On Graded S-Primary Ideals

Abstract

Let R be a commutative graded ring with unity, S be a multiplicative subset of homogeneous elements of R and P be a graded ideal of R such that $P\bigcap S=\varnothing$. In this article, we introduce the concept of graded S-primary ideals which is a generalization of graded primary ideals. We say that P is a graded S-primary ideal of R if there exists $s\in S$ such that for all $x,y\in h\left(R\right)$, if $xy\in P$, then $sx\in P$ or $sy\in Grad\left(P\right)$ (the graded radical of P). We investigate some basic properties of graded S-primary ideals.

1. Introduction

Throughout this article, G will be a group with the identity of e and R will be a commutative ring with a nonzero unity of 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. The elements of $R_g$ are called homogeneous of degree g. If $a\in R$, then a can be written uniquely as ${\displaystyle \sum _{g\in G}{a}_g}$, where $a_g$ is the component of a in $R_g$. The component $R_e$ is a subring of R and $1\in R_e$. The set of all homogeneous elements of R is ${\displaystyle h\left(R\right)=\bigcup _{g\in G}{R}_g}$. Let P be an ideal of a 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 $g\in G$. It is not necessary that every ideal of a graded ring is a graded ideal. For more details and terminology, look at [1,2].

Let P be a proper graded ideal of R. Then the graded radical of P is denoted by $Grad\left(P\right)$ and it is defined as written below:

$Grad\left(P\right)=\left\{{\displaystyle x=\sum _{g\in G}{x}_g\in R:\mathrm{for\; all} g\in G, \mathrm{there\; exists} n_g\in \mathbb{N} \mathrm{such\; that} x_g^{n_g}\in P}\right\}$.

Note that $Grad\left(P\right)$ is always a graded ideal of R (check [3]).

A proper graded ideal P of R is said to be a graded prime if $xy\in P$ implies that $x\in P$ or $y\in P$ where $x,y\in h\left(R\right)$ [3]. Graded prime ideals play a very important role in the Commutative Graded Rings Theory. There are several ways to generalize the concept of a graded prime ideal, for example, Refai and Al-Zoubi in [4] introduced the concept of graded primary ideals, a proper graded ideal P of R is said to be a graded primary ideal whenever $ab\in P$ where $a,b\in h\left(R\right)$, then either $a\in P$ or $b\in Grad\left(P\right)$.

Let $S\subseteq R$ be a multiplicative set and P be an ideal of R such that $P\bigcap S=\varnothing$. In [5], P is said to be a S-primary ideal of R if $s\in S$ exists such that for all $x,y\in R$, if $xy\in P$, then $sx\in P$ or $sy$ is in the radical of P.

Let R be a graded ring, $S\subseteq h\left(R\right)$ be a multiplicative set and P be a graded ideal of R such that $P\bigcap S=\varnothing$. In [6], P is said to be a graded S-prime ideal of R if $s\in S$ exists such that if $xy\in P$, then $sx\in P$ or $sy\in P$ where $x,y\in h\left(R\right)$. Also, several properties of graded S-prime ideals have been examined and investigated in [7]. In this article, motivated by [5], we introduce the concept of graded S-primary ideals. We say that P is a graded S-primary ideal of R if there exists $s\in S$ such that for all $x,y\in h\left(R\right)$, if $xy\in P$, then $sx\in P$ or $sy\in Grad\left(P\right)$. Clearly, every S-primary ideal is graded S-primary, we prove that the converse is not necessarily true (Example 1). It is also evident that every graded primary ideal that is disjoint with S is graded S-primary, we prove that the converse is not necessarily true (Example 2). Note that if S consists of units of $h\left(R\right)$, then the notions of graded S-primary and graded primary ideal coincide. We investigate some basic properties of graded S-primary ideals. Indeed, our results are motivated by the interesting results proved in [5,6,7].

2. Graded S-Primary Ideals

In this section, we introduce the concept of graded S-primary ideals. We investigate some basic properties of graded S-primary ideals.

Definition 1.

Let R be a graded ring, $S\subseteq h\left(R\right)$ be a multiplicative set and P be a graded ideal of R such that $P\bigcap S=\varnothing$. We say that P is a graded S-primary ideal of R if there exists $s\in S$ such that for all $x,y\in h\left(R\right)$, if $xy\in P$, then $sx\in P$ or $sy\in Grad\left(P\right)$.

Clearly, every S-primary ideal is graded S-primary, but the converse is not necessarily true, check the following example that is raised from ([7], Example 2.2):

Example 1.

