(K) N J(I) = . Assume that the bounded indices of K and L are M J(L) s and  t respectively. Clearly, J(K) and J(L) are both nil. Given any  a n ∈ J(I), then as+t = 0 and bs+t = 0. So there exist m1 , m2 ∈ M mb and n1 , n2 ∈ N such that  2(s+t) a n mb  s+t  s+t a n a n = mb mb  s+t   s+t  a n1 a n2 = m bs+t m2 bs+t  1   0 n1 0 n2 = m1 0 m2 0 = 0. Therefore J(I) is nil of bounded index. Clearly, I/J(I) ∼ = K/J(K) ⊕ L/J(L) is strongly π-regular. For any x ∈ I, we have a positive integer l such that
l
x + J(I) T /J(I)
l+1
= x + J(I) T /J(I)
kl+1
= x + J(I) T /J(I) ,
l where k = 2(s + t). So we have y + J(I) ∈ I/J(I) such that x + J(I) =
kl+1
x + J(I) y + J(I) . Hence xl − xkl+1 y ∈ J(I), and then xl −
k xkl+1 y = 0. Thus, we can find some z ∈ T such that xkl = xkl+1 z. Therefore strongly π-regular fromProposition 13.1.15.  I is  p a n a n = 0 for some p ∈ N. One easily Let ∈ I such that mb mb   p  p a n3 a n for some m3 ∈ M, n3 ∈ N . So ap = 0 check that = mb m3 b p in K and bp = 0 in L. Hence as = 0 and bt = 0. Analogously to the  2(s+t) a n = 0. This implies that I preceding discussion, we claim that mb is of bounded index, as asserted. 

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Let A = B = k[x]/(x2 ) = {a + bt | a, b ∈ k, t2 = 0} where k is a field of characteristic 2. Take M = N = k made into an A-module by α ∗ (a + bt) = αa with α, a, b ∈ k. Then A and B are both strongly πregular rings. Assume that (a + bt)n = 0 in A Then (a + bt)2n = 0, hence a2n = ((a + bt)2 )n = 0. So a = 0. Therefore (a + bt)2 = a2 = 0. That is, A = B is a strongly π-regular ring of bounded index 2. Then with the zero pairings, all the conditions in Theorem 13.1.21 are satisfied. 

 A 0 Corollary 13.1.22. The triangular matrix ring is a strongly M B π-regular rings of bounded index if and only if so are A and B. Proof. In light of Theorem 13.1.21, we choose N = 0. Then the result follows.  By induction, we deduce that the ring of all upper (lower) triangular matrices over any strongly π-regular ideal of bounded index is also a strongly π-regular ideal of bounded index. Corollary 13.1.23. Let T = (A, B, M, N, ψ, φ) with zero pairings. If A and B are regular rings of bounded index, then T is a strongly π-regular ring. Proof. Since A and B are regular rings of bounded index, it follows from [217, Theorem 7.15] that A and B are strongly π-regular. Hence we obtain the result by Theorem 13.1.21.  Let D be a division ring and let R = {(x1 , · · · , xn , y, y, · · · ) | xi ∈ Mi (D), n ∈ N, y ∈ D} where y is treated as a scalar matrix of proper size when multiplied with xi . By [424, Example 2.3], R is a strongly πregular ring not of bounded index, while its Jacobson radical is nilpotent. For Morita context over strongly π-regular ideals with nilpotent Jacobson radicals, we now observe the following fact. Theorem 13.1.24. Let T = (A, B, M, N, ψ, φ) with zero pairings, and let I = (K, L, M, N, ψ, φ), where K and L are ideals of A and B respectively. Then I is a strongly π-regular ideal with nilpotent Jacobson radical if and only if so are K and L. Proof. One direction is obvious. Conversely, assume that K and L are strongly π-regular ideals with nilpotent Jacobson radicals. As in the proof

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of Theorem 13.1.21, J(I) =



J(K) N M J(L)

, J(K)s = 0 and J(L)t = 0

  2(s+t) a n a n ∈ J(I), we see that = 0. mb mb Hence J(I)2(s+t) = 0. As in the proof of Theorem 13.1.21, I is a strongly π-regular ideal, as asserted.  for some s, t ∈ N. Given any





Corollary 13.1.25. Let T = (A, B, M, N, ψ, φ) with zero pairings. If A and B are right (left) artinian, then T is strongly π-regular. Proof. Since A and B are right (left) artinian, they are strongly π-regular rings with nilpotent Jacobson radicals. We obtain the result from Theorem 13.1.24.  Corollary 13.1.26. Let T = (A, B, M, N, ψ, φ) with zero pairings. If A and B are regular P.I. rings, then T is a strongly π-regular ring. Proof. Since A is a regular ring, we claim that every projective right A-module has the finite exchange property. By [372, Corollary 4.12], A is a strongly π-regular ring. Likewise, B is strongly π-regular. Clearly, J(A) = 0 and J(B) = 0. Thus the result follows from Theorem 13.1.24.  13.2

Stable Ideals

In this section, we introduce the concept of the stable ideal and give several characterizations. We develop some techniques to deal with diagonal reduction of square matrices over such an ideal. We prove that every regular square matrix over a stable exchange ideal is the product of an idempotent matrix and an invertible matrix, and admits a diagonal reduction. Definition 13.2.1. Let I be an ideal of a ring R. We say that I is stable provided that aR + bR = R with a ∈ I, b ∈ R implies that there exists some y ∈ R such that a + by ∈ U (R). If R has stable range one, then every ideal of R is stable. We also see that J(R) is stable for an arbitrary ring R. Thus, the concept of the stable ideal is a non-trivial generalization of the stable range one.

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Proposition 13.2.2. Let I be an exchange ideal of a ring R. Then the following are equivalent: (1) I is stable. (2) Every regular element in I is unit-regular. Proof. (1) ⇒ (2) Given any regular element x ∈ I, we have x = xyx for some y ∈ R. Since I is stable, from xy + (1 − xy) = 1, we can find w ∈ R such that y + w(1
− xy) = u ∈ U (R) by Lemma 4.1.2. Thus, x = xyx = x y + w(1 − xy) x = xux, as asserted. (2) ⇒ (1) Assume ax + b = 1 with a ∈ I and x, b ∈ R. Then b ∈ 1 + I. Since I is an exchange ideal, there exists e = e2 ∈ R such that e = bs and 1−e = (1−b)t, where s, t ∈ R. Thus, (1−e)axt+e = (1−e)(1−b)t+e = 1. Clearly, (1 − e)a is regular in I. So we may assume (1 − e)a = f u with f = f 2 ∈ R and u ∈ U (R). Hence, f uxt + e = 1 and so
a + bs (1 − f )u − a = (1 − e)a + e(1 − f )u = f u + e(1 − f )u
−1 = 1 + f uxt(1 − f ) u ∈ U (R), as required.



Corollary 13.2.3. Let I be an exchange ideal of a ring R. Then the following are equivalent: (1) I is stable. (2) For any regular a, b ∈ I, aR = bR implies that there exists u ∈ U (R) such that a = bu. Proof. (1)⇒(2) From aR = bR, we have a = bx and b = ay for some x, y ∈ R. Since b is regular, there exists c ∈ R such that b = bcb. Thus, we have a = b(cbx) and b = ay. Hence, b = b(cbx)y. Since (cbx)y + (1 − cbxy) = 1 with cbx ∈ I, we can find some z ∈ R such that cbx + (1 −
cbxy)z = u ∈ U (R). Consequently, a = bx = b(cbx) = b cbx + (1 − cbxy)z = bu. (2) ⇒ (1) Given any regular element x ∈ I, we have x = xyx for some y ∈ R. Set e = xy. Then e = e2 ∈ I. As xR = eR, x = eu for some u ∈ U (R) and so euy + (1 − xy) = 1. Therefore, x + (1 − xy)(1 − e)u =
−1 1 + euy(1 − e) u ∈ U (R). Hence, y + z(1
− xy) = v ∈ U (R) for some z ∈ R. Therefore x = xyx = x y + z(1 − xy) x = xvx, as required. 

Corollary 13.2.4. Let I be an exchange ideal of a ring R. Then the following are equivalent:

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(1) I is stable. (2) For any idempotents e, f ∈ I, eR ∼ = f R implies that there exists some u ∈ U (R) such that e = uf u−1 . (3) For any idempotents e, f ∈ I, eR ∼ = f R implies that (1−e)R ∼ = (1−f )R. Proof. (1) ⇒ (2) Let e, f ∈ I be idempotents such that ϕ : eR ∼ = f R. Then we have some a ∈ eRf and b ∈ f Re such that e = ab and f = ba. As b ∈ I is regular, it follows by Proposition 13.2.2 that b = bvb for some v ∈ U (R). Let u = (1 − ab − vb)v(1 − ba − bv). We can check that u ∈ U (R) and e = uf u−1 , as required. (2) ⇒ (3) Assume eR ∼ = f R with idempotents e, f ∈ I. Then there exists some u ∈ U (R) such that e = uf u−1 . Let a = (1 − e)u(1 − f ) and b = (1 − f )u−1 (1 − e). Then 1 − e = ab and 1 − f = ba. This implies that (1 − e)R ∼ = (1 − f )R. (3) ⇒ (1) Given any regular x ∈ I, there exists y ∈ I such that x = xyx. Then we have R = yxR ⊕ ker(x) = xR ⊕ (1 − xy)R. Since α : yxR ∼ = xyR with idempotents yx, xy ∈ I, we have ∼ (1 − xy)R. β : ker(x) = (1 − yx)R =

Let u : R → R be the map given by u(s + t) = α−1 (s) + β −1 (t) for any s ∈ xR and t ∈ (1 − xy)R. It is easy to verify that x = xux with u ∈ U (R), as required.  Lemma 13.2.5. Let I be an ideal of a ring R. If eRe has stable range one for any e = e2 ∈ I, then I is stable.

Proof. Suppose eR ∼ = f R with idempotents e, f ∈ I. We have right R-module decompositions R = eR ⊕ (1 − e)R = f R ⊕ (1 − f )R. Since EndR (eR) ∼ = eRe has stable range one, we can find some C ⊆ R such that R = C ⊕ (1 − e)R = C ⊕ (1 − f )R by Theorem 1.1.5. Therefore, (1 − e)R ∼  = (1 − f )R. Corollary 13.2.4 yields the result. For any ring R, we have that {I E R | I is a B-ideal } ⊆ {I E R | eRe has stable range one for any e = e2 ∈ I} ⊆ {I ER | I is stable }. Let R be
a regular ring, and let a ∈ R. In view of [217, Corollary 7.3], i EndR (aR) = ∞ if and only if for all positive integers n, aR contains direct sums of n + 1 nonzero pairwise isomorphism submodules. Furthermore, one sees that
i EndR (aR) ≤ n(< ∞) if and only if for any nonzero finitely generated projective right R-submodule Q of aR, (n + 1)Q 6. aR. Let S be the multiplicative semigroup generated by all idempotents of R. The depth ∆(R) of R is the least positive integer n such that each element of S can

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be written as a product of n idempotents if such an n exists, otherwise
∆(R) = ∞(cf.
[244]). According to [244, Proposition 1.1], ∆ EndR (aR) ≤ i EndR (aR) for any a ∈ R. Let
I = {a ∈ R | i EndR (aR) < ∞}. As is known, I is an ideal of R. Similarly to Example 13.1.20, we prove that I is a strongly π-regular ideal of R. Further, we get the following. Proposition 13.2.6. Every strongly π-regular ideal of a ring R is a stable exchange ideal. Proof. Let I be a strongly π-regular ideal of R. For any x ∈ I, we have n ∈ N and y ∈ I such that xn = xn+1 y and xy = yx. As in the proof of Lemma 13.1.3, we have an idempotent e ∈ Rx such that 1 − e ∈ R(1 − x). Thus, I is an exchange ideal. For any idempotent e ∈ I, it is easy to verify that eRe is a strongly π-regular ring. According to Theorem 13.1.7, eRe has stable range one. Therefore I is stable from Lemma 13.2.5.  Proposition 13.2.7. Every exchange ideal of bounded index is stable. Proof. Assume I has index n. For any e = e2 ∈ I, if exe ∈ eRe is nilpotent, then exe ∈ I is also nilpotent. Thus, eRe is an exchange ring of bounded index. In view of [422, Corollary 4], eRe has stable range one. So the result follows from Lemma 13.2.5.  Proposition 13.2.8. Let I be an exchange ideal of a ring R. If all idempotents in I are central, then I is stable. Proof. Assume all idempotents in I are central. For any e = e2 ∈ I, if exe is an idempotent of eRe, then exe ∈ I is central in I. So eRe is an exchange ring with all idempotents central. From [422, Theorem 6], eRe has stable range one. Thus, we complete the proof by Lemma 13.2.5.  Let I be an ideal of a ring R. We say that I is a P I-ideal provided that there exists a monic polynomial identity f (x1 , · · · , xn ) ∈ Z[x1 , · · · , xn ] such that f (r1 , · · · , rn ) = 0 for all ri ∈ I, where f is monic if at least one of the words of highest degree in the support of f has coefficient 1. Clearly, every ideal of a commutative ring is a P I-ideal. We claim that every exchange P I-ideal is stable. Let I be an exchange P I-ideal of a ring R. Then eRe is an exchange P I-ring for any e = e2 ∈ I. By [422, Corollary 2], eRe has stable range one for any e = e2 ∈ I. In view of Lemma 13.2.5, we are through.

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We say that an ideal I of a ring R is left quasi-duo, provided that for any maximal left ideal J of R, JI ⊆ J. Obviously, every ideal of a left quasi-duo ring is left quasi-duo. Similarly to [419, Proposition 2.2], we can prove that an ideal I of a ring R is left quasi-duo if for any x ∈ I, there exists n(x) ∈ N such that Rxn(x) I ⊆ Rxx(n) . Let V be an infinite-dimensional vector space over a division ring D. Set D D  D D  R= ⊕ EndD (V ) and I = . 0 D 0 D

Then R is not a left quasi-duo ring which contains a left quasi-duo ideal I. Moreover, there exists an idempotent e ∈ I which is not central. A right quasi-duo ideal can be defined in a similar way. Proposition 13.2.9. Every left quasi-duo exchange ideal is stable.

Proof. Let I be a left quasi-duo exchange ideal of a ring R, and let e = e2 ∈ I. Set S = eRe/J(eRe). Then S is an exchange ring by [382, Theorem 29.2]. For any idempotent f ∈ S, there exists an idempotent g ∈ eRe such that f = g + J(eRe). Now g = g 2 ∈ eRe ⊆ I. For any x ∈ eRe, we easily check that (gx − gxg)2 = 0. Let R(gx − gxg) + N = R with N ⊆ R R. Since the union of every ascending chain of proper submodules of R R is proper, by Zorn’s Lemma, we may assume N is a maximal left ideal of R. Obviously, N (gx − gxg) + N = R. Since I is left quasi-duo, we see that R ⊆ N I + N ⊆ N ; hence, N = R. Therefore, gx − gxg ∈ J(R). Likewise, we have xg − gxg ∈ J(R) and so gx − xg ∈ J(R). Hence, gx − xg = e(gx − xg)e ∈ eJ(R)e ⊆ J(eRe). Thus, gx = xg in S. This implies that all idempotents in S are central. By [422, Theorem 6], S has stable range one, and so has eRe. By Lemma 13.2.5, the proposition is proved.  Immediately, we deduce that every left quasi-duo exchange ring has stable range one. The right quasi-duo property can be deduced by symmetry. Lemma 13.2.10. Let I be an ideal of a ring R. Then the following are equivalent : (1) I is stable. (2) For any (aij ) ∈ GL2 (R) with a11 ∈ I, (aij ) = [∗, ∗]B21 (∗)B12 (∗)B21 (∗). Proof. (1)⇒(2) Assume (aij ) ∈ GL2 (R) with a11 ∈ I. Then a11 x+a12 y = 1 for some x, y ∈ R. Hence, we have z ∈ R such that a11 +a12 yz = u ∈ U (R).

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It can be directly checked that
B21 − (a21 + a22 yz)u−1 (aij )B21 (yz)B12 (−u−1 a12 ) = [u, a22 − (a21 + a22 yz)u−1 a12 ]. Thus, (aij ) = [∗, ∗]B21 (∗)B12 (∗)B21 (∗), as required.

(2) ⇒ (1) Given ax + b = 1 with a ∈ I, we have



 a b = −1 x

  −1  x xa − 1 a b ∈ GL2 (R). Hence, = [∗, ∗]B21 (∗)B12 (∗)B21 (−y) 1 a −1 x for some y ∈ R. Consequently, a + by ∈ U (R), as asserted.  It is well known that if R has stable range one, then so does Mn (R) (cf. [386, Theorem 2.4]). Unfortunately, Vaserstein’s technique can not be applied to stable ideals. We need to develop some new methods. Theorem 13.2.11. Let I be an ideal of a ring R. If I is stable, then so is Mn (I) for any n ∈ N. A B 
Proof. Assume ∈ GL2 Mn (R) with A = (aij ) ∈ Mn (I), and C D B = (bij ), C = (cij ), D = (dij ) ∈ Mn (R). Since a11 R + · · · + a1n R + b11 R + · · · + b1n R = R with a11 ∈ I, we can find y2 , . . . , yn , z1 , . . . , zn ∈ R such that a11 + a12 y2 + · · · + a1n yn + b11 z1 + · · · b1n zn = u1 ∈ U (R). Thus, we have     u1 a12 · · · a1n b11 · · · b1n 1   a′ a · · · a b · · · b  y 1 2n 21 2n    21 22  2   . . . . .. . . ..  .. ..    . . . .   . . .  . . . .    . . AB     1  =  ∗ an2 · · · ann bn1 · · · bnn  . yn    CD    ∗ c12 · · · c1n d11 · · · d1n   z1 1     .. . . ..   .. . .   .. .. . . .. . . .  . . .   . . . ∗ cn2 · · · cnn dn1 · · · dnn zn 1 Hence,   u1 a′12 · · · a′1n b′11 · · · b′1n  0 a′ · · · a′ b ′ · · · b ′   22 2n 21 2n  . . . .. . . ..  .. .. . . ...      . . .    ∗ 0 AB ∗ 0  ′ ′ ′ ′  =  0 an2 · · · ann bn1 · · · bnn  ,   0 In CD ∗ In  ∗ c12 · · · c1n d11 · · · d1n    .. . . ..   .. .. . . .. . . . . . .  . ∗ cn2 · · · cnn dn1 · · · dnn

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′ ′ ′ where a′22 = a22 − a′21 u−1 1 a12 ∈ I. Since a22 R + · · · + a2n R + b21 R + · · · + b′2n R = R with a′22 ∈ I, we have y3′ , . . . , yn′ , z1′ , . . . , zn′ ∈ R such that a′22 + a′23 y3′ + · · · + a′2n yn′ + b′21 z1′ + · · · b′2n zn′ = u2 ∈ U (R). Therefore,   u1 a′′12 · · · a′′1n b′′11 · · · b′′1n  0 u · · · a′′ b′′ · · · b′′  2  2n 21 2n  . . . .. .. . . ..   . . .   . . . .  . . .  AB   [∗, ∗] B21 (∗)[∗, ∗] =  0 0 · · · a′′nn b′′n1 · · · b′′nn  .   CD  ∗ ∗ · · · c1n d11 · · · d1n    .. . . ..   .. .. . . .. . . . . . . .  ∗ ∗ · · · cnn dn1 · · · dnn

Finally, we arrive at   AB [∗, ∗] B21 (∗)[∗, ∗] CD  (n) (n) (n) u1 a12 · · · a1n b11 · · ·  0 u · · · a(n) b(n) · · · 2  2n 21 . . . .. . . . . . . . . .. . . .  (n) =  0 0 · · · un bn1 · · ·   ∗ ∗ · · · ∗ d11 · · ·  . . . .. . . . . ...  .. .. . . ∗ ∗ · · · ∗ dn1 · · ·

(n)  b1n (n) b2n   ..   .  (n)  = [∗, ∗]B (∗)B (∗), 21 12 bnn   d1n   ..  . 

dnn

where u1 , u2 , . . . , un ∈ U (R). Therefore,   AB = [∗, ∗]B21 (∗)B12 (∗)B21 (∗). CD

In view of Lemma 13.2.10, we conclude that Mn (I) is stable.



Corollary 13.2.12. Let I be an exchange ideal of a ring R. Then the following are equivalent: (1) I is stable. (2) For any n ∈ N and any regular A ∈ Mn (I), there exist E = E 2 ∈ Mn (R), U ∈ GLn (R) such that A = EU . Proof. (1) ⇒ (2) According to Theorem 13.2.11, Mn (I) is a stable ideal of Mn (R). By virtue of Proposition 13.2.2, every regular square matrix over I is unit-regular, as required.

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(2)⇒(1) Take n = 1. Then we see that every regular element in I is unit-regular, completing the proof from Proposition 13.2.2.  Let I be a stable ideal of a ring R. We claim that for any regular A ∈ Mm×n (I)(m ≤ n), there exists an E = E 2 ∈ Mm (R) and a right invertible U ∈ Mm×n (R) such that A = EU . Let A = (aij ) ∈ Mm×n (I)(m ≤ n) be regular. Then we have B = (bij ) ∈ Mn×m (I) such that A = ABA. Set A′ = (a′ij ) and B ′ = (b′ij ) ∈ Mn (R), where a′ij = aij (1 ≤ i ≤ m), a′ij = 0(m + 1 ≤ i ≤ n), b′ij = bij (1 ≤ j ≤ m) and b′ij = 0(m + 1 ≤ j ≤ n). Then A′ = A′ B ′ A′ in Mn (R). Since I is stable, so is Mn (I) by Theorem 13.2.11. From A′ B ′ + (In − A′ B ′ ) = In with A′ ∈ Mn (I), we can find some Y ∈ Mn (R) such that U ′
:= A′ + (In − A′ B ′ )Y ∈ GLn (R). Hence, A′ = A′ B ′ A′ + (In − A′ B ′ )Y A′ = A′ B ′ U ′ . Write U ′ = (u′ij ). Choose U = (uij ) ∈ Mm×n (R), where uij = u′ij (1 ≤ i ≤ m, 1 ≤ j ≤ n). Let E = AB. Then E = E 2 ∈ Mm (R). Furthermore,      A E0 U = . 0 0 0 ∗ Thus, A = EU and U ∈ Mm×n (R) is right invertible, and we are done. For any regular A ∈ Mm×n (I)(m ≥ n), by symmetry, there exists an idempotent E ∈ Mn (R) and a left invertible U ∈ Mm×n (R) such that A = U E. Example 13.2.13. Let V be a countably generated infinite-dimensional
vector space over a division D. Then every matrix (aij ) ∈ Mn EndD (V ) with all dimD (aij V ) < ∞ is the product of an idempotent matrix and an invertible matrix over EndD (V ). Proof. Let I = {x ∈ EndD (V ) | dimD (xV ) < ∞}. Obviously, I is an ideal of EndD (V ). Given any x ∈ I, since D is a division ring, we have the following splitting short exact sequences of right D-modules: 0 → ker(x) → V → xV

0 → xV

→ 0,

→ V → V /xV → 0.

∼ xV ⊕ ker(x) ∼ Then V = = (V /xV ) ⊕ xV . Hence, dimD (im(x)) = dimD (coker(x)) = ∞ because dimD (xV ) < ∞. We conclude that x ∈ EndD (V ) is unit-regular. Thus, the result follows from Corollary 13.2.12.  Example 13.2.14. Let R be a regular ring, and let (aij ) ∈ Mn (R). If

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each EndR (aij R) is unit-regular, then (aij ) is the product of an idempotent matrix and an invertible matrix over R. Proof. I = {a ∈ R | EndR (aR) is unit-regular}. Let x, y ∈ I and z ∈ R. Construct a map ϕ : xR → zxR given by ϕ(xr) = zxr for any r ∈ R. Then ϕ is a splitting R-epimorphism; hence, zxR ⊕ D ∼ = xR for some right Rmodule D. This implies that zxR .⊕ xR. Write xR = eR, xzR = f R for some idempotents e, f ∈ R. It is easy to verify that f R ⊕ (1 − f )eR = eR, and so xzR ⊆⊕ xR. According to Corollary 1.1.6, EndR (xz) and EndR (zx) are unit-regular. Thus, xz, zx ∈ I. Write (x+y)R = gR and xR+yR = hR for some idempotents g, h ∈ R. Then gR ⊕ (h − gh)R = hR, and so (x + y) ⊆⊕ xR + yR. As R is regular, we have a splitting exact sequence \ 0 → xR yR → xR ⊕ yR → xR + yR → 0,

T and so (xR + yR) ⊕ (xR yR) ∼ = xR ⊕ yR. This implies that (x + y)R .⊕ xR ⊕ yR. In view
of Lemma 1.1.9, EndR (xR ⊕ yR) is unit-regular, and so is EndR (x + y)R . Therefore, x + y ∈ I. Consequently, I is an ideal of R. Let e ∈ I be an idempotent. By hypothesis, eRe is a unit-regular ring. In view of Lemma 13.2.5, I is a stable ideal. Therefore we complete the proof by Corollary 13.2.12.  Example 13.2.15. Let R be a regular, right self-injective ring. Then every matrix (aij ) ∈ Mn (R) with all aij R directly finite is the product of an idempotent matrix and an invertible matrix over R.

Proof. Let H = {x ∈ R | xR is a directly finite right R-module }. Let x, y ∈ H and z ∈ R. As in the proof of Example 13.2.14, xzR ⊆⊕ xR, zxR .⊕ xR and (x + y)R .⊕ xR ⊕ yR. Since xR ⊕ yR is directly finite by [217, Corollary 9.20], it follows from [253, Corollary 1] that x + y ∈ H. Thus, H is an ideal of R. For any idempotent e ∈ H, eR is a directly finite right R-module, and so EndR (eR) ∼ = eRe is a directly finite ring. By [217, Proposition 9.8], eRe is a regular, right self-injective ring. Hence, eRe is unit-regular. In view of Lemma 13.2.5, H is stable, and the result follows by Corollary 13.2.12. 
T As usual, GLn (I) = GLn (R) In + Mn (I) . As in the proof of Lemma 13.2.10, I is a B-ideal if and only if for any (aij ) ∈ GL2 (I), (aij ) = [∗, ∗]B21 (∗)B12 (∗)B21 (∗) if and only if for any (aij ) ∈ GL2 (I), (aij ) = [∗, ∗]B12 (∗)B21 (∗)B12 (∗). In contradiction of this fact, we now derive the following.

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Proposition 13.2.16. Let I be an ideal of a ring R. Then the following are equivalent : (1) I is stable. (2) For any (aij )   u B12 (∗) v (3) For any (aij )   u B21 (∗) v

∈ GL2 (I), there exist u, v ∈ U (R) such that (aij ) = B21 (∗)B12 (∗). ∈ GL2 (I), there exist u, v ∈ U (R) such that (aij ) = B12 (∗)B21 (∗).

Proof. (1) ⇒ (2) Let (aij ) ∈ GL2 (I). Then a12 R + a11 R = R, and so u := a12 + a11 y ∈ U (R) for an element y ∈ R. Thus,       1y a11 u ∗ (aij ) = = B12 (∗)B21 (∗). 01 a21 a22 + a21 y ∗ Therefore,

(aij ) =



∗

∗



B12 (∗)B21 (∗)B12 (∗),

as desired. (2) ⇒ (1) Given ax + b =1 with a ∈ I,x, b ∈ R, then ax(1 − b) + (1 − (1 − ax)b a ax)b = 1. This implies that ∈ GL2 (I). By hypothesis, we −x(1 − b) 1 have some u, v ∈ U (R) such that     (1 − ax)b a u = B12 (∗)B21 (∗)B12 (∗). −x(1 − b) 1 v

Thus, there exists some y ∈ R such that a + (1 − ax)by = u ∈ U (R). By virtue
of Lemma 4.1.2, we have an element z ∈ R such that x + x − z(1 − ax) b = x(1 − b) + z(1 − ax)b ∈ U (R). By using Lemma 4.1.2 again, a + bt ∈ U (R) for some t ∈ R. Therefore I is stable. (2) ⇒ (3) For any A ∈ GL2 (I), then     1 1 A ∈ GL2 (I). 1 1

By hypothesis, there are u, v ∈ U (R) such that       1 1 v A = B12 (∗)B21 (∗)B12 (∗). 1 1 u        1 v 1 u Therefore we get A = B12 (∗)B21 (∗)B12 (∗) = 1 u 1 v B21 (∗)B12 (∗)B21 (∗), as asserted.

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(3) ⇒ (2) is proved in the same manner.



As a consequence, we deduce that an ideal I of a ring R is stable if and only if I op is a stable ideal of Rop . That is, the stable property for ideals is left-right symmetric. Thus, an ideal I of a ring R is stable if and only if Ra + Rb = R with a ∈ I, b ∈ R implies that there exists some z ∈ R such that a + zb ∈ U (R). Theorem 13.2.17. Let I be a stable exchange ideal of a ring R. Then every regular square matrix over I admits a diagonal reduction. Proof. Given any regular A ∈ Mn (I), by Corollary 13.2.12, there exist E = E 2 ∈ Mn (I), U ∈ GLn (R) such that A = EU . Now we have a E splitting exact sequence 0 → ker(E) → nR −→ E(nR) → 0 of right Rmodules. Thus, E(nR) is a finitely generated projective right R-module. As E ∈ Mn (I), we see that E(nR) ∈ F P (I). In view of Lemma 13.1.8, we can find idempotents e1 , . . . , en ∈ I such that E(nR) ∼ = e1 R ⊕ · · · ⊕ en R ∼ = diag(e1 , . . . , en )(nR) as right R-modules. As in the proof of Theorem 7.3.10, EMn (R) ∼ = diag(e1 , . . . , en )Mn (R). According to Theorem 13.2.11 and [9, Theorem 1.4], Mn (I) is a stable exchange ideal. In view of Corollary 13.2.4, we can find V ∈ GLn (R) such that V EV −1 = diag(e1 , . . . , en ). Therefore, V AU −1 V −1 = diag(e1 , . . . , en ), as asserted.  Corollary 13.2.18. Let I be a stable exchange ideal of a ring R. Then every regular n × n matrix (n ≥ 2) over I is the sum of two invertible matrices over R. Proof. Let A ∈ Mn (I)(n ≥ 2) be regular. By virtue of Theorem 13.2.17, we have U, V ∈ GLn (R) such that U AV = diag(a1 , . . . , an ), which can be expressed as     a1 1 −1   a2   −1         .. .. .. .  + . . .        an−1 1 −1 1

This completes the proof.

−1

an



By Proposition 13.2.6 and Theorem 13.2.17, we see that every regular square matrix over a strongly π-regular ideal admits a diagonal reduction.

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Let R be a regular ring, and let (aij ) ∈ Mn (R). If each i EndR (aij R <
∞. Then (aij ) admits a diagonal reduction. Let I = {a ∈ R | i EndR (aR) < ∞}. Then I is an ideal of R. By virtue of Lemma 13.1.19 and [217, Theorem 7.15], I is strongly π-regular. Clearly, A ∈ Mn (I), and we are through. Let I be an ideal of a ring R. If a ∈ I, then the set \  Ω = K is a right ideal of R K {1 − au | u ∈ U (R)} = ∅

is a non-empty inductive set. By using Zorn’s Lemma, there exist some right ideals J which are maximal in Ω. We denote the set of all such J’s by a . Proposition 13.2.19. Let I be an ideal of a ring R. Then the following are equivalent : (1) I is stable. (2) For each a ∈ I, J ∈ a implies that a ∈ J.

Proof. (1) ⇒ (2) Let a ∈ I, J ∈ a and a 6∈ J. Then J $ aR + J ⊆ R. If aR + J = R, then ax + b = 1 for some x ∈ R, b ∈ J. Since I is stable, there exists a y ∈ R such that v := a + by ∈ U (R). Hence, 1 − av −1 = byv −1 ∈ J. This implies that J 6∈ a , a contradiction. Consequently, T (aR + J) {1 − au | u ∈ U (R)} = ∅. Write ac + d = 1 − au for some c ∈ R, d ∈ J and u ∈ U (R). Then a(c + u) + d = 1. By hypothesis, we have a z ∈ R such that w := a+bz ∈ U (R). We infer that 1−aw−1 = bzw−1 ∈ J, a contradiction. Therefore a ∈ I, J ∈ a implies that a ∈ J. (2) ⇒ (1) Assume that I is not stable. Then there exist some a ∈ I, x, b ∈ R such that ax + b = 1, but a + by 6∈ U (R) for any y ∈ R. We infer T that bR {1 − au | u ∈ U (R)} = ∅. Let \  Ω = K is a right ideal of R bR ⊆ K and K {1 − au | u ∈ U (R)} = ∅ .

It is easy to verify that Ω is a non-empty inductive set. By using Zorn’s Lemma, there exists an ideal J of R which is maximal in Ω. As a result, bR ⊆ J ∈ a . Hence, a, b ∈ J, and so J = R. This gives a contradiction. Therefore I is stable.  Corollary 13.2.20. Let R be an integral domain. If J(R) = 0 and | U (R) | < ∞, then every nonzero ideal of R is not stable. Proof. Let 0 6= I be a stable ideal of R. Then there exists some 0 6= a ∈ I. Write U (R) = {u1 , · · · , un }. If J ∈ a , it follows from Proposition 13.2.19 T that a ∈ J. In addition, J {1 − aui | 1 ≤ i ≤ n} = ∅. If K is a right

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T ideal such that J $ K ⊆ R, then K {1 − aui | 1 ≤ i ≤ n} 6= ∅. Hence we can find some ui such that b := 1 − aui ∈ K, and so aui + b = 1. As a, b ∈ K, we deduce that K = R. Consequently, J is a maximal right ideal of R. Thus, \  a = J is a maximal right ideal of R J {1 − au | u ∈ U (R)} = ∅ . T In view of Proposition 13.2.19, a ∈ J, and then J∈a

a

n Y

i=1

(1 − aui ) ∈

\ {J | J is a maximal right ideal and J ∈ a }.

For any maximal right ideal J, J 6∈ a implies that there exists some ui such that 1 − aui ∈ J. Hence, n Y \ a (1 − aui ) ∈ {J | J is a maximal right ideal and J 6∈ a }. i=1

Therefore a

n Q

(1 − aui ) ∈ J(R). Clearly, all 1 − aui 6= 0. This implies that

i=1

a = 0 from which the contradiction arises. Consequently, every nonzero ideal is not stable.  Immediately, we see that every nonzero ideal of Z is not stable. Further, every nonzero ideal of F [x], where F is a finite field, is not stable. In view of Lemma 1.3.1, an exchange ring R has stable range one if and only if every regular element in R is unit-regular. We shall give an analogue for ideals. Lemma 13.2.21. Let x, y ∈ R. Then 1 + xy is unit-regular if and only if so is 1 + yx. Proof. Suppose that 1 + xy is unit-regular. Clearly, one checks that         1 0 10 1x 1 + xy 0 1 0 1 −x = . 0 1 + yx y1 01 0 1 −y 1 0 1   1 0 Hence ∈ M2 (R) is unit-regular. Assume that 0 1 + yx       1 0 1 0 c11 c12 1 0 = . 0 1 + yx 0 1 + yx c21 c22 0 1 + yx

Then we get 1 + yx = (1 + yx)c22 (1 + yx). Since        1 0 1 0 0 0 10 + = , 0 1 + yx 0 c22 0 1 − (1 + yx)c22 01

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 z11 z12 So we can find ∈ M2 (R) such that z21 z22      1 0 0 0 z11 z12 + ∈ GL2 (R). 0 1 + yx 0 1 − (1 + yx)c22 z21 z22   1 0
That is, ∈ GL2 (R). As a result, u := ∗ 1 + yx + 1 − (1 + yx)c22 z22
1 + yx + 1 − (1 + yx)c22 z22 ∈ U (R). Hence, 1 + yx = eu, where e = (1 + yx)c22 ∈ R is an idempotent. Therefore 1 + yx ∈ R is unit-regular. The converse is symmetric.  Lemma 13.2.22. An element a ∈ R is unit-regular if and only if diag(a, 1, · · · , 1) ∈ Mn (R) is unit-regular. Proof. Suppose that diag(a, 1, · · · , 1) ∈ Mn (R) is unit-regular. Then diag(a, 1, · · · , 1) ∈ Mn (R) is regular; hence, we have x ∈ R such that a = axa. Clearly, we have diag(a, 1, · · · , 1)diag(x, 1, · · · , 1) + diag(1 − ax, 0, · · · , 0) = In . Write diag(a, 1, · · · , 1) = EU , where E = E 2 ∈ Mn (R) and U ∈ GLn (R). Then EU diag(x, 1, · · · , 1) + diag(1 − ax, 0, · · · , 0) = In ; hence, EU diag(x, 1, · · · , 1)(In − E) + diag(1 − ax, 0, · · · , 0)(In − E) = In − E. We infer that E + diag(1 − ax, 0, · · · , 0)(In − E) = In − EU diag(x, 1, · · · , 1)(In − E). Furthermore, we get diag(a, 1, · · · ,
1) + diag(1 − ax, 0, · · · , 0)(In − E)U = In − EU diag(x, 1, · · · , 1)(In − E) U ∈ GLn (R). Let (yij ) := (In − E)U . Then   a + (1 − ax)y11 by12 · · · by1n  0 1 ··· 0     .. . . ..  ∈ GLn (R). ..  . .  . . 0 0 ··· 1 It follows that
u := a + (1 − ax)y11 ∈ U (R). Hence a = axa = ax a + (1 − ax)y11 = axu. This implies that a = au−1 a ∈ R is unit-regular. The converse is obvious.  Let Φ(R) = {x ∈ R | 1 + sxt ∈ R is unit-regular for all s, t ∈ R}, and let Ψ(R) be the ideal generated by the
set Φ(R). We list some examples of such ideals. Ψ(Z) = 0; Ψ Z ⊕ Z/2Z = 0 ⊕ Z/2Z; and     Z/2Z Z/2Z
Z/2Z Z/2Z Ψ = . 0 Z/2Z 0 Z/2Z

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The following proposition shows that Ψ(R) has a ”radical-like” property.

Proposition 13.2.23. Let R be a ring. Then Ψ Mn (R) = Mn Ψ(R) for all n ∈ N.
Proof. Given any A
∈ Ψ Mn (R) , we have A = A1 + · · · Am (m ∈ N) with each Ai ∈ Φ Mn (R) . Write A1 = (aij ). For any r ∈ R, we have     r 0 ··· 0 1 + a11 r 0 · · · 0  0 0 · · · 0   a21 r 1 · · · 0      In + A1  . . . .  =  .. .. . . ..  . . . . . . . .  . . . . 0 0 ··· 0

an1 r

0 ··· 1 

0

0 ··· 1

is unit-regular. Clearly, there exists a V ∈ GLn (R) such that     1 + a11 r 0 · · · 0 1 + a11 r 0 · · · 0  a21 r 1 · · · 0   0 1 ··· 0      .. .. . . ..  =  .. .. . . ..  V.  . . . .  . . . . 

an1 r

0 ··· 1

1 + a11 r 0 · · · 0 0 1 ··· 0  .. . . ..  ∈ Mn (R) is unit-regular, and then so is 1 + .. . . . . 0 0 ··· 1 a11 r ∈ R from Lemma 13.2.22. It follows by Lemma 13.2.21 that a11
∈ Φ(R). Likewise, we prove that each
aij ∈ Φ(R). Hence, A1 ∈ Mn Ψ(R) . Similarly,
A2 , · · · , Am ∈ Mn Ψ(R) .
Consequently,
A = A1 + · · · + Am ∈ Mn Ψ(R) . We infer that Ψ Mn (R) ⊆ M Ψ(R) . n
Given any (aij ) ∈ Mn Ψ(R) , then each aij ∈ Ψ(R). Write aij = b1 + · · · + bk with each bs ∈ Φ(R)(1 ≤ s ≤ k). For any (rij ) ∈ Mn (R), we have     1 + b1 r11 b1 r12 · · · b1 r1n b1 0 · · · 0   0 0 ··· 0 0 1 ··· 0      In +  . . . .  (rij ) =  . .. . . ..  . ..   .. .. . . ..  . .  . 0 0 ··· 1 0 0 ··· 0   Hence  

Clearly, there is a U ∈ GLn (R) such that    1 + b1 r11 b1 0 · · · 0   0 0 ··· 0 0
   U In +  . . . .  (rij ) =  ..   .. .. . . ..  . 0 0 0 ··· 0

0 ··· 1 ··· .. . . . . 0 ···

 0 0  ..  . . 1

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As 1 + b1 r11 ∈ R is unit-regular, by Lemma 13.2.22, we have   b1 0 · · · 0  0 0 ··· 0
  U In +  . . . .  (rij ) ∈ Mn (R)  .. .. . . ..  0 0 ··· 0

is unit-regular, and then so is





b1 0  In +  .  ..

0 ··· 0 ··· .. . . . .

 0 0  ..  (rij ). .

0 0 ··· 0

   0 bi 0 · · · 0  0 0 ··· 0 0

   ..  ∈ Ψ Mn (R) . Likewise,  .. .. . . ..  ∈ Ψ Mn (R)  . . . . . 0 0 ··· 0 0 0 ··· 0   a11 0 · · · 0  0 0 ··· 0
  for all i. We infer that  . . . .  ∈ Ψ Mn (R) . Analogously, . . . .  . . . .

b1 0  Thus  .  ..

0 ··· 0 ··· .. . . . .



0 0 ··· 0

0 a12 0 0  . .  .. ..

··· ··· .. .

 0 0  ..  , · · · .

0 0 ··· 0

 0 0 ··· 0 0 0 ··· 0 
  , . . . ..  ∈ Ψ Mn (R) . . . . . . . .  

0 0 · · · ann




Therefore (aij ) ∈ Ψ Mn (R) , and so Mn Ψ(R) ⊆ Ψ Mn (R) , as required.  Lemma 13.2.24. Ψ(R) is a B-ideal. Proof. Given ax + b = 1 with a ∈ 1 + Ψ(R) and x, b ∈ R, we can find c1 , · · · , cm ∈ Φ(R) such that a = 1+c1 +· · ·+cm . Hence (1+c1 )x+(c2 +· · ·+ cm )x+b = 1. As 1+c1 ∈ R is unit-regular, there exist e = e2
∈ R, u ∈ U (R) such that 1+c1 = eu. Hence eux(1−e)+ (c2 +· · ·+cm )x+b (1−e) = 1−e, and then

−1 (1 + c1 ) + (c2 + · · · + cm )x + b (1 − e)u = 1 + eux(1 − e) u ∈ U (R).

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By Lemma
4.1.2, we can find a z1 ∈ R such that u1 := x + z1 (c2 + · · · + cm )x + b ∈ U (R). Thus, −1 −1 (1 + z1 c2 )xu−1 1 + z1 (c3 + · · · + cm )xu1 + z1 bu1 = 1.

Clearly, 1 + z1 c2 ∈ R is unit-regular. Similarly, we have a z2 ∈ R such that and then

−1 −1 xu−1 1 + z2 z1 (c3 + · · · + cm )xu1 + z2 z1 bu1 ∈ U (R),

x + z2 z1 (c3 + · · · + cm )x + z2 z1 b ∈ U (R).

By iteration of this process, we have a z ∈ R such that x + zb ∈ U (R). Therefore Ψ(R) is a B-ideal.  Theorem 13.2.25. Let R be an exchange ring, and let (aij ) ∈ Mn (R) be regular. If each 1 + aij r ∈ R is unit-regular for all r ∈ R, then (aij ) admits a diagonal reduction.
Proof. In view of Lemma 13.2.21, A := (aij ) ∈ Mn Ψ(R) . By Lemma 13.2.24, Ψ(R) is a B-ideal; hence, it is stable. According to Theorem 13.2.17, there exist U, V ∈ GLn (R) such that U AV is a diagonal matrix, and so the proof is true.  Proposition 13.2.26. Let R be a regular ring. Then Φ(R) is an ideal which is maximal with respect to B-ideals. Proof. Obviously, Φ(R) = {x ∈ R | 1 + xr ∈ R is unit-regular for all r ∈ R}. Let
x, y ∈ Φ(R), s,
t ∈ R. Then sxt ∈
Φ(R). Write 1 + (x +
y)t = 1 + (x + y)t c 1 + (x + y)t . Then 1 + (x + y)t c + 1 − (1 + (x + y)t)c = 1. Clearly, x + y ∈ Ψ(R). According to Lemma 13.2.24, Ψ(R) is a B-ideal.
Hence, v := c + z 1 − (1 +
(x + y)t)c ∈
U (R) for a z ∈ R. Therefore 1 + (x + y)t = 1 + (x + y)t v 1 + (x + y)t . That is, 1 + (x + y)t ∈ R is unit-regular. Thus x + y ∈ Φ(R), and so Φ(R) is an ideal of R. We infer that Φ(R) = Ψ(R). Hence, Φ(R) is a B-ideal by Lemma 13.2.24. In fact, one easily checks that P Φ(R) = {I | I is a B-ideal} IER

= {x ∈ R | RxR is a B-ideal}.

Therefore Φ(R) is a maximal ideal with respect to B-ideals.



Immediately, we deduce that an ideal I of a regular ring R is a B-ideal if and only if for any x ∈ I, r ∈ R, 1 + xr is unit-regular. For diagonal reduction of matrices over regular ideals, we can derive the following.

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Proposition 13.2.27. Every triangular matrix over a regular ideal admits a diagonal reduction by elementary transformations. Proof. Let I be a regular ideal of a ring R, and let A ∈ Mk (I) be a triangular matrix. The result holds for k = 1. Assume that the result holds for m − 1(m ≥ 2). Without loss of generality, we may assume that A is a lower triangular matrix. Write   0  ..  B  .  A= ,  0 am1 · · · ak(m−1) b

where B ∈ M(m−1)×(m−1) (I) is a lower triangular matrix. By induction, there exist P, Q ∈ Em−1 (R) such that P BQ is a diagonal reduction. So A can be reduced to the matrix   0  ..  P BQ  .  D :=  .  0 c1 · · · c(m−1) b

Write



a 0 ··· 0  ..  C D= . 0 c1 c2 · · · c(m−1)

 0 ..  . . 0 b

As I is regular, there exists an idempotent e ∈ I such that aR = eR. Write a = ec and e = ad. Set   d 0 · · · 0 1 − dc  0 0     . ..   . U = I m−2 . .  .    0 0  −1 0 · · · 0 c Then U ∈ Em (R). Further,



e 0 ··· 0 0   . DU =  C  ..  0 t s1 · · · sm−2

 0 0  ..   . .  0 s

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Write s = ss′ s and s′ = s′ ss′ . If we add to the first column of DU its m-th column right multiplied by −s′ t, we may assume that ss′ t = 0. Set   s 0 · · · 0 ss′ − 1  0 0     . ..   . I V = m−2 . .  .    0 0  1 + s′ s 0 · · · 0 s′ Then V ∈ Em (R). Furthermore,  se − t ∗ ··· ∗  0   ..  V DU =  C .   0 (1 + s′ s)e s′ s1 · · · s′ sm−2

Thus, A can be reduced to the form  y y1 · · · ym−2 0   . C V DU =   ..  0 e 0 ··· 0

0 0 .. . 0 f

0 0 .. .



   .   0  s′ s



   ,   

where f = s′ s ∈ R is an
idempotent. Write (1 − f )e = (1 − f )eh(1 − f )e. Let z = y 1 − h(1 − f )e . By elementary transformations, V DU can be reduced to the form   z y1 · · · ym−2 0 0 0     . .  . ..  C .  .   0 0 e f 0 ··· 0 One easily checks that 

z0 ef






B21 (−e)B12 h(1 − f ) B21 − (1 − f )e B12 − h(1 − f ) = diag(∗, ∗).

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That is, we can find some K = (kij ) ∈ E2 (R) such that diag(∗, ∗). Therefore,



z0 ef



K =



 k11 0 · · · 0 k12  0 0     .  .  ..  V DU  .. Ik−2     0 0  k21 0 · · · 0 k22   ∗ ∗ ··· ∗ 0 0 0    .  .  ..  can be reduced to  .. C  by elementary transformations. By   0 0 0 0 ··· 0 ∗ induction, we obtain the result.  It follows from Proposition 13.2.27 that every n × n triangular matrix (n ≥ 2) over regular ideals is the sum of two invertible matrices. 13.3

General Comparability

Let I be a regular ideal of a ring R, and let A, B ∈ F P (I). We claim that A . B if and only if A .⊕ B. One direction is obvious. Assume now that A . B. In light of Lemma 13.1.8, there exist idempotents e1 , · · · , en ; f1 , · · · , fn ∈ I such that A ∼ = e1 R ⊕ · · · ⊕ en R and A ∼ = f1 R ⊕ · · ·⊕fn R. Let T = Mn (R), E = diag(e1 , · · · , en ) and F = diag(f1 , · · · , fn ). Clearly, R1×n is flat left R-module, and so ET . F T . Thus ϕ : ET ∼ =D where D ⊆ F T . Obviously, D = ϕ(E)T . As I is regular, so is Mn (I). Since ϕ(E) = ϕ(E)E ∈ Mn (I), we can find some Y ∈ T such that ϕ(E) = ϕ(E)Y ϕ(E). Set G = ϕ(E)Y . Then T =
GT ⊕ (In − G)T = D ⊕ (In −
G)T . T T Therefore F T = F T D ⊕ (In − G)T = D ⊕ F T (In − G)T , and N n×1 ⊕ N n×1 thus D .⊕ F T . Hence, ϕ(E)T R . FT R . Consequently, T

T

A .⊕ B, and we are through. We say that I satisfies general comparability provided that for any idempotent x, y ∈ I, there exists some e ∈ B(R) such that exR . eyR and (1 − e)yR . (1 − e)xR. In this section, we investigate equivalent characterizations for regular ideals satisfying general compara-

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bility. One easily checks that I satisfies general comparability if and only if for any idempotents e, f ∈ I, there exist s ∈ eRf, t ∈ f Re such that ust = ue and (1 − u)ts = (1 − u)f for some u ∈ B(R). It is clear from this fact that general comparability for regular ideals is left-right symmetric. If R is a regular ring satisfying general comparability, then every ideal of R satisfies general comparability. Lemma 13.3.1. Let I be a regular ideal of a ring R. Let A ∈ F P (I). If A⊕B ∼ = C ⊕ D, then there exists a refinement matrix A B



CD  A1 D1 . C1 B1

Proof. In view of Lemma 13.1.8, there are some idempotents e1 , · · · , en ∈ I such that A ∼ = diag(e1 , · · · , en )(nR). Hence, EndR (A) ∼ = diag(e1 , · · · , en )Mn (R)diag(e1 , · · · , en ). According to [382, Theorem 28.7] and Lemma 13.1.4, A has the finite exchange property. Clearly, A ⊕ B = ψ −1 (C) ⊕ ψ −1 (D). As in the proof of Lemma 6.3.7, we have C1 ⊆⊕ ψ −1 (C) and B1 ⊆⊕ ψ −1 (D) such that A ⊕ B = A ⊕ C1 ⊕ B1 . Hence, there exist right R-modules A1 and D1 such that ψ(C) = A1 ⊕ C1 and ψ(D) = D1 ⊕ B1 . As A ⊕ C1 ⊕ B1 = A1 ⊕ C1 ⊕ D1 ⊕ B1 , we get A ∼ = A1 ⊕ D1 . Therefore, we get a refinement matrix A B as desired.



CD  A1 D1 , C1 B1 

Theorem 13.3.2. Let I be a regular ideal of a ring R. Then the following are equivalent: (1) I satisfies general comparability. (2) For all A, B ∈ F P (I), there exists an e ∈ B(R) such that Ae . Be and B(1 − e) . A(1 − e). Proof. (2) ⇒ (1) Let x, y ∈ I be idempotents. Then xR = xRxR ⊆ xRI ⊆ xR; hence, xR = xRI ∈ F P (I). Likewise, yR ∈ F P (I). By hypothesis, there is e ∈ B(R) such that exR . eyR and (1 − e)yR . (1 − e)xR. (1) ⇒ (2) Let A, B ∈ F P (I). Then we have idempotents e′1 , · · · , e′n , e′′1 , · · · , e′′n ∈ I such that A = e′1 R ⊕ · · · ⊕ e′n R and B = e′′1 R ⊕ · · · ⊕ e′′n R. If n = 1, the result follows. Assume that the result

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holds for n − 1 (n ≥ 2). By virtue of Lemma 13.3.1, there are decompositions A = A1 ⊕A2 , B = B1 ⊕B2 with A1 , B1 .⊕ (n−1)R and A2 , B2 .⊕ R, where A1 , A2 , B1 , B2 ∈ F P (I). So we have f1 , f2 ∈ B(R) such that A1 f1 . B1 f1 , B1 (1−f1 ) . A1 (1−f1 ), A2 f2 . B2 f2 , B2 (1−f2 ) . A2 (1−f2 ). Set e1 = f1 f2 , e2 = (1 − f1 )(1 − f2 ). Then Ae1 . Be1 , Be2 . Ae2 . Set g1 = f1 (1 − f2 ), g2 = f2 (1 − f1 ). Then A1 g1 . B1 g1 , A2 g2 . B2 g2 , B1 g2 . A1 g2 , B2 g1 . A2 g1 . Now assume that B1 g 1 ∼ = A1 g1 ⊕ D1 , B2 g2 ∼ = A2 g2 ⊕ D2 , A1 g2 ∼ = B1 g2 ⊕ C1 , A2 g1 ∼ = B2 g1 ⊕ C2 for some right R-modules C1 , C2 , D1 and D2 . Clearly, C1 ⊕ C2 , D1 ⊕ D2 .⊕ (n − 1)R. Further, C1 ⊕ C2 , D1 ⊕ D2 ∈ F P (I). So there exists h ∈ B(R) such that (C1 ⊕C2 )h . (D1 ⊕D2 )h and (D1 ⊕D2 )(1−h) . (C1 ⊕C2 )(1−h). Set e3 = gh, e4 = g(1 − h). Then Ae3 = = . . ∼ =

A1 g1 h ⊕ A1 g2 h ⊕ A2 g1 h ⊕ A2 g2 h A1 g1 h ⊕ B1 g2 h ⊕ C1 h ⊕ B2 g1 h ⊕ C2 h ⊕ A2 g2 h B1 g1 h ⊕ B1 g2 h ⊕ B2 g1 h ⊕ B2 g2 h ⊕ (D1 ⊕ D2 )h (B1 g1 ⊕ B2 g2 ⊕ B1 g2 ⊕ B2 g1 )h Be3 .

Similarly, we see that Be4 = = . ∼ = ∼ =

B1 g1 (1 − h) ⊕ B1 g2 (1 − h) ⊕ B2 g1 (1 − h) ⊕ B2 g2 (1 − h) (A1 g1 ⊕ D1 ⊕ A2 g2 ⊕ D2 ⊕ B1 g2 ⊕ B2 g1 )(1 − h) (A1 g1 ⊕ B1 g2 ⊕ A2 g2 ⊕ B2 g1 ⊕ C1 ⊕ C2 )(1 − h) (A1 g1 ⊕ A1 g2 ⊕ A2 g2 ⊕ A2 g1 )(1 − h) Ae4 .

Let e = e1 + e3 . Then e ∈ B(R) with 1 − e = e2 + e4 . Moreover, Ae . Be and B(1 − e) . A(1 − e), as asserted.  Corollary 13.3.3. Let I be a regular ideal of a ring R, and let A ∈ F P (I). If I satisfies general comparability, then EndR (A) satisfies general comparability. Proof. Let E = EndR (A), and let x, y ∈ E be idempotents. In view of [54, Lemma 2], Mn (I) is regular. Therefore E is regular from Lemma 13.1.8. In addition, xA, yA ∈ F P (I). Thus, there exists e ∈ B(R) such that exA . eyA and (1 − e)yA . (1 − e)xA. Define a map u : A → A given by u(a) = ae for any a ∈ A. Then uxA . uyA and (1 − u)yA . (1 − u)xA, where u ∈ B(E). As in the proof of Corollary 6.4.5, uxE . uyE and

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(1 − u)yE . (1 − u)xE. This implies that E satisfies general comparability.  Theorem 13.3.4. Let I be a regular ideal of a ring R. If I satisfies general comparability, then so does Mn (I). Proof. By virtue of [54, Lemma 2], Mn (I) is regular. Let x, y ∈ Mn (I), we then have x(nR), y(nR) ∈ F P (I). In view of Theorem 13.3.2, there exists e ∈ B(R) such that x(nR)e . y(nR)e and y(nR)(1 − e) . x(nR)(1 − e). Set E = diag(e, · · · , e) ∈ Mn (R). Then

ExRn×1 . EyRn×1 and (In − E)yRn×1 . (In − E)xRn×1 . N Observing that Rn×1 R1×n ∼ = Mn (R), we get R

ExMn (R) . EyMn (R) and (In − E)yMn (R) . (In − E)xMn (R),
where E ∈ B Mn (R) . Therefore Mn (I) satisfies general comparability. 

Corollary 13.3.5. Let I be a regular ideal of a ring R. If every idempotent in I is central, then Mn (I) satisfies general comparability.

Proof. Let x, y ∈ I be idempotents. Then u := (1 − x)(1 − y) ∈ B(R). One easily checks that uxR . uyR and (1 − u)yR . (1 − u)xR, and then I satisfies general comparability. Therefore we complete the proof by Theorem 13.3.4.  Let I be a regular ideal of a ring R, let A, B ∈ F P (I), and let n ∈ N. If every idempotent in I is central, we claim that (1) If nA ∼ = nB, then A ∼ = B. (2) If nA . nB, then A . B. Suppose that nA ∼ = nB. Since A, B ∈ F P (I), by virtue of Corollary 13.3.5 and Theorem 13.3.2, we have an idempotent e ∈ R such that Ae .⊕ Be and B(1−e) .⊕ A(1−e). So there is a right R-module C such that Ae⊕C ∼ = Be, and hence, nA ⊕ nC ∼ = n(Ae ⊕ C) ⊕ nA(1 − e) ∼ = nBe ⊕ nA(1 − e) ∼ = nAe ⊕ nA(1 − e) ∼ = nA. We note that a regular ideal I is a B-ideal if and only if eRe is unit-regular for all idempotents e ∈ I. Thus, I is a B-ideal; hence, nA is directly finite. This implies that nC = 0, and so Ae ∼ = Be. Likewise, we have A(1 − e) ∼ = B(1 − e). Therefore A ∼ = B. That is, (1) holds. (2) is proved in the same manner.

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Following Ara et al. (cf. [16]), an ideal I of a ring R is separative if the following condition holds: For all A, B ∈ F P (I), 2A ∼ = A⊕B ∼ = 2B ⇒ A ∼ = B. Lemma 13.3.6. Let I be a regular ideal of a ring R. Then the following are equivalent: (1) I is separative. (2) For all A, B, C ∈ F P (I), A ⊕ 2C ∼ = B ⊕ 2C =⇒ A ⊕ C ∼ = B ⊕ C. Proof. In view of Lemma 13.3.1, V(I) is a refinement monoid, where V(I) denotes the monoid of all P ∈ F P (I). Therefore the result follows by [16, Lemma 2.1].  Lemma 13.3.7. Let I be a regular ideal of a ring R. Then the following are equivalent: (1) I is separative. (2) For any idempotents e ∈ I, A⊕2(eR) ∼ = B⊕2(eR) implies that A⊕eR ∼ = B ⊕ eR. (3) eRe is a separative ring for all idempotents e ∈ I. Proof. (1) ⇒ (3) Let e ∈ I be an idempotent, and let A, B ∈ F P (eRe). If 2A ∼ =A⊕B ∼ = 2B, then O O O O 2A eR ∼ eR ⊕ B eR ∼ eR. =A = 2B

eRe eRe eRe N N eRe One easily checks that A eR, B eR ∈ F P (I). By hypothesis, eRe eRe N ∼ B N eR. Clearly, eR N Re ∼ A eR = = eRe. Therefore we get eRe eRe
N
N R N N A∼ eR Re ∼ eR Re ∼ = A = B = B, as required. eRe

R

eRe

R

A ⊕ eR eR

B ⊕ eR eR   A1 D1 . C1 B1

(3) ⇒ (2) Let A ⊕ 2(eR) ∼ = B ⊕ 2(eR) with e ∈ I. In view of Lemma 13.3.1, there exists a refinement matrix

As C1 C1

in the proof of Lemma 6.3.7, we may assume that B1 .⊕ C1 , D1 . Write ∼ E ⊕ 2B1 ∼ B ⊕ D ∼ eR ∼ = B1 ⊕ F and D1 ∼ = B1 ⊕ E. Then = B1 ⊕ N N= 1 N 1 = ∼ F ⊕ 2B . This implies that E Re, B Re, F Re ∈ F P (eRe). = 1 1 R

Further,

E

O R

Re ⊕ 2B1

O R

Re ∼ =F

R

O R

Re ⊕ 2B1

R

O R

Re.

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According to Lemma 13.3.6, we get O O O O E Re ⊕ B1 Re ∼ Re ⊕ B1 Re. =F R

R

R

R

As E .⊕ eR, there exists an idempotent ef e ∈ eRe such that E ∼ = ef e(eR), and thus, N N N
N E Re eR ∼ eR = ef e eR Re R eRe RN eRe
∼ eR = ef e eRe eRe

∼ = ef e(eR) ∼ = E. N N N N Likewise, B1 Re eR ∼ Re eR ∼ = B1 and F = F . Therefore D1 ∼ = R eRe R eRe ∼ A1 ⊕ D1 = ∼ A1 ⊕ C1 ∼ E ⊕ B1 ∼ = F ⊕ B1 ∼ = C1 , and so A ⊕ eR = = B ⊕ eR. ∼ (2) ⇒ (1) Let A, B, C ∈ F P (I) such that A ⊕ 2C = B ⊕ 2C. In

view of Lemma 13.1.8, there exist idempotents e1 , · · · , en ∈ I such that C ∼ = e1 R ⊕ · · · ⊕ en R. Hence 2(e1 R) ⊕ 2(e2 R) ⊕ · · · ⊕ 2(en R) ⊕ A ∼ = 2(e1 R) ⊕ 2(e2 R) ⊕ · · · ⊕ 2(en R) ⊕ B. As e1 ∈ I, by assumption, 2(e2 R) ⊕ · · ·⊕2(en R)⊕e1 R⊕B ∼ = 2(e2 R)⊕· · ·⊕2(en R)⊕e1 R⊕C. By iteration of this process, we derive that (e1 R) ⊕ · · · ⊕ (en R) ⊕ B ∼ = (e1 R) ⊕ · · · ⊕ (en R) ⊕ C. That is, A ⊕ B ∼ = A ⊕ C. Therefore I is a separative ideal from Lemma 13.3.6.  Theorem 13.3.8. Every regular ideal satisfying general comparability is separative.

Proof. Given any idempotent e ∈ I, by Corollary 13.3.3, we prove that EndR (eR) satisfies general comparability. Hence eRe is an exchange ring satisfying generalized 1-comparability. In view of Theorem 6.4.11, eRe is a separative ring. Therefore I is a separative ideal from Lemma 13.3.7.  Example 13.3.9. Let V be an infinite-dimensional vector space over a division ring D, and let R be a subring of EndD (V ) which contains I = {x ∈ EndD (V ) | dimD (xV ) < ∞}. Then I is a separative ideal of R. Proof. Let S = EndD (V ). Given any idempotents x, y ∈ I, we have xS . yS or yS . xS because S is a regular ring satisfying the comparability axiom. Observing that xR = xI = xS and yS = yI = yR, so either xR . yR or yR . xR. Thus, I as an ideal of R satisfies general comparability. According to Theorem 13.3.8, I is separative of R, as asserted. 

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Let M be a right R-module, and let Z(M ) = {m ∈ M | r(m) ⊆e R}. If Z(M ) = 0, we say that M is nonsingular; otherwise, we say that M is singular. We say that an ideal I of a ring R is right injective if I is an injective right R-module. Lemma 13.3.10. Let I and J be regular ideals of a ring R. If I is right injective and J ⊆ I, then there exists e ∈ B(R) such that J ⊆e eRR ⊆ I. Proof. For any m ∈ Z(I), there exists some z ∈ R such that m = mzm. T Hence, r(m) = (1 − zm)R. As r(m) zmR = 0, we get zmR = 0; hence, m = mzm = 0. That is, Z(I) = 0, i.e., I is nonsingular. Thus, J ⊆ I and I is a nonsingular injective right R-module. According to [217, Proposition 9.1], there is a K ⊆⊕ I such that J ⊆e K. Since I is injective, the exact sequence 0 → I → R → R/I → 0 splits, and so I ⊆⊕ R. Thus, K ⊆⊕ R, and we have K = eR for an idempotent e ∈ I. Given any x ∈ R, we have xJ ⊆ J ⊆ eR. Hence (1 − e)xJ = 0. As I is regular, we can find some y ∈ R such that (1 − e)xe = (1 − e)xey(1 − e)xe. Set f = ey(1 − e)xe ∈ eR. Then f = f 2 . As f J = ey(1 − e)xeJ = 0, we see that J ⊆ (1 − f )R; hence, J ∩ f R = 0. Clearly, f R ⊆ eR. It follows from J ⊆e eR that f R = 0, and so f = 0. This implies that (1 − e)xe = 0. Thus, (1 − e)Re = 0. As I is regular, there exists some t ∈ R such that ex(1 − e) = ex(1 − e)tex(1 − e) ∈ R(1 − e)ReR = 0, and then eR(1 − e) = 0. This shows that e ∈ B(R), as required.  Theorem 13.3.11. Every regular, right injective ideal of a ring satisfies general comparability. Proof. Let I be a regular, right injective ideal of a ring R, and let g ∈ I be an idempotent. Then gR⊕(1−g)I = I. This implies that gR is an injective right R-module. Let x, y ∈ I be idempotents, and let A = xR, B = yR. Let X denote the collection of all triples (A′ , B ′ , f ′ ) such that A′ ⊆ A, B ′ ⊆ B and f ′ : A′ → B ′ is an isomorphism. Define a partial order on X by setting (A′ , B ′ , f ′ ) ≤ (A′′ , B ′′ , f ′′ ) whenever A′ ⊆ A′′ , B ′ ⊆ B ′′ and f ′′ is an extension of f ′ . Clearly, X 6= ∅. Given (A1 , B1 , f1 ) ≤ · · · ≤ (An , Bn , fn ) ⊆ ∞ ∞ S S · · · in X, we can construct an R-morphism f : Ai → Bi . For any x∈

∞ S

i=1

i=1

i=1

Ai , there exists a minimal index m ∈ N such that x ∈ Am . Then we

define f (x) = fm (x). One easily checks that f is well defined. In addition, ∞ ∞
S S Ai , Bi , f ∈ X. Thus, X is inductive. By using Zorn’s Lemma, we i=1

i=1

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have a maximal element (A0 , B0 , f0 ) ∈ X. By the above discussion, A and B are both injective right R-modules. Thus A contains an injective hull A′0 for A0 and B contains an injective hull B0′ for B0 . In view of [4, Lemma 18.9], f0 : A0 → B0 extends to an isomorphism f0′ : A′0 → B0′ . By maximality, we have (A0 , B0 , f0 ) = (A′0 , B0′ , f0′ ); hence, A0 and B0 are injective right R-modules. Therefore A = A0 ⊕ C and B = B0 ⊕ D for some right R-modules C and D. Set H = {r ∈ I | Cr = 0}. Then H is a regular ideal of R. By Lemma 13.3.10, H ⊆e eRR ⊆ I for some e ∈ B(R). Given any x ∈ eR, then x ∈ I. We claim that x ∈ H. For any c ∈ C,
we have cx ∈ C and H ⊆ r(cx) ∩ eR. Given any K ⊆ eR, if r(cx) ∩ eR ∩ K = 0, then H ∩ K = 0. Since H ⊆e eRR , we have K = 0. In other words, L := r(cx) ∩ eR ⊆e eRR . Assume that cx 6= 0. Since I is regular and e ∈ B(R), we have y ∈ R such that cx = cxycx. Set f = ycx. Then f = f 2 ∈ eR is an idempotent of R. One sees that f L = ycxL = 0, so L ⊆ (1 − f )R. Hence, L ∩ f R = 0. Since L ⊆e eR and f R ⊆ eR, we deduce that f R = 0, thus f = 0. This contradicts cx 6= 0. So we derive that Cx = 0; and then x ∈ H. Therefore H = eR, and so Ce = 0. This shows that Ae ∼ = A0 e ∼ = B0 e .⊕ Be. Assume that D(1 − e) 6= 0. Then we have 0 6= x ∈ D(1 − e). As D ⊆⊕ R, there exists a nonzero idempotent g ∈ I such that D ∼ = gR. This implies that g 6∈ eR = H. So we have some y ∈ C such that yg 6= 0. Clearly, there exists an R-epimorphism ψ : D ∼ = gR → ygR. As in the discussion above, we see that ygR is projective. Hence, we have an Rmonomorphism f : ygR → D.
Since ygR ⊆ C and f (ygR) ⊆ D, we get A0 ⊕ygR, B0 ⊕f (ygR), f0 ⊕f ∈ X, a contradiction. Therefore D(1−e) = 0, so we conclude that B(1 − e) ∼ = B0 (1 − e) ∼ = A0 (1 − e) .⊕ A(1 − e), as required.  Lemma 13.3.12. Let I be a regular ideal of a ring R. If I satisfies general comparability, then aR + bR = R with a ∈ 1 + Q(I), b ∈ R implies that there exists y ∈ R such that a + by ∈ U + (R). Proof. Let x ∈ 1 + Q(I) be regular. Then there exists y ∈ R such that x = xyx. From x ∈ 1 + Q(I), we have y ∈ 1 + Q(I); hence, 1 − xy, 1 − yx ∈ Q(I). Thus, there are some m ∈ N such that (1 − xy)m , (1 − yx)m ∈ I, and so 1 − xy, 1 − yx ∈ I. Hence, we have e(1 − xy)R . e(1 − yx)R and (1 − e)(1 − yx)R . (1 − e)(1 − xy)R, where e ∈ B(R). From e(1 − xy)R . e(1 − yx)R, we see that e(1 − xy)R is isomorphic to a direct summand of e(1−yx)R; whence, we have an injection

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ψ : e(1 − xy)R → e(1 − yx)R. Obviously, eR = exyR ⊕ e(1 − xy)R = eyxR ⊕ e(1 − yx)R with φ : exyR = exR ∼ = eyxR given by φ(exr) = y(exr) for any exr ∈ exR. Define u ∈ EndR (eR) so that u restricts to φ and u restricts to ψ. Then ex = xu(e)x, where u(e) ∈ eRe is left invertible. Analogously, it follows from (1 − e)(1
− yx)R . (1 − e)(1 − xy)R that there exists some v ∈ EndR (1 − e)R such that (1 − e)x = xv(1 − e)x, where v(1 − e) ∈ (1 − e)R(1 − e) is right invertible. Consequently, x = ex + (1 − e)x = xwx, where w = u(e) + v(1 − e) ∈ U + (R). Given αβ +γ = 1 with α ∈ 1+Q(I), β, γ ∈ R, then α(β +γ)+(1−α)γ = 1. Let c = β + γ and d = (1 − α)γ. As d ∈ Q(I), we have some n ∈ N such that dn ∈ I. Thus, we can find an idempotent f ∈ dn R such that 1 − f ∈ (1 − dn )R. Write f = ds and 1 − f = (1 − d)t. Then αct + f = 1, and so (1 − f )αct + f = 1. This implies that (1 − f )α ∈ 1 + Q(I) is regular. By the above discussion, (1 − f )α ∈ R is related unit-regular. Analogously to Corollary 6.1.4, there exists a z ∈ R such that α + (1 − α)γs(z − α) = α + ds(z − α) = α + f (z − α) = (1 − f )α + f z ∈ U + (R).

In view of Lemma 4.1.2, we have some k ∈ R such that
β + 1 + k(1 − α) γ = β + γ + k(1 − α)γ ∈ U + (R).

By using Lemma 4.1.2 again, there is some δ ∈ R such that α+βδ ∈ U + (R), as required.  From Lemma 13.3.12, we claim that every regular ideal satisfying general comparability is a P B-ideal.

Proposition 13.3.13. Let I be a regular ideal of a ring R. If I satisfies general comparability, then for any
A ∈ Mn (I), there exists an idempotent E ∈ Mn (I) and a U ∈ U + Mn (R) such that A = EU .

Proof. In view of [54, Lemma 2], Mn (I) is regular. Given any A ∈ Mn
(I), we have some B ∈ Mn (R) such that A = ABA. Clearly, A+(In −AB) B+ (In − AB)(In − B) = In . Since A + (In − AB) ∈ In + Mn (I), by Lemma 13.3.12, we can find some Y ∈ M
n (R) such that U := A + (In − AB) + (In − AB)(In − B)Y ∈ U + Mn (R) . Let E = AB. Then E = E 2 ∈ Mn (I).
Therefore we conclude that A = AB A+(In −AB)+(In −AB)(In −B)Y = EU . 

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Let A ∈ Mn (R). If Mn (R)AMn (R) is regular and right injective, then
there exists an idempotent E ∈ Mn (R) and a U ∈ U + Mn (R) such that A = EU . According to Theorem 13.3.11, Mn (R)AMn (R) satisfies general comparability. As A ∈ Mn (R)AMn (R), we are done by Proposition 13.3.13. Example 13.3.14. Let R be a regular ring, and let (aij ) ∈ Mn (R). If all aij R are injective, then there exists an idempotent E ∈ Mn (R) and a
U ∈ U + Mn (R) such that (aij ) = EU .

