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CS-MODULES AND ANNIHILATOR CONDITIONS
MAHMOUD A. KAMAL and AMANY M. MENSHAWY
Received 4 June 2002
We studyS-R-bimodulesSMR with the annihilator conditionS=lS(A)+lS(B) for any closed submoduleA, and a complementB ofA, inMR. Such annihila-tor condition has a direct connection with the CS-condition for MR. We make use of this to give a new characterization of CS-modules. Bimodules SMR for which rMlS(A)=A(for every closed submoduleAofMR) are also dealt with. Such modules are calledW∗-modules. We give the extra added annihilator condi-tions toW∗-modules to be equivalent to the continuous (quasicontinuous) mod-ules.
2000 Mathematics Subject Classification: 16D80.
1. Introduction. LetRand S be rings and letSMR be a bimodule. For any X≤MandT≤S, writelS(X)= {s∈S:sX=0}andrM(T )= {m∈M:T m= 0}. Letλ:S→End(MR)be the canonical ring homomorphism. For eachs∈S, we identifyλ(s)withs. A submoduleAis essential inM(denoted byA≤eM) ifA∩B≠0 for every nonzero submoduleB ofM. A submoduleA is closed inM if it has no proper essential extensions inM. A≤⊕ M signifies that A is a direct summand of M (or simply a summand). A module M is called a CS-module if every closed submodule ofM is a summand. The moduleM is continuous if it is a CS-module and satisfies condition (C2): ifAB≤M with
A≤⊕M, thenB≤⊕M. A generalization of condition (C2) is (GC2) (see [4]): ifAis a submodule ofMwithAM, thenA≤⊕M. The moduleMis quasicontinuous if it is a CS-module and satisfies condition (C3): ifA,B≤⊕M withA∩B=0,
then A⊕B≤⊕ M. It is known thatM is quasicontinuous if and only ifM =
A⊕BwheneverAandBare complements of each other inM(see [3, Theorem 2.8]).
Camillo et al. [1] have dealt with Ikeda-Nakayama rings that are related to continuous and quasicontinuous rings.
InSection 3, we study the bimodulesSMRwhich satisfy the following con-dition:
S=lS(A)+lS(B) (1.1)
for any two relative complements Aand B inMR. Such modules are clearly quasicontinuous modules, while there are quasicontinuous modules which do not satisfy condition (1.1). For example, considerRas a commutative integral domain with field of quotientsQand letM=Q⊕Q. InLemma 3.2, we give a necessary and sufficient condition for quasicontinuous modules to satisfy condition (1.1). In the case ofS =End(MR), every quasicontinuous module must have condition (1.1). As a generalization of this condition, we introduce the concept of W∗-modules (bimodules SMR for which A=rMlS(A) for ev-ery closed submoduleAofMR). It is clear that any bimodule with condition (1.1) is aW∗-module, while in general the converse is not true.Proposition 3.8 indicates when aW∗-module satisfies condition (1.1).
InSection 4, we discuss the equivalence betweenW∗-modules and contin-uous (quasicontincontin-uous) modules over an arbitrary ringS. Then we draw the consequences whenS is the endomorphism ring ofMR.
2. CS-modules and annihilator conditions. The proofs of the lemmas and propositions, presented in this section, are adaptations of the arguments in [4].
Lemma2.1. LetSMRbe a bimodule. If for every closed submoduleAofMR there exists a complementBofAinMRsuch thatS=lS(A)+lS(B), thenMRis a CS-module.
Proof. Let A be a closed submodule of MR. Then by assumption there exists a complementB of A in MR such thatS =lS(A)+lS(B). Write 1S = u+v, where u∈lS(A)and v∈lS(B). It follows thata=vafor all a∈A, b=ubfor allb∈B, andvB=uA=0. ThusB⊆rM(v)⊆rM(v2)andrM(v2)∩ A=0. SinceB is a complement ofA in MR, we have B =rM(v)=rM(v2).
Similarly, A=rM(u)=rM(u2). Now we show that(vu)M =0. Let vum= a+b, wherem∈M, a∈A, andb∈B. Noting that vu=uv, we have that (v2u2)m=(vu)(a+b)=0. Henceu2m∈rM(v2)=rM(v), and this gives that
u2vm=vu2m=0. Thenvm∈r
M(u2)=rM(u); and thusvum=uvm=0. So(vu)M∩(A+B)=0. SinceA+Bis essential inMR,(vu)M=0. SouM⊆ rM(v)=B and vM⊆rM(u)=Aand hence M=vM+uM=A+B=A⊕B. ThereforeAis a summand ofMR.
