GENERALIZED LIFTING MODULES
YONGDUO WANG AND NANQING DING
Received 6 March 2006; Accepted 12 March 2006
We introduce the concepts of lifting modules and (quasi-)discrete modules relative to a given left module. We also introduce the notion of SSRS-modules. It is shown that (1) if Mis an amply supplemented module and 0→N→N→N→0 an exact sequence, then MisN-lifting if and only if it isN-lifting andN-lifting; (2) ifMis a Noetherian module, thenMis lifting if and only ifMisR-lifting if and only ifMis an amply supplemented SSRS-module; and (3) letMbe an amply supplemented SSRS-module such that Rad(M) is finitely generated, thenM=K⊕K, whereK is a radical module andK is a lifting module.
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
1. Introduction and preliminaries
Extending modules and their generalizations have been studied by many authors (see [2,3,8,7]). The motivation of the present discussion is from [2,8], where the concepts of extending modules and (quasi-)continuous modules with respect to a given module and CESS-modules were studied, respectively. In this paper, we introduce the concepts of lifting modules and (quasi-)discrete modules relative to a given module and SSRS-modules. It is shown that (1) if 0→N→N→N→0 is an exact sequence andM an amply supplemented module, thenMisN-lifting if and only if it is bothN-lifting and N-lifting; (2) ifMis a Noetherian module, thenMis lifting if and only ifMisR-lifting if and only ifM is an amply supplemented SSRS-module; and (3) let M be an amply supplemented SSRS-module such that Rad(M) is finitely generated, thenM=K⊕K, whereKis a radical module andKis a lifting module.
Throughout this paper,Ris an associative ring with identity and all modules are unital leftR-modules. We useN≤Mto indicate thatNis a submodule ofM. As usual, Rad(M) and Soc(M) stand for the Jacobson radical and the socle of a moduleM, respectively.
LetMbe a module andS≤M.Sis called small inM(notationSM) ifM=S+T for any proper submoduleTofM. LetNandLbe submodules ofM,Nis called a supple-ment ofLinMifN+L=M, andNis minimal with respect to this property. Equivalently,
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International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 47390, Pages1–9
M=N+LandN∩LN.N is called a supplement submodule ifN is a supplement of some submodule ofM.M is called an amply supplemented module if for any two sub-modulesAandB of M withA+B=M,Bcontains a supplement of A.M is called a weakly supplemented module (see [5]) if for each submoduleAofMthere exists a sub-moduleBofM such thatM=A+B andA∩BM. LetB≤A≤M. IfA/BM/B, thenBis called a coessential submodule ofAandAis called a coessential extension ofBin M. A submoduleAofM is called coclosed ifAhas no proper coessential submodules in M. Following [5],Bis called ans-closure ofAinMifBis a coessential submodule ofA andBis coclosed inM.
LetMbe a module.Mis called a lifting module (or satisfies (D1)) (see [9]) if for
ev-ery submoduleA of M, there exists a direct summand K of M such thatK≤A and A/KM/K, equivalently,Mis amply supplemented and every supplement submodule ofMis a direct summand.Mis called discrete ifMis lifting and has the following condi-tion.
(D2) IfA≤M such thatM/Ais isomorphic to a direct summand ofM, thenAis a
summand ofM.
Mis called quasidiscrete ifMis lifting and has the following condition:
(D3) For each pair of direct summandsAandBofMwithA+B=M,A∩Bis a direct
summand ofM. For more details on these concepts, see [9]. Lemma 1.1 (see [12, 19.3]). LetMbe a module andK≤L≤M.
(1)LMif and only ifKMandL/KM/K.
(2) IfMis a module andφ:M→Ma homomorphism, thenφ(L)Mwhenever LM.
Lemma 1.2 (see Lemma 1.1 in [5]). LetMbe a weakly supplemented module andN≤M. Then the following statements are equivalent.
(1)Nis a supplement submodule ofM. (2)Nis coclosed inM.
(3) For allX≤N,XMimpliesXN.
Lemma 1.3 (see Proposition 1.5 in [5]). LetMbe an amply supplemented module. Then every submodule ofMhas ans-closure.
Lemma 1.4 (see [12, 41.7]). LetMbe an amply supplemented module. Then every coclosed submodule ofMis amply supplemented.
