Throughout this note, $R$ denotes an associative ring with identity and $\alpha$ denotes a nonzero endomorphism, unless specified otherwise. For a ring $R$, we denote by $nil(R)$ the set of all nilpotent elements of $R$ and $T_{n}(R)$ the $n$-by-$n$ upper triangular matrix ring over $R$. In [8], Nielsen introduced the notion of a McCoy ring. A ring $R$ is said to be right McCoy (resp., left McCoy) if for each pair of nonzero polynomials $f(x), g(x)\in R[x]$ with $f(x)g(x)=0$, there exists a nonzero element $r\in R$ with $f(x)r=0$ (resp., $rg(x)=0$). A ring $R$ is McCoy if it is both left and right McCoy. The name of the ring was given due to N. H. McCoy who proved in [7] that commutative rings always satisfy this condition. A ring $R$ is called weak McCoy if for each pair of nonzero polynomials $f(x)=\sum\limits_{i=0}^{m}a_{i}x^{i}$ and $g(x)=\sum\limits_{j=0}^{n}b_{j}x^{j}\in R[x]$ with $f(x)g(x)=0$, there exists a nonzero element $r\in R$ such that $a_{i}r \in nil(R)$ (resp., $rb_{j}\in nil(R)$). A ring $R$ is called weak McCoy if it is both right and left weak McCoy. Due to Rege and Chhawchharia [9], a ring $R$ is called Armendariz if for given $f(x)=\sum\limits_{i=0}^{m}a_{i}x^{i}$ and $g(x)=\sum\limits_{j=0}^{n}b_{j}x^{j}\in R[x]$, $f(x)g(x)=0$ implies that $a_{i}b_{j}=0$ for each $i, j$ (the converse is obviously true). It is well-known that every reduced ring (i.e., rings without nonzero nilpotent elements in $R$) is an Armendariz ring and every Armendariz ring is McCoy. Recall that if $\alpha$ is an endomorphism of a ring $R$, then the map $R[x]\rightarrow R[x]$ defined by $\sum\limits_{i=0}^{m}a_{i}x^{i}\rightarrow \sum\limits_{i=0}^{m}\alpha(a_{i})x^{i}$ is an endomorphism of the polynomial ring $R[x]$ and clearly this map extends $\alpha$. We shall also denote the extended map $R[x]\rightarrow R[x]$ by $\alpha$ and the image of $f(x)\in R[x]$ by $\alpha(f(x))$. For basic and other results on McCoy rings, see, e.g., [3, 8, 10, 11].

We consider the McCoy properties related to an endomorphism $\alpha$ of a ring $R$ and call them $\alpha$-McCoy rings. It is clear that every McCoy ring is an $\alpha$-McCoy ring, but we shall give an example to show that there exists an $\alpha$-McCoy ring which is not McCoy. A number of properties of this version are established. It is proved that a ring $R$ is a right $\alpha$-McCoy ring if and only if $R[x]$ is right $\alpha$-McCoy. Moreover, we show that a ring $R$ is right $\alpha$-McCoy if and only if $R[x]/(x^{n})$ is right $\alpha$-McCoy. For a right Ore ring $R$, if $\alpha$ is an endomorphism of $R$ with $Q(R)$ the classical right quotient ring of $R$. It is proved that $R$ is right $\alpha$-McCoy if and only if $Q(R)$ is right $\alpha$-McCoy. Moreover, a weak form of $\alpha$-McCoy rings is investigated in the last section. We show that in general weak $\alpha$-McCoy rings need not be $\alpha$-McCoy. It is proved that if $R$ is a right weak $\alpha$-McCoy ring, then the $n$-by-$n$ upper triangular matrix ring $T_{n}(R)$ is a right weak $\alpha$-McCoy ring. And hence some results on McCoy rings are generalized.

In this section, we relate the problem on the various McCoy properties of a ring $R$ to an endomorphism $\alpha$ of $R$. We begin with the following definition.

