1-absorbing and weakly 1-absorbing prime submodules of a module over a noncommutative ring

Abstract

In this study, we aim to introduce the concepts of 1-absorbing prime submodules and weakly 1-absorbing prime submodules of a unital module over a noncommutative ring with nonzero identity. This is a new class of submodules between prime submodules (weakly prime submodules) and 2-absorbing submodules (weakly 2-absorbing submodules). Let R be a noncommutative ring with a nonzero identity $1\ne 0$ and M an R-module. A proper submdule P of M is said to be a 1-absorbing prime submodule (weakly 1-absorbing prime submodule) if for all nonunits $x,y\in R$ and $m\in M$ with xRyRm $\subseteq$ P $(\left\{ 0\right\} \ne$ xRyRm $\subseteq P)$, then $xy\in (M:_{R}P)$ or $m\in P$. Various properties and characterizations of these classes of submodules are considered.

1 Introduction

In this article, we focus only on noncommutative rings with nonzero identity and nonzero unital left modules. Let R always denote such a ring and let M denote such an R-module. The concept of prime ideals and its generalizations have a significant place in noncommutative algebra since they are used in understanding the structure of rings. Recall that in a commutative ring a proper ideal I of R is said to be a prime ideal if whenever $xy\in I$ then $x\in I$ or $y\in I.$ In [1], Anderson and Smith introduced a notion of weakly prime ideal which is a generalization of prime ideals. A proper ideal I of R is called weakly prime ideal if $0\ne xy\in I$ for some elements $x,y\in R$ implies that $x\in I$ or $y\in I.$ It is clear that every prime ideal is weakly prime but the converse is not true in general. Afterwards, Badawi, in his celebrated paper [2], introduced the notion of 2-absorbing ideals and used them to characterize Dedekind domains. Recall from [2], that a nonzero proper ideal I of R is called 2-absorbing ideal if $xyz\in I$ for some $x,y,z\in R$ implies either $xy\in I$ or $xz\in I$ or $yz\in I$. Note that every prime ideal is also a 2-absorbing ideal. After this, over the past decades, 2-absorbing version of ideals and many generalizations of 2-absorbing ideals attracted considerable attention by many researchers. Badawi and Darani in [3] defined and studied the notion of weakly 2-absorbing ideals which is a generalization of weakly prime ideals. A proper ideal I of R is called a weakly 2-absorbing ideal if for each $x,y,z\in R$ with $0\ne xyz\in I$, then either $xy\in I$ or $xz\in I$ or $yz\in I$.

In 2010, Hirano et al. extended the notion of weakly prime ideals in rings, not necessarily commutative or with identity. According their celebrated paper [11], a proper ideal P of R is called a weakly prime ideal of a ring R if whenever $a,b\in R$ such that $\left\{ 0\right\} \ne aRb\subseteq P,$ then $a\in P$ or $b\in P.$ They also verified that P is weakly prime ideal if and only if whenever J, K are right ideals of R such that $\left\{ 0\right\} \ne JK\subseteq P$, then $J\subseteq P$ or $K\subseteq P.$ An ideal I of R is said to be proper if $I\ne R$. Recall that a proper ideal I of R is called 2-absorbing as in [6] if whenever $aRbRc\subseteq I$ for some $a,b,c\in R$, then $ab\in I$ or $bc\in I$ or $ac\in I$. Let I be a proper ideal of R. Recall from [7] that a proper ideal I of R is said to be a weakly 2-absorbing ideal of R if whenever $a,b,c\in R$ with $\left\{ 0\right\} \ne aRbRc\subseteq I$, then $ab\in I$ or $ac\in I$ or $bc\in I$. Note that a 2-absorbing ideal is a weakly 2-absorbing ideal. However, these are different concepts.

In 2011, Darani and Soheilnia [5] introduced the concept 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 (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 submodules are generalization of prime submodules. Moreover, it is obvious that 2-absorbing ideals are special cases of 2-absorbing submodules.

In [8] and [9] the notions of 2-absorbing and weakly 2-absorbing submodules of a module over a noncommutative ring were introduced. A proper submodule P of a module M over a noncommutative ring R with identity is said to be a 2-absorbing (weakly 2-absorbing) submodule of M if whenever $a,b\in R$ and $m\in M$ with $aRbRm\subseteq P$ $(\left\{ 0\right\} \ne aRbRm\subseteq P)$, then $ab\in (P:M)\subseteq P$ or $am\in P$ or $bm\in P$.

Recently, in [16], Yassine et al. introduced a 1-absorbing prime ideal. This type of ideal which is a generalization of prime ideals of a commutative ring with identity. A proper ideal I of R is called 1–absorbing prime ideal if whenever $xyz\in I$ for some nonunits $x,y,z\in R,$ then either $xy\in I$ or $z\in I$. Note that every prime ideal is 1-absorbing prime and every 1-absorbing prime ideal is 2-absorbing. The converses are not true. More currently in [12] Koç et al. defined weakly 1-absorbing prime ideals which is a generalization of 1-absorbing prime ideal. A proper ideal I of R is called weakly 1-absorbing prime ideal if $0\ne xyz\in I$ for some nonunits $x,y,z\in R$ implies that $xy\in I$ or $z\in I$. Following Yassine et al. [15] and Koç [12] in [10] we introduced 1-absorbing prime ideals and weakly 1-absorbing prime ideals in noncommutative rings. For a noncommutative ring R, whenever $xRyRz\subseteq I$ $(\left\{ 0\right\} \ne xRyRz\subseteq I)$ for some nonunits $x,y,z\in R$, then $xy\in I$ or $z\in I,$ then I is a 1-absorbing prime ideal (weakly 1 absorbing prime ideal). In [14] Ugurlu introduced the concept of a 1-absorbing prime submodule of a unital module over a commutative ring with a non-zero identity. Also in [4] Celikel introduced the notion of 1-absorbing primary submodules of a unital module over a commutative ring with a non-zero identity. In this paper, after introducing the notion of 1-absorbing and weakly 1-absorbing prime submodules of a unital left module over a noncommutative ring with nonzero identity, we examine the properties of the new classes. We show that many of the results of Ugurlu in [14] for 1-absorbing prime submodules of a unital module over a commutative ring with a non-zero identity are also valid for 1-absorbing prime submodules of a unital left module over a noncommutative ring with nonzero identity. For all nonunit elements $a,b\in R$ and $m\in M$, if $aRbRm\subseteq N$, $(\left\{ 0\right\} \ne aRbRm\subseteq N)$ either $ab\in (N:_{R}M)$ or $m\in N$, then N is called a 1-absorbing prime submodule (weakly 1-absorbing prime submodule) of M. Recall that a proper submodule N of the R-module M is a prime (weakly prime) submodule if $aRm\subseteq N(\left\{ 0\right\} \ne aRm\subseteq N)$ for $a\in R$ and $m\in M$ then $a\in (N:_{R}M)$ or $m\in N.$