2. Relative lifting modules
To define the concepts of relative lifting and (quasi-)discrete modules, we dualize the concepts of relative extending and (quasi-)continuous modules introduced in [8] in this section. We start with the following.
LetNandMbe modules. We define the family
$(N,M)=
A≤M| ∃X≤N,∃f ∈Hom(X,M), f(X)A fM(X)
Proposition 2.1. $(N,M) is closed under small submodules, isomorphic images, and co-essential extensions.
Proof. We only show that $(N,M) is closed under coessential extensions. LetA∈$(N,M), A≤A≤M, andA/AM/A. There existX≤Nandf ∈Hom(X,M) such thatf(X)≤ A andA/ f(X)M/ f(X) sinceA∈$(N,M). Note that A/AM/A, so A/ f(X)
M/ f(X) byLemma 1.1(1). ThusA∈$(N,M).
Lemma 2.2. LetA∈$(N,M) andAbe coclosed inM. ThenB∈$(N,M) for any submodule BofA.
Proof. There existX≤Nandf∈Hom(X,M) such thatf(X)≤AandA/ f(X)M/ f(X) by hypothesis. SinceAis coclosed inM, f(X)=A. LetBbe any submodule ofAand Y= f−1(B)≤X≤N. Then f|
Y:Y→Mis a homomorphism such that f|Y(Y)=Bfor
f(X)=A. ClearlyB/ f|Y(Y)M/ f|Y(Y). ThereforeB∈$(N,M). Lemma 2.3. LetC≤A≤B≤MandAbe a coessential submodule ofB. IfCis ans-closure ofA, then it is also ans-closure ofB.
Proof. It is clear byLemma 1.1(1).
Proposition 2.4. LetMbe an amply supplemented module. Then everyAin $(N,M) has ans-closureAin $(N,M).
Proof. SinceA∈$(N,M), there existX≤N and f ∈Hom(X,M) such thatA/ f(X) M/ f(X). Note thatMis amply supplemented, and so f(X) has ans-closureAinMby
Lemma 1.3. Thus A is also ans-closure of A byLemma 2.3. The verification forA∈ $(N,M) is analogous to that forB∈$(N,M) inLemma 2.2.
LetNbe a module. Consider the following conditions for a moduleM.
($(N,M)-D1) For every submoduleA∈$(N,M), there exists a direct summandKofM
such thatK≤AandA/KM/K.
($(N,M)-D2) IfA∈$(N,M) such thatM/Ais isomorphic to a direct summand ofM,
thenAis a direct summand ofM.
($(N,M)-D3) IfAandLare direct summands ofM withA∈$(N,M) andA+L=M,
thenA∩Lis a direct summand ofM.
Definition 2.5. LetN be a module. A moduleM is said to beN-lifting,N-discrete, or N-quasidiscrete ifM satisfies $(N,M)-D1, $(N,M)-D1 and $(N,M)-D2or $(N,M)-D1
and $(N,M)-D3, respectively.
One easily obtains the hierarchy:MisN-discrete⇒MisN-quasidiscrete⇒Mis N-lifting. Clearly, the notion of relative discreteness generalizes the concept of discreteness. For any moduleN, lifting modules areN-lifting. But the converse is not true as shown in the following examples.
((1, 2)is a supplement of(1, 1)) and is not a direct summand of it though it is amply supplemented.
Example 2.7. LetM be a module with zero socle andSa simple module. ThenMis S-lifting since $(S,M) is a family only containing all small submodules ofM. So all torsion-freeZ-modules areS-lifting for any simpleZ-moduleS(see [12, Exercise 21.17]). In par-ticular, ZZand ZQareS-lifting for any simpleZ-module, but each one is not a lifting
module.
Lemma 2.8. LetMbe a module. Then $(M,M)= {A|A≤M} =N∈R- Mod$(N,M), where
R-Mod denotes the category of leftR-module.
Proof. It is straight forward.
Proposition 2.9. LetMbe a module. ThenMis lifting or (quasi-)discrete if and only ifM isM-lifting orM-(quasi-)discrete if and only ifMisN-lifting orN-(quasi-)discrete for any moduleN.
Proof. It is clear byLemma 2.8.
