On $(\sigma,\delta)$-skew McCoy modules

TL;DR

This paper introduces and explores the properties of $(\sigma,\delta)$-skew McCoy modules, generalizing existing module concepts and analyzing their behavior in skew triangular matrix extensions.

Contribution

It defines $(\sigma,\delta)$-skew McCoy modules, establishes their properties, and links them to skew triangular matrix extensions, broadening the understanding of module theory.

Findings

$(\sigma,\delta)$-skew McCoy modules generalize McCoy and $\sigma$-skew McCoy modules.

Connections between $(\sigma,\delta)$-skew McCoyness and $(\sigma,\delta)$-compatibility are established.

Characterizations of $(\sigma,\delta)$-skew McCoy modules via skew polynomial extensions are provided.

Abstract

Let $(\sigma,\delta)$ be a quasi derivation of a ring $R$ and $M_R$ a right $R$-module. In this paper, we introduce the notion of $(\sigma,\delta)$-skew McCoy modules which extends the notion of McCoy modules and $\sigma$-skew McCoy modules. This concept can be regarded also as a generalization of $(\sigma,\delta)$-skew Armendariz modules. Some properties of this concept are established and some connections between $(\sigma,\delta)$-skew McCoyness and $(\sigma,\delta)$-compatible reduced modules are examined. Also, we study the property $(\sigma,\delta)$-skew McCoy of some skew triangular matrix extensions $V_n(M,\sigma)$, for any nonnegative integer $n\geq 2$. As a consequence, we obtain: (1) $M_R$ is $(\sigma,\delta)$-skew McCoy if and only if $M[x]/M[x](x^n)$ is $(\overline{\sigma},\overline{\delta})$-skew McCoy, and (2) $M_R$ is $\sigma$-skew McCoy if and only if $M[x;\sigma]/M[x;\sigma](x^n)$ is $\overline{\sigma}$-skew McCoy.

Equations

$$M[x;\sigma ,\delta ]:=\left\{\sum _{i=0}^n{m}_i{x}^i\mid n\ge 0,m_i\in M\right\};$$

$$R_{4}=\left\{\left(\begin{array}[]{ccccc}a&a_{12}&a_{13}&a_{14}\\ 0&a&a_{23}&a_{24}\\ 0&0&a&a_{34}\\ 0&0&0&a\\ \end{array}\right)\mid a,a_{ij}\in R\right\},$$

$$+m_p{f}_{p-2}^p(a_q)=0,$$

$$+m_{p-1}{f}_{p-2}^{p-1}(a_{q-1})+m_p{f}_{p-2}^p(a_{q-1})+m_{p-3}{\sigma }^{p-3}(a_q)+m_{p-2}{f}_{p-3}^{p-2}(a_q)$$

$$+m_{p-1}{f}_{p-3}^{p-1}(a_q)+m_p{f}_{p-3}^p(a_q)=0,$$

$$⋮$$

$$⋮$$

$$m_p{a}_q=0\Rightarrow m_p{\sigma }^p(a_{q-1})a_q=m_p{f}_{p-1}^p(a_q)a_q=0.$$

$$+m_{p-1}{f}_{p-2}^{p-1}(a_q)a_q+m_p{f}_{p-2}^p(a_q)a_q=0,$$

$$+m_{p-2}{\sigma }^{p-2}(a_q)a_{q-1}+m_{p-1}{f}_{p-2}^{p-1}(a_q)a_{q-1}+m_p{f}_{p-2}^p(a_q)a_{q-1}=0$$

$$0=m_p{\sigma }^p(a_{q-2})a_{q-1}=m_p{f}_{p-1}^p(a_{q-1})a_{q-1}=m_{p-2}{\sigma }^{p-2}(a_q)a_{q-1}$$

$$=m_{p-1}{f}_{p-2}^{p-1}(a_q)a_{q-1}=m_p{f}_{p-2}^p(a_q)a_{q-1}$$

$$0=(m_p{x}^p+m_{p-1}{x}^{p-1}+\cdots +m_{1}x+m_0)(a_q^2{x}^q+a_q{a}_{q-1}{x}^{q-1}+\cdots +a_q{a}_{1}x+a_q{a}_0)$$

$$=(m_{p-1}{x}^{p-1}+\cdots +m_{1}x+m_0)(a_q^2{x}^q+a_q{a}_{q-1}{x}^{q-1}+\cdots +a_q{a}_{1}x+a_q{a}_0).$$

$$V_{3}(R)=\left\{\left(\begin{array}[]{ccc}a&b&c\\ 0&a&b\\ 0&0&a\\ \end{array}\right)\mid a,b,c\in R\right\}.$$

$$S_{n}(R):=\left\{\left(\begin{array}[]{ccccc}a&a_{12}&a_{13}&\ldots&a_{1n}\\ 0&a&a_{23}&\ldots&a_{2n}\\ 0&0&a&\ldots&a_{3n}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\ldots&a\\ \end{array}\right)\mid a,a_{ij}\in R\right\}$$

$$S_{n}(M):=\left\{\left(\begin{array}[]{ccccc}m&m_{12}&m_{13}&\ldots&m_{1n}\\ 0&m&m_{23}&\ldots&m_{2n}\\ 0&0&m&\ldots&m_{3n}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\ldots&m\\ \end{array}\right)\mid m,m_{ij}\in M\right\}$$

$$V_{n}(R):=\left\{\left(\begin{array}[]{cccccc}a_{0}&a_{1}&a_{2}&a_{3}&\ldots&a_{n-1}\\ 0&a_{0}&a_{1}&a_{2}&\ldots&a_{n-2}\\ 0&0&a_{0}&a_{1}&\ldots&a_{n-3}\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&0&\ldots&a_{1}\\ 0&0&0&0&\ldots&a_{0}\\ \end{array}\right)\mid a_{0},a_{1},a_{2},\cdots,a_{n-1}\in R\right\}$$

$$V_{n}(M):=\left\{\left(\begin{array}[]{cccccc}m_{0}&m_{1}&m_{2}&m_{3}&\ldots&m_{n-1}\\ 0&m_{0}&m_{1}&m_{2}&\ldots&m_{n-2}\\ 0&0&m_{0}&m_{1}&\ldots&m_{n-3}\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&0&\ldots&m_{1}\\ 0&0&0&0&\ldots&m_{0}\\ \end{array}\right)\mid m_{0},m_{1},m_{2},\cdots,m_{n-1}\in M\right\}$$

$$V_{n}(R,\sigma):=\left\{\left(\begin{array}[]{cccccc}a_{0}&a_{1}&a_{2}&a_{3}&\ldots&a_{n-1}\\ 0&a_{0}&a_{1}&a_{2}&\ldots&a_{n-2}\\ 0&0&a_{0}&a_{1}&\ldots&a_{n-3}\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&0&\ldots&a_{1}\\ 0&0&0&0&\ldots&a_{0}\\ \end{array}\right)\mid a_{0},a_{2},\cdots,a_{n-1}\in R\right\}$$

