Graded Weakly 2-Absorbing Ideals over Non-Commutative Graded Rings

Abstract

Let G be a group and R be a G-graded ring. In this paper, we present and examine the concept of graded weakly 2-absorbing ideals as in generality of graded weakly prime ideals in a graded ring which is not commutative, and demonstrates that the symmetry is obtained as a lot of the outcomes in commutative graded rings remain in graded rings that are not commutative.

1. Introduction

During the whole of this article, the rings are not certainly expected to have unity except pointed out alternatively. Likewise, an ideal in a ring means a two-sided ideal. Let G be a group with identity e and R be a ring. Then R is called graded ring which denoted by ‘GR-ring’ if ${\displaystyle R=\underset{g\in G}{⨁}{R}_g}$ where $R_g{R}_h\subseteq R_{gh}$ for $g,h\in G$. The additive subgroup stood for $R_g$ where $g\in G$. We call the homogeneous of degree g for the components of $R_g$. If $a\in R$, then a can be represented by ${\displaystyle \sum _{g\in G}{a}_g}$, with $a_g$ being the element of a in $R_g$. In fact the additive subgroup $R_e$ is a sub-ring of R, if R has a unity 1, then $1\in R_e$. Let ${\displaystyle \bigcup _{g\in G}{R}_g}$ be the collection of all homogeneous elements of R which is denoted by $h\left(R\right)$. Assume P is an ideal of a graded ring R. If ${\displaystyle P=\underset{g\in G}{⨁}(P\cap R_g)}$, so, P is announced for a graded ideal, and denoted by ‘GR-I’, i.e., for $a\in P$, ${\displaystyle a=\sum _{g\in G}{a}_g}$ where $a_g\in P$ and $g\in G$. It is not necessary for every ‘GR-I’ to be a GR-ring ([1], Example 1.1). For more details and terminology, see [2,3].

The following abbreviations are used towards the end of this paper: ‘CGR-ring’ stand for commutative graded rings, ‘NCGR-ring’ for non-commutative graded rings, ‘GR-P’ for graded prime, ‘GR-PI’ for graded prime ideals, ‘PGR-PI’ for proper graded prime ideals, ‘PGR-I’ for proper graded ideals, ‘GR-WPI’ for graded weakly prime ideals, ‘GR-2-AI’ for graded 2-Absorbing ideals, ‘GR-W-2-AI’ for a graded weakly 2-Absorbing ideals, and ‘GR-CW-2-AI’ for a graded completely weakly 2-Absorbing ideals.

For ‘CGR-ring’, ‘GR-2-AI’, generalized from ‘GR-PI’, which were presented as well as examined within [4]. Remember from [5] that a ‘PGR-I’ P of a ‘CGR-ring’ R is estimated to be a ‘GR-WPI’ of R if $x,y\in h\left(R\right)$ and $0\ne xy\in P$, then either $x\in P$ or $y\in P$. Also from [4] a ‘PGR-I’ P of a ‘CGR-ring’ R is announced for a ‘GR-2-AI’ of R, where, $x,y,z\in h\left(R\right)$ along with $xyz\in P$, therefore, either $xy\in P$, $xz\in P$ or $yz\in P$. The idea of a ‘GR-W-2-AI’ of a ‘CGR-ring’ R was presented in [4]. A ‘PGR-I’ P of a ‘CGR-ring’ R is called a ‘GR-W-2-AI’ of R if given $x,y,z\in h\left(R\right)$ and $0\ne xyz\in P$, so one of $xy$, $xz$ or $yz$ be in P.

The ‘GR-PI’ over ‘NCGR-rings’ have been put in place and examined by Abu-Dawwas, Bataineh, and Al-Muanger in [6]. A ‘PGR-I’ P of R is expressed to be ‘GR-P’ for both of I and J were ‘GR-I’ of R where, $IJ\subseteq P$, therefore $I\subseteq P$ or $J\subseteq P$. As a summarization of ‘GR-PI’ over ‘NCGR-ring’, the concept of ‘GR-2-AI’ over ‘NCGR-ring’ has been reported and investigated by Abu-Dawwas, Shashan and Dagher in [7]. A ‘PGR-I’ P of R is said to be ‘GR-2-AI’ where $x,y,z\in h\left(R\right)$ so that $xRyRz\subseteq P$, then $xy\in P$, $yz\in P$ or $xz\in P$. Recently, ‘GR-WPI’ over ‘NCGR-rings’ have been brought up and served by Alshehry and Abu-Dawwas in [1]. A ‘PGR-I’ P of R is said to be ‘GR-WP’ if once I and J are ‘GR-I’ of R such that $0\ne IJ\subseteq P$, then $I\subseteq P$ or $J\subseteq P$.

Within this article, we are following [8] to introduce and investigate the concept of ‘GR-W-2-AI’ as a generalization of ‘GR-WPI’ in a ‘GR-ring’ which is non-commutative, and demonstrates that the symmetry is obtained as a lot of the outcomes in ‘CGR-ring’ still remain in ‘NCGR-ring’.

2. Graded Weakly 2-Absorbing Ideals

This section consists of an examination and studies of ‘GR-W-2-AI’. During the whole of this section, we are dealing with a ring R, that is an ‘NCGR-ring’, having unity except pointed out alternatively.

Definition 1.

Let R be a ‘GR-ring’. Assume that P is a ‘PGR-I’ of R. Then we call P being a ‘GR-W-2-AI’ when $0\ne xRyRz\subseteq P$ gives $xy\in P$, $yz\in P$ or $xz\in P$ for each $x,y,z\in h\left(R\right)$. If $0\ne xyz\in P$ implies $xy\in P$, $yz\in P$ or $xz\in P$ for all $x,y,z\in h\left(R\right)$, we call P to be ‘GR-CW-2-AI’.

