PII. S016117120320346X http://ijmms.hindawi.com © Hindawi Publishing Corp.
ON SOME PROPERTIES OF
⊕
-SUPPLEMENTED MODULES
A. IDELHADJ and R. TRIBAK
Received 15 March 2002
A moduleMis⊕-supplemented if every submodule ofMhas a supplement which is a direct summand ofM. In this paper, we show that a quotient of a⊕-supplemented module is not in general⊕-supplemented. We prove that over a commutative ring R, every finitely generated⊕-supplementedR-moduleMhaving dual Goldie dimen-sion less than or equal to three is a direct sum of local modules. It is also shown that a ringRis semisimple if and only if the class of⊕-supplementedR-modules coincides with the class of injectiveR-modules. The structure of⊕-supplemented modules over a commutative principal ideal ring is completely determined.
2000 Mathematics Subject Classification: 16D50, 16D60, 13E05, 13E15, 16L60, 16P20, 16D80.
1. Introduction. All rings considered in this paper will be associative with an identity element. Unless otherwise mentioned, all modules will be left uni-tary modules. LetRbe a ring andManR-module. LetAandPbe submodules ofM. The submodulePis called asupplementofAif it is minimal with respect to the propertyA+P=M. AnyL≤M which is the supplement of anN≤M
will be called asupplement submoduleofM. If every submoduleU ofMhas a supplement in M, we callM complemented. In [25, page 331], Zöschinger shows that over a discrete valuation ringR, every complementedR-module satisfies the following property(P): every submodule has a supplement which is a direct summand. He also remarked in [25, page 333] that every module of the formM(R/a1)× ··· ×(R/an), whereR is a commutative local ring andai(1≤i≤n)are ideals ofR, satisfies(P). In [12, page 95], Mohamed and Müller called a module⊕-supplemented if it satisfies property(P).
On the other hand, letU and V be submodules of a module M. The sub-moduleVis called a complement ofUinMifVis maximal with respect to the propertyV∩U=0. In [17] Smith and Tercan investigate the following property which they called(C11): every submodule ofMhas a complement which is a
direct summand ofM. So, it was natural to introduce a dual notion of(C11)
which we called(D11)(see [6,7]). It turns out that modules satisfying(D11)
are exactly the⊕-supplemented modules. A moduleM is called a completely
⊕-supplemented (see [5]) (orsatisfies (D+
11)in our terminology, see [6, 7]) if
Our paper is divided into four sections. The purpose ofSection 2is to an-swer the following natural question: is any factor module of a⊕-supplemented module⊕-supplemented? Some relevant counterexamples are given.
InSection 3we prove that, over a commutative ring, every finitely generated
⊕-supplemented module having dual Goldie dimension less than or equal to three is a direct sum of local modules.
Section 4describes the structure of⊕-supplemented modules over commu-tative principal ideal rings.
In the last section we determine the class of ringsRwith the property that every⊕-supplementedR-module is injective. These turn out to be the class of all left NoetherianV-rings (Proposition 5.3). It is also shown that a ringR
is semisimple if and only if the class of⊕-supplementedR-modules coincides with the class of injectiveR-modules (Proposition 5.5).
For an arbitrary moduleM, we will denote by Rad(M)the Jacobson radical ofM. The injective hull ofMwill be denoted by E(M). The annihilator ofM will be denoted by AnnR(M). A submoduleAofMis calledsmallinM(AM) ifA+B≠Mfor any proper submoduleBofM. A nonzero moduleHis called
hollow if every proper submodule is small inH and is calledlocalif the sum of all its proper submodules is also a proper submodule. We notice that a local module is just a cyclic hollow module.
2. Quotients of⊕-supplemented modules. By [23, corollary on page 45], every factor module of a complemented module is complemented. Now, let
Mbe a⊕-supplemented module. In this section we will answer the following natural question: is any factor module ofM⊕-supplemented?
First, we mention the following result, which we will use frequently in the sequel.
Proposition2.1[6, Proposition 1]. The following are equivalent for a mod-uleM:
(i) Mis⊕-supplemented;
(ii) for any submoduleNofM, there exists a direct summandKofMsuch thatM=N+KandN∩Kis small inK.
A commutative ringRis a valuation ring if it satisfies one of the following three equivalent conditions:
(i) for any two elementsaandb, eitheradividesborbdividesa; (ii) the ideals ofRare linearly ordered by inclusion;
(iii) Ris a local ring and every finitely generated ideal is principal.