Full text

1. Introduction

Throughout this paper, $R$ denotes an associative ring with unity and $M_{R}$ a right $R$-module. For a subset $X$ of a module $M_{R}$, $r_{R}(X)=\{a\in R|Xa=0\}$ and $\ell_{R}(X)=\{a\in R|aX=0\}$ will stand for the right and the left annihilator of $X$ in $R$ respectively. An Ore extension of a ring $R$ is denoted by $R[x;\sigma,\delta]$, where $\sigma$ is an endomorphism of $R$ and $\delta$ is a $\sigma$-derivation, i.e., $\delta\colon R\rightarrow R$ is an additive map such that $\delta(ab)=\sigma(a)\delta(b)+\delta(a)b$ for all $a,b\in R$ (the pair $(\sigma,\delta)$ is also called a quasi-derivation of $R$). Recall that elements of $R[x;\sigma,\delta]$ are polynomials in $x$ with coefficients written on the left. Multiplication in $R[x;\sigma,\delta]$ is given by the multiplication in $R$ and the condition $xa=\sigma(a)x+\delta(a)$, for all $a\in R$. In the next, $S$ will stand for the Ore extension $R[x;\sigma,\delta]$. On the other hand, we have a natural functor $-\otimes_{R}S$ from the category of right $R$-modules into the category of right $S$-modules. For a right $R$-module $M$, the right $S$-module $M\otimes_{R}S$ is called the induced module [16]. Since $R[x;\sigma,\delta]$ is a free left $R$-module, elements of $M\otimes_{R}S$ can be seen as polynomials in $x$ with coefficients in $M$ with natural addition and right $S$-module multiplication.

For any $0\leq i\leq j\;(i,j\in\mathbb{N})$, $f_{i}^{j}\in End(R,+)$ will denote the map which is the sum of all possible words in $\sigma,\delta$ built with $i$ factors of $\sigma$ and $j-i$ factors of $\delta$ (e.g., $f_{n}^{n}=\sigma^{n}$ and $f_{0}^{n}=\delta^{n},n\in\mathbb{N}$). We have $x^{j}a=\sum_{i=0}^{j}f_{i}^{j}(a)x^{i}$ for all $a\in R$, where $i,j$ are nonnegative integers with $j\geq i$ (see [14, Lemma 4.1]).

Following Lee and Zhou [15], we introduce the notation $M[x;\sigma,\delta]$ to write the $S$-module $M\otimes_{R}S$. Consider

[TABLE]

which is an $S$-module under an obvious addition and the action of monomials of $R[x;\sigma,\delta]$ on monomials in $M[x;\sigma,\delta]_{R[x;\sigma,\delta]}$ via $(mx^{j})(ax^{\ell})=m\sum_{i=0}^{j}f_{i}^{j}(a)x^{i+\ell}$ for all $a\in R$ and $j,\ell\in\mathbb{N}$. The $S$-module $M[x;\sigma,\delta]$ is called the skew polynomial extension related to the quasi-derivation $(\sigma,\delta)$.

A module $M_{R}$ is semicommutative, if for any $m\in M$ and $a\in R$, $ma=0$ implies $mRa=0$ [6]. Let $\sigma$ an endomorphism of $R$, $M_{R}$ is called an $\sigma$-semicommutative module [17] if, for any $m\in M$ and $a\in R$, $ma=0$ implies $mR\sigma(a)=0$. For a module $M_{R}$ and a quasi-derivation $(\sigma,\delta)$ of $R$, we say that $M_{R}$ is $\sigma$-compatible, if for each $m\in M$ and $a\in R$, we have $ma=0\Leftrightarrow m\sigma(a)=0$. Moreover, we say that $M_{R}$ is $\delta$-compatible, if for each $m\in M$ and $a\in R$, we have $ma=0\Rightarrow m\delta(a)$=0. If $M_{R}$ is both $\sigma$-compatible and $\delta$-compatible, we say that $M_{R}$ is $(\sigma,\delta)$-compatible (see [3]). In [17], a module $M_{R}$ is called $\sigma$-skew Armendariz, if $m(x)f(x)=0$ where $m(x)=\sum_{i=0}^{n}m_{i}x^{i}\in M[x;\sigma]$ and $f(x)=\sum_{j=0}^{m}a_{j}x^{j}\in R[x;\sigma]$ implies $m_{i}\sigma^{i}(a_{j})=0$ for all $i,j$. According to Lee and Zhou [15], $M_{R}$ is called $\sigma$-Armendariz, if it is $\sigma$-compatible and $\sigma$-skew Armendariz.

Following Alhevas and Moussavi [1], a module $M_{R}$ is called $(\sigma,\delta)$-skew Armendariz, if whenever $m(x)g(x)=0$ where $m(x)=\sum_{i=0}^{p}m_{i}x^{i}\in M[x;\sigma,\delta]$ and $g(x)=\sum_{j=0}^{q}b_{j}x^{j}\in R[x;\sigma,\delta]$, we have $m_{i}x^{i}b_{j}x^{j}=0$ for all $i,j$.

In this paper, we introduce the concept of $(\sigma,\delta)$-skew McCoy modules which is a generalization of McCoy modules and $\sigma$-skew McCoy modules. This concept can be regarded also as a generalization of $(\sigma,\delta)$-skew Armendariz modules and rings. We study connections between reduced modules, $(\sigma,\delta)$-compatible modules and $(\sigma,\delta)$-skew McCoy modules. Also, we show that $(\sigma,\delta)$-skew McCoyness passes from a module $M_{R}$ to its skew triangular matrix extension $V_{n}(M,\sigma)$. In this sens, we complete the definition of skew triangular matrix rings $V_{n}(R,\sigma)$ given by Isfahani [12], by introducing the notion of skew triangular matrix modules. Moreover, we give some results on $(\sigma,\delta)$-skew McCoyness for skew triangular matrix modules.

