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pISSN 1225-6951

### 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 $a,b,c∈R$ and $abc∈P$, then $ab∈P$ or $bc∈P$ or $ac∈P$. 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 $a,b∈R$ and $m∈M$ with $abm∈P(0≠abm∈P)$, then $abM⊆P$ or $am∈P$ or $bm∈P$. 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 $N≤M$, 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 $(N:RM)={r∈R:rM⊆N}.$

Following [9] a proper ideal P of the ring R is 2-absorbing if $aRbRc⊆P$ implies $ab∈P$ or $ac∈P$ or $bc∈P$ 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 $aRbRx⊆N$ implies $ab∈(N:RM)$ or$ax∈N$ or $bx∈N$ for $a,b∈R$ and $x∈M.$ 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 $M,AN⊆P$ implies either $N⊆P$ or $AM⊆P$. This is equivalent to $aRx⊆P$ implies $a∈(P:M)$ or $x∈P$ for $a∈R$ and $x∈M.$ 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 $0≠aRb⊆P$ implies $a∈P$ or $b∈P$ 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 $r∈R$ and $m∈M$ with $0≠rRm⊆N$ then either $m∈N$ or $r∈(N:RM)$.

### Remark 1.2.

Let p and q be two prime numbers. In the $ℤ$-module $ℤpq$, 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 $PN≠0.$ We show that N is prime. Let r ∈ R and m ∈ M be such that $rRm⊆N.$ If $rRm≠0,$ then $r∈(N:RM)$ or $m∈N$ since N is weakly prime. So assume rRm=0. First assume $rN≠0$, say $rN≠0$ for some $n∈N.$ Now $0≠rn∈rR(n+m)⊆N$ and N weakly prime, gives $r∈(N:RM)$ or $(n+m)∈N.$ Hence $r∈(N:RM)$ or $m∈N$ since $n∈N.$ So we can assume that rN=0. Now suppose that $Pm≠0$, say $sm≠0$ for $s∈P⊆(N:RM).$ We have $0≠sm∈(r+s)Rm⊆N.$ Hence $(r+s)∈(N:RM)$ or $m∈N.$ So $r∈(N:RM)$ or $m∈N.$ Hence we can assume that Pm=0. Since $PN≠0,$ there exists $t∈P$ and $n∈N$ such that $tn≠0.$ Now we have $0≠tn∈(r+t)R(n+m)⊆N.$ Again, since N is weakly prime, we get $(r+t)∈(N:RM)$ or $(m+n)∈N.$ Hence $r∈(N:RM)$ or $m∈N.$ 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 $0≠PD⊆N$, either $P⊆(N:RM)$ or $D⊆N.$

• (3) For any element $a∈R$ and $L≤M$, if $0≠aRL⊆N$, then $L⊆N$ or $a∈(N:RM)$.

• (4) For any right ideal I of R and $L≤M$, if $0≠IL⊆N$, then $L⊆N$ or $I⊆(N:RM).$

• (5)] For any element $a∈R$ and $L≤M$, if $0≠RaRL⊆N$, then $L⊆N$ or $a∈(N:RM).$

• (6)] For any element $a∈R$ and $L≤M$, if $0≠RaL⊆N$ then $L⊆N$ 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 $0≠PD⊆N$ with $x∈D−N.$ We show that $P⊆(N:RM)$. Let $r∈P$. Now $rRx⊆rD⊆N.$ If $0≠rRx$, then N weakly prime gives $r∈(N:RM)$. So assume that rRx=0. First suppose that $rD≠0$, say $rD≠0$ where $d∈D$. If $d∉N$, then since $0≠rRd⊆N$ and N weakly prime $r∈(N:RM)$. If $d∈N$, then $rR(d+x)=rRd⊆N$, 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 $Px≠0$, say $ax≠0$ where $a∈P$. Now $0≠aRx⊆N$ and N weakly prime gives $a∈(N:RM)$. As $(r+a)Rx=aRx⊆N$, we get $r∈(N:RM)$, so $P⊆(N:RM)$. Therefore, we can assume that Px=0. Since $PD≠0$, there exist $b∈P$ and $d1∈D$ such that $bd1≠0$. As $(N:RM)N=0$ (by Theorem 1.3) and $0≠b(d1+x)=bd1∈N$ we can divide the proof into the following two cases:

Case 1. $b∈(N:RM)$ and $(d1+x)∉N$. Since $0≠(r+b)R(d1+x)=bRd1⊆N$, we obtain $(r+b)∈(N:RM)$, so $r∈(N:RM)$. Hence $P⊆(N:RM).$

Case 2. $b∉(N:RM)$ and $(d1+x)∈N$. As $0≠bRd1⊆N$ we have $d1∈N$, so $x∈N$ which is a contradiction. Thus $P⊆(N:RM).$

(2) $⇒$ (1) Suppose that $0≠sRm⊆N$ where $s∈R$ and $m∈M$. Take I=Rs and D=Rm. Then $0≠ID⊆N$, so either $I⊆(N:RM)$ or $D⊆N$; hence either $r∈(N:RM)$ or $m∈N$. Thus N is weakly prime.

(2) $⇒$ (3) Let $a∈R$ and $L≤M$ such that $0≠aRL⊆N.$ Now $0≠RaL⊆N$ and from 2. $L⊆N$ or $a∈Ra⊆(N:RM)$.

(3) $⇒$ (2) Let P be a left ideal of R and D a submodule of M with $0≠PD⊆N.$ If $D⊆N,$ then we are done. So suppose $D⊈N.$ We will show that $P⊆(N:RM).$ Let $a∈P.$ Hence $aRD⊆N.$ If $aRD≠0$ then it follows from (3) that $a∈(N:RM)$ and we have $P⊆(N:RM).$ So suppose $aRD=0.$ Because $PD≠0,$ there exists $p∈P$ such that $pRD≠0.$ We now have $0≠pRD=(a+p)RD⊆N.$ 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 $ℤ$-module $ℤ/8ℤ$. 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 $0≠aRb⊆(N:RM).$ Since M is a faithful R module we have $0≠aRbM⊆N.$ It follows that $0≠(RaR)(RbR)M⊆N$. From Theorem 1.4 we have $(RaR)M⊆N$ or $(RbR)M⊆N.$ 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 $0≠aRm⊆N$ with $m∈M$ and $a∉(N:M).$ Since M is a multiplication module there is an ideal I of R such that Rm=IM, then $0≠RaIM⊆N$. Hence $0≠RaI⊆(N:M).$ 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 $Rm=IM⊆N$. Thus $m∈N$ 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 $M=ℤ×ℤ$ is an $R=ℤ×ℤ$-module and $N=2ℤ×{0}$ is a submodule of M. (N:M)=0 is a weakly prime ideal. We have $(0,0)≠(2,0)(1,1)∈2ℤ×{0}.$ 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 $0≠aRbRm⊆N$ and $am∉N$, then $bM⊆N$ for all $a;b∈R$ and $m∈M$.