Proof. Let I = {a ∈ R | aR is injective}. Let x, y ∈ I and z ∈ R. As in the proof of Example 13.2.14, we show that zxR .⊕ xR, xzR .⊕ xR and (x + y)R .⊕ xR ⊕ yR. As xR and yR are injective, so are zxR, xzR and (x + y)R. This implies that zx, xz, x + y ∈ I. As a result, we see that I is an ideal of R. For any idempotent e ∈ I, eR is injective. It follows from Theorem 13.3.11 that I satisfies general comparability. As (aij ) ∈ Mn (I), we complete the proof by Proposition 13.3.13.  Example 13.3.15. Let R be a regular ring, and let A ∈ Mn (R). If A(nR) is injective, then there exists an idempotent E ∈ Mn (R) and a
U ∈ U + Mn (R) such that A = EU .

Proof. Let S = Mn (R). Then S is regular. As in the proof of Example 13.3.14, J = {a ∈ S | aS is injective} is an ideal of S, and J satisfies general comparability. It is easy to check N that AMn (R) ∼ = A(nR) R R1×n as S-modules. For any right S-modules N C, D, an S-morphism f : C → A(nR) R R1×n and an S-monomorphism g : C → D, there is an R-monomorphism g ⊗ 1S as Rn×1 is a flat left N N S-module. Further, A(nR) ∼ = A(nR) R1×n Rn×1 is injective. Hence, R

S

we have an R-morphism h such that the following diagram N N A(nR) R1×n Rn×1 R

S

f ⊗ 1Rn×1 ↑ N n×1 C R

տh

g⊗1Rn×1

֌

D

S

N

Rn×1

S

commutates. One easily checks that the following diagram σ N N N N A(nR) R1×n ∼ = A(nR) R1×n Rn×1 R1×n R

f↑ C

R

τ

∼ =

S

R

↑ f ⊗ 1Rn×1 ⊗ 1R1×n N n×1 N 1×n C R R S

R

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θ N N commutates. In addition, D Rn×1 R1×n ∼ = D and θ g ⊗ 1Rn×1 ⊗ S R

1R1×n τ = g. Thus, we deduce that σ −1 h⊗1R1×n θ−1 ·g = f . This implies N that A(nR) R1×n is an injective S-module, and then so is AMn (R). As R

a result, A ∈ J. In view of Proposition 13.3.13, we obtain the result. 13.4



Separative Regular Ideals

The main purpose of this section is to give various equivalent characterizations for a regular ideal of a ring to be separative. Lemma 13.4.1. Let I be a regular ideal of a ring. Then the following are equivalent: (1) (2) (3) (4) (5)

I is separative. For all A, B, C ∈ FP(I), A ⊕ C ∼ = B ⊕ C with C . A, B ⇒ A ∼ = B. For all A, B, C ∈ FP(I), A ⊕ C ∼ = B ⊕ C with C ∝ A, B ⇒ A ∼ = B. For all A, B ∈ FP(I), 2A ∼ = 2B and 3A ∼ = 3B ⇒ A ∼ = B. For all A, B ∈ FP(I), nA ∼ = nB and (n + 1)A ∼ = (n + 1)B(n ∈ N) ⇒ A∼ = B. (6) For all A, B, C ∈ F P (I), A ⊕ C ∼ = B ⊕ C .⊕ R with C . A, B =⇒ A∼ = B. Proof. In view of Lemma 13.3.1, V(I) is a refinement monoid, where V(I) denotes the monoid of all P ∈ F P (I). The result now follows by [16, Lemma 2.1 and Proposition 2.8].  Proposition 13.4.2. Let I be a regular ideal of a ring. Then the following are equivalent: (1) I is separative. (2) For all C ∈ F P (I), A ⊕ C ∼ = B ⊕ C with C .⊕ A, B ⇒ A ∼ = B for any right R-modules A and B. (3) For all C ∈ F P (I), A ⊕ C ∼ = B ⊕ C with C ∝ A, B ⇒ A ∼ = B for any right R-modules A and B. (4) For all C ∈ F P (I), A ⊕ 2C ∼ = B ⊕ 2C ⇒ A ⊕ C ∼ = B ⊕ C for any right R-modules A and B. Proof. (1) ⇒ (2) Let C ∈ F P (I) such that A ⊕ C ∼ = B ⊕ C, where C .⊕ A, B. Clearly, C has the finite exchange property. As in the proof of

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Lemma 6.3.7, we have a refinement matrix A C



BC  D1 A1 B1 C1

∼ A1 ⊕ C1 ∼ such that C1 .⊕ A1 , B1 . Thus C = = B1 ⊕ C1 . Since C ∈ F P (I), one easily checks that C1 , A1 , B1 ∈ F P (I). It follows by Lemma 13.4.1 that A1 ∼ = B1 . Therefore A ∼ = D1 ⊕ A1 ∼ = D 1 ⊕ B1 ∼ = B, as desired. (2) ⇒ (3) Let C ∈ F P (I) such that A ⊕ C ∼ = B ⊕ C, where C ∝ A, B. Then we have k ∈ N such that C .⊕ kA, kB. By the finite exchange property of C, we have right R-module decomposition C = C1 ⊕ · · · ⊕ Ck with all Ci .⊕ A(1 ≤ i ≤ k). Clearly, each Ci .⊕ C .⊕ kB; hence, we have right R-module decompositions Ci = C1i ⊕ · · · ⊕ Cki for 1 ≤ i ≤ k, where each Cji .⊕ B(1 ≤ j ≤ k). Therefore we get k M

i,j=1


Cji ⊕ A ∼ =

k M

i,j=1


Cji ⊕ B

with all Cji . A, B. By hypothesis, we have A ∼ = B, as required. (3) ⇒ (4) is obvious. (4) ⇒ (1) is trivial by Lemma 13.3.7. ⊕



Lemma 13.4.3. Let R be an exchange ring, and let A = (aij ) ∈ GLn (R)(n ≥ 2). Then A can be reduced by elementary column operations to a matrix (bij ) such that R = b11 R ⊕ · · · ⊕ b1n R and each b1i ∈ a1i Ra1i . Proof. As A ∈ GLn (R), we get

n P

a1i R = R. In view of [382, Theorem

i=1

28.7], there exists a complete set of idempotents in R, {e1 , · · · , en }, such that each ei ∈ a1i R. Write ei = a1i ri , where ri ei = ri . Thus, we get     1 e1 a11 a12 · · · a1n    ∗  −r2 a11 1 ∗     A . = . . . .. . . . . . . .    . .  . . . . −rn a11

1

∗

∗ ··· ∗

Further, we get     1 −r1 a12 e1 a11 e2 a12 1   ∗   −r2 a11 ∗ 1     A = .  .. .. . . . .   .. ..    . . ∗ ∗ −rn a12 1 −rn a11 1

· · · a1n



  ..  . . .  ··· ∗ ..

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Repeat the above process, A can be reduced by elementary column operations to a matrix (bij ), where each b1i = ei a1i . Thus, b1i ∈ a1i Ra1i . Moren n n P P P over, 1 = ei = ei a1i ri = bi ri . Therefore R = b1 R ⊕ · · · ⊕ bn R, as

required.

i=1

i=1

i=1



Lemma 13.4.4. Let R be an exchange ring, and let A ∈ GLn (R)(n ≥ 2). Then A can be reduced by elementary row and column operations to a matrix whose 1, 1 entry d is regular, with dR = (1 − p)R, Rd = R(1 − q) and RpR = RqR = R, where p, q ∈ R are idempotents. Proof. As A = (aij ) ∈ GLn (R)(n ≥ 2), it follows from Lemma 13.4.3 that A can be reduced by elementary column operations to a matrix (bij ) such that R = b11 R ⊕ · · · ⊕ b1n R and each b1i ∈ a1i Ra1i . Thus, we have a complete set {e1 , · · · , en } of idempotents in R such that each ei b1i = b1i . Write n n n n
P P P P 1= b1i xi = ei b1j xi and set c11 = b1i . Then R = Rc11 R. i=1

i=1

j=1

i=1

So A can be reduced by elementary column operations to a matrix (cij ) such that each c1i = b1i ∈ a1i Ra1i (i ≥ 2) and Rc11 R = R. As in the proof of Proposition 7.1.12, there exists an idempotent e ∈ c11 R such that ReR = R. Write e = c11 t1 . This implies that 

1 −t1 c12  1  (cij )  ..  .





c11 (1 − e)c12   ∗ ∗   = . ..   .. . ∗ ∗ 1

··· ··· .. .

 c1n ∗   ..  . . 

··· ∗

According to Lemma 13.4.3, A can be reduced by elementary column operations to a matrix (dij ) such that R = d11 R ⊕ · · · ⊕ d1n R and d12 ∈ (1 − e)c12 R(1 − e)c12 . Hence, d12 R = (1 − g)R for some idempotents g ∈ R. This implies that d12 ∈ R is regular. Clearly, d12 ∈ Ra12 . Moreover, (1 − g)R = d12 R ⊆ (1 − e)R, and so e = eg. This implies that Re ⊆ Rg. Consequently, we deduce that RgR = R. Clearly, (dij ) can be reduced by   d22 d21 · · · d2n  d12 d11 · · · d1n    elementary row and column operations to a matrix  . . . . .  .. .. . . ..  dn2 dn1 · · · dnn Since R is an exchange ring, so is the opposite ring Rop . Applying the dis-

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 · · · don2 · · · don1   o can be reduced to a matrix (qij ), . . ..  . .  do2n do1n · · · donn o op o o where q12 ∈ R d12 is regular, q12 Rop = (1o − q o )Rop for an idempotent o op op o op op q ∈ R with R q R = R . Therefore A can be reduced by elementary row and column operations to a matrix whose 2, 1 entry q12 is regular, with Rq12 = R(1 − q) and RqR = R. As q12 ∈ R is regular, q12 R = (1 − p)R for some idempotent h. Further, (1 − p)R = q12 R ⊆ d12 R ⊆ (1 − g)R; hence, g = gp. Therefore RpR = R. Using elementary operations to move q12 to the 1, 1 position, we establish the result.  do22  do21  cussion above to Rop ,  .  ..

do12 do11 .. .

The following elementary fact is due to Ara et al. (cf. [17, Theorem 2.8]). Lemma 13.4.5. Every separative exchange ring is a GE-ring. Proof. Let R be a separative exchange ring. Clearly, R is a GE1 -ring. Assume that R is a GEn−1 -ring (n ≥ 2). Let A ∈ GLn (R). By virtue of Lemma 13.4.4, we may assume that the 1, 1 entry d of A is regular, with dR = (1 − p)R, Rd = R(1 − q), RpR = RqR = R, where p, q ∈ R are idempotents. As Rd = R(1−q), we have some s, t ∈ R such that d = s(1−q) and td = 1 − q. Hence, dtd = d(1 − q) = d, and so r(d) = (1 − td)R = qR. n P As RqR = R, we have y1 , · · · , yn ∈ A such that R = yj qR. Construct a map ϕ : n(qR) → R given by ϕ(r1 , · · · , rn ) =

n P

j=1 ⊕

j=1

yj rj for any (r1 , · · · , rn ) ∈

n(qR). This implies that ϕ splits; hence, R . n(qR). That is, R ∝ qR. Further, (1 − dt)R ∼ = R/dR ∼ = pR. As RpR = R, we also obtain R ∝ pR; whence, R ∝ (1 − dt)R. Clearly, R = tdR ⊕ r(d) = dR ⊕ (1 − dt)R with tdR ∼ = dR. As dR ∝ r(d), (1 − dt)R, we get r(d) ∼ = R/dR. Thus, d ∈ R is unit-regular. Hence, we have an idempotent e ∈ R and some u ∈ U (R) such that d = ue. Set U = diag(u, In−1 ). Then the 1, 1 entry of U −1 A is an idempotent e. Set (bij ) = U −1 A. Then ex1 +b12 x2 +· · ·+b1n xn = 1 for some xi ∈ R. This implies that ex1 (1−e)+b12x2 (1−e)+· · ·+b1n xn (1−e) = 1−e, and so v := e + b12 x2 (1 − e) + · · · + b1n xn (1 − e) = 1 − ex1 (1 − e) ∈ U (R).

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

 1  x2 (1 − e) 1    Let V =   . Then the 1, 1 entry of U −1 AV is v, and .. . .  .  . xn (1 − e) 1 therefore there exist E, F ∈ En (R) such that EU −1 AV F = diag(v, A′ ), where A′ ∈ GLn−1 (R). By the induction hypothesis, A′ can be reduced a diagonal matrix by elementary row and column operations. Therefore we conclude that R is a GE-ring.  Let R be a separative exchange ring. It follows from Lemma 13.4.5 that the natural homomorphism U (R) → K1 (R) is surjective. Theorem 13.4.6. Let I be a separative regular ideal of a ring R. Then every square matrix over I admits a diagonal reduction by elementary transformations. Proof. Let A = (aij ) ∈ Mn (I). In view of Lemma 13.1.19, there exists an idempotent e ∈ I such that all aij ∈ eRe(1 ≤ i, j ≤ n); hence A ∈ Mn (eRe). According to Lemma 13.3.7, eRe is a separative regular ring. By Lemma 7.3.1, there exist U, V ∈ GLn (eRe) such that U AV = diag(r1 , · · · , rn ) for some r1 , · · · , rn ∈ eRe. Using Lemma 13.4.5, we have s t m k Q Q Q Q U −1 = Ei diag(c1 , · · · , cn ) Fi , V −1 = Gi diag(d1 , · · · , dn ) Hi , i=1

i=1

i=1

i=1

where E1 , · · · , Es , G1 , · · · , Gm ∈ Mn (eRe) are elementary row operations, F1 , · · · , Ft , H1 , · · · , Hk ∈ Mn (eRe) are elementary column operations and c1 , · · · , cn , d1 , · · · , dn ∈ eRe. Since E1 ∈ Mn (eRe) is an elementary row operation, it must be 

 

e

 .  ..   0 e   ..  .    e 0   ..  .

e



e

  .   ..     e α     .. ; .       e     ..   .

e

    e    ..     .     ; or   β      ..     .   e 


α ∈ eRe, β ∈ U (eRe) . Let E = diag(e, · · · , e) ∈ Mn (R). Then E1 +

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      In − E is equal to       

 

1 ..

455

. 1−e e

e ..

. 1−e

..

.

            ,            

1 

  1     ..    .       β+1−e  . It is reduced to      ..    .   1  





1 ..

1

..

    1 α   .. , .    1  ..  .  1 

    1 1   .. , .    0 1 − 2e  ..  .  1

.



1

.



    1   ..   1 α    .    ..   . Clearly, (1 − 2e)−1 =  β +1−e .   , or    ..     1 .  ..  1 .  1 1 − 2e ∈ U (R), and thus E1 + In − E is the product of some elementary row operations. Similarly, all Ei + In − E, Gi + In − E ∈ Mn (R) are products of some elementary row operations and all Fi + In − E, Hi + In − E ∈ Mn (R) are products of some elementary column operations. Therefore s t Q Q U −1 + In − E = (Ei + In − E)diag(c1 , · · · , cn ) (Fi + In − E);

            

..

V

.

−1

+ In − E =

i=1 m Q

i=1

i=1 k Q

(Gi + In − E)diag(d1 , · · · , dn )

As a result, we deduce that

i=1

(Hi + In − E).

(U + In − E)A(V + In − E) = diag(r1 , · · · , rn ).

Clearly, (U + In − E)−1 = U −1 + In − E = K1 · · · Kp diag(c′1 , · · · , c′n ); (V + In − E)−1 = V −1 + In − E = diag(d′1 , · · · , d′n )L1 · · · Lq ,

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where K1 , · · · Kp ∈ Mn (R) are elementary row operations and L1 , · · · Lq ∈ Mn (R) are elementary column operations. Therefore we conclude that
−1 ′ ′ ′ ′ Kp−1 · · · K1−1 AL−1 q · · · L1 = diag c1 r1 d1 , · · · , cn rn dn . This means that A can be diagonalized using elementary transformations. 

Corollary 13.4.7. Let R be a regular ring, and let A ∈ Mn (R). If A(nR) is injective, then A admits a diagonal reduction using elementary transformations. Proof. Let S = Mn (R), and let J = {a ∈ S | aS is injective}. As in the proof of Example 13.3.15, J is an ideal satisfying general comparability. In addition, A ∈ J. In light of Theorem 13.3.8, J is separative. So the result follows from Theorem 13.4.6.  In [433, Theorem 3.2.2], Zhang observed that a regular ring R is separative if and only if for each a ∈ R satisfying a(1 − a)R . r(a), R/aR is unit-regular. Now we extend this result to regular ideals as follows. Theorem 13.4.8. Let I be a regular ideal of a ring R. Then the following are equivalent : (1) I is separative. (2) For each regular a ∈ 1 + I satisfying (a − a2 )R . r(a), R/aR is unitregular. Proof. (1) ⇒ (2) Let a ∈ 1 + I be regular and (a − a2 )R . r(a), R/aR. Then we can find some x ∈ R such that a = axa. Clearly, x ∈ 1 + I. As 1 − a ∈ I, 1 − a ∈ R is regular. So we have an element y ∈ R such that
1−a = (1−a)y(1−a). Hence r(a) = (1−xa)R and r(1−a) = 1−(1−a)y R. Let e = 1 − xa and f = 1 − (1 − a)y. Then e = e2 ∈ I and f = f 2 ∈ 1 + I. Thus, r(a) + r(1 − a) = f R + eR. Given any r, s ∈ R, we have f s + er = f (s + er) + (1 − f )er. Since (1 − f )er ∈ I, we can find some t ∈ R such that (1 − f )er = (1 − f )ert(1 − f )er. Let g = (1 − f )ert(1 − f ). Then g = g 2 ∈ I. Furthermore, f s + er = f (s + er) + ger. In addition, f g = gf = 0. Hence, f (s + er) = f (f s + er) and ger = g(f s + er). As a result, we deduce that f (s + er) + ger = (f + g)(f s + er). This implies that f R + eR ⊆ (f + g)R. Obviously, (f + g)R ⊆ f R + eR. So r(a) + r(1 − a) = (f + g)R. Since (f + g)2 = f + g, r(a) + r(1 − a) is a direct summand of R. Now we can find a right R-module C such that R = r(a) ⊕ r(1 − a) ⊕ C. It follows

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that aR = ar(1 − a) ⊕ aC = r(1 − a) ⊕ aC. Thus, R = aR ⊕ (1 − ax)R = r(1 − a) ⊕ aC ⊕ (1 − ax)R, whence, r(a) ⊕ C ∼ = R/aR ⊕ aC. Clearly, C ∼ = aC, and so r(a) ⊕ C ∼ = R/aR ⊕ C. Let ϕ : C → (a − a2 )C given by ϕ(c) = (a − a2 )c for any c ∈ C. Then ϕ is an R-epimorphism. If ϕ(c) = (a − a2 )c = 0, then ac ∈ r(1 − a) and (1 − a)c ∈ r(a); hence,
T c = (1 − a)c + ac ∈ r(a) + r(1 − a) C = 0. This implies that ϕ is an R-monomorphism, and then ϕ : C ∼ = (a − a2 )C. We infer that (a − a2 )C = (a − a2 )R. Consequently, we deduce that r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R. As 1 − a ∈ I, (a − a2 )R ∈ F P (I). It follows from r(a) ∈ F P (I) that R/aR ∈ F P (I). Hence, (a − a2 )R .⊕ r(a), R/aR. In view of Proposition 13.4.2, r(a) ∼ = R/aR, and therefore a ∈ R is unitregular. (2) ⇒ (1) Given A ⊕ C ∼ = B ⊕ C .⊕ R and C .⊕ A, B with A, B, C ∈ F P (I), we write R = A1 ⊕ C1 ⊕ D = A2 ⊕ C2 ⊕ D, where A1 ∼ = A, C1 ∼ = C2 ∼ = C and A2 ∼ = B. Let a ∈ R induce an endomorphism of RR , which is zero on A1 , an isomorphism from C1 onto C2 , and the identity on D. One checks that (1 − a)R = (1 − a)(A1 ⊕ C1 ); hence, a ∈ 1 + I. As in the proof of (1) ⇒ (2), we see that r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R. It is easy to verify that aR = C2 ⊕ D; hence, there exists an idempotent e ∈ R such that aR = eR. Thus, a ∈ 1 + I is regular. Since 1 − a ∈ I and I is regular, a(1 − a)R is projective. Let ϕ : C1 → a(1 − a)C1 given by ϕ(c1 ) = a(1 − a)c1 for any c1 ∈ C1 . It is easy to verify that ϕ is an R-epimorphism. As a(1 − a)C1 ∼ = a(1 − a)R, we see that a(1 − a)C1 is projective. So we can find a right R-module E such that C1 ∼ = a(1 − a)C1 ⊕ ⊕ ⊕ ∼ E, i.e., a(1 − a)R = a(1 − a)C1 . C1 . A1 = r(a). Likewise, we have a(1 − a)R .⊕ C1 ∼ = C2 .⊕ A2 = R/aR. Thus (a − a2 )R . r(a), R/aR. By hypothesis, a ∈ R is unit-regular, and then A ∼ = r(a) ∼ = R/aR ∼ = B. In view of Lemma 13.4.1, I is separative.  Corollary 13.4.9. Let I be a regular ideal of a ring R, and let n ≥ 2. Then the following are equivalent: (1) I is separative. (2) Each regular a ∈ 1 + I satisfying



n r(a) ∼ = (n + 1) R/aR = n R/aR and (n + 1) r(a) ∼ is unit-regular.



Proof. (1)⇒(2) Let a ∈ 1 +
I be regular, n r(a) ∼ = n R/aR and
(n + 1) r(a) ∼ = (n + 1) R/aR . Then r(a), R/aR ∈ F P (I). Since I is

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separative, it follows by Lemma 13.4.1 that r(a) ∼ = R/aR, and so a ∈ R is unit-regular. (2)⇒(1) Let a ∈ 1 + I be regular and (a − a2 )R . r(a), R/aR. Then there exist right R-modules C and D such that r(a) ∼ = (a − a2 )R ⊕ C 2 ∼ and R/aR = (a − a )R ⊕ D. As in the proof of Theorem 13.4.8, we have r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R. Thus 2r(a) ∼ = r(a) ⊕ (a − a2 )R ⊕ C ∼ = R/aR ⊕ (a − a2 )R ⊕ C ∼ = R/aR ⊕ r(a).

Furthermore, we see that


2 R/aR ∼ = R/aR ⊕ (a − a2 )R ⊕ D ∼ = r(a) ⊕ (a − a2 )R ⊕ D ∼ = r(a) ⊕ R/aR.

∼ ∼ Hence 2r(a) ∼ 2 R/aR . As a result, 3r(a) 2 R/aR ⊕ r(a) = =


= R/aR ⊕
∼ ∼ R/aR⊕r(a) =
3 R/aR . Likewise,
we conclude that n r(a) = n R/aR and (n + 1) r(a) ∼ = (n + 1) R/aR . By assumption, a ∈ R is unit-regular. According to Theorem 13.4.8, I is separative. 

As in the proof of Corollary 13.4.9, we deduce that a regular ideal I of a ring R is separative if and only if each regular a ∈ 1 + I satisfying
∼ nr(a) = n R/aR for all n ≥ 2 is unit-regular. Lemma 13.4.10. Let I be a regular ideal of a ring R. Then the following are equivalent : (1) I is separative. (2) Each regular a ∈ 1 + I satisfying

2r(a) ∼ = r(a) ⊕ R/aR ∼ = 2 R/aR

is unit-regular.




∼ ∼ Proof. (1)
⇒ (2) Let a ∈ 1 + I be regular and 2r(a) = r(a) ⊕ R/aR = ∼ 2 R/aR . Then r(a), R/aR ∈ F P (I). By hypothesis, we get r(a) = R/aR, and then a ∈ R is unit-regular. (2) ⇒ (1) Let a ∈ 1+I be regular and (a−a2 )R . r(a), R/aR. As in the proof in Theorem 13.4.8, (a − a2 )R .⊕ r(a), R/aR. Thus, we have right Rmodules C and D such that r(a) ∼ = (a−a2 )R⊕C and R/aR ∼ = (a−a2 )R⊕D. As in the proof of Theorem 13.4.8, we have r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R. Analogously to Corollary 13.4.9, we get 2r(a) ∼ = r(a) ⊕ R/aR ∼ =

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2 R/aR . By assumption, a ∈ R is unit-regular. So the result follows from Theorem 13.4.8.  Theorem 13.4.11. Let I be a regular ideal of a ring R. Then the following are equivalent : (1) I is separative. (2) For any idempotents e, f ∈ I, ∼ eR ⊕ f R ∼ eR ⊕ eR = = f R ⊕ f R =⇒ eR ∼ = f R. Proof. (1) ⇒ (2) For any idempotents e, f ∈ I, we see that eR, f R ∈ F P (I). Thus, eR ⊕ eR ∼ = eR ⊕ f R ∼ = f R ⊕ f R implies that eR ∼ = f R.
(2) ⇒ (1) Let a ∈ 1+I be regular and 2r(a) ∼ = r(a)⊕R/aR ∼ = 2 R/aR . Then we can find some x ∈ 1+I such that a = axa; hence, r(a) = (1−xa)R and R/aR ∼ = (1 − ax)R. Thus, we get (1 − xa)R ⊕ (1 − xa)R ∼ = (1 − xa)R ⊕ (1 − ax)R ∼ = (1 − ax)R ⊕ (1 − ax)R. By assumption, we have φ : (1 − ax)R ∼ = (1 − xa)R. Clearly, ϕ : axR = aR ∼ = xaR given by ϕ(ar) = xar for any r ∈ R. Construct a map θ : R = aR ⊕ (1 − ax)R = xaR ⊕ (1 − xa)R given by θ(s + t) = ϕ(s) + φ(t) for any s ∈ aR and t ∈ (1 − ax)R. Then θ(1) ∈ R is invertible. In addition, aθ(1)a = a, i.e., a ∈ 1 + I is unit-regular. According to Lemma 13.4.10, we establish the result.  An an immediate consequence, we derive that a regular ring R is separative if and only if for any idempotents e, f ∈ R, eR ⊕ eR ∼ = eR ⊕ f R ∼ = ∼ f R ⊕ f R =⇒ eR = f R. Corollary 13.4.12. Let I be a regular ideal of a ring R. Then the following are equivalent : (1) I is separative. (2) For any a ∈ I, RaR is separative. (3) For any a ∈ I, EndR (aR) is separative. Proof. (1) ⇒ (2) Let a ∈ I. For any idempotent e ∈ RaR, we have e ∈ I; hence, eRe is a separative regular ring. Therefore RaR is separative from Lemma 13.3.7. (2) ⇒ (1) For any idempotent e ∈ I, it suffices to show that eRe is separative. As I is a regular ideal, eRe is a regular ring. Given any idempotents f, g ∈ eRe such that f (eRe) ⊕ f (eRe) ∼ = f (eRe) ⊕ g(eRe) ∼ = g(eRe) ⊕ g(eRe),

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then

N eR ∼ eR = f (eRe) eRe eRe eRe N N N ⊕g(eRe) eR ∼ eR ⊕ g(eRe) eR. = g(eRe) eRe eRe eRe N ∼ eR; hence, It is easy to verify that eRe eR = f (eRe)

N

eR ⊕ f (eRe)

N

eRe

fR ⊕ fR ∼ = f R ⊕ gR ∼ = gR ⊕ gR with f, g ∈ I. As in the proof of Theorem 13.4.8, we have an idempotent h ∈ I such that f R + gR = hR. This implies that f, g ∈ RhR. Hence, f R, gR ∈ F P (RhR). By hypothesis, RhR is separative. Hence, f R ∼ = gR. Thus, we N N N ∼ get f (eR) Re = g(eR) Re. One easily checks that eR Re ∼ = eRe R

R

R

as right eRe-modules. Consequently, we deduce that f (eRe) ∼ = g(eRe). Therefore eRe is a separative regular ring by Theorem 13.4.11, as required. (1) ⇔ (3) is obvious by Lemma 13.3.7. 

Lemma 13.4.13. Let I be a regular right ideal of a ring R, and let A be a finitely generated projective right R-module. Then A = AI if and only if A .⊕ nI for some n ∈ N. Proof. Assume that A = AI and x1 , · · · , xm are generators for A. Then we n P can find some y1 , · · · , yn ∈ A such that each xi ∈ yi I. Thus, A = yj I. Construct a map ϕ : nI → A given by ϕ(r1 , · · · , rn ) =

n P

j=1

yj rj for any

j=1

(r1 , · · · , rn ) ∈ nI. Clearly, ϕ is an R-epimorphism. By the projectivity, ϕ splits; hence, A .⊕ nI. Conversely, assume that A .⊕ nI. Then we can find a right R-module N N N D such that A ⊕ D ∼ = nI; hence, A (R/I) ⊕ D (R/I) ∼ = (nI) (R/I). R R R N Since I is regular, I = I 2 , and so I (R/I) = 0. Thus, A/AI ∼ = R N A (R/I) = 0. Therefore we see that A = AI, as asserted.  R

In [16, Proposition 6.2], Ara et al. proved that a regular ring R is separative in and only if each a ∈ R satisfying Rr(a) = ℓ(a)R = R(1 − a)R (∗) is unit-regular. In view of [244, Lemma 1.2] and [245, Proposition 2.3], every nilpotent element in a regular ring R satisfies the condition (∗). Now we extend the result of Ara et al. as follows. Theorem 13.4.14. Let I be a regular ideal of a ring R. Then the following are equivalent :

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(1) I is separative. (2) Each regular a ∈ 1 + I satisfying is unit-regular.

R(1 − a)R = Rr(a) = ℓ(a)R

Proof. (1) ⇒ (2) Let a ∈ 1 + I be regular and R(1 − a)R = Rr(a) = ℓ(a)R. Assume that a = aca for some c ∈ R. Since R(1 − a)R = Rr(a) and 1 − a ∈ I is regular, we get (1 − a)R = (1 − a)Rr(a). Clearly, r(a) is a regular right ideal. In view of Lemma 13.4.13, we have (1 − a)R .⊕ mr(a) for some m ∈ N. As R(1 − a)R = ℓ(a)R = R(1 − ac)R, we have (1 − a)R = (1 − a)R(1 − ac)R, and so (1 − a)R .⊕ n(1 − ac)R for some n ∈ N. Clearly, (1 − ac)R ∼ = R/aR. Thus, (1 − a)R .⊕ n(R/aR). As in the proof of Theorem 13.4.8, r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R. As a(1 − a) ∈ I is regular, we have some z ∈ R such that a(1 − a) = a(1 − a)za(1 − a). By the regularity of 1 −
a, it is easy to verify that (1 − a)R = a(1 − a)R ⊕ (1 − a) − a(1 − a)z R. So we see that (a − a2 )R .⊕ (1 − a)R, whence, r(a) ⊕ (1 − a)R ∼ = R/aR ⊕ (1 − a)R. As 1 − a ∈ I, (1 − a)R ∈ F P (I). According to Proposition 13.4.2, we have that r(a) ∼ = R/aR, and therefore a ∈ R is unit-regular. (2) ⇒ (1) Suppose that A ⊕ C ∼ = B ⊕ C .⊕ R and C .⊕ A, B for some A, B, C ∈ F P (I). Write R = A1 ⊕ C1 ⊕ D = A2 ⊕ C2 ⊕ D, where A1 ∼ = A, C1 ∼ =C ∼ = C2 and A2 ∼ = B. Let a ∈ R induce an endomorphism of RR , which is zero on A1 , an isomorphism from C1 onto C2 , and the identity on D. As in the proof of Theorem 13.4.8, a ∈ 1 + I is regular. Let ϕ : A1 ⊕ C1 → (1 − a)(A1 ⊕ C1 ) be a right R-module given by ϕ(x) = (1 − a)x for any x ∈ A1 ⊕ C1 . Since (1 − a)(A1 ⊕ C1 ) = (1 − a)R is projective, (1 − a)R .⊕ A1 ⊕ C1 .⊕ 2A1 = 2r(a). Since I is regular, so is r(a). In view of Lemma 13.4.13, (1 − a)R = (1 − a)Rr(a). This yields R(1 − a)R = Rr(a). Assume that a = aca for some c ∈ R. Then (1 − a)R .⊕ A1 ⊕ C1 ∼ = A2 ⊕ C2 .⊕ 2A2 = 2(R/aR) ∼ = 2(1 − ac)R. Clearly, c ∈ 1 + I, and so (1 − ac)R is regular. By using Lemma 13.4.13 again, (1 − a)R = (1 − a)R(1 − ac)R, and then R(1 − a)R = ℓ(a)R. By assumption, a ∈ R is unit-regular; hence, r(a) ∼ = R/aR. Thus we have A∼  = B. This establishes the result from Lemma 13.4.1. Corollary 13.4.15. A regular ring R is unit-regular if and only if the following hold : (1) R is separative. (2) For all ideals I, R/I is directly finite.

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(3) For all ideals I, every unit lifts modulo I. Proof. Assume that the given conditions hold. Let a ∈ R. Then there exists x ∈ R such that a = axa. Choose I = Rr(a). Then xa = 1 in R/I. By Condition (2), we get ax = 1, thus, a ∈ U (R/I). By Condition (3), there exists an element u ∈ U (R) such that a − u ∈ I. Let b = u−1 a. Then R(1 − b)R = R(u − a)R ⊆ I = Rr(b). Write b = bcb. Then Rr(b) = R(1 − cb)R = R(1 − b)(1 − cb)R ⊆ R(1 − b)R, so I = Rr(b) = R(1 − b)R. Likewise, R(1 − b)R ⊇ ℓ(b)R. By Condition (2) again, R/ℓ(b) is directly finite. For any d
∈ r(b), we get bd = 0. Clearly, 1 − bc ∈ ℓ(b)R, i.e.,
bc ≡ 1 mod ℓ(b)R . Hence, cb ≡ 1 mod ℓ(b)R , i.e, 1 − cb ∈ ℓ(b)R. This shows that d = (1 − cb)d ∈ ℓ(b)R, and then Rr(b) ⊆ ℓ(b)R. Consequently, Rr(b) = ℓ(b)R = R(1 − b)R. By virtue of Theorem 13.4.14, b ∈ R is a unit-regular, and so is a ∈ R. Therefore R is unit-regular. Conversely, Conditions (1) and (2) are obvious. Let I be an ideal of R. Given any u ∈ U (R/I), then there exists some v ∈ R such that 1 − uv ∈ I. It follows from uv + (1 − uv) = 1 that w := u + (1 − uv)z ∈ U (R) for some z ∈ R. Obviously, w − u = (1 − uv)z ∈ I, i.e., u ≡ w(mod I). Therefore the proof is true.  Lemma 13.4.16. Let I be a separative exchange ideal of a ring R. If e = e2 ∈ R, then eIe is a separative ideal of eRe. Proof. Clearly, eIe is an exchange ideal of eRe by [9, Proposition 1.3]. Given any idempotent exe ∈ eIe, then (exe)(eRe)(exe) = (exe)R(exe). Since exe ∈ I, by [16, Lemma 4.1], (exe)R(exe) is a separative exchange ring. By [16, Lemma 4.1] again, we obtain the result.  Theorem 13.4.17. Let I be a regular ideal of a ring R. Then the following are equivalent: (1) I is separative. (2) Mn (I) is separative. Proof. (1) ⇒ (2) Suppose that A ⊕ C ∼ = B ⊕ C with C .⊕ A, B and
C, A, B ∈ F P Mn (I) . Then O O O O A Rn×1 ⊕ C Rn×1 ∼ Rn×1 ⊕ C Rn×1 =B Mn (R)

Mn (R)

with

C

O

Mn (R)

Rn×1 .⊕ A

Mn (R)

O

Mn (R)

Rn×1 , B

Mn (R)

O

Mn (R)

Rn×1 .

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Clearly, (A

O

Mn (R) m P

Given any

ai

i=1

ai

i=1

O

Rn×1 .

Mn (R)

N N (x1i , · · · , xni )T ∈ A Rn×1 , we have bij ∈ A, rij ∈ Mn (R)

Mn (I) such that m P

Rn×1 )I ⊆ A

463

ki m P N P N (x1i , · · · , xni )T = (bij rij ) (x1i , · · · , xni )T

=

i=1 j=1 ki m P P

i=1 j=1

bij

N

rij (x1i , · · · , xni )T .

ij T T Set (cij 1 , · · · , cn ) = rij (x1i , · · · , xni ) . Then m P

ai

i=1

That is, A

ki P m P n
N P N (x1i , · · · , xni )T = bij (0, · · · , 1, · · · , 0)T cij k i=1 j=1 k=1 N n×1 R )I. ⊆ (A

N

Mn (R)

Mn (R)

Rn×1 ∈ F P (I). Likewise, B

N

Rn×1 , C

Mn (R)

N

Mn (R)

Rn×1 ∈

N F P (I). Since I is separative, we deduce that B Rn×1 Mn (R) N C Rn×1 . Therefore we get Mn (R)

B∼ = B

O

Mn (R)

Rn×1


O R

R1×n ∼ = C

O

Mn (R)

Rn×1


O R

∼ =

R1×n ∼ = C,

as desired. (2) ⇒ (1) In view of [54, Lemma 2], Mn (I) is regular. Choose e = diag(1, 0, · · · , 0) ∈ Mn (R). Then I ∼ = eMn (I)e, and the result follows from Lemma 13.4.16.  Corollary 13.4.18. Let I be a regular ideal of a ring R. Then the following are equivalent: (1) I is separative. (2) For all P ∈ F P (I), EndR (P ) is separative. Proof. (1) ⇒ (2) Clearly, I is an exchange ideal. Since P ∈ F P (I), by Lemma 13.1.8, there exist idempotents e1 , · · · , en ∈ I such that P ∼ = e1 R ⊕ ∼ · · · ⊕ en R. Hence, EndR (P ) = diag(e1 , · · · , en )Mn (R)diag(e1 , · · · , en ). In view of Theorem 13.4.17, Mn (I) is separative. Thus, EndR (P ) is separative from Lemma 13.3.7.

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(2) ⇒ (1) Given any idempotent e ∈ I, one easily checks that eR ∈ F P (I). Hence eRe ∼ = EndR (eR) is a separative ring. Therefore we complete the proof by Lemma 13.3.7.  Let A be a right R-module, and let S = EndR (A). We use addR (A) to denote the category of direct summands of direct sums of finite copies of A. Clearly, A is a left S-R-bimodule. Let M be a right R-module, and let N be a right S-module. Then HomR (S AR , MR ) is a right S-module N and N AR is a right R-module. Let P(S) be the category of finitely S

generated projective right S-modules. We now derive the following.

Lemma 13.4.19. Let
A be a finitely generated right R-module.
Then there exist F : P EndR (A) → addR (A), G : addR (A) → P EndR (A) such that FG ∼ = IaddR (A) and GF ∼ = IP

EndR (A)


.


AR : P EndR (A) → addR (A) and G = S
HomR (AR , −) : addR (A) → P EndR (A) . Since F and G preserve direct sums, we see at once that they are functors between the desired categories. Set S = EndR (A). For any M ∈ addR (A), we define ϕM : HomR (AR , MR ) × AR → MR given by ϕM (f, a) = f (a) for any f ∈ HomR (AR , MR ), a ∈ A. Then there N exists a unique φM : HomR (AR , MR ) AR → MR such that the following Proof.

Let F = −

N

S

diagram

⊗

HomR (A, M ) × A → HomR (A, M ) ϕM ↓ M

IM

→

↓ φM

N

A

S

M

commutates, where IM is the identical map. As A is finitely generated, it is routine to check that φM is an R-isomorphism.
For any R-morphism f : M → N , it is easy to verify that ϕN f∗ ⊗ 1R = f ϕM . Therefore we show that FG ∼ = IaddR (A) . N For any B ∈ P(S), we define σB : BS → HomR (AR , B AR ) given by S N σB (b) = gb for any b ∈ BS , where gb : AR → B AR given by gb (a) = S

b ⊗ a for any a ∈ AR . Then σB is an S-morphism. Clearly, σB is an S-isomorphism when B is a free right S-module. There is a splitting Sepimorphism θ : nS → BS when B is a finitely generated projective right

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S-morphism. Clearly, the following diagram θ

→

nS σnS ↓

HomR (AR , nS

N

AR )

S

(θ⊗1A )∗

→

BS ↓ σB

HomR (AR , B

N

AR )

S

commutates. This implies that σB is a splitting R-epimorphism. There is a splitting S-monomorphism τ : B → nS when B is a finitely generated projective right S-morphism. Clearly, the following diagram BS σB ↓

HomR (AR , B

τ

→ N

AR )

S

(τ ⊗1A )∗

→

nS ↓ σnS

HomR (AR , nS

N

AR )

S

commutates. This implies that σB is a splitting R-monomorphism. Therefore τB is an R-isomorphism. For any R-morphism g : B → C, it is easy to verify that σC g = (g ⊗ 1A )∗ )σB . Therefore we get GF ∼  = IP(S) . Lemma 13.4.20. Let R be an exchange ring, let P ∈ F P (R), and let C ∈ addR (P ). If EndR (P ) is separative, then for any right R-modules A and B, A ⊕ C ∼ = B ⊕ C with C .⊕ A, B implies that A ∼ = B.

∼ B ⊕ C with C .⊕ A, B, by hypothesis, C ∈ F P (R). Proof. Given A ⊕ C = Hence, C has the finite exchange property. As in the proof of Lemma 13.3.1, there exists a refinement matrix AC   B D 1 B1 . C A1 C1

As in the proof of Lemma 6.3.7, we may assume that C1 .⊕ A1 , B1 . Obviously, C1 , A1 , B there exist
1 ∈ addR (C). According to Lemma 13.4.19,
F : P EndR (C) → addR (C), G : addR (C) → P EndR (C) such that
. FG ∼ = Iadd (C) and GF ∼ =I R

P EndR (C)

Therefore G(A1 )⊕G(C1 ) ∼ = G(B1 )⊕G(C1 ) with G(C1 ) .⊕ G(A1 ), G(B1 ). As EndR (C) is separative, we get G(A1 ) ∼ = G(B1 ). Thus, F G(A1 ) ∼ = F G(B1 ). By Lemma 13.4.19 again, A1 ∼ = B1 . Therefore A ∼ = D1 ⊕A1 ∼ = D1 ⊕B1 ∼ = B, as required. 

Lemma 13.4.21. Let R be an exchange ring, and let x, y ∈ R be idempotents. If EndR (xR) and EndR (yR) are separative, then so is EndR (xR ⊕ yR).

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Proof. Suppose that EndR (xR) and EndR (yR) are separative. Given
A⊕C ∼ = B⊕C with C .⊕ A, B, where A, B, C ∈ F P
EndR (xR⊕yR) . By Lemma 13.4.19, there exist F : P End
R (xR ⊕ yR) → addR (xR ⊕ yR), G : addR (xR ⊕ yR) → P EndR (xR ⊕ yR) such that
. FG ∼ = Iadd (xR⊕yR) and GF ∼ =I R

P EndR (xR⊕yR)

In addition, F and G preserve direct sums. Thus, F (A) ⊕ F (C) ∼ = F (B) ⊕ F (C) with F (C) .⊕ F (A), F (B). Clearly, F (C) has the finite exchange property. As in the proof of Lemma 13.3.1, we have some C1 ∈ addR (xR), (C) = C1 ⊕ C2 . Thus
C2 ∈ addR (yR)
such that F C1 ⊕ C2 ⊕F (A) ∼ = C1 ⊕ C2 ⊕F (B) with C1 .⊕ C2 ⊕F (A), C2 ⊕F (B). In view of Lemma 13.4.20, C2 ⊕ F(A) ∼ = C2 ⊕ F(B) with C2 .⊕ F (A), F (B). By using Lemma 13.4.20 again, we get F (A) ∼ = F (B). This implies ∼ ∼ that GF (A) = GF (B). Therefore A = B, and then we conclude that EndR (xR ⊕ yR) is a separative ring.  Let I be an ideal of a ring R. Then the set of all n × n lower triangular matrices over I is an ideal of the ring of all n × n lower triangular matrices over R, and is denoted by T Mn (I). In general, T Mn (I) is not a regular ideal of T Mn (R).

Theorem 13.4.22. Let I be a regular ideal of a ring R. Then the following are equivalent: (1) I is separative. (2) The ideal T Mn (I) of all n × n lower triangular matrices over I is separative. Proof. (2) ⇒ (1) Choose E = diag(1, 0, · · · , 0) ∈ Mn (R). Then I ∼ = ET Mn (I)E. In view of [9, Proposition 1.3], T Mn (I) is an exchange ideal. According to Theorem 13.4.17, I is separative. (1) ⇒ (2) It will suffice to prove the result for n = 2. For  any idempotent  g0 f ∈ T M2 (I), there are idempotents g, h ∈ I such that f = . Hence ∗h   gRg 0 f T M2 (R)f ∼ . = ∗ hRh   gRg 0 Choose e = diag(g, 0). Then e e ∼ = gRg and diag(g, h) − ∗ hRh  
gRg 0
e diag(g, h) − e ∼ = hRh. As I is regular, both gRg and ∗ hRh

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 gRg 0 hRh are regular rings. In view of [264, Proposition 2.2], is ∗ hRh an exchange ring. According to Lemma 13.4.21, it is separative, and then so is f T M2(R)f . Therefore we conclude that T M2 (I) is separative by [16, Lemma 4.1].  Similarly, we deduce that a regular ideal I of a ring R is separative if and only if so is the ideal of all n × n upper triangular matrices over I. Let T be the ring of a Morita context (A, B, M, N, ψ, φ). If A and B are separative exchange rings, then so is T . Let e =
diag(1A , 0). Then A ∼ = eT e
∼ and B = diag(1A , 1B ) − e T diag(1A , 1B ) − e . Therefore we are done by Lemma 13.4.21. Lemma 13.4.23. Let R be a regular ring. Then I = {a ∈ R | EndR (aR) is separative} is a separative ideal of R. Proof. Let I = {a ∈ R | EndR (aR) is separative}. Let x, y ∈ I and z ∈ R. As in the proof of Example 13.2.14, we show that zxR .⊕ xR, xzR .⊕ xR and (x + y)R .⊕ xR ⊕ yR. As x, y ∈ I, EndR (xR) and EndR (yR) are separative. This implies that EndR (zxR) and EndR (xzR) are separative by Lemma 13.3.7. Hence, zx, xz ∈ I. In view of Lemma 13.4.21, EndR (xR ⊕
yR) is separative, and then so is EndR (x + y)R by Lemma 13.3.7. This implies that x + y ∈ I. As a result, I is an ideal of R. For any idempotent e ∈ I, eRe is separative. According to Lemma 13.3.7, I is a separative ideal of R.  Theorem 13.4.24. Let R be a regular ring, and let (aij ) ∈ Mn (R). If all EndR (aij R) are separative, then (aij ) admits a diagonal reduction. Proof. Let I = {a ∈ R | EndR (aR) is separative}. In view of Lemma 13.4.23, I is a separative ideal. Since all EndR (aij R) are separative, we show that all aij ∈ I. By virtue of Lemma 13.1.9, there exists an idempotent e ∈ I such that all aij ∈ eRe. As eRe ∼ = EndR (eR), eRe is separative. According to Lemma 7.3.1, (aij ) ∈ Mn (eRe) admits a diagonal reduction, i.e., there exist some U ′ , V ′ ∈ GLn (eRe) such that U ′ AV ′ = diag(r1 , · · · , rn ) for r1 , · · · , rn ∈ eRe. Let E = diag(e, · · · , e) ∈ Mn (R). Then U := U ′ + In − E, V = V ′ + In − E ∈ GLn (R). Further, U AV = diag(r1 , · · · , rn ), as asserted. 

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Chapter 14

Diagonal Reduction

The main purpose of this chapter is to study the conditions under which every regular matrix over an exchange ring admits a diagonal reduction. We say that a ring R is an almost hermitian ring provided that every regular matrix admits a diagonal reduction. A rectangular matrix A ∈ Mm×n (R) is completed provided that A can be completed to an invertible matrix by
adding some columns or rows. We use coln (R) rown (R) to denote the set of all completed n × 1 (1 × n) matrices over a ring R. We will see that almost hermitian rings behave in much the same way as those with stable range condition by means of completed matrices.

14.1

Almost Hermitian Rings, I

Lemma 14.1.1. Let Q be a right R-module. Then mQ ∼ = K ⊕ I and nQ ∼ = I ⊕ C if and only if there exists a regular f : mQ → nQ such that K∼ = ker(f ), I ∼ = im(f ) and C ∼ = coker(f ). Proof. Suppose that mQ ∼ = K ⊕ I and nQ ∼ = I ⊕ C. Then mQ = K1 ⊕ I1 and nQ = I2 ⊕ C2 with K1 ∼ = K, C ∼ = C2 and ψ : I1 ∼ = I ∼ = I2 . Let p : mQ = K1 ⊕ I1 → I1 be the projection and q : I2 → I2 ⊕ C2 = nQ be the injection. Define f = qψp : mQ → nQ. Then f is regular. Furthermore, we have K ∼ = K1 ∼ = ker(f ), I ∼ = I2 = im(f ) and C ∼ = C2 ∼ = coker(f ). Conversely, assume that there exists a regular f : mQ → nQ such 469

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that K ∼ = ker(f ), I ∼ = im(f ) and C ∼ = coker(f ). Since f is regular, we have g : nQ → mQ such that f = f gf . It is easy to verify that mQ = gf (mQ) ⊕ (1 − gf )(mQ) and nQ = f g(nQ) ⊕ (1 − f g)(nQ); hence, C := coker(f ) ∼ = (1 − f g)(nQ), I := im(f ) = f (mQ) = f g(nQ) ∼ = gf (mQ) and K = ker(f ) = (1 − gf )(mQ). Therefore, mQ ∼ = K ⊕ I and nQ ∼ = I ⊕ C, as asserted.  Let Q be a right R-module, and let F (Q) be the set of all nQ for n ≥ 0. For each F1 , F2 ∈ F (Q), let mor F (Q)(F1 , F2 ) be the set of all R-morphisms from F1 to F2 , and again let ◦ be the usual composition of homomorphisms. Then (F (Q), mor F (Q), ◦) is a category. We denote it by F (Q). Let M be the set of all nonnegative integers. For each m, n ∈ M, let mor M (Q)(n, m) be the set of all m × n matrices over EndR (Q), and again let ◦ be the usual product of matrices. Then (M, mor M (Q), ◦) is also a category. We denote it by M(Q). Lemma 14.1.2. Let Q be a right R-module. Then there exist covariant functors F : F (Q) → M(Q) and G : M(Q) → F(Q) such that F G = IM(Q) and GF = IF (Q) are both identity functors. Proof. Let mQ ∈ F (Q). Define F ′ (mQ) = m ∈ M. For each f : nQ → mQ, if pi : mQ → Q denotes the projection and qj : Q → nQ denotes the injection (1 ≤ i ≤ m, 1 ≤ j ≤ n), then we define   p1 f q1 p1 f q2 · · · p1 f qn  p2 f q1 p2 f q2 · · · p2 f qn 
  F ′′ f =  . .. ..  ∈ Mm×n EndR (Q) . ..   .. . . . pm f q1 pm f q2 · · · pm f qn Then F = (F ′ , F ′′ ) : F (Q) → M(Q) is a covariant functor. Let m ∈ M. Define G′ (m) = mQ ∈ F (Q). For each B ∈ Mm×n (EndR Q), we define     a1 a1  ..  ′′ ′′  ..  G B : nQ → mQ such that G B  .  = B  .  ∈ mQ. Then G = an an (G′ , G′′ ) : M(Q) → F (Q) is also a covariant functor. One easily checks that F G and GF are both identity functors.  We say that
f : 2Q → Q admits a diagonal reduction if F (f ) ∈ M1×2 EndR (Q) admits a diagonal reduction. By Lemma 14.1.2, f : 2Q → Q admits a diagonal reduction if and only if f is equivalent to (g, 0), where g : Q → Q.

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Lemma 14.1.3. Let Q be a right R-module, and let f : 2Q → Q be regular. Then f admits a diagonal reduction if and only if there is a decomposition ker(f ) ∼ = K1 ⊕ K2 such that K1 ⊕ im(f ) ∼ =Q∼ = K2 .

Proof. Assume there is a decomposition ker(f ) ∼ = K1 ⊕ K2 such that K1 ⊕im(f ) ∼ =Q∼ = K2 . Because of the regularity of f , we have K1 ⊕im(f ) ∼ = Q∼ = im(f ) ⊕ coker(f ). From Lemma 14.1.1, there is a regular g : Q → Q such that ker(g) ∼ = K1 , im(g) ∼ = im(f ) and coker(g) ∼ = coker(f ). So we have (g, 0) : 2Q → Q such that ker(g, 0) = ker(g) ⊕ Q ∼ = K1 ⊕ K2 ∼ = ∼ ker(f ), im(g, 0) = im(g) = im(f ) and coker(g, 0) = coker(g) ∼ = coker(f ). By virtue of Lemma 7.2.1, f admits a diagonal reduction. Conversely, assume now that f : 2Q → Q admits a diagonal reduction (g, 0) with g : Q → Q. Then g is also regular; hence, Q ∼ = ker(g) ⊕ im(g). By Lemma 7.2.1 again, ker(f ) ∼ = ker(g, 0), im(f ) ∼ = im(g, 0) and coker(f ) ∼ = coker(g, 0). It is easy to check that ker(g, 0) ∼ = ker(g) ⊕ Q, im(g, 0) ∼ = im(g), coker(g, 0) ∼ = coker(g). Set K1 = ker(g) and K2 = Q. Then ker(f ) ∼ = K1 ⊕ K2 , K1 ⊕ im(f ) ∼  = ker(g) ⊕ im(g) ∼ =Q∼ = K2 , as required.

Lemma 14.1.4. Let Q be a right R-module, and let f : 2Q → Q be regular. Then the following hold: (1) 2Q ⊕ coker(f ) ∼ = Q ⊕ ker(f ). (2) f admits a diagonal reduction if and only if Q ⊕ coker(f ) ∼ = ker(f ).

Proof. (1) In view of Lemma 14.1.1, we have 2Q ∼ = K ⊕ I and Q ∼ = I ⊕ C, where K ∼ = ker(f ), I ∼ = im(f ) and C ∼ = coker(f ). Thus 2Q ⊕ coker(f ) ∼ = ∼ ∼ 2Q ⊕ C = (K ⊕ I) ⊕ C = (I ⊕ C) ⊕ K = Q ⊕ K = Q ⊕ ker(f ), as desired. (2) Assume that f admits a diagonal reduction. According to Lemma 14.1.3, there is a decomposition ker(f ) ∼ = K1 ⊕ K2 such that K1 ⊕ im(f ) ∼ = ∼ ∼ ∼ ∼ Q∼ ker(f ) K ⊕K K ⊕Q K ⊕(coker(f )⊕im(f )) = K2 . Therefore = = = = 1 2 1 1
∼ K1 ⊕ im(f ) ⊕ coker(f ) = coker(f ) ⊕ Q. Assume that Q ⊕ coker(f ) ∼ = ker(f ). Set K1 = coker(f ) and K2 = Q. Then ker(f ) = K1 ⊕ K2 and K1 ⊕ im(f ) ∼ =Q∼ = K2 . It follows by Lemma 14.1.3 that f admits a diagonal reduction.  Theorem 14.1.5. Let Q be a right R-module having the finite exchange property, and let E = EndR (Q). Then the following are equivalent: (1) (2) (3) (4)

E is an almost hermitian ring. Every 1 × 2 regular matrix over E admits a diagonal reduction. Every 2 × 1 regular matrix over E admits a diagonal reduction. For any A, B ∈ F P (R), 2Q ⊕ A ∼ = Q ⊕ B implies that Q ⊕ A ∼ = B.

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Proof. (1) ⇒ (2) is trivial. (2) ⇒ (4) Let A, B ∈ F P (R) such that 2Q⊕A ∼ = Q⊕B. As in the proof of Lemma 6.3.7, we have decompositions 2Q = N1 ⊕ N2 , A = A1 ⊕ A2 and N1 ⊕ A1 ∼ = Q, N2 ⊕ A2 ∼ = B. So 2Q = N1 ⊕ N2 and Q ∼ = N1 ⊕ A1 . Using Lemma 14.1.1, we get a regular homomorphism f : 2Q → Q such that ker(f ) ∼ = A1 . Since f admits a diagonal = N2 , im(f ) ∼ = N1 and coker(f ) ∼ reduction, by Lemma 14.1.4, Q ⊕ coker(f ) ∼ = ker(f ). So Q ⊕ A1 ∼ = N2 . ∼ ∼ Therefore Q ⊕ A = A2 ⊕ N2 = B. (4) ⇒ (1) Let f : nQ → mQ be regular. Then nQ ∼ = ker(f ) ⊕ I and mQ ∼ = I ⊕coker(f ) for a right R-module I. This implies that mQ⊕ker(f ) ∼ = nQ ⊕ coker(f ). Assume that m = n(≥ 2). Given any decompositions nQ ∼ = K⊕I ∼ I ⊕ C, then K has the finite exchange property by [382, Lemma = 28.1]. Hence, we have K = X1 ⊕ X2 , I = Y1 ⊕ Y2 such that X1 ⊕ Y1 ∼ =I and X2 ⊕ Y2 ∼ = C. Thus nQ ⊕ Y2 ∼ = nQ ⊕ X1 . By hypothesis, we get Q⊕Y2 ∼ = Q⊕X1 . Further, Q = R1 ⊕R2 , Y2 = C1 ⊕C2 such that R1 ⊕C1 ∼ =Q ∼ ∼ and R2 ⊕ C2 ∼ X . This implies that nQ R ⊕ (I ⊕ X ⊕ C ) (I ⊕ X = 1 = 2 2 2 = 2⊕ ∼ C2 ) ⊕ C1 . One easily checks that 2Q ⊕ (n − 2)Q ⊕ R1 = Q ⊕ I ⊕ X2 ⊕ C2 . By hypothesis, we get (n − 1)Q ⊕ R1 ∼ = I ⊕ X2 ⊕ C2 . Consequently, we have that R2 ∼ = R2 ⊕ 0 ⊕ · · · ⊕ 0, I ⊕ X2 ⊕ C2 ∼ = R1 ⊕ Q ⊕ · · · ⊕ Q, C1 ∼ = C1 ⊕ 0 ⊕ · · · ⊕ 0

with R2 ⊕ R1 ∼ = C1 ⊕ R1 ∼ = Q. Similarly to Proposition 7.2.2, f admits a diagonal reduction. If n > m, then mQ ⊕ ker(f ) ∼ = 2Q ⊕ (n − 2)Q ⊕ coker(f ). By hypothesis, ker(f ) ∼ (n − m)Q ⊕ coker(f ). If n < m, 2Q ⊕ (m − 2)Q ⊕ ker(f ) ∼ = = nQ ⊕ coker(f ). By hypothesis, coker(f ) ∼ = (m − n)Q ⊕ ker(f ). Similarly to Proposition 7.2.11 and Lemma 14.1.4, f admits a diagonal reduction, as required. (1) ⇒ (3) ⇒ (4) are proved in the same manner. 

According to Theorem 12.1.12 and Theorem 14.1.5, a separative exchange ring is an almost hermitian ring if and only if it satisfies the finite stable range condition. Corollary 14.1.6. Let Q be a right R-module having the finite exchange property, and let E = EndR (Q). Then the following are equivalent: (1) E is an almost hermitian ring. (2) For any right R-modules A and B, 2Q ⊕ A ∼ = Q ⊕ B implies that Q⊕A∼ = B.

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(3) Q = A1 ⊕ B1 and 2Q = A2 ⊕ B2 with A1 ∼ = A2 imply that Q ⊕ B1 ∼ = B2 . ∼ Q⊕B, Proof. (1) ⇒ (2) Given any right R-modules A and B with 2Q⊕A =

then we have decompositions

2Q = N1 ⊕ N2 , A = A1 ⊕ A2 and N1 ⊕ A1 ∼ = Q, N2 ⊕ A2 ∼ =B

because Q has the finite exchange property. So 2Q = N1 ⊕ N2 and Q ∼ = N1 ⊕ A1 . Using Lemma 14.1.1, we get a regular R-morphism f : 2Q → Q such that ker(f ) ∼ = N2 , im(f ) ∼ = N1 and coker(f ) ∼ = A1 . Since f admits a diagonal reduction, by Lemma 14.1.4, Q ⊕ coker(f ) ∼ = ker(f ). So Q ⊕ A1 ∼ = ∼ ∼ N2 , and therefore Q ⊕ A = A2 ⊕ N2 = B. (2) ⇒ (3) is trivial. (3) ⇒ (1) Let f : 2Q → Q be regular. In view of Lemma 14.1.4, 2Q ⊕ coker(f ) ∼ = Q ⊕ ker(f ). Obviously, we have decompositions 2Q = N1 ⊕ N2 , coker(f ) = A1 ⊕ A2 and N1 ⊕ A1 ∼ = Q, N2 ⊕ A2 ∼ = ker(f ). Thus ∼ 2Q = N1 ⊕ N2 and Q = N1 ⊕ A1 . By hypothesis, Q ⊕ A1 ∼ = N2 , whence Q ⊕ coker(f ) ∼ = Q ⊕ (A1 ⊕ A2 ) ∼ = N2 ⊕ A2 ∼ = ker(f ). It follows from Lemma 14.1.4 that f admits a diagonal reduction. According to Theorem 14.1.5, we establish the result.  Corollary 14.1.7. Let P1 , · · · , Pn be right R-modules having the finite exchange property. If each
EndR (Pi ) is an almost hermitian ring, then so is EndR P1 ⊕ · · · ⊕ Pn .

Proof. It will suffice to show that the result
holds for n = 2.
Given any ∼ right R-modules
M and N such that
2 P1 ⊕ P2 ⊕ M = P1 ⊕ P2 ⊕ N , then
2P1 ⊕ 2P2 ⊕ M ∼ = = P1 ⊕ P2 ⊕ N
. By Corollary 14.1.6, P1 ⊕ 2P2 ⊕ M ∼ P2 ⊕N . Moreover, we get P1 ⊕P2 ⊕M ∼ N . Clearly, P ⊕P has the finite = 1 2 exchange property (cf. [382, Lemma 28.1]). Using Corollary 14.1.6 again,
we prove that EndR P1 ⊕ P2 is an almost hermitian ring, as required. 

Theorem 14.1.8. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) There exists a complete orthogonal set of idempotents in R, {e1 , · · · , en }, such that all ei Rei are almost hermitian rings.

Proof. (1) ⇒ (2) is trivial. (2) ⇒ (1) Let {e1 , · · · , en } be a complete orthogonal set of idempotents of R. Then R ∼ = e1 R⊕· · ·⊕en R as right R-modules. It follows by Corollary 14.1.7 that R is an almost hermitian ring. 

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Let T be the ring of a Morita context (A, B, M, N, ψ, φ) where A and B are exchange rings. If A and B are almost hermitian rings, then so is T by Theorem 14.1.8. Let R be an exchange ring. By induction, we see that if R is an almost hermitian ring then so is the matrix ring Mn (R). In general, the converse is not true. But we can derive the following fact. Corollary 14.1.9. Let R be an exchange ring, and let n ∈ N. Then R is an almost hermitian ring if and only if so is the ring of all n × n lower triangular matrices over R. Proof. One direction is clear by Theorem 14.1.8, [264, Proposition 2.2] and induction. Conversely, it will suffice to show that the result holds for n = 2. Now assume that T = T M2(R) is an almost hermitian ring. Given  any  regs 2 ular (x, y) ∈ R , then we have s, t ∈ R such that (x, y) = (x, y) (x, y). t    x0 y0 Clearly, ∈ T 2 is regular as well. Therefore we have some 00 00    u11 0 u12 0  m11 v11 m12 v12   W :=    u21 0   u22 0   ∈ GL2 (T ) m21 v21

such that



m22 v22

    y0 ∗∗ 00 W = . 00 ∗∗ 00   u11 u12 It is easy to verify that (x, y) = (∗, 0). Set u21 u22   ′  ′ u12 0 u11 0 ′ ′   m′11 v11 m′12 v12     . W −1 =  ′  u′ 0 u 0  x0 00



21

′ m′21 v21

Then we check that 

u11 u12 u21 u22



=



u′11 u′12 u′21 u′22

22

′ m′22 v22

−1

∈ U (T ).

By virtue of Theorem 14.1.5, R is an almost hermitian ring.



Analogously, we deduce that a ring R is an almost hermitian, exchange ring if and only if so is the ring of all n × n upper triangular matrices over R.

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Theorem 14.1.10. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any regular x ∈ R2 , there exists y ∈ col2 (R) such that x = xyx. (3) For any regular x ∈ 2R, there exists y ∈ row2 (R) such that x = xyx. Proof. (1) ⇒ (2) Using Theorem 14.1.5 and Lemma 12.1.10, R satisfies the 2-stable range. It follows by Lemma 12.3.12 that for any regular x ∈ R2 , there exists a unimodular y ∈ 2R such that x = xyx. Assume that zy = 1 for some z ∈ R2 . Clearly, 2R = (yz)(2R) ⊕ (I2 − yz)(2R) = y(R) ⊕ ker(z). Obviously, we have an isomorphism φ : y(R) ∼ = zy(R) = R given by y(r) → zy(r) for any r ∈ R. By Theorem 14.1.5 again, we have ψ : R ∼ = ker(z). Now we construct an isomorphism ϕ : 2R = R ⊕ R → y(R) ⊕ ker(z) = 2R given by ϕ(s, t) = φ−1 (s) + ψ(t) for any s, t ∈ R. According to Lemma 14.1.2, we have F (ϕ) ∈ GL2 (R). One easily checks that y is the first column of F (ϕ), as required. (2) ⇒ (1) For any regular x ∈ R2 , there exists y ∈ 2R which is a column of some invertible matrix such that x = xyx. Assume that y is the first column of some U ∈ GL2 (R). Set e = xy. Then xU  = (e, xz), where z 1 −xz is the second column of U . Hence V := U ∈ GL2 (R) and xV 0 1 is of the form (e, 0). Therefore R is an almost hermitian ring by Theorem 14.1.5. (1) ⇔ (3) is analogous to the above consideration.  Corollary 14.1.11. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) R satisfies the 2-stable range condition and R cancels from 2R ∼ = R⊕K. (3) R is a U C1 -ring satisfying the 2-stable range condition. Proof. (1) ⇒ (2) is clear by Theorem 14.1.5 and Lemma 12.1.10. (2) ⇒ (3) is proved as in the proof of Theorem 11.4.1. (3) ⇒ (1) is obvious by Lemma 12.3.12 and Theorem 14.1.10.



We see, from Corollary 14.1.11, that every exchange ring having stable range one is an almost hermitian ring. Corollary 14.1.12. Let R be an exchange ring. Then the following are equivalent:

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(1) R is an almost hermitian ring. (2) For any regular x ∈ R2 , there exists y ∈ col2 (R) such that xy ∈ R is an idempotent. (3) For any regular x ∈ 2R, there exists y ∈ row2 (R) such that yx ∈ R is an idempotent. Proof. (1)⇒(2) For each regular x ∈ R2 , it follows by Theorem 14.1.10 that there exists y ∈2 R which is a column of some invertible matrix such that x = xyx. Hence xy ∈ R is an idempotent, as required. (2)⇒(1) Given any regular x ∈ 2R, there exists y ∈ R2 such that x = xyx and y = yxy. Because y ∈ R2 is also regular, we have an element u ∈ 2R which is the column of some invertible matrix U such that yu = e is an idempotent of R. Hence wu = 1 for some w ∈ R2 . From xy +(I2 −xy) = I2 , we have xe + (I2 − xy)u = u. So
x + (I2 − xy)u e + (I2 − xy)u(1 − e) = u.

One easily checks that

(I2 − xy)u(1 − e) = (I2 − xy)u(1 − e)w(I2 − xy)u(1 − e).

Set g = (1 − e)w(I2 −
xy)u(1 − e). Then e = e2 , g = g 2 and eg
= ge = 0. Since x + (I2 − xy)u e + (I2 − xy)ug = u, x + (I2 − xy)u e = ue and (I2 −
xy)ug = ug, whence u(e + g) = u. Therefore
u 1 − (1 − e)w(I2 − xy)ue y+(1+(1−e)w(I2 −xy)ue)(1−e)w(I −xy) u = u (1−(1−e)w(I 2 2−

xy)ue)e + (1 − e)w(I2 − xy)u =
u e + (1 − e)w(I2 − xy)u(1 − e) = u. Set v := y
+ 1+(1−e)w(I2 −xy)ue (1−e)w(I2 −xy). Then u 1−(1−e)w(I
2− xy)ue vu = u. From wu = 1, we deduce that 1 − (1 − e)w(I2 − xy)ue vu = 1. As
−1 1 − (1 − e)w(I2 − xy)ue = 1 + (1 − e)w(I2 − xy)ue,

we have vu = 1 + (1 − e)w(I2 − xy)ue. Assume (u, r) ∈ GL2 (R)
for some r ∈ 2R; hence, v(u, r) = (vu, vr) = 1 + (1 − e)w(I2 − xy)ue, vr . Set ! V =
−1
−1 1 + (1 − e)w(I2 − xy)ue − 1 + (1 − e)w(I2 − xy)ue vr (u, r) , 0 1 then vV = (1, 0). Therefore v is the first row of V −1 . Clearly, we have
x = x y + (1 + (1 − e)w(I2 − xy)ue)(1 − e)w(I2 − xy) x = xvx.

It follows by Theorem 14.1.10 that R is an almost hermitian ring. (1) ⇔ (3) Clearly, a ring R is an almost hermitian ring if and only if so is its opposite ring Rop . In view of Lemma 1.4.7, a ring R is an exchange ring if and only if so is its opposite ring Rop . Applying (1) ⇔ (2) to Rop , we obtain the result. 

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Almost Hermitian Rings, II

The aim of this section is to investigate further the stability of almost hermitian rings by means of completed rectangular matrices. Lemma 14.2.1. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any m, n ∈ N (m ≥ n + 1), M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = mR, nR ∼ = A2 implies that (m − n)R ⊕ B1 ∼ = B2 . (3) For any m, n ∈ N (m ≥ n + 1), mR = A1 ⊕ B1 , nR = A2 ⊕ B2 with A1 ∼ = A2 implies that B1 ∼ = (m − n)R ⊕ B2 .

∼ Proof. (1) ⇒ (2) Suppose that M = A1 ⊕ B1 = A2 ⊕ B2 with A1 = ∼ ∼ ∼ mR, nR = A2 . Then A1 = A11 ⊕ A12 ⊕ · · · ⊕ A1m and A2 = A21 ⊕ A22 ⊕ · · · ⊕ A2n , where each A1i ∼ =R∼ = A2j . As A11 ⊕ A12 ∼ = 2R, it follows by Corollary 14.1.6 that A12 ⊕ · · · ⊕ A1m ⊕ B1 ∼ = A22 ⊕ · · · ⊕ A2n ⊕ B2 . By iteration of this process, we get (m − n)R ⊕ B1 ∼ = B2 . (2) ⇒ (3) Suppose that mR = A1 ⊕ B1 , nR = A2 ⊕ B2 with A1 ∼ = A2 . ∼ ∼ ∼ Then mR ⊕ B2 = A1 ⊕ B1 ⊕ B2 = A2 ⊕ B2 ⊕ B1 = nR ⊕ B1 . By assumption, we see that (m − n)R ⊕ B2 ∼ = B1 . (3) ⇒ (1) Choose m = 2, n = 1. Then the result follows from Corollary 14.1.6.  Theorem 14.2.2. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mm×n (R), there exists a completed U ∈ Mn×m (R) such that X = XU X. (3) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mn×m (R), there exists a completed U ∈ Mm×n (R) such that X = XU X. Proof. (1) ⇒ (2) Let X ∈ Mm×n (R) be regular. Then there exists Y ∈ Mn×m (R) such that X = XY X. Clearly, mR = XY (mR) ⊕ (Im − XY )(mR) and nR = Y X(nR) ⊕ (In − Y X)(nR). Obviously, we have an R-isomorphism φ : XY (mR) ∼ = Y X(nR) given by XY (r) 7→ Y XY (r) for any r ∈ mR. By virtue of Lemma 14.2.1, we have ψ : (Im − XY )(mR) ∼ = (m−n)R⊕(In −Y X)(nR). Now we construct an R-isomorphism ϕ : mR = XY (mR)⊕(Im −XY )(nR) → Y X(nR)⊕(In −Y X)(nR)⊕(m−n)R = mR

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given by ϕ(s, t) = φ(s) + ψ(t) for any s ∈ XY (mR), t ∈ (Im − XY )(mR). Let {e1 , · · · , em } be a basis of mR, and let ϕ(e1 , · · · , em ) = (e1 , · · · , em )A for a matrix A. Then  A∈ GLm (R). Let U ∈ Mn×m (R) be the first block U of A. That is, A = . Then ∗


X 0m×(m−n) = X 0m×(m−n) A X 0m×(m−n) ,

and so X = XU X, as required. (2) ⇒ (1) Choose m = 2, n = 1. Then the result follows by Theorem 14.1.10. (1) ⇔ (3) is proved by symmetry.  Corollary 14.2.3. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) R is a U C-ring satisfying the 2-stable range condition. Proof. (1) ⇒ (2) In view of Lemma 14.1.3, R satisfies the 2-stable range condition. For any n ≥ 2 and any unimodular x ∈ Rn , we can find some y ∈n R such that xy = 1. Hence, x ∈ Rn is regular. By virtue of Theorem 14.2.2, there exists u ∈ coln (R) such that x = xux. This implies that 1 = xy = xuxy = xu. Assume that (u, ∗) ∈ GLn (R).  Then we can find ∗ v ∈ M(n−1)×n (R) and w ∈ Mn×(n−1) (R) such that = (u, w)−1 . It is v easy to verify that     x 1 ∗ (u, w) = . v 0 In−1   x This implies that ∈ GLn (R), and so x ∈ rown (R). Therefore R is a v U C-ring. (2) ⇒ (1) is trivial by Lemma 14.1.3.  Corollary 14.2.4. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mm×n (R),
X 0m×(m−n) ∈ Mm (R) is unit-regular.