2. $(\sigma,\delta)$-skew McCoy modules

Cui and Chen [7, 8], introduced both concepts of McCoy modules and $\sigma$-skew McCoy modules. A module $M_{R}$ is called McCoy if $m(x)g(x)=0$, where $m(x)=\sum_{i=0}^{p}m_{i}x^{i}\in M[x]$ and $g(x)=\sum_{j=0}^{q}b_{j}x^{j}\in R[x]\setminus\{0\}$ implies that there exists $a\in R\setminus\{0\}$ such that $m(x)a=0$. A module $M_{R}$ is called $\sigma$-skew McCoy if $m(x)g(x)=0$, where $m(x)=\sum_{i=0}^{p}m_{i}x^{i}\in M[x;\sigma]$ and $g(x)=\sum_{j=0}^{q}b_{j}x^{j}\in R[x;\sigma]\setminus\{0\}$ implies that there exists $a\in R\setminus\{0\}$ such that $m(x)a=0$. With the same manner, we introduce the concept of $(\sigma,\delta)$-skew McCoy modules which is a generalization of McCoy modules, $\sigma$-skew McCoy modules and $(\sigma,\delta)$-skew Armendariz modules.

Definition 2.1**.**

Let $M_{R}$ be a module and $M[x;\sigma,\delta]$ the corresponding $(\sigma,\delta)$-skew polynomial module over $R[x;\sigma,\delta]$.

$\mathbf{(1)}$* The module $M_{R}$ is called $(\sigma,\delta)$-skew McCoy if $m(x)g(x)=0$, where $m(x)=\sum_{i=0}^{p}m_{i}x^{i}\in M[x;\sigma,\delta]$ and $g(x)=\sum_{j=0}^{q}b_{j}x^{j}\in R[x;\sigma,\delta]\setminus\{0\}$, implies that there exists $a\in R\setminus\{0\}$ such that $m(x)a=0$ $($i.e., $\sum_{i=\ell}^{p}m_{i}f_{\ell}^{i}(a)=0$, for all $\ell=0,1,\cdots,p)$.*

$\mathbf{(2)}$* The ring $R$ is called $(\sigma,\delta)$-skew McCoy if $R$ is $(\sigma,\delta)$-skew McCoy as a right $R$-module.*

Remark 2.2**.**

$\mathbf{(1)}$* If $M_{R}$ is an $(\sigma,\delta)$-skew Armendariz module then it is $(\sigma,\delta)$-skew McCoy $($Proposition 2.4$)$. But the converse is not true $($Example 2.5$)$.*

$\mathbf{(2)}$* If $\sigma=id_{R}$ and $\delta=0$ we get the concept of McCoy module, if only $\delta=0$, we get the concept of $\sigma$-skew McCoy module.*

$\mathbf{(3)}$* A module $M_{R}$ is $(\sigma,\delta)$-skew McCoy if and only if for all $m(x)\in M[x;\sigma,\delta]$, $r_{R[x;\sigma,\delta]}(m(x))\neq 0\Rightarrow r_{R[x;\sigma,\delta]}(m(x))\cap R\neq 0.$*

An ideal $I$ of a ring $R$ is called $(\sigma,\delta)$-stable, if $\sigma(I)\subseteq I$ and $\delta(I)\subseteq I$.

Proposition 2.3**.**

$\mathbf{(1)}$* Let $I$ be a nonzero right ideal of $R$. If $I$ is $(\sigma,\delta)$-stable then $R/I$ is an $R$-module $(\sigma,\delta)$-skew McCoy.*

$\mathbf{(2)}$* For any index set $I$, if $M_{i}$ is an $(\sigma_{i},\delta_{i})$-skew McCoy as $R_{i}$-module for each $i\in I$, then $\prod_{i\in I}M_{i}$ is an $(\sigma,\delta)$-skew McCoy as $\prod_{i\in I}R_{i}$-module, where $(\sigma,\delta)=(\sigma_{i},\delta_{i})_{i\in I}$.*

$\mathbf{(3)}$* Every submodule of an $(\sigma,\delta)$-skew McCoy module is $(\sigma,\delta)$-skew McCoy. In particular, if $I$ is a right ideal of an $(\sigma,\delta)$-skew McCoy ring then $I_{R}$ is $(\sigma,\delta)$-skew McCoy module.*

$\mathbf{(4)}$* A module $M_{R}$ is $(\sigma,\delta)$-skew McCoy if and only if every finitely generated submodule of $M_{R}$ is $(\sigma,\delta)$-skew McCoy.*

Proof.

$\mathbf{(1)}$ Let $m(x)=\sum_{i=0}^{p}\overline{m}_{i}x^{i}\in(R/I)[x;\sigma,\delta]$, where $\overline{m}_{i}=r_{i}+I\in R/I$ for all $i=0,1,\cdots,p$ and $r$ an arbitrary nonzero element of $I$. We have $m(x)r=\sum_{i=0}^{p}(r_{i}+I)\sum_{\ell=0}^{i}f_{\ell}^{i}(r)x^{\ell}\in I[x;\sigma,\delta]$, because $f_{\ell}^{i}(r)\in I$ for all $\ell=0,1,\cdots,i$. Hence $m(x)r=\bar{0}$.

$\mathbf{(2)}$ Let $M=\prod_{i\in I}M_{i}$ and $R=\prod_{i\in I}R_{i}$ such that each $M_{i}$ is an $(\sigma_{i},\delta_{i})$-skew McCoy as $R_{i}$-module for all $i\in I$. Take $m(x)=(m_{i}(x))_{i\in I}\in M[x;\sigma,\delta]$ and $f(x)=(f_{i}(x))_{i\in I}\in R[x;\sigma,\delta]\setminus\{0\}$, where $m_{i}(x)=\sum_{s=0}^{p}m_{i}(s)x^{s}\in M_{i}[x;\sigma_{i},\delta_{i}]$ and $f_{i}(x)=\sum_{t=0}^{q}a_{i}(t)x^{t}\in R_{i}[x;\sigma_{i},\delta_{i}]$ for each $i\in I$. Suppose that $m(x)f(x)=0$, then $m_{i}(x)f_{i}(x)=0$ for each $i\in I$. Since $M_{i}$ is $(\sigma_{i},\delta_{i})$-skew McCoy, there exists $0\neq r_{i}\in R_{i}$ such that $m_{i}(x)r_{i}=0$ for each $i\in I$. Thus $m(x)r=0$ where $0\neq r=(r_{i})_{i\in I}\in R$.