Among many results in this paper, it is shown in Proposition 2.4 if N is a 1-absorbing prime submodule of M then $(N:_{R}M)$ is a 1-absorbing prime ideal of R. It is also proved in Corollary 2.7 that if M is an R-module and $N_{1}$, $N_{2}$ submodules of M with $N_{2}\subseteq N_{1},$ then $N_{1}$ is a 1-absorbing prime submodule of M if and only if $N_{1}/N_{2}$ is a 1-absorbing prime submodule of $M/N_{2}$. We also have the following charaterization of 1-absorbing prime submodules in Theorem 2.9. A proper submodule of an R-module M is a 1-absorbing prime submodule of M if $I_{1}I_{2}K\subseteq N$ for some proper ideals $I_{1},I_{2}$ of R and some submodule K of M, then either $I_{1}I_{2}\subseteq (N:_{R}M)$ or $K\subseteq N.$ If there exists a weakly 1-absorbing prime submodule N in the R-module M that is not a prime submodule, then we show in Theorem 2.10 that R is a local ring. If R is a local ring and N is a weakly 1-absorbing prime submodule that is not 1-absorbing prime, then we show in Proposition 3.10 that $(N:_{R}M)^{2}N=\left\{ 0\right\} ,$ and in particular, $(N:_{R}M)^{3}\subseteq$Ann(M). If R is a local ring and M is a multiplication module and N is a weakly 1-absorbing prime submodule of M that is not a 1-absorbing prime submodule, then $N^{3}=\left\{ 0\right\}$ (Proposition 3.11). It is shown in Theorem 3.14 that if N is a proper submodule of the R module M,  then N is a weakly 1-absorbing prime submodule of M if for any proper ideals $I_{1},I_{2}$ of R and a submodule K of M such that $\left\{ 0\right\} \ne I_{1}I_{2}K\subseteq N$ and N is free triple-zero with respect to $I_{1},I_{2},K$, we have either $I_{1} I_{2}\subseteq$ $(N:_{R}M)$ or $K\subseteq N.$

We have the following diagram which clarifies the place of 1-absorbing prime submodules and weakly 1-absorbing prime submodules. Here, the arrows in the diagram are irreversible. $\begin{array}{ccccc} \text {prime submodule} &{} \Rightarrow &{} 1\text {-absorbing prime} &{} \Rightarrow &{} 2\text {-absorbing}\\ \Downarrow &{} &{} \Downarrow &{} &{} \Downarrow \\ \text {weakly prime } &{} \Rightarrow &{} \text {weakly }1\text {-absorbing prime} &{} \Rightarrow &{} \text {weakly }2\text {-absorbing} \end{array}$

2 1-absorbing prime submodules

Definition 2.1

Let M be an R-module and N be a proper submodule of M. For all nonunit elements $a,b\in R$ and $m\in M$ if $aRbRm\subseteq N$ either $ab\in (N:_{R}M)$ or $m\in N$, then N is called 1-absorbing prime submodule of M.

Proposition 2.2

Prime submodules $\Rightarrow$ 1-absorbing prime submodules $\Rightarrow$ 2-absorbing submodules.

Proof

Let N be a prime submodule of M. Take nonunit elements $a,b\in R$ and $m\in M$ such that $aRbRm\subseteq N$. Now $abRm\subseteq N$ and since N is a prime submodule, $ab\in (N:_{R}M)$ or $m\in N$, as desired.

Suppose N is a 1-absorbing prime submodule of M. Take any $a,b\in R$ and $m\in M$ such that $aRbRm\subseteq N$. If a and b are nonunits, we have $ab\in (N:_{R}M)$ or $m\in N$ and we are done. If a is a unit element, then $abm\in N$ implies $bm\in N.$ If b is a unit element, then there exists $b^{\prime }\in R$ such that $b^{\prime }b=1$ and we have $am=ab^{\prime }bm\in N$, as desired. $\square$

Example 2.3

For field K the ring $R=\left\{ \left[ \begin{array}{ccc} a &{}\quad 0 &{}\quad b\\ 0 &{}\quad a &{}\quad c\\ 0 &{}\quad 0 &{}\quad a \end{array} \right] :a,b,c\in K\right\}$ is a local ring whose unique maximal ideal M has square zero. Consider the R-module M. Then every proper submodule is a 1-absorbing prime submodule of M. To see this, choose nonunits $x,y\in R$ and $m\in M$ such that $xRyRm\subseteq N$. Since $xRy\subseteq M^{2}=\left\{ 0\right\}$, we have $xy\in (N:_{R}M)$ which implies N is a 1-absorbing prime submodule of M.

