on lifting modules

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On lifting modulesDerya Keskin aa Department of Mathematics, Hacettepe University, Beytepe, Ankara, 06532,Turkey E-mail:Published online: 27 Jun 2007.

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COMMUNICATIONS IN ALGEBRA, 28(7), 3427-3440 (2000)

ON LIFTING MODULES

Derya Keskin Department of Mathematics, Hacettepe University,

06532 Beytepe, Ankara, Turkey e-mail : keskinQhacettepe.edu.tr

Abstract

Let R he a ring wit,h ident,it,y and let M = MI 6B. . . @ Mn he a finite direct sum of relatively projective R-modules Mi. Then it is proved t,hat M is lifting if and only if M is amply supplemented and Mi is lifting for all 1 _< i 5 n. Let M - M I @ . . . @ A[,, he a finite direct sum of R-modules Mi. We prove that M is (quasi-) discrete if and only if A[,,. . . M,, are relatively projective (quasi-) discrete modules. We also prove t,hat, for an amply supplemented R-module A4 = Ad1 :@ &I2 such that Afi and Ad2 have the finite exchange property, M is lifting if and only if Afl and i%.lz are lifting anc! relatively small projective R-modules and every co-closed submodule N of M with M = N + &Il = N + M2 is a direct summand of M. Finally, we prove that, for a ring R such that every direct sum of a lifting R-module and a simple R-module is lifting, every simple R-module is small M-projective for any lift,ing R-module M.

1 Preliminaries

Throughout this paper all rings will have a n identity and all modules will be unital right R-modules.

Let M be a module and S < M. S is called small in M (notation S << M) if M # S + T for any proper submodule T of M. Dually, a submodule E of M is called essential in M if, E n A # 0 for any non-zero submodule A of hl. Note t h a t if any submodule E of M is essential in a submodule K of M , then we also say t h a t K is a n essential extension of E. A module M is called

Copyright 0 2000 by Marcel Dekker, Inc. www.dekker.com

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hollow if every proper submodule of M is small in M, and it is called local if it is hollow and has a unique maximal submodule.

Let M be a module. Let N and L be submodules of M. N is called a supplement of L if it is minimal with the property M = N -t L, equivalently, M = N + L and N n L << N. N is called a supplement submodule if N is a supplement of some submodule of M. Following [Wi], M is called a supplemented module if every submodule of M has a supplement. On the other hand, the module M is amply supplemented if, for any submodules A, B of M with M = A + B there exists a supplement P of A such that P < B. Following [ Z ] , the module M is called a weakly supplemented module if for each submodule A of M there exists a submodule B of M sych that M = A + B a n d A n B ~ M .

Let A4 be a module and B .< A 5 M. If A/B << M / B , then B is called a co-essential submodule of A and A is c,alled a co-essential extension of B in M. A suhrnodule A of h l is called co-closed if A has no proper co-essential submodule. Dually, a submodule A of M is called closed if A has no proper essential extension in hl. Also, we will call B an s-closure of A in M, if B is a co-essential submodule of A and B is co-closed in M.

Let A4 he a module. M is called a lzfiing module (or satisfies ( D l ) ) , if for any suhrnodule N of M, there exists a direc,t summand K of ill such that h' 5 N and N/I< << M / K , equivalently, for every submodule iV of A4 there exist subrnoclules K, K1 of M such that A l = I<* @ K 1 . II' 5 N and N n I? << I{'. Dually, M is called an extending module (or satisfies (Cl)) if, for any submodule N of M , there exists a direct summand K of M such that N is essentid in K. It is easy to see that a n~odule M is extending if and only if every closed submodule of M is a direct summand of 114. Ey [AM, Proposition 4.81, the module M is lifting if and only if M is amply supplemented and every supplement submoclule of M is a direct summand. Note tha t in [MhI], amply supplemented modules are called supplemented modules, and note that every direct summand of a lifting module is lifting [MM, Lernms 4.7) .

We start with the following lemma which appears in [L, Chapter I , Propo- sition 2.11.

Lemma 1.1 Let M be a m o d d e and N 5 M . Consider the following conditions:

( 1 ) N is u supplem.ent submodule of M .

(2) N is co-closed in M .

(3) For all X < N , X < M implies X < N .

Then (1)=+(.2)-(3) hold. I f M is a weakly supplemented module then ( 3 ) a ( I ) holds.