Apparently, when R is a ‘CGR-rings’ having unity, then the concepts of ‘GR-W-2-A’ and ‘GR-CW-2-AI’ coincide. The following example demonstrates that this will not be the case for ‘NCGR-ring’.

Example 1.

Consider $R=M_2\left(\mathbb{Z}\right)$ (the ring of all $2\times 2$ matrices with integer entries) and $G={\mathbb{Z}}_4$. Then R is graded by $R_0=\left(\begin{array}{cc}\mathbb{Z}& 0\\ 0& \mathbb{Z}\end{array}\right)$, $R_2=\left(\begin{array}{cc}0& \mathbb{Z}\\ \mathbb{Z}& 0\end{array}\right)$ and $R_1=R_3=0$.

Deal with ‘GR-I’ $P=M_2\left(2\mathbb{Z}\right)$ of R. P is Clearly a ‘GR-PI’ of R and so a ‘GR-W-2-AI’ of R. On the other side, P is not a ‘GR-CW-2-AI’ of R since $n,m\in \mathbb{Z}$, $A=\left(\begin{array}{cc}2n+1& 0\\ 0& 2m\end{array}\right)$, $B=\left(\begin{array}{cc}0& 2n+1\\ 2n+1& 0\end{array}\right)$ and $C=\left(\begin{array}{cc}2n+1& 0\\ 0& 4m\end{array}\right)\in h\left(R\right)$ where $0\ne ABC\in P$, for each of $AB$, $AC$ and $BC\notin P$.

Undoubtedly, every ‘GR-2-AI’ of a ‘GR-ring’ is a ‘GR-W-2-AI’. In any ‘GR-ring’, $P=\left\{0\right\}$ is ‘GR-W-2-AI’.

Individually, it is not necessary for $P=\left\{0\right\}$ to be ‘GR-2-AI’, check the next example.

Example 2.

Suppose that $R=M_2\left({\mathbb{Z}}_8\right)$ along with $G={\mathbb{Z}}_4$. Hence R will be ‘GR-ring’ by $R_0=\left(\begin{array}{cc}{\mathbb{Z}}_8& 0\\ 0& {\mathbb{Z}}_8\end{array}\right)$, $R_2=\left(\begin{array}{cc}0& {\mathbb{Z}}_8\\ {\mathbb{Z}}_8& 0\end{array}\right)$ and $R_1=R_3=0$. Undeniably, $P=\left\{\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\right\}$ is not a ‘GR-2-AI’ of R since $A=\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)\in h\left(R\right)$ with $ARARA\subseteq P$ but $A.A\notin P$.

Lemma 1.

For a ‘GR-ring’ R. Assume that P is a ‘GR-WPI’ of R.

1.

If for both I and J are graded right (left) ideals of R where, $0\ne IJ\subseteq P$. Then it is either $I\subseteq P$ or $J\subseteq P$.

2.

If $0\ne xRyRz\subseteq P$ such that $x,y,z\in h\left(R\right)$, therefore each of x, y or $z\in P$.

Proof. 

• Assume that both I and J are graded right (left) ideals of R in order that $0\ne IJ\subseteq P$. Let $\left(I\right)$ and $\left(J\right)$ be the ‘GR-I’ generated by I and J respectively. Then $0\ne \left(I\right)\left(J\right)\subseteq P$, whence $I\subseteq \left(I\right)\subseteq P$ or $J\subseteq \left(J\right)\subseteq P$.
• Suppose that $x,y,z\in h\left(R\right)$ where $0\ne xRyRz\subseteq P$. That being $0\ne \left(Rx\right)RyRz\subseteq P$ which it comes from (1) that $x\in Rx\subseteq P$ or $0\ne RyRz\subseteq P$. By reiterating this, the result follows.

□

Proposition 1.

In the ‘GR-ring’ R. P is a ‘GR-W-2-AI’ of R, if it is a ‘GR-WPI’ of R.

Proof. 

Let $x,y,z\in h\left(R\right)$ where $0\ne xRyRz\subseteq P$. By Lemma 1, $x\in P$ or $y\in P$ or $z\in P$. Accordingly, $xy\in P$ or $yz\in P$ or $xz\in P$, and the result holds. □

Proposition 2.

If P and K are two distinct ‘GR-WPI’ of a ‘GR-ring’ R, then $P\bigcap K$ is a ‘GR-W-2-AI’ of R.

Proof. 

Assume that $P\bigcap K=\left\{0\right\}$, it seems that $P\bigcap K$ is a ‘GR-W-2-AI’ of R. Let $x_1,x_2,x_3\in h\left(R\right)$ where $0\ne x_{1}R{x}_{2}R{x}_3\subseteq P\bigcap K$. Then $0\ne x_{1}R{x}_{2}R{x}_3\subseteq P$ and $0\ne x_{1}R{x}_{2}R{x}_3\subseteq K$. By Lemma 1 we have $x_i\in P$ and $x_j\in K$ for some i and j, then $x_i{x}_j\in P\bigcap K$. As a result, $P\bigcap K$ is a ‘GR-W-2-AI’ of R. □

Consider the two ‘GR-rings’ R and T. For all $g\in G$, $R\times T$ is a graded by ${(R\times T)}_g=R_g\times T_g$. $P\times K$ is a ‘GR-I’ of $R\times T$ if and only if P is a ‘GR-I’ of R and K is a ‘GR-I’ of T. The following example reveals that one can find ‘GR-W-2-AI’ which is not ‘GR-WPI’. Unfortunately, these rings that are used are commutative, indeed, we could not find such an example consisting of a non-commutative ring.