Consider $R=\mathbb{Z}\left[i\right]$ and $G={\mathbb{Z}}_2$. Then R is G-graded by $R_0=\mathbb{Z}$ and $R_1=i\mathbb{Z}$. Consider the graded ideal $I=5R$ of R. We show that I is a graded prime ideal of R. Let $xy\in I$ for some $x,y\in h\left(R\right)$.

Case (1): $x,y\in R_0$. In this case, $x,y\in \mathbb{Z}$ such that 5 divides $xy$, and then either 5 divides x or 5 divides y as 5 is a prime, which implies that either $x\in I$ or $y\in I$.

Case (2): $x,y\in R_1$. In this case, $x=ia$ and $y=ib$ for some $a,b\in \mathbb{Z}$ such that 5 divides $xy=-ab$, and then 5 divides $ab$ in $\mathbb{Z}$, and again either 5 divides a or 5 divides b, which implies that either 5 divides $x=ia$ or 5 divides $y=ib$, and hence either $x\in I$ or $y\in I$.

Case (3): $x\in R_0$ and $y\in R_1$. In this case, $x\in \mathbb{Z}$ and $y=ib$ for some $b\in \mathbb{Z}$ such that 5 divides $xy=ixb$ in R, that is $ixb=5(\alpha +i\beta )$ for some $\alpha ,\beta \in \mathbb{Z}$, which gives that $xb=5\beta$, that is 5 divides $xb$ in $\mathbb{Z}$, and again either 5 divides x or 5 divides b, and then either 5 divides x or 5 divides $y=ib$ in R, and hence either $x\in I$ or $y\in I$.

So, I is a graded prime ideal of R. Consider the graded ideal $P=10R$ of R and the multiplicative subset $S=\left\{2^n:nisanon\text{-}negativeinteger\right\}$ of $h\left(R\right)$. We show that P is a graded S-prime ideal of R. Note that $P\bigcap S=\varnothing$. Let $xy\in P$ for some $x,y\in h\left(R\right)$. Then 10 divides $xy$ in R. Then $xy\in I$, and then $x\in I$ or $y\in I$ as I is graded prime, which implies that $2x\in P$ or $2y\in P$. Therefore, P is a graded S-prime ideal of R, and hence P is a graded S-primary ideal of R.

On the other hand, P is not an S-primary ideal of R since $3-i,3+i\in R$ with $(3-i)(3+i)\in P$, ${\left(s(3-i)\right)}^n\notin P$ and ${\left(s(3+i)\right)}^n\notin P$ for each $s\in S$ and positive integer n.

It is obvious that every graded primary ideal that is disjoint with S is graded S-primary, but the converse is not necessarily true, check the next example. In fact, if S consists of units of $h\left(R\right)$, then the notions of graded primary and graded S-primary ideals coincide. The next example is motivated by ([7], Example 2.3).

Example 2.

Consider $R=\mathbb{Z}\left[X\right]$ and $G=\mathbb{Z}$. Then R is G-graded by $R_j=\mathbb{Z}{X}^j$ for $j\ge 0$ and $R_j=\left\{0\right\}$ otherwise. Consider the graded ideal $P=9XR$ of R and the multiplicative subset $S=\left\{9^n:nisanon\text{-}negativeinteger\right\}$ of $h\left(R\right)$. We show that P is a graded S-prime ideal of R. Note that $P\bigcap S=\varnothing$. Let $f\left(X\right)g\left(X\right)\in P$ for some $f\left(X\right),g\left(X\right)\in h\left(R\right)$. Then X divides $f\left(X\right)g\left(X\right)$, and X divides $f\left(X\right)$ or X divides $g\left(X\right)$, which implies that $9f\left(X\right)\in P$ or $9g\left(X\right)\in P$. Therefore, P is a graded S-prime ideal of R, hence that P is a graded S-primary ideal of R. On the other hand, P is not a graded primary ideal of R since $9,X\in h\left(R\right)$ with $9.X\in P$, $9^n\notin P$ and $X^n\notin P$ for each positive integer n.

Proposition 1.

Let R be a graded ring and $S\subseteq h\left(R\right)$ be a multiplicative set. If P is a graded S-primary ideal of R, then $Grad\left(P\right)$ is a graded S-prime ideal of R.

Proof. 

Since $P\bigcap S=\varnothing$, $Grad\left(P\right)\bigcap S=\varnothing$. Let $x,y\in h\left(R\right)$ such that $xy\in Grad\left(P\right)$. Then ${\left(xy\right)}^n=x^n{y}^n\in P$ for some positive integer n, and then there exists $s\in S$ such that $s{x}^n\in P$ or $s{y}^n\in Grad\left(P\right)$, which implies that $sx\in Grad\left(P\right)$ or $sy\in Grad\left(Grad\right(P\left)\right)=Grad\left(P\right)$. Therefore, $Grad\left(P\right)$ is a graded S-prime ideal of R. □

The next lemma is inspired by Example 2.