Proposition 2.10. LetMbe an amply supplemented module. Then the condition $(N, M)-D1is inherited by coclosed submodules ofM.
Proof. LetMsatisfy $(N,M)-D1andHbe a coclosed submodule ofM.His amply
sup-plemented byLemma 1.4. For anyA∈$(N,H),Ahas ans-closureA∈$(N,H) inHby
Proposition 2.4. SinceA∈$(N,H)⊆$(N,M) andM satisfies $(N,M)-D1, there is a
di-rect summandKofMsuch thatK≤AandA/KM/K. ByLemma 1.2,A/KH/K. NowA=KsinceAis coclosed inH. ThusHsatisfies $(N,H)-D1.
Corollary 2.11. LetMbe an amply supplemented module. Then the condition $(N, M)-D1is inherited by direct summands ofM.
Proposition 2.12. LetMbe an amply supplemented module. Then $(N,M)-Di(i=2, 3)
is inherited by direct summands ofM.
Proof. (1) LetMsatisfy $(N,M)-D2andHbe a direct summand ofM. We will show that
Hsatisfies $(N,H)-D2.
LetA∈$(N,H)⊆$(N,M) andH/Ais isomorphic to a direct summand ofH. Since H is a direct summand ofM, there existsH≤Msuch thatM=H⊕H. ThusM/A= (H⊕H)/A(H/A)⊕H, and soM/Ais isomorphic to a direct summand ofM.Ais a direct summand ofMsinceMsatisfies $(N,M)-D2, and henceAis a direct summand of
H.
(2) LetA∈$(N,H)⊆$(N,M) andA,Lbe direct summands ofH withA+L=H. We will show that A∩L is a direct summand ofH. SinceH is a direct summand of M, there existsH≤M such thatM=H⊕H. ThusM=(A+L)⊕H=A+ (L⊕H). NowA∩(L⊕H) is a direct summand ofM sinceM satisfies $(N,M)-D
3. Note that
Proof. The proof follows fromLemma 2.2,Corollary 2.11, andProposition 2.12. Proposition 2.14. Let 0→N→N→N→0 be an exact sequence. Then $(N,M)∪ $(N,M)⊆$(N,M). Therefore, ifM isN-lifting (resp., (quasi-)discrete), thenM isN -lifting andN-lifting (resp., (quasi-)discrete).
Proof. Without loss of generality we can assume thatN≤N andN=N/N. By def-inition,N≤Nimplies $(N,M)⊆$(N,M). Next, letA
2∈$(N,M). Then there exist
X≤N=N/N and f ∈Hom(X,M) such thatA2/ f(X)M/ f(X). WriteX=Y/N, Y≤N and letδ:Y →Y/Nbe the canonical homomorphism. It is clear thatg= f δ∈ Hom(Y,M) andg(Y)=f(X), henceA2/g(Y)M/g(Y). ThusA2∈$(N,M). Therefore
$(N,M)∪$(N,M)⊆$(N,M). The rest is obvious.
Dual to [8, Proposition 2.7], we have the following.
Theorem 2.15. Let 0→N→N→N→0 be an exact sequence andMan amply supple-mented module. ThenMisN-lifting if and only if it is bothN-lifting andN-lifting. Proof. LetMbeN-lifting. Then it is bothN-lifting andN-lifting byProposition 2.14. Conversely suppose thatM is bothN-lifting and N-lifting. For any submoduleA∈ $(N,M),Ahas ans-closure A∈$(N,M) byProposition 2.4. SinceA∈$(N,M), there existX≤Nand f ∈Hom(X,M) such thatA/ f(X)M/ f(X). SinceAis coclosed inM, f(X)=A. WriteY =X∩N≤Nand f|Y:Y→M is a homomorphism, then f(Y)≤ f(X)=A. Let f(Y) be ans-closure of f(Y) inA(forAis amply supplemented). Thus we conclude that f(Y)/ f(Y)M/ f(Y) and f(Y)∈$(N,M). SinceM isN-lifting, there exists a direct summandKofMsuch that f(Y)/KM/K. It is easy to seef(Y) is coclosed inM, hencef(Y)=Kis a direct summand ofM. WriteM=f(Y)⊕K,K≤M and A=A∩M= f(Y)⊕(A∩K). Defineh:W=(X+N)/N→M byh(x+N)= π f(x), whereπ:A→A∩Kdenotes the canonical projection. It is clear thath(W)= A∩K, thus (A∩K)/h(W)M/h(W), and hence (A∩K)∈$(N,M). SinceM is N-lifting, there exists a direct summandK of M such that (A∩K)/KM/K. SinceA∩Kis coclosed in M,A∩K=K. NowA∩K is a direct summand ofK. ThusAis a direct summand ofM. It follows thatMisN-lifting. Corollary 2.16. Let M be an amply supplemented module. If M is Ni-lifting for i= 1, 2,...,nandN=niNi, thenMisN-lifting.