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(3) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mn×m (R),   X ∈ Mm (R) is unit-regular. 0(m−n)×n Proof. (1) ⇒ (2) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mm×n (R), it follows by Theorem 14.2.2 that there exists a completed U ∈   U Mn×m (R) such that X = XU X. Assume that ∈ GLm (R). Then we ∗ see that  

U
X 0m×(m−n) = X 0m×(m−n) X 0m×(m−n) , ∗
and therefore X 0m×(m−n) is unit-regular. (2) ⇒ (1) For any x ∈ 2 R, (x, 0) ∈ M2 (R) is unit-regular.  regular    Thus, u u we can find some ∈ GL2 (R) such that (x, 0) = (x, 0) (x, 0). It ∗ ∗ is easy to see that x = xux, where u ∈ row2 (R). According to Theorem 14.1.10, R is an almost hermitian ring. (1) ⇔ (3) is proved in the same manner.  We deduce that a regular ring  Ris an hermitian ring if and only if for ab any (a, b) ∈ R2 , we have that ∈ M2 (R) is unit-regular if and only 00     a a0 2 if for any ∈ R, we have that ∈ M2 (R) is unit-regular. b b0 Proposition 14.2.5. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mm×n (R), there exists a completed U ∈ Mn×m (R) such that XU ∈ Mm (R) is an idempotent. (3) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mn×m (R), there exists a completed U ∈ Mm×n (R) such that XU ∈ Mn (R) is an idempotent. Proof. (1) ⇒ (2) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mm×n (R), there exists a completed U ∈ Mn×m (R) such that X = XU X. Let XU = E. Then E = E 2 ∈ Mm (R), as desired. (2) ⇒ (1) For any regular x ∈ R2 , there exists a y ∈ 2 R such that x = xyx and y = yxy. By assumption, we have a u ∈ row2 (R) such that

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e := yu ∈ M2 (R) is an idempotent. Since
xy + (1 − xy) = 1, we have xe + (1 − xy)u = u, and so x +
(1 − xy)u e + (1 − xy)u(I2 − e) = u. One easily checks that x + (1 − xy)u e = ue and (1 − xy)u(I2 − e) = u(I2 − e).   u As u ∈ row2 (R), we see that ∈ GL2 (R) for some s ∈ M1×2 (R). Write s  −1
u = v t with v, t ∈ col2 (R). Obviously, we have u(I2 − e)vu(I2 − s e) = u(I2 − e). Let g = (I2 − e)vu(I2 − e). Then g = g 2 ∈ M2 (R)
and u(e + g) = u. Furthermore, we see that
u I2 − (I2 − e)v(1 − xy)ue y + (I2 +(I2 −e)v(1−xy)ue)(I 2 −e)v(1−xy) u = u (I2 −(I

2 −e)v(1−xy)ue)e+ (I2 − e)v(1 − xy)u = u e + (I2 − e)v(1 − xy)u(I − e) = u(e + g) = u. Let 2
w = y
+ I2 +(I2 −e)v(1−xy)ue (I2 −e)v(1−xy).
Then u I2 −(I2 −e)v(1− xy)ue wu = u, and so u I2 − (I2 − e)v(1 − xy)ue w = 1. This implies that    

u 10 I2 −(I2 −e)v(1−xy)ue w (I2 + (I2 − e)v(1 − xy)ue)t = ∈ s ∗1 GL2 (R). Hence, w ∈ col2 (R). In addition, x = xwx. In view of Theorem 14.1.10, R is an almost hermitian ring. (1) ⇒ (3) For any m, n ∈ N (m ≥ n+1) and any regular X ∈ Mn×m (R), there exists a completed U ∈ Mm×n (R) such that X = XU X. Let XU = E. Then E = E 2 ∈ Mm (R), as required. (3) ⇒ (1) is obvious by Corollary 14.1.12.  Corollary 14.2.6. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mm×n (R), there exists an idempotent E ∈ Mm (R) and a completed U ∈ Mm×n (R) such that X = EU . (3) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mn×m (R), there exists an idempotent E ∈ Mn (R) and a completed U ∈ Mn×m (R) such that X = EU . Proof. (1) ⇒ (2) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mm×n (R), it follows by Proposition 14.2.5 that there exists a completed V ∈ Mn×m (R) such that X = XV X. Assume that  −1
V = U ∗ , ∗ m×m where U ∈ Mm×n (R). Then V U = In . Let E = XV . Then X = X(V U ) = EU with E = E 2 ∈ Mm (R), as required.

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(2) ⇒ (1) For any regular x ∈ R2 , there exists y ∈ 2 R such that x = xyx and y = yxy. By assumption, there exists an idempotent e ∈ M2 (R) and a v ∈ col2 (R) such that y = ev. It follows from yx + (I2 − yx) = I2 that evx + (I2 − yx) = I2 ; hence, evx(I2 − e) + (I2 − yx)(I2 − e) = I2 − e. This implies that e+(I2 −yx)(I2 −e) = I2 −evx(I2 −e) ∈ GL2 (R). Furthermore, we see that u := = = ∈

y + (I2 − yx)(I2 − e)v ev + (I2 − yx)(I2
− e)v I2 − evx(I2 − e) v col2 (R).

As a result, xu = xy ∈ R is an idempotent. In view of Corollary 14.1.12, R is an almost hermitian ring. (1) ⇔ (3) is proved in the same manner.  As in the proof of Corollary 14.2.6, we deduce that an exchange ring R is an almost hermitian ring if and only if for any regular x ∈ R2 , there exists an idempotent e ∈ M2 (R) and some u ∈ col2 (R) such that x = eu if and only if for any regular x ∈ R2 , there exists an idempotent e ∈ R and a u ∈ row2 (R) such that x = eu. If X ∈ Mm×n (R) is regular, then there exists some Y ∈ Mn×m (R) such that X = XY X and Y = Y XY . We say such Y is a reflexive inverse of X, and denote it by X + . Now we characterize almost hermitian rings by means of reflexive inverses. Proposition 14.2.7. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mm×n (R), there exists a completed U ∈ Mn×m (R) such that X = XX + U = U X + X. (3) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mn×m (R), there exists a completed U ∈ Mm×n (R) such that X = XX + U = U X + X. Proof. (1) ⇒ (2) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mm×n (R), there exists some Y ∈ Mn×m (R) such that X = XY X and Y = Y XY . Since Y ∈ Mn×m (R) is regular, by Theorem 14.2.2, we have Y = Y V Y for some completed V ∈ Mm×n (R). Let U = (Im − XY − V Y )V (In − Y X − Y V ). One easily checks that (Im − XY − V Y )2 = Im and (In − Y X − Y V )2 = In . Hence U ∈ Mm×n (R) is completed. Further, we see that XY U = −XY V (In − Y X − Y V ) = XY X = X and U Y X = −(Im − XY − V Y )V Y X = XY V Y X = X, as required.

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(2) ⇒ (1) is obvious from Corollary 14.2.6. (1) ⇔ (3) is proved in the same manner.



Corollary 14.2.8. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mm×n (R), there exists a completed U ∈ Mn×m (R) such that X = XX + U. (3) For any m, n ∈ N (m ≥ n + 1) and any regular X ∈ Mn×m (R), there exists a completed U ∈ Mm×n (R) such that X = XX + U. Proof. (1) ⇒ (2) is clear by Proposition 14.2.7. (2) ⇒ (1) is obvious from Corollary 14.2.6. (1) ⇔ (3) is proved in the same manner.



Now we characterize almost hermitian rings by a class of stable properties similar to the n-stable range condition. Lemma 14.2.9. Suppose that ax + b = 1 with a ∈ Rn , x ∈ Then the following are equivalent:

n

R, b ∈ R.

(1) There exists y ∈ Rn such that a + by ∈ rown (R). (2) There exists z ∈ n R such that x + zb ∈ coln (R). Proof. (1) ⇒ (2) Suppose that u := a + by ∈ rown (R). Clearly,    −1 a b x xa − In ∈ GLn+1 (R). = −In x 1 a Thus, we see that      a b In 0 u b = ∈ GLn+1 (R). −In x y 1 xy − In x

Assume that

Then 

 −1 u = (v, t), v ∈ coln (R), t ∈ Mn×(n−1) (R). s

u b xy − In x



In −vb 0 1



=



u 0 xy − In x − (xy − In )vb



∈ GLn+1 (R).

So we can find some w ∈ Mn×(n−1) (R) such that      0 u 0 t = . w xy − In x − (xy − In )vb 0

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By elementary transformations, we have     u 0   s xy − In (x − (xy − In )vb, w)      0 −1 (s, 0) 0  0 −(s, 0)U   w     → 0 →   U 0 U w w   0 −In−1   = 0  ∈ GL2n (R), U w   u 0 where U = . This implies that xy − In (x − (xy − In )vb     u 0   ∈ GL2n (R). s xy − In (x − (xy − In )vb, w)  
u Since ∈ GLn (R), we get x − (xy − In )vb, w ∈ GLn (R). Therefore s x + zb ∈ coln (R), where z = −(xy − In )v ∈ n R. (2) ⇔ (1) The result follows by applying (1) ⇒ (2) to the opposite ring Rop .  Theorem 14.2.10. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any n ≥ 2, a(n R) + bR = R with a ∈ Rn , b ∈ R implies that there exists y ∈ Rn such that a + by ∈ rown (R). (3) For any n ≥ 2, Rn a + Rb = R with a ∈ n R, b ∈ R implies that there exists z ∈ n R such that a + zb ∈ coln (R). Proof. (1) ⇒ (2) For any n ≥ 2, a(n R) + bR = R with a ∈ Rn , b ∈ R implies that there exist x ∈ Rn , y ∈ R such that ax + by = 1. Since R is an exchange ring, by [382, Proposition 29.1], there exists an idempotent e ∈ R such that e = bys and 1 − e = (1 − by)t for some s, t ∈ R. This implies that axt + e = 1, and so (1 − e)a ∈ Rn is regular. According to Corollary 14.2.6, we have an idempotent f ∈ R such that (1 − e)a = f u, where u ∈ rown (R). It follows from (1 − e)axt + e = 1 that f uxt(1 − f ) + e(1 − f ) = 1 − f .

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Hence



a + bys (1 − f )u − a = a + e (1 − f )u − a = (1 − e)a + e(1 − f )u = f u + e(1 − f )u
= 1 − f uxt(1 − f ) u
−1 = 1 + f uxt(1 − f ) u ∈ rown (R).
Let z = ys (1 − f )u − a . Then a + bz ∈ rown (R). (2) ⇒ (1) For any regular a ∈ R2 , we have x ∈ 2 R such that a = axa. It follows from ax+(1−ax) = 1 that a+(1−ax)y ∈ row2 (R) for some y ∈ R2 . In view of Lemma 14.2.9, there exists z ∈ n R such that u := x+z(1−ax) ∈
col2 (R). Therefore we conclude that a = axa = a x + z(1 − ax) a = aua, and so the result follows by Theorem 14.1.10. (1) ⇔ (3) is obvious by Lemma 14.2.9 and (1) ⇔ (2).  Corollary 14.2.11. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any n ≥ 2 and any regular x ∈ n R, there exists a ∈ Rn such that In + xa ∈ GLn (R), x+ + a ∈ rown (R). (3) For any n ≥ 2 and any regular x ∈ Rn , there exists a ∈ n R such that 1 + xa ∈ U (R), x+ + a ∈ coln (R).

Proof. (1) ⇒ (2) Since x ∈ n R is regular, we can find x+ ∈ Rn such that x = xx+ x and x+ = x+ xx+ . It follows from x+ x+ (1 − x+ x) = 1 that there exists z ∈ R such that x+ + (1 − x+ x)z ∈ rown (R). Let a = (1 − x+ x)z. Then x+ + a ∈ rown (R). In addition, we have In + xa = In + x(1 − x+ x)z = In ∈ GLn (R). (2) ⇒ (1) For any regular x ∈ R2 , there exists some x+ ∈ 2 R such that x = xx+ x and x+ = x+ xx+ . By assumption, we have a ∈ R2 such that I2 + x+ a ∈ GL2 (R), x + a ∈ row2 (R). One easily checks that     1 + ax+ 0 1 0 1 a 0 I2 x+ I2 0 I2   + 1 + ax (1 + ax+ )a = x+ I2 + x+ a   + 1 + ax a(In + x+ a) = x+ I + x+ a   2   1 a 1 0 1 0 = . 0 I2 x+ I2 0 I2 + x+ a

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Thus, 1 + ax+ ∈ U (R). Let u = x + a. Then w := 1 + (u − x)x+ ∈ U (R); hence, w−1 ux+ + w−1 (1 − xx+ ) = 1. As w−1 u ∈ row2 (R), it follows from Lemma 14.2.9 that v := x+ + tw−1 (1 − xx+ ) ∈ col2 (R) for an
element 2 + + −1 + t ∈ R. As a result, x = xx x = x x + tw (1 − xx ) x = xvx. According to Theorem 14.1.10, R is an almost hermitian ring. (1) ⇔ (3) is proved in the same manner.  Corollary 14.2.12. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any n ≥ 2 and any idempotents e ∈ R, f ∈ Mn (R), ϕ : eR ∼ = f (nR) implies that there exists u ∈ coln (R) such that ϕ(e) = ue = f u.

Proof. (1) ⇒ (2) For any idempotents e ∈ R, f ∈ Mn (R), ϕ : eR ∼ = f (nR) implies that there exist some a ∈ M1×n (R), b ∈ Mn×1 (R) such that e = ab, f = ba, a = eaf and b = f be by Lemma 6.4.16. Since ab + (1 − ab) = 1, in view of Theorem 14.2.10, we can find an element z ∈
n R such that v := b + z(1 − ab) ∈ coln (R). Hence, b = b + z(1 − ab) ab = ve. Let u = (In − f − va)v(1 − e − av). It is easy to verify that (In − f − va)2 = In and (1 − e − av)2 = 1. This implies that u ∈ coln (R). Further, f u = −f vav(1 − e − av) = −f v(1 − e − av) = f ve = b. Also we have ue = −(In − f − va)vave = −(In − f − va)b = −b + bab + ve = b. As in the proof of Lemma 6.4.16, ϕ(e) = b = ue = f u, as required. (2) ⇒ (1) Let x ∈ R2 be regular. Then there exists some y ∈ 2 R such that x = xyx and y = yxy. Thus, we get an R-isomorphism ϕ : xyR ∼ = yx(2R) given by ϕ(xyr) = yxyr for any r ∈ R. By hypothesis, there exists some v ∈ col2 (R) such that y = yxv = vxy. This implies
that y + (In − yx)v(1 − xy) = v. Thus, x = xyx = x y + (I2 − yx)v(1 − xy) x = xvx. In view of Theorem 14.1.10, R is an almost hermitian ring.  As a consequence, we derive that an exchange ring R is an almost hermitian ring if and only if for any idempotents e ∈ R, f ∈ M2 (R), ϕ : eR ∼ = f (2R) implies that there exists u ∈ col2 (R) such that ϕ(e) = ue = f u.

Proposition 14.2.13. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any n ≥ 2, a(n R) + bR = dR with a ∈ Rn , b, d ∈ R implies that there exist y ∈ Rn , u ∈ rown (R) such that a + by = du.

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(3) For any n ≥ 2, Rn a + Rb = dR with a ∈ n R, b, d ∈ R implies that there exist z ∈ n R, u ∈ coln (R) such that a + zb = ud. (4) For any n ≥ 2, a(n R) + bR = dR with a ∈ Rn , b, d ∈ R implies that there exist y ∈ R, u ∈ coln (R) such that au + by = d. (5) For any n ≥ 2, Rn a + Rb = Rd with a ∈ n R, b, d ∈ R implies that there exist z ∈ R, u ∈ rown (R) such that ua + zb = d. Proof. (1) ⇒ (2) For any n ≥ 2, a(n R) + bR = dR with a ∈ Rn , b, d ∈ R implies that there exist x ∈ n R, y ∈ R such that ax + by = d. Furthermore, we have s ∈ Rn and t ∈ R such that a = ds and b = dt. Clearly, dsx + dty = d, and so dv = 0, where v = 1 − sx − ty. Since sx + ty + v = 1 with s ∈ Rn , x ∈ n R, ty + v ∈ R, it follows by Theorem 10.2.10 that we have z ∈ Rn such that u := s + (ty + v)z ∈ rown (R). This implies that du = ds + d(ty + v)z = a + byz, as required. (2) ⇒ (1) is obvious from Theorem 14.2.10. (1) ⇔ (3) is symmetric. (1) ⇒ (4) Let a(n R) + bR = dR with a ∈ Rn and b, d ∈ R(n ≥ 2). Then we have x ∈ n R, y ∈ R, s ∈ Rn and t ∈ R such that ax + by = d, a = ds and b = dt. Let v = 1 − sx − ty. Then s + (ty + v)z ∈ rown (R) for
some z ∈ Rn . So we can find some u ∈ coln (R) such
that s + (ty + v)z u = 1. As a result, we deduce that d = d s + (ty + v)z u = au + byu, as desired. (4) ⇒ (1) For any regular a ∈ R2 , we have x ∈ 2 R such that a = axa. Since ax + (1 − ax) = 1, we can find u ∈ col2 (R) and
y ∈ R such that au + (1 − ax)y = 1. Thus, ax = ax au + (1 − ax)y = axau = au, i.e., au ∈ R is an idempotent. Therefore R is an almost hermitian ring by Corollary 14.1.12. (1) ⇔ (5) is symmetric.  Corollary 14.2.14. Let R be an exchange ring. Then the following are equivalent: (1) R is an almost hermitian ring. (2) For any n ≥ 2, a(n R) = bR with a ∈ Rn , b ∈ R implies that there exists u ∈ rown (R) such that a = bu. (3) For any n ≥ 2, Rn a = Rb with a ∈ n R, b ∈ R implies that there exists u ∈ coln (R) such that a = ub. (4) For any n ≥ 2, a(n R) = bR with a ∈ Rn , b ∈ R implies that there exists u ∈ coln (R) such that au = b. (5) For any n ≥ 2, Rn a = Rb with a ∈ n R, b ∈ R implies that there exists u ∈ rown (R) such that ua = b.

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Proof. (1) ⇒ (2) is clear by Proposition 14.2.13. (2) ⇒ (1) For any regular a ∈ R2 , we have x ∈ 2 R such that a = axa. Hence, a(2 R) = axR. By assumption, we can find u ∈ row2 (R) such that a = (ax)u. As ax = (ax)2 ∈ R, we see that R is an almost hermitian ring analogous to Corollary 14.2.6. (1) ⇔ (3) ⇔ (4) ⇔ (5) are proved in the same manner.  Clearly, Proposition 14.2.13 is an analogue of Proposition 12.3.17. For regular rings of finite stable range, we can derive a Zabavsky’s result, which is of interest in its own right (cf. [427, Theorem 1]). Lemma 14.2.15. Let R be a regular ring. If R satisfies the n-stable range condition, then every (n+1)×1 matrix over R admits a diagonal reduction. Proof. Let f : R → (n + 1)R be an R-morphism. Then f is regular. In view of Proposition 7.2.11, R ⊕ coker(f ) ∼ = (n + 1)R ⊕ ker(f ). According to Corollary 12.1.7, coker(f ) ∼ = nR ⊕ ker(f ). By Proposition 7.2.11 again, f admits a diagonal reduction, as required.  Proposition 14.2.16. Let R be a regular ring. If R satisfies the n-stable range condition, then every k × m matrix over R where | k − m | = n admits a diagonal reduction. Proof. Let A ∈ Mk×m (R), where | k − m | = n. Without loss of generality, we may assume that m = n + k. The result holds for k = 1 by Lemma 14.2.15 and symmetry. Assume that the result holds for k − 1(k ≥ 2). As | (k − 1) − (m − 1) | = n, by hypothesis, every (k − 1) × (m − 1) matrix admits a diagonal reduction. In view of Lemma 14.2.15, A can be reduced to the matrix   0  ..  B  .   ,  0 ak1 · · · ak(m−1) b where B ∈ M(k−1)×(m−1) (R). By induction, there exist P ∈ GLk−1 (R), Q ∈ GLm−1 (R) such that P BQ is a diagonal reduction. So A can be reduced to the matrix   a 0 0 ··· 0  .. ..   C . D= . . 0 0 c1 c2 · · · c(m−1) b

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Write a = ec and e = ad, where e = e2 ∈ R. Set   d 0 · · · 0 1 − dc 0 0     .  .  ..  U =  .. Im−2 .   0 0  1 0 · · · 0 −c Then U ∈ GLm (R). Further,



e 0 ··· 0 0   . DU =  C  ..  0 t s1 · · · sm−2

 0 0  ..   . .  0 s

Write s = ss′ s and s′ = s′ ss′ . Assume that ss′ t = 0. Set   s 0 · · · 0 ss′ − 1  0 0     . ..   . I V = k−2 . .  .    0 0  ′ ′ 1 + s s 0 ··· 0 s Then V ∈ GLk (R). Furthermore,  se − t ∗ ··· ∗  0   .. V DU =  C .    0 (1 + s′ s)e s′ s1 · · · s′ sm−2 Thus, we may assume that



y y1 · · · ym−2 0   . V DU =  C  ..  0 e 0 ··· 0

0 0 .. .



0 0 .. .



   .   0  s′ s

   ,   0 f

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where f = s′ s ∈ R is an
idempotent. Write (1 − f )e = (1 − f )eh(1 − f )e. Set z = y 1 − h(1 − f )e . Then V DU can be reduced to the form   z y1 · · · ym−2 0 0 0     . .   . . C . .  .   0 0 e f 0 ··· 0 As in the proof of Proposition 13.2.27, A can be reduced to   ∗ ∗ ··· ∗ 0 0 0    .  .  .  . C . .  .   0 0 0 0 ··· 0 ∗ By induction, we obtain the result.

14.3



Strongly Separative Rings

In this section, we investigate diagonal reduction over strongly separative rings. We characterize strongly separative property for exchange rings with a kind of stability. Theorem 14.3.1. Let R be an exchange ring. Then the following are equivalent: (1) R is strongly separative. (2) All corners of R are almost hermitian rings. Proof. (1) ⇒ (2) Let e ∈ R be an idempotent. As in the proof of [16, Lemma 4.1], it is clear from [16, Lemma 1.5] that eRe is a strongly separative exchange ring. Given 2(eR)⊕B ∼ = eR⊕C, then eR⊕(eR⊕B) ∼ = eR⊕C. ∼ Thus, we have eR ⊕ B = C by Corollary 12.2.11. According to Theorem 14.1.5, eRe is an almost hermitian ring. (2) ⇒ (1) Let A, B, C ∈ F P (R) such that A ⊕ C ∼ = B ⊕ C, where C .⊕ A. As in the proof of Lemma 6.3.7, we have a refinement matrix A C



BC  D1 A1 B1 C1

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such that C1 .⊕ A1 . Hence, 2C1 .⊕ C1 ⊕ A1 ∼ = C .⊕ A, and then A∼ = 2C1 ⊕ E for a right R-module E. One easily checks that A ⊕ C1 ∼ = ∼ ∼ ∼ D1 ⊕ A1 ⊕ C1 ∼ D ⊕ C D ⊕ B ⊕ C B ⊕ C , and so 3C ⊕ E = 1 = 1 = 1 1 = 1 1 C1 ⊕ B. As C1 ∈ F P (R), there exist idempotents e1 , · · · , en ∈ R such that C1 ∼ = e1 R ⊕ · · · ⊕ en R. Thus, 3(e1 R) ⊕ 3(e2 R) ⊕ · · · ⊕ 3(en R) ⊕ E ∼ = e1 R ⊕ e2 R ⊕ · · · ⊕ en R ⊕ B. Since e1 Re1 is an almost hermitian ring, it follows from Theorem 14.1.5 that 2(e1 R) ⊕ 3(e2 R) ⊕ · · · ⊕ 3(en R) ⊕ E ∼ = e2 R ⊕ · · · ⊕ en R ⊕ B. By iteration of this process, we get 2(e1 R ⊕ e2 R ⊕ · · · ⊕ en R) ⊕ E ∼ = B, i.e., ∼ ∼ A = 2C1 ⊕ E = B. Therefore R is strongly separative.  Corollary 14.3.2. Let R be an exchange ring. Then the following are equivalent: (1) R is strongly separative. (2) For any finitely generated projective right R-module C and any right R-modules A and B, 2C ⊕ A ∼ = C ⊕ B implies that C ⊕ A ∼ = B. Proof. (2) ⇒ (1) is obvious from Lemma 12.2.7. (1) ⇒ (2) Let C ∈ F P (R) such that 2C ⊕ A ∼ = C ⊕ B. Then there exist idempotents e1 , · · · , en ∈ R such that C ∼ = e1 R ⊕ · · · ⊕ en R. So 2(e1 R) ⊕ 2(e2 R) ⊕ · · · ⊕ 2(en R) ⊕ A ∼ = e1 R ⊕ e2 R ⊕ · · · ⊕ en R ⊕ B. In view of Theorem 14.3.1, EndR (e1 R) is an almost hermitian ring. Using Corollary 14.1.6, we have e1 R ⊕ 2(e2 R) ⊕ · · · ⊕ 2(en R) ⊕ A ∼ = e2 R ⊕ · · · ⊕ en R ⊕ B. By iteration of this process, C ⊕ A ∼ = e1 R ⊕ e2 R ⊕ · · · ⊕ en R ⊕ A ∼ = B, as asserted.  According to Theorem 14.3.1 and Theorem 14.1.5, we deduce that an exchange ring R is strongly separative if and only if every regular 1 × 2 matrix over eRe admits a diagonal reduction for all idempotents e ∈ R. Now we get an analogue for separative exchange rings. Proposition 14.3.3. Let R be an exchange ring. Then the following are equivalent: (1) R is separative. (2) For any idempotent e ∈ R, every regular matrix in M2 (eRe) admits a diagonal reduction.

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Proof. (1) ⇒ (2) is obvious by Lemma 7.3.1. (2) ⇒ (1) In view of Lemma 13.1.8, it will suffice to show that for any idempotent e ∈ R and any A, B ∈ F P (R), 2(eR) ⊕ A ∼ = 2(eR) ⊕ B ∼ implies that eR ⊕ A = eR ⊕ B. Analogously to Lemma 6.3.7, there exist decompositions 2(eR) = C11 ⊕ C12 , A = C21 ⊕ C22 , 2(eR) ∼ = C11 ⊕ C21 and B∼ = 2(eR) ⊕ C21 . This = C12 ⊕ C22 . Thus, 2(eR) ⊕ C12 ∼ = C11 ⊕ C21 ⊕ C12 ∼ implies that O O O O 2(eR) Re ⊕ C12 Re ∼ Re ⊕ C21 Re, = 2(eR) R

R

R

R

and so 2(eRe)⊕C12 e ∼ = 2(eRe)⊕C21 e. Let S = eRe. Analogously to Lemma 6.3.7, there exist right S-module decompositions 2S = B11 ⊕ B12 , C12 e = B21 ⊕ B22 , 2S ∼ = B11 ⊕ B21 and C21 e ∼ = B12 ⊕ B22 . Thus, we get 2S = ′ K ⊕ K = I ⊕ C, where K = B12 , K ′ = B11 , I ∼ = B11 and C ∼ = B21 . Clearly, there exists a regular f ∈ M2 (S) such that K = ker(f ), K ′ ∼ = im(f ) ∼ = I and C ∼ = coker(f ). By hypothesis, f admits a diagonal reduction. According to Proposition 7.2.2, there exist right S-module decompositions K = K1 ⊕ K2 , I ∼ = I1 ⊕ I2 , C ∼ = C1 ⊕ C2 , where Kj ⊕ Ij ∼ = Cj ⊕ Ij ∼ = S for ∼ ∼ j = 1, 2. It follows that S ⊕ K = C1 ⊕ I1 ⊕ K1 ⊕ K2 = S ⊕ C1 ⊕ K2 ∼ = C2 ⊕ I2 ⊕ C1 ⊕ K2 ∼ = S ⊕ C. Therefore eRe ⊕ C12 e ∼ = eRe ⊕ C21 e, and so O O O O eRe eR ⊕ C12 e eR ∼ eR ⊕ C21 e eR. = eRe eRe

eRe

eRe

eRe

This implies that eR ⊕ C12 ∼ = eR ⊕ C21 . As a result, we deduce that ∼ eR ⊕ A = eR ⊕ B, as required.  As a consequence, we deduce that an ideal I of an exchange ring R is separative if and only if for any idempotent e ∈ I, every regular matrix in M2 (eRe) admits a diagonal reduction. Lemma 14.3.4. Let R be an exchange ring. Then the following are equivalent: (1) For any x ∈ Ur,2 (R), there exists some y ∈ Uc,2 (R) such that xy = 0. (2) For any x ∈ Uc,2 (R), there exists some y ∈ Ur,2 (R) such that yx = 0. (3) For any right R-module A, 2R ∼ = R ⊕ A implies that R .⊕ A.

Proof. (1) ⇒ (3) Given R ⊕ R ∼ = R ⊕ A, we then have a splitting exact sequence i

f

0 → A → R ⊕ R → R → 0.

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Thus, there exists a right R-morphism g : R → R ⊕ R such that f g = 1R . Assume that f (1, 0) = a, f (0, 1) = b and g(1) = (x, y). Then 1 = f g(1) = f (x, y) = ax + by; hence, aR + bR = R. By assumption, there exist u, v ∈ R such that au = bv and Ru + Rv = R. Construct a map ϕ : R → R ⊕ R given by ϕ(r) = (ur, −vr) for any r ∈ R. Since f ϕ = 0, there exists a right R-morphism φ : R → A such that iφ = ϕ. Clearly, su + tv = 1 for some s, t ∈ R. Construct a map ψ : R ⊕ R → R given by ψ(r1 , r2 ) = sr1 − tr2 for any (r1 , r2 ) ∈ R⊕R. It is easy to verify that ψϕ = 1R ; hence, ϕ : R → R⊕R is a splitting R-monomorphism. This implies that φ : R → A is a splitting R-monomorphism, and so R .⊕ A. (3) ⇒ (1) Suppose that aR + bR = R with a, b ∈ R. Then we have x, y ∈ R such that ax + by = 1. Define f : R ⊕ R → R given by f (r1 , r2 ) = ar1 +br2 for any (r1 , r2 ) ∈ R⊕R and g : R → R⊕R given by g(r) = (xr, yr) for any r ∈ R. As f g = 1R , we have a splitting exact sequence i

f

0 → ker(f ) ֒→ R ⊕ R → R → 0,

where i is the inclusion map. This implies that ker(f ) ⊕ R ∼ = R ⊕ R. By assumption, we get R .⊕ ker(f ), and so we have two R-morphisms ϕ : R → ker(f ) and φ : ker(f ) → R such that φϕ = 1R . Thus, iϕ : R → R ⊕ R. Assume that iϕ(1) = (u, v) ∈ R ⊕ R. Then 0 = f iϕ(1) = f (u, v) = au + bv. Clearly, there exists a map j : R ⊕ R → ker(f ) such that ji = 1ker(f ) . Hence φj : R ⊕ R → R. Assume that φj(1, 0) = s and φj(0, 1) = t. Then su + tv = φj(u, 0) + φj(0, v) = φj(u, v) = φjiϕ(1R ) = 1R . This shows that Ru + Rv = R, as asserted. (2) ⇒ (3) Given R ⊕ R ∼ = R ⊕ A, then we have a splitting exact sequence i

f

0 → R → R ⊕ R → A → 0. So we can find a right R-morphism j : R ⊕ R → R such that ji = 1R . Assume that j(1, 0) = x, j(0, 1) = y and i(1) = (a, b). Then 1 = ji(1) = j(a, b) = xa + yb, and so Ra + Rb = R. By assumption, there exist u, v ∈ R such that ua = vb and uR + vR = R. Construct a map ϕ : R ⊕ R → R given by ϕ(r1 , r2 ) = ur1 − vr2 for any (r1 , r2 ) ∈ R ⊕ R. For any r ∈ R, ϕi(r) = ϕ(ar, br) = uar − vbr = 0, so ϕi = 0. Thus, we can find a right R-morphism φ : A → R such that φf = ϕ. Obviously, we have s, t ∈ R such that us + vt = 1; hence, ϕ(sr, −tr) = usr + vtr = r for any r ∈ R. This means that ϕ : R ⊕ R → R is an R-epimorphism. Hence φ : A → R is an R-epimorphism. As a result, we get R .⊕ A. (3) ⇒ (2) Suppose that Ra + Rb = R with a, b ∈ R. Then we have x, y ∈ R such that xa + yb = 1. Define j : R ⊕ R → R given by j(r1 , r2 ) =

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xr1 + yr2 for any r1 , r2 ∈ R and i : R → R ⊕ R given by i(r) = (ar, br) for any r ∈ R. Then ji = 1R , and so the exact sequence i

σ

0 → R → R ⊕ R → coker(i) → 0

splits, where σ is the natural R-epimorphism. Thus we have R ⊕ R ∼ = R ⊕ coker(i). By hypothesis, we have R .⊕ coker(f ). So there exists an R-epimorphism ϕ : coker(f ) → R, and then ϕσ : R ⊕ R → R. Assume that ϕσ(1, 0) = u and ϕσ(0, 1) = −v. Then 0 = ϕσi(1) = ϕσ(a, b) = ua − vb, i.e., ua = vb. Clearly, ϕσ is a splitting R-epimorphism. Hence we can find τ : R → R ⊕ R such that ϕστ = 1R . Let τ (1) = (s, t) ∈ R ⊕ R. Then 1 = ϕστ (1) = ϕσ(s, t) = us − vt. Therefore uR + vR = R, as asserted.  Lemma 14.3.5. Let R be an exchange ring. Then the following are equivalent : (1) R is strongly separative. (2) For any right R-modules A and B, 2A ∼ = A ⊕ B .⊕ R implies that ⊕ A . B. Proof. (1) ⇒ (2) is trivial. (2) ⇒ (1) Let A and B be finitely generated projective right R-modules such that A ⊕ A ∼ = A ⊕ B. Then we can find some n ∈ N such that A .⊕ nR. Since R is an exchange ring, it follows by [3, Proposition 1.2] ∼ that there exist A1 , ·
· · , An .⊕ R such that A
= A1 ⊕ · · · ⊕ An . Hence ∼ A1 ⊕ A2 ⊕· · ·⊕An ⊕A = A1 ⊕ A2 ⊕· · ·⊕An ⊕B . Let C1 = A2 ⊕· · ·⊕An ⊕A and B1 = A2 ⊕ · · · ⊕ An ⊕ B. Then A1 ⊕ C1 ∼ = A1 ⊕ B1 with A1 .⊕ C1 . Analogously to [3, Lemma 2.7], we have a refinement matrix A1 B1



A1 A′1 B1′

C1

 C1′ , D1

where A′1 .⊕ C1′ . Clearly, A′1 ⊕ C1′ ∼ = A1 .⊕ R with A′1 .⊕ C1′ . Thus, we have = A′1 ⊕ B1′ ∼ a refinement matrix A′1

C1′

 C1′′ , D1′ where A′′1 . C1′′ . It follows from A′′1 ⊕ C1′′ ∼ = A′1 with A′′1 . C1′′ = A′′1 ⊕ B1′′ ∼ ′′ ′′ ′ ′ ∼ ⊕ ′′ ′′ ∼ that C1 ⊕ C1 = C1 ⊕ B1 . A1 ⊕ C1 = A1 . R. By assumption, we have C ′′ .⊕ B ′′ , and so B ′′ ∼ = C ′′ ⊕ E for a right R-module E. As a result, we A′1 B1′

1

1

1

1



A′′1 B1′′

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get A′1 ∼ = A′′1 ⊕ B1′′ ∼ = A′′1 ⊕ C1′′ ⊕ E ∼ = A′1 ⊕ E. Since A′1 .⊕ C1′ , we see that ′ ∼ ′ ′ ∼ ′′ C1 = C1 ⊕ E, whence C1 = C1 ⊕ D1′ ⊕ E ∼ =
B1′′ ⊕ D1′ ∼ = B1′ , and so C1 ∼ = B
1 . This means that A2 ⊕ A3 ⊕ · · · ⊕ An ⊕ A = A2 ⊕ A3 ⊕ · · · ⊕ An ⊕ B . By iteration of this process, we prove that A ∼ = B. Therefore, R is strongly separative, which concludes the proof.  Theorem 14.3.6. Let R be an exchange ring. Then the following are equivalent : (1) R is strongly separative. (2) For any idempotent e ∈ R and any x ∈ Ur,2 (eRe), there exists y ∈ Uc,2 (eRe) such that xy = 0. (3) For any idempotent e ∈ R and any x ∈ Uc,2 (eRe), there exists y ∈ Ur,2 (eRe) such that yx = 0. Proof. (1) ⇒ (2) Let e ∈ R be an idempotent, and let S = eRe. Then S is strongly separative by [16, Lemma 1.5]. Hence, S is an almost hermitian ring from Theorem 14.3.1. In view of Corollary S is a U C1 -ring.  14.1.11,  ab Let aS + bS = S. Then there exists a matrix ∈ GL2 (R). Write ∗∗   −1  ∗ u ab = . ∗ −v ∗∗ Then au = bv and Su + Sv = S, as required. (2) ⇒ (1) Suppose that A ⊕ A ∼ = A ⊕ B .⊕ R. Then we can find an idempotent e ∈ R such that A ∼ = eR. Hence eR ⊕ eR ∼ = eR ⊕ B, and N N N N so eR Re ⊕ eR Re ∼ eR Re ⊕ B Re. This implies that eRe ⊕ = R R R R N eRe ∼ = eRe ⊕ B Re as right eRe-modules. By Lemma 14.3.4, we get R N eRe .⊕ B Re. As a result, we have that R

A∼ = eR ∼ = eRe

O

eR .⊕ B

eRe

Since B .⊕ eR ⊕ eR, we see that B

N R

O

Re

R

Re


N

eRe


O

eR.

eRe

N eR ∼ eR ∼ = Be = B, and eRe

so A .⊕ B. According to Lemma 14.3.5, R is strongly separative. (1) ⇔ (3) is analogous.



By induction and Theorem 12.2.8, an exchange ring R is strongly separative if and only if for any idempotent e ∈ R and any unimodular row x over eRe, there exists a unimodular column y over eRe such that xy = 0 if

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and only if for any idempotent e ∈ R and any unimodular column x over eRe, there exists a unimodular row y over eRe such that yx = 0. Corollary 14.3.7. An exchange ring R is strongly separative if and only if for all finitely generated projective left R-modules A and B, 2A ∼ = A⊕B ∼ implies that A = B. Proof. Let R be a strongly separative exchange ring. In view of Lemma 1.4.7, Rop is an exchange ring. For any S o = eo Ro eo with idempotent eo ∈ Ro , S o ao + S o bo = S o with ao , bo ∈ S o implies that aS + bS = S. Clearly, e ∈ R is an idempotent. In view of Theorem 14.3.6, we can find u, v ∈ R such that au = bu and Su + Sv = S. Thus, uo ao = v o bo and uo S o + v o S o = S o . By Theorem 14.3.6 again, Rop is strongly separative. Clearly, the opposite ring Rop is strongly separative if and only if for all finitely generated projective left R-modules A and B, A⊕A ∼ = A⊕B implies that A ∼  = B. The converse is symmetric. The preceding result shows that an exchange ring R is strongly separative if and only if so is the opposite ring Rop . Furthermore, we deduce that an exchange ring R is strongly separative if and only if for any finitely generated projective left R-module C, 2C ⊕ A ∼ = C ⊕ B implies that C ⊕ A ∼ =B for any left R-modules A and B. From Theorem 14.3.6, we see that an exchange ring R is strongly separative if and only if for any idempotents e ∈ R, every x ∈ Ur,2 (eRe) can be completed by a right invertible matrix if and only if for any idempotents e ∈ R, every x ∈ Uc,2 (eRe) can be completed by a left invertible matrix. Further, we can derive the following. Corollary 14.3.8. An exchange ring R is strongly separative if and only if every corner of R is a U C1 -ring. Proof. If R is a separative exchange ring, it follows by Theorem 14.3.1 that every corner of R is an almost hermitian ring. According to Corollary 14.1.11, every corner is a U C1 -ring. Conversely, let S be a corner  ofR. ab Suppose that (a, b) ∈ Ur,2 (S). By assumption, there exists some ∈ ∗∗   c11 c12 GL2 (S). So we have s, t ∈ S and ∈ M2 (S) such that c21 c22      ab c11 c12 10 = . st c21 c22 0 1 2×2

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c12 c22



Thus ac12 + bc22 = 0 and sc12 + tc22 = 1. This implies that (a, b) =   c12 0, where ∈ Uc,2 (S). According to Theorem 14.3.6, R is strongly c22 separative.  Corollary 14.3.9. Let R be a regular ring. Then the following are equivalent: (1) R is strongly separative. T (2) For any idempotent e ∈ R, ℓ(a) ℓ(b) = 0 in eRe implies that there T exist u, v ∈ eRe such that au = bv, Tr(u) r(v) = 0. (2) For any idempotent e ∈ R, r(a) r(b) = 0 in eRe implies that there T exist u, v ∈ eRe such that ua = vb, ℓ(u) ℓ(v) = 0.

Proof. Let S be a corner of R. Then S is regular. If Sa + Sb = S T T with a, b ∈ S, then r(a) r(b) = 0. Now assume that r(a) r(b) = 0.   a Construct a map ϕ : S → S ⊕ S given by ϕ(r) = r for any r ∈ S. If b       0 a 0 ϕ(r) = , then r= , and so ar = br = 0. This implies that 0 b 0 T r ∈ r(a) r(b) = 0. Thus ϕ : S → S ⊕ S is an R-monomorphism. As S is regular, there exist some c, d ∈ S such that      
a a a = . cd b b b
Construct an S-morphism θ : S ⊕ S → S corresponding to c d . For any r ∈ S, it is easy to verify that      
a a a ϕθϕ(r) = r= r = ϕ(r). cd b b b

This implies that θϕ(r) = r, and so θϕ = 1R . Hence ca + db = 1, and then T Sa+Sb = S. This implies that r(a) r(b) = 0 if and only if (a, b) ∈ Uc,2 (S). T Applying this to the opposite ring Rop , we deduce that ℓ(a) ℓ(b) = 0 if and only if (a, b) ∈ Ur,2 (S). Therefore the result follows from Theorem 14.3.6. 

Example 14.3.10. There exists a regular ring R with the following properties: (1) R is directly infinite.

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(2) R satisfies the comparability axiom. (3) R is strongly separative. Proof. Let V be an infinite-dimensional vector space over a field F , and let T = EndF (V ). Let M = {x ∈ T | dimF (xV ) < ∞}. Then M is the maximal ideal of T . Set A = T /M and write x for x + M . Let {e1 , e2 , · · · , en , · · · } be a basis of V . Define a : V → V given by a(e1 ) = 0 and a(ei ) = ei−1 (i = 2, 3, · · · ) and b : V → V given by b(ei ) = ei+1 (i = 1, 2, · · · ). Then ab = 1V and (ba − 1)V = F e1 . We claim that, for every 0 6= p(x) ∈ F [x], p(a) ∈ U (A). Let 0 6= p(x) ∈ F [x]. As a ∈ U (A), we may assume that p(x) = a0 +a1 x+· · ·+an xn , ai ∈ F ∞
P and a0 6= 0. Thus, p(x) ∈ U F [[x]] , i.e., there exists a q(x) = bi xi such i=0

that p(x)q(x) = q(x)p(x) = 1. It is easy to verify that ai (em ) = 0 for all i > m. Define ϕ : V → V given by ϕ(em ) = q(a)(em ). Then ϕ ∈ T such that p(a)ϕ = ϕp(a) = 1V . Consequently, p(a) ∈ U (T ); moreover, p(a) ∈ U (A). Let K = {p(a)q(a)−1 | p(x), q(x) ∈ F [x], q(x) 6= 0}. Then K is a field, and K ⊆ A. Let R = {x ∈ T | x ∈ K}. Then R is a subring of T and R/M = K. As R/M and M are both regular, so is R. Clearly, ab = ba = 1 in A. Hence, b = a−1 ∈ K. This implies that a, b ∈ R. As ab = 1 and ba 6= 1, we know that R is directly infinite. In view of [217, Proposition 8.4], R satisfies the comparability axiom. Let e ∈ R be an idempotent. As F is a field, e ∈ M or 1 − e ∈ M . Let S = M + F . Then S is a unit-regular subring of R. If e ∈ M , then eRe = eSe is unit-regular. Hence, eRe is an almost hermitian ring. If 1 − e ∈ M , then e = 1; hence, eRe/eM e ∼ = R/M ∼ = F . Thus, eRe/eM e
2 is a field. For any (x, y) ∈ (eRe) , there exists some U ∈ E2 eRe/eM e such that (x, y)U = (∗, 0). So we can find some V ∈ E2 (eRe) such that (x, y)W = (a, b) for some a ∈ eRe, b ∈ eM e. Since eM e ⊆ eSe, there exists an idempotent f ∈ eSe and a w ∈ U (eSe) such that b = f w. Obviously, −1 f ∈ eRe is an idempotent and w ∈ U (eRe). Hence, (x, y)W
diag(e, w ) = −1 (a, f ), and so (x, y)W diag(e, w )B21 (−a) = (e−f )a, f . Write (e−f )a = (e − f )ax(e − f )a. Set g = (e − f )ax(e − f ). Then (x, y)W diag(e, w−1 )B21 (−a)



0e e0




B21 x(e − f ) B21 (−ga) = (f + g, 0).

Consequently, eRe is an almost hermitian ring. According to Theorem 14.3.1, R is strongly separative. 

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Strongly Separative Ideals

One easily checks that an ideal I of an exchange ring R is strongly separative if and only if for any idempotent e ∈ I, eRe is an almost hermitian ring if and only if for any idempotent e ∈ I and any x ∈ Ur,2 (eRe), there exists y ∈ Uc,2 (eRe) such that xy = 0. The main purpose of this section is to investigate the strongly separative property for ideals of an exchange ring. Theorem 14.4.1. Let I be a strongly separative ideal of an exchange ring R, and let a − a2 ∈ I be regular. If Rar(a2 ) = R(a − a2 )R,

then a ∈ R is unit-regular. Proof. Since a − a2 ∈ R is regular, we can find some z ∈ R such that a(1−a) = a(1−a)za(1−a). Let x = 1+(1−a)z(1−a) and y = 1+aza. Then a = axa and 1 − a = (1 − a)y(1 − a), i.e., a, 1 − a ∈ R are both regular. Let b = 1−a. Then
R = r(a)⊕xaR = r(b)⊕ybR. Since R is an exchange ring, so is EndR r(a) from [382, Theorem 29.2]. So there exist C ⊆ r(b), D ⊆ ybR such that R = r(a) ⊕ C ⊕ D, and thus, aR = aC + aD = aC ⊕ aD. Given any x ∈ C, we see that x ∈ r(b), and so x = ax ∈ aC. Further, we see that ax = x ∈ C. This implies that C ⊆ aC ⊆ C. Hence, aC = C. This implies that aR = C ⊕ aD. Clearly, D ∼ = aD. Construct a map ϕ : D → a(1 − a)D given by ϕ(d) = a(1 − a)d for any d ∈ D. Then ϕ is an R-epimorphism. If a(1 − a)d = 0 for some d ∈ D, then (1 − a)d
∈ r(a). As ad ∈ C ⊕ D, T we see that (1 − a)d = d − ad ∈ r(a) C ⊕ D = 0; hence, (1 − a)d = 0. As d ∈ D ⊆ ybR, we can find some z ∈ R such that d = ybz. This implies that bd = bybz = (1 − a)d = 0, whence bz = 0 and so d = ybz = 0. This implies that D ∼ = a(1 − a)D. Clearly, a(1 − a)D ∼ = a(1 − a)R; hence, D∼ = aD ∼ = a(1 − a)R. In addition, R = r(a) ⊕ C ⊕ D = (1 − ax)R ⊕ aR = (1 − ax)R ⊕ C ⊕ aD,

and so r(a) ⊕ D ∼ = R/aR ⊕ D. Hence, r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R.

From Rar(a2 ) = R(a − a2 )R, we get a(1 − a)R ⊆ Rar(a2 ) ⊆ Rr(a). Thus we can find some x1 , · · · , xn ∈ R; y1 , · · · , yn ∈ r(a) such that a(1 − a) = n P xi yi . As a(1 − a) ∈ R is regular, we may assume that each xi ∈ a(1 −

i=1

a)R. Construct a map ϕ : nr(a) → a(1 − a)R given by ϕ(z1 , · · · , zn ) =

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xi zi for any (z1 , · · · , zn ) ∈ nr(a). Then ϕ is a right R-morphism. For

any a(1 − a)r ∈ a(1 − a)R, we can find (y1 r, · · · , yn r) ∈ nr(a) such that ϕ(y1 r, · · · , yn r) = a(1 − a)r; hence, ϕ is a right R-epimorphism. As a(1 − a)R ∈ F P (I), ϕ splits; hence, a(1 − a)R ∝ r(a). Similarly to Proposition  13.4.2, we get r(a) ∼ = R/aR, and therefore a ∈ R is unit-regular. Corollary 14.4.2. Let I be a strongly separative ideal of an exchange ring R, and let a − a2 ∈ I be regular. If ℓ(a2 )aR = R(a − a2 )R,

then a ∈ R is unit-regular. Proof. As is well known, an ideal I of an exchange ring R is strongly separative if and only if for any idempotent e ∈ I, eRe is strongly separative. In view of Corollary 14.3.7, I op is a strongly separative ideal of Rop . Applying Theorem 14.4.1 to the exchange ring Rop , we see that ao ∈ Rop is unit-regular, and therefore the result follows.  Lemma 14.4.3. Let I be a strongly separative ideal of an exchange ring R, and let a − a2 ∈ I be regular. If \ \ \ \ RaR Rr(a) = RaR R(1 − a)R or RaR ℓ(a)R = RaR R(1 − a)R,

then a ∈ R is unit-regular.

Proof. As in the proof of Theorem 14.4.1, we get r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R. T T T If RaR Rr(a) = RaR R(1 − a)R, then a(1 − a)R ⊆ RaR R(1 − a)R ⊆ Rr(a). Further, we see that a(1 − a)R ∝ r(a). Since I is a strongly separative ideal, similarly to Proposition 13.4.2, r(a) ∼ R/aR, and therefore T T = a ∈ R is unit regular. If RaR ℓ(a)R = RaR R(1 − a)R, we obtain the result by symmetry.  T T T T If RaR Rr(a) = RaR R(1 − a)R or RaR ℓ(a) = RaR R(1 − a)R is a strongly separative regular ideal of an exchange ring R, then a ∈ R is unit-regular. Theorem 14.4.4. Let I be a regular ideal of an exchange ring R. Then the following are equivalent: (1) I is strongly separative.

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(2) Each a ∈ R satisfying Rr(a) = R(1 − a)R and a − a2 ∈ I is unit-regular. (3) Each a ∈ R satisfying ℓ(a)R = R(1 − a)R and a − a2 ∈ I is unit-regular. Proof. (1) ⇒ (2) Suppose that Rr(a) = R(1 − a)R and a − a2 ∈ I. Then \ \ RaR Rr(a) = RaR R(1 − a)R.

It follows from Lemma 14.4.3 that a ∈ R is unit-regular. (2) ⇒ (1) Suppose A, B, C ∈ F P (I) with A ⊕ C ∼ = B ⊕ C .⊕ R and C .⊕ A. Write R = A1 ⊕ C1 ⊕ D ∼ = A2 ⊕ C2 ⊕ D, where A1 ∼ = A, ∼ ∼ ∼ C1 = C2 = C and A2 = B. Choose a ∈ R to induce an endomorphism of RR , which is zero on A1 , an isomorphism from C1 onto C2 , and the identity on D. For any x ∈ r(a), r ∈ R, we have rx = r(1 − a)x ∈ R(1 − a)R; hence, Rr(a) ⊆ R(1 − a)R. Clearly, A ∼ = r(a). Thus, (1 − a)R ∼ = A2 ⊕ ⊕ ∼ C2 = A1 ⊕ C1 . 2r(a). So we can find a right R-module E such that (1 − a)R ⊕ E ∼ = 2r(a). This implies that O O O (1 − a)R R/Rr(a) ⊕ E R/Rr(a) ∼ R/Rr(a). = 2r(a) R

R

N

R

As a ∈ R is regular, we see that r(a) R/Rr(a) = 0; hence, (1 − R N a)R R/Rr(a) = 0. This implies tat (1 − a)R = (1 − a)Rr(a). Thus, R

R(1 − a)R ⊆ Rr(a); hence, Rr(a) = R(1 − a)R. Furthermore, we see that
a(1 − a) ∈ a(1 − a) A2 ⊕ C2 ∈ F P (I), and thus a(1 − a) ∈ I. By assumption, a ∈ R is unit regular, i.e., there exists an invertible u ∈ R such that a = aua. Construct a map ϕ : (1 − ua)R → (1 − au)R given by
ϕ (1 − ua)r = u(1 − au)r for any r ∈ R. For any
(1 − ua)r ∈ (1 − ua)R, (1 − ua)r = u(1 − au)u−1 r = ϕ (1 − au)u−1 r . Hence, ϕ is an Repimorphism. Likewise, ϕ is R-monomorphic. Thus, r(a) ∼ = R/aR, i.e., A∼ = B. Analogously to [16, Corollary 2.9], I is strongly separative. (1) ⇔ (3) is proved by symmetry.  Corollary 14.4.5. Let I be a regular ideal of an exchange ring R. Then the following are equivalent: (1) I is strongly separative.

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(2) Each a ∈ R satisfying \ \ RaR Rr(a) = RaR R(1 − a)R and a − a2 ∈ I is unit-regular. (3) Each a ∈ R satisfying \ \ RaR ℓ(a)R = RaR R(1 − a)R and a − a2 ∈ I is unit-regular.

Proof. (1) ⇒ (2) is clear from Lemma 14.4.3. (2) ⇒ (1) Assume that a ∈ R satisfies

Rr(a) = R(1 − a)R and a − a2 ∈ I. T T Then RaR Rr(a) = RaR R(1 − a)R. By assumption, a ∈ R is unitregular. According to Theorem 14.4.4, I is strongly separative. (1)⇔(3) is proved by symmetry.  As a result, we deduce that a regular ring R is strongly separative if T T and only if each a ∈ R satisfying RaR Rr(a) = RaR R(1 − a)R is unitT T regular if and only if each a ∈ R satisfying RaR ℓ(a)R = RaR R(1−a)R is unit-regular. Lemma 14.4.6. Let R be a regular ring. Then I = {a ∈ R | EndR (aR) is strongly separative} is a strongly separative ideal of R. Proof. Let I = {a ∈ R | EndR (aR) is strongly separative}. Let x, y ∈ I and z ∈ R. As in the proof of Example 13.2.14, we show that zxR .⊕ xR, xzR .⊕ xR and (x+y)R .⊕ xR⊕yR. Since EndR (xR) and EndR (yR) are strongly separative, so are EndR (zxR) and EndR (xzR) by [16, Lemma 1.5]. Hence, zx, xz ∈ I. Analogously to Lemma 13.4.21, EndR (xR ⊕ yR)
is strongly separative, and then so is EndR (x + y)R by [16, Lemma 1.5]. Thus, x + y ∈ I. So we see that I is an ideal of R. For any idempotent e ∈ I, eRe is strongly separative. By [16, Lemma 1.5] again, I is a strongly separative ideal.  Proposition 14.4.7. Let I be an ideal of a regular ring R. Then the following are equivalent: (1) I is strongly separative.
(2) For all a ∈ I, EndR (a − a2 )R is strongly separative.

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Proof. (1) ⇒ (2) Let a ∈ I. Then there exists an element x ∈ R such that
a− a2 = (a− a2 )x(a− a2 ). Let e = (a− a2 )x ∈ I. Then EndR (a− a2 )R ∼ = eRe; hence, it is strongly separative from [16, Lemma 1.5]. (2) ⇒ (1) Let J = {x ∈ R | EndR (xR) is strongly separative}. In view of Lemma 14.4.6, J is a strongly separative ideal of R. Given A ⊕ C ∼ = B⊕ C .⊕ R and C .⊕ A with A, B, C ∈ F P (I), we write R = A1 ⊕ C1 ⊕ D = A2 ⊕ C2 ⊕ D, where A1 ∼ = A, C1 ∼ = C2 ∼ = C and A2 ∼ = B. Let a ∈ R induce an endomorphism of RR , which is zero on A1 , an isomorphism from C1 onto C2 , and the identity on D. Then a ∈ 1 + I. As in the proof of Theorem
13.4.14, we show that Rr(a) = R(1−a)R. By hypothesis, EndR (a−a2 )R is strongly separative, and so a − a2 ∈ J. In view of Theorem 14.4.4, a ∈ R is unit-regular. This implies that A ∼ = r(a) ∼ = R/aR ∼ = B. Therefore I is strongly separative.  Let R be a regular ring, and let (aij ) ∈ Mm×n (R). If all EndR (aij R) are strongly separative, then we claim that (aij ) admits a diagonal reduction. Let I = {a ∈ R | EndR (aR) is strongly separative}. In view of Lemma 14.4.6, I is a strongly separative ideal. Obviously, each aij ∈ I. By virtue of Lemma 13.1.19, there exists an idempotent e ∈ I such that all aij ∈ eRe. As eRe ∼ = EndR (eR), eRe is strongly separative. In view of Theorem 14.3.1, (aij ) ∈ Mm×n (R) admits a diagonal reduction, i.e., there exist U ′ ∈ GLm (R), V ′ ∈ GLn (eRe) such that U ′ AV ′ is a diagonal matrix. Let E = diag(e, · · · , e) ∈ Mm (R) and F = diag(e, · · · , e) ∈ Mn (R). Then U := U ′ + Im − E ∈ GLm (R) and V = V ′ + In − E ∈ GLn (R). Then U AV is a diagonal matrix over R, and we are done. Lemma 14.4.8. Let R be a regular ring. Then the following are equivalent: (1) R is strongly separative. (2) Each a ∈ R satisfying


(a − a2 )R . r(a) and i EndR (aR) = ∞

is unit-regular. (3) Each a ∈ R satisfying


(a − a2 )R . R/aR and i EndR (aR) = ∞

is unit-regular.


Proof. (1) ⇒ (2) Let a ∈ R satisfy (a−a2 )R . r(a) and i EndR (aR) = ∞. As in the proof of Theorem 14.4.1, we get r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R.

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Since R is strongly separative, it follows that r(a) ∼ = R/aR, and therefore a ∈ R is unit-regular. (2) ⇒ (1) If R is of bounded index, then R is unit-regular by [217, Corollary 7.11].
Hence, R is strongly separative. Otherwise, I := {x ∈ R | i EndR (xR) < ∞} is an ideal of R. One easily checks that I is a Bideal. Given A ⊕ C ∼ = B ⊕ C .⊕ R and C .⊕ A with A, B, C ∈ F P (R), we write R = A1 ⊕ C1 ⊕ D = A2 ⊕ C2 ⊕ D, where A1 ∼ = A, C1 ∼ = C2 ∼ = C and ∼ A2 = B. Let a ∈ R induce an endomorphism of RR , which is zero on A1 , an isomorphism from C1 onto C2 , and the identity on D. As in the proof of Theorem 13.4.8, we see that (a − a2 )R .⊕ r(a). In addition, we get

r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R.
If i EndR (aR) < ∞, then a ∈ I; hence, (a − a2 )R ∈ F P (I). In light ∼ of Lemma 13.1.9, we get
r(a) = R/aR. This implies that a ∈ R is unitregular. If i EndR (aR) = ∞, by hypothesis, a ∈ R is unit-regular. In any case, r(a) ∼ = R/aR. Therefore A ∼ = r(a) ∼ = R/aR ∼ = B, as required. (1) ⇔ (3) is proved in the same manner. 

Theorem 14.4.9. Let I be an ideal of a regular ring R. Then the following are equivalent : (1) I is strongly separative. (2) Each a ∈ 1 + I satisfying


(a − a2 )R . r(a) and i EndR (aR) = ∞

is unit-regular. (3) Each a ∈ 1 + I satisfying


(a − a2 )R . R/aR and i EndR (aR) = ∞

is unit-regular.


Proof. (1) ⇒ (2) Let a ∈ 1+I satisfy (a−a2 )R . r(a) and i EndR (aR) = ∞. As in the proof of Theorem 14.4.1, we get r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R.

Clearly, (a − a2 )R ∈ F P (I). Similarly to Lemma 13.4.1, we get r(a) ∼ = R/aR, and then a ∈ R is unit-regular. (2) ⇒ (1) Let e = e2 ∈ I and S = eRe. It will suffice to show
that 2 S is strongly separative. Given (a − a )S . r(a) and i EndS (aS) = ∞ with a ∈ S, then a + 1 − e ∈ 1 + I. Since R is regular, we can find an element x ∈ eRe such that a = axa, and so r(a) = (e − xa)S. Assume

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that (a − a2 )S ⊕ D ∼ = (e − xa)S for a right S-module D. Then (a − N N N a2 )S eR ⊕ D eR ∼ eR. By the regularity again, we get = (e − xa)S S S S N (a − a2 )R ⊕ D eR ∼ = (e − xa)R. Let b = a + 1 − e. Then b − b2 = S

a − a2 , e − xa = 1 − (x + 1 −
e)b and b = b(x + 1 − e)b. Thus we see that (b − b2 )R . 1 − (x + 1 − e)b R ∼ = r(b).
Let f = ax + 1 − e. Then f = f 2 and EndR bR ∼ = f Rf . Assume that i(f Rf ) = n < ∞. Given any nilpotent c ∈ axSax, then c ∈ f (eSe)f ⊆ f
Sf , and so cn = 0. This implies that i(axSax) ≤ n
< ∞, i.e., i EndS (aS) < ∞. This gives a contradiction. So i EndR (bR) = ∞. By assumption, b ∈ R is unit-regular. Since b(x + 1 − e) + (e − ax) = 1, there exists some that b + (e −
ax)y = u ∈ U (R). Hence
−1y ∈ R such −1 b + (e − ax)y u = 1 = u b + (e − ax)y . Clearly,
(1 − e)u−1 e = 0, −1 −1 i.e., eu e = u e. As
a result, a + (e − ax)(eye) (eu−1 e) = e = (eu−1 e) a + (e − ax)(eye) . Thus a + (e − ax)(eye)
∈ U (eRe). One easily checks that a = axa = ax a + (e − ax)(eye) . Therefore a = axa =
−1 a a + (e − ax)(eye) a, i.e., a ∈ S is unit-regular. According to Lemma 14.4.8, S is strongly separative, as desired. (1) ⇔ (3) Use the similar proof as above.  Corollary 14.4.10. Let I be an ideal of a regular ring R. Then the following are equivalent: (1) I is strongly separative. (2) Each a ∈ 1 + I satisfying


2r(a) ∼ = r(a) ⊕ R/aR and i EndR (aR) = ∞

is unit-regular. (3) Each a ∈ 1 + I satisfying



r(a) ⊕ R/aR ∼ = 2 R/aR and i EndR (aR) = ∞

is unit-regular.

Proof. (1) ⇒ (2) is clear from [16, Lemma 5.1].
(2) ⇒ (1) Let a ∈ 1 + I such that (a − a2 )R . r(a) and i EndR (aR) = ∞. Then we have a right R-module C such that r(a) ∼ = (a − a2 )R ⊕ C. 2 2 Hence, 2r(a) ∼ = r(a)⊕(a−a )R⊕C ∼ = R/aR⊕(a−a )R⊕C ∼ = r(a)⊕R/aR. By assumption, a ∈ R is unit-regular. So the result follows from Theorem 14.4.9. (1) ⇔ (3) is analogous. 

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In particular, an ideal I of a regular ring R is strongly separative if and only if each a ∈ 1 + I satisfying 2r(a) ∼ if and = r(a) ⊕ R/aR is unit-regular
only if each a ∈ 1 + I satisfying r(a) ⊕ R/aR ∼ = 2 R/aR is unit-regular. Lemma 14.4.11. Let R be a regular ring. Then the following are equivalent : (1) R is strongly separative. (2) Each a ∈ R satisfying


Rr(a) = R(1 − a)R and i EndR (aR) = ∞

is unit-regular. (3) Each a ∈ R satisfying


ℓ(a)R = R(1 − a)R and i EndR (aR) = ∞

is unit-regular.

Proof. (1) ⇒ (2) is obvious from Theorem 14.4.4. (2) ⇒ (1) If R is of bounded index, then R is unit-regular by [217, Corollary 7.11].
Hence, R is strongly separative. Otherwise, I := {x ∈ R | i EndR (xR) < ∞} is an ideal of R. One easily checks that I is a Bideal. Given A ⊕ C ∼ = B ⊕ C .⊕ R and C .⊕ A with A, B, C ∈ F P (R), we write R = A1 ⊕ C1 ⊕ D = A2 ⊕ C2 ⊕ D, where A1 ∼ = A, C1 ∼ = C2 ∼ = C and A2 ∼ = B. Let a ∈ R induce an endomorphism of RR , which is zero on A1 , an isomorphism from C1 onto C2 , and the identity on D. As in the proof of Theorem 13.4.14, we see that Rr(a) = R(1 − a). In addition, we get r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R.


If i EndR (aR) < ∞, then a ∈ I; hence, (a − a2 )R ∈ F P (I). By virtue ∼ of Lemma 13.1.9, we get
r(a) = R/aR. This implies that a ∈ R is unitregular. If i EndR (aR) = ∞, by hypothesis, a ∈ R is unit-regular. In any case, r(a) ∼ = R/aR. Therefore A ∼ = r(a) ∼ = R/aR ∼ = B, as desired. (1) ⇔ (3) is proved in the same manner.  Theorem 14.4.12. Let I be an ideal of a regular ring R. Then the following are equivalent: (1) I is strongly separative. (2) Each a ∈ 1 + I satisfying


R(1 − a)R = Rr(a) and i EndR (aR) = ∞

is unit-regular.

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(3) Each a ∈ 1 + I satisfying


R(1 − a)R = ℓ(a)R and i EndR (aR) = ∞

is unit-regular.

Proof. (1)⇒(2) is clear by Theorem 14.4.4. (2) ⇒ (1) Let e = e2 ∈ I and S = eRe. It suffices to prove
that S is strongly separative. Given (e − a)S . r(a) and i EndS (aS) = ∞ with a ∈ S, then a + 1 − e ∈ 1 + I. Let b = a + 1 − e.
Similarly to Theorem 14.4.9, we have (1 − b)R ∝ r(b) and i EndR (bR) = ∞. By assumption, b ∈ R is unit-regular. Furthermore, we see that a ∈ eRe is unit-regular. According to Lemma 14.4.11, S is strongly separative. As a result, I is strongly separative. (1) ⇔ (3) is proved in the same manner.  As a consequence, we deduce that an ideal I of a regular ring R is strongly separative if and only if each a ∈ 1+I satisfying Rr(a) = R(1−a)R is unit-regular if and only if each a ∈ 1 + I satisfying ℓ(a)R = R(1 − a)R is unit-regular. Let I be an ideal of an exchange ring R. As is well known, I is strongly separative if and only if for any idempotent e ∈ I, eRe is strongly separative. It follows from Theorem 14.3.1 that I is strongly separative if and only if for any idempotent e ∈ I, eRe is an almost hermitian ring. A natural question to ask is whether the matrix over a strongly ideal admits  separative   a F F 0F diagonal reduction. Let F be a field, let R = , and let I = . 0 F 0 0 Then I is an ideal of the exchange ring R. For any idempotent e ∈ I, eRe is strongly separative. Thus, I is a strongly separative ideal. One easily   01 checks that ∈ M2 (I) can not be reduced to a diagonal matrix. Thus, 00 the answer to the preceding problem is negative. Let I be a strongly separative regular ideal of a ring R, and let (aij ) ∈ Mm×n (I). In view of Lemma 13.1.19, there exists an idempotent e ∈ I such that each aij ∈ eRe; hence, (aij ) ∈ Mm×n (eRe). Clearly, eRe is strongly separative. According to Theorem 14.3.1, we have U ′ ∈ GLm (eRe) and V ′ ∈ GLn (eRe) such that U ′ AV ′ is a diagonal matrix over eRe. Set U = U ′ + diag(1 − e, · · · , 1 − e)m×m and V = V ′ + diag(1 − e, · · · , 1 − e)n×n. Then U ∈ GLm (R) and V ∈ GLn (R). Further, U AV is a diagonal matrix. That is, (aij ) admits a diagonal reduction. Thus, we conclude that every matrix over strongly separative regular ideals admits a diagonal reduction.

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Lemma 14.4.13. Let I be a regular ideal of a ring R. Then the following are equivalent: (1) Every 2 × 1 matrix over I admits a diagonal reduction. (2) For all n ∈ N, every n × 1 matrix over I admits a diagonal reduction. Proof. (1) ⇒ (2) Clearly, the result holds for n = 1, 2. Assume that the result holds for n = k(k ≥ 2). Let a1 , · · · , ak , ak+1 ∈ R. Then there exists a Q ∈ GLk (R) such that Q(a1 , · · · , ak )T = (b, 0, · · · , 0)T . Hence,   Q0 (a1 , · · · , ak , ak+1 )T = (b, 0, · · · , 0, ak+1 )T . 0 1

By hypothesis, there exists some (cij ) ∈ GL2 (R) such that (cij )(b, ak+1 )T = (d, 0)T . Therefore,   c11 0 · · · 0 c12  0 1 ··· 0 0      .. .. . . .. ..  Q 0  . . . . .  (a1 , · · · , ak , ak+1 )T = (d, 0, · · · , 0, 0)T ,    . . . . .  0 1  .. .. . . .. ..  c21 0 · · · 0 c22 as required. (2) ⇒ (1) is trivial.



Proposition 14.4.14. Let I be a regular ideal of a ring R. Then the following are equivalent: (1) (2) (3) (4)

Every matrix over I admits a diagonal reduction. Every 1 × 2 matrix over I admits a diagonal reduction. Every 2 × 1 matrix over I admits a diagonal reduction. 2R = A1 ⊕ B1 and R = A2 ⊕ B2 with A1 ∼ = A2 , A1 , A2 ∈ F P (I) imply that B1 ∼ . = R ⊕ B2  a0 (5) For any a, b ∈ I, ∈ M2 (R) is unit-regular. b0 Proof. (1) ⇒ (3) is trivial. (3) ⇒ (1) Let A ∈ Mk×m (R). The result holds for m = 1 by Lemma 14.4.13. Assume that the result holds for m − 1(m ≥ 2). In view of Lemma 14.4.13, A can be reduced to the matrix   0  ..  B  .   ,  0 ak1 · · · ak(m−1) b

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where B ∈ M(k−1)×(m−1) (R). By induction, A can be reduced to the matrix   a 0 0 ··· 0  .. ..   C . D= . . 0 0 c1 c2 · · · c(m−1) b As I is regular, we may writea = ec and e = ad,where e = e2 ∈ I. Set d 0 · · · 0 1 − dc 0 0     . ..   . U = I m−2 . .  .   0 0  1 0 · · · 0 −c Then U ∈ GLm (R), and that   e 0 0 ··· 0 0 0    .  .  ..  DU =  .. C .   0 0 t s1 · · · sm−2 s ′ ′ ′ ′ Write s = ss s and s = s  ss . We may assume thatss′ t = 0. Set s 0 · · · 0 ss′ − 1  0 0      . .   . . Ik−2 V = . . .    0 0  1 + s′ s 0 · · · 0 s′ Then V ∈ GLk (R). Furthermore,   se − t 0 ∗ ··· ∗  0 0     .. ..   V DU =  C . . .     0 0  (1 + s′ s)e s′ s1 · · · s′ sm−2 s′ s Thus, we may assume that   y y1 · · · ym−2 0 0 0     . .   . . C V DU =  . . ,   0 0 e f 0 ··· 0

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where f = s′ s ∈ R is an
idempotent. Write (1 − f )e = (1 − f )eh(1 − f )e. Set z = y 1 − h(1 − f )e . Then V DU can be reduced to the form   z y1 · · · ym−2 0 0 0    . ..    . C . .  .   0 0 e f 0 ··· 0

As in the proof of Proposition 13.2.27, we can find some K = (kij ) ∈   z0 GL2 (R) such that K = diag(∗, ∗). Therefore, ef   k11 0 · · · 0 k12  0 0      . .  ..  V DU  .. Ik−2     0 0  k21 0 · · · 0 k22 can be reduced to



∗ ∗ ··· ∗ 0   .  . C  .  0 0 0 ··· 0

0 0 .. . 0 ∗



   .   