Proof. Let $a,b∈R$ and $m∈M$. Assume that $0≠aRbRm⊆N$ and $am∉N.$ Now we have $0≠(RaR)(RbR)m⊆N$. From Theorem 1.4 we have $RaRM⊆N$ or $RbRm⊆N.$ Since $am∉N$ we have $RbRm⊆N$. Because $0≠aRbRm$ we must have $0≠bRm⊆N$. Now, since N is weakly prime we get $bM⊆N$ or $m∈N.$ Since $am∉N,$ we must have $bM⊆N$ 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 $x∉P$. Then $rRx⊆P$. Suppose $rRx≠0$. Hence $r∈(P:M)$ because P is weakly prime and $x∉P$. 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 $r∈(0:Rx)$, $rRx=0⊆P$ and so $r∈(P:Rx)$. If $r∈(P:M)$ then $rM⊆P$, and this implies $rRx⊆rM⊆P.$ Hence $r∈(P:Rx)$ and therefore $(P:Rx)=(P:M)∪(0:Rx).(2)⇒(3)$ Is obvious. $(3)⇒(1)$ Suppose that $0≠rRx⊆P$ with $r∈R$ and $x∈M−P$. 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 $M=M1⊕M2$ and $N⊆M1⊕M2$. 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 $N=Q⊕M2$ be a weakly prime submodule of M. Let $0≠rRq⊆Q$ , $q∉Q$ . Then $(q,0)∉Q⊕M2$, while $0≠rR(q,0)⊆Q⊕M2$. Since $N=Q⊕M2$ is a weakly prime submodule of M we have $r∈(M1⊕M2:Q⊕M2).$ Hence $rM1⊆Q$ and Q is a weakly prime submodule of M1. Now, suppose $r∈R$, $x∈M1$ such that rRx=0, but $x∉Q$, $r∉(Q:M1).$ Assume that $rM2≠0,$ so there esists $m∈M2$ such that $rm≠0.$ Thus $(0,0)≠rR(x,m)=(rRx,rRm)=(0,rRm)⊆Q⊕M2=N.$ N is a weakly prime submodule of M, so either $(x,m)∈Q⊕M2$ or $r∈(Q⊕M2:M1⊕M2).$ Thus either $x∈Q$ or $r∈(Q:M1)$ which is a contradiction with hypothesis, hence $rM2=0.(⇐)$ Let $r∈R$ and $(x,y)∈M.$ Assume $(0,0)≠rR(x,y)⊆Q⊕M2$, so if $rRx≠0,$ then $x∈Q$ or $r∈(Q:M1)$, since Q is a weakly prime submodule of M1. Thus either $(x,y)∈Q⊕M2=N$ or $r∈(N:M).$ If rRx=0, suppose $x∉Q$, $r∉(Q,M1).$ Then by hypothesis $rM2=0$ and so $rRy⊆rM2=0.$ Hence $rR(x,y)=(0,0)$ which is a contradiction. Thus either $x∈Q$ or $r∈(Q:M1)$ and hence either $(x,y)∈Q⊕M2=N$ or $r∈(Q⊕M2:M1⊕M2).$

### 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 $M1⊕M2.$

Proof. Let $r∈R$ and $(x,y)∈M.$ If $(0,0)≠rR(x,y)⊆(0)⊕M2,$ then rRx=0 and $rRy⊆M2.$ Since (0) is a prime submodule of M1, either x=0 or $r∈((0):M1).$ Hence either $(x,y)=(0,y)∈(0)⊕M2$ or $r∈((0)⊕M2:M1⊕M2),$ that is $(0)⊕M2$ is a weakly prime submodule of $M1⊕M2.$

### Proposition 1.13.

Let M1 and M2 be R-modules. If $U⊕W$ is a weakly prime submodule of $M1⊕M2,$ 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 $M=ℤ⊕ℤ$ is a $ℤ$-module and consider the submodule $N=pℤ⊕{0}$ of M. $pℤ$ is a prime submodule of the $ℤ$-module $ℤ$ and hence also a weakly prime submodule and {0} is a weakly prime submodule of the $ℤ$ module $ℤ$. $N=pℤ⊕{0}$ is not weakly prime since $(0,0)≠p(1,0)∈pℤ⊕{0}$ but $p∉(pℤ⊕{0}:ℤℤ⊕ℤ)$ and $(1,0)∉pℤ⊕{0}.$

