Endo-Noetherian rings

Abstract

A commutative ring R is said to be endo-Noetherian if the chain of annihilators ${ ann}(a_{1})\subseteq$${ ann}(a_{2})\subseteq \cdots$ stabilizes for each sequence $(a_{k})_{k}$ of elements of R (Ndiaye and Gueye in Int J Appl Math 86:871–881, 2013). In this paper, we give equivalent conditions for the power series (resp., polynomial) rings over an Armendariz ring to be endo-Noetherian. We also study several properties of an endo-Noetherian ring. The concept of the amalgamated duplication of R along an ideal $I, R\bowtie I$ is studied.

1 Introduction

All the rings considered are commutative with identity. An R-module M is called endo-Noetherian if it satisfies the ascending chain condition on endomorphic kernels, i.e. the chain $Ker(f_{1}) \subseteq Ker(f_{2}) \subseteq \cdots$ stabilizes for each sequence $(f_k)_{k}$ of endomorphisms of M [17]. A commutative ring R is said to be endo-Noetherian if it is endo-Noetherian as an R-module. Equivalently, R is endo-Noetherian if the chain of annihilators ${ ann}(a_{1})\subseteq { ann}(a_{2})\subseteq \cdots$ stabilizes for each sequence $(a_{k})_{k}$ of elements of R. The class of endo-Noetherian rings includes Noetherian rings and iso-Noetherian rings (Recall from [7] that an R-module M is iso-Noetherian if for every ascending chain $M_{1}\subseteq M_{2}\subseteq \cdots$ of submodules of M, there exists an index $n\ge 1$ such that $M_{n}\simeq M_{i}$ for every $i \ge n$. We say that a ring R is iso-Noetherian if the R-module R is iso-Noetherian). The purpose of this paper is to characterize when the power series (resp., polynomial) rings over Armendariz rings are endo-Noetherian.

In the first part of this paper, we study several properties of an endo-Noetherian ring. We show that if R is a commutative ring and S a multiplicative subset of R consisting of regular elements, then R is endo-Noetherian if and only if $S^{-1}R$ is endo-Noetherian. Also, we construct an example of a commutative ring R and a multiplicative subset S of R such that $R_S$ is Noetherian (so endo-Noetherian) but R is not endo-Noetherian. We study when the direct product of rings is endo-Noetherian. We prove that if $(R_i)_{i\in I}$ is a family of commutative rings and $R=\prod _{i\in I} R_{i}$ the product of $R_i,$ then R is endo-Noetherian if and only if I is a finite set and for each $i\in I, R_{i}$ is endo-Noetherian. Let $(*)$ be the following property: every increasing sequence of the form $(I:a_{1})$$\subseteq (I:a_{2}) \subseteq \cdots$ is stationary, where I is an ideal of R and $(a_{k})_k$ a sequence of elements of R. Note that if R satisfies the property $(*),$ then R satisfies the accr condition (we say that R satisfies the accr condition if the ascending chain of colon ideals of the form $(I:a)\subseteq (I:a^2)\subseteq (I:a^3) \subseteq \cdots$ is stationary for each $a\in R$ and every ideal I of R). We show that R satisfies the property $(*)$ if and only if R / I is an endo-Noetherian ring for each ideal I of R. We also study the amalgamated duplication of R along an ideal $I, R\bowtie I$ to be endo-Noetherian. The amalgamated duplication of a ring R along an ideal I is a ring that is defined as the following subring with unit element (1, 1) of $R \times R$:

$$\begin{aligned} R\bowtie I = \{(r, r + i) | r \in R, i \in I\}. \end{aligned}$$

This construction has been studied, in the general case, and from the different points of view of pullbacks, by D’Anna and Fontana [5]. According to [6], an element of R is called a regular element if it is not a zerodivisor. Also, we say that I is a regular ideal if it contains a regular element of R. We prove that if I is a regular ideal of R,  then R is endo-Noetherian if and only if $R\bowtie I$ is endo-Noetherian. In the second part of this paper, we give equivalent conditions for the power series (resp., polynomial) rings over an Armendariz ring to be endo-Noetherian. First, we need to collect some necessary notions. Recall that a ring R is called an Armendariz ring if whenever the polynomials $f= \sum _{i=0}^{m}a_{i}X^{i}$ and $g= \sum _{i=0}^{n}b_{i}X^{i}\in R[X]$ satisfy $fg=0$ we have $a_{i}b_{j}=0$ for every i and j [14]. Also, R is said power serieswise Armendariz if for each $f=\sum _{i\ge 0}a_iX^i$ and $g=\sum _{i\ge 0}b_iX^i \in R[\![X]\!]$ such that $fg=0,$ then $a_ib_j=0$ for each i and j. We prove that if R is an Armendariz commutative ring, then the following are equivalent:

• 1. R[X] is endo-Noetherian.

• 2. R satisfies the acc on annihilators of finite subset.

• 3. R satisfies the acc on annihilators of finitely generated ideals of R.

Finally, if R is a power serieswise Armendariz ring, then the following statements are equivalent:

• 1. $R[\![X]\!]$ is endo-Noetherian.

• 2. R satisfies the acc on annihilators of countably subset.

• 3. R satisfies the acc on annihilators of countably generated ideals of R.

