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### Article

Kyungpook Mathematical Journal 2021; 61(1): 33-48

Published online March 31, 2021

### OnWeakly Prime andWeakly 2-absorbing Modules over Non-commutative Rings

Nico J. Groenewald

Department of Mathematics, Nelson Mandela University, Port Elizabeth, South Africa
e-mail : nico.groenewald@mandela.ac.za

Received: March 22, 2020; Revised: September 8, 2020; Accepted: October 8, 2020

Most of the research on weakly prime and weakly 2-absorbing modules is for modules over commutative rings. Only scatterd results about these notions with regard to non-commutative rings are available. The motivation of this paper is to show that many results for the commutative case also hold in the non-commutative case. Let R be a non-commutative ring with identity. We define the notions of a weakly prime and a weakly 2-absorbing submodules of R and show that in the case that R commutative, the definition of a weakly 2-absorbing submodule coincides with the original definition of A. Darani and F. Soheilnia. We give an example to show that in general these two notions are different. The notion of a weakly m-system is introduced and the weakly prime radical is characterized interms of weakly m-systems. Many properties of weakly prime submodules and weakly 2-absorbing submodules are proved which are similar to the results for commutative rings. Amongst these results we show that for a proper submodule Ni of an Ri-module Mi, for i=1,2, if N1×N2 is a weakly 2-absorbing submodule of M1×M2, then Ni is a weakly 2-absorbing submodule of Mi for i=1,2

Keywords: 2-absorbing submodule, weakly 2-absorbing submodule, prime submodule, weakly prime submodule, weakly prime radical.

In 2007 Badawi [3] introduced the concept of 2-absorbing ideals of commutative rings with identity, which is a generalization of prime ideals, and investigated some properties. He defined a 2-absorbing ideal P of a commutative ring R with identity to be a proper ideal of R such that if a,b,cR and abcP, then abP or bcP or acP. In 2011, Darani and Soheilnia [7] introduced the concepts of 2-absorbing and weakly 2-absorbing submodules of modules over commutative rings with identities. A proper submodule P of a module M over a commutative ring R with identity is said to be a 2-absorbing submodule (weakly 2-absorbing submodule) of M if whenever a,bR and mM with abmP(0abmP), then abMP or amP or bmP. One can see that 2-absorbing and weakly 2-absorbing submodules are generalizations of prime submodules. Moreover, it is obvious that 2-absorbing ideals are special cases of 2-absorbing submodules.

Throughout this paper, all rings are associative with identity elements (not necessarily commutative) and modules are unitary left modules. Let R be a ring and M be an R-module. We write NM, if N is a submodule of M. In recent years the study of the absorbing properties of rings and modules, and related notions, have been topics of interest in ring and module theory. In [11] the notion of 2-absorbing modules over non-commutative rings was introduced. In this paper we study the notion of weakly prime and weakly 2-absorbing modules over non-commutative rings. We prove basic properties of weakly 2-absorbing submodules. In particular, we show that If R is a commutative ring then the notion of a weakly 2-absorbing submodule coincides with that of the original definition introduced by Darani and Soheilnia in [7]. For an R-module M and a submodule N of M we have (N:RM)={rR:rMN}.

Following [9] a proper ideal P of the ring R is 2-absorbing if aRbRcP implies abP or acP or bcP for a,b and c elements of R. Following [11] a proper submodule N of the R-module M is a 2-absorbing submodule of M if aRbRxN implies ab(N:RM) oraxN or bxN for a,bR and xM. From [8] a proper submodule P of M is called a prime submodule of M if, for every ideal A of R and every submodule N of M,ANP implies either NP or AMP. This is equivalent to aRxP implies a(P:M) or xP for aR and xM. It is clear that a submodule N of an R-module M is prime if and only P=(N:RM) is a prime ideal of R.

From [10] A proper ideal P of the ring R is weakly prime if 0aRbP implies aP or bP for a and b elements of R.

### Definition 1.1.

([1, Definition 3.3]) Let M be a left R-module. A proper submodule N of M is called a weakly prime submodule of M if whenever rR and mM with 0rRmN then either mN or r(N:RM).

### Remark 1.2.

Let p and q be two prime numbers. In the -module pq, the submodule (0) is weakly prime, but not prime.

Compare the next Theorem with [2, Corollary 2.3].

### Theorem 1.3.

Let R be a ring, and M an R-module and N a weakly prime submodule of M. If N is not a prime submodule of M, then for any subset P of R such that P(N:RM) we have PN=0. In particular (N:RM)N=0.

Proof. Suppose P is a subset of R such that P(N:RM). Suppose PN0. We show that N is prime. Let r ∈ R and m ∈ M be such that rRmN. If rRm0, then r(N:RM) or mN since N is weakly prime. So assume rRm=0. First assume rN0, say rN0 for some nN. Now 0rnrR(n+m)N and N weakly prime, gives r(N:RM) or (n+m)N. Hence r(N:RM) or mN since nN. So we can assume that rN=0. Now suppose that Pm0, say sm0 for sP(N:RM). We have 0sm(r+s)RmN. Hence (r+s)(N:RM) or mN. So r(N:RM) or mN. Hence we can assume that Pm=0. Since PN0, there exists tP and nN such that tn0. Now we have 0tn(r+t)R(n+m)N. Again, since N is weakly prime, we get (r+t)(N:RM) or (m+n)N. Hence r(N:RM) or mN. Thus N is a prime submodule.