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ON LIFTING MODULES 3429

Proof . (1)*(2) Let N be a supplement of L in M. Then M = N + L and N is minimal with this property. Let K 5 N and N / K << M / K . Then M = K + L. By the minimality of N , N = K. Hence N is co-closed in M . (2)=+(3) Let N be a co-closed submodule of M. Let X 5 N and X <( M . Assume N = X + Y and Y 5 N. Now we want to show that N / Y << M / Y . Let M/Y = N/Y+H/Y with Y < H 5 M. Then M = N+H = X+Y+H = X + H implies M = H. Therefore N / Y << M / Y . Since N is co-closed N = Y. Thus X << N. (3)*(1) Let M be a weakly supplemented module. Assume that for all X 5 N , X << M implies X << N . Since M is weakly supplemented there exists a submodule L of M such that M = N + L and N n L << M. By assumption, N n L << N . Hence N is a supplement of L in M.

Lemma 1.2 Let M be a module and let M = A + B and M = ( A n B ) + C T h e n M = ( B n C ) + A = ( A n C ) + B .

Proof . ( A n C ) + B = [ ( A n C ) + ( A n B ) ] + B = [ A n ( C + ( A n B ) ) ] + B = A + B = M . ( B n C ) t - A = [ ( B n C ) + ( A n B ) ] + A = [ B n ( C + ( A n B ) ) ] + A - B + A = M .

Note that Lemma 1.2 is Lemma 2.2 in [R] , and it can also be found in [L, Chapter 1, Lemma 1.11.

Lemma 1.3 Let hf be n module and let A l = A + B, B < C and C / B << A!/B. Then ( A n C ) / ( A n B) << M/(A n B ) .

Proof . Let Al / (AnB) = ( A n C ) / ( A n B ) + X / ( , 4 n B ) with A n B < X 5 121. Then Af = (AnC)+X By Lemma 1.2, M = C t ( A n X ) . Since C / B << M,IB A! = B + ( A 11 X). Again by Lemma 1.2, M = X + ( A n B) . Hence M --- X. Thus ( A n C j / ( A n B ) << M / ( A n B). 0

Lemma 1.4 (1) Let M be a module and let B < C submodules of ild such that C / B is a supplement submodt~le of M / B and B is rz supplement submodule of M. Then C is a supplement submodule of M .

( 2 ) Let M be a weakly supplemented module artd B < C submodules of A4 such that C / B is co-closed in M / B and B is co-closed in M . Then C is co-closed in M .

Proof . (1) Let C / B be a supplement of C'/B in M / B and let B be a supplement of Bf in M. Then hd/B = C/B+C1/B , C / B n C 1 / B <( C/B. Also M = B+Bf and BnBf << B. Clearly BnB' << C. M = (CnC')+B1 and M = C+C1 imply M = C+(BfnC ' ) (Lemma 1.2.). Now, C = Cn(B+B1) = B + ( C n B f ) and ( C n C t ) / B << C / B . By Lemma 1.3, ( C n C ' n B f ) / ( B n B ' ) ex C / ( B n B t ) . Also B n B' << C. Therefore (C n C' n B') << C. Thus C is a supplement of B' n C' in M.

( 2 ) It is clear by Lemma 1.1 and (1).

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Proposition 1.5 Let M be an amply supplemented module. Then every submodule of M has an s-closure.

Proof. Let A < M. Since M is amply supplemented there exists a submodule B of M such t h a t B is minimal with the property M = A+ B. Again since M is arnply supplemented there exists a submodule C of M such tha t C < A, M = C + B and C n B < C. Now we show tha t A / C << hl/C. Let C S X < M . I f M = X + A , t h e n M = C + ( X n B ) + A = A + ( X n B ) . B y the minitnality of B , X n B = B. Therefore M = X , a cont,radiction. Thus n/l # A + X. Hence A/C << M/C. C is co-closed as C is a supplement submodule (Lemma 1.1). So C is a n s-closure of A in M.

Lemma 1.6 Let hf be an R-module. Suppose A and B are submodules of h4. B is an s-closure of A i n M if and only if B is a minimal co-essential submodule of A.

Proof. Let B be an s-closure of A in M. Then AIB << MlB and B is co-closed in M. We show t h a t B is minimal co-essential submodule of A. Let X 5 B and AIX << MIX. Then B/X << MIX. Because B is co-closed in M , B = X . Thus B is a minimal co-essential suhmodule of A. Conversely, let B be a minimal co-essential sul>module of A. Then AlB << M/B. Now we show t h a t B is co-closed in M. Let X 5 B and BIX << M / X . Let MIX = A1.Y + ClX with X _< C 5 M. Then A f = A + C , and so A1 = A + (C .+ B). Since All3 < MIB A4 = C + B. Since B/X < R1l.X lLf = C .t X =. C . So, A / S << M / X . By the ~ninilnality of B , B - X. Therefore B is co-closecl in M. U

Lemma 1.7 Let hl be an R-module. Then M is amply suppkmented if and only if M as weakly supplemented and every submodule of A4 has a7~ s-closure in M .

