S-Ideals: A Unified Framework for Ideal Structures via Multiplicatively Closed Subsets

1. Introduction

Throughout this paper, all rings are assumed to be commutative with identity. For such a ring R, we denote by $U\left(R\right)$ the set of units, by $\mathrm{reg}\left(R\right)$ the set of regular (i.e., non zero-divisor) elements, by $\mathrm{zd}\left(R\right)$ the set of zero divisors, by $N\left(R\right)$ the set of nilpotent elements, and by $J\left(R\right)$ the Jacobson radical of R. If A is a subset of R and I is an ideal of R, we define the ideal quotient

$$(I:A)=\{r\in R\mid rA\subseteq I\}.$$

In particular, the annihilator of A is defined by $\mathrm{Ann}\left(A\right):=(0:A)$.

Multiplicatively closed subsets of a commutative ring have long played a fundamental role in commutative algebra, particularly in the localization of rings and modules. Moreover, they play a central role in the structure theory of ideals, offering a flexible framework for defining and analyzing diverse classes of ideals. For instance, the m.c.s. of regular elements of a ring R is used to define r-ideals, where a proper ideal I of R satisfies the condition: if $ab\in I$ and $a\in \mathrm{reg}\left(R\right)$, then $b\in I$.

We aim in this paper to provide a general framework for defining ideals using arbitrary multiplicatively closed subsets of a ring. Specifically, we introduce the notion of S-ideals. For an arbitrary multiplicatively closed subset S, a proper ideal I of a ring R is called an S-ideal if for all $a,b\in R$, the condition $ab\in I$ and $a\in S$ implies $b\in I$. Equivalently, I is an S-ideal if and only if $S^{-1}I\cap R=I$.

In this work, we explore the theory of S-ideals in depth, examining their foundational properties, characterizations, relationships to other types of ideals, and their behavior under various ring constructions and extensions. Beyond r-ideals, which are defined with respect to the multiplicatively closed set $S=\mathrm{reg}\left(R\right)$, we show in this paper that several well-known classes of ideals naturally arise as S-ideals for suitable choices of multiplicatively closed subsets $S\subseteq R$. For instance, every P-primary ideal of R is an S-ideal when $S=R\setminus P$. More generally, any ideal that admits a primary decomposition can be viewed as an S-ideal where S is the complement of the finite union of the associated prime ideals.

Moreover, $z_0$-ideals, which are central in the theory of rings with zero-divisors, fit naturally within the S-ideal framework as well. Recall that an ideal $I\subseteq R$ is a $z_0$-ideal if, for each element $a\in I$, the intersection $P_a$ of all minimal prime ideals containing a is also contained in I. We show that these ideals can be considered as S-ideals with respect to the multiplicatively closed subset $S=R\setminus \bigcup \{\mathfrak{p}\mid \mathfrak{p}\mathrm{minimal}\mathrm{prime}\mathrm{of}R\}$. These examples demonstrate how varying the choice of multiplicatively closed subsets enables a unified and generalized approach to studying ideal structures across various classes.

In addition to connecting familiar ideal classes to suitable multiplicatively closed subsets, we explore how certain specific forms of S lead to meaningful characterizations of S-ideals. For example, when $S=1+J$ for an ideal J of R, or when S is the set of powers of a fixed non-unit element, we provide explicit descriptions of the resulting S-ideals. Similarly, when S consists of primitive polynomials in a polynomial ring over R, or other carefully structured multiplicatively closed subsets, we are able to determine necessary and sufficient conditions for an ideal of R to be an S-ideal. These cases illustrate the flexibility of the S-ideal framework and highlight its potential for unifying diverse ideal-theoretic behaviors under a common generalization.

Section 2 of the paper is devoted to presenting several fundamental properties and characterizations of S-ideals. In addition to foundational results highlighted earlier, Proposition 1 provides a concise summary of key properties and includes a wide range of illustrative examples. One of the notable results we establish is that every maximal S-ideal of a ring R is prime (Proposition 4). Furthermore, we generalize the classical Prime Avoidance Lemma to the broader setting of multiplicatively closed subsets, offering a new version tailored to S-ideals (Corollary 1). When $S\subseteq \mathrm{reg}\left(R\right)$, we explore connections between S-ideals and von Neumann regularity in the ring of fractions $S^{-1}R$. In particular, we provide two distinct characterizations of this regularity: one in terms of semiprime S-ideals and the other via minimal prime S-ideals (Theorems 2 and 3). We also identify rings where $\left\{0\right\}$ is the only S-ideal and provide a precise characterization in Proposition 6. For an arbitrary multiplicatively closed set S, Theorem 5 describes the structure of nonzero S-ideals within GCD-domains.

In the final section, we examine the behavior of S-ideals under various ring-theoretic constructions and extensions. These include localizations (Theorem 9), homomorphic images or quotient rings (Proposition 9), and direct products of rings (Proposition 10). We also study S-ideals in more advanced constructions such as idealization rings (Proposition 13) and the amalgamation of rings along an ideal, with key results captured in Theorems 10 and 11.

2. S-Ideals and Their Foundational Properties

In this section, we develop the foundational theory of S-ideals in commutative rings. We provide key definitions, explore characterizations, and relate S-ideals to known classes such as prime, semiprime and r-ideals. Several properties are examined with respect to different types of multiplicatively closed subsets. We also characterize when the localization $S^{-1}R$ is von Neumann regular in terms of S-ideals. These results unify and extend existing notions like r-ideals under a broader framework.

Definition 1. 

Let I be a proper ideal of a ring R, and let S be a multiplicatively closed subset of R. We say that I is an S-ideal if for all $a,b\in \mathbb{R}$, the conditions $ab\in I$ and $a\in S$ imply that $b\in I$.

If I is an S-ideal of a ring R, then $I\cap S=\varnothing$. Indeed if there is $s\in I\cap S$, then $s·1\in I$ and $s\in S$ imply $1\in I$, a contradiction. However, the converse is not true in general. For example, choose $S=\{2^k:k\ge 0\}\subseteq \mathbb{Z}$ and $I=6\mathbb{Z}$. Then $I\cap S=\varnothing$ but I is not an S-ideal of $\mathbb{Z}$ since $2·3=6\in I$ and $2\in S$ but $3\notin I$. It is clear that if $S\subseteq T$ are multiplicatively closed subsets of a ring R, then any T-ideal of R is an S-ideal.

We begin with the following elementary result, whose proof is straightforward and is left to the reader.

Proposition 1. 

Let S be a multiplicatively closed subset of a ring R.

• The class of $reg\left(R\right)$-ideals is the same as the class of r-ideals.
• Every proper ideal in a ring R is an S-ideal if and only if $S=U\left(R\right)$.
• The ideal $\left\{0\right\}$ of R is an S-ideal if and only if $S\subseteq reg\left(R\right)$.
• If I is an S-ideal of R, then so is $\sqrt{I}$.
• The intersection of any family of S-ideals is an S-ideal.
• If I is an S-ideal and $A\subseteq R\setminus I$, then $(I:A)$ is an S-ideal. In particular, if $S\subseteq reg\left(R\right)$, then $Ann\left(A\right)$ is always an S-ideal.
• If $S\subseteq reg\left(R\right)$, then every von Neumann regular ideal is an S-ideal.
• If I is a proper ideal of R, then I is an S-ideal if and only if for ideals K and J of R, $KJ\subseteq I$ and $K\cap S\ne \varnothing$ imply $J\subseteq I$.
• If K is an ideal of R with $K\cap S\ne \varnothing$ and I, J are S-ideals such that $IK=JK$, then $I=J$.
• If I and J are ideals of R such that $J\cap S\ne \varnothing$ and $IJ$ is an S-ideal, then $I=IJ$.
• A prime (resp. primary) ideal P of R is an S-ideal if and only if $P\cap S=\varnothing$.

By using Zorn’s Lemma, one can prove that if I is an ideal of a ring R such that $I\cap S=⌀$, then there exists a prime ideal P of R such that $I\subseteq P$ and $P\cap S=⌀$. Thus, every ideal disjoint from S is contained in a prime S-ideal.

Remark 1. 

Any ideal I of a ring R is an S-ideal with respect to some multiplicatively closed subset of R. Indeed, the set $S:=\{a\in R\mid ab\in I$$⟹b\in I$$\forall b\in R\}$ is the largest multiplicatively closed set for which I is an S-ideal.

Let R be a commutative ring and S a multiplicatively closed subset of R. For any ideal I of R, the S-compoment of I is defined as:

$$I_S:=\{r\in R\mid \exists s\in S\mathrm{such}\mathrm{that}sr\in I\}.$$

It is clear that $I_S$ is an ideal of R containing I and it is well-known that $I_S=S^{-1}I\cap R$,8]

Proposition 2. 

Let S be a multiplicatively closed subset of a ring R and I be an ideal of R disjoint with S. Then $I_S$ is the smallest S-ideal of R containing I.

Proof. 