Example 3.

Let $R={\mathbb{Z}}_2\left[i\right]$, $T={\mathbb{Z}}_4\left[i\right]$, and $G={\mathbb{Z}}_2$. Then $R_0={\mathbb{Z}}_2$ and $R_1=i{\mathbb{Z}}_2$ are the grades at that point of R. As well, T is a graded by $T_0={\mathbb{Z}}_4$ and $T_1=i{\mathbb{Z}}_4$. In order that, $R\times T$ is a graded by ${(R\times T)}_j=R_j\times T_j$ for all $j=0,1$. Therefore, $\left\{0\right\}$ is a ‘GR-I’ of R and $2T$ is a ‘GR-I’ of T as $2\in h\left(T\right)$, so $P=\left\{0\right\}\times 2T$ is a ‘GR-I’ of $R\times T$. Since $x=(0,1),y=(1,2)\in h(R\times T)$ with $(0,0)\ne xy=(0,2)\in P$, $x\notin P$ and $y\notin P$. Then P is not a ‘GR-WPI’ of $R\times T$. Individually, P is ‘GR-2-AI’ and hence a ‘GR-W-2-AI’ of $R\times T$.

Theorem 1.

Let R be a ‘GR-ring’. Suppose that P is a ‘PGR-I’ of R. Assume that for graded left ideals $E,F$ and G of R such that $0\ne EFG\subseteq P$, since $EG\subseteq P$, $FG\subseteq P$ or $EF\subseteq P$. Then P is a ‘GR-W-2-AI’ of R.

Proof. 

Suppose that $x,y,z\in h\left(R\right)$ where $0\ne xRyRz\subseteq P$. therefore, $RxRyRzR\subseteq P$, and as a consequence, since R has a unity, $0\ne xRyRz=1.xR.1.yR.1.z.1\subseteq \left(RxR\right)\left(RyR\right)\left(RzR\right)\subseteq P$. By assumption, we have $xy\in \left(RxR\right)\left(RyR\right)\subseteq P$ or $yz\in \left(RyR\right)\left(RzR\right)\subseteq P$ or $xz\in \left(RxR\right)\left(RzR\right)\subseteq P$. Accordingly, P will be ‘GR-W-2-AI’. □

Theorem 2.

Theorem 1 still true if graded left ideals are replaced by graded right ideals.

Let R be a ‘GR-ring’ and K is a ‘GR-I’ of R, then $R/K$ is a graded by ${(R/K)}_g=(R_g+K)/K$ for any $g\in G$. For P as an ideal of R and K is a ‘GR-I’ of R such that $K\subseteq P$, then P is a ‘GR-I’ of R if and only if $P/K$ is a ‘GR-I’ of $R/K$.

Proposition 3.

For a graded ring R. Assume that P is a ‘GR-W-2-AI’ of R. Let $K\subseteq P$, if K is a ‘GR-I’ of R, then $P/K$ is a ‘GR-W-2-AI’ of $R/K$.

Proof. 

Let $x+K,y+K,z+K\in h(R/K)$ with $0+K\ne (x+K)(R/K)(y+K)(R/K)(z+K)\subseteq P/K$. Hence $x,y,z\in h\left(R\right)$ with $0\ne xRyRz\subseteq P$. Because P is ‘GR-W-2-AI’, for $xy\in P$, $yz\in P$ or $xz\in P$, therefore, $(x+K)(y+K)\in P/K$ or $(y+K)(z+K)\in P/K$ or $(x+K)(z+K)\in P/K$. So, $P/K$ is a ‘GR-W-2-AI’ of $R/K$. □

Proposition 4.

For a graded ring R. Let $K\subseteq P$ is a ‘PGR-I’ of a ‘GR-ring’ R. Then P is a ‘GR-W-2-AI’ of R, if K is a ‘GR-W-2-AI’ of R and $P/K$ is a ‘GR-W-2-AI’ of $R/K$.

Proof. 

Suppose that $x,y,z\in h\left(R\right)$ with $0\ne xRyRz\subseteq P$. Therefore, $x+K,y+K,z+K\in h(R/K)$ such that $(x+K)(R/K)(y+K)(R/K)(z+K)\subseteq P/K$. If $0\ne xRyRz\subseteq K$, then $xy\in K\subseteq P$ or $yz\in K\subseteq P$ or $xz\in K\subseteq P$ since K is a ‘GR-W-2-AI’ of R. If $xRyRz⊈K$, then $0+K\ne (x+K)(R/K)(y+K)(R/K)(z+K)\subseteq P/K$. Since $P/K$ is a ‘GR-W-2-AI’ of $R/K$, $(x+K)(y+K)\in P/K$ or $(y+K)(z+K)\in P/K$ or $(x+K)(z+K)\in P/K$, that yields that $xy\in P$, $yz\in P$ or $xz\in P$. Therefore, P is a ‘GR-W-2-AI’ of R. □

For two ‘GR-rings’ S and T. We call $f:S\to T$ to be graded homomorphism f is ring homomorphism and $f\left(S_g\right)\subseteq T_g$ for every $g\in G$.

Proposition 5.

Let S and T be two ‘GR-rings’ and $f:S\to T$ be graded homomorphism. Then $Ker\left(f\right)$ is a ‘GR-I’ of S.

Proof. 

