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Vol. 8, No. 3, 2015, 417-430

ISSN 1307-5543 – www.ejpam.com

Classical 2-Absorbing Submodules of Modules over

Commutative Rings

Hojjat Mostafanasab

1,∗

, Ünsal Tekir

2

and Kür¸sat Hakan Oral

3

1Department of Mathematics and Applications, University of Mohaghegh Ardabili, P. O. Box 179,

Ardabil, Iran

2Department of Mathematics, Marmara University, Ziverbey, Goztepe, Istanbul 34722, Turkey 3 Department of Mathematics, Yildiz Technical University, Davutpasa Campus, Esenler, Istanbul,

Turkey

Abstract. In this article, all rings are commutative with nonzero identity. Let M be anR-module.

A proper submodule N of M is called aclassical prime submodule, if for eachm M and elements

a,bR,a bmN implies thatamNor bmN. We introduce the concept of "classical 2-absorbing submodules" as a generalization of "classical prime submodules". We say that a proper submoduleN

of M is a classical 2-absorbing submodule if whenevera,b,cRand mM witha bcmN, then

a bmNoracmNor bcmN.

2010 Mathematics Subject Classifications: 13A15, 13C99, 13F05

Key Words and Phrases: Classical prime submodule, Classical 2-absorbing submodule

1. Introduction

Throughout this paper, we assume that all rings are commutative with 1 #= 0. Let R be a commutative ring and M be an R-module. A proper submodule N of M is said to be a prime submodule, if for each element aR and mM, amN implies that mN or a ∈ (N :R M) ={rR | r MN}. A proper submodule N of M is called aclassical prime submodule, if for each mM and a,bR, a bmN implies that amN or bmN. This notion of classical prime submodules has been extensively studied by Behboodi in[9, 10]

(see also,[11], in which, the notion of “weakly prime submodules” is investigated). For more information on weakly prime submodules, the reader is referred to[3, 4, 12].

Badawi gave a generalization of prime ideals in[5]and said such ideals 2-absorbing ideals. A proper ideal I ofR is a2-absorbing ideal of R if whenever a,b,cR and a bcI, then a bI or acI or bcI. He proved that I is a 2-absorbing ideal of R if and only if ∗Corresponding author.

Email addresses:h.mostafanasab@gmail.com(H. Mostafanasab),utekir@marmara.edu.tr(Ü. Tekir), khoral@yildiz.edu.tr(K. Hakan Oral)

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whenever I1, I2, I3 are ideals of R with I1I2I3I, then I1I2I or I1I3I or I2I3I. Anderson and Badawi[2]generalized the notion of 2-absorbing ideals ton-absorbing ideals. A proper ideal I ofRis called ann-absorbing(resp. a strongly n-absorbing)idealif whenever x1· · ·xn+1 ∈ I for x1, . . . ,xn+1 ∈ R (resp. I1. . .In+1 ⊆ I for ideals I1, . . . ,In+1 of R), then there are nof the xi’s (resp. n of the Ii’s) whose product is in I. The reader is referred to

[6–8]for more concepts related to 2-absorbing ideals. Yousefian Darani and Soheilnia in[13]

extended 2-absorbing ideals to 2-absorbing submodules. A proper submoduleN ofMis called a 2-absorbing submodule of M if whenever a bmN for some a,bR and mM, then amN or bmN or a b ∈!N:RM

"

. Generally, a proper submodule N ofM is called an n-absorbing submodule if whenever a1. . .anmN fora1, . . .anRand mM, then either a1. . .an∈(N :RM)or there aren−1 ofai’s whose product withmis inN, see[14]. Several authors investigated properties of 2-absorbing submodules, for example[15].

In this paper we introduce the definition of classical 2-absorbing submodules. A proper sub-moduleN of anR-moduleM is calledclassical 2-absorbing submoduleif whenevera,b,cR andmM with a bcmN, then a bmN or acmN or bcmN. Clearly, every classi-cal prime submodule is a classiclassi-cal 2-absorbing submodule. We show that every Noetherian R-moduleM contains a finite number of minimal classical 2-absorbing submodules (Theorem 3). Further, we give the relationship between classical 2-absorbing submodules, classical prime submodules and 2-absorbing submodules (Proposition 2, Proposition 7). Moreover, we charac-terize classical 2-absorbing submodules in (Theorem 2, Theorem 4). In (Theorem 7, Theorem 8) we investigate classical 2-absorbing submodules of a finite direct product of modules.