Let $a,b\in R$ such that $ab\in I_S$ and $a\in S$. Then there is $s\in S$ such that $\left(sa\right)b=s\left(ab\right)\in I$. Hence, $b\in I_S$ and $I_S$ is an S-ideal of R. Now, let J be an S-ideal such that $I\subseteq J$. Take $r\in I_S$ so that there is $s\in S$such that $sr\in I\subseteq J$. Since J is an S-ideal and $s\in S$, then $r\in J$. Hence $I_S\subseteq J$, and $I_S$ is the smallest S-ideal containing I. □

Let I be an idealof a ring R and S be a multiplicatively closed subset of R. Following [11], the S-radical of I is defined by

$$\sqrt[S]{I}=\{a\in R:s{a}^n\in I\mathrm{for}\mathrm{some}s\in S\mathrm{and}n\in \mathbb{N}\},$$

Which is an ideal of R containing $\sqrt{I}$. We can easily conclude that $\sqrt[S]{I}={\left(\sqrt{I}\right)}_S$ and so $\sqrt[S]{I}$ is an S-ideal of R.

For a multiplicatively closed subset S of a ring R, the following theorem provides a useful characterization of S-ideals in R.

Theorem 1. 

Let R be a ring, I a proper ideal of R, and S a multiplicatively closed subset of R. The following are equivalent:

• I is an S-ideal of R.
• $(I:s)=I$ for every $s\in S$.
• $I_S=S^{-1}I\cap R=I$.

Proof. $\left(1\right)⟹\left(2\right)$ Assume I is an S-ideal and let $s\in S$. If $b\in (I:s)$. Then $sb\in I$ and so $b\in I$ as I is an S-ideal. Thus, $(I:s)\subseteq I$ and so $(I:s)=I$ as the reverse containment is obvious.

$\left(2\right)⟹\left(3\right)$ Assume that for every $s\in S$, $(I:s)=I$ and let $b\in I_S$ so that there exists $s\in S$ such that $sb\in I$. Then $b\in (I:s)=I$. Therefore, $I_S\subseteq I$ and since clearly $I\subseteq I_S$, we get $I_S=I$.

$\left(3\right)⟹\left(1\right)$ Suppose $I_S=I$ and let $a,b\in R$ such that $ab\in I$ and $a\in S$. Then $b\in I_S=I$ and I is an S-ideal of R. □

Recall that for an ideal I of a ring R, $Min\left(I\right)$ denotes the set of minimal prime ideals of R that contain I. The following proposition shows that minimal prime ideals over an S-ideal are themselves S-ideals.

Proposition 3. 

Let S be a multiplicatively closed subset of a ring R and $P\in Min\left(I\right)$, where I is an S-ideal of R. Then P is an S-ideal.

Proof. 

Let $a,b\in R$ such that $ab\in P$ and $a\in S$. by [7] [Theorem 2.1], there exist $x\notin P$ and $n\in \mathbb{N}$ such that $x{\left(ab\right)}^n=x{a}^n{b}^n\in I$ . Since $a^n\in S$ and I is an S-ideal, we infer that $x{b}^n\in I\subseteq P$. It follows that $b^n\in P$ as $x\notin P$. Thus $b\in P$ and P is is an S-ideal of R. □

By Proposition 1 (3), we conclude that whenever $S\subseteq reg\left(R\right)$, then any minimal prime ideal of R is an S-ideal.

It is important to note that the conclusion of the preceding proposition may fail if the prime ideal P is not a minimal prime over I, as demonstrated in the following example.

Example 1. 

Consider the non-prime ideal $I=〈x^2〉$ in $\mathbb{Z}\left[x\right]$. Then the prime ideal $P=〈2,x〉$ is not minimal over I. Consider the multiplicatively closed subset $S=\left\{f\in \mathbb{Z}\left[x\right]:f\left(0\right)\ne 0\right\}$ of $\mathbb{Z}\left[x\right]$. Then P is not an S-ideal since clearly, $P\cap S\ne \varnothing$. We prove that I is an S-ideal. Let $f={\sum }_{j=0}^m{a}_j{x}^j\in S$ and $g={\sum }_{i=0}^n{b}_i{x}^i\in \mathbb{Z}\left[x\right]$ such that $fg\in I=〈x^2〉$. Then $a_0{b}_0=0$, $a_0{b}_1+a_1{b}_0=0$ and $a_0\ne 0$ imply $b_0=b_1=0$. Therefore, $g\in I$ and I is an S-ideal of $\mathbb{Z}\left[x\right]$.

Next, we prove that maximal S-ideals of a ring R are prime.

Proposition 4. 

Let S be a multiplicatively closed subset of a ring R. Then any maximal S-ideal of R is prime.

Proof. 

Suppose I is a maximal S-ideal of R. Let $a,b\in R$ such that $ab\in I$ and $a\notin I$. Then the ideal $(I:a)$ is proper in R and is an S-ideal by Proposition 1 (6). Since $I\subseteq (I:a)$ and I is a maximal S-ideal, then $b\in (I:a)=I$ and I is a prime ideal of R. □

Let $B\subseteq {\bigcup }_{i\in I}{A}_i$, where B and each $A_i$ are subsets of a ring R. Recall that this inclusion is said to be irreducible if no $A_i$ can be removed from the union; that is, for every $i\in I$, we have $B⊈{\bigcup }_{j\in I\setminus \left\{i\right\}}{A}_j$.

In the following, we generalize [13] [Theorem 3.8] concerning irreducible coverings of ideals by replacing the set of regular elements with an arbitrary multiplicatively closed set.

Proposition 5. 

Let $I\subseteq {\bigcup }_{i=1}^n{J}_i$ be an irreducible inclusion of ideals in a ring R and let $S\subseteq R$ be a multiplicatively closed set. Suppose that $J_1$ is an S-ideal and that $J_i\cap S\ne \varnothing$ for all $i=2,\dots ,n$. Then $I\subseteq J_1$.

Proof. 

Since the inclusion is irreducible, we have $I⊈{\bigcup }_{i=2}^n{J}_i$, so there exists an element $a\in I\setminus {\bigcup }_{i=2}^n{J}_i$. In particular, $a\in J_1$. Now take any $x\in I\cap {\bigcap }_{i=2}^n{J}_i$; then $x+a\in I$, but $x+a\notin {\bigcup }_{i=2}^n{J}_i$ because otherwise we would have $a\in {\bigcup }_{i=2}^n{J}_i$, contradicting the choice of a. Therefore, $x+a\in J_1$, and since $a\in J_1$, it follows that $x\in J_1$. Hence, $I\cap {\bigcap }_{i=2}^n{J}_i\subseteq J_1$, and so $I·{\prod }_{i=2}^n{J}_i\subseteq J_1$. Now, since each $J_i\cap S\ne \varnothing$, we conclude that ${\prod }_{i=2}^n{J}_i\cap S\ne \varnothing$. Since $J_1$ is an S-ideal, then by Proposition 1 (8), we conclude that $I\subseteq J_1$, as required. □

The following corollary is an interesting variant of the Prime Avoidance Lemma, generalized to an arbitrary multiplicatively closed set.

Corollary 1. 

Let $Q\subseteq {\bigcup }_{i=1}^n{P}_i$, where Q and each $P_i$ are ideals of a ring R, and assume the inclusion is irreducible. Let S be a multiplicatively closed subset of R. Suppose that $P_1$ is an S-ideal and that $P_i\cap S\ne \varnothing$ for all $i\ge 2$. Then $Q\subseteq P_1$.

In particular, if in the above irreducible inclusion, $S\subseteq reg\left(R\right)$ and $P_1\in Min\left(R\right)$, then $Q=P_1$ and $Q\in Min\left(R\right)$.

In the following theorem, we provide a characterization of the von Neumann regularity of the localization $S^{-1}R$ for an arbitrary multiplicatively closed subset $S\subseteq reg\left(R\right)$. This result extends the corresponding characterization for the case $S=reg\left(R\right)$, which appears in [13] [Proposition 3.6] in the literature concerning r-ideals.

Theorem 2. 

Let R be a commutative ring and $S\subseteq reg\left(R\right)$ a multiplicatively closed subset. Then $S^{-1}R$ is a von Neumann regular ring if and only if every proper S-ideal of R is semiprime.

Proof. 

⇒) Suppose $S^{-1}R$ is a von Neumann regular ring. Let I be a proper S-ideal of R, and let $x\in R$ such that $x^2\in I$. Then ${\displaystyle \frac{x^2}{1}}\in S^{-1}I$ and since $S^{-1}R$ is von Neumann regular, there exists ${\displaystyle \frac{y}{s}}\in S^{-1}R$ such that ${\displaystyle \frac{x}{1}}={\left({\displaystyle \frac{x}{1}}\right)}^2·{\displaystyle \frac{y}{s}}$. Thus, ${\displaystyle \frac{x}{1}}\in S^{-1}I$, which implies there exists $s\in S$ such that $sx\in I$. Since I is an S-ideal and $s\in S$, it follows that $x\in I$. Hence I is semiprime.