Definition 2.1  An endomorphism $\alpha$ of a ring $R$ is called right (resp., left) McCoy, if for each pair of nonzero polynomials $f(x)=\sum\limits_{i=0}^{m}a_{i}x^{i}$ and $g(x)=\sum\limits_{j=0}^{n}b_{j}x^{j}\in R[x]$ with $\alpha(f(x))g(x)=0$ (resp., $f(x)\alpha(g(x))=0$), there exists a nonzero element $r\in R$ such that $\alpha(f(x))r=0$ (resp., $r\alpha(g(x))=0$). A ring $R$ is called right (resp., left) $\alpha$-McCoy if there exists a right (resp., left) McCoy endomorphism $\alpha$ of $R$. $R$ is an $\alpha$-McCoy ring if it is both right and left $\alpha$-McCoy.

It is clear that every right McCoy ring is right $\alpha$-McCoy. However, we can give the following example to show that there exists a McCoy endomorphism $\alpha$ of a ring $S$ such that $S$ is not a McCoy ring.

Example 2.2  Let $\mathbb{Z}$ be the ring of integers. Consider the ring

$ S = \left\{ {\left( {\begin{array}{*{20}c} a \;\; b \\ 0 \;\; c \\ \end{array} } \right)\left| {a, b, c \in \mathbb{Z}} \right.} \right\}. $

Let $\alpha: S\rightarrow S$ be an endomorphism defined by $\alpha \left( {\left( {\begin{array}{*{10}c} a \;\; b \\ 0 \;\; c \\ \end{array} } \right)} \right) = \left( {\begin{array}{*{20}c} a \;\; 0 \\ 0 \;\; 0 \\ \end{array} } \right). $ If $f(x)=\sum\limits_{i=0}^{n} \left( \begin{array}{cc} a_{i} \;\; b_{i} \\ 0 \;\; c_{i} \\ \end{array} \right)x^{i}$ and $g(x)=\sum\limits_{j=0}^{m} \left( \begin{array}{cc} d_j \;\; e_j \\ 0 \;\; f_j \\ \end{array} \right)x^{j}$ are nonzero polynomials in $S[x]$ such that $\alpha(f(x))g(x)=0$. Then we have

$ \begin{eqnarray*} \alpha \left( {f\left( x \right)} \right)g\left( x \right)\;\;=\;\; \sum\limits_{k = 0}^{m + n} {\left( {\sum\limits_{i + j = k} {\left( {\begin{array}{*{20}c} {a_i } \;\; 0 \\ 0 \;\; 0 \\ \end{array}} \right)\left( {\begin{array}{*{20}c} {d_j } \;\; {e_j } \\ 0 \;\; {f_j } \\ \end{array}} \right)} } \right)} x^k \\ \;\;=\;\; \sum\limits_{k = 0}^{m + n} {\left( {\sum\limits_{i + j = k} {\left( {\begin{array}{*{20}c} {a_i \;\; d_j } {a_i \;\; e_j } \\ 0 \;\;\;\; 0 \\ \end{array}} \right)} } \right)} x^k = 0. \end{eqnarray*} $

This implies that

$ \sum\limits_{k = 0}^{n + m}(\sum\limits_{i+j=k}a_{i}d_{j})x^{k}=0, \sum\limits_{k = 0}^{n + m}(\sum\limits_{i+j=k}a_{i}e_{j})x^{k}=0. $

Let $f_{1}(x)=\sum\limits_{i=0}^{n}a_{i}x^{i}$, $g_{1}(x)=\sum\limits_{j=0}^{m}d_{j}x^{j}$ and $g_{2}(x)=\sum\limits_{j=0}^{m}e_{j}x^{j}$. Then we have $f_1(x)g_{1}(x)=f_1(x)g_{2}(x)=0$. Since every reduced ring is an Armendariz ring, it follows that $ a_{i}d_{j}=a_{i}e_{j}=0 $ for each $i, j$. If $a_{i}=0$, then we are done. If $a_{i}\neq 0$, then we have $d_{j}=e_{j}=0$. Now if we let

$ r=\left(\begin{array}{cc} 0 \;\; 0 \\ 0 \;\; f_{j} \\ \end{array} \right) $

for some $f_{j}\neq 0$, then $r\neq 0$ and $\alpha(f(x))r=0$. This shows that the endomorphism $\alpha$ of $S$ is right McCoy. Similarly, we can prove that the endomorphism $\alpha$ of $S$ is left McCoy. But $S$ is neither left nor right McCoy by [10, Theorem 2.1].