Example2.2. LetRbe a commutative local ring which is not a valuation ring and letn≥2. By [21, Theorem 2], there exists a finitely presented indecompos-able moduleM=R(n)/Kwhich cannot be generated by fewer thannelements. By [6, Corollary 1],R(n)is⊕-supplemented. HoweverMis not⊕-supplemented [6, Proposition 2].
Thedual Goldie dimension of anR-module, denoted by corank(RM), was introduced by Varadarajan in [19]. IfM=0, the corank ofMis defined as 0. LetM≠0 andkan integer greater than or equal to one. If there is an epimor-phismf:M→ki=1Ni, where eachNi≠0, we say that the corank(RM)≥k.
If corank(RM)≥kand corank(RM)k+1, then we define corank(RM)=k. If the corank(RM)≥kfor everyk≥1, we say that the corank(RM)= ∞. It was shown in [14,19] that the corank(RM) <∞if and only if there is an epimor-phismf:M→ki=1Hi, whereHiis hollow and ker(f )is small inM.
As in [20], a moduleMhas theexchange propertyif for any moduleG, where
G=M ⊕C= ⊕i∈IDi (2.1)
withM M, there are submodulesDi≤Disuch thatG=M ⊕(⊕i∈IDi). Before proceeding any further, we consider another example (note that the module considered is decomposable).
Example2.3. LetR be a commutative local ring which is not a valuation ring. Letaandbbe elements ofR, neither of them divides the other. By taking a suitable quotient ring, we may assume(a)∩(b)=0 andam=bm=0, where
mis the maximal ideal ofR. LetFbe a free module with generatorsx1,x2, and
x3. LetKbe the submodule generated byax1−bx2and letM=F/K. Thus,
M=Rx1⊕Rx2⊕Rx3
Rax1−bx2
=Rx1+Rx2
⊕Rx3. (2.2)
Suppose that M is⊕-supplemented. There exist submodulesHand N ofM such thatM=H⊕N,Rx1+N=M, andRx1∩Nis small inN(Proposition 2.1).
By the proof of [21, Theorem 2],Rx1+Rx2is an indecomposable module which
cannot be generated by fewer than 2 elements. Thus corank(Rx1+Rx2)=2
by [14, Proposition 1.7]. Hence corank(M)=3. Since HM/N and M/N Rx1/(N∩Rx1), we get that H is a local direct summand of M and hence
corank(N)=2 (see [14, Corollary 1.9]). SinceR is a commutative local ring, EndR(Rx3)is a local ring by [4, Theorem 4.1]. Since Rx3 has the exchange
property [20, Proposition 1], there are submodulesH ≤HandN ≤N such thatM=Rx3⊕H ⊕N. ThereforeRx1+Rx2H ⊕N. ThusH ⊕N is
inde-composable. HenceN =0 orH =0. But corank(M)=3 and corank(N)=2, soM=Rx3⊕NandNRx1+Rx2is indecomposable. Sincex1,x2∈M, there
areα,β∈Randy1,y2∈Nsuch thatx1=αx3+y1andx2=βx3+y2. Hence
x1−αx3∈Nandx2−βx3∈N. ButM=Rx3⊕[R(x1−αx3)+R(x2−βx3)].
there exists α ∈R such that x3−α x1∈N. Note thatαx1−ααx3∈N
and (1−αα)x3∈ N∩Rx3. Thus (1−αα)x3= 0, that is, (1−αα)x3 ∈
R(ax1−bx2). Hence 1−αα=0. Soαis invertible andα−1=α. Note that
ax1−αx3
−bx2−βx3
=bβ−aαx3. (2.3)
Thusa(x1−αx3)−b(x2−βx3)≠0. Otherwise,(bβ−aα)x3∈R(ax1−bx2),
which givesbβ=aαand thena=bβα, which is a contradiction. Since(bβ− aα)x3∈N∩Rx3, thenN∩Rx3≠0, which is a contradiction. It follows that
Mis not⊕-supplemented. ButRx1⊕Rx2⊕Rx3is completely⊕-supplemented
[6, Corollary 2].
These examples show that a factor module of a⊕-supplemented module is not in general⊕-supplemented.
Proposition 2.5 deals with a special case of factor modules of ⊕ -supple-mented modules. First we prove the following lemma.