⇐) Conversely, suppose every proper S-ideal of R is semiprime. Let ${\displaystyle \frac{x}{s}}\in S^{-1}R$. Let $I=\left(x^2\right)$ and consider the ideal $I_S$. If $I\cap S\ne \varnothing$, then $r{x}^2\in S$ for some $r\in R$. Thus, ${\displaystyle \frac{x}{s}}={\displaystyle \frac{r{x}^3}{sr{x}^2}}$ is a unit in $S^{-1}R$ and so is a von Neumann regular element in $S^{-1}R$. Suppose $I\cap S=\varnothing$. Then $I_S$ is an S-ideal of R with $x^2\in I\subseteq I_S$. By assumption, $I_S$ is semiprime in R and so $x\in I_S$. Hence, there exists $s\in S$ such that $sx\in I$, i.e., $sx=x^{2}r$ for some $r\in R$. Then in $S^{-1}R$, ${\displaystyle \frac{x}{s}}={\displaystyle \frac{x^{2}r}{s^2}}={\left({\displaystyle \frac{x}{s}}\right)}^2·{\displaystyle \frac{r}{1}}$ and so again ${\displaystyle \frac{x}{s}}$ is a von Neumann regular element in $S^{-1}R$. Hence $S^{-1}R$ is a von Neumann regular ring. □

It is well known that in a von Neumann regular ring, for any prime ideal P containing an element x, the annihilator of x is contained in P. We recall the result established in [13] [Proposition 3.5], where the author characterizes the von Neumann regularity of the total ring of quotients $Q\left(R\right)$ in terms of r-ideals. In the following theorem, we extend this result to a more general setting by replacing $reg\left(R\right)$ with an arbitrary multiplicatively closed subset $S\subseteq reg\left(R\right)$.

Theorem 3. 

Let R be a reduced ring, and let $S\subseteq reg\left(R\right)$ be a multiplicatively closed subset. Then the $S^{-}R$ is a von Neumann regular ring if and only if every prime S-ideal P of R is a minimal prime ideal of R.

Proof. 

$\Rightarrow )$ Suppose that $S^{-}R$ is a von Neumann regular ring. Let P be a prime S-ideal of R. Then $P\cap S=⌀$ and so $S^{-1}P$ is a prime ideal of $S^{-}R$. Assume, for contradiction, that $P\notin Min\left(R\right)$. Then there exists an element $a\in P$ such that ${Ann}_R\left(a\right)⊈P$. So there exists $r\in {Ann}_R\left(a\right)$ with $r\notin P$. Therefore, ${\displaystyle \frac{r}{1}}\notin S^{-1}P$, but ${\displaystyle \frac{r}{1}}·{\displaystyle \frac{a}{1}}={\displaystyle \frac{0}{1}}$ in $S^{-}R$, i.e., ${\displaystyle \frac{r}{1}}\in {Ann}_{R_{S^{-}R}}\left({\displaystyle \frac{r}{1}}\right)$. Hence, the annihilator of ${\displaystyle \frac{a}{1}}$ is not contained in $S^{-1}P$, which contradicts the assumption that $S^{-}R$ is von Neumann regular. Thus, $P\in Min\left(R\right)$.

$\Leftarrow )$ Conversely, assume that every prime S-ideal P of R is a minimal prime ideal. Let M be a maximal ideal of $S^{-}R$. Then $M=S^{-1}P$ for some prime ideal P of R disjoint from S. This prime ideal P is an S-ideal and so by assumption, $P\in Min\left(R\right)$. Hence, $M=S^{-1}P\in Min\left(S^{-}R\right)$. Therefore, every maximal ideal of $S^{-}R$ is minimal, and so $S^{-}R$ is von Neumann regular. □

Let S be a multiplicatively closed subset of a ring R. The saturation of S is the set $S^{*}=\{r\in R:{\displaystyle \frac{r}{1}}$ is a unit in $S^{-1}R\}$. It is clear that $S^{*}$ is a multiplicatively closed subset of R and that $S\subseteq S^{*}$. Moreover, it is well known that $S^{*}=\{x\in R:xy\in S$ for some $y\in R$}, see [8]. The set S is called saturated if $S^{*}=S$.

In the following proposition, we characterize rings in which $\left\{0\right\}$ is the only S-ideal where S is saturated.

Proposition 6. 

Let S be a saturated multiplicatively closed subset of a ring R. Then the following are equivalent:

• The zero ideal is the only S-ideal of R.
• R is a domain and $S=R\setminus \left\{0\right\}$.

Proof. 

$⟹)$ Suppose $\left\{0\right\}$ is the only ideal of R. It is well known that there is a prime ideal P of R with $P\cap S=\varnothing$. Thus, P is an S-ideal of R by Proposition 1 (11). By assumption, $\left\{0\right\}=P$ is a prime ideal and so R is a domain. Now, Suppose there exists $0\ne x\notin R\setminus S$. Since S is saturated, then $〈x〉\cap S=\varnothing$ and so by Proposition 2, ${〈x〉}_S$ is an S-ideal of R. Therefore, ${〈x〉}_S=\left\{0\right\}$ which is a contradiction as $0\ne x\in$${〈x〉}_S$. Therefore, $S=R\setminus \left\{0\right\}$.

$⟸)$ We prove this implication without the assumption that S is saturated. Suppose R is a domain and $S=R\setminus \left\{0\right\}$. Then $S\subseteq reg\left(R\right)$ and so $\left\{0\right\}$ is an S-ideal by Proposition 1 (3). If I is any non-zero ideal of R, then $I\cap S\ne \varnothing$ and so I is not an S-ideal of R. Thus, $\left\{0\right\}$ is the only S-ideal of R. □

If S is not saturated in Proposition 1, then we may find a ring R such that $\left\{0\right\}$ is the only S-ideal of R but $S\ne R-\left\{0\right\}$. For example, consider the multiplicatively closed subset $S=D-\left\{0,-1\right\}$ of any domain D. If I is an S-ideal of D, then $I\cap S=\varnothing$ and so $I\subseteq \left\{0,-1\right\}$. Therefore, $I=\left\{0\right\}$ and $\left\{0\right\}$ is the only S-ideal of R. Note that S is not saturated as $1=(-1)(-1)\in S$ but $-1\notin S$.

Note also that if R is a non domain and $S=R\setminus \left\{0\right\}$, then R has no S-ideals. Indeed, $\left\{0\right\}$ is not an S-ideal by Proposition 1 (3) and any nonzero ideal I of R is also not an S-ideal since $I\cap S\ne \varnothing$.

Next, we characterize non-zero principal S-ideals of any GCD-domain.

Theorem 4. 

Let R be a GCD-domain, S a multiplicatively closed subset of R and $I=nR$ be a proper non-zero ideal of R. Then I is an S-ideal of R if and only if $gcd(n,s)=1$ for all $s\in S$.

Proof. 

$⟹)$ Suppose $I=nR$ is a non-zero S-ideal of R. Assume, for contradiction, that there is a non unit element $d\in R$ such that $gcd(n,s_0)=d$ for some $s_0\in S$. Then there exists an irreducible element p of R such that $p\mid n$ and $p\mid s_0$. Write $s_0=tp$ for some $t\in R$. Then $s_0·\left({\displaystyle \frac{n}{p}}\right)=tp·\left({\displaystyle \frac{n}{p}}\right)=tn\in I$ and $s_0\in S$. But, $n/p\notin I=n\mathbb{Z}$ (since $p\mid n$ and p is not a unit), contradicting the assumption that I is an S-ideal of R. Therefore, $gcd(n,s)=1$ for all $s\in S$.

$⟸)$ Suppose $gcd(n,s)=1$ for all $s\in S$ and let $a,b\in R$ such that $ab\in I$ and $a\in S$. Then $ab=tn$ for some $t\in R$ and $gcd(a,n)=1$. Since R is a GCD-domain, then $n\mid b$ and so $b\in I$. Thus, I is an S-ideal of R. □

Moreover, we consider in the following theorem the non-principal case of Theorem 4. We start by the following lemma.

Lemma 1. 

Let R be a GCD-domain, S a multiplicatively closed subset of R and I be a non-zero proper ideal of R. Then for every $s\in S$, there exists an element $r\in I$ such that $gcd(r,s)=1$.

Proof. 

Assume, for contradiction, that there exists $s_0\in S$ such that every element $r\in I$ satisfies $gcd(r,s_0)\ne 1$, say, $d_r=gcd(r,s_0)$ is a nonunit for all $r\in I$. Let $A=\left\{d_r:r\in I\right\}$ equipped with the divisibility partially ordered: $d_{r_1}⪯d_{r_2}$ if and only if $d_{r_1}\mid d_{r_2}$. Then A is non-empty and every chain $\left\{d_{r_i}:i\in \Lambda \right\}$in A has a lower bound $d=gcd\left\{r_i:i\in \Lambda \right\}$. By Zorn’s Lemma, A has a minimal element $d=gcd(r,s_0)$ for some $r\in I$. Let $r^{\prime }={\displaystyle \frac{r}{d}}$. Then $gcd(r^{\prime },s_0)=gcd({\displaystyle \frac{r}{d}},s_0)=gcd({\displaystyle \frac{r}{d}},{\displaystyle \frac{d{s}_0}{d}})$. Since $gcd({\displaystyle \frac{r}{d}},{\displaystyle \frac{s_0}{d}})=1$, we have $gcd(r^{\prime },s_0)=gcd({\displaystyle \frac{r}{d}},d)=d^{\prime }$. If $d^{\prime }\ne 1$, then $d^{\prime }\mid d$ and since d is minimal in A, then $d^{\prime }$ must be a unit and $gcd(r^{\prime },s_0)=1$. Now, $s_0{r}^{\prime }={\displaystyle \frac{s_0}{d}}r\in I$ and $s_0\in S$ imply $r^{\prime }\in I$ since I is an S-ideal of R. This contradicts the assumption that $gcd(r,s_0)\ne 1$ for every $0\ne r\in I$. Therefore, for every $s\in S$, there is $r\in I$ such that $gcd(r,s)=1$ as needed. □

Theorem 5. 

