Rings with uniformly S-w-Noetherian spectrum

TL;DR

This paper introduces and characterizes rings with uniformly S-w-Noetherian spectrum, providing several equivalent conditions and extending classical results to this new class of rings.

Contribution

It introduces the concept of rings with uniformly S-w-Noetherian spectrum and establishes multiple equivalent characterizations, including extensions to polynomial and localization rings.

Findings

A ring has uniformly S-w-Noetherian spectrum iff each radical ideal is radically S-w-finite.

Such rings have stationary ascending chains of radical w-ideals.

The classical case is extended to include countably generated ideals and polynomial rings.

Abstract

In this paper, the notion of rings with uniformly S-w-Noetherian spectrum is introduced. Several characterizations of rings with uniformly S-w-Noetherian spectrum are given. Actually, we show that a ring R has uniformly S-w-Noetherian spectrum with respect to some s 2 S if and only if each ascending chain of radical w-ideals of R is stationary with respect to s 2 S, if and only if each radical (prime) (w-)ideal of R is radically S-w-finite with respect to s, if and only if each countably generated ideal of R is radically S-w-finite with respect to s, if and only if R[X] has uniformly S-w-Noetherian spectrum, if and only if RfXg has uniformly S-Noetherian spectrum. Beside, we show a ring R has Noetherian (resp., uniformly S-Noetherian) spectrum with respect to s if and only if each countably generated ideal of R is radically finite (S-finite with respect to s), which is a new result in the classical case.

Equations

$$\mathcal{I}=\{\beta \mid (s{r}_{\beta }{)}^n\notin \sqrt{⟨r_{\alpha }\mid \alpha <\beta ⟩}\text{mbox}foralln\ge 1\}$$

$$(s{r}_{{\beta }_1}{)}^{n_1}\in ⟨r_{\alpha }\mid \alpha <{\beta }_1⟩.$$

$$(s{r}_{{\beta }_{2,i_k}}{)}^{n_2}\in ⟨r_{\alpha }\mid \alpha <{\beta }_2⟩$$

$$(s{r}_{{\beta }_1}{)}^{n_1{n}_2{m}_2}=(s\sum _{{\beta }_{2,i}\in \mathcal{I}}{k}_i{r}_{{\beta }_{2,i}}+s\sum _{{\beta }_{2,i}\notin \mathcal{I}}{k}_i{r}_{{\beta }_{2,i}}{)}^{n_2{m}_2}=(s\sum _{{\beta }_{2,i}\notin \mathcal{I}}{k}_i{r}_{{\beta }_{2,i}}{)}^{n_2{m}_2}+M,$$

$${\mathcal{I}}^{\prime }=\{{\beta }_1<{\beta }_2<{\beta }_3<\cdots \}\subseteq \mathcal{I}.$$

$$s⟨r_{\beta }\mid \beta \in {\mathcal{I}}^{\prime }⟩\subseteq \sqrt{⟨r_{\beta }\mid \beta \in {\mathcal{I}}^{\prime \prime }⟩}.$$

$$S^{\ast }=\{s\in R\mid s_1=s{s}_2\text{ for some }{s}_1,s_2\in S\}.$$

$$\textstyle{\boxed{$u$\text{-}$S$\text{-Noetherian spectrum}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$

$$s_k{I}_k\subseteq \sqrt{(J_k{)}_{w_k}}.$$

$$(s_1,\dots ,s_n)I\subseteq \sqrt{(\prod _{k=1}^n{J}_k{)}_w}.$$

$$s_k{I}_k\subseteq \sqrt{(J_k{)}_{w_k}}.$$

$$sI\subseteq \sqrt{(\bigcup _{i=1}^k{I}_i{)}_w}=\sqrt{\bigcup _{i=1}^k{I}_i}=\bigcup _{i=1}^k\sqrt{I_i}=\bigcup _{i=1}^k{I}_i=I_k.$$

$$\mathcal{I}=\{\beta \mid J(s{r}_{\beta }{)}^n⊈\sqrt{⟨r_{\alpha }\mid \alpha <\beta ⟩}\text{mbox}foralln\ge 1\text{mbox}andallJ\in \mathrm{GV}(R)\}$$

$$(J_{1}s{r}_{{\beta }_1}{)}^{n_1}\subseteq ⟨r_{\alpha }\mid \alpha <{\beta }_1⟩.$$

$$(J_{2}s{r}_{{\beta }_{2,i_k}}{)}^{n_2}\subseteq ⟨r_{\alpha }\mid \alpha <{\beta }_2⟩$$

$$(j_{1,l}{j}_{2,l^{\prime }}s{r}_{{\beta }_1}{)}^{n_1{n}_2{m}_2}$$

$$=(j_{1,l}{j}_{2,l^{\prime }}s\sum _{{\beta }_{2,i}\notin \mathcal{I}}{k}_i{r}_{{\beta }_{2,i}}{)}^{n_2{m}_2}+M,$$

$${\mathcal{I}}^{\prime }=\{{\beta }_1<{\beta }_2<{\beta }_3<\cdots \}\subseteq \mathcal{I}.$$

$$s⟨r_{\beta }\mid \beta \in {\mathcal{I}}^{\prime }⟩\subseteq \sqrt{⟨r_{\beta }\mid \beta \in {\mathcal{I}}^{\prime \prime }{⟩}_w}.$$

$$S_w=\{f\in R[x]|c(f)\in \mathrm{GV}(R)\},$$

$$s(P+Rx)\subseteq \sqrt{(I_x{)}_W},s(P+Ry)\subseteq \sqrt{(I_y{)}_W}$$

$$s^2(P+Rx)(P+Ry)$$

$$=\sqrt{(I_x{)}_W(I_y{)}_W}$$

$$\subseteq \sqrt{((I_x{)}_W(I_y{)}_W{)}_W}$$

$$\subseteq \sqrt{(I_x{I}_y{)}_W}.$$

$$s(P+(a){)}_W\subseteq \sqrt{(B+(a){)}_W}.$$

$$a^{m}g=fq+r\text{and}\mathrm{deg}(r)<\mathrm{deg}(f).$$

$$(ag{)}^m=fq{g}^{m-1}+r{g}^{m-1}\in ((f)+M{)}_W,$$

$$s^2{P}^2$$

$$\subseteq s^{2}P\sqrt{(B+(a){)}_W}$$

$$\subseteq s(\sqrt{P(B+(a))}{)}_W$$

$$\subseteq s(\sqrt{B+aP}{)}_W$$

$$\subseteq s\sqrt{(B+((f)+M{)}_W{)}_W}$$

$$\displaystyle\subseteq\sqrt{\bigl{(}B+(f)+sM\bigr{)}_{W}}$$

$$\displaystyle\subseteq\sqrt{\bigl{(}B+(f)+KR[X]\bigr{)}_{W}}.$$

Full text

Beside, we show a ring $R$ has Noetherian (resp., uniformly $S$-Noetherian) spectrum with respect to $s$ if and only if each countably generated ideal of $R$ is radically finite ($S$-finite with respect to $s$), which is a new result in the classical case.

1. Introduction

Throughout this paper, $R$ is always a commutative ring with identity. For a subset $U$ of an $R$-module $M$, we denote by $\langle U\rangle$ the $R$-submodule of $M$ generated by $U$. A subset $S$ of $R$ is called a multiplicative subset of $R$ if $1\in S$ and $s_{1}s_{2}\in S$ for any $s_{1}\in S$, $s_{2}\in S$.