Lemma 1.

Let R be an integral domain. Suppose that R is a graded ring, $a,b\in h\left(R\right)$ such that $Ra$ is a nonzero graded prime ideal of R and $Rb$ is a graded primary ideal of R. If $Rb⊈Ra$ and $S=\left\{b^n:nisanon\text{-}negativeinteger\right\}$. Then $P=Rab$ is a graded S-prime ideal of R which is not graded primary.

Proof. 

Firstly, we show that $a\notin P$. If $a\in P$, then $a=rab$ for some $r\in R$, and then $a(1-rb)=0$, which implies that $a=0$ or $1=rb$, and then $Ra=\left\{0\right\}$ or b is a unit, which is a contradiction in both cases. Secondly, we show that $P\bigcap S=\varnothing$. If $x\in P\bigcap S$, then $x=b^n\in Ra$ for some non-negative integer n, and then $b\in Ra$ as $Ra$ is graded prime, and so $Rb\subseteq Ra$, which is a contradiction. Now, let $x,y\in h\left(R\right)$ such that $xy\in P$. Then $xy\in Ra$, and then $x\in Ra$ or $y\in Ra$, so $s=b\in S$ such that $sx\in P$ or $sy\in P$. Therefore, P is a graded S-prime ideal of R. On the other hand, $a,b\in h\left(R\right)$ such that $ab\in P$ and $a\notin P$. If $b\in Grad\left(P\right)$, then $b^n\in P$ for some positive integer n, which yields that $b^n\in P\bigcap S$, which is a contradiction. Therefore, P is not a graded primary ideal of R. □

Remark 1.

In Example 2, $\langle X\rangle$ is a nonzero graded prime ideal of R and $\langle 9\rangle$ is a graded primary ideal of R with $\langle 9\rangle ⊈\langle X\rangle$. So, by Lemma 1, $P=\langle 9X\rangle =9XR$ is a graded S-prime ideal of R which is not graded primary, where $S=\left\{9^n:nisanon\text{-}negativeinteger\right\}$.

Proposition 2.

Let R be a graded ring, $S\subseteq h\left(R\right)$ be a multiplicative set and P be a graded ideal of R such that $P\bigcap S=\varnothing$. Then P is a graded S-primary ideal of R if and only if $(P:s)$ is a graded primary ideal of R for some $s\in S$.

Proof. 

Suppose that P is a graded S-primary ideal of R. Then there exists $s\in S$ such that whenever $x,y\in h\left(R\right)$ with $xy\in P$, then either $sx\in P$ or $sy\in Grad\left(P\right)$. We show that $Grad\left((P:s)\right)=Grad\left((P:s^n)\right)$ for all positive integer n. Let n be a positive integer. Then $(P:s)\subseteq (P:s^n)$, and then $Grad\left((P:s)\right)\subseteq Grad\left((P:s^n)\right)$. Let $x\in Grad\left((P:s^n)\right)$. Then $x_g\in Grad\left((P:s^n)\right)$ for all $g\in G$ as the graded radical is a graded ideal, and then there exists a positive integer k such that $x_g^k{s}^n\in P$ for all $g\in G$. If $s^{n+1}\in Grad\left(P\right)$, then $s^{(n+1)m}\in P\bigcap S$ for some positive integer m, which is a contradiction. So, $s{x}_g^k\in P$ for all $g\in G$, and hence $x_g\in Grad\left((P:s)\right)$ for all $g\in G$, so $x\in Grad\left(\right(P:s\left)\right)$. Therefore, $Grad\left((P:s)\right)=Grad\left((P:s^n)\right)$. Now, let $x,y\in h\left(R\right)$ such that $xy\in (P:s)$. Then $sxy\in P$, and then $s^{2}x\in P$ or $sy\in Grad\left(P\right)$. If $s^{2}x\in P$, then as $s^3\notin Grad\left(P\right)$, we have $sx\in P$, which means that $x\in (P:s)$. If $sy\in Grad\left(P\right)$, then ${\left(sy\right)}^n=s^n{y}^n\in P$ for some positive integer n, and then $y\in Grad\left((P:s^n)\right)=Grad\left((P:s)\right)$. Hence, $(P:s)$ is a graded primary ideal of R. Conversely, assume that $(P:s)$ is a graded primary ideal of R for some $s\in S$. Let $x,y\in h\left(R\right)$ such that $xy\in P\subseteq (P:s)$. Then $x\in (P:s)$ or $y\in Grad\left(\right(P:s\left)\right)$. Therefore, either $sx\in P$ or $sy\in Grad\left(P\right)$. This shows that P is a graded S-primary ideal of R. □

Proposition 3.