Proposition 2.4

If N is a 1-absorbing prime submodule of M then we have the following:

• 1. $(N:_{R}M)$ is a 1-absorbing prime ideal of R.

• 2. (N : Rm) is a 1-absorbing prime ideal of R for every $m\in M\backslash N$.

Proof

Let N be a 1-absorbing prime submodule of M.

1. Choose nonunits $a,b,c\in R$ such that $aRbRc\subseteq (N:_{R}M).$ For all $m\in M,$ then $aRbRcm\subseteq N$. By our hypothesis, $ab\in (N:_{R}M)$ or $cm\in N$. If $ab\in (N:_{R}M),$ then we done. So suppose $ab\notin (N:_{R}M).$ Hence $cm\in N$ for all $m\in M.$ This implies that $c\in (N:_{R}M)$. Consequently, $(N:_{R}M)$ is 1-absorbing prime ideal of R.

2. Choose nonunits $a,b,c\in R$ such that $aRbRc\subseteq (N:Rm).$ Hence $aRbRcRm\subseteq N$ and therefore $aRbRcrm\subseteq N$ for all $r\in R.$ By our hypothesis or $ab\in (N:_{R}M)$ or $crm\in N$ for all $r\in R.$ Thus $ab\in (N:_{R}M)\subseteq (N:Rm)$ or $c\in (N:Rm).$ Consequently, (N : Rm) is 1-absorbing prime ideal of R. $\square$

The converse of the above proposition is not true in general.

Example 2.5

Let p be a fixed prime integer. Then $\mathbb {Z}(p^{\infty })=\{a\in Q/\mathbb {Z}:a=r/p^{n}+\mathbb {Z}$ for some $r\in \mathbb {Z}$ and $n\ge 0\}$ is a nonzero submodule of $Q/\mathbb {Z}$. Let $G_{t}=\{a\in Q/\mathbb {Z}:a=r/p^{t}+\mathbb {Z}$ for some $r\in \mathbb {Z}\}$ for all $t\ge 0$. It is well known that each proper submodule of $\mathbb {Z}(p^{\infty })$ is equal to $G_{t}$ for some $t\ge 0$. $G_{t}$ is not a 1-absorbing prime submodule of $\mathbb {Z}(p^{\infty })$ since for $p^{2}(1/p^{t+2}+\mathbb {Z})\in G_{t}$ we have $(1/p^{t+2}+Z)\notin G_{t}$ and $p^{2}\notin (G_{t}:_{\mathbb {Z} }\mathbb {Z}(p^{\infty }))=\left\{ 0\right\}$. We can see that $(G_{t}:_{\mathbb {Z}}\mathbb {Z}(p^{\infty }))=\left\{ 0\right\}$ is a 1-absorbing prime ideal of $\mathbb {Z}$ for all $t\ge 0$.

Note that from the above remark we have that some modules do not have any 1-absorbing prime submodules. Since each proper submodule of $\mathbb {Z} (p^{\infty })$ is equal to $G_{t}$ for some $t\ge 0$, so $\mathbb {Z}(p^{\infty })$ does not have any 1-absorbing prime submodule.

Proposition 2.6

Let $M_{1}$ and $M_{2}$ be R-modules and $f:M_{1} \rightarrow M_{2}$ be a module homomorphism. Then the following statements hold:

• 1. If $N_{2}$ is a 1-absorbing prime submodule of $M_{2}$, then $f^{-1}(N_{2})$ is a 1-absorbing prime submodule of $M_{1}$.

• 2. Let f be an epimorphism. If $N_{1}$ is a 1-absorbing prime submodule of $M_{1}$ containing $\ker (f)$, then $f(N_{1})$ is a 1-absorbing prime submodule of $M_{2}$.

Proof

1. Suppose that a, b are nonunit elements of R, $m_{1}\in M_{1}$ and $aRbRm_{1}\subseteq f^{-1}(N_{2})$. Then $aRbRf(m_{1})\subseteq N_{2}.$ Since $N_{2}$ is a 1-absorbing prime submodule, we have either $ab\in (N_{2}:_{R}M_{2})$ or $f(m_{1})\in N_{2}$. Here, we show that $(N_{2}:_{R} M_{2})\subseteq (f^{-1}(N_{2}):M_{1})$. Let $r\in (N_{2}:M_{2})$. Then $rM_{2}\subseteq N_{2}$ which implies that $rf^{-1}(M_{2})\subseteq f^{-1}(N_{2})$, i.e., $rM_{1}\subseteq f^{-1}(N_{2})$. Thus $r\in (f^{-1} (N_{2}):M_{1})$. Hence $ab\in$ $(f^{-1}(N_{2}):M_{1})$ or $m_{1}\in f^{-1}(N_{2})$. Hence $f^{-1}(N_{2})$ is a 1-absorbing prime submodule of $M_{1}$.

2. Suppose that are nonunit elements a and b of R, $m_{2}\in M_{2}$ and $aRbRm_{2}\in f(N_{1})$. Since f is an epimorphism, there exists $m_{1}\in M_{1}$ such that $f(m_{1})=m_{2}$. Since $\ker (f)\subseteq N_{1}$, $aRbRm_{1}\subseteq N_{1}$. Hence $ab\in (N_{1}:M_{1})$ or $m_{1}\in N_{1}$. Here, we show that $(N_{1}:M_{1})\subseteq (f(N_{1}):M_{2})$. Let $r\in (N_{1}:M_{1})$. Then $rM_{1}\subseteq N_{1}$ which implies that $rf(M_{1} )\subseteq f(N_{1})$. Since f is onto, we conclude that $rM_{2}\subseteq f(N_{1})$, that is, $r\in (f(N_{1}):M_{2})$. Thus $ab\in (f(N_{1}):M_{2})$ or $m_{2}\in f(N_{1})$, as desired $\square$

As a consequence of Proposition 2.6, we have the following result.

