Graded 1-Absorbing Prime Ideals over Non-Commutative Graded Rings

Abstract

In this article, we define and study graded 1-absorbing prime ideals and graded weakly 1-absorbing prime ideals in non-commutative graded rings as a new class of graded ideals that lies between graded prime ideals (graded weakly prime ideals) and graded 2-absorbing ideals (graded weakly 2-absorbing ideals). Let G be a group and let R be a non-commutative G-graded ring with nonzero unity. Let P be a proper graded ideal of R. We then say that P is a graded 1-absorbing prime ideal (a graded weakly 1-absorbing prime ideal) of R if, for each nonunit homogeneous element $r,s,t\in R$ with $rRsRt\subseteq P$ ($\left\{0\right\}\ne rRsRt\subseteq P$), either $rs\in P$ or $t\in P$. We present a number of properties and characterizations of these graded ideals.

1. Introduction

1.1. Motivation

The area of graded ring theory has long been a major research field because of its uses in algebraic geometry, combinatorial algebra, and theoretical physics. One of the main topics in this subject is the grouping and description of graded ideals. Those are indispensable in gaining insight into the graded ring structure. Among these, graded prime ideals and their generalizations have been extensively investigated.

The matter of n-absorbing ideals and their modifications is a case of a much deeper insight into the nature of rings with respect to the factorization properties. For graded commutative rings, a graded n-absorbing ideal is bounded between graded prime ideals and graded primary ideals. It is, however, found that the structures of graded ideals in non-commutative graded rings have been much less understood, and plenty of important questions have generally remained unresolved.

This article introduces and studies graded 1-absorbing prime ideals and graded weakly 1-absorbing prime ideals in non-commutative graded rings. These are new types of graded ideals that naturally extend the definitions of graded prime and graded weakly prime ideals, remaining intimately related to graded 2-absorbing and graded weakly 2-absorbing ideals. The interest in this refinement stems from the necessity to refine the hierarchy of graded ideals in order to demonstrate the behavior concerning ideal absorption in non-commutative settings. This entails the study of their essential properties and characterizations in the hopes of providing a meaningful result toward a view of the ideal structure in graded rings.

In addition, within the non-commutative context, further complications arise, rendering the study of graded ideals even more compelling. The rules governing multiplication follow a straightforward course in commutative cases, while in non-commutativity, one must worry about the ordering of elements and the varying interactions in the ring. The introduction of graded 1-absorbing prime ideals and their weak variations provides a starting point for continuing research on the structural features pertaining to non-commutative graded rings.

1.2. Preliminaries

Let G be a group and R be a ring with nonzero unity 1. Then, R is called G-graded if ${\displaystyle R=\underset{g\in G}{⨁}{R}_g}$ with $R_g{R}_h\subseteq R_{gh}$ for all $g,h\in G$, where $R_g$ is an additive subgroup of R for all $g\in G$, where $R_g{R}_h$ consists of all finite sums of elements $a_g{b}_h$ with $a_g\in R_g$ and $b_h\in R_h$. We denote this by $G\left(R\right)$. The elements of $R_g$ are called homogeneous of degree g. If $a\in R$, then a can be written uniquely as ${\displaystyle a=\sum _{g\in G}{a}_g}$, where $a_g$ is the component of a in $R_g$ and $a_g=0$ except for finitely many. The additive subgroup $R_e$ is in fact a subring of R and $1\in R_e$. The set of all homogeneous elements of R is ${\displaystyle \bigcup _{g\in G}{R}_g}$ and is denoted by $h\left(R\right)$. For more terminology, see [1,2]. Let P be an ideal of a G-graded ring R. Then, P is called a graded ideal if ${\displaystyle P=\underset{g\in G}{⨁}(P\cap R_g)}$; i.e., for $a\in P$, ${\displaystyle a=\sum _{g\in G}{a}_g}$ where $a_g\in P$ for all $g\in G$. An ideal of a graded ring is not necessarily a graded ideal (see [3], Example 1.1).

We only consider non-commutative graded rings with nonzero unity in this article. Such a graded ring will always be indicated by R. Many varieties of graded ideals, including graded prime, graded primary, graded maximal, etc., have been produced over a long period of time. While defining a graded ring, each of them is important. Since they are used to comprehend the structure of graded rings, the idea of graded prime ideals and its generalizations play a crucial role in non-commutative graded algebra.

Recall that in a commutative graded ring, a proper graded ideal P of R is said to be a graded prime ideal if whenever $x,y\in h\left(R\right)$ with $xy\in P$, either $x\in P$ or $y\in P$ [4]. In [5], Atani introduced the notion of graded weakly prime ideals which is a generalization of graded prime ideals. A proper graded ideal P of R is called a graded weakly prime ideal if $0\ne xy\in P$ for some elements $x,y\in h\left(R\right)$ implies that either $x\in P$ or $y\in P$. It is clear that every graded prime ideal is graded weakly prime but the converse is not true in general. Afterwards, Al-Zoubi et al., in their celebrated article [6], introduced the notion of graded 2-absorbing ideals (graded weakly 2-absorbing ideals). Recall from [6] that a nonzero proper graded ideal P of R is called a graded 2-absorbing ideal (graded weakly 2-absorbing ideal) if $xyz\in P$ ($0\ne xyz\in P$) for some $x,y,z\in h\left(R\right)$ implies $xy\in P$ or $xz\in P$ or $yz\in P$. Note that every graded prime ideal is also a graded 2-absorbing ideal (graded weakly 2-absorbing ideal). Following this, many researchers paid close attention to the graded 2-absorbing (graded weakly 2-absorbing) version of graded ideals as well as numerous generalizations of graded 2-absorbing ideals (graded weakly 2-absorbing ideals). Recently, in [7], Abu-Dawwas et al. introduced a graded 1-absorbing prime ideal. This type of graded ideal is a generalization of graded prime ideals. A proper graded ideal P of R is called a graded 1-absorbing prime ideal if whenever $xyz\in P$ for some nonunits $x,y,z\in h\left(R\right)$, either $xy\in P$ or $z\in P$. Note that every graded prime ideal is a graded 1-absorbing prime and every graded 1-absorbing prime ideal is a graded 2-absorbing ideal. The converses are not true. More currently, in [8], Tekir et al. defined graded weakly 1-absorbing prime ideals which is a generalization of the graded 1-absorbing prime ideal. A proper graded ideal P of R is called a graded weakly 1-absorbing prime if $0\ne xyz\in P$ for some nonunits $x,y,z\in h\left(R\right)$ implies that either $xy\in P$ or $z\in P$.

