Generalizations of Principally Injective Ringsu

JOURNAL OF ALGEBRA 206, 706᎐721Ž. 1998 ARTICLE NO. JA977403

Generalizations of Principally Injective RingsU

Stanley S. Page

The Uni¨ersity of British Columbia, Vancou¨er, British Columbia, V6T 1Z2, Canada E-mail: [email protected]

and

Yiqiang Zhou

Memorial Uni¨ersity of Newfoundland, St. John’s, Newfoundland, A1C 5S7, Canada E-mail: [email protected]

Communicated by Kent R. Fuller

Received December 20, 1997

A ring R is said to be right P-injective if every homomorphism of a principal right ideal to R is given by left multiplication by an element of R. This is equivalent to saying that lrŽ.a s Ra for every a g R, where l and r are the left and right annihilators, respectively. We generalize this to only requiring that for each 0 / a g R, lrŽ.a contains Ra as a direct summand. Such rings are called right AP-injective rings. Even more generally, if for each 0 / a g R there exists an n ) 0 with ann/ 0 such that Ra is not small in lrŽan., R will be called a right QGP-injective ring. Among the results for right QGP-injective rings we are able to show that the radical is contained in the right singular ideal and is the singular ideal with a mild additional assumption. We show that the right socle is contained in the left socle for semiperfect right QGP-injective rings. We give a decomposition of a right QGP-injective ring, with one additional assumption, into a semisimple ring and a ring with square zero right socle. In the third section we explore, among other things, matrix rings which are AP-injective, giving necessary and sufficient conditions for a matrix ring to be an AP-injective ring. ᮊ 1998 Academic Press

INTRODUCTION

Throughout we assume R is a ring with identity and modules are unitary right R-modules. Let JRŽ.be the Jacobson radical of R and ZM ŽR .be

* The research was supported by NSERC Grants A80204 and OGP0194196.

706

0021-8693r98 $25.00 Copyright ᮊ 1998 by Academic Press All rights of reproduction in any form reserved. PRINCIPALLY INJECTIVE RINGS 707 the singular submodule of an R-module M. The Baer criterion for testing when an R-module M is injective naturally leads one to consider the case when one can extend any homomorphism from any principal right ideal of R to M to a homomorphism of R to M. A module M is said to be principally injective, or P-injective for short, if every map from any principal right ideal to M extends to a map of R to M, i.e., is given by left multiplication by an element of M. This is equivalent to saying that lrMRŽ.a s Ma for all a g R, where l and r are the left and right annihilators, respectively. When the ring is P-injective as a right module, the ring is said to be a right P-injective ring. Many results which are known for right self-injective rings, such as the equality of the Jacobson radical with the right singular ideal, and in some cases equality of the right and left socle of R, seewx 2, 6 , were shown to hold for right P-injective rings in w 7, 9x . The second author began to study the rings for which one could write each homomorphism from a principal right ideal to R as the sum of two homomorphisms in a non-trivial way, with the first term a map which did extend to all of R and the second term one which could not be extended. More precisely, suppose the module M has the property that for all

0 / a g R, lrMRŽ.a s Ma [ Xa, where this is a direct sum over the endomorphism of ring of M. Such modules are called AŽ. almost P-injective modules. Many of the results ofwx 9 were obtainable for the class of right AP-injective rings. This suggested the possibility one could obtain these same results if for all 0 / a g R, Ra is not small in lrŽ.a as a left R-module. Indeed, this turns out to be the case and is the subject of Section 2 where we consider the QŽ. quasi P-injective rings and even a more general setting, QGP-injective rings. In Section 3 we extend the results on maximal right ideals of Camillowx 3 on commutative rings andw 9, 10x for right P-injective rings to AP-injective rings. Also, we study the categorical aspects of AP-injective rings and give necessary and sufficient conditions for a matrix ring over a right AP-injective ring to be right AP-injective. Section 1 gives the notation and relation with annihilators which are needed in Sections 2 and 3 as well as examples to illustrate the concepts.

1. NOTIONS AND EXAMPLES

For a right R-module M, we let S s EndŽ.M and then we have an Ž.S, R -bimodule M. For X : M and A : R, the right annihilator of X in R is rRŽ.X s Ä4r g R : Xr s 0 and the left annihilator of A in M is l M Ž.A s Ä4x g M : xA s 0 . For singletons Äx4Äand a4, we abbreviate to rRMŽ.x and l Ž.a . We may omit the subscripts R and M if there is no chance for ambiguity. For an R-module N and a submodule P of N,we 708 PAGE AND ZHOU will identify HomŽ.N, M with the set of maps in Hom Ž.P, M that can be extended to N, and hence HomŽ.N, M becomes a left S-submodule of HomŽ.P, M . In particular, for an element a g R,Hom Ž.R, M will be regarded as a left S-submodule of HomŽ.aR, M .

DEFINITION 1.1. Given a module MR we let S s EndŽ.M . The module M is said to be almost principally injectië Ž.or AP-injectië for short if, for any a g R, there exists an S-submodule XaMof M such that lrŽŽ..Ra s Ma [ Xa as left S-modules. The module M is called almost general principally injectië Ž.or AGP-injectië for short if, for any 0 / a g R, there exists a positive integer n s naŽ.and an S-submodule Xa of M such that n n n a / 0 and lrMRŽŽa .. s Ma [ Xaas left S-modules. In each case the module Xaamay not be unique, but we take one such X for each a and form the S-module bMŽ.s ÝaaX . We call bMŽ.an index bound of M and the set of those XaRan index set of M.If R is an AP-injectiveŽ resp. AGP-injective.Ž module, then we call R a right AP-injectië resp. right AGP-injectië. ring. The concepts of AP-injective modules and AGP-injective modules are explained by the following lemma.