Remark 2.2. The converse of Lemma 2.1is not true. For example, there are torsion-free CS-modules over commutative integral domains, which do not satisfy the given condition inLemma 2.1.
Lemma2.3. LetSMRbe a bimodule, whereSMis faithful, and letMR=A⊕B. If the projectionfofMontoAalongBis given byf (m)=smfor somes∈S, and allm∈M, thenS=lS(A)+lS(B).
For any submodulesAand BofMR and anyt∈S, defineαt :A+B→M, a+b→ta(see [4]).
Proposition2.4. LetSMRbe a bimodule such thatSM is faithful. The fol-lowing are equivalent:
(1) MR is CS and for anyf2=f ∈End(MR), there existss ∈S such that f (m)=sm, for allm∈MR;
(2) for every closed submoduleAofMR, there exists a complementBofAin MRsuch thatS=lS(A)+lS(B);
(3) for every closed submoduleAofMR, there exists a complementBofAin MRsuch thatS=lS(A)⊕lS(B);
(4) for every closed submoduleAofMR, there exists a complementBofAin MRsuch that for everyt∈S, the diagram
0 A+B M
M
αt (2.1)
can be extended byλ(s), for somes∈S.
Proof. (1)⇒(2). Let A be a closed submodule of MR. Since MR is a CS-module, there existsf2=f∈End(MR)such thatA=f M. By (1), there exists
s∈S such thatf (m)=sm, for allm∈MR. Hence(s2−s)M=(f2−f )M=0. SinceSMis faithful, it follows thatsis an idempotent inS. Now we have
lS(A)=lS(f M)=lS(sM)=lS(s)=S(1−S). (2.2)
Similarly,lS(B)=SS, whereB=:(1−f )M. ThusS=lS(A)+lS(B).
(2)⇒(1). It is clear byLemma 2.1thatMRis CS. Now letf2=f∈End(MR),
and denoteA=f (M). By (2), there exists a complementB ofA inMR such thatS =lS(A)+lS(B). The argument of the proof ofLemma 2.1shows that M=A⊕B. Letπbe the projection ofMontoAalongB. Then
lS(A)=lS(πM)= {s∈S:sπ=0} (2.3)
(by consideringsthe homomorphism given by left multiplication bys) and
Let 1=s+s, wheres∈lS(A)ands∈lS(B). Thussπ=0 ands(1−π)=0. It follows that 0=s(1−π)=(1−s)(1−π)=1−π−s. Thereforef (m)=
π(m)=smfor allm∈M.
(2)⇒(3). From the argument in the proof ofLemma 2.1, we haveM=A⊕B. SinceSM is faithful, we have 0=lS(M)=lS(A+B)=lS(A)lS(B)and hence S=lS(A)⊕lS(B).
(3)⇒(4). LetAbe a closed submodule ofMR. By (3), there exists a complement B of A such that S =lS(A)⊕lS(B). Write t=u+v, where u∈ lS(A) and v∈lS(B). Thenαt(a+b)=ta=(u+v)a=va=v(a+b)=λ(v)(a+b).
(4)⇒(2). LetAbe a closed submodule ofMR. By (4), there exists a complement BofAinMRsatisfying diagram (2.1). By (4), there existss∈S such thatλ(s) extends αt. Thus, for alla∈A and b∈B, ta=αt(a+b)=λ(s)(a+b)= s(a+b). It follows that(1−s)a+(−s)b =0, for all a∈A and b ∈B. So 1−s∈lS(A) and −s ∈lS(B)and hence 1 =(1−s)−(−s)∈ lS(A)+lS(B). ThereforeS=lS(A)+lS(B).
Corollary2.5. The following are equivalent for a bimoduleSMRwithS= End(MR):
(1) MRis a CS-module;
(2) for every closed submoduleAofMR, there exists a complementBofAin MRsuch thatS=lS(A)+lS(B);
(3) for every closed submoduleAofMR, there exists a complementBofAin MRsuch thatS=lS(A)⊕lS(B);
(4) for every closed submoduleAofMR, there exists a complementB ofA inMRsuch that for everyt∈S, diagram (2.1) can be extended by some g:M→M.
Proposition2.6. LetSbe the center ofEnd(MR). The following are equiv-alent:
(1) for every closed submoduleAofMR, there exists a complementBofAin MRsuch thatS=lS(A)+lS(B);
(2) MRis CS and every idempotent ofEnd(MR)is central; (3) MRis CS and every closed submodule ofMRis fully invariant.
Proof. (1)(2) byProposition 2.4.