Proof. T h e necessity is clear from definitions and Proposition 1.5. Conversely, suppose t h a t h4 is weakly supplemented and every subrnodule of M has an s- closure in M. Let A, B 5 M with 114 = A+B. Since M is weakly supplemented there exists a subrnodule C of M such t h a t M = ( A n B ) +C and A n B n C < A1. By Lemma 1.2, M = '4 + ( B n C ) . Let D be a n s-closure of B n C in hi. Then h l = A + D and A n D << M. Since D is co-closed in M A n D D (Lemma 1.1). Thus D is a supplement of A in M and D 5 B. I t follows t h a t M is amply supplemented.

Proposition 1.8 Let M = MI @ M2 be a weakly supplemented module. Sup- pose that for every co-closed submodule N of M such that either M = N + MI or M = N+M2, N is a direct summand of M . Let K be a co-closed submodule of M such that every submodule of M / K has an s-closure i n M/K. Then K is a direct summand of M.

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ON LIFTING MODULES 343 1

Proof. Consider the submodule ( K + M l ) / K of MIK. Then by hypothesis, there exists a co-closed submodule N / K of M / K such that N / K < ( K + M l ) / K and ( K + Ml) /N << M / N . By Lemma 1.4(2), N is co-closed in M. Moreover M = ( K + M I ) + M2 implies that M = N + Mz. By hypothesis, M = N $ N' for some submodule N' of M.

Next K = N n ( K + N') and M = N + ( K + N') . Thus M/K = N / K $

( K + N 1 ) / K , and so ( K + N 1 ) / K is co-closed in M / K . Again by Lemma 1.4(2), K + N' is co-closed in M. Moreover, M = ( K + M I ) + N' = ( K + N') + MI. Therefore K 4- N' is a direct summand of M by hypothesis. There exists a submodule K' of M such that M = ( K + N') % K'. Hence N' = ( K + N') n (N' + K') and N n ( K + N') n (N' + K' ) = K .n (N' + K') = 0. So M = K $ ( N 1 + K ' ) . 0

Corollary 1.9 Let M = M1 $ M2 be an amply supplemented module. Suppose that for every co-closed submodule N of M such that either M = N + MI or M = N + M2, N is a direct summand of A{. Then every co-closed su.bmodule of M 2s a direct summand.

Proof. Let K be a co-closed submodule of A6. It is easy to see that M / K is amply supplemented. By Proposit~on 1.5, every submodule of M / K has an s-closure in A l / K . Therefore K is a direct summand of M hy Proposition 1.8. CJ

2 Lifting modules

Theorem 2.1 Let ill = MI $ hlz be cln amply supplemented ntoduie. Then the followzng statements are equzvalent.

( 1 ) M is lifting.

( 2 ) Every co-closed submodule K of 1Lf such thul either M = K + MI or M = I( + M2 is a direct summand of M .

(3) Every co-closed subrnodule K of M such that either (K+M1)/K <( M/IC or ( K + M 2 ) / K << M / K or M = K+Ml = K + M2 is a direct summand of M.

Proof. (1)-(2) Follows from Lemma 1.1 and Corollary 1.9. (2)*(3) Clear. (3)==+(2) Let K be a co-closed submodule of M such that hl = K + Mz, the case M = K + MI being analogous. Since M / K is amply supplemented ( K + M l ) / K has an s-closure in M I K , that is, there exists a submodule N of M such that K 5 N 5 K + MI, (K+Ml) /N << M/N and N / K is co-closed in M / K . Also, N + Ml = I< + MI and ( N + Ml) /N << M/N. Since N / K is co- closed in M / K and K is co-closed in M N is co-closed in M (Lemma 1.4(2)).

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By hypothesis, there exists a submodule N' of M such that M = N $ N'. We have M = K + M2 = ( K + N') + M2 = ( K + N') + MI. Note that M = K+N1+N and K = ( K + N 1 ) f l N , and so M / K = N / K $ ( K + N r ) / K . Therefore ( K + N1) /K is co-closed in M / K . Again by Lemma. 1.4(2), K + N' is co-closed in M. By hypothesis, K + N' is a direct summand of M. Since K + N' = K $ N' K is a direct summand of M.