Let R be a graded ring and $S\subseteq h\left(R\right)$ be a multiplicative set. Suppose that P is a graded primary ideal of R with $P\bigcap S=\varnothing$. Then for any $s\in S$, $sP$ is a graded S-primary ideal of R. Moreover, If $P\ne \left\{0\right\}$ and ${\displaystyle \bigcap _{n=1}^{\infty }R{s}^n=\left\{0\right\}}$, then $sP$ is not a graded primary ideal of R.

Proof. 

Let $s\in S$ and $I=sP$. As $I\subseteq P$ and $P\bigcap S=\varnothing$, it follows that $I\bigcap S=\varnothing$. Since P is a graded primary ideal of R with $Grad\left(P\right)\bigcap S=\varnothing$, we get that $(I:s)=P$. Consequently, $(I:s)$ is a graded primary ideal of R. Therefore, we obtain from Proposition 2 that $I=sP$ is a graded S-primary ideal of R. Moreover, assume that $P\ne \left\{0\right\}$ and ${\displaystyle \bigcap _{n=1}^{\infty }R{s}^n=\left\{0\right\}}$. if $P=sP$, then $P=s^{n}P$ for each $n\ge 1$. From ${\displaystyle \bigcap _{n=1}^{\infty }R{s}^n=\left\{0\right\}}$, it follows that $P=\left\{0\right\}$, which is a contradiction. In consequence, $P\ne sP$. So, there exists $x\in P-sP$, and then $x_g\notin sP$ for some $g\in G$. Note that $x_g\in P$ as P is a graded ideal. Hence, $s{x}_g\in sP=I$ with $x_g\notin I$ and $s\notin Grad\left(I\right)$. Therefore, $I=sP$ is not a graded primary ideal of R. □

Proposition 4.

Allow R to be a graded ring and $S\subseteq h\left(R\right)$ be a multiplicative set. Suppose that $n\ge 1$, $i\in \{1,\dots ,n\}$ and $P_i$ is a graded ideal of R with $P_i\bigcap S=\varnothing$. If $P_i$ is a graded S-primary ideal of R for each i with $Grad\left(P_i\right)=Grad\left(P_j\right)$ for all $i,j\in \{1,\dots ,n\}$, then ${\displaystyle \bigcap _{i=1}^n{P}_i}$ is a graded S-primary ideal of R.

Proof. 

Since $P_i$ is a graded S-primary ideal of R, there exists $s_i\in S$ to this extent for all $x,y\in h\left(R\right)$ with $xy\in P_i$, we have either $s_{i}x\in P_i$ or $s_{i}y\in Grad\left(P_i\right)$. Let ${\displaystyle s=\prod _{i=1}^n{s}_i}$. Then $s\in S$. Assume that $x,y\in h\left(R\right)$ in such a way ${\displaystyle xy\in \bigcap _{i=1}^n{P}_i}$ and ${\displaystyle sx\notin \bigcap _{i=1}^n{P}_i}$. Then $sx\notin P_k$ for some $1\le k\le n$, and then $s_{k}x\notin P_k$. Seeing as $xy\in P_k$, $s_{k}y\in Grad\left(P_k\right)$. Therefore, $sy\in Grad\left(P_k\right)$. By assumption, $Grad\left(P_1\right)=Grad\left(P_i\right)$ for all $1\le i\le n$. Thus ${\displaystyle sy\in Grad\left(P_1\right)=\bigcap _{i=1}^{n}Grad\left(P_i\right)=Grad\left({\displaystyle \bigcap _{i=1}^n{P}_i}\right)}$. Therefore, ${\displaystyle \bigcap _{i=1}^n{P}_i}$ is a graded S-primary ideal of R.  □

Recall that if R is a G-graded ring and $S\subseteq h\left(R\right)$ is a multiplicative set, then $S^{-1}R$ is a G-graded ring with ${\left(S^{-1}R\right)}_g=\left\{\frac{a}{s},a\in R_h,s\in S\cap R_{h{g}^{-1}}\right\}$ for all $g\in G$. In addition, if I is a graded ideal of R, then $S^{-1}I$ is a graded ideal of $S^{-1}R$ [2].

Lemma 2.

Let R be a graded ring and P be a graded ideal of R. If P is a graded prime ideal of R, then $S^{-1}P$ is a graded prime ideal of $S^{-1}R$.

Proof. 