Corollary 2.17. LetMbe an amply supplemented module. ThenMis lifting if and only ifMisN-lifting andM/N-lifting for every submoduleNofMif and only ifMisN-lifting andM/N-lifting for some submoduleNofM.
Recall that a moduleM is said to be distributive ifN∩(K+L)=(N∩K) + (N∩L) for all submodulesN,K,LofM. A moduleMhas SSP (see [4]) if the sum of any pair of direct summands ofMis a direct summand ofM.
Proof. We only need to show thatMsatisfies $(N,M)-D3whenMsatisfies $(N,M)-D3
and $(N,M)-D
3byTheorem 2.15. LetA∈$(N,M) andA,Hbe direct summands ofM
withA+H=M. We know thatA=A1⊕A2, whereA1∈$(N,M),A2∈$(N,M) from
the proof ofTheorem 2.15. SinceMis a distributive module with SSP,A1∩HandA2∩H
are direct summands ofM. This implies thatA∩H is a direct summand ofM. ThusM
satisfies $(N,M)-D3.
3. SSRS-modules
In [2], a module is called a CESS-module if every complement with essential socle is a direct summand. As a dual of CESS-modules, the concept of SSRS-modules is given in this section. It is proven that: (1) letM be an amply supplemented SSRS-module such that Rad(M) is finitely generated, thenM=K⊕K, whereKis a radical module andK is a lifting module; (2) letMbe a finitely generated amply supplemented module, thenM is an SSRS-module if and only ifM/Kis a lifting module for every coclosed submodule KofM.
Definition 3.1. A module is called an SSRS-module if every supplement with small radical is a direct summand.
Lifting modules are SSRS-modules, but the converse is not true. For example,ZZis an
SSRS-module which is not a lifting module.
Proposition 3.2. LetMbe an SSRS-module. Then any direct summand ofMis an SSRS-module.
Proof. LetKbe a direct summand ofM andNa supplement submodule ofK such that Rad(N)N. Let N be a supplement of Lin K, that is, N+L=K and N∩LN. Since K is a direct summand ofM, there existsK≤M such thatM=K⊕K. Note thatM=N+ (L⊕K) andN∩(L⊕K)=N∩LN. ThereforeN is a supplement of L⊕KinM. ThusNis a direct summand ofMsinceM is an SSRS-module. SoNis a
direct summand ofK. The proof is complete.
Proposition 3.3. LetM be a weakly supplemented SSRS-module andK a coclosed sub-module ofM. ThenKis an SSRS-module.
Proof. It follows from the assumption and [4, Lemma 2.6(3)].
Proposition 3.4. LetMbe an amply supplemented module. ThenMis an SSRS-module if and only if for every submoduleNwith small radical, there exists a direct summandKofM such thatK≤NandN/KM/K.
Proof. “⇐.” LetNbe a supplement submodule with small radical. By assumption, there exists a direct summandKofMsuch thatK≤N andN/KM/K. SinceNis coclosed inM,N=K. ThusNis a direct summand ofM.
Rad(N)NandNis a supplement submodule byLemma 1.2. ThereforeNis a direct
summand ofMby assumption. This completes the proof.
Corollary 3.5. LetM be an amply supplemented SSRS-module. Then every simple sub-module ofMis either a direct summand or a small submodule ofM.
Proposition 3.6. LetMbe an amply supplemented module. ThenMis an SSRS-module if and only if for every submoduleNofM, everys-closure ofNwith small radical is a lifting module and a direct summand ofM.
Proof. It is straight forward.
Proposition 3.7. Let M be an amply supplemented SSRS-module. ThenM=K⊕K, whereKis semisimple andKhas small socle.