Proposition 2.2 Let M = MI $ M2 be a module. Let N be a subrnodule of A4 such that ( N + Ml)/Ml << M/Ml and M = N + M2. Then (IN+ MI) /N << MIN.

Proof. Let N be a subrnodule of M such that ( N -I. M1)/Ml < MIMI and

hi = N+M2. We have M/Ml 2 M2 4 M2/(M2nN) 2 (M2+N)/N = MIN. In the above map (N + Ml)/Ml is mapped onto ( N + Ml) /N. Hence the Proposition follows.

Let MI and he modules. The ~llodule MI is small Mz-projective if every hoinomorphism f : MI 4 h.lz/A, where A is a submodule of M2 and I m f << M2/A, can he lifted to a hoinomorphism cp : M I ---t M2. The modules MI and h/lz are .relatively small projective if Mi is small Mj-projective, for every i , f {1,2), i # j. It is clear that if the module MI is hf2--prqjective then 12.1, is small h12-projective.

Lemma 2.3 Let R be a righ.t hereditary riug. Then an injective right R- module X is small AT-pro3ect.ive for all right R-modules i2.1. In particulnr, the Prufer p-group Z ( p m ) is s m d hi-projective for every Z-module M .

Proof. Let A4 be any right R-module, let N 5 M and let p be any homo- !norphisin from X to M/N such that Imcp < M / N . Since I m p is injective I m p = 0. Therefore p = 0.

Lemma 2.4 Let MI and M2 be modules and M = Ml $ Mz. The following statements are equivalent.

( 1 ) A41 is small M2-projective.

(2) For every submodule N of M such that ( N + Ml) /N << M / N , there exists a submodule N' of N such that M = N' $ Mz.

Proof. (1)*(2) Let N be a submodule of M such that ( N + Ml) /N << M/N. Then M = N + M2. Consider the homomorphism g : Ml -4 MIN, ml H m~ + N and the epimorphism f : MZ -+ MIN, m2 c-) m2 + N. Since Img = ( N i- Ml)/N Img << M/N. So, there exists a homomorphism cp : MI -4 Mz such that f p = g. Define N' = { a - p(a) I a E MI ). Clearly N' is a submodule of N and M = N' &I M2. (2)=+(1) Let A be a submodule of Mz, f : MI -+ M2/A a homomorphism

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such that I m f << M 2 / A and .rr : Mz -+ M 2 j A the natural epimorphism. Define N = { a + b E MI @ M2 1 f ( a ) = -n(b) ,a E Ml,b E M z ) . Clearly A < N and M = N + M2. Let I m f = X / A with X < M2. Consider the homomorphism h : M 2 / A -+ M / N , m 2 + A 6 m2 + N . Then h ( X / A ) = ( X + N ) / N , and so ( X + N ) / N << M / N . Since ( N + M l ) / N 5 ( X + N ) / N ( N + h l l ) / N <( M / N . Hence there exists a submodule N' of N such that M = N' @ M2 by hypothesis. Consider the canonical projection a : N' $

M2 ---+ M2. Then the homomorphism f can be lifted to the homomorphism alM, : M1 -+ M2. Thus MI is small M2-projective. O

The following lemma is well-known from [Wi, 41.14).

Lemma 2.5 Let hll and M2 be modules and M = MI @ M2. The following statements are equivalent.

( 2 ) For every subrnodule N of M such that M = N + M2, there exists a subm,odule N' of N such that A4 = N ' 0

Proposi t ion 2.6 Let A4, and M2 be modzrles and let A 4 = Ail, @ h12 be an amply supplemented module. Then the followmg statements are equzualent.

( 1 ) For every (co-closed) suhmodule N of A4 such that ( N + Ml)/Ml << M/M, and A 4 = N + M2. there exists (1 submodule N' of N such that A i = N' $ hI2.

( 2 ) For every (co-closed) subrnodule N of A4 s.uch that ( N + M l ) / N << M / N , there exists a submodule :V' of N such that M = N' 69 M2.

(3) For every co-closed submodule N of M such that (Ai + MI ) / M I << MIMI a n d M = N + M 2 , M = N $ M z .

( 4 ) MI is small M2-projective.