Let $x,y\in h\left(R\right)$ and $s_1,s_2\in S$ in such wise $\frac{x}{s_1}\frac{y}{s_2}\in S^{-1}P$. Then there exists $s_3\in S$ such that $s_{3}xy\in P$, and $s_{3}x\in P$ or $y\in P$. If $s_{3}x\in P$, subsequently $\frac{x}{s_1}=\frac{s_{3}x}{s_3{s}_1}\in S^{-1}P$. If $y\in P$, then $\frac{y}{s_2}\in S^{-1}P$. Thereupon, $S^{-1}P$ is a graded prime ideal of $S^{-1}R$.  □

By ([4], Lemma 1.8), if P is a graded primary ideal of R, then $Q=Grad\left(P\right)$ is a graded prime ideal of R, and we say that P is a graded Q-primary ideal of R.

Lemma 3.

Allow R to be a graded ring and P be a graded ideal of R. If P is a graded Q-primary ideal of R, then $S^{-1}P$ is a graded $S^{-1}Q$-primary ideal of $S^{-1}R$.

Proof. 

Let $x,y\in h\left(R\right)$ and $s_1,s_2\in S$ such that $\frac{x}{s_1}\frac{y}{s_2}\in S^{-1}P$. Then there exists $s_3\in S$ such that $s_{3}xy\in P$, then $s_{3}x\in P$ or $y\in Grad\left(P\right)$. If $s_{3}x\in P$, then $\frac{x}{s_1}=\frac{s_{3}x}{s_3{s}_1}\in S^{-1}P$. If $y\in Grad\left(P\right)$, then $\frac{y}{s_2}\in S^{-1}Grad\left(P\right)=Grad\left(S^{-1}P\right)$ by ([8], Proposition 3.11 (v)). Therefore, $S^{-1}P$ is a graded primary ideal of $S^{-1}R$. Note that, $Grad\left(S^{-1}P\right)=S^{-1}Grad\left(P\right)=S^{-1}Q$ which is a graded prime ideal of $S^{-1}R$ by Lemma 2. Thereupon, $S^{-1}P$ is a graded $S^{-1}Q$-primary ideal of $S^{-1}R$.  □

Proposition 5.

Let R be a graded ring and $S\subseteq h\left(R\right)$ be a multiplicative set. Suppose that P is a graded ideal of R with $P\bigcap S=\varnothing$. Then P is a graded S-primary ideal of R if and only if $S^{-1}P$ is a graded primary ideal of $S^{-1}R$ and $SP=(P:s)$ for some $s\in S$.

Proof. 

Suppose that P is a graded S-primary ideal of R. Then there exists $s\in S$ in such a manner for all $x,y\in h\left(R\right)$ with $xy\in P$, we have either $sx\in P$ or $sy\in Grad\left(P\right)$. Considering $P\bigcap S=\varnothing$, $S^{-1}P\ne S^{-1}R$ by ([8], Proposition 3.11 (ii)). Allow $x,y\in h\left(R\right)$ and $s_1,s_2\in S$ such that $\frac{x}{s_1}\frac{y}{s_2}\in S^{-1}P$. Then there exists $s_3\in S$ such that $s_{3}xy\in P$, and then $s{s}_{3}x\in P$ or $sy\in Grad\left(P\right)$. If $s{s}_{3}x\in P$, then $\frac{x}{s_1}=\frac{s{s}_{3}x}{s{s}_3{s}_1}\in S^{-1}P$. If $sy\in Grad\left(P\right)$, then $\frac{y}{s_2}=\frac{sy}{s{s}_2}\in S^{-1}Grad\left(P\right)=Grad\left(S^{-1}P\right)$ by ([8], Proposition 3.11 (v)). Thus, $S^{-1}P$ is a graded primary ideal of $S^{-1}R$. Now, by Proposition 2, $(P:s)$ is a graded primary ideal of R for some $s\in S$. Clearly, $(P:s)\bigcap S=\varnothing$. On that account, $S\left(\right(P:s\left)\right)=(P:s)$ by Lemma 3. Also, by ([8], Corollary 3.15), $S^{-1}(P:s)=\left(S^{-1}P{:}_{S^{-1}R}\frac{s}{1}\right)$. Since $\frac{s}{1}\in U\left(S^{-1}R\right)$, $S^{-1}(P:s)=S^{-1}P$, and $S\left(\right(P:s\left)\right)=SP$, accordingly $SP=(P:s)$. Contrarily, if $S^{-1}P$ is a graded $S^{-1}Q$-primary ideal of $S^{-1}R$, then $SP$ is a graded Q-primary ideal of R. Hence, we get that $(P:s)$ is a graded primary ideal of R for some $s\in S$. Thence, we obtain by Proposition 2 that P is a graded S-primary ideal of R.  □

Theorem 1.

