On graded 2-absorbing $I_{e}$-prime submodules of graded modules over graded commutative rings

1.   Introduction and preliminaries

Throughout this paper all rings are commutative with identity and all modules are unitary.

Badawi in ^[15] introduced the concept of 2-absorbing ideals of commutative rings. The notion of 2-absorbing ideals was extended to 2-absorbing submodules in ^[17] and ^[24]. Recently, Farshadifar in ^[18] introduced and studied the concept of 2-absorbing $I$-prime submodules.

Refai and Al-Zoubi in ^[25] introduced the concept of graded primary ideal. The concept of graded 2-absorbing ideal was introduced and studied by Al-Zoubi, Abu-Dawwas and Ceken in ^[4]. The concept of graded prime submodule was introduced and studied by many authors, see for example ^{[1,2,10,11,12,14,23]}. The concept of graded 2-absorbing submodule, generalizations of graded prime submodule, was introduced by Al-Zoubi and Abu-Dawwas in ^[3] and studied in ^[7,8]. Then many generalizations of graded 2-absorbing submodules were studied such as graded 2-absorbing primary (see^[16]), graded weakly 2-absorbing primary (see ^[6]) and graded classical 2-absorbing (see ^[5]). Recently, Alghueiri and Al-Zoubi in ^[13] introduced the concept of graded $I_{e}$-prime submodule over a commutative ring as a new generalization of graded prime submodule. Here, we introduce the concept of graded 2-absorbing $I_{e}$-prime submodule as a new generalization of a graded 2-absorbing prime submodule on the one hand and a generalization of a graded $I_{e}$-prime submodule on other hand.

First, we recall some basic properties of graded rings and modules which will be used in the sequel. We refer to ^{[19,20,21,22]} for these basic properties and more information on graded rings and modules.

Let $G$ be an abelian multiplicative group with identity element $e.$ A ring $R$ is called a graded ring (or $G$-graded ring) if there exist additive subetaoups $R_{h}$ of $R$ indexed by the elements $h\in G$ such that $R = \oplus _{h\in G}R_{h}$ and $R_{g}R_{h}\subseteq R_{gh}$ for all $g, h\in G$. The non-zero elements of $R_{h}$ are said to be homogeneous of degree $h$ and all the homogeneous elements are denoted by $h(R)$, i.e., $h(R) = \cup _{h\in G}R_{h}$. If $a\in R$, then $a$ can be written uniquely as $\sum_{h\in G}a_{h}$, where $a_{g}$ is called a homogeneous component of $a$ in $R_{h}$. Moreover, $R_{e}$ is a subring of $R$ and $1\in R_{e}$ (see ^[22]). Let $R = \oplus _{h\in G}R_{h}$ be a $G$-graded ring. An ideal $J$ of $R$ is said to be a graded ideal if $J = \sum_{h\in G}(J\cap R_{h}): = \sum_{h\in G}J_{h}$ (see ^[22]).

Let $R = \oplus _{h\in G}R_{h}$ be a $G$-graded ring. A left $R$-module $M$ is said to be a graded $R$-module (or $G$-graded $R$-module) if there exists a family of additive subetaoups $\{M_{h}\}_{h\in G}$ of $M$ such that $M = \oplus _{h\in G}M_{h}$ and $R_{g}M_{h}\subseteq M_{gh}$ for all $g, h\in G$. Also if an element of $M$ belongs to $\cup _{h\in G}M_{h} = h(M)$, then it is called a homogeneous. Note that $M_{h}$ is an $R_{e}$-module for every $h\in G$. Let $R = \oplus _{h\in G}R_{h}$ be a $G$-graded ring. A submodule $N$ of $M$ is said to be a graded submodule of $M$ if $N = \oplus _{h\in G}(N\cap M_{h}): = \oplus _{h\in G}N_{h}$. In this case, $N_{h}$ is called the $h$ -component of $N$. Moreover, $M/N$ becomes a $G$-graded $R$-module with $h$ -component $(M/N)_{h}: = (M_{h}+N)/N$ for $h\in G$ (see ^[22]).

2.   Results

Definition 2.1. Let $R$ be a $G$-graded ring, $M$ a graded $R$-module, $I = \oplus _{g\in G}I_{g}$ a graded ideal of $R$, $N = \oplus _{g\in G}N_{g}$ a graded submodule of $M$ and $g\in G.$

(i) We say that $N_{g}$ is a $g$-$2$-absorbing $I_{e}$-prime submodule of the $R_{e}$-module $M_{g}$, if $N_{g}\not = M_{g}$; and whenever $r_{e}, s_{e}\in R_{e}$ and $m_{g}\in M_{g}$ with $r_{e}s_{e}m_{g}\in N_{g}\backslash I_{e}N_{g}$, implies either $r_{e}s_{e}\in (N_{g}:_{R_{e}}M_{g})$ or $r_{e}m_{g}\in N_{g}$ or $s_{e}m_{g}\in N_{g}.$