LEMMA 1.2. Let MR be a module, S s EndŽ.M , and a g R.

Ž.1 If lrMR Ž Ž..a s Ma [ X for some X : M as left S-modules, then we ha¨e Hom RRŽ.aR, M s Hom Ž.R, M [ ⌫ as left S-modules, where ⌫ s Ä f g Hom RŽ.Ž.aR, M : fag X 4;

Ž.2 If Hom RR ŽaR, M .s Hom ŽR, M .[ ⌫ as left S-modules, then lrMRŽŽ..a s Ma [ X as left S-modules, where X s Ä faŽ.: f g ⌫4;

Ž.3 Ma is a summand of lrMRŽ Ž..a as left S-modules iff Hom RŽ.R, M is a summand of Hom RŽ.aR, M as left S-modules.

Proof. The map ␪ : lrMRŽŽ..a ª Hom R ŽaR, M .with ␪ Ž.m s ␭mis an S-isomorphism, where ␭mm: aR ª M is defined by ␭ Ž.ar s mr. Also, ␪Ž.Ma s Hom ŽR, M .. So, the lemma follows. Moreover, SŽ.Ma is non- small in lrMRŽŽ..a if and only if HomŽ.R, M is non-small in HomŽ.aR, M .

From Lemma 1.2, the following lemma follows.

LEMMA 1.3. For a g R and a module MRMR, lrŽŽ..a s Ma if and only if e¨ery R-homomorphism of aR into M extends to R.

Recall that a module MR is called P-injectiveŽ. principally injective if every R-homomorphism from a principal right ideal to M is given by left multiplication by an element of M, while a module NR is said to be GP-injectiveŽ. generally principally injective if, for any 0 / a g R, there PRINCIPALLY INJECTIVE RINGS 709 exists a positive integer n such that an / 0 and any R-homomorphism of anR into N extends to R. Therefore, by Lemma 1.3, we have

LEMMA 1.4.Ž. 1 A module MR is P-injecti¨e iff M is AP-injecti¨ewithan Ž.or each index bound b Ž.Ž. M s 0;

Ž.2 A module MR is GP-injecti¨e iff M is AGP-injecti¨e with an Ž or each.Ž index bound b M .Žs 0.. We fix the following notation. If N is a submodule of M, we write NM< to indicate that N is a direct summand of M. For an Ž.R, R -bimodule M, we let R A M be the trivial extension of R and M, i.e., R A M s R [ M as an abelian group, with the following multiplication:

Ž.Ž.Žr, xs, y s rs, ry q xs ..

For a subset I of R and a subset X of M, we let I A X s ÄŽ.r, x g R A M : r g I, x g X 4. EXAMPLES 1.5.Ž. 1 A commutative AP-injective ring R with an index bound bRŽ.s JR Ž.s Soc Ž.R , which has infinite Goldie dimension and is not GP-injective. ϱ Let Q s Ł is1 Fii, where each F s Z44, be the full product of rings Z and let R be the subring of Q generated by ϱ 2 F and 1 . Then [is1 iQ R s [i 2 FiQq Z1 . We next show that, for any a g R, there exists Xa : R such that lrŽ.a s Ra [ Xa asŽ. left R-modules, and thus R is AP-injective. Suppose a s x q n1Qig R with x g[2 F and n g Ä40, 1, 2, 3 . If n s 1or 3, then a is invertible in R and we choose Xa s Ž.0; If n s 2, then lrŽ.a r Ž.a ϱ 2 F 2Z1 Ž ϱ 2 F . Ra and so we choose X s s [is1 iQq s [is1 i[ a ϱϱ2 F ;If n 0, then a 2 F and so Ra< ϱ 2 F , implying s [is1 iis g [is1 [is1 i ϱ 2 F Ra Y for some Y R, and lrŽ.a r Ž.a Ž ϱ 2 F . [is1 ias [ a: s s [is1 i[ 2Z1Qa . In this case, we choose X s Ya[ 2Z1Q . Hence, R is AP-injective and bRŽ. Ž ϱ 2 F .Ž2Z1 . Clearly, bR.JRŽ.SocŽR.. Note s [is1 iQ[ s s 2 that, in the last case, a s 0 and Ra / lrŽ.a . Hence R is not GP-injective. Clearly, the Goldie dimension of R is infinite. Ž.2 A finite commutative ring which is AP-injective with an index bound bRŽ.such that 0 - Soc Ž.R - bR Ž.- JR Ž., but not GP-injective.

Let R s Z444A Ž.Z [ Z be the trivial extension of Z4and the Z 4-module Z44[ Z . We leave for the readers to check that R is AP-injective with an index bound bRŽ.s Ž.0 A ŽZ44[ Z .. Note that SocŽ.R s Ž.0 A Ž2Z4 [ 2Z44.Ž.Žand JRs 2Z A Z4[ Z4 .. Therefore, 0 - SocŽ.Ž.R - bR- JRŽ.. To see R is not GP-injective, we consider a s Ž.0, 1, 0 . Note that 2 a s 0 and lrŽ.a s Ž.0 A ŽZ44[ Z ./ Ž.0 A ŽZ4[ Ž..0 s Ra. Therefore, R is not GP-injective. 710 PAGE AND ZHOU

Ž.3 A noncommutative right AP-injective ring which is not a right GP-injective. Let C be a noncommutative division subring of a division ring D such that the C-vector space C D has dimension ) 1. Let R s C A D be the trivial extension of C and the C-module D. Then R is not commutative. Let 0 / a s Ž.c, d g R.If c / 0, then a is invertible in R and so we can let Xa s Ž.0. If c s 0, then lr Ž.a s Ž.0 A D and Ra s Ž.0 A Cd. Write D s Cd [ D1 as a left C-vector space and let Xa s Ž.0 A D1. Then 2 lrŽ.a s Ra [ Xa. Therefore, R is right AP-injective. Note that a s 0 and lrŽ.a / Ra. Thus, R is not right GP-injective.