Proof. For Soc(M), there exists a direct summandK ofMsuch that Soc(M)/KM/K byProposition 3.4. It is easy to see thatKis semisimple. SinceKis a direct summand of M, there existsK≤Msuch thatM=K⊕K. Note that Soc(M)=Soc(K)⊕Soc(K). So Soc(M)/K=(K⊕Soc(K))/KM/K=(K⊕K)/K. Thus Soc(K)K. Recall that a moduleMis called a radical module if Rad(M)=M. Dual to [2, Theorem 2.6], we have the following.
Theorem 3.8. LetMbe an amply supplemented SSRS-module such that Rad(M) is finitely generated. ThenM=K⊕K, whereKis a radical module andKis a lifting module. Proof. Rad(Rad(M))Rad(M) since Rad(M) is finitely generated. There exists a di-rect summandK of M such that Rad(M)/K M/K byProposition 3.4. SinceK is a direct summand ofM, there existsK≤Msuch thatM=K⊕K. Note that Rad(M)= Rad(K)⊕Rad(K). ThereforeK=K∩Rad(M)=Rad(K) and Rad(M)/K =(Rad(K)⊕ Rad(K))/KM/K=(K⊕K)/K. Thus Rad(K)=Kand Rad(K)K.
Next, we show thatKis a lifting module.Kis amply supplemented since it is a direct summand ofM. So we only prove that every supplement submodule ofKis a direct summand ofK. LetNbe a supplement submodule ofK. ByLemma 1.2and Rad(K) K, we know that Rad(N)N.Nis a direct summand ofKsinceKis an SSRS-module
byProposition 3.2. The proof is complete.
Corollary 3.9. LetMbe an amply supplemented module with small radical. ThenM is an SSRS-module if and only ifMis a lifting module.
Theorem 3.10. LetM be a finitely generated amply supplemented module. Then the fol-lowing statements are equivalent.
(1)Mis an SSRS-module. (2)Mis a lifting module.
(3)M/Kis a lifting module for every coclosed submoduleKofM. Proof. (1)⇔(2) follows fromCorollary 3.9.
(3)⇒(1) is clear.
M/KandKis coclosed inM. Rad(A)AsinceMis finitely generated andAis coclosed inM.A is a direct summand ofM by assumption. ThusA/K is a direct summand of
M/K.
Lemma 3.11. LetMbe a module. Then the following statements are equivalent.
(1) For every cyclic submoduleNofM, there exists a direct summandKofMsuch that K≤NandN/KM/K.
(2) For every finitely generated submoduleNofM, there exists a direct summandKof Msuch thatK≤NandN/KM/K.
Proof. See [12, 41.13].
Corollary 3.12. LetMbe a Noetherian module. Then the following statements are equiv-alent.
(1)MisR-lifting.
(2)MisF-lifting, for any free moduleF. (3)Mis lifting.
(4)Mis an amply supplemented SSRS-module.
Proof. It is easy to see that $(R,M) and $(F,M) are closed under cyclic submodules. The rest follows immediately fromTheorem 3.10andLemma 3.11. Corollary 3.13. LetRbe a left perfect (semiperfect) ring. Then every SSRS-module (finitely generated SSRS-module) is a lifting module.
Proof. It follows from the fact that every module over a left perfect ring has small radical,
[11, Theorems 1.6 and 1.7] andCorollary 3.9.
A moduleMis uniserial (see [6]) if its submodules are linearly ordered by inclusion and it is serial if it is a direct sum of uniserial submodules. A ringRis right (left) serial if the right (left)R-moduleRR(RR) is serial and it is serial if it is both right and left serial.
Corollary 3.14. The following statements are equivalent for a ringRwith radicalJ. (1)Ris an artinian serial ring andJ2=0.
(2)Ris a left semiperfct ring and every finitely generated module is an SSRS-module. (3)Ris a left perfect ring and every module is an SSRS-module.
Proof. It holds by [6, Theorem 3.15], [10, Theorem 1 and Proposition 2.13], and
Corol-lary 3.13.
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Yongduo Wang: Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou 730050, China
E-mail address:[email protected]
Nanqing Ding: Department of Mathematics, Nanjing University, Nanjing 210093, China