By induction, A ∈ Mk×m (I) admits a diagonal reduction. (1) ⇔ (2) is symmetric. (2) ⇒ (4) Suppose that 2R = A1 ⊕ B1 , R = A2 ⊕ B2 with A1 ∼ = A2 , A1 , A2 ∈ F P (I). Then there is a regular homomorphism f : 2R → R such that ker f ∼ = B1 , Imf ∼ = A1 and coker f ∼ = B2 . As A1 = A1 I, we deduce that Imf = (Imf )I ⊆ I. Thus, f admits a diagonal reduction. By Lemma 14.1.4, R ⊕ coker f ∼ = ker f . So B1 ∼ = R ⊕ B2 . (4) ⇒ (5) For any a, b ∈ I, there exists Y ∈ I 2 such that X =   a XY X, where X = . Clearly, 2R = XY (2R) ⊕ (I2 − XY )(2R) b and R = Y X(R) ⊕ (1 − Y X)(R). Obviously, we have an R-isomorphism φ : XY (2R) ∼ = Y X(R) given by XY (r) 7→ Y XY (r) for any r ∈ 2R. By hypothesis, we have ψ : (I2 − XY )(2R) ∼ = R ⊕ (1 − Y X)(R). Now we construct an R-isomorphism ϕ : 2R = XY (2R) ⊕ (I2 − XY )(2R) →

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Y X(R) ⊕ (1 − Y X)(R) ⊕ R = 2R given by ϕ(s, t) = φ(s) + ψ(t) for any s ∈ XY (2R), t ∈ (I2 − XY )(2R). Let {e1 , e2 } be a basis of 2R, and let ϕ(e1 , e2 ) = (e1 , e2 )U for a matrix U . Then U ∈ GL2 (R). It is easy to verify that


X 02×1 = X 02×1 U X 02×1 ,

as required. (5) ⇒ (3) For any regular x ∈ 2 I, there exists y ∈ R2 which is a row of some invertible matrix such that x = xyx. Assume  that y is the first row of e some U ∈ GL2 (R). Set e = yx. Then U x = , where z is the second zx     1 0 e row of U . Hence V := U ∈ GL2 (R) and V x is of the form . −zx 1 0 Therefore the proof is true. 

As an immediate consequence, we derive that every matrix over regular right (left) hermitian rings admits a diagonal reduction. That is, every right (left) hermitian exchange ring is an almost hermitian ring. Corollary 14.4.15. Every matrix over a stable regular ideal admits a diagonal reduction. Proof. Let (a, b) ∈ I 2 . Since I is stable, by virtue of Proposition 13.2.2, we can findw1 , w2 ∈ U(R) such that aw1 = f, bw2 = e are idempotents.
1 f −1 Let Q1 = . Then (a, b)diag(w1 , w2 )Q1 = (1 − e)f, e . As −f 1 (1 − e)f ∈ I, it follows from Proposition 13.2.2 that there exists a u ∈ U (R) such that f1 = (1 − e)f u is an idempotent. Set  g = f1 (1  − e). Let Q2 = 1−g 1 diag(u, 1), Q3 = diag(1 − f1 e, 1 + f1 e) and Q4 = . Then −g 1     ab 0∗ diag(w1 , w2 )Q1 Q2 Q3 Q4 = . cd ∗∗

One easily checks that Q1 , Q2 , Q3 , Q4 ∈ E2 (R). According to Proposition 14.4.14, we obtain the result.  Let R be regular, and let (aij ) ∈ Mm×n (R). If for any r ∈ R, each 1 + aij r ∈ R is unit-regular, then A admits a diagonal reduction. In view of
Lemma 13.2.21, (aij ) ∈ Mm×n Ψ(R) . According to Lemma 13.2.24, Ψ(R) is a B-ideal; hence, it is stable. Clearly, Ψ(R) is regular, and we are done by Corollary 14.4.15.

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Chapter 15

Clean Properties, I

An element in a ring R is clean provided that it is the sum of an idempotent and an invertible element. A ring R is clean provided that every element in R is clean. In [63, Theorem 1], Camillo and Khurana proved that every element in a unit-regular ring is clean. In [338, Theorem], Nicholson and Varadarajan proved that every countable linear transformation over a division ring is clean. This shows that clean elements may not be unitregular even in a regular ring. Many authors have investigated cleanness in different ways, cf. [63-65], [337-341] and [368]. The problem of deciding the cleanness of a single element is considerably harder. In [278], Khurana   ab and Lam proved that many matrices ∈ M2 (Z) are unit-regular but 00       12 5 12 7 13 5 not clean, e.g., , , , etc. The main purpose of this 0 0 0 0 0 0 chapter is to investigate cleanness as it relates to stable range condition.

15.1

Cleanness in Separative Ideals

The following simple fact is due to Nicholson (cf. [334, Lemma 2.8]). Lemma 15.1.1. Let A be a quasi-projective right R-module. If A1 ⊆⊕ A and A = A1 + B, then there exists some A2 ⊆ B such that A = A1 ⊕ A2 . 511

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Proof. Write A = A1 ⊕ C. Then there exists a projection p : A → A1 such that A1 = pA. Let ϕ : A → A/B be the natural map. As A is quasi-projective, there exists a map α : A → A such that the following diagram α

A ← A p↓ ↓ϕ ϕ |A1

A1 ։ A/B commutates. Hence, ϕ = ϕ|A1 pα, i.e., ϕpα = ϕ. Clearly, p = p2 ∈ EndR (A). Let γ = p + pα(1 − p). Then γ = γ 2 ∈ EndR (A). Let A2 = ker(γ). Then A2 = (1 − γ)(A) = (1 − pα)(1 − p)(A) ⊆ ker(ϕ) = B. T For any a ∈ A1 A2 , we see that a = p(a) and γ(a) = 0. Thus, a = T (1 − γ)(a) + γ(a) = (1 − pa)(1 − p)p(a) = 0. This implies that A1 A2 = 0. For any a ∈ A, a = γ(a) + (1 − γ)(a) ∈ pA + (1 − γ)(A) = A1 + A2 . This implies that A = A1 + A2 . Therefore A = A1 ⊕ A2 , as asserted.  Lemma 15.1.2. Let R be a ring, and let a ∈ R be regular. If

(1) aR/ar(a2 ) is projective;
(2) ar(a2 ) ∼ = R/ r(a) + aR ;

then a ∈ R is clean. Proof. Suppose that Conditions (1) and (2) hold. Since aR/ar(a2 ) is projective, we have a splitting exact sequence 0 → ar(a2 ) ֒→ aR → aR/ar(a2 ) → 0.

Thus, there exists a right R-module Z such that aR = ar(a2 )⊕Z. Since aR
is projective, so is ar(a2 ). Thus, we see that R/ aR + r(a) is projective, and so
0 → aR + r(a) ֒→ R → R/ aR + r(a) → 0
splits. Hence, we have a right R-module Y such that R = aR + r(a) ⊕ Y , and so, aR + r(a) is projective. As a ∈ R is regular, there exists a right R-module aR + r(a) = aR +
T E such
that R = aR ⊕ E.
TThis implies that r(a) aR⊕E = aR⊕ aR+r(a) E, i.e., aR ⊆⊕ aR+r(a). By Lemma 15.1.1, we can find a right R-module X
⊆ r(a) such that aR + r(a) = T aR ⊕ X. Hence, r(a) = r(a) aR ⊕ X = K ⊕ X, where K = ar(a
2 ). By assumption, we have a right R-isomorphism φ : Y ∼ = R/ r(a) + aR ∼ = 2 ar(a ) = K. So there is an isomorphism α : X ⊕ Y → X ⊕ K given by α(x+y) = x+φ(y) for any x ∈ X and y ∈ Y . Let h : R = K ⊕X ⊕Y ⊕Z →

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K ⊕ X ⊕ Y ⊕ Z = R given by h(k + x + y + z) = α−1 (x + k) + y for any k ∈ K, x ∈ X, y ∈ Y, z ∈ Z. Let v : R = K ⊕ X ⊕ Y ⊕ Z → K ⊕ X ⊕ Y ⊕ Z = R given by v(k + x + y + z) = α(x + y) + φ−1 (k) for any k ∈ K, x ∈ X, y ∈ Y, z ∈ Z. For any k ∈ K, x ∈
X, y ∈ Y, z ∈ Z, we have
hvhv(k + x + y + z)
= hvh α(x + y) + φ−1 (k) = hv x + y + φ−1 (k) = h α(x+y)+αφ−1 (k) = x+y+φ−1 (k) = hv(k+x+y+z). Let e = hv. Then e ∈ EndR (R) is an idempotent. Assume that (a − hv)(k + x + y + z) = 0 for some k ∈ K, x ∈ X, y ∈ Y, z ∈ Z. Then a(y + z) = hv(k + x + y + z) = x+ y + φ−1 (k) ∈ aR ∩(X ⊕ Y ) = 0, and then x = −y − φ−1 (k) ∈ X ∩Y = 0. It follows from a(y + z) = 0 that y + z ∈ r(a)∩(Y ⊕ Z) = 0; hence y + z = 0. This implies that y = −z ∈ Y ∩ Z = 0, and then y = z = 0. Furthermore, we get φ−1 (k) = −y = 0, and so k = 0. This means that a − hv : R → R is a monomorphism. Given any
k ∈ K, x ∈ X, y ∈ Y, z ∈ Z, we have k + z ∈ K ⊕ Z = aR = a r(a)⊕Y ⊕Z = a(Y ⊕Z). So we can find some y ′ ∈ Y and z ′ ∈ Z such that k + z = a(y ′ + z ′ ). Choose x′ = −x and k ′ = −φ(y + y ′ ). It
is easy to verify that (a−hv)(k ′ +x′ +y ′ +z ′ ) = a(y ′ +z ′ )− x′ +y ′ +φ−1 (k ′ ) = k+x+y +z. This means that a − hv : R → R is an epimorphism. Therefore a − hv is an isomorphism. Let e = hv. Then a is the sum of an idempotent e and a unit a − hv, and therefore a ∈ R is clean.  Theorem 15.1.3. Let I be a separative ideal of an exchange ring R, and let a − a2 ∈ I be regular. If
(1) aR/ar(a2 ) and R/ aR + r(a) are projective; (2) Rar(a2 ) = ℓ(a2 )aR = R(a − a2 )R; then a ∈ R is clean. Proof. Suppose that Conditions (1) and (2) hold. Since aR/ar(a2 ) is projective, analogously to Lemma 15.1.2, there exists a right R-module Z
such that aR = ar(a2 ) ⊕ Z. Since R/ aR + r(a) is projective,
0 → aR + r(a) ֒→ R → R/ aR + r(a) → 0
splits. Thus, we have a right R-module Y such that R = aR + r(a) ⊕ Y . Hence, we can construct X, Y, K, Z as in the proof of Lemma 15.1.2 such that R = aR ⊕ X ⊕ Y, aR = K ⊕ Z, r(a) = K ⊕ X and aR + r(a) = aR ⊕ X, where K = ar(a2 ). Construct C and D as in the proof of Theorem 14.4.1. Then R = C ⊕ aD ⊕ X ⊕ Y = r(a) ⊕ C ⊕ D = C ⊕ D ⊕ X ⊕ K. This implies that a(1 − a)R ⊕ K ∼ = a(1 − a)R ⊕ Y.

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From Rar(a2 ) = R(a−a2 )R, we get a(1−a)R ⊆ Rar(a2 ). Thus we can find n P some x1 , · · · , xn ∈ R; y1 , · · · , yn ∈ ar(a2 ) such that a(1 − a) = xi yi . As i=1

a(1 − a) ∈ R is regular, we may assume that each xi ∈ a(1 − a)R. Construct n
P a map ϕ : n ar(a2 ) → a(1 − a)R given by ϕ(z1 , · · · , zn ) = xi zi for i=1
any (z1 , · · · , zn ) ∈ n ar(a2 ) . Clearly, ϕ is a right R-morphism.
For any a(1 − a)r ∈ a(1 − a)R, we can find (y1 r, · · · , yn r) ∈ n ar(a2 ) such that ϕ(y1 r, · · · , yn r) = a(1 − a)r; hence, ϕ is a right R-epimorphism. Now a(1 − a)R ∈ F P (I), whence ϕ splits. This implies that a(1 − a)R ∝ ar(a
2 ). Write aR + r(a) = gR for an idempotent g ∈ R. Then R/ aR + r(a) ∼ = (1 − g)R. As a − a2 ∈ R is regular, there exists an idempotent e ∈ R such that (a − a2 )R = eR. This implies that e ∈ ReR = ℓ(a2 )aR. Thus, we n P have some x1 , · · · , xn ∈ ℓ(a2 )a and r1 , · · · , rn ∈ R such that e = xi ri . i=1

Write xi = si a for some si ∈ ℓ(a2 ). Then si a2 = 0. Write g = ar + b, where r ∈ R, b ∈ r(a). Then xi g = si a(ar + b) = si a2 r = 0. This implies that xi = xi (1 − g). Let f = diag(1 − g, · · · , 1 − g) ∈ Mn (R), and     r1 r1  ..   ..  let c = e(x1 , · · · , xn )f, d = f  .  e. Then e = (x1 , · · · , xn )  .  = rn rn cd, c = ecf, d = f dc. In
view of Lemma 6.4.16, eR .⊕ f
(nR). Hence, a(1 − a)R .⊕ n (1 − g)R , i.e., a(1 − a)R ∝ R/ aR + r(a) . Similarly to
Proposition 13.4.2, we have K ∼ = Y , and thus, ar(a2 ) ∼ = R/ r(a) + aR . According to Lemma 15.1.2, a ∈ R is clean.  Corollary 15.1.4. Let I be a separative ideal of a regular ring R. Then each a ∈ R satisfying Rar(a2 ) = ℓ(a2 )aR = R(a − a2 )R ⊆ I is clean. Proof. Let a ∈ R satisfy Rar(a2 ) = ℓ(a2 )aR = R(a − a2 )R ⊆ I. Then T aR, r(a), aR + r(a) and K = ar(a2 ) = aR r(a) are direct summands of RR . In view of Lemma 15.1.1, there exists X ⊆ r(a) such that aR + r(a) = aR ⊕ X. In addition, we have right R-modules Y and Z such that R = aR ⊕ X ⊕ Y and aR = K ⊕ Z.
Hence R = K ⊕ X ⊕ Y ⊕ Z. 2 ∼ Clearly, aR/ar(a ) = Z and R/ aR + r(a) ∼ = Y . Thus, aR/ar(a2 ) and
R/ aR + r(a) are projective. According to Theorem 15.1.3, a ∈ R is clean.  Let R be a separative regular ring. It follows from Corollary 15.1.4 that each a ∈ R satisfying Rar(a2 ) = ℓ(a2 )aR = R(a − a2 )R is clean.

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Example 15.1.5. Let V be an infinite-dimensional vector space over a
field F , and let R = M3 EndF (V ) . Then R is weakly stable by Example 5.2.4 and Theorem 5.1.6. According   to Theorem 5.2.9, R is a separative 010 regular ring. Choose a =  0 0 0  ∈ R. Clearly, Rar(a2 ) ⊆ Ra(1 − 001 a2 )R = Ra(1 − a)(1 + a)R ⊆ R(a − a2 )R. As a2 = a3 , it follows that R(a − a2 )R = Ra(1 − a)R ⊆ Rar(a2 ). As a result, Rar(a2 ) = R(a − a2 )R. Likewise, we prove that ℓ(a2 )aR = R(a − a2 )R. Therefore we conclude that a ∈ R satisfies Rar(a2 ) = ℓ(a2 )aR = R(a − a2 )R. According to Corollary     100 −1 1 0 15.1.4, a ∈ R is clean. In fact, we have a =  0 1 0  +  0 −1 0 .  000 0 0 1 Example 15.1.6. Let V be an infinite-dimensional vector space over a field F , and let R = EndF (V ). Then R ∼ = 2R ∼ = 3R, so there exist orthogonal idempotents e1 , e2 , e3 ∈ R such that each ei R ∼ = R and e1 + e2 + e3 = 1. Clearly, we can find eij ∈ ei Rej such that ei = eij eji whenever i 6= j. Let a = e21 + e32 . Then e3 R ⊆ r(a). Given any x ∈ r(a), we have (e21 +e32 )x = 0. Hence e1 x = e12 (e21 +e32 )x = 0. Similarly, e2 x = 0. Thus we get x = (e1 + e2 + e3 )x = e3 x. So r(a) = e3 R, and then r(a) = e3 R ⊆ (e21 + e32 )e23 R ⊆ aR. Likewise, we have ℓ(a) = Re1 ⊆ Re12 (e21 + e32 ) ⊆ Ra. One easily checks that a = (e2 + e3 + e21 )(e1 + e3 + e32 )(e1 + e2 ) is the product of 3 idempotents. In light of [245, Proposition 2.3], we have Rr(a) = ℓ(a)R = R(1 −
a)R. It follows that R(a − a2 )R ⊆ R(1 − T a)R = Rr(a) = R r(a) aR = Rar(a2 ) ⊆ R(a − a2 )R. This implies that R(a − a2 )R = Rar(a2 ). Analogously, we have R(a − a2 )R = ℓ(a2 )aR. Therefore Rar(a2 ) = ℓ(a2 )aR = R(a − a2 )R. Since R is weakly stable, it follows by Corollary 15.1.4 that a ∈ R is clean. In fact, a = 1 + (a − 1) with (a − 1)−1 = −1 − e21 − e31 − e32 ∈ U (R). In this case, a2 6= a3 .  Proposition 15.1.7. A regular ring
R is unit-regular if and only if for each a ∈ R, ar(a2 ) ∼ = R/ aR + r(a) .

Proof. Let R be a regular ring, and let a ∈ R. Then aR, r(a), aR + r(a) and K = ar(a2 ) are direct summands of RR . In view of Lemma 15.1.1, there exists X ⊆ r(a) such that aR + r(a) = aR ⊕ X. Further, we have right R-modules Y and Z such that R = aR ⊕ X ⊕ Y and aR
= K ⊕ Z. T Thus, R = K ⊕ X ⊕ Y ⊕ Z. In addition, r(a) = r(a) aR ⊕ X = K ⊕ X. As in the proof of Theorem 15.1.3, we have a(1 − a)R ⊕ ar(a2 ) ∼ = a(1 −

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a)R ⊕ R/ aR ⊕ r(a) . If R
is unit-regular, it follows by Corollary 1.1.7 that ar(a2 ) ∼ = R/ aR + r(a)
. Conversely, we have r(a) = X
⊕ ar(a2 ) and R/aR ∼ = X ⊕ R/ aR + r(a) . Thus ar(a2 ) ∼ = R/ aR + r(a) implies that ∼ r(a) = R/aR; hence, a ∈ R is unit-regular, and therefore the proof is true.  Theorem 15.1.8. Let I be a strongly separative ideal of an exchange ring R, and let a − a2 ∈ I be regular. If
(1) aR/ar(a2 ) and R/ aR + r(a) are projective; (2) Rar(a2 ) = R(a − a2 )R or ℓ(a2 )aR = R(a − a2 )R; then a ∈ R is clean. Proof. Construct K, Y as in the proof of Theorem 15.1.3, we see that a(1 − a)R ⊕ K ∼ = a(1 − a)R ⊕ Y.

If Rar(a2 ) = R(a−a2 )R, then a(1−a)R ∝ ar(a2 ). If ℓ(a2 )aR = R(a−a2 )R,
then a(1 − a)R ∝ R/ aR + r(a) . Similarly
to Lemma 12.2.7, we have that K ∼ = Y , and thus, ar(a2 ) ∼ = R/ r(a) + aR . According to Lemma 15.1.2, a ∈ R is clean.  Corollary 15.1.9. Let I be a strongly separative ideal of a regular ring R. Then each a ∈ R satisfying Rar(a2 ) = R(a − a2 )R ⊆ I or ℓ(a2 )aR = R(a − a2 )R ⊆ I is clean. Proof. Let a ∈ R satisfy Rar(a2 ) = R(a − a2 )R ⊆ I or ℓ(a2 )aR = R(a − a2 )R ⊆ I. As in the proof of Corollary 15.1.4, aR/ar(a2 ) is projective. According to Theorem 15.1.8, a ∈ R is clean.  Let R be a strongly separative regular ring. Then each a ∈ R satisfying Rar(a2 ) = R(a − a2 )R or ℓ(a2 )aR = R(a − a2 )R is clean. Theorem 15.1.10. Let R
be an exchange ring, and let a − a2 ∈ R be regular. If EndR (a − a2 )R has stable range one, then a ∈ R is clean.

Proof. As in the proof of Theorem 14.4.1, we can find C ⊆ r(b) and D ⊆ ybR such that R = r(a) ⊕ C ⊕ D. In addition, aR = C ⊕ aD with D ∼ = aD ∼ = a(1−a)R. Obviously, R = aR⊕(1−ax)R = r(a)⊕xaR. As EndR (aR) is an exchange ring, there exist X1 ⊆ r(a), Y1 ⊆ xaR such that R = aR⊕X1 ⊕Y1 . It is easy to verify that r(a) = r(a) ∩ (X1 ⊕ aR ⊕ Y1 ) = X1 ⊕ X2 , where X2 = r(a) ∩ (aR ⊕ Y1 ). Likewise, we have a right R-module Y2 such that xaR = Y1 ⊕ Y2 . As a result, we deduce that R = X1 ⊕ X2 ⊕ C ⊕ D =

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∼ C ⊕ aD ⊕ X1 ⊕ Y1 . This
implies that a(1 − a)R ⊕ X2 = a(1 − a)R ⊕ Y1 . 2 Since EndR (a − a )R has stable range one, by Theorem 1.1.5, we get a right R-isomorphism ψ : X2 → Y1 . Construct a right R-isomorphism k : X1 ⊕ X2 → X1 ⊕ Y1 given by k(x1 + x2 ) = x1 + ψ(x2 ) for any x1 ∈ X1 , x2 ∈ X2 . Let h : R = X1 ⊕X2 ⊕Y1 ⊕Y2 → X1 ⊕Y1 ⊕X2 ⊕Y2 = R given by h(x1 +x2 +y1 +z) = k(x1 +x2 )+y1 for any x1 ∈ X1 , x2 ∈ X2 , y1 ∈ Y1 , z ∈ Z. Let v : R = X1 ⊕ Y1 ⊕ X2 ⊕ Y2 → X1 ⊕ X2 ⊕ Y1 ⊕ Y2 = R given by v(x1 + y1 + x2 + z) = k −1 (x1 + y1 ) + ψ(x2 ) for any x1 ∈ X1 , y1 ∈ Y1 , x2 ∈ X2 , y2 ∈ Y2 . Analogously to Lemma 15.1.2, we conclude that hvhv = hv and that a − hv ∈ R is an isomorphism, and therefore the result follows. Corollary 15.1.11. Let I be an ideal of a regular ring R, and let a ∈ I. If I is a B-ideal, then a + b ∈ R is clean for all idempotents b ∈ R. Proof. Let b ∈ R be an idempotent. Then (a + b) − (a + b)2 = (a − a2 ) − ab − ba ∈ I; hence, (a + b) − (a + b)2 ∈ I is regular.
Thus, we can find
x ∈ R such that (a + b) − (a +
b)2 = (a + b) − (a + b)2 x (a + b) −
(a + b)2 . Let f = (a + b) − (a + b)2 x. Then EndR ((a + b) − (a + b)2 )R = f Rf . Given bc + d = f with b, c, d ∈ f Rf , then (b + 1 − f )(c + 1 − f ) + d = 1 with b + 1 − f ∈ 1 + I. Since I is a B-ideal, there exists z ∈ R such that u := b+1−f +dz ∈ U (R). Thus, (b+1−f +dz)u−1 = 1;
hence, (1−f )u−1 f = 0.
This implies that u−1
f = f u−1 f , and so b+d(f zf ) f u−1 f = f . Clearly,
−1 f u f b + d(f zf ) = f . This implies that b + d(f zf ) ∈ U (f Rf ), i.e., f Rf has stable range one. According to Theorem 15.1.10, a + b ∈ R is clean.  Let R be a regular ring, and let a ∈ R. We then have the following variations. (1) If RaR is a B-ideal, then a + b is clean for all idempotents b ∈ R. (2) If RaR is of bounded index, then a + b ∈ R is clean for all idempotents b ∈ R. As is well known, an ideal I of a regular ring R has stable range if and only if eRe is unit-regular for all idempotents e ∈ I. For any idempotent e ∈ RaR, eRe is a regular ring of bounded index. In view of [217, Corollary 7.11], eRe has stable range one. Thus, RaR is a B-ideal.
(3) If i EndR (aR) < ∞,
then a + b is clean for all idempotents b ∈ R. Let I = {x ∈ R | i EndR (aR) < ∞}. Then I is an ideal of R. Furthermore, I is a B-ideal, and we are done. (4) If EndR (aR) is unit-regular, then a + b is clean for all idempotents b ∈ R. As in the proof of Example 13.2.14, I = {a ∈ R | EndR (aR) is unit-regular} is an ideal of R, as required. (5) Let R be a right self-injective, regular ring, and let a ∈ R. If aR is directly

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finite, then we claim that a + b is clean for all idempotents b ∈ R. Let I = {x ∈ R | xR is directly finite}. By virtue of [217, Corollary 9.21], I is an ideal of R. For any idempotent e ∈ I, we have that EndR (eR) ∼ = eRe is unit-regular from [217, Theorem 9.17]. This implies that I is a B-ideal. According to Corollary 15.1.11, a + b is clean for all idempotents b ∈ R. Proposition 15.1.12. Let I be an ideal of a regular ring R. Then the following are equivalent: (1) I is a B-ideal.
(2) For each a ∈ I, EndR (a − a2 )R has stable range one.

Proof. (1) ⇒ (2) Let a ∈ I. Then there exists an idempotent x ∈ R such that a − a2
= (a − a2 )x(a − a2 ). Let e = (a − a2 )x ∈ I. Then EndR (a − a2 )R ∼ = eRe; hence, it has stable range one. (2) ⇒ (1) Let ax + c = 1 with a ∈ 1 + I, x, c ∈ R.
Choose b = 1 − a ∈ I. Then b − b2 = a − a2 ; hence, EndR (a − a2 )R has stable range one. According to Theorem 15.1.10, there exists an idempotent e ∈ R and a unit u ∈ R such that a = e + u. Construct h, X1 , X2 , Y1 and Y2 as in Theorem 15.1.10. We see that \

\
au−1 − 1 a = eu−1 a ∈ hR aR ⊆ X1 ⊕ Y1 X2 ⊕ Y2 = 0; hence, a = au−1 a. Let f = au−1 . Then f ux + c = 1, and thus f ux(1 − f ) + c(1 − f ) = 1
− f . This implies that a + c(1 − f )u = f u + c(1 − f )u = 1 − f ux(1 − f ) u ∈ U (R), and therefore I is a B-ideal. 

As a consequence, we deduce that a regular
ring R is unit-regular if 2 and only if for each a ∈ R, End (a − a )R is unit-regular. In fact, R
2 EndR (a − a )R is unit-regular implies that a ∈ R is unit-regular. The counterexample below shows that the converse is not true. Example 15.1.13. Let Z3 [[t]] be the ring of formal power
series over the field Z3 in an indeterminate t, and let R = EndZ3 Z3 [[t]] . Clearly, R is regular. Define a : Z3 [[t]] → Z3 [[t]] given by a(f ) = 2f for any f ∈ Z3 [[t]]. Then
we see that a = a−1 ∈ U (R) is unit-regular. Obviously, EndR (a − a2 )R ∼ = R. Define b : Z3 [[t]] → Z3 [[t]] given by b(f ) = tf for any f ∈ Z3 [[t]]. Then b ∈ R is injective; it is a splitting monomorphism. Hence b ∈ R is left invertible. If b ∈ R is unit-regular, then there exists some u ∈ U (R) such that b = bub; hence, ub = 1. This implies that b ∈ U (R),
and then b is surjective. This is impossible. As a result, EndR (a − a2 )R is not unit-regular. 

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Regular Rings in which 2 is Invertible

Theorem 15.2.1. Let I be a separative ideal of a regular ring R in which 2 is invertible, and let a − a3 ∈ I. If Rar(a2 ) = ℓ(a2 )aR = R(a − a3 )R, then a ∈ R is clean. Proof. Since R is regular, we can construct X, Y, K, Z as in the proof of Lemma 15.1.2 such that R = aR ⊕ X ⊕ Y, aR = K ⊕ Z, r(a) = K ⊕ X and aR + r(a) = aR ⊕ X, where K = ar(a2 ). Since R is a regular
ring, we have a right R-module D such that
R= T r(a)+r(1−a)+r(1+a) ⊕D. Given any r ∈ r(a) r(1−a)+r(1+a) , then ar = 0 and r = y +z with y = ay, z = −az; hence, ar = ay +az = y −z = 0. This implies that y = z, and so 2y = 0. As 2 is
invertible, we get y = z = 0. T This implies that r(a) r(1 − a) + r(1 + a) = 0. Given any r ∈ r(1 −
T a) r(a)+r(1+a) , then r = y +z with r = ar, ay = 0 and z = −az. This implies that 0 = ay = a(r −z) = r +z; hence, r = −z. Thus, z = az = −az, and so 2az = 0. As a result,
az = 0, and then r =T−z = az = 0. That
is, T r(1 − a) r(a) + r(1 + a) = 0. Likewise, r(1 + a)
r(a) + r(1 − a) = 0. Thus, we have that R = r(a) ⊕ r(1 − a) ⊕ r(1 + a) ⊕ D. Construct a map ϕ : D → a(1 − a)(1 + a)D given by ϕ(d) = a(1 − a)(1 + a)d for any d ∈ D. If a(1 − a)(1 + a)d = 0, then a(1 − a)d ∈ r(1 + a), a(1 + a)d ∈ r(1 − a) and (1 − a)(1 + a)d ∈ r(a). It is easy to verify that d = (1 − a2 )d + a2 d 1 = (1 − a)(1 + a)d + 12 a(1 + a)d
T− 2 a(1 − a)d ∈ r(a) + r(1 − a) + r(1 + a) D,

and so d = 0. This implies that ϕ : D → (a−a3 )D is a right R-isomorphism. Consequently, D ∼ =D∼ = (a − a3 ))D = (a − a3 )R. Hence, R= = = =

r(a) ⊕ r(1 − a) ⊕ r(1 + a) ⊕ D K ⊕ X ⊕ r(1 − a) ⊕ r(1 + a) ⊕ D aR ⊕ X ⊕ Y r(1 − a) ⊕ r(1 + a) ⊕ aD ⊕ X ⊕ Y,

and so (a − a3 )R ⊕ K ∼ = (a − a3 )R ⊕ Y. Since Rar(a2 ) = R(a − a2 )R, analogously to Theorem 15.1.3, we show that 2 (a − a2 )R ∝ ar(a
). Write aR + r(a) = gR 3for an idempotent g ∈ R.2 Then ∼ R/ aR + r(a) = (1 − g)R. Write (a − a )R = eR, where e = e ∈ R.

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This implies that e ∈ ReR = ℓ(a
2 )aR. Similarly to Theorem 15.1.3, we see that (a − a3 )R ∝ R/ aR + r(a) . Similarly
to Proposition 13.4.2, we have K ∼ = Y , and thus, ar(a2 ) ∼ = R/ r(a) + aR . According to Lemma 15.1.2, a ∈ R is clean.  Let R be a separative regular ring in which 2 is invertible. It follows by Theorem 15.2.1 that each a ∈ R satisfying Rar(a2 ) = ℓ(a2 )aR = R(a−a3 )R is clean. Now we characterize separative ideals of regular rings in which 2 is invertible. Lemma 15.2.2. Let R be a regular ring in which 2 is invertible, and let a ∈ R. Then r(a) ⊕ (a − a3 )R ∼ = R/aR ⊕ (a − a3 )R.

Proof. Construct K, Y as in the proof of Theorem 15.2.1. Then (a−a3 )R⊕ K∼ = (a − a3 )R ⊕ Y. Furthermore, r(a) = K ⊕ X and R/aR ∼ = X ⊕ Y ; hence, 3 3 ∼ r(a) ⊕ (a − a )R = R/aR ⊕ (a − a )R, as asserted.  Theorem 15.2.3. Let I be an ideal of a regular ring R in which 2 is invertible. Then the following are equivalent : (1) I is separative. (2) Each a ∈ 1 + I satisfying

R(1 − a2 )R = Rr(a) = ℓ(a)R

is unit-regular. Proof. (1) ⇒ (2) Suppose that a ∈ 1 + I satisfies R(1 − a2 )R = Rr(a)
= ℓ(a)R. Construct a map ϕ : (1 − a2 )R → (a − a3 )R given by ϕ (1 − a2 )r = (a − a3 )r for any r ∈ R. Clearly, ϕ is an R-epimorphism. As a − a3 ∈ R is regular, (a − a2 )R is projective. Thus, ϕ splits. So we can find a right R-module D such that (1−a2 )R ∼ = (a−a3 )R⊕D. In view of Lemma 15.2.2, 2 2 r(a)⊕(1−a )R ∼ = R/aR⊕(1−a )R. Since a ∈ R is regular, there exists some x ∈ R such that a = axa; hence, R(1−a2 )R = ℓ(a)R = R(1−ax)R. In view of [217, Corollary 2.23], (1 − a2 )R ∝ (1 − ax)R, i.e., (1 − a2 )R ∝ R/aR. Also we have (1 − a2 )R ∝ r(a). Clearly, r(a), R/aR ∈ F P (I). As I is separative, we get r(a) ∼ = R/aR, and then a ∈ 1 + I is unit-regular. (2) ⇒ (1) Given A ⊕ C ∼ = B ⊕ C .⊕ R and C .⊕ A, B with A, B, C ∈ F P (I), we write R = A1 ⊕ C1 ⊕ D = A2 ⊕ C2 ⊕ D, where A1 ∼ = A, C1 ∼ = C2 ∼ = C and A2 ∼ = B. Let a ∈ R induce an endomorphism of RR , which is zero on A1 , an isomorphism from C1 onto C2 , and the identity on

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D. Clearly, a ∈ 1 + I is
regular. In addition, (1 − a2 )R ∼ = (1 − a2 ) A1 ⊕ C1
; hence, (1 − a2 ) A1 ⊕ C1 is projective. Let ϕ : A1 ⊕ C1 → (1 − a2 ) A1 ⊕ C1 given by ϕ(a1 + c1 ) = (1 − a2 )(a1 + c1 ) for any a1 ∈ A1 , c1 ∈ C1 . Then ψ splits; hence,
there exists a right R-module A1 ⊕ C1 ∼ =
⊕ E such that 2 2 ⊕ (1−a ) A1 ⊕C1 ⊕E. Thus, (1−a ) A1 ⊕C1 . A1 ⊕C1 . 2A1 ∼ = 2r(a). Hence (1 − a2 )R ∝ r(a). Likewise, (1 − a2 )R ∝ R/aR. In view of [217, Corollary 2.23], we get R(1 − a2 )R ⊆ Rr(a) and R(1 − a2 )R ⊆ ℓ(a)R. On the other hand, Rr(a) ⊆ R(1 − a2 )R and l(a)R ⊆ R(1 − a2 )R. Thus, R(1 − a2 )R = Rr(a) = ℓ(a)R. By hypothesis, a ∈ R is unit-regular, and then A ∼  = r(a) ∼ = R/aR ∼ = B. Therefore I is separative. Corollary 15.2.4. Let I be an ideal of a regular ring R in which 2 is invertible. If each a ∈ 1 + I satisfying R(1 − a2 )R = Rr(a2 ) = ℓ(a2 )R is unit-regular, then I is separative. Proof. Suppose that a ∈ 1 + I satisfies R(1 − a2)R = Rr(a) = ℓ(a)R. Then R(1 − a2 )R = Rr(a) ⊆ Rr(a2 ) ⊆ R(1 − a2 )R, and so R(1 − a2 )R = Rr(a2 ). Similarly, we see that R(1 − a2 )R = ℓ(a2 )R. As a result, a ∈ R is unitregular. According to Theorem 15.2.3, I is separative.  Let I be a separative ideal of a regular ring R, and let a ∈ 1 + I. If R(1 − a2 )R = Rr(a2 ) = ℓ(a2 )R, it follows by Theorem 13.4.14 that a2 is unit-regular. Corollary 15.2.5. Let R be a regular ring in which 2 is invertible. Then the following are equivalent : (1) R is separative. (2) Each a ∈ R satisfying


R(1 − a2 )R = Rr(a) = ℓ(a)R and i EndR (aR) = ∞

is unit-regular.

Proof. (1) ⇒ (2) is trivial by
Theorem 15.2.3. (2) ⇒ (1) If i EndR (aR) < ∞ for all a ∈ R, then R has stable
range one; hence, it is separative. Otherwise, I := {a ∈ R | i EndR (aR) < ∞} is an ideal of
R. Let a ∈ R such that R(1 − a2 )R = Rr(a) = ℓ(a)R. If
i EndR (aR) = ∞, then a ∈ R is unit-regular. If i EndR (aR) < ∞, then a ∈ I. Clearly, I is a B-ideal, and so it is stable from Lemma 13.2.5.

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According to Proposition 13.2.2, a ∈ I is unit-regular. Therefore R is separative from Theorem 15.2.3.  Lemma 15.2.6. Let I be an ideal of a regular ring R in which 2 is invertible. Then the following are equivalent : (1) I is separative. (2) Each a ∈ 1 + I satisfying (a − a3 )R . r(a), R/aR is unit-regular. (3) Each a ∈ 1 + I satisfying (a − a3 )R ∝ r(a), R/aR is unit-regular. Proof. (1) ⇒ (3) Suppose that (a − a3 )R ∝ r(a), R/aR with a ∈ 1 + I. In view of Lemma 15.2.2, r(a) ⊕ (a − a3 )R ∼ = R/aR ⊕ (a − a3 )R with 3 3 r(a), R/aR, (a − a )R ∈ F P (I). As (a − a )R ∝ r(a), R/aR, we obtain r(a) ∼ = R/aR, and so a ∈ 1 + I is unit-regular. (3) ⇒ (2) is trivial. (2) ⇒ (1) Given A ⊕ C ∼ = B ⊕ C . R and C . A, B with A, B, C ∈ F P (I), we write R = A1 ⊕ C1 ⊕ D = A2 ⊕ C2 ⊕ D, where A1 ∼ = A, C1 ∼ = C2 ∼ = C and A2 ∼ = B. Let a ∈ R induce an endomorphism of RR , which is zero on A1 , an isomorphism from C1 onto C2 , and the identity
on D. Then a ∈ 1 + I. It is easy to see that (a − a3 )R ∼ = (a − a3 ) A1 ⊕ C1 ∼ = (a − a3 )C1 ; hence, (a − a3 )C1 is projective. Let ϕ : C1 → (a − a3 )C1 given by ϕ(c1 ) = (a − a3 )(c1 ) for any c1 ∈ C1 . Then ψ splits, and so we have a right R-module E such that C1 ∼ = (a − a3 )C1 ⊕ E. This implies 3 that (a − a )C1 . C1 . A1 ∼ = r(a). Hence (a − a3 ) . r(a). Likewise, (a − a3 )R . R/aR. By hypothesis, a ∈ R is unit-regular. Therefore A∼  = r(a) ∼ = R/aR ∼ = B, as required. Theorem 15.2.7. Let I be an ideal of a regular ring R in which 2 is invertible. Then the following are equivalent : (1) I is separative. (2) Each a ∈ 1 + I satisfying \ \ \ R(1 − a2 )R RaR = Rr(a) ℓ(a)R RaR is unit-regular.

T T T Proof. (1) ⇒ (2) Suppose that R(1 − a2 )R RaR = Rr(a) ℓ(a)R RaR T with a ∈ 1 + I. Then R(a − a3 )R ⊆ R(1 − a2 )R RaR ⊆ Rr(a); hence, (a − a3 )R ⊆ Rr(a). In view of [217, Corollary 2.23], (a − a3 )R ∝ r(a). Likewise, we get (a − a3 )R ∝ R/aR. By virtue of Lemma 15.2.6, a ∈ 1 + I is unit-regular.

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(2) ⇒ (1) Suppose that R(1 − a2 )R = Rr(a) = ℓ(a)R with a ∈ 1 + I. Then T T T R(1 − a2 )R RaR ⊆ Rr(a) ℓ(a)R RaR T ⊆ R(1 − a2 )R RaR; T T T hence, R(1 − a2 )R RaR = Rr(a) ℓ(a)R RaR. By assumption, a ∈ 1 + I is unit-regular. Therefore we complete the proof using Theorem 15.2.3.  Corollary 15.2.8. Let I be an ideal of a regular ring R in which 2 is invertible. If each a ∈ 1 + I satisfying \ \ \ R(1 − a2 )R RaR = Rr(a2 ) ℓ(a2 )R RaR

is unit-regular, then I is separative. T T T Proof. Given R(1 − a2 )R RaR = Rr(a) ℓ(a)R RaR with a ∈ 1 + I, then T T T R(1 − a2 )R RaR ⊆ Rr(a) ℓ(a)R RaR T T ⊆ Rr(a2 ) ℓ(a2 )R RaR T ⊆ R(1 − a2 )R RaR; T T T hence, R(1 − a2 )R RaR = Rr(a2 ) ℓ(a2 )R RaR. This implies that a ∈ 1 + I is unit-regular. According to Theorem 15.2.7, I is separative.  Corollary 15.2.9. Let I be an ideal of a regular ring R in which 2 is invertible. Then the following are equivalent : (1) I is separative. (2) Each a ∈ 1 + I satisfying


R(1 − a2 )R = Rr(a) = ℓ(a)R and i EndR (aR) = ∞

is unit-regular. (3) Each a ∈ 1 + I satisfying \ \ \
R(1 − a2 )R RaR = Rr(a) ℓ(a)R RaR and i EndR (aR) = ∞ is unit-regular.

Proof. (1) ⇒ (3) is obvious from Theorem 15.2.7. (3) ⇒ (2) is similar to the proof of Theorem 15.2.7. (2) ⇒ (1) Let e = e2 ∈ I and S = eRe. By [16, Lemma 4.1], it suffices to prove that S
is separative. Assume that S(e − a2 )S = Sr(a) = ℓ(a)S and i EndS (aS) = ∞. Let b = a + 1 − e. Then 1 − b2 = e − a2 ,

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whence R(1 − b2 )R = R(e − a2 )R ⊆ RSr(a)R ⊆ Rr(b) ⊆ R(1 − b2 )R. 2 Thus, R(1 − b2 )R = Rr(b). Likewise, R(1
− b )R = ℓ(b)R. As in the proof of Theorem 14.4.9, we get i EndR (bR) = ∞. By hypothesis, b ∈ R is unit-regular. Similarly to Theorem 14.4.9, a ∈ S is unit-regular. According to Corollary 15.2.5, S is separative, as desired.  Now we extend this corollary to general regular rings in which 2 may not be an invertible element. Theorem 15.2.10. Let I be an ideal of a regular ring R. Then the following are equivalent: (1) I is separative. (2) Each a ∈
1 + I satisfying Rr(a) = ℓ(a)R = R(1 − a)R and i EndR (aR) = ∞ is unit-regular.

Proof. (1)⇒(2) is clear from Lemma 13.4.10.
(2) ⇒ (1) Let J = {x ∈ R | i EndR (xR) < ∞}. If J = R, then R is a regular ring of bounded index; hence, it has stable range one. This implies that I is separative. Otherwise, J is an ideal of R. Furthermore, J is a B-ideal. Let a ∈ 1 + I satisfy

R(1 − a)R = Rr(a) = ℓ(a)R.

If i EndR (aR) = ∞, then a ∈ 1 + I is unit-regular. If i EndR (aR) < ∞, then a ∈ J. In view of Proposition 13.2.2, a ∈ 1 + I is unit-regular. According to Theorem 13.4.14, I is separative, as asserted.  Corollary 15.2.11. Let I be a ideal of a regular ring R. Then the following are equivalent : (1) I is separative. (2) Each a ∈ 1 +
I satisfying Rr(a)aR = Raℓ(a)R = RaR(1 − a)R and i EndR (aR) = ∞ is unit-regular.

Proof. (1) ⇒ (2)
Let a ∈ 1 + I satisfy Rr(a)aR = Raℓ(a)R = RaR(1 − a)R and i EndR (aR) = ∞. Then Ra(1 − a)R ⊆ RaR(1 − a)R = Rr(a)aR ⊆ Rr(a), and so a(1 − a)R ⊆ Rr(a). In view of [217, Corollary 2.23], (a − a2 )R ∝ r(a). Write a = aca. Then R(1 − ac)R = ℓ(a)R, and so (a− a2 )R ⊆ Ra(1 − a)R ⊆ RaR(1 − a)R = Raℓ(a)R ⊆ ℓ(a)R = R(1 − ac)R. By [217, Corollary 2.23] again, (a − a2 )R ∝ (1 − ac)R. That is, (a − a2 )R ∝ R/aR. Clearly, (a − a2 )R ∈ F P (I). As in the proof of Theorem 14.4.1, r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R.

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Therefore r(a) ∼ = R/aR, and so a ∈ 1 + I is unit-regular. (2) ⇒ (1) Let
a ∈ 1 + I satisfy Rr(a) = ℓ(a)RT= R(1 − a)R and i EndR (aR) = ∞. Then RaR(1 − a)R = RaR R(1 − a)R = T Rr(a) RaR = Rr(a)aR. Likewise, we have RaR(1 − a)R = Raℓ(a)R. By hypothesis, a ∈ 1 + I is unit-regular. According to Theorem 15.2.10, we establish the result.  Example 15.2.12. Let F be a field, let Rn = Mn (F ) for n = 1, 2, · · · , ∞ Q and let S be the F -subalgebra of Rn (the product of all Rn ) generated by 1 and

∞ L

n=1

n=1

Rn (the direct sum of all Rn ) . Set R = F ⊕ S. Choose

a = (1F , 0) and b = (0, 1S ). Then a, b ∈ R are idempotents, and satisfy Rr(a) = ℓ(a)R = R(1 − a)R and Rr(b) = ℓ(b)R = R(1 − b)R, respectively. Since every primitive factor ring of S is artinian, one
easily checks that R is a separative regular ring. Clearly, i EndR (aR) = i(F ) = 1, while
i EndR (bR) = i(S) = ∞. This shows that Theorem 15.2.10 is a nontrivial generalization of Theorem 13.4.14.  Theorem 15.2.13. Let I be an ideal of a regular ring R. Then the following are equivalent: (1) I is separative. (2) Each a ∈ 1 + I satisfying R(1 − a)R = Rr(a) = ℓ(a)R,
2r(a) ∼ = r(a) ⊕ R/aR ∼ = 2 R/aR

is unit-regular. (3) Each a ∈ 1 + I satisfying

is unit-regular.

Rr(a)aR = Raℓ(a)R = RaR(1 − a)R,
2r(a) ∼ = r(a) ⊕ R/aR ∼ = 2 R/aR

Proof. (1) ⇒ (3) is obvious from Lemma 13.4.10. (3) ⇒ (2) is analogous to Corollary 15.2.11. (2) ⇒ (1) Given A⊕ C ∼ = B ⊕ C .⊕ R, C . A, B with A, B, C ∈ F P (I), we write R = A1 ⊕ C1 ⊕ D = A2 ⊕ C2 ⊕ D, where A1 ∼ = A, C1 ∼ = C2 ∼ =C ∼ and A2 = B. Let a ∈ R induce an endomorphism of RR , which is zero on A1 , an isomorphism from C1 onto C2 , and the identity on D. As in the proof of Theorem 13.4.14, R(1 − a)R = Rr(a) = ℓ(a)R with a ∈ 1 + I. As

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in the proof of Theorem 13.4.8, we get r(a) ⊕ (a − a2 )R ∼ = R/aR ⊕ (a − a2 )R,

where (a−a2 )R ∝ r(a), R/aR. Similarly to Corollary 13.4.9, we get 2r(a) ∼ =
∼ r(a) ⊕ R/aR = 2 R/aR . By hypothesis, a ∈ 1 + I is unit-regular. Hence, A∼  = r(a) ∼ = R/aR ∼ = B. Therefore I is separative. Corollary 15.2.14. Let I be an ideal of a regular ring R. Then the following are equivalent: (1) I is separative. (2) Each a ∈ 1 + I satisfying


R(1 − a)R = Rr(a) = ℓ(a)R, i EndR (aR)
= ∞, 2r(a) ∼ = r(a) ⊕ R/aR ∼ = 2 R/aR

is unit-regular. (3) Each a ∈ 1 + I satisfying


Rr(a)aR = Raℓ(a)R = RaR(1 − a)R, i EndR (aR) = ∞,
∼ ∼ 2r(a) = r(a) ⊕ R/aR = 2 R/aR

is unit-regular.

Proof. (1) ⇒ (3) and (3) ⇒ (2) are both clear.
(2) ⇒ (1) Let J = {x ∈ R | i EndR (xR) < ∞}. Analogously to Theorem 15.2.10, we show that I is separative by Theorem 15.2.13.  Analogously to Corollary 15.2.14, it follows by Corollary 15.2.9 that a regular ring R in which 2 is invertible is separative if and
only if each a ∈ R satisfying R(1 − a
2 )R = Rr(a) = ℓ(a)R, i EndR (aR) = ∞, 2r(a) ∼ = r(a) ⊕ R/aR ∼ 2 R/aR is unit-regular if and only if each
a ∈ R satisfying = T T T R(1 − a2 )R RaR = Rr(a) ℓ(a)R RaR, i EndR (aR) = ∞, 2r(a) ∼ =
r(a) ⊕ R/aR ∼ = 2 R/aR is unit-regular.

Corollary 15.2.15. Let I be an ideal of a regular ring R. Then the following are equivalent:

(1) I is separative.

∼ ∼ (2) For
any idempotents e, f
∈ I, 2 eR = eR⊕f R = 2 f R , i (1−e)R(1− e) = i (1 − f )R(1 − f ) = ∞ =⇒ eR ∼ = f R. Proof. (1) ⇒ (2) is obvious.
(2) ⇒ (1)
Let a ∈ 1 + I such that 2r(a) ∼ = r(a) ⊕ R/aR ∼ = 2 R/aR and i EndR (aR) = ∞. Then we can find some x ∈ 1+I such that a = axa and

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x = xax. Thus, we see that (1−xa)R⊕(1−xa)R ∼ = (1−xa)R⊕(1−ax)R ∼ = (1 − ax)R ⊕ (1 − ax)R. In addition, we have

i (1 − (1 − ax))R(1 − (1 − ax)) = i EndR (aR) = ∞.

m Assume that = 0 for some r ∈ R, then
mi(xaRxa) = n < ∞. If (axrax)
n xa(xra)xa = 0, and so xa(xra)xa = 0. Thus, (xra)n = 0. This implies that (axrax)n = a(xra)n x = 0, and so i(axRax) < ∞. This gives a
contradiction. Hence, i (1 − (1 − xa))R(1 − (1 − xa)) = ∞. By hypothesis, we get ϕ : (1 − ax)R ∼ = (1 − xa)R. Clearly, φ : axR = aR ∼ = xaR given by φ(ar) = xar for any r ∈ R. Define u ∈ EndR (R) so that u restricts to φ : aR → xaR and u restricts to ϕ : (1 − ax)R → (1 − xa)R. Then u(1) ∈ U (R). In addition, a = au(1)a, i.e., a ∈ R is unit-regular. Therefore we complete the proof by Corollary 15.2.14. 

Similarly, we derive that a regular ring R is unit-regular if and only
if ∼ for any idempotents e, f ∈ R, (1 − e)R (1 − f )R, i (1 − e)R(1 − e) = =
∼ i (1 − f )R(1 − f ) = ∞ =⇒ eR = f R. 15.3

Cleanness in Exchange Rings

In [334, Proposition 1.8], Nicholson proved that every exchange ring with all idempotents central is clean. By an argument of Yu [424, Example 3.10], there is an exchange ring R with artinian primitive factors, while that has noncentral idempotents. Now we investigate cleanness for exchange rings with artinian primitive factors. We start by recording the following elementary property. Lemma 15.3.1. Let D be a division ring, and let n ∈ N. Then for any γ ∈ Mn (D), there exists α = α2 ∈ Mn (D) such that γ − α ∈ GLn (D). That is, Mn (D) is clean. Proof. For any 0 6= γ ∈ D, γ = 0 + γ, where 0 = 02 ∈ D, γ ∈ D∗ . Obviously, 0 = 1 + (−1), where 1 = 12 ∈ D, −1 ∈ D∗ . Thus, the result holds for n = 1. In other words, D is clean. As in the proof of Proposition 13.1.17, we complete the proof by induction.  Theorem 15.3.2. Every exchange ring with artinian primitive factors is clean.

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Proof. Let R be an exchange ring with artinian primitive factors. Assume that R is not a clean ring. Then we can find some x ∈ R such that x 6= e+u for any e = e2 ∈ R and u ∈ U (R). Let Ω be the set of all the two-sided ideals A of R such that x + A ∈ R/A cannot be written as the sum of an idempotent and a unit in R/A. Obviously, Ω 6= ∅. Given any ascending chain A1 ⊆ A2 ⊆ · · · ⊆ Ak ⊆ · · · in Ω, we set ∞ S M = An . Then M is an ideal of R. If M is not in Ω, then there n=1

exist g + M = (g + M )2 ∈ R/M, w + M ∈ U (R/M ) such that x + M = (g + M ) + (w + M ). Hence x − (g + w), g − g 2 , ws − 1, sw − 1 ∈ M for some s ∈ R, and so we can find positive integers ni (1 ≤ i ≤ 4) such that x − (g + w) ∈ An1 , g − g 2 ∈ An2 , ws − 1 ∈ An3 , sw − 1 ∈ An4 . Let p = max{n1 , n2 , n3 , n4 }. Then we see that x−(g+w), g−g 2 , ws−1, sw−1 ∈ Ap . This contradicts the choice of Ap , so M ∈ Ω. Thus, Ω is inductive. By using Zorn’s Lemma, we have an ideal Q of R such that Q is maximal in Ω. Let S = R/Q. The maximality of Q ∈ Ω implies that S is indecomposable as a ring. If J(S) 6= 0, we may assume that J(S) = N/Q with Q $ N . By the maximality of Q, x is the sum of an idempotent and a unit modulo N.
From S/J(S) ∼ = R/N , we have (x + Q) + J(S) = (f + Q) + J(S) + (v +

2 Q) + J(S) with (f +
Q) + J(S) = (f + Q) + J(S) ∈ S/J(S), (v + Q) + J(S) ∈ U S/J(S) . Since S/J(S) is an exchange ring, we may assume that f + Q = (f + Q)2 ∈ R/Q. On the other hand, units lift modulo the Jacobson radical of S. We may assume that v + Q ∈ U (S). Thus
x + Q = (f + Q) + (v + Q) + (r + Q) for some r + Q ∈ J(S). Clearly, (v + Q) + (r + Q) ∈ U (S), and it yields a contradiction. This implies that J(S) = 0, so S is an indecomposable ring with J(S) = 0. Since R is an exchange ring with artinian primitive factors, by virtue of [424, Lemma 3.7], S is simple artinian. That is, S ∼ = Mm (D) for some m ∈ N. In view of Lemma 15.3.1, S is clean. Thus, x can be seen as the sum of an idempotent and a unit modulo Q. This gives a contradiction, whence R is a clean ring.  Corollary 15.3.3. Every exchange ring of bounded index is clean. Proof. Let R be an exchange ring of bounded index. As in the proof of Corollary 1.3.14, R is an exchange ring with artinian primitive factors, to which we can apply Theorem 15.3.2.  Corollary 15.3.4. Every exchange ring with all idempotents central is

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clean. Proof. Let R be an exchange ring with all idempotents central. As in the proof of Corollary 1.3.15, R/J(R) is of bounded index; hence, it is clean by Corollary 15.3.3. Since R is an exchange ring, every idempotent lifts modulo J(R). Also we see that every unit lifts modulo J(R). We conclude that R is clean.  In [424, Theorem 3.6], Yu proved that R is strongly π-regular if R is an exchange ring with artinian primitive factors and R/J(R) is homomorphically semiprimitive. Recall that a subset I of a ring R is right T -nilpotent in case for every sequence a1 , a2 , · · · in I there is an n such that an · · · a1 = 0. Now we extend Yu’s result as follows. Theorem 15.3.5. Let R be an exchange ring with artinian primitive factors. If the Jacobson radical of any homomorphic image of R is right T -nilpotent or locally nilpotent, then R is a strongly π-regular ring. Proof. Assume that R is not a strongly π-regular ring. Then there exists some x ∈ R such that xn 6∈ xn+1 R for any n ∈ N. Set Ω = {A E R | xn 6= xn+1 y (mod A) for any n ∈ N and any y ∈ R}.

Clearly, Ω is a nonempty inductive set. Just as in Theorem 15.3.2 we get an ideal Q which is maximal in Ω. Let S = R/Q. We then see that S is indecomposable as a ring. Assume that J(R/Q) is right T -nilpotent. If Q is a prime ideal of R, then R/Q is a prime ring. Assume that 0 6= a ∈ J(R/Q). Then there exist x1 , x2 , · · · , xn , · · · ∈ R/Q such that ax1 a 6= 0, ax2 ax1 a 6= 0, · · · , axn a · · · ax1 a 6= 0, · · · . Since J(R/Q) is right T -nilpotent, we obtain a contradiction. Thus J(R/Q) = 0. Using [424, Lemma 3.7], we see that R/Q is simple artinian, whence it is strongly π-regular. This contradicts the choice of Q. So we may assume that Q is not prime. Consequently, we have two ideals K and L of R such that Q $ K, L and KL ⊆ Q. By the maximality of Q, there are positive integers k, l and s, t ∈ R such that xk = xk+l+1 s (mod K) and xl = xk+l+1 t (mod L). It is easy to verify that (xk − xk+l+1 s)(xl − xk+l+1 t) ∈ KL ⊆ Q. Hence we have some z ∈ R such that xk+l = xk+l+1 z (mod Q), a contradiction. Assume that J(R/Q) is locally nilpotent. If J(S) 6= 0, by the maximality of Q, there exists a positive integer m such that (x + Q)m =

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(x + Q)m+1 w (mod J(S)) for some w ∈ R. Write a for x + Q. Then j := am − am+1 w satisfies j ∈ J(S). Since J(S) is locally nilpotent, there is a positive integer i such that (SjS)i = 0. One easily computes that, for k ≥ 1, am − am+k wk = j + ajw + a2 jw2 + · · · + ak−1 jwk−1 ∈ SjS.

Consequently (am − ami+1 v)i = 0 for some v ∈ R. Therefore we claim that xmi = xmi+1 q for some q ∈ R, a contradiction. Thus J(S) = 0. By [424, Lemma 3.7], we have that R/Q is simple artinian. This contradicts the choice of Q. Therefore we conclude that R is a strongly π-regular ring.  

 F F . Then R shows 0 F that the generalization of [424, Theorem 3.6] to Theorem 15.3.5 is nontrivial.  Example 15.3.6. Let F be a field, and let R =

Corollary 15.3.7. Let R be an commutative exchange ring. Then the following are equivalent: (1) R is a strongly π-regular ring. (2) The Jacobson radical of any homomorphic image of R is nil.



Recall that a right R-module M has the exchange property, if for every L module A and any two decompositions A = M ′ ⊕ N = Ai , where M ′ ∼ = i∈I
L M , there are submodules A′i ⊆ Ai such that M = M ′ ⊕ A′i . Following i∈I

Stock [372], we say a ring R is a right P -exchange ring if every projective right R-module has the exchange property. Clearly, every regular ring is a right P -exchange ring. But the converse is not true (cf. [372, Example 4.6]). Now we investigate right P -exchange rings with artinian primitive factors. Corollary 15.3.8. Let R be a right P -exchange ring with artinian primitive factors. Then R is strongly π-regular. Proof. Let Q be an ideal of R. Since R is a right P -exchange ring, so is R/Q by [372, Proposition 4.1]. Further, J(R/Q) is right T -nilpotent. By virtue of Theorem 15.3.5, the result follows. 

Let F be a field, let Rn = Mn (F ) for n = 1, 2, · · · , and let R be the F Q L subalgebra of Rn generated by 1 and Rn . By an argument of Goodearl

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(cf. [217]), R is a right P -exchange ring with artinian primitive factors, while the index of R is ∞. So Corollary 15.3.8 is a nontrivial generalization of [372, Theorem 4.11]. Yu [424, Example 3.10] gave an example of a right quasi-duo and right P -exchange ring with noncentral idempotents. So the generalization from [372, Theorem 4.8] to the following result, which was also proven by Yu (cf. [419, Theorem 4.4]), is nontrivial. Proposition 15.3.9. Let R be a right quasi-duo ring. Then the following are equivalent :

(1) R is a right P -exchange ring. (2) R/J(R) is regular and J(R) is right T -nilpotent.

Proof. (2) ⇒ (1) follows from [372, Theorem 4.2]. (1) ⇒ (2) In view of [372, Proposition 4.1], J(R) is right T -nilpotent. Since R is a right quasi-duo and right P -exchange ring, then so is R/J(R) from [372, Proposition 4.1]. In light of [419, Corollary 2.4], R/J(R) is reduced, and so it has artinian primitive factors from [424, Theorem 3.11]. By virtue of Corollary 15.3.8, R/J(R) is strongly π-regular. According to [424, Theorem 3.8], we conclude that R/J(R) is regular.  Corollary 15.3.10. The following are equivalent:

(1) R is abelian regular. (2) R is a reduced and right P -exchange ring.

Proof. (1) ⇒ (2) is trivial. (2) ⇒ (1) Since R is reduced, it is abelian. According to [419, Proposition 4.3], R is a right quasi-duo ring. By virtue of Proposition 15.3.9, J(R) is right T -nilpotent, and so J(R) = 0. By using Proposition 15.3.9 again, R is regular, as required.  Lemma 15.3.11. Let D be a division ring, and let A ∈ Mn (D)(n ≥ 2). Then there exists some U ∈ GLn (D) such that A ± U ∈ GLn (D).

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Proof. Since D is a division ring, there exist U, V ∈ GLn (D) such that   1D   1D     ..   .     U AV =  . 1D      0    ..   .  0 n×n

Assume that n is an even number. Let   1D 1D  1 0   D   ..   1D .     ..  . W = . 0       1D 0     .. ..   . . 1D 0 n×n Then W ∈ GLn (D). Clearly,  0  −1 1  D D   −1D    U AV − W =      

On the other hand,



−1D .. ..

. . 1D −1D 0 .. .. . . −1D

2 1 1  D D   1D    U AV + W =      

0 1D

.. ..

. . 1D 1D 0 .. .. . . 1D 0



       ∈ GLn (D).                    

n×n

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

 2 1D 1 1   D D    1D 1D     ..   . 1   D U AV + W =  .  ..     1D .     .. ..   . . 1D 1D n×n

As n is an even number, one easily checks that U AV + W ∈ GLn (D). Assume that n is anodd number. Let  1D 1D  −1 0   D    ..   1D .     .   .. 0 W = .      1D 0     .. ..   . . 1D 0

Then



0 1 1  D D   −1D    U AV − W =      

On the other hand,



n×n

−1D .. ..

. . 1D −1D 0 .. .. . .

2  −1 1  D D   1D    U AV + W =      

−1D

0



       ∈ GLn (D).      

1D .. ..

. . 1D 1D 0 .. .. . . 1D 0

             

n×n

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

 2 1D  −1 1   D D    1D 1D     ..   1D .  U AV + W =  .   ..     1D .     .. ..   . . 1D 1D n×n

As n is an odd number, one easily checks that U AV + W ∈ GLn (D). In any case, we can find some W ∈ GLn (D) such that U AV ± W ∈ GLn (D), and so A ± U −1 W V −1 ∈ GLn (D).  Theorem 15.3.12. Let R be an exchange ring with artinian primitive factors. Then for any a ∈ R, there exists an idempotent e ∈ R and a u ∈ U (R) such that a ± eu ∈ U (R). Proof. Assume that there exists some a ∈ R such that a + eu 6∈ U (R) or a − eu 6∈ U (R) for any idempotent e ∈ R and any u ∈ U (R). Let Ω = {I E R | a + eu 6∈ U (R/I) or a − eu 6∈ U (R/I) for any idempotent e ∈ R/I and any u ∈ U (R/I)}. Similarly to Theorem 15.3.2, we see that Ω is a nonempty inductive set. By using Zorn’s Lemma, there exists an ideal Q which is maximal in Ω. Since every idempotent lifts modulo J(R/Q), analogously to Theorem 15.3.2, we prove that R/Q/J(R/Q) is an indecomposable ring. In view of [424, Lemma 3.7], R/Q/J(R/Q) is a simple artinian ring, and so R/Q/J(R/Q) ∼ = Mn (D) for a natural number n. If n = 1, then R/Q/J(R/Q) is a division ring, so we can find
an idempotent e ∈ R/Q/J(R/Q) such that a ± e ∈ U R/Q/J(R/Q) . If
a ± v ∈ U R/Q/J(R/Q) for some n ≥ 2, it follows by Lemma 15.3.11 that
v ∈ U R/Q/J(R/Q) . Since every unit and every idempotent lift modulo the Jacobson radical, we get a contradiction. As a result, we can find an idempotent e ∈ R and some u ∈ U (R) such that a ± eu ∈ U (R).  Corollary 15.3.13. Let R be an exchange ring with artinian primitive factors. Then for any a ∈ R, 2a is the sum of two units. Proof. For any a ∈ R, we can find an idempotent e ∈ R and some u ∈ U (R) such that a±eu ∈ U (R) by Theorem 15.3.12. Let v = a+eu and w = a−eu. Then 2a = (a + eu) + (a − eu) = v + w with v, w ∈ U (R), as required. 

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Let R be an exchange ring of bounded index. Then for any A ∈ Mn (R), 2A is the sum of two invertible matrices. This result also holds for any exchange ring with all idempotents central. Corollary 15.3.14. Let R be a strongly π-regular ring. Then for any a ∈ R, 2a is the sum of two units. Let a ∈ R, and let S be the subring generated by the set ∞ P {1R , a, a2 , · · · , an , · · · }. That is, S = Zai . Then S is commutative. Proof.

i=0

Let f : S → R be the inclusion. In view of [57, Corollary 1.10], we can find a commutative strongly π-regular subring T of R such that S ⊆ T ⊆ R. As T is a commutative strongly π-regular ring, it is an exchange ring with all idempotents central. By virtue of Corollary 15.3.13, we can find u, v ∈ U (T ) such that 2a = u + v. Therefore we have u, v ∈ U (R) such that 2a = u + v, as desired.  As an immediate consequence of Corollary 15.3.14, we derive the known result: every element in a strongly π-regular ring is generated by two units if 2 is invertible. Let A be an artinian right R-module, and let σ ∈ EndR (A). Then we claim that there exist α, β ∈ AutR (A) such that 2σ = α + β.
By [68, Corollary 6], EndR (A) is semilocal; hence, EndR (A)/J EndR (A) is a right artinian ring, and we are done by Corollary 15.3.14. 15.4

Exchange P B-Ideals

We now investigate the analogue of cleanness as it relates to exchange P Bideals. Lemma 15.4.1. Let I be a P B-ideal of a ring R. Then for any idempotent e ∈ I, M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = eR ∼ = A2 implies that there exist ideals I, J and right R-morphisms ϕ : B1 → B2 , ψ : B2 → B1 such that I♮J, ϕI ψI = 1B1 /B1 I and ψJ ϕJ = 1B2 /B2 J . Proof. Let e ∈ I be an idempotent. Given any right R-module decompositions M = A1 ⊕ B1 = A2 ⊕ B2 with A1 ∼ = eR ∼ = A2 , let ρ′i : M → ′ Ai , ρi : M → Bi be the projections. Let τi : Ai → M, τi : Bi → M be the injections. Let ιi : Ai ∼ = eR, πi = ιi ρ′i and σi = τi′ ι−1 i . Then 1eR = π2 σ2 = π2 (σ1 π1 + τ1 ρ1 )σ2 = (π2 σ1 )(π1 σ2 ) + (π2 τ1 ρ1 σ2 ) = ax + b.

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Identifying eRe with EndR (eR). In view of Lemma 9.4.1, eRe is a P B −1 corner. Thus, we have y ∈ eRe and u ∈ eRe p such that u = π2 (σ1 + τ1 ρ1 σ2 y). Similarly to Lemma 9.2.1, there exists some v ∈ eRe such that R(e−uv)R(e−vu)R and R(u−uvu)R
are nilpotent. Let p = uv and q = vu. Clearly, u ≡ uvu mod R(u − uvu)R . Hence, p, q ∈ R/R(u − uvu)R are idempotents. Letting n be the nilpotency index of p − p2 , we get   n   n X
n X
k
n n 0 = p−p2 = pn−k −p2 = pn −pn+1 (−1)k−1 pk−1 . k k k=0 k=1  
n Pn n k−1 Let p′ = pn (−1) pk−1 . It is easy to verify that p′ ∈ k=1 k
eRe is an idempotent. In addition, p ≡ p′ mod R(u − uvu)R . Likewise,
we have an idempotent q ′ ∈ eRe such that q ≡ q ′ mod R(u − uvu)R . Let α = σ1 + τ1 ρ1 σ2 y, D1 = ker(p′ π1 ), D2 = ker(q ′ π2 ) and E = αp′ (eR). Analogously to Theorem 8.2.1, we get M = E ⊕ D1 = E ⊕ D2 . Let C1 = σ1 (e − p′ )(eR) and C2 = σ2 (e − q ′ )(eR). Then D1 = C1 ⊕ B1 and D2 = C2 ⊕ B2 . Therefore M = E ⊕ C1 ⊕ B1 = E ⊕ C2 ⊕ B2 .

Let I = R(e − p′ )R and J = R(e − q ′ )R. It
is easy to verify that IJ ⊆ m R(e − p)R(e − q)R + R(u − uvu)R. Hence, IJ = 0 for some m ∈ N, i.e., I♮J. Furthermore, we see that C1 I = C1 and C2 J = C2 . Clearly, we have an isomorphism σ : C1 ⊕ B1 ∼ = C2 ⊕ B2 . Let η1 : B1 → B1 ⊕ C2 be the injection, θ1 : C2 ⊕ B2 → B2 be the projection, and let η2 : B2 → C2 ⊕ B2 be the injection, θ2 : C1 ⊕ B1 → B1 be the projection. Choose ϕ = θ1 ση1 and ψ = θ2 σ −1 η2 . Then ϕ : B1 → B2 , ψ : B2 → B1 . Clearly, η1

σ

η2

σ−1

θ

ϕ : B1 → C1 ⊕ B1 → C2 ⊕ B2 →1 B2 , θ

2 ψ : B2 → C2 ⊕ B2 → C1 ⊕ B1 → B1 .

One easily checks that

ϕI ψI = θ1 ση1 I θ2 σ −1 η2 I = (θ1 )I (σ)I (η1 )I (θ2 )I (σ −1 )I (η2 )I .


For any c ∈ C1 , b1 ∈ B1 , we see that
c ∈ C1 = C1 I ⊆ C1 ⊕ B1
I; hence, (η1 )I (θ2
)I c + b1 + (C ⊕ B1 )I
= (η1 )I (θ2 )I b1 + (C1 ⊕ B1 )I = b1 + C1 ⊕ B1 I = c + b1 + C1 ⊕ B1 I. This implies that (η1 )I (θ2 )I =

, whence ϕI ψI = 1B /B I . Likewise, ψJ ϕJ = 1B /B J , 1 1 1 2 2 C1 ⊕B1 / C1 ⊕B1 I

as required.



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Theorem 15.4.2. Let I be an exchange P B-ideal of a ring R. Then for any regular a ∈ I, there exists an idempotent e ∈ R and a u ∈ Rp−1 such that a = e + u. Proof. Let a ∈ I be regular. Then we have some x ∈ I such that a = axa and x = xax, and so R = aR ⊕ (1 − ax)R = xR ⊕ r(a). Clearly, aR has the finite exchange property, and so there exist right R-modules X1 , Y1 such that R = aR ⊕ X1 ⊕ Y1 with X1 ⊆ r(a) and Y1 ⊆ xR. It is easy to verify that r(a) = X1 ⊕ X2 , where X2 = r(a) ∩ (aR ⊕ Y1 ). Likewise, we have a right R-module Y2 such that xR = Y1 ⊕ Y2 . Thus R = X1 ⊕ X2 ⊕ Y1 ⊕ Y2 , and so R = aR ⊕ X1 ⊕ Y1 = xR ⊕ X1 ⊕ X2 with aR ∼ = xR. Since aR ⊕ Y1 ∼ = xR ⊕ X2 , in view of Lemma 15.4.1, there exist two ideals I1 and I2 and two right R-morphisms ϕ : Y1 → X2 and ψ : X2 → Y1 such that I1 ♮I2 , ϕI1 ψI1 = 1X2 /X2 I1 and ψI2 ϕI2 = 1Y1 /Y1 I2 . Let h : R = X1 ⊕ X2 ⊕ Y1 ⊕ Y2 → X1 ⊕ X2 ⊕ Y1 ⊕ Y2 = R given by h(x1 + x2 + y1 + y2 ) = x1 + ψ(x2 ) + y1 for any x1 ∈ X1 , x2 ∈ X2 , y1 ∈ Y1 , y2 ∈ Y2 . Let v : R = X1 ⊕ X2 ⊕ Y1 ⊕ Y2 → X1 ⊕ X2 ⊕ Y1 ⊕ Y2 = R given by v(x1 + y1 + x2 + y2 ) = x1 + ϕ(y1 ) + ψ(x2 ) for any x1 ∈ X1 , x2 ∈ X2 , y1 ∈ Y1 , y2 ∈ Y2 . For any x1 ∈ X1 , x2 ∈ X2 , y1 ∈ Y1 , y2 ∈ Y2 , we have
(hv)I1 (hv)I1 x1 + x2 + y1 + y2
= hI1 vI1 hI1 x1 + ϕ(y1 ) + ψ(x2 )
= hI1 vI1 x1 + ψϕ(y1 ) + ψ(x2 )
= hI1 x1 + ϕ(y1 ) + x2 = x1 + ψϕ(y1 ) + ψ(x2 )
= (hv)I1 x1 + x2 + y1 + y2 .