• 4. For each sequence $(f_k=\sum _{j\ge 0}a_{k,j}X^j)_k$ of elements of $R[\![X]\!]$ such that ${ ann}_{R[\![X]\!]}(f_1) \subseteq { ann}_{R[\![X]\!]}(f_2)$$\subseteq \cdots ,$ there exists an $n\in \mathbb {N}$ such that for each $k\ge n, \cap _{j\ge 0}{} { ann}_R(a_{k,j}) = \cap _{j\ge 0}{} { ann}_R(a_{n,j})$.

2 General properties

Let R be a commutative ring. According to [17], the ring R is said to be endo-Noetherian if the increasing sequence ${ ann}(a_{1})\subseteq { ann}(a_{2})\subseteq \cdots$ stabilizes for any sequence $(a_{k})_k$ of elements of R. In this section, we study several properties of the endo-Noetherian ring.

Example 2.1

• 1. Every Noetherian ring is an endo-Noetherian ring.

• 2. The converse of (1) is not true in general. Let R be a commutative ring with identity and M an R-module. The idealization of M in R (or trivial extension of R by M) is a commutative ring $R(+)M=\lbrace (r,m) \;|\;r\in R$ and $m\in M \rbrace$ under the usual addition and the multiplication defined as $(r_{1},m_{1})(r_{2},m_{2}) = (r_{1}r_{2},r_{1}m_{2}+r_{2}m_{1})$ for all $(r_{1},m_{1}),(r_{2},m_{2})\in R(+)M$. Now, let K be a field and V a K-vector space, then $K(+)V$ is an endo-Noetherian ring, because for each $(k,m)\in K\times V\backslash (0,0),$

  $$\begin{aligned} { ann}_{K(+)V}((k,m)) =\left\{ \begin{array}{ll} (0,0) &{}\quad {\texttt {if}}\quad k\ne 0\\ 0(+)V &{}\quad {\texttt {if}}\quad k= 0 \end{array}\right. \end{aligned}$$

  Note that if V has an infinite dimensional, then by [2, Theorem 4.8], $K(+)V$ is not Noetherian.

• 3. Recall from [7] that an R-module M is iso-Noetherian if for every ascending chain $M_{1}\subseteq M_{2}\subseteq \cdots$ of submodules of M, there exists an index $n\ge 1$ such that $M_{n}\simeq M_{i}$ for every $i \ge n$. We say that a ring R is iso-Noetherian if the R-module R is iso-Noetherian. By [7, Lemma 5.4(1)], every iso-Noetherian ring is endo-Noetherian.

Recall that a commutative ring R is said to be strongly Hopfian if the chain of annihilators ${ ann}(a)\subseteq$${ ann}(a^2)\subseteq \cdots$ is stationary for each $a\in R$ [3, 9, 12, 13]. Note that every endo-Noetherian ring satisfies the strongly Hopfian property. In Example 2.7, we provide a strongly Hopfian ring which is not endo-Noetherian.

Proposition 2.2

Let R be a commutative ring and S a multiplicative subset of R consists of regular elements. Then, the following assertions are equivalent:

• 1. R is endo-Noetherian.

• 2. $S^{-1}R$ is endo-Noetherian.

Proof

$(1) \Rightarrow (2)$. Let $(\frac{a_{k}}{s_{k}})_k$ be a sequence of elements of $S^{-1}R$ such that ${ ann}_{S^{-1}R}(\frac{a_{1}}{s_{1}}) \subseteq { ann}_{S^{-1}A}(\frac{a_{2}}{s_{2}}) \subseteq \cdots$. We will show that ${ ann}_{R}(a_{k}) \subseteq { ann}_{R}(a_{k+1})$ for each $k\in \mathbb {N}$. Let $b\in { ann}_{R}(a_{k})$. Since $\frac{b}{1}$$\in { ann}_{S^{-1}R}(\frac{a_{k}}{s_{k}}), \frac{b}{1} \in$${ ann}_{S^{-1}R}(\frac{a_{k+1}}{s_{k+1}})$. This implies that $b\in { ann}_{R}(a_{k+1}),$ because S consists of regular elements.

Now, as R is endo-Noetherian, there exists a positive integer n such that for each $k\ge n$, ${ ann}(a_{k})={ ann}(a_{n})$. Let $k\ge n$ and let $\frac{b}{t}\in { ann}(\frac{a_{k}}{s_{k}})$. Then, $ba_{k}=0;$ so $ba_{n}=0$ which implies that $\frac{b}{t} \in$${ ann}(\frac{a_{n}}{s_{n}})$. Thus, the chain ${ ann}_{S^{-1}R}(\frac{a_{1}}{s_{1}}) \subseteq$${ ann}_{S^{-1}R}(\frac{a_{2}}{s_{2}}) \subseteq \cdots$ is stationary, and hence the ring $S^{-1}R$ satisfies the endo-Noetherian property.