2. RIGHT QGP-INJECTIVE RINGS

In this section we consider concepts more general than AP- and AGP- injective rings. As motivation we note that equality of JRŽ.and ZR ŽP .was proved by Kohwx 6 for self-injective rings and extended to P-injective rings and GP-injective rings by Nicholson and Yousifwx 9 and Nam, Kim, and Kimwx 7 , respectively. We will see in Example 2.19 that there are rings satisfying the assumptions in the next theorem but which are not right AGP-injective. Moreover, many results hold in this more general setting and therefore, we make the following definitions. DEFINITION 2.1. Let R be a ring and a g R. We say that a is a right QP-injective element of R if lrŽ.a s Ra q Xaawith a f X , the sum being as left ideals. The ring R is called a right QP-injective ring if every nonzero element of R is a QP-injective element of R, while R is said to be a right QGP-injective ring if every nonzero element of R has a nonzero power which is a QP-injective element. When a g R is a QP-injective element of R and we write lrŽ.a s Ra q Xa, it will be understood that a f Xaaand X is a left ideal of R.

We note that the condition that a is not in Xa is equivalent to Ra is not small in lrŽ.a . There is the slightly stronger condition which we will use later that requires Ra l Xa be small in Ra. It is easy to see that Z is a QGP-injective ring but does not satisfy the assumption in the second part of the following result:

THEOREM 2.2. Let R be a right QGP-injecti¨e ring. Then JŽ. R : ZR ŽR .. n nn n If for each 0 / a g R, lrŽa . s Ra q Xaann with Ra l X nil and a f X a n for some n ) 0 and some left ideal XaRn of R, then JŽ. R s ZR Ž ..

Proof. Let a g JRŽ..If a f ZR ŽR ., then r Ž.a l I s 0 for some 0 / I : RR. Take an s g I such that as / 0. Then there exists 0 / u g asR such that lrŽ.u s Ru q Xuuwith u not in X . Write u s asc for some PRINCIPALLY INJECTIVE RINGS 711 c g R.If t g rŽ.asc , then asct s 0, implying ct g r Ž.Ž.as s r s , since r Ž.a l I s 0. Hence, Ž.sc t s 0 and so t g r Ž.sc . This shows that r Žasc .s r Ž.sc . Note that sc g lrŽ.sc s lr Žasc .s R Ž asc .q Xu. Write sc s dasc q x. Then 1 1 Ž.1 y da sc s x, and so sc s Ž.1 y day x. Then u s asc s aŽ.1 y day x g Xu, a contradiction. For the second statement of the theorem, let b g ZRŽ.R . Then, for each c g R, cb g ZRŽ.R and so 1 y cb / 0. There exists l ) 0 such that l u s Ž.1 y cb / 0 and Ru q Xu s lrŽ.u . Note that u s 1 y ¨ for some ¨ g ZRŽ.R . Since r Ž.Ž.¨ l r 1 y ¨ s 0, we have r Ž.1 y ¨ s 0. Then R s lrŽ.u s Ru q Xuuwith Ru l X s I a nil left ideal. Then we have g q x s 2 1 for g g Ru and x g Xu. Since g y g sygx g I we can use the arguments in Fuller and Andersonwx 1, p. 301 on lifting idempotents to see n n 1 that there is a t g R and n ) 1 such that gt s tg and g s g q t. From n n this it follows that e s g t is idempotent and in Ru. Next we have Rg q I s Ru. Now we claim Re q I s Rg q I. To see this we note that 2 knn1 n 2 n 2 g y g g I and so Rg q I s Rg q I. Since g s gtq s ggŽ . ts ggt n n n n s иии s g g t s g e we have the claim. This means that u s se q i for some s g R and i in I. Recall that u s 1 y ¨ with ¨ g ZRŽ.R . Since r Ž.¨ is right essential, if e / 1, there is a non-zero elementŽ.Ž. 1 y erg 1 y eR l rŽ.¨ . Then Ž 1 y ¨ .Ž1 y er .s Ž1 y er .s i Ž1 y er .. But then Ž 1 y i .Ž1 y er. s 0, with 1 y i a unit. This is a contradiction. So, e s 1 and hence u is left invertible. The result follows. We note that the first part of the theorem remains true if it is only assumed that each 0 / aR contains a nonzero right QP-injective element.

COROLLARY 2.3. If R is a right AGP-injecti¨e ring, then JŽ. R s ZR ŽR ..

For a ring R, JRŽ.is nilpotent if R is right selfinjective and RrSocŽ.RR has the ACC on right annihilators by Armendariz and Parkwx 2 . The same result holds for right P-injective ringswx 9, Theorem 2.2 . The following lemma iswx 9, Lemma 2.1 .