(ii) We say that $N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M$, if $N\not = M$; and whenever $r_{h}, s_{\lambda }\in h(R)$ and $m_{\alpha }\in h(M)$ with $r_{h}s_{\lambda }m_{\alpha }\in N\backslash I_{e}N$, implies either $r_{h}s_{\lambda }\in (N:_{R}M)$ or $r_{h}m_{\alpha }\in N$ or $s_{\lambda }m_{\alpha }\in N.$

Proposition 2.2. Let $R$ be a $G$-graded ring, $M$ a graded $R$-module, $I = \oplus _{g\in G}I_{g}$ a graded ideal of $R$ and $N = \oplus _{g\in G}N_{g}$ a graded submodule of $M$. If $N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M$, then for any $g\in G$ with $N_{g}\neq M_{g}$, $N_{g}$ is a $g$-$2$-absorbing $I_{e}$-prime submodule of the $R_{e}$-module $M_{g}$.

Proof. Let $r_{e}, s_{e}\in R_{e}$ and $m_{g}\in M_{g}$ such that $r_{e}s_{e}m_{g}\in N_{g}\backslash I_{e}N_{g}$, so $r_{e}s_{e}m_{g}\in N\backslash I_{e}N$ and then either $r_{e}s_{e}\in (N:_{R}M)$ or $r_{e}m_{g}\in N$ or $s_{e}m_{g}\in N$ as $N$ is a graded $2$ -absorbing $I_{e}$-prime submodule of $M$. Since $M_{g}\subseteq M$ and $N_{g} = N\cap M_{g}$, we conclude that either $r_{e}s_{e}\in (N_{g}:_{R_{e}}M_{g})$ or $r_{e}m_{g}\in N_{g}$ or $s_{e}m_{g}\in N_{g}.$ Therefore, $N_{g}$ is a $g$-$2$-absorbing $I_{e}$-prime submodule of $M_{g}$.

Recall from ^[3] that a proper graded submodule $N$ of a graded $R$ -module $M$ is said to be a graded weakly 2-absorbing submodule of $M$ if whenever $r_{g}, s_{h}\in h(R)$ and $m_{\lambda }\in h(M)$ with $0\neq r_{g}s_{h}m_{\lambda}\in N$, then either $r_{g}m_{\lambda }\in N$ or $s_{h}m_{\lambda }\in N$ or $r_{g}s_{h}\in (N:_{R}M)$.

Remark 2.3. Let $R$ be a $G$-graded ring, $M$ a graded $R$-module and $I = \oplus _{g\in G}I_{g}$ a graded ideal of $R$. If $I = (0)$, then the notion of graded $2$-absorbing $I_{e}$-prime submodule is exactly the notion of graded weakly $2$-absorbing submodule.

Recall from ^[3] that a proper graded submodule $N$ of a graded $R$ -module $M$ is said to be a graded 2-absorbing submodule of $M$ if whenever $r_{g}, s_{h}\in h(R)$ and $m_{\lambda }\in h(M)$ with $r_{g}s_{h}m_{\lambda }\in N$, then either $r_{g}m_{\lambda }\in N$ or $s_{h}m_{\lambda }\in N$ or $r_{g}s_{h}\in (N:_{R}M)$.

It is easy to see that every graded $2$-absorbing submodule is a graded $2$-absorbing $I_{e}$-prime submodule. The following example shows that the converse is not true in general.

Example 2.4. Let $G = \mathbb{Z} _{2}$ and $R = \mathbb{Z}$ be a $G$-graded ring with $R_{0} = \mathbb{Z}$ and $R_{1} = \{0\}.$ Let $M = \mathbb{Z} _{12}$ be a graded $R$-module with $M_{0} = \mathbb{Z} _{12}$ and $M_{1} = \{\bar{0}\}.$ Now, consider the graded submodule $N = (\bar{0})$ of $M$, then $N$ is not a graded $2$-absorbing submodule of $M$ since $\bar{2}\cdot \bar{2}\cdot \bar{3}\in N$ and neither $\bar{2}\cdot \bar{3}\in N$ nor $\bar{2}\cdot \bar{2}\in (N:_{ \mathbb{Z} } \mathbb{Z} _{12}).$ However, for any graded ideal $I = \oplus _{g\in G}I_{g}$ of $R$, $N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M$.

Let $R$ be a $G$-graded ring, $M$ a graded $R$-module and $I = \oplus _{g\in G}I_{g}$ a graded ideal of $R$. Recall from ^[13] that a proper graded submodule $N$ of $M$ is said to be a graded $I_{e}$-prime submodule of $M$ if whenever $r_{h}\in h(R)$ and $m_{\lambda }\in h(M)$ with $r_{h}m_{\lambda }\in N-I_{e}N$, implies either $m_{\lambda }\in N$ or $r_{h}\in (N:_{R}M).$

It is easy to see that every graded $I_{e}$-prime submodule is a graded $2$-absorbing $I_{e}$-prime submodule. The following example shows that the converse is not true in general.