$(2) \Rightarrow (1)$. Assume that $S^{-1}R$ is endo-Noetherian and let $(a_k)_k$ be a sequence of elements of R such that ${ ann}(a_1)$$\subseteq { ann}(a_2) \subseteq \cdots$. Since S consists of regular elements, for each $k\in \mathbb {N}^*$, ${ ann}_{S^{-1}R}(\frac{a_{k}}{1}) \subseteq { ann}_{S^{-1}R}(\frac{a_{k+1}}{1})$. Thus by hypothesis, there exists a nonzero positive integer n such that for each $k\ge n$, ${ ann}_{S^{-1}R}(\frac{a_{n}}{1}) = { ann}_{S^{-1}R}(\frac{a_{k}}{1})$. Now, let $k\ge n,$ and $\alpha \in { ann}(a_k),$ then $\frac{\alpha }{1} \in$${ ann}_{S^{-1}R}(\frac{a_{k}}{1}) = { ann}_{S^{-1}R}(\frac{a_{n}}{1});$ so $\alpha a_{n}=0,$ because S consists of regular elements. This implies that ${ ann}(a_n) = { ann}(a_k)$ for each $k\ge n$. Hence, R is endo-Noetherian. $\square$

We next construct an example of a commutative ring R and a multiplicative subset S of R such that $R_S$ is Noetherian (so endo-Noetherian) but R is not endo-Noetherian. To prove this, we need to collect some necessary notions. Let R be a commutative ring and S be a multiplicative subset of R. Recall from [1] (or [10]) that an ideal I of R is called S-finite (resp., S-principal), if $sI\subseteq J\subseteq I$ for some finitely generated (resp., principal) ideal J of R and some $s\in S$. Also, R is called S-Noetherian if each ideal of R is S-finite. In [1], the authors showed that R is S-Noetherian if and only if every prime ideal is S-finite.

Example 2.3

Let F be a field, $\mathbf{X}=\{X_i \,|\, i \in \mathbb {N}\}$ a set of indeterminates over F, I the ideal of $F[\mathbf{X}]$ generated by the set $\{X_0^jX_j \,|\, j\ge 1\}$ and $R=F[\mathbf{X}]/I$. For each $i \in \mathbb {N}$, let $\overline{X_i}$ denote the homomorphic image of $X_i$ in R. Let $S=\{\overline{X_0}^i\mid i\in \mathbb {N}\}$ and ${\overline{\mathbf{X}}}=\{\overline{X_i} \,|\, i \in \mathbb {N} \}$. Then, S is a multiplicative subset of R.

• 1. R is not strongly Hopfian. Indeed, let $a=\overline{X_0},$ then for each $n\in \mathbb {N}^*, \overline{X_n}\in$${ ann}(a^n){\setminus } { ann}(a^{n-1})$. So R is not endo-Noetherian.

• 2. Let P be the prime ideal of R generated by the set $\{\overline{\mathbf{X}}\}$. Then, P is the only prime ideal of R. Since $\overline{X_0}\in P$, $\overline{X_0}P\subseteq$$(\overline{X_0})\subseteq P;$ so P is an S-finite ideal of R. This implies that R is an S-Noetherian ring, and hence $R_S$ is Noetherian [1, Proposition 2(f)]. In particular, R is an S-Noetherian ring which is not Noetherian.

Let R be a commutative ring and S a multiplicative subset of R. Our next result gives a sufficient condition for the localization of $R, R_S$ satisfies the endo-Noetherian property implies that R is endo-Noetherian.

Proposition 2.4

Let R be a quasi-local ring with maximal ideals $M_{1},\ldots ,M_{n}$. If $(R{\setminus } M_{i})^{-1}R$ is an endo-Noetherian ring for each $1\le i\le n,$ then R is endo-Noetherian.

Proof

Let $(a_{k})_k$ be a sequence of elements of R such that ${ ann}_{R}(a_{1}) \subseteq { ann}_{R}(a_{2})\subseteq \cdots$. It is easy to show that ${ ann}_{(R{\setminus } M_{i})^{-1}R}(\frac{a_{1}}{1})$$\subseteq { ann}_{(R{\setminus } M_{i})^{-1}R}(\frac{a_{2}}{1})$$\subseteq \cdots$ for each $1\le i\le n$. Note that for each $1\le i\le n$, there exists a positive integer $m_i$ such that for each $k\ge m_i, { ann}_{(R{\setminus } M_{i})^{-1}R}(\frac{a_{k}}{1})$$= { ann}_{(R{\setminus } M_{i})^{-1}R}(\frac{a_{m_i}}{1})$. Let $m=\max \{m_i\mid 1\le i\le n\}$. Then, for each $1\le i\le n$ and each $k\ge m, { ann}_{(R{\setminus } M_{i})^{-1}R}(\frac{a_{k}}{1})$$= { ann}_{(R{\setminus } M_{i})^{-1}R}(\frac{a_{m}}{1})$.

Now, we will prove that for each $k\ge m, { ann}_{R}(a_{k})= { ann}_{R}(a_{m})$. Let $k\ge m$ and $b\in { ann}_{R}(a_{k})$. Then $ba_k=0,$ and for each $1\le i\le n, \frac{b}{1}\frac{a_k}{1}= 0$ in $(R{\setminus } M_{i})^{-1}R;$ so for each $1\le i\le n$, $\frac{b}{1}\frac{a_m}{1}= 0$. This implies that for each $1\le i\le n,$ there exists an $s_{i}\in R{\setminus } M_{i}$ such that $s_{i}ba_{m}=0$. As the ideal generated by $\{s_{1},\ldots ,s_{n}\}$ is equal to R,  we deduce that $ba_{m}=0,$ and hence R is endo-Noetherian. $\square$

Theorem 2.5

Let $(R_i)_{i\in I}$ be a family of commutative rings and $R=\prod _{i\in I} R_{i}$ the product of $R_i$. Then, R is endo-Noetherian if and only if I is finite and for each $i\in I$, $R_{i}$ is endo-Noetherian.