Corollary 2.7

Let M be an R-module and $N_{1}$, $N_{2}$ be submodules of M with $N_{2}\subseteq N_{1}$. Then $N_{1}$ is a 1-absorbing prime submodule of M if and only if $N_{1}/N_{2}$ is a 1-absorbing prime submodule of $M/N_{2}$.

Proof

Suppose that $N_{1}$ is a 1-absorbing primary submodule of M. Consider the canonical epimorphism $f:M\rightarrow M/N_{2}$ in Proposition 2.6. Then $N_{1}/N_{2}$ is a 1-absorbing prime submodule of $M/N_{2}$. Conversely, let a and b are nonunit elements of R, $m\in M$ such that $aRbRm\subseteq N_{1}$. Hence $aRbR(m+N_{2})\subseteq N_{1}/N_{2}.$ Since $N_{1}/N_{2}$ is a 1-absorbing prime submodule of $M/N_{2}$, it implies either $ab\in (N_{1}/N_{2}:_{R}M/N_{2})$ or $m+N_{2}\in M/N_{2}$. Therefore $ab\in (N_{1}:_{R}M)$ or $m\in N_{1}$. Thus $N_{1}$ is a 1-absorbing prime submodule of M. $\square$

Let $M_{1}$ be $R_{1}$-module and $M_{2}$ be $R_{2}$-module where $R_{1}$ and $R_{2}$ are noncommutative rings with identity. Let $R=R_{1}\times R_{2}$ and $M=M_{1}\times M_{2}$. Then M is an R-module and every submodule of M is of the form $N=N_{1}\times N_{2}$ for some submodules $N_{1},N_{2}$ of $M_{1},M_{2}$, respectively.

Proposition 2.8

Let $M_{1}$ be $R_{1}$-module and $M_{2}$ be $R_{2}$-module where $R_{1}$ and $R_{2}$ are noncommutative rings with identity. Let $R=R_{1}\times R_{2}$ and $M=M_{1}\times M_{2}$. Suppose that $N_{1}$ is a proper submodule of $M_{1}$. If $N=N_{1}\times M_{2}$ is a 1-absorbing prime submodule of the R-module M, then $N_{1}$ is a 1-absorbing prime submodule of M.

Proof

Suppose that $N=N_{1}\times M_{2}$ is a 1-absorbing prime submodule of M. Put $M^{\prime }=M/(\{0\}\times M_{2})$ and $N^{\prime }=N/(\{0\}\times N_{2})$. From Corollary 2.7, $N^{\prime }$ is a 1-absorbing prime submodule of $M^{\prime }$. Since $M^{\prime }\cong M_{1}$ and $N^{\prime }\cong N_{1}$, we conclude the result. $\square$

Next we give several characterizations of 1-absorbing prime submodules of an R-module.

Theorem 2.9

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

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

• (2) If a, b are nonunit elements of R such that $ab\notin (N:_{R}M)$, then $(N:_{M}aRbR)\subseteq N$.

• (3) If a, b are nonunit elements of R, and K is a submodule of M with $aRbK\subseteq N$, then $ab\in (N:_{R}M)$ or $K\subseteq N.$

• (4) If $I_{1}I_{2}K\subseteq N$ for some proper ideals $I_{1},I_{2}$ of R and some submodule K of M, then either $I_{1}I_{2}\subseteq (N:_{R}M)$ or $K\subseteq N.$

Proof

$(1)\Rightarrow (2)$ Suppose that a, b are nonunit elements of R such that $ab\notin (N:_{R}M)$. Let $m\in (N:_{M}aRbR)$. Hence $aRbRm\subseteq N$. Since N is 1-absorbing prime submodule and $ab\notin (N:_{R}M)$, we have $m\in N$, and so $(N:_{M}aRbR)\subseteq N$.

$(2)\Rightarrow (3)$ Suppose that $ab\notin (N:_{R}M)$. Since $aRbRK\subseteq aRbK\subseteq N$, we have $K\subseteq (N:_{M}aRbR)\subseteq N$ by (2). $(3)\Rightarrow (4)$ Suppose $I_{1}I_{2}K\subseteq N$ for some proper ideals $I_{1},I_{2}$ of R and some submodule K of M. Assume on the contrary that neither $I_{1}I_{2}\subseteq (N:_{R}M)$ nor $K\subseteq N$. Then there exist nonunit elements $a\in I_{1},b\in I_{2}$ with $ab\notin (N:_{R}M)$. Thus $aRbK\subseteq N$, which contradicts (3).

$(4)\Rightarrow (1)$ Let $a,b\in R$ be nonunit elements, $m\in M$ and $aRbRm\subseteq N$. Put $I_{1}=RaR,I_{2}=RbR,K=Rm$. Now $RaRRbRRm\subseteq RaRbRm\subseteq N.$ Thus $ab\in RaRRbR\subseteq (N:_{R}M)$ or $m\in Rm\subseteq N$ and we are done. $\square$

Theorem 2.10

Let M be an R-module. If N is a 1-absorbing prime submodule of M that is not a prime submodule, then R is a local ring.

Proof

Suppose that N is a 1-absorbing prime submodule of M that is not a prime submodule. Then there exist a nonunit $r\in R$ and $m\in M$ such that $rRm\subseteq N$ but $r\notin (N:_{R}M)$ and $m\notin N$. Choose a nonunit element $s\in R$. Hence we have that $rRsRm\subseteq rRm\subseteq N$ and $m\notin N$. Since N is 1-absorbing prime, $rs\in (N:_{R}M)$. Let us take a unit element $u\in R$. We claim that $s+u$ is a unit element of R. To see this, assume $s+u$ is a nonunit. Then $rR(s+u)Rm\subseteq rRm\subseteq N$. As N is 1-absorbing prime, $r(s+u)\in (N:_{R}M)$. This means that $ru\in (N:_{R}M)$, i.e., $r\in (N:_{R}M)$, which is a contradiction. Thus for any nonunit element s and unit element u in R, we have $s+u$ is a unit element. From [9, Lemma 4.1], we have that R is a local ring. $\square$

Corollary 2.11

Let M be an R-module where R is not a local ring. Then a proper submodule N of M is a 1-absorbing prime submodule if and only if N is a prime submodule of M.