Compare (1) (2) of the next Theorem with [2, Theorem 2.4]

### Theorem 1.4.

Let N be a proper submodule of a left R-module M. Then the following are equivalent:

• (1) N is a weakly prime submodule of M.

• (2) For a left ideal P of R and submodule D of M with 0PDN, either P(N:RM) or DN.

• (3) For any element aR and LM, if 0aRLN, then LN or a(N:RM).

• (4) For any right ideal I of R and LM, if 0ILN, then LN or I(N:RM).

• (5)] For any element aR and LM, if 0RaRLN, then LN or a(N:RM).

• (6)] For any element aR and LM, if 0RaLN then LN or a(N:RM).

Proof.

(1) (2) Suppose that N is a weakly prime submodule of M. If N is prime, then the result is clear from [5, Proposition 1.1]. So we can assume that N is weakly prime that is not prime. Let 0PDN with xDN. We show that P(N:RM). Let rP. Now rRxrDN. If 0rRx, then N weakly prime gives r(N:RM). So assume that rRx=0. First suppose that rD0, say rD0 where dD. If dN, then since 0rRdN and N weakly prime r(N:RM). If dN, then rR(d+x)=rRdN, so r(N:RM) or (d+x)N. Thus, r(N:RM); hence P(N:RM). So we can assume that rD=0. Suppose that Px0, say ax0 where aP. Now 0aRxN and N weakly prime gives a(N:RM). As (r+a)Rx=aRxN, we get r(N:RM), so P(N:RM). Therefore, we can assume that Px=0. Since PD0, there exist bP and d1D such that bd10. As (N:RM)N=0 (by Theorem 1.3) and 0b(d1+x)=bd1N we can divide the proof into the following two cases:

Case 1. b(N:RM) and (d1+x)N. Since 0(r+b)R(d1+x)=bRd1N, we obtain (r+b)(N:RM), so r(N:RM). Hence P(N:RM).

Case 2. b(N:RM) and (d1+x)N. As 0bRd1N we have d1N, so xN which is a contradiction. Thus P(N:RM).

(2) (1) Suppose that 0sRmN where sR and mM. Take I=Rs and D=Rm. Then 0IDN, so either I(N:RM) or DN; hence either r(N:RM) or mN. Thus N is weakly prime.

(2) (3) Let aR and LM such that 0aRLN. Now 0RaLN and from 2. LN or aRa(N:RM).

(3) (2) Let P be a left ideal of R and D a submodule of M with 0PDN. If DN, then we are done. So suppose DN. We will show that P(N:RM). Let aP. Hence aRDN. If aRD0 then it follows from (3) that a(N:RM) and we have P(N:RM). So suppose aRD=0. Because PD0, there exists pP such that pRD0. We now have 0pRD=(a+p)RDN. It follows from (3) that (a+p)(N:RM) and we have P(N:RM).

(3) (4) (5) (6) is now easy to see.

### Remark 1.5.

From [5] we know that if N is a prime submodule of an R-module M, then (N:RM) is a prime ideal of R. Suppose that N is weakly prime which is not prime. Contrary to what happens for a prime submodules, the ideal (N:RM) is not, in general, a weakly prime ideal of R. For example, let M denote the cyclic -module /8. Take N={0}. Certainly N is a weakly prime submodule of M, but (N:RM)=8Z is not a weakly prime ideal of R, but we have the following result:

### Proposition 1.6.

Let R be a ring with identity M a faithful R-module, and N a weakly prime submodule of M. Then (N:RM) is a weakly prime ideal of R.

Proof. Assume that M is a faithful R module and let 0aRb(N:RM). Since M is a faithful R module we have 0aRbMN. It follows that 0(RaR)(RbR)MN. From Theorem 1.4 we have (RaR)MN or (RbR)MN. Hence a (N:RM) or b (N:RM) and it follows that (N:RM) is a weakly prime ideal.

From [12] we have that M is a multiplication module over a non-commutative ring if and only if (N:M)M=N for each submodule N of M.

### Proposition 1.7.

Let M be a multiplication R-module. If (N:M) is a weakly prime ideal of R, then N is a weakly prime submodule of M.

Proof. Let 0aRmN with mM and a(N:M). Since M is a multiplication module there is an ideal I of R such that Rm=IM, then 0RaIMN. Hence 0RaI(N:M). Since (N:M) is a weakly prime ideal of R, we have Ra(N:M) or I(N:M). Since a(N:M), we have I(N:M). Hence Rm=IMN. Thus mN and N is a weakly prime submodule of M.

### Remark 1.8.

The converse of Proposition 1.7 is not true in general. Suppose that M=× is an R=×-module and N=2×{0} is a submodule of M. (N:M)=0 is a weakly prime ideal. We have (0,0)(2,0)(1,1)2×{0}. Now, neither (2,0)(N:M) nor (1,1)N. Hence N is not weakly prime. Notice that M is not a multiplication module.

### Lemma 1.9.

Let R be a ring, M an R-module and N a weakly prime submodule of M. If 0aRbRmN and amN, then bMN for all a;bR and mM.