Early in 1968, Ohm and Pendleton [4] introduced a ring $R$ with Noetherian spectrum, i.e., the prime spectrum ${\rm Spec}(R)$ of $R$ is Noetherian. Since the open subsets of ${\rm Spec}(R)$ and the radical ideals of $R$ correspond to each other one-to-one in order, a ring $R$ has the Noetherian spectrum if and only if the ascending chain condition holds for the radical ideals of $R$. In 1985, Ribenboim [5] showed that showed that a ring $R$ has the Noetherian spectrum if and only if every (prime) ideal of $R$ is radically of finite type (or radically finite), i.e., an ideal $I$ satisfies that $\sqrt{I}=\sqrt{J}$ for some finitely generated subideal $J$ of $I$, where $\sqrt{I}:=\{r\in R\mid r^{n}\in I\ \mbox{for some}\ n\geq 1\}$. Some more works on rings with Noetherian spectrum can be found in [1, 2, 7].

In 2014, Kim, Kwon and Rhee [3] called an integral domain $R$ having strong Mori spectrum (or $w$-Noetherian spectrum) if the induced topology on $w\mbox{-Spec}(R)$ by the Zariski topology on ${\rm Spec}(R)$ is Noetherian, or equivalently, it satisfies the descending chain condition on the sets of the form $W(I):=\{\,P\in w\mbox{-Spec}(R)\mid I\subseteq P\,\}$ where $I$ runs over $w$-ideals of $R$. Moreover, several characterizations of integral domains with strong Mori spectrum were given (see [3, Theorem 2.6]). Recently, Samir and Ahmed [6] led into the notion of rings with uniformly $S$-Noetherian spectrum. An ideal $I$ of $R$ is said to be radically $S$-finite (with respect to $s$) if $sI\subseteq\sqrt{J}$ for some finitely generated subideal $J$ of $I$ and some $s\in S$. A ring $R$ is said to have an uniformly $S$-Noetherian spectrum if there is $s\in S$ such that every ideal of $R$ is radically $S$-finite with respect to $s$. Similar characterizations of rings with uniformly $S$-Noetherian spectrum to classical cases can be found in [6, Theorem 3.11, Proposition 3.16]. Subsequently, the authors [12] investigated rings with uniforly $S$-Noetherian spectrum under various constructions,including flat overrings, Serre’s conjecture rings, Nagata rings, Anderson rings, and semigroup rings.

Since the $w$-operations and multiplicative subsets play a different methods to study rings and modules. It is very interesting to unify them together. For example, the uniformly $S$-$w$-Noetherian rings introduced in [11] are generalizations of both $w$-Noetherian rings and uniformly $S$-Noetherian rings. The main purpose of this paper is to generalize integral domains with strong Mori spectrum and rings with uniformly $S$-Noetherian spectrum. Actually, we introduce the notion of rings with uniformly $S$-$w$-Noetherian spectrum. An ideal $I$ of $R$ is said to be *radically $S$-$w$-finite * (with respect to $s$) if $sI\subseteq\sqrt{F_{w}}$ for some finitely generated subideal $F$ of $I$ and some $s\in S$. A ring $R$ is said to have uniformly $S$-$w$-Noetherian spectrum (with respect to $s$) if there exists $s\in S$ such that every ideal of $R$ is radically $S$-$w$-finite with respect to $s$. We characterize rings with uniformly $S$-$w$-Noetherian spectrum via various aspects:

Theorem 1.1**.**

$($=Theorem 3.8+Theorem 3.12$)$ Let $R$ be a ring, $S$ a multiplicative subset of $R$ and $s\in S$. Then the following statements are equivalent.

1. (1)

$R$* has uniformly $S$-$w$-Noetherian spectrum with respect to $s$.* 2. (2)

Each ascending chain of radical $w$-ideals of $R$ is stationary with respect to $s\in S$. 3. (3)

Each radical $(w$-$)$ideal of $R$ is radically $S$-$w$-finite with respect to $s$. 4. (4)

Each prime $(w$-$)$ideal of $R$ is radically $S$-$w$-finite with respect to $s$. 5. (5)

Each countably generated ideal of $R$ is radically $S$-$w$-finite with respect to $s$. 6. (6)

$R[X]$* has uniformly $S$-$w$-Noetherian spectrum with respect to $s$.* 7. (7)

$R\{X\}$* has uniformly $S$-Noetherian spectrum with respect to $s$.*

Beside, we show a ring $R$ has Noetherian (resp., uniformly $S$-Noetherian) spectrum with respect to $s$ if and only if each countably generated ideal of $R$ is radically finite ($S$-finite with respect to $s$), which is a new result in the classical case.

As our work involves $w$-operations, we recall the related notions from [9]. Let $R$ be a ring and $J$ a finitely generated ideal of $R$. Then $J$ is called a ${\rm GV}$-ideal if the natural homomorphism $R\rightarrow{\rm Hom}_{R}(J,R)$ is an isomorphism. The set of all ${\rm GV}$-ideals is denoted by ${\rm GV}(R)$. Let $M$ be an $R$-module. Define

${\rm tor_{\rm GV}}(M):=\{x\in M|Jx=0$, for some $J\in{\rm GV}(R)\}.$

An $R$-module $M$ is said to be ${\rm GV}$-torsion (resp. ${\rm GV}$-torsion-free) if ${\rm tor_{\rm GV}}(M)=M$ (resp. ${\rm tor_{\rm GV}}(M)=0$). A ${\rm GV}$-torsion-free module $M$ is called a $w$-module if ${\rm Ext}_{R}^{1}(R/J,M)=0$ for any $J\in{\rm GV}(R)$, and the $w$-envelope of $M$ is given by

$M_{w}:=\{x\in E(M)|Jx\subseteq M$, for some $J\in{\rm GV}(R)\},$

where $E(M)$ is the injective envelope of $M$. Therefore, $M$ is a $w$-module if and only if $M_{w}=M$. A ${\rm DW}$ ring $R$ is a ring for which every $R$-module is a $w$-module. Examples of ${\rm DW}$ rings include semihereditary rings and commutative rings with krull dimension $\leq 1$. The set of all prime $w$-ideals is denoted by $w$-${\rm Spec}(R)$. A maximal $w$-ideal is an ideal of $R$ which is maximal among the $w$-submodules of $R$. The set of all maximal $w$-ideals is denoted by $w$-${\rm Max}(R)$. All maximal $w$-ideals are prime ideals , i.e., $w$-${\rm Max}(R)\subseteq w$-${\rm Spec}(R)$ (see [9, Theorem 6.2.14]).

2. Rings with uniformly $S$-Noetherian spectrum

Recently, Samir and Hamed [6] introduced the notion of rings with uniformly $S$-Noetherian spectrum. An ideal $I$ of $R$ as being radically $S$-finite (with respect to $s$) if $sI\subseteq\sqrt{J}\subseteq\sqrt{I}$ for some finitely generated subideal $J$ of $I$ and some $s\in S$. A ring $R$ is said to have an uniforly $S$-Noetherian spectrum (abbreviated to be $u$-$S$-Noetherian spectrum) if there is $s\in S$ such that every ideal of $R$ is radically $S$-finite with respect to $s$.