Hence (hv)2I1 = (hv)I1 . Likewise, we verify that (hv)2I2 = (hv)I2 . Let T T
2 f = hv(1). Then f 2 − f ∈ I1 I2 . As I1 I2 ⊆ I1 I2 , we see that
n f − f 2 = 0 for some n ∈ N. That is, f n (1 − f )n = 0. Working in the polynomial ring Z[x], we have  2n 
X 2n 1 = x + (1 − x) = x2n−k (1 − x)k . k k=0






2n x2n−k (1 − x)k . Then g(x) ≡ 0 mod xn and g(x) ≡ k=0
k
2 n n 1 mod (1−x)n . Hence g(x) ≡ g(x) mod x (1−x) . Put e = g(f ). Then
T 2 n n 2 e ≡ e mod f (1 − f
) , i.e., e = e . Since f (1 − f ) ∈ I1 I2 , we have T e ≡ f 2n mod I1
I2 . Furthermore, we see that f k = f k+1 + (1 − f )f k = T T
k+1 f mod I1 I2 for all k. Therefore e ≡ f mod I1 I2 . Let g(x) =

n P

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Let a − hv = u. Construct a right R-morphism φ : R = im(a) ⊕ X1 ⊕ Y1 → R; z + x1 + y1 7→ xz + x1 + ϕ(y1 ), ∀z ∈ im(a), x1 ∈ X1 , y1 ∈ Y1 . Then


φI1 (a − hv)I1 x1 + y1 + x2 + y2
= φI1 a(y1 + y2 ) − kl(x1 + y1 ) − ψI1 (x2 )
= φI1 a(y1 + y2 ) − x1 − ψI1 ϕI1 (y1 ) − ψI1 (x2 ) = xa(y1 + y2 ) − x1 − ϕI1 ψI1 ϕI1 (y1 ) − ϕI1 ψI1 (x2 ) = y1 + y2 − x1 − ϕI1 (y1 ) − x2 , ∀x1 ∈ X1 , y1 ∈ Y1 , x2 ∈ X2 , y2 ∈ Y2 .

Furthermore, we get
φI1 (a − hv)I1 φI1 (a − hv)I1 x1 + y1 + x2 + y2
= φI1 (a − hv)I1 y1 + y2 − x1 − ϕI1 (y1 ) − x2 = y1 + y2 + x1 − ϕI1 (y1 ) + ϕI1 (y1 ) + x2 = x1 + x2 + y1 + y2 , ∀x1 ∈ X1 , y1 ∈ Y1 , x2 ∈ X2 , y2 ∈ Y2 . This implies
that φI1 (a − hv)I1 φI1 (a − hv)I1 = 1R/I1 , and so (a − hv)I1 ∈ EndR R/I1 is left invertible. Thus, we have an element w ∈ R such that
1 + I1 = (w + I1 ) u + I1 . Hence, 1 − wu ∈ I1 , i.e., wu ≡ 1 (mod I1 ).
Given any t ∈ aR, x1 ∈ X1 , y1 ∈ Y1 , we have t ∈ aR = a Y1 ⊕ Y2 . So we can find some y1′ ∈ Y1 and y2′ ∈ Y2 such that t = a(y1′ + y2′ ). Choose x′1 = −x1 and x′2 = −ϕ(y1 + y1′ ). It is easy to verify that
uI2 x′1 + x′2 + y1′ + y2′
= a(y1′ + y2′ ) − x′1 + y1′ + ψ(x′2 )
= t − − x1 + y1′ − y1 − y1′ = t + x1 + y1 . This means that uI2 : R/I2 → R/I2 is a right R-epimorphism. Hence, 1 + I2 = uI2 (v + I2 ) =
uv + I2 for an element v ∈ R. Thus, 1 − uv ∈ I2 , i.e., uv ≡
1 mod I2 . It follows from I1 ♮I2 that u ∈ Rp−1 . As e ≡ T T f mod I1 I2 , we have some r ∈ I1 I2 such that f = e + r. Hence,
2
2m T
2 a = e + u + r. Clearly, RrR ⊆ I1 I2 ⊆ I1 I2 , and then RrR =0 for some m ∈ N. One easily checks that u′ := u + r ∈ Rp−1 and a = e + u′ , as required.  Let R be an exchange P B-ring, and let A ∈ Mn (R) be regular. As a consequence of Theorem 9.2.5 and Theorem 15.4.2, there exists an idempotent E ∈ Mn (R) and an element U ∈ Mn (R)−1 p such that A = E + U .

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Corollary 15.4.3. Let R be a ring, and let a ∈ R be regular. If RaR is an exchange P B-ideal, then there exists an idempotent e ∈ R and a u ∈ Rp−1 such that a = e + u. Proof. As a ∈ RaR is regular, we obtain the result from Theorem 15.4.2. Corollary 15.4.4. Let R be an exchange ring, and let a ∈ R be regular.
If EndR aR is a P B-ring, then there exists an idempotent e ∈ R and a u ∈ Rp−1 such that a = e + u. Proof. Since a ∈ R is regular, there
is some x ∈ R such that a = axa. Let e = ax. Then eRe ∼ = EndR aR ; hence, eRe is an exchange P B-ring. According
to Theorem 15.4.2, there exists an idempotent f ∈ eRe and a −1 u ∈ eRe p such that ae = f + u. Hence, a = ae + a(1 − e) = (f − 1 + e) + u + 1 − e + a(1 − e).
−1 Clearly, f − 1 + e = (f − 1 + e)2 ∈ R. As u ∈ eRe p , we can find an element v ∈ eRe such that (e
− uv)♮(e − vu). One easily checks that (e − uv) + u + 1 − e + a(1 − e) (v + 1 − e) = 1 + a(1 − e); hence,


(e − uv) 1 − a(1 − e) + u + 1 − e + a(1 − e) (v + 1 − e) 1 − a(1 − e) = 1. This implies that




1 − u + 1 − e + a(1 − e) (v + 1 − e) 1 − a(1 − e) = (e − uv) 1 − a(1 − e) . Likewise, we get




1 − 1 − va(1 − e) (v + 1 − e) u + 1 − e + a(1 − e) = 1 − va(1 − e) (e − vu).

Thus,


1 − (u + 1 − e + a(1 − e))(v + 1 − e)(1 − a(1 − e))
♮ 1 − (1 − va(1 − e))(v + 1 − e)(u + 1 − e + a(1 − e)) ,

and so u + 1 − e + a(1 − e) ∈ Rp−1 . Therefore the result follows.



Proposition 15.4.5. Let I be an exchange P B-ideal of a ring R. Then for any regular a ∈ I, there exist u, v ∈ Rp−1 such that 2a = u + v. Proof. Let a ∈ I be regular. Construct h, v, e, u, φ, I1 and I2 as in the proof of Theorem 15.4.2. Then a − e ∈ Rq−1 . Further, we see that
φI1 (a + hv)I1 φI1 (a + hv)I1 x1 + y1 + x2 + y
2 = φI1 (a + hv)I1 y1 + y2 + x1 + ϕI (y1 ) + x2 = x1 + x2 + y1 + y2 , ∀x1 ∈ X1 , y1 ∈ Y1 , x2 ∈ X2 , y2 ∈ Y2 .

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This implies
that φI1 (a + hv)I1 φI1 (a + hv)I1 = 1R/I1 , and so (a + hv)I1 ∈ EndR R/I1 is left invertible. Set u′ = a +
hv. Given any t ∈ aR, x1 ∈ X1 , y1 ∈ Y1 , we have t ∈ aR = a Y1 ⊕ Y2 . So we can find some y1′ ∈ Y1 and y2′ ∈ Y2 such that t = a(y1′ + y2′ ). Choose x′1 = x1 and x′2 = ϕ(y1 − y1′ ). It is easy to verify that
u′I2 x′1 + x′2 + y1′ + y2′
= a(y1′ + y2′ ) + x′1 + y1′ + ψ(x′2 )
= t + x1 + y1′ + y1 − y1′ = t + x1 + y1 .
This means that u′I2 ∈ EndR R/I2 is a right R-epimorphism. Analogously to Theorem 15.4.2, we see that a+ e ∈ Rp−1 . Therefore 2a = (a− e)+ (a+ e) is the sum of two elements in Rp−1 .  Let R be an exchange QB-ring, and let A ∈ Mn (R) be regular. Analogously, we deduce that A is the sum of an idempotent E ∈ Mn (R) and some U ∈ Mn (R)−1 such that A = E + U , and that 2A is the sum of two q ∗ matrices in Mn (R)−1 q . Let A be an extremally rich C -algebra of real rank zero. Then we claim that every regular element in A is the sum of two extreme partial isometries. Using [16, Theorem 7.2], A is an exchange ring. In view of [25, Proposition 9.1], A is a QB-ring. Let a ∈ A be regular. Then there exist two extreme partial isometries u, v ∈ A such that 2a = u + v. Clearly, 2 ∈ U (A). Therefore a = 2−1 u + 2−1 v, and we are done.

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Chapter 16

Clean Properties, II

In this chapter, we investigate cleanness and its extension for endomorphism rings over a countably generated right vector space and some square matrices over several rings with stable range condition. In Section 16.1, we prove a theorem of Nicholson and Varadarajan. In Section 16.2, we investigate other decompositions of endomorphisms over a countably generated right vector space over a division ring. In Section 16.3, we study the cleanness of square matrices over several commutative rings with stable range condition. Finally, in Section 16.4, we extend the study of cleanness to strong cleanness.

16.1

Infinite-Dimensional Linear Transformations, I

The main purpose of this section is to prove that every linear transformation over a countably generated right vector space over a division ring is the sum of an idempotent and an automorphism. This was firstly proved by Nicholson and Varadarajan [338, Theorem]. In [431, Theorem], Zelinsky proved that every linear transformation over a division ring is the sum of two automorphisms with the exception of the identity transformation on a space of two elements. For countably linear transformations, we will prove Zelinsky’s result by means of the decompositions of matrices. Let V be a countably generated infinite-dimensional right vector space 541

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over a division ring D, and let {x1 , x2 , x3 , · · · } be a basis of V . We say that σ : V → V is a shift operator provided that σ(xi ) = xi+1 for all i ∈ N. Lemma 16.1.1. Let V be a countably generated infinite-dimensional right vector space over a division ring D, and let σ ∈ EndD (V ) be a shift operator. Then there exists α = α2 ∈ EndD (V ) such that σ − α ∈ AutD (V ). Proof. Let E= 

E L  and let A =  0  .. .

0 E L .. .

0 0 E .. .



0 −1D 0 1D



,L =



0 1D 0 0



,

 ··· ···  . Then A is a column-finite matrix over D. It ···  .. .

is easy to verify that E 2 = E, LE + EL = L and L2 = 0.    E2 0 0 ··· E 0 2   LE + EL E 0 ···   L E  L2 LE + EL E2 ··· A2 =  =0 L   0 L2 LE + EL · · ·    .. .. .. .. .. .. . . . . . .

Let

Hence we get  0 ··· 0 ···  = A. E ···  .. . . . .

 K 0 0 ···    0 K 0 ··· 0 1D   K= and B =  0 0 K · · ·  . 1D −1D   .. .. .. . . . . . . Then B is a column-finite matrix over D. In addition, we have   −1 K 0 0 ··· −1  0 K 0 ···   B −1 =  0 . 0 K −1 · · ·    .. .. .. . . . . . .   0 0 Let Q = . Then 1D 0   Q 0 0 ···  L Q 0 ···   A + B =  0 L Q ···.   .. .. .. . . . . . . 

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Let {x1 , x2 , x3 , · · · } be a basis of V . Define α, β ∈ EndD (V ) given by α(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )A, β(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )B. Then α = α2 ∈ EndD (V ), β ∈ AutD (V ). Furthermore, we see that (α + β)(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )(A + B)   Q 0 0 ···  L Q 0 ···   = (x1 , x2 , x3 , · · · )  0 L Q · · ·    .. .. .. . . . . . . = σ(x1 , x2 , x3 , · · · ),

and therefore σ = α + β, as required.



Lemma 16.1.2. Let V be a countably generated right vector space over a division ring D, and let γ ∈ EndD (V ) be such that V is spanned by {y, γ(y), γ 2 (y), · · · }. Then there exists α = α2 ∈ EndD (V ) such that γ − α ∈ AutD (V ). Proof. We may assume that V 6= 0. If γ n (y) 6∈ yD + γ(y)D + γ 2 (y)D + · · · + γ n−1 (y)D for all n ∈ N, then {y, γ(y), γ 2 (y), · · · } is a basis of V . Since γ is a shift operator with respect to this basis, by virtue of Lemma 16.1.1, we have α = α2 ∈ EndD (V ) such that γ − α ∈ AutD (V ). Assume that there exists some n ∈ N such that γ n (y) ∈ yD + γ(y)D + 2 γ (y)D + · · · + γ n−1 (y)D. If n is minimal with this property, then {y, γ(y), γ 2 (y), · · · , γ n−1 (y)} is a basis of V ; hence, V ∼ = nD, and so EndD (V ) ∼ = Mn (D). By Lemma 15.3.1, we have α = α2 ∈ EndD (V ) such that γ − α ∈ AutD (V ), and we are done.  Lemma 16.1.3. Let V be a countably generated right vector space over a division ring D, let γ ∈ EndD (V ), and let U be a γ-invariant vector subspace of V . Assume that V = U + K where K = yD + γ(y)D + γ 2(y)D + · · · for some y ∈ V . If there exists π = π 2 ∈ EndD (U ) such that γ|U − π ∈ AutD (U ), then there exists α = α2 ∈ EndD (V ) such that γ − α ∈ AutD (V ) and α|U = π. Proof. Clearly, there exists a vector subspace M such that V = M ⊕ U . As U is γ-invariant, we get γ : V /U → V /U defined by γ(v) = γ(v). Let θ : V → V be given by θ(m + u) = m for any m ∈ M, u ∈ U . Then

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θ(V ) = M and θ2 = θ. This induces a D-isomorphism θ0 : V /U → M . Let ζ = θ0 γθ0−1 ∈ EndD (M ). Then ζθ0 = θ0 γ, and so ζθ0 (v) = θ0 γ(v) for any v ∈ V . This implies that ζθ(v) = θγ(v). For any m ∈ M , ζ(m) = ζθ(m) = θγ(m). Hence θζ(m) = θ2 γ(m) = θγ(m). It follows that γ(m) − ζ(m) ∈ ker(θ) = U for all m ∈ M . By assumption,
we see that
{y, γ(y), γ 2 (y), · · · } spans V /U , and then {θ0 (y), θ0 γ(y)
, θ0 γ 2 (y)
, · · · } spans M . It is easy to verify that θ0 γ
n (y) =
ζ n θ0 (y) for all n ∈ N. This means that {θ0 (y), ζ θ0 (y) , ζ 2 θ0 (y) , · · · } spans M . By virtue of Lemma 16.1.2, there exists α1 = α21 ∈ EndD (M ) such that ζ − α1 ∈ AutD (M ). By hypothesis, there exists α2 = α22 ∈ EndD (U ) such that γ|U − α2 ∈ AutD (U ). Define α : V → V given by α(m + u) = α1 (m) + α2 (u) for any m ∈ M, u ∈ U . Then α = α2 ∈ EndD (V ). Let β1 = ζ − α1 and β2 = γ|U −
α2 . Let β : V → V be given by β(m + u) = β1 (m) + β2 (u) + γ(m) − ζ(m) for any m ∈ M, u ∈ U . One easily checks that (α + β)(m + u)
= α1 (m) + α2 (u) +
β1 (m) + β2 (u) +
γ(m) − ζ(m)
= α1 (m) + β1 (m) + α2 (u) +
β2 (u) + γ(m) − ζ(m) = ζ(m) + γ(u) + γ(m) − ζ(m) = γ(u) + γ(m) = γ(m + u). for any m ∈ M, u ∈ U . Thus, γ − α = β. If β(m + u) = 0, then β1 (m) = T −β2 (u) − γ(m) + ζ(m) ∈ M U = 0; hence, β1 (m) = 0. This implies that m = 0. Furthermore, we get −β2 (u) = 0, and so u = 0. Thus β is a D-monomorphism. Obviously, U ⊆ im(β). It suffices to show that M ⊆ im(β). If m ∈ M , we can find m0 ∈ M such that m = β1 (m0 ). Also we have u0 ∈ U such that −(γ(m0 ) − ζ(m0 )) = β2 (u0 ). It is easy to check that β(m0 + u0 )
= β1 (m0 ) + β2 (u0 ) + γ(m0 ) − ζ(m0 ) = m. That is, β is a D-epimorphism, and so γ − α = β ∈ AutD (V ). This yields the result. 

Theorem 16.1.4. Every endomorphism over a countably generated right vector space over a division ring is the sum of an idempotent endomorphism and an automorphism.

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Proof. Let V be a countably generated right vector space over a division ring D, and let γ ∈ EndD (V ). Set Ω = {(U, σ, π) | UD ⊆ V is γ − invariant, γ |U −σ = π, σ = σ 2 ∈ EndD (U ), π ∈ AutD (U )}. Clearly, Ω is not empty. Partially order Ω by writing (U, σ, π) ≤ (U ′ , σ ′ , π ′ ) if U ⊆ U ′ , σ = σ ′ |U ∞ S , π = π ′ |U . Given (U1 , σ1 , π1 ) ≤ (U2 , σ2 , π2 ) ≤ · · · in Ω, then Ui is a γ-invariant vector subspace of V . Define σ :

σ(xi ) = σi (xi ) for any xi ∈ Ui and π :

∞ S

Ui →

i=1 ∞ S

∞ S

i=1

Ui given by

i=1

∞ S

Ui →

∞ [

∞ [

Ui , π ∈ AutD Ui .

i=1

Ui given by π(xi ) =

i=1

πi (xi ) for any xi ∈ Ui . It is easy to verify that σ and π are well defined. Furthermore, we see that γ |∞ S

Ui

i=1

Thus,

∞ S

i=1

−σ = π, σ = σ 2 ∈ EndD

i=1

i=1


Ui , σ, π ∈ Ω, and so Ω is inductive. By using Zorn’s Lemma,

there exists (W, α, β) ∈ Ω which is maximal in Ω. If W 6= V , then can find some y ∈ V − W . Let K = yD + γ(y)D + γ 2 (y)D + · · · , and W0 = W + K. Obviously, W0 is γ-invariant. In view of Lemma 16.1.3, see that W0 ∈ Ω, a contradiction. Thus, V = W , as desired.

we let we 

Lemma 16.1.5. Let V be a countably generated infinite-dimensional right vector space over a division ring D, and let σ ∈ EndD (V ) be a shift operator. Then there exist α, β ∈ AutD (V ) such that σ = α + β, α2 = β 2 = 1V . Proof. Let C= 

C L  and let A =  0  .. .

 0 C L .. .

1D 0 0 −1D 0 0 C .. .

is easy to verify that



,L =



0 1D 0 0



,E =



1D 0 0 1D



,

 ··· ···  . Then A is a column-finite matrix over D. It ···  .. .

C 2 = E, LC = −CL =



0 −1D 0 0



and L2 = 0.

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Hence, we get 

Let

rings

C2 0 0 2  LC + CL C 0  2 2  2 L LC + CL C A =  2 0 L LC + CL  .. .. .. . . .

K=



−1D 1D

0 0 0 C2 .. .

  ··· E ···  0  ··· =0  ···  . .. .. .



K  0 0  and B =  0 1D  .. .

0 K 0 .. .

0 0 K .. .

0 E 0 .. .

0 0 E .. .

 ··· ···  . ···  .. .

 ··· ···  . ···  .. .

Then B is a column-finite matrix over D. In addition, we have  2    K 0 0 ··· E 0 0 ···  0 K2 0 · · ·   0 E 0 · · ·      B2 =  0 0 K 2 · · ·  =  0 0 E · · ·  .     .. .. .. . . .. .. .. . . . . . . . . . .   0 0 Let Q = . Then 1D 0   Q 0 0 ···  L Q 0 ···   A + B =  0 L Q ···.   .. .. .. . . . . . .

Let {x1 , x2 , x3 , · · · } be a basis of V . Define α, β ∈ EndD (V ) given by α(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )A, β(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )B.

Then α2 = β 2 = 1V . Furthermore, we see that

(α + β)(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )(A + B)   Q 0 0 ···  L Q 0 ···   = (x1 , x2 , x3 , · · · )  0 L Q · · ·    .. .. .. . . . . . . = σ(x1 , x2 , x3 , · · · ),

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and therefore σ = α + β.



Lemma 16.1.6. Let D be a division ring, and let γ ∈ Mn (D). If | D | 6= 2, then there exists α ∈ GLn (R) such that γ − α ∈ GLn (D). Proof. Let x ∈ D. If x 6= 0, then we choose u 6∈ {0, x}. Hence, x − u ∈ D∗ . If x = 0, then x − 1D ∈ D∗ . In any case, every element in D is the sum of two units. Assume that every element in Mn (D) is the sum of two invertible matrices. Given any γ ∈ Mn+1 (D), we have γ11 ∈ D, γ12 ∈ M1×n (D), γ21 ∈ Mn×1 (D) and γ22 ∈ Mn×n (D) such that   γ11 γ12 γ= . γ21 γ22

Thus, we can find u ∈ D∗ such that γ11 − u = v ∈ D∗ . Further, we have U ∈ GLn (D) such that (γ22 − γ21 u−1 γ12 ) − U = V ∈ GLn (D). Thus,   v γ12 γ − diag(u, U ) = . γ21 U + γ21 v −1 γ12 Obviously, diag(u, U ) ∈ GLn+1 (D). Moreover,   −1  −1 v + v −1 γ12 V −1 γ21 v −1 −v −1 γ12 V −1 v γ12 . = −V −1 γ21 v −1 V −1 γ21 U + γ21 v −1 γ12 Thus, the result holds for n+1. By induction, we show that γ−α ∈ GLn (D) for some α ∈ GLn (D), and therefore the proof is true.  Lemma 16.1.7. Let V be a countably generated right vector space over a division ring D, and let γ ∈ EndD (V ) be such that V is spanned by {y, γ(y), γ 2 (y), · · · }. If | D | 6= 2, then there exists α ∈ AutD (V ) such that γ − α ∈ AutD (V ).

Proof. We may assume that V 6= 0. If γ n (y) 6∈ yD + γ(y)D + γ 2(y)D + · · ·+ γ n−1 (y)D for all n ∈ N, then {y, γ(y), γ 2 (y), · · · } is a basis of V . Since γ is a shift operator with respect to this basis, it follows from Lemma 16.1.5 that there exists α ∈ AutD (V ) such that γ − α ∈ AutD (V ). Assume that there exists some n ∈ N such that γ n (y) ∈ yD +γ(y)D +γ 2(y)D +· · ·+γ n−1 (y)D. If n is minimal with this property, then {y, γ(y), γ 2 (y), · · · , γ n−1 (y)} is a basis of V ; hence, V ∼ = nD, and so EndD (V ) ∼ = Mn (D). According to Lemma 16.1.6, the result follows.  Lemma 16.1.8. Let V be a countably generated right vector space over a division ring D where | D | 6= 2, let γ ∈ EndD (V ), let U be a γ-invariant

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vector subspace of V , and let V = U + K where K = yD + γ(y)D + γ 2 (y)D + · · · for some y ∈ V . If there exists π ∈ AutD (U ) such that γ |U −π ∈ AutD (U ), then there exists α ∈ AutD (V ) such that γ − α ∈ AutD (V ), α| U = π. Proof. Construct M, θ, θ0 , ζ as in the proof of Lemma 16.1.3. Then γ(m) − ζ(m)
∈ ker(θ) In addition,
= U for all m ∈ M . 2 {θ0 (y), ζ θ0 (y) , ζ θ0 (y) , · · · } spans M . In view of Lemma 16.1.7, ζ = π0 + σ0 , π0 , σ0 ∈ AutD (M ). By hypothesis, we have γ| U = π + σ, π, σ ∈ AutD (U ). Define α : V → V given by α(m+ u) = π0 (m)+ π(u) for any m ∈
M, u ∈ U and β : V → V given by β(m+u) = σ0 (m)+ σ(u)+γ(m)−ζ(m) for any m ∈ M, u ∈ U. Clearly, α ∈ AutD (V ) and α |U = π. As in the proof of Lemma 16.1.3, we have (α + β)(m + u) = γ(m + u). for any m ∈ M, u ∈ U . Hence, γ = α + β. If β(m + u) = 0, then T σ0 (m) = −σ(u) − γ(m) + ζ(m) ∈ M U = 0; hence, m = 0. This implies that u = 0. Obviously, U ⊆ im(β). It suffices to show that M ⊆ im(β). If m ∈ M , we can find m0 ∈ M such that m = σ0 (m0 ). Also we have u0 ∈ U such that ζ(m0 ) − γ(m0 ) = σ(u0 ). Then β(m0 + u0 ) = m. Therefore β ∈ AutD (V ), as required.  Theorem 16.1.9. Every endomorphism over a countably generated right vector space over a division ring with more than two elements is the sum of two automorphisms. Proof. Let V be a countably generated right vector space over a division ring D, where | D | 6= 2. Let γ ∈ EndD (V ). Set Ω = {(U, σ, π) | UD ⊆ V is γ − invariant, γ |U = σ + π, σ, π ∈ AutD (U )}. Clearly, Ω is not empty. Partially order Ω by writing (U, σ, π) ≤ (U ′ , σ ′ , π ′ ) if U ⊆ U ′ , σ = σ ′ |U , π = π ′ |U . ∞ S Given (U1 , σ1 , π1 ) ≤ (U2 , σ2 , π2 ) ≤ · · · in Ω, then Ui is a γ-invariant vector subspace of V . Define σ :

for any xi ∈ Ui and π :

∞ S

i=1

∞ S

Ui →

i=1 ∞ S

Ui →

∞ S

i=1

Ui given by σ(xi ) = σi (xi )

i=1

Ui given by π(xi ) = πi (xi ) for any

i=1

xi ∈ Ui . It is easy to verify that σ and π are well defined. Furthermore, we see that ∞ [
∞ −σ = π, σ, π ∈ AutD Ui . γ |S Ui

Thus,

∞ S

i=1




i=1

i=1

Ui , σ, π ∈ Ω, and so Ω is inductive. By using Zorn’s Lemma,

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there exists (W, α, β) ∈ Ω which is maximal in Ω. If W 6= V , then we have some y ∈ V − W . Let K = yD + γ(y)D + γ 2 (y)D + · · · , and let W0 = W + K. Then W0 is γ-invariant. In view of Lemma 16.1.8, W0 ∈ Ω, a contradiction. Thus, V = W , as needed.  16.2

Infinite-Dimensional Linear Transformations, II

If a ring R satisfies the Goodearl-Menal condition, then R satisfies the property (∗): for any x, y ∈ R, there exists a u ∈ U (R) such that x + u, y − u−1 ∈ U (R). Now we investigate such property for endomorphism rings of a countably generated right vector space over a division ring. Lemma 16.2.1. Let V be a countably generated infinite-dimensional right vector space over a division ring D, and let σ ∈ EndD (V ) be a shift operator. Then there exists α ∈ AutD (V ) such that σ + α, σ − α−1 ∈ AutD (V ). Proof. Let    0 0 1D 0 0 0 C =  1D 0 0  , L =  0 0 0  , 00 0 0 1D 0 



C L  and let A =  0  .. .

0 C L .. .

0 0 C .. .

 ··· ···  . Then A is a column-finite matrix over D. It ···  .. .

is easy to verify that C 3 = L2 = 0. Let    0 00 C 0 0 ··· L 0 0  0 C 0 ···    B =  0 0 C ···,D =  0 L 0    .. .. .. .. .. .. . . . . . . . . . Then



C L  A=0  .. .

0 C L .. .

0 0 C .. .

 ··· ···  . ···  .. .

 ··· ···  = B + D. ···  .. .

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Let {x1 , x2 , x3 , · · · } be a basis of V . Define ε, δ ∈ EndD (V ) given by σ(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )A, ε(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )B, δ(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )D. Then (ε + δ)(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )(B + D)  C 0 0 L C 0  = (x1 , x2 , x3 , · · · )  0 L C  .. .. .. . . . = σ(x1 , x2 , x3 , · · · ),

 ··· ···  ···  .. .

and then σ = ε + δ. Clearly, ε3 = δ 2 = 0. Thus, σ = (δ + 1) + (ε − 1) and σ = (δ − 1) + (ε + 1). Let α = −(δ + 1). Then α−1 = δ − 1. It is easy to verify that (ε − 1)−1 = −ε2 − ε − 1 and (ε + 1)−1 = ε2 − ε + 1. Therefore σ + α, σ − α−1 ∈ AutD (V ), as asserted.  We note that | D | = 2 if and only if D ∼ = Z/2Z and | D | = 3 if and only if D ∼ = Z/3Z. In addition, Z/2Z and Z/3Z do not satisfy the property (∗). Let R = Z/3Z. Choose x = 2. We see that there is not any u ∈ U (R) such that x + u, x − u−1 ∈ U (R). Lemma 16.2.2. Let V be a countably generated right vector space over a division ring D, and let γ ∈ EndD (V ) be such that V is spanned by {y, γ(y), γ 2 (y), · · · }. If | D | 6= 2, 3, then there exists α ∈ AutD (V ) such that γ + α, γ − α−1 ∈ AutD (V ). Proof. We may assume that V 6= 0. If γ n (y) 6∈ yD + γ(y)D + γ 2(y)D + · · ·+ γ n−1 (y)D for all n ∈ N, then the result follows by Lemma 16.2.1. Assume that there exists some n ∈ N such that γ n (y) ∈ yD +γ(y)D +γ 2(y)D +· · ·+ γ n−1 (y)D. If n is minimal with this property, then EndD (V ) ∼ = Mn (D). As | D | 6= 2, 3, it follows by Lemma 2.4.5 that Mn (D) satisfies the GoodearlMenal condition. For any γ ∈ M2 (Z/2Z), we verify that there exists α ∈ AutD (V ) such that γ + α, γ − α−1 ∈ AutD (V ), as required.  Lemma 16.2.3. Let V be a countably generated right vector space over a division ring D where | D | 6= 2, 3, let γ ∈ EndD (V ), let U be a γinvariant vector subspace of V , and let V = U + K where K = yD +

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γ(y)D + γ 2 (y)D + · · · for some y ∈ V . If there exists π ∈ AutD (U ) such that γ| U + π, γ| U − π −1 ∈ AutD (V ), then there exists α ∈ AutD (V ) such that γ + α, γ − α−1 ∈ AutD (V ) and α| U = π. Proof. Construct M, θ, θ0 , ζ as in the proof of Lemma 16.1.3. Then γ(m) − ζ(m)
∈ ker(θ) In addition,
= U for all m ∈ M . 2 {θ0 (y), ζ θ0 (y) , ζ θ0 (y) , · · · } spans M . As dimD (M ) = dimD (V /U ), it follows by Lemma 16.2.2 that there exists α1 ∈ AutD (M ) such that ζ + α1 , ζ − α−1 1 ∈ AutD (M ). By assumption, there exists α2 ∈ AutD (U ) such that γ|

U

+ α2 , γ|

U

− α−1 2 ∈ AutD (U ).

Define α : V → V given by α(m + u) = α1 (m) + α2 (u) for any m ∈ M, u ∈ U . Let β1 = ζ + α1 and β2 = γ| U +
α2 . Let β : V → V given by β(m + u) = β1 (m) + β2 (u) + γ(m) − ζ(m) for any m ∈ M, u ∈ U . One easily checks that α ∈ AutD (V ). Furthermore, we see that (−α + β)(m + u)
= −α1 (m) − α2 (u) +
β1 (m) + β2 (u) + γ(m)
− ζ(m) = − α1 (m) + β1 (m) + − α2 (u) + β2 (u) + γ(m) − ζ(m) = γ(u) + γ(m) = γ(m + u). for any m ∈ M, u ∈ U . Thus, γ + α = β. As in the proof of Lemma 16.1.3, we show that β is a D-epimorphism, and so γ + α = β ∈ AutD (V ). Let σ1 = ζ − α−1 and σ2 = γ| U − α
−1 1 2 . Let δ : V → V given by δ(m + u) = σ1 (m) + σ2 (u) + γ(m) − ζ(m) for any m ∈ M, u ∈ U . Then (α−1 + δ)(m + u)
−1 = α−1 1 (m) + α2 (u)
+ σ1 (m) + σ2 (u) +
γ(m) − ζ(m) −1 = α−1 1 (m) + σ1 (m) + α2 (u) + σ2 (u) + γ(m) − ζ(m) = γ(u) + γ(m) = γ(m + u).

for any m ∈ M, u ∈ U . So γ − α−1 = δ. Analogously, we see that δ ∈ AutD (V ). Therefore the result follows.  Theorem 16.2.4. Let V be a countably generated right vector space over a division ring D, and let γ ∈ EndD (V ). If | D | 6= 2, 3, then there exists α ∈ AutD (U ) such that γ + α, γ − α−1 ∈ AutD (U ). Proof. Let Ω = {(U, σ) | UD ⊆ V is γ − invariant, σ, γ| U + σ, γ| U − σ −1 ∈ AutD (U )}. Partially order Ω by writing (U, σ) ≤ (U ′ , σ ′ ) if U ⊆ U ′ , σ =

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σ′ |

U.

Given (U1 , σ1 ) ≤ (U2 , σ2 ) ≤ · · · in Ω, then

vector subspace of V . Define σ :

i=1

for any xi ∈ Ui . Then γ |∞ S

i=1

and thus,

∞ S

i=1

Ui

∞ S

+σ, γ | ∞ S

i=1

Ui

Ui →

∞ S

∞ S

Ui is a γ-invariant

i=1

Ui given by σ(xi ) = σi (xi )

i=1

−σ −1 ∈ AutD

∞ [

i=1


Ui ,


Ui , σ ∈ Ω; hence, Ω is an inductive nonempty set. By using

Zorn’s Lemma, there exists (W, α, β) ∈ Ω which is maximal in Ω. If W 6= V , then we can find some y ∈ V − W . Let K = yD + γ(y)D + γ 2 (y)D + · · · , and let W0 = W + K. Obviously, W0 is γ-invariant. In view of Lemma 16.2.3, W0 ∈ Ω, a contradiction. Therefore V = W , and we are done.  Let V be a countably generated right vector space over a division ring D, and let γ ∈ EndD (V ). If |D| 6= 2, 3, we claim that there exists α ∈ AutD (V ) such that x − α, x − α−1 ∈ AutD (V ). The proof is similar to the preceding discussion from Lemma 16.1.5. For a semilocal ring R, R satisfies the Goodearl-Menal condition if and only if for any γ ∈ R, there exists α ∈ U (R) such that γ + α, γ − α−1 ∈ U (R). This should be contracted to the following simple fact: a semilocal ring R satisfies unit 1-stable range if and only if for any γ ∈ R, there exists α ∈ U (R) such that γ−α, γ−α−1 ∈ U (R). Let R be a strongly π-regular ring. Then for any x ∈ R there exists an idempotent e ∈ R such that x2 + xe + 1 ∈ U (R) (cf. [80, Corollary 13]). For countably linear transformations over vector spaces, we can derive the following. Corollary 16.2.5. Let V be a countably generated right vector space over a division ring D, and let γ ∈ EndD (V ). If |D| 6= 2, 3, then there exists α ∈ AutD (V ) such that γ 2 + γα + 1 ∈ AutD (V ). Proof. In view of Theorem 16.2.4, there exists some ε ∈ AutD (V ) such that −γ + ε, −γ − ε−1 ∈ AutD (V ). Let −γ + ε = α and −γ − ε−1 = σ. Then γε+1 = −σε, and so γ(γ +α)+1 = −σε. Therefore γ 2 +γα+1 ∈ AutD (V ), as asserted.  Lemma 16.2.6. Let V be a countably generated infinite-dimensional right vector space over a division ring D, and let σ ∈ EndD (V ) be a shift operator. Then there exists α ∈ AutD (V ) such that σ 2 − α2 ∈ AutD (V ).

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Proof. Let

and let



0  1D A=  0 1  D 1D  0 E=  0 0 

1D 0 0 0 0 1D 0 0

0 0 0 1D 0 0 1D 0

rings

553

   0 0 0 −1D 0   0   , B =  0 0 0 −1D  ,   1D 00 0 0  0 00 0 0  0 0   ∈ M4 (D), 0  1D

   A 0 0 ··· −E 0 0 · · ·  0 A 0 ···  −B −E 0 · · ·      X =  0 0 A · · ·  , Y =  0 −B −E · · ·  ,     .. .. .. . . .. .. .. . . . . . . . . . .  −1    A 0 0 ··· E 0 0 ···  0 A−1 0 · · ·   0 E 0 ···     Z= 0 , I = −1   0 0 E ··· 0 A ···    .. .. . . .. .. .. .. . . . . . . . . . .

be column-finite matrices over D. One easily checks that XZ = ZX = I and Y (−2I − Y ) = (−2I − Y )Y = I. Let {x1 , x2 , x3 , · · · } be a basis of V . Define α, β, φ, ϕ ∈ EndD (V ) given by α(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )X, β(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )Y, φ(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )Z, ϕ(x1 , x2 , x3 , · · · ) = (x1 , x2 , x3 , · · · )(−2I − Y ). Then αφ = φα = 1V and βϕ = ϕβ = 1V . That is, α, β ∈ AutD (V ). In addition, we see that (α2 + β)(x1 , x2 , x3 , · · · )
= (x1 , x2 , x3 , · · · ) X 2 + Y     2 −E 0 0 · · · A 0 0 ···  0 A2 0 · · ·   −B −E 0 · · · 
    = (x1 , x2 , x3 , · · · )  0 0 A2 · · ·  +  0 −B −E · · ·      .. .. .. . . .. .. .. . . . . . . . . . .  2  A −E 0 0 ···  −B A2 − E 0 ···   2 = (x1 , x2 , x3 , · · · )  , 0 −B A − E ···   .. .. .. .. . . . .

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

 1D 0 0 0  0 1D 0 0 
2  where A2 =   1D 0 1D 0 . Thus, we have α + β (x1 , x2 , x3 , · · · ) = 0 1D 0 1D σ 2 (x1 , x2 , x3 , · · · ), and therefore σ 2 − α2 = β ∈ AutD (V ).  Lemma 16.2.7. Let V be a countably generated right vector space over a division ring D. If | D | 6= 2, 3, then for any γ ∈ Mn (D), there exists α ∈ GLn (D) such that γ 2 − α2 ∈ AutD (V ). Proof. Suppose that | D | 6= 2, 3. Then | D | ≥ 4. We find some u ∈ D such that u 6∈ {0, 1D , −1D }; hence, u2 6= 1D . So we have some v ∈ D∗ such that 1D = u2 + v. Let a ∈ D. If a 6= 1, then a − 12D ∈ D∗ . For any a ∈ D, we conclude that a − u2 ∈ D∗ for some u ∈ D∗ . For any 2 γ ∈ Mk (D), assume that there  exists α ∈ Mk (D) such that γ − α ∈ γ11 γ12 GLk (D). Given γ = ∈ Mk+1 (D) with γ11 ∈ D, γ22 ∈ Mk (D), γ21 γ22 then we find u1 , v1 ∈ D∗ such that γ11 = u21 + v1 . By assumption, there exist u2 , v2 ∈ GLk (D) such that γ22 − γ21 v1−1 γ12 = u22 + v2 . Hence γ =   2  v1 γ12 u1 + . One easily checks that γ21 v2 + γ21 v1−1 γ12 u2 

−1 v1 γ12 γ21 v2 + γ21 v1−1 γ12  −1  v1 + v1−1 γ12 v2−1 γ21 v1−1 −v1−1 γ12 v2−1 = −v2−1 γ21 v1−1 v2−1 ∈ GLk+1 (D).

By induction, we obtain the result.



Lemma 16.2.8. Let V be a countably generated right vector space over a division ring D, and let γ ∈ EndD (V ) be such that V is spanned by {y, γ(y), γ 2 (y), · · · } for some y ∈ V . If | D | 6= 2, 3, then there exists α ∈ AutD (V ) such that γ 2 − α2 ∈ AutD (V ). Proof. We may assume that V 6= 0. If γ n (y) 6∈ yD + γ(y)D + γ 2 (y)D + · · · + γ n−1 (y)D for all n ∈ N, then γ is a shift operator with respect to the basis {y, γ(y), γ 2 (y), · · · }. If there exists some n ∈ N such that γ n (y) ∈ yD + γ(y)D + γ 2(y)D + · · ·+ γ n−1 (y)D and n is minimal with this property, then EndD (V ) ∼ = Mn (D). Analogously to Lemma 16.1.2, we complete the proof by Lemma 16.2.6 and Lemma 16.2.7. 

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Lemma 16.2.9. Let V be a countably generated right vector space over a division ring D, let γ ∈ EndD (V ), let U be a γ-invariant vector subspace of V , let V = U + K where K = yD + γ(y)D + · · · for some y ∈ V , and let |D| 6= 2, 3. If there exist π, σ ∈ AutD (U ) such that γ 2 | U = π 2 + σ, then there exist α, β ∈ AutD (V ) such that γ 2 = α2 + β, α| U = π, β| U = σ. Proof. Clearly, we have a vector subspace M such that V = M ⊕ U and γ : V /U → V /U defined by γ(v) = γ(v). Let θ : V → V given by θ(m + u) = m for any m ∈ M, u ∈ U . Then θ(V ) = M and θ2 = θ. This induces a D-isomorphism θ0 : V /U → M . Let δ = θ0 γθ0−1 ∈ EndD (M ). Then δθ0 = θ0 γ, and so δθ0 (v) = θ0 γ(v) for any v ∈ V . This induces that δθ(v) = θγ(v). For any m ∈ M , δ(m) = δθ(m) = θγ(m). Furthermore, we get δ 2 (m) = δθγ(m) = θγ 2 (m). Hence θδ 2 (m) = θ2 γ 2 (m) = θγ 2 (m). This implies that γ 2 (m) − δ 2 (m) ∈ ker(θ) = U for all
m ∈ M .
Analogously to Lemma 16.1.3, {θ0 (y), δ θ0 (y) , δ 2 θ0 (y) , · · · } spans M . In view of Lemma 16.2.8, we have δ 2 = π02 + σ0 , π0 , σ0 ∈ AutD (M ). By hypothesis, we have γ2|

U

= π 2 + σ, π, σ ∈ AutD (U ).

Define α : V → V given by α(m + u) = π0 (m) + π(u) for any m ∈ M, u
∈ U and β : V → V given by β(m + u) = σ0 (m) + σ(u) + γ 2 (m) − δ 2 (m) for any m ∈ M, u ∈ U. Obviously, α| U = π and β| U = σ. In addition, we have (α2 + β)(m + u)
= π02 (m) + π 2 (u) + σ0 (m) + σ(u) + γ 2 (m) − δ 2 (m)
= δ 2 (m) + γ 2 (u) + γ 2 (m) − δ 2 (m) = γ 2 (m + u). for any m ∈ M, u ∈ U . As a result, we get γ 2 = α2 + β. Clearly, α ∈ AutD (V ). If β(m + u) = 0, then σ0 (m) = −σ(u) − γ 2 (m) + δ 2 (m) ∈ T M U = 0; hence, σ0 (m) = 0. This implies that m = 0. Furthermore, we have −σ(u) = 0, and so u = 0. Thus β is a D-monomorphism. Clearly, U ⊆ im(β). If m ∈ M , there exists some m0 ∈ M such that m = σ0 (m0 ). Also we have u0 ∈ U such that δ 2 (m0 )− γ 2 (m0 ) = σ(u0 ). One easily checks that β(m0 + u0 )
= σ0 (m0 ) + σ(u0 ) + γ 2 (m0 ) − δ 2 (m0 ) = m.

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That is, β is a D-epimorphism. Therefore β ∈ AutD (V ), as required.



Theorem 16.2.10. Let V be a countably generated right vector space over a division ring D, and let γ ∈ EndD (V ). If |D| 6= 2, 3, then there exists α ∈ AutD (V ) such that γ 2 − α2 ∈ AutD (V ). Proof. Let Ω = {(U, σ, π) | UD ⊆ V is γ − invariant, γ 2 | U = σ 2 + π, σ, π ∈ AutD (U )}. Clearly, Ω is not empty. Partially order Ω by writing (U, σ, π) ≤ (U ′ , σ ′ , π ′ ) if U ⊆ U ′ , σ = σ ′ | U , π = π ′ | U . Analogously to Theorem 16.1.4, there exists (W, α, β) ∈ Ω which is maximal in Ω. If W 6= V , then we have some y ∈ V − W . Let K = yD + γ(y)D + γ 2 (y)D + · · · , and let W0 = W + K. Obviously, W0 is γ-invariant. According to Lemma 16.2.9, we can show that W0 ∈ Ω. This gives a contradiction, and then V = W . So the proof is true.  Corollary 16.2.11. Let V be an infinite-dimensional vector space over a division ring D. If 2, 3 ∈ D∗ , then for any γ ∈ EndD (V ), there exists α ∈ AutD (V ) such that γ 2 − α2 ∈ AutD (V ). Proof. As 2, 3 ∈ D∗ , | D | 6= 2, 3. Thus, the result follows from Theorem 16.2.10.  Let V be an infinite-dimensional vector space over Z/5Z. It follows from Corollary 16.2.11 that for any γ ∈ EndZ/5Z (V ), there exists α ∈ AutZ/5Z (V ) such that γ 2 − α2 ∈ AutZ/5Z (V ). Let V be an infinite-dimensional vector space over any infinite field F . Then for any γ ∈ EndF (V ), there exists α ∈ AutF (V ) such that γ 2 − α2 ∈ AutF (V ). We note that the requirement that 2, 3 ∈ D∗ in Corollary 16.2.11 is not necessary. Example 16.2.12. Let D = {0, e, a, b} be a set. Define operations by the following tables: + 0 e a b

0 0 e a b

e e 0 b a

a a b 0 e

b b a e 0

,

× 0 e a b

0 0 0 0 0

e 0 e a b

a 0 a b e

b 0 b e a

.

Then D is a field with | D | = 4. Let V be an infinite-dimensional right vector space over D. In view of Theorem 16.2.10, we prove that for any γ ∈ EndD (V ), there exists α ∈ AutD (V ) such that γ 2 − α2 ∈ AutD (V ). In

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this case, 2 6∈ D∗ . 16.3



Clean Matrices

Throughout this section, R is a commutative ring with  a1    0 use a1 , a2 , · · · , an to denote the n × n matrix  .  ..

an identity. We  a2 an 0 ··· 0   .. . . ..  . We . .  .

0 0 ··· 0 say that A ∈ Mn (R) is e-clean provided that there exists an idempotent E ∈ Mn (R) such that A − E ∈ GLn (R)anddetE = e. In [278], Khurana and Lam proved that many matrices a, b ∈ M2 (Z) are unit-regular, while they are not 0-clean. The main purpose  of this sectionis to get a general criterion of e-cleanness for the matrix a1 , a2 , · · · , an . As an application, we prove that 0-cleanness and unit-regularity for any such n×n matrix over a Dedekind domain coincide for all n ≥ 3. We use Ur (R) to denote the sets of all unimodular rows over R. We begin with a simple fact. Lemma 16.3.1. Let E = (eij ) ∈ Mn (R) be an idempotent, and let e = detE. If (ei1 , · · · , ein ) ∈ Ur (R), then eeii = e, eeij = 0 (j 6= i). Proof. Assume that (en1 , · · · , enn ) ∈ Ur (R). Then there are some xi such that en1 x1 + · · · + enn xn = 1. Let f = 1 − e. Over the factor ring R/f R, E 2 has determinant 1. As E = E , we get E = In . This implies that eni ∈ f R for any i(6= n). Therefore, eenj = 0 (j 6= n).
Obviously, enn xn ≡ 1 mod en1 R + · · · + en(n−1) R . Thus, 

e11  e21  E= .  .. 0

 · · · e1n · · · e2n 
 mod en1 R + · · · + en(n−1) R . . . ..   . . 0 · · · enn

e12 e22 .. .

Since E = E 2 , we see that enn = enn 2 . This implies that enn = 1, i.e., 1 − enn ∈ en1 R + · · · + en(n−1) R. Hence, we get e = eenn . The general case is proved in a similar fashion.  Theorem 16.3.2. Let   a1 , · · · , an ∈ R, and let e ∈ R be an idempotent. Then a1 , a2 , · · · , an is e-clean if and only if the following hold:

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(1) There exist x1 , · · · , xn ∈ R such that a1 x1 + · · · + an xn ∈ R is e-clean. (2) ex2 = · · · = exn = 0.
(3) x1 ≡ 1 mod x2 R + · · · + xn R .   Proof. Suppose that a1 , a2 , · · · , an is e-clean. Then we have an idempotent matrix E   = (eij ) ∈ Mn (R) and a U = (uij ) ∈ GLn (R) such that a1 , a2 , · · · , an = E + U and detE = e. Thus, 

a1 − e11 a2 − e12  −e21 −e22   .. ..  . . −en1 −en2

 · · · an − e1n · · · −e2n    = U. .. ..  . . · · · −enn

This implies that a1 a2 −e21 −e22 . .. .. . −e −e n1 n2

· · · an · · · −e2n n . + (−1) detE = detU. .. . .. · · · −enn

Hence, a1 A11 +a2 A12 +· · ·+an A1n = e+(−1)n+1u, where u = detU ∈ U (R) and each A1i is the algebraic complement of E corresponding to e1i (1 ≤ i ≤ n). Let x1 = A11 , x2 = A12 , · · · , xn = A1n . Then a1 x1 +a2 x2 +· · ·+an xn = e + (−1)n+1 u is e-clean. As E ∈ Mn (R) is an idempotent with detE = e, in view of Lemma 16.3.1, we get eeii = e, eeij = 0 (1 ≤ i 6= j ≤ n). This implies that eA12 = · · · = eA1n = 0; hence, ex2 = · · · = exn = 0. Clearly, we have e21 A11 + e22 A12 + · · · + e2n A1n = 0, and thus,
e21 A11 ≡ 0 mod x2 R + · · · + xn R .

On the other hand, u11 A11 + u12 A12 + · · · + u1n A1n = (−1)n+1 u, and thus,
u11 A11 ≡ (−1)n+1 u mod x2 R + · · · + xn R .

As u ∈ U (R), we deduce that


e21 ≡ 0 mod x2 R + · · · + xn R .

Similarly, we show that


e31 , · · · , en1 ≡ 0 mod x2 R + · · · + xn R .

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Since E 2 = E, we see that    e11 e12 · · · e1n e11 e12 · · · e1n  0 e22 · · · e2n   0 e22 · · · e2n      . . . .  .  . .  .. .. . . ..   .. .. . . . ..  0 e · · · enn 0 en2 · · · enn  n2  e11 e12 · · · e1n  0 e22 · · · e2n 
  ≡ . . . ..  mod x2 R + · · · + xn R . . . .  . . . . 

This implies that 

As a result, Hence,

  e22 · · · e2n e22 · · · e2n  .. . . ..   .. . . ..   . . .  . . .  en2 · · · enn e · · · enn   n2 e22 · · · e2n
  ≡  ... . . . ...  mod x2 R + · · · + xn R . en2 · · · enn


A211 = A11 mod x2 R + · · · + xn R .


u11 A211 ≡ u11 A11 mod x2 R + · · · + xn R .

Therefore we get That is,

0 en2 · · · enn


A11 ≡ 1 mod x2 R + · · · + xn R .
x1 ≡ 1 mod x2 R + · · · + xn R .

Conversely, assume that Conditions (1), (2) and (3) hold. By Condition (1), we can find some x1 , · · · , xn ∈ R such that a1 x1 + · · · + ai xi + · · · + an xn is e-clean. Let c1 = x1 and ci = −xi (2 ≤ i ≤ n). Then a1 c1 − · · · − ai ci − · · · − an cn is e-clean. By Condition (3), we can find some k2 , · · · , kn ∈ R such that c1 = 1 + k2 c2 + · · · + kn cn . Let     e 1 k2 · · · kn 1 −k2 · · · −kn   1   c2 1  1     E = (eij ) =  .   . . . . .. ..   ..     .. 1 cn 1 1

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By Condition (2), it is easy to verify that E = E 2 ∈ Mn (R) and detE = e. Let   a1 · · · an 0 ··· 0    U = (uij ) =  . . .  − E.  .. . . ..  0 ··· 0 Then a1 − e11 a2 − e12 · · · an − e1n −e21 −e22 · · · −e2n detU = .. .. .. .. . . . . −e −en2 · · · −enn n1 a1 a2 · · · an −e21 −e22 · · · −e2n n = . .. . . .. + (−1) detE .. . . . −e −e · · · −e n1 n2 nn
= (−1)n−1 a1 A11 + · · · + an A1n + (−1)n e, where A11 , · · · , A1n are algebraic complements of E corresponding to e11 , · · · , e1n respectively. Obviously,   1 + (e − c1 ) (e − c1 )k2 (e − c1 )k3 · · · (e − c1 )kn  c2 1 + k2 c2 k3 c2 ··· kn c2     c3 k2 c3 1 + k3 c3 · · · kn c3  E= .   .. .. .. .. ..   . . . . .

cn k2 cn k3 cn · · · 1 + kn cn It is easy to see that A11 = 1 + k2 c2 + · · · + kn cn = c1 . Furthermore, we see that each A1i = −ci (2 ≤ i ≤ n). Clearly, there is some u ∈ U (R) such that a1 A11 + · · · + a1n A1n = a1 c1 − · · · − ai ci − · · · − an cn = e + u. Thus, detU = (−1)n−1 (e + u) + (−1)n e = (−1)n−1 u ∈ U (R), and then U ∈ GLn (R). Therefore A is e-clean, as asserted. 

  Corollary 16.3.3. Let a1 , · · · , an ∈ R (n ∈ N). If a , a , · · · , a is 1 2 n  0-clean, then so is a1 u1 , a2 u2 , · · · , an un for any u1 , · · · , un ∈ U (R).   Proof. Assume that a1 , a2 , · · · , an is 0-clean. According to Theorem 16.3.2, there exist x1 , x2 , · · · , xn ∈ R such that a1 x1 + · · · + an xn = u ∈
U (R) and x1 ≡ 1 mod x2 R + · · · + xn R . Thus, we deduce that (a1 u1 )x1 + a2 (u1 x2 ) + · · · + an (u1 xn ) = u1 u ∈ U (R). In addition,
x1 ≡ 1 mod (u1 x2 )R + · · · + (u1 xn )R .

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By using Theorem 16.3.2 again, we have an idempotent E ∈ Mn (R) and a U ∈ GLn (R) such that   a1 u 1 a2 · · · an  0 0 ··· 0     . .. . . ..  = E + U and detE = 0.  .. . . .  0

0 ··· 0

Therefore we conclude that 

a1 u 1 , a2 u 2 , · · · , an u n 



1

 u−1 2  + ..  .

as desired.

u−1 n









1

 u−1 2  = ..  . 

1

  u2   U  ..   .

un

u−1 n





1

  u2   E  ..   .

un

   

  , 



  As immediate consequences, we a1 , a2 , · · · , an is 1-clean  derive that  if and only if a1 ∈ 1 + U (R), and a1 , a2 , · · · , an is 0-clean if and only if the following hold: (1) There exist x1 , · · · , xn ∈ R such
that a1 x1 + · · · + an xn ∈ U (R). (2) x1 ≡ 1 mod x2 R + · · · + xn R .

  Example   16.3.4. Let us show that 12, 5, 3 ∈ M3 (Z) is clean, while 12, 5 ∈ M2 (Z) is not clean. Since 12 × (−2) + 5 × 2+ 3 × 5 =  1 and −2 ≡ 1 (mod 2R + 5R), it follows by Theorem 16.3.2 that 12, 5, 3 ∈ M3 (Z) is   0-clean. In fact, we have the decomposition: 12, 5, 3 = E + U, where     3 8 −2 9 −3 5 E =  −2 −7 2  , U =  2 7 −2  −5 −20 6

2

5 20 −6

with E = E , detE = 0 and detU = 1.  Clearly, 12 ∈ 6 1 + U (Z), and so 12, 5 ∈ M2 (Z) is not 1-clean. If   12, 5 ∈ M2 (Z) is 0-clean, then there exist some x1 , x2 ∈ N such that 12x1 + 5x2 = ±1 and x1 ≡ 1 (mod x2 Z). Write x1 + x2 y = 1. Then (12y − 5)x2 = 13 or 11. Hence, 12y−5 = ±1, ±11 or ±13. So 12y = −8, −6, 4, 6, 16

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  or 18,  a contradiction. Thus, 12, 5 ∈ M2 (Z)is not0-clean. Therefore  12, 5 ∈ M2 (Z) is not clean. In this case, 12, 5 ∈ M2 (Z) is unit     10 12 5 regular since 12, 5 = .  00 −5 −2

  Let a1 , · · · , an , an+1 ∈ R (n ∈ N).  If a1 , a2 , · · · , an ∈ Mn (R) is eclean, then so is a1 , a2 , · · · , an+1 ∈ Mn+1 (R). Example 16.3.4 shows that the converse is not true. Note that Theorem 16.3.2 illustrates the process ofcomputing ”clean decompositions” of numerical examples. For  instance, 9, 7 ∈ M2 (Z) is clean, and       4 3 54 9, 7 = + . −4 −3 43 Lemma 16.3.5. Let a1 , a2 , · · · , an+1 ∈ R (n ∈ N). If (a2 , · · · , an+1 ) ∈  Ur (R), then a1 , a2 , · · · , an+1 ∈ Mn+1 (R) is 0-clean.

Proof. Since (a2 , · · · , an+1 ) ∈ Ur (R), there are x2 , · · · , xn+1 ∈ R such that a2 x2 + · · · + an+1 xn+1 = 1. Thus, a1 × 0 + a2 x2 + · · · + an+1 xn+1 = 1. It is easy to see that
0 ≡ 1 mod x2 R + · · · + xn+1 R . According to Theorem 16.3.2, we obtain the result.



Theorem 16.3.6. Let a1 , a2 , · · · , an+1 ∈ R (n ∈ N). If R satisfies the n-stable range condition, then the following are equivalent:   (1) a1 , a2 , · · · , an+1 is 0-clean. (2) (a1 , a2 , · · · , an+1 ) ∈ Ur (R). Proof. (1) ⇒ (2) By virtue of Theorem 16.3.2, there exist x1 , · · · , xn+1 ∈ R such that a1 x1 + · · · + an+1 xn+1 = u ∈ U (R); hence, a1 x1 u−1 + · · · + an+1 xn+1 u−1 = 1. That is, (a1 , a2 , · · · , an+1 ) ∈ Ur (R). (2) ⇒ (1) Since (a1 , a2 , · · · , an+1 ) ∈ Ur (R), there exist c2 , · · · , cn+1 ∈ R such that  (a2 + a1 c2 , · · · , an+1 + a1 cn+1 )∈ Ur (R). In view of Lemma 16.3.5, a1 , a2 + a1 c2 , · · · , an+1 + a1 cn+1 ∈ Mn+1 (R) is 0-clean. Thus, we have an idempotent E ∈ Mn+1 (R) and some U ∈ GLn+1 (R) such that  a1 , a2 + a1 c2 , · · · , an+1 + a1 cn+1 = E + U and detE = 0. Let   1 c2 · · · cn+1 0 1 ··· 0    Q=. . . .  ∈ GLn+1 (R).  .. .. . . ..  0 0 ··· 1

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    Then, Q−1 a1 , a2 , · · · , an+1 Q = a1 , a2 + a1 c2 , · · · , an+1 + a1 cn+1 = E + U. Therefore a1 , a2 , · · · , an+1 = QEQ−1 + QU Q−1 . In addition, QEQ−1 ∈ Mn+1 (R) is an idempotent matrix, detQEQ−1 = 0 and QU Q−1 ∈ GLn+1 (R). Therefore the result follows.  Corollary 16.3.7. Let R be a Dedekind domain, and let a1 , · · · , an ∈ R (n ≥ 3). Then the following are equivalent:   (1) a1 , a2 , · · · , an  is 0-clean. (2) a1 , a2 , · · · , an 6= 0 is unit-regular. (3) (a1 , · · · , an ) ∈ Ur (R). Proof. (1) ⇔ (3) Since R is a Dedekind domain, it follows by Example 12.1.14 that R satisfies the 2-stable range condition, and so it is clear from Theorem 16.3.6.   (2) ⇒ (3) Let a1 , a2 , · · · , an 6= 0 be unit-regular. Then there exists an E = (eij ) ∈ M ij ) ∈ GLn (R) such that  idempotent  n (R) and a U = (u −1 a1 , a2 , · · · , an = EU, i.e., a that 1 , a2 , · · · , an U  =E. This implies  eij = 0 for i = 2, ·
· · , n. Thus, a1 , a2
, · · · , an = e11 , e12 , · · · , e1n U ; hence, a1 , · · · , an = e11 u11 , · · · , u1n . Clearly, e11 = e211 ∈ R, and then, e11 = 1. Therefore we get (a1 , · · · , an ) = (u11 , · · · , u1n ) ∈ Ur (R). (3) ⇒ (2) Since (a1 , · · · , an−1 , an ) ∈ Ur (R), there are some x1 , · · · , xn ∈ R such that a1 x1 + · · · + an xn = 1. As R satisfies the 2-stable range condition, we have bi , ci (3 ≤ i ≤ n) such that (a1 + a3 b3 + · · · + an bn , a2 + a3 c3 + · · · +
an cn ) ∈ Ur (R). Thus, a1 + a3 b3 + · · · + an bn )x + (a2 + a3 c3 + · · · + an cn y = 1 for some x, y ∈ R. One easily checks that   a1 a2 a3 · · · an−1 an  −y x 0 ··· 0 0       −b3 −c3 1 · · · 0 0   . .. . . . ..  ..  ∈ GLn (R).  . . .. .  . .  .    −bn−1 −cn−1 0 · · · 1 0  −bn −cn 0 · · · 0 1 Therefore

  a1 1 0 ··· 0  −y     0 0 ··· 0 a1 , a2 , · · · , an =  . . . .   −b3  .. .. . . ..   .  .. 0 0 ··· 0 −bn 

a2 x −c3 .. . −cn

··· ··· ··· .. . ···

 an 0   0  , ..  .  1

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as desired.



The following result should contracted to the fact that the problem  be  of deciding the cleanness of a, b ∈ M2 (Z) is considerably harder (cf. [278]).   Corollary 16.3.8. Let a1 , · · · , an ∈ Z (n ≥ 3). Then a1 , a2 , · · · , an ∈ Mn (Z) is clean if and only if a1 = 0 or a1 = 2 or (a1 , · · · , an ) ∈ Ur (R).   Proof. If a1 , a2 , · · · , an ∈ Mn (Z) is 1-clean, then we can find an idempotent E ∈ Mn (Z) and some U = (uij ) ∈ GLn (Z) such that  a1 , a2 , · · · , an = E + U and detE = 1. Thus, E = diag(1, · · · , 1) ∈ Mn (Z). This implies that uij = 0 (i 6= 1, j), uii = −1 (2 ≤ i ≤ n). Hence, a1 − 1 ∈ U (Z), i.e., a1 = 0, 2. Therefore we conclude that  a1 , a2 , · · · , an ∈ Mn (Z) is 1-clean if and only if either a1 = 0 or a1 = 2. Consequently, the result follows from Corollary 16.3.7.  We say that 0 6= A ∈ Mn (R) has that there exist some  rank 1 provided  P, Q ∈ GLn (R) such that P AQ = a1 , · · · , an for some a1 , · · · , an ∈ R.

Corollary 16.3.9. Let R be a Dedekind domain, and let A ∈ Mn (R) (n ≥ 3). If A has rank 1, then the following are equivalent: (1) A is 0-clean. (2) A is unit-regular.

Proof. (1) ⇒ (2) As A has rank 1, there exist some P, Q ∈ GLn (R) such that P AQ = [[a1 , · · · , an ]] for some a1 , · · · , an ∈ R. Thus,     P AP −1 = a1 , · · · , an Q−1 P −1 = b1 , · · · , bn   for some b1 , · · · , bn ∈ R. This b1 , · · · , bn is 0-clean. Ac implies that  cording to Corollary 16.3.7, b1 , · · · , bn is unit-regular. Therefore A is unit-regular.   (2) ⇒ (1) As in the precedingdiscussion,P AP −1 = b1 , · · · , bn 6= 0 for some b1 , · · · , b b1 , · · · , bn is unit-regular. In view of n ∈ R. Thus,  Corollary 16.3.7, b1 , · · · , bn is 0-clean, and therefore so is A. 

It is clear that every polynomial ring over a field is not clean. Furthermore, [417, Example 3.3] shows that every polynomial ring over a commutative ring is not semiclean. We end this section by noting that Theorem 16.3.2 provides an explicit way to represent such a matrix over polynomial rings as the sum of an idempotent matrix and an invertible matrix.

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Example 16.3.10. Let 1 + xy, x2 , y ∈ M3 Z[x, y] . Obviously, we have (1 + xy) · (1 − xy) + x2 · y + y · x2 (−1 + y) = 1. In addition, 1 − xy = 1 + y · (−x) + x2 (−1 + y) · 0. Thus, we say     1 −x 0 0 00 1x0 E = 0 1 0 −y 1 00 1 0 0 0 1 x2 (1 − y) 0 1  xy −x(1 − xy) 0 = −y 1 − xy 0  . 2 3 x (1 − y) x (1 − y) 1
Then E = E 2 ∈ M3 Z[x, y] and detE = 0. Let 

001



   1 + xy x2 y 1 x(1 + x − xy) y U = 0 0 0 − E =  y −1 + xy 0 . 2 3 0 0 0 −x (1 − y) −x (1 − y) −1
Then U ∈ GL3 Z[x, y] and detU = 1. This proves that   1 + xy, x2 , y     1 x(1 + x − xy) y xy −x(1 − xy) 0 = y −1 + xy 0  −y 1 − xy 0  +  2 3 2 3 −x (1 − y) −x (1 − y) −1 x (1 − y) x (1 − y) 1 is clean.



From the proceeding discussion, we see that it is hard to determine the cleanness of 2 × 2 matrices over an integral domain. We record a result due to Rajeswari and Aziz [358].  ab ∈ M2 (Z). Then (a) A is 1-clean cd if and only if detA − trA = 0 or − 2; and (b) A is 0-clean if and only if one of the following hold: Proposition 16.3.11. Let A =



ad − bc = ±1. ad − bc − a + bx = ±1 is solvable. ad − bc − d + bx = ±1 is solvable. The Diophantine equation (−b)x2 + (a − d)xy + cy 2 + bx + ad − bc −
a − 1 y = 0 has a solution (x, y) such that y 6= 0 and y | x − x2 . (5) The Diophantine equation (−b)x2 + (a − d)xy + cy 2 + bx + ad − bc −
a + 1 y = 0 has a solution (x, y) such that y 6= 0 and y | x − x2 . (1) (2) (3) (4)

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Proof. (a) It is easy to verify that A is 1-clean ⇔ there exists a U ∈ GL2 (Z) such that A = I2 + U a − 1 −b ∈ U (Z) ⇔ −c d − 1 ⇔ detA − trA = 0 or − 2.

 ab is 0-clean. Then there exists an cd idempotent E ∈ M2 (Z)  suchthat  A− E ∈ GL2 (Z), where detE = 00 00 10 0. Let Q = { , , }. One easily checks that E ∈ 00 ∗1 ∗0     x y 00 Q or , yz = x − x2 , where y 6= 0. If E = , then z 1−x 00   0 0 ad − bc = ±1. If E = for a w ∈ R, then ad − bc − a + bx = ±1 w1   1 0 is a solution w. If E = for a w ∈ R, then ad − bc − d + bx = ±1 w0   x y is a solution w. Otherwise, E = , yz = x − x2 , where y 6= 0. z 1−x   a − x b − y a−x b−y ∈ U (Z). Thus, ∈ GL2 (Z). Hence, c−z d−1+x c − z d − 1 + x This implies that u := (ad − bc) + (a − d)x + bz + cy −
a ∈ U (Z). As a result, (a − d)xy + b(x − x2 ) + cy 2 + ad − bc −
a − u y = 0. That is, 2 2 (−b)x + (a − d)xy + cy + bx + ad − bc − a − u = 0 has a solution (x, y) such that y 6= 0 and y | x − x2 . As u = ±1, we see that one of Conditions (3) and (4) hold. If one of Conditions (1), (2) and (3) holds, then A − E ∈ GL2 (Z) where E ∈ Q is an idempotent. Thus, A ∈ GL2 (Z) is clean. Suppose that
there exists a u = ±1 such that (−b)x2 +(a−d)xy+cy 2 +bx+ ad−bc−a+u y = 0 has a solution (x, y) such 6= 0 and y | x − x2 . Then, x − x2 = yz for  that y  x y a z ∈ R. Choose E = . Then E = E 2 . It is easy to verify that z 1−x a − x b − y = ad−a+ax−dx+x−x2 −bc+bz+yc−yz = det(A−E) = c − z d − 1 + x (a − d)x + bz + yc +
ad − bc − a. This implies that ydet(A − E) =
(a − d)x + bz +yc+ad−bc−a y = (−b)x2 +(a−d)xy +cy 2 +bx+ ad−bc−a y = −uy. As y 6= 0, we get det(A − E) = −u ∈ U (Z). Therefore A − E ∈ GL2 (Z), as required.  (b) Suppose that A =



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It is possible for a matrix to be 1-clean as well as 0-clean. Con   7 11 7 11 sider which is both 1-clean as well as 0-clean because = 1 3 1 3         6 11 10 0 −2 7 13 + = + . 1 2 01 0 1 1 2 16.4

Strong Cleanness

We say that a ∈ R is strongly clean provided that there exists an idempotent e ∈ R and a unit u ∈ R such that a = e + u and eu = ue. A ring R is strongly clean in the case when every element in R is strongly clean. As in the proof of Proposition 13.1.17, every strongly π-regular ring is strongly clean. The main purpose of this section is to investigate strongly clean 2×2 matrices over a ring. Now we begin with a simple fact which is due to Chen et al. ([180, Lemma 1.5]) Lemma 16.4.1. Let R be an integral domain. Then A  is an  ∈ M2 (R) a b idempotent if and only if A = 0 or A = I2 or A = where c 1−a bc = a − a2 in R. Proof. Let A ∈ M2 (R) be an idempotent. Then (detA)2 = detA. Hence, detA = 1 or detA = 0. If detA = 1, then A ∈ GL  2 (R), and so A = I2 . If ab detA = 0, then ad − bc = 0, where A = . As A2 = A, we deduce cd that a(a + d) = a, b(a + d) = b, c(a + d) = c and d(a + d) = d. If A 6= 0, then a + d = 1. Thus, d = 1 − a, and so bc = a − a2 , as required. The converse is obvious.    Proposition 16.4.2. Let R be an integral domain. Then [a, b] ∈ M2 (R) is strongly clean if and only if either a ∈ U (R) or 1 − a ∈ U (R). Proof.

If a ∈ U (R), then       ab 0 −a−1 b a (1 + a−1 )b = + . 00 0 1 0 −1

If 1 − a ∈ U (R), then 

ab 00



=



10 01



+



a−1 b 0 −1



.

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  In these [a, b] ∈ M2 (R) is strongly clean. Conversely, assume that  cases,  A := [a, b] is strongly clean. If I2 − A is invertible, then 1 − a ∈ GL2 (R). Thus, we may assume that I2 − A is a nonunit. Obviously, A is a nonunit. Write A = E + U withE = E 2 , U ∈ GL2 (R) and EU = U E. In view x y of Lemma 16.4.1, E = , where zy = x − x2 . In addition, U = z 1−x   ∗ ∗ . It follows from EA = AE that az = 0, thus we may assume −z x − 1 that z = 0. We infer that x(1 − x) = 0, and so x = 0 or  x =  1. As 0y U ∈ GL2 (R), x 6= 1. This implies that x = 0; hence, E = . Since 01   a b−y A−E = ∈ GL2 (R), we get a ∈ U (R), as asserted.  0 −1   Corollary 16.4.3. Let a, b ∈ Z. Then [a, b] ∈ M2 (Z) is strongly clean if and only if a = 0, ±1, 2.

Proof. Clearly, a ∈ U (Z) or 1 − a ∈ U (Z) if and only if a = 0, ±1, 2. Therefore we complete the proof by Proposition 16.4.2.    According to Corollary 16.4.3, [9, 7] ∈ M2 (Z) is not strongly clean. But it is clean from Theorem 16.3.2. Thus, every strongly clean ring is clean, but the converse is not true. For several kinds of 2 × 2 matrices over Z, we can derive accurate characterizations. Let A = (aij ) ∈ M2 (R), sA = a11 − a22 and tA = (trA)2 − 4detA. Now we extend [180, Lemma 1.6] as follows. Lemma 16.4.4. Let R be an integral domain, and let A ∈ M2 (R). If A is strongly clean in M2 (R), then either A ∈ GL2 (R), or I2 − A ∈ GL2 (R), or there exists q ∈ R such that s2A = tA q 2 . Proof. Suppose that A and I2 − A are nonunits. In view of Lemma 16.4.1, there exist a, b, c ∈ R such that A = E + (A − E), where   a b E = E2 = , bc = a − a2 , EA = AE and A − E ∈ GL2 (R). c 1−a As EA = AE, we get sA b = a12 (2a − 1) and sA c = a21 (2a − 1). Hence, s2A (a − a2 ) = s2A bc = a12 a21 (2a − 1)2 . That is, (s2A + 4a12 a21 )(a − a2 ) = a12 a21 . Obviously, tA = s2A + 4a12 a21 . Therefore s2A = tA − 4a12 a21 = tA − 4tA (a − a2 ) = tA (1 − 2a)2 , as required. 