(2)⇒(3). LetAbe a closed submodule ofM. By CS,Ais a direct summand of MR. ThenA=f (M)for somef2=f∈End(MR). For anyg∈EndR(M), since
fis central by (2),g(A)=g(f (M))=f (g(M))⊆f (M)=A. This shows thatA is a fully invariant submodule ofM.
(3)⇒(2). Letf ,g∈EndR(M)withf2=f. Thereforef (M)is a closed
3. Condition (1.1) and its generalizations. The next lemma is clear.
Lemma3.1. The following are equivalent for a bimoduleSMR:
(1) S=lS(A)+lS(B)for any two relative complementsAandBofMR; (2) for any submodulesAandBofMRwithA∩B=0,S=lS(A)+lS(B).
We say that a bimodule SMR has condition (1.1) if it satisfies one of the equivalent conditions ofLemma 3.1.
The next lemma follows from [4, Lemma 3].
Lemma3.2. LetSMRbe a bimodule such thatSMis faithful. Then the follow-ing are equivalent:
(1) Mhas condition (1.1);
(2) Mis quasicontinuous and every idempotent inEnd(MR)is a left multipli-cation by an element ofS.
Remark3.3[4, Theorem 8]. In the case of S=End(MR), it is clear from Lemma 3.2that an R-moduleMis quasicontinuous if and only ifMhas condi-tion (1.1).
Proposition 3.4. Let SMR be a bimodule which satisfies condition (1.1). ThenA=rMlS(A)for all closed submodulesAofMR.
Proof. LetAbe a closed submodule ofMRandBa submodule ofrMlS(A) such thatA∩B=0. By Zorn’s lemma, there exists a complementC ofA in MRwithB⊆C. By condition (1.1), we haveS =lS(A)+lS(C)⊆lS(A)+lS(B), soS=lS(A)+lS(B). SincelS(A)=lSrMlS(A)≤lS(B), it follows thatS=lS(B) and henceB=0. This shows thatA≤erMlS(A). SinceAis a closed submodule ofMR, we haveA=rMlS(A).
A bimoduleSMRis called aW∗-module ifA=rMlS(A)for every closed sub-moduleAofMR. It is clear byProposition 3.4that every bimoduleSMR with condition (1.1) is aW∗-module. But there are bimodules which areW∗-modules and do not satisfy condition (1.1). For example, letS=R=F F
0F
, whereF is any field and letM=RRR. It is clear thatMisW∗-module. ButMRis not qua-sicontinuous, and henceMdoes not satisfy condition (1.1).
Lemma3.5. The following are equivalent for a bimoduleSMR: (1) A≤er
MlS(A)for all submodulesAofMR; (2) SMRis aW∗-module.
Proof. (1)⇒(2). This implication is obvious.
(2)⇒(1). LetAbe a submodule ofMRandCa maximal essential extension of AinMR. We have by(2)thatA≤eC=rMlS(C). SincerMlS(A)≤rMlS(C), we haveA≤erMlS(A).
Proof. LetTbe a maximal left ideal ofS. SinceT⊆lSrM(T ), we have either lSrM(T )=T orlSrM(T )=S. IflSrM(T )=S, thenrM(T )=0. If lSrM(T )=T, letN be a nonzero submodule ofrM(T ). ThenT=lSrM(T )⊆lS(N)⊆S, and the maximality ofT yieldsT=lS(N). It follows thatrM(T )=rMlS(N). Since MisW∗-module, we have byLemma 3.5thatN≤erM(T ). ThereforerM(T )is uniform.
Corollary3.7. LetSMRbe aW∗-module, where every maximal left ideal of Sis a left annihilator. ThenrM(T )is uniform for every maximal left idealTofS.
Proof. LetTbe a maximal left ideal ofS. FromProposition 3.6, it is enough to show thatrM(T )≠0. LetrM(T )=0. By assumption,T=lSrM(T )=lS(0)=S, which contradicts the maximality ofT.
Proposition3.8. The following are equivalent for a bimoduleSMR: (1) SMR is aW∗-module andlS(A)+lS(B)is a left annihilator for any two
relative complementsAandBinMR; (2) SMRhas condition (1.1).
Proof. (1)⇒(2). LetAandBbe two relative complements inMR. Then by (1), S=lS(0)=lS(A∩B)=lS(rMlS(A)∩rMlS(B))=lSrM(lS(A)+lS(B))=lS(A)+ lS(B). ThereforeMhas condition (1.1).
(2)⇒(1). This implication is obvious.
4. The relation between W∗-modules and (quasi-) continuous modules. The following is an immediate consequence ofProposition 3.8.