Let R be a ring, S a multiplicatively closed subset of R and I a proper ideal of R. If $Rs+I=R$ for all $s\in S$, then I is an S-ideal of R. The converse is true if I is non-zero and R is a GCD-domain.

Proof. 

Let $a,b\in R$ such that $ab\in I$ and $a\in S$. Since by assumption, $Ra+I=R$, there exist $r\in R$ and $i\in I$ such that $1=ra+i$. Multiplying both sides by b, we get $b=bra+bi\in I$. Therefore, I is an S-ideal of R as required. Now, assume R is a GCD-domain and I is a non-zero S-ideal of R. If $s\in S$, then by Lemma 1, there exists an element $r\in I$ such that $gcd(r,s)=1$. Thus, there exist $x,y\in R$ such that $1=xs+yr\in Rs+I$. Therefore, $Rs+I=R$ for every $s\in S$ as needed. □

Corollary 2. 

Let R be a ring, $0\ne m\in R$ a non-unit, $S=\{m^n\mid n\ge 0\}$ and I a proper ideal of R. If $Rm+I=R$, then I is an S-ideal of R. The converse is true if I is non-zero and R is a GCD-domain.

Corollary 3. 

Let $S=\{m^k:k\ge 0\}\subseteq \mathbb{Z}$ where $m\in \mathbb{Z}\setminus \{0,\pm 1\}$. Then a proper ideal $I=n\mathbb{Z}$ of $\mathbb{Z}$ is an S-ideal if and only if $gcd(m,n)=1$.

If we consider the case where m is a prime integer in Corollary 3, then we achieve an example of a ring in which every ideal disjoint with S is an S-ideal.

Corollary 4. 

Let $S=\{p^k:k\ge 0\}\subseteq \mathbb{Z}$ where p is a prime integer and let $I=n\mathbb{Z}$ be a proper ideal of $\mathbb{Z}$. The following are equivalent:

• I is an S-ideal.
• $I\cap S=\varnothing$.
• $gcd(p,n)=1$.

Proof. 

The proof is straightforward and is left to the reader. □

The converse of Theorem 5 does not hold in general, as demonstrated by the following example.

Example 2. 

Let k be a field and consider the ring $R=K[x,y]$. Define the multiplicatively closed subset $S=\{x^k:k\ge 0\}\subseteq R$ where $m=x\in R$ is a non-zero non-unit element. Let $I=\left(y\right)$ be the ideal of R generated by y. First, we verify that I is an S-ideal. Let $a\in S$ and $b\in R$ such that $ab\in I$. Since $a=x^k$ for some $k\ge 0$, we have

$$x^{k}b\in I=\left(y\right)\Rightarrow x^{k}b=yf\mathrm{for}\mathrm{some}f\in R.$$

Since $x^k$ and y are coprime in the UFD R, then the factor y must divide b and so $b\in \left(y\right)=I$. Therefore, I is an S-ideal of R. On the other hand, $Rm+I=(x,y)\ne R$. Note that R is not a GCD-domain since there is no greatest common divisor of x and y in R.

Problem 1. 

Can the converse of Theorem 1 be extended to a broader class of commutative rings? Specifically, can one identify ring-theoretic conditions weaker than being a GCD-domain under which the converse implication also holds?

Proposition 7. 

Let R be a ring, P a prime ideal of R, and let $S=R\setminus P$. Then an ideal I of R is an S-ideal if and only if I is P-primary.

Proof. 

Suppose I is an S-ideal and $a,b\in R$ such that $ab\in I$ and $a\notin P$. Then $a\in S$ and by the S-ideal property, it follows that $b\in I$. Therefore, I is P-primary. Conversely, suppose I is P-primary. Let $a,b\in R$ such that $ab\in I$ and $a\in S$. Then $a\notin P=\sqrt{I}$ and hence $b\in I$ as I is P-primary. □

The following theorem is a generalization of the preceding proposition, where the multiplicatively closed set S is defined by excluding the union of finitely many prime ideals rather than a single prime ideal.

Theorem 6. 

Let R be a commutative ring with identity, and let $P_1,\dots ,P_n$ be prime ideals of R. Define the multiplicatively closed set $S:=R\setminus {\bigcup }_{i=1}^n{P}_i$ . Then any S-ideal I of R has a primary decomposition $I={\bigcap }_{i=1}^n{Q}_i$ where

$$Q_i=I_{P_i}:=\{r\in R:\exists s_i\notin P_i\mathrm{such}\mathrm{that}{s}_{i}r\in I\}.$$

Proof. 

Assume I is an S-ideal, where $S=R\setminus {\bigcup }_{i=1}^n{P}_i$ and let $Q_i=I_{P_i}=\{r\in R:\exists s_i\notin P_i$ such that $s_{i}r\in I\}$. Then $Q_i$ is an $(R\setminus P_i)$-ideal for each i by Proposition 2. Therefore, $Q_i$ is a $P_i$-primary ideal of R by Proposition 7. If $r\in I$, then $1·r=r\in I$ with $1\notin P_i$. Thus, $r\in I_{P_i}$ for all i and so $I\subseteq {\bigcap }_{i=1}^n{Q}_i$. Conversely, Let $r\in {\bigcap }_{i=1}^n{Q}_i$. Then for each i, there exists $s_i\notin P_i$ such that $s_{i}r\in I$. Let $J=〈s_1,s_2,\dots ,s_n〉=\{t_1{s}_1+\dots +t_n{s}_n\mid t_i\in R\}$ be the ideal generated by all of the $s_i$. We show that $J\cap S\ne \varnothing$. Suppose, to the contrary, that $J\subseteq {\bigcup }_{i=1}^n{P}_i$. Then by Prime Avoidance Lemma, $J\subseteq P_k$ for some $k\in \left\{1,2,\dots ,n\right\}$. But $s_k\in J$ and $s_k\notin P_k$, a contradiction. Hence, there exists $s\in J\cap S$. Since $s\in J$, we can write $s=t_1{s}_1+\dots +t_n{s}_n$ so that

$$sr=t_1\left(s_{1}r\right)+\dots +t_n\left(s_{n}r\right)\in I.$$

Since $s\in S$, it follows that $r\in I_S=I$. Therefore, ${\bigcap }_{i=1}^n{Q}_i\subseteq I$ and the equality $I={\bigcap }_{i=1}^n{Q}_i$ holds. □

In the following theorem we describe the S-ideals of a ring R with respect to the multiplicatively closed subset $S=1+J$ for some proper ideal J of R.

Theorem 7. 

Let R be a ring, J a proper ideal of R and set $S=1+J$. Let I be an ideal of R. If $J\subseteq I$, then I is an S-ideal of R. Moreover, The converse holds if $J\subseteq J\left(R\right)$ and I is maximal among proper S-ideals of R.

Proof. 

Suppose $J\subseteq I$. Take any $s=1+j\in S$ and $r\in R$ such that $sr\in I$. Then $sr=r+jr\in I$, and since $jr\in I$ (as $j\in J\subseteq I$), it follows that $r=sr-jr\in I$. Thus, I is an S-ideal of R. Conversely, suppose $J\subseteq J\left(R\right)$ and I is maximal among proper S-ideals of R. Then $1+J\subseteq U\left(R\right)$, and thus every proper ideal of R is an S-ideal. Suppose there is $j\in J\setminus I$. Then the ideal $I+Rj$ is strictly larger than I and still a proper ideal. Indeed, Suppose for contradiction $I+Rj=R$. Then $i+rj=1$for some $r\in R$ and $i\in I$ and so $i=1-rj\in U\left(R\right)$, a contradiction. Thus, $I+Rj$ is an S-ideal which contradicts the maximality of I. Therefore, every $j\in J$ lies in I and $J\subseteq I$. □

However, the converse of Theorem 7 need not be true in general as we can see in the following example:

Example 3. 

Let $R=\mathbb{Z}$ and let $J=2\mathbb{Z}$, so that $S=1+J=\{1+2k\mid k\in \mathbb{Z}\}=\mathbb{Z}\setminus 2\mathbb{Z}$. Then $I=4\mathbb{Z}$ is a $2\mathbb{Z}$-primary ideal of $\mathbb{Z}$ and so it is an S-ideal by Proposition 7. However, $J=2\mathbb{Z}⊈4\mathbb{Z}=I$.