2. Characterizations of Classical 2-Absorbing Submodules

First of all we give a module which has no classical 2-absorbing submodule. Example 1. Let p be a fixed prime integer and!0=!∪{0}. Then

E!p":=

#

α∈"/#|α= r

pn +#for some r∈#and n∈!0

$

is a nonzero submodule of the#-module"/#. For each t ∈!0, set

Gt:=

#

α"/#|α= r

pt +#for some r∈#

$

.

Notice that for each t∈!0, Gtis a submodule of E!p"generated by p1t+#for each t∈!0. Each

proper submodule of E!p"is equal to Gifor some i∈!0(see,[17, Example 7.10]). However, no Gtis a classical2-absorbing submodule of E

!

p". Indeed, pt1+3+#∈E

!

p". Then

p3%p1t+3 +#

&

= p1t +#∈Gtbut p2

% 1

pt+3+#

&

= pt1+1+#∈/Gt.

Theorem 1. Let f :MM)be an epimorphism of R-modules.

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(ii) If N is a classical 2-absorbing submodule of M containing Ker(f), then f(N)is a classical 2-absorbing submodule of M).

Proof. (i)Since f is epimorphism, f−1(N))is a proper submodule ofM. Leta,b,cRand mM such thata bcmf−1(N)). Thena bc f(m)∈N). Hencea b f(m)∈N)orac f(m)∈N) or bc f(m)N), and thusa bmf−1(N))oracmf−1(N))or bcmf−1(N)). So, f−1(N)) is a classical 2-absorbing submodule ofM.

(ii)Let a,b,cRand m) ∈M) be such that a bcm) ∈ f(N). By assumption there exists mM such thatm)= f(m)and so f(a bcm) f(N). SinceKer(f)N, we havea bcmN. It implies thata bmN or acmN or bcmN. Hence a bm)∈ f(N)or acm) ∈ f(N)or bcm)∈ f(N). Consequently f(N)is a classical 2-absorbing submodule ofM).

As an immediate consequence of Theorem 1 we have the following corollary.

Corollary 1. Let M be an R-module and L N be submodules of M . Then N is a classical 2-absorbing submodule of M if and only if N/L is a classical 2-absorbing submodule of M/L. Proposition 1. Let M be an R-module and N1, N2 be classical prime submodules of M . Then N1N2 is a classical 2-absorbing submodule of M .

Proof. Let for somea,b,cRandmM,a bcmN1N2. Since N1 is a classical prime submodule, then we may assume that amN1. Likewise, assume that bmN2. Hence a bmN1N2which impliesN1N2 is a classical 2-absorbing submodule.

Proposition 2. Let N be a proper submodule of an R-module M .

(i) If N is a 2-absorbing submodule of M , then N is a classical 2-absorbing submodule of M .

(ii) N is a classical prime submodule of M if and only if N is a 2-absorbing submodule of M and(N:RM)is a prime ideal of R.

Proof. (i)Assume thatNis a 2-absorbing submodule ofM. Leta,b,cRandmM such that a bcmN. Therefore eitheracmN or bcmN ora b∈(N :M). The first two cases lead us to the claim. In the third case we have that a bmN. ConsequentlyN is a classical 2-absorbing submodule.

(ii)It is evident that if N is classical prime, then it is 2-absorbing. Also,[3, Lemma 2.1]

implies that(N:R M)is a prime ideal ofR. Assume thatN is a 2-absorbing submodule of M and(N:RM)is a prime ideal ofR. Leta bmNfor somea,bRandmM such that neither amN nor bmN. Thena b∈(N :R M)and so either a∈(N :R M)or b∈(N :R M).This contradiction shows thatN is classical prime.

he following example shows that the converse of Proposition 2(i)is not true.

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Let M be anR-module andN a submodule ofM. For everyaR, {mM |amN}is denoted by(N :Ra). It is easy to see that(N:M a)is a submodule ofM containingN.

Theorem 2. Let M be an R-module and N be a proper submodule of M . The following conditions are equivalent:

(i) N is classical 2-absorbing;

(ii) For every a,b,cR,(N:Ma bc) = (N:M a b)∪(N:Mac)∪(N :M bc);

(iii) For every a,bR and mM with a bm/N ,(N:Ra bm) = (N:Ram)∪(N:R bm); (iv) For every a,bR and mM with a bm/N ,(N:Ra bm) = (N:Ram)or

(N :Ra bm) = (N:R bm);

(v) For every a,bR and every ideal I of R and mM with a bI mN , either a bmN or aI mN or bI mN ;

(vi) For every aR and every ideal I of R and mM with aI m#⊆N ,(N:RaI m) = (N:Ram) or(N:RaI m) = (N:R I m);

(vii) For every aR and every ideals I, J of R and mM with aI J mN , either aI mN or aJ mN or I J mN ;

(viii) For every ideals I , J of R and mM with I J m#⊆N ,(N :RI J m) = (N:RI m)or

(N :RI J m) = (N:RJ m);

(ix) For every ideals I , J , K of R and mM with I J K mN , either I J mN or I K mN or J K mN ;

(x) For every mM\N ,(N:Rm)is a 2-absorbing ideal of R.