These concepts have been expanded by numerous authors to a non-commutative graded ring setup. In 2018, Abu-Dawwas et al. extended the notion of graded prime ideals to graded rings, not necessarily commutative or with unity. According to their celebrated article [9], a proper graded ideal P of R is called a graded prime ideal if whenever $a,b\in h\left(R\right)$ such that $aRb\subseteq P$, either $a\in P$ or $b\in P$. They also verified that P is a graded prime ideal if and only if whenever $I,J$ are graded right ideals of R such that $IJ\subseteq P$, either $I\subseteq P$ or $J\subseteq P$. In 2021, Alshehry et al. extended the notion of graded weakly prime ideals to graded rings, not necessarily commutative or with unity. According to their celebrated article [3], a proper graded ideal P of R is called a graded weakly prime ideal if whenever $a,b\in h\left(R\right)$ such that $\left\{0\right\}\ne aRb\subseteq P$, either $a\in P$ or $b\in P$. They also verified that P is a graded weakly prime ideal if and only if whenever $I,J$ are graded right ideals of R such that $\left\{0\right\}\ne IJ\subseteq P$, either $I\subseteq P$ or $J\subseteq P$. In 2022, Alshehry et al. extended the notion of graded 2-absorbing (graded weakly 2-absorbing) ideals to graded rings, not necessarily commutative or with unity. According to their celebrated article [10], a proper graded ideal P of R is called graded 2-absorbing (graded weakly 2-absorbing) if whenever $aRbRc\subseteq P$ ($\left\{0\right\}\ne aRbRc\subseteq P$) for some $a,b,c\in h\left(R\right)$, $ab\in P$ or $bc\in P$ or $ac\in P$. It should be noted that a graded 2-absorbing ideal is a graded weakly 2-absorbing ideal. These are distinct ideas, though. Now, following Groenewald in [11], we introduce and examine graded 1-absorbing prime ideals and graded weakly 1-absorbing prime ideals in non-commutative graded rings. For a non-commutative graded ring R, whenever $rRsRt\subseteq P$ ($\left\{0\right\}\ne rRsRt\subseteq P$) for some nonunits $r,s,t\in h\left(R\right)$, then either $rs\in P$ or $t\in P$, and then P is a graded 1-absorbing prime ideal (graded weakly 1-absorbing prime ideal). We give many properties and characterizations of these graded ideals.

2. Graded 1-Absorbing Prime Ideals

In this section, we introduce and examine the concept of graded 1-absorbing prime ideals over non-commutative graded rings. From [7], we have the fact that a proper graded ideal P of a commutative graded ring R is called a graded 1-absorbing prime ideal if whenever $xyz\in P$ for some nonunits $x,y,z\in h\left(R\right)$, either $xy\in P$ or $z\in P$. In what follows, R is a non-commutative graded ring with nonzero unity unless indicated otherwise. We have the following:

Definition 1.

Let R be a graded ring. A proper graded ideal P of R is called graded 1-absorbing prime if for all nonunit elements $r,s,t\in h\left(R\right)$ such that $rRsRt\subseteq P$, we have either $rs\in P$ or $t\in P$.

For commutative graded rings, the two concepts of graded 1-absorbing prime ideals coincide. We now give an example to show that for non-commutative graded rings, it is not the same case:

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 G-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 $R_1=R_3=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$. Since $P=M_2\left(2\mathbb{Z}\right)$ is a graded prime ideal of R, P is a graded 1-absorbing prime ideal of R. On the other hand, $A=\left(\begin{array}{cc}3& 0\\ 0& 0\end{array}\right)$, $B=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$, and $C=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)\in h\left(R\right)$ are nonunit elements such that $ABC=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\in P$, $AB=\left(\begin{array}{cc}3& 0\\ 0& 0\end{array}\right)\notin P$, and $C\notin P$.

Lemma 1.

Let P be a graded 1-absorbing prime ideal of a graded ring R. If J is a proper graded ideal of R and $r,s\in h\left(R\right)$ are nonunit elements such that $rRsJ\subseteq P$, then either $rs\in P$ or $J\subseteq P$.

Proof. 

Suppose that $rs\notin P$ and $t\in J$. Then, $t_g\in J$ for all $g\in G$ as J is a graded ideal. Let $g\in G$. Then, $t_g$ is a nonunit element as J is proper. Now, $rRsR{t}_g\subseteq rRsJ\subseteq P$, which implies that $t_g\in P$ as P is a graded 1-absorbing prime and $rs\notin P$. So, $t_g\in P$ for all $g\in G$, and hence ${\displaystyle t=\sum _{g\in G}{t}_g\in P}$. Thus, $J\subseteq P$. □

Theorem 1.

Suppose that P is a proper graded ideal of a graded ring R. Then, P is a graded 1-absorbing prime ideal of R if and only if whenever $IJK\subseteq P$, for some proper graded ideals $I,J,K$ of R, either $IJ\subseteq P$ or $K\subseteq P$.

Proof. 

Suppose that P is a graded 1-absorbing prime ideal of R and $IJK\subseteq P$ for some proper graded ideals $I,J,K$ of R such that $IJ⊈P$. Then, there are nonunit elements $r\in I$ and $s\in J$ such that $rs\notin P$, and then there are $g,h\in G$ such that $r_g{s}_h\notin P$. Note that as $I,J$ are graded ideals, $r_g\in I$ and $s_h\in J$. Since $r_{g}R{s}_{h}K\subseteq P$ and $r_g{s}_h\notin P$, it follows from Lemma 1 that $K\subseteq P$. Conversely, suppose that $rRsRt\subseteq P$ for some nonunit elements $r,s,t\in h\left(R\right)$ and $rs\notin P$. Assume that $I=RrR,J=RsR$, and $K=RtR$. Then $I,J,K$ are proper graded ideals of R with $IJK\subseteq P$ and $IJ⊈P$. Hence, $K=RtR\subseteq P$, and thus $t\in P$. □

Let R and S be two G-graded rings. Then, a ring homomorphism $f:R\to S$ is said to be a graded ring homomorphism if $f\left(R_g\right)\subseteq S_g$ for all $g\in G$ [1].