Proof. Let a,bR and mM. Assume that 0aRbRmN and amN. Now we have 0(RaR)(RbR)mN. From Theorem 1.4 we have RaRMN or RbRmN. Since amN we have RbRmN. Because 0aRbRm we must have 0bRmN. Now, since N is weakly prime we get bMN or mN. Since amN, we must have bMN and we are done.

The following result gives characterizations of weakly prime submodules.

### Theorem 1.10.

Let M be an R-module. The following asserations are equivalent:

• (1) P is a weakly prime submodule of M.

• (2) (P:Rx)=(P:M) ∪ (0:Rx) for any x ∈ M-P.

• (3) (P:Rx)=(P:M) or (P:Rx)=(0:Rx) for any x ∈ M-P.

Proof. (1)(2) Let r(P:Rx) and xP. Then rRxP. Suppose rRx0. Hence r(P:M) because P is weakly prime and xP. If rRx=0, then r(0:Rx). Thus (P:Rx)(P:M)(0:Rx). Now if r(P:M)(0:Rx) then either r(P:M) or r(0:Rx). Hence, when r(0:Rx), rRx=0P and so r(P:Rx). If r(P:M) then rMP, and this implies rRxrMP. Hence r(P:Rx) and therefore (P:Rx)=(P:M)(0:Rx).(2)(3) Is obvious. (3)(1) Suppose that 0rRxP with rR and xMP. Then r(P:Rx) and r(0:Rx). It follows from (3) that r(P:Rx)=(P:M), as required.

### Proposition 1.11.

Let M1 and M2 be unitary R-modules over a ring R. Let M=M1M2 and NM1M2. Then the following are satisfied:

• (1) N=Q ⊕ M2 is a weakly prime submodule of M if and only if Q is a weakly prime submodule of M1 and r ∈ R, x ∈ M1 with rRx=0, but x ∉ Q, r∉(Q:M1) implies rM2=0.

• (2) N=M1 ⊕ Q is a weakly prime submodule of M if and only if Q is a weakly prime submodule of M2 and r ∈ R, x∈ M2 with rRx=0, but x ∉ Q, r∉(Q:M2) implies rM1=0.

Proof. We will prove (1) and the proof of (2) will be similar. () Let N=QM2 be a weakly prime submodule of M. Let 0rRqQ , qQ . Then (q,0)QM2, while 0rR(q,0)QM2. Since N=QM2 is a weakly prime submodule of M we have r(M1M2:QM2). Hence rM1Q and Q is a weakly prime submodule of M1. Now, suppose rR, xM1 such that rRx=0, but xQ, r(Q:M1). Assume that rM20, so there esists mM2 such that rm0. Thus (0,0)rR(x,m)=(rRx,rRm)=(0,rRm)QM2=N. N is a weakly prime submodule of M, so either (x,m)QM2 or r(QM2:M1M2). Thus either xQ or r(Q:M1) which is a contradiction with hypothesis, hence rM2=0.() Let rR and (x,y)M. Assume (0,0)rR(x,y)QM2, so if rRx0, then xQ or r(Q:M1), since Q is a weakly prime submodule of M1. Thus either (x,y)QM2=N or r(N:M). If rRx=0, suppose xQ, r(Q,M1). Then by hypothesis rM2=0 and so rRyrM2=0. Hence rR(x,y)=(0,0) which is a contradiction. Thus either xQ or r(Q:M1) and hence either (x,y)QM2=N or r(QM2:M1M2).

### Remark 1.12.

Let M1 and M2 be R-modules. If (0) is a prime submodule of M1, then (0)M2 is a weakly prime submodule of M1M2.

Proof. Let rR and (x,y)M. If (0,0)rR(x,y)(0)M2, then rRx=0 and rRyM2. Since (0) is a prime submodule of M1, either x=0 or r((0):M1). Hence either (x,y)=(0,y)(0)M2 or r((0)M2:M1M2), that is (0)M2 is a weakly prime submodule of M1M2.

### Proposition 1.13.

Let M1 and M2 be R-modules. If UW is a weakly prime submodule of M1M2, then U and W are weakly prime submodules of M1 and M2 respectively.

Proof. The proof is straight forward so it is omitted.

### Remark 1.14.

The converse of Proposition 1.13 is not true in general as the following example shows.

### Example 1.15.

Suppose M= is a -module and consider the submodule N=p{0} of M. p is a prime submodule of the -module and hence also a weakly prime submodule and {0} is a weakly prime submodule of the module . N=p{0} is not weakly prime since (0,0)p(1,0)p{0} but p(p{0}:) and (1,0)p{0}.

We begin this section with the definition of weakly m-systems.

### Definition 2.1

Let R be a ring and M be an R-module. A nonempty set SM\{0} is called a weakly m-system if, for each ideal A of R, and for all submodules K,LM, if (K+L)S, (K+AM)S, and AL0 then (K+AL)S.

### Proposition 2.2.

Let M be an R-module. Then a submodule P of M is weakly prime if and only if M\P is a weakly m-system.