Apparently, $Ker\left(f\right)$ is an ideal of S. Assume that $x\in Ker\left(f\right)$. Hence $x\in S$ such that $f\left(x\right)=0$. Now, ${\displaystyle x=\sum _{g\in G}{x}_g}$, with $x_g\in S_g$ for all $g\in G$, which lead to $f\left(x_g\right)\in f\left(S_g\right)\subseteq T_g$ for all $g\in G$. As a result, for $g\in G$, $f\left(x_g\right)\in h\left(T\right)$ with ${\displaystyle 0=f\left(x\right)=f\left({\displaystyle \sum _{g\in G}{x}_g}\right)=\sum _{g\in G}f\left(x_g\right)}$, which yields that $f\left(x_g\right)=0$ for all $g\in G$ along with $\left\{0\right\}$ is a ‘GR-I’. Therefore, $x_g\in Ker\left(f\right)$ for any $g\in G$, and then $Ker\left(f\right)$ is a ‘GR-I’ of S. □

Theorem 3.

For the two ‘GR-rings’ S and T and $f:S\to T$ be surjective graded homomorphism.

1.

$f\left(P\right)$ will be a ‘GR-W-2-AI’ of T, if P is a ‘GR-W-2-AI’ of S and $Ker\left(f\right)\subseteq P$.

2.

$f^{-1}\left(I\right)$ will be a ‘GR-W-2-AI’ of S, if I is a ‘GR-W-2-AI’ of T and $Ker\left(f\right)$ is a ‘GR-W-2-AI’ of R.

Proof. 

• Let $f\left(P\right)$ be a ‘GR-I’ of T. Because P is a ‘GR-W-2-AI’ of R and $Ker\left(f\right)\subseteq P$, Proposition 3 shows that $P/Ker\left(f\right)$ is a ‘GR-W-2-AI’ of $S/Ker\left(f\right)$. The result holds Since $S/Ker\left(f\right)$ is isomorphic to T.
• Assume that $f^{-1}\left(I\right)$ is a ‘GR-I’ of S. Let $K=f^{-1}\left(I\right)$. Then $Ker\left(f\right)\subseteq K$. We observe that $K/Ker\left(f\right)$ is a ‘GW-2-AI’ of $S/Ker\left(f\right)$, since $S/Ker\left(f\right)$ is isomorphic to T. Because $Ker\left(f\right)$ is a ‘GR-W-2-AI’ of S and $K/Ker\left(f\right)$ is a ‘GR-W-2-AI’ of $S/Ker\left(f\right)$, Proposition 4 states that $K=f^{-1}\left(I\right)$ is a ‘GR-W-2-AI’ of S.

□

Motivated by Theorem 1, we observe the next question.

Question 1.

IfPis a ‘GR-W-2-AI’ ofRthat is not a ‘GR-2-AI’ and$0\ne EFK\subseteq P$for some ‘GR-I’E,FandKofR. Does it indicate that$EF\subseteq P$or$EK\subseteq P$or$FK\subseteq P$?

We will give a partial answer through the coming discussions. Motivated by ([4], Definition 3.3), we introduce the following:

Definition 2.

Assume that R is a ‘GR-ring’, $g\in G$ and P is a ‘GR-I’ of R with $P_g\ne R_g$.

1.

If for each $x,y,z\in R_g$ where $x{R}_{e}y{R}_{e}z\subseteq P$, then P is said to be a ‘GR-2-AI’ of R, therefore, $xy\in P$, $yz\in P$ or $xz\in P$.

2.

If for each $x,y,z\in R_g$ where $0\ne x{R}_{e}y{R}_{e}z\subseteq P$, then P is said to be a ‘GR-W-2-AI’ of R, therefore, $xy\in P$, $yz\in P$ or $xz\in P$.

3.

For $x,y,z\in R_g$, let P is a ‘GR-W-2-AI’ of R and. We denote ‘GR-3-Z’ for $(x,y,z)$ which is the graded-triple-zero of P if $x{R}_{e}y{R}_{e}z=0$, such that $xy\notin P$, $yz\notin P$ and $xz\notin P$.

Note that if P is ‘GR-W-2-AI’ which is not ‘GR-2-AI’, then P involves a ‘GR-3-Z’ $(x,y,z)$ for $x,y,z\in R_g$.

Proposition 6.

Assume that $x{R}_{e}y,K_g\subseteq P$ for any $x,y\in R_g$ and some graded left ideal K of R, and that P is a ‘GR-W-2-AI’ of R. Let $(x,y,z)$ is not a ‘GR-3-z’ of P for every $z\in K_g$. If $xy\notin P$, then $x{K}_g\subseteq P$ or $y{K}_g\subseteq P$.

Proof. 

Consider that $x{K}_g⊈P$ along with $y{K}_g⊈P$. Then there exist $r,s\in K_g$ such that $xr\notin P$ and $ys\notin P$. Since $x{R}_{e}y{R}_{e}r\subseteq x{R}_{e}y{K}_g\subseteq P$ and since $(x,y,r)$ is not a GR-3-Z of P and $xy\notin P$, $xr\notin P$, we obtain that $yr\in P$. Also, since $x{R}_{e}y{R}_{e}s\subseteq x{R}_{e}y{K}_g\subseteq P$ and since $(x,y,s)$ is not a GR-3-Z of P and $xy\notin P$, $ys\notin P$, we obtain that $xs\in P$. Now, since $x{R}_{e}y{R}_e(r+s)\subseteq x{R}_{e}y{K}_g\subseteq P$ and since $(x,y,r+s)$ is not a GR-3-Z of P and $xy\notin P$, we get $x(r+s)\in P$ or $y(r+s)\in P$. If $x(r+s)\in P$, then since $xs\in P$, $xr\in P$, a contradiction. If $y(r+s)\in P$, then since $yr\in P$, $ys\in P$, a contradiction. Hence, $x{K}_g\subseteq P$ or $y{K}_g\subseteq P$. □

Definition 3.