Theorem 2.

Let $f:R\to S$ be a graded ring epimorphism such that $f\left(r\right)$ is a nonunit in S for every nonunit element r in R.

1.

If K is a graded 1-absorbing prime ideal of S, then $f^{-1}\left(K\right)$ is a graded 1-absorbing prime ideal of R.

2.

If P is a graded 1-absorbing prime ideal of R with $Ker\left(f\right)\subseteq P$, then $f\left(P\right)$ is a graded 1-absorbing prime ideal of S.

Proof. 

1.

Suppose that $rRsRt\subseteq f^{-1}\left(K\right)$ for some nonunit elements $r,s,t\in h\left(R\right)$. Then, $f\left(rRsRt\right)=f\left(r\right)Sf\left(s\right)Sf\left(t\right)\subseteq K$, which means that $f\left(r\right)f\left(s\right)\in K$ or $f\left(t\right)\in K$. It follows that $rs\in f^{-1}\left(K\right)$ or $t\in f^{-1}\left(K\right)$. Hence, $f^{-1}\left(K\right)$ is a graded 1-absorbing prime ideal of R.

2.

Suppose that $xSySz\subseteq f\left(P\right)$ for some nonunit elements $x,y,z\in h\left(S\right)$. Since f is surjective, there exist nonunit elements $r,s,t\in h\left(R\right)$ such that $x=f\left(r\right)$, $y=f\left(s\right)$, and $z=f\left(t\right)$. Therefore, $f\left(rRsRt\right)=f\left(r\right)Sf\left(s\right)Sf\left(t\right)=xSySz\subseteq f\left(P\right)$. Since $Ker\left(f\right)\subseteq P$, we conclude that $rRsRt\subseteq P$. Thus, $rs\in P$ or $t\in P$, and so $xy\in f\left(P\right)$ or $z\in f\left(P\right)$. Hence, $f\left(P\right)$ is a graded 1-absorbing prime ideal of S.

□

Let R be a G-graded ring and P be a graded ideal of R. Then, $R/P$ is a G-graded ring by ${(R/P)}_g=(R_g+P)/P$, for all $g\in G$ [1].

Corollary 1.

Suppose that P and K are proper graded ideals of a graded ring R with $P\subseteq K$ and $U(R/P)=\{r+P:r\in U(R\left)\right\}$. Then, K is a graded 1-absorbing prime ideal of R if and only if $K/P$ is a graded 1-absorbing prime ideal of $R/P$.

Proof. 

Define $f:R\to R/P$ by $f\left(r\right)=r+P$. Then, f is a graded ring epimorphism, $f\left(1_R\right)=1_R+P=1_{R/P}$, $f\left(r\right)$ is a nonunit for every nonunit $r\in R$, and $Ker\left(f\right)=P\subseteq K$. Suppose that K is a graded 1-absorbing prime ideal of R. Then, $f\left(K\right)=K/P$ is a graded 1-absorbing prime ideal of $R/P$ by Theorem 2 (2). Conversely, $f^{-1}(K/P)=K$ is a graded 1-absorbing prime ideal of R by Theorem 2 (1). □

Definition 2.

Let R be a graded ring, $g\in G$, and P be a graded ideal of R such that $P_g\ne R_g$. Then,

1.

P is called a g-1-absorbing prime if for all nonunit elements $r,s,t\in R_g$ such that $rRsRt\subseteq P$, either $rs\in P$ or $t\in P$.

2.

P is called a g-prime if for all nonunit elements $r,s\in R_g$ such that $rRs\subseteq P$, either $r\in P$ or $s\in P$.

Theorem 3.

Let R be a graded ring. Suppose that R has a g-1-absorbing prime ideal that is not g-prime. If $\alpha ,\beta \in R_g$ such that α is a nonunit element and β is a unit element, then $\alpha +\beta$ is a unit element.

Proof. 

Let P be a g-1-absorbing prime ideal of R that is not g-prime. Then, there are nonunit elements $r,s\in R_g$ such that $rRs\subseteq P$ and $r,s\notin P$. Since $\alpha RrRs\subseteq P$, it follows that $\alpha r\in P$. Also, $(\alpha +\beta )RrRs\subseteq P$. If $\alpha +\beta$ is a nonunit element, then $(\alpha +\beta )r\in P$. Hence, $\beta r\in P$, and then since $\beta$ is a unit element, we have $r\in P$, which is a contradiction. Hence, $\alpha +\beta$ is a unit element. □

Theorem 4.

Let R be a graded ring such that $R_e$ is not a local ring. Then, every e-1-absorbing prime ideal of R is e-prime.

Proof. 

Suppose that R has an e-1-absorbing prime ideal that is not e-prime. Then, by Theorem 3, for $\alpha ,\beta \in R_e$ such that $\alpha$ is a nonunit element and $\beta$ as a unit element, we have $\alpha +\beta$ as a unit element, and then by ([11], Lemma 4.1), $R_e$ is a local ring, which is a contradiction. □

A proper graded ideal M of a graded ring R is said to be a graded maximal ideal if whenever P is a graded ideal of R such that $M\subseteq P\subseteq R$, either $P=M$ or $P=R$. A graded ring R is said to a graded local ring if it has a unique graded maximal ideal [1].

Lemma 2.

Let R be a graded local ring with unique graded maximal ideal M. Then, P is a graded 1-absorbing prime ideal of R if and only if P is a graded prime ideal of R or $M^2\subseteq P⊊M$.

Proof. 