Proof. Suppose S=M\P. Let A be an ideal in R and K and L be submodules of M such that (K+L)S, (K+AM)S and AL0. If (K+AL)S= then K+ALP. Hence ALP and since P is weakly prime, and AL0, LP or AMP. It follows that (K+L)S= or (K+AM)S=, a contradiction. Therefore, S is a weakly m-system in M. Conversely, let S=M\P be a weakly m-system in M. Suppose ALP and AL0, where A is an ideal of R and L is a submodule M. If LP and AMP, then LS and AMS. Thus, ALS, a contradiction. Therefore, P is a weakly prime submodule of M.

The following proposition offers several characterizations of a weakly m-system S when it is the complement of a submodule.

### Proposition 2.3.

Let R be a ring and M be an R-module. Let P be a proper submodule of M, and let S:=M\P. Then the following statements are equivalent:

• (1) P is weakly prime;

• (2) S is a weakly m-system;

• (3) for each left ideal AR, and for every submodule LM, if LS, AMS and AL0 then ALS;

• (4) for each ideal AR, and for every mM, if RmS, AMS= and AL0, then ARmS;

• (5) for each a∈ R, and for each mM, if RmS, aMS and aRm0, then aRmS.

Proof. (1) (2) follows from Proposition 2.2. (2) (3) (4) (5) is clear (5) (1). Suppose a ∈ R and m ∈ M with 0aRmP. If RmP and aMP, then RmS and aMS and aRm0. From (5) aRmS. Hence aRmP a contradiction. Hence RmP or aMP and P is weakly prime.

### Proposition 2.4.

Let M be an R-module, SM be a weakly m-system, and let P be a submodule of M maximal with respect to the property that P is disjoint from S. Then P is a weakly prime submodule.

Proof. Suppose 0ALP, where A is an ideal of R and LM. If LP and AMP, then by the maximal property of P, we have, (P+L)S and (P+AM)S. Thus, since S is a weakly m-system (P+AL)S and it follows that PS, a contradiction. Thus, P must be a weakly prime submodule.

Next we need a generalization of the notion of N for any submodule N of M. We adopt the following:

### Definition 2.5.

Let R be a ring and M be an R-module. For a submodule N of M, if there is a weakly prime submodule containing N, then we define N:={mM : every weakly m-system containing m meets N}. If there is no weakly prime submodule containing N, then we put N=M.

### Theorem 2.6.

Let M be an R-module and NM. Then either N=M or N equals the intersection of all the weakly prime submodules of M containing N.

Proof. Suppose that NM. This means that {P|P is a weakly prime submodule of M and NP}. We first prove that N{P|P is a weakly prime submodule of M and NP}. Let mN and P be any weakly prime submodule of M containing N. Consider the m-system M\P. This m-system cannot contain m, for otherwise it meets N and hence also P. Therefore, we have mP. Conversely, assume mN. Then, by Definition 2.5, there exists an m-system S containing m which is disjoint from N. By Zorn's Lemma, there exists a submodule PN which is maximal with respect to being disjoint from S. By Proposition 2.4, P is a weakly prime submodule of M, and we have mP, as desired.

### 3. Weakly 2-absorbing Submodules

From [11] we have the following:

### Definition 3.1.

Let P be a proper ideal of a ring R. Then P is a 2-absorbing ideal of R if aRbRP implies abP or bc ∈ P or ac ∈ P for all a;b;cR.

### Definition 3.2.

Let R be a ring and N be a proper submodule of an R-module M. Then N is 2-absorbing submodule of M if aRbRmN implies abMN i.e. ab(N:RM) or amN or bmN for all a;bR and mM.

### Remark 3.3.

If R is a commutative ring then this notion of a 2-absorbing submodule coincides with that of Darani and Soheilnia [7].

We now have the following:

### Definition 3.4.

Let R be a ring and N be a proper submodule of an R-module M. Then N is a weakly 2-absorbing submodule of M if 0aRbRmN implies abMN i.e. ab(N:RM) or amN or bmN for all a;bR and mM.

### Remark 3.5.

Every 2-absorbing submodule is weakly 2-absorbing but the converse does not necessarily hold. For example consider the case where R=,M=/30 and N=0. Then 2.3.(5+30)=0N while 2.3(N:RM), 2.(5+30Z)N and 3.(5+30Z)N. Therefore N is not 2-absorbing while it is weakly 2-absorbing.

### Proposition 3.6.

Let x ∈ M and a ∈ R. Then if annl(x)(Rx:M), the submodule Rx is 2-absorbing if and only if Rx is weakly 2- absorbing.

Proof. Let Rx be a weakly 2-absorbing submodule of M and suppose r,s ∈ R and m ∈ M with rRsRmRx. Since Rx is a weakly 2-absorbing submodule, we may assume rRsRm=0, otherwise Rx is 2-absorbing. Now rRsR(x+m)Rx. If rRsR(x+m)0 then we have rs(Rx:M) or r(x+m)Rx or s(x+m)Rx, as Rx is a weakly 2-absorbing submodule. Hence rs(Rx:M) or rm ∈ Rx or sm ∈ Rx. Now let rRsR(x+m)=0. Then rRsRm=0 implies rRsRx=0. Hence rsannl(x)(Rx:M). Thus Rx is 2-absorbing.

### Proposition 3.7.

Let R be a ring and N be a proper submodule of an R-module M. If N is weakly prime, then it is weakly 2-absorbing.

Proof. Assume N is a weakly prime submodule of the R-module M and 0aRbRmN for all a;bR and mM. Suppose amN. It now follows from Proposition 1.9 that bMN and consequently abMN. Hence N is weakly 2-absorbing.