Proof. (i)⇒(ii)Suppose thatN is a classical 2-absorbing submodule ofM. Let

m ∈ !N:Ma bc". Then a bcmN. Hence a bmN or acmN or bcmN. Therefore m∈!N:M a b

"

orm∈!N:M ac

"

orm∈!N:M bc

"

. Consequently,

!

N:M a bc"=!N:M a b"∪!N:M ac"∪!N:M bc".

(ii)(iii)Leta bm/ N for some a,bRandmM. Assume that x ∈(N :R a bm). Then a b x mN, and so m ∈ (N :M a b x). Since a bm/ N, m/ (N :M a b). Thus by part (i), m (N :M a x) or m (N :M b x), whence x (N :R am) or x (N :R bm). Therefore (N:Ra bm) = (N:Ram)∪(N:R bm).

(iii)⇒(i v)By the fact that if an ideal (a subgroup) is the union of two ideals (two subgroups), then it is equal to one of them.

(i v)(v)Let for somea,bR, an idealIofRandmM,a bI mN. HenceI⊆(N:Ra bm). If a bmN, then we are done. Assume that a bm/ N. Therefore by part(i v)we have that I⊆(N:Ram)orI⊆(N:R bm), i.e.,aI mNor bI mN.

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(i x)⇒(i)Is trivial.

(i x)(x)Straightforward.

Corollary 2. Let R be a ring and I be a proper ideal of R.

(i) RI is a classical 2-absorbing submodule of R if and only if I is a 2-absorbing ideal of R.

(ii) Every proper ideal of R is 2-absorbing if and only if for every R-module M and every proper submodule N of M , N is a classical 2-absorbing submodule of M .

Proof. (i)LetI be a classical 2-absorbing submodule ofR. Then by Theorem 2,(I:R1) =I is a 2-absorbing ideal ofR. For the converse see part(i)of Proposition 2.

(ii)Assume that every proper ideal ofRis 2-absorbing. LetNbe a proper submodule of an R-moduleM. Since for everymM\N,(N :Rm)is a proper ideal ofR, then it is a 2-absorbing ideal ofR. Hence by Theorem 2, N is a classical 2-absorbing submodule of M. We have the converse immediately by part(i).

Proposition 3. Let M be an R-module and(Ki |iI

)

be a chain of classical 2-absorbing sub-modules of M . TheniIKi is a classical 2-absorbing submodule of M .

Proof. Suppose thata bcm∈ ∩iIKi for somea,b,cRandmM. Assume that

a bm ∈ ∩/ iIKi and acm ∈ ∩/ iIKi. Then there are t, lI where a bm/ Kt and acm/ Kl. Hence, for everyKsKt and everyKdKl we have thata bm/Ks andacm/Kd. Thus, for every submoduleKhsuch thatKhKtandKhKl we getbcmKh. Hencebcm∈ ∩iIKi.

A classical 2-absorbing submodule of M is calledminimal, if for any classical 2-absorbing submoduleK ofM such thatKN, thenK=N. Let Lbe a classical 2-absorbing submodule ofM. Set

Γ=(K|Kis a classical 2-absorbing submodule ofM andKL).

If (Ki:iI) is any chain in Γ, then iIKi is in Γ, by Proposition 3. By Zorn’s Lemma, Γ contains a minimal member which is clearly a minimal classical 2-absorbing submodule ofM. Thus, every classical 2 -absorbing submodule of M contains a minimal classical 2-absorbing submodule ofM. IfMis a finitely generated, then it is clear thatMcontains a minimal classical 2-absorbing submodule.

Theorem 3. Let M be a Noetherian R-module. Then M contains a finite number of minimal classical2-absorbing submodules.

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and M/(T+J K m) have only finitely many minimal classical 2-absorbing submodules. Sup-pose P/T be a minimal classical 2-absorbing submodule ofM/T. So I J K m T P, which implies that I J mP or I K mP or J K mP. Thus P/(T+I J m) is a minimal classical 2-absorbing submodule of M/(T+I J m) or P/(T+I K m)is a minimal classical 2-absorbing submodule of M/(T+I K m) or P/(T+J K m)is a minimal classical 2-absorbing submodule ofM/(T+J K m). Thus, there are only a finite number of possibilities for the submodule P. This is a contradiction.