According to [1], an endomorphism $\alpha$ of a ring $R$ is called right (resp., left) reversible if whenever $ab=0$ for $a, b\in R$, $b\alpha(a)=0$ (resp., $\alpha(b)a=0)$. A ring $R$ is called right (resp., left) $\alpha$-reversible if there exists a right (resp., left) reversible endomorphism $\alpha$ of $R$. $R$ is $\alpha$-reversible if it is both left and right $\alpha$-reversible.

Note 2.3  It is well-known that every reversible ring is a McCoy ring. Based on this fact, one may suspect that every left (resp., right) $\alpha$-reversible ring is McCoy. But this is not true by Example 2.2 and [1, Example 2.2]. In general, we do not know if every $\alpha$-reversible ring is $\alpha$-McCoy. In fact, Example 2.2 shows that a right $\alpha$-reversible ring can be $\alpha$-McCoy.

The next proposition gives more examples of right $\alpha$-McCoy rings.

Proposition 2.4  Let $R$ be a ring and $\alpha$ an endomorphism of $R$. Then $R$ is a right $\alpha$-McCoy ring if and only if $R[x]$ is a right $\alpha$-McCoy ring.

Proof  Assume that $R$ is a right $\alpha$-McCoy ring. Let $p(y)=f_{0}+f_{1}y+\cdots+f_{m}y^{m}$, $q(y)=g_{0}+g_{1}y+\cdots+g_{n}y^{n}$ be in $R[x][y]$ with $\alpha(p(y))q(y)=0$. We also let

$ f_{i}=a_{i_{0}}+a_{i_{1}}x+\cdots+a_{w_{i}}x^{w_{i}}, g_{j}=b_{j_{0}}+b_{j_{1}}x+\cdots+b_{v_{j}}x^{v_{j}} $

for each $0\leq i \leq m$ and $0\leq j \leq n$, where $a_{i_{0}}, a_{i_{1}}, \cdots, a_{w_{i}}, b_{j_{0}}, b_{j_{1}}, \cdots, b_{v_{j}}\in R$. We claim that $R[x]$ is right $\alpha$-McCoy. Take a positive integer $k$ such that $k>\max\{{\rm deg}(f_{i}), {\rm deg}(g_{j})\}$ for any $0\leq i \leq m$ and $0\leq j \leq n$, where the degree is as polynomials in $R[x]$ and the degree of zero polynomial is take to be zero. Then

$ p(x^{k})=f_{0}+f_{1}x^{k}+\cdots+f_{m}x^{mk}, q(x^{k})=g_{0}+g_{1}x^{k}+\cdots+g_{n}x^{nk}\in R[x], $

and hence the set of coefficients of the $f_{i}^{, }$s (resp., $g_{j}^{, }$s) equals the set of coefficients of $p(x^{k})$ (resp., $q(x^{k})$). Since $\alpha(p(y))q(y)=0$, we have $\alpha(p(x^{k}))q(x^{k})=0$. It follows that there exists $0\neq r \in R\subseteq R[x]$ such that $\alpha(p(x^{k}))r=0$. This implies that $\alpha(p(y))r=0$, and so $R[x]$ is right $\alpha$-McCoy. Conversely, suppose that $f(y)=\sum\limits_{i=0}^{m}a_{i}y^{i}$, $g(y)=\sum\limits_{j=0}^{n}b_{j}y^{j} \in R[y] \backslash \{0\}$ such that $\alpha(f(y))g(y)=0$. Since $R[x]$ is right $\alpha$-McCoy, there exists $0\neq r(x)\in R[x]$ such that $\alpha(f(y))r(x)=0$. This shows that $\alpha(a_{i})r(x)=0$ for each $i$. It follows from $0\neq r(x)$ that there exists $0\neq r_{j}\in R$ such that $\alpha(a_{i})r_{j}=0$ for each $i$. Therefore, $\alpha(f(y))r_{j}=0$ and so $R$ is right $\alpha$-McCoy.

Corollary 2.5  A ring $R$ is a right McCoy ring if and only if $R[x]$ is right McCoy.

Let $R$ be a ring and $\triangle$ a multiplicative monoid in $R$ consisting of central regular elements, and let $\triangle^{-1}R$ =$\{u^{-1}a|u\in \triangle, a\in R\}$, then $\triangle^{-1}R$ is a ring. For an endomorphism $\alpha$ of $R$ with $\alpha(\Delta)\subseteq \Delta$, the induced map $\bar{\alpha}: \triangle^{-1}R\rightarrow \triangle^{-1}R$ defined by $\bar{\alpha}(u^{-1}a)=\alpha(u)^{-1}\alpha(a)$ is also an endomorphism. We have the following result for the right $\alpha$-McCoy property.