Let R be a graded ring, S be a multiplicative subset of $h\left(R\right)$ and P be a graded ideal of R such that $P\bigcap S=\varnothing$. Thus the following statements are equivalent:

1. 

P is a graded S-primary ideal of R.

2. 

$(P:s)$ is a graded primary ideal of R for some $s\in S$.

3. 

$S^{-1}P$ is a graded primary ideal of $S^{-1}R$ and $SP=(P:s)$ for some $s\in S$.

Proof. 

It follows from Propositions 2 and 5. □

Proposition 6.

Let R be a graded ring, S be a multiplicative subset of $h\left(R\right)$ and P be a graded ideal of R such that $P\bigcap S=\varnothing$. If P is a graded S-primary ideal of R, then the ascending sequence of graded ideals $(P:sr)\subseteq (P:s{r}^2)\subseteq (P:s{r}^3)\subseteq \dots$ is stationary for some $s\in S$ and for all $r\in h\left(R\right)$.

Proof. 

By Proposition 2, $(P:s)$ is a graded primary ideal of R for some $s\in S$. Let $r\in h\left(R\right)$. Suppose that $r\notin Grad\left(\right(P:s\left)\right)$. As $(P:s)$ is a graded primary ideal of R, it follows that for all positive integer n, $(P:s{r}^n)=(P:s)$. Assume that $r\in Grad\left(\right(P:s\left)\right)$. Then $s{r}^k\in P$ for some positive integer k. Hence, for all $j\ge k$, $(P:s{r}^j)=R$.  □

Proposition 7.

Let R be a graded ring, S be a multiplicative subset of $h\left(R\right)$ and P be a graded ideal of R such that $P\bigcap S=\varnothing$. If P is a graded S-primary ideal of R, then the ascending sequence of graded ideals $(P:r)\subseteq (P:r^2)\subseteq (P:r^3)\subseteq \dots$ is S-stationary for all $r\in h\left(R\right)$.

Proof. 

Let $r\in h\left(R\right)$. Now, there exists positive integer n such that for all $j\ge n$, $(P:s{r}^j)=(P:s{r}^n)$ for some $s\in S$ by Proposition 6. Let $j\ge n$ and $a\in (P:r^j)$. Then $sa{r}^j\in P$ so, $a\in (P:s{r}^j)=(P:s{r}^n)$. This implies that $sa\in (P:r^n)$. This proves that $s(P:r^j)\subseteq (P:r^n)$ for all $j\ge n$. Wherefore, the ascending sequence of graded ideals $(P:r)\subseteq (P:r^2)\subseteq (P:r^3)\subseteq ...$ is S-stationary for all $r\in h\left(R\right)$. □

Remark 2.

Let R be a graded ring that is not graded local, $S=U\left(R\right)$, $X_1,X_2$ be two distinct graded maximal ideals of R and $P=X_1\bigcap X_2$. Presume $r\in h\left(R\right)$. Then for any positive integer n, $(P:r^n)=(X_1:r^n)\bigcap (X_2:r^n)$. For $i=1,2$, if $r\in X_i$, then $(X_i:r^n)=R$ for all positive integer n, and if $r\notin X_i$, then $(X_i:r^n)=X_i$ for all positive integer n. As a result, the ascending sequence of graded ideals $(P:r)\subseteq (P:r^2)\subseteq (P:r^3)\subseteq \dots$ is stationary, but P is not a graded primary ideal of R.

Let R be a G-graded ring. Then R is said to be a graded von Neumann regular ring if for each $a\in R_g$ ($g\in G$), there exists $x\in R_{g^{-1}}$ such that $a=a^{2}x$ [9].

Proposition 8.

Let R be a graded von Neumann regular ring and I be a graded ideal of R. Then $Grad\left(I\right)=I$.

Proof. 

Clearly, $I\subseteq Grad\left(I\right)$. Let $a\in Grad\left(I\right)$. Then $a_g\in Grad\left(I\right)$ for all $g\in G$ as $Grad\left(I\right)$ is a graded ideal. Suppose that $g\in G$. Then $a_g^n\in I$ for some positive integer n. Since R is graded von Neumann regular, there exists $x\in R_{g^{-1}}$ such that $a_g=a_g^{2}x$. Hence, $R{a}_g=R{a}_g^2$. So, $R{a}_g=R{a}_g^n\subseteq I$ and so, $a_g\in I$ for all $g\in G$, and hence $a\in I$. This proves that $Grad\left(I\right)\subseteq I$ and so, $I=Grad\left(I\right)$.  □

Corollary 1.

Let R be a graded von Neumann regular ring and I be a graded ideal of R. Then I is a graded prime ideal of R if and only if I is a graded primary ideal of R.

Proof. 