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 $S⊆M\{0}$ is called a weakly m-system if, for each ideal A of R, and for all submodules $K,L⊆M$, if $(K+L)∩S≠∅$, $(K+AM)∩S≠∅,$ and $AL≠0$ 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 $AL≠0.$ If $(K+AL)∩S=∅$ then $K+AL⊆P$. Hence $AL⊆P$ and since P is weakly prime, and $AL≠0$, $L⊆P$ or $AM⊆P$. 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 $AL⊆P$ and $AL≠0,$ where A is an ideal of R and L is a submodule M. If $L⊈P$ and $AM⊈P$, then $L∩S≠∅$ and $AM∩S≠∅$. Thus, $AL∩S≠∅$, 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 $A⊆R$, and for every submodule $L≦M$, if $L∩S≠∅$, $AM∩S≠∅$ and $AL≠0$ then $AL∩S≠∅$;

• (4) for each ideal $A⊆R$, and for every $m∈M$, if $Rm∩S≠∅$, $AM∩S=∅$ and $AL≠0$, then $ARm∩S≠∅$;

• (5) for each a∈ R, and for each $m∈M$, if $Rm∩S≠∅$, $aM∩S≠∅$ and $aRm≠0$, then $aRm∩S≠∅.$

Proof. (1) $⇔$ (2) follows from Proposition 2.2. (2) $⇒$ (3) $⇒$ (4) $⇒$ (5) is clear (5) $⇒$ (1). Suppose a ∈ R and m ∈ M with $0≠aRm⊆P.$ If $Rm⊈P$ and $aM⊈P,$ then $Rm∩S≠∅$ and $aM∩S≠∅$ and $aRm≠0.$ From (5) $aRm∩S≠∅.$ Hence $aRm⊈P$ a contradiction. Hence $Rm⊆P$ or $aM⊆P$ and P is weakly prime.

### Proposition 2.4.

Let M be an R-module, $S⊆M$ 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 $0≠AL⊆P$, where A is an ideal of R and $L≦M$. If $L⊈P$ and $AM⊈P$, 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 $P∩S≠∅$, 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 $√N:={m∈M$ : 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 $N≤M$. Then either $√N=M$ or $√N$ equals the intersection of all the weakly prime submodules of M containing N.

Proof. Suppose that $√N≠M$. This means that {P|P is a weakly prime submodule of M and $N⊆P}≠∅$. We first prove that $√N⊆{P|P$ is a weakly prime submodule of M and $N⊆P}$. Let $m∈√N$ 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 $m∈P$. Conversely, assume $m∉√N$. 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 $P⊇N$ 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 $m∉P$, 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 $aRbR⊆P$ implies $ab∈P$ or bc ∈ P or ac ∈ P for all $a;b;c∈R.$

### 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 $aRbRm⊆N$ implies $abM⊆N$ i.e. $ab∈(N:RM)$ or $am∈N$ or $bm∈N$ for all $a;b∈R$ and $m∈M$.

### 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 $0≠aRbRm⊆N$ implies $abM⊆N$ i.e. $ab∈(N:RM)$ or $am∈N$ or $bm∈N$ for all $a;b∈R$ and $m∈M$.

### Remark 3.5.

Every 2-absorbing submodule is weakly 2-absorbing but the converse does not necessarily hold. For example consider the case where $R=ℤ,M=ℤ/30ℤ$ and N=0. Then $2.3.(5+30ℤ)=0∈N$ 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 ann$l(x)⊆(Rx:M)$, 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 $rRsRm⊆Rx$. 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 $rs∈ ann l(x)⊆(Rx:M).$ 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 $0≠aRbRm⊆N$ for all $a;b∈R$ and $m∈M$. Suppose $am∉N.$ It now follows from Proposition 1.9 that $bM⊆N$ and consequently $abM⊆N$. 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 $N∩K$ is a weakly prime submodule of M so that $N∩K$ 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 $N∩K$ is a proper submodule of M. Next, let a,b∈ R and m ∈ M be such that $0≠aRbRm⊆N∩K$ but $am∉N∩K$ and $ab∉(N∩K:M)$. 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 $0≠aRbRm⊆N∩K⊆N$ and $am∉N$, it follows from Proposition 1.9 that $bM⊆N$. 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 $bM⊆N$ and then bm∈ N. Since $0≠aRbRm⊆K$ 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 $aM⊆RaRM⊆K$ or $bm∈RbRm⊆K$ If $aM⊆K$, then $abM⊆aM⊆K$ which contradicts $ab∉(K:RM)$. Thus bm∈ K. Hence $bm∈N∩K$. The proof of Case(3) is similar to that of Case(2). Hence $N∩K$ 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, $ab∉(N:RM)$, $am∉N$ and $bm∉N.$