$\mathbf{(3)}$ and $(4)$ are obvious. ∎

Proposition 2.4**.**

If $M_{R}$ is an $(\sigma,\delta)$-skew Armendariz module then it is $(\sigma,\delta)$-skew McCoy.

Proof.

Let $m(x)=\sum_{i=0}^{p}m_{i}x^{i}\in M[x;\sigma,\delta]$ and $g(x)=\sum_{j=0}^{q}b_{j}x^{j}\in R[x;\sigma,\delta]\setminus\{0\}$. Suppose that $m(x)g(x)=0$, then $m_{i}x^{i}b_{j}x^{j}=0$ for all $i,j$. Since $g(x)\neq 0$ then $b_{j_{0}}\neq 0$ for some $j_{0}\in\{0,1,\cdots,p\}$. Thus $m_{i}x^{i}b_{j_{0}}x^{j_{0}}=0$ for all $i$. On the other hand $m_{i}x^{i}b_{j_{0}}x^{j_{0}}=\sum_{\ell=0}^{p}(\sum_{i=\ell}^{p}m_{i}f_{\ell}^{i}(b_{j_{0}}))x^{\ell+j_{0}}=0$, and so $\sum_{i=\ell}^{p}m_{i}f_{\ell}^{i}(b_{j_{0}})=0$ for all $\ell=0,1,\cdots,p$. Thus $m(x)b_{j_{0}}=0$, therefore $M_{R}$ is $(\sigma,\delta)$-skew McCoy. ∎

By the next example, we see that the converse of Proposition 2.4 does not hold.

Example 2.5**.**

Let $R$ be a reduced ring. Consider the ring

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Since $R$ is reduced then it is right McCoy and so $R_{4}$ is right McCoy, by [18, Proposition 2.1]. But $R_{4}$ is not Armendariz by [13, Example 3].

A module $(\sigma,\delta)$-skew McCoy need not to be McCoy by [8, Example 2.3(2)]. Also, the following example shows that, there exists a module which is McCoy but not $(\sigma,\delta)$-skew McCoy.

Example 2.6**.**

Let $\mathbb{Z}_{2}$ be the ring of integers modulo $2$, and consider the ring $R=\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$ with the usual addition and multiplication. Let $\sigma$ be an endomorphism of $R$ defined by $\sigma((a,b))=(b,a)$ and $\delta$ an $\sigma$-derivation of $R$ defined by $\delta((a,b))=(a,b)-\sigma((a,b))$. The ring $R$ is commutative reduced then it is McCoy. However, for $p(x)=(1,0)x$ and $q(x)=(1,1)+(1,0)x\in R[x;\sigma,\delta]$. We have $p(x)q(x)=0$, but $p(x)(a,b)\neq 0$ for any $0\neq(a,b)\in R$. Therefore, $R$ is not $(\sigma,\delta)$-skew McCoy. Also, $R$ is not $(\sigma,\delta)$-compatible, because $(0,1)(1,0)=(0,0)$, but $(0,1)\sigma((1,0))=(0,1)^{2}\neq(0,0)$ and $(0,1)\delta((1,0))=(0,1)(1,1)=(0,1)\neq(0,0)$.

Lemma 2.7**.**

Let $M_{R}$ be an $(\sigma,\delta)$-compatible module. For any $m\in M_{R}$, $a\in R$ and nonnegative integers $i,j$. We have the following:

$\mathbf{(1)}$* $ma=0\Rightarrow m\sigma^{i}(a)=m\delta^{j}(a)=0$.*

$\mathbf{(2)}$* $ma=0\Rightarrow m\sigma^{i}(\delta^{j}(a))=m\delta^{i}(\sigma^{j}(a))=0$.*

Proof.

The verification is straightforward. ∎

If $M_{R}$ is an $(\sigma,\delta)$-compatible module then $ma=0\Rightarrow mf_{i}^{j}(a)=0$ for any nonnegative integers $i,j$ such that $i\geq j$, where $m\in M_{R}$ and $a\in R$. For a subset $U$ of $M_{R}$ and $(\sigma,\delta)$ a quasi-derivation of $R$, the set of all skew polynomials with coefficients in $U$ is denoted by $U[x;\sigma,\delta]$.

Lemma 2.8**.**

Let $M_{R}$ be a module and $(\sigma,\delta)$ a quasi-derivation of $R$. The following are equivalent:

$\mathbf{(1)}$* For any $U\subseteq M[x;\sigma,\delta]$, $(r_{R[x;\sigma,\delta]}(U)\cap R)[x;\sigma,\delta]=r_{R[x;\sigma,\delta]}(U)$.*

$\mathbf{(2)}$* For any $m(x)=\sum_{i=0}^{p}m_{i}x^{i}\in M[x;\sigma,\delta]$ and $f(x)=\sum_{j=0}^{q}a_{j}x^{j}\in R[x;\sigma,\delta]$. If $m(x)f(x)=0$ implies $\sum_{\ell=i}^{p}m_{\ell}f_{i}^{\ell}(a_{j})=0$ for all $i,j$.*

Proof.