Proposition 2.6  Let $R$ be a ring with an endomorphism $\alpha$. If $R$ is right $\alpha$-McCoy, then $\triangle^{-1}R$ is right $\alpha$-McCoy.

Proof  Assume that $R$ is right $\alpha$-McCoy and let

$ f(x)=\sum\limits_{i=0}^{m}u_{i}^{-1}a_{i}x^{i}, g(x)=\sum\limits_{j=0}^{n}v_{j}^{-1}b_{j}x^{j} \in \bigtriangleup^{-1}R[x] $

with $\alpha(f(x))g(x)=0$. Then we have

$ F(x)=(u_{m}u_{m-1}\cdots u_{0})f(x), G(x)=(v_{n}v_{n-1}\cdots v_{0})g(x)\in R[x]. $

Since $R$ is right $\alpha$-McCoy and $\alpha(F(x))G(x)=0$, this implies that there exists a nonzero $r \in R$ such that $\alpha(u_{m}u_{m-1}\cdots u_{0}u_{i}^{-1}a_{i})r=0$ for all $i, j$, and so $\alpha(a_{i})r=0$ since $\triangle$ is a multiplicative monoid in $R$ consisting of central regular elements and $u_{i}, v_{j}\in \bigtriangleup $ for all $i, j$. It follows that $\alpha(u_{i}^{-1}a_{i})r=\alpha(u_{i})^{-1}\alpha(a_{i})r=0$ for all $i, j$. This shows that $\triangle^{-1}R$ is right $\alpha$-McCoy.

The ring of Laurent polynomials in $x$, with coefficients in a ring $R$, consists of all formal sum $\sum\limits_{i=k}^{n}m_{i}x^{i}$ with obvious addition and multiplication, where $m_{i}\in R$ and $k, n$ are (possibly negative) integers. We denote this ring by $R[x; x^{-1}]$. For an endomorphism $\alpha$ of a ring $R$, the map $\bar{\alpha}: R[x; x^{-1}]\rightarrow R[x; x^{-1}]$ defined by $\bar{\alpha}(\sum\limits_{i=k}^{n}a_{i}x^{i})=\sum\limits_{i=k}^{n}\alpha(a_{i})x^{i}$ extends $\alpha$ and is also an endomorphism of $R[x; x^{-1}]$.

Corollary 2.7  Let $R$ be a ring. If $R$ is a right $\alpha$-McCoy ring, then $R[x; x^{-1}]$ is right $\alpha$-McCoy.

Proof  Let $\triangle=\{{1, x, x^{2}, \cdots}\}$, then clearly $\triangle$ is a multipicatively closed subset of $R[x]$. Since $R[x; x^{-1}]\cong \triangle^{-1}R[x]$, it follows directly from Proposition 2.6 that $R[x; x^{-1}]$ is right $\alpha$-McCoy.

According to [2], an endomorphism $\alpha$ of a ring $R$ is called semicommutative if $ab=0$ implies that $aR\alpha(b)=0$ for all $a, b\in R$. A ring $R$ is called $\alpha$-semicommutative if there exists a semicommutative endomorphism $\alpha$ of $R$. Recall from [3] that a ring $R$ is said to be right linearly McCoy if given nonzero linear polynomials $f(x), g(x)\in R[x]$ with $f(x)g(x)=0$, there exists a nonzero element $r\in R$ with $f(x)r=0$. We can define linearly $\alpha$-McCoy rings similarly. It was proved in [3, Proposition 5.3] that every semicommutative ring is right linearly McCoy. The next example gives an example of right linearly $\alpha$-McCoy rings which is not $\alpha$-semicommutative.

Example 2.8  Let $R=\mathbb{Z}_{2}\oplus \mathbb{Z}_{2}$, where $\mathbb{Z}_{2}$ is the ring of integers modulo 2. Then $R$ is a right linearly $\alpha$-McCoy ring since $R$ is a commutative reduced ring. Let $\alpha: R\rightarrow R$ be an endomorphism defined by $\alpha((a, b))=(b, a)$. For $(1, 0), (0, 1)\in R$, we have $(1, 0)(0, 1)=0$ but $(1, 0)(1, 1)\alpha(0, 1)\neq 0$. It follows that $R$ is not $\alpha$-semicommutative.