Compare the following theorem with that of [7, Theorem 2.3(ii)].

### Theorem 3.8.

The intersection of each pair of weakly prime submodules of an R-module M is a weakly 2-absorbing submodule of M.

Proof. Let N and K be two weakly prime submodules of M. If N=K, then NK is a weakly prime submodule of M so that NK is a weakly 2-absorbing submodule of M. Assume that N and K are distinct. Since N and K are proper submodules of M, it follows that NK is a proper submodule of M. Next, let a,b∈ R and m ∈ M be such that 0aRbRmNK but amNK and ab(NK:M). Then, we can conclude that (a) am ∉ N or am ∉ K, and (b) ab(N:RM) or ab(K:RM). These two conditions give 4 cases:

• (1) am ∉ N and ab ∉ (N:RM);

• (2) am ∉ N and ab ∉ (K:RM);

• (3) am ∉ K and ab ∉ (N:RM);

• (4) am ∉ K and ab ∉ (K:RM).

We first consider Case(1). Since 0aRbRmNKN and amN, it follows from Proposition 1.9 that bMN. This is a contradiction because ab(N:RM). Hence Case(1) does not occur. Similarly, Case(4) is not possible. Next, Case(2) is considered. Again, we obtain that bMN and then bm∈ N. Since 0aRbRmK it follows that 0(RaR)(RbR)m)K. Hence, from the fact that K is weakly prime and from Theorem 1.4 it follows that aMRaRMK or bmRbRmK If aMK, then abMaMK which contradicts ab(K:RM). Thus bm∈ K. Hence bmNK. The proof of Case(3) is similar to that of Case(2). Hence NK is a weakly 2-absorbing submodule of M.

### Definition 3.9.

Let N be a weakly 2-absorbing submodule of M. (a,b,m) is called a triple-zero of N if aRbRm=0, ab(N:RM), amN and bmN.

The following result is an analogue of [6, Theorem 1].

### Theorem 3.10.

Let N be weakly 2-absorbing submodule of M and (a,b,m) be a triple-zero of N for some a,b ∈ R and m∈ M. Then the followings hold.

• (1) aRbN=a(N:RM)m=b(N:RM)m=0.

• (2) a(N:RM)N=b(N:RM)N=(N:RM)bN=(N:RM)bm=(N:RM)2m=0.

Proof. Suppose that (a,b,m) is a triple-zero of N for some a,b ∈ R and m ∈ M.

(1) Assume that aRbN0. Then there is an element n ∈ N such that aRbRn0. Now aRbR(m+n)=aRbRm+aRbRn=aRbRn0 since aRbRm=0 because (a,b,m) is a triple-zero of N. Since 0aRbR(m+n)N and N weakly 2-absorbing we have ab(N:RM) or a(m+n)N or b(m+n)N. Since (a,b,m) is a triple-zero of N,ab(N:RM). Hence a(m+n)N or b(m+n)N and consequently am ∈ N or bm ∈ N a contradiction. Hence aRbN=0. Now, we suppose that a(N:RM)m0. Thus there exists an element r(N:RM) such that arm0. Hence aR(r+b)rm=aRrRm+aRbRm=aRrRm. Since 0armaRrRmN and N weakly 2-absorbing we have a(r+b)(N:RM) or amN or (r+b)mN. Hence ab(N:RM) or amN or bm ∈ N a contradiction since (a,b,m) is a triple-zero of N. Similarly, it can be easily seen that b(N:RM)m=0.

(2) Assume that a(N:RM)N0. Then there are r(N:RM), n ∈ N such that arn0. By (1), we get a(b+r)(m+n)=abm+abn+arm+arn=arn0. Now 0aR(b+r)R(m+n)N. Therefore, we have a(b+r)(N:RM) or a(m+n)N or (b+r)(m+n)N and we obtain ab(N:RM) or am ∈ N or bm ∈ N, a contradiction. Hence a(N:RM)N=0. In a similar way, we get b(N:RM)N=0. Now, we suppose that (N:RM)bN0. Then there are r(N:RM), nN such that rbn0. Now, from above (a+r)b(n+m)=abn+abm+rbn+rbm=rbn0. Hence 0(a+r)RbR(n+m)N and since N is weakly 2-absorbing (a+r)b(N:RM) or (a+r)(n+m)N or b(n+m)N. Hence ab(N:RM) or amN or bm ∈ N a contradiction since (a,b,m) is a triple-zero of N. Now, we suppose that (N:RM)bm0. Then there is r(N:RM) such that rbm0. Hence 0rbm=(a+r)bm(a+r)RbRm=rRbRmN. Since N is weakly 2-absorbing, we have (a+r)b(N:RM) or (a+r)mN or bm ∈ N. Therefore ab(N:RM) or am ∈ N or bm ∈ N a contradiction since (a,b,m) is a triple-zero of N. Hence (N:RM)bm=0. Lastly, we show that (N:RM)2m=0. Let (N:RM)2m0. Thus there exist r,s(N:RM) where rsm0. By (1), we get (a+r)(b+s)m=rsm0. Thus we have 0(a+r)R(b+s)RmN. Hence (a+r)(b+s)(N:RM) or (a+r)mN or (b+s)mN.

