S-Noetherian generalized power series rings

Abstract

Let $R$ be a ring with identity, $(M,\leq)$ a commutative positive strictly ordered monoid and $\omega_{m}$ an automorphism for each $m\in M$. The skew generalized power series ring $R[[M,\omega]]$ is a common generalization of (skew) polynomial rings, (skew) power series rings, (skew) Laurent polynomial rings, (skew) group rings, and Mal’cev Neumann Laurent series rings. If $S\subset R$ is a multiplicative set, then $R$ is called right $S$-Noetherian, if for each ideal $I$ of $R$, $Is\subseteq J\subseteq I$ for some $s\in S$ and some finitely generated right ideal $J$. Unifying and generalizing a number of known results, we study transfers of $S$-Noetherian property to the ring $R[[M,\omega]]$. We also show that the ring $R[[M,\omega]]$ is left Noetherian if and only if $R$ is left Noetherian and $M$ is finitely generated. Generalizing a result of Anderson and Dumitrescu, we show that,when $S\subseteq R$ is a $\sigma$-anti-Archimedean multiplicative set with $\sigma$ an automorphism of $R$, then $R$ is right $S$-Noetherian if and only if the skew polynomial ring $R[x;\sigma]$ is right $S$-Noetherian.

Department of Pure Mathematics, Faculty of Mathematical Sciences,
Tarbiat Modares University, Tehran, Iran, P.O. Box: 14115-134.

Keywords: $S$-Noetherian ring, skew generalized power series ring; right archimedean ring; skew Laurent series ring; skew polynomial ring.

Subject Classification: $16P40;16D15;16D40;16D70;16S36$

1 Introduction

Throughout this paper, $R$ is a ring (not necessary commutative) with identity. In [3], the authors introduced the concept of “almost finitely generated” to study Querr$\acute{e}$’s characterization of divisorial ideals in integrally closed polynomial rings. Later, Anderson and Dumitrescu [1] abstracted this notion to any commutative ring and defined a general concept of Noetherian rings. They call $R$ an $S$-Noetherian ring if each ideal of $R$ is $S$-finite, i.e., for each ideal $I$ of $R$, there exist an $s\in S$ and a finitely generated ideal $J$ of $R$ such that $Is\subseteq J\subseteq I$. By [1, Proposition 2(a)], any integral domain $R$ is ($R\setminus\{0\}$)-Noetherian; so an $S$-Noetherian ring is not generally Noetherian. Also, $M$ is said to be $S$-finite if there exist an $s\in S$ and a finitely generated $R$-submodule $F$ of $M$ such that $sM\subseteq F.$ Also, $M$ is called $S$-Noetherian if each submodule of $M$ is $S$-finite. In [1], the authors gave a number of $S$-variants of well-known results for Noetherian rings: $S$-versions of Cohen’s result, the Eakin-Nagata theorem, and the Hilbert basis theorem under an additional condition. More precisely, in [1, Propositions 9 and 10], the authors showed that, if $S$ is an anti-Archimedean subset of an $S$-Noetherian ring $R$, then the polynomial ring $R[X_{1},\cdots,X_{n}]$ is also an $S$-Noetherian ring; and if $S$ is an anti-Archimedean subset of an $S$-Noetherian ring $R$ consisting of nonzero divisors, then the power series ring $R[[X_{1},\cdots,X_{n}]]$ is an $S$-Noetherian ring. Note that if $S$ is a set of units of $R$, then the results above are nothing but the Hilbert basis theorem and a well-known fact that $R[[X]]$ is Noetherian if $R$ is Noetherian. In [16, Theorem 2.3], Liu generalized this result to the ring of generalized power series as follows: If $S$ is an anti-Archimedean subset of a ring $R$ consisting of nonzero divisors and ($\Gamma,\leq$) is a positive strictly ordered monoid (defined in Secion 4), then $R[[M,\leq]]$ is $S$-Noetherian if and only if $R$ is $S$-Noetherian and $\Gamma$ is finitely generated. Note that this recovers the result for the Noetherian case shown in [6, , Theorem 4.3] when $S$ is a set of units. Also, the authors in [14] study on transfers of the $S$-Noetherian property to the constructions $D+(X_{1},\cdots,X_{n})E[X_{1},\cdots,X_{n}]$ and $D+(X_{1},\cdots,X_{n})E[[X_{1},\cdots,X_{n}]]$ and Nagata’s idealization is studied in [15].

The authors in [8, Theorem 7.7, page(65)] proved that $R[M]$ is Noetherian if and only if $R$ is Noetherian and $M$ is finitely generated. Brookfield [6] proved that if $(M,\leq)$ is a commutative positively ordered monoid, then $R[[M,\leq]]$ is right Noetherian if and only if $R$ is right Noetherian and $M$ is finitely generated.

Ribenboim [22] and Varadarajan [26], have carried out an extensive study of rings of generalized power series. They investigated conditions under which a ring of generalized power series $R[[M,\leq]]$ is Noetherian, where $R$ is a commutative ring with identity and $(M,\leq)$ is a strictly ordered monoid.

In this paper we obtain results pertaining to Noetherian nature of generalized power series rings. These considerably strengthen earlier results of Ribenboim [22], Varadarajan [26], Brookfield [6], D. D. Anderson, and T. Dumitrescu[1], D. D. Anderson, B. G. Kang, and M. H. Park [2] , D. D. Anderson, D. J. Kwak, M. Zafrullah [3] on this topic.

More precisely, we show that, if $S$ is an $\sigma$-anti-Archimedean multiplicative subset of an $S$-Noetherian ring $R$ with an automorphism $\sigma$, then the skew polynomial ring $R[x;\sigma]$ is also an $S$-Noetherian ring; and if $(M,\leq)$ is a commutative positively ordered monoid and $\omega_{m}$ is an automorphism over $R$ for every $m\in M$, then the skew generalized power series ring $R[[M,\omega]]$ is right Noetherian if and only if $R$ is right Noetherian and $M$ is finitely generated. When $(M,\leq)$ is a commutative positive strictly ordered monoid and $\omega_{m}$ is an automorphism for each $m\in M$, we unify and generalize the above mentioned results, and study transfers of $S$-Noetherian property to the skew generalized power series ring $R[[M,\omega]]$.

2 S-Noetherian property on skew polynomial rings

If $R$ is a commutative ring and $S$ is a multiplicative subset of $R$, in [1], the authors proved that the necessary condition for the ring of fractions $R_{S}$ to be a Noetherian ring is that $R$ be an $S$-Noetherian ring. In noncommutative rings, the situation is more complicated. In fact, if $S$ is a right (resp., left) permutable and right (resp., left) reversible (i.e $S$ is right (resp., left) denominator set), then $R$ has a ring of fraction $RS^{-1}$ (resp., $S^{-1}R$). In this situation, denominator sets (both left and right denominator sets) act like a multiplicatively closed sets in the commutative case. Our interest in this note is multiplicatively closed subsets (i.e. denominator subsets) in noncommutative rings. First we define the notion of $S$-Noetherian rings for noncommutative rings.

Definition 2.1.

Let $R$ be a ring and $S$ a multiplicative subset of $R$. An ideal $I$ of $R$ is called right $S$-finite (resp., $S$-principal), if there exists a finitely generated (resp., principal) right ideal $J$ of $R$ and some $s\in S$ such that $Is\subseteq J\subseteq I$.