Proof

The “only if” part assume that $R=\prod _{i\in I} R_{i}$ is endo-Noetherian. If I is not finite, then the sequence ${ ann}_R((0,1,1,\ldots )) \subseteq { ann}_R((0,0,1,1,\ldots ))$$\subseteq \cdots$ does not stabilize. It is easy to prove that if $R=\prod _{i=1}^n R_{i}$ is endo-Noetherian, then for each $1\le i\le n, R_i$ is endo-Noetherian.

For “if” part, let $R_1,\ldots ,R_n$ be endo-Noetherian rings. We will show that $R=\prod _{i=1}^n R_{i}$ is endo-Noetherian. To prove this, it sufficient to show the proof in the case $n=2$ and conclude by induction on n.

Let $R=A\times B$ and $(a_{1},b_{1}),\ldots \in R$ such that ${ ann}_{R}((a_{1},b_{1})) \subseteq \cdots$. It is easy to show that ${ ann}_{A}(a_{1}) \subseteq { ann}_{A}(a_{2})$$\subseteq \cdots$ and ${ ann}_{B}(b_{1}) \subseteq$${ ann}_{B}(b_{2})\cdots$. Since A and B are endo-Noetherian rings, there exists a positive integer n such that for each $k\ge n, { ann}_{A}(a_{k}) = { ann}_{A}(a_{n})$ and ${ ann}_{B}(b_{k}) = { ann}_{B}(b_{n})$. This implies that for each $k\ge n$, ${ ann}_{R}((a_{k},b_{k})) = { ann}_{R}((a_{n},b_{n})),$ and hence R is endo-Noetherian. $\square$

Corollary 2.6

Let $(R_i)_{1\le i\le n}$ be a finite family of commutative rings and $R=\prod _{i=1}^n R_{i}$. Then, R is endo-Noetherian if and only if $R_i$ is endo-Noetherian for each $1\le i\le n$.

Example 2.7

Let $(F_i)_{i\in I}$ be an infinite family of fields and $R=\prod _{i\in I} F_{i}$. Since R is a zero-dimensional ring, by [12, Remark 1.20], R is strongly Hopfian. But by Theorem 2.5, R is not an endo-Noetherian ring.

Remark 2.8

Note that if R is endo-Noetherian and I an ideal of R, then R / I is not endo-Noetherian in general. In fact, let K be a field and $R = K[X_{0},\ldots ],$ then R is endo-Noetherian. Let I be the ideal of R generated by the elements $X_{0}X_{1},X_{0}^{2}X_{2},\cdots$. Then, R / I is not strongly Hopfian, because $\overline{X_n}\in$${ ann}_{R/I}(\overline{X_0^n}){\setminus } { ann}_{R/I}(\overline{X_0^{n-1}})$. In particular, R / I is not endo-Noetherian.

Let R be a commutative ring and let $(*)$ be the following property: every increasing sequence of the form $(I:a_{1})$$\subseteq (I:a_{2}) \subseteq \cdots$ is stationary, where I is an ideal of R and $(a_{k})_k$ a sequence of elements of R. Note that if R satisfies the property $(*),$ then R satisfies the accr condition (we say that R satisfies the accr condition if the ascending chain of colon ideals of the form $(I:a)\subseteq (I:a^2)\subseteq (I:a^3) \subseteq \cdots$ is stationary for each $a\in R$ and every ideal I of R). It is easy to show that if R satisfies the property $(*),$ then R is an endo-Noetherian ring and the converse does not hold in general. It suffices to take an integral domain which is not satisfied accr. Indeed, let F be a field, $\mathbf{X}=\{X_i \,|\, i \in \mathbb {N}\}$ a set of indeterminates over F and $R=F[\mathbf{X}]$. Let Y be an indeterminate over R. Since R is not Noetherian, by [16, Theorem 2], $R[\![Y]\!]$ does not satisfy accr, in particular $R[\![Y]\!]$ does not satisfy $(*)$. Note that $R[\![Y]\!]$ is endo-Noetherian, because it is an integral domain.

Theorem 2.9

The following assertions are equivalent for a commutative ring R.

• 1. R / I is an endo-Noetherian ring for each ideal I of R.

• 2. R satisfies the property $(*)$.

Proof

$(1) \Rightarrow (2)$. Let I be an ideal of R and $a_{1},a_{2},\ldots \in R$ such that $(I:a_{1})\subseteq (I:a_{2})$$\subseteq \cdots$. We show that ${ ann}_{R/I}(\overline{a_{1}})$$\subseteq { ann}_{R/I}(\overline{a_{2}}) \subseteq \cdots$. Let $\overline{\alpha } \in { ann}_{R/I}(\overline{a_{k}}),$ then $\alpha a_{k} \in I;$ so $\alpha a_{k+1}\in I$. Therefore, $\overline{\alpha } \in { ann}_{R/I}(\overline{a_{k+1}})$.

