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It is proved that the aforementioned property is not closed under direct sums


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Vol. 19 (2018), No. 2, pp. 945–952 DOI: 10.18514/MMN.2018.2276


YELIZ KARA Received 14 March, 2017

Abstract. We elaborate the class ofP C-modules in which every projection invariant submod- ules has projection invariant closures in this study. We provide examples that the class ofP C- modules does not belong to the system of generalization of extending modules. Moreover, we clarify direct sums and direct summands properties for the former class of modules. It is proved that the aforementioned property is not closed under direct sums. Thereupon, we cope with when the direct sums ofP C-modules enjoy with the property.

2010Mathematics Subject Classification: 16D10; 16D50

Keywords: complement submodule, extending module, projection invariant submodule


Let R be a ring with unity and M a unitary right R-module. A submodule K of M is called a complement in M if K has no proper essential extension in M. Recall that a moduleM is calledextending(orCS) if every submoduleN ofM is essential in a direct summandM; or every complement submodule ofM is a direct summand ofM [4,15]. A submoduleN ofM is called aprojection (fully) invariant iff .N /N for all f2Df 2End.MR/(f 2End.MR/). It is well known that every fully invariant submodule is projection invariant (see, [5, page 50]). There are many examples of projection invariant submodules in different algebraic structures.

Motivating on this class of submodules, a moduleM is said to be-extending[3], if every projection invariant submodule ofM is essential in a direct summand ofM. It is shown that the class of-extending modules is closed under direct sums, but not direct summands (see, [3, Example 5.5], [14, Example 4]). Hence it is investigated some special subclasses of-extending modules. To this end, a moduleM is called strongly-extending[9], if every projection invariant submodule of M is essential in a fully invariant direct summand of M. It is proved that the class of strongly -extending modules is a proper subclass of-extending modules.

LetN be a submodule ofM. A submoduleT ofM is called a closure(or es- sential closure) ofN inM, ifN e T cM. It is renowned that, every submodule has a closure (see, [15, Proposition 2.5]). Consequently, we concern with closure

c 2018 Miskolc University Press


properties of projection invariant submodules. Thereby, we call a moduleM ispro- jection invariant closure module(denoted,P C-module), if every projection invariant submodule ofM has a projection invariant closure inM. We observe indecompos- able modules, uniform modules and nonsingular modules are the examples ofP C- modules. Moreover it is shown that theP C condition is more general than strongly -extending property. Although strongly-extending property implies-extending condition, we provide by counter examples that the classes ofP C-modules and- extending modules are incomparable. Apart from that, we delve into the direct sums and direct summands properties of P C-modules. We prove that projection invari- ant direct summands ofP C-modules areP C-modules. Even so the aforementioned property is not closed under direct sums. Therefore we show when the direct sums of P C-modules isP C-module. To this end, a moduleM hascomplement sum prop- erty, CSP [7], if the sum of every pair of complements of M is a complement of M. Furthermore we get the hang of CSP condition which is not Morita invariant.

Additionally, we characterize that the extending and quasi-continuous conditions are equivalent for a module with CSP.

For notation, we useMn.R/andRnfor the fulln-by-nmatrix ring overRand the direct sum ofncopies ofRfor any positive integern, respectively. For a nonempty subset X ofM, X M, X e M, X c M, X d M andX Ep M denote the submodule ofM, the essential submodule ofM, the complement submodule ofM, the direct summand ofM and the projection invariant submodule ofM, respectively.

For unknown terminology and notation, see [1,4,10,11].


We locate P C-modules with the other renowned classes of modules (e.g., - extending, strongly-extending) in this section. Observe that indecomposable mod- ules and uniform modules are the examples ofP C-modules. Furthermore, nonsin- gular modules have the foregoing property as shown in the first result.

Lemma 1. Every nonsingular module is aP C-module.

Proof. LetMRbe a nonsingular module andX EpM. ThenX eT cM for some submoduleT ofM. SinceMRis nonsingular andX EpM,T is a projection invariant submodule ofM by [3, Lemma 2.3]. HenceM is aP C-module.