A ring $R$ is said to be right $S$-Noetherian (resp., $S$-PRIR), if each right ideal of $R$ is right $S$-finite (resp., $S$-principal). This definition can be done similarly for left side ideals.

Also, we say that an $R$-module $M$ is right (or left) $S$-finite if $Ms\subseteq F$ (resp., $sM\subseteq F$) for some $s\in S$ and a finitely generated submodule $F$ of $M$. A module $M$ is called right (or left) $S$-Noetherian if each submodule of $M$ is a right (or left) $S$-finite module.

The author in [1] justified the definition of $S$-Noetherian for commutative rings by proving some interesting properties of $S$-Noetherian ring. For example, they showed that if $R$ is $S$-Noetherian, then the ring of fractions $R_{S}$ is Noetherian and they found the conditions for the reverse of this proposition.

Given rings $R,T$, an ideal $J$ of $T$ is said to be extended, if there exists an ideal $I$ of $R$ such that $\varphi(I)=J$ where $\varphi:R\longrightarrow T$ is a ring monomorphism. Also, a ring $R$ is von Neumann regular if for every $a\in R$ there exists an $x$ in $R$ such that $a=axa$. The center of a ring $R$ is denoted by $Cent(R)$.

Proposition 2.2.

Let $R$ be a ring, $S\subseteq R$ a multiplicative set and $I$ a right ideal of $R$.

1) If $R$ is von Neumman regular, $S$ a denominator set and $I\cap S\neq\emptyset$, then $I$ is right $S$-principal.

2) If $S\subseteq T$ are right denominator subsets of $R$ and $R$ is right $S$-Noetherian(resp., $S$-PRIR), then $R$ is right $T$-Noetherian(resp., $T$-PRIR).

3) If $R$ is von Neumman regular and $S$ a denominator set, then $R$ is right $S$-Noetherian (resp., $S$-PRIR) if and only if $R$ is right Noetherian (resp., PRIR).

4) If $S$ is a denominator set and $R$ is right $S$-Noetherian (resp., $S$-PRIR), then $RS^{-1}$ is right Noetherian.

5) If $S$ is central in $R$, then the conditions 1-4 and those of [1, Proposition 2] follow.

Proof.

1) Let $S\subseteq R$ be a denominator set, $R$ a von Neumman regular ring and $I$ a right ideal of $R$. Then for each $s\in I\cap S$, one can see that $Is\subseteq Rs=s\frac{1}{s}Rs$, where $\frac{1}{s}$ is the inverse of $s$ in $RS^{-1}$. It is sufficient to see that $\frac{1}{s}Rs\subseteq R$. For each $s\in S$, there exists $a\in R$ such that $sas=s$, so $sa=s\frac{1}{s}=1$ (in $RS^{-1}$). Thus $sa=1$ and hence $a=\frac{1}{s}$. Therefore $\frac{1}{s}\in R$ and $Rs\subseteq R$, so $\frac{1}{s}Rs\subseteq R$.

2) Let $S\subseteq T$ be denominator subsets of $R$. If $R$ is right $S$-Noetherian (resp., $S$-PRIR), then for each right ideal of $R$, there exists $s\in S$ such that $Is\subseteq J\subseteq I$ for some finitely generated (resp., principal) right ideal of $R$. Since $s\in S$, $S\subseteq T$, $s\in T$ which means that $R$ is right $T$-Noetherian (resp., $T$-PRIR).

3) Assume that $R$ is a right Noetherian (resp., PRIR) ring. Each right ideal of $R$ is finitely generated (resp., principal). So for each $s\in S$, one can see that $Is\subseteq I$. Hence $R$ is right $S$-Noetherian (resp., $S$-PRIR). On the other hand, assume that $R$ is right $S$-Noetherian (resp., $S$-PRIR), so there exists $s\in S$ such that $Is\subseteq J\subseteq I$ for some finitely generated (resp., principal) right ideal of $R$. Also suppose that $sts=s$ for some $t\in R$. So $Is\subseteq I$. Also, $It\subseteq I$, so $Its\subseteq Is=Ists$. So $Its.\frac{1}{s}\subseteq Ists.\frac{1}{s}$. Hence $It\subseteq Ist$. Also $Is\subseteq I$ yields that $Ist\subseteq It\subseteq Ist$. So $Ist=It$. Thus $Ists=Its$ which means that $Is=Ists=Its$. However $Its=I\frac{1}{s}sts=I\frac{1}{s}s=I$. So $Is=I$. Thus $I=Is\subseteq J\subseteq I$ and hence $I=J$, and since $J$ is a finitely generated (resp., principal) right ideal of $R$, so is $I$.

4) This proof is an inspiration from [4, proposition 3.11 part (i)]. First, we claim that each ideal of $RS^{-1}$ is extended. Let a right ideal $J$ of ring of fraction $RS^{-1}$ and $\frac{x}{s}=b\in J$. So $\frac{x}{1}=\frac{x}{s}.\frac{s}{1}\in J.\frac{s}{1}\subseteq J$. So $\frac{x}{1}\in J$. Hence, $\varphi^{-1}(\frac{x}{1})\in\varphi^{-1}(J)$ which means that $x\in\varphi^{-1}(J)$. Thus, $\varphi(x)\in\varphi(\varphi^{-1}(J))$, so $\frac{x}{1}\in\varphi(\varphi^{-1}(J))$. So $\frac{x}{1}.\frac{s}{s}=\frac{x.s}{s}.\frac{s}{s}=\frac{xs}{s}\in\varphi(\varphi^{-1}(J))$. Note that $\varphi(\varphi^{-1}(J))$ is an ideal of $RS^{-1}$ and $s\in U(RS^{-1})$, so we have

$$\frac{xs}{s}.\frac{1}{s}=\frac{x}{s}\in\varphi(\varphi^{-1}(J))\frac{1}{s}\subseteq\varphi(\varphi^{-1}(J)).$$

So $b=\frac{x}{s}\in\varphi(\varphi^{-1}(J))$ which implies $J\subseteq\varphi(\varphi^{-1}(J))$. On the other hand, $\varphi(\varphi^{-1}(J))\subseteq J$ holds for each ideal of $RS^{-1}$. Thus $J=\varphi(\varphi^{-1}(J))$ and $J$ is an extended ideal of $RS^{-1}$.

Let a right ideal $K$ of ring of fraction $RS^{-1}$. Since $R$ is right $S$-Noetherian there exists $s\in S$ and a finitely generated (resp., principal) right ideal $W$ of $R$ such that $\varphi^{-1}(K)s\subseteq W\subseteq\varphi^{-1}(K)$. So $\varphi(\varphi^{-1}(K)s)\subseteq\varphi(W)\subseteq\varphi(\varphi^{-1}(K))$. We know that $\varphi(\varphi^{-1}(K)s)=\varphi(\varphi^{-1}(K))\varphi(s)$. Also, $\varphi(s)\in U(RS^{-1})$ and $\varphi(\varphi^{-1}(K))=K$. So $K\subseteq\varphi(W)\subseteq K$. So $K=\varphi(W)$. Since $W$ is finitely generated, $\varphi(W)$ is finitely generated. So $K$ is finitely generated which means that $RS^{-1}$ is right Noetherian.

5) The proof is straightforward by [1, Proposition 2]. ∎

Now we generalize a theorem of D.D. Anderson and Tiberiu Dumitrescu [1, Proposition 9], for commutative polynomial ring $R[x]$, in a more general setting. We show that if $R$ is a right (or left) $S$-Noetherian ring with an automorphism $\sigma$, then $R[x;\sigma]$ is a right (or left) $S$-Noetherian ring.