Apply Proposition 8.  □

Theorem 2.

Let R be a graded ring, S be a multiplicative subset of $h\left(R\right)$ and P be a graded ideal of R such that $P\bigcap S=\varnothing$. Suppose that $S^{-1}R$ is graded von Neumann regular. Then P is a graded S-prime ideal of R if and only if P is a graded S-primary ideal of R.

Proof. 

Suppose that P is a graded S-primary ideal of R. By Proposition 5, $S^{-1}P$ is a graded primary ideal of $S^{-1}R$ and $SP=(P:s)$ for some $s\in S$. Since $S^{-1}R$ is graded von Neumann regular, we get that $S^{-1}P$ is a graded prime ideal of $S^{-1}R$ by Corollary 1. Thereupon, $SP$ is a graded prime ideal of R. As $SP=(P:s)$, we obtain that $(P:s)$ is a graded prime ideal of R for some $s\in S$. Therefore, it follows from ([7], Proposition 2.4) that P is a graded S-prime ideal of R. The converse is clear.  □

3. Graded Strongly S-Primary Ideals

In this section, we introduce and study the concept of graded strongly S-primary ideals. We examine some basic properties of graded strongly S-primary ideals.

Definition 2.

1. 

Let R be a graded ring and P be a graded primary ideal of R. Then P is said to be a graded strongly primary ideal of R if ${\left(Grad\left(P\right)\right)}^n\subseteq P$ for some $n\in \mathbb{N}$.

2. 

Let R be a graded ring, $S\subseteq h\left(R\right)$ be a multiplicative set and P be a graded S-primary ideal of R. Then P is said to be a graded strongly S-primary ideal of R if there exist $s^{\prime }\in S$ and $n\in \mathbb{N}$ such that $s^{\prime }{\left(Grad\left(P\right)\right)}^n\subseteq P$.

Proposition 9.

Let R be a graded ring and $S\subseteq h\left(R\right)$ be a multiplicative set. If P is a graded S-prime ideal of R, then P is a graded strongly S-primary ideal of R.

Proof. 

Since P is a graded S-prime ideal of R, $(P:s)$ is a graded prime ideal of R for some $s\in S$ by ([7], Proposition 2.4), and then $s\left(Grad\right(P\left)\right)\subseteq s\left(Grad\right((P:s)\left)\right)=s(P:s)\subseteq P$. Therefore, P is a graded strongly S-primary ideal of R.  □

Proposition 10.

Allow R to be a graded ring, $S\subseteq h\left(R\right)$ be a multiplicative set and P be a graded ideal of R such that $P\bigcap S=\varnothing$. Then P is a graded strongly S-primary ideal of R if and only if $(P:s)$ is a graded strongly primary ideal of R for some $s\in S$.

Proof. 

Suppose that P is a graded strongly S-primary ideal of R. Then there exist $s,s^{\prime }\in S$ and $n\in \mathbb{N}$ such that for all $x,y\in h\left(R\right)$ with $xy\in P$, we have either $sx\in P$ or $sy\in Grad\left(P\right)$ and $s^{\prime }{\left(Grad\left(P\right)\right)}^n\subseteq P$. Note that $s{s}^{\prime }\in S$, for all $x,y\in h\left(R\right)$ with $xy\in P$, we have either $s{s}^{\prime }x\in P$ or $s{s}^{\prime }y\in Grad\left(P\right)$ and $s{s}^{\prime }{\left(Grad\left(P\right)\right)}^n\subseteq P$. Hence, on replacing $s,s^{\prime }$ by $s{s}^{\prime }$, we can assume without loss of generality that $s=s^{\prime }$. Now, $(P:s)$ is a graded primary ideal of R by Proposition 2. Let $r\in Grad\left(\right(P:s\left)\right)$. Then $s{r}^m\in P$ for some $m\in \mathbb{N}$. Hence, $sr\in Grad\left(P\right)$. This implies that $s.Grad\left(\right(P:s\left)\right)\subseteq Grad\left(P\right)$. Take that $I=(P:s)$. Then $s^{n+1}{\left(Grad\left(I\right)\right)}^n\subseteq s{\left(Grad\left(P\right)\right)}^n\subseteq P\subseteq (P:s)$. As $s^{n+1}\notin Grad\left((P:s)\right)$ and $(P:s)$ is a graded primary ideal of R, we get that ${\left(Grad\left(I\right)\right)}^n\subseteq (P:s)=I$. This proves that $(P:s)$ is a graded strongly primary ideal of R. Contrariwise, take that $I=(P:s)$. Now, P is a graded S-primary ideal of R by Proposition 2 and there exists $n\in \mathbb{N}$ such that ${\left(Grad\left(I\right)\right)}^n\subseteq I=(P:s)$. As $P\subseteq I$, we get that ${\left(Grad\left(P\right)\right)}^n\subseteq {\left(Grad\left(I\right)\right)}^n\subseteq (P:s)$. This implies that $s{\left(Grad\left(P\right)\right)}^n\subseteq P$ and so, P is a graded strongly S-primary ideal of R.  □

Proposition 11.