Lemma2.4. LetMbe a nonzero module and letUbe a submodule ofMsuch thatf (U)≤Ufor eachf∈EndR(M). IfM=M1⊕M2, thenU=U∩M1⊕U∩M2.
Proof. Letπi:M→Mi(i=1,2) denote the canonical projections. Letx be an element of U. Thenx=π1(x)+π2(x). By hypothesis,πi(U)≤U for i=1,2. Thusπi(x)∈U∩Mifori=1,2. HenceU≤U∩M1⊕U∩M2. It follows
thatU=U∩M1⊕U∩M2.
Proposition 2.5. LetM be a nonzero module and let U be a submodule ofM such thatf (U)≤Ufor eachf∈EndR(M). IfM is⊕-supplemented, then M/U is⊕-supplemented. If, moreover,U is a direct summand ofM, thenU is also⊕-supplemented.
Proof. Suppose thatM is ⊕-supplemented. LetL be a submodule of M
which containsU. There exist submodulesNandN ofMsuch thatM=N⊕N,
M=L+N, andL∩N is small inN (Proposition 2.1). By [23, Lemma 1.2(d)],
(N+U)/U is a supplement ofL/UinM/U. Now applyLemma 2.4to get that
U=U∩N⊕U∩N. Thus,
(N+U)∩(N +U)≤(N+U+N)∩U+(N+U+U)∩N. (2.4)
Hence,
(N+U)∩(N +U)≤U+(N+U∩N+U∩N)∩N. (2.5)
It follows that (N+U)∩(N +U) ≤U and ((N+U)/U)⊕((N +U)/U)= M/U. Then(N+U)/U is a direct summand ofM/U. Consequently,M/U is
⊕-supplemented.
M=K⊕K,M=V+K, andV∩KK(Proposition 2.1). ThusU=V+U∩K. ButU=U∩K⊕U∩K (Lemma 2.4), henceU∩K is a direct summand ofU. Moreover,V∩(U∩K)=V∩Kis small inK. Then,V∩(U∩K)is small inU∩K
by [23, Lemma 1.1(b)]. ThereforeU∩Kis a supplement ofV inUand it is a direct summand ofU. ThusUis⊕-supplemented.
Corollary2.6. LetM be anR-module and P(M)the sum of all its radi-cal submodules. IfM is⊕-supplemented, thenM/P(M)is ⊕-supplemented. If, moreover,P(M)is a direct summand ofM, thenP(M)is also⊕-supplemented.
Proof. ByProposition 2.5, it suffices to prove that f (P(M))≤P(M)for eachf∈EndR(M). LetNbe a radical submodule ofM and letf be an endo-morphism ofMandgits restriction toN. By [1, Proposition 9.14],g(Rad(N))≤ Rad(f (N)). But Rad(N) =N and f (N) =g(N), hence f (N)≤ Rad(f (N)). Thus, Rad(f (N))=f (N). This implies that f (N)≤P(M), and the corollary is proved.
We recall that a moduleMis called semi-Artinian if every nonzero quotient module ofMhas nonzero socle. For a moduleRM, we define
Sa(M)=
U≤M Usemi-Artinian
U. (2.6)
By [18, Chapter VIII, Section 2, Corollary 2.2], ifRis a left Noetherian ring and
RM a semi-Artinian left R-module, thenM is the sum of its submodules of finite length.
IfRis a commutative Noetherian ring andMis anR-module, then Sa(M)= L(M), the sum of all Artinian submodules ofM.
Corollary2.7. LetM be a ⊕-supplementedR-module. ThenM/Sa(M)is ⊕-supplemented. If, moreover,Sa(M)is a direct summand ofM, thenSa(M)is also⊕-supplemented.
Proof. ByProposition 2.5, it suffices to prove thatf (Sa(M))≤Sa(M)for eachf∈EndR(M). LetU be a semi-Artinian submodule ofMand letf be an endomorphism ofMandgits restriction toU. ThusU/Ker(g)g(U). Hence
f (U)U/Ker(g). But it is easy to check thatU/Ker(g) is a semi-Artinian module. Therefore,f (U)is semi-Artinian.
Remark2.8. LetMbe a⊕-supplemented module. It is clear thatM/Rad(M)
andM/Soc(M)are also⊕-supplemented (seeProposition 2.5and [1, Proposi-tions 9.14 and 9.8]).