In [2] the authors defined the notion of anti-Archimedean multiplication set. Now we introduce the notion of $\sigma$-anti-Archimedean multiplication set:

Definition 2.3.

Let $R$ be a ring with an automorphism $\sigma$ and $S$ a multiplicative set. Then $R$ is called left $\sigma$-anti-Archimedean over $S$, if there exists $s\in S$, such that

$$(\bigcap_{l\geq 1,k_{i}\geq 0}R\sigma^{k_{1}}(s)\sigma^{k_{2}}(s)\cdots\sigma^{k_{l}}(s)\big)\cap S\neq\emptyset.$$

Theorem 2.4.

Let $R$ be a ring with an automorphism $\sigma$ and $S\subseteq R$ a $\sigma$-anti-Archimedean multiplicative set. Then $R$ is right (or left) $S$-Noetherian if and only if $R[x;\sigma]$ is right (or left) $S$-Noetherian.

Proof.

($\Rightarrow$) We prove the theorem for the right version. The proof of left version is similar. First, we claim that if $D$ is a finitely generated $R$-module and $R$ is a right $S$-Noetherian ring, then $D$ is a right $S$-Noetherian module. For this claim, assume that $D$ is a finitely generated right $R$-module. So there exists a finitely generated free right $R$-module $F$ and a surjective homomorphism $\pi:F\longrightarrow D$. We show that $D$ is a right $S$-Noetherian $R$-module. For this, let $N:=\pi^{-1}(T)$, for a submodule $T$ of $D$. We have $N\simeq I_{1}\oplus I_{2}\cdots\oplus I_{l}$, for some right ideals $I_{i}$ of $R$, $1\leq i\leq l$. Since $R$ is a right $S$-Noetherian ring, there exists $s_{i}\in S$ such that $I_{i}s_{i}\subseteq J_{i}$ for a finitely generated ideals $J_{i}$ of $R$, $1\leq i\leq l$. Now take $s^{\prime}:=s_{1}s_{2}\cdots s_{l}\in S$, we show that $Ns^{\prime}\subseteq K$ for a finitely generated $R$-submodule $K$ of $F$. One can see that $Ns_{1}=I_{1}s_{1}\oplus I_{2}s_{1}\oplus\cdots\oplus I_{l}s_{1}$. Since $I_{i}$ is a right ideal of $R$ so we have $I_{i}s_{1}\subseteq I_{i}$ for $i\neq 1$ and $I_{1}s_{1}\subseteq J_{1}$, for a finitely generated right ideal $J_{1}$ of $R$. So we have $Ns_{1}\subseteq J_{1}\oplus I_{2}\oplus I_{3}\cdots\oplus I_{l}$. Continuing in this way, $Ns_{1}s_{2}\cdots s_{l}\subseteq J_{1}\oplus J_{2}\oplus\cdots\oplus J_{l}\simeq K$, where $J_{i}$ is a finitely generated right ideal of $R$,$1\leq i\leq l$, and hence $K$ is a finitely generated $R$-submodule of $F$. Thus $Ns^{\prime}\subseteq K$ and hence $F$ is a right $S$-Noetherian $R$-module. Next, since $T=\pi(N)$ and $Ns^{\prime}\subseteq K$, we have $\pi(Ns^{\prime})=\pi(N)s^{\prime}=Ts^{\prime}\subseteq\pi(K)$. We know that $K$ is finitely generated in $F$, so $\pi(K)$ is finitely generated $R$-submodule of $D$. Thus, $Ts^{\prime}\subseteq\pi(K)$ which means that $T$ is $S$-finite. Since $T$ is an arbitrary $R$-submodule of $D$, $D$ is a right $S$-Noetherian module.

Now, we prove that $A:=R[x;\sigma]$ is a right $S$-Noetherian ring. Let $I$ be right ideal of $A$ and suppose that

$$J=\{r_{i}\in R|r_{i}\,\,\textrm{is a leading coefficient of any polynomial in}\,\,I\}\cup\{0\}.$$

It is easy to see that $J$ is a right ideal. Since $R$ is right $S$-Noetherian, $Js\subseteq(a_{1}R+a_{2}R+\cdots+a_{n}R)$ for some $s\in S$ and $a_{i}\in J$. So there exist polynomials $f_{i}\in I$ with $f_{i}=a_{i,n_{i}}x^{n_{i}}+\cdots+a_{0,i}$. Let $d=max\{n_{i}\}$. Assume that $T$ is the set of all polynomials in $I$ with degree less than $d$. Obviously, $T$ is a finitely generated right $R$-submodule of $A$. So by the first claim, $T$ is right $S$-Noetherian. Hence there exist $t\in S$, $g_{i}\in T$ for $1\leq i\leq m$ such that $Tt\subseteq(g_{1}R+g_{2}R+\cdots+g_{m}R)$. Let $h(x)=\sum_{i=1}^{z}b_{i}x^{i}\in I$, so $b_{z}\in J$ which means that $b_{z}\in(a_{1}R+a_{2}R+\cdots+a_{n}R)$. Thus $h\sigma^{-z}(s)$ can be written as follows:

$$h\sigma^{-z}(s)=v^{(1)}+w^{(1)}+q^{(1)},$$

where $v^{(1)}\in(f_{1}A+f_{2}A+\cdots+f_{n}A)$, $w^{(1)}\in\{f\in A|d+1\leq deg(f)\leq z-1\}$ and $q^{(1)}\in T$. Continuing in this way and multiplying $\sigma^{-z+1}(s),\sigma^{-z+2}(s),\cdots,\sigma^{-1-d}(s)$ from right side respectively, so there exists some $v\in(f_{1}A+f_{2}A+\cdots+f_{n}A)$, $w\in T$ such that

$$h\sigma^{-z}(s)\sigma^{-z+1}(s)\cdots\sigma^{-d-1}(s)=v+w.$$

Assume that $s_{i}=\sigma^{-z+i}$ and multipling $t$ from right side, then $hs_{1}s_{2}\cdots s_{z-d}t=vt+wt$. But $wt\in Tt$, so $wt\in(g_{1}R+g_{2}R+\cdots+g_{m}R)\subseteq(g_{1}A+g_{2}A+\cdots+g_{m}A)$. Hence,

$$hs_{1}s_{2}\cdots s_{z-d}t\in(f_{1}A+f_{2}A+\cdots+f_{n}A+g_{1}A+\cdots+g_{m}A).$$

Since $s_{i}$’s and $t$ are independent from the choice of $h\in I$, we have

$$Is_{1}s_{2}\cdots s_{z-d}t\subseteq(f_{1}A+f_{2}A+\cdots+f_{n}A+g_{1}A+\cdots+g_{m}A).$$

Finally, since $s_{1}s_{2}\cdots s_{z-d}t\in S$, the ideal $I$ is $S$-finite and because $I$ was chosen an arbitrary right ideal of $A$, hence $A$ is a right $S$-Noetherian ring.