Proof . (1)*(2) Let N be a submodule of M such that ( N + M l ) / N << M I N . Since M / M 1 is amply supplemented there exists a submodule X of M such that MI < X, MIIW = X/M1+ ( N + M I ) / M I and ( X n ( N + M 1 ) ) / M l << X I M I . Now M = N + X = N + ( X n ( M l + M 2 ) ) = ( N + M 1 ) + ( X n M 2 ) . Since ( N + M l ) j N << M / N M = ( X n M 2 ) + N . By Lemma 1.2, M = M 2 + ( X n N ) . Also, ( N n X ) + MI = X n ( N + M I ) implies that ( ( N n X ) + M 1 ) / M l << X I M I . Therefore ( ( N n X) + Ml) /Ml << M / M l . By hypothesis, there exists a submodule N' of M such that N' 5 N n X and M = N' $ Mz. (2 )=+(1) Clear by Proposition 2.2. (3)*(1) Let N be a submodule of M such that ( N + M l ) / M l << M/Ml and M = N + M2. Since M is amply supplemented M2 has a supplement

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N' such that N' 5 N . Clearly (N' + M l ) / M l < ( N + M l ) / M l , and so ( N ' + M l ) / M l << M / M 1 . B y hypothesis, M = N 1 @ M2. (1)*(3) Let N be a co-closed submodule of M such that ( N + M l ) / M l << M / M 1 and M = N + M2+ By hypothesis, there exists a submodule N' of N such that M = N' $ Mz. We have N = N' @ ( N n M 2 ) Hence ( ( N n hlz) + Mj)/Ml << M / M 1 . Considering the obvious epimorphism M / M l -+ hI2 we get N n M2 < M2 and hence N n Mz < M . Considering the obvious map hl -+ M / N 1 we get ( ( N n M z ) + N f ) / N ' = N / N 1 << M / N 1 . As N is co-closed N' = N. (2)-(4) It is Lemma 2.4. 0

Lemma 2.7 Let MI be any module and M2 a lifting'module and let M = A[, @ M2. If MI is small M2-projective, then every co-closed submodule N of M such that ( N + M l ) / N << M / N is a direct summand.

Proof. Let N he a co-closed suh~i~oclule of M such that ( K + M 1 ) / N << AI /N . B y Lemma 2.4, there exists a submodule N' of N such that M = N' 83 h f 2 . Clearly h l / N 1 is lifting, and since N is ro-closed in h f N /N ' is co-clos~d in M / N 1 . Therefore N / N f is a direct summand of M I N ' . Hence N is a direct stlimnand of 1 U . n

Theorem 2.8 Let A i l and h.12 be l ~ f t f r q modules and let 111 = AI1 tn hI2 he an 1mp1y supplemented module If one of the follow~ng condhons holds, thew h i 1s lzftmg

( I ) Ad, is small Mz-prqjective and every co-closed subnrodnie N of A4 such that h f = N + MI is a direct summand.

(2) MI and hl2 are relatively small projective and every co-closed submodule N of M such that M = N + A41 = N + M2 is a direct summand.

(3) &I2 is M I -projective and MI is small M2-projective.

( 4 ) MI is semisimple and small M2-projective.

Proof. ( 1 ) and ( 2 ) follow from Theorem 2.1 and Lemma 2.7. (3) Let N be a co-closed submodule of M such that M = N + M I . By

Lemma 2.5, there exists a submodule N' of N such that M = N 1 $ M I . Since M / N 1 is lifting and N / N 1 is co-closed in M / N 1 N / N 1 is a direct summand of M / N 1 . Therefore N is a direct summand of M . Now (3 ) follows from (1) .

( 4 ) follows from (3 ) . 0

Corollary 2.9 Let M = MI $. . . $ M,, be a finite direct sum of relatively projective modules Mi. Then M is lifting 2f and only if M is amply supplemented and Mi is lifting for all 1 5 i 5 n.

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Proof . The necessity is clear. Conversely, suppose that M is amply supplemented and all Mi are lifting. By induction on n , it is enough t o prove that M is lifting when n = 2. This follows from Theorem 2.8. 0

Note that Corollary 2.9 dualizes [DHSW, Proposition 7.101.

Corol la ry 2.10 Let R be a right hereditary ring. Let MI and M2 be R- modules such that MI is injective and M = MI @ M2 is amply supplemented. Then M is lifting i f and only if MI and M2 are lifting and every co-closed submodule N of M such that M = N -t MI is a direct summand.

Proof . By Lemma 2.3 and Theorein 2.8(1). O

Let M be a module. Consider the following conditions:

(D2) If N is a submodule of M such that M / N is isomorphic t o a direct summand of M, then N is a direct sulninand of M.