Example 2.5. Let $G = \mathbb{Z} _{2}$ and $R = \mathbb{Z}$ be a $G$-graded ring with $R_{0} = \mathbb{Z}$ and $R_{1} = \{0\}.$ Let $M = \mathbb{Z}$ be a graded $R$-module with $M_{0} = \mathbb{Z}$ and $M_{1} = \{0\}.$ Now, consider the graded ideal $I = 2 \mathbb{Z}$ of $R$ and the graded submodule $N = 4 \mathbb{Z}$ of $M$. Then $N$ is not a graded $I_{e}$-prime submodule of $M$ since $2\cdot 2\in 4 \mathbb{Z} \backslash 8 \mathbb{Z}$ and neither $2\in 4 \mathbb{Z}$ nor $2\in (4 \mathbb{Z} :_{ \mathbb{Z} } \mathbb{Z}).$ However, easy computations show that $N$ is a graded $2$-absorbing submodule of $M$ and then a graded $2$-absorbing $I_{e}$-prime$.$

Let $R$ be a $G$-graded ring, $M$ a graded $R$-module, $N = \oplus _{g\in G}N_{g}$ a graded submodule of $M$ and $g\in G.$ Recall from ^[3] that $N_{g}$ is said to be a $g$-2-absorbing submodule of the $R_{e}$-module $M_{g}$ if $N_{g}\neq M_{g}$; and whenever $r, s\in R_{e}$ and $m\in M_{g}$ with $rsm\in N_{g}$, then either $rs\in (N_{g}:_{R_{e}}M_{g})$ or $rm\in N_{g}$ or $sm\in N_{g}.$

Theorem 2.6. Let $R$ be a $G$-graded ring, $M$ a graded $R$-module, $I = \oplus _{g\in G}I_{g}$ a graded ideal of $R$ and $N = \oplus _{g\in G}N_{g}$ a graded $2$-absorbing $I_{e}$-prime submodule of $M$. Then for any $g\in G$ with $N_{g}\neq M_{g},$ either $N_{g}$ is $g$-$2$-absorbing submodule of the $R_{e}$-module $M_{g}$ or $(N_{g}:_{R_{e}}M_{g})^{2}N_{g}\subseteq I_{e}N_{g}$.