($\Leftarrow$) Let $I$ be a right ideal of $R$. Suppose that

$$J=\{f\in A|\,\,\textrm{the leading coefficient of}\,f\,\textrm{is in}\,I\}.$$

Then $J$ is a right ideal of $A$. Since $A$ is right $S$-Noetherian, there exists $s\in S$ such that $Js\subseteq K\subseteq J$, where $K$ is a finitely generated right ideal of $A$. Suppose that $K=(f_{1}A+f_{2}A+\cdots+f_{l}A)$. Let $r\in I$, then there exists some $f\in J$ such that $fs=\sum a_{i}f_{i}$. So if $r_{i}$ is the leading coefficient of $f_{i}$, $1\leq i\leq l$, then $rs\in(r_{1}R+r_{2}R\cdots+r_{l}R)$. So $Is\subseteq(r_{1}R+r_{2}R+\cdots+r_{l}R)$. Also, $K\subseteq J$, so each leading coefficient of $K$ is a leading coefficient of $J$. So $(r_{1}R+r_{2}R+\cdots+r_{l}R)\subseteq I$ and hence $I$ is right $S$-finite and $R$ is right $S$-Noetherian. ∎

We have the following generalization of a theorem of D.D. Anderson and Tiberiu Dumitrescu [1, Proposition 9].

Corollary 2.5.

Let $R$ be a (not necessarily commutative) ring and $S\subseteq R$ an anti-Archimedean multiplicative set. If $R$ is $S$-Noetherian then so is the polynomial ring $R[X_{1},X_{2},\cdots,X_{n}]$.

3 Noetherian Skew Generalized Power Series Rings

Throughout this section, $(M,\leq)$ is assumed to be a strictly ordered commutative monoid. The pair $(M,\leq)$ is called an ordered monoid with order $\leq$, if for every $m,m^{\prime},n\in M$, $m\leq m^{\prime}$ implies that $nm\leq nm^{\prime}$ and $mn\leq m^{\prime}n$. Also, an ordered monoid $(M,\leq)$ is said to be strictly orderd if for every $m,m^{\prime},n\in M$, $m<m^{\prime}$ implies that $nm<nm^{\prime}$ and $mn<m^{\prime}n$. Let $(M,\leq)$ be a partially ordered set. The set $(M,\leq)$ is called Artinian if every strictly decreasing sequence of elements of $M$ stablized, and also $(M,\leq)$ is called narrow if the number of incomparable elements in every subset of $M$ is finite. Thus, we can conclude that $(M,\leq)$ is Artinian and narrow if and only if every nonempty subset of $M$ has at least one but only a finite number of minimal elements.

The author in [24] introduced the ring of generalized power series $R[[M]]$ for a strictly ordered monoid $M$ and a ring $R$ consisting of all functions from $M$ to $R$ whose support is Artinian and narrow with the pointwise addition and the convolution multiplication. There are a lot of interesting examples of rings in this form (e.g., Elliott and Ribenboim, [7]; Ribenboim,[23]) and it was extensively studied by many authors, recently.

In [21], the authors defined a “twisted” version of the mentioned construction and study on ascending chain condition for its principal ideals. Now we recall the construction of the skew generalized power series ring introduced in [21]. Let $R$ be a ring, $(M,\leq)$ a strictly ordered monoid, and $\omega:M\rightarrow End(R)$ a monoid homomorphism. For $m\in M$, let $\omega_{m}$ denote the image of $m$ under $\omega$, that is $\omega_{m}=\omega(m)$. Let $A$ be the set of all functions $f:M\rightarrow R$ such that the support $\supp(f)=\{m\in M|f(m)\neq 0\}$ is Artinian and narrow. Then for any $m\in M$ and $f,g\in A$ the set

$$\chi_{m}(f,g)=\{(u,v)\in\supp(f)\times\supp(g):m=uv\}$$

is finite. Thus one can define the product $fg:M\rightarrow R$ of $f,g\in A$ as follows:

$$fg(m)=\sum_{(u,v)\in\chi_{m}(f,g)}f(u)\omega_{u}(g(v)),$$

(by convention, a sum over the empty set is $0$). Now, the set $A$ with pointwise addition and the defined multiplication is a ring, and called the ring of skew generalized power series with coefficients in $R$ and exponents in $M$. To simplify, take $A$ as a formal series $\sum\limits_{m\in M}r_{m}x^{m},$ where $r_{m}=f(m)\in R$. This ring can be denoted either by $R[[M^{\leq},\omega]]$ or by $R[[M,\omega]]$ (see [18] and [19]).

For every $r\in R$ and $m\in M$ we can defined the maps $c_{r},e_{m}:M\longrightarrow R$ by

$$c_{r}(x)=\begin{cases}r\quad;x=1\\ 0\quad;\text{Otherwise}\end{cases},e_{m}(x)=\begin{cases}1\quad;x=m\\ 0\quad;\text{Otherwise}\end{cases}$$ (3.1)

where $x\in M$. By way of illustration, $c_{r}(x)$ and $e_{m}(x)$ are like $r$ and $x^{m}$ in usual polynomial ring $R[x]$, respectively.

The following proposition which is proved in [11, Theorem 2.1], can characterize all Artinian and narrow sets.

Proposition 3.1.

Let $(M,\leq)$ be an ordered set. Then the following conditions are equivalent

(1) $(M,\leq)$ is Artinian and narrow.

(2) For any sequence $(m_{n})_{n\in\mathbb{N}}$ of elements of $M$ there exist indices $n_{1}<n_{2}<n_{3}<\cdots$ such that $m_{n_{1}}\leq m_{n_{2}}\leq m_{n_{3}}\leq\cdots$ .

(3) For any sequence $(m_{n})_{n\in N}$ of elements of $M$ there exist indices $i<j$ such that $m_{i}\leq m_{j}$.

The author in [6] introduced the concept of a lower set. A lower set of $L$ is a subset $I\subseteq L$ such that $x\leq y\in I$ implies $x\in I$ for all $x,y\in L$, (which we denoted by $\Downarrow L$ for the set of lower sets of $L$ ordered by inclusion). In this concept, we can ignore the condition narrow by lower set, indeed it is proved that if $L$ is a partially ordered set, then $\Downarrow L$ is Artinian if and only if $L$ is Artinian and narrow. He also showed that if $\alpha:K\longrightarrow L$ is strictly increasing map between partially ordered sets, then if $L$ satisfies Artinian (or Noetherian) property, then so is $K$. Moreover, if $\alpha$ is surjective and $\Downarrow K$ satisfies Artinian (or Noetherian) property, then so does $\Downarrow L$.

An ordered monoid $(M,\leq)$ is called positively ordered if $m\geq 0$ for all $m\in M$. In this condition, $m\preceq m^{\prime}$ implies $m\leq m^{\prime}$ for all $m,m^{\prime}\in M$. Now, according to [6, in section 4] we have

$$R[[M,\omega,\leq]]=\{f\in R[[M,\omega]]\quad|\Downarrow(\supp(f),\leq)\quad is\quad\text{Artinian}\}.$$ (3.2)

If $\Downarrow(M,\leq)$ is Artinian, $R[[M,\omega,\leq]]=R[[M,\omega]]$. For instance, $\Downarrow(\mathbb{F},\preccurlyeq)$ and $\Downarrow(\mathbb{F}^{n},\preccurlyeq)$ are Artinian, and so $R[[\mathbb{F},\omega,\preccurlyeq]]=R[[\mathbb{F},\omega]]$ and $R[[\mathbb{F}^{n},\omega,\preccurlyeq]]=R[[\mathbb{F}^{n},\omega]]$ such that $\mathbb{F}$ be a free monoid.