Now, since R / I is endo-Noetherian, there exists a positive integer n such that ${ ann}_{R/I}(\overline{a_{k}}) = { ann}_{R/I}(\overline{a_{n}})$ for each $k\ge n$. We will prove that $(I:a_{k}) = (I:a_{n})$ for each $k\ge n$. Let $b\in (I:a_{k})$. Then, $\overline{b}\overline{a_{k}} = \overline{0}$ in R / I. Therefore, $\overline{b}\in { ann}_{R/I}(\overline{a_{k}})$$\subseteq { ann}_{R/I}(\overline{a_{n}});$ so $ba_{n}\in I$.

$(2) \Rightarrow (1)$. Let $a_{1},a_{2},\ldots \in R$ such that ${ ann}_{R/I}(\overline{a_{1}}) \subseteq$${ ann}_{R/I}(\overline{a_{2}}) \subseteq \cdots$. We show that $(I:a_{1}) \subseteq (I:a_{2})\subseteq \cdots$. Let $b\in (I:a_{k})$. Then, $\overline{b} \in { ann}_{R/I}(\overline{a_{k}})$$\subseteq { ann}_{R/I}(\overline{a_{k+1}})$. Thus, $b\in (I:a_{k+1})$.

Now, by hypothesis, there exists a positive integer n such that $(I:a_{k}) = (I:a_{n})$ for each $k\ge n$. This implies that ${ ann}_{R/I}(\overline{a_{k}}) = { ann}_{R/I}(\overline{a_{n}})$ for each $k\ge n$. $\square$

Proposition 2.10

Let R, T be two commutative rings and let $f:R\rightarrow T$ be a ring homomorphism. If R satisfies $(*)$, then f(R) satisfies $(*)$. In particular, f(R) is an endo-Noetherian ring.

Proof

Let I be an ideal of f(R) and $a_{1},a_{2},\ldots \in R$ such that $(I:_{f(R)}f(a_{1})) \subseteq (I:_{f(R)}f(a_{2})) \subseteq \cdots$. We show that $(f^{-1}(I):_{R}a_{1}) \subseteq$$(f^{-1}(I):_{R}a_{2}) \subseteq \cdots$. Let $x\in R$ such that $xa_{k} \in f^{-1}(I);$ so $f(x) \in (I:_{f(R)}f(a_{k}))$. Therefore, $f(x)\in (I:_{f(R)}f(a_{k+1})),$ and hence $xa_{k+1}\in f^{-1}(I)$.

Now, since R satisfies $(*),$ there exists a positive integer n such that $(f^{-1}(I):_{R}a_{k}) = (f^{-1}(I):_{R}a_{n})$ for each $k\ge n$. We show that $(I:_{f(R)}f(a_{k})) = (I:_{f(R)}f(a_{n}))$ for each $k\ge n$. Let $y\in f(R)$ such that $yf(a_{k}) \in I$. There exists $x\in R$ such that $y=f(x)$. Then, $f(xa_{k})\in I$. This implies that $xa_{k} \in f^{-1}(I)$. Thus, $xa_{n}\in f^{-1}(I)$. Hence, $yf(a_{n})\in I$. $\square$

Recall the following definitions: A ring R is called a zero divisor ring (a ZD-ring), if the set of zero divisors on the R-module R / I is a finite union of prime ideals for each ideal I of R. A Laskerian ring is one for which every ideal is a finite intersection of primary ideals, and finally, a ring R is strongly Laskerian if it is Laskerian and for each ideal I of R there exists $k\in \mathbb {N}$ such that $(\sqrt{I})^k$$\subseteq I$. Clearly, every Laskerian ring is a ZD-ring. Heinzer and Ohm [11], proved that a polynomial ring R[X] is a ZD-ring if and only if R is Noetherian, and hence the conditions Noetherian, Strongly Laskerian, Laskerian, and ZD are equivalent in R[X]. Similarly, Gilmer and Heinzer [8] proved that a power series ring $R[\![X]\!]$ is Laskerian if and only if R is Noetherian (but R need not be Noetherian if $R[\![X]\!]$ is a ZD-ring). We remark that Lu [16] gave a very simple proof to show that if R[X] (resp. $R[\![X]\!])$ satisfies accr, then R is Noetherian. Using this result, we prove the following proposition.

Proposition 2.11

R[X] (resp., $R[\![X]\!])$ satisfies $(*)$ if and only if R is Noetherian.

Proof

Since R[X] satisfies $(*)$, R[X] satisfies accr; so by [16, Theorem 2], R is Noetherian. The other implication is obvious. $\square$

Remark 2.12

By Proposition 2.11, the conditions $(*),$ Noetherian, Strongly Laskerian, Laskerian, and ZD are equivalent in R[X]. This shows that the property $(*)$ does not passe to the polynomial (resp., formal power series) rings. Indeed, we will construct a commutative ring R satisfies the condition $(*),$ but R is not Noetherian. Let K be a field and V an infinite-dimensional K-vector space. By Example 2.1(2), $R=K(+)M$ is not Noetherian. On the other hand, by [2, Corollary 3.4], every proper ideal of R has the form $I=0(+)N$ for some submodule N of M. Since $R/I=(K(+)M)/(0(+)N) \simeq K(+)(M/N),$ by Example 2.1(2), R / I is endo-Noetherian for every ideal I of R. Thus by Theorem 2.9, R satisfies $(*)$.