LEMMA 2.4. Let R be a ring and let I be an ideal of R such that RrI satisfies the ACC on right annihilators of principal left ideals. If y1,..., yn,... are in lŽ.I , then there exists n G 1 such that rŽ.Ž.ym иии y1 s r yn иии y1 for all m G n. THEOREM 2.5. Let I be an ideal of R such that RrI satisfies the ACC on right annihilators of principal left ideals. Let R be a right QGP-injecti¨e ring. Then lŽ.I l J Ž R . is right T-nilpotent. Proof. Suppose lŽ.I l JR Ž .is not right T-nilpotent. Then there exists a sequence a12, a ,... in lŽ.I l JR Ž .such that an иии aa21/ 0 for all n ) 0. By Lemma 2.4 there exists s 1 such that rŽ.Ž.a иии a r a иии a for 1 G n 1 s s1 1 712 PAGE AND ZHOU all n s . Take b a иии a . By our assumption, there exists n ) 0 G 11s s1 11 n1 such that c11s b / 0 and lrŽ.c1s Rc 1q X 1for some X1with c 1f X 1. Suppose a иии abn1 0 for some n ) s . Then b n1y1 rŽ.a иии a ns1q11s 1 1 g n 1 s rŽ.b , implying c b n1 0, a contradiction. Therefore, a иии ac/ 0 11s 1s ns1q11 for all n ) s12. Again by Lemma 2.4, there exists s ) s1such that rŽan иии a .Žr a иии a .for all n s . Let b a иии a . Then bc/ s121q1 s ssq122G s ss21q121 n2 0. Therefore, there exists n22) 0 such that c s Ž.bc212/ 0 and lrŽ.c s Rc22q X for some X2with c 22f X . It follows from the choice of b2that a иии ac/ 0 for all n ) s . In the same manner, a simple induction ns2q12 2 ensures that there exist two sequences in lŽ.I l JR Ž ., Ä4b12, b ,... and Ä4c12, c , . . . , and a sequence Än1, n 2, . . . 4 of positive integers, such that: Ž. n1 Ž.n i 1 c11s b and ciiis bcy1 for i ) 1;

Ž.2 Each ci / 0;

Ž.3 For each i, lrŽ.ciiis Rc q X for some X iwith c iif X .

We now have an ascending chain of right annihilators, rŽ.c12: r Ž.c : иии Ž.Ž. . Thus, by Lemma 2.4, r ckq1 s r ck for some k ) 0. It follows that Rckq1 q Xkq1 s Rckkq X and so, for some r g R and x g Xkq1, ck s n n 1 Ž.kq 1 Ž.kq 1y rckq1 q x s rbkq1 ckkq x s rbq1 ckk bq1 ck q x. Therefore, we n 1 Ž.k q 1 y havew 1 y rbkq 1 ckk bq 1 xw ckks x and so c s 1 y n 1 1 n Ž.kq 1y y Ž.kq 1 rbkq1 ckk bq1 x x. Thus, ckq1 s bckq1 kkg X q1. This is a contradiction and thus lŽ.I l JR Ž .is right T-nilpotent.

PROPOSITION 2.6. Let R be a right QGP-injecti¨e ring. Then SocŽ.RR : rŽ.J , where J s JR Ž .. Proof. Let tR : R be simple. Suppose jt / 0 for some j g J. Then 2 rŽ.jt s r Ž.t . Since R is right QGP-injective and Ž.jt s 0, there is a left ideal Xjt such that Ž.jt f Xjt and lr Ž.jt s Rjt Ž.q Xjt. Note that t g lr Ž.jt . y1 Write t s rjt q x where x is in x jt. ThenŽ. 1 y rj t s x,sot s Ž.1 y rj x g Xjt. This means that jt g Xjt, a contradiction. COROLLARY 2.7. Let R be semiperfect and right QGP-injecti¨e. Then SocŽ.RRR: Soc Ž.R and Z Ž. R R : ZR Ž. Rs JR Ž..

Proof. Since R is semiperfect, we have ZRŽ.RRq ZR Ž.: JR Ž.byw 11, Lemma 2x . The inclusion JRŽ.: ZR ŽR .is by Theorem 2.2. The first part is by Proposition 2.6 A ring R is called right Kasch if every simple right R-module embeds in R, or equivalently if lŽ.M / 0 for every maximal right ideal M of R. Analogously, we can define a left Kasch ring. COROLLARY 2.8. If R is a semiperfect right QGP-injecti¨e right Kasch ring such that Ra is an essential submodule of lrŽ.a for all a g R, then Soc ŽRR . is left essential in R, SocŽ.RRR s Soc Ž.R , and J Ž. R s ZR Ž. RRs ZR Ž.. PRINCIPALLY INJECTIVE RINGS 713

Proof. Since R is right Kasch, we see that lrŽ.b l l ŽŽJR ../ 0 for any 0 / b g R. By the assumptions, this implies that SocŽ.RR is an essential left ideal of R. Thus, by Corollary 2.7, SocŽ.RRRs Soc Ž.R . We have JRŽ.: ZR ŽRR .since l ŽŽ..JR s Soc ŽR .is essential in RR. Now the last statement of the corollary follows from Corollary 2.7. Remark 2.9. It should be noted that with the hypothesis of Corollary 2.8 results similar to those of Theorem 2.3 ofwx 9 can be obtained. COROLLARY 2.10. Let I be an ideal of R and suppose that RrI satisfies the ACC on right annihilators. If R is right QGP-injecti¨e, then lŽ.I l JR Ž . is nilpotent. In particular, if I s SocŽ.RorIRRs Soc Ž.R , then J Ž. R is nilpotent. Proof. It follows from Theorem 2.5 and a well-known result of Fisher wx4 that <ŽŽl I .l JR Ž ..q I

COROLLARY 2.11. Suppose that RrSocŽ.RR satisfies the ACC on right annihilators. If R is right AGP-injectiëŽ in particular, right GP-injectiëor right P-injectië.Ž., then J R is nilpotent.

A module M is said to satisfy Ž.C2 if for any two submodules X and Y of M with X ( YM<<, we have XMwx8 . We say M satisfies Ž.) if, for any two submodules X and Y of M with X : Y and X ( YM<<, we have XM. We say M satisfies Ž.)) if for any two simple submodules X and Y of M with X ( YM<<, we have XM. ␴ 2 LEMMA 2.12. Let a, b g R such that aR ( bR s eR, where e s e . Then there exists an idempotent f g R such that af s a and rŽ.a s r Ž.f . Proof. This is by the proof ofw 9, Theorem 1.2Ž. 1x .