$(1)\Rightarrow(2)$. Let $m(x)=\sum_{i=0}^{p}m_{i}x^{i}\in M[x;\sigma,\delta]$ and $f(x)=\sum_{j=0}^{q}a_{j}x^{j}\in R[x;\sigma,\delta]$. If $m(x)f(x)=0$, we have $f(x)\in r_{R[x;\sigma,\delta]}(m(x))=(r_{R[x;\sigma,\delta]}(m(x))\cap R)[x;\sigma,\delta]$. Then $a_{j}\in r_{R[x;\sigma,\delta]}(m(x))$ for all $j$, so that $m(x)a_{j}=0$ for all $j$. But $m(x)a_{j}=0\Leftrightarrow\sum_{\ell=i}^{p}m_{\ell}f_{i}^{\ell}(a_{j})=0$ for all $0\leq i\leq p$. Thus $\sum_{\ell=i}^{p}m_{\ell}f_{i}^{\ell}(a_{j})=0$ for all $i,j$. $(2)\Rightarrow(1)$. Let $U\subseteq M[x;\sigma,\delta]$, we have always $(r_{R[x;\sigma,\delta]}(U)\cap R)[x;\sigma,\delta]\subseteq r_{R[x;\sigma,\delta]}(U)$. Conversely, let $f(x)\in r_{R[x;\sigma,\delta]}(U)$ then by $(2)$, we have $Ua_{j}=0$ for all $j$ and so $a_{j}\in r_{R[x;\sigma,\delta]}(U)\cap R$. Therefore $f(x)\in(r_{R[x;\sigma,\delta]}(U)\cap R)[x;\sigma,\delta]$. ∎

Theorem 2.9** (McCoy’s Theorem for module extensions).**

Let $M_{R}$ be a module and $N$ a nonzero submodule of $M[x;\sigma,\delta]$. If one of the equivalent conditions of Lemma 2.8 is satisfied. Then $r_{R[x;\sigma,\delta]}(N)\neq 0$ implies $r_{R}(N)\neq 0$.

Proof.

Suppose that $r_{R[x;\sigma,\delta]}(N)\neq 0$, then there exists $0\neq f(x)=\sum_{i=0}^{p}a_{i}x^{i}\in r_{R[x;\sigma,\delta]}(N)$. But $r_{R[x;\sigma,\delta]}(N)=(r_{R[x;\sigma,\delta]}(N)\cap R)[x;\sigma,\delta]$ by Lemma 2.8. Therefore all $a_{i}$ are in $r_{R[x;\sigma,\delta]}(N)$, so $a_{i}\in r_{R}(N)$ for all $i$. Since $f(x)\neq 0$ then there exists $i_{0}\in\{0,1,\cdots p\}$ such that $0\neq a_{i_{0}}\in r_{R}(N)$. So that $r_{R}(N)\neq 0$. ∎

Definition 2.10**.**

Let $M_{R}$ be a module and $\sigma$ an endomorphism of $R$. We say that $M_{R}$ satisfies the condition $(\mathcal{C_{\sigma}})$ if whenever $m\sigma(a)=0$ with $m\in M$ and $a\in R$, then $ma=0$.

Proposition 2.11**.**

Let $m(x)=\sum_{i=0}^{p}m_{i}x^{i}\in M[x;\sigma,\delta]$ and $f(x)=\sum_{j=0}^{q}a_{j}x^{j}$ $\in R[x;\sigma,\delta]$ such that $m(x)f(x)=0$. If one of the following conditions hold:

$\mathbf{(a)}$* $M_{R}$ is $(\sigma,\delta)$-skew Armendariz and satisfy the condition $(\mathcal{C_{\sigma}})$.*

$\mathbf{(b)}$* $M_{R}$ is reduced and $(\sigma,\delta)$-compatible. Then $m_{i}a_{j}=0$ for all $i,j$.*

Proof.

$\mathbf{(a)}$ Since $M_{R}$ is $(\sigma,\delta)$-skew Armendariz then from $m(x)f(x)=0$, we get $m_{i}x^{i}a_{j}x^{j}=0$ for all $i,j$. But $m_{i}x^{i}a_{j}x^{j}=m_{i}\sum_{\ell=0}^{i}f_{\ell}^{i}(a_{j})x^{j+\ell}=m_{i}\sigma^{i}(a_{j})x^{i+j}+Q(x)=0$ where $Q(x)$ is a polynomial in $M[x;\sigma,\delta]$ of degree strictly less than $i+j$. Thus $m_{i}\sigma^{i}(a_{j})=0$, therefore $m_{i}a_{j}=0$ for all $i,j$.

$\mathbf{(b)}$ We will use freely the fact that, if $ma=0$ then $m\sigma^{i}(a)=m\delta^{j}(a)=mf_{i}^{j}(a)=0$ for any nonnegative integers $i,j$ with $j\geq i$. From $m(x)f(x)=0$, we have the following system of equations:

[TABLE]

[TABLE]

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[TABLE]

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[TABLE]

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From equation $(0)$, we have $m_{p}a_{q}=0$ by $\sigma$-compatibility. Multiplying equation $(1)$ on the right hand by $a_{q}$, we get

[TABLE]

Since $M_{R}$ is semicommutative, then

[TABLE]

By Lemma 2.12, equation $(1^{\prime})$ gives $m_{p-1}a_{q}=0$. Also, by $(\sigma,\delta)$-compatibility, equation $(1)$ implies $m_{p}\sigma^{p}(a_{q-1})=0$, because $m_{p}a_{q}=m_{p-1}a_{q}=0$. Thus $m_{p}a_{q-1}=0$.

Summarizing at this point, we have

[TABLE]

Now, multiplying equation $(2)$ on the right hand by $a_{q}$, we get

[TABLE]

[TABLE]

With the same manner as above, equation $(2^{\prime})$ gives $m_{p-2}\sigma^{p-2}(a_{q})a_{q}=0$ and thus $m_{p-2}a_{q}=0\;(\beta)$. Also, multiplying equation $(2)$ on the right hand by $a_{q-1}$, we get

[TABLE]

[TABLE]

Equations $(\alpha)$ and $(\beta)$ implies

[TABLE]

[TABLE]

Hence, equation $(2^{\prime\prime})$ gives $m_{p-1}\sigma^{p-1}(a_{q-1})a_{q-1}=0$ and by Lemma 2.12, we get $m_{p-1}a_{q-1}=0\;(\gamma)$. Now, by equations $(\alpha)$,$(\beta)$ and $(\gamma)$, we get $m_{p-1}\sigma^{p-1}(a_{q-1})=m_{p}f_{p-1}^{p}(a_{q-1})=m_{p-2}\sigma^{p-2}(a_{q})=m_{p-1}f_{p-2}^{p-1}(a_{q})=m_{p}f_{p-2}^{p}(a_{q})=0$. Therefore equation $(2)$ implies $m_{p}\sigma^{p}(a_{q-2})=0$, so that $m_{p}a_{q-2}=0$.

