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 $a,b,c\in R$ and $abc\in P$, then $ab\in P$ or $bc\in P$ or $ac\in P$. 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 $a,b\in R$ and $m\in M$ with $abm\in P(0\ne abm\in P)$, then $abM\subseteq P$ or $am\in P$ or $bm\in P$. 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 $N\le M$, 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 $(N{:}_{R}M)=\{r\in R:rM\subseteq N\}.$

Following [9] a proper ideal P of the ring R is 2-absorbing if $aRbRc\subseteq P$ implies $ab\in P$ or $ac\in P$ or $bc\in P$ 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 $aRbRx\subseteq N$ implies $ab\in (N{:}_{R}M)$ or$ax\in N$ or $bx\in N$ for $a,b\in R$ and $x\in M.$ 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 $M,AN\subseteq P$ implies either $N\subseteq P$ or $AM\subseteq P$. This is equivalent to $aRx\subseteq P$ implies $a\in (P:M)$ or $x\in P$ for $a\in R$ and $x\in M.$ It is clear that a submodule N of an R-module M is prime if and only $P=(N{:}_{R}M)$ is a prime ideal of R.

From [10] A proper ideal P of the ring R is weakly prime if $0\ne aRb\subseteq P$ implies $a\in P$ or $b\in P$ 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 $r\in R$ and $m\in M$ with $0\ne rRm\subseteq N$ then either $m\in N$ or $r\in (N{:}_{R}M)$.

Remark 1.2.

Let p and q be two prime numbers. In the $\mathbb{Z}$-module $\mathbb{Z}pq$, 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 $P\subseteq (N{:}_{R}M)$ we have PN=0. In particular $(N{:}_{R}M)N=0.$

Proof. Suppose P is a subset of R such that $P\subseteq (N{:}_{R}M).$ Suppose $PN\ne 0.$ We show that N is prime. Let r ∈ R and m ∈ M be such that $rRm\subseteq N.$ If $rRm\ne 0,$ then $r\in (N{:}_{R}M)$ or $m\in N$ since N is weakly prime. So assume rRm=0. First assume $rN\ne 0$, say $rN\ne 0$ for some $n\in N.$ Now $0\ne rn\in rR(n+m)\subseteq N$ and N weakly prime, gives $r\in (N{:}_{R}M)$ or $(n+m)\in N.$ Hence $r\in (N{:}_{R}M)$ or $m\in N$ since $n\in N.$ So we can assume that rN=0. Now suppose that $Pm\ne 0$, say $sm\ne 0$ for $s\in P\subseteq (N{:}_{R}M).$ We have $0\ne sm\in (r+s)Rm\subseteq N.$ Hence $(r+s)\in (N{:}_{R}M)$ or $m\in N.$ So $r\in (N{:}_{R}M)$ or $m\in N.$ Hence we can assume that Pm=0. Since $PN\ne 0,$ there exists $t\in P$ and $n\in N$ such that $tn\ne 0.$ Now we have $0\ne tn\in (r+t)R(n+m)\subseteq N.$ Again, since N is weakly prime, we get $(r+t)\in (N{:}_{R}M)$ or $(m+n)\in N.$ Hence $r\in (N{:}_{R}M)$ or $m\in N.$ Thus N is a prime submodule.

Compare (1) $\iff$ (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 $0\ne PD\subseteq N$, either $P\subseteq (N{:}_{R}M)$ or $D\subseteq N.$

• (3) For any element $a\in R$ and $L\le M$, if $0\ne aRL\subseteq N$, then $L\subseteq N$ or $a\in (N{:}_{R}M)$.

• (4) For any right ideal I of R and $L\le M$, if $0\ne IL\subseteq N$, then $L\subseteq N$ or $I\subseteq (N{:}_{R}M).$

• (5)] For any element $a\in R$ and $L\le M$, if $0\ne RaRL\subseteq N$, then $L\subseteq N$ or $a\in (N{:}_{R}M).$

• (6)] For any element $a\in R$ and $L\le M$, if $0\ne RaL\subseteq N$ then $L\subseteq N$ or $a\in (N{:}_{R}M)$.

Proof.