Now we give a generalization of a result [6, Theorem 4.3] of G. Brookfield:

Theorem 3.2.

Let $R$ be a ring, $(M,\leq)$ a positive strictly ordered monoid and $\omega_{m}$ an automorphism of $R$ with $\omega_{m}\omega_{n}=\omega_{n}\omega_{m}$ for each $m,n\in M$. Then $R[[M,\omega]]$ is left Noetherian if and only if $R$ is left Noetherian and $M$ is finitely generated.

Proof.

$\Leftarrow$) In the first place, we claim that if $\varphi:(N,\leq)\rightarrow(M,\leq)$ is a surjective strict monoid homomorphism, induces a surjective ring homomorphism $\varphi^{*}:R[[N,\omega,\leq]]\rightarrow R[[M,\omega,\leq]]$. Since $\varphi$ is strict, $\varphi^{-1}(x)$ is antichain in $(N,\leq)$ for all $x\in M$. Thus, if $f\in R[[N,\omega,\leq]]$ then $\varphi^{-1}(x)\cap\supp(f)$ is finite and we can define $\varphi^{*}(f)=f^{*}$, where $f^{*}(x)=\sum_{x^{\prime}\in\varphi^{-1}(x)}f(x^{\prime})$ for $x\in M$. We show that $\varphi^{*}$ is a ring homomorphism. One can see that

$$(fg)^{*}(m)=\sum_{m^{\prime}\in\varphi^{-1}(m)}(fg)(m^{\prime})=\sum_{xy=m}\sum_{\begin{subarray}{c}x^{\prime}y^{\prime}=m^{\prime}\\ m^{\prime}\in\varphi^{-1}(m)\end{subarray}}\bigg(f(x^{\prime})\alpha_{x^{\prime}}(g(y^{\prime}))\bigg).$$ (3.3)

On the other hand

$$(f^{*}g^{*})(m)= \big(\varphi^{*}(f)\varphi^{*}(g)\big)(m)=\sum_{xy=m}\bigg(\varphi^{*}(f(x))\alpha_{x}\big(\varphi^{*}(g(y)\big)\bigg)$$

$$= \sum_{xy=m}\bigg(\sum_{x^{\prime}\in\varphi^{-1}(x)}f(x^{\prime})\bigg)\alpha_{x}\bigg(\sum_{y^{\prime}\in\varphi^{-1}(y)}g(y^{\prime}))\bigg)$$

$$= \sum_{xy=m}\sum_{x^{\prime}\in\varphi^{-1}(x)}\sum_{y^{\prime}\in\varphi^{-1}(y)}\bigg(f(x^{\prime})\alpha_{x^{\prime}}(g(y^{\prime}))\bigg).$$

Since $\varphi^{-1}$ is a homomorphism, $\varphi^{-1}(x)\varphi^{-1}(y)=\varphi^{-1}(xy)$ and so $\varphi^{-1}(m)=\varphi^{-1}(x)\varphi^{-1}(y)$. So

$$(f^{*}g^{*})(m)=\sum_{xy=m}\sum_{\begin{subarray}{c}m^{\prime}=x^{\prime}y^{\prime}\\ m^{\prime}\in\varphi^{-1}(m)\end{subarray}}\bigg(f(x^{\prime})\alpha_{x^{\prime}}(g(y^{\prime}))\bigg).$$ (3.4)

By equations 3.3 and 3.4 we see that $(fg)^{*}(m)=(f^{*}g^{*})(m)$. We have also

$$(f+g)^{*}(x)= \sum_{x^{\prime}\in\varphi^{-1}(x)}(f+g)(x^{\prime})=\sum_{x^{\prime}\in\varphi^{-1}(x)}(f(x^{\prime})+g(x^{\prime}))$$

$$= \sum_{x^{\prime}\in\varphi^{-1}(x)}f(x^{\prime})+\sum_{x^{\prime}\in\varphi^{-1}(x)}g(x^{\prime})=f^{*}(x)+g^{*}(x).$$

Thus $\varphi^{*}:R[[N,\omega,\leq]]\rightarrow R[[M,\omega,\leq]]$ is a ring homomorphism. Now, we show that $\varphi^{*}$ is surjective. Suppose that $f\in R[[M,\omega,\leq]]$, where $\{f(n)\}_{n\in M}$ are the coefficients of $f$ in $R$. For every $n\in M$, the set $\varphi^{-1}(n)$ is nonempty and finite, say $\varphi^{-1}(n)=\{m_{1},m_{2},\dots,m_{k}\}$, where $k$ and all the $m_{i}$ depends on $n$. We define the function $g\in R[[N,\omega,\leq]]$ as follows

$$g(m_{j})=\begin{cases}f(n)\quad;j=1\\ 0\quad;\text{otherwise}.\end{cases}$$ (3.5)

Notice that $g$ is independent of $n$, since if $n\neq n^{\prime}$, then $\varphi^{-1}(n)\cap\varphi^{-1}(n^{\prime})=\emptyset$. Also, for each $n\in M$ we have

$$\varphi^{*}(g)(n)=\sum_{m\in\varphi^{-1}(n)}g(m)=\sum_{j=1}^{k}g(m_{j})=g(m_{1})=f(n).$$

This means that $\varphi^{*}(g)=f$, and hence $\varphi^{*}$ is surjective. So we proved the claim. It is well-known that there is an strict monoid surjection $\varphi:(\mathbb{F}^{n},\preccurlyeq)\rightarrow(M,\preccurlyeq)$ for some $n\in\mathbb{N}$. Also, the identity map $(M,\preccurlyeq)\to(M,\leq)$ is a surjection. So the composition of these two maps is a surjection and by [6, Lemma 2.1]. Hence $R[[M,w,\leq]]$ is a homomorphic image of the ring $R[[\mathbb{F}^{n},\omega,\preceq]]$. Since $R[[\mathbb{F}^{n},\omega,\preceq]]=R[[\mathbb{F}^{n},\omega]]$ and $R[[\mathbb{F}^{n},\omega]]$ is Noetherian, its projection $R[[M,w,\leq]]$ is also Noetherian. Moreover, we show that $R[[M,\omega,\leq]]=R[[M,\omega]]$. If $R[[M,\omega,\leq]]$ is left Noetherian, then $\Downarrow(M,\preccurlyeq)$ is Artinian. By applying [6, Lemma 2.1(2)] to the identity map $(M,\preccurlyeq)\to(M,\leq)$, one can see that $\Downarrow(M,\leq)$ is Artinian. Thus $R[[M,\omega,\leq]]=R[[M,\omega]]$.

$\Rightarrow$) The method of this part is inspired from [6, Theorem 4.3]. The trivial case of $M$ is obvious. By [6, Lemmas 3.1 and 3.2], $M$ is strict and $\preccurlyeq$ is a partial order on $M$.

Suppose $T=R[[M,\omega,\leq]]$ is left Noetherian. One can see that $M$ is finitely generated similar to the proof of [6, Theorem 4.3]. Hence we have to prove that $R$ is Noetherian similar to the proof of ([25, Theorem 5.2(i)], [26, Theorem 3.1(i)]). Let $I_{T}=\{f\in T\mid\omega_{x}(f(y))\in I;x,y\in M\}$. It is easy to see that $I_{T}$ is a left ideal of $T$. So for each ideal $I$ of $R$, there is a correspondent ideal in $T$. Also if $I\subset J$ ,then $I_{T}\subset J_{T}$. Hence if there exists a nonstabilized ascending chain in $R$, then there is one in $T$. But this is impossible, so $R$ is left Noetherian. ∎

In Theorem 3.2 if we set $\sigma$ the identity homomorphism then we have:

Corollary 3.3.