Summarizing at this point, we have $m_{i}a_{j}=0$ with $i+j\in\{p+q,p+q-1,p+q-2\}$. Continuing this procedure yields $m_{i}a_{j}=0$ for all $i,j$. ∎

Lemma 2.12**.**

Let $M_{R}$ be an $(\sigma,\delta)$-compatible module, if $ma^{2}=0$ implies $ma=0$ for any $m\in M$ and $a\in R$. Then

$\mathbf{(1)}$* $m\sigma(a)a=0$ implies $ma=m\sigma(a)=0$.*

$\mathbf{(2)}$* $ma\sigma(a)=0$ implies $ma=m\sigma(a)=0$.*

Proof.

The proof is straightforward. ∎

According to Lee and Zhou [15], a module $M_{R}$ is called $\sigma$-reduced, if for any $m\in M$ and $a\in R$. We have

1. $\mathbf{(1)}$

$ma=0$ implies $mR\cap Ma=0$. 2. $\mathbf{(2)}$

$ma=0$ if and only if $m\sigma(a)=0$.

The module $M_{R}$ is called reduced if $M_{R}$ is $id_{R}$-reduced.

Lemma 2.13** ([15, Lemma 1.2]).**

The following are equivalent for a module $M_{R}$:

1. $\mathbf{(1)}$

$M_{R}$* is $\sigma$-reduced.* 2. $\mathbf{(2)}$

The following three conditions hold: For any $m\in M$ and $a\in R$,

1. $\mathbf{(a)}$

$ma=0$* implies $mRa=mR\sigma(a)=0$.* 2. $\mathbf{(b)}$

$ma\sigma(a)=0$* implies $ma=0$.* 3. $\mathbf{(c)}$

$ma^{2}=0$* implies $ma=0$.*

By Lemma 2.13, a module $M_{R}$ is reduced if and only if it is semicommutative with $ma^{2}=0$ implies $ma=0$ for any $m\in M$ and $a\in R$.

Corollary 2.14** ([1, Theorem 2.19]).**

Every $(\sigma,\delta)$-compatible and reduced module is $(\sigma,\delta)$-skew Armendariz.

Proof.

Clearly from Proposition 2.11(b). ∎

Let $M_{R}$ be a module and $(\sigma,\delta)$ a quasi derivation of $R$. We say that $M_{R}$ satisfies the condition $(*)$, if for any $m(x)\in M[x;\sigma,\delta]$ and $f(x)\in R[x;\sigma,\delta]$, $m(x)f(x)=0$ implies $m(x)Rf(x)=0$. A module $M_{R}$ which satisfies the condition $(*)$ is semicommutative. But the converse is not true, by the next example.

Example 2.15**.**

Take the ring $R=\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$ with $(\sigma,\delta)$ as considered in Example 2.6. Since $R$ is commutative then the module $R_{R}$ is semicommutative. However, it does not satisfy the condition $(*)$. For $p(x)=(1,0)x$ and $q(x)=(1,1)+(1,0)x\in R[x;\sigma,\delta]$. We have $p(x)q(x)=0$, but $p(x)(1,0)q(x)=(1,0)+(1,0)x\neq 0$. Thus $p(x)Rq(x)\neq 0$.

Theorem 2.16**.**

If a module $M_{R}$ is $(\sigma,\delta)$-compatible and reduced, then it satisfies the condition $(*)$.

Proof.

Let $m(x)=\sum_{i=0}^{p}m_{i}x^{i}\in M[x;\sigma,\delta]$ and $f(x)=\sum_{j=0}^{q}a_{j}x^{j}\in R[x;\sigma,\delta]$, such that $m(x)f(x)=0$. By Proposition 2.11(b) and semicommutativity of $M_{R}$, we have $m_{i}Ra_{j}=0$ for all $i$ and $j$. Moreover, compatibility implies $m_{i}f_{k}^{\ell}(Ra_{j})=0$ for all $i,j,k,\ell$. Therefore $m(x)Rf(x)=0$. ∎

Since the ring $R=\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$ is reduced, then from Example 2.15, we can see that the condition “$(\sigma,\delta)$-compatible” in Theorem 2.16 is not superfluous.

Proposition 2.17**.**

Let $M_{R}$ be an $(\sigma,\delta)$-compatible module which satisfies $(*)$. Suppose that for any $m(x)=\sum_{i=0}^{p}m_{i}x^{i}\in M[x;\sigma,\delta]$ and $f(x)=\sum_{j=0}^{q}a_{j}x^{j}\in R[x;\sigma,\delta]\setminus\{0\}$, $m(x)f(x)=0$. Then $m_{i}a_{q}^{p+1}=0$ for all $i=0,1,\cdots,p$.

Proof.

Let $m(x)=\sum_{i=0}^{p}m_{i}x^{i}\in M[x;\sigma,\delta]$ and $f(x)=\sum_{j=0}^{q}a_{j}x^{j}\in R[x;\sigma,\delta]\setminus\{0\}$, such that $m(x)f(x)=0$. We can suppose that $a_{q}\neq 0$. From $m(x)f(x)=0$, we get $m_{p}\sigma^{p}(a_{q})=0$. Since $M_{R}$ is $(\sigma,\delta)$-compatible, we have $m_{p}a_{q}=0$ which implies $m_{p}x^{p}a_{q}=0$. Since $m(x)f(x)=0$ implies $m(x)a_{q}f(x)=0$. Then

[TABLE]

[TABLE]

If we put $f^{\prime}(x)=a_{q}f(x)$ and $m^{\prime}(x)=\sum_{i=0}^{p-1}m_{i}x^{i}$ then we get $m_{p-1}a_{q}^{2}=0$. Continuing this procedure yields $m_{i}a_{q}^{p+1-i}=0$ for all $i=0,1,\cdots,p$. Consequently $m_{i}a_{q}^{p+1}=0$ for all $i=0,1,\cdots,p$. ∎

Corollary 2.18**.**

Let $M_{R}$ be an $(\sigma,\delta)$-compatible module over a reduced ring $R$. If $M_{R}$ satisfies $(*)$, then it is $(\sigma,\delta)$-skew McCoy.

Proof.