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 $aRbN≠0$. Then there is an element n ∈ N such that $aRbRn≠0.$ Now $aRbR(m+n)=aRbRm+aRbRn=aRbRn≠0$ since aRbRm=0 because (a,b,m) is a triple-zero of N. Since $0≠aRbR(m+n)⊆N$ 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 $a(N:RM)m≠0$. Thus there exists an element $r∈(N:RM)$ such that $arm≠0$. Hence $aR(r+b)rm=aRrRm+aRbRm=aRrRm$. Since $0≠arm∈aRrRm⊆N$ and N weakly 2-absorbing we have $a(r+b)∈(N:RM)$ or $am∈N$ or $(r+b)m∈N$. Hence $ab∈(N:RM)$ or $am∈N$ 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 $a(N:RM)N≠0$. Then there are $r∈(N:RM)$, n ∈ N such that $arn≠0$. By (1), we get $a(b+r)(m+n)=abm+abn+arm+arn=arn≠0$. Now $0≠aR(b+r)R(m+n)⊆N$. 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 $(N:RM)bN≠0$. Then there are $r∈(N:RM)$, $n∈N$ such that $rbn≠0$. Now, from above $(a+r)b(n+m)=abn+abm+rbn+rbm=rbn≠0$. 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 $am∈N$ or bm ∈ N a contradiction since (a,b,m) is a triple-zero of N. Now, we suppose that $(N:RM)bm≠0.$ Then there is $r∈(N:RM)$ such that $rbm≠0$. Hence $0≠rbm=(a+r)bm∈(a+r)RbRm=rRbRm⊆N$. Since N is weakly 2-absorbing, we have $(a+r)b∈(N:RM)$ or $(a+r)m∈N$ 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 $(N:RM)2m≠0$. Thus there exist $r,s∈(N:RM)$ where $rsm≠0$. By (1), we get $(a+r)(b+s)m=rsm≠0$. Thus we have $0≠(a+r)R(b+s)Rm⊆N$. Hence $(a+r)(b+s)∈(N:RM)$ or $(a+r)m∈N$ or $(b+s)m∈N$.

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, $(N:RM)3⊆Ann(M)$.

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 $(N:RM)2N≠0$. Thus there exist $r,s∈(N:RM)$ and n ∈ N with $rsn≠0$. By Theorem 3.10, we get $(a+r)(b+s)(n+m)=rsn≠0$. 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 $0≠IJK⊆L$ for some ideals I,J,K of R. Does it imply that $IJ⊆L$ or $IK⊆L$ or $JK⊆L$?

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 $0≠I1I2K⊆N$ 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 $a∈I1,b∈I2$ 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 $aRbK⊆N$for 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 $aK⊆N$ or $bK⊆N$.

Proof. Assume that $aK⊈N$ and $bK⊈N$. Then there are x,y ∈ K such that $ax∉N$ and $by∉N$. 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 $ax∉N$. In a similar way, $ay∈N$. 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 $ay∈N$, 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 $aK⊆N$ or $bK⊆N$.

Let N be a weakly 2-absorbing submodule of an R-module M and $I1I2K⊆N$ 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 $0≠IJK⊆N$ 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 $IK⊆N$ or $JK⊆N$.