Recall that an ideal I of a ring R is called a $z_0$-ideal if for every $a\in I$, the intersection $P_a$ of all minimal prime ideals containing a is contained in I, i.e., $P_a\subseteq I$. In [13] [Theorem 2.19], it is proved that every $z_0$-ideal of a ring R is an r-ideal. More generally, we next determine a multiplicatively closed subset S such that every $z_0$-ideal is an S-ideal.

Proposition 8. 

Let R be a commutative ring and define the multiplicatively closed subset

$$S=R\setminus \bigcup \{P\mid P\in \mathrm{Min}(R\left)\right\}.$$

Then every $z_0$-ideal of R is an S-ideal.

Proof. 

Suppose I is a $z_0$-ideal and let $ab\in I$ with $a\in S$. Since $ab\in I$, by the definition of $z_0$-ideals, we have $P_{ab}\subseteq I$. Also, since $a\in S$, it follows that $a\notin P$ for any $P\in \mathrm{Min}\left(R\right)$, hence $a\notin P_{ab}$. Therefore, clearly $b\in P_{ab}\subseteq I$, and so I is an S-ideal. □

As the following example indicates, the converse of Proposition 8 does not necessarily hold.

Example 4. 

Let $R=\mathbb{Z}\left[x\right]/\left(x^2\right)$, and let $I=\left(0\right)$ be the zero ideal of R. We know that the unique minimal prime ideal of R is $\mathrm{Min}\left(R\right)=\left\{\right(x\left)\right\}$. Then

$$S=R\setminus \bigcup \{P\mid P\in \mathrm{Min}(R\left)\right\}=R\setminus \left(x\right)=\{a+bx\in R\mid a,b\in \mathbb{Z},a\ne 0\}.$$

Since clearly $S\subseteq \mathrm{reg}\left(R\right)$, then $I=\left(0\right)$ is an S-ideal by Proposition 1 (3). However, if we take $a=0\in I$, then $P_a=\left(x\right)$ is not contained in I. Therefore, I is not a $z_0$-ideal of R.

In [13] [Definition 3.11], the concept of r-multiplicatively closed sets (briefly, r-m.c.s) was introduced using the original m.c.s $S=\mathrm{reg}\left(R\right)$. To allow broader applications, we now introduce S-multiplicatively closed sets (briefly, S-m.c.s), where S is any multiplicatively closed set of R.

Definition 2. 

Let S be a subset of a ring R. We say that a subset T of R is an S-multiplicatively closed subset (briefly, S-m.c.s) if T contains at least one element $s\ne 1$ from S and for all $s\in S\cap T$ and $t\in T$, we have $st\in T$. We say that T is S-saturated if it is an S-m.c.s and for all $a,b\in R$, $ab\in T$ implies $a\in T$ and $b\in T$.

Remark 2. 

Similar to the case of r-m.c.s T where we may assume for practical purposes that $reg\left(R\right)\subseteq T$,13], we may similarly assume that any S-m.c.s T of R contains S. This is because the set $T^{\prime }=T\cup S\cup \{st:s\in S,t\in T\}$ is again an S-m.c.s larger than T containing S and clearly for any ideal I of R, $I\cap T=\varnothing$ if and only if $I\cap T^{\prime }=\varnothing$.

Lemma 2. 

Let $\{I_i:i\in \Delta \}$ be a family of S-ideals of R. Then the set $T=R\setminus {\bigcup }_{i\in \Delta }{I}_i$ is an S-saturated m.c.s.

Proof. 

Since each $I_i$ is an S-ideal, $I_i\cap S=\varnothing$. Hence, $S\subseteq T$. Now, suppose $s\in S\cap T$ and $t\in T$. Then for all i, $s,t\notin I_i$ and since $I_i$ is an S-ideal, then $st\notin I_i$. Thus, $st\in T$, proving that T is an S-mcs. To show thatT is S-saturated, assume $ab\in T$. Then $ab\notin I_i$ for all $i\in \Delta$ and so $a,b\notin I_i$ for all $i\in \Delta$. Thus, $a,b\in T$. □

We now present an analogue of [13] [Proposition 3.14] by replacing $reg\left(R\right)$ with an arbitrary S.

Theorem 8. 

Let S be a subset of a ring R. Then a subset T of R is an S-saturated m.c.s of R such that $S\subseteq T$ if and only if $T=R\setminus {\bigcup }_{I\in \mathcal{A}}I$ for some family $\mathcal{A}$ of S-ideals of R.

Proof. 

⇒) Assume T is an S-saturated m.c.s. of R and $S\subseteq T$. Let

$$\mathcal{A}=\{I\subseteq R:I\mathrm{is}\mathrm{an}S-\mathrm{ideal}\mathrm{and}I\cap T=\varnothing \}.$$

By the definition of $\mathcal{A}$, we have clearly, $T\subseteq R\setminus {\bigcup }_{I\in \mathcal{A}}I$. To prove the reverse inclusion, let $r\in R\setminus {\bigcup }_{I\in \mathcal{A}}I$ and suppose that $r\notin T$. If $〈r〉\cap T\ne \varnothing$, say $ar\in T$ for some $a\in R$, then T being S-saturated implies $r\in T$, a contradiction. Therefore, $〈r〉\cap T=\varnothing$ and so $〈r〉\cap S=\varnothing$ as $S\subseteq T$. By Proposition 2, the ideal ${〈r〉}_S$ is an S-ideal and ${〈r〉}_S\cap T\ne \varnothing$. Thus, $r\in {〈r〉}_S\subseteq {\bigcup }_{I\in \mathcal{A}}I$, a contradiction. Therefore, $r\in T$ and the equality$T=R\setminus {\bigcup }_{I\in \mathcal{A}}I$ holds.

⇐) Lemma 2. □

3. S-Ideals in Ring Extensions and Constructions

In this section, we investigate the behavior of S-ideals under various ring-theoretic constructions and extensions. We examine how the concept of S-ideals is preserved or transformed when passing to localized rings (rings of fractions), quotient rings, and finite Cartesian products. In addition, we study the extension of S-ideals to more structured rings, including polynomial rings, idealization rings, and amalgamated algebras. This analysis provides insight into the structural robustness of S-ideals and illustrates their applicability across a wide range of ring-theoretic settings.

If S and T are multiplicatively closed subsets of a ring R, then

$$T^{-1}S:=\left\{{\displaystyle \frac{s}{t}}:s\in S,t\in T\right\}$$

is multiplicatively closed subsets of $T^{-1}R$. In particular, ${\left\{1\right\}}^{-1}S=\left\{{\displaystyle \frac{s}{1}}:s\in S\right\}$ is multiplicatively closed in $T^{-1}R$.

Theorem 9. 

Let I a proper ideal of a ring R and let $S,T$ be multiplicatively closed subsets of R. If I is an S-ideal of R and $I\cap T=\varnothing$, then $T^{-1}I$ is an $T^{-1}S$-ideal of $T^{-1}R$.

Proof. 

Suppose there is ${\displaystyle \frac{s}{t}}\in T^{-1}S\cap T^{-1}I$. Then there exists $t^{\prime }\in T$ such that $t^{\prime }s$$\in I$ and I being an S-ideal implies $t^{\prime }$$\in I\cap T$, a contradiction. Thus, $T^{-1}S\cap T^{-1}I=\varnothing$. Let ${\displaystyle \frac{a}{t_1}}\in T^{-1}S$ and ${\displaystyle \frac{b}{t_2}}\in T^{-1}R$, and suppose $\left({\displaystyle \frac{a}{t_1}}\right)\left({\displaystyle \frac{b}{t_2}}\right)={\displaystyle \frac{ab}{t_1{t}_2}}\in T^{-1}I.$ Then there exists $t_3\in T$ such that $t_{3}ab\in I$. Since I is an S-ideal and $a\in S$, then $b{t}_3\in I$ and so ${\displaystyle \frac{b}{t_2}}={\displaystyle \frac{b{t}_3}{t_2{t}_3}}\in T^{-1}I$. Hence, $T^{-1}I$ is an $T^{-1}S$-ideal of $T^{-1}R$. □

In particular, if $S=T$, then all elements of $T^{-1}S$ are units in $T^{-1}R$. Thus, Proposition 1 (2) implies the following corollary:

Corollary 5. 

Let S be a multiplicatively closed subset of a ring R. Then any proper ideal of $S^{-1}R$ is an $S^{-1}S$-ideal.

However, the converse of Theorem 9 is not true in general.

Example 5. 