(1) $\Rightarrow$ (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 $0\ne PD\subseteq N$ with $x\in D-N.$ We show that $P\subseteq (N{:}_{R}M)$. Let $r\in P$. Now $rRx\subseteq rD\subseteq N.$ If $0\ne rRx$, then N weakly prime gives $r\in (N{:}_{R}M)$. So assume that rRx=0. First suppose that $rD\ne 0$, say $rD\ne 0$ where $d\in D$. If $d\notin N$, then since $0\ne rRd\subseteq N$ and N weakly prime $r\in (N{:}_{R}M)$. If $d\in N$, then $rR(d+x)=rRd\subseteq N$, so $r\in (N{:}_{R}M)$ or $(d+x)\in N$. Thus, $r\in (N{:}_{R}M)$; hence $P\subseteq (N{:}_{R}M)$. So we can assume that rD=0. Suppose that $Px\ne 0$, say $ax\ne 0$ where $a\in P$. Now $0\ne aRx\subseteq N$ and N weakly prime gives $a\in (N{:}_{R}M)$. As $(r+a)Rx=aRx\subseteq N$, we get $r\in (N{:}_{R}M)$, so $P\subseteq (N{:}_{R}M)$. Therefore, we can assume that Px=0. Since $PD\ne 0$, there exist $b\in P$ and $d_1\in D$ such that $b{d}_1\ne 0$. As $(N{:}_{R}M)N=0$ (by Theorem 1.3) and $0\ne b(d_1+x)=b{d}_1\in N$ we can divide the proof into the following two cases:

Case 1. $b\in (N{:}_{R}M)$ and $(d_1+x)\notin N$. Since $0\ne (r+b)R(d_1+x)=bR{d}_1\subseteq N$, we obtain $(r+b)\in (N{:}_{R}M)$, so $r\in (N{:}_{R}M)$. Hence $P\subseteq (N{:}_{R}M).$

Case 2. $b\notin (N{:}_{R}M)$ and $(d_1+x)\in N$. As $0\ne bR{d}_1\subseteq N$ we have $d_1\in N$, so $x\in N$ which is a contradiction. Thus $P\subseteq (N{:}_{R}M).$

(2) $\Rightarrow$ (1) Suppose that $0\ne sRm\subseteq N$ where $s\in R$ and $m\in M$. Take I=Rs and D=Rm. Then $0\ne ID\subseteq N$, so either $I\subseteq (N{:}_{R}M)$ or $D\subseteq N$; hence either $r\in (N{:}_{R}M)$ or $m\in N$. Thus N is weakly prime.

(2) $\Rightarrow$ (3) Let $a\in R$ and $L\le M$ such that $0\ne aRL\subseteq N.$ Now $0\ne RaL\subseteq N$ and from 2. $L\subseteq N$ or $a\in Ra\subseteq (N{:}_{R}M)$.

(3) $\Rightarrow$ (2) Let P be a left ideal of R and D a submodule of M with $0\ne PD\subseteq N.$ If $D\subseteq N,$ then we are done. So suppose $D⊈N.$ We will show that $P\subseteq (N{:}_{R}M).$ Let $a\in P.$ Hence $aRD\subseteq N.$ If $aRD\ne 0$ then it follows from (3) that $a\in (N{:}_{R}M)$ and we have $P\subseteq (N{:}_{R}M).$ So suppose $aRD=0.$ Because $PD\ne 0,$ there exists $p\in P$ such that $pRD\ne 0.$ We now have $0\ne pRD=(a+p)RD\subseteq N.$ It follows from (3) that $(a+p)\in (N{:}_{R}M)$ and we have $P\subseteq (N{:}_{R}M).$

(3) $\iff$ (4) $\iff$ (5) $\iff$ (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 $(N{:}_{R}M)$ 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 $(N{:}_{R}M)$ is not, in general, a weakly prime ideal of R. For example, let M denote the cyclic $\mathbb{Z}$-module $\mathbb{Z}/8\mathbb{Z}$. Take $N=\left\{0\right\}$. Certainly N is a weakly prime submodule of M, but $(N{:}_{R}M)=8Z$ 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 $(N{:}_{R}M)$ is a weakly prime ideal of R.