Proof. Let N be a weakly 2-absorbing submodule of an R-module M and $0≠IJK⊆N$ 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 $IK⊆N$ or $JK⊆N$. Assume $IK⊈N$ and $JK⊈N$. Then $a1K⊈N$ and $a2K⊈N$ where a1 ∈ I and a2 ∈ J. From Lemma 4.2 $a1a2∈(N:RM)$ since $a1Ra2K⊆IJK⊆N$ and $a1K⊈N$ and $a2K⊈N$. By our assumption, there are $b1∈I$ and $b2∈J$ such that $b1b2∉(N:RM)$. By Lemma 4.2, we get $b1K⊆N$ or $b2K⊆N$ since $b1Rb2K⊆IJK⊆N$ and $b1b2∉(N:RM)$. We have the following cases: Case (1) $b1K⊆N$ and $b2K⊈N$: Since $a1Rb2K⊆IJK⊆N$ and $a1K⊈N$ and $b2K⊈N$ it follows from Lemma 4.2 that $a1b2∈(N:RM).$ Since $b1K⊆N$ and $a1K⊈N$, we conclude $(a1+b1)K⊈N$. On the other hand since $(a1+b1)Rb2K⊆N$ and neither $(a1+b1)K⊆N$ nor $b2K⊆N$, 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) $b2K⊆N$ and $b1K⊈N:$ By a similar argument to case (1) we get a contradiction. Case (3) $b1K⊆N$ and $b2K⊆N:b2K⊆N$ and $a2K⊈N$ gives $(a2+b2)K⊈N.$ But $a1R(a2+b2)K⊆N$ and neither $a1K⊆N$ nor $(a2+b2)K⊆N$, 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 $(a1+b1)Ra2K⊆N$ and neither $a2K⊆N$ nor $(a1+b1)K⊆N$, 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 $(a1+b1)R(a2+b2)K⊆N$ and neither $(a1+b1)K⊆N$ nor $(a2+b2)K⊆N$, 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 $IK⊆N$ or $JK⊆N.$

### Proposition 5.1.

Let $R=R1×R2$ and $M=M1×M2$ where M1 is an R1 module and $0≠M2$ 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 $a,b∈R1$ and $x∈M1$ such that $aR1bR1x⊆N1.$ For every $0≠y∈M2$ we have $(a,1)(b,1)(x,y)=(abx,y)≠(0,0).$ Now $(0,0)≠(abx,y)∈aR1bR1x×1R11R1y⊆N1×M2.$ 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 $ax∈N1$ or $bx∈N1.$

### Proposition 5.2.

Let $R=R1×R2$ and $M=M1×M2$ where $0≠M1$ 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 $0≠M2$ is an R2module. Let $N1≠M1.$ 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 $0≠aR1cR1x⊆N1$ and $bR2dR2y=0.$ Since N1 is a weakly prime submodule of M1 we get $a∈(N1:M1)$ or $c∈(N1:M1)$ or $x∈N1.$ 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 $0≠M1$ is an R1 module and $0≠M2$ is an R2 module. If $N=N1×N2$ is a weakly 2-absorbing submodule of M, $N1≠M1$, and $N2≠M2$, then N1 and N2 are weakly prime submodules of M1 and M2 respectively.

Proof. Let $0≠rRx⊆N1$, where $r∈R1$ and $x∈M1$. Consider $z∈M2\N2$. 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 $z∈M2\N2$, $(r,1)(x,z)∉N$. 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 $a1,a2∈R1$ and $m∈M1$ such that $0≠a1R1a2R1m⊆N1$. Clearly, $(0,0)≠(a1,1)(R1×R2)(a2,1)(R1×R2)(m,m2)$ for any $m2∈N2.$ Hence $(0,0)≠(a1,1)(R1×R2)(a2,1)(R1×R2)(m,m2)⊆a1R1a2R1m×1R21R2m2⊆N1×N2.$ 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 $a1m∈N1$ or $a2m∈N1$. 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 $M=ℤ×ℤ$ is an $R=ℤ×ℤ$-module and $N=pℤ×{0}$ is a submodule of M where $pℤ$ is a prime submodule and hence a weakly 2-absorbimg submodule of the $ℤ$ module $ℤ$ and {0} is weakly 2-absorbing.submodule of the $ℤ$ module $ℤ$. Then $(N:M)=0$. Assume that $(0,0)≠(p,1)(1,0)(1,1)∈pℤ×{0}$. 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 $a1,a2∈R$ and m ∈ M1 such that $0≠a1Ra2Rm⊆N1$. 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 $a1m∈N1$ or $a2m∈N1$. Hence N1 is a weakly 2-absorbing submodule of M1