Let R be a ‘GR-ring’ $g\in G$ and P be a ‘GR-W-2-AI’ of R. Assume that $A_g{B}_g{K}_g\subseteq P$ for some ‘GR-I’ $A,B$ and K of R. If $(x,y,z)$ is not a ‘GR-3-Z’ of P for every $x\in A_g,y\in B_g$ and $z\in K_g$. We can state P as being a free ‘GR-3-Z’ respecting $ABK$. The next proposition is clear.

Proposition 7.

Let P is a ‘GR-W-2-AI’ of R. Presume that $A_g{B}_g{K}_g\subseteq P$ and P to be a free ‘GR-3-Z’ in respect to $ABK$, for some ‘GR-I’ $A,B$ and K of R. If $x\in A_g,y\in B_g$ and $z\in K_g$, then $xy\in P$, $xz\in P$ or $yz\in P$.

Theorem 4.

Infer that P is a ‘GR-W-2-AI’ of R. Lets take $0\ne A_g{B}_g{K}_g\subseteq P$ and P to be a free ‘GR-3-Z’ in respect to $ABK$, for some ‘GR-I’ $A,B$ and K of R. Then $A_g{K}_g\subseteq P$, $B_g{K}_g\subseteq P$ or $A_g{B}_g\subseteq P$.

Proof. 

Suppose that $A_g{K}_g⊈P$, $B_g{K}_g⊈P$ and $A_g{B}_g⊈P$. There exist $x\in A_g$ and $y\in B_g$ where $x{K}_g⊈P$ and $y{K}_g⊈P$. Now, $x{R}_{e}y{K}_g\subseteq A_g{B}_g{K}_g\subseteq P$. Since $x{K}_g⊈P$ and $y{K}_g⊈P$, it comes from Proposition 6 that $xy\in P$. Because $A_g{B}_g⊈P$, there are $a\in A_g$ and $b\in B_g$ where $ab\notin P$. Since $a{R}_{e}b{K}_g\subseteq A_g{B}_g{K}_g\subseteq P$ and $ab\notin P$, it comes from Proposition 6 that $a{K}_g\subseteq P$ or $b{K}_g\subseteq P$.

Case (1): $a{K}_g\subseteq P$ and $b{K}_g⊈P$. Since $x{R}_{e}b{K}_g\subseteq A_g{B}_g{K}_g\subseteq P$ and $x{K}_g⊈P$ and $b{K}_g⊈P$, it follows from Proposition 6 that $xb\in P$. Since $a{K}_g\subseteq P$ and $x{K}_g⊈P$, we obtain that $(x+a)K_g⊈P$. On the other hand, since $(x+a)R_{e}b{K}_g\subseteq P$ and neither $(x+a)K_g\subseteq P$ nor $b{K}_g\subseteq P$, we have that $(x+a)b\in P$ by Proposition 6, and hence $ab\in P$, which is not true.

Case (2): $b{K}_g\subseteq P$ and $a{K}_g⊈P$. Using an analogous assertion to case (1), we will have an inconsistency.

Case (3): $a{K}_g\subseteq P$ and $b{K}_g\subseteq P$. Since $b{K}_g\subseteq P$ and $y{K}_g⊈P$, $(y+b)K_g⊈P$. But $x{R}_e(y+b)K_g\subseteq P$ and neither $x{K}_g\subseteq P$ nor $(y+b)K_g\subseteq P$, and hence $x(y+b)\subseteq P$ by Proposition 6. Since $xy\in P$ and $(xy+xb)\in P$, we have that $xb\in P$. Since $(x+a)R_{e}y{K}_g\subseteq P$ and neither $y{K}_g\subseteq P$ nor $(x+a)K_g\subseteq P$, we conclude that $(x+a)y\in P$ by Proposition 6, and hence $ax\in P$. Since $(x+a)R_e(y+b)K_g\subseteq P$ and neither $(x+a)K_g\subseteq P$ nor $(y+b)K_g\subseteq P$, we have $(x+a)(y+b)\in P$ by Proposition 6. But $xy,xb,ay\in P$, so $ab\in P$, a contradiction. Consequently, $A_g{K}_g\subseteq P$ or $B_g{K}_g\subseteq P$ or $A_g{B}_g\subseteq P$. □

Lemma 2.

For a ‘GR-ring’ R. Assume that P is a ‘GR-W-2-AI’ and $(x,y,z)$ is a ‘GR-3-Z’ of P for some $x,y,z\in R_g$. Then

1.

$x{R}_{e}y{P}_g=\left\{0\right\}$,

2.

$P_{g}y{R}_{e}z=\left\{0\right\}$,

3.

$x{P}_{g}z=\left\{0\right\}$,

4.

$P_g^{2}z=\left\{0\right\}$,

5.

$x{P}_g^2=\left\{0\right\}$,

6.

$P_{g}y{P}_g=\left\{0\right\}$.

Proof. 

