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### EXTENSIONS OF WEAKLY SUPPLEMENTED MODULES

RAFAIL ALIZADE and ENGIN BÜYÜKA ¸SIK

Abstract

It is shown that weakly supplemented modules need not be closed under extension (i.e. ifUand M/U are weakly supplemented thenMneed not be weakly supplemented). We prove that, if U has a weak supplement inMthenMis weakly supplemented. For a commutative ringR, we prove thatRis semilocal if and only if every direct product of simpleR-modules is weakly supplemented.

1. Introduction

Throughout,R is a commutative ring with identity andM is a unital leftR- module. ByNM, we mean thatN is a submodule of M. A submodule LMis said to beessentialinM, denoted asLM, ifLN =0 for every nonzero submoduleNM. A submodule S of M is called small (inM), denoted asS M, ifM =S+Lfor every proper submoduleLofM. By RadM we denote the sum of all small submodules ofM or, equivalently the intersection of all maximal submodules ofM. A ringRis said to besemilocal ifR/RadR is semisimple. By ([7] Proposition 20.2) R is semilocal if and only ifRhas only finitely many maximal ideals. A moduleMissupplemented (see [12]), if every submoduleN ofM has asupplement, i.e. a submoduleK minimal with respect toN+K =M.Kis a supplement ofNinMif and only ifN+K=M andNKK(see [12]). IfN+K =MandNKM, thenK is called aweak supplementofN (see, [14] and [8]).M is aweakly supplementedmodule if every submodule ofM has a weak supplement. By we denote the set of all maximal ideals ofR. LetRbe a domain andM be an R-module. The submoduleT (M)= {m∈M |rm=0 for some 0=rR}

is called thetorsionsubmodule ofM, and ifM = T (M)thenM is called a torsion module.

LetRbe a Dedekind domain andᒍ ∈. The submoduleT(M)= {m ∈ M | ᒍnm = 0 for some n > 0} is called the ᒍ-primary part of M, and T (M)=

T(M)(see Proposition 10.6.9 in [3]).

Received February 28, 2007.

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A classMof modules is said to beclosed under extensionifU, M/UM impliesMM. In this case we say thatM is an extension ofU byM/U.

Let R be a noetherian local ring. Rudlof proved that an R-module M is weakly supplemented if and only if it is an extension of a supplemented module by a supplemented module (see Theorem 3.1 in [10]). He also proved that over a noetherian ring every extension of a supplemented module by a supplemented module is weakly supplemented (see Proposition 3.6 in [10]). In general a weakly supplemented modules need not be an extension of a supplemented module by a supplemented module. For example theZ-moduleQis weakly supplemented andQdoes not contain any supplemented submodule (see [13], Theorem 3.1).

In this paper we show that the class of weakly supplemented modules need not be closed under extensions, that is ifUandM/Uare weakly supplemented for some submoduleU ofM thenM need not be weakly supplemented. But ifU has a weak supplement inM we show thatM is weakly supplemented.

We prove that a commutative ringR is semilocal if and only if every direct product of simpleR-modules is weakly supplemented. LetR be a Dedekind domain. We obtain that anR-moduleM is weakly supplemented if and and only ifT (M)andM/T (M)are weakly supplemented andT (M)has a weak supplement inM. IfM is a torsionR-module with RadM M then every submodule ofM is weakly supplemented.

2. Extensions of weakly supplemented modules

A submoduleN of a module M is calledclosed in M if N K for some KM impliesK = N. A submoduleN of M is calledcoclosed inM if N/K M/K for someKM impliesK =N.

Theorem2.1. Let0→LMN →0be a short exact sequence. If LandN are weakly supplemented andLhas a weak supplement inM then M is weakly supplemented.

IfLis coclosed inM then the converse holds, that is ifM is weakly sup- plemented thenLandN are weakly supplemented.