3. Some properties of finitely generated ⊕-supplemented modules. A module M is called supplemented if for any two submodulesA and B with
The proof of the next result is taken from [6, Lemma 2], but is given for the sake of completeness.
Lemma3.1. LetMbe a⊕-supplementedR-module. IfMcontains a maximal submodule, thenMcontains a local direct summand.
Proof. LetLbe a maximal submodule ofM. SinceM is⊕-supplemented, there exists a direct summandK ofM such that Kis a supplement ofL in
M. Then for any proper submoduleX ofK, Xis contained inLsinceLis a maximal submodule andL+Xis a proper submodule ofMby minimality of
K. HenceX≤L∩KandXis small inKby [12, Lemma 4.5]. ThusKis a hollow module, and the lemma is proved.
Proposition 3.2. If M is a⊕-supplemented module such that Rad(M) is small inM, thenMcan be written as an irredundant sum of local direct sum-mands ofM.
Proof. Since Rad(M)is small inM,Mcontains a maximal submodule and henceMcontains a local direct summand byLemma 3.1. LetNbe the sum of all local direct summands ofM. IfNis a proper submodule ofM, then there exists a maximal submoduleLofMsuch thatN≤L(see [8, Proposition 9 and Theorem 8]). LetPbe a direct summand ofMsuch thatP is a supplement of
LinM. Note thatPis a local module (see the proof ofLemma 3.1) and hence it is contained inN, soM=L+P≤L+N=L. This is a contradiction. Hence we haveN=M. Now letM=i∈ILiwhere eachLiis a local direct summand ofM. Then,
M
Rad(M)=
i∈I L
i+Rad(M) Rad(M)
(3.1)
and each
Li+Rad(M) Rad(M)
Li
Li∩Rad(M) (3.2)
is simple by [23, Lemma 1.1(c)]. Hence
M
Rad(M)=
k∈K L
k+Rad(M) Rad(M)
(3.3)
for some subsetK⊆I. ThusM=k∈KLksince Rad(M)is small inM. Clearly, the sumk∈KLkis irredundant.
Proof. ByProposition 3.2,M=H1+H2+···+Hn, where eachHiis a local direct summand ofMand the sumni=1Hi is irredundant. By [16, Corollary
4.6],Mis supplemented. Thereforen=corank(M)by [14, Proposition 1.7] and [19, Lemma 2.36 and Theorem 2.39].
Remark3.4. (i) The moduleM=(Rx1+Rx2)⊕Rx3inExample 2.3is not
⊕-supplemented. On the other hand,Mcan be written as follows:M=(Rx1+
Rx2)⊕R(x1−x3);M=(Rx1+Rx2)⊕R(x2−x3); andM=R(x1−x3)+R(x2−
x3)+Rx3. ThereforeMis an irredundant sum of local direct summands ofM.
However,Mis not⊕-supplemented.
(ii) In the same example, we have thatK=Rx1+Rx2is an indecomposable
direct summand of
M=Rx1⊕Rx2⊕Rx3
Rax1−bx2
=Rx1+Rx2⊕Rx3. (3.4)
ThenK is not an irredundant sum of local direct summands. This example shows that, in general, a direct summand of a module which is written as an irredundant sum of local direct summands does not have the same property.
Proposition 3.5. Let M be a finitely generated ⊕-supplemented module such thatk=corank(M)≤2. ThenMis a direct sum of local modules.
Proof. It is clear that ifk=1, then M is a local module. Now suppose thatk=2. SinceMis⊕-supplemented,Mcontains a local direct summandH
(Lemma 3.1). LetKbe a submodule ofMsuch thatM=H⊕K. By [14, Corollary 1.9], we have corank(K)=1 and henceKis a local module (see [19, Proposition 1.11]). ThusMis a direct sum of local modules, as required.
Our next objective is to prove that over a commutative ring, ifMis a finitely generated⊕-supplemented module with corank(M)=3, thenMis a direct sum of local modules. We first prove the following generalization of [11, Lemma 2.3].
Lemma3.6. LetL1,L2,...,Lnbe indecomposable direct summands of a mod-uleMsuch thatEndR(Li)is a local ring for eachi(1≤i≤n). IfLiLjfor all i≠j, thenni=1Liis direct and is a direct summand ofM.