Proof. Let $g\in G$ with $N_{g}\neq M_{g}.$ Then $N_{g}$ is a $g$-$2$ -absorbing $I_{e}$-prime submodule of the $R_{e}$-module $M_{g}$ by Proposition 2.2. Suppose that $(N_{g}:_{R_{e}}M_{g})^{2}N_{g}\not\subseteq I_{e}N_{g}$. Now, let $r_{e}, s_{e}\in R_{e}$ and $m_{g}\in M_{g}$ such that $r_{e}s_{e}m_{g}\in N_{g}$. If $r_{e}s_{e}m_{g}\not\in I_{e}N_{g}$, then either $r_{e}s_{e}\in (N_{g}:_{R_{e}}M_{g})$ or $r_{e}m_{g}\in N_{g}$ or $s_{e}m_{g}\in N_{g}$ as $N_{g}$ is a $g$-$2$-absorbing $I_{e}$-prime submodule of the $R_{e}$-module $M_{g}.$ So now we can assume that $r_{e}s_{e}m_{g}\in I_{e}N_{g}$. First, suppose that $r_{e}s_{e}N_{g}\not\subseteq I_{e}N_{g}$, so there exists $n_{g}\in N_{g}$ such that $r_{e}s_{e}n_{g}\not\in I_{e}N_{g}$ and it follows that $r_{e}s_{e}(m_{g}+n_{g})\in N_{g}\backslash I_{e}N_{g}$. Then we get either $r_{e}s_{e}\in (N_{g}:_{R_{e}}M_{g})$ or $r_{e}(m_{g}+n_{g})\in N_{g}$ or $s_{e}(m_{g}+n_{g})\in N_{g}$ as $N_{g}$ is a $g$-$2$-absorbing $I_{e}$-prime submodule of $M_{g}$. Hence, either $r_{e}s_{e}\in (N_{g}:_{R_{e}}M_{g})$ or $r_{e}m_{g}\in N_{g}$ or $s_{e}m_{g}\in N_{g}$. Now, we may assume that $r_{e}s_{e}N_{g}\subseteq I_{e}N_{g}$. If $r_{e}(N_{g}:_{R_{e}}M_{g})m_{g} \not\subseteq I_{e}N_{g}$, then there exists $t_{e}\in (N_{g}:_{R_{e}}M_{g})$ such that $r_{e}t_{e}m_{g}\not\in I_{e}N_{g}$. This yields that $r_{e}(s_{e}+t_{e})m_{g}\in N_{g}\backslash I_{e}N_{g}$ and then we have either $r_{e}(s_{e}+t_{e})\in (N_{g}:_{R_{e}}M_{g})$ or $r_{e}m_{g}\in N_{g}$ or $(s_{e}+t_{e})m_{g}\in N_{g}$ as $N_{g}$ is a $g$-$2$-absorbing $I_{e}$ -prime submodule of the $R_{e}$-module $M_{g}$. Thus, either $r_{e}s_{e}\in (N_{g}:_{R_{e}}M_{g})$ or $r_{e}m_{g}\in N_{g}$ or $s_{e}m_{g}\in N_{g}$. We get the same result if $s_{e}(N_{g}:_{R_{e}}M_{g})m_{g}\not\subseteq I_{e}N_{g},$ so assume that $r_{e}(N_{g}:_{R_{e}}M_{g})m_{g}\subseteq I_{e}N_{g}$ and $s_{e}(N_{g}:_{R_{e}}M_{g})m_{g}\subseteq I_{e}N_{g}$. Now, since $(N_{g}:_{R_{e}}M_{g})^{2}N_{g}\not\subseteq I_{e}N_{g}$, there exist $r_{e}^{\prime }, s_{e}^{\prime }\in (N_{g}:_{R_{e}}M_{g})$ and $n_{g}^{\prime }\in N_{g}$ with $r_{e}^{\prime }s_{e}^{\prime }n_{g}^{\prime }\not\in I_{e}N_{g}$. If $r_{e}s_{e}^{\prime }n_{g}^{\prime }\not\in I_{e}N_{g}$, then $r_{e}(s_{e}+s_{e}^{\prime })(m_{g}+n_{g}^{\prime })\in N_{g}\backslash I_{e}N_{g}$ implies that either $r_{e}(s_{e}+s_{e}^{\prime })\in (N_{g}:_{R_{e}}M_{g})$ or $r_{e}(m_{g}+n_{g}^{\prime })\in N_{g}$ or $(s_{e}+s_{e}^{\prime })(m_{g}+n_{g}^{\prime })\in N_{g}$. Hence, either $r_{e}s_{e}\in (N_{g}:_{R_{e}}M_{g})$ or $r_{e}m_{g}\in N_{g}$ or $s_{e}m_{g}\in N_{g}$. Now, assume that $r_{e}s_{e}^{\prime }n_{g}^{\prime }\in I_{e}N_{g}$. Similarly, assume that $r_{e}^{\prime }s_{e}^{\prime }m_{g}\in I_{e}N_{g}$ and $r_{e}^{\prime }s_{e}n_{g}^{\prime }\in I_{e}N_{g}$. Then from $(r_{e}+r_{e}^{\prime })(s_{e}+s_{e}^{\prime })(m_{g}+n_{g}^{\prime })\in N_{g}\backslash I_{e}N_{g}$, we get $(r_{e}+r_{e}^{\prime })(s_{e}+s_{e}^{\prime })\in (N_{g}:_{R_{e}}M_{g})$ or $(r_{e}+r_{e}^{\prime })(m_{g}+n_{g}^{\prime })\in N_{g}$ or $(s_{e}+s_{e}^{\prime })(m_{g}+n_{g}^{\prime })\in N_{g}$ and it follows that either $r_{e}s_{e}\in (N_{g}:_{R_{e}}M_{g})$ or $r_{e}m_{g}\in N_{g}$ or $s_{e}m_{g}\in N_{g}$. Therefore, $N_{g}$ is a $g$-$2$-absorbing submodule of the $R_{e}$-module $M_{g}$.

Theorem 2.7. Let $R$ be a $G$-graded ring, $M$ a graded $R$-module, $I = \oplus _{g\in G}I_{g}$ a graded ideal of $R$, $N$ a graded $2$-absorbing $I_{e}$-prime submodule of $M$ and $K = \oplus _{\lambda\in G}K_{\lambda}$ a graded submodule of $M$. If $r_{g}, s_{h}\in h(R)$ and $\lambda \in G$ with $r_{g}s_{h}K_{\lambda }\subseteq N$ and $2r_{g}s_{h}K_{\lambda }\not\subseteq I_{e}N$, then either $r_{g}s_{h}\in (N:_{R}M)$ or $r_{g}K_{\lambda }\subseteq N$ or $s_{h}K_{\lambda }\subseteq N$.

Proof. Suppose that $r_{g}s_{h}\not\in (N:_{R}M)$. Now, let $k_{\lambda _{1}}\in K_{\lambda }$. If $r_{g}s_{h}k_{\lambda _{1}}\not\in I_{e}N$, then either $r_{g}k_{\lambda _{1}}\in N$ or $s_{h}k_{\lambda _{1}}\in N$ as $N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M$ and $r_{g}s_{h}\not\in (N:_{R}M)$, which yields that $k_{\lambda _{1}}\in (N:_{M}r_{g})\cup (N:_{M}s_{h})$. Now, we can assume that $r_{g}s_{h}k_{\lambda _{1}}\in I_{e}N$. Since $2r_{g}s_{h}K_{\lambda }\not\subseteq I_{e}N$, there exists $k_{\lambda _{2}}\in K_{\lambda }$ such that $2r_{g}s_{h}k_{\lambda _{2}}\not\in I_{e}N$ and then $r_{g}s_{h}k_{\lambda _{2}}\in N\backslash I_{e}N$. Hence, we get either $r_{g}k_{\lambda _{2}}\in N$ or $s_{h}k_{\lambda _{2}}\in N$ as $N$ is a graded $2$-absorbing $I_{e}$ -prime and $r_{g}s_{h}\not\in (N:_{R}M)$. Also, $r_{g}s_{h}(k_{\lambda _{1}}+k_{\lambda _{2}})\in N\backslash I_{e}N$ implies either $r_{g}(k_{\lambda _{1}}+k_{\lambda _{2}})\in N$ or $s_{h}(k_{\lambda _{1}}+k_{\lambda _{2}})\in N$. Hence, we consider three cases.