Let $m(x)=\sum_{i=0}^{p}m_{i}x^{i}\in M[x;\sigma,\delta]$ and $f(x)=\sum_{j=0}^{q}a_{j}x^{j}\in R[x;\sigma,\delta]\setminus\{0\}$, such that $m(x)f(x)=0$. We can suppose that $a_{q}\neq 0$. By Proposition 2.17, we have $m_{i}a_{q}^{p+1}=0$ for all $i=0,1,\cdots,p$. Since $M_{R}$ is $(\sigma,\delta)$-compatible, we get $m_{i}x^{i}a_{q}^{p+1}=m_{i}\sum_{\ell=0}^{i}f_{\ell}^{i}(a_{q}^{p+1})x^{\ell}=0$ for all $i$. Hence $m(x)a_{q}^{p+1}=0$ where $a_{q}^{p+1}\neq 0$, because $R$ is reduced. Consequently $M_{R}$ is $(\sigma,\delta)$-skew McCoy. ∎

Example 2.19**.**

Consider a ring of polynomials over $\mathbb{Z}_{2}$, $R=\mathbb{Z}_{2}[x]$. Let $\sigma\colon R\rightarrow R$ be an endomorphism defined by $\sigma(f(x))=f(0)$. Then

$\mathbf{(1)}$* $R$ is not $\sigma$-compatible. Let $f=\overline{1}+x$, $g=x\in R$, we have $fg=(\overline{1}+x)x\neq 0$, however $f\sigma(g)=(\overline{1}+x)\sigma(x)=0$.*

$\mathbf{(2)}$* $R$ is $\sigma$-skew Armendariz [10, Example 5].*

From Example 2.19, we see that the ring $R=\mathbb{Z}_{2}[x]$ is $\sigma$-skew McCoy because it is $\sigma$-skew Armendariz, but it is not $\sigma$-compatible. Thus the $(\sigma,\delta)$-compatibility condition is not essential to obtain $(\sigma,\delta)$-skew McCoyness.

Example 2.20** ([5, Example 2.5]).**

Let $R$ be a ring, $\sigma$ an endomorphism of $R$ and $\delta$ be a $\sigma$-derivation of $R$. Suppose that $R$ is $\sigma$-rigid. Consider the ring

[TABLE]

The ring $V_{3}(R)$ is $(\overline{\sigma},\overline{\delta})$-skew McCoy, reduced and $(\overline{\sigma},\overline{\delta})$-compatible, and by Theorem 2.16, it satisfies the condition $(*)$.

3. $(\sigma,\delta)$-skew McCoyness of some matrix extensions

For a nonnegative integer $n\geq 2$, let $R$ be a ring and $M$ a right $R$-module. Consider

[TABLE]

and

[TABLE]

Clearly, $S_{n}(M)$ is a right $S_{n}(R)$-module under the usual matrix addition operation and the following scalar product operation. For $U=(u_{ij})\in S_{n}(M)$ and $A=(a_{ij})\in S_{n}(R)$, $UA=(m_{ij})\in S_{n}(M)$ with $m_{ij}=\sum_{k=1}^{n}u_{ik}a_{kj}$ for all $i,j$. A quasi derivation $(\sigma,\delta)$ of $R$ can be extended to a quasi derivation $(\overline{\sigma},\overline{\delta})$ of $S_{n}(R)$ as follows: $\overline{\sigma}((a_{ij}))=(\sigma(a_{ij}))$ and $\overline{\delta}((a_{ij}))=(\delta(a_{ij}))$. We can easily verify that $\overline{\delta}$ is a $\overline{\sigma}$-derivation of $S_{n}(R)$.

Theorem 3.1**.**

A module $M_{R}$ is $(\sigma,\delta)$-skew McCoy if and only if $S_{n}(M)$ is $(\overline{\sigma},\overline{\delta})$-skew McCoy as an $S_{n}(R)$-module for any nonnegative integer $n\geq 2$.

Proof.

The proof is similar to [4, Theorem 14]. ∎

Now, for $n\geq 2$. Consider

[TABLE]

and

[TABLE]

With the same method as above, $V_{n}(M)$ is a right $V_{n}(R)$-module, and a quasi derivation $(\sigma,\delta)$ of $R$ can be extended to a quasi derivation $(\overline{\sigma},\overline{\delta})$ of $V_{n}(R)$. Note that $V_{n}(M)\cong M[x]/M[x](x^{n})$ where $M[x](x^{n})$ is a submodule of $M[x]$ generated by $x^{n}$ and $V_{n}(R)\cong R[x]/(x^{n})$ where $(x^{n})$ is an ideal of $R[x]$ generated by $x^{n}$.

Proposition 3.2**.**

A module $M_{R}$ is $(\sigma,\delta)$-skew McCoy if and only if $V_{n}(M)$ is $(\overline{\sigma},\overline{\delta})$-skew McCoy as an $V_{n}(R)$-module for any nonnegative integer $n\geq 2$.

Proof.

The proof is similar to that of [4, Theorem 14] or [8, Proposition 2.27]. ∎

Corollary 3.3**.**

For a nonnegative integer $n\geq 2$, we have:

$\mathbf{(1)}$* $M_{R}$ is $(\sigma,\delta)$-skew McCoy if and only if $M[x]/M[x](x^{n})$ is $(\overline{\sigma},\overline{\delta})$-skew McCoy.*

$\mathbf{(2)}$* $R$ is $(\sigma,\delta)$-skew McCoy if and only if $R[x]/(x^{n})$ is $(\overline{\sigma},\overline{\delta})$-skew McCoy.*

$\mathbf{(3)}$* $R$ is McCoy if and only if $R[x]/(x^{n})$ is McCoy.*

In the next, we define skew triangular matrix modules $V_{n}(M,\sigma)$, based on the definition of skew triangular matrix rings $V_{n}(R,\sigma)$ given by Isfahani [12]. Let $\sigma$ be an endomorphism of a ring $R$ and $M_{R}$ a right $R$-module. For $n\geq 2$. Consider

[TABLE]

and

[TABLE]

Clearly $V_{n}(M,\sigma)$ is a right $V_{n}(R,\sigma)$-module under the usual matrix addition operation and the following scalar product operation.

[TABLE]

[TABLE]

$c_{i}=m_{0}\sigma^{0}(a_{i})+m_{1}\sigma^{1}(a_{i-1})+m_{2}\sigma^{2}(a_{i-2})+\cdots+m_{i}\sigma^{i}(a_{0})$ for each $0\leq i\leq n-1$.