Let $R=\mathbb{Z}$, $S=\{3^n:n\ge 0\}$, $T=\mathbb{Z}\setminus \left\{0\right\}$ and $I=6\mathbb{Z}$. Then $T^{-1}R=\mathbb{Q}$, $T^{-1}I=6\mathbb{Q}$ and $T^{-1}S=\left\{{\displaystyle \frac{3^n}{m}}:n\ge 0,m\in \mathbb{Z}\setminus \left\{0\right\}\right\}$. We show that $T^{-1}I$ is a $T^{-1}S$-ideal in $\mathbb{Q}$: Let $x\in \mathbb{Q}$ and suppose there exists $s\in T^{-1}S$ such that $sx\in 6\mathbb{Q}$, that is, $sx=6q$ for some $q\in \mathbb{Q}.$ Since $s\in T^{-1}S$, write $s={\displaystyle \frac{3^n}{m}}$ for some $n\ge 0$, $m\in \mathbb{Z}\setminus \left\{0\right\}$. Then ${\displaystyle \frac{3^n}{m}}x=6q$ and so $x={\displaystyle \frac{6mq}{3^n}}\in 6\mathbb{Q}=T^{-1}I$. Therefore, $T^{-1}I$ is a $T^{-1}S$-ideal in $\mathbb{Q}$. On the other hand, by Corollary 3, $I=6\mathbb{Z}$ is not an S-ideal in $\mathbb{Z}$ since $gcd(6,2)\ne 1$.

Proposition 9. 

Let $f:R_1\to R_2$ be a ring homomorphism and S be a multiplicatively closed subset of $R_1$. Then the following statements hold.

• If J is an $f\left(S\right)$-ideal of $R_2$, then $f^{-1}\left(J\right)$ is an S-ideal of $R_1$.
• If f is an epimorphism and I is an S-ideal of $R_1$ containing $Ker\left(f\right)$, then $f\left(I\right)$ is an $f\left(S\right)$-ideal of $R_2.$

Proof. (1) Let $a,b\in R_1$ with $a\in S$ and $ab\in f^{-1}\left(J\right).$ Then $f\left(ab\right)=f\left(a\right)f\left(b\right)\in J$ with $f\left(a\right)\in f\left(S\right)$ and since J is an $f\left(S\right)$-ideal of $R_2$, $f\left(b\right)\in J$. Thus, $b\in f^{-1}\left(J\right)$.

(2) Let $a,b\in R_2$ such that $a\in f\left(S\right)$ and $ab\in f\left(I\right)$. Write $a=f\left(s\right)$ for some $s\in S$ and since f is onto, there is $y\in R_1$ such that $b=f\left(y\right)$. Since $f\left(s\right)f\left(y\right)\in f\left(I\right)$ and $Ker\left(f\right)\subseteq I$, we have $sy\in I$ and so $y\in I$ as I is an S-ideal of $R_1$. Therefore, $b=f\left(y\right)\in f\left(I\right)$ and $f\left(I\right)$ is an $f\left(S\right)$-ideal of $R_2$. □

Let S be a multiplicatively closed subset of a ring R and I be an ideal of R disjoint with S. If we denote $r+I\in R/I$ by $\overline{r}$, then $\overline{S}=\{\overline{s}:s\in S\}$ is a multiplicatively closed subset of $R/I$. In view of Proposition 9, we conclude the following result for $\overline{S}$-ideals of $R/I$.

Corollary 6. 

Let S be a multiplicatively closed subset of a ring R and I, J are two ideals of R with $I\subseteq J$ . Then J is an S-ideal of R if and only if $J/I$ is an $\overline{S}$-ideal of $R/I$.

Proof. 

We apply the canonical epimorphism $\pi :R\to R/I$ in Proposition 9. □

Corollary 7. 

Let n be a positive integer and let $\overline{S}$ be a multiplicatively closed subset of the ring ${\mathbb{Z}}_n$. Then a proper ideal $I=〈\overline{m}〉$ of ${\mathbb{Z}}_n$ is an $\overline{S}$-ideal if and only $gcd(m,s)=1$ for every $s\in S$.

Proof. 

Consider the canonical epimorphism $\pi :\mathbb{Z}\to \mathbb{Z}/n\mathbb{Z}$ in Proposition 4 and use Theorem 5. □

Let ${\left\{R_i\right\}}_{i\in \Lambda }$ be a family of rings and for each $i\in \Lambda$, let $S_i$ be a multiplicatively closed subset of $R_i$. The following proposition investigates $\left({\prod }_{i\in \Lambda }{S}_i\right)$-ideals in the Cartesian product ring ${\prod }_{i\in \Lambda }{R}_i$.

Proposition 10. 

Let ${\left\{R_i\right\}}_{i\in \Lambda }$ be any family of rings and for each $i\in \Lambda$, let $S_i\subseteq R_i$ be a multiplicatively closed subset and $I_i\subseteq R_i$ be an ideal. Define:

$$R=\prod _{i\in \Lambda }{R}_i,S=\prod _{i\in \Lambda }{S}_i,I=\prod _{i\in \Lambda }{I}_i.$$

Then I is an S-ideal of R if and only if $I_i$ is an $S_i$-ideal of $R_i$ for each $i\in \Lambda$.

Proof. 

$\Rightarrow )$ Assume $I={\prod }_{i\in \Lambda }{I}_i$ is an S-ideal in $R={\prod }_{i\in \Lambda }{R}_i$. Fix $j\in \Lambda$ and suppose $s_j\in S_j$, $r_j\in R_j$ with $s_j{r}_j\in I_j$. Define elements in R as follows:

$$s=\left(s_i\right)\in S\mathrm{where}{s}_i=\left\{\begin{array}{cc}{s}_j\hfill & \mathrm{if}i=j,\hfill \\ 1\hfill & \mathrm{otherwise},\hfill \end{array}\right.r=\left(r_i\right)\in R\mathrm{where}{r}_i=\left\{\begin{array}{cc}{r}_j\hfill & \mathrm{if}i=j,\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Then $sr\in I$ because $s_j{r}_j\in I_j$ and $s_i{r}_i=0\in I_i$ for $i\ne j$. Since I is an S-ideal of R, it follows that $r\in I$, hence $r_j\in I_j$. Therefore, $I_j$ is an $S_j$-ideal of $R_j$ for all $j\in \Lambda$.

$\Leftarrow )$ Assume each $I_i$ is an $S_i$-ideal in $R_i$. Let $s=\left(s_i\right)\in S$, $r=\left(r_i\right)\in R$ such that $sr=\left(s_i{r}_i\right)\in I={\prod }_{i\in \Lambda }{I}_i$. Then $s_i{r}_i\in I_i$ for all $i\in \Lambda$. Since each $I_i$ is an $S_i$-ideal and $s_i\in S_i$, it follows that $r_i\in I_i$ for all $i\in \Lambda$. Thus, $r\in I$ and hence, I is an S-ideal in R. □

Let R be a commutative ring with identity and $S\subseteq R$ a multiplicatively closed subset. If I is an S-ideal of R, then it is easy to verify that $I\left[x\right]$, the ideal generated by I in $R\left[x\right]$, is also an S-ideal in $R\left[x\right]$. That is, the S-ideal property is preserved under extension to the polynomial ring.

However, in general, the set $S\left[x\right]\subseteq R\left[x\right]$, consisting of polynomials with coefficients in S, is not a multiplicatively closed subset of $R\left[x\right]$. For example, if $S=\{1,3\}$ in ${\mathbb{Z}}_6$, then $f\left(x\right)=3x+1$ and $g\left(x\right)=x+3$ are in $S\left[x\right]$ but $f\left(x\right)g\left(x\right)=3x^2+4x+3\notin S\left[x\right]$.

To address this issue, we consider instead the set of monomials of the form $s{x}^n$ with $s\in S$ and $n\ge 0$. Define:

$$S^{\prime }=\left\{s{x}^n:s\in S,n\ge 0\right\}\subseteq R\left[x\right].$$

It is straightforward to check that $S^{\prime }$ is a multiplicatively closed subset of $R\left[x\right]$.

In the following proposition, we characterize the $S^{\prime }$-ideals of $R\left[x\right]$ in terms of S-ideals of R.

Proposition 11. 

Let R be a commutative ring with identity, S a multiplicatively closed subset of R, and let

$$S^{\prime }=\left\{s{x}^n:s\in S,n\ge 0\right\}\subseteq R\left[x\right].$$

Then for any ideal I of R, we have:

$$I\mathrm{is}\mathrm{an}S-\mathrm{ideal}\mathrm{in}R\iff I\left[x\right]\mathrm{is}\mathrm{an}{S}^{\prime }-\mathrm{ideal}\mathrm{in}R\left[x\right].$$

Proof. 

$\Rightarrow )$ Suppose I is an S-ideal in R. Let $f\left(x\right)\in R\left[x\right]$ and $s{x}^n\in S^{\prime }$ with $s\in S$, $n\ge 0$ such that $\left(s{x}^n\right)f\left(x\right)\in I\left[x\right]$. Write $f\left(x\right)={\sum }_{j=0}^d{b}_j{x}^j$. Then: $\left(s{x}^n\right)f\left(x\right)={\sum }_{j=0}^{d}s{b}_j{x}^{n+j}\in I\left[x\right]$. So for all j, $s{b}_j\in I$. Since I is an S-ideal and $s\in S$, we get $b_j\in I$ for all j, so $f\left(x\right)\in I\left[x\right]$. Hence, $I\left[x\right]$ is an $S^{\prime }$-ideal.