Case 1: $r_{g}k_{\lambda _{2}}\in N$ and $s_{h}k_{\lambda _{2}}\in N$. Then $r_{g}(k_{\lambda _{1}}+k_{\lambda _{2}})\in N$ or $s_{h}(k_{\lambda _{1}}+k_{\lambda _{2}})\in N$ implies either $r_{g}k_{\lambda _{1}}\in N$ or $s_{h}k_{\lambda _{1}}\in N$.

Case 2: $r_{g}k_{\lambda _{2}}\in N$ and $s_{h}k_{\lambda _{2}}\not\in N$. Assume that $r_{g}k_{\lambda _{1}}\not\in N$. Then $r_{g}(k_{\lambda _{1}}+k_{\lambda _{2}})\not\in N$ and so $s_{h}(k_{\lambda _{1}}+k_{\lambda _{2}})\in N$. Thus, $r_{g}(k_{\lambda _{1}}+2k_{\lambda _{2}})\not\in N$ and $s_{h}(k_{\lambda _{1}}+2k_{\lambda _{2}})\not\in N$. Now, we get $r_{g}s_{h}(k_{\lambda _{1}}+2k_{\lambda _{2}})\in I_{e}N$ as $N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M$ and $r_{g}s_{h}\not\in (N:_{R}M)$, \ and so $2r_{g}s_{h}k_{\lambda _{2}}\in I_{e}N$, a contradiction. Thus, $r_{g}k_{\lambda _{1}}\in N$.

Case 3: $r_{g}k_{\lambda _{2}}\not\in N$ and $s_{h}k_{\lambda _{2}}\in N$. Then the proof is similar to that of Case 2. Therefore, $K_{\lambda }\subseteq (N:_{M}r_{g})\cup (N:_{M}s_{h})$ and then either $r_{g}K_{\lambda }\subseteq N$ or $s_{h}K_{\lambda }\subseteq N$.

Theorem 2.8. Let $R$ be a $G$-graded ring, $M$ a graded $R$-module, $I = \oplus _{g\in G}I_{g}$ a graded ideal of $R$ and $N$ a graded $2$-absorbing $I_{e}$-prime submodule of $M.$ Let $J = \oplus _{h\in G}J_{h}$ be a graded ideal of $R$ and $K = \oplus _{\lambda\in G}K_{\lambda}$ a graded submodule of $M$. If $r_{g}\in h(R)$ and $h, \lambda \in G$ with $r_{g}J_{h}K_{\lambda }\subseteq N$ and $4r_{g}J_{h}K_{\lambda }\not\subseteq I_{e}N$, then either $r_{g}J_{h}\subseteq (N:_{R}M)$ or $r_{g}K_{\lambda }\subseteq N$ or $J_{h}K_{\lambda }\subseteq N$.