We recall from[5]that if I is a 2-absorbing ideal of a ringR, then either-I=Pwhere P is a prime ideal ofR or-I= P1P2 where P1, P2 are the only distinct minimal prime ideals ofI.

Corollary 3. Let N be a classical 2-absorbing submodule of an R-module M . Suppose that mM\N and*(N:Rm) =P where P is a prime ideal of R and(N :R m)#= P. Then for each x ∈*(N:Rm)\(N :R m), (N :R x m)is a prime ideal of R containing P. Furthermore, either

(N:Rx m)⊆(N:R y m)or(N:R y m)⊆(N:Rx m)for every x,y

*

(N:Rm)\(N:Rm). Proof. By Theorem 2 and[5, Theorem 2.5].

Corollary 4. Let N be a classical 2-absorbing submodule of an R-module M . Suppose that mM\N and*(N:Rm) =P1∩P2 where P1 and P2 are the only nonzero distinct prime ideals of R that are minimal over (N :R m). Then for each x ∈ *(N:Rm)\(N :R m), (N :R x m)

is a prime ideal of R containing P1 and P2. Furthermore, either (N :R x m) ⊆ (N :R y m) or

(N:R y m)⊆(N :Rx m)for every x,y

*

(N:Rm)\(N:Rm). Proof. By Theorem 2 and[5, Theorem 2.6].

AnR-moduleM is called amultiplication moduleif every submoduleN ofM has the form I M for some idealI ofR. LetN andK be submodules of a multiplicationR-moduleM with N =I1MandK=I2M for some idealsI1andI2ofR. The product ofNandKdenoted byN K is defined byN K=I1I2M. Then by[1, Theorem 3.4], the product ofN andKis independent of presentations ofN andK.

Proposition 4. Let M be a multiplication R-module and N be a proper submodule of M . The following conditions are equivalent:

(i) N is a classical 2-absorbing submodule of M ;

(ii) If N1N2N3mN for some submodules N1, N2, N3of M and mM , then either N1N2mN or N1N3mN or N2N3mN .

Proof. (i) ⇒(ii)Let N1N2N3mN for some submodules N1, N2, N3 of M and mM. Since M is multiplication, there are ideals I1, I2, I3 ofRsuch that N1 = I1M, N2 = I2M and N3 = I3M. Therefore I1I2I3mN, and so either I1I2mN or I1I3mN or I2I3mN. HenceN1N2mN or N1N3mN orN2N3mN.

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In [16], Quartararoet al. said that a commutative ring R is au-ring providedR has the property that an ideal contained in a finite union of ideals must be contained in one of those ideals; and aum-ring is a ringRwith the property that anR-module which is equal to a finite union of submodules must be equal to one of them. They show that every B´ezout ring is a u-ring. Moreover, they proved that every Prüfer domain is au-domain. Also, any ring which contains an infinite field as a subring is au-ring,[17, Exercise 3.63].

Theorem 4. Let R be a um-ring, M be an R-module and N be a proper submodule of M . The following conditions are equivalent:

(i) N is classical 2-absorbing;

(ii) For every a,b,cR,(N:Ma bc) = (N:M a b)or(N :M a bc) = (N:M ac)or

(N :M a bc) = (N:M bc);

(iii) For every a,b,cR and every submodule L of M , a bc LN implies that a b LN or ac LN or bc LN ;

(iv) For every a,bR and every submodule L of M with a b L#⊆N ,(N:Ra b L) = (N :Ra L)or (N :Ra b L) = (N:Rb L);

(v) For every a,bR, every ideal I of R and every submodule L of M , a bI LN implies that a b LN or aI LN or bI LN ;

(vi) For every aR, every ideal I of R and every submodule L of M with aI L#⊆N ,(N:RaI L) = (N:Ra L)or(N:RaI L) = (N:RI L);

(vii) For every aR, every ideals I , J of R and every submodule L of M , aI J LN implies that aI LN or aJ LN or I J LN ;

(viii) For every ideals I , J of R and every submodule L of M with I J L#⊆N ,(N:RI J L) = (N:RI L) or(N:RI J L) = (N:RJ L);

(ix) For every ideals I , J , K of R and every submodule L of M , I J K LN implies that I J LN or I K LN or J K LN ;

(x) For every submodule L of M not contained in N ,(N:RL)is a 2-absorbing ideal of R. Proof. Similar to the proof of Theorem 2.