$\Leftarrow )$ Suppose $I\left[x\right]$ is an $S^{\prime }$-ideal. Let $r\in R$ and $s\in S$ with $sr\in I$. Then $s{x}^0·r=\left(sr\right)\in I\Rightarrow \left(s{x}^0\right)\left(r\right)\in I\left[x\right]$. Since $s{x}^0\in S^{\prime }$ and $r\in R\left[x\right]$, the assumption implies $r\in I\left[x\right]$, i.e., $r\in I$. Thus, I is an S-ideal. □

We now study S-ideals of $R\left[x\right]$ with respect to other types of multiplicatively closed subsets, beyond those inherited directly from R. This broader approach allows us to explore more refined structures within the polynomial ring.

For a polynomial $f\in R\left[x\right]$, the content of f, denoted $C\left(f\right)$, is the ideal of R generated by its coefficients. A polynomial $f\in R\left[x\right]$ is said to be primitive if its content $C\left(f\right)=R$. It is proved in [12] [Proposition 33.1] that $T=\{f\in R[x]:C(f)=R\}$ is a complement of a union of prime ideals of $R\left[x\right]$ and so it is a (regular) multiplicatively closed subset of $R\left[x\right]$. The ring of fractions of $R\left[x\right]$ at T is denoted by $R\left(x\right)=T^{-1}R\left[x\right]$, and it is called the Nagata ring of R. By [12] [Proposition 33.1], for any ideal I of R, $T^{-1}\left(I\left[x\right]\right)\cap R\left[x\right]=I\left[x\right]$ and so $I\left[x\right]$ is an T-ideal of $R\left[x\right]$.

In particular, if we consider the multiplicatively subset $N=\{f\in R[x]:f$ is monic$\}\subseteq T$, then also $I\left[x\right]$ is an N-ideal of $R\left[x\right]$ for every ideal I of R. The ring of fractions of $R\left[x\right]$ at N is denoted by $R〈x〉=N^{-1}R\left[x\right]$ and it is well-known in the literature, see [12].

While $I\left[x\right]$ is an S-ideal of $R\left[x\right]$ for every ideal $I\subseteq R$ when $S=T$ or $S=N$ as defined above, this property does not necessarily hold for arbitrary multiplicatively closed subsets $S\subseteq R\left[x\right]$ that contain T or N.

Proposition 12. 

Let S be a multiplicatively closed subset of a ring R. Define

$$T:=\{f\in R[x]:C(f)\cap S\ne \varnothing \}$$

Then J is an S-ideal of R if and only if $J\left[x\right]$ is an T-ideal of $R\left[x\right]$.

Proof. 

$⟹)$ Suppose J is an S-ideal of R, and let $g\left(x\right),f\left(x\right)\in R\left[x\right]$ such that $g\left(x\right)\in T$ and $g\left(x\right)f\left(x\right)\in J\left[x\right]$. Since $g\left(x\right)\in T$, there exists $s\in S$ such that $s\in C\left(g\right)$. Write $f={\sum }_{j=0}^m{b}_j{x}^j$ and $g={\sum }_{i=0}^n{a}_i{x}^i.$ Then $gf={\sum }_{k=0}^{m+n}{c}_k{x}^k$, with $c_k={\sum }_{i+j=k}{a}_i{b}_j.$ Since $gf\in J\left[x\right]$, we have $c_k\in J$ for all k. In particular, for each j, $a_i{b}_j\in J$for all i. Let $s\in C\left(g\right)=〈a_0,\dots ,a_n〉$, so there exist $r_0,\dots ,r_n\in R$ such that $s={\sum }_{i=0}^n{r}_i{a}_i$. Then for each j, we compute

$$s{b}_j=\left(\sum _{i=0}^n{r}_i{a}_i\right)b_j=\sum _{i=0}^n{r}_i{a}_i{b}_j\in J,$$

since each $a_i{b}_j\in J$. Thus, $s{b}_j\in J$ for all j, and since $s\in S$ and J is an S-ideal, it follows that $b_j\in J$. Hence, all coefficients of f lie in J and $f\in J\left[x\right]$. Therefore, $J\left[x\right]$ is a T-ideal.

$⟸)$ Suppose $J\left[x\right]$ is a T-ideal and let $a,b\in R$ with $ab\in J$ and $a\notin S$. Define $f\left(x\right)=b$, $g\left(x\right)=a$, so $g\left(x\right)f\left(x\right)=ab\in J\left[x\right]$ and $g\left(x\right)\in T$. Then by assumption, $b=f\left(x\right)\in J\left[x\right]$ and so $b\in J$. Therefore, J is an S-ideal of R. □

The following corollaries are special cases of the preceding proposition.

Corollary 8. 

Let P be a prime ideal of a ring R, and define

$$T:=\{f\in R[x]:\mathrm{the}\mathrm{constant}\mathrm{term}\mathrm{of}f\notin P\}.$$

Then I is an S-ideal of R with $S:=R\setminus P$ if and only if $I\left[x\right]$ is an T-ideal of $R\left[x\right]$.

Corollary 9. 

Let I be an ideal of ring R. Then I is an $(R\setminus \{0\left\}\right)$-ideal of R if and only if $I\left[x\right]$ is an $\left(R\right[x]\setminus \{0\left\}\right)$-ideal of $R\left[x\right]$.

Corollary 10. 

Let $S\subseteq$ be a multiplicatively closed subset of a commutative ring R, and define

$$T=\{f\in R[x]:\mathrm{the}\mathrm{leading}\mathrm{coefficient}\mathrm{of}f\mathrm{lies}\mathrm{in}S\}.$$

Then I is an S-ideal of R with if and only if $I\left[x\right]$ is an T-ideal of $R\left[x\right]$.

Recall that the idealization of an R-module M denoted by $R(+)M$ is the commutative ring $R\times M$ with coordinate-wise addition and multiplication defined as $(r_1,m_1)(r_2,m_2)=(r_1{r}_2,r_1{m}_2+r_2{m}_1)$. For an ideal I of R and a submodule N of M, $I(+)N$ is an ideal of $R(+)M$ if and only if $IM\subseteq N$. If S is a multiplicatively closed subset of R, then clearly the set $S(+)K=\left\{(s,m):s\in S,m\in K\right\}$ is a multiplicatively closed subsets of the ring $R(+)M$ for any submodule K of M.

Definition 3. 

Let R be a ring and let N be a proper submodule of an R-module M. Let S be a multiplicatively closed subset of R. Then N is called an S-submodule of M if whenever $s\in S$ and $m\in M$, $sm\in N$ implies $m\in N$.

Equivalently, N is an S-submodule of M if and only if $S^{-1}N\cap M=N$.

Next, for a submodule K of an R-module M, we determine the relation between S-ideals of R and $S(+)K$-ideals of $R(+)M$.

Proposition 13. 

Let $N,K$ be submodules of an R-module M, S be a multiplicatively closed subset of R and I be an ideal of R where $IM\subseteq N$. Then $I(+)N$ is an $S(+)K$-ideal of $R(+)M$ if and only if I is an S-ideal of R and N is an S-submodule of M.

Proof. 

$⟹)$ Suppose $I(+)N$ is an $S(+)K$-ideal of $R(+)M$. Let $a,b\in R$ such that $ab\in I$ and $a\in S$. Then $(a,0)(b,0)\in I(+)N$ with $(a,0)\in S(+)K$ and by assumption, $(b,0)\in I(+)N$. Thus, $b\in I$ and I is an S-ideal of R. Now, let $s\in S$ and $m\in M$ such that $sm\in N$. Then $(s,0)(0,m)=(0,sm)\in I(+)N$ with $(s,0)\in S(+)K$. Again by assumption, $(0,m)\in I(+)N$ and so $m\in N$. Thus, N is an S-submodule of M.

$⟸)$ Suppose I is an S-ideal of R and N is an S-submodule of M. Let $(r_1,m_1),(r_2,m_2)\in R(+)M$ such that $(r_1,m_1)(r_2,m_2)\in I(+)N$ and $(r_1,m_1)\in S(+)K$. Then $r_1{r}_2\in I$ and $r_1\in S$ and I being an S-ideal implies $r_2\in I$. Since $r_1{m}_2+r_2{m}_1\in N$ and $r_2{m}_1\in IM\subseteq N$, then $r_1{m}_2\in N$. Since N is an S-submodule of M and $r_1\in S$, then $m_2\in N$. Therefore, $(r_2,m_2)\in I(+)N$ and $I(+)N$ is an $S(+)K$-ideal of $R(+)M$. □

The converse of the proposition 13 is not true in general. That is, even if I is an S-ideal of R, the ideal $I(+)N$ of $R(+)M$ need not be an $S(+)K$-ideal if N is not an S-submodule of M. We demonstrate this with the following example:

Example 6. 