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Proposition 16.4.5.



x+1 x y x



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∈ M2 (Z) is strongly clean if and only if

(1) x = −1, y = −3, 0, ±1; (2) x = 0; (3) x = 1, y = 0, ±1, 3.   x+1 x Proof. Let A = . If Condition (1) holds, then x = 0. Clearly, y x       10 0 0 1 0 = + , y0 −y 1 2y −1

thus A is strongly clean. If x = 1, y = ±1, 3 or x = −1, y = ±1, −3 hold, then A ∈ GL2 (Z) or I2 − A ∈ GL2 (Z); hence, A is strongly clean. In addition, we have these strongly clean expressions:       21 11 10 = + and 01 00 01       0 −1 1 −1 −1 0 = + . 0 −1 0 0 0 −1

Thus, A is strongly clean provided that   x = ±1, y = 0, as required. x+1 x Conversely, assume that ∈ M2 (Z) is strongly clean. If x 6= y x ±1, then A and I2 − A are nonunits. In view of Lemma 16.4.4, we have q ∈ Z such that 1 = (1 + 4xy)q 2 . This implies that 1 + 4xy = 1; hence, x = 0 or y = 0. If y = 0, then we have   a b E= and U ∈ M2 (R) c 1−a such that



x+1 x 0 x



= E + U, E = E 2 , EU = U E.

2 It is easy to verify  that c = 0, and  so a = a. This implies that a = 0, 1. If x+1 ∗ a = 0, then U = ∈ GL2 (Z), and thus x2 − 1 = ±1. Hence, 0 x−1   x∗ x = 0. If a = 1, then U = ∈ GL2 (Z), and thus x = ±1. This gives 0x a contradiction. Thus, we will further consider the case x = ±1. If x = 1 and y 6= 0, ±1, 3, then A and I2 − A are nonunits, and so we have some a, b, c ∈ R such that     21 a b = +U y1 c 1−a

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is a strongly clean expression. It follows from       21 a b a b 21 = y1 c 1−a c 1−a y1

1 6∈ Z for all that (4y + 1)a2 − (4y + 1)a + y = 0. Thus, a = 21 ± 2√4y+1 y ∈ Z − {0}, a contradiction. If x = −1 and y 6= −3, 0, ±1, then A and I2 − A are nonunits, and so we have some a, b, c ∈ R such that     0 −1 a b = +U y −1 c 1−a

is a strongly clean expression. It follows from       0 −1 a b a b 0 −1 = y −1 c 1−a c 1−a y −1

1 that (4y − 1)a2 − (4y − 1)a + y = 0. Thus, a = 21 ± 2√4y−1 6∈ Z for all y ∈ Z − {0}, a contradiction. Therefore the proof is true. 



 x−1 x Analogously, we prove that ∈ M2 (Z) is strongly clean if and y x only if (1) x = −3, y = −7; (2) x = −1, y = −7, −5, −3, −1; (3) x = 0; (4) x = 1, y = 0, ±1; (5) x = 2, y = 0; and (6) x = 3, y = 3. Recall that a ring R is called an ID-ring provided that every idempotent matrix over R admits an idempotent diagonal reduction (cf. [267] and [268]). The following classes of commutative rings are ID-rings: (1) πregular rings; (2) semilocal rings; (3) artinian rings; and (4) polynomial rings in one variable over principal ideal rings. We say that a ring R is connected if it has no nontrivial idempotents. Theorem 16.4.6. Let R be a connected ID-ring. Then A ∈ M2 (R) is strongly clean if and only  if either A  ∈ GL2 (R), or I2 − A ∈ GL2 (R), or A 1 + w1 0 is similar to a matrix , where w1 , w2 ∈ U (R). 0 w2 Proof. If either A or clean. For  I2 − A is invertible,  then  A is strongly  1 + w1 0 10 w1 0 any w1 , w2 ∈ U (R), = + is strongly clean. 0 w2 00 0 w2 Thus, one direction is obvious. Conversely, assume that A ∈ M2 (R) is strongly clean. Then there exists an idempotent E ∈ M2 (R) and an invertible U ∈ M2 (R) such that A = E + U with EU = U E. Suppose that A and I2 − A are nonunits. Since

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E admits an idempotent diagonal matrix, we can find V, K ∈ GL2 (R) such that E = V F K, where F = F 2 = diag(f1 , f2 ). Set Y = V F V −1 , and X = I2 −E+Y . Then Y E = E and Y = EK −1 V −1 , where X −1 = I2 +E−Y . So EY = Y . As EV F = V F , we deduce that XV F = (I2 − E + Y )V F = V F . Further, EXV = Y V = V F . Thus, EXV = XV F . Let J = (XV )−1 . Then JEJ −1 = diag(f1 , f2 ). As diag(f1 , f2 ) ∈ M2 (R) is an idempotent, we deduce that f1 = 1, f2 = 0 or f1= 0,f2 = 1. Therefore there  exists  1 0 10 −1 −1 H ∈ GL2 (R) such that HEH = . Thus, HAH = + 00 00 HU H −1 . Set W = (wij ) := HU H −1 . It follows from EU = U E that     10 10 W = W ; hence, w12 = w21 = 0 and w11 , w22 ∈ U (R). 00 00   1 + w11 0 Therefore A is similar to , as required.  0 w22 Corollary 16.4.7. Let R be a local ring. Then A ∈ M2 (R) is strongly clean if and only if either A ∈ GL2 (R), or I2 −A ∈ GL2 (R), or A is similar to a diagonal matrix. Proof. Obviously, R is connected. In view of Theorem 13.2.17, R is an ID-ring. Therefore we complete the proof by Theorem 16.4.6.  Corollary 16.4.8. Let R be a commutative ring over which every nonzero finitely generated projective R-module is free. Then A ∈ M2 (R) is strongly clean if and only  if eitherA ∈ GL2 (R), or I2 −A ∈ GL2 (R), or A is similar 1 + w1 0 to a matrix , where w1 , w2 ∈ U (R). 0 w2 Proof. Let 0 6= e ∈ R. Then eR ∼ = R, and so e = 1. Thus, R is connected. According to Proposition 11.4.9, R is an ID-ring, and we are done by Theorem 16.4.6. 

Example 16.4.9. Let A ∈ M2 (Z). Then A ∈ M2 (Z) is strongly clean if and only if A ∈ GL2 (Z), GL2 (Z),orA is similar to one of the  orI2− A ∈    00 0 0 20 2 0 matrices in the set , , , . 01 0 −1 01 0 −1

Proof. Clearly, every nonzero finitely generated projective Z-module is free, and so the result follows by Corollary 16.4.8.  Let p, q ∈ Z be two distinct prime numbers, and let R = { m n ∈

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Q | (m, n) = 1, (n, pq) = 1}. Then all maximal ideals of R are pR and qR. Thus, R is a commutative semilocal ring. Hence, every nonzero finitely generated projective R-module is free. Thus, A ∈ M2 (R) is strongly clean if and only if either A ∈GL2 (R), or I2 − A ∈ GL2 (R), or A is similar 1 + w1 0 to a matrix , where w1 , w2 ∈ U (R). In this case, R is not 0 w2 local. Let K be a principal ideal domain. By using the Quillen-Suslin Theorem, every nonzero finitely generated
projective K[x]-module is free. Thus, by Corollary 16.4.8, A ∈ M2 K[x] is strongly clean if and only if

either A ∈ GL2 K[x] , or I2 − A ∈ GL2 K[x] , or A is similar to a ma  1 + w1 (x) 0 trix , where w1 (0), w2 (0) ∈ U (K) and the derivative 0 w2 (x) (n)

(n)

w1 , w2

are nilpotent for all n ∈ N.

Lemma 16.4.10. Let R be a commutative local ring. Then A ∈ is  M2 (R)  a b an idempotent if and only if either A = 0, or A = I2 , or A = c 1−a where bc = a − a2 in R. Proof. One direction is clear. Let A ∈ M2 (R) be an idempotent. Then
detA I2 −detA = 0. As R is local, either detA ∈ U (R) or 1−detA ∈ U (R).   ab Thus, detA = 1 or detA = 0. If detA = 1, then A = I2 . Write A = . cd If detA = 0, as in the proof of Lemma 16.4.1, ad − bc = 0, a(a +
d) = a, b(a + d) = b, c(a + d) = c and d(a + d) = d. If A 6∈ J M2 (R) , then one of the entries
of A is in U (R); hence, a + d = 1. Thus, bc = a − a2 . If A ∈ J M2 (R) , then I2 − A ∈ GL2 (R). This implies that A = 0, as required.  Lemma 16.4.11. Let R be a local ring, and let A ∈ M2 (R). Then  either  0λ A ∈ GL2 (R), or I2 − A ∈ GL2 (R), or A is similar to a matrix , 1µ where λ ∈ J(R), µ ∈ 1 + J(R).   ab Proof. Let A = . Case I. c ∈ U (R). Then one easily checks that cd   0λ −1 −1 −1 [c, 1]B12 (−ac )AB12 (ac )[c , 1] = for some λ, µ ∈ R. If λ ∈ 1µ U (R), then A ∈ GL2 (R). If λ ∈ J(R), 1 − µ ∈ U (R), [c, 1]B12 (−ac−1 )(I2 −

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1 −λ −1 1 − µ



573

 1 −λ A)B12 (ac )[c , 1] = = B21 (−1) ; hence, 0 1−λ−µ   0λ I2 − A ∈ GL2 (R). Otherwise, A is similar to a matrix , where 1µ     01 01 λ ∈ J(R), µ ∈ 1 + J(R). Case II. b ∈ U (R). Then A = 10 10   dc . Case III. c, b ∈ J(R), a − d ∈ U (R). Then B21 (−1)AB21 (1) = ba   a+b b . Case IV. c, b ∈ J(R), a, d ∈ U (R). Then a−d+b−c b−d   a b B21 (−ca−1 )A = ; hence, A ∈ GL2 (R). Case V. c, b, a, d ∈ 0 d − ca−1 b J(R). Then I2 − A ∈ GL2 (R). In any case, the proof is true.  −1



rings



We now give a new simple way to look at the following result which is well known (cf. [51, Proposition 30] and [416, Theorem 7]). Theorem 16.4.12. Let R be a commutative local ring. Then the following are equivalent: (1) M2 (R) is strongly clean. (2) For any w ∈ J(R), the equation x2 + x = w is solvable in R. (3) For any w1 , w2 ∈ J(R), the equation x2 + (1 + w1 )x + w2 = 0 is solvable in R.   0 w1 (4) For any w1 , w2 ∈ J(R), ∈ M2 (R) is strongly clean. 1 1 + w2   0 −w(1 + 4w)−1 Proof. (1) ⇒ (2) Let w ∈ J(R). Choose A = (aij ) = . 1 1 Then A and I2 − A are nonunits. By hypothesis  and Lemma 16.4.10, a b there exists a nontrivial idempotent E = ∈ M2 (R) such that c 1−a A − E ∈ GL2 (R) and EA = AE. As in the proof of Lemma 16.4.4, we get (s2A + 4a12 a21 )(a − a2 ) = a12 a21 . That is,
1 − 4w(1 + 4w)−1 (a − a2 ) = −w(1 + 4w)−1 .
−1 Hence, a − a2 = − 1 − 4w(1 + 4w)−1 w(1 + 4w)−1 = −w. Set b = −a. 2 Then b + b = w, as desired. (2) ⇒ (3) For any w1 , w2 ∈ J(R), there exists some w0 ∈ J(R) such that (1 + w1 )−1 = 1 + w0 . By hypothesis, we can find a ∈ R such that
2 a2 +a = −(1+w0 )2 w2 . Set b = (1+w0 )−1 a. Then (1+w0 )b +(1+w0)b =

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−(1+w0 )2 w2 . Hence, b2 +(1+w0 )−1 b = −w2 , and so b2 +(1+w1 )b+w2 = 0. Therefore x2 + (1 + w1 )x = 0 is solvable in R.  + w2  0 a12 (3) ⇒ (4) Let A = , where a12 ∈ J(R), a22 ∈ 1 + J(R). Let 1 a22 a11 = 0 and a21 = 1. Clearly, A and I2 − A are nonunits. We see that sA = −a22 ∈ U (R) and tA = s2A + 4a12 a21 ∈ U (R). 2 Let w = −t−1 A a12 a21 ∈ J(R). By hypothesis, x + x = w is solvable, 2 and so is x − x = w. If x 6∈ U (R), then x ∈ J(R), and so 1 − x ∈ U (R). Observing that (1 − x)2 − (1 − x) = w is solvable, we may assume that x2 − x = w has a solution a ∈ U (R) without loss of generality. Set −1 2 b = s−1 A a12 (2a − 1) and c
= sA a21 (2a − 1).
Then sA−1bc = a12 a21 (2a − 2 2 1) = a12 a21 4(a − a) + 1 = a12 a21 4w + 1 = a12 a21 tA (tA − 4a12 a21 ) =   a b 2 2 2 2 2 a12 a21 t−1 s = −s w = −s (a − a). So bc = a − a . Let E = . A A A A c 1−a Then E = E 2 ∈ M2 (R) by Lemma 16.4.10. As a21 b = a12 c, we see that a11 a + a21 b = a11 a + a12 c and a12 c + a22 (1 − a) = a21 b + a22 (1 − a). Since sA b = a12 (2a − 1) and sA c = a21 (2a − 1), we get a12 a + a22 b = a11 b + a12 (1 − a) and a11 c + a21 (1 − a) = a21 a + a22 c. Thus, we have EA = AE. It suffices to show that A − E ∈ GL2 (R). Observe that det(A − E) = (a11 − a)(a22 + a − 1) − (a12 − b)(a21 − c) = detA + sA a − a11 + 2a12 c ∈ U (R). Therefore A ∈ M2 (R) is strongly clean, as asserted. (4) ⇒ (1) Let B ∈ M2 (R). If B is invertible or I2 − B is invertible, then B ∈ M2 (R) is strongly clean. Now assume that B and I2 − B are nonunits. In view of Lemma  16.4.11, there exists a U ∈ GL2 (R) such 0 a12 that A := U BU −1 = , where a12 ∈ J(R), a22 ∈ 1 + J(R). By 1 a22 hypothesis, A is strongly clean, and then so is B. Therefore M2 (R) is strongly clean.  Corollary 16.4.13. Let R be a commutative local ring. If the following are equivalent:

1 2

∈ R, then

(1) M2 (R) is strongly clean. (2) For any w ∈ J(R), the equation x2 − 1 = w is solvable in R. (3) For any A ∈ M2 (R), A ∈ GL2 (R) or I2 − A ∈ GL2 (R) or there exists q ∈ R such that s2A = tA q 2 .

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Proof. (1) ⇒ (3) Analogously to the proof of Lemma 16.4.4, it follows from Lemma 16.4.10.
  0 − 14 1 − (1 + w)−1 ∈ (3) ⇒ (2) Let w ∈ J(R), and let A = 1 1 M2 (R). Then A and I2 − A are nonunits. By
hypothesis, there exists a q ∈ R such that 1 = 1 − 41 (1 − (1 + w)−1 ) q 2 . As a result, q 2 − 1 =
−1 1 − 14 (1 − (1 + w)−1 ) − 1 = w, as desired. (2) ⇒ (1) For any w ∈ J(R), we can find some a ∈ R such that 2 a2 − 1 = 4w. Set b = a−1 2 . Then a = 2b + 1, and so 4(b + b) = 4w; hence, 2 x + x = w has a solution of b ∈ R. According to Theorem 16.4.12, M2 (R) is strongly clean.  The condition 12 ∈
R in Corollary 16.4.13 is necessary. Let Z4 =
{0, 1, 2, 3}. Then J Z4 = {0, 2}. According to Theorem 16.4.12, M2 Z4 is strongly clean. But the equation x2 − 1 = 2 is not solvable in Z4 . Let e ∈ R be an idempotent. If R is strongly clean, then so is eRe (cf. [184, Theorem 2.4]). This raises a natural question: Is strong cleanness Morita invariant? Let p ∈ Z be prime, and let Z(p) , { m n | m, n ∈ Z, p ∤ n}, the ring of integers localized at the prime ideal (p). Then Z(p) is a local integral domain. In addition, J(Z(p) ) = { pm n | m, n ∈ Z, p ∤ n}. Clearly, p ∈ J(Z(p) ). Observe that x2 +x+p = 0 is not solvable in Z(p) . According to Theorem 16.4.12, M2 (Z(p) ) is not strongly clean. Further, Z(p) is strongly clean. Thus, the answer to the  question Let R = M2 (Z), and    is  negative.  23 10 13 let a = ∈ R. Then a = + is strongly clean. Choose 13 01 12       11 33 ZZ e= . If eae = ∈ = eRe is strongly clean, then 3 ∈ Z 00 00 0 0 is clean. This gives a contradiction. Therefore eae ∈ eRe is not strongly clean. This means that the preceding result can not be extended to a single element. A long standing question asks whether strongly clean rings have stable range one (cf. [336]). So as to construct a natural subclass of strongly clean rings which have stable range one, we replace U (R) by J(R) and parallel to introduce the concept of strongly J-cleanness. We say that an element a ∈ R is strongly J-clean provided that there exists an idempotent e ∈ R and an element w ∈ J(R) such that a = e + w and ew = we. A ring R is strongly J-clean in the case where every element in R is strongly J-clean. Proposition 16.4.14. Every strongly J-clean element is strongly clean.

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Proof. Let x ∈ R be strongly J-clean. Then there exists an idempotent e ∈ R and a w ∈ J(R) such that x = e + w and ew = we. Hence, x = (1 − e) + (2e − 1 + w). As (2e − 1)2 = 1, we see that x − (1 − e) ∈ U (R). Thus, x ∈ R is strongly clean.  Let R = { m n | m, n ∈ Z, 3 ∤ n}. Then R is a local ring with the Jacobson radical J(R) = 3R. As every local ring is strongly clean, so is
2 R. Obviously, 12 ∈ R satisfies 21 − 21 = 12 6∈ J(R); hence, 21 ∈ R is not strongly J-clean. This implies that the converse of Proposition 16.4.14 is not true, that is, the definition of strongly J-cleanness is different from that of strong cleanness. An element e ∈ R/J(R) is said to lift strongly modulo J(R) provided that there exists an idempotent f ∈ R such that e − f ∈ J(R) and ef = f e. Proposition 16.4.15. Let R be a ring. Then the following are equivalent: (1) R is a strongly J-clean ring. (2) R/J(R) is Boolean and each idempotent lifts strongly modulo J(R). (3) R/J(R) is Boolean and R is strongly clean. Proof. (1) ⇒ (2) Clearly, R/J(R) is Boolean. Let e ∈ R/J(R) be idempotent. By hypothesis, there exists an idempotent f ∈ R such that e − f ∈ J(R) and ef = f e. Thus, e lifts strongly modulo J(R), as required. (2) ⇒ (1) For any x ∈ R, then x ∈ R/J(R) is an idempotent. By assumption, we can find an idempotent e ∈ R such that x − e ∈ J(R) and ex = xe. Set w = x−e. Then x = e+w, w ∈ J(R) and ew = we. Therefore R is a strongly J-clean ring. (1) ⇒ (3) is obvious from Proposition 16.4.14. (3) ⇒ (1) As R/J(R) is Boolean, (2)2 = 2, and so 2 ∈ J(R). For any x ∈ R, by hypothesis, there exists an idempotent e ∈ R such that u := x − e ∈ U (R) and ex = xe. As u = u2 , we deduce that u ∈ 1 + J(R). Thus,
x = (1 − e) + 2e + (x − e − 1) is a strong J-clean decomposition, as required.



Since every Boolean ring has stable range one, it follows from Proposition 16.4.15 that every strongly J-clean ring has stable range one. Thus, for any finitely generated projective right modules A,B and C over a strongly J-clean ring, A ⊕ B ∼ = A ⊕ C implies that B ∼ = C.

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Recall that a ring R is a uniquely clean ring provided that each element in R has a unique representation as the sum of an idempotent and a unit (cf. [340]). In [3], Anderson and Camillo asked whether the homomorphic image of a uniquely clean ring is uniquely clean. Nicholson and Zhou gave an affirmative answer to this question (cf. [340, Theorem 22]). Now we include a proof of this fact in a different way. Corollary 16.4.16. Let R be a ring. Then the following are equivalent: (1) R is a uniquely clean ring. (2) R is strongly J-clean with all idempotents central. (3) Every homomorphic image of R is a uniquely clean ring. Proof. (1) ⇒ (2) Let R be a uniquely clean ring. In view of [340, Theorem 20], every idempotent in R is central. For any x ∈ R, by using [340, Theorem 20] again, there exists a unique idempotent e ∈ R such that e − a ∈ J(R). Hence, a = e + w for a w ∈ J(R). In addition, ew = we. Consequently, R is strongly J-clean. (2) ⇒ (1) Clearly, R is strongly clean. Let x ∈ R. Write x = e1 + u1 = e1 + u2 , e1 = e21 , e2 = e22 and u1 , u2 ∈ U (R). Then x = e1 + u1 = e2 + u2 . According to Proposition 16.4.15, R/J(R) is Boolean; hence, u1 = u2 = 1. This implies that e1 = e2 , and so e1 − e2 ∈ J(R). Further, we deduce that e2 (1 − e1 ) = (e1 − e2 )(e1 − 1) ∈ J(R). As every idempotent in R is central, e2 (1 − e1 ) ∈ R is an idempotent, thus, e2 (1 − e1 ) = 0. It follows that e2 = e2 e1 . Likewise, e1 = e1 e2 . As a result, we derive that e1 = e2 , and then u1 = u2 . Therefore R is a uniquely clean ring. (3) ⇒ (1) is trivial. (1) ⇒ (3) Since R is a uniquely clean ring, it follows from (1) ⇒ (2) that R is a strongly J-clean ring with all idempotents central. As R is clean, every idempotent lifts modulo any ideal of R. Thus, every idempotent in the homomorphic image of R is central as well. It is easy to verify that every homomorphic image of R is strongly J-clean again. According to (2) ⇒ (1), we conclude that every homomorphic image of R is a uniquely clean ring, as asserted.  Immediately, we deduce that for a commutative uniquely  clean  ring Z2 Z2 ∼ R, R/M = Z2 for all maximal ideals M of R. Let R = . By 0 Z2 [340, Example 21], R is not uniquely clean. But it is strongly J-clean (cf. Corollary 16.4.24 below). So we get { uniquely clean rings } $ { strongly

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J-clean rings } $ { strongly clean rings }. Example 16.4.17. Here are some examples and non-examples. (1) Any quotient of a strongly J-clean ring is strongly J-clean. (2) Any direct product of strongly J-clean rings is strongly J-clean. (3) Every nil clean local ring, i.e., a local ring in which each element is the sum of an idempotent and a nilpotent element, is strongly J-clean. (4) R[x] is never a strongly J-clean ring as it is not clean. (5) Every power series ring over a strongly J-clean ring is strongly J-clean. But Z2 [[x]] is not nil clean. (6) R[x]/(xn ) is strongly J-clean if and only if so is R. (7) If R is commutative, then R is strongly J-clean if and only if so is the   ab ring T = { | a, b ∈ R}.  0a Lemma 16.4.18. Let R be a ring, and let a = e + w be a strongly J-clean decomposition of a in R. Then ℓ(a) ⊆ ℓ(e) and r(a) ⊆ r(e). Proof. Let r ∈ ℓ(a). Then ra = 0. Write a = e + w, e = e2 , w ∈ J(R) and ew = we. Thus re = −rw; hence, re = −rwe = −rew. It follows that re(1 + w) = 0, and so re = 0. That is, r ∈ ℓ(e). Therefore ℓ(a) ⊆ ℓ(e). A similar argument shows that r(a) ⊆ r(e).  Proposition 16.4.19. Let R be a ring, and let f ∈ R be an idempotent. Then a ∈ f Rf is strongly J-clean in R if and only if a is strongly J-clean in f Rf . Proof. Suppose that a = e + w, e = e2 ∈ f Rf, w ∈ J(f Rf ) and ew = we. Obviously, w ∈ f J(R)f ⊆ J(R). Hence, a ∈ f Rf is strongly J-clean in R. Conversely, suppose that a = e + w, e = e2 ∈ R, w ∈ J(R) and ew = we. T T As a ∈ f Rf , we see that 1 − f ∈ ℓ(a) r(a) ⊆ ℓ(e) r(e) = R(1 − T e) (1 − e)R = (1 − e)R(1 − e). Hence, ef = e = f e. We observe that a = f ef + f wf , (f ef )2 = f ef , f wf ∈ f J(R)f = J(f Rf ). Furthermore, f ef · f wf = f ewf = f wef = f wf · f ef . Therefore we obtain the result. Corollary 16.4.20. Let R be a ring, and let e ∈ R be an idempotent. If R is strongly J-clean, then so is eRe. Proof. Let a ∈ eRe. As R is strongly J-clean, we see that a ∈ eRe is strongly J-clean in R. According to Proposition 16.4.19, a ∈ eRe is strongly J-clean in eRe, as asserted. 

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The strongly J-clean condition possesses many nice properties, e.g., 2 is in the Jacobson radical for each strongly J-clean ring and a ring is Boolean if and only if it is strongly J-clean with the zero Jacobson radical. We will consider the strong J-cleanness of the ring of upper triangular matrices over a local ring. These results also hold for the ring of all lower triangular matrices in a similar fashion. Lemma 16.4.21. Let R be a local ring. Then the following are equivalent: (1) R is strongly J-clean. (2) R is uniquely clean. (3) R/J(R) ∼ = Z2 . Proof. (1) ⇒ (3) Assume that R is strongly J-clean. It follows from Proposition 16.4.15 that R/J(R) is Boolean. On the other hand, R/J(R) is a field. Therefore R/J(R) ∼ = Z2 . (3) ⇒ (1) Suppose that R/J(R) ∼ = Z2 . For any x ∈ R, x ∈ J(R) or x − 1 ∈ J(R). Hence, x ∈ R is strongly J-clean, as desired. (2) ⇔ (3) follows from [340, Theorem 15].  Let a ∈ R. Then la : R → R and ra : R → R denote, respectively, the abelian group endomorphisms given by la (r) = ar and ra (r) = ra for all r ∈ R. Thus, la − rb is an abelian group endomorphism such that (la − rb )(r) = ar − rb for any r ∈ R. Following Diesl, a local ring R is bleached provided that for any a ∈ U (R), b ∈ J(R), la −rb , lb −ra are both surjective. The class of bleached local rings contains many familiar examples, e.g., commutative local rings, local rings with nil Jacobson radicals, local rings for which some power of each element of their Jacobson radicals is central (cf. [50, Example 13]). Lemma 16.4.22. Let R be a local ring and suppose that A = (aij ) ∈ T Mn (R). Then for any set {eii } of idempotents in R such that eii = ejj whenever laii − rajj is not a surjective abelian group endomorphism of R, there exists an idempotent E ∈ T Mn(R) such that AE = EA and Eii = eii for every i ∈ {1, · · · , n}. Proof. See [50, Lemma 7].



Theorem 16.4.23. Let R be a local ring, and let n ≥ 2. Then the following are equivalent: (1) T Mn(R) is strongly J-clean.

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(2) R is bleached and R/J(R) ∼ = Z2 . Proof. (1) ⇒ (2) Let e = diag(1, 0, · · · , 0) ∈ T Mn (R). Then R ∼ = eT Mn(R)e. According to Corollary 16.4.20, R is strongly J-clean. It follows from Lemma 16.4.21 that R/J(R) ∼ = Z2 . By Corollary 16.4.20 again, T M2 (R) is strongly J-clean. Let a ∈ U (R) and b ∈ J(R). We will show that la − rb : R → R is surjective. For any  v ∈ R, it suffices to find some av x ∈ R such that ax − xb = v. Let r = . As R/J(R) ∼ = Z2 , we see 0b that a ∈ 1+J(R). The idempotents in R are 0 and 1. Thus, the idempotent in T M2 (R) is one of the following: 

0x 00

       0x 1x 1x , , , or . 01 00 01

Since T M2 (R) is strongly J-clean, there exists an idempotent e ∈ TM2 (R) 
1x such that r − e ∈ J T M2 (R) and er = re. This implies that e = ; 00
otherwise, r − e 6∈ J T M2 (R) . As er = re, we deduce that ax − xb = v. Thus, la − rb : R → R is surjective. Analogously, lb − ra : R → R is surjective. Therefore R is bleached. (2) ⇒ (1) Let A = (aij ) ∈ T Mn (R). We need to construct an idempo
tent E ∈ T Mn(R) such that EA = AE and such that
A−E ∈ J T Mn (R) . S As R/J(R) ∼ 1 + J(R) . Begin by construct= Z2 , we see that R = J(R) ing the main diagonal of E. Set eii = 0 if aii ∈ J(R), and set eii = 1 otherwise. Thus, aii − eii ∈ J(R) for every i. If eii 6= ejj , then it must be the case (without loss of generality) that aii ∈ U (R) and ajj ∈ J(R). As R is bleached, laii − rajj : R → R is surjective. According to Lemma 16.4.22, there exists an idempotent E ∈ T Mn (R) such that AE = EA
and Eii = eii for every i ∈ {1, · · · , n}. In addition, A − E ∈ J T Mn (R) . Therefore the result follows.  Corollary 16.4.24. Let R be a commutative local ring. Then the following are equivalent: (1) T Mn(R) is strongly J-clean. (2) R/J(R) ∼ = Z2 . Proof. (1) ⇒ (2) is obvious from Theorem 16.4.23. (2) ⇒ (1) As R is a commutative local ring, it is bleached. Therefore the result follows from Theorem 16.4.23. 

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Corollary 16.4.25. Let R be a local ring. Then T M2 (R) is strongly J-clean if and only if (1) R/J(R) ∼ = Z2 .
(2) For any a ∈ 1+J(R), b ∈ J(R) and v ∈ R, there exists P ∈ U T M2 (R)     av a0 such that P −1 P = . 0b 0b Proof. Suppose that T M2(R) is strongly J-clean. By virtue of Theorem 16.4.23, R is bleached and R/J(R) ∼ = Z2 . Let a ∈ 1 + J(R), b ∈ J(R) and v ∈ R. Then la − rb : R → R is surjective. Hence, there exists x ∈ R such that ax − xb = v. It is easy to verify that    −1   1x av 1x a0 = , 01 0b 01 0b as required.



 av Conversely, assume that Conditions (1) and (2) hold. Let ∈ 0b T M2 (R).   If either a ∈ J(R), b ∈ J(R) or a ∈ 1 + J(R), b ∈ 1 + J(R), then av ∈ T M2(R) is strongly J-clean. If a ∈ 1+J(R), b ∈ J(R), then there 0b    
av a0 exists some P ∈ U T M2 (R) such that P −1 P = . In view of 0b 0b Lemma 16.4.21, R is strongly J-clean. Thus, we have idempotents e, f ∈ R and v, w ∈J(R)such that = f + w2, ew1 = w1 e, f w2 = w2 f .  a=  e + w1 , b  av e0 w1 0 Therefore = P P −1 + P P −1 . If a ∈ J(R) and 0b 0f 0 w2       av 10 1 − a −v b ∈ 1 + J(R), then = − . As 1 − a ∈ 1 + 0b 01 0 1−b J(R) and 1 − b ∈ J(R), by exists  the preceding  discussion,   there   P ∈
1 − a v e 0 w 0 1 U T M2 (R) such that P −1 P = + , where 0 1−b 0f 0 w2 e = e2 , f =f 2 ∈ R;  w1 , w2 ∈ J(R),  ew  1 = w1 e and  f w2 = w2 f . As a result, a v 1 − e 0 −w 0 1 P −1 P = + , as asserted.  0b 0 1−f 0 −w2 Let R be a local ring. It follows from Corollary 16.4.25 that T M2
(R) is strongly J-clean if
and only if for any A ∈ T M2(R),
A ∈ J T M2 (R) or I2 − A ∈ J T M2 (R) or there exists P ∈ U T M2 (R) such that P −1 AP =   1 + w1 0 , where w1 , w2 ∈ J(R). This fact could be extended to the 0 w2

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ring of full matrices over a local ring. Proposition 16.4.26. Let R be a local
ring. Then A ∈ M2 (R)
is strongly J-clean if and only if A ∈ J M2 (R) , or I2 − A ∈ J M2 (R) , or A is   1 + w1 0 similar to a matrix , where w1 , w2 ∈ J(R). 0 w2
Proof. If either A or I2 − A is in J M2 (R) , then A is strongly J-clean.       1 + w1 0 10 w1 0 For any w1 , w2 ∈ J(R), = + is strongly 0 w2 00 0 w2 J-clean. Thus, one direction is obvious. Conversely, assume that A ∈ M2 (R) is strongly J-clean. Then there
exists an idempotent E ∈ M2 (R) and a W ∈ J M2 (R) such that A =
E + W with EW = W E. Suppose that A and I2 − A are not in J M2 (R) . As in the proof a H ∈ GL2 (R) such that  of  Theorem 16.4.6, there  exists  1 0 1 0 HEH −1 = . Thus, HAH −1 = + HW H −1 . Set V = (vij ) := 00 00     10 10 HW H −1 . It follows from EW = W E that V =V ; hence, 00 00   1 + v11 0 v12 = v21 = 0 and v11 , v22 ∈ J(R). Therefore A is similar to , 0 v22 as required.  Though it is hard to determine if a 2 × 2 matrix over local rings is strongly J-clean, surprisingly, we can derive the following. Corollary 16.4.27. Let R be a commutative local ring. Then M2 (R) is not strongly J-clean.   11 Proof. Let A = ∈ M2 (R). Then A ∈ GL2 (R) and I2 −A ∈ GL2 (R). 10

Thus, A 6∈ J M2 (R) and I2 − A 6∈ J M2 (R) . Suppose that A ∈ M2 (R) is stronglyJ-clean.  In view of Proposition 16.4.26, A is similar to a diagonal α0 matrix , where α ∈ 1 + J(R), β ∈ J(R). Hence, −1 = detA = αβ ∈ 0β J(R). This gives a contradiction. Therefore A ∈ M2 (R) is not strongly J-clean.  Recall that a matrix A ∈ M2 (R) is called singular if A is not invertible. A singular matrix A ∈ M2 (R) is called purely singular if I2 − A is

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singular. Let R be a local ring, and let A ∈ M2 (R) be purely singular. Then A ∈ M 2 (R) is strongly J-clean if and only if A is similar to a matrix  1 + w1 0 , where w1 , w2 ∈ J(R). 0 w2 Proposition 16.4.28. Let R be a local ring, and let A ∈ M2 (R). Then the following are equivalent: (1) A ∈ M2 (R) is strongly clean. (2) A ∈ GL2 (R) or I2 − A ∈ GL2 (R) or A ∈ M2 (R) is strongly J-clean. Proof. (1) ⇒ (2) Let A ∈ M2 (R). If A is not a purely singular matrix, then either A ∈ GL2 (R) or I2 − A ∈ GL2 (R). If A is a purely singular matrix, in view 2], there exist w1 , w2 ∈ J(R) such that A  of [416, Lemma  1 + w1 0 is similar to . Hence, we have some P ∈ GL2 (R) such that 0 w    2  10 w1 0 P −1 AP = + , and so 00 0 w2     10 w1 0 A=P P −1 + P P −1 , 00 0 w2 as required. (2) ⇒ (1) is clear from Proposition 16.4.14.



Let p be prime, and let M = { pmn | m ∈ Z, n ∈ N} be an additive
cp = endZ Zp∞ subgroup of Q with subgroup Z. Denote Zp∞ = M/Z. Let Z be the ring of p-adic integers.

b p ) is strongly Corollary 16.4.29. Every purely singular matrix in M2 (Z J-clean.

b p is commutative local. In view of [180, Theorem 2.4], Proof. Clearly, Z b M2 (Zp ) is strongly clean. Therefore we complete the proof from Proposition 16.4.28.  Recall that a ring R is strongly nil clean provided that for any x ∈ R there exists an idempotent e ∈ R and a nilpotent w ∈ R such that x = e + w and ew = we. Clearly, every commutative strongly nil clean ring ∞ ` is strongly J-clean. But the converse is not true. Consider R = Z2n . n=1

For each n ∈ N, Z2n is a local ring with the maximal ideal 2Z2n . Thus, Z2n /2Z2n ∼ = Z2 . Hence, Z2n is strongly J-clean by Lemma 16.4.21, and so

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R is strongly J-clean. Since the element r = (0, 2, 2, · · · ) ∈ R is certainly not strongly nil clean, R is not strongly nil clean. As in the proof of Proposition 16.4.28, we deduce that every purely singular 2 × 2 matrix over a commutative local ring R is strongly nil clean if and only if J(R) is nil. As every idempotent matrix over local rings admits a diagonal reduction, it follows that each purely singular 2 × 2 matrix A over a commutative local artinian ring satisfies these conditions: A = E + W, E is similar to a diagonal matrix, W ∈ M2 (R) is nilpotent and E and W commutate. It is worth noting that such a decomposition over a field is called the JordanChevalley decomposition in Lie algebra theory. Lemma 16.4.30. Let R be a commutative ring, and let A = (aij ) ∈ M2 (R). If a21 ∈ U (R) and the equation x2 − trA · x + detA = 0 has two roots x1 , x2 ∈ R such that x1 − x2 ∈ U (R), then A is similar to diag(x1 , x2 ). Proof. As x1 , x2 ∈ R are distinct roots of x2 − trA · x + detA = 0, we get x21 − trA · x1 + detA = 0 and x22 − trA · x2 + detA = 0. Hence, (x1 − x2 )(x1 + x2 −  trA) = 0. This implies that trA = x1 + x2 and detA = x1 x2 . Set    x1 − a22 x2 − a22 a a22 − x2 21 −1 P = . Then P −1 = a21 (x1 − x2 )−1 a21 a −a21 x1 − a22  21  x1 and P −1 AP = , as required.  x2 Theorem 16.4.31. Let R be a commutative local ring. Then the following are equivalent: (1) A ∈ M2 (R) is
strongly J-clean.
(2) A ∈ J M2 (R) or I2 − A ∈ J M2 (R) or the equation x2 − trA · x + detA = 0 has a root in J(R) and a root in 1 + J(R). Proof. (1) ⇒ (2) Let A ∈ M2 (R)
be strongly J-clean. Suppose that A, I2 −A ∈ M2 (R) are not in J M2 (R) . In view of Proposition 16.4.26, A is   λ0 similar to the matrix B = ∈ M2 (R), where λ ∈ J(R), µ ∈ 1 + J(R). 0µ Thus, x2 − trA · x + detA = det(xI2 − A) = det(xI2 − B) = (x − λ)(x − µ). Hence, x2 − trA · x + detA = 0 has a root λ ∈ J(R) and a root µ ∈ 1 + J(R). (2) ⇒
(1) Let A ∈ M2 (R). If either A ∈ M2 (R) or I2 − A ∈ M2 (R) is in J M2 (R) , then A ∈ M2 (R) is strongly J-clean. Otherwise, it follows by hypothesis that the equation x2 − trA · x + detA = 0 has a root x1 ∈ J(R) and a root x2 ∈ 1 + J(R). Clearly, x1 − x2 ∈ U (R). Similarly to Lemma 16.4.30, we deduce that trA = x1 + x2 and detA = x1 x2 . As detA ∈

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J(R), A 6∈ GL2 (R). Obviously, det(I2 − A) = 1 − trA + detA ∈ J(R), and so I2 − A 6∈ GL2 (R). According to Lemma 16.4.11,  there  are some 0λ λ ∈ J(R), µ ∈ 1 + J(R) such that A is similar to B = . Obviously, 1µ x2 −trB·x+detB = det(xI2 −B) = det(xI2 −A) = x2 −trA·x+detA; and so x2 −trB ·x+detB = 0 has a root in 1+J(R) and a root in J(R). According   α0 −1 to Lemma 14.4.30, there exists a P ∈ GL2 (R) such that P BP = 0β     00 α 0 for some α ∈ J(R), β ∈ 1 + J(R). Thus, P −1 BP = + 01 0 β−1 is a strongly J-clean expression. Therefore A ∈ M2 (R) is strongly J-clean, as desired.  Corollary 16.4.32. Let R be a commutative local ring, and let A ∈ M2 (R). Then the following are equivalent: (1) A ∈ M2 (R) is
strongly J-clean.
(2) A ∈ J M2 (R) or I2 − A ∈ J M2 (R) , or trA ∈ 1 + J(R) and the equation x2 −
x = − detA tr 2 A has a root in
J(R). (3) A ∈ J M2 (R) or I2 − A ∈ J M2 (R) , or trA ∈ 1 + J(R), detA ∈ J(R) detA and the equation x2 − x = tr2 A−4detA is solvable.
Proof. (1) ⇒ (2) Assume that A, I2 − A ∈ M2 (R) are not in J M2 (R) . In   λ0 view of Proposition 16.4.26, A is similar to , where λ ∈ J(R), µ ∈ 0µ 1 + J(R). Thus, trA = λ + µ ∈ 1 + J(R), detA = λµ. Clearly, y 2 − (λ + µ)y + λµ = 0 has a root in J(R). Thus, so does the equation (λ + µ)−1 y 2 − y = −(λ + µ)−1 λµ. Set x = (λ + µ)−1 y. Then (λ + µ)x2 − (λ + µ)x = −(λ + µ)−1 λµ. We infer that x2 − x = −(λ + µ)−2 λµ. Therefore x2 − x = − detA tr 2 A has a root in J(R). (2) ⇒ (3) Suppose that trA ∈ 1+J(R) and the equation x2 −x = − detA tr 2 A has a root a ∈ J(R). Then detA ∈ J(R). It is easy to verify that
2
detA
a(2a − 1)−1 − a(2a − 1)−1 = =

=

tr 2 A· 4(a2 −a)+1 detA
tr 2 A· −4(trA)−2 detA+1 detA tr 2 A−4detA .

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detA Therefore the equation x2 − x = tr2 A−4detA is solvable.
(3) ⇒ (1) Assume that A and I2 − A are not in J M2 (R) . Then detA has trA ∈ 1 + J(R), detA ∈ J(R) and the equation x2 − x = tr2 A−4detA a root a ∈ R. Clearly, b := 1 − a
∈ R is a root of this equation. One
2 easily checks that a(2a − 1)−1 trA − trA · a(2a − 1)−1 trA + detA = 2

2

−a) 2 − tr4(aA·(a 2 −a)+1 + detA = 0. Thus, the equation x − trA · x + detA = 0 has roots a(2a − 1)−1 trA and b(2b − 1)−1 trA. Since ab ∈ J(R), we see that a + b = 1 and either a ∈ J(R) or b ∈ J(R). Thus, x2 − trA · x + detA = 0 has a root in 1 + J(R) and a root in J(R). By using Theorem 16.4.31, we obtain the result. 


Clearly, the equation x2 − x = − 92 has a root 23 ∈ J Z(2) . It follows  
1 1 ∈ M2 Z(2) is strongly J-clean. As from Corollary 16.4.32 that 2 −9 0 an immediate consequence, we can derive the following result (cf. [306, Theorem 2.6]). Corollary 16.4.33. Let R be a commutative local ring, and let A ∈ M2 (R). Then the following are equivalent: (1) A ∈ M2 (R) is strongly clean. (2) A ∈ GL2 (R) or I2 −A ∈ GL2 (R) or the equation x2 −trA·x+detA = 0 has a root in J(R) and a root in 1 + J(R). detA (3) A ∈ GL2 (R) or I2 − A ∈ GL2 (R) or the equation x2 − x = tr2 A−4detA is solvable. Proof. (1) ⇒ (2) If A 6∈ GL2 (R) and I2 − A 6∈ GL2 (R), it follows from Proposition 16.4.28 that A ∈ M By virtue of
2 (R) is strongly J-clean.
Theorem 16.4.31, A ∈ J M2 (R) or I2 − A ∈ J M2 (R) or the equation x2 − trA · x +
detA = 0 has a root in J(R) and a root in 1 + J(R). If
A ∈ J M2 (R) , then I2 − A ∈ GL2 (R). If I2 − A ∈ J M2 (R) , then A ∈ GL2 (R), as desired. (2) ⇒ (3) By hypothesis, the equation x2 − trA · x + detA = 0 has a root a ∈ J(R) and a root b ∈ 1 + J(R). Then trA = a + b and detA = ab. One detA easily checks that the equation x2 − x = tr2 A−4detA has a root a(a − b)−1 ∈ R, as desired. (3) ⇒ (1) If A 6∈ GL2 (R) and I2 − A 6∈ GL2 (R), then the equation detA x2 − x = tr2 A−4detA is solvable. In view of Lemma 16.4.11, A is similar   0λ to a matrix , where λ ∈ J(R), µ ∈ 1 + J(R). This implies that 1µ

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trA = µ ∈ 1 + J(R) and detA = −λ ∈ J(R). In light of Corollary 16.4.32, A ∈ M2 (R) is strongly J-clean, as needed.  Proposition 16.4.34. Let R be a commutative local ring. If the following are equivalent:

1 2

∈ R, then

(1) A ∈ M2 (R) is
strongly J-clean.
(2) A ∈ J M2 (R) or I2 − A ∈ J M2 (R) , or trA ∈ 1 + J(R), detA ∈ J(R) and tr2 A − 4detA = u2 for some u ∈ R.
Proof. (1) ⇒
(2) According to Corollary 16.4.32, A ∈ J M2 (R) or I2 − A ∈ J M2 (R) , or trA ∈ 1 + J(R), detA ∈ J(R) and the equation x2 − detA is solvable. If a ∈ R is the root of the equation, then x = tr2 A−4detA 2

tr A ∈ U (R). Therefore tr2 A − 4detA = (2a − 1)2 = 4(a2 − a) + 1 = tr2 A−4detA
2 trA · (2a − 1)−1 , as desired. (2) ⇒ (1) If trA ∈ 1 + J(R), detA ∈ J(R) and tr2 A − 4detA = u2 for detA some u ∈ R, then u ∈ U (R) and the equation x2 − x = tr2 A−4detA has a 1 −1 root 2 u (trA + u). In light of Corollary 16.4.32, A ∈ M2 (R) is strongly J-clean. 

Corollary 16.4.35. Let R be a commutative local ring, and let A ∈ M2 (R). If 21 ∈ R, then the following are equivalent: (1) A ∈ M2 (R) is strongly clean. (2) A ∈ GL2 (R) or I2 −A ∈ GL2 (R) or tr2 A−4detA = u2 for some u ∈ R. Proof. (1) ⇒ (2) If A 6∈ GL2 (R) and I2 − A 6∈ GL2 (R), then A ∈ M2 (R) is strongly J-clean by Proposition 16.4.28. According to Proposition 16.4.34, tr2 A − 4detA = u2 for some u ∈ R. (2) ⇒ (1) If A 6∈ GL2 (R) and I2 − A6∈ GL 2 (R), it follows from Lemma 0λ 16.4.11 that A is similar to a matrix , where λ ∈ J(R), µ ∈ 1 + 1µ J(R). Hence, trA = µ ∈ 1 + J(R) and detA = −λ ∈ J(R). By virtue of Proposition 16.4.34, A ∈ M2 (R) is strongly J-clean, as required. 

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Chapter 17

Abelian Rings and Exchange

A ring R is said to be an abelian ring provided that every idempotent in R is central. A ring R is abelian if and only if every direct summand of RR has a unique complementary direct summand, i.e., RR = A ⊕ B = A ⊕ C =⇒ B = C(cf. [418, Proposition 1.3]). The main purpose of this chapter is to investigate abelian rings related to stable range condition. In Section 17.1, we study various kinds of separation properties of medium spaces. In Section 17.2, we give fundamental element-wise properties of abelian exchange rings. In Section 17.3, we develop the K0 property via medium spaces for abelian exchange rings. Finally, in Section 17.4, we study the K0 groups of abelian exchange rings by means of their Pierce spaces.

17.1

Medium Spaces

Define the maximal spectrum of R as the set of all maximal ideals of R which we denote as M ax(R). Define the prime spectrum of R as the set of all prime ideals of R which we denote as Spec(R). We record two simple facts which will be used frequently. Lemma 17.1.1. Let P be an ideal of a ring R. Then the following are equivalent: (1) P ∈ Spec(R). 589

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(2) For any a, b ∈ R, aRb ⊆ P implies that a ∈ P or b ∈ P . Proof. (1) ⇒ (2) Suppose that aRb ⊆ P, a, b ∈ R. Choose I = RaR and J = RbR. Then IJ ⊆ P ; hence, I ⊆ P or J ⊆ P . Consequently, a ∈ P or b ∈ P. (2) ⇒ (1) Let I and J be ideals of R. If IJ ⊆ P and I * P , then there exists some a ∈ I, but a 6∈ P . For any b ∈ J, aRb ⊆ IJ ⊆ P . By hypothesis, a ∈ P or b ∈ P . Hence, b ∈ P , and so J ⊆ P . That is, IJ ⊆ P implies that I ⊆ P or J ⊆ P . Therefore P ∈ Spec(R).  Proposition 17.1.2. Let R be a ring. Then M ax(R) ⊆ Spec(R). Proof. Let P ∈ M ax(R) and aRb ⊆ P . If a 6∈ P , then RaR+P = R. Write n n P P xi ayi + c = 1, where xi , yi ∈ R, c ∈ P . Hence, b = xi ayi b + cb ∈ i=1

i=1

RaRb + P ⊆ P , and so either a ∈ P or b ∈ P . According to Lemma 17.1.1, P ∈ Spec(R), as desired.  S One easily verifies that M ax(Z×Z) $ M ax(Z×Z) {0×Z} $ Spec(Z× Z). In this section, we are concern with sets between maximal spectrums and prime spectrums. Suppose that M ax(R) ⊆ Ξ(R) ⊆ Spec(R). Let I be an ideal of R, and let E(I) = {P
∈ Ξ(R) | I 6⊆ P }. Then E(R) = S P
Ξ(R), E(0) = ∅, E(I) ∩ E(J) = E IJ and E(Ii ) = E Ii . So Ξ(R) i

i

is a topological space, where {E(I) | I E R} is the collection of its open sets. Let V (I) = Ξ(R) − E(I). Then V (I) = {P ∈ Ξ(R) ⊆ P }. In
| IT S addition, V (0) = Ξ(R), V (R) = ∅, V (I) V (J) = V IJ and V (Ii ) = i P
V Ii . Let a ∈ R, and let V (a) := V (RaR). It is easy to verify that i T V (I) = V (a). Let S and T be two sets. We always use S ⊔ T to denote a∈I

the set S ∪ T with S ∩ T = ∅.

Lemma 17.1.3. Ξ(R) is a compact space. S P
P Proof. Suppose that E(Ii ) = Ξ(R). Then E Ii = Ξ(R). If R 6= Ii , i i i P then the set Ω = {I E R | Ii ⊆ I} is non-empty. By using Zorn’s i P Lemma, we have some J ∈ M ax(R) such that Ii ⊆ J. So J ∈ Ξ(R), i P
P while J 6∈ E Ii . This implies that Ii = R, and then there are i

r1 ∈ I1 , · · · , rn ∈ In such that

n P

i=1

i

ri = 1. As a result, R =

n P

i=1

Ii . Further,

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i=1

E(Ii ) = E

n P

i=1


Ii = Ξ(R). Therefore Ξ(R) is a compact space.

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Let R = F [x1 , x2 ] be the polynomial ring over a field F , and let (x1 ) and (x2 ) be the principal ideals in R. Then (x1 ) and (x2 ) are prime but S not maximal since (xi ) $ (x1 , x2 ). Choose Ξ(R) = M ax(R) {(x1 )}. Then M ax(R) $ Ξ(R) $ Spec(R). In view of Lemma 17.1.3, Ξ(R) is a compact topological space that lies between M ax(R) and Spec(R). An element a ∈ R is nearly clean provided that there exists an idempotent u ∈ R and a full element u ∈ R such that a = e + u. We say that R is nearly clean in the case where every element in R is nearly clean. For commutative rings, the exchange property, the cleanness and the nearly cleanness coincide. Now we characterize the nearly clean ring R by means of the separation property of the topological space Ξ(R). Lemma 17.1.4. Let R be an abelian ring. Then (1) a ∈ R is nearly clean if and only if there exists an idempotent e ∈ R such that V (a − 1) ⊆ V (e) ⊆ Ξ(R) − V (a); and (2) −a ∈ R is nearly clean if and only if there exists an idempotent e ∈ R such that V (a + 1) ⊆ V (e) ⊆ Ξ(R) − V (a). Proof. (1) Let a ∈ R be nearly clean. Then there exists an idempotent e ∈ R and a full u ∈ R such that a = e + u. For any P ∈ V (a − 1), we have P 6∈ V (1 − e); otherwise, u = (a − 1) + (1 − e) ∈ P . Clearly, Ξ(R) = F V (e) V (1 − e), and so P ∈ V (e). We infer that V (a − 1) ⊆ V (e). If P ∈ Ξ(R) and P 6∈ Ξ(R) − V (a), then P ∈ V (a); hence, P 6∈ V (e); otherwise, u = a − e ∈ P . This implies that V (e) ⊆ Ξ(R) − V (a). Consequently, V (a − 1) ⊆ V (e) ⊆ Ξ(R) − V (a). Conversely, assume that there exists an idempotent e ∈ R such that V (a − 1) ⊆ V (e) ⊆ Ξ(R) − V (a). Let u = a − e. If RuR 6= R, by using Zorn’s Lemma, we have a maximal ideal P such that RuR ⊆ P $ R. In view of Proposition 17.1.2, P ∈ Spec(R). As eR(1 − e) ⊆ P , we get e ∈ P or 1 − e ∈ P . Hence, P ∈ V (e) or P ∈ V (1 − e). If P ∈ V (e), then a = e + u ∈ P , whence, P ∈ V (a). This gives a contradiction. If P ∈ V (1 − e), then a − 1 = (e − 1) + u ∈ P , whence, P ∈ V (a − 1). This implies that P ∈ V (e), a contradiction. Therefore a − e ∈ R is full, as required. (2) By (1), −a ∈ R is nearly clean if and only if there exists an idempotent e ∈ R such that V (−a − 1) ⊆ V (e) ⊆ Ξ(R) − V (−a). Obviously, V (−a − 1) = V (a + 1) and V (−a) = V (a), and therefore the result follows. 

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A topological space X is said to be strongly zero-dimensional provided that any two disjoint closed sets are separated by clopen sets, that is, if A and B are disjoint closed sets, then there exist disjoint clopen sets C1 , C2 such that A ⊆ C1 and B ⊆ C2 (cf. [309]). Let X be strongly zerodimensional. It follows from Urysohn’s Lemma that for any two disjoint closed sets A and B, there exists a continuous function f : X → [0, 1] such that f (x) = 0, x ∈ A; f (x) = 1, x ∈ B. It follows from Tietze’s Theorem that for any closed set A and a continuous map f : A → R, there exists a continuous map g : X → R such that g |A = f . In [367, Theorem 2.7], Samei proved that a commutative semiprimitive Gelfand ring R is clean if and only if M ax(R) is strongly zero-dimensional. In [309, Theorem 2.6], Lu and Yu extended these results and showed that a commutative ring is clean if and only if Spec(R) is strongly zero-dimensional. Theorem 17.1.5. Let R be an abelian ring, and let M ax(R) ⊆ Ξ(R) ⊆ Spec(R). Then the following are equivalent: (1) R is nearly clean. (2) For any a ∈ R, there exists an idempotent e ∈ R such that V (a) ⊆ V (e) and V (1 − a) ⊆ V (1 − e). (3) For any disjoint closed sets A and B of Ξ(R), there exists an idempotent e ∈ R such that A ⊆ V (e) and B ⊆ V (1 − e). In this case, Ξ(R) is strongly zero-dimensional. Proof. (1) ⇒ (3) Let A and B be disjoint closed sets of Ξ(R). Then T A B = ∅. Clearly, there exist two ideals I and J such that A = V (I) T and B = V (J); hence, V (I) V (J) = ∅. If I + J 6= R, then there exists a maximal ideal P of R such that I + J ⊆ P $ R. Hence, P ∈ V (I + J) = T V (I) V (J), a contradiction. This implies that I + J = R, and so a+ b = 1 for some a ∈ I and b ∈ J. As a ∈ R is nearly clean, it follows from Lemma 17.1.4 that there exists an idempotent e ∈ R such that V (a − 1) ⊆ V (1 − e) ⊆ Ξ(R) − V (a). It is easy to check that B = V (J) ⊆ V (b) = V (a − 1) ⊆ V (1 − e) ⊆ Ξ(R) − V (a) ⊆ Ξ(R) − V (I) = Ξ(R) − A. Clearly, B ⊆ V (1 − e). As V (1 − e) ⊆ Ξ(R) − A, we see that A ⊆ V (e), as required.

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T (3) ⇒ (2) For any a ∈ R, V (a) V (1 − a) = ∅. That is, V (a) and V (a − 1) are disjoint closed sets of Ξ(R). By hypothesis, there exists an idempotent e ∈ R such that A ⊆ V (e) and B ⊆ V (1 − e). (2) ⇒ (1) For any a ∈ R, we can find an idempotent e ∈ R such that V (a) ⊆ V (1 − e) and V (a − 1) ⊆ V (e). If P ∈ Ξ(R) and P 6∈ Ξ(R) − V (a), then P ∈ V (a), and so P ∈ V (1 − e). This implies that P 6∈ V (e). Hence, V (e) ⊆ Ξ(R) − V (a). As a result, V (a − 1) ⊆ V (e) ⊆ Ξ(R) − V (a). According to Lemma 17.1.4, a ∈ R is nearly clean, as desired.



Corollary 17.1.6. Let R be an abelian ring. Then the following are equivalent: (1) R is nearly clean. (2) For any disjoint compact sets A and B of M ax(R), there exists an idempotent e ∈ R such that A ⊆ V (e) and B ⊆ V (1 − e). Proof. (1) ⇒ (2) Let P, Q ∈ M ax(R), P 6= Q. Then there exist a ∈ P, b ∈ Q such that a + b = 1. As R is nearly clean, we can find an idempotent e ∈ R and a full u ∈ R such that a = e + u. This implies that e 6∈ P . Clearly, b = 1 − a = (1 − e) − u, and so 1 − e 6∈ Q. Consequently, 1 − e ∈ P and e ∈ Q. Thus, P ∈ V (1 − e) and Q ∈ V (e). One easily checks that F M ax(R) = V (e) V (1 − e). Hence, V (e) and V (1 − e) are clopen sets. That is, M ax(R) is an Hausdorff space. In view of Lemma 17.1.3, M ax(R) is compact. So every compact subset of M ax(R) is closed. Any disjoint compact subsets A and B of M ax(R) are closed. According to Theorem 17.1.5, there exists an idempotent e ∈ R such that A ⊆ V (e) and B ⊆ V (1 − e). (2) ⇒ (1) As M ax(R) is compact, every closed set is compact. Thus the result follows from Theorem 17.1.5.  Corollary 17.1.7. Let R be an abelian exchange ring. Then for any P ∈ M ax(R) and any closed set B of M ax(R) which does not contain P , there exist clopen sets U and V of M ax(R) such that P ∈ U and B ⊆ V ; hence, M ax(R) is a T3 -space. Proof. In view of Corollary 15.3.4, R is nearly clean. Let P ∈ M ax(R), and T let B be a closed set of M ax(R). If P 6∈ B, then {P } B = ∅. As M ax(R) is compact, every closed set is compact. Thus, B is compact. Clearly, {P } is compact, and we are through by Corollary 17.1.6. 

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T Let Λ(R) = {P | P ∈ Ξ(R)}. Then Λ(R) is an ideal that lies between T P (R) and BM (R), where P (R) = {P | P ∈ Spec(R)} is the prime radical T of R and BM (R) = {P | P ∈ M ax(R)} is the Brown-McCoy radical of R. Ξ(R) is called a medium space in the case where Λ(R) is nil. Lemma 17.1.8. If x ∈ Λ(R), then R(1 + x)R = R. Proof. Let x ∈ Λ(R). If R(1 + x)R 6= R, then there exists an ideal P ∈ M ax(R) such that R(1 + x)R ⊆ P by Zorn’s Lemma. This proves that 1 + x ∈ P ; hence, x 6∈ P . As P ∈ M ax(R) ⊆ Ξ(R), we get a contradiction. Therefore R(1 + x)R = R.  Proposition 17.1.9. Let R be an abelian exchange ring, and let M ax(R) ⊆ Ξ(R) ⊆ Spec(R). If J(R) is nil, then Ξ(R) is a medium space. Proof. Let a ∈ Λ(R). For any b ∈ R, it follows from Lemma 17.1.8 that R(1 + ab)R = R. Let c = 1 + ab. Then there exist x1 , · · · , xm ∈ R such that x1 cR + · · · + xm cR = R. By [382, Proposition 29.1], there are orthogonal m P idempotents g1 ∈ x1 cR, · · · , gm ∈ xm cR such that gi = 1. Set gi = xi cyi i=1

with yi = yi gi . Let ei = cyi xi . Then e2i = ei and ei R ∼ = gi R. Since R = e1 R ⊕ (1 − e1 )R = e2 R ⊕ (1 − e2 )R, we can find some right R-modules A1 , A′1 , B1 and B1′ such that R = e1 R ⊕ A1 ⊕ A′1 , A1 ⊕ B1 = e2 R and A′1 ⊕ B1′ = (1 − e2 )R. Clearly, there exists an idempotent f1 ∈ R such that f1 R = e1 R ⊕ A1 , and so f1 ∈ e1 R + e2 R. This
implies that f1 ∈ cR. Obviously, e1 R .⊕ f1 R. Further, e1 R ∼ = B1 ⊕ B1′ . = R/ A1 ⊕ A′1 ∼ ⊕ ⊕ ∼ Hence, e2 R . A1 ⊕ e1 R = f1 R. Thus, e1 R, e2 R . f1 R, f1 ∈ cR. Since R = f1 R ⊕ (1 − f1 )R = e3 R ⊕ (1 − e3 )R, we can find some right R-modules A2 , A′2 , B2 and B2′ such that R = f1 R ⊕ A2 ⊕ A′2 , A2 ⊕ B2 = e3 R and A′2 ⊕ B2′ = (1 − e3 )R. Clearly, there exists an idempotent f2 ∈ R such that f2 R = f1 R ⊕ A2 , and so f2 ∈ f1 R + e3 R. This implies that f2 ∈ cR. Obviously, f1 R .⊕ f2 R.
′ Further, f1 R ∼ = B2 ⊕ B2′ . Hence, e3 R .⊕ A2 ⊕ f1 R ∼ = f2 R. = R/ A2 ⊕ A2 ∼ ⊕ Thus, e1 R, e2 R, e3 R . f2 R, f2 ∈ cR. By iteration of this process, we can find an idempotent f ∈ cR such that e1 R, · · · , en R .⊕ f R. So we see that gi R .⊕ f R. In view of Lemma 6.1.2, n P there exist a ∈ gi Rf, b ∈ f Rgi such that gi = ab ∈ Rf R. As gi = 1, i=1

we get Rf R = R. Since R is an abelian ring, we get f = 1. This implies that c = 1 + ab ∈ U (R), and so a ∈ J(R). Since J(R) is nil, so is Λ(R). Consequently, Ξ(R) is a medium space. 

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Let R be an abelian strongly π-regular ring. Then each topological space between the maximal spectrum M ax(R) and the prime spectrum Spec(R) is a medium space. For any x ∈ J(R), there exists some m ∈ N such that xm = xm+1 y for some y ∈ R; hence, xm (1 − xy) = 0. Clearly, 1 − xy ∈ R is invertible. This implies that xm = 0. That is, J(R) is nil, and we are through by Proposition 17.1.9. Lemma 17.1.10. Let R be an abelian ring, and let Ξ(R) be a medium space. Then A is a clopen subset of Ξ(R) if and only if there exists a unique idempotent e ∈ R such that A = V (e). Proof. One direction is obvious. Conversely, assume that A is an clopen subset of Ξ(R). Then we can find two ideals I and J of R such that A = V (I) and Ξ(R) = V (I) ⊔ V (J). Thus, V (IJ) = Ξ(R); hence, IJ ⊆ Λ(R). T On the other hand, V (I) V (J) = ∅, and so V (I + J) = ∅. If I + J 6= R, then there exists a maximal ideal P of R such that I + J ⊆ P $ R. Clearly, P ∈ Ξ(R), and so P ∈ V (I + J). This gives a contradiction. Hence I + J = R. Write 1 = a1 + a2 , a1 ∈ I, a2 ∈ J. Then a1 a2 ∈ Λ(R). By hypothesis, Λ(R) is nil. Clearly, a1 a2 = a1 (1 − a1 ) = (1 − a1 )a1 = a2 a1 . So we can find  some n ∈ N such that an1 (1 − a1 )n = (a1 a2 )n = 0. Let n P 2n f (x) = x2n−i (1 − x)i . Then f (x) ≡ 0 (mod xn ). It follows from i i=0  2n  X
n 2n f (x) + x2n−i (1 − x)i = x + (1 − x) = 1 i i=n+1




that f (x) ≡ 1 mod (1 − x)n . Thus, f (x) 1 − f (x) ≡ 0 mod xn (1 − x)n . Let e1 = f (a1 ). Then e1 (1 − e1 ) = 0, and so e1 ∈ R is an idempotent. Let m e2 = 1 − e1 . Then e2 = e22 and e1 e2 = 0. Clearly, am 1 = (a1 + a2 )a1 ≡ m+1 2n a1 (mod IJ) for any m ∈ N. Thus, e1 ≡ a1 ≡ · · · ≡ a1 (mod IJ). Likewise, e2 ≡ a2 (mod IJ). Hence, we see that e1 ∈ I and e2 ∈ J. So we have Ξ(R) = V (e1 ) ⊔ V (e2 ), V (I) ⊆ V (e1 ), V (J) ⊆ V (e2 ). As Ξ(R) = V (I) ⊔ V (J), we deduce that V (I) = V (e1 ) and V (J) = V (e2 ). Assume that A = V (e). Then V (e1 ) = V (e), and so V (1 − e1 ) = V (1 − e). If P ∈ V (e1 ), then we have P ∈ V (e), and so e1 − e ∈ P . If P 6∈ V (e1 ), then P ∈ V (1 − e1 ); hence, P ∈ V (1 − e). Thus, e1 − e = (1 − e) − (1 − e1 ) ∈ P . It follows that e1 − e ∈ Ξ(R). This implies that (e1 − e)2p+1 = e1 − e = 0 for some p ∈ N, and so e1 = e. Therefore the result follows. 

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Theorem 17.1.11. Let R be an abelian ring. If Ξ(R) is a medium space, then the following are equivalent: (1) R is nearly clean. (2) Ξ(R) is strongly zero-dimensional. Proof. (1) ⇒ (2) is trivial by Theorem 17.1.5. T (2) ⇒ (1) For any a ∈ R, we see that V (a) V (a−1) = ∅. That is, V (a) and V (a−1) are disjoint closed sets of Ξ(R). Thus, we can find two distinct clopen sets U and V such that V (a) ⊆ U and V (a − 1) ⊆ V . According to Lemma 17.1.10, we have idempotents e, f ∈ R such that U = V (e) and T V = V (f ). As V (e) V (f ) = V (eR + f R) = ∅, it is easy to check that eR + f R = R. Hence, (1 − e)R ⊆ (1 − e)f R ⊆ f R. If P ∈ Ξ(R) and P 6∈ Ξ(R) − V (a), then P ∈ V (a), and so P ∈ V (e). This implies that P 6∈ V (1 − e). Obviously, V (f ) ⊆ V (1 − e). Hence, P 6∈ V (f ), and so V (f ) ⊆ Ξ(R) − V (a). As a result, V (a − 1) ⊆ V (f ) ⊆ Ξ(R) − V (a). According to Lemma 17.1.4, a ∈ R is nearly clean, as desired.



Lemma 17.1.12. Let R be an abelian ring. Then the following are equivalent: (1) R is nearly clean. (2) R/P (R) is nearly clean. (3) Spec(R) is strongly zero-dimensional. Proof. (1) ⇒ (3) is obvious by Theorem 17.1.5. (3) ⇒ (1) Given any x ∈ P (R), we claim that x is nilpotent. If not, T we choose S = {1, x, x2 , · · · } and let Ω = {I | I E R such that I S = ∅}. Clearly, Ω is a nonempty inductive set. By using Zorn’s Lemma, there exists an ideal Q which is maximal in Ω. If Q is not prime, then we can find some a, b ∈ R such that aRb ⊆ Q, while a, b 6∈ Q. By this construction, we have m, n ∈ N such that xm ∈ Q + RaR and xn ∈ Q + RbR. Hence, xm+n ∈ Q + RaRbR ⊆ Q, a contradiction. We infer that Q ∈ Spec(R), and so x ∈ Q. This also gives a contradiction. Consequently, we conclude that P (R) is nil. Hence, Spec(R) is a medium space. Therefore R is nearly clean from Theorem 17.1.11. (1) ⇒ (2) is obvious. (2) ⇒ (1) As in the preceding discussion, P (R) is nil. For any a ∈ R, there exists an idempotent e ∈ R/P (R) and a full u ∈ R/P (R) such that a − e − u ∈ P (R). Further, a = e + u + s for some s ∈ P (R). Since

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idempotents lift modulo any nil ideal, and so we may assume that e = e2 ∈ n n P P R. Write 1 = si uti + r for some r ∈ P (R). Then 1 = si (u + s)ti + t i=1

for some t ∈ P (R). Hence,

n P

i=1

i=1

−1

si (u + s)ti (1 − t)

= 1. We infer that

u + s ∈ R is full. Therefore R is nearly clean.



Recall that a ring R is a pm (Gelfand) ring provided that each prime ideal is contained in exactly one maximal ideal. Any ring of the form C(X), the ring of continuous real valued functions on a (completely regular) topological space X, is a pm ring. In [3, Corollary 4], Anderson and Camillo proved that every commutative clean ring is always a pm ring. Now we extend this result as follows. Theorem 17.1.13. Let R be an abelian ring. Then the following are equivalent: (1) R is nearly clean. (2) R is a pm ring and M ax(R) is strongly zero-dimensional. Proof. (1) ⇒ (2) In view of Theorem 17.1.5, M ax(R) is strongly zerodimensional. Let I ∈ Spec(R). If there exist P, Q ∈ M ax(R) such that T I ⊆ P Q, then P 6= Q. As in the proof of Corollary 17.1.6, P ∈ V (1 − e) and Q ∈ V (e). Clearly, e(1 − e) ∈ I, hence, e ∈ I or 1 − e ∈ I. This implies that e ∈ P or 1 − e ∈ Q, which are absurd. Therefore R is a pm ring. (2) ⇒ (1) As R is a pm ring, there exists a map ϕ : Spec(R) → M ax(R), ϕ(P ) = M , where M is the unique maximal ideal such that P ⊆ M . For any I E R, we let VS (I) := {P ∈ Spec(R) | I ⊆ P }
and VM (I) := {P ∈ M ax(R) | I ⊆ P }. It is easy to check that ϕ VS (I) = VM (I). For any disjoint closed sets A, B ⊆ Spec(R), there exist two ideals I and J of R such that A = VS (I) and B = VS (J). Hence, ϕ(A) and ϕ(B) are both T closed. As VS (I) VS (J) = ∅, we see that VS (I + J) = ∅; hence, I + J = R. T Thus, we infer that VM (I) VM (J) = VM (I + J) = VM (R) = ∅. In other words, ϕ(A) and ϕ(B) are disjoint closed sets of M ax(R). Since M ax(R) is strongly zero-dimensional, we can find disjoint clopen sets U, V ⊆ M ax(R) such that VM (I) ⊆ U, VM (J) ⊆ V . Clearly, A ⊆ ϕ← ϕ(A) ⊆ ϕ← (U ) and B ⊆ ϕ← ϕ(B) ⊆ ϕ← (V ). As is well known, ϕ is continuous; hence, ϕ← (U ) T and ϕ← (V ) are clopen. For any P ∈ ϕ← (U ) ϕ← (V ), there exists a unique T M ∈ M ax(R) such that P ⊆ M . Hence, M ∈ U V , a contradiction. This T implies that ϕ← (U ) ϕ← (V ) = ∅. Consequently, Spec(R) is strongly zerodimensional. Therefore R is nearly clean by Lemma 17.1.12. 

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Corollary 17.1.14. Let R be a commutative ring. Then the following are equivalent: (1) R is clean. (2) Spec(R) is strongly zero-dimensional. (3) R is a pm ring and M ax(R) is strongly zero-dimensional. Proof. It is an immediate consequence of Lemma 17.1.12 and Theorem 17.1.13.  In [367], Samei investigated clean elements in commutative rings. He proved that if an ∈ R is clean then so is a, where R is a reduced commutative ring. In a commutative ring R, the cleanness of an element a may not imply that of a2 . An obvious example is 2 ∈ Z. In the following, we find some criteria under which a2 ∈ R is nearly clean, where R is an abelian ring. Theorem 17.1.15. Let R be an abelian ring. Then the following are equivalent: (1) a2 ∈ R is nearly clean. (2) a, −a ∈ R are nearly clean.

Proof. (1) ⇒ (2) Since a2 ∈ R is nearly clean, it follows from Lemma 17.1.4 that there exists an idempotent e ∈ R such that V (a2 − 1) ⊆ V (e) ⊆ Spec(R) − V (a2 ).