Proposition 2.12

Let $\left\{ N_{i}:i\in \Delta \right\}$ be a chain of 1-absorbing prime submodules of the R-module M. Then ${\textstyle \bigcap \nolimits _{i\in \Delta }} N_{i}$ is a 1-absorbing prime submodule of M.

Proof

Let $\left\{ N_{i}:i\in \Delta \right\}$ be a chain of 1-absorbing prime submodules of M. Take nonunit elements $a,b\in R$ and $m\in M$ such that $aRbRm\subseteq {\textstyle \bigcap \nolimits _{i\in \Delta }} N_{i}$. Assume that $m\notin {\textstyle \bigcap \nolimits _{i\in \Delta }} N_{i}$, so there exists $i\in \Delta$ such that $m\notin N_{i}$. Since $N_{i}$ is 1-absorbing prime, we conclude $ab\in (N_{i}:M)$. For any $j\in \Delta$, we have $N_{i}\subseteq N_{j}$ or $N_{j}\subseteq N_{i}$. Without loss of generality, if $N_{i}\subseteq N_{j}$ then $(N_{i}:M)\subseteq (N_{j}:M)$, that is, $ab\in (N_{j}:M)$. If $N_{j}\subseteq N_{i}$, then $ab\in (N_{j}:M)$ since $m\notin N_{j}$ and $N_{j}$ is 1-absorbing prime. Hence we have $ab\in {\textstyle \bigcap \nolimits _{i\in \Delta }} \left\{ (N_{i}:M):i\in \Delta \right\} =(( {\textstyle \bigcap \nolimits _{i\in \Delta }} N_{i}:i\in \Delta ):M).$ $\square$

Definition 2.13

Let M be an R-module and N be a proper submodule of M. Let P be a 1-absorbing prime submodule of M such that $N\subseteq P$. If there does not exist a 1-absorbing prime submodule $P^{\prime }$ such that $N\subseteq P^{\prime }\subset P$, then P is called a minimal 1-absorbing prime submodule over N.

Proposition 2.14

Let M be an R-module and N be a proper submodule of M. If P is a 1-absorbing prime submodule of M such that $N\subseteq P$, then there exists a minimal 1-absorbing prime submodule over N that is contained in P.

Proof

Let $\Lambda =\left\{ P_{i}:P_{i}\text { is a submodule of }M\text { such that }N\subseteq P_{i}\subseteq P\right\} .$ Since $N\subseteq P,$ we have $\Lambda \ne \emptyset .$ Consider $(\Lambda ,\supseteq )$. Let us take a chain $\left\{ N_{i}:i\in \Delta \right\}$ in $\Lambda$. Since by Proposition 2.12, $\cap _{i\in \Delta }N_{i}$ is a 1-absorbing prime submodule of M,  there exists a maximal element $K\in \Lambda$ by applying Zorn’s Lemma. Then K is 1-absorbing prime and $N\subseteq K\subseteq P$. Now we will show that K is a minimal 1-absorbing prime submodule over N. On the contrary, assume that there exists a 1-absorbing prime submodule $K^{\prime }$ such that $N\subseteq K^{\prime }\subseteq K$. Then $K^{\prime } \in \Lambda$ and $K\subseteq K^{\prime }$. This implies $K=K^{\prime }$. Consequently, K is a minimal 1-absorbing prime submodule of N. $\square$

Corollary 2.15

Let M be an R-module. Every 1-absorbing prime submodule of M contains at least one minimal 1-absorbing prime submodule of M.

3 Weakly 1-absorbing prime submodules

Definition 3.1

Let R be a ring and N be a proper submodule of an R-module M. Then N is a weakly 1-absorbing prime submodule of M if $\left\{ 0\right\} \ne aRbRm\subseteq N$ implies $abM\subseteq N$ i.e. $ab\in (N:_{R}M)$ or $m\in N$ for nonunits $a,b\in R$ and $m\in M$.

Remark 3.2

1.:

Every 1–absorbing prime submodule is weakly 1-absorbing prime but the converse does not necessarily hold. For example consider the case where $R=\mathbb {Z},$ $M=\mathbb {Z}/30\mathbb {Z}$ and $N=\left\{ 0\right\}$. Then $2\cdot 3\cdot (5+30\mathbb {Z})=0\in N$ while $2\cdot 3\notin (N:_{R}M),$ $(5+30Z)\notin N$. Therefore N is not 1-absorbing prime while it is weakly 1-absorbing prime.

2.:

Every weakly prime submodule is weakly 1 absorbing prime but the converse does not necessarily hold. Let $M=\mathbb {Z}_{12}$ be a module over $\mathbb {Z}$ and $W=\{\overline{0},\overline{4},\overline{8}\}$ be a proper submodule of M. Let $r,s\in \mathbb {Z}$ and $m\in M.$ Now $(W:M)=4\mathbb {Z}.$ Therfore, for $0\ne rsm\in W,$ we get $m=\overline{4}$ or $m=\overline{8}$ which are elements in W or $rs\in 4\mathbb {Z}.$ So W is a weakly 1 absorbing prime submodule. W is not weakly prime since $\overline{0} \ne 2\cdot \overline{2}\in W$ with $\overline{2}\notin W$ and $2\notin (W:M).$

Question. Suppose that L is a weakly 1-absorbing prime submodule of an R-module M and $\left\{ 0\right\} \ne IJK\subseteq L$ for some ideals I, J of R and a submodule K of M. Does it imply that $IJ\subseteq (M:_{R}L)$ or $K\subseteq L$?