They gave several characterizations of rings with uniforly $S$-Noetherian spectrum. Actually, they showed that

Theorem 2.1**.**

[6, Theorem 3.11, Proposition 3.16]** Let $R$ be a commutative ring, $S$ a multiplicative subset of $R$ and $s\in S$. Then the following statements are equivalent.

1. (1)

$R$* has uniformly $S$-Noetherian spectrum with respect to $s$.* 2. (2)

Each ascending chain of radical ideals of $R$ is stationary with respect to $s\in S$. 3. (3)

Each prime $($radical$)$ ideal of $R$ is radically finite with respect to $s\in S$.

It is well-known that a ring $R$ is a Noetherian ring (resp., PIR) if and only if every countably generated ideal of $R$ is finitely $($resp., principally$)$ generated. The author in [11] obtained that a ring $R$ is a uniformly $S$-Noetherian ring (resp., uniformly $S$-PIR) if and only if there is $s\in S$ such that every countably generated ideal of $R$ is $S$-finite $($resp., $S$-principal$)$ with respect to $s$ under a mild condition. The main aim in this section is to extend this result to rings with uniformly $S$-Noetherian spectrum.

Theorem 2.2**.**

Let $R$ be a commutative ring, $S$ a multiplicative subset of $R$ and $s\in S$. Then $R$ has uniformly $S$-Noetherian spectrum with respect to $s$ if and only if each countably generated ideal of $R$ is radically $S$-finite with respect to $s$.

Proof.

Suppose $R$ has uniformly $S$-Noetherian spectrum with respect to $s$. Then, trivially, each countably generated ideal of $R$ is radically $S$-finite with respect to $s$.

On the other hand, let $I=\langle r_{1},r_{2},\cdots\rangle$ be a countably generated ideal of $R$. Set $I_{n}=\sqrt{\langle r_{1},r_{2},\cdots,r_{n}\rangle}$. Then $\{I_{n}\mid n\geq 1\}$ is an ascending chain of radical ideals of $R$. So there is a positive integer $k$ such that $sI_{n}\subseteq I_{k}$ for all $n\geq k$. Note that $\sqrt{I}=\bigcup\limits_{n\geq 1}I_{n}$. Indeed, for any $r\in\sqrt{I}$, we have $r^{m}\in I$ for some $m\geq 1$ and so $r^{m}\in\langle r_{1},r_{2},\cdots,r_{n}\rangle$ for some $n\geq 1$. Hence $r\in I_{n}$, that is, $\sqrt{I}=\bigcup\limits_{n\geq 1}I_{n}$. Now, we claim that $s\sqrt{I}\subseteq I_{k}$. Indeed, let $r\in\sqrt{I}$. Then $r\in I_{n}$ for some $n\geq 1$. So $sr\in sI_{n}\subseteq I_{k}$. Hence $s\sqrt{I}\subseteq I_{k}$. Consequently, $I$ is radically $S$-finite with respect to $s$.

$(4)\Rightarrow(3)$ Let $I=\langle r_{\alpha}\mid\alpha<\kappa\rangle$ be an ideal of $R$, where $\kappa$ is a cardinal. Consider the index set

[TABLE]

We claim that for any $\boldsymbol{r_{\beta^{\prime}}\in I}$, we have $\boldsymbol{sr_{\beta^{\prime}}\in\sqrt{\langle r_{\beta}\mid\beta\in\mathcal{I}\rangle}}$. Since the first ordinal is in $\mathcal{I}$, we have $\mathcal{I}$ is nonempty. Let $r_{\beta^{\prime}}\in I$. If $\beta^{\prime}\in\mathcal{I}$, we are done. Otherwise, there exists $n_{1}\geq 1$ such that

[TABLE]

So $(sr_{\beta_{1}})^{n_{1}}=\sum k_{i}r_{\beta_{2,i}}$ with each $\beta_{2,i}<\beta_{1}$ and each $k_{i}\in R$. If all such $\beta_{2,i}$s belong to $\mathcal{I}$, we are done. Otherwise, let $\beta_{2,i_{1}},\cdots\beta_{2,i_{m_{2}}}$ be the finite set of all $\beta_{2,i}$s which are not in $\mathcal{I}$. Then, for all $k=1,\dots,m_{2}$, we have

[TABLE]

for some $n_{2}\geq 1$ and some $\beta_{2}<\beta_{1}$. Hence

[TABLE]

where $M$ is a multiple of $\sum\limits_{\beta_{2,i}\in\mathcal{I}}k_{i}r_{\beta_{2,i}}$. So $M$ is in $\langle r_{\beta}\mid\beta\in\mathcal{I}\rangle$. Note that $(s\sum\limits_{\beta_{2,i}\not\in\mathcal{I}}k_{i}r_{\beta_{2,i}})^{n_{2}m_{2}}$ is the linear combination of $(sk_{i}r_{\beta_{2,i}})^{n_{2}}$, and so is in $\langle r_{\alpha}\mid\alpha<\beta_{2}\rangle$. Hence $(sr_{\beta_{1}})^{n_{1}n_{2}m_{2}}$ is in $\langle r_{\beta}\mid\beta\in\mathcal{I}\ \mbox{or}\ \beta<\beta_{2}\rangle$. Continuing this process, there exists an sequence $r_{\beta_{1}},r_{\beta_{2}},r_{\beta_{3}},\dots$ with $\beta_{1}>\beta_{2}>\beta_{3}>\dots$. Note that $\{\beta_{i}\mid i\geq 1\}$ is a subset of the well-ordered set $\kappa$. So $\{\beta_{i}\mid i\geq 1\}$ has a minimal element, which is a contradiction. Consequently, the claim holds.

If $\mathcal{I}$ is finite, then $I$ is radically $S$-finite with respect to $s$, completing the proof. Now assume that $\mathcal{I}$ is infinite. Consider a countable subset

[TABLE]

Since $\langle r_{\beta}\mid\beta\in\mathcal{I}^{\prime}\rangle$ is radically $S$-finite with respect to $s$ by assumption, there exists a finite subset $\mathcal{I}^{\prime\prime}=\{\beta_{1},\beta_{2},\dots,\beta_{k}\}\subseteq\mathcal{I}^{\prime}$ such that

[TABLE]

So $(s\beta_{k+1})^{n_{k+1}}\in\langle r_{\beta}\mid\beta\in\mathcal{I}^{\prime\prime}\rangle$ for some $n_{k+1}>1.$ Consequently, $\beta_{k+1}\not\in\mathcal{I}$, which is a contradiction.

Consequently, $I$ is radically $S$-finite with respect to $s$, and it follows that $R$ has the uniformly $S$-Noetherian spectrum property with respect to $s$. ∎

Corollary 2.3**.**

Let $R$ be a commutative ring. Then $R$ has Noetherian spectrum if and only if each countably generated ideal of $R$ is radically finite.

Proof.

Take $S=\{1\}$ in Theorem 2.2. ∎

3. Rings with uniformly $S$-$w$-Noetherian spectrum

In 2014, Kim, Kwon, and Rhee [3] defined an integral domain having strong Mori spectrum. The related notions can be trivially extend to commutative rings with zero divisors, which can be called rings with $w$-Noetherian spectrum. An ideal $I$ of $R$ is said to be *radically $w$-finite *, if $I=\sqrt{F_{w}}$ for some finitely generated subideal $F$ of $I$. And a ring $R$ is said to have $w$-Noetherian spectrum if every ideal of $R$ is radically $w$-finite.