PROPOSITION 2.13.Ž. 1 If, for each 0 / a g R, there exists a left ideal Xa of R such that lrŽ.a s Ra q Xaa with Ra l X nil and small in Ra, then RR satisfiesŽ. C2 .

Ž.2 If, for each 0 / a g R, there exists n ) 0 and a left ideal Xan of R n n nn such that a / 0 and lrŽa . s Ra q Xaann with Ra l X nil and small in n Ra , then RR satisfies Ž.) and Ž)) .. ␴ 2 Proof. Ž.1 Let a, b g R such that aR ( bR s eR, where e s e .We 2 want to show that aR s gR for some g s g . By Lemma 2.12, there exists an idempotent f such that af s a and rŽ.a s r Ž.f . Then f g lr Ž.f s lr Ž.a . We have lrŽ.a s Ra q Xaafor some left ideal X of R where I s Ra l X a is small in Ra and nil. Set f s ra q x, where r g R and x g Xa. Then 714 PAGE AND ZHOU

ra s raf s rara q rax. Thus, ra y rara s rax g Ra l Xa s I, showing that ra y rara is nilpotent. Letting g s ra we see that Rg q I s Ra as ag y a g I. But I is small in Ra,so Rg s Ra. As in the proof of Theorem 2.2 we see that for some idempotent e, Ra s Re q I s Re, the last equality because I is small in Ra. From this we see that Re is a direct summand of Rf, hence, Ra is a direct summand of Rf. Then Rf s Ra [ Y for some Y and f s sa q y for some y g Y. Then a y asa s ay s 0. This shows that 2 aR s Ž.as R and Ž.as s as. Ž.2 To show RR satisfies Ž) ., let a, b g R such that a g bR s eR, 2 ␴ where e s e , and that aR ( bR. We want to show that aR s gR for some 2 g s g . There exists c g R such that ␴ Ž.ac s e and hence Ž.ac R s aR,so n we may assume that ␴ Ž.a s e. There exists n ) 0 such that a / 0 and n nn lrŽa . s Ra q Xaaannnfor some left ideal X of R with Ra l X nil and n 0 i 1 i small in Ra . We write a s e and we have ␴ ŽaRq .Ž.s ␴ aaRs ii i i 1 ea R s aR for i s 0, 1, . . . , n y 1. This implies that aRa< y Rm i 1 i i 1 i ␴ ŽaRq .<␴ ŽaR. m aRaRq < for i s 1,...,n y 1. Therefore, we have 2 n n 1 aR< eR m aRaR< m иии m aRa< y R. So, to show aR< R, it suffices to nn␴ n1 ␴␴␴ show that aRR< . We note that aR( aRy ( иии ( aR ( eR, i.e., ␤ nn n aR( eR and ␤Ž.a s e, where ␤ s ␴ . Now we can use the proof ofŽ. 1 to conclude that anR is a direct summand of R. ␴ To show RR satisfies Ž.)) , let a, b g R such that aR ( bR s eR, where 2 e s e and eR is a minimal right ideal of R. We want to show that 2 n n aR s gR for some g s g . There exists n ) 0 such that a / 0 and a is 2 ޑސ-injective. Since Ž.aR s 0oraR s gR for an idempotent g, we may assume n s 1. Thus, we can proceed as in the proof ofŽ. 1 . For a module M we let EMŽ.be the injective hull of M. The module M is said to be weakly injective if, for any finitely generated submodule N : EMŽ., we have N : X ( M for some X : EM Ž.wx5. LEMMA 2.14. Let M be an f. g. module. If M is weakly injecti¨e and satisfies Ž.) , then M is injecti¨e. Proof. Let x g EMŽ.. Then there exists X : EM Ž.such that M q xR : X ( M. Hence X satisfies Ž.) , and so MX< . This implies that M s X, so x g M. The following corollary follows immediately from Lemma 2.14 and Proposition 2.13.

COROLLARY 2.15. If R is as in Proposition 2.13Ž. 2 and RR is weakly injecti¨e, then R is right self-injecti¨e. The following result was proved inwx 9, Theorem 1.4 for a semiperfect right P-injective ring R. PRINCIPALLY INJECTIVE RINGS 715

THEOREM 2.16. If R is as in Proposition 2.13Ž. 2 and is semiperfect, then

R ( R12= R , where R 1 is semisimple and eëry simple right ideal of R2 is nilpotent. Proof. The arguments of the proof ofwx 9, Theorem 1.4 will work when one uses Proposition 2.13Ž. 1 instead ofwx 9, Theorem 1.2 . A ring R is called Dedekind finite if u¨ s 1in R implies ü s 1. PROPOSITION 2.17. Suppose that, for any 0 / a g R, there exists n ) 0 n n n and a left ideal Xan : JŽ. R such that a / 0 and lrŽa . s Ra q Xan. Then the following are equiälent: Ž.1 RrJ Ž R . is Dedekind finite;

Ž.2 E¨ery monomorphism RRRª R is an isomorphism; Ž.3 R is Dedekind finite.