This section is devoted to studying the above question for modules over noncommutative rings.

Proposition 3.3

Let $x\in M$ and $a\in R$. Then if $\hbox {ann}_{l}(x)\subseteq (Rx:M)$, the submodule Rx is 1-absorbing prime if and only if Rx is weakly 1-absorbing prime.

Proof

If Rx is 1-absorbing prime then it is clear that Rx is weakly 1- absorbing prime. Let Rx be a weakly 1-absorbing prime submodule of M and suppose $r,s\in R$ are nonunits and $m\in M$ with $rRsRm\subseteq Rx$. Since Rx is a weakly 1-absorbing prime submodule, we may assume $rRsRm=\left\{ 0\right\}$, otherwise Rx is 1-absorbing prime. Now $rRsR(x+m)\subseteq Rx$. If $rRsR(x+m)\ne \left\{ 0\right\}$ then we have $rs\in (Rx:M)$ or $(x+m)\in Rx$, as Rx is a weakly 1-absorbing prime submodule. Hence $rs\in (Rx:M)$ or $m\in Rx$. Now let $rRsR(x+m)=\left\{ 0\right\}$. Then $rRsRm=\left\{ 0\right\}$ implies $rRsRx=\left\{ 0\right\}$. Hence $rs\in$ann$_{l}(x)\subseteq (Rx:M).$ Thus Rx is 1-absorbing prime. $\square$

Proposition 3.4

Let R be a ring and N be a proper submodule of an R-module M.

• 1. If N is weakly prime, then it is weakly 1-absorbing prime.

• 2. If N is a weakly 1-absorbing prime submodule of M, then it is a weakly 2-absorbing submodule.

Proof

1. Assume N is a weakly prime submodule of the R-module M and $\left\{ 0\right\} \ne aRbRm\subseteq N$ for nonunits $a,b\in R$ and $m\in M$. Suppose $m\notin N.$ Since $abRm\subseteq N$ and $m\notin N,$ we have $ab\in (N:_{R}M).$ Hence N is weakly 1-absorbing prime. 2. Assume N is a weakly 1-absorbing prime submodule of the R-module M and $\left\{ 0\right\} \ne aRbRm\subseteq N$ for $a,b\in R$ and $m\in M$. If a is a unit, then it is easy to see that $bm\in N.$ If b is a unit, then there exists $b^{\prime }\in R$ such that $ab^{\prime }bm=am\in N.$ If both a and b are nonunits, then $ab\in (N:_{R}M)$ since N is a weakly 1-absorbing prime submodule of the R-module M. Hence N is weakly 2-absorbing. $\square$

Proposition 3.5

Let N be a weakly 1-absorbing prime submodule of an R-module M. Assume that K is a submodule of M with $N\subsetneqq K$. Then N is a weakly 1-absorbing prime submodule of K.

Proof

Let $a,b\in R$ be nonunits and $k\in K$ with $\left\{ 0\right\} \ne aRbRk\subseteq N$. Then $ab\in (N:_{R}M)$ or $k\in N$ as N is a weakly 1-absorbing prime submodule of M. Thus $ab\in (N:_{R}K)$ or $k\in K$ since $(N:_{R}M)\subseteq (N:_{R}K)$ and $N\subseteq K.$ $\square$

Proposition 3.6

Let N, K be submodules of an R-module M with $K\subseteq N$. If N is a weakly 1-absorbing prime submodule of M, then N/K is a weakly 1-absorbing prime submodule of M/K. The converse is true when K is a weakly 1-absorbing prime submodule.

Proof

Assume that N is a weakly 1-absorbing prime submodule of M. Let $a,b\in R$ be nonunits and $m+K\in M/K$ where $\left\{ 0_{M/K}\right\} \ne aRbR(m+K)\subseteq N/K$. Since $aRbR(m+K)\ne \left\{ 0_{M/K}\right\}$, we get $aRbRm\subseteq N$ and $aRbPm\nsubseteq K$. If $aRbRm=\left\{ 0\right\}$, we obtain $aRbRm+K=\left\{ 0_{M/K}\right\}$. So $aRbRm\ne \left\{ 0\right\}$. Thus $ab\in (N:_{R}M)$ or $m\in N$ as N is weakly 1-absorbing prime. Consequently, we get $ab\in (N/K:_{R}M/K)$ or $m+K\in N/K$. Conversely, let K be a weakly 1-absorbing prime submodule. Assume that N/K is a weakly 1-absorbing prime submodule of M/K. Let $a,b\in R$ be nonunits and $m\in M$ where $\left\{ 0\right\} \ne aRbRm\subseteq N$. Then we have $aRbRm+K\subseteq N/K$. If $aRbRm+K=\left\{ 0_{M/K}\right\}$, then $aRbRm\subseteq K$. Thus $ab\in (K:_{R}M)$ or $m\in K$, since K is weakly 1-absorbing prime. Therefore, $ab\in (N:_{R}M)$ or $m\in N$, since $K\subseteq N$. Let $aRbRm+K=aRbR(m+K)\ne \left\{ 0_{M/K}\right\}$. Then $ab\in (N/K:_{R}M/K)$ or $m+K\in N/K$. Thus $ab\in (N:_{R}M)$ or $m\in N$. $\square$

Definition 3.7

Let N be a weakly 1-absorbing prime submodule of an R-module M. For nonunits $a,b\in R$ and $m\in M,$ (a, b, m) is called a triple-zero of N if $aRbRm=0$, $ab\notin (N:_{R}M)$ and $m\notin N$.