Next, we will generalize integral domains with strong Mori spectrum and rings with uniformly $S$-Noetherian spectrum.

Definition 3.1**.**

Let $R$ be a ring, and $S$ a multiplicative subset of $R$.

1. (1)

An ideal $I$ of $R$ is said to be radically $S$-$w$-finite $($abbreviated to be $u$-$S$-$w$-Noetherian spectrum$)($with respect to $s)$ if $sI\subseteq\sqrt{F_{w}}$ for some finitely generated subideal $F$ of $I$ and some $s\in S$. 2. (2)

A ring $R$ is said to have uniformly $S$-$w$-Noetherian spectrum $($abbreviated to be $u$-$S$-$w$-Noetherian spectrum$)($with respect to $s)$ if there exists $s\in S$ such that every ideal of $R$ is radically $S$-$w$-finite with respect to $s$.

If $R$ is a ${\rm DW}$-ring, then rings with uniformly $S$-$w$-Noetherian spectrum have uniformly $S$-Noetherian spectrum. It is clear that if $S_{1}$ is a multiplicative subset of $R$ contained in $S_{2}$, then rings with uniformly $S_{1}$-$w$-Noetherian spectrum also have uniformly $S_{2}$-$w$-Noetherian spectrum. The saturation $S^{\ast}$ of a multiplicative subset $S$ of $R$ is defined as

[TABLE]

A multiplicative subset $S$ of $R$ is called saturated if $S=S^{\ast}$. Note that $S^{\ast}$ is always a saturated multiplicative subset containing $S$. Clearly, rings with uniformly $S^{\ast}$-$w$-Noetherian spectrum are exactly those with uniformly $S$-$w$-Noetherian spectrum.

We trivially have the following implications:

[TABLE]

Rings with uniformly $S$-$w$-Noetherian spectrum may not have uniformly $S$-Noetherian spectrum.

Example 3.2**.**

Let $R=k[x_{1},x_{2},\cdots]$ with $k$ a field. Set $S=\{1\}.$ Then $R$ has uniformly $S$-$w$-Noetherian spectrum as $R$ is a UFD. However, $R$ does not have uniformly $S$-Noetherian spectrum, since the prime ideals $\langle x_{1}\rangle\subseteq\langle x_{1},x_{2}\rangle\subseteq\cdots$ is not stationary.

Let $R$ be a ring and $S$ a multiplicative subset of $R$. For any $s\in S$, there is a multiplicative subset $S_{s}=\{1,s,s^{2},....\}$ of $S$. We denote by $M_{s}$ the localization of $M$ at $S_{s}$. Certainly, $M_{s}\cong M\otimes_{R}R_{s}$. A multiplicative subset of $R$ is said to be regular if it consists of non-zero-divisors.

Proposition 3.3**.**

Let $R$ be a ring, $S$ a regular multiplicative subset of $R$ and $s\in S$. Suppose $R$ has uniformly $S$-$w$-Noetherian spectrum with respect to $s$. Then $R_{s}$ has $w$-Noetherian spectrum.

Proof.

Let $I_{s}$ be an ideal of $R_{s}$ with $I$ an ideal of $R$. Then there is a finitely generated ideal $F$ of $I$ such that if $sI\subseteq\sqrt{F_{w}}=(\sqrt{F})_{w}$ (Note that $\sqrt{I_{w}}=(\sqrt{I})_{w}$ for any ideal $I$ of $R$). For any $r\in I$, we have $sr\in\sqrt{F}_{w}$ Hence there is an ideal $J\in{\rm GV}(R)$ such that $Jsr\in\sqrt{F}$. Hence $J_{s}\frac{r}{1}\in\sqrt{F}_{s}=\sqrt{F_{s}}$. Since $J_{s}\in{\rm GV}(R_{s})$ by [10, Lemma 4.3(1)], $\frac{r}{1}\in\sqrt{(F_{s})_{w}}.$ Hence $I_{s}\subseteq\sqrt{(F_{s})_{w}}$. Consequently, $R_{s}$ has $w$-Noetherian spectrum. ∎

Proposition 3.4**.**

Let $n\geq 2$ be an integer, $R_{1},\dots,R_{n}$ be rings and $S_{1},\dots,S_{n}$ be multiplicative subsets of $R_{1},\dots,R_{n}$, respectively. Then the following statements are equivalent.

1. (1)

$R_{k}$* has $S_{k}$-$w$-Noetherian spectrum for all $k=1,\dots,n$.* 2. (2)

$\prod_{k=1}^{n}R_{k}$* has \Bigl{(}\prod_{k=1}^{n}S_{k}\Bigr{)}-$w$-Noetherian spectrum.*

Proof.

Write $R:=\prod_{k=1}^{n}R_{k}$ and $S:=\prod_{k=1}^{n}S_{k}$. Let $w$ be the $w$-operation of $R$, and $w_{k}$ be the $w$-operation of $R_{k}$ for each $k=1,\dots,n$.

$(1)\Rightarrow(2)$ Let $I$ be an ideal of $R$. Then for each $k=1,\dots,n$, we can find an ideal $I_{k}$ of $R_{k}$ such that $I=\prod_{k=1}^{n}I_{k}$. Since each $R_{k}$ has $S_{k}$-$w_{k}$-Noetherian spectrum, there exist an element $s_{k}\in S_{k}$ and a finitely generated subideal $J_{k}$ of $I_{k}$ such that

[TABLE]

Note that $\prod_{k=1}^{n}\sqrt{(J_{k})_{w_{k}}}=\sqrt{(\prod_{k=1}^{n}J_{k})_{w}}$; so

[TABLE]

Hence $I$ is radically $S$-$w$-finite. Thus $R$ has $S$-$w$-Noetherian spectrum.

(2) $\Rightarrow$ (1) Suppose that $R$ has $S$-$w$-Noetherian spectrum and for each $k\in\{1,\dots,n\}$, let $I_{k}$ be an ideal of $R_{k}$. Then $I:=\prod_{k=1}^{n}I_{k}$ is an ideal of $R$; so we can find an element $(s_{1},\dots,s_{n})\in S$ and a finitely generated subideal $J$ of $I$ such that $(s_{1},\dots,s_{n})I\subseteq\sqrt{J_{w}}$. Note that for each $k=1,\dots,n$, there exists a finitely generated ideal $J_{k}$ of $R_{k}$ such that $J=\prod_{k=1}^{n}J_{k}$. Hence for each $k\in\{1,\dots,n\}$,

[TABLE]

Thus for each $k=1,\dots,n$, $I_{k}$ is radically $S_{k}$-$w_{k}$-finite, which means that $R_{k}$ has $S_{k}$-$w_{k}$-Noetherian spectrum. ∎

Rings with uniformly $S$-$w$-Noetherian spectrum may not have uniformly $w$-Noetherian spectrum.

Example 3.5**.**

Let $R_{1}$ be a ring with $w$-Noetherian spectrum, and $R_{2}$ be a ring without $w$-Noetherian spectrum. Set $R=R_{1}\times R_{2}$, and $S=\{1\}\times\{0,1\}$. Then $R$ has uniformly $S$-$w$-Noetherian spectrum but does not have $w$-Noetherian spectrum by Proposition 3.4.