• Assume that $x{R}_{e}y{P}_g\ne \left\{0\right\}$. Then there exist $r\in R_e$ and $p\in P_g$ such that $0\ne xryp$. Now, $xry(p+z)=xryp+xryz=xryp\ne 0$. Hence, $0\ne x{R}_{e}y{R}_e(p+z)\subseteq P$. We have $x(p+z)\in P$ or $y(p+z)\in P$, since P is ‘GR-W-2-AI’. Thus $xz\in P$ or $yz\in P$ is a contradiction.
• Suppose that $P_{g}y{R}_{e}z\ne \left\{0\right\}$. Then there exist $r\in R_e$ and $p\in P_g$ such that $0\ne pyrz$. Now, $(x+p)yrz=xyrz+pyrz=pyrz\ne 0$. Hence, $0\ne (x+p)R_{e}y{R}_{e}z\subseteq P$. If P is ‘GR-W-2-AI’ We have $(x+p)y\in P$ or $(x+p)z\in P$. As a result, $xy\in P$ or $xz\in P$ is a contradiction.
• Suppose that $x{P}_{g}z\ne \left\{0\right\}$. However, there exists $p\in P_g$ for which $0\ne xpz$. Now, $x(y+p)z=xyz+xpz=xpz\ne 0$. Hence, $0\ne x{R}_e(y+p)R_{e}z\subseteq P$. We have $x(y+p)\in P$ or $(y+p)z\in P$. Because P is ‘GR-W-2-AI’. Hence, $xy\in P$ or $yz\in P$ is a contradiction.
• Suppose that $P_g^{2}z\ne \left\{0\right\}$. Moreover, there exist $p,q\in P_g$ in which $0\ne pqz$. Now, $(x+p)(y+q)z=xyz+xqz+pyz+pqz=pqz\ne 0$ by (2) and (3). Hence, $0\ne (x+p)R_e(y+q)R_{e}z\subseteq P$. We have $(x+p)z\in P$ or $(y+q)z\in P$ or $(x+p)(y+q)\in P$. Because P is ‘GR-W-2-AI’. Hence, $xz\in P$ or $yz\in P$ or $xy\in P$ is a contradiction.
• Suppose that $x{P}_g^2\ne \left\{0\right\}$. Moreover, there exist $p,q\in P_g$, where, $0\ne xpq$. Now, by (1) and (3), $x(y+p)(z+q)=xyz+xyq+xpz+xpq=xpq\ne 0$. As a result, $0\ne x{R}_e(y+p)R_e(z+q)\subseteq P$. We have $x(y+p)\in P$, $x(z+q)\in P$ or $(y+p)(z+q)\in P$. Because, P is ‘GR-W-2-AI’. Hence, $xy\in P$, $xz\in P$ or $yz\in P$ is a contradiction.
• Suppose that $P_{g}y{P}_g\ne \left\{0\right\}$. Then there exist $p,q\in P_g$ such that $0\ne pyq$. Now, by (1) and (2), $(x+p)y(z+q)=xyz+xyq+pyz+pyq=pyq\ne 0$. Hence, $0\ne (x+p)R_{e}y{R}_e(z+q)\subseteq P$. We have $(x+p)y\in P$ or $y(z+q)\in P$ or $(x+p)(z+q)\in P$. Because P is ‘GR-W-2-AI’. As a result, $xy\in P$, $yz\in P$ or $xz\in P$ is a contradiction.

□

The following theorem is a consequence result from Lemma 2.

Theorem 5.

Let R be a ‘GR-ring’, $g\in G$ and P be a ‘GR-I’ of R such that $P_g^3\ne \left\{0\right\}$. Then P is ‘GR-W-2-AI’ if and only if P is ‘GR-2-AI’.

Proof. 

Assume that P is a ‘GR-W-2-AI’ that is not the same as a ‘GR-2-AI’ of R. For some $x,y,z\in R_g$. Let P has a ‘GR-3-Z’, say $(x,y,z)$. Therefore, if $P_g^3\ne \left\{0\right\}$, there exist $p,q,r\in P_g$ where $pqr\ne 0$, and then $(x+p)(y+q)(z+r)=pqr\ne 0$. As a result, $0\ne (x+p)R_e(y+q)R_e(z+r)\subseteq P$. We have either $(x+p)(y+q)\in P$, $(x+p)(z+r)\in P$ or $(y+q)(z+r)\in P$. Because P is ‘GR-W-2-AI’, and thus either $xy\in P$, $xz\in P$ or $yz\in P$ which is a contradiction. Hence, P is a ‘GR-2-AI’ of R. The contrary is self-evident. □

Corollary 1.

Assume R to be a ‘GR-ring’. If P is a ‘GR-W-2-AI’ of R and it is not ‘GR-2-AI’, then $P_g^3=\left\{0\right\}$.

Allow $R$ to be a ‘GR-ring’ and M to be an R-module. Then M is considered to be a graded if for any $g\in G$, $M={⨁}_{g\in G}{M}_g$ with $R_g{M}_h\subseteq M_{gh}$, where $M_g$ is an additive subgroup of M. The components of $M_g$ are known as homogeneous of degree g.

For any $g\in G$ It is obvious that $M_g$ is an $R_e$-submodule of M. The set of all homogeneous components of M is ${\bigcup }_{g\in G}{M}_g$ and is denoted by $h\left(M\right)$. Let N be an R-submodule which is a graded R-module M, and denoted by ‘GR-R’-submodule.

If $N={⨁}_{g\in G}(N\cap M_g)$, or equivalently, $x={\sum }_{g\in G}{x}_g\in N$, i.e., $x_g\in N$ for any $g\in G$. Then N is said to be graded R-submodule.

It is well known that an R-submodule of a ‘GR-R’-module does not need to be graded. For more terminology see [2,3].