Proof. Without restriction of generality we will assume thatLM. Let Sbe a weak supplement ofLinMi.e.L+S=M andLSM. Then we have,

M/(LS)=L/(LS)S/(LS)

L/LSis weakly supplemented as a factor module ofL. On the other hand, S/(LS)∼=M/L∼=N is weakly supplemented. ThenM/(LS)is weakly supplemented as a sum of weakly supplemented modules (see [8] Proposi- tion 2.5). ThereforeMis weakly supplemented by ([8], Proposition 2.2 (4)).

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Suppose thatL is coclosed. Then LS L by ([5], Lemma 1.1) i.e.

Lis a supplement ofS in M. Therefore Lis weakly supplemented by ([8], Proposition 2.2 (5)).

Proposition2.2.LetRbe a semilocal ring (not necessarily commutative) andMbe anR-module. SupposeUMsuch thatM/Uis finitely generated.

IfU is weakly supplemented thenM is weakly supplemented.

Proof. SupposeM/U is generated by

m1+U, m2+U, . . . , mn+U.

For the submoduleK = Rm1+Rm2+ · · · +Rmn we haveU +K = M. ThenM is weakly supplemented by ([8], Proposition 2.5).

The following well known lemma is given for completeness.

Lemma2.3. LetM be a module andU be a finitely generated submodule ofM contained inRadM. ThenU is small inM.

A module M is said to be locally noetherianif every finitely generated submodule ofM is noetherian.

Proposition2.4.LetM be a locally noetherian module andX⊆ RadM. SupposeM/Xis finitely generated. IfXandM/Xare weakly supplemented thenM is weakly supplemented.

Proof. SinceM/Xfinitely generated,X+L=M for some finitely gen- erated submoduleLofM. ThenXLX ⊆ RadM is finitely generated, becauseLis finitely generated andM is locally noetherian. SoXLM. ThusLis a weak supplement ofXinM. ThereforeMis weakly supplemented by Theorem 2.1.

We shall give an example in order to prove that the class of weakly supple- mented modules need be closed under extensions. The following lemmas will be useful to present this example.

Lemma2.5 ([1], Lemma 4.4). Let R be a Dedekind domain. For anR- moduleM the following are equivalent:

(1) M is injective, (2) M is divisible,

(3) M =P M for every maximal idealP ofR, (4) M does not contain any maximal submodule.

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Note that ifM is divisible module over a Dedekind domain then RadM = M. Hence ifNis a module with RadN =0 thenNdoes not contain divisible submodule.

Lemma2.6. LetRbe a domain anda maximal ideal ofR. Then for every-primaryR-moduleM,M/RadM is semisimple.

M. We will show thatM = M for every \ {ᒍ}. LetxM, thenᒍnx=0 for somen∈N. Sincen+ᒎ=R, we have 1=p+qfor somep∈ᒍnandq ∈ᒎ. So we getx =px+qx=qx ∈ᒎM, henceM = ᒎM. Therefore RadM =

M = ᒍM. Then sinceR/ᒍis a fieldM/RadM =M/Mis semisimpleR/ᒍ-module, and so it is semisimple as anR-module.

Corollary2.7. LetRbe a Dedekind domain andM a torsionR-module, thenM/RadMis semisimple.

Proof. SinceRis a Dedekind domain andMa torsionR-module, we have M =

T(M).

∼= ⊕[T(M)/RadT(M)] is semisimple by Lemma 2.6, and by Theorem 9.6 in [2].

Lemma2.8. LetRbe a Dedekind domain andKbe the field of quotients ofR. ThenRKis weakly supplemented.

Proof. SinceRis a Dedekind domain andK/Ris a torsionR-module, we haveK/R∼=

PTP(K/R)soK/R is supplemented by Theorem 2.4 and Theorem 3.1 in [13]. SinceR is finitely generated and RadK = Kwe have RK. ThereforeKis weakly supplemented by Proposition 2.2 (4) in [8].