Proof. Assume that M is a faithful R module and let $0\ne aRb\subseteq (N{:}_{R}M).$ Since M is a faithful R module we have $0\ne aRbM\subseteq N.$ It follows that $0\ne (RaR)(RbR)M\subseteq N$. From Theorem 1.4 we have $(RaR)M\subseteq N$ or $(RbR)M\subseteq N.$ Hence $a\in$ $(N{:}_{R}M)$ or $b\in$ $(N{:}_{R}M)$ and it follows that $(N{:}_{R}M)$ 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 $0\ne aRm\subseteq N$ with $m\in M$ and $a\notin (N:M).$ Since M is a multiplication module there is an ideal I of R such that Rm=IM, then $0\ne RaIM\subseteq N$. Hence $0\ne RaI\subseteq (N:M).$ Since (N:M) is a weakly prime ideal of R, we have $Ra\subseteq (N:M)$ or $I\subseteq (N:M).$ Since $a\notin (N:M),$ we have $I\subseteq (N:M).$ Hence $Rm=IM\subseteq N$. Thus $m\in N$ 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 $M=\mathbb{Z}\times \mathbb{Z}$ is an $R=\mathbb{Z}\times \mathbb{Z}$-module and $N=2\mathbb{Z}\times \left\{0\right\}$ is a submodule of M. (N:M)=0 is a weakly prime ideal. We have $(0,0)\ne (2,0)(1,1)\in 2\mathbb{Z}\times \left\{0\right\}.$ Now, neither $(2,0)\in (N:M)$ nor $(1,1)\in N.$ 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 $0\ne aRbRm\subseteq N$ and $am\notin N$, then $bM\subseteq N$ for all $a;b\in R$ and $m\in M$.

Proof. Let $a,b\in R$ and $m\in M$. Assume that $0\ne aRbRm\subseteq N$ and $am\notin N.$ Now we have $0\ne (RaR)(RbR)m\subseteq N$. From Theorem 1.4 we have $RaRM\subseteq N$ or $RbRm\subseteq N.$ Since $am\notin N$ we have $RbRm\subseteq N$. Because $0\ne aRbRm$ we must have $0\ne bRm\subseteq N$. Now, since N is weakly prime we get $bM\subseteq N$ or $m\in N.$ Since $am\notin N,$ we must have $bM\subseteq N$ 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. $(1)\Rightarrow (2)$ Let $r\in (P:Rx)$ and $x\notin P$. Then $rRx\subseteq P$. Suppose $rRx\ne 0$. Hence $r\in (P:M)$ because P is weakly prime and $x\notin P$. If $rRx=0$, then $r\in (0:Rx)$. Thus $(P:Rx)\subseteq (P:M)\cup (0:Rx).$ Now if $r\in (P:M)\cup (0:Rx)$ then either $r\in (P:M)$ or $r\in (0:Rx)$. Hence, when $r\in (0:Rx)$, $rRx=0\subseteq P$ and so $r\in (P:Rx)$. If $r\in (P:M)$ then $rM\subseteq P$, and this implies $rRx\subseteq rM\subseteq P.$ Hence $r\in (P:Rx)$ and therefore $(P:Rx)=(P:M)\cup (0:Rx).(2)\Rightarrow (3)$ Is obvious. $(3)\Rightarrow (1)$ Suppose that $0\ne rRx\subseteq P$ with $r\in R$ and $x\in M-P$. Then $r\in (P:Rx)$ and $r\notin (0:Rx)$. It follows from (3) that $r\in (P:Rx)=(P:M)$, as required.

Proposition 1.11.