Proof. We use induction overn. Assume thatL1+L2+···+Ln−1is a direct
sum and is a direct summand ofMand letL=L1⊕L2⊕···⊕Ln−1. There exists
every indecomposable direct summand ofL is isomorphic to one of theLi, 1≤i≤n−1. It follows thatLnis isomorphic to one of theLi, 1≤i≤n−1, which is a contradiction. ThereforeL =Land M=Ln⊕L⊕N, that is,M= L1⊕L2⊕···⊕Ln−1⊕Ln⊕N , and the lemma is proved.
Corollary3.7. Suppose thatR is commutative or left Noetherian. LetL1,
L2,...,Lnbe hollow local direct summands of a moduleM. IfLiLjfor alli≠j, thenni=1Liis direct and is a direct summand ofM.
Proof. This is a consequence of [4, Theorems 4.1 and 4.2] andLemma 3.6.
Proposition3.8. Suppose thatRis a commutative ring. LetMbe a finitely generated⊕-supplemented module such that all the hollow direct summands of Mare isomorphic. ThenMis a direct sum of hollow local modules.
Proof. ByProposition 3.2, we can writeM=H1+H2+ ··· +Hnas an ir-redundant sum of hollow local direct summands. By hypothesis,H1H2
··· Hn. Thus,
AnnRH1=AnnRH2= ··· =AnnRHn. (3.5)
Hence,
AnnR(M)= n
i=1
AnnRHi=AnnRHi for eachi (1≤i≤n). (3.6)
Therefore all hollow local direct summands ofMare isomorphic toR/I, where
I=AnnR(M). LetHbe a local submodule ofMsuch thatHis not small inM. SinceMis⊕-supplemented, there exist submodulesNandN ofMsuch that
H+N=M,N ⊕N=M, andH∩N is small inN (Proposition 2.1). It follows thatN M/NH/(H∩N). Hence,N is a local module. This implies that AnnR(N)=I and AnnR(H/(H∩N))=I. Thus, the set{r∈R|r x∈N} =I, whereH=Rx. Lety∈H∩N. There existsα∈Rwithy=αx. Soα∈I and hencey=0 sinceI⊆AnnR(H). ThereforeH∩N=0 andM=H⊕N. It follows that every nonsmall local submodule ofM is a direct summand ofM. Note that corank(M) <∞(Corollary 3.3). Applying [23, corollary on page 45] and [8, Proposition 9], we get thatMis a direct sum of local modules.
Corollary3.9. LetR be a commutative ring andM a finitely generated ⊕-supplemented module withcorank(M)=3. ThenM is a direct sum of local modules.
Proof. LetF0be an irredundant set of representatives of the local direct
summands ofM(F0is not empty byLemma 3.1). ByCorollary 3.7, Card(F0)≤3.
If Card(F0)=3, then M is a direct sum of local modules (Corollary 3.7). If
M=L1⊕L2⊕L3(Corollary 3.7). But corank(M)=3. Therefore corank(L3)=1
(see [14, Corollary 1.9]) and henceL3is a local module. If Card(F0)=1, then
Mis a direct sum of local modules byProposition 3.8.
Remark3.10. (i) IfMis a finitely generated⊕-supplemented module with corank(M)≤2, thenM is completely⊕-supplemented (see [6, Proposition 6] andProposition 3.5).
(ii) IfR is a commutative ring andM a finitely generated⊕-supplemented module with corank(M)=3, thenM is completely ⊕-supplemented (see [6, Corollary 6] andCorollary 3.9).
4. ⊕-supplemented modules over commutative principal ideal rings. In this section, the structure of⊕-supplemented modules over a principal ideal ring is completely determined.
LetR be a commutative Noetherian ring. LetΩbe the set of all maximal ideals ofR. As in [24, page 53], ifm∈ΩandMis anR-module, we denote the
m-local component ofMbyKm(M)= {x∈M|x=0orthe only maximal ideal over AnnR(x)ism}. We callM m-local ifKm(M)=Mor, equivalently, ifmis the only maximal ideal over eachp∈Ass(M). In this case,mis anRm-module by the following operation:(r /s)x:=r x withx=sx (r∈R,s∈R\m). The submodules ofMoverRand overRmare identical.
ForK(M)= {x∈M|Rxis complemented}, we always have a decomposi-tionK(M)= ⊕m∈ΩKm(M)and for a complemented moduleM, we haveM= K(M)[24, Theorems 2.3 and 2.5].
A principal ideal ring is calledspecial if it has only one prime idealp≠R
andpis nilpotent [22, page 245].