Lemma2.9. LetRbe a Dedekind domain and{ᒍi}i∈Ibe an infinite collec- tion of distinct maximal ideals ofR. LetM =

i∈I(R/i)be the direct product of the simpleR-modulesR/i andT = T (M)be the torsion submodule of M. Then the following hold,

(1) M/T is divisible, thereforeM/T ∼=K(J )for some index setJ, (2) RadM =0.

Proof. (1) Letᒍbe a maximal ideal ofR. Thenᒍ(M/T )=(M+T )/T. Now ifis not one of the ideals{ᒍi}i∈IthenM+T =Mand so(M/T )=

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M/T. Suppose ᒍ ∈ {ᒍi}i∈I, sayᒍ = ᒍj for some jI, thenᒍM = M(j) whereM(j)consists of those elements ofM whosej th coordinate is zero.

LetM(j)be the submodule ofM whose all coordinates exceptjth are zero.

ClearlyM(j)T. ThenM = M(j)+M(j)⊆ᒍM +T, soM +T =M and henceᒍ(M/T )=M/T. Therefore by Lemma 2.5M/T is divisible, and since it is torsion-free we haveM/T ∼=K(J ).

(2)M/M(j)∼=R/jis a simple module, soM(j)is a maximal submodule ofM for everyjI. Then we get RadM

j∈IM(j)=0.

Theorem2.10.For a commutative ringR, the following are equivalent.

(1) Ris semilocal,

(2) Every direct product of simpleR-modules is semisimple,

(3) Every direct product of simpleR-modules is weakly supplemented.

Proof. (1)(2)Let ᒍ1,2, . . . ,n be the maximal ideals of R. Then RadR=ᒍ1∩ᒍ2∩· · ·∩ᒍn=ᒍ1.2. . .n. LetMbe a direct product of simple R-modules. Since every simpleR-module is isomorphic to one of the simple modulesR/j, j = 1, . . . , n, we have ᒍ1.2. . .nM = 0. So thatM is an R/RadR-module. By the hypothesisR/RadRis semisimple, and soM is a semisimpleR/RadR-module. ThereforeM is a semisimpleR-module.

(2)(3)Obvious.

(3)(1)Let M =

(R/). From the proof of Lemma 2.9(2) we have Rad(M) = 0. Since M is weakly supplemented,M is semisimple by Corollary 2.3 in [8]. So thatM =

(R/)∼=

I(R/)for some index setsI. In this case(1+ᒍ)Mcan have only finitely many nonzero components in the last decomposition. Thereforeis finite, i.e. Rhas only finitely many maximal ideals. HenceRis semilocal.

Example2.11. LetRandM be as in Lemma 2.9 andT = ⊕i∈I(R/i) be the torsion submodule of M. Note that T is semisimple, so it is weakly supplemented. LetN be a submodule ofM such thatN/T ∼=K. ThenN/T is weakly supplemented by Lemma 2.8. Note that RadN =0 by Lemma 2.9 and N is not semisimple becauseN/T ∼= K is not semisimple. Hence by Corollary 2.3 in [8],N is not weakly supplemented.

Remark2.12. In Theorem 2.1 the hypothesis thatLhas a weak supplement can not be omitted. LetT andN be as in Example 2.11. ThenT has no weak supplement inN. Otherwise we would haveT +A=M andTANfor some submoduleAofN. Since RadN =0 we haveTA=0. So the sum T +A=M is a direct sum andN/T ∼=Ais divisible, a contradiction.

The proof of the following lemma is standard.

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Lemma 2.13 (see [6], Exercise 6.34). Let R be a domain andM be an R-module. Then the torsion submoduleT (M)ofM is closed inM.

Note that over a Dedekind domain a submodule is closed if and only if it is coclosed (see [13], Satz 3.4).

Proposition2.14.LetRbe a Dedekind domain andM be anR-module.

Then the following holds.

(1) IfMis weakly supplemented thenT (M)andM/T (M)are weakly sup- plemented. If T (M) has a weak supplement in M then the converse holds.