Let M₁ and M₂ be unitary R-modules over a ring R. Let $M=M_1\oplus M_2$ and $N\subseteq M_1\oplus M_2$. 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. $(\Rightarrow )$ Let $N=Q\oplus M_2$ be a weakly prime submodule of M. Let $0\ne rRq\subseteq Q$ , $q\notin Q$ . Then $(q,0)\notin Q\oplus M_2$, while $0\ne rR(q,0)\subseteq Q\oplus M_2$. Since $N=Q\oplus M_2$ is a weakly prime submodule of M we have $r\in (M_1\oplus M_2:Q\oplus M_2).$ Hence $r{M}_1\subseteq Q$ and Q is a weakly prime submodule of M₁. Now, suppose $r\in R$, $x\in M_1$ such that rRx=0, but $x\notin Q$, $r\notin (Q:M_1).$ Assume that $r{M}_2\ne 0,$ so there esists $m\in M_2$ such that $rm\ne 0.$ Thus $(0,0)\ne rR(x,m)=(rRx,rRm)=(0,rRm)\subseteq Q\oplus M_2=N.$ N is a weakly prime submodule of M, so either $(x,m)\in Q\oplus M_2$ or $r\in (Q\oplus M_2:M_1\oplus M_2).$ Thus either $x\in Q$ or $r\in (Q:M_1)$ which is a contradiction with hypothesis, hence $r{M}_2=0.(\Leftarrow )$ Let $r\in R$ and $(x,y)\in M.$ Assume $(0,0)\ne rR(x,y)\subseteq Q\oplus M_2$, so if $rRx\ne 0,$ then $x\in Q$ or $r\in (Q:M_1)$, since Q is a weakly prime submodule of M₁. Thus either $(x,y)\in Q\oplus M_2=N$ or $r\in (N:M).$ If rRx=0, suppose $x\notin Q$, $r\notin (Q,M_1).$ Then by hypothesis $r{M}_2=0$ and so $rRy\subseteq r{M}_2=0.$ Hence $rR(x,y)=(0,0)$ which is a contradiction. Thus either $x\in Q$ or $r\in (Q:M_1)$ and hence either $(x,y)\in Q\oplus M_2=N$ or $r\in (Q\oplus M_2:M_1\oplus M_2).$

Remark 1.12.

Let M₁ and M₂ be R-modules. If (0) is a prime submodule of M₁, then $(0)\oplus M_2$ is a weakly prime submodule of $M_1\oplus M_2.$

Proof. Let $r\in R$ and $(x,y)\in M.$ If $(0,0)\ne rR(x,y)\subseteq (0)\oplus M_2,$ then rRx=0 and $rRy\subseteq M_2.$ Since (0) is a prime submodule of M₁, either x=0 or $r\in ((0):M_1).$ Hence either $(x,y)=(0,y)\in (0)\oplus M_2$ or $r\in ((0)\oplus M_2:M_1\oplus M_2),$ that is $(0)\oplus M_2$ is a weakly prime submodule of $M_1\oplus M_2.$

Proposition 1.13.

Let M₁ and M₂ be R-modules. If $U\oplus W$ is a weakly prime submodule of $M_1\oplus M_2,$ 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 $M=\mathbb{Z}\oplus \mathbb{Z}$ is a $\mathbb{Z}$-module and consider the submodule $N=p\mathbb{Z}\oplus \left\{0\right\}$ of M. $p\mathbb{Z}$ is a prime submodule of the $\mathbb{Z}$-module $\mathbb{Z}$ and hence also a weakly prime submodule and {0} is a weakly prime submodule of the $\mathbb{Z}$ module $\mathbb{Z}$. $N=p\mathbb{Z}\oplus \left\{0\right\}$ is not weakly prime since $(0,0)\ne p(1,0)\in p\mathbb{Z}\oplus \left\{0\right\}$ but $p\notin (p\mathbb{Z}\oplus \{0\}{:}_{\mathbb{Z}}\mathbb{Z}\oplus \mathbb{Z})$ and $(1,0)\notin p\mathbb{Z}\oplus \left\{0\right\}.$

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 $S\subseteq M\backslash \left\{0\right\}$ is called a weakly m-system if, for each ideal A of R, and for all submodules $K,L\subseteq M$, if $(K+L)\cap S\ne \varnothing$, $(K+AM)\cap S\ne \varnothing ,$ and $AL\ne 0$ then $(K+AL)\cap S\ne \varnothing$.

Proposition 2.2.

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

Proof. Suppose $S=M\backslash P$. Let A be an ideal in R and K and L be submodules of M such that $(K+L)\cap S\ne \varnothing$, $(K+AM)\cap S\ne \varnothing$ and $AL\ne 0.$ If $(K+AL)\cap S=\varnothing$ then $K+AL\subseteq P$. Hence $AL\subseteq P$ and since P is weakly prime, and $AL\ne 0$, $L\subseteq P$ or $AM\subseteq P$. It follows that $(K+L)\cap S=\varnothing$ or $(K+AM)\cap S=\varnothing$, a contradiction. Therefore, S is a weakly m-system in M. Conversely, let $S=M\backslash P$ be a weakly m-system in M. Suppose $AL\subseteq P$ and $AL\ne 0,$ where A is an ideal of R and L is a submodule M. If $L⊈P$ and $AM⊈P$, then $L\cap S\ne \varnothing$ and $AM\cap S\ne \varnothing$. Thus, $AL\cap S\ne \varnothing$, 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 $S:=M\backslash P$. Then the following statements are equivalent:

• (1) P is weakly prime;

• (2) S is a weakly m-system;

• (3) for each left ideal $A\subseteq R$, and for every submodule $L\leqq M$, if $L\cap S\ne \varnothing$, $AM\cap S\ne \varnothing$ and $AL\ne 0$ then $AL\cap S\ne \varnothing$;

• (4) for each ideal $A\subseteq R$, and for every $m\in M$, if $Rm\cap S\ne \varnothing$, $AM\cap S=\varnothing$ and $AL\ne 0$, then $ARm\cap S\ne \varnothing$;

• (5) for each a∈ R, and for each $m\in M$, if $Rm\cap S\ne \varnothing$, $aM\cap S\ne \varnothing$ and $aRm\ne 0$, then $aRm\cap S\ne \varnothing .$

Proof. (1) $\iff$ (2) follows from Proposition 2.2. (2) $\Rightarrow$ (3) $\Rightarrow$ (4) $\Rightarrow$ (5) is clear (5) $\Rightarrow$ (1). Suppose a ∈ R and m ∈ M with $0\ne aRm\subseteq P.$ If $Rm⊈P$ and $aM⊈P,$ then $Rm\cap S\ne \varnothing$ and $aM\cap S\ne \varnothing$ and $aRm\ne 0.$ From (5) $aRm\cap S\ne \varnothing .$ Hence $aRm⊈P$ a contradiction. Hence $Rm\subseteq P$ or $aM\subseteq P$ and P is weakly prime.

Proposition 2.4.

Let M be an R-module, $S\subseteq M$ 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 $0\ne AL\subseteq P$, where A is an ideal of R and $L\leqq M$. If $L⊈P$ and $AM⊈P$, then by the maximal property of P, we have, $(P+L)\cap S\ne \varnothing$ and $(P+AM)\cap S\ne \varnothing$. Thus, since S is a weakly m-system $(P+AL)\cap S\ne \varnothing$ and it follows that $P\cap S\ne \varnothing$, a contradiction. Thus, P must be a weakly prime submodule.

Next we need a generalization of the notion of $\surd N$ 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 $\surd N:=\{m\in M$ : every weakly m-system containing m meets N}. If there is no weakly prime submodule containing N, then we put $\surd N=M.$

Theorem 2.6.

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

Proof. Suppose that $\surd N\ne M$. This means that {P|P is a weakly prime submodule of M and $N\subseteq P\}\ne \varnothing$. We first prove that $\surd N\subseteq \left\{P\right|P$ is a weakly prime submodule of M and $N\subseteq P\}$. Let $m\in \surd N$ and P be any weakly prime submodule of M containing N. Consider the m-system $M\backslash P$. This m-system cannot contain m, for otherwise it meets N and hence also P. Therefore, we have $m\in P$. Conversely, assume $m\notin \surd N$. 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 $P\supseteq N$ 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 $m\notin P$, as desired.

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 $aRbR\subseteq P$ implies $ab\in P$ or bc ∈ P or ac ∈ P for all $a;b;c\in R.$

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 $aRbRm\subseteq N$ implies $abM\subseteq N$ i.e. $ab\in (N{:}_{R}M)$ or $am\in N$ or $bm\in N$ for all $a;b\in R$ and $m\in M$.

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 $0\ne aRbRm\subseteq N$ implies $abM\subseteq N$ i.e. $ab\in (N{:}_{R}M)$ or $am\in N$ or $bm\in N$ for all $a;b\in R$ and $m\in M$.

Remark 3.5.

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

Proposition 3.6.