Suppose that P is a graded 1-absorbing prime ideal of R that is not graded prime. Clearly, $P⊊M$. Since P is not graded prime, there are $r,s\in h\left(M\right)-P$ such that $rRs\subseteq P$. Let $a,b\in M$. Then, $a_g,b_h\in M$ for all $g,h\in G$ as M is a graded ideal. Also, $\left(a_{g}R{b}_{h}R\right)RrRs\subseteq P$, which implies that $\left(a_{g}R{b}_{h}R\right)r\subseteq P$ for all $g,h\in G$. Again, we have $a_g{b}_h\in P$ for all $g,h\in G$. Thus, $ab\in P$, and hence $M^2\subseteq P$. Conversely, if P is a graded prime ideal, then P is a graded 1-absorbing prime ideal. Suppose that $M^2\subseteq P⊊M$. Then, clearly, P is proper. Let $r,s,t\in h\left(M\right)$ such that $rRsRt\subseteq P$. Then, $rs\in M^2\subseteq P$. Therefore, P is a graded 1-absorbing prime ideal of R. □

Proposition 1.

Let R be a graded local ring with unique graded maximal ideal M. Then, R has a graded 1-absorbing prime ideal that is not graded prime if and only if $M^2\ne M$.

Proof. 

Suppose that R has a graded 1-absorbing prime ideal P of R that is not graded prime. Then, by Lemma 2, $M^2\subseteq P⊊M$, and hence $M^2\ne M$. Conversely, it is clear that $M^2⊊M$, so $M^2$ is proper and there exists $m\in M-M^2$, and then as M is a graded ideal, there exists $g\in G$ such that $m_g\in M-M^2$. But $m_{g}R{m}_g\subseteq M^2$, so $M^2$ is not a graded prime ideal of R. Let $r,s,t\in h\left(M\right)$ such that $rRsRt\subseteq M^2$. Then, $rs\in M^2$, and hence $M^2$ is a graded 1-absorbing prime ideal of R. □

Proposition 2.

Let R be a graded local ring. If P and K are two graded 1-absorbing prime ideals of R that are not graded prime, then $P\bigcap K$ and $P+K$ are graded 1-absorbing prime ideals of R.

Proof. 

Let M be a unique graded maximal ideal of R. Then, by Lemma 2, $M^2\subseteq P\bigcap K$. Let $r,s,t\in h\left(M\right)$ such that $rRsRt\subseteq P\bigcap K$. Then, $rs\in M^2\subseteq P\bigcap K$, and hence $P\bigcap K$ is a graded 1-absorbing prime ideal of R. Similarly, $P+K$ is a graded 1-absorbing prime ideal of R. □

A graded ring R is said to be graded prime if $\left\{0\right\}$ is a graded prime ideal of R [4].

Theorem 5.

Let R be a graded local ring with unique graded maximal ideal M. Then, $\left\{0\right\}$ is a graded 1-absorbing prime ideal of R if and only if R is graded prime or $M^2=\left\{0\right\}$.

Proof. 

The result holds directly from Lemma 2. □

Assume that R is a G-graded ring and M is an R-module. Then, M is said to be G-graded if ${\displaystyle M=\underset{g\in G}{⨁}{M}_g}$ with $R_g{M}_h\subseteq M_{gh}$ for all $g,h\in G$, where $M_g$ is an additive subgroup of M for all $g\in G$. The elements of $M_g$ are called homogeneous of degree g. It is clear that $M_g$ is an $R_e$-submodule of M for all $g\in G$. We assume that ${\displaystyle h\left(M\right)=\bigcup _{g\in G}{M}_g}$. Let N be an R-submodule of a graded R-module M. Then, N is said to be a graded R-submodule if ${\displaystyle N=\underset{g\in G}{⨁}(N\bigcap M_g)}$; i.e., for $x\in N$, ${\displaystyle x=\sum _{g\in G}{x}_g}$, where $x_g\in N$ for all $g\in G$. It is known that an R-submodule of a graded R-module is not necessarily graded. Let M be an R-module. The idealization $R(+)M=\left\{\right(r,m):r\in R,m\in M\}$ of M is a ring with componentwise addition and multiplication; $(x,m_1)+(y,m_2)=(x+y,m_1+m_2)$ and $(x,m_1)(y,m_2)=(xy,x{m}_2+y{m}_1)$ for each $x,y\in R$ and $m_1,m_2\in M$. By ([11], Remark 3.1), $(r,m)$ is a nonunit in $R(+)M$ if and only if r is a nonunit in R. Let G be an abelian group and M be a G-graded R-module. Then, $X=R(+)M$ is G-graded by $X_g=R_g(+)M_g$ for all $g\in G$ [12]. Moreover, if P is an ideal of R and N is an R submodule of M such that $PM\subseteq N$, then $P(+)N$ is a graded ideal of $R(+)M$ if and only if P is a graded ideal of R and N is a graded R-submodule of M ([12], Proposition 3.3).

Theorem 6.

Let M be a graded R-module and P be a proper graded ideal of R. Then, $P(+)M$ is a graded 1-absorbing prime ideal of $R(+)M$ if and only if P is a graded 1-absorbing prime ideal of R.

Proof. 

Suppose that $P(+)M$ is a graded 1-absorbing prime ideal of $X=R(+)M$ and let $rRsRt\subseteq P$ for some nonunits $r,s,t\in h\left(R\right)$. Then, $(r,0),(s,0),(t,0)$ are nonunits in $h\left(R\right(+\left)M\right)$ with $(r,0)X(s,0)X(t,0)\subseteq P(+)M$, and then $(r,0)(s,0)=(rs,0)\in P(+)M$ or $(t,0)\in P(+)M$. Hence, $rs\in P$ or $t\in P$. Thus, P is a graded 1-absorbing prime ideal of R. Conversely, let $(r,m)X(s,n)X(t,p)\subseteq P(+)M$ for some nonunits $r,s,t\in h\left(R\right)$ and $m,n,p\in h\left(M\right)$. Then, $rRsRt\subseteq P$, and then $rs\in P$ or $t\in P$. If $rs\in P$, then $(r,m)(s,n)=(rs,rn+sm)\in P(+)M$, and if $t\in P$, then $(t,p)\in P(+)M$. Hence, $P(+)M$ is a graded 1-absorbing prime ideal of $R(+)M$. □

3. Graded Weakly 1-Absorbing Prime Ideals

In this section, we introduce and examine the concept of graded weakly 1-absorbing prime ideals over non-commutative graded rings.