[6, Theorem 4.3] Let $R$ be a ring and $(M,\leq)$ a positive strictly ordered monoid. Then $R[[M,\leq]]$ is left Noetherian if and only if $R$ is left Noetherian and $M$ is finitely generated.

Finally, we conclude the following result which connects the results of previous sections.

Corollary 3.4.

Let $R$ be an $S$-Noetherian von Neumman regular ring and $S$ a denominator set. Assume that $(M,\leq)$ is a finitely generated positive strictly ordered monoid and $\omega_{m}$ an automorphism of $R$ with $\omega_{m}\omega_{n}=\omega_{n}\omega_{m}$ for each $m,n\in M$. Then $(S^{-1}R)[[M,\omega]]$ is a left Noetherian ring.

Proof.

The ring $S^{-1}R$ is Noetherian by Theorem 2.2. Since $(M,\leq)$ is a positive strictly ordered monoid and $\omega_{m}$ is an automorphism for all $m\in M$, $(S^{-1}R)[[M,\omega]]$ is a Noetherian ring by Theorem 3.2. ∎

4 S-Noetherian property of generalized skew power series rings

Recall that a ring is called right duo (resp., left duo) if all of its right (resp., left) ideals are two-sided. Also, a right and left duo ring is called a duo ring. We know that if a ring is duo, then every prime ideal is completely prime. It is known that a power series ring over a duo ring need not be duo (on either side).

Lemma 4.1.

Let $R$ be a duo ring and $S\subset R$ a denominator set. If $s\in S$, $r\in R$ then there exists $s_{1}\in S$ such that $srs_{1}=rss_{1}$.

Proof.

Let $s\in S$ and $r\in R$. Since $R$ is duo, there exist $s^{\prime}\in S$ such that $sr=rs^{\prime}$, so $\frac{1}{s}.\frac{sr}{1}=\frac{1}{s}.\frac{rs^{\prime}}{1}$. Hence $\frac{r}{1}=\frac{rs^{\prime}}{s}=\frac{r}{1}.\frac{s^{\prime}}{s}$. Thus $\frac{r}{1}(1-\frac{s^{\prime}}{s})=0$, which means that $\frac{r(s-s^{\prime})}{s}=0_{S^{-1}R}$. So $r(s-s^{\prime})s_{1}=0_{R}$. So $rss_{1}=rs^{\prime}s_{1}$ and since $rs^{\prime}=sr$ we have $srs_{1}=rss_{1}$. ∎

In the previous result, it is easy to see that if $s\in S$, $r\in R$, then there exists $s_{1}\in S$ such that $s_{1}sr=s_{1}rs$. We will use this point in the proposition below.

Proposition 4.2.

Let $R$ be a duo ring, $S\subseteq R$ a denominator set and $M$ an $S$-finite $R$-module. Then $M$ is $S$-Noetherian if and only if $PM$ is an $S$-finite submodule, for each $S$-disjoint prime ideal $P$ of $R$.

Proof.

The “only if” part is clear. For the converse, assume that $PM$ is $S$-finite for each $P$ prime ideal of $R$ with $P\cap S=\emptyset$. Since $M$ is $S$-finite, $wM\subseteq F$ for some $w\in S$ and some finitely generated submodule $F$. If $M$ is not $S$-Noetherian, the set $\mathfrak{F}$ of all non-$S$-finite submodules of $M$ is not empty. So $\mathfrak{F}$ has a maximal element like $N$ by Zorn’s lemma. We claim that $P=[N:M]:=\{r\in R\mid rM\subseteq N\}$ is a prime ideal of $R$ and is disjoint from $S$. Suppose to the contrary that $P\cap S\neq\emptyset$ and $s\in P\cap S$. Then we have

$$swN\subseteq swM\subseteq sF\subseteq sM\subseteq N.$$

So $swN\subseteq sF\subseteq N$ and $N$ becomes $S$-finite. This contradiction shows that $P\cap S=\emptyset$. Now suppose that $P$ is not a prime ideal of $R$. So $P$ is not completely prime. So there exist $a,b\in R\setminus P$ and $ab\in P$. So $N+aM$ is $S$-finite, hence $s(N+aM)\subseteq(R(n_{1}+am_{1})+\cdots+R(n_{p}+am_{p}))$ for some $s\in S$, $n_{i}\in N$ and $m_{i}\in M$. Also $[N:a]$ is $S$-finite. So $t[N:a]\subseteq(Rq_{1}+Rq_{2}+\cdots+Rq_{k})$ for some $t\in S$ and $q_{j}\in[N:a]$. Since $R$ is duo and $S$ is a denominator set in $R$, there exists $s^{\prime\prime}\in S$ such that $s^{\prime\prime}at=s^{\prime\prime}ta$ by Theorem 4.1. Also $s(N+aM)\subseteq(R(n_{1}+am_{1})+\cdots+R(n_{p}+am_{p}))$. Thus $sx=\sum r_{i}n_{i}+r_{i}am_{i}$. This means that $sx=\sum r_{i}n_{i}+a\sum r^{\prime}_{i}m_{i}$ for some $r^{\prime}_{i}\in R$. Since $sx,\sum r_{i}n_{i}\in N$, we have $\sum r^{\prime}_{i}m_{i}\in[N:a]$. So

$$s^{\prime\prime}tsx=s^{\prime\prime}t\sum r_{i}n_{i}+s^{\prime\prime}t\sum ar^{\prime}_{i}m_{i}=\sum s^{\prime\prime}tr_{i}n_{i}+s^{\prime\prime}at\sum r^{\prime}_{i}m_{i}=\sum s^{\prime\prime}tr_{i}n_{i}+s^{\prime\prime}a\sum c_{j}q_{j}.$$

So $s^{\prime\prime}tsx=\sum s^{\prime\prime}tr_{i}n_{i}+\sum c^{\prime}s^{\prime\prime}_{j}aq_{j}$ for some $c^{\prime}_{j}\in R$. Hence $s^{\prime\prime}tsx\in(Rn_{1}+\cdots+Rn_{p}+Rs^{\prime\prime}aq_{1}+\cdots+Rs^{\prime\prime}aq_{k})$. So $s^{\prime\prime}tsN\subseteq(Rn_{1}+\cdots+Rn_{p}+Rs^{\prime\prime}aq_{1}+\cdots+Rs^{\prime\prime}aq_{k})\subseteq N$. Thus $N$ is $S$-finite and this contradicts to the fact that $N$ is maximal in $\mathfrak{F}$. Therefore $P$ is a prime ideal of $R$. Moreover $P=[N:M]\subseteq[N:F]\subseteq[N:wM]=[P:w]=P$. Hence $[N:F]=P$. Let $F=(Rf_{1}+Rf_{2}+\cdots+Rf_{k})$. Since $R$ is a duo ring, $P=[N:\sum Rf_{i}]=\bigcap[N:f_{i}]$. So $P=[N:f_{i}]$ for some $f_{i}\in\{f_{1},f_{2},\cdots,f_{k}\}.$ One can show that $tN\subseteq(Rn_{1}+Rn_{2}+\cdots+Rn_{l})+PM$ for some $t\in S$ and $n_{i}\in N$ as above or in similar way as that employed in [1, Proposition 4]. Since $PM$ is $S$-finite, $vPM\subseteq G\subseteq PM\subseteq N$ for some $v\in S$ and a finitely generated submodule $G$ of $M$. So

$$vtN\subseteq v(Rn_{1}+Rn_{2}\cdots+Rn_{l})+vPM\subseteq(Rn^{\prime}_{1}+Rn^{\prime}_{2}\cdots+Rn^{\prime}_{l})+G\subseteq N$$

for some $n^{\prime}_{i}\in N$. So $N$ becomes $S$-finite which is a contradiction. So $M$ is $S$-Noetherian. ∎

Lemma 4.3.