Assume M to be an $bi$-R-module. The idealization (trivial extension) $R⋉M=\left\{\right(r,m):r\in R,m\in M\}$ of M is a ring with component wise addition defined by: $(x,m_1)+(y,m_2)=(x+y,m_1+m_2)$ and multiplication is defined by: $(x,m_1)(y,m_2)=(xy,x{m}_2+m_{1}y)$ for each $x,y\in R$ and $m_1,m_2\in M$. Let G be an Abelian group and M be a ‘GR-R’-module. Then for any $g\in G$, $X=R⋉M$ is a graded by $X_g=R_g⨁M_g$ [9].

Theorem 6.

Let R be a GR-ring with unity, M be a GR-$bi$-R-module and P be a ‘P-GR-I’ of R. Hence, $P⋉M$ is a ‘GR-2-AI’ of $R⋉M$ if and only if P is a ‘GR-2-AI’ of R.

Proof. 

For some $x,y,z\in h\left(R\right)$. Assume that $P⋉M$ is a ‘GR-2-AI’ of $R⋉M$ and $xRyRz\subseteq P$. Then $(x,0),(y,0),(z,0)\in h(R⋉M)$ with $(x,0)R⋉M(y,0)R⋉M(z,0)\subseteq P⋉M$, and then $(x,0)(y,0)=(xy,0)\subseteq P⋉M$, $(x,0)(z,0)=(xz,0)\subseteq P⋉M$ or $(y,0)(z,0)=(yz,0)\subseteq P⋉M$. A a result, $xy\in P$, $xz\in P$ or $yz\in P$, as required. In the opposite case, let $(x,m)R⋉M(y,n)R⋉M(z,p)\subseteq P⋉M$ for some $(x,m),(y,n),(z,p)\in h(R⋉M)$. Therefore, $x,y,z\in h\left(R\right)$ with $xRyRz\subseteq P$, we obtain $xy\in P$, $xz\in P$ or $yz\in P$. If $xy\in P$ true, then $(x,m)(y,n)=(xy,xn+ym)\subseteq P⋉M$. Similarly, if $xz\in P$, then $(x,m)(z,p)\in P⋉M$, and if $yz\in P$, then $(y,n)(z,p)\in P⋉M$, and so on, this completes the proof. □

Theorem 7.

Let R be a ‘GR-ring’ with unity, M to be a ‘GR-bi-R’-module and P to be a ‘PGR-I’ of R. If $P⋉M$ is a ‘GR-W-2-AI’ of $R⋉M$, then P is a ‘GR-W-2-AI’ of R.

Proof. 

For $x,y,z\in h\left(R\right)$, let $0\ne xRyRz\subseteq P$. Then $(0,0)\ne (x,0)R⋉M(y,0)R⋉M(z,0)\subseteq P⋉M$, and then $(xy,0)\in P⋉M$, $(xz,0)\in P⋉M$ or $(yz,0)\in P⋉M$. As a result, $xy\in P$, $xz\in P$ or $yz\in P$. So, P is ‘GR-W-2-AI’. □

Theorem 8.

Let R be a ‘GR-ring’ with unity, M be a ‘GR-bi-R’-module, $g\in G$ and P to be a ‘GR-I’ of R with $P_g\ne R_g$. Hence $P⋉M$ is a ‘GR-W-2-AI’ of $R⋉M$ if and only if P is a ‘GR-W-2-AI’ of R and for every ‘GR-3-Z’, $(x,y,z)$ of P we got $x{R}_{e}y{R}_e{M}_g=M_g{R}_{e}y{R}_{e}z=x{M}_{g}z=0$.

Proof. 

Assume that $P⋉M$ is a ‘GR-W-2-AI’ of $R⋉M$. Let $0\ne x{R}_{e}y{R}_{e}z\subseteq P$, with $x,y,z\in R_g$. Then $(0,0)\ne (x,0)R_e⋉M_e(y,0)R_e⋉M_e(z,0)\subseteq P⋉M$, and then $(xy,0)\in P⋉M$ or $(xz,0)\in P⋉M$ or $(yz,0)\in P⋉M$. As a result, $xy\in P$, $xz\in P$ or $yz\in P$. So, P is ‘GR-W-2-AI’. Preduse that $(x,y,z)$ is a ‘GR-3-Z’ of P. Assume that $x{R}_{e}y{R}_e{M}_g\ne 0$. Hence there exist $r,s\in R_e$ and $m\in M_g$ such that $xrysm\ne 0$, and then $(0,0)\ne (xrysz,xrysm)=(x,0)(r,0)(y,0)(s,0)(z,m)\in (x,0)R_e⋉M_e(y,0)R_e⋉M_e(z,m)\subseteq x{R}_{e}y{R}_{e}z⋉M_g=0⋉M_g\subseteq P⋉M$. However, $(x,0)(y,0)\notin P⋉M$ and $(x,0)(z,m)\notin P⋉M$ and $(y,0)(z,m)\notin P⋉M$, which contradicting the statement that $P⋉M$ is a ‘GR-W-2-AI’. If $M_g{R}_{e}y{R}_{e}z\ne 0$, hence, there exist $n\in M_g$ and $r,s\in R_e$ such that $nrysz\ne 0$. As above, we have $(0,0)\ne (xrysz,nrysz)=(x,n)(r,0)(y,0)(s,0)(z,0)\in (x,n)R_e⋉M_e(y,0)R_e⋉M_e(z,0)\subseteq x{R}_{e}y{R}_{e}z⋉M_g=0⋉M_g\subseteq P⋉M$. however, there is a contradiction between $(x,n)(y,0)\notin P⋉M$, $(x,n)(z,0)\notin P⋉M$ and $(y,0)(z,0)\notin P⋉M$. If $x{M}_{g}z\ne 0$, then there exists $t\in M_g$ where, $xtz\ne 0$. At the present, $(0,0)\ne (xyz,xtz)=(x,0)(1,0)(y,t)(1,0)(z,0)\in (x,0)R_e⋉M_e(y,t)R_e⋉M_e(z,0)\subseteq x{R}_{e}y{R}_{e}z⋉M_g=0⋉M_g\subseteq P⋉M$. However, there is a contradiction between $(x,0)(y,t)\notin P⋉M$ and $(x,0)(z,0)\notin P⋉M$ and $(y,t)(z,0)\notin P⋉M$. Conversely, suppose that $(0,0)\ne (x,n)R_e⋉M_e(y,m)R_e⋉M_e(z,t)\subseteq P⋉M$ for $(x,n),(y,m),(z,t)\in R_g⋉M_g$. Then $x,y,z\in R_g$ with $x{R}_{e}y{R}_{e}z\subseteq P$.