Definition 3.

Let R be a graded ring. A proper graded ideal P of R is called graded weakly 1-absorbing prime if for all nonunit elements $r,s,t\in h\left(R\right)$ such that $\left\{0\right\}\ne rRsRt\subseteq P$, we have either $rs\in P$ or $t\in P$.

Clearly, every graded 1-absorbing prime ideal is graded weakly 1-absorbing prime. However, the next example shows that a graded weakly 1-absorbing prime ideal is not necessarily graded 1-absorbing prime:

Example 2.

Consider $R=M_2\left({\mathbb{Z}}_{30}\right)$ and $G={\mathbb{Z}}_4$. Then, R is G-graded by $R_0=\left(\begin{array}{cc}{\mathbb{Z}}_{30}& 0\\ 0& {\mathbb{Z}}_{30}\end{array}\right)$, $R_2=\left(\begin{array}{cc}0& {\mathbb{Z}}_{30}\\ {\mathbb{Z}}_{30}& 0\end{array}\right)$, and $R_1=R_3=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$. Clearly, $P=\left\{\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\right\}$ is a graded weakly 1-absorbing prime ideal of R. On the other hand, P is not graded 1-absorbing prime, since $A=\left(\begin{array}{cc}2& 0\\ 0& 0\end{array}\right)$, $B=\left(\begin{array}{cc}3& 0\\ 0& 0\end{array}\right)$, and $C=\left(\begin{array}{cc}5& 0\\ 0& 0\end{array}\right)\in h\left(R\right)$ are nonunit elements such that $ARBRC=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\in P$, $AB=\left(\begin{array}{cc}6& 0\\ 0& 0\end{array}\right)\notin P$, and $C\notin P$.

Proposition 3.

Let R be a graded ring and K be a graded weakly 1-absorbing prime ideal of R. If P is a graded ideal of R with $P\subseteq K$, then $K/P$ is a graded weakly 1-absorbing prime ideal of $R/P$.

Proof. 

Suppose that $\{0+P\}\ne (r+P)(R/P)(s+P)(R/P)(t+P)\subseteq K/P$ for some nonunits $r+P,s+P,t+P\in h(R/P)$. Then, $r,s,t\in h\left(R\right)$ are nonunits such that $\left\{0\right\}\ne rRsRt\subseteq K$, and then either $rs\in K$ or $t\in K$, which implies that either $(r+P)(s+P)\in K/P$ or $t+P\in K/P$. Hence, $K/P$ is a graded weakly 1-absorbing prime ideal of $R/P$. □

Proposition 4.

Let $P\subseteq K$ be proper graded ideals of a graded ring R such that $U(R/P)=\{r+P:r\in U(R\left)\right\}$. If P is a graded weakly 1-absorbing prime ideal of R and $K/P$ is a graded weakly 1-absorbing prime ideal of $R/P$, then K is a graded weakly 1-absorbing prime ideal of R.

Proof. 

Let $r,s,t\in h\left(R\right)$ be nonunits such that $\left\{0\right\}\ne rRsRt\subseteq K$. If $rRsRt\subseteq P$, then either $rs\in P\subseteq K$ or $t\in P\subseteq K$. Suppose that $rRsRt⊈P$. Then, $r+P,s+P,t+P\in h(R/P)$ are nonunits such that $\{0+P\}\ne (r+P)(R/P)(s+P)(R/P)(t+P)\subseteq K/P$, and then either $(r+P)(s+P)\in K/P$ or $t+P\in K/P$, which implies that either $rs\in K$ or $t\in K$. Hence, K is a graded weakly 1-absorbing prime ideal of R. □

Let $f:R\to S$ be a graded ring homomorphism. Then, by ([10], Proposition 5), $Ker\left(f\right)$ is a graded ideal of S.

Theorem 7.

Let $f:R\to S$ be a graded ring epimorphism.

1.

If P is a graded weakly 1-absorbing prime ideal of R and $Ker\left(f\right)\subseteq P$, then $f\left(P\right)$ is a graded weakly 1-absorbing prime ideal of S.

2.

If Q is a graded weakly 1-absorbing prime ideal of S, $f\left(r\right)$ is a nonunit in S for every nonunit r in R, and $Ker\left(f\right)$ is a graded weakly 1-absorbing prime ideal of R, then $f^{-1}\left(Q\right)$ is a graded weakly 1-absorbing prime ideal of R.

Proof. 

1.

Since P is a graded weakly 1-absorbing prime ideal of R and $Ker\left(f\right)\subseteq P$, we conclude that $P/Ker\left(f\right)$ is a graded weakly 1-absorbing prime ideal of $R/Ker\left(f\right)$ by Proposition 3, and then the result follows since $R/Ker\left(f\right)$ is isomorphic to S.

2.

Let $P=f^{-1}\left(Q\right)$. Then, $Ker\left(f\right)\subseteq P$. Since $R/Ker\left(f\right)$ is isomorphic to S, we conclude that $P/Ker\left(f\right)$ is a graded weakly 1-absorbing prime ideal of $R/Ker\left(f\right)$. Since $Ker\left(f\right)$ is a graded weakly 1-absorbing prime ideal of R and $P/Ker\left(f\right)$ is a graded weakly 1-absorbing prime ideal of $R/Ker\left(f\right)$, we conclude that $P=f^{-1}\left(Q\right)$ is a graded weakly 1-absorbing prime ideal of R by Proposition 4.

□

Definition 4.

Let R be a graded ring, $g\in G$, and P be a graded ideal of R such that $P_g\ne R_g$. Then, P is called g-weakly 1-absorbing prime if for all nonunit elements $r,s,t\in R_g$ such that $\left\{0\right\}\ne rRsRt\subseteq P$, either $rs\in P$ or $t\in P$.