LetMZ DZ=Zp for any primep. It is clear that MZ is not nonsingular, but it is aP C-module. It follows that the converse of above lemma is not true. Moreover, there are examples which show that theP C condition does not imply uniform or in- decomposable properties. For example, letMZD.Z˚Z/Z. It is renowned thatMZ is nonsingular, and hence it is aP C-module by Lemma1. HoweverMZ is neither uniform nor indecomposable. The next result identifies the connection between the classes of strongly-extending modules andP C-modules.


Lemma 2. IfMRis a strongly-extending module, thenMR hasP C condition.

But the converse need not to be true.

Proof. Let MR be a strongly -extending module and N Ep M. Then N e

Dd M for some fully invariant direct summandD ofM. It follows that N e Dc M whereDEp M. ThereforeMRis aP C-module. On the other hand, for the converse, letR be a domain which is not right Ore. SinceRR is a prime ring, every nonzero ideal ofRRis essential inRR. ThusRRis aP C-module, becauseRis an indecomposableR-module. HoweverRRis not uniform, soRis not-extending and hence it is not strongly-extending by [9, Corollary 2.4].

It is interpreted that every strongly -extending module is -extending in [9].

In connection with the above lemma one might ask that are there any implications betweenP C-modules and-extending modules? Now, we get across the following counter examples for the former question.

Example1. .i /ConsiderMZDQ

i2IAi whereAi DZfori2I. Note that the Specker groupMZis not-extending by [5]. On the other hand,MZis a nonsingular by [6, Proposition 1.22]. ThereuponMZis a PC-module by Lemma1.

.i i /LetMZDZ˚.Z=Zp/for any primep. Then MZ is-extending which is not strongly -extending by [9, Corollary 2.4 .iv/».i /]. Hence MZ is not a PC-module by Proposition1(i).

Note that a ringRis calledAbelianif every idempotent ofRis central. The next fact provides the implications between the classes of-extending modules, strongly -extending modules andP C-modules under some additional conditions.

Proposition 1. .i /LetMRbe a PC-module. ThenMRis a-extending module if and only ifMRis a strongly-extending module.

.i i /LetSDEnd.MR/be an Abelian ring. IfMR is a-extending, thenMR is a PC-module.

.i i i /IfMRbe a PC-module and every projection invariant essentially closed sub- module ofM is a fully invariant direct summand, thenMRis strongly-extending.

Proof. .i /LetMRbe a-extending module andX EpM. ThenXeKdM for some direct summand K of M. Since MR is a P C-module, K is projection invariant inM. HenceK is a fully invariant submodule ofM by [5, page 50]. Thus M is a strongly-extending module. The converse follows by [9, Corollary 2.4].

.i i /LetXEpM. ThenX eKdM for some submoduleKofM. SinceSis Abelian, it can be easily seen that every direct summand ofM is projection invariant inM. ThereforeM is aP C-module.

.i i i /LetM be a P C-module and X Ep M. Then there exists a projection in- variant submodule K ofM such thatX eK c M. By hypothesis, K is a fully invariant direct summand, soM is strongly-extending.


Notice thatMZD.Z=Zp/˚Q is aP C-module for any prime p. Indeed, it is clear from [8, Example 2.14],MZ is a-extending module with an Abelian endo- morphism ring. Thereby,MZ is aP C-module by Proposition1.i i /. This example and Example 1.i i / show that any submodules of a P C-module need not to be a P C-module, in general.

One might wonder whetherP C-condition belongs to the system of generalization of extending modules or not. Hence we supply examples which simplify that there is no implication between the classes ofP C-modules and extending modules.

Example2. .i /LetRD

Z Z 0 Z

be the upper triangular 2-by-2 matrix ring over Z. It is follows thatRRis not extending by [15, Example 3.84]. SinceRRis nonsin- gular, it is aP C-module by Lemma1.

.i i /LetS3be the symmetric group on the letterf1; 2; 3gandZ3the ring of integers modulo 3. LetRDZ3ŒS3be the group ring of the group S3overZ3. ThenR is right self-injective. Thus RR is extending, and hence it is -extending. However RRis not a strongly-extending module by [2, Example 1.1] and [9, Corollary 2.4].