(b) Let $a1Ra2Rm=0$ with $a1a2∉(N1:M1)$ and $a1m∉N1$ and $a2m∉N1$ for $a1,a2∈R$ and $m∈M1$. Suppose $a1Ra2M2≠0$. Hence there exists $m2∈M2$ such that $a1Ra2M2≠0$ 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 $a1m∈N1$ or $a2m∈N1$ a contradiction. Hence $a1Ra2M2=0$.

(2) $⇒$ (1).

Let $a1,a2∈R$ and $(m1,m2)∈M1×M2$ such that $(0,0)≠(a1,a1)(R×R)(a2,a2)(R×R)(m1,m2)⊆N1×M2.$ If $0≠a1Ra2Rm1$ then $0≠a1Ra2Rm1⊆N1$ and N1 a weakly 2-absorbing submodule of M1 gives $a1a2∈(N1:M1)$ or $a1m1∈N1$ or $a2m2∈N1$. 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 $a1Ra2Rm2≠0$ and therefore $a1Ra2M2≠0$. By b. $a1a2∈(N1:M1)$ or $a1m1∈N1$ or $a2m2∈N1$. 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 $m1∈N1$ such that $m1≠0$. To show that N2=M2 or N3=M3. Assume that $N2≠M2$ and $N3≠M3$. Thus there exist $m2∈M2$ and $m3∈M3$ such that $m2∉N2$ and $m3∉N3$. 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 $m2∈N2$ or $m3∈N3$ 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 $N={(0,0,0)}$.

1. A. E. Ashour and M. Hamoda. Weakly primary submodules over non-commutative rings, J. Progr. Res. Math. 7(2016), 917-927.
2. S. E. Atani and F. Farzalipour. On weakly prime submodules, Tamkang J. Math. 38(2007), 247-252.
3. A. Badawi. On 2-absorbing ideals of commutative rings, Bull. Aust. Math. Soc. 75(2007), 417-429.
4. A. Badawi and A. Y. Darani. On weakly 2-absorbing ideals of commutative rings, Houston J. Math. 39(2013), 441-452.
5. M. Behboodi. On the prime radical and Baer's lower nilradical of modules, Acta Math. Hungar. 122(2009), 293-306.
6. A. Y. Darani, F. Soheilnia, U. Tekir, and G. Ulucak. On weakly 2-absorbing primary submodules of modules over commutative rings, J. Korean Math. Soc. 54(5)(2017), 1505-1519.
7. A. Y. Darani and F. Soheilnia. 2-absorbing and weakly 2-absorbing submodules, Thai J. Math. 9(2011), 577-584.
8. J. Dauns. Prime modules, J. Reine Angew. Math. 298(1978), 156-181.
9. N. J. Groenewald. On 2-absorbing ideals of non-commutative rings, JP J. Algebra Number Theory Appl. 40(5)(2018), 855-867.
10. N. J. Groenewald. On weakly and strongly 2-absorbing ideals of non-commutative rings, submitted.
11. N. J. Groenewald and B. T. Nguyen. On 2-absorbing modules over noncommutative rings, Int. Electron. J. Algebra 25(2019), 212-223.
12. A. A. Tuganbaev. Multiplication modules over noncommutative rings, Sb. Math. 194(12)(2003), 1837-1864.