(D3) For every direct summands K , L of M with M = K + L, K I? L is a direct su~ninand of M.

If the module M is lifting and has (D2) ((U3)) then it is called a discrete(qwrr.si- discrete) ~notlule (See, [MWI]).

Lemma 2.11 Let h1, and Af2 be relatzvely project~ve quasi-dzscrete modules Then ,If -= MI $ nilz r s a lzftzng module

Proof. [I;, Theorein 81. 0

Propos i t i on 2.12 Let M = MI 69 M2 where Dll and M2 are relatively projective and M2 is a quasi-discrete module. Let K , L be direct sumrnnnds of M such that M = K + L. Suppose further that M = K + M2. Then h' n L is a direct summand of Ad.

Proof . [I<, Proposition 121. 0

T h e o r e m 2.13 Let M be a module such that M =#MI @ . . . @ h4,, is a finite direct sum of modules Mi ( 1 < i 5 n). Then M is quasi-discrete if and only if M I , . . . , Mn are relatively projective quasi-discrete modules.

Proof . Necessity follows from [MM, Lemmas 4.7 and 4.231. Conversely, suppose that MI, . . . , M, are relatively projective quasi-discrete. By induction on n , it is sufficient t o prove the case n = 2. Thus suppose tha t M = MI $ M2. By Lemma 2.11, M is lifting. Now we prove that M has (D3). Let K , L be direct summands of M with M = K + L. Then M/K is lifting. By Proposition 1.5, there exists a submodule Kl of M containing K such that K1 /K 5 (K+ Ml) /K , K I / K is co-closed in M / K and (K+ Ml)/Kl << M/KI.

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3436 KESKIN

Since M / K is lifting there exists a submodule Kz of M containing K such that M / K = K l / K $ K z / K . Clearly M = K 2 + MI and moreover Kt and K2 are direct summands of M . By Proposition 2.12, Kz n L is a direct summand of .M. On the other hand, (K l + M2) + ( K + M I ) = M implies that Ad = K, + '21,. Again by Proposition 2.12, K n L = Kl n (K2 n L) is a direct summand of M. I t follows that M is quasi-discrete. 0

Note that Theorem 2.13 is dual to [ M M , Corollary 2.141,

Corollary 2.14 A finite direct sum MI @ . . - $ M, is quasi-discrete if and only if Mi $ Mj is quasi-discrete for all 1 5 i < j < n.

Corollary 2.15 [ M M , Corollary 4.501 Let M be a module such that M - MI @ . . . $ M, is a finite direct sum of hollow modules A.li (1 < i 5 n ) . Then M is quasi-discrete if and only i f MI , . . . , M,, are relatively projectzve modules.

Example 2.16 Let M = Z(py) $ . . + @ Z(p;P) where for every 1 5 i < j 5 n p, # P, and for every 1 L 1 5 n. p, zs a p r m e mteger Ther, ;M zs qcinsl-

dzscrefe

Theorem 2.17 Let A1 be n module such that 111 = MI (5 . . . 5) Ad,, is n fintte direct sum o j moddes A4, ( 1 < i < n ) . Then A l is discrete ,if und only i j MI, . . . , A4,, are relntively projectioe discrete modules.

Proof. B y Theorem 2.13 and [Mhl, Theorem 4 15 and Theorem 5 21. 17

Corollary 2.18 A ,finite direct sum MI $ . . . % Ad, is discrete if and oidy if M, 63 Mj is discrete for all 1 5 i < 3 5 n.

3 Lifting modules with summands satisfying the finite exchange property

A module M is said t o have the finite exchange property if, for every finite index set I , whenever M @ N = @icrAi for modules N and Ai, i E I, then M $ N = M @ ($t,lB,) for submodules Bi of Ai, i E I . Discrete modules have this property (See, [MM, Corollary 4.19 and Corollary 5.51). Any direct summand of a module with the finite exchange property has the finite exchange property [ M M , Lemma 3.201.

Lemma 3.1 Let MI and Mz be modules and M = MI @ Mz. Let K be a direct summand of M such that ( K + M l ) / K <( M / K . If I( has the finite exchange property, then there exists a submodule B of M2 such that M = K @ B.

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Proof . Suppose that K has the finite exchange property. Then M = K@A%B for some A 5 hJl and B 5 M2. As ( K + M l ) / K << M / K , we can conclude tha t M = K @ B. 0

The following lemma is known from [SC, Lemma 101. We give the proof for the completeness.