Consequently, ab(N:RM) or am ∈ N or bm ∈ N a contradiction, since (a,b,m) is a triple-zero of N. Therefore (N:M)2m=0.

The following result is an analogue of [6, Lemma 1].

### Proposition 3.11.

Assume that N is a weakly 2-absorbing submodule of an R-module M that is not 2-absorbing. Then (N:RM)2N=0. In particular, (N:RM)3Ann(M).

Proof. Suppose that N is a weakly 2-absorbing submodule of an R-module M that is not 2-absorbing. Then there is a triple-zero (a,b,m) of N for some a,b ∈ R and m ∈ M. Assume that (N:RM)2N0. Thus there exist r,s(N:RM) and n ∈ N with rsn0. By Theorem 3.10, we get (a+r)(b+s)(n+m)=rsn0. Then we have 0(a+r)R(b+s)R(n+m)N. Since N is weakly 2-absorbing we have (a+r)(b+s)(N:RM) or (a+r)(n+m)N or (b+s)(n+m)N and so ab(N:RM) or am ∈ N or bm ∈ N, which is a contradiction.

Thus (N:RM)2N=0. We get (N:RM)3((N:RM)2N:M)=(0:M)=Ann(M).

### 4. On a Question from Badawi and Yousefian

In [4], the authors asked the following question:

Question. Suppose that L is a weakly 2-absorbing ideal of a ring R and 0IJKL for some ideals I,J,K of R. Does it imply that IJL or IKL or JKL?

This section is devoted to studying the above question and its generalization in modules over non-commutative rings.

### Definition 4.1.

Let N be a weakly 2-absorbing submodule of an R-module M and let 0I1I2KN for some ideals I1,I2 of R and some submodule K of M. N is called free triple-zero in regard to I1 ,I2,K if (a,b,m) is not a triple-zero of N for every aI1,bI2 and m ∈ K.

The following result and its proof are analogous of [6, Lemma 2].

### Lemma 4.2.

Let N be a weakly 2-absorbing submodule of M. Assume that aRbKNfor some a,b ∈ R and some submodule K of M where (a,b,m) is not a triple-zero of N for every m ∈ K. If ab(N:RM), then aKN or bKN.

Proof. Assume that aKN and bKN. Then there are x,y ∈ K such that axN and byN. We get bx ∈ N since N is a weakly 2-absorbing submodule, (a,b,x) is not a triple-zero of N, ab(N:RM) and axN. In a similar way, ayN. Now, aRbR(x+y)N and since (a,b,x+y) is not a triple-zero of N and ab(N:RM) we have a(x+y) ∈ N or b(x+y) ∈ N. Assume that a(x+y)=(ax+ay) ∈ N. As ayN, we get ax ∈ N, a contradiction. Assume that b(x+y)=(bx+by) ∈ N. As bx ∈ N, we get by ∈ N, a contradiction again. Hence we obtain that aKN or bKN.

Let N be a weakly 2-absorbing submodule of an R-module M and I1I2KN for some for some ideals I1,I2 of R and some submodule K of M where N is free triple-zero in regard to I1,I2,K. Note that if a ∈ I1,b ∈ I2 and m ∈ K, then ab(N:RM) or am ∈ N or bm ∈ N.

The following result and its proof are analogous of [6, Theorem 1] and its proof.

### Theorem 4.3.

Assume that N is a weakly 2-absorbing submodule of an R-module M and 0IJKN for some ideals I,J of R and some submodule K of M where N is free triple-zero in regard to I,J,K. Then IJ(N:RM) or IKN or JKN.

Proof. Let N be a weakly 2-absorbing submodule of an R-module M and 0IJKN for some ideals I,J of R and some submodule K of M where N is free triple-zero in regard to I,J,K. Suppose IJ(N:RM). We show that IKN or JKN. Assume IKN and JKN. Then a1KN and a2KN where a1 ∈ I and a2 ∈ J. From Lemma 4.2 a1a2(N:RM) since a1Ra2KIJKN and a1KN and a2KN. By our assumption, there are b1I and b2J such that b1b2(N:RM). By Lemma 4.2, we get b1KN or b2KN since b1Rb2KIJKN and b1b2(N:RM). We have the following cases: Case (1) b1KN and b2KN: Since a1Rb2KIJKN and a1KN and b2KN it follows from Lemma 4.2 that a1b2(N:RM). Since b1KN and a1KN, we conclude (a1+b1)KN. On the other hand since (a1+b1)Rb2KN and neither (a1+b1)KN nor b2KN, we get that (a1+b1)b2(N:RM) by Lemma 4.2. But, because (a1+b1)b2=(a1b2+b1b2)(N:RM) and (a1+b1)b2(N:RM), we get b1b2(N:RM) which is a contradiction. Case (2) b2KN and b1KN: By a similar argument to case (1) we get a contradiction. Case (3) b1KN and b2KN:b2KN and a2KN gives (a2+b2)KN. But a1R(a2+b2)KN and neither a1KN nor (a2+b2)KN, hence a1(a2+b2)(N:RM) by Lemma 4.2. Since a1a2(N:RM) and (a1a2+a1b2)(N:RM), we have a1b2(N:RM). Since (a1+b1)Ra2KN and neither a2KN nor (a1+b1)KN, we conclude (a1+b1)a2(N:RM) by Lemma 4.2. But (a1+b1)a2=a1a2+b1a2, so (a1a2+b1a2)(N:RM) and since a1a2(N:RM), we get b1a2(N:RM). Now, since (a1+b1)R(a2+b2)KN and neither (a1+b1)KN nor (a2+b2)KN, we have (a1+b1)(a2+b2)=(a1a2+a1b2+b1a2+b1b2)(N:RM) by Lemma 4.2. But a1a2,a1b2,b1a2(N:RM), so b1b2(N:RM) which is a contradiction. Consequently IKN or JKN.