In [4], D’Anna and Fontana considered a different type of construction obtained involving a ring R and an ideal $I \subseteq R$ that is denoted by $R\bowtie I,$ called amalgamated duplication, and it is defined as the following subring of $R\times R:$

$$\begin{aligned} R\bowtie I=\{(r,r+i) \mid r\in R, i\in I\}. \end{aligned}$$

According to [6], an element of R is called a regular element if it is not a zerodivisor. Also, we say that I is a regular ideal if it contains a regular element of R. We end this section by the following result.

Theorem 2.13

Let R be a commutative ring and I a regular ideal of R. Then, the following assertions are equivalent:

• 1. R is endo-Noetherian.

• 2. $R\times R$ is endo-Noetherian.

• 3. $R\bowtie I$ is endo-Noetherian.

Proof

$(1) \Rightarrow (2)$. It follows from Theorem 2.5.

$(2) \Rightarrow (3)$. Let $(r_{1},r_{1}+i_{1})$, $(r_{2},r_{2}+i_{2}),\ldots \in R\bowtie I$ such that ${ ann}_{R\bowtie I}((r_{1},r_{1}+i_{1})) \subseteq { ann}_{R\bowtie I}((r_{2},r_{2}+i_{2})) \subseteq \cdots$. We will show that ${ ann}_{R\times R}((r_{k},r_{k}+i_{k})) \subseteq { ann}_{R\times R}((r_{k+1},r_{k+1}+i_{k+1}))$ for each $k\ge 1$. Let $(a,b)\,\in { ann}_{R\times R}((r_{k},r_{k}+i_{k}))$ and let i be a regular element of I. We have $(i,i)(a,b)= (ia,ia+i(b-a))\in$$R\bowtie I$. Moreover, $(i,i)(a,b)(r_{k},r_{k}+i_{k}) = (0,0),$ then $(i,i)(a,b)\in { ann}_{R\bowtie I}((r_{k+1},r_{k+1}+i_{k+1}))$. Thus, $(a,b)\in { ann}_{R\times R}((r_{k+1},r_{k+1}+i_{k+1})),$ because (i, i) is regular in $R\times R$. Now, by hypothesis, there exists a positive integer n such that ${ ann}_{R\times R}((r_{k},r_{k}+i_{k})) = { ann}_{R\times R}((r_{n},r_{n}+i_{n}))$ for each $k\ge n$. This implies that for each $k\ge n, (R\bowtie I)\cap { ann}_{R\times R}((r_{k},r_{k}+i_{k})) = ( R\bowtie I)\bigcap { ann}_{R\times R}((r_{n},r_{n}+i_{n}));$ so ${ ann}_{R\bowtie I}((r_{k},r_{k}+i_{k})) = { ann}_{R\bowtie I}((r_{n},r_{n}+i_{n}))$ for each $k\ge n$.

$(3) \Rightarrow (1)$. Note that this implication is always true and does not need the assumption that I contains a regular element.

Let $r_{1},r_{2},\ldots \in R$ such that ${ ann}_{R}(r_{1})$$\subseteq { ann}_{R}(r_{1}) \subseteq \cdots$. We will prove that ${ ann}_{R\bowtie I}((r_{k},r_{k})) \subseteq { ann}_{R\bowtie I}((r_{k+1},r_{k+1}))$ for each $k\ge 1$. Let $(\alpha ,\alpha + i)\in R\bowtie I$ such that $(\alpha ,\alpha + i)(r_{k},r_{k})$$=(0,0)$. Then, $\alpha$ and $\alpha + i$ are in ${ ann}_{R}(r_{k});$ so $(\alpha ,\alpha + i)\in { ann}_{R\bowtie I}((r_{k+1},r_{k+1}))$. Since $R\bowtie I$ is endo-Noetherian, there exists a positive integer n such that ${ ann}_{R\bowtie I}((r_{k},r_{k})) = { ann}_{R\bowtie I}((r_{n},r_{n}))$ for each $k\ge n$. Now, we show that ${ ann}_{R}(r_{k}) = { ann}_{R}(r_{n})$ for each $k\ge n$. Let $b\in { ann}_{R}(r_{k})$. Since $(b,b)\in { ann}_{R\bowtie I}((r_{k},r_{k}))$, $(b,b)(r_{n},r_{n})=(0,0)$ thus $br_{n}=0,$ and hence $b\in { ann}_{R}(r_{n})$. $\square$

3 The case of Armendariz and power serieswise Armendariz rings

Recall that a ring R is called an Armendariz ring if whenever the polynomials $f= \sum _{i=0}^{m}a_{i}X^{i}$ and $g= \sum _{i=0}^{n}b_{i}X^{i}\in R[X]$ satisfy $fg=0$ we have $a_{i}b_{j}=0$ for every i and j [14]. We next give a necessary and sufficient condition for the polynomial ring R[X] to be endo-Noetherian, where R is an Armendariz ring.

Theorem 3.1

Let R be an Armendariz commutative ring. Then, the following statements are equivalent:

• 1. R[X] is endo-Noetherian.

• 2. R satisfies the acc on annihilators of finite subset.

• 3. R satisfies the acc on annihilators of finitely generated ideals of R.