L e m m a 3.2 Let K and K' be modules, let M = K $ K' and let T be a submodule of M with the finite exchange property. If M = N' @ T , for some N' < K'. then K has the finite exchange property.

Proof . Suppose that M = N' @ T , for some N' < K'. Then K' = K' n (N' $ T ) = N' @ (K' n T ) , M = K $ K ' = K $ N' $ (K' n T ) and T = T n ( K @ N' $ (Ii" n T ) ) = (K' n T ) @ ( ( K @ N') n T ) . Thus, it is easy t o see tha t M = ((K $ N') n T ) @ K' and we can conclude that K is isomorphic t o ( K 6 N') n T, which is a direct summand of T. Therefore K has the finite exchange property.

P ropos i t i on 3.3 Let A f l be a module with the finite exchange property and let M2 be any module. .If M = MI 8 M2 is lifting, th,en MI is small M2 -projectiae.

Proof . Let K bt. a co-closed submodule of 111 such that M = I< -t M2 and ( I< + A f , ) / M , (< i l I /Af , . By Proposition 2.2. (K + M 1 ) / K << Ag/I<. Since 111 is lifting ,I1 = K @ K' for some submodule K'. Because All has the finite exchange property, M = M, @ A 3 B, for some A < li, B 5 K'. Since (I< + AI,)/K < Al/K M = K % B , and since (K +- hfl)/Ml << MIA11 Ad= B @ M 1 . So K f = B a n d A = O . Nowwehave M = M1$K1 . ByLemma 3.2, K has the finite exchange property. Therefore there exists a submodule X of M2 such that M = K $ X by Lemma 3.1. So. M = X $ M I . Hence ,Y = i& and M = K @ M2. Thus MI is small M2-projective by Proposition 2.6. 0

T h e o r e m 3.4 Let M1 and M2 be modules with the ,finite exchange property and let A4 = MI % M2 be amply supplemented. Then M is lifting i f and only 2f MI and M2 are lifting and relatively small projective and every co-closed submodule N of M such that M = N + MI = N + M2 is a direct summand of M .

Proof . By Theorem 2.8 and Proposition 3.3. 0

Corol la ry 3.5 Let MI and M2 be discrete modules and let M = MI @ M2 be amply supplemented. Then M is lifting if and only if MI and M2 are relatively small projective and every co-closed submodule N of M such that M = N + MI = N + M2 is a direct summand of M .

Proo f . By Theorem 3.4.

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3438 KESKIN

Lemma 3.6 Let R be any commutative domain and let K be the field of fractions of R. Let V be an9 non-zero vector space over the field K . Then the R-module V has (02).

Proof. By [GW, Proposition 6.121, V is an injective R-module. Note that V is extending by [MM, Proposition 2.11. Let U be any submodule of V such that V/U is isomorphic to a direct summand of V . Then V/U is a torsion-free R-module. Since U is closed in the extending module V U is a direct summand of V. It follows that V has (D2).

Now any direct sum of two discrete modules need not be amply supplemented as we see in the following example.

Example 3.7 Let R be a discrete valuation ring with field of fractions K . Assume that KR is not quasi-projective. Then the R.-module K @ K is not amply supplemented.

Proof. By Lemma 3.6, KR has (D2). Since KR is hollow KH is lifting. and so it is discrete. By Theorem 2.17 and Lemma 3.6, K 6 k' is not lifting. Let B 2 I( @ K and A he a supplement of B in K @ li. Since K gi IC is clivisible Ir's ri' = (K Q) K)re = Ar + BI. < Ar + B, for every 0 # 7, E R. Thus ilr = A, ar!d A is divisible, hence injec,tive. Therefore A is a, direct suminand of K a f i . Hence K @ K is not amply suppleinentecl by [MM, Proposition 4.81.

Theorem 3.8 Let A l l be a semmmple module and let M2 he a module w ~ t h the finzte exchange property, and let M - A l l F7 MZ be a m p l y supplemented. Then M zs lzfting if and only zf A4 zs lzfting and MI u small I\&-projectrue.

Proof. By Theorem 2.8 and Proposition 3.3.

In particular, Theorem 3.8 characterizes when the direct sum of a semisimple module and,a discrete module is lifting.

Corollary 3.9 Let R be a right perfect ring. Let MI be an R-module with the finite exchange property, Mz a projective R-module and M = Ml $ M2. Then M is lifting if and only if MI is lifting and small ib!2-p~ojective.