Now assume thatRRis aP C-module. ThenRRis not-extending by Proposition 1(i), a contradiction. ThereforeRRis not aP C-module.

Recall from [12], a moduleM is calledUC-module, if every submodule ofM has a unique closure. Following the idea in [12], it is natural to think ofunique pro- jection invariant closure module(denoted,UP C-module) in which every projection invariant submodule has a unique projection invariant closure.

Observe that every nonsingular andP C-module are the examples ofUP C-modules.

However basically the results and their proofs arise out of just simple motifications of the result in [12]. To this end, Example1.i i /and Example2.i i /identify that pro- jection invariant submodules need not to have a unique projection invariant closure.


In this section, our main goal is to deal with the direct summand and direct sum properties ofP C-module.

Proposition 2. LetM be aP C-module such thatM DM1˚M2whereM1and M2are projection invariant submodules ofM. ThenM1andM2areP C-modules.

Proof. LetM DM1˚M2for some projection invariantM1; M2M and letX1

be a projection invariant submodule ofM1. ThenX1˚M2 is a projection invariant submodule ofM by [3, Lemma 4.13]. HenceX1˚M2e Kc M for some pro- jection invariant submoduleK ofM. SinceKEp M,K D.K\M1/˚.K\M2/ whereK1DK\M1EpM1andK2DK\M2EpM2by [5, page 50]. Hence we obtainX1DM1\.X1˚M2/e K\M1DK1. It is easy to see thatK1c M. It follows from [10, Proposition 6.24 (1)] that K1c M1. Therefore M1 is aP C- module. Similarly, it can be seen thatM2is aP C-module.


Example1.i i /examines that the direct sums ofP C-modules need not to beP C- module. To this end, we determine when the class ofP C-modules is closed under direct sums.

Proposition 3. LetM DM1˚M2 be an extending module for someM1; M2 M. IfM1andM2are PC-modules, thenM is a PC-module.

Proof. LetY EpM. ThenY D.Y\M1/˚.Y\M2/whereY1DY\M1EpM1

andY2DY \M2Ep M2 by [5, page 50]. Hence there exist projection invariant submodulesK1ofM1andK2ofM2such thatX1eK1cM1andX2eK2c M2. It follows thatM1andM2are extending by [11, Proposition 2.7]. ThusK1and K2 are direct summands ofM1 andM2, respectively. ConsequentlyK1˚K2 is a direct summand of M, so K1˚K2 is a complement in M. ThereforeX DX1˚ X2eK1˚K2cM such thatK1˚K2EpM. TherebyM is aP C-module.

Corollary 1. LetM DM1˚M2be a semisimple (or uniform, or injective) module for someM1; M2M. IfM1andM2are PC-modules, thenM is a PC-module.

Proof. It is a consequence of Proposition3.

Recall that a moduleMRhas thesummand sum property, SSP, if for allD1; D2d

MR,D1CD2d M. Motivating SSP definition on complement submodules, M hascomplement sum property, CSP[7], if for all K1; K2cMR,K1CK2c M. Even though the authors in [7] proved that CSP condition implies SSP, we give the following example which shows that the reverse implication of the former property is not true, in general.

Example3. LetM be theZ-moduleZ˚.Z=Zp/as in Example1(ii). Note that HomZ.Z;Z=Zp/DZ=Zpand HomZ.Z=Zp;Z/D0.

ThusM has SSP by [15, Exercise 2.41]. On the other hand, we claimMZdoes not have CSP. Assume the contrary. SinceZandZ=ZpareP C-modules,MZis aP C- module by Theorem 3, a contradiction (see, Example1(ii)). ThereforeM does not satisfy CSP.

Now we compose some useful properties of modules with CSP which might help us to consider being Morita invariant property as well as the application to the full matrix rings for the former condition. Let us begin with an easy fact and an example for extending case.

Lemma 3. SSP and CSP conditions are coincide for an extending module.

Proof. It is routine to check.

Example4. LetKbe a field andRRD

K K 0 K

. ThenRRis an extending module which does not satisfyC3. HenceRRdoes not have SSP. It follows from Lemma3 thatRRdoes not have CSP.