Let $P$ be a prime ideal of $R$. We call a ring $R$ has uniformly $P$-$w$-Noetherian spectrum provided that $R$ has uniformly $(R-P)$-$w$-Noetherian spectrum.

Proposition 3.6**.**

The following statements are equivalent for a ring $R$.

1. (1)

$R$* has $w$-Noetherian spectrum.* 2. (2)

$R$* has uniformly $P$-$w$-Noetherian spectrum for any $P\in\mathrm{Spec}(R)$.* 3. (3)

$R$* has uniformly $M$-$w$-Noetherian spectrum for any $M\in\mathrm{Max}(R)$.* 4. (4)

$R$* has uniformly $P$-$w$-Noetherian spectrum for any $P\in w\mbox{-}\mathrm{Spec}(R)$.* 5. (5)

$R$* has uniformly $M$-$w$-Noetherian spectrum for any $M\in w\mbox{-}\mathrm{Max}(R)$.*

Proof.

$(1)\Rightarrow(2)\Rightarrow(3)$ and $(1)\Rightarrow(4)\Rightarrow(5)$: These implications are trivial.

$(3)\Rightarrow(1)$: Let $I$ be an ideal of $R$. By (3), for any $M\in\mathrm{Max}(R)$, there exists an $s_{M}\in R\setminus M$ such that $s_{M}I\subseteq\sqrt{(F_{M})_{w}}$ for some finitely generated subideal $F_{M}$ of $I$. Let $K$ be the ideal of $R$ generated by the set $\{s_{M}\mid M\in\mathrm{Max}(R)\}$. Then $K=R$. Hence $1=s_{M_{1}}r_{1}+\dots+s_{M_{n}}r_{n}$ for some $r_{1},\dots,r_{n}\in R$. Set $F:=\sum_{i=1}^{n}F_{M_{i}}$. Then $F$ is a finitely generated subideal of $I$. Furthermore, $I=1\cdot I=(s_{M_{1}}r_{1}+\dots+s_{M_{n}}r_{n})I\subseteq s_{M_{1}}I+\dots+s_{M_{n}}I\subseteq\sum_{i=1}^{n}\sqrt{(F_{M_{i}})_{w}}\subseteq\sqrt{\sum_{i=1}^{n}(F_{M_{i}})_{w}}\subseteq\sqrt{F_{w}}\subseteq\sqrt{I_{w}}$. Hence $R$ has $w$-Noetherian spectrum.

$(5)\Rightarrow(1)$: Let $I$ be an ideal of $R$. By (5), for any $M\in w\mbox{-}\mathrm{Max}(R)$, there exists an $s_{M}\in R\setminus M$ such that $s_{M}I\subseteq\sqrt{(F_{M})_{w}}$ for some finitely generated subideal $F_{M}$ of $I$. Let $K$ be the ideal of $R$ generated by the set $\{s_{M}\mid M\in w\mbox{-}\mathrm{Max}(R)\}$. Then $K_{w}=R$. Hence there exists $\{s_{M_{1}},\dots,s_{M_{n}}\}$ such that $J:=\langle s_{M_{1}},\dots,s_{M_{n}}\rangle\in{\rm GV}(R)$. Set $F:=\sum_{i=1}^{n}F_{M_{i}}$. Then $F$ is a finitely generated subideal of $I$. Furthermore, $JI=s_{M_{1}}I+\dots+s_{M_{n}}I\subseteq\sum_{i=1}^{n}\sqrt{(F_{M_{i}})_{w}}\subseteq\sqrt{\sum_{i=1}^{n}(F_{M_{i}})_{w}}\subseteq\sqrt{F_{w}}\subseteq\sqrt{I_{w}}$. Hence $I_{w}\subseteq\sqrt{F_{w}}\subseteq\sqrt{I_{w}}$ , implying that $R$ has $w$-Noetherian spectrum. ∎

Lemma 3.7**.**

Let $R$ be a ring, $S$ a multiplicative subset of $R$ and $s\in S$. Let $I$ and $K$ be two ideals of $R$ such that $I_{w}=K_{w}$. Then $I$ is radically $S$-$w$-finite with respect to $s$ if and only if so is $K$.

Proof.

Suppose $I$ is radically $S$-$w$-finite with respect to $s$. $sI\subseteq\sqrt{F_{w}}$ for some finitely generated subideal $F=\langle f_{1},\cdots,f_{n}\rangle$ of $I$. Since $I_{w}=K_{w}$, there is $J\in{\rm GV}(R)$ such that $JF\subseteq K.$ Hence $sK\subseteq sK_{w}=sI_{w}\subseteq(sI)_{w}\subseteq(\sqrt{F_{w}})_{w}=\sqrt{F_{w}}=\sqrt{(J_{w}F_{w})_{w}}=\sqrt{(JF)_{w}},$ by [9, [Proposition 6.2.3(5), Proposition 6.2.3]. Since $JF$ is finitely generated, $K$ is radically $S$-$w$-finite with respect to $s$. The converse is similarly true. ∎

Theorem 3.8**.**

Let $R$ be a ring, $S$ a multiplicative subset of $R$ and $s\in S$. Then the following statements are equivalent.

1. (1)

$R$* has uniformly $S$-$w$-Noetherian spectrum with respect to $s$.* 2. (2)

Each ascending chain of radical $w$-ideals of $R$ is stationary with respect to $s\in S$. 3. (3)

Each radical $(w$-$)$ideal of $R$ is radically $S$-$w$-finite with respect to $s$. 4. (4)

Each prime $(w$-$)$ideal of $R$ is radically $S$-$w$-finite with respect to $s$. 5. (5)

Each countably generated ideal of $R$ is radically $S$-$w$-finite with respect to $s$.

Proof.

The equivalence of $w$-ideals and ideals in $(3),(4)$ can be deduced by Lemma 3.7.

$(1)\Rightarrow(3)\Rightarrow(4)$ Trivial.

$(4)\Rightarrow(3)$ On contrary, assume that there exists a radical $w$-ideal $K$ of $R$ which is not radically $S$-$w$-finite with respect to $s$. Set

$\mathcal{F}=\{$radical $w$-ideals of $R$ that are not radically $S$-$w$-finite type with respect to $s\}.$

Since $K\in\mathcal{F}$, we have $\mathcal{F}\neq\emptyset$. Ordered $\mathcal{F}$ by inclusion. Let $\{I_{\lambda}\}_{\lambda\in\Lambda}$ be a chain in $\mathcal{F}$ and set $I=\bigcup_{\lambda\in\Lambda}I_{\lambda}.$ Then $I$ is a radical $w$-ideal of $R.$ Indeed, let $r\in\sqrt{I}$, then $r^{k}\in I=\bigcup_{\lambda\in\Lambda}I_{\lambda}$ for some positive integer $k.$ So $r^{k}\in I_{\lambda}$ for some $\lambda\in\Lambda.$ Thus $r\in\sqrt{I_{\lambda}}=I_{\lambda}\subseteq I.$ And note that the directed unions of $w$-ideals is also a $w$-ideal (see [9, Proposition 6.1.19]). We claim that $I\in\mathcal{F}$. Assume that $I\notin\mathcal{F}.$ There exists a finitely generated subideal $J$ of $I$ such that $sI\subseteq\sqrt{J_{w}}$. Since $J$ is finitely generated, there exists $\lambda\in\Lambda$ such that $J\subseteq I_{\lambda}$, and so $sI\subseteq sI_{\lambda}\subseteq\sqrt{J_{w}}\subseteq\sqrt{(I_{\lambda})_{w}}$, a contradiction. Hence, by Zorn’s Lemma, there is a maximal element $P$ of $\mathcal{F}$.