Proof. By [MM, Theorem 4.411, M is amply supplemented and A12 is discrete. Therefore Mz has the finite exchange property. Now it is clear from Theorem 2.8 and Proposition 3.3. U

4 Rings for which every direct sum of a lifting module and a simple module is lifting

Let R be a ring. Consider the following (*) property:

(*) MI @ M2 is a lifting R-module whenever A l l is a simple .R-module and M2

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ON LIFTlNG MODULES 3439

is a lifting R-module. A module M is called a V-module if every simple module is M-injective.

The ring R is said t o be a right V-ring if RR is a V-module. Let R be a right V-ring and let M be a lifting R-module. Then M is

semisimple since the Jacobson radical of M is equal t o 0 [DHSW, 2.131. There- fore every right V-ring satisfies (*).

Proposition 4.1 Let R satisfy (*). Let ~LI' be a lifting R-module with local endomorphisrn ring. Then A4 is a simple R.-module or M is a local R-module with unique maximal submodule S which zs the socle of M

Proof. Suppose that M is not simple. Let 0 f m E h.1 such that A4 # mR. Let N be any essential submodule of mR. Suppose that N # mR. Let P be a maximal submodule of mR such that N 5 P. Let U = mRjP. Consider the incli~sion mapping f : U --+ M j P and the natural hnmomurphism .ir :

IM -+ M / P . Since R satisfies (*) U 8 hf is lifting. The mapplng f is not an epimorphisrn because M # mR. By [I<, Lemma 11, f can he lifted to a homomorphis~n g : U ---t hl. Since TI/ = ,f # 0 it follows tha.t ,9 f 0, and hence 0 # g ( U ) = V is a simple submodule of M . Clearly V n P = 0 and mR < V 3 P. It follows that nzR = nzR,i\ (V 8 P) = P @ ( m R r l V) and hence mR n I/ $. 0. that is, nzR n V = I/. Hence ml3 = V @ P. But P is essential in mR, so P = mR and 1) = 0, a contradiction. Thus nzR = N for any essential submodule N of mR. Therefore m R is semisimple. Let S denote the socle of M . We have proved that M = mR for all m E M \ S . Hence S is a maximal submoduie of M. Since M is an indecomposable lifting module it is hollow. Hence M is a local R-module.

Corollary 4.2 Let R satisfy (*). Then every lifting R-module with local en- domorphism ring has an essential socle.

A semiperfect ring need not satisfy (*). Consider the right uniserial ring R which is semiperfect with unique composition series 0 C U C V C R. Then R @ V/U is not lifting by [I<, Corollary 21. Thus this ring does not satisfy (*).

Theorem 4.3 Let R satisfy (*). Let M be a lifting R-module. Then every simple R-module is small M-projective.

Proof. Let S be a simple R-module. Then S @ M is lifting. Hence by Propo- sition 3.3, S is small M-projective. 0

Acknowledgments I would like to express my gratefulness t,o the referee and Professor R. Wisbauer for valuable suggestions which improved the pre- sentation of the paper.

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REFERENCES

[DHSW ] N. V. Dung, D. V. Huynh, P. F. Srnith and R. Wishauer, Extend- zng modules, Pitman Research Notes in Mathematics Series, Longman, Harlow, (1994).

[GW ] 1i.R. Goodearl and R.B. Warfield, An introduction to noncommudative Noetherzan rings, London Math. Soc. Student Texts 16, Cambridge Univ. Press, Cambridge, (1989).

[K ) D. Ileskin, Finite direct sums of (Dl)-modules, Turkish Journal of Math. 22(1), (1998), 85-91.

[L 1 C. Lornp, On dm1 Goldie dimension, M.Sc. Thesis, Glasgow University. (1996).

JMM ] S. H. Moharnerl and B. J Muller, Cont~twous and dtscrete rnodtiles, London Math. Soc LNS 147 Camht~dge [Tnlv. Press, C'drnhridg~, (1900)

(33. 1 E. R.eiter, A dual to the Golclie ascending chain conclitim on the direct sum of submodules, Bull. Cal. Math. Soc., 73, (1981), 55-63.

[SC ] C. Santa-Clara, Extending modules with i4ject,ive or sernisimple summands, Journal of Pure and Applied Algebw, 127, (1998), 193-203.

[Wi ] R. Wisbauer, Foundations of module and ring theory, Gordon and Breach, Philadelphia, (1991).

[Z ] H. Zoschinger, Minimax-Moduln, J. Algebra, 102, (1986)) 1-32.

Received: April 1999

Revised: January 2000

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on lifting modules

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