Theorem4.1. LetRbe a commutative local principal ideal ring (not neces-sarily a domain) with maximal idealm.
(i)Ifmis nilpotent, then everyR-module is⊕-supplemented.
(ii)Ifmis not nilpotent, thenRis a domain andRMis a⊕-supplemented R-module if and only ifMRa⊕Qb⊕(Q/R)c⊕B(1,...,n), whereQis the quotient field ofRandB(1,...,n)denotes the direct sum of arbitrarily many copies of R/m,...,R/mn, for some positive integern.
Proof. (i) Suppose that m is nilpotent. By [1, Theorem 15.20], R is an Artinian principal ideal ring. Thus, everyR-module is⊕-supplemented by [7, Theorem 1.1].
(ii) Suppose thatmis not nilpotent. ThenRis not a special principal ideal ring. By [22, Chapter IV, Section 15, Theorem 33],Ris a principal ideal domain and the result follows from [12, Proposition A.7].
The proof of the following result can be found in [7, Proposition 2.1].
(i) Mis⊕-supplemented;
(ii) M=K(M)andKm(M)is⊕-supplemented for allm∈Ω.
Corollary4.3. LetRbe a commutative principal ideal ring (not necessarily a domain) andManR-module. The following conditions are equivalent:
(i) Mis⊕-supplemented;
(ii) (1) the ringR/pis local for allp∈Ass(M);
(2) ifm∈Ωsuch thatmRmis not nilpotent, thenKm(M)Ram⊕Q(Rm)b ⊕[Q(Rm)/Rm]c⊕Bm(1,...,nm)(inMod-Rm), whereQ(Rm)is the quotient field ofRmandBm(1,...,nm)denotes the direct sum of ar-bitrarily many copies ofRm/mRm,...,Rm/(mRm)nm, for some pos-itive integernm.
Proof. SeeProposition 4.2, [13, Proposition 2.2(b)], andTheorem 4.1.
Proposition4.4(see [7, Corollary 2.2]). LetRbe a commutative Noetherian ring andManR-module. The following assertions are equivalent:
(i) Mis completely⊕-supplemented;
(ii) M=K(M)andKm(M)is completely⊕-supplemented for allm∈Ω.
Corollary4.5. LetRbe a commutative principal ideal ring (not necessarily a domain) andM anR-module. ThenMis⊕-supplemented if and only ifM is completely⊕-supplemented.
Proof. ByProposition 4.4and the proof ofTheorem 4.1, it suffices to prove the result for anR-moduleMover a local principal ideal domainRwith maxi-mal idealm≠0. IfMis⊕-supplemented, thenMRa⊕Qb⊕(Q/R)c⊕B(1,...,
n), whereQis the quotient field ofRandB(1,...,n)denotes the direct sum of arbitrarily many copies ofR/m,...,R/mn(Theorem 4.1). By [7, Theorem 2.1], Qb⊕(Q/R)c and Ra⊕B(1,...,n)both are⊕-supplemented. By [6, Corollary 2],Ra⊕B(1,...,n)is completely⊕-supplemented. Now consider the module Qb⊕(Q/R)c. SinceQandQ/Rare injective, EndR(Q)and EndR(Q/R)are local rings (see [1, Lemma 25.4]). By [1, Corollary 12.7] and [12, Proposition A.7],Qb⊕
(Q/R)cis completely⊕-supplemented. HenceQb⊕(Q/R)c⊕Ra⊕B(1,...,n)is completely⊕-supplemented (see [7, Corollary 2.1]).
5. Some rings whose modules are⊕-supplemented. A ringR is called a
leftV-ring if every simple leftR-module is injective. The ring Ris called an
SSI-ring if every semisimple leftR-module is injective.
Lemma5.1. LetMbe a module withRad(M)=0. ThenMis⊕-supplemented if and only ifMis semisimple.
Proof. This is clear by [19, Proposition 3.3].
Proof. By [3, page 236, Theorem (Villamayor)], for every left R-module, Rad(M)=0. Therefore, every⊕-supplementedR-module is semisimple (Lemma 5.1).
Proposition5.3. LetRbe a ring. The following statements are equivalent:
(i) every⊕-supplementedR-module is injective;
(ii) Ris a left NoetherianV-ring.
Proof. (i)⇒(ii). Since every semisimpleR-module is⊕-supplemented, every semisimpleR-module is injective. ThusRis anSSI-ring. By [3, Proposition 1],
Ris a left NoetherianV-ring.