Proposition 5. Let R be a um-ring and N be a proper submodule of an R-module M . Then N is a classical 2-absorbing submodule of M if and only if N is a 3-absorbing submodule of M and

(N:RM)is a 2-absorbing ideal of R.

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and so either a1a2 ∈(N :R M)or a1a3 ∈(N :R M) or a2a3 ∈(N :R M). This contradiction shows thatN is classical 2-absorbing.

Proposition 6. Let M be an R-module and N be a classical 2-absorbing submodule of M . The following conditions hold:

(i) For every a,b,cR and mM ,(N:Ra bcm) = (N:Ra bm)∪(N:Racm)∪(N:R bcm); (ii) If R is a u-ring, then for every a,b,cR and mM , (N :R a bcm) = (N :R a bm) or

(N :Ra bcm) = (N :Racm)or(N:Ra bcm) = (N:R bcm).

Proof. (i)Leta,b,cRandmM. Suppose that r∈(N :Ra bcm). Then a bc(r m)∈N. So, eithera b(r m)∈N or ac(r m)∈N or bc(r m)∈N. Therefore, either r ∈(N :R a bm)or r∈(N:Racm)or r∈(N:R bcm). Consequently

(N:Ra bcm) = (N:Ra bm)∪(N:Racm)∪(N:R bcm). (ii)Use part(i).

Proposition 7. Let R be a um-ring, M be a multiplication R-module and N be a proper submodule of M . The following conditions are equivalent:

(i) N is a classical 2-absorbing submodule of M ;

(ii) If N1N2N3N4N for some submodules N1, N2, N3, N4 of M , then either N1N2N4N or N1N3N4N or N2N3N4N ;

(iii) If N1N2N3N for some submodules N1, N2, N3of M , then either N1N2N or N1N3N or N2N3N ;

(iv) N is a 2-absorbing submodule of M ;

(v) (N :RM)is a 2-absorbing ideal of R.

Proof. (i)⇒(ii)LetN1N2N3N4N for some submodulesN1, N2, N3,N4 ofM. SinceM is multiplication, there are ideals I1, I2, I3 ofRsuch thatN1=I1M,N2=I2M andN3=I3M. ThereforeI1I2I3N4N, and soI1I2N4N orI1I3N4Nor I2I3N4N. Thus by Theorem 4, eitherN1N2N4N orN1N3N4Nor N2N3N4N.

(ii)(iii)Is easy.

(iii)⇒(i v)Suppose that I1I2KN for some idealsI1,I2ofRand some submoduleK ofM. It is sufficient to setN1:=I1M,N2:=I2M andN3=K in part(iii).

(i v)(i)By part(i)of Proposition 2. (i v)⇒(v)By[15, Theorem 2.3].

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Definition 1. Let R be a um-ring, M be an R-module and S be a subset of M\ {0}. If for all ideals I , J , Q of R and all submodules K, L of M ,(K+I J L)S#=.and(K+IQ L)S#=.and

(K+JQ L)S #=.implies(K+I JQ L)S#=., then the subset S is called classical 2-absorbing m-closed.

Proposition 8. Let R be a um-ring, M be R-module and N a submodule of M . Then N is a classical 2-absorbing submodule if and only if M\N is a classical 2-absorbing m-closed.

Proof. Suppose that N is a classical 2-absorbing submodule of M and I, J, Q are ideals ofRandK, L are submodules of M such that(K+I J L)∩S #=.and(K+IQ L)∩S #=.and (K+JQ L)S #= .whereS = M\N. Assume that(K+I JQ L)S =.. ThenK+I JQ LN and soKN andI JQ LN. Since N is a classical 2-absorbing submodule, we getI J LN or IQ LN or JQ LN. If I J LN, then we get(K+I J L)∩S =., sinceKN. This is a contradiction. By the other cases we get similar contradictions. Now for the converse suppose thatS=M\N is a classical 2-absorbing m-closed and assume thatI JQ LN for some ideals I, J, Q ofR and submodule L of M. Then we get for submodule K = (0), K+I JQ LN. Thus (K+I JQ L)∩S = .. SinceS is a classical 2-absorbing m-closed, (K+I J L)∩S = .or (K+IQ L)S =.or(K+JQ L)S =.. Hence I J LN or IQ LN orJQ LN. SoN is a classical 2-absorbing submodule.