Definition 5.

Suppose that P is a graded weakly 1-absorbing prime ideal of a graded ring R and $\alpha ,\beta ,\gamma \in h\left(R\right)$ are nonunit elements.

1.

We say that $(\alpha ,\beta ,\gamma )$ is a homogeneous triple-zero (htz for short) of P if $\alpha R\beta R\gamma =\left\{0\right\}$, $\alpha \beta \notin P$, and $\gamma \notin P$.

2.

Suppose that $IJK\subseteq P$ for some graded ideals $I,J,K$ of R. Then, we say that P is a free htz with respect to $IJK$ if $(\alpha ,\beta ,\gamma )$ is not an htz of P, for every $\alpha \in I\bigcap h\left(R\right),\beta \in J\bigcap h\left(R\right)$, and $\gamma \in K\bigcap h\left(R\right)$.

Definition 6.

Suppose that P is a g-weakly 1-absorbing prime ideal of a graded ring R and $\alpha ,\beta ,\gamma \in R_g$ are nonunit elements.

1.

We say that $(\alpha ,\beta ,\gamma )$ is a g-triple-zero (g-tz for short) of P if $\alpha R\beta R\gamma =\left\{0\right\}$, $\alpha \beta \notin P$, and $\gamma \notin P$.

2.

Suppose that $IJK\subseteq P$ for some graded ideals $I,J,K$ of R. Then, we say that P is a free g-tz with respect to $IJK$ if $(\alpha ,\beta ,\gamma )$ is not a g-tz of P, for every $\alpha \in I_g,\beta \in J_g$, and $\gamma \in K_g$.

Clearly, if P is a graded weakly 1-absorbing prime ideal that is not graded 1-absorbing prime, then there exists an htz of P. Also, if P is a g-weakly 1-absorbing prime ideal that is not g-1-absorbing prime, then there exists a g-tz of P. So, we introduce the following example that is motivated by Example 2:

Example 3.

Consider $R=M_2\left({\mathbb{Z}}_{30}\right)$ and $G={\mathbb{Z}}_4$. Then, R is G-graded by $R_0=\left(\begin{array}{cc}{\mathbb{Z}}_{30}& 0\\ 0& {\mathbb{Z}}_{30}\end{array}\right)$, $R_2=\left(\begin{array}{cc}0& {\mathbb{Z}}_{30}\\ {\mathbb{Z}}_{30}& 0\end{array}\right)$, and $R_1=R_3=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$. Then, by Example 2, $P=\left\{\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\right\}$ is a graded weakly 1-absorbing prime ideal of R that is not graded 1-absorbing prime, and P is a 0-weakly 1-absorbing prime ideal of R that is not 0-1-absorbing prime. Note that $A=\left(\begin{array}{cc}2& 0\\ 0& 0\end{array}\right)$, $B=\left(\begin{array}{cc}3& 0\\ 0& 0\end{array}\right)$, $C=\left(\begin{array}{cc}5& 0\\ 0& 0\end{array}\right)\in R_0\subseteq h\left(R\right)$ are nonunit elements such that $ARBRC=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$, $AB=\left(\begin{array}{cc}6& 0\\ 0& 0\end{array}\right)\notin P$ and $C\notin P$. Hence, $(A,B,C)$ are htzs of P, and $(A,B,C)$ is a 0-tz of P.

Lemma 3.

Let P be a g-weakly 1-absorbing prime ideal of a graded ring R and suppose that $(\alpha ,\beta ,\gamma )$ is a g-tz of P for some nonunit elements $\alpha ,\beta ,\gamma \in R_g$. Assume that the sum of every two nonunit elements of $R_g$ is nonunit. Then,

1.

$\alpha R_g\beta P_g=\left\{0\right\}$.

2.

$P_g\beta R_g\gamma =\left\{0\right\}$.

3.

$\alpha P_g\gamma =\left\{0\right\}$.

4.

$P_g^2\gamma =\left\{0\right\}$.

5.

$\alpha P_g^2=\left\{0\right\}$.

6.

$P_g\beta P_g=\left\{0\right\}$.

Proof. 

1.

Suppose that $\alpha R_g\beta P_g\ne \left\{0\right\}$. Then, there exist $r\in R_g$ and $p\in P_g$ such that $0\ne \alpha r\beta p$. Now, $\alpha r\beta (p+\gamma )=\alpha r\beta p+\alpha r\beta \gamma =\alpha r\beta p\ne 0$. Hence, $\left\{0\right\}\ne \alpha R\beta R(p+\gamma )\subseteq P$. Since $\alpha \beta \notin P$, we have $(p+\gamma )\in P$ and consequently $\gamma \in P$, which is a contradiction.

2.

Suppose that $P_g\beta R_g\gamma \ne \left\{0\right\}$. Then, there exist $r\in R_g$ and $p\in P_g$ such that $0\ne p\beta r\gamma$. Now, $(\alpha +p)\beta r\gamma =\alpha \beta r\gamma +p\beta r\gamma =p\beta r\gamma \ne 0$. Hence, $\left\{0\right\}\ne (\alpha +p)R\beta R\gamma \subseteq P$. Since $\gamma \notin P$, we have $(\alpha +p)\beta \in P$ and then $\alpha \beta \in P$, which is a contradiction.

3.

Suppose that $\alpha P_g\gamma \ne \left\{0\right\}$. Then, there exists $p\in P_g$ such that $0\ne \alpha p\gamma$. Now, $\alpha (\beta +p)\gamma =\alpha \beta \gamma +\alpha p\gamma =\alpha p\gamma \ne 0$. Hence, $\left\{0\right\}\ne \alpha R(\beta +p)R\gamma \subseteq P$. Since $\gamma \notin P$, we have $\alpha (\beta +p)\in P$ and then $\alpha \beta \in P$, which is a contradiction.

4.