Consider the ring $R=\mathbb{Z}$ and the R-module $M={\mathbb{Z}}_6$ and let $S=\mathbb{Z}\setminus 2\mathbb{Z}$. Define the ideal $I=\left\{0\right\}$ of $\mathbb{Z}$, and the submodule $N=\{\overline{0},\overline{3}\}$ of ${\mathbb{Z}}_6$. Then $IM=0\subseteq N$ and I is an S-ideal of $\mathbb{Z}$. However, N is not an S-submodule of M since if we take $s=3\in S$ and $m=\overline{2}\in {\mathbb{Z}}_6$, then $sm\in N$, but $m=\overline{2}\notin N$. Consequently, although I is an S-ideal, the ideal $I(+)N$ fails to be an $S(+)M$-ideal of $R(+)M$ since $(3,\overline{1})\in S(+)M$ and $(3,\overline{1})(0,\overline{2})=(0,\overline{0})\in I(+)N$ but $(0,\overline{2})\notin I(+)N$. This confirms that the converse of the theorem is not valid when N is not an S-submodule.

Let R and $R^{\prime }$ be two rings, J be an ideal of $R^{\prime }$ and $f:R\to R^{\prime }$ be a ring homomorphism. The set $R{\bowtie }^{f}J=\left\{(r,f(r)+j):r\in R,j\in J\right\}$ is a subring of $R\times R^{\prime }$ called the amalgamation of R and $R^{\prime }$ along J with respect to f. In particular, if $I{d}_R:R\to R$ is the identity homomorphism on R, then $R\bowtie J=R{\bowtie }^{I{d}_R}J=\left\{(r,r+j):r\in R,j\in J\right\}$ is the amalgamated duplication of a ring along an ideal J. Many properties of this ring have been investigated and analyzed over the last two decades, see for example [5,6].

Let I be an ideal of R and K be an ideal of $f\left(R\right)+J$. Then $I{\bowtie }^{f}J=\left\{(i,f(i)+j):i\in I,j\in J\right\}$ and ${\overline{K}}^f=\{(a,f\left(a\right)+j):a\in R$, $j\in J$, $f\left(a\right)+j\in K\}$ are ideals of $R{\bowtie }^{f}J$,6]. For a multiplicatively closed subset S of R, one can easily verify that $S{\bowtie }^{f}J=\left\{(s,f(s)+j):s\in S,j\in J\right\}$ and $W=\left\{(s,f(s\left)\right):s\in S\right\})$ are multiplicatively closed subsets of $R{\bowtie }^{f}J$. Let T be a multiplicatively closed subset of $R^{\prime }$. Then clearly, the set ${\overline{T}}^f=\{(s,f\left(s\right)+j):s\in R,$ $j\in J,$$f\left(s\right)+j\in T\}$ is also a multiplicatively closed subset of $R{\bowtie }^{f}J$.

Next, for a multiplicatively closed subset S of a ring R, we determine when the ideal $I{\bowtie }^{f}J$ is an $\left(S{\bowtie }^{f}J\right)$-ideal in $R{\bowtie }^{f}J$.

Theorem 10. 

Consider the amalgamation of rings R and $R^{\prime }$ along the ideals J of $R^{\prime }$ with respect to a homomorphism f. Let S be a multiplicatively closed subset of R and I be an ideal of R. The following statements are equivalent.

• $I{\bowtie }^{f}J$ is an $\left(S{\bowtie }^{f}J\right)$-ideal of $R{\bowtie }^{f}J$.
• $I{\bowtie }^{f}J$ is an W-ideal of $R{\bowtie }^{f}J$.
• I is a S-ideal of R.

Proof. 

Note that clearly, I is proper in R of and only if $I{\bowtie }^{f}J$ is proper in $R{\bowtie }^{f}J$.

$\left(1\right)⟹\left(2\right)$ Clear as $W\subseteq S{\bowtie }^{f}J$.

$\left(2\right)⟹\left(3\right)$ Suppose $I{\bowtie }^{f}J$ is an $\left(S{\bowtie }^{f}J\right)$-ideal of $R{\bowtie }^{f}J$. Let $a,b\in R$ such that $ab\in I$ and $s\in S$. Then $(a,f\left(a\right))(b,f\left(b\right))\in I{\bowtie }^{f}J$ with $(a,f(a\left)\right)\in W$. By assumption, $(b,f\left(b\right))\in I{\bowtie }^{f}J$ and so $b\in I$. Therefore, I is an S-ideal of R.

$\left(3\right)⟹\left(1\right)$ Suppose I is an S-ideal of R. Let $(a,f\left(a\right)+j_1)(b,f\left(b\right)+j_2)=(ab,(f\left(a\right)+j_1)(f\left(b\right)+j_2))\in I{\bowtie }^{f}J$ for $(a,f\left(a\right)+j_1),(b,f\left(b\right)+j_1)\in R{\bowtie }^{f}J$ where $(a,f\left(a\right)+j_1)\in S{\bowtie }^{f}J$. Then $ab\in I$ and $a\in S$ and as I is an S-ideal, then $b\in I$. Thus, $(b,f\left(b\right)+j_2)\in I{\bowtie }^{f}J$ and $I{\bowtie }^{f}J$ is an $\left(S{\bowtie }^{f}J\right)$-ideal of $R{\bowtie }^{f}J$. □

Corollary 11. 

Consider the amalgamation of rings R and $R^{\prime }$ along the ideal J of $R^{\prime }$ with respect to a homomorphism f. Let S be a multiplicatively closed subset of R. The $\left(S{\bowtie }^{f}J\right)$-ideals of $R{\bowtie }^{f}J$ containing $\left\{0\right\}\times J$ are of the form $I{\bowtie }^{f}J$ where I is an S-ideal of $R.$

Proof. 

From Theorem 10, $I{\bowtie }^{f}J$ is an $\left(S{\bowtie }^{f}J\right)$-ideal of $R{\bowtie }^{f}J$ for any S-ideal I of R. Let K be an $\left(S{\bowtie }^{f}J\right)$-ideal of $R{\bowtie }^{f}J$ containing $\left\{0\right\}\times J.$ Consider the surjective homomorphism $\phi :R{\bowtie }^{f}J\to R$ defined by $\phi (a,f(a)+j)=a$ for all $(a,f\left(a\right)+j)\in R{\bowtie }^{f}J$. Then $Ker\left(\phi \right)=\left\{0\right\}\times J\subseteq K$ and so $I:=\phi \left(K\right)$ is an S-ideal of R by Proposition 9. Since $\left\{0\right\}\times J\subseteq K$, we conclude that $K=I{\bowtie }^{f}J$ as needed. □

In the following theorem, we clarify the relation between T-ideals of $R^{\prime }$ and ${\overline{T}}^f$-ideal of $R{\bowtie }^{f}J$.

Theorem 11. 

Consider the amalgamation of rings R and $R^{\prime }$ along the ideals J of $R^{\prime }$ with respect to an epimorphism f. Let K be an ideal of $R^{\prime }$ and T be a multiplicatively closed subset of $R^{\prime }$. Then ${\overline{K}}^f$ is an ${\overline{T}}^f$-ideal of $R{\bowtie }^{f}J$ if and only if K is an T-ideal of $R^{\prime }$.

Proof. 

Note that clearly, K is proper in $R^{\prime }$ if and only if ${\overline{K}}^f$ is proper in $R{\bowtie }^{f}J$.

Suppose ${\overline{K}}^f$ is an ${\overline{T}}^f$-ideal of $R{\bowtie }^{f}J$. Let $a^{\prime }$,$b^{\prime }\in R^{\prime }$ such that $a^{\prime }{b}^{\prime }\in K$ and $a^{\prime }\in T$. Choose $a,b\in R$ such that $f\left(a\right)=a^{\prime }$ and $f\left(b\right)=b^{\prime }$. Then $(a,a^{\prime }),(b,b^{\prime })\in R{\bowtie }^{f}J$ with $(a,a^{\prime })(b,b^{\prime })=(ab,a^{\prime }{b}^{\prime })\in {\overline{K}}^f$ and $(a,a^{\prime })\in {\overline{T}}^f$. By assumption, we have $(b,b^{\prime })\in {\overline{K}}^f$ and so $b^{\prime }\in K$. Therefore, K is an T-ideal of $R^{\prime }$. conversely, suppose K is an T-ideal of $R^{\prime }$. Let $(a,f\left(a\right)+j_1)(b,f\left(b\right)+j_2)=(ab,(f\left(a\right)+j_1)(f\left(b\right)+j_2))\in {\overline{K}}^f$ for $(a,f\left(a\right)+j_1),(b,f\left(b\right)+j_1)\in R{\bowtie }^{f}J$ where $(a,f\left(a\right)+j_1)\in {\overline{T}}^f$. Then $(f\left(a\right)+j_1)(f\left(b\right)+j_2)\in K$ with $(f\left(a\right)+j_1)\in T$ and so $(f\left(b\right)+j_2)\in K$ as K is an T-ideal of $R^{\prime }$. It follows that $(b,f\left(b\right)+j_2)\in {\overline{K}}^f$ and ${\overline{K}}^f$ is an ${\overline{T}}^f$-ideal of $R{\bowtie }^{f}J$. □

Corollary 12. 

Let R,I, J, K, S and T be as in Theorems 10 and 11. Then

• $I\bowtie J$ is an $(S\bowtie J)$-ideal of $R\bowtie J$ if and only if I is an S-ideal of R.
• $\overline{K}$ is a $\overline{T}$-ideal of $R\bowtie J$ if and only if K is a T-ideal of R.