Clearly, V (a − 1) ⊆ V (a2 − 1) and V (a) ⊆ V (a2 ). Thus, we get V (a − 1) ⊆ V (e) ⊆ Spec(R) − V (a). By using Lemma 17.1.4 again, a ∈ R is nearly clean. As (−a)2 ∈ R is nearly clean, so is −a ∈ R by a similar argument. (2) ⇒ (1) Let a, −a ∈ R be nearly clean. By virtue of Lemma 17.1.4, we can find two clopen sets U, W of Spec(R) such that V (a − 1) ⊆ U ⊆ Spec(R) − V (a) and V (a + 1) ⊆ W ⊆ Spec(R) − V (a). Thus, V (a − 1)

[

V (a + 1) ⊆ U

[

W ⊆ Spec(R) − V (a). S It is easy to verify that V (a2 − 1) = V (a − 1) V (a + 1) and V (a2 ) = V (a), and thus we deduce that [ V (a2 − 1) ⊆ U W ⊆ Spec(R) − V (a2 ).

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By using Lemma 17.1.10, we find an idempotent e ∈ R such that U V (e). According to Lemma 17.1.4, a2 ∈ R is clean.

599

S

V = 

Corollary 17.1.16. Let R be an abelian ring. Then the following are equivalent: (1) a2 ∈ R is nearly clean. (2) For any idempotent e ∈ R, a + e ∈ R is nearly clean. Proof. (1) ⇒ (2) Let a2 ∈ R be nearly clean. Then a, −a ∈ R are both nearly clean from Theorem 17.1.15. Thus, we have idempotents e1 , e2 ∈ R and full u1 , u2 ∈
R such that a = e1 + u
1 and −a = e2 + u2 . Let T S T U = V (e) V (e1 ) V (1 − e) V (1 − e2 ) . Let P ∈ V (a + e − 1). Clearly, P ∈ V (e) or P ∈ V (e − 1). If P ∈ V (e), then a − 1 ∈ P . Hence, e1 − 1 6∈ P ; otherwise, u1 ∈ P , a contradiction. This implies that e1 ∈ P . T Thus, P ∈ V (e) V (e1 ). If P ∈ V (e − 1), then a ∈ P . This implies that T e2 6∈ P , and so e2 − 1 ∈ P . Hence, P ∈ V (e − 1) V (e2 − 1). Therefore we get V (a + e − 1) ⊆ U . If P 6∈ Spec(R) − V (a + e), then P ∈ V (a + e). T If P ∈ V (e) V (e1 ), then u1 = (a + e) − e − e1 ∈ P , a contradiction. If T P ∈ V (1 − e) V (1 − e2 ), then u2 = (1 − e2 ) − (1 − e) − (a + e) ∈ P , a contradiction. Hence, U ⊆ Spec(R) − V (a + e). Thus, we see that V (a + e − 1) ⊆ U ⊆ Spec(R) − V (a + e). As U is a clopen subset of Spec(R), it follows from Lemma 17.1.10 that U = V (f ) for an idempotent f ∈ R. In view of Lemma 17.1.4, a + e ∈ R is nearly clean. (2) ⇒ (1) By hypothesis, a, a + 1 ∈ R are both nearly clean. Thus, we can find an idempotent e ∈ R and a full u ∈ R such that a + 1 = e + u; hence, −a = (1 − e) + (−u) is nearly clean. It follows from Theorem 17.1.15 that a2 ∈ R is nearly clean.  Example 17.1.17. Let F be a field of characteristic 0 and R = F [x, y] be the additive group of polynomials in two indeterminates. Define multiplication in R by requiring that multiplication be distributive, that ax = xa, ya = ay for all a ∈ F , that the product of x and y be the polynomial xy as usual, but that the product of y and x be the polynomial xy + 1. Then R is a simple ring which has no zero divisors. In addition, R is not a division ring. If a = 1, then a ∈ R is nearly clean. If a 6= 1, then a − 1 ∈ R is full. Hence, a ∈ R is nearly clean. Clearly, R has no non-trivial idempotents. Thus, R is an abelian nearly clean ring. 

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Let Lat(R) be the lattice of all right ideals of a ring R. We say that R is T T a D-ring provided that Lat(R) is distribute, i.e., A (B + C) = (A B) + T (A C) for all A, B, C ∈ Lat(R). We note that every D-ring is abelian (cf. [371, Corollary 2]). Let e ∈ R be an idempotent, and let a ∈ R.
Set T f = e − (1 − e)ae. Then eR, f R ∈ Lat(R). Hence, eR = eR f R +
T eR (1 − f )R , and so f eR ⊆ eR. We infer that ef e = f e = f . This implies that ae = eae. Likewise, ea = eae. Thus ea = ae, and we are done. 17.2

Abelian Exchange Rings

Lemma 17.2.1. A ring R is a local ring if and only if it is an indecomposable, abelian exchange ring. Proof. If R is a local ring, then it is an indecomposable, abelian exchange ring. Conversely, assume that R is an indecomposable, abelian exchange ring. For any x ∈ R, by Lemma 1.4.7, we can find an idempotent e ∈ R such that e ∈ xR and 1 − e ∈ (1 − x)R. Since R is indecomposable as a ring, all central idempotents in R are 0 and 1. So we may assume e = 0 or 1. Thus, x ∈ R is right invertible or 1 − x ∈ R is right invertible. If ab = 1, then ba ∈ B(R); hence, ba = ba(ab) = a(ba)b = (ab)2 = 1. That is, R is directly finite. Therefore either x or 1 − x is invertible, and so R is a local ring by [4, Proposition 15.1.5].  Construct R as in Example 17.1.17. If R is an exchange ring, it follows from Lemma 17.2.1 that R is local. As J(R) = 0, R is a division ring. This gives a contradiction. Therefore R is not an exchange ring. This implies that { abelian exchange rings } $ { abelian nearly clean rings }. The main purpose of this section is to investigate abelian exchange rings. Let S(R) be the nonempty set of all ideals of a ring R generated by central idempotents. By Zorn’s Lemma, S(R) contains maximal elements. If P is a maximal element of the set S(R), we say that R/P is a Pierce stalk of R. Now we begin with several element-wise characterizations of abelian exchange rings. Theorem 17.2.2. Let R be a ring. Then the following are equivalent : (1) R is an abelian exchange ring. (2) For any x ∈ R, there exists e ∈ B(R) such that e ∈ xR and 1 − e ∈ (1 − x)R.

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(3) For any x ∈ R, there exist e ∈ B(R), u ∈ U (R) such that x = e + u. (4) For any x, y ∈ R, there exists e ∈ B(R) such that 1 + x(y − e) ∈ U (R). (5) aR + bR = R with a, b ∈ R implies that there exists e ∈ B(R) such that a + be ∈ U (R). Proof. (1) ⇒ (5) Assume that aR + bR = R with a, b ∈ R and c ∈ R. Let f (X1 , X2 , Y1 , Y2 ) = 1 − (X1 + X2 Y1 )Y2 , g(X1 , X2 , Y1 , Y2 ) = 1 − Y2 (X1 + X2 Y1 ), h(X3 , Y2 ) = X3 Y2 − Y2 X3 and k(Y1 ) = Y1 − Y12 be the polynomials in noncommutative indeterminate X1 , X2 , X3 , Y1 and Y2 . Let R/P be an arbitrary Pierce stalk of R. Then we can find some x, y ∈ R such that ax + by = 1 in R/P . In view of [382, Proposition 32.2], R/P is a local ring. If a 6∈ J(R/P ), then a + b × 0 ∈ U (R/P ). If a ∈ J(R/P ), it follows from a(x − y) + (a + b)y = 1 that a + b × 1 ∈ U (R/P ). In any case, we can find an idempotent t ∈ R/P such that a + bt ∈ U (R/P ) and tc = ct. Thus, we have a w ∈ R/P such that (a + bt)w = 1 = w(a + bt). This means that f (a, b, t, w) = 0, g(a, b, t, w) = 0, h(c, t) = 0 and k(t) = 0. In view of [382, Lemma 11.4], there exist some e, d ∈ R such that f (a, b, e, d) = 0, g(a, b, e, d) = 0, h(c, e) = 0 and k(e) = 0. This implies that we can find a central idempotent e ∈ R such that a + be ∈ U (R), as asserted. (5) ⇒ (4) Given any x, y ∈ R, since (−x)R + (1 + xy)R = R, we can find some e ∈ B(R) such that (−x)e + (1 + xy) ∈ U (R), i.e., 1 + x(y − e) ∈ U (R), as required. (4) ⇒ (3) Given any x ∈ R, we have e ∈ B(R) such that v := 1 + 1 × (−x − f ) ∈ U (R). Therefore, x = e + u where e = 1 − f and u = −v, as desired. (3) ⇒ (2) Given any x = x2 ∈ R, there exists a central idempotent e and a unit u such that x = e + u. So e + u = e + eu + ue + u2 and u = 1 − 2e. Hence, x = 1 − e is central. Thus, R is a clean ring with all idempotents central. For any a ∈ R, write a = f + v, where f = f 2 , v ∈ U (R). Set g = v(1 − f )v −1 . Then g = g 2 ∈ R. Clearly, we see that
(a − g)v = f + v − v(1 − f )v −1 v = v 2 + f v − v + vf = a2 − a.
It is easy to
verify that g = a 1 − (a − 1)v −1 ∈ ag and 1 − g = (1 − a) av −1 + 1 ∈ (1 − a)R, as desired. (2) ⇒ (1) Clearly, R is an exchange ring. For any idempotent x ∈ R, there exists some e ∈ B(R) such that e ∈ xR and 1 − e ∈ (1 − x)R. This implies that e − x = (1 − x)e − x(1 − e) ∈ (x − x2 )R, and so x = e ∈ B(R). Therefore R is an abelian ring. 

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The preceding proof indeed shows that every clean ring is an exchange ring. But the converse is not true. Let F be a field, and let A = F [[x]]. Let K be the field of fractions of A. Define R = {r ∈ EndF (A) | ∃ q ∈ K and ∃ n > 0 with r(a) = qa for all a ∈ (xn )}. Then R is an exchange ring but not a clean ring (cf. [66]). Corollary 17.2.3. If R is a reduced exchange ring, then for any x, y ∈ R, there exists a central idempotent e such that 1 + x(y − e) ∈ U (R). Proof. Let R be a reduced exchange ring, and let x ∈ R. For each idempotent e ∈ R, we see that ex(1 − e) is a nilpotent element. Since R is reduced, ex(1 − e) = 0 and so ex = exe. Similarly, xe = exe. Thus, ex = xe, i.e., e is central. This implies that R is abelian. By virtue of Theorem 17.2.2, we see that for any x, y ∈ R, there exists a central idempotent e such that 1 + x(y − e) ∈ U (R).  Let R be an abelian regular ring. Then, for any x, y ∈ R, there exists e ∈ B(R) such that 1 + x(y − e) ∈ U (R). Let R be a ring in which 2 is invertible. If every element in R is the sum of an idempotent and a central unit, we claim that R is an abelian exchange ring. Given any idempotent e ∈ R, we have an idempotent f and a central unit u such that e = f + u. So f + u = e = e2 = f + f u + uf + u2 , whence f = 2−1 (1 − u) is central. Thus, e is central. This means that R is abelian. According to Theorem 17.2.2, R is an abelian exchange ring, and we are through. Corollary 17.2.4. Let R be a strongly π-regular ring. Then, for any x ∈ R, there exists an idempotent e such that x2 + xe + 1 ∈ U (R). Given any x ∈ R, let S be the subring generated by the set ∞ P {1R , x, x , · · · , xn , · · · }. That is, S = Zxi . Then S is commutative. Proof.

2

i=0

Let f : S → R be the inclusion. In view of [57, Corollary 1.10], we can find a commutative strongly π-regular subring T of R such that S ⊆ T ⊆ R. According to Theorem 17.2.2, there exists an idempotent e ∈ T ⊆ R such that 1 + (−x)(−x − e) ∈ U (T ); hence, x2 + xe + 1 = 1 + x(x − e) ∈ U (T ) ⊆ U (R), as asserted. 

Wenote that  the central condition in Theorem 17.2.2 is necessary. Let Z2 Z2 R = . Then R is an exchange ring. One directly checks that 0 Z2 aR + bR = R with a, b ∈ R implies that there exists an idempotent e ∈ R

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such that a + be ∈ U (R). In this case, R is not abelian. Lemma 17.2.5. Let R be an abelian exchange ring. Then R/P is a local ring for any prime ideal P of R. Proof. Let P ∈ Spec(R). As R is an exchange ring, so is R/P . In view of [382, Theorem 29.2], every idempotent lifts modulo P . Hence, R/P is abelian. For any idempotent e ∈ B(R/P ), we see that e(R/P )(1 − e) = 0. As P is a prime ideal of R, R/P is a prime ring. Thus, e = 0 or e = 1. That is, R/P is indecomposable. According to Lemma 17.2.1, R/P is a local ring, as asserted.  Yu [422, Remark 5] claimed that it seems unlikely that exchange rings with all idempotents central would have artinian primitive factors. As an immediate consequence of Lemma 17.2.5, we claim that every abelian exchange ring has artinian primitive factors. Let R be an abelian exchange ring, and let P be a primitive ideal of R. Since every primitive ideal is prime, by Lemma 17.2.5, R/P is a local ring. Since P is a primitive ideal, we see that J(R/P ) = 0 and so R/P is a division ring. Therefore R/P is artinian, and we are done. A ring R is fully idempotent provided that I = I 2 for every right ideal I of R. Theorem 17.2.6. Let R be an exchange ring. If R is fully idempotent, then the following are equivalent : (1) R is an abelian ring. (2) R/P is local for any prime ideal P of R. Proof. (1) ⇒ (2) is obvious from Lemma 17.2.5. (2) ⇒ (1) Assume there exists an idempotent e ∈ R which is not central. Let Ω be the set of ideals A of R such that (e + A)(x + A) 6= (x + A)(e + A) for some x ∈ R. Clearly, Ω 6= ∅. Given an ascending chain A1 ⊆ A2 ⊆ · · · ⊆ An ⊆ · · · in Ω, let M = ∞ S Ai . Then M is an ideal of R. Assume M is not in Ω. Then (e + M )(x +

i=1

M ) = (x+M )(e+M ) for any x ∈ R. So ex−xe ∈ M , whence ex−xe ∈ Am for some positive integer m, i.e., (e + Am )(x + Am ) = (x + Am )(e + Am ) in R/Am . This contradiction shows that M is in Ω. By Zorn’s Lemma, we can find an ideal Q of R which is maximal in Ω. If Q is prime, then R/Q is local; hence, it is abelian. Thus, (e + Q)(x + Q) = (x + Q)(e + Q) for

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any x ∈ R, which is a contradiction. This implies that Q is not a prime ideal of R. Thus, we can find two ideals K and L of R such that Q $ K, L, and KL ⊆ Q. By the maximality of Q, (e + K)(x + K) = (x + K)(e + K) and (e + L)(x + L) = (x + L)(e + L) for any x ∈ R. This implies that ex − xe ∈ K ∩ L = KL ⊆ Q, and so (e + Q)(x + Q) = (x + Q)(e + Q) for any x ∈ R. This gives a contradiction. Therefore, every idempotent in R is central, whence the result.  Corollary 17.2.7. A regular ring R is an abelian ring if and only if R/P is a division ring for any prime ideal P of R. Proof. As is well known, every regular ring is a fully idempotent and exchange ring. If R is an abelian regular ring, by Theorem 17.2.6, R/P is local for any P ∈ Spec(R). It follows from J(R/P ) = 0 that R/P is a division ring. The converse is obvious by Theorem 17.2.6.  Lemma 17.2.8. Let R be an abelian exchange ring, and let A, B be finitely generated projective right R-modules. Then there exist orthogonal idempotents e1 , · · · , en ∈ R and nonnegative integers tij such that A∼ = t11 (e1 R) ⊕ · · · ⊕ t1n (en R) and B ∼ = t21 (e1 R) ⊕ · · · ⊕ t2n (en R). Proof. According to Lemma 13.1.8, we have idempotents f1 , · · · , fn ∈ R such that A ∼ = s11 (f1 R) ⊕ · · · ⊕ s1n (fn R), B ∼ = s21 (f1 R) ⊕ · · · ⊕ s2n (fn R) for some nonnegative integers sij . We now induct on n. If n = 1, we have an idempotent f1 ∈ R such that A ∼ = s11 (f1 R), B ∼ = s21 (f1 R). Assume that there exist orthogonal idempotents e1 , · · · , em ∈ R such that A∼ = t11 (e1 R) ⊕ · · · ⊕ t1m (em R), B ∼ = t21 (e1 R) ⊕ · · · ⊕ t2m (em R),

if n = k(k ∈ N). Let n = k + 1(k ∈ N). Then we have

A∼ = s11 (f1 R) ⊕ · · · ⊕ s1k (fk R) ⊕ s1(k+1) (fk+1 R), B∼ = s21 (f1 R) ⊕ · · · ⊕ s2k (fk R) ⊕ s2(k+1) (fk+1 R).

By the induction hypothesis, we have orthogonal idempotents e1 , · · · , em ∈ R such that A∼ = t11 (e1 R) ⊕ · · · ⊕ t1m (em R) ⊕ s1(k+1) (fk+1 R), B∼ = t21 (e1 R) ⊕ · · · ⊕ t2m (em R) ⊕ s2(k+1) (fk+1 R) for some nonnegative integers tij . Since R is abelian, we have fk+1 (1 − e1 − · · · − em ) = (1 − e1 − · · · − em )fk+1 . Hence A∼ = t11 (e1 R) ⊕ · · · ⊕ t1m (em R) ⊕ s1(k+1) (ek+1 R), B∼ = t21 (e1 R) ⊕ · · · ⊕ t2m (em R) ⊕ s2(k+1) (ek+1 R),

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where ek+1 = fk+1 (1 − e1 − · · · − em ). Clearly, e1 , · · · , ek , ek+1 ∈ R are orthogonal idempotents. Therefore we establish the result.  Let Z+ = {0, 1, 2, · · · }. Then Z+ is a topological space in which all its subsets are open sets. Let A be a finitely generated projective right R-module, and let P ∈ Spec(R). In view of Lemma 17.2.5, R/P is local. Thus, there exists a unique n ∈ Z+ such that A/AP ∼ = n(R/P ). Denote n by rankR/P (A/AP ). Proposition 17.2.9. Let R be an abelian exchange ring, and let A, B be finitely generated projective right R-modules. If Ξ(R) is a medium space, then A ∼ = B if and only if rankR/P (A/AP ) = rankR/P (B/BP ) for all P ∈ Ξ(R).

Proof. If A ∼ = B, then rankR/P (A/AP ) = rankR/P (B/BP ) for all P ∈ Ξ(R). Conversely, suppose that A ≇ B. By virtue of Lemma 17.2.8, there exist orthogonal idempotents e1 , · · · , en and nonnegative integers tij such that A∼ = t11 (e1 R) ⊕ · · · ⊕ t1n (en R) and B ∼ = t21 (e1 R) ⊕ · · · ⊕ t2n (en R). Thus, there is some j such that ej 6= 0 and t1j 6= t2j . By hypothesis, Λ(R) is nil. Hence, ej 6∈ Λ(R), and so there exists some P ∈ Ξ(R) such that ej 6∈ P . If i 6= j, then ei ej = 0 ∈ P ; hence, ei ∈ P . This implies that A/AP ∼ = t1j (R/P ) and B/BP ∼ = t2j (R/P ). Further, A/AP ≇ B/BP , a contradiction. Therefore the result follows.  So as to construct more examples of abelian exchange rings, we recall the concept of ideal-extensions. Let R be a ring with an identity and S be a ring (not necessarily unitary), and let S be an R-R-bimodule in which (s1 s2 )r = s1 (s2 r), r(s1 s2 ) = (rs1 )s2 and (s1 r)s2 = s1 (rs2 ) for all s1 , s2 ∈ S, r ∈ R. Construct I(R; S) = {(r, s) | r ∈ R, s ∈ S}. Define (r1 , s1 ) + (r2 , s2 ) = (r1 + r2 , s1 + s2 ); (r1 , s1 )(r2 , s2 ) = (r1 r2 , s1 s2 + r1 s2 + s1 r2 ). Then I(R; S) is a ring with an identity (1, 0). We say I(R; S) is an idealextension of R by S. Proposition 17.2.10. The ideal-extension I(R; S) is an abelian exchange ring if the following conditions are satisfied: (1) R is an abelian exchange ring.

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(2) If e = e2 ∈ R, then es = se for all s ∈ S. (3) If s ∈ S, then there exists some s′ ∈ S such that ss′ = s′ s and s + s′ + ss′ = 0. If S contains no nonzero idempotents, then the preceding conditions are necessary. Proof. Suppose that Conditions (1), (2) and (3) hold. Let (r, s) ∈ I(R; S). By Condition (1), we can find a central idempotent e ∈ R such that r = e + u, u ∈ U (R). Hence, (r, s) = (e, 0) + (u, s). Clearly, (e, 0)2 = (e, 0). By Condition (2), we see that (e, 0) ∈ I(R; S) is a central idempotent. By Condition (3), there exists some y ∈ R such that u−1 sy = yu−1 s, y +u−1 s+ u−1 sy = 0. Hence, (u, s)−1 = (u−1 , yu−1 ). Therefore I(R; S) is an abelian exchange ring from Theorem 17.2.2. Suppose that I(R; S) is an abelian exchange ring and S contains no nonzero idempotents. It is directly verified that R is an abelian exchange ring. Let e = e2 ∈ R and s ∈ S. Then (e, 0) ∈ I(R; S) is an idempotent; hence, (e, 0)(0, s) = (0, s)(e, 0). Thus, es = se and so Condition (2) holds. Let s ∈ S. Write (1, s) = (f, y) + (u,
v) where (f, y) ∈ I(R; S) is a central idempotent and (u, v) ∈ U I(R; S) . From 1 − f = u ∈ U (R), we deduce that f = 0. It follows from (0, y)2 = (0, y) that y 2 = y. By hypothesis,
y = 0. This implies that s = y + v = v; hence, (1, s) ∈ U I(R; S) . Thus there is some (c, s′ ) ∈ I(R; S) such that (1, s)(c, s′ ) = (1, 0) = (c, s′ )(1, s). This implies that c = 1; hence, s + s′ + ss′ = s + s′ + s′ s = 0, as asserted. Example 17.2.11. Let R be an abelian exchange ring. Then Sn = {(aij ) ∈ T Mn (R) | a11 = · · · = ann } is an abelian exchange ring. Proof. Let T = {(aij ) ∈ T Mn (R) | a11 = · · · = ann = 0}. Then Sn ∼ = I(R; T ). One easily checks that Conditions (1), (2) and (3) in Proposition 17.2.10 are satisfied. Therefore we complete the proof by Proposition 17.2.10.  Let R be a ring and α : R → R be a ring endomorphism, and let R[[x, α]] denote the ring of skew formal series over R, that is all formal power series in x with coefficients from R with multiplication defined by xr = α(r)x for all r ∈ R. Example 17.2.12. Let R be a ring, and let α : R → R be a ring endomorphism. Then R[[x, α]] is an abelian exchange ring if and only if R is an abelian exchange ring and α(e) = e for all e = e2 ∈ R.

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Proof. Let T = R[[x, α]] · x. Then R[[x, α]] ∼ = I R, T ). Clearly, Condition (3) holds. Condition (2) is satisfied if and only if α(e) = e for all e = e2 ∈ R. In addition, T contains no nonzero idempotents. Thus, the result follows from Proposition 17.2.10.  Let R = Z4 ⊕ Z
4 and α : R → R given by α(a, b) = (b, a). Clearly, Z4 is local with J Z4 = {0, 2}. Hence, R is an abelian exchange ring. But R[[x, α]] is not an abelian exchange ring from Example 17.2.12. In addition, we notice that the center of any abelian exchange ring is also an abelian exchange ring from Theorem 17.2.2. But the converse is not true. Example 17.2.13. Let D be a division ring, and let S =

∞ Q

Di , where

i=1

Di = D for all i. Define α : S → S by α(a1 , a2 , · · · ) = (a1 , a1 , a2 , · · · ). Then α is injective but not onto. Let R = S[x, α] be a skew polynomial ring. Then the center C(R) = {(a, a, a, · · · ) | a ∈ C(D)} ∼ = C(D), and so C(R) is an abelian exchange ring. As α(0, 1, 1, · · · ) = (0, 0, 1, · · · ), it follows from Example 17.2.12 that R is not an abelian exchange ring. In fact, 0 is the only idempotent contained in xR and 1 6∈ (1 − x)R.  Example 17.2.14. Let R be a ring. Then the following are equivalent: (1) (2) (3) (4) (5)

R is an abelian exchange ring. R[[x]] is an abelian exchange ring. R[[x]]/(x2 ) is an abelian exchange ring. R[[x]]/(xn ) is an abelian exchange ring for some n ≥ 2. R[[x]]/(xn ) is an abelian exchange ring for all n ≥ 2.

Proof. (1) ⇒ (2) Choose α : R → R by α(a) = a. Then R[[x]] is an abelian exchange ring by Example 17.2.12. (2) ⇒ (5) is clear from Theorem 17.2.2. (5) ⇒ (4) ⇒ (3) are trivial. n−1 P (3) ⇒ (1) Let S = R[[x]]/(xn )(n ≥ 2). Then S = { ai xi | xn = i=0

0, a0 , · · · , an−1 ∈ R}. Construct a map ϕ : S → R given by ϕ

n−1 P i=0


ai xi =

a0 . Then ϕ is a ring epimorphism, and so R ∼ = S/kerϕ. According to Theorem 17.2.2, R is an abelian exchange ring.  Example 17.2.15. Let R be a ring. Then the following are equivalent:

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(1) R is an abelian exchange ring. (2) R[x]/(x2 ) is an abelian exchange ring. Proof. It is straightforward to check that R[x]/(x2 ) ∼ = R[[x]]/(x2 ). Therefore we complete the proof from Example 17.2.14.  17.3

K0 -Groups

Let R be a ring with an identity 1. Two right R-modules A, B ∈ F P (R) are said to be stably isomorphic provided that A ⊕ nR ∼ = B ⊕ nR for some n ∈ N. Clearly, stable isomorphism defines an equivalence on F P (R). We use [A] to denote the equivalence class of any A ∈ F P (R). Let K0 (R)+ = {[A] | A ∈ F P (R)}. We define an operation + on K0 (R)+ according to the rule [A] + [B] = [A ⊕ B]. Then K0 (R)+ becomes an abelian semigroup with a zero element [0]. Let K0 (R) be the abelian group generated by K0 (R)+ . Let R be a Dedekind domain, and let Cl(R) be the abelian group whose generators are the isomorphism classes hIi of nonzero ideals I of R with the operation hIihJi = hIJi.
We say that Cl(R) is the ideal class group of R, √ e.g., Cl Z 21 (1 + −23) ∼ = Z3 . As is well known, K0 (R) ∼ = Z[R] ⊕ Cl(R) (cf. [368]). Let R be an abelian exchange ring. Then K0 (R) is the abelian group generated by {[eR] | e = e2 ∈ R} because every finitely generated projective right R-module is isomorphic to a direct sum of principal right ideals of R. Now we investigate such abelian groups for abelian exchange rings. Lemma 17.3.1. Let R be an exchange ring, and let A be a finitely generated projective right R-module. Then there is a complete set of idempotents {e1 , · · · , en } in R and nonnegative integers tij such that A ∼ = t11 (e1 R) ⊕ · · · ⊕ t1n (en R). Proof. According to Lemma 13.1.8, we have idempotents f1 , · · · , fn ∈ R such that A ∼ = f1 R ⊕ · · · ⊕ fn R. Let Ii = fi R and Ji = (1 − fi )R. Then R = Ii ⊕ Ji . As R = I1 ⊕ J1 = I2 ⊕ J2 , we get a refinement matrix I2 J2



I1 h11 R h21 R

J1

 h12 R , h22 R

where {h11 , h12 , h21 , h22 } is a complete set of idempotents of R. In addition, I1 ∼ = h11 R ⊕ h21 R, I2 ∼ = h21 R ⊕ h22 R. As R = h11 R ⊕ h21 R ⊕ h12 R ⊕ h22 R =

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I3 ⊕ J3 , we get a refinement matrix I3 J3



h11 R

h21 R

h12 R

k11 R k21 R

k12 R k22 R

k13 R k23 R

h22 R

 k14 R , k24 R

where {kij } is a complete set of idempotents of R. In addition, I1 ∼ = k11 R ⊕ k21 R ⊕ k12 R ⊕ k22 R, I2 ∼ = k11 R ⊕ k21 R ⊕ k13 R ⊕ k23 R and I3 ∼ = k11 R ⊕ k12 R ⊕ k13 R ⊕ k14 R. By iteration of this process, the result follows.  Lemma 17.3.2. Let R be an abelian exchange ring, and let Ξ(R) be a medium space. If e, f ∈ R are idempotents, then E(f R) = E(gR) if and only if f = g. Proof. One direction is obvious. Now assume that G G

Ξ(R) = E(f R) E (1 − f )R = E(gR) E (1 − g)R .



If E(f R) = E(gR), then E (1
− f )R = E (1
− g)R . For any P ∈ Ξ(R), we have either P 6∈ E (1−f )R = E (1−g)R or P 6∈ E(f R) = E(gR). So 1 − f, 1 − g ∈ P or f, g ∈ P . In any case, we have g − f ∈ P , and therefore g − f ∈ Λ(R). As Λ(R) is nil, there exists k ∈ N such that (g − f )2k+1 = 0. Since g and f are central, we see that 2k+1

(g − f )

= (g − f ) g + f + gf

2k−1 X

i

(−1)

i=1



2k i






= g − f; 

hence, g = f .

Lemma 17.3.3. Let R be an abelian exchange ring, let Ξ(R) be a medium n F space. If Ξ(R) = E(Ii ), then there exists a unique complete set of i=1

idempotents, {e1 , · · · , en }, such that each E(Ii ) = E(ei R). n n
F P Proof. Since Ξ(R) = E(Ii ) = E Ii , analogously to Lemma 17.1.3, we have R = E(I1 ) ∩ E whence I1 ∩

n P

n P

i=1

i=1

i=1

Ii . So we can find ei ∈ Ii such that 1 =


Ii = ∅, we see that I1 ∩

i=2 n P

i=2




Ii ⊆ Λ(R). From

e1 −e21

n P

i=2

n P

ei . As

i=1


Ii ⊆ P for all P ∈ Ξ(R),

= e1 e2 +· · ·+e1 en ∈ I1 ∩

n P

i=2


Ii ,

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we have some m1 ∈ N such that (e1 − e21 )m1 = 0. Likewise, we have some mi such that (ei − e2i )mi = 0 for all i = 2, · · · , n. If i 6= j, then we see that n X ei ej ∈ Ii ∩ Ij ⊆ Ii ∩ ( Ij ). j=1 j6=i

Since Ii ∩(

n P

j=1

Ij ) ⊆ P for all P ∈ Ξ(R), we have that ei ej ∈ Λ(R), so we can

j6=i

find nij ∈ N such that (ei ej )nij = 0. Let m = max{nij , mk | 1 ≤ i, j, k ≤ n, i 6= j}. Then (ei − e2i )m = 0 for all i = 1, · · · , n and (ei ej )m = 0(i 6= j).   m P 2m ei2m−r (1 − ei )r . It follows from ei (1 − ei ) = (1 − ei )ei Let fi = r r=0   2m P 2m that 1 − fi = e2m−r (1 − ei )r . As a result, i r r=m fi (1 − fi ) =

m P

r=0

= 0;



2m r



e2m−r (1 i

r

− ei )


2m P

r=m



2m r



e2m−r (1 − ei )r i




that is, fi = fi2 . In addition, we have fi fj = (fi fj )m = 0(i 6= j). So {f1 , · · · , fn } is a set of orthogonal idempotents of  R. For  any P ∈ Ξ(R), we m P 2m have P ∈ Spec(R). If f1 , · · · , fn ∈ P , then e2m−r (1 − ei )r ∈ P i r r=0 for all i = 1, · · · , n; hence,  n m  X X 2m r=0

i=1

This implies that

m P

r=0



2m r

Let y =

r=1



n P

i=1

r

2m r



n P

i=1

i=1

e2m−r ( i


e2m−r (1 − ei )r ∈ P. i


e2m−r (1 − ei )r ∈ P ; and then i

 X m  n X 2m r=0

m P



r

X
e2m−r ( ej )r ∈ P. i

n P

j=1 j6=i

j6=i

n
P ej )r . Then e2m + y ∈ P , where i i=1

y ∈ Λ(R). Furthermore, we have z ∈ Λ(R) such that

n P

i=1

ei


2m

+y +z ∈ P .

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Hence 1 + y + z ∈ P . By virtue of Lemma 17.1.8, R(1 + y + z)R = R, and so P = R, a contradiction. As a result, we can find some i such that fi R 6⊆ P , n S and so P ∈ E(fi R). That is, Ξ(R) = E(fi R). It follows by fi R ⊆ Ii i=1

that E(fi R) ⊆ E(Ii ). Hence

E(R) = Ξ(R) = and so R =

n P

i=1

fi R =

n P

i=1

n G

E(fi R) = E

i=1




n X i=1


fi R ,

fi R. This implies that

n P

fi = 1. That

i=1

is, {f1 , · · · , fn } is a complete set of idempotents. One easily checks that E(fi R) = E(Ii ) for all i = 1, · · · , n. Assume that g1 , · · · , gn ∈ R is a complete set of idempotents such that E(gi R) = E(Ii ) for all i = 1, · · · , n. It follows from Lemma 17.3.2 that g1 = f1 . Analogously, we have gi = fi for all i = 2, · · · , n. Consequently, there exists a unique complete set of idempotents, {f1 , · · · , fn }, such that each E(Ii ) = E(fi R).  Lemma 17.3.4. Let R be an abelian exchange ring, and let A be a finitely generated projective right R-module. If Ξ(R) is a medium space, then the rule fA (P ) = rankR/P (A/AP ) defines a continuous map fA : Ξ(R) → Z+ . Conversely, any continuous map of Ξ(R) into Z+ arises in this manner. Proof. We may assume that A 6= 0. It follows from Lemma 17.3.1 that there exists a complete set of idempotents {e1 , · · · , ek } in R and nonnegative integers ti such that A ∼ = t1 (e1 R) ⊕ · · · ⊕ tk (ek R). Let n1 , · · · , nm be P ei . Then there exists distinct numbers of all ti . For each nj , set gj = ti =nj

a complete set of idempotents in R, {g1 , g2 , · · · , gm }, such that m M A∼ nj (gj R). = j=1

By hypothesis, Λ(R) is nil. Hence, each gj 6∈ Λ(R), and so there exists some P ∈ Ξ(R) such that gj 6∈ P . If i 6= j, then gi gj = 0 ∈ P ; hence, gi ∈ P . This implies that {P ∈ Ξ(R) | fA (P ) = nj } = {P ∈ Ξ(R) | gj 6∈ P }.

S S S For any open set {n} ⊆ Z+ , we have fA−1 {n} = fA−1 {n} , so it n n n
suffices to prove that fA−1 {n} is an open set. Observe that
fA−1 {n} = {P ∈ Ξ(R) | fA (P ) = n}  E(gj R) if n = nj ; = ∅ if n 6∈ {n1 , · · · , nm }.

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This implies that fA : Ξ(R) → Z+ is continuous. Let f : Ξ(R) → Z+ be continuous. In view of Lemma 17.1.3, we may assume that im(f ) = {n1 , · · · , nk } ⊆ Z+ . Set Xi = {P ∈ Ξ(R) | f (P ) = ni }. Then Xi is a clopen set of Ξ(R). By virtue of Lemma 17.3.3, there exists a unique complete set of idempotents, {e1 , · · · , ek }, such that each k L Xi = E(ei R). Set B = ni (ei R). Then rankR/P (B/BP ) = f (P ) for all i=1

P ∈ Ξ(R). Therefore f = fB , as required.



Theorem 17.3.5. Let R be an abelian exchange ring, and let Ξ(R) be a medium space. Then there exists a semigroup isomorphism ϕ : K0 (R)+ →
+ {f : Ξ(R) → Z | f is continuous} given by ϕ [A] = fA for any [A] ∈ K0 (R)+ , where fA (P ) = rankR/P (A/AP ) for any P ∈ Ξ(R). Proof. Define a semigroup morphism ϕ : K0 (R)+ → {f : Ξ(R) → Z+ |
f is continuous} given by ϕ [A] = fA for any [A] ∈ K0 (R)+ , where fA (P ) = rankR/P (A/AP ) for any P ∈ Ξ(R). Clearly, ϕ is well defined. If ϕ([A]) = ϕ([B]), then rankP/P (A/AP ) = rankR/P (B/BP ) for all P ∈ Ξ(R). In view of Proposition 17.2.9, A ∼ = B, and so [A] = [B]. That is, ϕ is a monomorphism. For any g ∈ {f : Ξ(R) → Z+ | f is continuous}, it follows from Lemma 17.3.4 that there exists some B ∈ F P (R) such that g = fB = ϕ([B]). Therefore ϕ is an epimorphism, as desired.  Let Ξ(R) be a medium space. For any g, h ∈ {f : Ξ(R) → Z | f is continuous }, we define (g + h)(P ) = g(P ) + h(P ) for all P ∈ Ξ(P ). Then {f : Ξ(R) → Z | f is continuous} is an abelian group. Corollary 17.3.6. Let R be an abelian exchange ring, and let Ξ(R) be a medium space. Then there exists a group isomorphism K0 (R) ∼ = {f : Ξ(R) → Z | f is continuous}. Proof. According to Theorem 17.3.5, there exists a semigroup isomorphism
ϕ : K0 (R)+ → {f : Ξ(R) → Z+ | f is continuous} given by ϕ [A] = fA for any [A] ∈ K0 (R)+ . Let H be the abelian group generated by {f : Ξ(R) → Z+ | f is continuous}. Then there exists a group isomorphism K0 (R) ∼ = H. We note that every continuous f : Ξ(R) → Z+ can be extended to a continuous fˆ : Ξ(R) → Z. Construct a group morphism φ : H → {f : Ξ(R) → Z | f is continuous} given by φ(f − g) = fˆ − gˆ. It is easy to verify that φ is a monomorphism. If h : Ξ(R) → Z is continuous,

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then we can define two continuous maps h+ , h− : Ξ(R) → Z+ given by  h(P ) if h(P ) ≥ 0; h+ (P ) = 0 if h(P ) < 0; and

h− (P ) =



0 if h(P ) ≥ 0; −h(P ) if h(P ) < 0.

One easily checks that h = φ(h+ −h− ), and so φ is an epimorphism. Hence, H∼ = {f : Ξ(R) → Z | f is continuous }, and therefore the result follows.  Let R be an abelian exchange ring. Then we claim that K0 (R) ∼ = {f : Spec(R) → Z | f is continuous}. As in the proof of Lemma 17.1.12, Spec(R) is a medium space, and we are done by Corollary 17.3.6. Let p be a prime and put R = Z(p) , the localization of integers at (p). Then R is a local ring with M ax(R) = {pZ(p) }. Hence it is an abelian exchange ring, and so K0 (R) ∼ = {f : Spec(R) → Z | f is continuous} ∼ = Z, where Spec(R) = {{0}, pZ(p)}. Observe that since M ax(R) is a single point topological space, the maps from M ax(R) to Z are defined by giving the image of pZ(p) and hence K0 (R) ∼ = Z ∼ = {f : M ax(R) → Z | f is continuous}. In this case, T {P | P ∈ M ax(R)} = pZ(p) ; hence, the intersection of ideals in M ax(R) is not nil. Thus the condition of Λ(R) being nil is not necessary. Corollary 17.3.7. Let R be an abelian exchange ring, and let M ax(R) ⊆ Ξ(R) ⊆ Spec(R). If J(R) is nil, then K0 (R) ∼ = {f : Ξ(R) → Z | f is continuous }. Proof. In light of Proposition 17.1.9, Ξ(R) is a medium space. Therefore the result follows from Corollary 17.3.6.  Let R be an abelian strongly π-regular ring, and let M ax(R) ⊆ Ξ(R) ⊆ Spec(R). Then J(R) is nil. It follows from Corollary 17.3.7 that K0 (R) ∼ = {f : Ξ(R) → Z | f is continuous}. Let R be a commutative ring over which every module has a maximal submodule, and let M ax(R) ⊆ Ξ(R) ⊆ Spec(R). Then we claim that K0 (R) ∼ = {f : Ξ(R) → Z | f is continuous}. Koifman proved that every commutative ring over which every module has a maximal submodule is a right P -exchange ring. Since R is commutative, by [372, Proposition 4.1], J(R) is right T -nilpotent, and the result follows from Corollary 17.3.7. One observes that the set of continuous maps from those topological spaces to Z does not depend (up to isomorphy) on the given set as long

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as the intersection of all the ideals in it is nil. The impression one can obtain from this observation is that it could even be enough to consider only the maps from M ax(R). This fact remains unclear, but we can derive the following. Corollary 17.3.8. Let R be an abelian exchange ring. Then K0 (R) ∼ = {f :
M ax R/J(R) → Z | f is continuous}.

Proof. Let S = R/J(R). Since R is an abelian exchange ring, so is S. As J(S) = 0, it follows from Corollary 17.3.7 that K0 (S) ∼ = {f : M ax(S) → Z | f is continuous}. In view of [9, Theorem 3.5], there exists a surjective map K0 (ϕ) : K0 (R) → K0 (S), where ϕ : R → S is Given  the natural projection.  any [P ] − [Q] ∈ K0 (ϕ), we have P/P J(R) = Q/QJ(R) . Since R is an exchange ring, there exists some m ∈ N such that P ′ /P ′ J(R) ∼ = Q′ /Q′ J(R), ′ ∼ m ′ ∼ m ′ ∼ ′ where P = P ⊕R and Q = Q⊕R . Clearly, P = Q , and then [P ] = [Q]. This implies that K0 (R) ∼  = K0 (S), and therefore the proof is true.

Corollary 17.3.9. Let R be an abelian exchange ring,
and let M ax(R) ⊆ Ξ(R) ⊆ Spec(R). Then K0 (R) ∼ {f : Spec R/Λ(R) → Z | f is continu= ous }. Proof. Let S = R/Λ(R). Since R is an abelian exchange ring, so is S. Choose Ξ(S) = Spec(S). As in the proof of Lemma 17.1.12, Λ(S) is nil. By virtue of Corollary 17.3.6, K0 (S) ∼ = {f : Spec(S) → Z | f is continuous}. In view of [2,
Theorem 3.5], there exists a surjective map K0 (ϕ) : K0 (R) → K0 R/Λ(R) , where ϕ : R → R/Λ(R) is the natural projection. Given any    [P ] − [Q] ∈ K0 (ϕ), we have P/P Λ(R) = Q/QΛ(R) . So we can find some m ∈ N such that P ′ /P ′ Λ(R) ∼ = P ⊕ Rm = Q′ /Q′ Λ(R), where P ′ ∼ m ′ ∼ and Q = Q ⊕ R . In view of Lemma 1.4.11, we have right R-module decompositions P′ ∼ = P1′ ⊕ P2′ , Q′ ∼ = Q′1 ⊕ Q′2 , P1′ ∼ = Q′1 , P2′ = P2′ Λ(R) and Q′2 = Q′2 Λ(R). Since R is an exchange ring, there exists idempotents e1 , · · · , en ∈ R such n L that P2′ ∼ ei R. As P2′ = P2′ Λ(R), we deduce that ei R = ei RΛ(R) = i=1

for all i. Hence each ei ∈ Λ(R). According to Lemma 17.1.8, we have R(1 − ei )R = R; hence, each ei = 0. It follows that P2′ = 0. Likewise, Q′2 = 0. So we get P ′ ∼ = Q′ .
As a result, we prove that [P ] = [Q]. ∼ Therefore K0 (R) = K0 R/Λ(R) , as required. 

Theorem 17.3.10. Let R be an abelian exchange ring, and let M ax(R) ⊆

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Ξ(R) ⊆ Spec(R). If J(R) is nil, then K0 (R) ∼ = {f : M ax R/Λ(R) → Z | f is continuous}.
Proof. Analogously to Corollary 17.3.9, we have K0 (R) ∼ = K0 R/Λ(R) . Let S = R/Λ(R). Then S is an abelian exchange ring. Let a ∈ J(S). For any x, y ∈ R, we have that 1 − xay ∈ S is invertible. So we can find some z ∈ R such that 1 − z(1 − xay), 1 − (1 − xay)z ∈ Λ(R). This implies that 1 − z(1 − xay), 1 − (1 − xay)z ∈ J(R). As a result, 1 − xay ∈ R is invertible, and then a ∈ J(R). So an = 0 for some n ∈ N. Furthermore, we see that a = 0. According to Corollary 17.3.7, K0 (S) ∼ = {f : M ax(S) →
Z | ∼ f is continuous}. Therefore we prove that K0 (R) = {f : M ax R/Λ(R) → Z | f is continuous}.  Corollary 17.3.11. Let R be an abelian right P -exchange ring,
and let M ax(R) ⊆ Ξ(R) ⊆ Spec(R). Then K0 (R) ∼ = {f : M ax R/Λ(R) → Z | f is continuous}. Proof. In view of [372, Proposition 4.1], J(R) is right T -nilpotent, and the proof is complete by Theorem 17.3.10. 

17.4

Pierce Spaces

Many authors have investigated Pierce stalks of a ring R (cf. [60-61]). The study of Pierce stalks for noncommutative rings was developed by Burgess and Stephenson in their paper [58]. Subsequently, Burgess and Goursaud in their paper [56] developed a technique and successfully used it to compute K0 (R) in the case where R is a regular, biregular ring of bounded index. For any commutative ring R, K0 (R) ∼ = {f : Spec(R) → Z | f is continuous} ⊕xk0 R, where xk0 R = ker(rank) (cf. [368, Proposition 25]). The main purpose of this section is to give an analogue of this result. If R/P is a Pierce stalk of R, then P is called a Pierce ideal of R. Let P ier(R) be the set of all Pierce ideals of R. Let V(I) = {P ∈ P ier(R) | I ⊆ P , where I is an ideal generated by central idempotents of R}. Lemma 17.4.1. The P ier(R) is a topological space with closed sets V(I), where all ideals I are generated by central idempotents. Proof. Clearly, V(R) = ∅ and V(0) = P ier(R). Let I and J be generated by the sets of central idempotents {ei } and {fj }, respectively. Then IJ

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is generated by the set of central idempotents {ei fj }. Given any P ∈ S V(I) V(J), we have that I ⊆ P or J ⊆ P ; hence, IJ ⊆ P . Thus, S P ∈ V(IJ), and so V(I) V(J) ⊆ V(IJ). Given any P ∈ V(IJ), then IJ ⊆ P . Assume that I P and J P . Then we can find central idempotents e ∈ I, f ∈ J such that e, f 6∈ P . By the maximality of P , we obtain P + eR = R and P + f R = R. As a result, P + ef R = R, and then ef 6∈ P . But we have ef ∈ IJ ⊆ P ; this is a contradiction. S Hence I ⊆ P or J ⊆ P , and so P ∈ V(I) V(J). This means that S S V(IJ) ⊆ V(I) V(J). Therefore we conclude that V(I) V(J) = V(IJ). If {Iλ |λ ∈ Λ} is a set of ideals generated by central idempotents of R, then P Iλ is an ideal generated by central idempotents. One easily checks that λ∈Λ T P
V(Iλ ) = V Iλ , as required.  λ

λ

P ier(R) is said to be the Pierce space of a ring R. Let E(I) = P ier(R)− V(I). Then P ier(R) is a topological space where {E(I) | I is an ideal generated by central idempotents of R} is the collection of its open sets. Let Z be a topological space with discrete topology. For any g, h ∈ {f : P ier(R) → Z | f is continuous}, we define (g + h)(P ) = g(P ) + h(P ), (−f )(P ) = −f (P ) and 0(P ) = 0 for all P ∈ P ier(R). Then {f : P ier(R) → Z | f is continuous} is an abelian group. Theorem 17.4.2. Let R be a ring. Then the following are equivalent: (1) B(R) = {0, 1}. (2) P ier(R) is a connected space. (3) {f : P ier(R) → Z | f is continuous} ∼ = Z. Proof. (1) ⇒ (3) Since B(R) = {0, 1}, P ier(R) = {{0}}. Construct a map
ϕ : {f : P ier(R) → Z | f is continuous} → Z given by ϕ(f ) = f {0} for any f ∈ {f : P ier(R) → Z | f is continuous}. For any f, g ∈ {f : P ier(R) → Z | f is continuous}, one easily checks that ϕ(f + g) = ϕ(f ) + ϕ(g), ϕ(−f ) = −ϕ(f ), ϕ(0) = 0. Thus, ϕ is a group homomorphism. If ϕ(f ) = 0, then f (P ) = 0 for any P ∈ P ier(R); hence, f = 0. This means that ϕ is monomorphic. Given
any n ∈ Z, we have a continuous map h : P ier(R) → Z given by h {0} = n. As ϕ(h) = n, we see that ϕ is epimorphic, and therefore {f : P ier(R) → Z | f is continuous} ∼ = Z.

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(3) ⇒ (2) Assume that P ier(R) is not a connected space. Then there F exist open sets K and L of P ier(R) such that P ier(R) = K L. For any m, n ∈ Z, we define fm,n : P ier(R) → Z given by  m if P ∈ K; fm,n (P ) = n if P ∈ L.
S −1
S S −1 For any open set {aλ } ⊆ Z, we see that fm,n {aλ } = fm,n {aλ } , where

λ

λ

−1 fm,n

λ


 ∅ if aλ 6= m, n; {aλ } = K if aλ = m;  L if aλ = n.

Thus, fm,n is continuous, and then fm,n ∈ {f : P ier(R) → Z | f is continuous}. Construct a map ϕ : Z ⊕ Z → {f : P ier(R) → Z | f is continuous} given by ϕ(m, n) = fm,n for any (m, n) ∈ Z ⊕ Z. Clearly, ϕ is a group monomorphism. As {f : P ier(R) → Z | f is continuous} ∼ = Z, we get a group monomorphism φ : Z ⊕ Z → Z. Since every subgroup of a finitely generated free abelian group of rank s is free of rank ≤ s, we deduce that Z ⊕ Z ∼ = Z or 0, a contradiction. Therefore P ier(R) is a connected space. (2) ⇒ (1) If B(R) 6= {0, 1}, then there exists a non-trivial central idempotent e ∈ R. If P ∈ P ier(R), it follows by [382, Remark 11.1] that either e ∈ P or 1 − e ∈ P , i.e., e 6∈ P or 1 − e 6∈ P . This implies that
eR P or (1 − e)R P . That is, P ∈
E(eR) or P ∈TE (1 − e)R
. S Therefore, P ier(R) = E(eR) E (1 − e)R . If P ∈ E(eR) E (1 − e)R , then eR P and (1 − e)R P , i.e., e 6∈ P and 1 − e 6∈ P . In view of [382, Remark 11.1], 1 − e ∈ P and e ∈ P . This implies
that T 1 = (1 − e) + e ∈ P ; this gives a contradiction.
So E(eR) E (1 − e)R = ∅, F and then P ier(R) = E(eR) E (1 − e)R . Hence P ier(R) is not a connected space, a contradiction. Therefore B(R) = {0, 1}. 

√ Example 17.4.3. Let R = Z[ −5]. Then {f : P ier(R) → Z | f is continuous} ∼ = Z and K0 (R) ∼ = Z ⊕ Z/2Z.

Proof. Clearly, R is a Dedekind domain; hence B(R) = {0, 1}. In view of Theorem 17.4.2, {f : P ier(R) → Z | f is continuous} ∼ = Z. By virtue of [368, Example 4.1], K0 (R) ∼  = Z ⊕ Z/2Z. 

F F 0 F P ier(R) → Z | f is continuous} ∼ = Z and K0 (R) ∼ = Z ⊕ Z. Example 17.4.4. Let F be a field, and let R =



. Then {f :

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   00 1F 0 Proof. It is easy to verify that B(R) = { , }. In view of 00 0 1F Theorem 17.4.2, {f : P ier(R) → Z | f is continuous} ∼ = Z.   0F Clearly, J(R) = ; hence, R/J(R) ∼ Thus, = F ⊕ F. 0 0
K0 R/J(R) ∼ = Z ⊕ Z. Let π : R → R/J(R) be the natural epimorphism. Then K0 π is monomorphic. Since R is an exchange ring, it follows from
[9, Theorem 3.5] that K0 π is epimorphic. Therefore K0 (R) ∼ = K0 R/J(R) ∼ = Z ⊕ Z.  Proposition 17.4.5. Let R be a ring. If |B(R)| < ∞, then {f : P ier(R) → Z | f is continuous } is a finitely generated free abelian group. Proof. As | B(R) | is finite, so is | P ier(R) |. Write P ier(R) = {P1 , · · · , Pn }. Construct a map ϕ : {f : P ier(R) → Z | f is continuous} →
f 7→ f (P1 ), · · · , f (Pn ) .

n L 1

Z;

One easily checks that

ϕ(f + g) = ϕ(f ) + ϕ(g), ϕ(−f ) = −ϕ(f ), ϕ(0) = 0. Thus, ϕ is a group homomorphism. If ϕ(f ) = 0, then f (P1 ) = · · · =
f (Pn ) = 0, and so f P ier(R) = 0. That is, f = 0; hence, ϕ is monomorn L phic. For any (m1 , · · · , mn ) ∈ Z, we construct a map h : P ier(R) → Z 1

given by

For any open set where

   m1 if P .. h(P ) = .   mn if P

= P1 ; .. . = Pn .



S S S {aλ } ⊆ Z, we see that h−1 {aλ } = h−1 {aλ } , λ

h−1

 ∅   
 {P1 } {aλ } = ..   .   {Pn }

λ

if aλ 6∈ {m1 , · · · , mn }; if aλ = m1 ; .. . if aλ = mn .

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Clearly, {Pi } =

j=1

619

E(Pj ) is open. Thus, h is continuous, and then f ∈

j6=i

{f : P ier(R) → Z | f is continuous}. In addition, ϕ(h) = (m1 , · · · , mn ). Therefore, h is epimorphic, and so the result follows.  Lemma 17.4.6. P ier(R) is a compact space. S P
Proof. Suppose that E(Iλ ) = P ier(R). Then E Iλ = P ier(R). λ P λ Assume that R 6= Iλ . Let λ

Ω = {I | I ⊇

X

Iλ is an ideal generated by central idempotents in R}.

λ

Clearly, Ω 6= ∅. Given any I1 ⊆ I2 ⊆ · · · ⊆ In ⊆ · · · in Ω, it is easy ∞ S to verify that Ii ∈ Ω. By using Zorn’s Lemma, we have an ideal J i=1

which is maximal in Ω. Since J is
generated by central idempotents in P R, J ∈ P ier(R), while J 6∈ E Iλ . This gives a contradiction. Hence, P λ

λ

Iλ = R, and then there exist r1 ∈ I1 , · · · , rn ∈ In such that

This implies that R =

n P

Ii . So

n P

ri = 1.

i=1

i=1 n [

i=1

E(Ii ) = E

n X
Ii = E(R) = P ier(R), i=1



and therefore P ier(R) is a compact space.

Lemma 17.4.7. Let I and J be ideals generated by central idempotents F in a ring R. If P ier(R) = E(I) E(J), then there exists a unique
central idempotent e ∈ R such that E(I) = E(eR) and E(J) = E (1 − e)R . F S Proof. Since P ier(R) = E(I) E(J), we have P ier(R) = E(I) E(J) = E(I + J). As in the proof in Lemma 17.4.6, we get I + J = R. On the T other hand, E(IJ) = E(I) E(J) = ∅. For any P ∈ P ier(R), we have T P 6∈ E(IJ); hence, IJ ⊆ P . Thus, IJ ⊆ P . In view of [382, P ∈P ier(R)

Remark 11.2], IJ = 0. Write 1 = e1 + e2 , where e1 ∈ I, e2 ∈ J. Then e1 e2 = e1 (1−e1 ) = (1−e2 )e2 ∈ IJ, and therefore e1 e2 = 0, e21 = e1 , e22 = e2 . For any x ∈ R, we see that e1 x − e1 xe1 = e1 x(1 − e1 ) = e1 xe2 ∈ IJ; hence, e1 x = e1 xe1 . Likewise, xe1 = e1 xe1 . Thus, e1 x = xe1 , and so

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e1 ∈ B(R). This implies that e2 ∈ B(R). As in the proof in Theorem F 17.4.2, P ier(R) = E(e1 R) E(e2 R). Since e1 R ⊆ I and e2 R ⊆ J, we have E(e1 R) ⊆ E(I) and E(e2 R) ⊆ E(J), so we conclude that E(I) = E(e1 R) and E(J) = E(e2 R). Suppose f is an another central idempotent in R such that E(I) =
E(f R) and E(J) = E (1 − f )R . For any P ∈ P ier(R), we have either
P ∈ E(e1 R) = E(f R) or P ∈ E(e2 R) = E (1 − f )R . Thus, e1 , f ∈ P or T P. 1−e1 , 1−f ∈ P . Hence, e1 −f ∈ P . This implies that e1 −f ∈ In view of [382, Remark 11.2], we obtain e1 = f , as required.

P ∈P ier(R)



Theorem 17.4.8. Let R be a ring. Then there exists a group homomorphism θ : {f : P ier(R) → Z | f is continuous } → K0 (R). Proof. Given then
Sany f ∈ {f : P ier(R) → ZS | f
is continuous}, S f P ier(R) = {nλ }. Hence P ier(R) = f −1 {nλ } = f −1 ({nλ }). In λ λ λ
view of Lemma 17.4.6, we can find n1 , · · · , nr ∈ Z such that f P ier(R) = {n1 , · · · , nr }. Obviously, {ni } is both open and closed in Z. Since f is con
tinuous, f −1 {ni } is open and closed in P ier(R). In view of Lemma 17.4.7,
there is a unique central idempotent ei ∈ R such that f −1 {ni } = E(ei R) T for each i. If i 6= j, then E(ei R) E(ej R) = ∅, and so E(ei ej R) = ∅. As in the proof in Lemma 17.4.7, ei ej = 0. Next,

E

r X i=1

r r G
G
ei R = E(ei R) = f −1 {ni } = P ier(R).

This implies that R =

i=1

r P

i=1

ei R. Thus, 1 =

i=1

r P

i=1

ei si for some si ∈ R. On

multiplying each side by ei , we see that ei = ei si , and so 1 =

r P

ei .

i=1

Such a set {e1 , · · · , er } we call the complete set of central idempotents r P corresponding to f . Put θ(f ) = ni [ei R] ∈ K0 (R). Then θ : {f : i=1

P ier(R) → Z | f is continuous} → K0 (R) is a map. For any g, h ∈ {f : P ier(R) → Z | f is continuous}, there exist unique complete sets {g1 , · · · , gs } of central idempotents corresponding to g and {h1 , · · · , ht } of central idempotents corresponding to h. Write E(gi R) =

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g −1 {mi } and E(hj R) = h−1 {nj } . Then θ(g) + θ(h) = = = =

s P

t P

mi [gi R] +

nj [hj R]

i=1 s P

t t s  P  P   P mi g i ( hj )R + nj hj ( gi )R

i=1 s P

t P

i=1 s P

j=1

j=1

mi

t L

j=1



g i hj R +

j=1

t P

nj

j=1

s L

i=1

g i hj R

i=1

(mi + nj )[gi hj R].



i=1 j=1


Suppose (g + h) P ier(R) = {c1 , · · · , cp }. Then θ(g) + θ(h) =

= =

p P

k=1 p P

k=1 p P

k=1

Clearly, {

P

mi +nj =c1

g i hj , · · · ,

P

P

ck

[gi hj R]

mi +nj =ck

ck



g i hj R

mi +nj =ck

 ck (

mi +nj =cp

L

P

mi +nj =ck



 gi hj )R .

gi hj } is the complete set of central

idempotents corresponding to f +g. So θ(f +g) = θ(f )+θ(g), and therefore θ is a group homomorphism.  Corollary 17.4.9. Let R be a commutative ring. Then Z is isomorphic to a direct summand of {f : P ier(R) → Z | f is continuous }. Proof. For any m ∈ Z, we define fm (P ) = m for any P ∈ P ier(R). Obviously, fm ∈ {f : P ier(R) → Z | f is continuous}. Construct a group homomorphism φ : Z → {f : P ier(R) → Z | f is continuous} given by φ(m) = fm for any m ∈ Z. Since R is a commutative ring, there exists a maximal ideal M of R; hence, R/M is a field. Let ϕ : R → R/M be the natural
epimorphism. Then ψ : K0 (R/M ) = Z[R/M ] ∼ = Z, where ψ m[R/M ] = m for any m[R/M ] ∈ K0 (R/M ). By virtue of Theorem 17.4.8, there exists a group homomorphism θ : {f : P ier(R) → Z | f is continuous} → K0 (R). For any m

∈ Z, ψ(K0 ϕ)θφ(m) = ψ(K0 ϕ)θ(fm ) = ψ(K0 ϕ) m[R] = ψ m[R/M ] = m. Thus, ψ(K0 ϕ)θφ = 1Z , and so φ : Z → {f : P ier(R) → Z | f is continuous} is a splitting monomorphism. Therefore {f : P ier(R) → Z | f is continuous} ∼ = Z ⊕ {f : P ier(R) → Z | f is continuous}/im(φ), as required. 

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In view of Theorem 17.1.5, every space between the maximal spectrum and the prime spectrum of an abelian exchange ring is strongly zerodimensional. A natural question asks if the Pierce space of such a ring is a space between two spectrums. Let p be a prime and put R = Z(p) . Then  R is local; hence, it is an abelian exchange ring. Obviously, P ier(R) = {0} and M ax(R) = {pR}. Thus, M ax(R) P ier(R). Hence, P ier(R) is not a medium space. Let e ∈ B(R), and let V(e) := V(eR). We extend Lemma 17.1.10 as follows. Lemma 17.4.10. Let R be an abelian ring. Then A is a clopen subset of P ier(R) if and only if there exists a unique idempotent e ∈ R such that A = V(e). Proof. If A = V(e) for an idempotent e ∈ R, then P ier(R) = A ⊔ V(1−e). Thus, A is an open subset of P ier(R); hence, A is clopen. Conversely, assume that A is an clopen subset of P ier(R). Then we have two ideals I and J, which are generated by central idempotents of R, such that A = V(I) and P ier(R) = V(I) ⊔ V(J). Thus, V(IJ) = P ier(R); hence, IJ ⊆ T T P . In view of [382, Remark 11.2], IJ = 0. V(I) V(J) = ∅, P ∈P ier(R)

and so V(I + J) = ∅, i.e., E(I + J) = P ier(R). As in the proof of Lemma 17.4.6, we get I + J = R. Write 1 = e1 + e2 , e1 ∈ I and e2 ∈ J. Then e1 e2 = 0, e1 = e21 and e2 = e22 . So we have P ier(R) = V(e1 ) ⊔ V(e2 ), V(I) ⊆ V(e1 ), V(J) ⊆ V(e2 ).

It follows from P ier(R) = V(I) ⊔ V(J) that V(I) = V(e1 ) and V(J) = V(e2 ). Assume that A = V(e). Then V(e1 ) = V(e), and so V(1 − e1 ) = V(1 − e). If P ∈ V(e1 ), then P ∈ V(e), and so e1 − e ∈ P . If P 6∈ V(e1 ), then P ∈ V(1−e1 ); hence, P ∈ V(1−e). Thus, e1 −e = (1−e)−(1−e1 ) ∈ P . T It follows that e1 − e ∈ P . In view of [382, Remark 11.2], e1 = e. P ∈P ier(R)

Therefore the proof is true.



Theorem 17.4.11. Let R be an abelian exchange ring. Then P ier(R) is strongly zero-dimensional. T Proof. Let A and B be disjoint closed sets of P ier(R). Then A B = ∅. Clearly, there exists two ideals I and J, which are generated by central T idempotents, such that A = V(I) and B = V(J); hence, V(I) V(J) = ∅. This implies that V(I + J) = ∅, i.e., E(I + J) = P ier(R). As in the proof of Lemma 17.4.6, we get I + J = R, and so a + b = 1 for some a ∈ I and b ∈ J. Write a = e + u, where e ∈ R is an idempotent and u ∈ U (R).

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For any P ∈ V(J), we have J ⊆ P ; hence, b ∈ P . As b = 1 − a = (1 − e) − u, we see that 1 − e 6∈ P . In view of [382, Remark 11.1], e ∈ P , i.e., eR ⊆ P . This implies that P ∈ V(e). Hence, V(J) ⊆ V(e). For any P ∈ V(I), we have I ⊆ P ; hence, a ∈ P . As a = e + u, we see that e 6∈ P . By using [382, Remark 11.1] again, 1 − e ∈ P , i.e., (1 − e)R ⊆ P . This implies that P ∈ V(1 − e). Hence, V(I) ⊆ V(1 − e). Choose U = V(e). Then B = V(J) ⊆ U ⊆ P ier(R) − V(1 − e) ⊆ P ier(R) − V(I) = P ier(R) − A. Clearly, B ⊆ U . Let V = P ier(R) − U . Then V is a clopen set of P ier(R). T As U ⊆ P ier(R)−A, we see that A ⊆ V . In addition, U V = ∅. Therefore P ier(R) is strongly zero-dimensional.  In view of [382, Proposition 32.2], all Pierce stalks of an abelian exchange ring are local. Thus, we see that an exchange ring is abelian if and only if it is a subring of direct products of local rings. Let A be a finitely generated projective right module over an abelian exchange ring R. If P ∈ P ier(R), then A/AP is a finitely generated free R-module; hence, A/AP ∼ = n(R/P ) for a nonnegative number n. We denote such an n by rankP (A). Lemma 17.4.12. Let R be an abelian exchange ring, and let A and B be finitely generated projective right R-modules. Then the following are equivalent: (1) A ∼ = B. (2) rankP (A) = rankP (B) for all P ∈ P ier(R). (3) A/AP ∼ = B/BP for all P ∈ P ier(R). Proof. (1) ⇒ (2) ⇒ (3) are trivial. (3) ⇒ (1) Since A and B are finitely generated projective right Rmodules, we can find E = E 2 , F =
F 2 ∈ Mn (R)
such that A ∼ = E(nR) 6 B. and B ∼ = F n(R/P ) . Assume that A ∼ = = F (nR). Hence, E n(R/P ) ∼ Then E(nR) 6∼ F (nR). Let Ω = {P | P is an ideal of R generated by =

∼ central idempotents of R such that E n(R/P ) = 6 F n(R/P ) }. Clearly, 0 ∈ Ω; hence, Ω 6= ∅. Given P1 ⊆ P2 ⊆ · · · ⊆ Pn ⊆ · · · in Ω, we have ∞ S that Q := Pi ∈ Ω. Clearly, Q is an ideal of R generated by central i=1

idempotents of R. If Q 6∈ Ω, then E n(R/Q) ∼ = F n(R/Q) ; hence, EMn (R/Q) ∼ = F Mn (R/Q). Thus, we can find some X, Y ∈ Mn (R) such

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that E = XY , F = Y X, X = XY X

and Y = Y XY .

That is, E ≡ XY, F ≡ Y X, X ≡ XY X, Y ≡ Y XY mod As a result, we can find an ideal P ∈ Ω such that


Mn (Q) .


E ≡ XY, F ≡ Y X, X ≡ XY X, Y ≡ Y XY mod Mn (P ) .


It follows
that EMn (R/P ) ∼ = F Mn (R/P ), and so E n(R/P ) ∼ = F n(R/P ) . This gives a contradiction. In other words, Ω is inductive. By using Zorn’s Lemma, we have an ideal M of R, which is maximal in Ω. Since A/AM 6∼ = B/BM , R/M is not a Pierce stalk of R. In view of [382, Remark 11.1], there exists a central idempotent e ∈ R such that M + eR and M + (1 − e)R are both ideals of R generated by central idempotents of R, and that M + eR and M + (1 − e)R properly contain the ideal M . By the maximal of M , we see that

A/A M + eR
∼ = B/B M + eR ,
A/A M + (1 − e)R ∼ = B/B M + (1 − e)R .

Clearly, we have that



M + eR M + (1 − e)R = M = M + (1 − e)R M + eR .

According to Lemma 1.4.12, A/AM ∼ = B/BM , a contradiction. This shows ∼ that A = B. 

Let R be an abelian exchange ring, let A be a finitely generated projective right R-module, and let n be a positive integer. If A/AP can be generated by n elements for all P ∈ P ier(R), then we claim that A can be generated by n elements. Let P ∈ P ier(R). Since A/AP is a projective right R-module generated by n elements, A/AP .⊕ (nR)/(nR)P . In view of Lemma 17.4.12, A .⊕ nR. Therefore A can be generated by n elements. Lemma 17.4.13. Let R be an abelian exchange ring. Then there exists a semigroup homomorphism ϕ :
K0 (R)+ → {f : P ier(R) → Z+ | f is continuous } given by ϕ [M ] = fM for any [M ] ∈ K0 (R)+ , where fM (P ) = rankP M for any P ∈ P ier(R). Proof. Let [M ] ∈ K0 (R)+ . Since R is an exchange ring, by Lemma 13.1.8, n L there exist idempotents e1 , · · · , en ∈ R such that M ∼ ei R. Define = i=1

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fM : P ier(R) → Z+ given by fM (P ) = rankP M for any P ∈ P ier(R). n P Clearly, fM = fei R . It will suffice to prove that each fei R is continuous. i=1S
S
S For any open set {nλ } ⊆ Z+ , we have fe−1 {nλ } = fe−1 {nλ } , so it iR iR λ λ λ
−1 will suffice to prove that f {n } is an open set. Clearly, rankP (ei R) + λ ei R
rankP (1 − ei )R = 1 for any P ∈ P ier(R); hence,
fe−1 {nλ } iR = {P ∈ P ier(R) | fei R (P ) = nλ }  if ∅ = {P ∈ P ier(R) | rankP (ei R) = 0} if  {P ∈ P ier(R) | rankP (ei R) = 1} if

nλ = 6 0, 1; nλ = 0; nλ = 1.

If P ∈ P ier(R) such that rankP (ei R) = 0, then ei R/ei RP = 0; hence, ei ∈ ei R = ei P ⊆ P . In view of [382,
Remark 11.1], 1−ei 6∈ P . That is, (1− ei )R 6⊆ P , and so
P ∈ E (1−ei )R . Thus, {P ∈
P ier(R) | rankP (ei R) = 0} ⊆ E (1 − ei )R . Given any P ∈ E (1 − ei )R , then (1 − ei )R 6⊆ P . We infer that 1 − ei 6∈ P . In view of [382, Remark 11.1], ei ∈ P . This proves
that ei R/ei RP = 0; hence, rankP (ei R) = 0. Therefore
E (1 − ei )R ⊆ −1 {P ∈ P ier(R)|rank P (ei R) = 0}. Therefore, fei R {nλ } are respectively
∅, E (1 −
ei )R and E(ei R) which are all open sets. This implies that fe−1 {nj } is an open set. Therefore fM is continuous, and so fM ∈ {f : iR P ier(R) → Z+ | f is continuous}. One easily checks that ϕ is a semigroup homomorphism, as required.  Theorem 17.4.14. Let R be an abelian exchange ring. Then K0 (R) ∼ = {f : P ier(R) → Z | f is continuous}. Proof. In light of Lemma 17.4.13, there exists a group homomorphism ϕ :
K0 (R) → {f : P ier(R) → Z | f is continuous} given by ϕ [M ] − [N ] = fM − fN for any [M ] − N ∈ K0 (R), where fM (P ) = rankP M for any P ∈ P ier(R). In view of Theorem 17.4.8, there exists a group homomorphism θ : {f : P ier(R) → Z | f is continuous} → K0 (R). For any f ∈ {f : P ier(R) → Z | f is continuous}, it will suffice to prove that ϕθ(f ) = f . Let {f1 , · · · , fn } be the complete set of central idempotents corresponding to f . For any P ∈ P ier(R), there exists i ∈ {1, · · · , n} such that P ∈ E(fi R), and that rankP (fi R) ≤ 1. As in the proof of Lemma 17.4.13, we
obtain {P ∈ P ier(R) | rankP (fi R) = 0} =
P F E (1 − fi )R = E ( fk )R = E(fk R). Hence, rankP (fi R) 6= 0, and k6=i

k6=i

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so rankP (fi R) = 1. Furthermore, rankP (fj R) = 0(j 6= i), and then n n


P P ϕθ(f ) (P ) = ϕ mi [fi R] (P ) = mi ϕ [fi R] (P ) =

n P

i=1

i=1

mi rankP (fi R) = mi = f (P ).

i=1

As a result, we get

ϕθ = 1{f :P ier(R)→Z

is continuous} . It follows from Lemma 17.4.12 that ker(ϕ) = {[A] − [B] | fA = fB } = {[A] − [B] | fA (P ) = fB (P ) for all P ∈ P ier(R)} = {[A] − [B] | rankP A = rankP B for all P ∈ P ier(R)} = 0. Therefore K0 (R) ∼ = {f : P ier(R) → Z | f is continuous}, as asserted.  | f

For example, Z6 is an abelian π-regular ring. In view of Theorem 17.4.14, K0 (Z6 ) ∼ = {f : P ier(Z6 ) → Z | f is continuous } ∼ = Z ⊕ Z. Corollary 17.4.15. Let R be an abelian exchange ring. If | P ier(R) | = ∞, then every finitely generated free abelian group is isomorphic to a direct summand of {f : P ier(R) → Z | f is continuous}. Proof. Given any n ∈ N, we can find {P1 , · · · , Pn } ⊆ P ier(R). For distinct Pierce ideals Pi and Pj , we can find a e ∈ B(R) such that e ∈ Pj and e 6∈ Pi . Thus, Pi + eR = R. This implies that Pi + Pj = R. By using the Chinese n n T L Remainder Theorem, we have R/ Pk ∼ R/Pk . According to [382, = k=1

k=1

Proposition 32.2], all Pierce stalks of R are local; hence, K0 (R/Pk ) ∼ = Z. n n
L T Thus, K0 R/ Pk ∼ Z. We see in [9, Theorem 3.5] that there exists an = k=1

1

epimorphism K0 (R) → K0 R/

n T

k=1


Pk . By Theorem 17.4.14, there exists

an epimorphism ϕ : {f : P ier(R) → Z | f is continuous} →

n L 1

Z. As

n L 1

Z

is a free abelian group, we get {f : P ier(R) → Z | f is continuous} ∼ = n
L Z ⊕ kerϕ, as desired.  1

Let F be a field, N be the set of positive integers, Fi = F (i ∈ N), Q and let R = Fi . Then R is an abelian exchange ring. Choose i∈N

P1 = {(0, x2 , x3 , x4 , · · · ) | x2 , x3 , x4 , · · · ∈ F }, P2 = {(x1 , 0, x3 , x4 , · · · ) | x1 , x3 , x4 , · · · ∈ F }, P3 = {(x1 , x2 , 0, x4 , · · · ) | x1 , x2 , x4 , · · · ∈ F }, · · · . Then | P ier(R) | = ∞ as {P1 , P2 , P3 , · · · } ⊆ P ier(R).