Proposition 9. Let R be a um-ring, M be an R-module, N a submodule of M and S=M\N . The following conditions are equivalent:

(i) N is a classical 2-absorbing submodule of M ;

(ii) S is a classical 2-absorbing m-closed;

(iii) For every ideals I , J , Q of R and every submodule L of M , if I J LS#=.and IQ LS#=.

and JQ LS#=., then I JQ LS#=.;

(iv) For every ideals I , J , Q of R and every mM , if I J mS #= . and IQmS #= . and JQmS#=., then I JQmS#=..

Proof. It follows from the previous Proposition, Theorem 2 and Theorem 4.

Theorem 5. Let R be a um-ring, M be an R-module and S be a classical 2-absorbing m-closed. Then the set of all submodules of M which are disjoint from S has at least one maximal element. Any such maximal element is a classical 2-absorbing submodule.

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Theorem 6. Let R be a um-ring and M be an R-module.

(i) IfF is a flatR-module andN is a classical 2-absorbing submodule of M such that FN#=FM, thenFN is a classical 2-absorbing submodule ofFM.

(ii) Suppose thatFis a faithfully flatR-module. ThenNis a classical 2-absorbing submodule ofM if and only ifFN is a classical 2-absorbing submodule ofFM.

Proof. (i)Leta,b,cR. Then we get by Theorem 4,!N:Ma bc"=!N:M a b"or

!

N:M a bc"= !N:Mac" or!N:M a bc"=!N:M bc". Assume that !N:M a bc"= !N:Ma b". Then by[4, Lemma 3.2],

!

FN:FMa bc"=F!N :M a bc"=F!N:Ma b"=!FN:FMa b". Again Theorem 4 implies thatFN is a classical 2-absorbing submodule ofFM.

(ii) Let N be a classical 2-absorbing submodule of M and assume that FN = FM. Then 0 FN →⊆ FM → 0 is an exact sequence. Since F is a faithfully flat module, 0N→⊆ M→0 is an exact sequence. SoN=M, which is a contradiction. So

FN#=FM. ThenFN is a classical 2-absorbing submodule by(1). Now for conversely, let FN be a classical 2-absorbing submodule of FM. We have FN #= FM and so N #=M. Leta,b,cR. Then!FN:FM a bc"=!FN:FM a b"or

!

FN:FMa bc

"

=!FN :FM ac

"

or!FN:FMa bc

"

=!FN :FM bc

"

by Theorem 4. Assume that!FN:FM a bc"=!FN:FMa b". Hence

F⊗!N:Ma b

"

=!FN:FMa b

"

=!FN:FMa bc

"

=F⊗!N:M a bc

"

.

So 0 F ⊗!N:Ma b" →⊆ F ⊗!N :M a bc" → 0 is an exact sequence. Since F is a faith-fully flat module, 0!N:Ma b

" ⊆

→!N:Ma bc

"

→0 is an exact sequence which implies that

!

N:M a b"=!N:M a bc". ConsequentlyN is a classical 2-absorbing submodule ofM by The-orem 4.

Corollary 5. Let R be a um-ring, M be an R-module and X be an indeterminate. If N is a classical 2-absorbing submodule of M , then N[X]is a classical 2-absorbing submodule of M[X].

Proof. Assume that N is a classical 2-absorbing submodule of M. Notice that R[X] is a flat R-module. So by Theorem 6,R[X]⊗N 0 N[X] is a classical 2-absorbing submodule of R[X]M0M[X].

For anR-moduleM, the set of zero-divisors ofM is denoted byZR(M).

Proposition 10. Let M be an R-module, N be a submodule and S be a multiplicative subset of R. (i) If N is a classical 2-absorbing submodule of M such that!N:RM

"

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(ii) If S−1N is a classical 2-absorbing submodule of S−1M such that ZR(M/N)∩S=., then N is a classical 2-absorbing submodule of M .

Proof. (i)LetNbe a classical 2-absorbing submodule ofM and!N:RM"∩S=.. Suppose that a1

s1

a2 s2

a3

s3

m s4 ∈S

−1N. Then there exist nN andsS such that a1

s1

a2 s2

a3

s3

m s4 =

n

s. Therefore there exists ans)∈Ssuch thats)sa1a2a3m=s)s1s2s3s4nN. Soa1a2a3(sm)∈Nfors∗=s)s. Since N is a classical 2-absorbing submodule we get a1a2(sm) N or a1a3(sm) N or a2a3(sm)∈N. Thus a1a2m

s1s2s4 =

a1a2(sm)

s1s2s4s∗ ∈S−1N or as11sa33sm4S−1N or a2a3m

s2s3s4S−1N.