Proof. Suppose that $r_{g}J_{h}\not\subseteq (N:_{R}M)$ and $r_{g}K_{\lambda }\not\subseteq N.$ Now, since $r_{g}J_{h}\not\subseteq (N:_{R}M),$ there exists $j_{h_{1}}\in J_{h}$ such that $r_{g}j_{h_{1}}\not \in (N:_{R}M).$ Also, since $4r_{g}J_{h}K_{\lambda }\not\subseteq I_{e}N$, there exists $j_{h_{2}}\in J_{h}$ such that $4r_{g}j_{h_{2}}K_{\lambda }\not\subseteq I_{e}N$\ and then $2r_{g}j_{h_{2}}K_{\lambda }\not\subseteq I_{e}N.$ Now, let $j_{h}\in J_{h}$, if $2r_{g}j_{h}K_{\lambda }\not\subseteq I_{e}N$, then by Theorem 2.7, we get $j_{h}\in ((N:_{R}M):_{R}r_{g})\cup (N:_{R}K_{\lambda })$ as $N$ is a graded $2$ -absorbing $I_{e}$-prime submodule of $M.$ So we can assume that $2r_{g}j_{h}K_{\lambda }\subseteq I_{e}N.$ If $4r_{g}j_{h_{1}}K_{\lambda }\not\subseteq I_{e}N$, then $2r_{g}j_{h_{1}}K_{\lambda }\not\subseteq I_{e}N$. Thus $j_{h_{1}}K_{\lambda }\subseteq N$ by Theorem 2.7 as $N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M$. So, $2r_{g}(j_{h}+j_{h_{1}})K_{\lambda }\not\subseteq I_{e}N$ implies that $j_{h}+j_{h_{1}}\in ((N:_{R}M):_{R}r_{g})\cup (N:_{R}K_{\lambda }).$ Assume that $j_{h}+j_{h_{1}}\in ((N:_{R}M):_{R}r_{g})\backslash (N:_{R}K_{\lambda })$ then consider $2r_{g}(j_{h}+j_{h_{1}}+j_{h_{1}})K_{\lambda } = 2r_{g}j_{h}K_{\lambda }+4r_{g}j_{h_{1}}K_{\lambda }\not\subseteq I_{e}N$, which yields that $j_{h}+j_{h_{1}}+j_{h_{1}}\in ((N:_{R}M):_{R}r_{g})\cup (N:_{R}K_{\lambda }).$ But $j_{h_{1}}K_{\lambda }\subseteq N$ and $(j_{h}+j_{h_{1}})K_{\lambda }\not\subseteq N$ implies that $(j_{h}+j_{h_{1}}+j_{h_{1}})K_{\lambda }\not\subseteq N,$ also $r_{g}j_{h_{1}}\not\in (N:_{R}M)$ and $r_{g}(j_{h}+j_{h_{1}})\in (N:_{R}M)$ implies that $r_{g}(j_{h}+j_{h_{1}}+j_{h_{1}})\not\in (N:_{R}M),$ a contradiction. Hence, $j_{h}+j_{h_{1}}\in (N:_{R}K_{\lambda })$. Thus $j_{h}K_{\lambda }\subseteq N$ since $j_{h_{1}}K_{\lambda }\subseteq N.$ Similarly, if $r_{g}j_{h_{2}}\not\in (N:_{R}M),$ then we get the result in the same manner. So now we can assume that $r_{g}j_{h_{2}}\in (N:_{R}M)$ and $4r_{g}j_{h_{1}}K_{\lambda }\subseteq I_{e}N.$ Thus, $4r_{g}(j_{h_{1}}+j_{h_{2}})K_{\lambda }\not\subseteq I_{e}N$, then $2r_{g}(j_{h_{1}}+j_{h_{2}})K_{\lambda }\not\subseteq I_{e}N.$ It follows that $(j_{h_{1}}+j_{h_{2}})K_{\lambda }\subseteq N$ by Theorem 2.7 as $N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M$ and $r_{g}(j_{h_{1}}+j_{h_{2}})\not\in (N:_{R}M).$ So, $2r_{g}(j_{h}+(j_{h_{1}}+j_{h_{2}}))K_{\lambda }\not\subseteq I_{e}N$ implies that $j_{h}+(j_{h_{1}}+j_{h_{2}})\in ((N:_{R}M):_{R}r_{g})\cup (N:_{R}K_{\lambda }).$ Assume that $j_{h}+(j_{h_{1}}+j_{h_{2}})\in ((N:_{R}M):_{R}r_{g})\backslash (N:_{R}K_{\lambda })$ then consider $2r_{g}(j_{h}+2(j_{h_{1}}+j_{h_{2}}))K_{ \lambda } = 2r_{g}j_{h}K_{\lambda }+4r_{g}(j_{h_{1}}+j_{h_{2}})K_{\lambda }\not\subseteq I_{e}N$, which yields that $j_{h}+2(j_{h_{1}}+j_{h_{2}})\in ((N:_{R}M):_{R}r_{g})\cup (N:_{R}K_{\lambda }).$ But $(j_{h_{1}}+j_{h_{2}})K_{\lambda }\subseteq N$ and $(j_{h}+(j_{h_{1}}+j_{h_{2}}))K_{\lambda }\not\subseteq N$ implies that $(j_{h}+2(j_{h_{1}}+j_{h_{2}}))K_{\lambda }\not\subseteq N,$ also $r_{g}(j_{h_{1}}+j_{h_{2}})\not\in (N:_{R}M)$ and $r_{g}(j_{h}+(j_{h_{1}}+j_{h_{2}}))\in (N:_{R}M)$ implies that $r_{g}(j_{h}+2(j_{h_{1}}+j_{h_{2}}))\not\in (N:_{R}M),$ a contradiction. Hence, $j_{h}+(j_{h_{1}}+j_{h_{2}})\in (N:_{R}K_{\lambda })$. Thus $j_{h}K_{\lambda }\subseteq N$ since $(j_{h_{1}}+j_{h_{2}})K_{\lambda }\subseteq N.$ Therefore, $J_{h}\subseteq ((N:_{R}M):_{R}r_{g})\cup (N:_{R}K_{\lambda })$ and then $r_{g}J_{h}\subseteq (N:_{R}M)$ or $J_{h}K_{\lambda }\subseteq N$, but $r_{g}J_{h}\not\subseteq (N:_{R}M)$, so $J_{h}K_{\lambda }\subseteq N.$

Theorem 2.9. Let $R$ be a $G$-graded ring, $M$ a graded $R$-module, $I = \oplus _{g\in G}I_{g}$ be a graded ideal of $R$ and $N$ a proper graded submodule of $M.$ Then the following statements are equivalent:

(i) $N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M$.