Let x ∈ M and a ∈ R. Then if ann${}_l(x)\subseteq (Rx:M)$, 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 $rRsRm\subseteq Rx$. Since Rx is a weakly 2-absorbing submodule, we may assume rRsRm=0, otherwise Rx is 2-absorbing. Now $rRsR(x+m)\subseteq Rx$. If $rRsR(x+m)\ne 0$ then we have $rs\in (Rx:M)$ or $r(x+m)\in Rx$ or $s(x+m)\in Rx$, as Rx is a weakly 2-absorbing submodule. Hence $rs\in (Rx:M)$ or rm ∈ Rx or sm ∈ Rx. Now let rRsR(x+m)=0. Then rRsRm=0 implies rRsRx=0. Hence $rs\in ann{}_l(x)\subseteq (Rx:M).$ 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 $0\ne aRbRm\subseteq N$ for all $a;b\in R$ and $m\in M$. Suppose $am\notin N.$ It now follows from Proposition 1.9 that $bM\subseteq N$ and consequently $abM\subseteq N$. 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 $N\cap K$ is a weakly prime submodule of M so that $N\cap K$ 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 $N\cap K$ is a proper submodule of M. Next, let a,b∈ R and m ∈ M be such that $0\ne aRbRm\subseteq N\cap K$ but $am\notin N\cap K$ and $ab\notin (N\cap K:M)$. Then, we can conclude that (a) am ∉ N or am ∉ K, and (b) $ab\notin (N{:}_{R}M)$ or $ab\notin (K{:}_{R}M).$ 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 $0\ne aRbRm\subseteq N\cap K\subseteq N$ and $am\notin N$, it follows from Proposition 1.9 that $bM\subseteq N$. This is a contradiction because $ab\notin (N{:}_{R}M)$. Hence Case(1) does not occur. Similarly, Case(4) is not possible. Next, Case(2) is considered. Again, we obtain that $bM\subseteq N$ and then bm∈ N. Since $0\ne aRbRm\subseteq K$ it follows that $0\ne (RaR)(RbR)m)\subseteq K$. Hence, from the fact that K is weakly prime and from Theorem 1.4 it follows that $aM\subseteq RaRM\subseteq K$ or $bm\in RbRm\subseteq K$ If $aM\subseteq K$, then $abM\subseteq aM\subseteq K$ which contradicts $ab\notin (K{:}_{R}M)$. Thus bm∈ K. Hence $bm\in N\cap K$. The proof of Case(3) is similar to that of Case(2). Hence $N\cap K$ 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, $ab\notin (N{:}_{R}M)$, $am\notin N$ and $bm\notin N.$

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 $aRbN\ne 0$. Then there is an element n ∈ N such that $aRbRn\ne 0.$ Now $aRbR(m+n)=aRbRm+aRbRn=aRbRn\ne 0$ since aRbRm=0 because (a,b,m) is a triple-zero of N. Since $0\ne aRbR(m+n)\subseteq N$ and N weakly 2-absorbing we have $ab\in (N{:}_{R}M)$ or $a(m+n)\in N$ or $b(m+n)\in N$. Since (a,b,m) is a triple-zero of $N,ab\notin (N{:}_{R}M).$ Hence $a(m+n)\in N$ or $b(m+n)\in N$ and consequently am ∈ N or bm ∈ N a contradiction. Hence aRbN=0. Now, we suppose that $a(N{:}_{R}M)m\ne 0$. Thus there exists an element $r\in (N{:}_{R}M)$ such that $arm\ne 0$. Hence $aR(r+b)rm=aRrRm+aRbRm=aRrRm$. Since $0\ne arm\in aRrRm\subseteq N$ and N weakly 2-absorbing we have $a(r+b)\in (N{:}_{R}M)$ or $am\in N$ or $(r+b)m\in N$. Hence $ab\in (N{:}_{R}M)$ or $am\in N$ or bm ∈ N a contradiction since (a,b,m) is a triple-zero of N. Similarly, it can be easily seen that $b(N{:}_{R}M)m=0$.