Suppose that $P_g^2\gamma \ne \left\{0\right\}$. Then, there exist $p,q\in P_g$ such that $0\ne pq\gamma$. Now, $(\alpha +p)(\beta +q)\gamma =\alpha \beta \gamma +\alpha q\gamma +p\beta \gamma +pq\gamma =pq\gamma \ne 0$ by (2) and (3). Hence, $\left\{0\right\}\ne (\alpha +p)R(\beta +q)R\gamma \subseteq P$. Since $\gamma \notin P$, we have $(\alpha +p)(\beta +q)\in P$ and then $\alpha \beta \in P$, which is a contradiction.

5.

Suppose that $\alpha P_g^2\ne \left\{0\right\}$. Then, there exist $p,q\in P_g$ such that $0\ne \alpha pq$. Now, $\alpha (\beta +p)(\gamma +q)=\alpha \beta \gamma +\alpha \beta q+\alpha p\gamma +\alpha pq=\alpha pq\ne 0$ by (1) and (3). Hence, $\left\{0\right\}\ne \alpha R(\beta +p)R(\gamma +q)\subseteq P$. So we have $\alpha (\beta +p)\in P$ or $(\gamma +q)\in P$. Hence, $\alpha \beta \in P$ or $\gamma \in P$, which is a contradiction.

6.

Suppose that $P_g\beta P_g\ne \left\{0\right\}$. Then, there exist $p,q\in P_g$ such that $0\ne p\beta q$. Now, $(\alpha +p)\beta (\gamma +q)=\alpha \beta \gamma +\alpha \beta q+p\beta \gamma +p\beta q=p\beta q\ne 0$ by (1) and (2). Hence, $\left\{0\right\}\ne (\alpha +p)R\beta R(\gamma +q)\subseteq P$. So we have $(\alpha +p)\beta \in P$ or $(\gamma +q)\in P$. Hence, $\alpha \beta \in P$ or $\gamma \in P$, which is a contradiction.

□

Theorem 8.

Let R be a graded ring and $g\in G$ such that the sum of every two nonunit elements of $R_g$ is a nonunit. Suppose that P is a graded ideal of R such that $P_g^3\ne \left\{0\right\}$. Then, P is a g-weakly 1-absorbing prime ideal of R if and only if P is a g-1-absorbing prime ideal of R.

Proof. 

Suppose that P is a g-weakly 1-absorbing prime ideal of R that is not a g-1-absorbing prime. Then, P has a g-tz $(\alpha ,\beta ,\gamma )$ for some nonunits $\alpha ,\beta ,\gamma \in R_g$. Assume that $pqr\ne 0$ for some $p,q,r\in P_g$. Then, by Lemma 3, we have $(\alpha +p)(\beta +q)(\gamma +r)=pqr\ne 0$. Hence, $\left\{0\right\}\ne (\alpha +p)R(\beta +q)R(\gamma +r)\subseteq P$. So, we have either $(\alpha +p)(\beta +q)\subseteq P$ or $(\gamma +r)\in P$, and thus either $\alpha \beta \in P$ or $\gamma \in P$, which is a contradiction. Hence, P is a g-1-absorbing prime ideal of R. The converse is clear. □

Corollary 2.

Let R be a graded ring and $g\in G$ such that the sum of every two nonunit elements of $R_g$ is a nonunit. If P is a g-weakly 1-absorbing prime ideal of R that is not g-1-absorbing prime, then $P_g^3=\left\{0\right\}$.

Theorem 9.

Let R be a graded ring and $g\in G$ such that the sum of every two nonunit elements of $R_g$ is a nonunit. Suppose that P is a g-weakly 1-absorbing prime ideal of R such that $\left\{0\right\}\ne P_g=P_g^2$. Then, P is a g-1-absorbing prime ideal of R.

Proof. 

Now, $P_g^3=P_g{P}_g^2=P_g^2=P_g\ne \left\{0\right\}$, and then it follows from Theorem 8 that P is a g-1-absorbing prime ideal of R. □

Proposition 5.

Let R be a graded ring and $g\in G$ such that the sum of every two nonunit elements of $R_g$ is a nonunit. Suppose that $a\in R_g$ such that $l\left(a\right)=\{r\in R:ra=0\}=\left\{0\right\}$. Then, the graded left ideal $Ra$ is g-1-absorbing prime if and only if $Ra$ is a g-weakly 1-absorbing prime left ideal of R.

Proof. 

Suppose that $Ra$ is a g-weakly 1-absorbing prime left ideal of R that is not g-1-absorbing prime. Then, $Ra$ has a g-tz $(\alpha ,\beta ,\gamma )$ for some nonunit elements $\alpha ,\beta ,\gamma \in R_g$. Now, $\alpha R\beta R(\gamma +a)=\alpha R\beta R\gamma +\alpha R\beta Ra=\alpha R\beta Ra$. If $\alpha R\beta Ra=\left\{0\right\}$, then $\alpha R\beta R\subseteq l\left(a\right)=\left\{0\right\}$, and then $\alpha \beta =0\in Ra$, which is a contradiction. Hence, $\alpha R\beta Ra\ne \left\{0\right\}$, and so $\left\{0\right\}\ne \alpha R\beta R(\gamma +a)\subseteq Ra$, which implies that $(\gamma +a)\in Ra$, and then $\gamma \in Ra$, which is a contradiction. Therefore, $Ra$ is g-1-absorbing prime. The converse is clear. □

A G-graded ring R is said to be a cross-product if $R_g$ contains a unit for all $g\in G$ [1]. Let S and T be two G-graded rings. Then, $R=S\times T$ is a G-graded ring by $R_g=S_g\times T_g$ for all $g\in G$ [1].

Theorem 10.

Let S and T be two graded rings such that T is a cross-product. Let $R=S\times T$ and P be a proper graded ideal of S. If $P\times T$ is a graded weakly 1-absorbing prime ideal of R, then P is a graded 1-absorbing prime ideal of S.

Proof. 