Proof. Ž.1 « Ž.2 . Let f : RRRª R be monic and a s fŽ.1 . Then a / 0 n and rŽ.a s 0. By hypothesis, there exists n ) 0 such that a / 0 and n n n lrŽa . s Ra q Xaannfor some X : JRŽ.. Since r Ž.a s 0, r Ža . s 0. n n Therefore, R s Ra q Xan. Write 1 s ba q x for some b g R and x g n n y1 Xan. ByŽ. 1 , we have 1 s abq y for some y g JRŽ.. Then abŽ.1 y y n 1 1 s 1 and so faŽ y bŽ.1 y y y . s 1, showing that f is an isomorphism. Ž.2 « Ž.3 . Let ab s 1. Define f : RRRª R by frŽ.s br. Then f is monic and, byŽ. 2 , 1 s bc for some c. Therefore, c s Ž.Ž.ab c s abc s a. Ž.3 « Ž.1 . Let ab s 1in RrJRŽ.. Then ab s 1 q x for some x g JRŽ. 1 1 and hence 1 s Ž.1 q xaby .By3,1 Ž.Ž.s b 1 q xay . It follows that 1 s y1 bŽ.1 q xas b1a s ba. As in Proposition 2.17, the proof ofŽ. 2 « Ž.3 « Ž.1 does not need any assumptions on R.

COROLLARY 2.18. If R is right AGP-injectië with an index bound bŽ. R : JŽ.Ž R in particular, right GP-injectiëorP-injectië ., then the following are equiälent: Ž.1 R is Dedekind finite;

Ž.2 E¨ery monomorphism RRRª R is isomorphism; Ž.3 RrJ Ž R . is Dedekind finite. EXAMPLE 2.19. A finite commutative ring R which satisfies the assumptions in Proposition 2.13Ž. 2 but is not AGP-injective.

Let R s Z88A 2Z be the trivial extension of Z8and the Z 8-module 2Z 8. For a s Ž.n, x g R,if n s 1 or 3 or 5, then a is invertible in R. Thus, lrŽ.a s Ra. For a1s Ž.Ž.4, 0 , lr a 1s 4Z 88A 4Z and Ra 18s 4Z A Ž.0 ; For a22s Ž.4, 2 or Ž.Ž. 4, 6 , lr a s 4Z8A 2Z8and Ra2s ÄŽ.Ž.Ž.Ž0, 0 , 0, 4 , 4, 2 , 4, 716 PAGE AND ZHOU

6.4 ; For a33883s Ž.Ž.4, 4 , lr a s 4Z A 4Z and Ra s ÄŽ.Ž.0, 0 , 4, 44 ; For a4s Ž.Ž.0, 4 , lr a488s 4Z A 4Z and Ra 4s Ž.0 A 4Z 8. Thus, Raii

3. RIGHT AP-INJECTIVE RINGS

In this section some results on maximal right ideals of commutative P-injective rings by Camillowx 3 and of right P-injective rings by Nicholson and Yousifwx 9, 10 are extended to right AP-injective rings which admit an index set Ä4Xaa: a g R of ideals such that X bs Xbafor all a, b g R. Necessary and sufficient conditions are obtained for a matrix ring to be right AP-injective. The proofs of Lemmas 3.1 and 3.2 and of Corollary 3.3, are adaptations of arguments inwx 9, 10 and they are included for completeness.

LEMMA 3.1. Gi¨en a setÄ4 Xa : a g R of left ideals of R, the following are equi¨alent:

Ž.1 lr Ž.a s Ra [ Xa for all a g R;

Ž.2 lwbR l rŽ.a x s Ž.Xab: b l q Ra and Ž. Xab: b l l Ra : l Ž.b for all a, b g R, whereŽ. Xab: b l s Ä4x g R : xb g Xab . Proof. Ž.1 « Ž.2 . Let x g lwbR l rŽ.a x. Then rŽ.ab : r Ž.xb and so xb g lrŽ.xb : lr Ž.ab s Rab [ Xab. Write xb s rab q y, where r g R and y g Xab. Then Ž.x y ra b s y g Xab and hence x y ra g Ž.Xab: b l. It follows that x g Ž.Xab: b l q Ra. Obviously, Ra : lwbR l rŽ.a x.If y g Ž.Xab: b l, then yb g Xab : lrŽ.ab . Let bs g bR l rŽ.a . Then abs s 0. Hence ybs s 0 since yb g lrŽ.ab . This shows that y g lwbR l rŽ.a x.We have proved that lwbR l rŽ.a x s Ž.Xab: b l q Ra.If ra g Ž.Xab: b l l Ra, then rab g Xab l Rab, showing that rab s 0. Hence ra g lŽ.b . Ž.2 « Ž.1 . Let b s 1.

LEMMA 3.2. Let R be a right AP-injecti¨e ring with an index setÄ Xa, a g R4 of ideals such that Xabs X ba. If 0 / uR is a uniform right ideal of R, define

Mu s Ä4x g R : rŽ.x l uR / 0. Ž. Then Mu is the unique maximal left ideal of R which contains Ýag RaulX : u . PRINCIPALLY INJECTIVE RINGS 717

Proof. It is easy to see that Muais a left ideal. Let y g Ž.X u: u l. Then yu g Xau. Thus yua g Xau l Rua s Xua l Rua since Xaus X ua is an ideal. Then yua s 0 and so y g Mu if ua / 0. If ua s 0 then lrŽ.ua s 0, and so Xaus X ua s 0. This shows that yu s 0 and hence y g Mu. There- fore, Ž.Xau: u l: M u for all a g R. Now if a f Mu, then rŽ.a l uR s 0. By Lemma 3.1, we have R s lrŽŽa .l uR .s ŽXau: u . l q Ra. Then R s Muuq Ra, showing that M is a maximal left ideal. Ž. Let L be a left ideal of R such that Ýag RaulX : u : L / M u. Then, as above, R s Ž.Xau: u l q Ra for any a g L y Mu. Therefore, L s R. COROLLARY 3.3. Let R be a right AP-injecti¨e ring with an index set Ä4Xaa: a g R of ideals such that X bs Xba for all a, b g R. If R is right finitely dimensional, then

Ž.1 M is a maximal left ideal of R iff M s Mu for some uniform right ideal uR; Ž.2 RrJ Ž R . is semisimple artinian.