Theorem 1. LetRbe a ring such thatRDReRandSDeRefor somee2De2 R. ThenMRhas CSP if and only if the rightS-module moduleM ehas CSP.

Proof. It is clear from [15, Lemma 2.76 and Proposition 2.77 (ii)].

Corollary 2. LetRbe a ring such thatRDReRfor somee2De2R. ThenRR

has CSP if and only if the righteRe-moduleResatisfies CSP.

Proof. It is clear from Theorem1.

Theorem 2. Mn.R/has CSP condition if and only if the free rightR-moduleRn has CSP condition.

Proof. Note thatMn.R/DMn.R/eMn.R/wheree is the matrix unit with 1 in the.1; 1/-th position and zero elsewhere. Now apply the Theorem1and Corollary2

to get the theorem.

As an application of Theorem2, it can be easily seen that M2.Z/does not have CSP condition. Indeed, it is renowned thatMZD.Z˚Z/Z is an extending module which does not satisfy SSP by [15, Example 2.82]. HenceMZ does not have CSP by Lemma3. ThusM2.Z/does not have CSP by Theorem2. This example explains that CSP condition is not Morita invariant. Now we proceed our main aim of this section.

Theorem 3. LetM be a rightR-module with CSP such thatM DM1˚M2 for someM1; M2M. IfM1andM2areP C-modules, thenM is a PC-module.

Proof. LetXEpM. ThenXD.X\M1/˚.X\M2/whereX1DX\M1Ep M1andX2DX\M2EpM2by [5, page 50]. Hence there exist projection invariant submodulesK1ofM1andK2ofM2such thatX1eK1cM1andX2eK2c M2. ThusX eK1˚K2M whereK1˚K2EpM. Note thatK1˚K2cM

by CSP condition. ThusM is aP C-module.

Recall from [13], a module M has.Pn/condition if for every submodule K of M such thatT is a direct sumT1˚ ˚Tnof complementsTi .1i n/inM, every homomorphism˛1WT !M can be lifted to a homomorphism˛2WM !M. In [13], the authors proved that ifM satisfies.Pn/ thenM satisfies.Pn 1/for all n2. Moreover, they engage a characterization of quasi-continuous modules in terms of .Pn/ conditions for every positive integer n. Realize that .P1/ does not imply.P2/(see, [13, Example 10]). The following result spells out.P1/and.P2/ conditions are equivalent for a module with CSP.

Proposition 4. LetM be a rightR-module with CSP. ThenM has.P2/condition if and only ifM has.P1/condition.

Proof. LetM be a right R-module with CSP. IfM has.P2/, thenM has.P1/ by [13, page 341]. Conversely, let M has.P1/. Consider K1; K2 c M with ˛W


K1˚K2!M homomorphism. Since M has CSP, K1˚K2c M. Hence ˛W K1˚K2!M can be lifted to a homomorphism WM !M by the condition of .P1/. ThereforeM has.P2/condition.

We strengthen the characterization of quasi-continuous modules, which is presen- ted in [13], for a module with CSP condition.

Theorem 4. The followings are equivalent for a moduleMRwith CSP.

.1/ M is quasi-continuous.

.2/ M has.Pn/for every positive integern.

.3/ M has.Pn/for some integern2.

.4/ M has.P2/.

.5/ M has.P1/.

.6/ M is extending.

Proof. .1/,.2/,.3/,.4/Clear from [13, Theorem 4].

.4/,.5/It follows from Proposition4.

.5/,.6/ Let M has .P1/. Then M has.P2/ by Proposition 4. Hence M is extending from the fact of.4/,.1/. Conversely, let Kc M and˛WK !M be a homomorphism. Since M is extending,K is a direct summand ofM. Consider gD where WM !K is projection andWK !M is inclusion. It is easy to

check thatgjKD˛, henceM has.P1/.


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Author’s address

Yeliz Kara

Uludag University, Department of Mathematics, G¨or¨ukle, 16059 Bursa, Turkey E-mail address:yelizkara@uludag.edu.tr



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