Note that if N is a weakly 1-absorbing prime submodule of M and there is no triple-zero of N, then N is a 1-absorbing prime submodule of M.

Proposition 3.8

Let N be a weakly 1-absorbing prime submodule of M and K be a proper submodule of M with $K\subseteq N$. Then for any nonunits $a,b\in R$ and $m\in M$, (a, b, m) is a triple-zero of N if and only if $(a,b,m+K)$ is a triple-zero of N/K.

Proof

Let (a, b, m) be a triple-zero of N for some nonunits $a,b\in R$ and $m\in M$. Then $aRbRm=\left\{ 0\right\}$, $ab\notin (N:_{R}M)$ and $m\notin N$. By Proposition 3.6, we get that N/K is a weakly 1-absorbing prime submodule of M/K. Thus $aRbR(m+K)=K,ab\notin (N/K:M/K)$ and $(m+K)\notin N/K$. Hence $(a,b,m+K)$ is a triple-zero of N/K. Conversely, assume that $(a,b,m+K)$ is a triple-zero of N/K. Suppose that $aRbRm\ne \left\{ 0\right\}$. Then $aRbRm\subseteq N$ since $aRbR(m+K)=K$. Thus $ab\in (N:_{R}M)$ or $m\in N$ as N is weakly 1-absorbing prime, a contradiction. So it must be $aRbRm=\left\{ 0\right\}$. Consequently, (a, b, m) is a triple-zero of N. $\square$

Theorem 3.9

Let R be a local ring and let N be weakly 1-absorbing prime submodule of M and (a, b, m) be a triple-zero of N for some nonunits $a,b\in R$ and $m\in M$. Then the following hold.

• 1. $aRbN=a(N:_{R}M)m=b(N:_{R}M)m=\left\{ 0\right\} .$

• 2. $a(N:_{R}M)N=b(N:_{R}M)N=(N:_{R}M)bN=(N:_{R}M)bm=(N:_{R}M)^{2}m=\left\{ 0\right\} .$

Proof

Suppose that (a, b, m) is a triple-zero of N for some nonunits $a,b\in R$ and $m\in M$. 1. Assume that $aRbN\ne \left\{ 0\right\}$. Then there is an element $n\in N$ such that $aRbRn\ne \left\{ 0\right\} .$ Now $aRbR(m+n)=aRbRm+aRbRn=aRbRn\ne \left\{ 0\right\}$ since $aRbRm=\left\{ 0\right\} .$ Since $\left\{ 0\right\} \ne aRbR(m+n)\subseteq N\$and N a weakly 1-absorbing prime submodule, we have $ab\in (N:_{R}M)$ or $(m+n)\in N$. Hence $ab\in (N:_{R}M)$ or $m\in N$. This is a contradiction since (a, b, m) is a triple zero of N. Hence $aRbN=\left\{ 0\right\} .$ Now, we suppose that $a(N:_{R}M)m\ne \left\{ 0\right\}$. Thus there exists an element $r\in (N:_{R}M)$ such that $arm\ne 0$. Hence $aR(r+b)Rm=aRrRm+aRbRm=aRrRm$. Hence $\left\{ 0\right\} \ne aR(r+b)Rm\subseteq N.$ Since R is local, the set of nonunit elements of R is an ideal of R. Therefore $(r+b)$ is a nonunit. Since N is a weakly 1-absorbing prime submodule, we have $a(r+b)\in (N:_{R}M)$ or $m\in N$. Consequently, $ab\in (N:_{R}M)$ or $m\in N$ a contradiction since (a, b, m) is a triple-zero of N. Hence $a(N:_{R}M)m=\left\{ 0\right\} .$ Similarly, we can proof that $b(N:_{R}M)m=\left\{ 0\right\}$. 2. Assume that $a(N:_{R}M)N\ne \left\{ 0\right\}$. Then there are $r\in (N:_{R}M),$ $n\in N$ such that $arn\ne 0$. By (1), we get $a(b+r)(m+n)=abm+abn+arm+arn=arn\ne 0.$ Now $\left\{ 0\right\} \ne aR(b+r)R(m+n)\subseteq N.$ Again, since R is a local ring $(b+r)$ is a nonunit and since N is 1-absorbing prime, we have $a(b+r)\in (N:_{R}M)$ or $(m+n)\in N$. Hence we obtain $ab\in (N:_{R}M)$ or $m\in N$ a contradiction. Hence $a(N:_{R}M)N=\left\{ 0\right\} .$ In a similar way we get $b(N:_{R}M)N=\left\{ 0\right\} .$ Now, we suppose that $(N:_{R} M)bN\ne \left\{ 0\right\} .$ 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 $\left\{ 0\right\} \ne (a+r)RbR(n+m)\subseteq N$ and since N is weakly 1-absorbing and $(a+r)$ is a nonunit, we have $(a+r)b\in (N:_{R}M)$ or $(n+m)\in N.$ Hence $ab\in (N:_{R}M)$ or $m\in N$ a contradiction since (a, b, m) is a triple-zero of N. Now, we suppose that $(N:_{R} M)bm\ne \left\{ 0\right\} .$ 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 a weakly 1-absorbing prime submodule, we have $(a+r)b\in (N:_{R}M)$ or $m\in N$. Therefore $ab\in (N:_{R}M)$ or $m\in N$ a contradiction since (a, b, m) is a triple-zero of N. Hence $(N:_{R}M)bm=\left\{ 0\right\} .$ Lastly, we show that $(N:_{R}M)^{2}m=\left\{ 0\right\} .$ Let $(N:_{R} M)^{2}m\ne \left\{ 0\right\}$. 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 $\left\{ 0\right\} \ne (a+r)R(b+s)Rm\subseteq N$. Since R is a local ring, $(a+r)$ and $(b+s)$ are nonunits. Hence $(a+r)(b+s)\in (N:_{R}M)$ or $m\in N$. Consequently $ab\in (N:_{R}M)$ or $m\in N$ a contradiction since (a, b, m) is a triple-zero of N. Therefore $(N:_{R}M)^{2}m=\left\{ 0\right\}$. $\square$

Proposition 3.10

Let R be a local ring. Assume that N is a weakly 1-absorbing prime submodule of an R-module M that is not 1-absorbing prime. Then $(N:_{R}M)^{2}N=\left\{ 0\right\}$. In particular, $(N:_{R}M)^{3}\subseteq$Ann(M).