Proof

$(1) \Rightarrow (2)$. Let $S_{1}, S_{2},\ldots$ be a sequence of finite subset such that ${ ann}_{R}(S_{1}) \subseteq { ann}_{R}(S_{2})$$\subseteq \cdots$. Put $S_{i} = \lbrace a_{0,i},a_{1,i},\ldots ,a_{n_{i},i} \rbrace$ for each i and let $P_{i}$$= \sum _{j=0}^{n_{i}}a_{j,i}X^{j}$. We will show that ${ ann}_{R[X]}(P_{1}) \subseteq { ann}_{R[X]}(P_{2}) \subseteq \cdots$. Let $k\in \mathbb {N}^*$ and let $f = \sum _{j=0}^{d}b_{j}X^{j}$ such that $fP_{k}=0$. Since R is an Armendariz ring, for each $0\le j\le d$ and $0\le l\le n_{k}$, $b_{j}a_{l,k}=0$. Thus for each $0\le j\le d$, $b_{j} \in$${ ann}_{R}(S_{k}),$ which implies that for each $0\le j\le d$, $b_{j} \in { ann}_{R}(S_{k+1});$ so $b_{j}a_{l,k+1}=0$ for each $0\le j\le d$ and $0\le l\le n_{k+1}$. This implies that $fP_{k+1}=0$ and we conclude that ${ ann}_{R[X]}(P_{1}) \subseteq$${ ann}_{R[X]}(P_{2}) \subseteq \cdots$. Now, since R[X] is endo-Noetherian, there exists a positive integer n such that ${ ann}_{R[X]}(P_{k}) = { ann}_{R[X]}(P_{n})$ for each $k\ge n$. It is easy to show that ${ ann}_{R}(S_{k})= { ann}_{R}(S_{n})$ for each $k\ge n$. Hence, R satisfies the acc on annihilators of finite subset.

$(2)\Rightarrow (1)$ Let $(P_k)_k$ be a sequence of elements of R[X] such that ${ ann}_{R[X]}(P_{1}) \subseteq { ann}_{R[X]}(P_{2})$$\subseteq \cdots$. For each $i\ge 1,$ let $P_{i} = \sum _{j=0}^{n_{i}}a_{j,i}X^{j}$ and put $S_{i}=\lbrace a_{0,i},a_{1,i},\ldots ,a_{n_{i},i} \rbrace$. We will prove that ${ ann}_{R}(S_{k})\subseteq { ann}_{R}(S_{k+1})$ for each $k\ge 1$. Let $\alpha \in R$ such that $\alpha S_{k}=0$. Then by construction of $S_k,$ we have $\alpha P_{k}=0;$ so $\alpha P_{k+1}=0$ which implies that $\alpha S_{k+1}=0$. Since R satisfies the acc on annihilators of finite subset, there exists a positive integer n such that ${ ann}_{R}(S_{k}) = { ann}_{R}(S_{n})$ for each $k\ge n$. Now, let $Q=\sum _{i=0}^rq_{i}X^{i}\in R[X]$ such that $QP_{k}=0$. Then $q_{i}S_{k} = 0$ for each $0\le i\le r;$ so $q_{i}S_{n}=0$ for each $0\le i\le r$. This implies that $QP_{n}=0,$ and hence ${ ann}_{R[X]}(P_{n}) = { ann}_{R[X]}(P_{k})$ for each $k\ge n$.

$(2) \Leftrightarrow (3)$. It follows from the fact that for each finitely generated ideal $I=(a_1,\ldots ,a_n)$ of $R, { ann}_R(I)= { ann}_R(S)= \cap _{k=1}^n{ ann}_R(a_k),$ where $S=\{a_k\mid 1\le k\le n\}$. $\square$

Let R be a ring. According to [12], R is said power serieswise Armendariz if for each $f=\sum _{i\ge 0}a_iX^i$ and $g=\sum _{i\ge 0}b_iX^i \in R[\![X]\!]$ such that $fg=0,$ then $a_ib_j=0$ for each i and j. Next, we characterize when the power series rings over power serieswise Armendariz rings are endo-Noetherian. First, we need to recall some results. Note that those results can be found in [12] or [15].

Notation 3.2

Let R be a commutative ring. We note ${ ann}_R(2^R) =\{{ ann}_R(U)\mid U\subseteq R\},$ and ${ ann}_{R[\![X]\!]}(2^{R[\![X]\!]})$$=\{{ ann}_{R[\![X]\!]}(V) \mid V\subseteq R[\![X]\!]\}$. If $f\in R[\![X]\!],$ we note $C_f$ the set of all coefficients of f and for $V\subseteq R[\![X]\!],$ we note $C_V=\cup _{f\in V}C_f$.

Remark that ${ ann}_{R[\![X]\!]}(V)\cap R= { ann}_{R}(V)= { ann}_{R}(C_V)$ for $V\subseteq R[\![X]\!]$. So we can define the application

$$\begin{aligned} \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} \psi &{} : &{} { ann}_{R[\![X]\!]}(2^{R[\![X]\!]}) &{} \rightarrow &{} { ann}_R(2^R) \\ &{} &{} I &{} \mapsto &{} I\cap R. \end{array} \end{aligned}$$

Then, $\psi$ is surjective.