Suppose that $xSySz\subseteq P$ for some nonunit elements $x,y,z\in h\left(S\right)$. Then, $x\in S_g,y\in S_h$, and $z\in S_k$ for some $g,h,k\in G$. Choose unit elements $a\in T_g,b\in T_h$, and $c\in T_k$. Then, $\left\{\right(0,0\left)\right\}\ne (x,a)R(y,b)R(z,c)=(xSySz,aTbTc)\subseteq P\times T$, and so either $(x,a)(y,b)=(xy,ab)\in P\times T$ or $(z,c)\in P\times T$. Hence, either $xy\in P$ or $z\in P$. Thus, P is a graded 1-absorbing prime ideal of S. □

A graded ring R is said to be a graded division ring if every nonzero homogeneous element of R is a unit [1].

Theorem 11.

Let S and T be two graded rings that are cross-products but not graded division rings. Let $R=S\times T$ and P be a nonzero proper graded ideal of R. If P is a graded weakly 1-absorbing prime ideal of R, then $P=P_1\times T$ for some graded prime ideal $P_1$ of S or $P=S\times P_2$ for some graded prime ideal $P_2$ of T.

Proof. 

Now, $P=P_1\times P_2$ for some graded ideals $P_1$ of S and $P_2$ of T. Since P is nonzero, $P_1\ne \left\{0\right\}$ or $P_2\ne \left\{0\right\}$. Without loss of generality, we may assume that $P_1\ne \left\{0\right\}$. Then, there exists $0\ne x\in P_1$, and then $x_g\ne 0$, for some $g\in G$. Note that $x_g\in P_1$ as $P_1$ is a graded ideal. Choose a unit element $t\in T_g$, where $\left\{(0,0)\right\}\ne (1,0)R(1,0)R(x_g,t)=(1S1S{x}_g,0)\subseteq P$; we conclude $(1,0)\in P$ or $(x_g,t)\in P$. Then, we have $P_1=S$ or $P_2=T$. Assume that $P_1=S$. Now, we will show that $P_2$ is a graded prime ideal of T. Let $yTz\subseteq P_2$ for some $y,z\in h\left(T\right)$. If y or z is a unit element, then we have $y\in P_2$ or $z\in P_2$. So, assume that $y,z$ are nonunit elements in T. Since S is not a graded division ring, there exists a nonzero nonunit $s\in h\left(S\right)$. Note that $y\in T_g,z\in T_h$, and $s\in S_k$ for some $g,h,k\in G$, so choose unit elements $\alpha \in S_g,\beta \in S_h$, and $\gamma \in T_k$. This implies that $\left\{\right(0,0\left)\right\}\ne (s,\gamma )R(\alpha ,y)R(\beta ,z)=(sS\alpha S\beta ,\gamma TyTz)\subseteq P$. We conclude that $(s,\gamma )(\alpha ,y)=(s\alpha ,\gamma y)\in P$ or $(\beta ,z)\in P$. Thus, we get $y\in P_2$ or $z\in P_2$, and so $P_2$ is a graded prime ideal of T. In another case, one can similarly show that $P=P_1\times T$ and $P_1$ is a graded prime ideal of S. □

Theorem 12.

Let S, T, and L be graded rings such that they are cross-products. Let $R=S\times T\times L$ and $H=S\times T$. If $P=P_1\times P_2\times P_3$ is a nonzero graded weakly 1-absorbing prime ideal of R, then $P_1\times P_2$ is a graded 1-absorbing prime ideal of H or $P_3$ is a graded 1-absorbing prime ideal of L.

Proof. 

Since P is nonzero, there exists $(0,0,0)\ne (a,b,c)\in P$, and then $a_g\ne 0$ or $b_h\ne 0$ or $c_k\ne 0$ for some $g,h,k\in G$. Choose unit elements $t_1\in T_g,l_1\in L_g,s_1\in S_h,l_2\in L_h,s_2\in S_k$, and $t_2\in T_k$. So, $\left\{(0,0,0)\right\}\ne (a_g,t_1,l_1)R(s_1,b_h,l_2)R(s_2,t_2,c_k)=(a_{g}S{s}_{1}S{s}_2,t_{1}T{b}_{h}T{t}_2,l_{1}L{l}_{2}L{c}_k)\subseteq P$, and then either $(a_g{s}_1,t_1{b}_h,l_1{l}_1)\in P$ or $(s_2,t_2,c_k)\in P$. Hence, either $P_1=S$ and $P_2=T$, or $P_3=L$, and so $P=P_1\times P_2\times L$ or $P=S\times T\times P_3$. Thus, by Theorem 10, $P_1\times P_2$ is a graded 1-absorbing prime ideal of H or $P_3$ is a graded 1-absorbing prime ideal of L. □

Proposition 6.

Let R be a graded local ring with unique graded maximal ideal M. If $M^3=\left\{0\right\}$, then every proper graded ideal of R is a graded weakly 1-absorbing prime ideal of R.

Proof. 

Clearly, $\left\{0\right\}$ is a graded weakly 1-absorbing prime ideal of R. Let P be a nonzero proper graded ideal of R. Assume that P is not a graded weakly 1-absorbing prime ideal. Then, there exist nonunit elements $r,s,t\in h\left(R\right)$ such that $\left\{0\right\}\ne rRsRt\subseteq P$ with $rs\notin P$ and $t\notin P$. Since $r,s,t$ are nonunits, they are elements of M. So, $rRsRt\subseteq M^3=\left\{0\right\}$, which is a contradiction. □

4. Conclusions

In this article, we introduced and examined graded 1-absorbing prime ideals and graded weakly 1-absorbing prime ideals in non-commutative graded rings as a new class of graded ideals between graded prime ideals (graded weakly prime ideals) and graded 2-absorbing ideals (graded weakly 2-absorbing ideals). Let G be a group and R be a non-commutative G-graded ring with nonzero unity. A proper graded ideal P of R is said to be a graded 1-absorbing prime ideal (graded weakly 1-absorbing prime ideal) if for all nonunit homogeneous elements $r,s,t\in R$ with $rRsRt\subseteq P$ ($\left\{0\right\}\ne rRsRt\subseteq P$), we have either $rs\in P$ or $t\in P$. We gave many properties and characterizations of these graded ideals. As a proposal for future work, we will study non-commutative graded rings in which every proper graded ideal is graded weakly 1-absorbing prime.