(2) IfRadT (M)M thenM is weakly supplemented if and only ifT (M) has a weak supplement inMandM/T (M)is weakly supplemented.

(3) Suppose M is torsion. Then M is weakly supplemented if RadM is weakly supplemented and has a weak supplement inM.

(4) SupposeM/RadM is finitely generated andRadM M. ThenM is weakly supplemented ifRadM is weakly supplemented.

Proof. (1) Suppose M is weakly supplemented. Then T (M) is a weak supplement inM. SinceT (M) is also coclosed it is a supplement inM by ([5], Lemma 1.1). Then T (M) and M/T (M) are weakly supplemented by Proposition 2.2(5) in [8].

IfT (M)has a weak supplement thenM is weakly supplemented by The- orem 2.1.

(2)T (M)/RadT (M) is semisimple by Lemma 2.7 so it is weakly sup- plemented. ThenT (M)is weakly supplemented by Proposition 2.2(4) in [8].

Then the proof is clear by (1).

(3) By Lemma 2.7M/RadM is semisimple. Then the proof is clear by Theorem 2.1.

(4) SupposeM/RadM is generated by

Then for the finitely generated submoduleK=Rm1+Rm2+ · · · +Rmnwe have RadM+K = M andK∩RadM is finitely generated asK is finitely generated, soK∩RadM M by Lemma 2.3 i.e.Kis a weak supplement of RadM inM.

By ([2] Proposition 9.15) Rad(M/RadM) = 0, and since RadM M, M/RadM is torsion. Therefore M/RadM is semisimple by Lemma 2.7.

HenceMis weakly supplemented by Theorem 2.1.

A moduleMis called coatomic if every proper submodule ofMis contained in a maximal submodule ofM. Over a commutative noetherian ring every

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submodule of a coatomic module is coatomic (see Lemma 1.1 in [15]). Note that coatomic modules have small radicals.

Proposition2.15.Let R be a Dedekind domain and M be a torsionR- module. IfRadM Mthen every submodule ofM is weakly supplemented.

Proof. The module M/RadM is semisimple by Lemma 2.7. Since RadM M, every submodule of M is contained in a maximal submod- ule i.e.M is coatomic. LetN be a submodule ofM. ThenN is coatomic so RadN N, and sinceN is torsion, N/RadN is semisimple. HenceN is weakly supplemented by Proposition 2.2(4) in [8].

A domain R is said to be one-dimensional if R/I is artinian for every nonzero idealIofR. One-dimensional domains are proper generalizations of Dedekind domains.

Lemma2.16. LetRbe a ring,I RandM be anR-module. IfIM has a weak supplementK inM, thenK is a weak supplement ofInM inM for everyn1.

Proof. By hypothesisIM+K =M. Then we haveI2M +IK = IM, soI2M +IK +K = IM+Kwhich givesI2M +K = M. Continuing in this way we get:

InM +K=MandInMKIMKM. This means thatKis a weak supplement ofInMinM.

Proposition2.17. Let Rbe a one-dimensional domain andM be anR- module. Suppose thatIis a nonzero ideal ofR. IfInMis weakly supplemented andIkM has a weak supplement inM for somek n, then M is weakly supplemented.

Proof. Since R is a domain In = 0. So R/In is an artinian ring be- causeR is one-dimensional. ThenM/InM is a supplementedR/In-module by Theorem 24.25 in [7] and Theorem 4.41 in [9]. HenceM/InMis a weakly supplementedR-module. By Lemma 2.16,InMhas a weak supplement inM. Therefore by Theorem 2.1,M is weakly supplemented.

Corollary 2.18.Let R be a one-dimensional domain and M be anR- module. IfrMis weakly supplemented for some0 =rRand has a weak supplement inM thenM is weakly supplemented.

Acknowledgements.The authors would like to thank the referee for the valuable suggestions and comments.

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