(ii)Assume thatS−1Nis a classical 2-absorbing submodule ofS−1MandZR(M/N)S=.. Leta,b,cRandmM such thata bcmN. Then a

11b1cm1 ∈S−1N. Therefore a11bm1 ∈S−1N or a

1c1m1 ∈S−1N or 1b1cm1 ∈S−1N. We may assume that a11bm1 ∈S−1N. So there existsuS such thatua bmN. ButZR(M/N)∩S=., whence a bmN. ConsequentlyN is a classical 2-absorbing submodule ofM.

Let Ri be a commutative ring with identity and Mi be an Ri-module, for i = 1, 2. Let R=RR2. ThenM = MM2 is anR-module and each submodule ofM is in the form of N =N1×N2 for some submodulesN1ofM1 andN2 ofM2.

Theorem 7. Let R=RR2 be a decomposable ring and M=MM2 be an R-module where M1 is an R1-module and M2 is an R2-module. Suppose that N=N1×N2is a proper submodule of M . Then the following conditions are equivalent:

(i) N is a classical 2-absorbing submodule of M ;

(ii) Either N1 =M1 and N2 is a classical 2-absorbing submodule of M2or N2= M2 and N1is a classical 2-absorbing submodule of M1 or N1, N2 are classical prime submodules of M1, M2,respectively.

Proof. (i)⇒(ii)Suppose thatN is a classical 2-absorbing submodule ofM such that N2=M2. From our hypothesis,Nis proper, soN1#=M1. SetM)={0}M×M

2. HenceN

)= N

{0}×M2

is a classical 2-absorbing submodule of M) by Corollary 1. Also observe that M) ∼= M1 and N)∼=N1. ThusN1is a classical 2-absorbing submodule of M1. Suppose thatN1#=M1 and N2#=M2. We show thatN1is a classical prime submodule ofM1. Since N2#=M2, there exists m2M2\N2. Leta bm1N1 for somea,bR1andm1M1. Thus

(a, 1)(b, 1)(1, 0)(m1,m2) = (a bm1, 0)∈N=N1×N2.

So either(a, 1)(1, 0)(m1,m2) = (am1, 0)∈N or (b, 1)(1, 0)(m1,m2) = (bm1, 0)∈N. Hence either am1N1 or bm1N1 which shows that N1 is a classical prime submodule of M1. Similarly we can show thatN2is a classical prime submodule of M2.

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Lemma 1. Let R=R1×R2×· · ·×Rnbe a decomposable ring and M =M1×M2×· · ·×Mnbe an R-module where for every1in, Miis an Ri-module, respectively. A proper submodule N of M is a classical prime submodule of M if and only if N =×n

i=1Nisuch that for some k∈{1, 2, . . . ,n}, Nkis a classical prime submodule of Mk, and Ni=Mi for every i∈{1, 2, . . . ,n}\{k}.

Proof. (⇒) LetN be a classical prime submodule of M. We knowN =×n

i=1Ni where for every 1in,Ni is a submodule ofMi, respectively. Assume thatNris a proper submodule ofMrandNs is a proper submodule ofMsfor some 1r <sn. So, there are mrMr\Nr andmsMs\Ns. Since

(0, . . . , 0, r-th

+,-.

1R

r , 0, . . . , 0)(0, . . . , 0,

s-th

+,-.

1R

s , 0, . . . , 0)(0, . . . , 0,

r-th

+,-.

mr , 0, . . . , 0, s-th

+,-.

ms , 0, . . . , 0) =(0, . . . , 0)∈N,

then either

(0, . . . , 0, r-th

+,-.

1Rr , 0, . . . , 0)(0, . . . , 0, r-th

+,-.

mr , 0, . . . , 0, s-th

+,-.

ms , 0, . . . , 0) =(0, . . . , 0,

r-th

+,-.

mr , 0, . . . , 0)∈N or

(0, . . . , 0, s-th

+,-.

1R

s , 0, . . . , 0)(0, . . . , 0,

r-th

+,-.

mr , 0, . . . , 0, s-th

+,-.

ms , 0, . . . , 0) =(0, . . . , 0,

s-th

+,-.

ms , 0, . . . , 0)N,

which is a contradiction. Hence exactly one of theNi’s is proper, sayNk. Now, we show that Nk is a classical prime submodule of Mk. Let a bmkNk for some a,bRk and mkMk. Therefore

(0, . . . , 0, k-th

+,-.

a , 0, . . . , 0)(0, . . . , 0, k-th

+,-.

b , 0, . . . , 0)(0, . . . , 0, k-th

+,-.

mk , 0, . . . , 0)

=(0, . . . , 0, k-th

+ ,- .

a bmk, 0, . . . , 0)∈N, and so

(0, . . . , 0, k-th

+,-.

a , 0, . . . , 0)(0, . . . , 0, k-th

+,-.

mk , 0, . . . , 0) = (0, . . . , 0, k-th

+,-.

amk, 0, . . . , 0)∈N or

(0, . . . , 0, k-th

+,-.

b , 0, . . . , 0)(0, . . . , 0, k-th

+,-.

mk , 0, . . . , 0) = (0, . . . , 0, k-th

+,-.