Proof. Let uR12[ uR[ иии [ uRn be a right essential ideal of R with each uRi a uniform right ideal. Ž.2 For any x n M , we have rŽ.x uR/ 0 for i 1,...,n. g Fis1 uii l s Then rŽ.x is an essential right ideal of R. Thus, x g ZRŽ.R . By Corollary 2.3, x g JRŽ.. Ž.1 By Lemma 3.2, we only need to show that every maximal left ideal M of R is of the form Mu for some u g R such that uR is a uniform right ideal. If not, then M / M . There exists m M such that rŽ.m uR u1 g l 1 s 0. It follows that rŽ.Ž.mu r u . Then u lr Ž.mu Rmu X . Thus, 11: 1g 1s 1[ mu1 there exists r R such thatŽ. 1 rm u X . This implies that 1 rm g y 1 g mu1 y Ž.X : u M . Let m rm.If1 m M for all i, then, from g mu111 lu: 11s y g u i the proof ofŽ. 2 , 1 y m1 g JRŽ.: M, implying 1 g M, a contradiction. So, say 1 m M . Then rŽ.1 m uR 0. Thus,Ž. 1 muR uR y 1 f u2 y 12l s y 122( is a uniform right ideal and so M / Mu for u s Ž.1 y mu12. As above, XX X 1 y m g Mu for some m g M. Then rŽ1 y m . l uR / 0. This implies thatŽ 1 mX .Ž1 m . M . Therefore, there exists m M such that y y 1 g u2 g 1 m M M . By a simple induction, there exists m M such that y g uu12l g 1 m M for i 1,...,n. By the proof ofŽ. 2 , 1 m JR Ž . M,a y g u i s y g : contradiction.

COROLLARY 3.4. Let R be a right AP-injecti¨e ring with an index set Ä4Xaa: a g R of ideals such that X bs Xba for all a, b g R. If R is left Kasch and R has enough uniform right ideals, then M is a maximal left ideal of R iff

M s Mu for some uniform right ideal uR. Proof. One direction is by Lemma 3.2. Let M be a maximal left ideal. Then rŽ.M / 0 since R is left Kasch. Choose a uniform right ideal 718 PAGE AND ZHOU uR : rŽ.M .If x g Mu, then 0 / r Ž.x l uR : r Ž.M . Thus, R / lrw Ž.x l uRx = lrŽ.M = M, implying lrw Ž.x l uRx s M. So, x g M. By Lemma 3.2, Mu s M. LEMMA 3.5. Let R be a right AP-injecti¨e ring with an index bound bŽ. R . If cR [ dR and Rc [ Rd are direct for c, d g R, then lŽ.c q l Ž.d s R, where c s c q bŽ. R and d s d q bRŽ.. Proof. The correspondence Ž.c q dRª R given by Ž.c q dr¬ cr is a Ž.Ž. well-defined map. Thus, c g lr c q d s Rcq d [ Xcqd. There exists r g R such that c y rcŽ.Ž.Ž.q d g bR. Then 1 y rcy rd s 0. This implies that 1 y r g lŽ.c and yr g l Ž.d . Thus, R s lŽ.c q l Ž.d . A ring R is called right duo if every right ideal of R is an ideal.

THEOREM 3.6. Let R be a right AP-injecti¨e ring with an index set Ä4Xaa: a g R of ideals such that X bs Xba for all a, b g R. Suppose R is right duo. Then the following are equi¨alent: Ž.1 R is right finitely dimensional;

Ž.2 R has a finite number of maximal left ideals and b Ž R .s ÝaaXis right finitely dimensional.

Proof. Ž.1 « Ž.2 . By Corollary 3.3, RrJRŽ. is semisimple artinian. Since R is right duo, RrJRŽ.is right duo. It follows that RrJRŽ.is a direct sum of division rings. Therefore,Ž. 2 holds. Ž.2 « Ž.1 . Suppose R contains as a right ideal an infinite direct sum aR12[ aR[ иии with each aRi / 0. Since bRŽ.as right R-module has finite Goldie dimension, there exists a natural number n such that bRŽ.l иии wxaRnn[ aRq1 [ s 0. Note that each aRi is an ideal. It follows that иии Rann[ Ra q1 [ is direct. For each i G n, there exists a maximal left ideal Miiiisuch that lŽ.a : M . By Ž. 2 , we have M s Mjfor some i / j with i G n and j G n. Then lŽ.aijiq l Ž.a : M , a contradiction to Lemma 3.5. Letting bRŽ.s 0, we obtainwx 10, Theorem 1.1 . When R is commutative, Theorem 3.6 yields the following result:

COROLLARY 3.7. Let R be a commutati¨eAP-injecti¨e ring with an index bound bŽ. R . Then R has finite Goldie dimension iff R has a finite number of maximal ideals and bŽ. R is of finite Goldie dimension. Example 1.5Ž. 1 shows that, in Theorem 3.6 or Corollary 3.7, the condition that R has a finite number of maximalŽ. left ideals alone does not always imply that R has finite Goldie dimension. We now take up the question of when a matrix ring is an AP-injective ring. PRINCIPALLY INJECTIVE RINGS 719

THEOREM 3.8. The following are equi¨alent for two natural numbers n and m with 1 F m F n:

Ž.1 For any A s Žaij .g M n Ž. R with aij s 0 whene¨er i ) m, MRAn Ž. is a direct summand of lrŽ.A as a left Mn Ž. R -module;

Ž.2 For any A g MnŽ. R such that A has at most m nonzero rows, MnnŽ. R A is a direct summand of lrŽ.A as a left M Ž. R -module; m Ž.3 Hom RR ŽR , R.Ž is a direct summand of Hom I, R. as a left R-module for any n-generated R-submodule of R m.

m Proof. Ž.1 « Ž.3 . Let I s aR1 q иии qaRn : R and write Ž.a1,...,an s Ž.Že1,...,emn,0,...,0 A, where A g MR., the sequence e1,...em is the standard basis of the R-module R mmover R, and 0 is the 0 element of R .