(ii)⇒(i). LetM be a ⊕-supplementedR-module. SinceR is a left V-ring,M
is semisimple (Corollary 5.2). ThusM is an injectiveR-module (see [3, Prop-osition 1]).
Corollary5.4. LetRbe a commutative ring. The following are equivalent:
(i) every⊕-supplementedR-module is injective;
(ii) Ris semisimple.
Proof. (i)⇒(ii). It is a consequence of Proposition 5.3 and [3, page 236, Proposition 1 and its first corollary].
(ii)⇒(i) This application is obvious.
Proposition5.5. The following assertions are equivalent for a ringR:
(i) for everyR-moduleM,Mis⊕-supplemented if and only ifMis injective;
(ii) Ris semisimple.
Proof. (i)⇒(ii). Suppose thatRsatisfies the stated condition. ByProposition 5.3,Ris a left NoetherianV-ring. Now, letM be an injectiveR-module. Then
Mis⊕-supplemented and, sinceRis aV-ring,Mis semisimple (Corollary 5.2). ThereforeRis a semisimple ring.
(ii)⇒(i). It is easy to show that everyR-module is⊕-supplemented and every
R-module is injective.
Remark5.6. IfRis a commutative local Noetherian ring having an injective hollow radicalR-moduleH, then theR-moduleM=H(N)is injective. However Mis not⊕-supplemented (see [7, Remark 2.1(3)]). For example, ifRis a local Dedekind domain with quotient field K, then K(N) is an injective R-module which is not⊕-supplemented.
Our next objective is to determine the class of commutative Noetherian rings
Rwith the property that every injectiveR-module is⊕-supplemented. First we prove the following lemma.
Proof. By [10, Theorem 15.9], every injectiveR-module is projective. Since
Ris left perfect, every projectiveR-module is⊕-supplemented (see [6, Propo-sition 13]) and the result is proved.
Proposition5.8. For a commutative Noetherian ringR, the following state-ments are equivalent:
(i) every injectiveR-module is⊕-supplemented;
(ii) Ris Artinian andE(R/m)is a localR-module for each maximal idealm ofR;
(iii) Ris Artinian andR/Imis a quasi-Frobenius ring for each maximal ideal mofR, whereIm=AnnR(E(R/m)).
Proof. (i)⇒(ii). By [15, page 53, corollary of Theorem 2.32] and [10, Corol-lary 3.86], it suffices to prove thatE(R/p)is a finitely generatedR-module for each prime idealp ofR. SinceE(R/p)is indecomposable (see [15, page 53, corollary of Theorem 2.32]) andE(R/p)is⊕-supplemented,E(R/p)is hollow [6, Proposition 2]. ByRemark 5.6,E(R/p)is not radical. Thus,E(R/p)is a local
R-module.
(ii)⇒(iii). Letmbe a maximal ideal ofR. SinceE(R/m)is a localR-module,
E(R/m) R/Im where Im =AnnR(E(R/m)). Thus, R/Im is an injectiveR -module. By [9, Theorem 203],R/Imis an injective(R/Im)-module, that is, the ring R/Im is self-injective. Since R/Im is an Artinian ring, R/Im is a quasi-Frobenius ring, and the result is proved.
(iii)⇒(i). LetMbe an injectiveR-module. By [15, Theorem 4.5], we can write
M = ⊕i∈IE(R/mi)where the mi are maximal ideals ofR. Now, E(R/mi) is an(R/Imi)-module and the(R/Imi)-submodules ofE(R/mi)are the same as
theR-submodules ofE(R/mi), thereforeR(E(R/mi))is⊕-supplemented (see Lemma 5.7and [9, Theorem 203]). By [6, Proposition 2],E(R/mi)(i∈I) is a hollowR-module. By [1, Corollary 15.21], Rad(E(R/mi))is small inE(R/mi). Thus,E(R/mi)(i∈I) is a localR-module. It follows by [1, Corollary 15.21] and [6, Corollary 2] thatMis⊕-supplemented.
Proposition5.9. Letpbe a prime ideal of a commutative Noetherian ring Rsuch thatE(R/p)is hollow. Then there is a maximal idealmofRsuch that
(i) mis the only maximal ideal overp;
(ii) E(R/p)has the structure of anRm-module;
(iii) the submodules ofE(R/p)overRand overRmare identical.