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If R is an abelian exchange ring, we claim that Z is isomorphic to a direct summand of K0 (R). Clearly, there exists an inclusion map φ : Z[R] ֒→ K0 (R). Let ϕ : R → R/P be a natural epimorphism, where P ∈ P ier(R). Then K0 ϕ : K0 (R) → K0 (R/P ). In view of [382, Proposition 32.2], R/P is local. Hence, K0 (R/P ) = Z[R/P ]. Since R and R/P have invariant basis numbers, we have a group isomorphism ψ : Z[R/P ] → Z[R]

given by ψ n[R/P ] = n[R] for any n ∈ Z. It is easy to check that ψ K0 ϕ φ = 1Z[R] , and so φ is a split monomorphism. Therefore K0 (R) ∼ = Z[R] ⊕ K0 (R)/Z[R] ∼ = Z ⊕ K0 (R)/Z[R], and we are done. Corollary 17.4.16. Let R be an abelian exchange ring. Then the following are equivalent: (1) (2) (3) (4)

R is a local ring. K0 (R) ∼ = Z. {f : P ier(R) → Z | f is continuous} ∼ = Z. Z ⊕ Z is not isomorphic to any direct summand of K0 (R).

Proof. (1) ⇒ (4) If Z ⊕ Z is isomorphic to a direct summand of K0 (R), we have an abelian group G such that Z ⊕ Z ⊕ G ∼ = K0 (R) ∼ = Z. According to ∼ Corollary 1.4.16, Z ⊕ G = 0, a contradiction. Thus, Z ⊕ Z is not isomorphic to any direct summand of K0 (R). (4) ⇒ (3) If there exists a non-trivial central idempotent e ∈ R, then R = eRe ⊕ (1 − e)R(1 − e). In addition, eRe and (1 − e)R(1 − e) are both abelian exchange rings. By the previous observation, K0 (eRe) ∼ = Z ⊕ K0 (eRe)/Z[eRe] and
 
K0 (1 − e)R(1 − e) ∼ = Z ⊕ K0 (1 − e)R(1 − e) /Z (1 − e)R(1 − e) . As a result, we see that
K0 (R) ∼ = K0 (eRe) ⊕ K0 (1 − e)R(1 − e)
∼ = Z ⊕ Z ⊕ K0 (eRe)/Z[eRe] ⊕ K0 (1 − e)R(1 − e) , which gives a contradiction. So B(R) = {0, 1}. According to Theorem 17.4.2, {f : P ier(R) → Z | f is continuous} ∼ = Z. (3) ⇒ (1) By virtue of Theorem 17.4.2, B(R) = {0, 1}. For any x ∈ R, we see that either x ∈ R or 1 − x is right invertible. Write xy = 1. Then yx ∈ B(R), and so yx = 1. Hence, x ∈ U (R). Write (1−x)z = 1. Similarly, we see that 1 − x ∈ U (R). Therefore R is a local ring from [4, Proposition 15.1.5]. (2) ⇔ (3) follows from Theorem 17.4.14. 

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Bibliography

[1] [2] [3]

[4] [5] [6] [7] [8] [9] [10]

M.S. Ahn and D.D. Anderson, Weakly clean rings and almost clean rings, Rocky Mountain J. Math., 36(2006), 783–798. H. Al–Ezeh, Exchange P F –rings and almost P P –rings, Mat. Vesnik., 42 (1990), 77–80. D.D. Anderson and V.P. Camillo, Commutative rings whose elements are a sum of a unit and idempotent, Comm. Algebra, 30(2002), 3327–3336. F. Anderson and K. Fuller, Rings and Categories of Modules, Springer, Berlin, 1973. P. Ara, Stable range of ℵ0 -continuous regular rings, Lect. Notes Math., 1328, 1988, 1–7. P. Ara, ℵ0 -continuous regular rings, J. Algebra, 109(1987), 115– 126. P. Ara, K-theory for Rickart C ∗ -algebras, K-Theory, 5(1991), 281– 292. P. Ara, Strongly π–regular rings have stable range one, Proc. Amer. Math. Soc., 124(1996), 3293–3298. P. Ara, Extensions of exchange rings, J. Algebra, 197(1997), 409– 423. P. Ara, Stability properties of exchange rings, International Symposium on Ring Theory, MA: Birkhaeuser, Trends in Mathematics, 2001, 23–42. 629

rings

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[11] [12] [13]

[14] [15]

[16]

[17]

[18] [19]

[20] [21]

[22]

[23]

[24]

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P. Ara, The exchange property for purely infinite simple rings, Proc. Amer. Math. Soc., 132(2004), 2543–2547. P. Ara; M. G´ omez Lozano and M. Siles Molina, Local rings of exchange rings, Comm. Algebra, 26 (1998), 4191–4205. P. Ara and K.R. Goodearl, The almost isomorphism relation for simple regular rings, Publications Matem` atiques, 36(1992), 369– 388. P. Ara and K.R. Goodearl, Stable rank of corner rings, Proc. Amer. Math. Soc., 133(2005), 379–386. P. Ara; K.R. Goodearl; K.C. O’Meara and E. Pardo, Diagonalisation of matrices over regular rings, Linear Algebra Appl., 265(1997), 146–163. P. Ara; K.R. Goodearl; K.C. O’Meara and E. Pardo, Separative cancellation for projective modules over exchange rings, Israel J. Math., 105(1998), 105–137. P. Ara; K.R. Goodearl; K.C. O’Meara and E. Pardo, K1 of separative exchange rings and C ∗ –algebras with real rank zero, Pacific J. Math., 195(2000), 261–275. P. Ara; K.R. Goodearl and E. Pardo, K0 of purely infinite simple regular rings, K-Theory, 26(2002), 69–100. P. Ara; K.R. Goodearl; E. Pardo and D.V. Tyukavkin, Ktheoretically simple von Neumann regular rings, J. Algebra, 174(1995), 659–677. P. Ara; K.C. O’Meara and F. Perera, Stable finiteness of group rings in arbitrary characteristic, Adv. Math., 170(2002), 224–238. P. Ara; K.C. O’Meara and F. Perera, Gromov translation algebras over discrete trees are exchange rings, Trans. Amer. Math. Soc., 356(2003), 2067–2079. P. Ara; K.C. O’Meara and D.V. Tyukavkin, Cancelation of projective modules over regular rings with comparability, J. Pure Appl. Algebra, 107(1996), 19–38. P. Ara and E. Pardo, Refinement monoids with weak comparability and applications to regular rings and C ∗ –algebras, Proc. Amer. Math. Soc., 124 (1996), 715–720. P. Ara; E. Pardo and F. Perera, The structure of countably generated projective modules over regular rings, J. Algebra, 226(2000), 161–190.

rings

December 8, 2010

10:20

World Scientific Book - 9in x 6in

Bibliography

[25] [26] [27] [28] [29] [30]

[31] [32] [33]

[34]

[35]

[36] [37] [38] [39] [40]

rings

631

P. Ara; G.K. Pedersen and F. Perera, An infinite analogue of rings with stable rank one, J. Algebra, 230(2000), 608-655. P. Ara; G.K. Pedersen and F. Perera, A closure operation in rings, Internat. J. Math., 12(2001), 791–812. P. Ara; G.K. Pedersen and F. Perera, Extensions and pullbacks in QB-rings, Algebra Represent. Theory, 8(2005), 75–97. E.P. Armendariz; G.F. Birkenmeier and J.K. Park, Rings containing ideals with bounded index, Comm. Algebra, 30(2002), 787–801. E.P. Armendariz and J.W. Fisher, Regular P.I.-rings, Proc. Amer. Math. Soc., 39(1973), 247–251. E.P. Armendariz; J.W. Fisher and R.L. Snider, On injective and surjective endomorphisms of finitely generated modules, Comm. Algebra, 6(1978), 659–672. D.M. Arnold, Finite Rank Torsion Free Abelian Groups and Rings, Lect. Notes Math., 931, 1982. F. Azarpanah, When is C(X) a clean ring? Acta Math. Hungar., 94(2002), 53–58. G. Baccella, Von Neummann regularity of V –rings with artinian primitive factor rings, Proc. Amer. Math. Soc, 103(1988), 747– 749. G. Baccella, Exchange property and the natural preorder between simple modules over semi-artinian rings, J. Algebra, 253(2002), 133–166. G. Baccella and A. Ciampella, K0 of semiartinian unit– regular rings, In: Abelian groups, module theory, and topology (Padua,1997), Dekker, New York, 1998, 69–78. A. Badawi, On semicommutative π–regular rings, Comm. Algebra, 22 (1994), 151–157. A. Badawi, On abelian π–regular rings, Comm. Algebra, 25(1997), 1009–1021. C.S. Ballantine, Products of idempotent matrices, Linear Algebra Appl., 19 (1978), 81–86. B. Banascewski, Gelfand and exchange rings: their spectra in point– free topology, Arabian J. Sci. Engin., 25(2000), 3–22. B. Banascewski, Ring theory and pointfree topology, Topology Appl., 137(2004), 21–37.

December 8, 2010

10:20

632

[41] [42]

[43]

[44] [45]

[46] [47] [48] [49] [50] [51]

[52] [53] [54] [55] [56]

World Scientific Book - 9in x 6in

Rings Related to Stable Range Conditions

H. Bass and R. Guralnick, Projective modules with free multiples and powers, Proc. Amer. Math. Soc., 96(1986), 207–208. K.I. Beidar; K.C. O’Meara and R.M. Raphael, On uniform digitalization of matrices over regular rings and one-accessible regular algebras, Comm. Algebra, 32(2004), 3543–3562. A.J. Berrick and M.E. Keating, An Introduction to Rings and Modules with K-Theory in View, London: Cambridge Univer. Press, 2000. G.F. Birkenmeier, A survey of regularity conditions and the simplicity of prime factor rings, Vietnam J. Math., 23(1995), 29–38. G.F. Birkenmeier; J.Y. Kim and J.K. Park, Regularity conditions and the simplicity of prime factor rings, J. Pure Appl. Algebra, 115(1997), 213–230. B. Blackadar, The stable rank of full corners in C ∗ -algebras, Proc. Amer. Math. Soc., 132(2004), 2945–2950. B. Blackadar, A stable cancellation theorem for simple C ∗ -algebras, Proc. London Math. Soc., 47(1983), 285–302. B. Blackadar, Rational C ∗ -algebras and nonstable K-theory, Rocky Mountain J. Math., 20(1990), 285–316. G. Borooah, Clean Conditions and Incidence Rings, Ph.D. Thesis, University of California, Berkeley, 2005. G. Borooah; A.J. Diesl and T.J. Dorsey, Strongly clean triangular matrix rings over local rings, J. Algebra, 312(2007), 773–797. G. Borooah; A.J. Diesl and T.J. Dorsey, Strongly clean matrix rings over commutative local rings, J. Pure Appl. Algebra, 212(2008), 281–296. G. Brookfield, Monoids and Categories of Noetherian Modules, Ph.D. Thesis, University of California, 1997. G. Brookfield, Direct sum cancellation of noetherian modules, J. Algebra, 200 (1998), 207–224. B. Brown and N.H. McCoy, The maximal regular ideal of a ring, Proc. Amer. Math. Soc., 1(1950), 165–171. H.H. Brungs, Left euclidean rings, Pacific J. Math., 45(1973), 27– 33. W.D. Burgess and J.-M. Goursaud, K0 of a ring whose central localizations are simple artinian, Canad. Math. Bull., 25(1982), 344–347.

rings

December 8, 2010

10:20

World Scientific Book - 9in x 6in

Bibliography

[57] [58] [59] [60] [61] [62] [63] [64] [65]

[66] [67] [68] [69] [70] [71] [72] [73] [74]

rings

633

W.D. Burgess and P. Menal, On strongly π–regular rings and homomorphisms into them, Comm. Algebra, 16(1988), 1701–1725. W.D. Burgess and W. Stephenson, Pierce sheaves of noncommutative rings, Comm. Algebra, 4(1976), 51–75. W.D. Burgess and W. Stephenson, An analogue of the Pierce sheaf for noncommutative rings, Comm. Algebra, 6(1978), 863–886. W.D. Burgess and W. Stephenson, Rings all of whose Pierce stalks are local, Canad. Math. Bull., 22(1979), 159–164. D.G. Burkholder, Modules with local pierce stalks, J. Algebra, 145(1992), 339–348. G. C˘ alug˘ areanu, Abelian groups with semilocal endomorphism rings, Comm. Algebra, 30(2002), 4105–4111. V.P. Camillo and D. Khurana, A characterization of unit regular rings, Comm. Algebra, 29(2001), 2293–2295. V.P. Camillo; D. Khurana; T.Y. Lam; W.K. Nicholson and Y. Zhou, Continuous modules are clean, J. Algebra, 304(2006), 94–111. V. Camillo and J. J. Sim´on, The Nicholson–Varadarajan theorem on clean linear transformations, Glasg. Math. J., 44(2002), 365369. V. Camillo and H.P. Yu, Exchange rings, units and idempotents, Comm. Algebra, 22(1994), 4737–4749. V. Camillo and H.P. Yu, Stable range one for rings with many idempotents, Trans. Amer. Math. Soc., 347(1995), 3141–3147. R. Camps and W. Dicks, On semilocal rings, J. Algebra, 81(1993), 203–211. R. Camps and P. Menal, Power cancellation for artinian modules, Comm. Algebra, 19(1991), 2081–2095. R. Camps and P. Menal, The power-substitution property for rings of continuous functions, J. Algebra, 161(1993), 455–466. M.J. Canfell, Completion of diagrams by automorphisms and Bass’ first stable range condition, J. Algebra, 176(1995), 480–503. J.J. Carmona; J. Cufi and P. Menal, On the unit-1-stable rank of rings of analytic functions, Publ. Mat., 36(1992), 439–447. M. Chacron, Algebraic algebras over π–regular rings, J. Algebra, 24(1973), 71–79. H. Chen, Related comparability over regular rings, Algebra Colloq., 3(1996), 277–282.

December 8, 2010

10:20

634

[75] [76] [77] [78] [79] [80] [81] [82] [83] [84] [85] [86] [87] [88] [89] [90] [91] [92]

World Scientific Book - 9in x 6in

Rings Related to Stable Range Conditions

H. Chen, Elements in one-sided unit-regular rings, Comm. Algebra, 25 (1997), 2517–2529. H. Chen, On related unit regular elements, Algebra Colloq., 4(1997), 323–328. H. Chen, Comparability of modules over regular rings, Comm. Algebra, 25 (1997), 3531–3543. H. Chen, Exchange rings, related comparability and powersubstitution, Comm. Algebra, 26(1998), 3383–3401. H. Chen, Rings with stable range conditions, Comm. Algebra, 26 (1998), 3653–3668. H. Chen, On exchange rings with all idempotents central, Algebra Colloq., 6 (1999), 45–50. H. Chen, Exchange rings with artinian primitive factors, Algebra Represent. Theory, 2(1999), 201–207. H. Chen, Related comparability over exchange rings, Comm. Algebra, 27 (1999), 4209–4216. H. Chen, Rings with many idempotents, Int. J. Math. Math. Sci., 22 (1999), 547–558. H. Chen, On m-fold stable exchange rings, Comm. Algebra, 27(1999), 5639–5647. H. Chen, Exchange rings satisfying unit 1-stable range, Kyushu J. Math., 54 (2000), 1–6. H. Chen, Generalized stable exchange rings, Southeast Asian Bull. Math., 24 (2000), 19–24. H. Chen, On generalized stable rings, Comm. Algebra, 28(2000), 1907–1917. H. Chen, On stable range conditions, Comm. Algebra, 28(2000), 3913–3924. H. Chen, Elements in exchange rings with related comparability, Internat. J. Math. Math. Sci., 23(2000), 639–644. H. Chen, Power-substitution, exchange rings and unit-regularity, Comm. Algebra, 28(2000), 5223–5233. H. Chen, Regular rings with finite stable range, Comm. Algebra, 29(2001),157–166. H. Chen, Units, idempotents, and stable range conditions, Comm. Algebra, 29(2001), 703–717.

rings

December 8, 2010

10:20

World Scientific Book - 9in x 6in

Bibliography

[93] [94] [95] [96] [97] [98] [99] [100] [101] [102] [103] [104] [105] [106] [107] [108] [109] [110]

rings

635

H. Chen, Regular rings with generalized s-comparability, Southeast Asian Bull. Math., 25(2001), 15–21. H. Chen, Stable ranges for Morita contexts, Southeast Asian Bull. Math., 25 (2001), 209–216. H. Chen, Idempotents in exchange rings, New Zealand J. Math., 30(2001), 103–110. H. Chen, Exchange rings having stable range one, Int. J. Math. Math. Sci., 25(2001), 763–770. H. Chen, Stable rings generated by their units, Int. J. Math. Math. Sci., 26(2001), 759–764. H. Chen, Morita contexts with many units, Comm. Algebra, 30(2002), 1499–1512. H. Chen, Morita contexts with generalized stable conditions, Comm. Algebra, 30(2002), 2699–2713. H. Chen, On partial unit-regularity, Kyungpook Math. J., 42(2002), 13–19. H. Chen, On partially one-sided unit-regularity, Comm. Algebra, 30(2002), 5781–5793. H. Chen, On directly infinitely stable rings, J. Appl. Algebra Discrete Struct., 1(2003), 13–26. H. Chen, Strongly π-regular Morita contexts, Bull. Korean Math. Soc., 40(2003), 91–99. H. Chen, On exchange QB-rings, Comm. Algebra, 31(2003), 831– 841. H. Chen, Extensions of rings with many units, Comm. Algebra, 31(2003), 2403–2416. H. Chen, On strongly stable rings, Comm. Algebra, 31(2003), 2771– 2789. H. Chen, Partial cancellation of modules, Acta Math. Hungar., 100(2003), 205–214. H. Chen, Generalized stable regular rings, Comm. Algebra, 31(2003), 4899–4910. H. Chen, Separativity of endomorphism rings of modules with finite exchange rings, Chin. J. Contemp. Math., 24(2003), 283–290. H. Chen, Exchange rings satisfying power ideal-cancellation, Sci. China, Series A: Mathematics, 46(2003), 804–814.

December 8, 2010

10:20

636

[111] [112] [113]

[114] [115] [116] [117] [118] [119] [120] [121] [122] [123] [124] [125] [126] [127] [128]

World Scientific Book - 9in x 6in

Rings Related to Stable Range Conditions

H. Chen, Quasi-separative modules with finite exchange property, Algebra Colloq., 10(2003), 535–541. H. Chen, On unit 1-stable range, J. Appl. Algebra Discrete Struct., 1(2003), 189–196. H. Chen, Exchange rings with stable range conditions, Recent Research on Pure and Applied Algebra, Nova Science Publishers Inc. New York, 2003, 47-58. H. Chen, Pseudo-similarity in semigroups, Semigroup Forum, 68(2004)59–63. H. Chen, Exchange rings over which every regular matrix admits a diagonal reduction, J. Algebra Appl., 3(2004), 207–217. H. Chen, Extensions of unit-regular rings, Comm. Algebra, 32 (2004), 2359–2364. H. Chen, Full elements in regular rings, Taiwanese J. Math., 8(2004), 203–209. H. Chen, Extensions of unit 1-stable range, Comm. Algebra, 32(2004), 3071–3085. H. Chen, One-sided unit-regular ideals of regular rings, Taiwanese J. Math., 8(2004), 761–770. H. Chen, Self-cancellation of modules having the finite exchange property, Chin. Ann. Math., 26B(2005), 111–118. H. Chen, On quasi-rank functions of a non-unital regular ring, J. Appl. Algebra Discrete Struct., 3(2005), 41–53. H. Chen, On exchange ideals, Bull. Korean Math. Soc., 42(2005), 295–305. H. Chen, Comparability for ideals of regular rings, Sci. China, Series A: Mathematics, 48(2005), 757–768. H. Chen, Full elements in one-sided unit-regular rings, Comm. Algebra, 33(2005), 4115–4123. H. Chen, K0 of exchange rings with all idempotents central, Comm. Algebra, 33(2005), 4219–4227. H. Chen, A note on potent elements, Kyungpook Math. J., 45(2005), 519–526. H. Chen, A note on quasi-injective modules, J. Appl. Algebra Discrete Struct., 4(2006), 13–21. H. Chen, Pseudo-similarity and stable power-substitution, Adv. Algebra & Analysis, 1(2006), 7–12.

rings

December 8, 2010

10:20

World Scientific Book - 9in x 6in

Bibliography

[129] [130] [131] [132] [133] [134] [135] [136] [137] [138] [139] [140] [141] [142] [143] [144] [145] [146] [147]

rings

637

H. Chen, Regular rings, idempotents and products of one-sided units, Comm. Algebra, 34(2006), 737–741. H. Chen, Separative ideals, clean elements and unit-regularity, Comm. Algebra, 34(2006), 911–921. H. Chen, Diagonal reductions of matrices over exchange ideals, Czechoslovak Math. Math., 56(2006),9–18. H. Chen, Regular elements which is the sum of an idempotent and a left cancelable element, Taiwanese J. Math., 10(2006), 881–890. H. Chen, On QB∞ -rings, Comm. Algebra, 34(2006), 2057–2068. H. Chen, Separative cancellation over regular rings, Comm. Algebra, 34(2006), 2597–2606. H. Chen, On regular QB-ideals, Taiwanese J. Math., 10(2006), 1261–1269. H. Chen, Extensions of exchange QB-rings, Algebra Colloq., 13(2006), 667–674. H. Chen, On simple factorization of invertible matrices, Linear & Multilinear Algebra, 55(2007), 81-92. H. Chen, Exchange rings generated by their units, Acta Math. Sin., 23B(2007), 357–364. H. Chen, Separativity of regular rings in which 2 is invertible, Comm. Algebra, 35(2007), 1661–1673. H. Chen, Exchange rings with stable range one, Czechoslovak Math. J., 57(132) (2007), 579-590. H. Chen, Strong separativity over exchange rings, Czechoslovak Math. J., 58(133)(2008), 417–428. H. Chen, Generalized stable exchange rings, Algebra Colloq., 15(2008), 193–198. H. Chen, Exchange rings having stable range one, Chin. J. Contemp. Math., 29(2008), 95–102. H. Chen, On medium stable rings, Chin. Ann. Math., 29A(2008), 755–764. H. Chen, 2 × 2 Invertible matrices over weakly stable rings, J. Korean Math. Soc., 46(2009), 257–269. H. Chen, Clean matrices over commutative rings, Czechoslovak Math. J., 59(2009), 145–158. H. Chen, Cancellation of small projectives over exchange rings, Comm. Algebra, 37(2009), 2145-2158.

December 8, 2010

10:20

638

[148] [149] [150] [151] [152] [153] [154] [155] [156] [157] [158] [159] [160] [161] [162] [163] [164] [165] [166]

World Scientific Book - 9in x 6in

Rings Related to Stable Range Conditions

H. Chen, On separative refinement monoids, Bull. Korean Math. Soc., 46(2009), 489-498. H. Chen, Diagonalization of regular matrices over exchange rings, Internat. Electronic J. Algebra, 6(2009), 107-114. H. Chen, Generalized stable rings and regularity, Algebra Colloq., to appear. H. Chen, Decompositions of linear transformations over division rings, Algebra Colloq., to appear. H. Chen, Topological space and nearly clean rings, Chin. Ann. Math.(A), to appear. H. Chen, Exchange rings with many units, preprint, 2010. H. Chen and M. Chen, Generalized ideal-stable regular rings, Comm. Algebra, 31(2003), 4989–5001. H. Chen and M. Chen, On unit-regular ideas, New York J. Math., 9(2003), 295–302. H. Chen and M. Chen, Simple extensions of regular rings, Southeast Asian Bull. Math., 28(2004), 235–240. H. Chen and M. Chen, Whitehead groups of module extensions, Acta Math. Sci., 24B(2004), 171–174. H. Chen and M. Chen, On products of three triangular matrices over associative rings, Linear Algebra Appl., 387(2004), 297–311. H. Chen and M. Chen, Unit 1-stable range for ideals, Int. J. Math. Math. Sci., 46(2004), 2477–2482. H. Chen and M. Chen, On m-fold stable ideals, J. Algebra Appl., 3(2004), 239–246. H. Chen and M. Chen, On strongly π-regular ideals, J. Pure Appl. Algebra, 195(2005), 21–32. H. Chen and M. Chen, On exchange qb-ideals, Bull. Korean Math. Soc., 42 (2005), 45–51. H. Chen and M. Chen, The symmetry of unit ideal stable range conditions, Arch. Math. (Brno), 41(2005), 181–186. H. Chen and M. Chen, Extensions of GM -rings, Czechoslovak Math. J., 55(2005), 273–281. H. Chen and M. Chen, Ideal-comparability over regular rings, Acta Math. Sin., 22B(2006), 417–424. H. Chen; M. Chen and J. Li, T -stable ideals of regular rings, Advance in Algebra, World Scientific, 2003, 291-300.

rings

December 8, 2010

10:20

World Scientific Book - 9in x 6in

Bibliography

[167] [168] [169] [170] [171] [172]

[173] [174] [175] [176] [177] [178]

[179]

[180] [181] [182]

rings

639

H. Chen and F. Li, Separativity over regular rings, Algebra Colloq., 6(1999), 321–328. H. Chen and F. Li, Rings with many unit-regular elements, Chin. J. Contemp. Math., 21(2000), 33–38. H. Chen and F. Li, Separativity of modules with finite exchange property, Sci. China, Series A: Mathematics, 43(2000),795–803. H. Chen and F. Li, Comparability and s-comparability of modules over exchange rings, Acta Math. Sin., 17B:3(2001), 481–488. H. Chen and F. Li, Exchange rings having ideal-stable range one, Sci. China, 44(2001), 580–586. H. Chen and F. Li, Whitehead groups of exchange rings with artinian primitive factors, Int. J. Math. Math. Sci., 26(2001), 393– 398. H. Chen and F. Li, On ideals of regular rings, Acta Math. Sin., 18(2002), 347–356. H. Chen and F. Li, Exchange rings satisfying related comparability, Collect. Math., 53(2002), 157–164. H. Chen and F. Li, Exchange rings satisfying the n-stable range condition I, Algebra Colloq., 9(2002), 351–360. H. Chen and F. Li, Exchange rings satisfying the n-stable range conditions II, Algebra Colloq., 10(2003), 1–8. H.V. Chen and A.Y.M. Chin, A note on regular rings with stable range one, Internat. J. Math. Math. Sci., 31(2002), 449–450. J. Chen; Z. Wang and Y. Zhou, Rings in which elements are uniquely the sum of an idempotent and a unit that commutate, J. Pure Appl. Algebra, 213(2009), 215–223. J. Chen; X. Yang and Y. Zhou, When is the 2 × 2 matrix ring over a commutative local ring strongly clean? J. Algebra, 301(2006), 280–293. J. Chen; X. Yang and Y. Zhou, On strongly clean matrix and triangular matrix rings, Comm. Algebra, 34(2006), 3659–3674. J. Chen and Y. Zhou, Strongly clean power series rings, Proc. Edinb. Math. Soc., 50(2007), 73–85. J. Chen; W.K. Nicholson and Y. Zhou, Group rings in which every element is uniquely the sum of a unit and an idempotent, J. Algebra, 306(2006), 453–460.

December 8, 2010

10:20

640

[183] [184] [185] [186] [187] [188] [189]

[190] [191] [192] [193] [194]

[195] [196] [197] [198] [199]

World Scientific Book - 9in x 6in

Rings Related to Stable Range Conditions

W. Chen, P -clean rings, Int. J. Math. Math. Sci., 2006(2006), 1–7. W. Chen, A question on strongly clean rings, Comm. Algebra, 34(2006), 2347–2350. A.Y.M. Chin, A note on strongly π-regular rings, Acta Math. Hungar., 102 (2004), 337–341. W.S. Ching, Linear equations over commutative rings, Linear Algebra Appl., 18(1977), 257–266. M. Compa˜ ny Cabezos, M.G˜omez Lozano and M. Siles Molina, Exchange Morita rings, Comm. Algebra, 29 (2001), 907–925. I.G. Connell, Some ring theoretic Schr¨oder-Bernstein theorems, Trans. Amer. Math. Soc., 132(1968), 335–351. G.E. Cooke, A weakening of the euclidean property for integral domains and applications to algebraic number theory I, J. Reine Angew. Math., 282(1976), 133–156. G. Corach and A.R. Larotonda, Stable range in Banach algebras, J. Pure Appl. Algebra, 32(1984), 289–300. D.L. Costa, Zero-dimensionality and the GE2 of polynomial rings, J. Pure Appl. Algebra, 50(1988), 223–229. F. Couchot, Indecomposable modules and Gelfand rings, Comm. Algebra, 35 (2007), 231–241. P. Crawley and B. J´onsson, Refinements for infinite direct decompositions of algebraic systems, Pacific J. Math., 14(1964), 797–855. S. Dhompongsa and H. Tansee, The exchange property of modules with the finite exchange property, Comm. Algebra, 31(2003), 859– 868. T.J. Dorsey, Cleanness and Strong Cleanness of Rings of Matrices, Ph.D. Thesis, University of California, Berkeley, 2006. G. Ehrlich, Unit-regular rings, Portugal. Math., 27(1969), 209–212. G. Ehrlich, Units and one-sided units in regular rings, Trans. Amer. Math. Soc., 216(1976), 81–90. E.W. Ellers and H. Ishibashi, Factorization of transformations over a local rings, Linear Algebra Appl., 85(1987), 17–27. D.R. Estes and R.M. Guralnick, Module equivalences: local to global when primitive polynomials represent units, J. Algebra, 77(1982), 138–157.

rings

December 8, 2010

10:20

World Scientific Book - 9in x 6in

Bibliography

[200] [201] [202] [203]

[204] [205] [206]

[207] [208] [209] [210] [211]

[212] [213] [214] [215] [216]

rings

641

D.R. Estes and R.M. Guralnick, A stable range for quadratic forms over commutative rings, J. Pure Appl. Algebra, 120(1997), 255-280. D. Estes and J. Ohm, Stable range in commutative rings, Comm. Algebra, 7(1967), 343–362. E.G. Evans, Jr., Krull–Schmidt and cancellation over local rings, Pacific J. Math., 46(1973), 115–121. A. Facchini, Module Theory: Endomorphism Rings and Direct Sum Decompositions in Some Classes of Modules, Birkh¨auser, Basel– Boston Berlin, 1998. C. Faith, Dedekind finite rings and a theorem of Kaplansky, Comm. Algebra, 31(2003), 4175–4187. L. Fan and X. Yang, Strongly clean property and stable range one, Comm. Algebra, 37(2009), 2021–2029. J.W. Fisher and R.L. Snider, On the von Neumann regularity of rings with regular prime factor rings, Pacific J. Math., 54(1974), 135–144. J.W. Fisher and R.L. Snider, Rings generated by their units, J. Algebra, 42 (1976), 363–368. L. Fuchs, On a substitution property of modules, Monatsh. Math., 75(1971), 198–204. L. Fuchs, The cancellation property of modules, Lect. Notes Math., 246 (1972), 191–212. R. G¨ obel and W. May, Cancellation of direct sums of countable abelian p-groups, Proc. Amer. Math. Soc., 131(2003), 2705–2710. A. Ghorbani and A. Haghany, Duality for weakly co-hopfian and generalized hopfian modules, Comm. Algebra, 31(2003), 2811– 2817. L. Gillman and M. Henriksen, Some remarks about elementary divisor rings, Trans. Amer. Math. Soc., 82(1956), 362–365. M.A. Gonzalez-Barroso and E. Pardo, Structure of nonunital purely infinite simple rings, Comm. Algebra, 34(2006), 617–624. K.R. Goodearl, Direct sum properties of quasi-injective modules, Bull. Amer. Math. Soc., 82(1976), 108–110. K.R. Goodearl, Power-cancellation of groups and modules, Pacific J. Math., 64(1976), 387–411. K.R. Goodearl, The state space of K0 of a ring, Lect. Notes Math., 734 (1979), 91–117.

December 8, 2010

10:20

642

[217] [218] [219] [220] [221] [222] [223]

[224] [225] [226] [227] [228] [229] [230] [231] [232]

[233]

World Scientific Book - 9in x 6in

Rings Related to Stable Range Conditions

K.R. Goodearl, Von Neumann Regular Rings, Pitman, London, San Francisco, Melbourne, 1979. K.R. Goodearl, Directly finite ℵ0 -continuous regular rings, Pacific J. Math., 100(1982), 105–122. K.R. Goodearl, Cancellation of low-rank vector bundles, Pacific J. Math., 113 (1984), 289–302. K.R. Goodearl, Surjective endomorphisms of finitely generated modules, Com. Algebra, 15(1987), 589–609. K.R. Goodearl, K0 of regular rings with bounded index of nilpotence, Contemp. Math., 171(1994), 173–199. K.R. Goodearl, Torsion in K0 of unit–regular rings, Proc. Edinburgh Math. Soc., 38(1995), 331–341. K.R. Goodearl, von Neumann regular rings and direct sum decomposition problems, Facchini A. and Menini C.(eds.), Abelian Groups and Modules, Dordrecht: Kluwer Academic Publishers, Math. Appl., Dordredt, 343, 1995, 249–255. K.R. Goodearl and P. Menal, Stable range one for rings with many units, J. Pure Appl. Algebra, 54(1988), 261–287. K.R. Goodearl and J. Moncasi, Cancellation of finitely generated modules over regular rings, Osaka J. Math., 26(1989), 679–685. K.R. Goodearl and R.B. Warfield, Jr., Algebras over zero– dimensional ring, Math. Ann., 223(1976), 157–168. K.R. Goodearl and R.B. Warfield, Jr., State spaces of K0 of noetherian rings, J. Algebra, 71(1981), 322–378. R.M. Guralnick, Roth’s theorem and decomposition of modules, Linear Algebra Appl., 39(1980), 155–165. R.M. Guralnick, Matrix Equivalence and isomorphism of modules, Linear Algebra Appl., 43(1982), 125–136. R.M. Guralnick, Power cancellation of modules, Pacific J. Math., 124 (1986), 131–144. R.M. Guralnick and C. Lanski, Pseudosimilarity and cancellation of modules, Linear Algebra Appl., 47(1982), 111–115. R.M. Guralnick; L.S. Levy and R.B. Warfield, Jr., Cancellation counterexamples in Krull dimension 1, Proc. Amer. Math. Soc., 109(1990), 323–326. W.H. Gustafson, Roth’s theorems over commutative rings, Linear Algebra Appl., 23(1979), 245–251.

rings

December 8, 2010

10:20

World Scientific Book - 9in x 6in

Bibliography

[234] [235] [236] [237]

[238]

[239] [240]

[241] [242] [243] [244]

[245] [246]

[247] [248]

rings

643

A.W. Hager and C.M. Kimber, Clean rings of continuous functions, Algebra Universalis, 56(2007), 77–92. A. Haghany, Morita contexts and torsion theories, Math. Japonica, 42 (1995), 137–142. A. Haghany, Hopficity and co-hopficity for Morita contexts, Comm. Algebra, 27(1999), 477–492. F.J. Hall; R.E. Hartwig; I.J. Katz and M. Newman, Pseudosimilarity and partial unit regularity, Czechoslovak Math. J., 33(1983), 361–372. J. Han and W.K. Nicholson, Notes on clean rings. In Proceedings of the 31st Symposium on Ring Theory and Representation Theory and Japan–Korea Ring Theory and Representation Theory Seminar (Osaka, 1998), Shinshu Univ., Matsumoto 1999, 23–32. J. Han and W.K. Nicholson, Extensions of clean rings, Comm. Algebra, 29(2001), 2589–2595. K. Hanada; Y. Kuratomi and K. Oshiro, On direct sums of extending modules and internal exchange property, J. Algebra, 250(2002), 115–133. D. Handelman, K1 of noncommutative von Neumann regular rings, J. Pure Appl. Algebra, 8(1976), 105–118. D. Handelman, Perspectivity and cancellation in regular rings, J. Algebra, 48 (1977), 1–16. D. Handelman, Stable range in AW ∗ algebras, Proc. Amer. Math. Soc., 76(1979), 241–249. J. Hannah and K.C. O’Meara, Depth of idempotent-generated subsemigroups of a regular ring, Proc. London Math. Soc., 59(1989), 464–482. J. Hannah and K.C. O’Meara, Products of idempotents in regular rings II, J. Algebra, 123(1989), 223–239. J. Hannah and K.C. O’Meara, Products of simultaneously triangulable idempotent matrices, Linear Algebra Appl., 149(1991), 185– 190. M. Harada, On the exchange property in a direct sum of indecomposable modules, Osaka J. Math., 12 (1975), 719–736. M. Harada and T. Ishii, On perfect rings and the exchange property, Osaka J. Math., 12(1975), 483–491.

December 8, 2010

10:20

644

[249] [250] [251] [252] [253] [254] [255] [256] [257] [258] [259] [260] [261] [262]

[263]

[264] [265]

World Scientific Book - 9in x 6in

Rings Related to Stable Range Conditions

R.E. Hartwig, Roth’s equivalence problem in unit-regular rings, Proc. Amer. Math. Soc., 59(1976), 39–44. R.E. Hartwig, How to partially order regular elements, Math. Japonica, 25(1980), 1–13. R.E. Hartwig, On a theorem of Flanders, Proc. Amer. Math. Soc., 85 (1982), 310–312. R.E. Hartwig and F.J. Hall, Pseudo-similarity for matrices over a field, Proc. Amer. Math. Soc., 71(1978), 6–10. R.E. Hartwig and J. Luh, On finite regular rings, Pacific J. Math., 69(1977), 73–95. R.E. Hartwig and J. Luh, A note on the group structure of unit regular ring elements, Pacific J. Math., 71(1977), 449-461. R.E. Hartwig and M.S. Putcha, Semisimilarity for matrices over a division ring, Linear Algebra Appl., 39(1981), 125–132. M. Henriksen, On a class of regular rings that are elementary divisor rings, Arch. Math., 24(1973), 133–141. M. Henriksen, Two classes of rings generated by their units, J. Algebra, 31(1974), 182–193. R.H. Herman and L.N. Vaserstein, The stable range of C ∗ -algebras, Invent. Math., 77(1984), 553–555. Y. Hirano, Some studies on strongly π–regular rings, Math. J. Okayama Univ., 20(1978), 141–149. Y. Hirano, Some characterizations of π–regular rings of bounded index, Math. J. Okayama Univ., 32 (1990), 97–101. Y. Hirano and J.K. Park, On self-injective strongly π-regular rings, Comm. Algebra, 21(1993), 85-91. C.Y. Hong; N.K. Kim; T.K. Kwak and Y. Lee, On weak π-regularity of rings whose prime ideals are maximal, J. Pure Appl. Algebra, 146(2000), 35–44. C.Y. Hong; N.K. Kim and T.K. Kwak, On the maximality of prime ideals in exchange rings, Comm. Korean Math. Soc., 17(2002), 409–422. C.Y. Hong; N.K. Kim and Y. Lee, Exchange rings and their extensions, J. Pure Appl. Algebra, 179(2003), 117–126. C. Huh; N.K. Kim and Y. Lee, On exchange rings with primitive factor rings artinian, Comm. Algebra, 28(2000), 4989–4993.

rings

December 8, 2010

10:20

World Scientific Book - 9in x 6in

Bibliography

[266] [267] [268] [269] [270] [271] [272] [273] [274] [275] [276] [277] [278] [279] [280]

[281] [282]

rings

645

C. Huh; N.K. Kim and Y. Lee, Examples of strongly π-regular rings, J. Pure Appl. Algebra, 189(2004), 195–210. D. Huylebrouck, Equivalence of von Neumann regular and idempotent matrices, Czechoslovak Math. J., 41(1991), 44–50. D. Huylebrouck and J. van Geel, Diagonalisation of idempotent matrices, J. Algebra, 105(1987), 196–206. M. Janjic, A commutativity condition for π-regular rings, Facta Univ. Ser. Math. Inform., 11(1996), 9–10. J.A. Jeong; G.H. Park and D.Y. Shin, Stable rank and real rank of graph C ∗ -algebras, Pacific J. Math., 200(2001), 331–343. J. Kado, Note on exchange property, Math. J. Okayama Univ., 18 (1975/1976), 153–157. W. van der Kallen, The K2 of rings with many units, Am. scient. ´ Norm. Sup., 10(1977), 473–515. Ec. W. van der Kallen; H. Maazen and J. Stienstra, A presentation for some K2 (n, R), Bull. Amer. Math. Soc., 81(1975), 934–936. H. Kambara, On directly finite regular rings, Osaka J. Math., 27(1990), 629–654. I. Kaplansky, Elementary divisors and modules, Trans. Amer. Math. Soc., 66(1949), 464–491. W.P. Keigher, On the ring of Hurwitz series, Comm. Algebra, 25(1997), 1845–1859. F. Keune, The K2 of a 1-fold stable rings, Lect. Notes Math., 1046(1984), 193–228. D. Khurana and T.Y. Lam, Clean matrices and unit-regular matrices, J. Algebra, 280(2004), 683–698. D. Khurana and T.Y. Lam, Rings with internal cancellation, J. Algebra, 284 (2005), 203–235. N.K. Kim, On exchange rings, Proceedings of the 31st Symposium on Ring Theory and Representation Theory and Japan–Korea Ring Theory and Representation Theory Seminar (Osaka,1998), Shinshu Univ. Matsumoto, 1999, 93–97. N.K. Kim and Y. Lee, On right quasi–duo rings which are π–regular, Bull. Korean Math. Soc., 37(2000), 217–227. B. Kirkwood and B.R. McDonald, The symplectic group over a ring with one in its stable range, Pacific J. Math., 92(1981), 111–125.

December 8, 2010

10:20

646

[283] [284] [285]

[286] [287] [288] [289] [290] [291] [292] [293] [294]

[295]

[296] [297] [298]

World Scientific Book - 9in x 6in

Rings Related to Stable Range Conditions

A.A. Klein, The sum of nil one–sided ideals of bounded index of a ring, Israel J.Math., 88 (1994), 25–30. M. Kutami, ℵ0 -continuous directly finite projective modules over regular rings, Math. J. Okayama Univ., 25(1983), 157–163. M. Kutami, On projective modules over directly finite regular rings satisfying the comparability axiom, Osaka J. Math., 22(1985), 815– 819. M. Kutami, On unit–regular rings satisfying s–comparability, Osaka J. Math., 33(1996), 983–995. M. Kutami, The property special (DF ) for unit–regular rings, Math. J. Okayama Univ., 38(1996), 1–11 (1998). M. Kutami, On regular rings with s–comparability, Comm. Algebra, 27(1999), 2917–2933. M. Kutami, Regular rings with comparability and some related properties, Comm. Algebra, 30(2002), 3337–3349. M. Kutami, On von Neumann regular rings with weak comparability, J. Algebra, 265(2003), 285–298. M. Kutami, A construction of unit-regular rings which satisfy (DF ), Comm. Algebra, 32(2004), 1509–1517. M. Kutami, On regular rings with the property (DF ), Advances in Ring Theory, World Sci. Publ., Hackensack, NJ, 2005, 128–138. M. Kutami, On von Neumann regular rings with weak comparability II, Comm. Algebra, 33(2005), 3137–3147. M. Kutami and I. Inoue, The property (DF ) for regular rings whose primitive factor rings are artinian, Math. J. Okayama Univ., 35(1993), 169–179. M. Kutami and K. Oshiro, An example of a ring whose projective modules have the exchange property, Osaka J. Math., 17(1980), 415–420. M. Kutami and H. Tsunashima, Unit-regular rings satisfying weak comparability, Comm. Algebra, 29(2001), 1131–1140. T.Y. Lam, A First Course in Noncommutative Rings, SpringerVerlag, 1991. T.Y. Lam, Modules with isomorphic multiples and rings with isomorphic matrix rings, a survey, L’Enseignement Math´ematique, Geneva, 1999.

rings

December 8, 2010

10:20

World Scientific Book - 9in x 6in

Bibliography

[299] [300] [301] [302] [303] [304] [305] [306] [307] [308] [309] [310] [311] [312] [313] [314] [315]

[316]

rings

647

T.Y. Lam, A crash course on stable range, cancellation, substitution and exchange, J. Algebra Appl., 3(2004), 301–343. T.Y. Lam and A.S. Dugas, Quasi-duo rings and stable range descent, J. Pure Appl. Algebra, 195(2005), 243–259. T.Y. Lam and W. Murray, Unit regular elements in corner rings, Bull. Hong Kong Math. Soc., 1(1997), 61–65. T.K. Lee and Y. Zhou, A class of exchange rings, Glasgow Math. J., 50(2008), 509–522. Y. Lee and C. Huh, A note on π–regular rings, Kyungpook Math.J., 38(1998), 157–161. Y. Lee; C.O. Kim and H.K. Kim, On strongly π-regular rings, Kyungpook Math. J., 37(1997), 291–295. Q. Li, Comparability, K0 and Its State Space, Ph.D. Thesis, Nanjing University, 2002. Y. Li, Strongly clean matrix rings over local rings, J. Algebra, 312(2007), 397–404. T.M. Lok and K.P. Shum, On matrix rings over unit-regular rings, Beitr. Algebra Geom., 44(2003), 179–188. D. Lu; Q. Li and W. Tong, Comparability, stability, and completion of ideals, Comm. Algebra, 32(2004), 2617–2634. D. Lu and W. Yu, On prime spectrums of commutative rings, Comm. Algebra, 34(2006), 2667–2672. X. Lu and H. Qin, K0 of exchange rings with stable range 1, Vietnam J. Math., 34(2006), 171–178. X. Lu and H. Qin, Boolean algebras, generalized abelian rings, and Grothendieck groups, Comm. Algebra, 34(2006), 641–659. G. Marks, A criterion for unit-regularity, Acta Math. Hungra., 111(2006), 311–312. J.C. McConnell and J.C. Robson, Noncommutative Noetherian Rings, New York, NY: John Wiley, 1988. B.R. McDonald, GL2 of rings with many units, Comm. Algebra, 8(1980), 869–888. B.R. McDonald and W. C. Waterhouse, Projective modules over rings with many units, Proc. Amer. Math. Soc., 83(1981), 455– 458. W.W. McGovern, A characterization of commutative clean rings, Int. J. Math. Game Theory Algebra, 15(2006), 403–413.

December 8, 2010

10:20

648

[317] [318] [319] [320] [321] [322]

[323] [324] [325]

[326]

[327] [328] [329] [330]

[331]

[332]

World Scientific Book - 9in x 6in

Rings Related to Stable Range Conditions

W.W. McGovern, B´ezout rings with almost stable range 1, J. Pure Appl. Algebra, 212(2008), 340–348. P. Menal, On π–regular rings whose primitive factor rings are artinian, J.Pure Appl. Algebra, 20(1981), 71–78. P. Menal, Cancellation modules over regular rings, Lect. Notes Math., 1328 (1988), 187–208. P. Menal and J. Moncasi, On regular rings with stable range 2, J. Pure Appl. Algebra, 24(1982), 25–40. P. Menal and J. Moncasi, K1 of von Neumann regular rings, J. Pure Appl. Algebra, 33(1984), 295–312. P. Menal and J. Moncasi, Lifting units in self-injective rings and an index theory for Rickart C ∗ -algebras, Pacific J. Math., 126(1987), 295-329. P. Menal and L.N. Vaserstein, On the structure of GL2 over stable range one rings, J. Pure Appl. Algebra, 64(1990), 149–162. S.H. Mohamed and B.J. M¨ uller, Continuous modules have the exchange property, Abelian groups theory, Perth, 1987, 285–289. S.H. Mohamed and B.J. M¨ uller, On the exchange property for quasi-continuous modules, Abelian groups and modules, Padova, 1994, 367–372. S.H. Mohamed and B.J. M¨ uller, ℵ–exchange rings, Abelian Groups, Module Theory, and Topology (Padua 1997), Dekker, New York, 1998, 311–317. J. Moncasi, A regular rings whose K0 is not a Riesz group, Comm. Algebra,13(1985), 125-131. G.S. Monk, A characterization of exchange rings, Proc. Amer. Math. Soc., 35(1972), 349–353. R. Mortini and R. Rupp, Totally reducible elements in rings of analytic functions, Comm. Algebra, 20(1992), 1705–1713. K.R. Nagarajan; M. Paul Devasahayam and T. Soundararajan, Products of three triangular matrices, Linear Algebra Appl., 292(1999), 61–71. K.R. Nagarajan; M. Paul Devasahayam and T. Soundararajan, Products of three triangular matrices over commutative rings, Linear Algebra Appl., 348(2002), 1-6. W.K. Nicholson, I–rings, Trans. Amer. Math. Soc., 207(1975), 361–373.

rings

December 8, 2010

10:20

World Scientific Book - 9in x 6in

Bibliography

[333] [334] [335] [336] [337] [338] [339] [340]

[341] [342] [343]

[344] [345] [346] [347] [348] [349]

rings

649

W.K. Nicholson, Semiregular modules and rings, Canad. J. Math., 28(1976), 1105–1120. W.K. Nicholson, Lifting idempotents and exchange rings, Trans. Amer. Math. Soc., 229(1977), 269–278. W.K. Nicholson, On exchange rings, Comm. Algebra, 25(1997), 1917–1918. W.K. Nicholson, Strongly clean rings and Fitting’s lemma, Comm. Algebra, 27(1999), 3583–3592. W.K. Nicholson, Clean rings: a survey, Advances in Ring Theory, World Sci. Publ., Hackensack, NJ, 2005, 181–198. W.K. Nicholson and K. Varadarajan, Countable linear transformations are clean, Proc. Amer. Math. Soc., 126(1998), 61–64. W.K. Nicholson; K. Varadarajan and Y. Zhou, Clean endomorphism rings, Arch. Math., 83(2004), 340–343. W.K. Nicholson and Y. Zhou, Rings in which elements are uniquely the sum of an idempotent and a unit, Glasg. Math. J., 46(2004), 227–236. W.K. Nicholson and Y. Zhou, Clean general rings, J. Algebra, 291(2005), 297–311. W.K. Nicholson and Y. Zhou, Strongly lifting, J. Algebra, 285(2005), 795–818. W.K. Nicholson and Y. Zhou, Endomorphisms that are the sum of a unit and a root of a fixed polynomial, Canad. Math. Bull., 49(2006), 265–269. P.P. Nielsen, Abelian exchange modules, Comm. Algebra, 33(2005), 1107–1118. P.P. Nielsen, Countable exchange and full exchange rings, Comm. Algebra, 35 (2007), 3–23. M. Ohori, On strongly π–regular rings and periodic rings, Math. J. Okayama Univ., 27(1985), 49–52. K.C. O’Meara, Simple regular rings satisfying weak comparability, J. Algebra, 141(1991), 162–186. K.C. O’Meara, The exchange property for row and column-finite matrix rings, J. Algebra, 268(2003), 744–749. K.C. O’Meara and R.M. Raphael, Uniform diagonalisation of matrices over regular rings, Algebra Univers., 45(2001), 383–405.

December 8, 2010

10:20

650

[350]

[351]

[352] [353]

[354] [355] [356] [357] [358] [359] [360] [361] [362] [363] [364] [365] [366]

World Scientific Book - 9in x 6in

Rings Related to Stable Range Conditions

K.C. O’Meara and C. Vinsonhaler, Separative cancellation and multiple isomorphism in torsion-free abelian groups, J. Algebra, 221(1999), 536–550. K.C. O’Meara and C. Vinsonhaler, On approximately simultaneously diagonalizable matrices, Linear Algebra Appl., 412(2006), 39– 74. K. Oshiro, Projective modules over von Neumann regular rings have the finite exchange property, Osaka J. Math., 20(1983), 695–699. K. Oshiro and S.T. Rizvi, The exchange property of quasicontinuous modules with the finite exchange property, Osaka J. Math., 33(1996), 217–234. E. Pardo, Comparability, separativity, and exchange rings, Comm. Algebra, 24(1996), 2915–2929. E. Pardo, Metric completions of ordered groups and K0 of exchange rings, Trans. Amer. Math. Soc., 350(1998), 913–933. F. Perera, Lifting units modulo exchange ideals and C ∗ –algebras with real rank zero, J. Reine Angew. Math., 522(2000), 51–62. A.M. Rahimi, Some Results on Stable Range in Commutative Rings, Ph.D. Thesis, The University of Texas at Arlington, 1993. K.N. Rajeswari and R. Aziz, A note on clean matrices in M2 (Z), Internat. J. Algebra, 3(2009), 241–248. R. Raphael, Rings which are generated by their units, J. Algebra, 28(1974), 199–205. M.A. Rieffel, Dimension and stable rank in the K-theory of C ∗ algebras, Proc. London Math. Soc., 46(1983), 301–333. A.G. Robertson, Stable range in C ∗ -algebras, Math. Proc. Camb. Phil. Soc., 87(1980), 413–418. O.M. Romaniv, Elementary reduction of matrices over right 2euclidean rings, Ukrainian Math. J., 56(2004), 2028–2034. J. Rosenberg, Algebraic K–Theory and Its Applications, Grad. Texts in Math. Springer–Verlag, New York, 147, 1994. J.J. Rotman, An Introduction to Homological Algebra, Academic Press, New York, San Francisco, London, 1979. D.E. Rush, B´ezout domains with stable range 1, J. Pure Appl. Algebra, 158(2001), 309–324. A. S´ aez-Schwedt, Feedback cyclization for rings with finite stable range, Linear Algebra Appl., 427(2007), 234–241.

rings

December 8, 2010

10:20

World Scientific Book - 9in x 6in

Bibliography

[367] [368] [369] [370]

[371] [372] [373]

[374] [375] [376] [377] [378] [379]

[380] [381] [382] [383]

rings

651

K. Samei, Clean elements in commutative reduced rings, Comm. Algebra, 32(2004), 3479–3486. J.R. Silvester, Introduction to Algebraic K-Theory, Chapman and Hall, London and New York, 1981. J.R. Silvester, On the GLn of a semilocal ring, Lect. Notes Math., 966(1982), 244-260. A.K. Srivastava, Rings Generated by Properties of Direct Sums of Modules and on Rings Generated by Units, Ph.D. Thesis, Ohio University, 2007. W. Stephenson, Modules whose lattice of submodules is distributive, Proc. London Math. Soc., 28(1974), 291–310. J. Stock, On rings whose projective modules have the exchange property, J. Algebra, 103(1986), 437–453. T. Sudo, Stable rank of the reduced C ∗ -algebras of non-amenable Lie groups of type I, Proc. Amer. Math. Soc., 125(1997), 3647– 3654. R.G. Swan, A cancellation theorem for projective modules in the metastable range, Invent. Math., 27(1974), 23–43. K. Tanabe, Some conditions for nil ideals and π–regular ideals, Comm. Algebra, 27 (1999), 2573–2581. A. Toms, Cancellation does not imply stable rank one, Bull. London Math. Soc., 38(2006), 1005–1008. M. Tousi and S. Yassemi, Tensor products of clean rings, Glasg. Math. J., 47(2005), 501–503. A.A. Tuganbaev, Quasi–projective modules with the finite exchange property, Usp. Mat. Nauk, 54(1999), 191–192. A.A. Tuganbaev, Modules with the exchange property and exchange rings, Handbook of Algebra, North–Holland, Amsterdam, 2000, 439–459. A.A. Tuganbaev, Semiregular, weakly regular, and π–regular rings, J. Math. Sci., 109(2002), 1509–1588. A.A. Tuganbaev, Rings and modules with exchange properties, J. Math. Sci., 110(2002), 2348–2431. A.A. Tuganbaev, Rings Close to Regular, Kluwer Academic Publishers, Dordrecht, Boston, London, 2002. D.V. Tyukavkin, Regular rings of bounded index, Mat. Zametki, 54(1993), 87–93.

December 8, 2010

10:20

652

[384] [385] [386] [387]

[388]

[389] [390] [391] [392] [393] [394] [395] [396]

[397]

[398] [399]

World Scientific Book - 9in x 6in

Rings Related to Stable Range Conditions

K. Varadarajan, Study of Hopficity in certain classes of rings, Comm. Algebra, 28(2000), 771–783. L.N. Vaserstein, The stable range of rings and the dimension of topological spaces, Functional Anal. Appl., 5(1971), 102–110. L.N. Vaserstein, Bass’s first stable range condition, J. Pure Appl. Algebra, 34(1984), 319–330. L.N. Vaserstein and E. Wheland, Commutators and companion matrices over rings of stable rank 1, Linear Algebra Appl., 142(1990), 263-277. L.N. Vaserstein and E. Wheland, Factorization of invertible matrices over rings of stable rank one, J. Austral. Math. Soc., 48(1990), 455–460. F.D. Veldkamp, Projective planes over rings of stable rank 2, Geom. Dedicata, 11(1981), 285–308. Z. Wang and J. Chen, On two open problems about strongly clean rings, Bull. Austral. Math. Soc., 70(2004), 279–282. R. Ware, Endomorphism rings of projective modules, Trans. Amer. Math. Soc., 155(1971), 233–256. R.B. Warfield, Jr., Exchange rings and decompositions of modules, Math. Ann., 199(1972), 31–36. R.B. Warfield, Jr., Stable equivalence of matrices and resolutions, Comm. Algebra, 17(1978), 1811–1828. R.B. Warfield, Jr., Cancellation of modules and groups and stable range of endomorphism rings, Pacific J. Math., 91 (1980), 457–485. F. Wehrung, Injective positively ordered monoids I, J. Pure Appl. Algebra, 83(1992), 43–82. F. Wehrung, Restricted injectivity, transfer property and decompositions of separative positively ordered monoids, Comm. Algebra, 22(1994), 1747–1781. F. Wehrung, Semilattices of finitely generated ideals of exchange rings with finite stable rank, Trans. Amer. Math. Soc., 356(2003), 1957–1970. J. Wei, Exchange rings with weakly stable range one, Vietnam J. Math., 32(2004), 441–449. J. Wei, Power-substitution and exchange rings, Advances in Ring Theory, World Sci. Publ., Hackensack, NJ, 2005, 251–262.

rings

December 8, 2010

10:20

World Scientific Book - 9in x 6in

Bibliography

[400] [401] [402] [403] [404] [405] [406] [407] [408] [409] [410] [411] [412] [413] [414] [415] [416] [417] [418]

rings

653

J. Wei, Unit-regularity and stable range conditions, Comm. Algebra, 33(2005), 1937–1946. T. Wu, Exchange rings with primitive factor rings artinian, Algebra Colloq., 3 (1996), 225–230. T. Wu, On the stable range of endomorphism rings of quasiprojective modules, Comm. Algebra, 26(1998), 139–146. T. Wu, On endomorphism rings of regular modules, Acta Math. Sin., 14B (1998), 385–390. T. Wu, The power–substitution condition of endomorphism rings of quasi–projective modules, Comm. Algebra, 28 (2000), 407–418. T. Wu, Exchange rings with general ℵ0 -comparability, Comm. Algebra, 29(2001), 815–828. T. Wu, On exchange rings with bounded index of nilpotence, Comm. Algebra, 29(2001), 3089–3098. T. Wu, Further results on finitely generated projective modules, Acta Math. Sin., 18B(2002), 225-228. T. Wu and P. Chen, On finitely generated projective modules and exchange rings, Algebra Colloq., 9(2002), 433–444. T. Wu and W. Tong, Finitely generated projective modules over exchange rings, Manuscr. Math., 86(1995), 149–157. T. Wu and W. Tong, Stable range condition and cancellation of modules, Pitman Res. Notes Math. 346, 1996, 98–104. T. Wu and Y. Xu, On the stable range condition of exchange rings, Comm. Algebra, 25(1997), 2355–2363. G. Xiao and W. Tong, n-clean rings and weakly unit stable range ring, Comm. Algebra, 33(2005), 1501–1517. G. Xiao and W. Tong, n-clean rings, Algebra Colloq., 13(2006), 599–606. K. Yamagata, The exchange property and direct sums of indecomposable injective modules, Pacific J. Math., 55 (1974), 301–317. X. Yang and Y. Zhou, Some families of strongly clean rings, Linear Algebra Appl., 425 (2007), 119-129. X. Yang and Y. Zhou, Strong cleanness of the 2 × 2 matrix ring over a general local ring, J. Algebra, 320(2008), 2280-2290. Y. Ye, Semiclean rings, Comm. Algebra, 31(2003), 5609–5625. H.P. Yu, On modules for which the finite exchange property implies the exchange property, Comm. Algebra, 22(1994), 3887–3901.

December 8, 2010

10:20

654

[419] [420] [421] [422] [423] [424] [425] [426] [427] [428] [429] [430] [431] [432] [433] [434] [435] [436] [437]

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Rings Related to Stable Range Conditions

H.P. Yu, On quasi-duo rings, Glasg. Math. J., 37 (1995), 21–31. H.P. Yu, On strongly π–regular rings of stable range one, Bull. Austral. Math. Soc., 51(1995), 433–437. H.P. Yu, On ℵ0 –quasi–continuous exchange rings, Comm. Algebra, 23(1995), 2187–2197. H.P. Yu, Stable range one for exchange rings, J. Pure Appl. Algebra, 98(1995), 105–109. H.P. Yu, On Krull–Schmidt and a problem of Matlis, Comm. Algebra, 24(1996), 2851–2858. H.P. Yu, On the structure of exchange rings, Comm. Algebra, 25(1997), 661–670. B.V. Zabavsky, Elementary reduction of matrices over adequate domain, Math. Stud., 17(2002), 115–116. B.V. Zabavsky, Reduction of matrices over B´ezout rings of stable rank not high than 2, Ukrainian Math. J., 55(2003), 665–670. B.V. Zabavsky, Diagonalization of matrices, Math. Stud., 23(2005), 3–10. B.V. Zabavskij, On noncommutative rings with elementary divisors, Ukr. Mat. Zh., 42(1990), 847–850. B.V. Zabavskij, On prime unit regular rings of elementary divisors, Math. Stud., 2(1993), 21–22. B.V. Zabavskij and O.M. Romaniv, Rings with elementary reduction of matrices, Ukrainian Math. J., 52(2000), 1872–1881. D. Zelinsky, Every linear transformation is a sum of nonsingular ones, Proc. Amer. Math. Soc., 5(1954), 627–630. G. Zhang; W. Tong and F. Wang, Spectrum of a noncommutative ring, Comm. Algebra, 34(2006), 2795–2810. H. Zhang, On Cancellation Property, Ph.D. Thesis, Nanjing University, 2004. H. Zhang and W. Tong, The cancellation range of rings, Arch. Math., 85(2005), 327–334. H. Zhang and W. Tong, Weakly stable modules, Comm. Algebra, 34(2006), 983–989. H. Zhang and W. Tong, The cancellation property of projective modules, Algebra Colloq., 13(2006), 617–622. S. Zhang, Diagonalizing projections in multiplier algebras and in matrices over a C ∗ -algebra, Pacific J. Math., 145(1990), 181–200.

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Bibliography

[438] [439]

[440] [441] [442] [443]

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W. Zhang, Extensions of generalized stable rings, Bull. Korean Math. Soc., 46(2009), 1091–1097. B. Zimmermann–Huisgen, Rings whose right modules are direct sums of indecomposable modules, Proc. Amer. Math. Soc., 77(1979), 191–197. B. Zimmermann–Huisgen, Classes of modules with the exchange property, J. Algebra, 88(1984), 416–434. B. Zimmermann–Huisgen and W. Zimmermann, Algebraically compact rings and modules, Math. Z., 161(1978), 81–93. B. Zimmermann–Huisgen and W. Zimmermann, Classes of modules with the exchange property, J. Algebra, 88 (1984), 416–434. B. Zimmermann–Huisgen, Exchanging torsion modules over Dedekind domain, Arch. Math., 55(1990), 241–246.

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rings

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Index

1-dimensional complex, 69 2-comparability, 30 2-stable range condition, 332, 371 C ∗ -algebra, 64, 141, 243, 315 extremally rich, 242, 540 Mn -unique, 185 P -space, 322 P B-corner, 269, 282 R-algebra, 300, 316 Z-algebra, 387 m-fold stable, 69, 80, 299 n-stable range condition, 215, 361, 368, 394, 487 s-comparability, 174, 179

cokernel, 19 comparability, 119 comparability axiom, 144, 145, 149, 150, 227 completable, 351 completion of diagram, 251 corner, 216 depth, 423 diagonal power-reduction, 321 diagonal reduction, 10, 208, 222, 431, 470, 487, 507 diagonal refinement, 211 Diophantine equation, 565 direct summand, 34, 170, 204, 217, 319, 410, 464, 515, 626, 627 distribute lattice, 600 domain, 19 algebraically closed, 58, 296, 299 B´ ezout, 298 Dedekind, 39, 296, 373, 563, 608 integral, 39, 339, 342 integrally closed, 296 polynomial, 591 principal ideal, 22, 353, 572 Drazin inverse, 306

algebra AW ∗ , 65 algebraic, 63, 410 finite dimensional, 63 algebraic integer, 298 artinian primitive factors, 26, 27, 56, 93, 109, 115, 527 atomless, 176 bimodule, 48, 82, 124 bounded index, 27, 32, 57, 115, 418, 424, 517, 528, 535 Brown-McCoy radical, 594

element, 19 e-clean, 557 clean, 138, 511 Fredholm, 411 full, 198

cancellation, 20, 362 center, 311, 607 centrally orthogonal, 190 657

rings

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Rings Related to Stable Range Conditions

maximal, 600 nilpotent, 27, 408, 417, 596 quasi-invertible, 235 regular, 9, 125, 138, 153, 164, 165, 207, 246, 272, 278, 311, 408, 422, 433, 540 related unit-regular, 152 semiclean, 564 strongly π-regular, 305, 406 unit π-regular, 308, 309 unit-regular, 9, 19, 308, 408, 433, 456, 479, 501, 505, 521 elementary transformation, 146 envelope, 261 essential submodule, 262 extreme partial isometry, 540 finite exchange property, 1, 20, 127, 302 finite rank, 386 finite stable range, 383 fully invariant, 129, 257 general ℵ0 -comparability, 181 general comparability, 151, 181, 440 generalized s-comparability, 181, 184 generator, 213 Goodearl-Menal condition, 58, 61, 69, 95, 111, 112 group, 69 K0 , 608 K1 , 65 abelian, 130, 386 cyclic, 130 ideal class, 608 nonreduced, 130 reduced torsion, 130 special linear, 336 torsion-free, 387 group inverse, 161 homomorphic image, 55, 373, 529, 577 Hurwitz series, 89

ideal B, 274, 409, 423, 517 P B, 282, 448, 535 abelian, 189 exchange, 385 injective, 446 maximal, 92, 343, 589 nil, 415, 596 nilpotent, 271 PI, 424 Pierce, 615 prime, 589 primitive, 27 principal, 22, 46, 87, 591 quasi-duo, 425 regular, 191, 417, 442, 499 separative, 456 stable, 421 strongly π-regular, 406 strongly π ∼regular, 416 strongly separative, 498, 506 ideal-extension, 605 idempotent, 8, 182, 567 full, 399 orthogonal, 141 perfect, 283 index, 417 integral, 296 Jacobson radical, 331, 386, 415, 418, 420, 421, 528, 529, 534, 576, 579 linear transformation, 541, 552 locally nilpotent, 268, 529 matrix cyclic, 228 idempotent, 570 invertible, 10, 143 left invertible, 495 purely singular, 584 rank 1, 564 rectangular, 351, 356, 469 regular, 123, 140, 217, 271, 321, 405, 471, 491 right invertible, 495

rings

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Index

simple, 223 strongly J-clean, 582 triangular, 10, 142, 198 maximal spectrum, 595 medium stable, 113, 116 module hereditarily projective, 39 artinian, 7, 55, 61 cyclic, 77 directly finite, 128 directly infinite, 128 finitely generated projective, 80, 249, 358 injective, 446 nonsingular, 446 power-free, 356 quasi-injective, 84, 251, 254, 262, 290 quasi-projective, 23, 44, 253, 257, 258, 260, 290, 414 semisimple, 7, 84, 205, 206 simple, 80 stably free, 317, 356 monic polynomial, 296 monoid, 169 conical, 174 ordered-separative, 173 refinement, 172, 179 separative, 172 simple, 174 Morita context, 48, 103, 197, 376, 474 Morita equivalent, 6 Morita invariant, 6, 50, 77, 119, 151, 166, 185, 240, 268, 315 norm, 325 order-unit, 176 perspectivity, 34 Pierce stalk, 623 polynomial identity, 424 power-cancellation, 291, 307, 311 power-substitution, 292, 294, 304, 307, 310, 315 prime radical, 594

rings

659

prime spectrum, 595 primitive criterion, 89, 301 primitive polynomial, 88, 301 progenerator, 213 pseudo reducible, 270 pseudo substitution, 244 pseudo-orthogonal, 265 pseudosimilar, 242 reducible, 237 reflexive inverses, 481 related comparability, 151, 153, 160, 165, 169, 198 related unit, 152, 164, 168 right T -nilpotent, 529, 613 ring D-, 600 GE, 200, 334 GEn , 200 ID-, 570 P B, 265, 269, 538 QB, 235, 240, 251, 257, 540 U -irreducible, 58 U C, 345 ω-stage euclidean, 325, 327, 339 abelian, 589, 592, 595, 596, 599, 603, 622 almost hermitian, 469, 477, 479, 481, 483, 498 artinian, 27, 535 B´ezout, 87, 336 birregular, 615 bleached, 579 Boolean, 136, 165 clean, 138, 511 commutative, 10, 87, 557 directly finite, 30, 92, 335, 429 directly infinite, 30, 496 division, 27, 152, 547, 549 euclidean, 325 exchange, 1, 21, 24, 28, 30 extension, 296 fully idempotent, 603 generalized stable, 193, 194, 207, 208, 219 hereditary, 39

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Rings Related to Stable Range Conditions

hermitian, 10 homomorphically semiprimitive, 529 indecomposable, 600 integral extension, 296 local, 18, 573, 623 medium stable, 111 nearly clean, 596 noetherian, 87, 317 one-sided unit-regular, 125, 151 opposite, 3, 495 P.I., 421 perfect, 165 power series, 89 prime, 265 principal ideal, 570 purely infinite, 141, 233 quasi-duo, 92, 115, 531 reduced, 602 regular, 9, 22, 496, 502, 518, 522 right N F , 196 right P -exchange, 530, 613 right q, 208 right self-injective, 185, 289, 429 self-injective, 208 semihereditary, 39 semilocal, 7, 55, 93, 105, 535, 572 semiperfect, 18 semiprime, 265 semiregular, 28, 103 semisimple, 7 separative, 110, 131, 171, 249, 371, 444, 459, 460, 490 simple, 389 skew formal series, 606 skew polynomial, 607 stably euclidean, 325, 327, 331, 350 strongly π-regular, 91, 311, 530, 567, 595, 613 strongly J-clean, 575 strongly clean, 567 strongly separative, 10, 378, 382, 489, 494, 495, 516 strongly stable, 97 triangular matrix, 48, 82, 83, 105, 474

UC, 353, 478 uniquely clean, 577 unit π-regular, 308, 311 unit-regular, 9, 22, 34, 429, 515, 518 weakly stable, 119, 124, 127, 132, 141, 222, 227 zero-dimensional, 87, 88 set clopen, 592 closed, 592 complete orthogonal, 103, 113, 321, 608, 611 open, 616 shift operator, 542 small projectives, 30 space compact, 591 Hausdorff, 593 medium, 596 strongly zero-dimensional, 592, 622 topological, 316, 590, 616 splitting epimorphism, 246 splitting monomorphism, 246 stable power-substitution, 315, 319, 320 stable range one, 1, 9, 21, 30, 33, 140, 250, 311, 399 subdirect product, 9, 285, 319 submodule fully invariant, 288 maximal, 613 superfluous, 413 submonoid, 174 subring, 318, 335, 386, 497 substitution, 2, 75 transpose, 3 trivial extension, 106 unimodular column, 494 unimodular row, 212, 237, 265, 345, 362, 494, 557 unit, 57, 58, 69, 87, 88, 528, 534, 535

rings

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Index

unit 1-stable range, 41, 45, 49, 51, 55, 95 vector space, 199, 240 countably generated, 428, 541, 549 infinite-dimensional, 23, 128, 147, 157, 198, 215, 227, 267, 288, 308, 371, 406, 497, 515, 556 weak substitution, 166, 362 Weyl-algebra, 19 zero divisor, 38, 87, 599

rings

661