Let $R$ be a ring with an endomorphism $\sigma$. If $R[[x;\sigma]]$ is a duo ring, then $\sigma$ is surjective.

Proof.

Suppose that $a\in R$. Since $R[[x;\sigma]]$ is a duo ring we have $ax=xf$ such that $f=\sum_{i=0}^{\infty}f_{i}x^{i}$. So $xf=x\sum_{i=0}^{\infty}f_{i}x^{i}=\sum_{i=0}^{\infty}\sigma(f_{i})x^{i+1}$. Now, since $ax=xf$, $\sigma(f_{i})=0$ for all $i\neq 0$ and $\sigma(f_{0})=a$. Thus, for each $a\in R$ there exists $f_{0}\in R$ such that $a=\sigma(f_{0})$. ∎

Theorem 4.4.

Let $R$ be a ring, $S\subseteq R$ a $\sigma$-anti-Archimedean denominator set (consisting nonzero devisors) and $\sigma_{1},\cdots,\sigma_{n}$ are monomorphisms of $R$ with $\sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i}$, for each $i,j$. Assume that $R[[X_{1},\cdots,X_{n};\sigma_{1},\cdots,\sigma_{n}]]$ is a duo ring. If $R$ is $S$-Noetherian, then the ring $R[[X_{1},\cdots,X_{n};\sigma_{1},\cdots,\sigma_{n}]]$ is also $S$-Noetherian.

Proof.

We use the method in [1, Proposition 10] employed by Anderson and Dumitrescu. As $S$ is $\sigma$-anti-Archimedean in every ring containing $R$ as a subring, we shall prove the case $n=1$, so we assume that $T=R[[x;\sigma]]$ is duo and $\sigma$ is an automorphism of $R$. It is enough to prove that every prime ideal $P$ of $T$ is $S$-finite. Let $\pi:T\rightarrow R$ the $R$-algebra homomorphism sending $x$ to zero and $P^{\prime}=\pi(P)$. Since $R$ is $S$-Noetherian, there exists $s\in S$ such that $sP^{\prime}\subseteq(Rg_{1}(0)+Rg_{2}(0)+\cdots+Rg_{k}(0))$ for some $g_{i}\in P$. If $x\in P$, then $P=(TP^{\prime}+Tx)$. If $g_{i}(x)=\sum a_{i}x^{i}$, then $g_{i}(x)=\sum x^{i}\sigma^{-i}(a_{i})\in(TP^{\prime}+Tx)$. So $sP\subseteq(TP^{\prime}+Tx)=(Tg_{1}+\cdots+Tg_{k})\subseteq P$. This means that $P$ is $S$-finite. Let $x\notin P$ and $f\in P$. So $sf(0)=\sum d_{0,j}g_{j}(0)$ for some $d_{0,j}\in R$. So $xf_{1}=sf-\sum d_{0,j}g_{j}\in P$ for some $f_{1}\in T$. Considering $x\notin P$, $f_{1}\in P$. So $sf_{1}=\sum d_{1,j}g_{j}+xf_{2}$ for some $f_{2}\in T$. Hence $\sigma(s)sf=\sum\sigma(s)d_{0,j}g_{j}+x\sum d_{1,j}g_{j}+x^{2}f_{2}$. Also $f_{2}\in P$, since $x\notin P$ and $sf_{1}-\sum d_{1,j}g_{j}\in P$. In this way, one can see that for each $L\geq 0$,

$$\big(\prod_{l=0}^{L}\sigma^{l}(s)\big)f=\sum_{i=0}^{L}x^{i}\sum_{j=1}^{k}(\prod_{l=i+1}^{L}\sigma^{l}(s))d_{i,j}g_{j}+x^{L+1}f_{L+1}.$$

Since $S\cap\big(\bigcap_{l\geq 1,i_{j}\in\mathbb{N}\cup\{0\}}\sigma^{i_{1}}(s)\cdots\sigma^{i_{l}}(s)R\big)\neq\emptyset$, there exists $t\in R$ such that $\frac{t}{\sigma^{i_{1}}(s)\cdots\sigma^{i_{k}}(s)}\in R$ for each $i_{j}\in\mathbb{N}\cup\{0\}$, $k\in\mathbb{N}$. Moreover

$$tf=\sum_{j}\sum_{i}\Big(\frac{ts\sigma^{-i}(d_{ij})}{\prod_{l}\sigma^{l}(s)}\Big)x^{i}g_{j}.$$

So $tf=\sum_{j}h_{j}g_{j}$ where $h_{j}=\sum_{i}\frac{ts\sigma^{-i}(d_{i,j})}{\prod_{l}\sigma^{l}(s)}x^{i}$. So $tf\in(Tg_{1}+Tg_{2}+\cdots+Tg_{k})$. Hence $tP\subseteq(Tg_{1}+Tg_{2}+\cdots+Tg_{k})$. Since $g_{i}\in P$, $(Tg_{1}+Tg_{2}+\cdots+Tg_{k})\subseteq P$. Thus $R[[x;\sigma]]$ is an $S$-Noetherian ring. ∎

The following proposition which is proved in [1], is the corollary of the above theorem.

Corollary 4.5.

[1, Proposition 10] Let $R$ be a commutative ring and $S\subseteq R$ an anti-Archimedean multiplicative set of $R$. If $R$ is $S$-Noetherian, then so is $R[[X_{1},\cdots,X_{n}]]$.

A ring R is called strongly regular if every principal right (or left) ideal is generated by a central idempotent. A ring is said to be left self injective if it is injective as a left module over itself. Hirano in [12, Theorem 4] shows that if $R$ is a self-injective strongly regular ring, then $R[[x]]$ is a duo ring.

We have the following generalization of a theorem of D.D. Anderson and Tiberiu Dumitrescu [1, Proposition 10].

Theorem 4.6.

Let $R$ be a duo ring with an automorphism $\sigma$ and $S\subseteq R$ a $\sigma$-anti-Archimedean denominator set (consisting nonzero devisors). If $R$ is $S$-Noetherian, then so is the skew power series ring $R[[x;\sigma]]$.

Proof.

We can prove this theorem in a similar way as in Theorem 4.4. Consider the notations in the proof of Theorem 4.4. Let $x\in P$. Since $\sigma$ is bijective, $P$ is $S$-finite. Let $x\notin P$ and $f\in P$, so $xf_{1}=sf-\sum d_{0i}g_{i}\in P$. Note that for each $h\in R[[x;\sigma]]$ and $I$ is a left ideal of $R[[x;\sigma]]$, $xh\in I$ yields that $xR[[x;\sigma]]h\in I$. So $f_{1}\in P$. The rest of the proof is similar to what we did in Theorem 4.4. ∎

The following corollary is a generalization of the case $n=1$ in [1, Proposition 10] for the category of duo rings.