### Proposition 5.1.

Let R=R1×R2 and M=M1×M2 where M1 is an R1 module and 0M2 is an R2 module. If N1 is a proper submodule of M1 then the following statements are equivalent:

• (1) N1 is a 2-absorbing submodule of M1;

• (2) N1 × M2 is a 2-absorbing submodule of M1 × M2;

• (3) N1 × M2 is a weakly 2-absorbing submodule of M1 × M2.

Proof. (1) (2) follows from [11, Theorem 2.5]. (2) (3) is clear. We show (3) (1) Let a,bR1 and xM1 such that aR1bR1xN1. For every 0yM2 we have (a,1)(b,1)(x,y)=(abx,y)(0,0). Now (0,0)(abx,y)aR1bR1x×1R11R1yN1×M2. Since N1×M2 is a weakly 2-absorbing submodule of M1×M2, we get (a,1)(b,1)(M1×M2:N1×M2) or (a,1)(x,y)N1×M2 or (b,1)(x,y)N1×M2. Hence ab(N1:M1) or axN1 or bxN1.

### Proposition 5.2.

Let R=R1×R2 and M=M1×M2 where 0M1 is an R1 module and M2 is an R2module. If N2 is a proper submodule of M2 then the following statements are equivalent:

• (1) N2 is a 2-absorbing submodule of M1;

• (2) M1 × N2 is a 2-absorbing submodule of M1 × M2;

• (3) M1 × N2 is a weakly 2-absorbing submodule of M1 × M2.

Proof. Similar to Proposition 5.1.

### Proposition 5.3.

Let R=R1×R2 and M=M1×M2 where M1 is an R1 module and 0M2 is an R2module. Let N1M1. If N1 is a weakly prime submodule of M1 and 0 a prime submodule of M2 then N1×{0} is a weakly 2-absorbing submodule of M1×M2.

Proof. Assume (0,0)(a,b)R(c,d)R(x,y)N1×{0} where (a,b)R,(c,d)R and (x,y)M. Hence 0aR1cR1xN1 and bR2dR2y=0. Since N1 is a weakly prime submodule of M1 we get a(N1:M1) or c(N1:M1) or xN1. Also, since 0 is a prime submodule of M2 and bR2dR2y=0 we have b(0:M2) or d(0:M2) or y=0. In any of the above cases we have (a,b)(c,d)(N1×{0}:M) or (a,b)(x,y)N1×{0} or (c,d)(x,y)N1×{0}.

### Proposition 5.4.

Let R=R1×R2 and M=M1×M2 where 0M1 is an R1 module and 0M2 is an R2 module. If N=N1×N2 is a weakly 2-absorbing submodule of M, N1M1, and N2M2, then N1 and N2 are weakly prime submodules of M1 and M2 respectively.

Proof. Let 0rRxN1, where rR1 and xM1. Consider zM2\N2. Then (0,0)(1,0)R(r,1)R(x,z)N and as N is weakly 2-absorbing, (1,0)(r,1)(N:M) or (r,1)(x,z)N or (1,0)(x,z)N. Note that since zM2\N2, (r,1)(x,z)N. Thus (1,0)(r,1)(N:M)=(N1:M1)×(N2:M2) or (1,0)(x,z)N. Therefore, r(N1:M1) or x ∈ N1. This shows that N1 is a weakly prime submodule of M1. Similarly we can show that N2 is a weakly prime submodule of M2.

### Proposition 5.5.

Let Ni be a proper submodule of an Ri-module Mi, for i=1,2.

If N1×N2 is a weakly 2-absorbing submodule of M1×M2, then

• (1) N1 is a weakly 2-absorbing submodule of M1,

• (2) N2 is a weakly 2-absorbing submodule of M2.

Proof.

• (1) Suppose that N1×N2 is a weakly 2-absorbing submodule of M1×M2. Let a1,a2R1 and mM1 such that 0a1R1a2R1mN1. Clearly, (0,0)(a1,1)(R1×R2)(a2,1)(R1×R2)(m,m2) for any m2N2. Hence (0,0)(a1,1)(R1×R2)(a2,1)(R1×R2)(m,m2)a1R1a2R1m×1R21R2m2N1×N2. Since N1×N2 is a weakly 2-absorbing submodule of M1×M2, (a1,1)(a2,1)(N1×N2:M1×M2) or (a1,1)(m,m2)N1×N2 or (a2,1)(m,m2)N1×N2. Consequently a1a2(N1:M1) or a1mN1 or a2mN1. Hence N1 is a weakly 2-absorbing submodule of M1.

• (2) This follows as in part (1).