We claim that the maximal element $P$ of $\mathcal{F}$ is a prime ideal of $R.$ Indeed, on contrary, assume that $P$ is not a prime ideal of $R.$ There exist $a,b\in R\setminus P$ such that $ab\in P.$ We put $I:=P+aR$ and $K:=P+bR$. Since $IK=P^{2}+(aR)P+(bR)P+(ab)R\subseteq P,$ we have $\sqrt{IK}\subseteq\sqrt{P}=P.$ Also $P\subseteq I\cap K\subseteq\sqrt{I\cap K}=\sqrt{IK},$ thus $P=\sqrt{IK}$. On the other hand, $P\subsetneq I\subseteq\sqrt{I}$ and $P\subsetneq K\subseteq\sqrt{K}.$ Then by maximality of $P,$ there exist finitely generated subideals $I^{\prime}$ of $I$ and $K^{\prime}$ of $K$ such that $s\sqrt{I}\subseteq\sqrt{(I^{\prime})_{w}}$ and $s\sqrt{K}\subseteq\sqrt{(K^{\prime})_{w}}$. Let $x\in\sqrt{I}$ and $y\in\sqrt{K}$. Then $sx\in\sqrt{(I^{\prime})_{w}}$ and $sy\in\sqrt{(K^{\prime})_{w}}$, which implies that $(sx)^{k}\in(I^{\prime})_{w}$ and $(sy)^{k^{\prime}}\in(K^{\prime})_{w}$ for some positive integers $k,k^{\prime}$. Thus there exists $L_{1},L_{2}\in{\rm GV}(R)$ such that $(L_{1}L_{2}sxy)^{k+k^{\prime}}\subseteq I^{\prime}K^{\prime}$, so $L_{1}L_{2}sxy\in\sqrt{I^{\prime}K^{\prime}}$. Hence $sxy\in(\sqrt{I^{\prime}K^{\prime}})_{w}=\sqrt{(I^{\prime}K^{\prime})_{w}}$ This shows that $s\sqrt{I}\sqrt{K}\subseteq\sqrt{(I^{\prime}K^{\prime})_{w}}$. Thus $sP\subseteq s\sqrt{IK}=s\sqrt{I}\sqrt{K}\subseteq\sqrt{(I^{\prime}K^{\prime})_{w}},$ which contradicts the maximality of $P$. So $P$ is a prime ideal of $R$ which is not radically $S$-$w$-finite with respect to $s$, a contradiction.

$(3)\Rightarrow(1)$ Let $I$ be an ideal of $R$. By (3), $\sqrt{I}$ is radically $S$-$w$-finite with respect to $s$, then $s\sqrt{I}\subseteq\sqrt{(a_{1},\dots,a_{n})_{w}}$ for some $a_{1},\dots,a_{n}\in\sqrt{I}$. So for all $1\leq i\leq n$, $a_{i}^{r_{i}}\in I$ for some positive integer $r_{i}$. For each $1\leq i\leq n$, let $b_{i}=a_{i}^{r_{i}}$ and let $F:=(b_{1},\dots,b_{n})$. Then $F$ is a finitely generated subideal of $I$. Moreover, $sI\subseteq s\sqrt{I}\subseteq\sqrt{(a_{1},\dots,a_{n})_{w}}\subseteq\sqrt{F_{w}}\subset\sqrt{I_{w}}$.

$(1)\Rightarrow(2)$ Let $I_{1}\subseteq I_{2}\subseteq I_{3}\subseteq\cdots$ be ascending chain of radical $w$-ideals of $R$. Let $I=\bigcup\limits_{i=1}^{\infty}I_{i}$. Then $I$ is radically $S$-$w$-finite with respect to $s$, that is, $sI\subseteq\sqrt{J_{w}}$ for some finitely generated subideal $J$ of $I$. Assume that $J\subseteq\bigcup\limits_{i=1}^{k}I_{i}$. Then

[TABLE]

Consequently, $sI_{n}\subseteq I_{k}$ for any $n\geq k.$

$(2)\Rightarrow(5)$ Let $I=\langle r_{1},r_{2},\cdots\rangle$ be a countably generated ideal of $R$. Set $I_{n}=\sqrt{\langle r_{1},r_{2},\cdots,r_{n}\rangle}$. Then $\{(I_{n})_{w}\mid n\geq 1\}$ is an ascending chain of radical $w$-ideals of $R$ (Note that $(\sqrt{I})_{w}=\sqrt{I_{w}}$ for each ideal $I$ of $R$). So there is a positive integer $k$ such that $s(I_{n})_{w}\subseteq(I_{k})_{w}$ for all $n\geq k$. Note that $\sqrt{I_{w}}=\sqrt{I}_{w}=\bigcup\limits_{n\geq 1}(I_{n})_{w}$. Now, we claim that $s\sqrt{I_{w}}\subseteq(I_{k})_{w}$. Indeed, let $r\in\sqrt{I_{w}}$. Then $r\in(I_{n})_{w}$ for some $n\geq 1$. So $sr\in s(I_{n})_{w}\subseteq(I_{k})_{w}$. Hence $sI\subseteq sI_{w}\subseteq s\sqrt{I_{w}}\subseteq(I_{k})_{w}$. Consequently, $I_{w}$ is radically $S$-$w$-finite with respect to $s$.

$(5)\Rightarrow(1)$ Let $I=\langle r_{\alpha}\mid\alpha<\kappa\rangle$ be an ideal of $R$, where $\kappa$ is a cardinal. Consider the index set

[TABLE]

We claim that for any $\boldsymbol{r_{\beta^{\prime}}\in I}$, we have $\boldsymbol{Jsr_{\beta^{\prime}}\in\sqrt{\langle r_{\beta}\mid\beta\in\mathcal{I}\rangle}}$ for some $\boldsymbol{J\in{\rm GV}(R)}$. Since the first ordinal is in $\mathcal{I}$, we have $\mathcal{I}$ is nonempty. Let $r_{\beta^{\prime}}\in I$. If $\beta^{\prime}\in\mathcal{I}$, we are done. Otherwise, there exists $n_{1}\geq 1$ and $J_{1}=\langle j_{1,1},\dots,j_{1,l_{1}}\rangle\in{\rm GV}(R)$ such that

[TABLE]

So $(j_{1,l}sr_{\beta_{1}})^{n_{1}}=\sum k_{i}r_{\beta_{2,i}}$ with each $\beta_{2,i}<\beta_{1}$, each $k_{i}\in R$ and each $l=1,\dots,l_{1}$. If all such $\beta_{2,i}$s belong to $\mathcal{I}$, we are done. Otherwise, let $\beta_{2,i_{1}},\cdots\beta_{2,i_{m_{2}}}$ be the finite set of all $\beta_{2,i}$s which are not in $\mathcal{I}$. Then, for all $k=1,\dots,m_{2}$, there exists $J_{2}=\langle j_{2,1},\dots,j_{2,l_{2}}\rangle\in{\rm GV}(R)$ such that