(2) Assume that $a(N{:}_{R}M)N\ne 0$. Then there are $r\in (N{:}_{R}M)$, n ∈ N such that $arn\ne 0$. By (1), we get $a(b+r)(m+n)=abm+abn+arm+arn=arn\ne 0$. Now $0\ne aR(b+r)R(m+n)\subseteq N$. Therefore, we have $a(b+r)\in (N{:}_{R}M)$ or $a(m+n)\in N$ or $(b+r)(m+n)\in N$ and we obtain $ab\in (N{:}_{R}M)$ or am ∈ N or bm ∈ N, a contradiction. Hence $a(N{:}_{R}M)N=0$. In a similar way, we get $b(N{:}_{R}M)N=0$. Now, we suppose that $(N{:}_{R}M)bN\ne 0$. Then there are $r\in (N{:}_{R}M)$, $n\in N$ such that $rbn\ne 0$. Now, from above $(a+r)b(n+m)=abn+abm+rbn+rbm=rbn\ne 0$. Hence $0\ne (a+r)RbR(n+m)\subseteq N$ and since N is weakly 2-absorbing $(a+r)b\in (N{:}_{R}M)$ or $(a+r)(n+m)\in N$ or $b(n+m)\in N$. Hence $ab\in (N{:}_{R}M)$ or $am\in N$ or bm ∈ N a contradiction since (a,b,m) is a triple-zero of N. Now, we suppose that $(N{:}_{R}M)bm\ne 0.$ Then there is $r\in (N{:}_{R}M)$ such that $rbm\ne 0$. Hence $0\ne rbm=(a+r)bm\in (a+r)RbRm=rRbRm\subseteq N$. Since N is weakly 2-absorbing, we have $(a+r)b\in (N{:}_{R}M)$ or $(a+r)m\in N$ or bm ∈ N. Therefore $ab\in (N{:}_{R}M)$ or am ∈ N or bm ∈ N a contradiction since (a,b,m) is a triple-zero of N. Hence $(N{:}_{R}M)bm=0$. Lastly, we show that ${(N{:}_{R}M)}^{2}m=0$. Let ${(N{:}_{R}M)}^{2}m\ne 0$. Thus there exist $r,s\in (N{:}_{R}M)$ where $rsm\ne 0$. By (1), we get $(a+r)(b+s)m=rsm\ne 0$. Thus we have $0\ne (a+r)R(b+s)Rm\subseteq N$. Hence $(a+r)(b+s)\in (N{:}_{R}M)$ or $(a+r)m\in N$ or $(b+s)m\in N$.

Consequently, $ab\in (N{:}_{R}M)$ or am ∈ N or bm ∈ N a contradiction, since (a,b,m) is a triple-zero of N. Therefore ${(N:M)}^{2}m=0$.

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 ${(N{:}_{R}M)}^{2}N=0$. In particular, ${(N{:}_{R}M)}^3\subseteq Ann(M)$.

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 ${(N{:}_{R}M)}^{2}N\ne 0$. Thus there exist $r,s\in (N{:}_{R}M)$ and n ∈ N with $rsn\ne 0$. By Theorem 3.10, we get $(a+r)(b+s)(n+m)=rsn\ne 0$. Then we have $0\ne (a+r)R(b+s)R(n+m)\subseteq N$. Since N is weakly 2-absorbing we have $(a+r)(b+s)\in (N{:}_{R}M)$ or $(a+r)(n+m)\in N$ or $(b+s)(n+m)\in N$ and so $ab\in (N{:}_{R}M)$ or am ∈ N or bm ∈ N, which is a contradiction.

Thus ${(N{:}_{R}M)}^{2}N=0.$ We get ${(N{:}_{R}M)}^3\subseteq ({(N{:}_{R}M)}^{2}N:M)=(0:M)=Ann(M).$

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

Question. Suppose that L is a weakly 2-absorbing ideal of a ring R and $0\ne IJK\subseteq L$ for some ideals I,J,K of R. Does it imply that $IJ\subseteq L$ or $IK\subseteq L$ or $JK\subseteq L$?

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 $0\ne I_1{I}_{2}K\subseteq N$ for some ideals $I_1,I_2$ 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 $a\in I_1,b\in I_2$ 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 $aRbK\subseteq N$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 $ab\notin (N{:}_{R}M)$, then $aK\subseteq N$ or $bK\subseteq N$.

Proof. Assume that $aK⊈N$ and $bK⊈N$. Then there are x,y ∈ K such that $ax\notin N$ and $by\notin N$. We get bx ∈ N since N is a weakly 2-absorbing submodule, (a,b,x) is not a triple-zero of N, $ab\notin (N{:}_{R}M)$ and $ax\notin N$. In a similar way, $ay\in N$. Now, $aRbR(x+y)\subseteq N$ and since (a,b,x+y) is not a triple-zero of N and $ab\notin (N{:}_{R}M)$ we have a(x+y) ∈ N or b(x+y) ∈ N. Assume that a(x+y)=(ax+ay) ∈ N. As $ay\in N$, 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 $aK\subseteq N$ or $bK\subseteq N$.

Let N be a weakly 2-absorbing submodule of an R-module M and $I_1{I}_{2}K\subseteq N$ 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 $ab\in (N{:}_{R}M)$ 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 $0\ne IJK\subseteq N$ 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 $IJ\subseteq (N{:}_{R}M)$ or $IK\subseteq N$ or $JK\subseteq N$.