The converse of the above proposition is no true in general:

### Example 5.6.

Suppose that M=× is an R=×-module and N=p×{0} is a submodule of M where p is a prime submodule and hence a weakly 2-absorbimg submodule of the module and {0} is weakly 2-absorbing.submodule of the module . Then (N:M)=0. Assume that (0,0)(p,1)(1,0)(1,1)p×{0}. Then neither (p,1)(1,0)(N:M) nor (p,1)(1,1)N nor (1,0)(1,1)N. Hence N is not weakly 2-absorbing.

### Proposition 5.7.

Let Ni be a proper submodule of an R-module Mi, for i=1,2 Then the following conditions are equivalent:

• (1) N1 △ M2 is a weakly 2-absorbing submodule of M1 △ M2;

• (2)

• (a) N1 is a weakly 2-absorbing submodule of M1;

• (b) For each a1,a2 ∈ R and m ∈ M1 such that a1Ra2Rm=0 if a1a2∉(N1:M1) and a1m ∉ N1 and a2m ∉ N1 then a1Ra2M2=0.

Proof. (1) (2).

(a) Suppose N1×M2 is a weakly 2-absorbing submodule of M1×M2. Let a1,a2R and m ∈ M1 such that 0a1Ra2RmN1. Now (0,0)(a1,0)(R×R)(a2,0)(R×R)(m,0)N1×M2. Hence N1×M2 a weakly 2-absorbing submodule of M1×M2 gives (a1a2,0)=(a1,0)(a2,0)(N1×M2:M1×M2) or (a1,0)(m,0)N1×M2 or (a2,0)(m,0)N1×M2. Consequently a1a2(N1:M1) or a1mN1 or a2mN1. Hence N1 is a weakly 2-absorbing submodule of M1

(b) Let a1Ra2Rm=0 with a1a2(N1:M1) and a1mN1 and a2mN1 for a1,a2R and mM1. Suppose a1Ra2M20. Hence there exists m2M2 such that a1Ra2M20 and therefore (0,0)a1Ra2(m,m2)a1Ra2Rm×a1Ra2Rm2=(a1,1)(R×R)(a2,1)(R×R)(m,m2)N1×M2. Since N1×M2 is a weakly 2-absorbing submodule of M1×M2 we have (a1,1)(a2,1)(N1×M2:M1×M2) or (a1,1)(m,m2)N1×M2 or (a2,1)(m,m2)N1×M2. Hence a1a2(N1:M1) or a1mN1 or a2mN1 a contradiction. Hence a1Ra2M2=0.

(2) (1).

Let a1,a2R and (m1,m2)M1×M2 such that (0,0)(a1,a1)(R×R)(a2,a2)(R×R)(m1,m2)N1×M2. If 0a1Ra2Rm1 then 0a1Ra2Rm1N1 and N1 a weakly 2-absorbing submodule of M1 gives a1a2(N1:M1) or a1m1N1 or a2m2N1. Hence (a1,a1)(a2,a2)(N1×M2:M1×M2) or (a1,a1)(m1,m2)N1×M2 or (a2,a2)(m1,m2)N1×M2. Thus N1×M2 is a weakly 2-absorbing submodule of M1×M2. If a1Ra2Rm1=0, then a1Ra2Rm20 and therefore a1Ra2M20. By b. a1a2(N1:M1) or a1m1N1 or a2m2N1. Thus (a1,a1)(a2,a2)(N1×M2:M1×M2) or (a1,a1)(m1,m2)N1×M2 or (a2,a2)(m1,m2)N1×M2. Hence N1×M2 is a weakly 2-absorbing submodule of M1×M2.

### Proposition 5.8.

Let Ni be a submodule of an Ri-module Mi, for i=1,2,3. If N is a weakly 2-absorbing submodule of M1×M2×M3, then N={(0,0,0)} or N is a 2-absorbing submodule of M1×M2×M3.

Proof. Suppose that N is a weakly 2-absorbing submodule of M1×M2×M3 that is not 2-absorbing . We will show that N={(0,0,0)}. Now suppose that N1×N2×N3 {0}×{0}×{0}. Thus Ni{0}, for some i=1,2,3. We claim that N1{0}. There exists m1N1 such that m10. To show that N2=M2 or N3=M3. Assume that N2M2 and N3M3. Thus there exist m2M2 and m3M3 such that m2N2 and m3N3. Since (1,0,1)(1,1,0)(m1,m2,m3)=(m1,0,0)(0,0,0), we have (0,0,0)(1,0,1)(R1×R2×R3)(1,1,0)(R1×R2×R3)(m1,m2,m3)N1×N2×N3. Now, because N1×N2×N3 is a weakly 2-absorbing submodule of M1×M2×M3, we have (1,0,1)(1,1,0)(N1×N2×N3:M1×M2×M3) or (1,0,1)(m1,m2,m3)N1×N2×N3 or (1,1,0)(m1,m2,m3)N1×N2×N3. Hence m2N2 or m3N3 a contradiction. Therefore N=N1×M2×N3 or N=N1×N2×M3. If N=N1×M2×N3, then (0,1,0)(N1×M2×N3:M1×M2×M3). By Proposition 3.11, {0}×M2×{0}=(0,1,0)2N(N:N1×M2×N3)={(0,0,0)}, which is a contradiction. Hence N={(0,0,0)}.

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