[TABLE]

for some $n_{2}\geq 1$ and some $\beta_{2}<\beta_{1}$. Hence for all $l=1,\dots,l_{1}$ and $l^{\prime}=1,\dots,l_{2}$, we have

[TABLE]

where $M$ is a multiple of $\sum\limits_{\beta_{2,i}\in\mathcal{I}}k_{i}r_{\beta_{2,i}}$. So $M$ is in $\langle r_{\beta}\mid\beta\in\mathcal{I}\rangle$. Note that $(j_{1,l}j_{2,l^{\prime}}s\sum\limits_{\beta_{2,i}\not\in\mathcal{I}}k_{i}r_{\beta_{2,i}})^{n_{2}m_{2}}$ is the linear combination of $(j_{1,l}j_{2,l^{\prime}}sk_{i}r_{\beta_{2,i}})^{n_{2}}$, and so is in $\langle r_{\alpha}\mid\alpha<\beta_{2}\rangle$. Hence $(J_{1}J_{2}sr_{\beta_{1}})^{n_{1}n_{2}m_{2}}$ is contained in $\langle r_{\beta}\mid\beta\in\mathcal{I}\ \mbox{or}\ \beta<\beta_{2}\rangle$. Continuing this process, there exists an sequence $r_{\beta_{1}},r_{\beta_{2}},r_{\beta_{3}},\dots$ with $\beta_{1}>\beta_{2}>\beta_{3}>\dots$. Note that $\{\beta_{i}\mid i\geq 1\}$ is a subset of the well-ordered set $\kappa$. So $\{\beta_{i}\mid i\geq 1\}$ has a minimal element, which is a contradiction. Consequently, the claim holds.

If $\mathcal{I}$ is finite, then $I$ is radically $S$-$w$-finite with respect to $s$, completing the proof. Now assume that $\mathcal{I}$ is infinite. Consider a countable subset

[TABLE]

Since $\langle r_{\beta}\mid\beta\in\mathcal{I}^{\prime}\rangle$ is radically $S$-$w$-finite with respect to $s$ by assumption, there exists a finite subset $\mathcal{I}^{\prime\prime}=\{\beta_{1},\beta_{2},\dots,\beta_{k}\}\subseteq\mathcal{I}^{\prime}$ such that

[TABLE]

So $(Js\beta_{k+1})^{n_{k+1}}\subseteq\langle r_{\beta}\mid\beta\in\mathcal{I}^{\prime\prime}\rangle$ for some $n_{k+1}>1$ and some $J\in{\rm GV}(R)$. Consequently, $\beta_{k+1}\not\in\mathcal{I}$, which is a contradiction.

Consequently, $I$ is radically $S$-$w$-finite with respect to $s$, and it follows that $R$ has the uniformly $S$-$w$-Noetherian spectrum property with respect to $s$. ∎

Corollary 3.9**.**

Let $R$ be a ring. Then the following statements are equivalent.

1. (1)

$R$* has $w$-Noetherian spectrum.* 2. (2)

Each ascending chain of radical $w$-ideals of $R$ is stationary. 3. (3)

Each radical $(w$-$)$ideal of $R$ is radically $w$-finite. 4. (4)

Each prime $(w$-$)$ideal of $R$ is radically $w$-finite. 5. (5)

Each countably generated ideal of $R$ is radically $w$-finite.

Proof.

Take $S=\{1\}$ in Theorem 3.8. ∎

To build a connection between the $w$-operation case and ordinal case, Wang and Kim [8] introduced the notion of the $w$-Nagata ring, $R\{x\}$, of $R$. It is a localization of $R[X]$ at the multiplicative closed set

[TABLE]

where $c(f)$ is the content of $f$. Note that $R\{x\}$ can naturally be seen as an extension of $R$. The $w$-Nagata module $M\{x\}$ of an $R$-module $M$ is defined as $M\{x\}=M[x]_{S_{w}}\cong M\bigotimes_{R}R\{x\}$.

Lemma 3.10**.**

Let $I$ be an ideal of $R$. Then $\sqrt{I}\{X\}=\sqrt{I\{X\}}.$

Proof.

Let $\frac{f}{g}\in\sqrt{I}\{X\}$ with $f=a_{n}X^{n}+\cdots+a_{0}\in\sqrt{I}[X]$ and $c(g)\in{\rm GV}(R)$. Then there exists $m\geq 1$ such that $a_{i}^{m}\in I$ for each $i=0,\cdots,n$. So $f^{m(n+1)}\in I[x]$. Hence $(\frac{f}{g})^{m(n+1)}\in I\{X\}.$ Consequently, $\frac{f}{g}\in\sqrt{I\{X\}}$. Hence$\sqrt{I}\{X\}\subseteq\sqrt{I\{X\}}$.

On the other hand, then $a_{n}^{m}\in$ obviously. Let $\frac{f}{g}\in\sqrt{I\{X\}}$. Then there exists $m\geq 1$ such that $(\frac{f}{g})^{m}\in I\{X\}$ with $f^{m}\in I[X]$ and $c(g^{m})\in{\rm GV}(R)$. Since $c(g^{m})\subseteq c(g)$, we have $c(g)\in{\rm GV}(R).$ Assume $f=a_{n}X^{n}+\cdots+a_{0}.$ We will show each $a_{i}\in\sqrt{I}$ by induction on $n$. If $n=0$, it is trivial. For general case, let $f_{1}=f-a_{n}X^{n}$. Then $f^{m}=\sum\limits_{i=0}^{m}C_{m}^{i}a_{n}^{m-i}X^{n(m-i)}f_{1}^{i}=a_{n}^{m}X^{nm}+\cdots\in I[X].$ Then $a_{n}^{m}\in I$, and thus $a_{n}\in\sqrt{I}$. Note that $f_{1}^{2m}\in I[x]$ and $\deg(f_{1})<\deg(f)$. One can show that each $a_{i}\in\sqrt{I}$ by induction. So $f\in\sqrt{I}[x]$. Hence$\sqrt{I}\{X\}\supseteq\sqrt{I\{X\}}$.

∎

Proposition 3.11**.**

Let $R$ be a ring, $S$ a multiplicative subset of $R$ and $s\in S$. Then an ideal $I$ of $R$ is radically $S$-$w$-finite with respect to $s$ if and only if $I\{X\}$ is radically $S$-finite with respect to $s$.

Proof.

Suppose $I$ is radically $S$-$w$-finite with respect to $s$. Then there exists a finitely generated subideal $F$ of $I$ such that $sI\subseteq\sqrt{F_{w}}$. Then $sI\{X\}\subseteq\sqrt{F_{w}}\{X\}=\sqrt{F_{w}\{X\}}=\sqrt{F\{X\}}$ by Lemma 3.10. Hence $I\{X\}$ is radically $S$-$w$-finite with respect to $s$.

On the other hand, suppose $I\{X\}$ is radically $S$-finite with respect to $s$. Then $sI\{X\}\subseteq\sqrt{A_{S_{w}}}$ for some finitely generated subideal $A$ of $I[X]$. So $c(A)$ is a finitely generated subideal of $I$. Since $A\subseteq c(A)[\textsc{X}]\subseteq I[X]$, we have $sI\{X\}\subseteq\sqrt{A_{S_{w}}}\subseteq\sqrt{c(A)\{X\}}=\sqrt{c(A)}\{X\}$ by Lemma 3.10. Consequently, we have $sI\subseteq(sI)_{w}\subseteq(\sqrt{c(A)}))_{w}=\sqrt{c(A)_{w}}$. Hence, $I$ is radically $S$-$w$-finite with respect to $s$. ∎

Theorem 3.12**.**

Let $R$ be a ring, $S$ a multiplicative subset of $R$ and $s\in S$. Then the following statements are equivalent.