Proof. Let N be a weakly 2-absorbing submodule of an R-module M and $0\ne IJK\subseteq N$ 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 $IJ⊈(N{:}_{R}M)$. We show that $IK\subseteq N$ or $JK\subseteq N$. Assume $IK⊈N$ and $JK⊈N$. Then $a_{1}K⊈N$ and $a_{2}K⊈N$ where a₁ ∈ I and a₂ ∈ J. From Lemma 4.2 $a_1{a}_2\in (N{:}_{R}M)$ since $a_{1}R{a}_{2}K\subseteq IJK\subseteq N$ and $a_{1}K⊈N$ and $a_{2}K⊈N$. By our assumption, there are $b_1\in I$ and $b_2\in J$ such that $b_1{b}_2\notin (N{:}_{R}M)$. By Lemma 4.2, we get $b_{1}K\subseteq N$ or $b_{2}K\subseteq N$ since $b_{1}R{b}_{2}K\subseteq IJK\subseteq N$ and $b_1{b}_2\notin (N{:}_{R}M)$. We have the following cases: Case (1) $b_{1}K\subseteq N$ and $b_{2}K⊈N$: Since $a_{1}R{b}_{2}K\subseteq IJK\subseteq N$ and $a_{1}K⊈N$ and $b_{2}K⊈N$ it follows from Lemma 4.2 that $a_1{b}_2\in (N{:}_{R}M)\text{.}$ Since $b_{1}K\subseteq N$ and $a_{1}K⊈N$, we conclude $(a_1+b_1)K⊈N$. On the other hand since $(a_1+b_1)R{b}_{2}K\subseteq N$ and neither $(a_1+b_1)K\subseteq N$ nor $b_{2}K\subseteq N$, we get that $(a_1+b_1)b_2\in (N{:}_{R}M)$ by Lemma 4.2. But, because $(a_1+b_1)b_2=(a_1{b}_2+b_1{b}_2)\in (N{:}_{R}M)$ and $(a_1+b_1)b_2\in (N{:}_{R}M)$, we get $b_1{b}_2\in (N{:}_{R}M)$ which is a contradiction. Case (2) $b_{2}K\subseteq N$ and $b_{1}K⊈N:$ By a similar argument to case (1) we get a contradiction. Case (3) $b_{1}K\subseteq N$ and $b_{2}K\subseteq N:b_{2}K\subseteq N$ and $a_{2}K⊈N$ gives $(a_2+b_2)K⊈N\text{.}$ But $a_{1}R(a_2+b_2)K\subseteq N$ and neither $a_{1}K\subseteq N$ nor $(a_2+b_2)K\subseteq N$, hence $a_1(a_2+b_2)\in (N{:}_{R}M)$ by Lemma 4.2. Since $a_1{a}_2\in (N{:}_{R}M)$ and $(a_1{a}_2+a_1{b}_2)\in (N{:}_{R}M)$, we have $a_1{b}_2\in (N{:}_{R}M)$. Since $(a_1+b_1)R{a}_{2}K\subseteq N$ and neither $a_{2}K\subseteq N$ nor $(a_1+b_1)K\subseteq N$, we conclude $(a_1+b_1)a_2\in (N{:}_{R}M)$ by Lemma 4.2. But $(a_1+b_1)a_2=a_1{a}_2+b_1{a}_2$, so $(a_1{a}_2+b_1{a}_2)\in (N{:}_{R}M)$ and since $a_1{a}_2\in (N{:}_{R}M)$, we get $b_1{a}_2\in (N{:}_{R}M)$. Now, since $(a_1+b_1)R(a_2+b_2)K\subseteq N$ and neither $(a_1+b_1)K\subseteq N$ nor $(a_2+b_2)K\subseteq N$, we have $(a_1+b_{1)}(a_2+b_2)=(a_1{a}_2+a_1{b}_2+b_1{a}_2+b_1{b}_2)\in (N{:}_{R}M)$ by Lemma 4.2. But $a_1{a}_2,a_1{b}_2,b_1{a}_2\in (N{:}_{R}M)$, so $b_1{b}_2\in (N{:}_{R}M)$ which is a contradiction. Consequently $IK\subseteq N$ or $JK\subseteq N\text{.}$