(ii) $N/I_{e}N$ is a graded weakly $2$-absorbing submodule of $M/I_{e}N.$

Proof. $(i)\Rightarrow (ii)$ Suppose that $N$ is a graded 2-absorbing $I_{e}$-prime submodule of $M$. Now, let $r_{g}, s_{h}\in h(R)$ and $(m_{\lambda }+I_{e}N)\in h(M/I_{e}N)$ with $0_{M/I_{e}N}\not = (r_{g}s_{h}m_{ \lambda }+I_{e}N)\in N/I_{e}N$, this yields that $r_{g}s_{h}m_{\lambda }\in N\backslash I_{e}N.$ Hence, either $r_{g}m_{\lambda }\in N$ or $s_{h}m_{\lambda }\in N$ or $r_{g}s_{h}M\subseteq N$ as $N$ is a graded $2$ -absorbing $I_{e}$-prime submodule of $M$. Then either $(r_{g}m_{\lambda }+I_{e}N)\in N/I_{e}N$ or $(s_{h}m_{\lambda }+I_{e}N)\in N/I_{e}N$ or $r_{g}s_{h}(M/I_{e}N)\subseteq N/I_{e}N.$ Therefore, $N/I_{e}N$ is a graded weakly $2$-absorbing submodule of $M/I_{e}N.$

$(i)\Rightarrow (ii)$ Suppose that $N/I_{e}N$ is a graded weakly $2$ -absorbing submodule of $M/I_{e}N.$ Let $r_{g}, s_{h}\in h(R)$ and $m_{\lambda }\in h(M)$ such that $r_{g}s_{h}m_{\lambda }\in N\backslash I_{e}N.$ This follows that $\ 0_{M/I_{e}N}\not = (r_{g}s_{h}m_{\lambda }+I_{e}N) = r_{g}s_{h}(m_{\lambda }+I_{e}N)\in N/I_{e}N.$ Thus, either $r_{g}s_{h}\in (N/I_{e}N:_{R}M/I_{e}N)$ or $(r_{g}m_{\lambda }+I_{e}N)\in N/I_{e}N$ or $(s_{h}m_{\lambda }+I_{e}N)\in N/I_{e}N$\ and then either $r_{g}s_{h}\in (N:_{R}M)$ or $r_{g}m_{\lambda }\in N$ or $s_{h}m_{\lambda }\in N.$ Therefore, $N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M$.

Recall from ^[9] that a graded zero-divisor on a graded $R$-module $M$ is an element $r_{g}\in h(R)$ for which there exists $m_{h}\in h(M)$ such that $m_{h}\not = 0$ but $r_{g}m_{h} = 0$. The set of all graded zero-divisors on $M$ is denoted by $G$-$Zdv_{R}(M)$.

The following result studies the behavior of graded $2$-absorbing $I_{e}$ -prime submodules under localization.

Theorem 2.10. Let $R$ be a $G$-graded ring, $M$ a graded $R$-module, $S\subseteq h(R)$ be a multiplicatively closed subset of $R$ and $I = \oplus _{g\in G}I_{g}$ a graded ideal of $R$.

(i) If $N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M$ with $(N:_{R}M)\cap S = \emptyset$, then $S^{-1}N$ is a graded $2$-absorbing $I_{e}$ -prime submodule of $S^{-1}M$.

(ii) If $S^{-1}N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $S^{-1}M$ with $S\cap G$-$Zdv_{R}(M/N) = \emptyset$, then $N$ is a graded $2$ -absorbing $I_{e}$-prime submodule of $M$.

Proof. $(i)$ Since $(N:_{R}M)\cap S = \emptyset,$ $S^{-1}N$ is a proper graded submodule of $S^{-1}M.$\ Let $\frac{r_{g}}{s_{1}}, \frac{s_{h}}{s_{2}} \in h(S^{-1}R)$ and $\frac{m_{\lambda }}{s_{3}}\in h(S^{-1}M)$ such that $\frac{r_{g}}{s_{1}}\frac{s_{h}}{s_{2}}\frac{m_{\lambda }}{s_{3}}\in S^{-1}N\backslash I_{e}S^{-1}N.$ Then there exists $t\in S$ such that $tr_{g}s_{h}m_{\lambda }\in N\backslash I_{e}N$ which yields that either $tr_{g}m_{\lambda }\in N$ or $ts_{h}m_{\lambda }\in N$ or $r_{g}s_{h}\in (N:_{R}M)$ as $N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M.$ Hence, either $\frac{r_{g}m_{\lambda }}{s_{1}s_{3}} = \frac{tr_{g}m_{\lambda } }{ts_{1}s_{3}}\in S^{-1}N$ or $\frac{s_{h}m_{\lambda }}{s_{2}s_{3}} = \frac{ ts_{h}m_{\lambda }}{ts_{2}s_{3}}\in S^{-1}N$ or $\frac{r_{g}s_{h}}{ s_{1}s_{2}}\in S^{-1}(N:_{R}M) = (S^{-1}N:_{S^{-1}R}S^{-1}M).$ Therefore, $S^{-1}N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $S^{-1}M$.