Also, let $U\subseteq R,$ then ${ ann}_R(U)[\![X]\!]$$\subseteq { ann}_{R[\![X]\!]}(U)$. Conversely, let $f=\sum _{i\ge 0}a_iX^i$ such that $fU=0,$ then $a_iU=0$ for each $i\in \mathbb {N};$ so $f\in { ann}_R(U)[\![X]\!]$. Then, ${ ann}_R(U)[\![X]\!] = { ann}_{R[\![X]\!]}(U)$. Thus, we can define the application

$$\begin{aligned} \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} \phi &{} : &{} { ann}_R(2^R) &{} \rightarrow &{} { ann}_{R[\![X]\!]}(2^{R[\![X]\!]}) \\ &{} &{} J &{} \mapsto &{} J[\![X]\!]. \end{array} \end{aligned}$$

Then, $\phi$ is injective.

Proposition 3.3

[15] Let R be a ring. The following are equivalent:

• 1. R is power serieswise Armendariz.

• 2. $\phi$ is surjective.

• 3. $\psi$ is injective.

Now, we are ready to prove the main result of this section. Note that the proof is inspired by [12, Proposition 2.28].

Theorem 3.4

Let R be a power serieswise Armendariz ring. Then, the following statements are equivalent:

• 1. $R[\![X]\!]$ is endo-Noetherian.

• 2. R satisfies the acc on annihilators of countably subset.

• 3. R satisfies the acc on annihilators of countably generated ideals of R.

• 4. For each sequence $(f_k=\sum _{j\ge 0}a_{k,j}X^j)_k$ of elements of $R[\![X]\!]$ such that ${ ann}_{R[\![X]\!]}(f_1) \subseteq { ann}_{R[\![X]\!]}(f_2)$$\subseteq \cdots ,$ there exists an $n\in \mathbb {N}$ such that for each $k\ge n, \cap _{j\ge 0}{} { ann}_R(a_{k,j}) = \cap _{j\ge 0}{} { ann}_R(a_{n,j})$.

Proof

$(1) \Rightarrow (2)$. Let $S_{1}, S_{2},\ldots$ be a sequence of countably subset of R such that ${ ann}_{R}(S_{1}) \subseteq { ann}_{R}(S_{2}) \subseteq \cdots$. For each $i\in \mathbb {N},$ let $S_{i} = \{ a_{j,i}\mid j\in \mathbb {N}\}$ and put $P_{i} = \sum _{j\ge 0}a_{j,i}X^{j} \in R[\![X]\!]$. In the same way of the proof of Theorem 3.1, we can conclude that ${ ann}(S_n)= { ann}(S_{n+1})= { ann}(S_{n+2})=\cdots$ for some $n\in \mathbb {N}^*$. So R satisfies the acc on annihilators of countably subset.

$(2) \Rightarrow (1)$. Let $(P_k)_k$ be a sequence of elements of $R[\![X]\!]$ such that ${ ann}_{R[\![X]\!]}(P_{1}) \subseteq$${ ann}_{R[\![X]\!]}(P_{2}) \subseteq \cdots$. For each $i\ge 1,$ let $P_{i} = \sum _{j\ge 0}a_{j,i}X^{j}$ and put $S_{i}= \{a_{j,i} \mid j\in \mathbb {N}\}$. Now, in the same way of the proof of Theorem 3.1, we can conclude that ${ ann}_{R[\![X]\!]}(P_n)= { ann}_{R[\![X]\!]}(P_{n+1})= { ann}_{R[\![X]\!]}(P_{n+2})=\cdots$ for some $n\in \mathbb {N}^*$. Hence, $R[\![X]\!]$ is an endo-Noetherian ring.

$(2) \Leftrightarrow (3)$. Obvious.

$(1) \Leftrightarrow (4)$. By Proposition 3.3, R is power serieswise Armendariz if and only if $\phi$ is surjective if and only if $\psi$ is injective. In that case, $\phi \circ \psi =id$ and $\psi \circ \phi =id$. Then for each $f\in R[\![X]\!]$, ${ ann}_{R[\![X]\!]}(f)= \phi \circ \psi ({ ann}_{R[\![X]\!]}(f))= \phi ({ ann}_R(f))= { ann}_R(f)[\![X]\!]$. It is easy to prove that for each $f=\sum _{j\ge 0}a_{j}X^j\in R[\![X]\!], { ann}_R(f)= \cap _{j\ge 0}{} { ann}_R(a_j)$. Now, $R[\![X]\!]$ is endo-Noetherian if and only if for each sequence $(f_k=\sum _{i\ge 0}a_{k,j}X^j)_k$ of elements of $R[\![X]\!]$ such that ${ ann}_{R[\![X]\!]}(f_1)$$\subseteq { ann}_{R[\![X]\!]}(f_2) \subseteq \cdots ,$ there exists an $n\in \mathbb {N}$ such that for each $k\ge n$, ${ ann}_{R[\![X]\!]}(f_k)= { ann}_{R[\![X]\!]}(f_n)$ if and only if ${ ann}_{R}(f_k)[\![X]\!]= { ann}_{R}(f_n)[\![X]\!]$ if and only if $\cap _{j\ge 0}{} { ann}_R(a_{k,j}) = \cap _{j\ge 0}{} { ann}_R(a_{n,j})$. $\square$