We denote elements of $V_{n}(R,\sigma)$ by $(a_{0},a_{1},\cdots,a_{n-1})$ and elements of $V_{n}(M,\sigma)$ by $(m_{0},m_{1},\cdots,m_{n-1})$. There is a ring isomorphism $\varphi\colon R[x;\sigma]/(x^{n})\rightarrow V_{n}(R,\sigma)$ given by $\varphi(a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n-1}x^{n-1}+(x^{n}))=(a_{0},a_{1},a_{2},\cdots,a_{n-1})$, and an abelian group isomorphism $\phi\colon M[x,\sigma]/M[x,\sigma](x^{n})\rightarrow V_{n}(M,\sigma)$ given by $\phi(m_{0}+m_{1}x+m_{2}x^{2}+\cdots+m_{n-1}x^{n-1}+(x^{n}))=(m_{0},m_{1},m_{2},\cdots,m_{n-1})$ such that

[TABLE]

for any $N(x)=m_{0}+m_{1}x+m_{2}x^{2}+\cdots+m_{n-1}x^{n-1}+(x^{n})\in M[x,\sigma]/M[x,\sigma](x^{n})$ and $A(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n-1}x^{n-1}+(x^{n})\in R[x;\sigma]/(x^{n})$. The endomorphism $\sigma$ of $R$ can be extended to $V_{n}(R,\sigma)$ and $R[x;\sigma]$, and we will denote it in both cases by $\overline{\sigma}$.

Theorem 3.4**.**

A module $M_{R}$ is $\sigma$-skew McCoy if and only if $V_{n}(M,\sigma)$ is $\overline{\sigma}$-skew McCoy as an $V_{n}(R,\sigma)$-module for any nonnegative integer $n\geq 2$.

Proof.

We shall adapt the proof of [4, Theorem 14] to this situation. Note that $V_{n}(R,\sigma)[x,\overline{\sigma}]\cong V_{n}(R[x,\sigma],\overline{\sigma})$ and $V_{n}(M,\sigma)[x,\overline{\sigma}]\cong V_{n}(M[x,\sigma],\overline{\sigma})$. We only prove when $n=2$, because other cases can be proved with the same manner. Suppose that $M_{R}$ is $\sigma$-skew McCoy. Let $0\neq m(x)\in V_{2}(M,\sigma)[x,\overline{\sigma}]$ and $0\neq f(x)\in V_{2}(R,\sigma)[x,\overline{\sigma}]$ such that $m(x)f(x)=0$, where

[TABLE]

[TABLE]

Then \left(\begin{array}[]{cc}\alpha_{11}&\alpha_{12}\\ 0&\alpha_{11}\\ \end{array}\right)\left(\begin{array}[]{cc}\beta_{11}&\beta_{12}\\ 0&\beta_{11}\\ \end{array}\right)=0, which gives $\alpha_{11}\beta_{11}=0$ and $\alpha_{11}\beta_{12}+\alpha_{12}\overline{\sigma}(\beta_{11})=0$ in $M[x;\sigma]$. If $\alpha_{11}\neq 0$, then there exists $0\neq\beta\in\{\beta_{11},\beta_{12}\}$ such that $\alpha_{11}\beta=0$. Since $M_{R}$ is $\sigma$-skew McCoy then there exists $0\neq c\in R$ which satisfies $\alpha_{11}c=0$, thus \left(\begin{array}[]{cc}\alpha_{11}&\alpha_{12}\\ 0&\alpha_{11}\\ \end{array}\right)\left(\begin{array}[]{cc}0&c\\ 0&0\\ \end{array}\right)=\left(\begin{array}[]{cc}0&\alpha_{11}c\\ 0&0\\ \end{array}\right)=0. If $\alpha_{11}=0$ then \left(\begin{array}[]{cc}0&\alpha_{12}\\ 0&0\\ \end{array}\right)\left(\begin{array}[]{cc}0&c\\ 0&0\\ \end{array}\right)=0, for any $0\neq c\in R$. Therefore, $V_{2}(M,\sigma)$ is $\overline{\sigma}$-skew McCoy.

Conversely, suppose that $V_{2}(M,\sigma)$ is an $\overline{\sigma}$-skew McCoy module. Let $0\neq m(x)=m_{0}+m_{1}x+\cdots+m_{p}x^{p}\in M[x;\sigma]$ and $0\neq f(x)=a_{0}+a_{1}x+\cdots+a_{q}x^{q}\in R[x;\sigma]$, such that $m(x)f(x)=0$. Then \left(\begin{array}[]{cc}m(x)&0\\ 0&m(x)\\ \end{array}\right)\left(\begin{array}[]{cc}f(x)&0\\ 0&f(x)\\ \end{array}\right)=\left(\begin{array}[]{cc}m(x)f(x)&0\\ 0&m(x)f(x)\\ \end{array}\right)=0, so there exists 0\neq\left(\begin{array}[]{cc}a&b\\ 0&a\\ \end{array}\right)\in V_{2}(R,\sigma) such that \left(\begin{array}[]{cc}m(x)&0\\ 0&m(x)\\ \end{array}\right)\left(\begin{array}[]{cc}a&b\\ 0&a\\ \end{array}\right)=0, because $V_{2}(M,\sigma)$ is $\overline{\sigma}$-skew McCoy. Thus $m(x)a=m(x)b=0$, where $a\neq 0$ or $b\neq 0$. Therefore, $M_{R}$ is $\sigma$-skew McCoy. ∎

Corollary 3.5**.**

For a nonnegative integer $n\geq 2$, we have:

$\mathbf{(1)}$* $M_{R}$ is $\sigma$-skew McCoy if and only if $M[x;\sigma]/M[x;\sigma](x^{n})$ is $\overline{\sigma}$-skew McCoy.*

$\mathbf{(2)}$* $R$ is $\sigma$-skew McCoy if and only if $R[x;\sigma]/(x^{n})$ is $\overline{\sigma}$-skew McCoy.*

$\mathbf{(3)}$* $M_{R}$ is McCoy if and only if $M[x]/M[x](x^{n})$ is McCoy.*

$\mathbf{(4)}$* $R$ is McCoy if and only if $R[x]/(x^{n})$ is McCoy.*