$(ii)$ Let $r_{g}, s_{h}\in h(R)$ and $m_{\lambda }\in h(M)$ such that $r_{g}s_{h}m_{\lambda }\in N\backslash I_{e}N$. Then $\frac{r_{g}}{1}\frac{ s_{h}}{1}\frac{m_{\lambda }}{1}\in S^{-1}N\backslash I_{e}S^{-1}N$. Since $S^{-1}N$ is a graded $2$-absorbing $I_{e}$-prime submodule of $S^{-1}M$, either $\frac{r_{g}}{1}\frac{m_{\lambda }}{1}\in S^{-1}N$ or $\frac{s_{h}}{1} \frac{m_{\lambda }}{1}\in S^{-1}N$ or $\frac{r_{g}}{1}\frac{s_{h}}{1}\in (S^{-1}N:_{S^{-1}R}S^{-1}M)$. If $\frac{r_{g}m_{\lambda }}{1}\in S^{-1}N$, then there exists $t_{1}\in S$ such that $t_{1}r_{g}m_{\lambda }\in N$. This yields that $r_{g}m_{\lambda }\in N$ since $S\cap G$-$Zdv_{R}(M/N) = \emptyset$. Similarly, if $\frac{s_{h}m_{\lambda }}{1}\in S^{-1}N$, then there exists $t_{2}\in S$ such that $t_{2}s_{h}m_{\lambda }\in N$. This yields that $s_{h}m_{\lambda }\in N$ since $S\cap G$-$Zdv_{R}(M/N) = \emptyset$. Now, if $\frac{r_{g}s_{h}}{1}\in (S^{-1}N:_{S^{-1}R}S^{-1}M) = S^{-1}(N:_{R}M)$, then there exists $t_{3}\in S$ such that $t_{3}r_{g}s_{h}M\subseteq N$ and hence $r_{g}s_{h}\in (N:_{R}M)$ since $S\cap G$-$Zdv_{R}(M/N) = \emptyset$. Therefore, $N$ is a graded $2$ -absorbing $I_{e}$-prime submodule of $M$.

Proposition 2.11. Let $R$ be a $G$-graded ring, $M_{1}$ and $M_{2}$ be two graded $R$-modules, $I = \oplus _{g\in G}I_{g}$ a graded ideal of $R$ and $N_{1}$ and $N_{2}$ be two graded submodules of $M_{1}$ and $M_{2}$, respectively. Then:

(i) If $N_{1}$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M_{1},$ then $N_{1}\times M_{2}$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M_{1}\times M_{2}.$

(ii) If $N_{2}$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M_{2},$ then $M_{1}\times N_{2}$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M_{1}\times M_{2}.$

Proof. $(i)$ Suppose that $N_{1}$ is a graded $I_{e}$-prime submodule of $M_{1}.$ Now, let $r_{g}, s_{h}\in h(R)$ and $(m_{_{\lambda 1}}, m_{_{\lambda 2}})\in h(M_{1}\times M_{2})$ such that $r_{g}s_{h}(m_{_{\lambda 1}}, m_{_{\lambda 2}}) = (r_{g}s_{h}m_{_{\lambda 1}}, r_{g}s_{h}m_{\lambda 2})\in (N_{1}\times M_{2})\backslash I_{e}(N_{1}\times M_{2}) = (N_{1}\backslash I_{e}N_{1})\times (M_{2}\backslash I_{e}M_{2}),$ which follows that $r_{g}s_{h}m_{_{\lambda 1}}\in N_{1}\backslash I_{e}N_{1}.$ Hence, either $r_{g}m_{_{\lambda 1}}\in N_{1}$ or $s_{h}m_{\lambda 1}\in N_{1}$ or $r_{g}s_{h}M_{1}\subseteq N_{1}$ and then either $r_{g}(m_{\lambda 1}, m_{\lambda 2})\in N_{1}\times M_{2}$ or $s_{h}(m_{\lambda 1}, m_{\lambda 2})\in N_{1}\times M_{2}$ or $r_{g}s_{h}(M_{1}\times M_{2})\subseteq N_{1}\times M_{2}.$ Therefore, $N_{1}\times M_{2}$ is a graded $2$-absorbing $I_{e}$-prime submodule of $M_{1}\times M_{2}.$

$(ii)$ The proof is similar to that in part $(i).$

Acknowledgments

We would like to thanks the honorable reviewers for their valuable comments and suggestions, which are really helpful to enrich the quality of our paper. Furthermore, we are grateful to the journal authority for their proper judgements and kind consideration

Conflict of interest

The authors declare that they have no any competing interests