Case (1):$x{R}_{e}y{R}_{e}z\ne 0$. Since P is GR-W-2-AI, it might be $xy\in P$, $xz\in P$ or $yz\in P$. Hence, $(x,n)(y,m)\in P⋉M$, $(x,n)(z,t)\in P⋉M$ or $(y,m)(z,t)\in P⋉M$, as desired.

Case (2):$x{R}_{e}y{R}_{e}z\ne 0$. If $xy\notin P$, $xz\notin P$ and $yz\notin P$, then $(x,y,z)$ is a ‘GR-3-Z’ of P and by assumption $x{R}_{e}y{R}_e{M}_g=M_g{R}_{e}y{R}_{e}z=x{M}_{g}z=0$. Now, $(x,n)R_e⋉M_e(y,m)R_e⋉M_e(z,t)\subseteq \left(x{R}_{e}y{R}_{e}z,M_g{R}_{e}y{R}_{e}z+x{M}_{g}z+x{R}_{e}y{R}_e{M}_g\right)=(0,0)$, a contradiction. □

Question 2.

As a proposal for future work, we think it will be worthy to study non-commutative graded rings such that every ‘GR-I’ is ‘GR-W-2-AI’. What kind of results will be achieved?

The following abbreviations are used throw this Article: ‘GR-SW-2-AI’ for the graded strongly weakly 2-absorbing ideals.

On the other hand, we present the idea of ‘GR-SW-2-AI’, and examine ‘GR-rings’ in which every ‘GR-I’ is ‘GR-SW-2-AI’.

Definition 4.

Let R be a ‘GR-ring’ and P to be a ‘PGR-I’ of R. If $A,B$ and C are ‘GR-I’ of R where $0\ne ABC\subseteq P$. So, $AC\subseteq P$, $BC\subseteq P$ or $AB\subseteq P$. Then P is said to be a ‘GR-SW-2-AI’ of R.

Proposition 8.

Let P be a ‘PGR-I’ of R. Then P is a ‘GR-SW-2-AI’ of R if and only if for any ‘GR-I’ $A,B$ and C of R such that $P\subseteq A$ (or $P\subseteq B$ or $P\subseteq C$), $0\ne ABC\subseteq P$ implies that $AB\subseteq P$, $AC\subseteq P$ or $BC\subseteq P$.

Proof. 

The result holds by the above definition If P is a ‘GR-SW-2-AI’ of R. Conversely, let $K,B$ and C be ‘GR-I’ of R where, $0\ne KBC\subseteq P$. Hence $A=K+P$ is a GR-I of R such that $0\ne ABC\subseteq P$, and then by assumption, $AB\subseteq P$ or $AC\subseteq P$ or $BC\subseteq P$. As a result, $KB\subseteq P$, $KC\subseteq P$ or $BC\subseteq P$. Hence, P becomes a ‘GR-SW-2-AI’ of R. □

Proposition 9.

Let R be a ‘GR-ring’. Then every ‘GR-I’ of R is ‘GR-SW-2-AI’ if and only if for any ‘GR-I’ $I,J$ and K of R, $IJ=IJK$, $IK=IJK$, $JK=IJK$ or $IJK=0$.

Proof. 

Suppose that every ‘GR-I’ of R is ‘GR-SW-2-AI’. Let $I,J$ and K be ‘GR-I’ of R. If $IJK\ne R$, then $IJK$ is ‘GR-SW-2-AI’. Suppose that $IJK\ne 0$. Then $0\ne IJK\subseteq IJK$ and $IJ\subseteq IJK$, $IK\subseteq IJK$ or $JK\subseteq IJK$ and hence $IJ=IJK$, $IK=IJK$ or $JK=IJK$. If $IJK=R$, then $I=J=K=R$. Conversely, let P be a PGR-I of R, $0\ne IJK\subseteq P$ for some ‘GR-I’ $I,J$ and K of R. Then $IJ=IJK\subseteq P$ or $IK=IJK\subseteq P$ or $JK=IJK\subseteq P$. Hence, P is a ‘GR-SW-2-AI’ of R. □

Corollary 2.

Assume R to be a ‘GR-ring’ where every ‘GR-I’ of R is ‘GR-SW-2-AI’. Then $I^3=I^2$ or $I^3=0$ for every ‘GR-I’ of R.

3. Conclusions

In this study, we introduced and examined the concept of Gr-W-2-AI over non-commutative graded rings, several results were achieved. As a proposal for further work on the topic, we are going to examine the concept of Gr-W-1-AI over non-commutative graded rings.