1. (1)

$R$* has uniformly $S$-$w$-Noetherian spectrum.* 2. (2)

$R[X]$* has uniformly $S$-$w$-Noetherian spectrum.* 3. (3)

$R\{X\}$* has uniformly $S$-Noetherian spectrum.*

Proof.

We denote by the $w$-operation of $R$ by $w$, and the $w$-operation of $R[x]$ by $W$, respectively.

$(1)\Rightarrow(2)$ Suppose that $R$ has the uniformly $S$-$w$-Noetherian spectrum property with respect to some $s\in S$. We claim that $R[X]$ also has the uniformly $S$-$w$-Noetherian spectrum property with respect to $s$.

Let $P$ be a $w$-ideal of $R[X]$ that is maximal among those that are not radically $S$-$w$-finite with respect to $s$. Then $P$ is a prime ideal of $R[X]$. Indeed, suppose $xy\in P$ with $x\notin P$ and $y\notin P$. Then both $P+Rx$ and $P+Ry$ are $S$-$w$-finite with respect to $s$, implying

[TABLE]

for some finitely generated ideals $I_{x}$ and $I_{y}$ of $P+Rx$ and $P+Ry$, respectively. It follows that

[TABLE]

Thus, $s^{2}P\subseteq\sqrt{(I_{x}I_{y})_{W}}$, leading to a contradiction since $sP\subseteq\sqrt{(I_{x}I_{y})_{W}}$.

Set $\mathfrak{p}=P\cap R$. Then $\mathfrak{p}$ is a prime $w$-ideal of $R$. So $s\mathfrak{p}\subseteq\sqrt{K_{w}}$ for some finitely generated subideal $K$ of $\mathfrak{p}$. Set $M=(\mathfrak{p}R[x])_{W}$. Then $sM=s(\mathfrak{p}R[x])_{W}\subseteq(\sqrt{K}R[x])_{W}$ by [9, Proposition 6.2.3]. So $M$ is a radically $S$-$w$-finite $w$-ideal with respect to $s$. Note $M=(\mathfrak{p}R[x])_{W}=((P\cap R)R[x])_{W}\subseteq P_{W}=P.$ Hence $M\subsetneq P.$

Let $f$ be an element of least degree in $P\setminus M$. Let $n$ be the degree of $f$ and let $a$ be the coefficient of $X^{n}$ in $f$. Then $n\geq 1$ and $a\notin P$. Indeed, if $n=0$, then $f=a\in P\cap R\subseteq M$. This contradicts the choice of $f$. Hence $n\geq 1$. Next, suppose to the contrary that $a\in P$. Then $f-aX^{n}\in P$. Since $\deg(f-aX^{n})<\deg(f)$, $f-aX^{n}\in M$ by the minimality of $n$. Also, note that $a\in P\cap R$; so $aX^{n}\in(P\cap R)R[X]\subseteq M$. Therefore $f=(f-aX^{n})+aX^{n}\in M$, which is a contradiction. Hence $a\notin P$.

Since $a\notin P$ and $P$ is a $w$-ideal of $R[x]$, $P\subsetneq(P+(a))_{W}$. So by the maximality of $P$, $(P+(a))_{W}$ is a $w$-finite $w$-ideal of $R[X]$. Therefore there exists a finitely generated ideal $F$ of $R[X]$ such that $s(P+(a))_{W}\subseteq\sqrt{F_{W}}$, where $F=B+(a)$ for some finitely generated subideal $B$ of $P$, that is,

[TABLE]

Let $g\in P$. If $g\in M$, then $ag\in((f)+M)_{W}$. Now suppose that $g\in P\setminus M$. Let $m$ be the degree of $g$. Then $m\geq n$. Notice the minimality of $n$. By a routine iterative calculation, one can find suitable elements $q,r\in R[X]$ such that

[TABLE]

Therefore $r=a^{m}g-fq\in P$. By the minimality of $n$, we have $r\in M$. Also, note that

[TABLE]

so $ag\in\sqrt{((f)+M)_{W}}$. Hence $aP\subseteq\sqrt{((f)+M)_{W}}$.

Now, we obtain

[TABLE]

Hence sP\subseteq\sqrt{\bigl{(}B+(f)+KR[X]\bigr{)}_{W}} implying $P$ is a $S$-$w$-finite ideal of $R[X]$ with respect to $s$, as $B+(f)+KR[X]$ is a finitely generated subideal of $P$. This is a contradiction to the fact that $P$ is not an $S$-$w$-finite ideal of $R[X]$ with respect to $s$. Thus $R[X]$ has $S$-$w$-Noetherian spectrum with respect to $s$.

$(2)\Rightarrow(1)$ Let $I$ be an ideal of $R$. Then $IR[X]$ is an ideal of $R[X]$. Since $R[X]$ has uniformly $S$-$w$-Noetherian spectrum with respect to $s$, $IR[X]$ is a radically $S$-$w$-finite ideal of $R[X]$ with respect to $s$. So there is a finitely generated subideal $K$ of $I$ such that

[TABLE]

Therefore, it follows by follows by [9, Proposition 6.5.2] that

[TABLE]

Hence $I$ is a $S$-$w$-finite ideal of $R$ with respect to $s$. Thus $R$ has $S$-$w$-Noetherian spectrum with respect to $s$.

$(2)\Rightarrow(3)$: Let $I$ be an ideal of $R\{x\}.$ Then there exists an ideal $K$ of $R[X]$ such that $I=K_{S_{w}}$. Since $R[X]$ has $w$-Noetherian spectrum, there exists a finitely generated subideal $L$ of $K$ such that $sK\subseteq\sqrt{L_{W}}$. So for any $f\in K$ there is $J\in{\rm GV}(R)$ and a positive integer $n\geq 1$ such that $(Jsf)^{n}\subseteq L$. Since $(J^{n})_{S_{w}}=R\{x\}$, we have $(s\frac{f}{1})^{n}\in L_{S_{w}}$. Hence $sI\subseteq\sqrt{L_{S_{w}}}$. Consequently, $R\{x\}$ has uniformly $S$-Noetherian spectrum with respect to $s$.

$(3)\Rightarrow(1)$: Suppose $R\{x\}$ has uniformly $S$-Noetherian spectrum with respect to $s$. Let $I$ be an ideal of $R$. Then $I\{x\}$ is radically $S$-finite with respect to $s$ as an $R\{x\}$-ideal. So $I$ is radically $S$-$w$-finite with respect to $s$ by Proposition 3.11. Hence, $R$ is has uniformly $S$-$w$-Noetherian spectrum. ∎

Corollary 3.13**.**

Let $R$ be a ring. Then the following statements are equivalent.

1. (1)

$R$* has $w$-Noetherian spectrum.* 2. (2)

$R[X]$* has $w$-Noetherian spectrum.* 3. (3)

$R\{X\}$* has Noetherian spectrum.*

Proof.

Take $S=\{1\}$ in Theorem 3.12 ∎