Then byŽ. 1 we have lr ŽA .s MRAnA Ž . [ X for some left ideal XAof MRnŽ.. Let

␣ a иии ␣ a ¡ Ž.1 Ž.n ¦ 0 иии 0 ⌫ s~¥␣ g Hom RAŽ.I, R : ..g X . .. ¢§00 иии 0

It can be verified that ⌫ is a left R-submodule of Hom RŽ.I, R . We claim m that Hom RRŽ.I, R s Hom ŽR , R. [ ⌫ as left R-modules. In fact, for ␣ g Hom RŽ.I, R , write

␣ Ž.a12␣ Ž.a иии ␣ Ž.an 00иии 0 B s ...... 000иии 0

Then B g lrŽ.A and hence B s Ž.cAij q Ž.dij , where Ž.cijg MR n Ž.and Ž. ␤ mmm␤Ž. ␥ dijg X A. Let : R ª R by Ýis1 eriis Ý is11criiand : I ª R by ␥ Ž.nn ␤ Ž.m␣ ␤ ␥ Ýis1 ariis Ý is11drii. Then g Hom RR , R and s q . Note that

d11 иии d1n 10 иии 0 0 иии 0 00 иии 0 .. . ..s .. .Ž.dijg X A...... 00 иии 0 000 иии 0 m So, ␥ g ⌫. Therefore, we have Hom RRŽ.I, R s Hom ŽR , R. q ⌫. Sup- m n pose ␥ g ⌫ l Hom RŽ R , R.Ž.. Then there exists c1,..., xm,0,...,0 g R 720 PAGE AND ZHOU such that

Ž.␥ Ž.a1 ,...,␥ Ž.an s Ž.c1 ,...,cm ,0,...,0 A. Therefore,

␥ Ž.a1 иии ␥ Ž.an c1 иии cm 0 иии 0 0 иии 0 0 иии 00иии 0 A ..s ...... 00 иии 0 00 иии 00иии 0

g MRAnAŽ. l X , m implying ␥ s 0. Thus, Hom RRŽ.I, R s Hom ŽR , R. [ ⌫. Ž.3 « Ž.1 . Suppose A s Ž.Ž.aijg MR n with aij s 0 whenever i ) m. Let I s aR1 q иии qaRn , where Ž.Ža1,...,an s e1,...,em,0,...,0 .A.By m Ž.3 , we have Hom RRŽ.ŽI, R s Hom R , R. [ ⌫ for some left R-submodule ⌫ of Hom RŽ.I, R . Let ␣ a ␣ a иии ␣ a ¡ 11Ž. 12 Ž.1 Ž.n ¦ ␣21Ž.a ␣ 22 Ž.a иии ␣2 Ž.an XAis~¥.. .: ␣ g ⌫, i s 1,...,n . .. . ␣ Ž.a ␣ Ž.a иии ␣ Ž.a ¢§0n 1 n 2 nn

Then XAnis a left ideal of MRŽ.. We claim that lrŽ.A s MRAn Ž. [ XA as left MRnŽ.-modules. It can easily be checked that XA : lrŽ.A .If Y s Ž.yij g lr Ž.A , then Ab Ž.ij ¬ Yb Ž.ij is an MRn Ž.-homomorphism from Ž. Ž. n AMnn R to YM R and induces an R-homomorphism fi: Ý js1 arjj¬ n Ý js1 yrij jfrom I to R for each 1 F i F n. Write f is g iq h i, where m giRg Hom Ž R , R.Žand h ig ⌫. Then, for each i, there exists ci1,...,cin. n g R such that ŽŽ.gai 1 ,..., gain Ž ..Žs c i1,...,cAin .. Therefore,

ha11Ž.иии ha1 Ž.n

ha21Ž.иии ha2 Ž.n Y s Ž.yijs Ž.cA ij qg..MRAnŽ.q XA. .. haŽ.иии ha Ž. 0n 1 nn Therefore, lrŽ.A s MRAnA Ž. q X . Finally, let C g MRAnAŽ. l X . Then, for some Ždijn .g MR Ž.and some ␣i g ⌫ Ž.i s 1,...,n ,

␣11Ž.a иии ␣1 Ž.an .. C ss..Ž.dAij . ␣ a иии ␣ a 0nŽ.1 nn Ž. PRINCIPALLY INJECTIVE RINGS 721

Then, for each i, ŽŽ.␣i a1 ,...,␣in Ž..Ža s d i1,...,dAin ., which shows that m ␣iRg Hom Ž R , R.Žl ⌫. Thus, each ␣is 0 and hence C s 0. So, lr A.s MRAnAŽ. [ X . Ž.2 « Ž.1 . This is obvious. Ž.1 « Ž.2 . For any A g MRn Ž .such that A has at most m nonzero rows, there exists an invertible matrix U g MRnŽ.such that the last n y m rows of UA are 0. Thus, the implication follows.

COROLLARY 3.9. The following are equi¨alent for a natural number n:

Ž.1 Mn Ž R . is right AP-injecti¨e; n Ž.2 Hom RR ŽR , R.Ž is a direct summand of Hom I, R. as left R-modules for any n-generated R-submodule I of R n.

REFERENCES

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