Moreover, as anRm-module,E(R/p)is isomorphic to an injective envelope of Rm/S−1pwhereS=R\m.
Proof. Suppose thatE(R/p) is hollow. Since [13, Proposition 1.1] gives that E(R/p)is m-local for somem∈Ω, mis the only maximal ideal over
E(R/p)is indecomposable as anR-module and itsRm-submodules are also R-submodules so thatE(R/p)is also indecomposable as anRm-module. Since AssR(E(R/p))= {p}, there is an elementx∈E(R/p)such that AnnR(x)=p. But it is easy to check that AnnRm(x)=S−1pwithS=R\mandS−1pis a prime
ideal ofRm. ThenE(R/p)is isomorphic to an injective envelope ofRm/S−1p by [15, page 53, Corollary of Theorem 2.32].
Proposition5.10. Letpbe a prime ideal of a commutative Noetherian ring R. Then the following are equivalent:
(i) E(R/p)is hollow local;
(ii) pis maximal andRpis a quasi-Frobenius ring.
Proof. (i)⇒(ii). Suppose thatE(R/p) is hollow local. By Proposition 5.9,
E(R/p) is m-local for some maximal ideal m of R and as an Rm-module, E(Rm/S−1p)is hollow local, where S =R\m. SinceRm is Noetherian local, Rm is Artinian by [9, Theorem 207]. HenceS−1p is a maximal ideal ofRm. ThusS−1p=S−1m. Thereforep=mis maximal. Moreover, by [15, page 47,
Corollary 2], AnnRm(E(Rm/S−1m))=0. Then E(Rm/S−1m)Rm. So Rm is
self-injective. ThereforeRmis a quasi-Frobenius ring.
(ii)⇒(i). Suppose that p is maximal and Rp is a quasi-Frobenius ring. Put E=E(R/p). By [15, Proposition 4.23],E(R/p)=∞n=1AnnE(pn). ThenEisp
-local. ThusE is anRp-module and the submodules ofE overR and overRp are identical. The proof ofProposition 5.9shows that, as anRp-module,E is isomorphic toE(Rp/pRp), wherepRpdenotes the unique maximal ideal ofRp. On the other hand, sinceRpis a self-injective Artinian local ring,E(Rp/pRp), as anRp-module, is isomorphic toRp(see [10, Theorem 15.27]). HenceE(Rp/pRp) is a localRp-module. Consequently,Eis a localR-module.
Lemma 5.11. Let R be a commutative ring. If R is Noetherian and Rm is quasi-Frobenius for every maximal idealmofR, thenRis quasi-Frobenius.
Proof. Letmbe a maximal ideal ofR. SinceRm is quasi-Frobenius, then Rmis Artinian and somRm, the maximal ideal ofRm, is a minimal prime ideal. Thereforemis a minimal prime ideal ofR. The ringRis Noetherian and every prime ideal is maximal, henceRis Artinian. LetR=R1× ··· ×Rt where each Riis Artinian and local. Since eachRiis a localization ofR, thenRiis quasi-Frobenius for eachi=1,...,t. It is not difficult to see that a finite product of rings is quasi-Frobenius if and only if each factor is quasi-Frobenius (see [10, Theorem 15.27]). HenceR=R1×···×Rtis quasi-Frobenius.
Theorem5.12. For a commutative Noetherian ringR, the following state-ments are equivalent:
(i) every injectiveR-module is⊕-supplemented;
Proof. (i)⇒(ii). It is a consequence of Propositions5.8and5.10. (ii)⇒(iii). It is clear byLemma 5.11.
(iii)⇒(i). SeeLemma 5.7.
Proposition5.13. For aV-ring, the following statements are equivalent:
(i) Ris semisimple;
(ii) everyR-module is⊕-supplemented.
Proof. (i)⇒(ii). It is obvious.
(ii)⇒(i). Suppose that everyR-module is⊕-supplemented. ByCorollary 5.2, everyR-module is semisimple. ThusRis semisimple, as required.
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A. Idelhadj: Département de Mathématiques, Faculté des Sciences de Tétouan, Uni-versité Abdelmalek Essaâdi, B.P 21.21 Tétouan, Morocco
E-mail address:[email protected]
R. Tribak: Département de Mathématiques, Faculté des Sciences de Tétouan, Univer-sité Abdelmalek Essaâdi, B.P 21.21 Tétouan, Morocco