Corollary 4.7.

Let $R$ be a duo ring and $S\subseteq R$ an anti-Archimedean denominator set (consisting nonzero devisors) of $R$. If $R$ is $S$-Noetherian, then so is the power series ring $R[[x]]$.

Now we extend the last result for the skew generalized power series ring $R[[M,\omega]]$.

Theorem 4.8.

Let $R$ be a duo ring, $(M,\leq)$ a positive strictly ordered commutative monoid and $\omega_{m}$ a monomorphism of $R$ with $\omega_{m}\omega_{n}=\omega_{n}\omega_{m}$ for each $m,n\in M$. Assume that $S\subset R$ is an $\omega_{m}$-anti-Archimedean denominator set (consisting nonzero devisors) of $R$ and $R[[M,\omega]]$ be a duo ring. Then $R[[M,\omega]]$ is left (or right) $S$-Noetherian if and only if $R$ is left (or right) $S$-Noetherian and $M$ is finitely generated.

Proof.

($\Leftarrow$) We use the method of G. Brookfeild employed in [6]. We know that the surjective homomorphism $\varphi:\mathbb{F}^{n}\longrightarrow M$ (where $\mathbb{F}$ is a free monoid) induces a projection

$$\varphi^{*}:R[[\mathbb{F}^{n},(\omega,\preceq)]]\longrightarrow R[[M,(\omega,\leq)]]$$

and $R[[M,(\omega,\leq)]]=R[[M,\omega]]$ by [6, Theorem 4.3]. Moreover, since $R[[\mathbb{F}^{n},(\omega,\preceq)]]$ is $S$-Noetherian, so is $R[[M,\omega]]$ by [16, Lemma 2.2] for noncommutative version.

($\Rightarrow$) Let $A:=R[[M,\omega]]$ be $S$-Noetherian. Let $\{m_{n}|n\in\mathbb{N}\}$ be an infinite sequence in $M$. Let $I=(Ae_{m_{1}}+Ae_{m_{2}}+\cdots)$. Since $A$ is $S$-Noetherian, there exists $s\in S$ such that $c_{s}I\subseteq J\subseteq I$ for $J$ finitely generated ideal of $A$. So $c_{s}I\subseteq(Ae_{m_{i_{1}}}+Ae_{m_{i_{2}}}+\cdots+Ae_{m_{i_{k}}})$ for some $k\in\mathbb{N}$. So $c_{s}e_{m_{l}}=\sum_{t=0}^{k}f_{t}e_{m_{it}}$ for some $l\neq i_{t}$. So $m_{l}\in\bigcup_{t=0}^{k}\supp(f_{t}e_{m_{i_{t}}})$ for each $m\in M$, $(f_{t}e_{m_{i_{t}}})(m)=\sum_{m^{\prime}m^{\prime\prime}=m}f_{t}(m^{\prime})\omega_{m^{\prime}}(e_{m_{i_{t}}}(m^{\prime\prime}))$. So $m_{l}\in\bigcup_{t=0}^{k}\big\{\supp(f_{t})+\supp(\omega_{m^{\prime}}(e_{m_{i_{t}}}(m^{\prime\prime})))\big\}$. There exists $m_{1}\in M$ such that $m_{1}m_{it}=m$ for some $0\leq t\leq L$. So

$$(f_{t}e_{m_{i_{t}}})(m)=f_{t}(m_{1})\omega_{m_{1}}(e_{m_{i_{t}}}(m_{i_{t}}))=f_{t}(m_{1}).$$

Thus $m_{1}\in\supp(f_{t})$ and $m_{1}m_{i_{t}}\in\supp(f_{t}e_{m_{i_{t}}})$ for some $0\leq t\leq L$. So for each $m\in\supp(\omega_{m^{\prime}}(e_{m_{i_{t}}}(m^{\prime})))$, $m_{i_{t}}\preceq m$ for some $0\leq t\leq L$. Since $m_{l}\in\supp(f_{t}e_{m_{i_{t}}})$, $m_{i_{t}}\preceq m_{l}$ for some $0\leq t\leq L$. Since $M$ is positive strictly ordered monoid, $M$ is finitely generated by [6, Lemma 3.3].

Let $I$ be an ideal of $R$, so $AI$ is an ideal of $A$. So there exists $s\in S$ such that $c_{s}AI\subseteq J\subseteq AI$ for some $J$ finitely generated ideal of $A$. Set

$$T=\{f(\pi(f))|f\in c_{s}AI\}.$$

We claim that $T=sI$. Let $t\in T$, so $t=h(\pi(h))$ and $h=c_{s}g$ for some $g\in AI$. So $t=sg(\pi(sg))$. This means that $t\in sI$ considering the fact that

$$I=\{f(\pi(f))|f\in AI\}.$$

So $T\subseteq sI$. Now let $i\in I$, so $i\in AI$. Since $si(m)=0$ for $m\neq 1$, $si\in T$. Thus $sI\subseteq T$. Hence $sI=T$. But $sI=T\subseteq J^{\prime}\subseteq I$ where $J^{\prime}=\{f(\pi(f))|f\in J\}$. Let $J=(Aj_{1}+Aj_{2}+\cdots+Aj_{p})$. So it is easy to show that $J^{\prime}=(Rj_{1}(\pi(j_{1}))+Rj_{2}(\pi(j_{2}))+\cdots+Rj_{p}(\pi(j_{p})))$. So $J^{\prime}$ is finitely generated in $R$. Hence $I$ is $S$-finite and $R$ is left $S$-Noetherian. ∎

Recall from [5], that a ring $R$ is right (left) $\aleph_{0}$-injective provided any homomorphism from a countably generated right (left) ideal of $R$ into $R$ extends to a right (left) $R$-module endomorphism of $R$. By an $\aleph_{0}$-injective ring we mean a right and left $\aleph_{0}$-injective ring.

Corollary 4.9.

Let $R$ be an strongly regular and an $\aleph_{0}$-injective ring with automorphisms $\sigma_{1},\sigma_{2}$ such that $\sigma_{1}\sigma_{2}=\sigma_{2}\sigma_{1}$. Assume that $S\subset R$ is an $\sigma_{1}$, $\sigma_{2}$ anti-Archimedean denominator set (consisting nonzero devisors). If $R$ is left (or right) $S$-Noetherian, then so is $R[[x,y;\sigma_{1},\sigma_{2}]]$.

Proof.

Assume that $R$ is an strongly regular and $\aleph_{0}$-injective ring. Then by [20], $A=R[[x;\sigma_{1}]]$ is duo ring and $S$-left Noetherian ring. Then $A[[y;\sigma_{2}]]$ is left $S$-Noetherian ring. ∎

The following corollary is a generalization of the case $n=2$ of in [1, Proposition 10] for the category of duo rings.

Corollary 4.10.

Let $R$ be an strongly regular self-injective ring and $S\subseteq R$ an anti-Archimedean denominator set (consisting nonzero devisors) of $R$. If $R$ is left (or right) $S$-Noetherian, then so is $R[[x,y]]$.

Corollary 4.11.

Let $R$ be a duo ring, $S\subseteq R$ an anti-Archimedean denominator set (consisting nonzero devisors) of $R$. Assume that $R[[M]]$ is a duo ring. Then $R[[M]]$ is left (or right) $S$-Noetherian if and only if $R$ is left (or right) $S$-Noetherian and $M$ is finite generated.