Proof

Suppose that N is a weakly 1-absorbing prime submodule of an R-module M that is not 1-absorbing prime. Then there is a triple-zero (a, b, m) of N for some nonunits $a,b\in R$ and $m\in M$. Assume that $(N:_{R}M)^{2} N\ne \left\{ 0\right\}$. Thus there exist $r,s\in (N:_{R}M)$ and $n\in N$ with $rsn\ne 0$. By Theorem 3.9, we get $(a+r)(b+s)(n+m)=rsn\ne 0$. Then we have $\left\{ 0\right\} \ne (a+r)R(b+s)R(n+m)\subseteq N.$ Since N is weakly 1-absorbing prime, we have $(a+r)(b+s)\in (N:_{R}M)$ or $(n+m)\in N$ and so $ab\in (N:_{R}M)$ or $m\in N$ which is a contradiction. Thus $(N:_{R}M)^{2}N=\left\{ 0\right\} .$ We get $(N:_{R}M)^{3}\subseteq ((N:_{R}M)^{2}N:M)=(\left\{ 0\right\} :M)=$Ann(M). $\square$

Proposition 3.11

Let R be a local ring and let M be a multiplication R-module and N be a weakly 1-absorbing prime submodule of M that is not a 1-absorbing prime submodule. Then $N^{3}=\left\{ 0\right\}$.

Proof

We have that $(N:_{R}M)M=N$ since M is a multiplication module. Then $N^{3}=(N:_{R}M)^{3}M=(N:_{R}M)^{2}N=\left\{ 0\right\}$. Consequently, $N^{3}=\left\{ 0\right\}$. $\square$

Definition 3.12

Let N be a weakly 1-absorbing prime submodule of an R-module M and let $a,b\in R$ be nonunits. Let $\left\{ 0\right\} \ne I_{1}I_{2}K\subseteq N$ for some ideals $I_{1},I_{2}$ of R and some submodule K of M. N is called free triple-zero in regard to $I_{1},I_{2},K$ if (a, b, m) is not a triple-zero of N for every $a\in I_{1},b\in I_{2}$ and $m\in K$.

Lemma 3.13

Let N be a weakly 1-absorbing prime submodule of M. Assume that $aRbK\subseteq N$ for some nonunits $a,b\in R$ and some submodule K of M where (a, b, m) is not a triple-zero of N for every $m\in K$. If $ab\notin (N:_{R}M),$ then $K\subseteq N$.

Proof

Suppose that $aRbK\subseteq N$, but $ab\notin (N:_{R}M)$ and $K\nsubseteq N$. Then there exists an element $k\in K\backslash N$. But (a, b, k) is not a triple-zero of N and $aRbRk\subseteq N$ and $ab\notin (N:_{R}M)$ and $k\notin N$, a contradiction. Hence $K\subseteq N.$ $\square$

Let N be a weakly 1-absorbing prime submodule of an R-module M and $I_{1}I_{2}K\subseteq N$ for some for some ideals $I_{1},I_{2}$ of R and some submodule K of M where N is free triple-zero in regard to $I_{1},I_{2},K.$ Note that if $a\in I_{1},b\in I_{2}$ and $m\in K$, then $ab\in (N:_{R}M)$ or $m\in N.$

Theorem 3.14

Suppose that N is a proper submodule of the R module M. Then the following statements are equivalent.

• 1. N is a weakly 1-absorbing prime submodule of M.

• 2. For any proper ideals $I_{1},I_{2}$ of R and a submodule K of M such that $\left\{ 0\right\} \ne I_{1}I_{2}K\subseteq N$ and N is free triple-zero with respect to $I_{1},I_{2},K$, we have either $I_{1} I_{2}\subseteq$ $(N:_{R}M)$ or $K\subseteq N.$

Proof

$(1)\Rightarrow (2)$ Suppose that N is a weakly 1-absorbing prime submodule of M and $\left\{ 0\right\} \ne I_{1}I_{2}K\subseteq N$ for proper ideals $I_{1},I_{2}$ of R and a submodule K of M such that N is free triple-zero with respect to $I_{1},I_{2},K$. Then there are nonunit elements $a\in I_{1}$ and $b\in I_{2}$ such that $ab\notin (N:_{R}M)$. Since $aRbK\subseteq N$, $ab\notin (N:_{R}M)$ and (a, b, k) is not a triple-zero of N for every $k\in K$, it follows from Lemma 3.13 that $K\subseteq N$. $(2)\Rightarrow (1)$ Suppose that $\left\{ 0\right\} \ne aRbRm\subseteq N$ for some nonunit elements $a,b\in R$ and $m\in M.$ Suppose $ab\notin (N:_{R} M)$. Let $I_{1}=RaR,$ $I_{2}=RbR$ and $K=Rm.$ Now, $\left\{ 0\right\} \ne RaRRbRRm\subseteq RN\subseteq N.$ Hence $\left\{ 0\right\} \ne I_{1} I_{2}K\subseteq N$ and $I_{1}I_{2}\nsubseteq$ $(N:_{R}M).$ From (2), it follows that $m\in Rm=K\subseteq N$ and we are done. $\square$