Proposition4.1. LetSMR be a bimodule withS=End(MR). Then the fol-lowing are equivalent:
(1) SMR is aW∗-module andlS(A)+lS(B)is a left annihilator for any two
relative complementsAandBofMR; (2) MRis quasicontinuous.
Proposition 4.2. LetSMR be a bimodule, whereSM is faithful. Then the following are equivalent:
(1) SMRis aW∗-module,lS(A)+lS(B)is an annihilator for any two relative complementsAandBofMR, andMRhas GC2;
(2) MR is a continuous module and every idempotent inEnd(MR)is a left multiplication by an element ofS.
Proof. (1)⇒(2). We have byProposition 3.8 that MR has condition (1.1). Therefore, byLemma 3.2,MR is a quasicontinuous module. Lets∈End(MR) be a monomorphism, withsM≤eM. By GC2it follows thatsM=M. Then by [3,
Lemma 3.14],MRis a continuous module. The rest of the proof of (2) follows fromLemma 3.2.
Corollary4.3. LetSMRbe a bimodule withS=End(MR). Then the follow-ing are equivalent:
(1) SMRis aW∗-module,lS(A)+lS(B)is an annihilator for any two relative complementsAandBofMR, andMRhas GC2;
(2) MRis a continuous module.
In particular, ifMRis of finite uniform dimension, thenSis semiperfect.
Proof. It is clear that every monomorphismf ∈End(MR) is an isomor-phism (due to GC2andMof finite uniform dimension). Hence,Msatisfies the
assumptions in Camps and Dicks [2, Theorem 5], and so End(MR)is semilocal. Therefore by using [3, Proposition 3.5 and Lemma 3.7], idempotents ofS/J(S) lift to idempotents ofS, and thusSis semiperfect.
Lemma4.4. LetSMR be a bimodule such that every finitely generated left ideal ofSis a left annihilator of a subset ofMR, and every closed submodule of MRis a right annihilator of a finite subset ofS. ThenMhas condition (1.1).
Proof. LetA1and A2be complements of each other inMR. Then by as-sumption, we haveAi=rM(Yi)for some finite subsetsYi of S. Again by as-sumption,SYi=lS(Ki)for some subsets Ki in MR, where i=1,2. Now S = lS(A1∩A2)=lS(rM(Y1)∩rM(Y2))=lSrM(SY1+SY2)=SY1+SY2(due to the
as-sumption and sinceSY1+SY2is finitely generated). HenceS=lS(K1)+lS(K2)=
lSrMlS(K1)+lSrMlS(K2)=lSrM(Y1)+lSrM(Y2)=lS(A1)+lS(A2). ThereforeM
satisfies condition (1.1).
Lemma4.5. LetSMRbe a bimodule and let every idempotent inEnd(MR)be a left multiplication by an element ofS. IfMRis a CS-module, then every closed submodule ofMRis a right annihilator of a finite subset ofS.
Proof. LetAbe a closed submodule ofMR. Then by CS, there existsf2= f∈End(MR)such that A=rM(1−f )= {m∈M:(1−s)m=0} =rM(1−s), where(1−s)∈S.
The following corollary is an immediate consequence of Lemmas4.4and4.5.
Corollary 4.6. Let SMR be a bimodule, where S =End(MR). Let every finitely generated left ideal of S be a left annihilator of a subset of M. Then the following are equivalent:
(1) every closed submodule ofMis a right annihilator of a finite subset ofS; (2) Mis a CS-module.
Theorem4.7. LetSMRbe a bimodule, whereS=End(MR). Let every finitely generated left ideal ofSbe a left annihilator of a subset ofM. Then the following are equivalent:
Proof. By Lemmas 4.4and 4.5, we have that M has condition (1.1) . By Remark 3.3,Mis quasicontinuous. To show thatMis continuous, by [3, Lemma 3.14], it is enough to show that every essential monomorphisms∈S is an isomorphism. Lets∈S be a monomorphism, withsM≤eM. By assumption, SS =lS(X)for some subsetX ofM. It follows thatX=0 and henceSS=s. Thensis a split monomorphism, and thereforesM=M.
References
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Uni-versity Press, Cambridge, 1990.
[4] R. Wisbauer, M. F. Yousif, and Y. Zhou,Ikeda-Nakayama modules, Beiträge Algebra Geom.43(2002), no. 1, 111–119.
Mahmoud A. Kamal: Department of Mathematics, Faculty of Education, Ain Shams University, Roxy, Cairo, Egypt
E-mail address:[email protected]