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Theorem 8. Let R=R1×R2×· · ·×Rn(2n<∞)be a decomposable ring and

M = M1×M2 ×· · ·× Mn be an R-module where for every 1 i n, Mi is an Ri-module, respectively. For a proper submodule N of M the following conditions are equivalent:

(i) N is a classical 2-absorbing submodule of M ;

(ii) Either N = ×nt=1Nt such that for some k ∈ {1, 2, . . . ,n}, Nk is a classical 2-absorbing submodule of Mk, and Nt=Mt for every t∈{1, 2, . . . ,n}\{k}or Nnt=1Ntsuch that for some k,m∈{1, 2, . . . ,n}, Nkis a classical prime submodule of Mk, Nm is a classical prime submodule of Mm, and Nt=Mt for every t∈{1, 2, . . . ,n}\{k,m}.

Proof. We argue induction on n. For n = 2 the result holds by Theorem 7. Then let 3n<∞and suppose that the result is valid whenK=M1×· · ·×Mn1. We show that the result holds whenM=K×Mn. By Theorem 7,Nis a classical 2-absorbing submodule of Mif and only if eitherN=L×Mnfor some classical 2-absorbing submoduleLofK orN=K×Ln for some classical 2-absorbing submodule Ln of Mn or N = L×Ln for some classical prime submodule LofK and some classical prime submoduleLn ofMn. Notice that by Lemma 1, a proper submodule LofKis a classical prime submodule of K if and only if Lnt=11Nt such that for somek∈{1, 2, . . . ,n−1}, Nk is a classical prime submodule of Mk, andNt =Mt for everyt∈{1, 2, . . . ,n−1}\{k}. Consequently we reach the claim.

References

[1] R. Ameri. On the prime submodules of multiplication modules, International Journal of Mathematics and Mathematical Sciences, 27, 1715–1724. 2003.

[2] D. F. Anderson and A. Badawi.On n-absorbing ideals of commutative rings, Comunications in Algebra, 39, 1646–1672. 2011.

[3] A. Azizi.On prime and weakly prime submodules, Vietnam Journal of Mathematics, 36(3), 315–325. 2008.

[4] A. Azizi.Weakly prime submodules and prime submodules, Glasgow Mathematical Journal, 48, 343–346. 2006.

[5] A. Badawi.On2-absorbing ideals of commutative rings, Bulletin of the Australian Mathe-matical Society, 75, 417–429. 2007.

[6] A. Badawi, Ü. Tekir, and E. Yetkin.On2-absorbing primary ideals in commutative rings, Bulletin of the Korean Mathematical Society, 51(4), 1163–1173. 2014.

[7] A. Badawi, E. Yetkin, and Ü. Tekir.On weakly2-absorbing primary ideals of commutative rings, Journal of the Korean Mathematical Society, 52(1), 97–111. 2015.

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[9] M. Behboodi. A generalization of Bear’s lower nilradical for modules, Journal of Algebra and its Applications, 6(2), 337-353. 2007.

[10] M. Behboodi. On weakly prime radical of modules and semi-compatible modules, Acta Mathematical Hungarica, 113(3), 239-250. 2006.

[11] M. Behboodi and H. Koohy. Weakly prime modules, Vietnam Journal of Mathematics, 32(2), 185-195. 2004.

[12] M. Behboodi and S. H. Shojaee. On chains of classical prime submodules and dimension theory of modules, Bulletin of the Iranian Mathematical Society, 36(1), 149–166. 2010.

[13] A. Y. Darani and F. Soheilnia. 2-absorbing and weakly2-absorbing submodules, Thai Jour-nal of Mathematics, 9(3), 577–584. 2011.

[14] A. Y. Darani and F. Soheilnia.On n-absorbing submodules, Mathematical Communications, 17, 547-557. 2012.

[15] Sh. Payrovi and S. Babaei.On 2-absorbing submodules, Algebra Colloquium, 19, 913–920. 2012.

[16] P. Quartararo and H. S. Butts. Finite unions of ideals and modules, Proceedings of the American Mathematical Society, 52, 91-96. 1975.

References

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