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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.

Proof. RadM =

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).

Then M/RadM =[⊕T(M)]/[⊕RadT(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

m1+RadM, m2+RadM, . . . , mn+RadM

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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REFERENCES

1. Alizade, R., Bilhan, G., and Smith, P. F.,Modules whose maximal submodules have supple- ments, Comm. Algebra 29:6 (2001), 2389–2405.

2. Anderson, F. W., and Fuller, K. R.,Rings and Categories of Modules, Springer, New York, 1992.

3. Cohn, P. M.,Basic Algebra: Groups, Rings and Fields, Springer, London, 2002.

4. Kaplansky, I., Infinite Abelian Groups, Ann Arbor, Michigan: Michigan University Press, 1965.

5. Keskin, D.,On lifting modules, Comm. Algebra 28:7 (2000), 3427–3440.

6. Lam, T. Y.,Lectures on Modules and Rings, Springer, New York, 1999.

7. Lam, T. Y.,A first course in Noncommutative Rings, Springer, New York, 1999.

8. Lomp, C.,On semilocal modules and rings, Comm. Algebra 27:4 (1999), 1921–1935.

9. Mohamed, S. H., and Müller, B. J.,Continuous and Discrete Modules, Cambridge University Press 1990.

10. Rudlof, P.,On the structure of couniform and complemented modules, J. Pure Appl. Alg. 74 (1991), 281–305.

11. Santa-Clara, C., and Smith, P. F.,Direct product of simple modules over Dedekind domains, Arch. Math. (Basel) 82 (2004), 8–12.

12. Wisbauer, R.,Foundations of Modules and Rings, Gordon and Breach, 1991.

13. Zöschinger, H.,Komplementierte Moduln über Dedekindringen, J. Algebra. 29 (1974), 42–56.

14. Zöschinger, H.,Invarianten wesentlicher Überdeckungen, Math. Ann. 237 (1978), 193–202.

15. Zöschinger, H.,Koatomare Moduln, Math. Z. 170 (1980), 221–232.

DEPARTMENT OF MATHEMATICS IZMIR INSTITUTE OF TECHNOLOGY GÜLBAHÇEKÖYÜ

35430 URLA IZMIR TURKEY

E-mail:rafailalizade@iyte.edu.tr, enginbuyukasik@iyte.edu.tr

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