Let R be a graded ring and $S\subseteq h\left(R\right)$ be a multiplicative set. Suppose that $n\ge 1$, $i\in \{1,\dots ,n\}$ and $P_i$ is a graded ideal of R with $P_i\bigcap S=\varnothing$. If $P_i$ is a graded strongly S-primary ideal of R for each i with $Grad\left(P_i\right)=Grad\left(P_j\right)$ for all $i,j\in \{1,\dots ,n\}$, then ${\displaystyle \bigcap _{i=1}^n{P}_i}$ is a graded strongly S-primary ideal of R.

Proof. 

It is already verified that ${\displaystyle \bigcap _{i=1}^n{P}_i}$ is a graded S-primary ideal of R by Proposition 4. Now, for each $i\in \{1,...,n\}$, there exist $s_i\in S$ and a positive integer $k_i$ such that $s_i{\left(Grad\left(P_i\right)\right)}^{k_i}\subseteq P_i$. As $Grad\left({\displaystyle \bigcap _{i=1}^n{P}_i}\right)=Grad\left(P_j\right)$ for all $j\in \{1,...,n\}$, it follows that $s{\left(Grad\left(I\right)\right)}^k\subseteq I$, where ${\displaystyle s=\prod _{i=1}^n{s}_i}$, ${\displaystyle I=\bigcap _{i=1}^n{P}_i}$ and $k=max\{t_1,...,t_n\}$. This proves that ${\displaystyle \bigcap _{i=1}^n{P}_i}$ is a graded strongly S-primary ideal of R. □

Proposition 12.

Let R be a graded ring and $S\subseteq h\left(R\right)$ be a multiplicative set. Intend that P is a graded ideal of R with $P\bigcap S=\varnothing$. Then P is a graded strongly S-primary ideal of R if and only if $S^{-1}P$ is a graded strongly primary ideal of $S^{-1}R$ and $SP=(P:s)$ for some $s\in S$.

Proof. 

Suppose that P is a graded strongly S-primary ideal of R. Then there exist $s\in S$ and $n\in \mathbb{N}$ such that for all $x,y\in h\left(R\right)$ with $xy\in P$, we have either $sx\in P$ or $sy\in Grad\left(P\right)$ and $s{\left(Grad\left(P\right)\right)}^n\subseteq P$. It is already verified that $S^{-1}P$ is a graded primary ideal of $S^{-1}R$ and $SP=(P:s)$ for some $s\in S$ by Proposition 5. Now, as $\frac{s}{1}\in U\left(S^{-1}R\right)$, it follows from ([8], Proposition 3.11 (v)) that ${\left(Grad\left(S^{-1}P\right)\right)}^n=S^{-1}\left(s{\left(Grad\left(P\right)\right)}^n\right)\subseteq S^{-1}P$. Hence, $S^{-1}P$ is a graded strongly primary ideal of $S^{-1}R$. Again, if $S^{-1}P$ is a graded strongly $S^{-1}Q$-primary ideal of $S^{-1}R$, then $SP$ is a graded strongly Q-primary ideal of R. Hence, we get that $(P:s)$ is a graded strongly primary ideal of R for some $s\in S$. Therefore, we obtain by Proposition 10 that P is a graded strongly S-primary ideal of R. □

Theorem 3.

Allow R to be a graded ring, S to be a multiplicative subset of $h\left(R\right)$ and P to be a graded ideal of R such that $P\bigcap S=\varnothing$. Then the following statements are equivalent:

1. 

P is a graded strongly S-primary ideal of R.

2. 

$(P:s)$ is a graded strongly primary ideal of R for some $s\in S$.

3. 

$S^{-1}P$ is a graded strongly primary ideal of $S^{-1}R$ and $SP=(P:s)$ for some $s\in S$.

Proof. 

It follows from Propositions 10 and 12. □

4. Conclusions

In this study, we introduced the concept of graded S-primary ideals which is a generalization of graded primary ideals. Furthermore, we introduced the concept of graded strongly S-primary ideals. We investigated some basic properties of graded S-primary ideals and graded strongly S-primary ideals. As a proposal to further the work on the topic, we are going to study the concepts of graded S-absorbing and graded S-absorbing primary ideals as a generalization of the concepts of graded absorbing and graded absorbing primary ideals.