Let $A(R, \alpha)$ be the subset $\{x^{-i}rx^{i}|r\in R, i\geq 0\}$ of the skew Laurent polynomial ring $R[x, x^{-1}; \alpha]$, where $\alpha: R\rightarrow R$ is an injective ring endomorphism of a ring $R$ (see [5] for more details). Elements of $R[x, x^{-1}; \alpha]$ are finite sums of elements of the form $x^{-i}rx^{i}$ where $r\in R$ and $i$ is a non-negative integer. Multiplication is subject to $xr=\alpha(r)x$ and $rx^{-1}=x^{-1}\alpha(r)$ for all $r\in R$. Note that for each $j\geq 0$, $x^{-i}rx^{i}=x^{-(i+j)}\alpha^{j}(r)x^{(i+j)}$. It follows that the set $A(R, \alpha)$ of all such elements forms a subring of $R[x, x^{-1}; \alpha]$ with

$ \begin{eqnarray*} x^{-i}rx^{i}+x^{-j}sx^{j}=x^{-(i+j)}(\alpha^{j}(r)+\alpha^{i}(s))x^{(i+j)}, \\ (x^{-i}rx^{i})(x^{-j}sx^{j})=x^{-(i+j)}(\alpha^{j}(r)\alpha^{i}(s))x^{(i+j)}\end{eqnarray*} $

for $r, s\in R$ and $i, j\geq 0$. Note that $\alpha$ is actually an automorphism of $A(R, \alpha)$.

Proposition 2.9  If $R$ is an $\alpha$-rigid ring, then $A(R, \alpha)$ is right $\alpha$-McCoy.

Proof  It follows directly from the fact that $A(R, \alpha)$ is an $\alpha$-rigid ring by [4] and that every $\alpha$-rigid ring is right $\alpha$-McCoy.

Proposition 2.10  Let $R$ be a ring and $\alpha$ an endomorphism of $R$. Then $R$ is a right $\alpha$-McCoy ring if and only if $R[x]/(x^{n})$ is a right $\alpha$-McCoy ring, where $(x^{n}) $ is the ideal generated by $x^{n}$.

Proof  Assume that $R$ is right $\alpha$-McCoy and we denote the element $\bar{x}$ in $R[x]/(x^{n})$ by $u$. Then

$ R[x]/(x^{n})=R[u]=R+Ru+\cdots+Ru^{n-1}, $ $

where $u$ commutes with elements of $R$ and $u^{n}=0$. Let $f(y)=\sum\limits_{i=0}^{p}f_{i}y^{i}$ and $g(y)=\sum\limits_{j=0}^{q}g_{j}y^{j}$ be nonzero polynomials in $R[u][y]$ with $\alpha(f(y))g(y)=0$, where

$ f_{i}=\sum\limits_{s=0}^{n-1}a_{is}u^{s}, ~~ g_{j}=\sum\limits_{t=0}^{n-1}b_{jt}u^{t}. $

Moreover, if we let $k_{s}(y)=\sum\limits_{i=0}^{p}a_{is}y^{i}$, $h_{t}(y)=\sum\limits_{j=0}^{q}b_{jt}y^{j}$. Then we have

$ \begin{array}{l} 0\;\; = \;\;\alpha (f(y))g(y) = (\sum\limits_{i = 0}^p \alpha ({f_i}){y^i})(\sum\limits_{j = 0}^q {{g_j}} {y^j})\\ \;\;\;\;\;\; = \;\;(\sum\limits_{i = 0}^p {\sum\limits_{s = 0}^{n - 1} \alpha } ({a_{is}}){u^s}{y^i})(\sum\limits_{j = 0}^q {\sum\limits_{t = 0}^{n - 1} {{b_{jt}}} } {u^t}{y^j})\\ \;\;\;\;\;\; = \;\;\sum\limits_{s = 0}^{n - 1} {(\sum\limits_{i = 0}^p \alpha ({a_{is}}){y^i})} \sum\limits_{t = 0}^{n - 1} {(\sum\limits_{j = 0}^q {{b_{jt}}} {y^j}){u^{s + t}} = (\sum\limits_{s = 0}^{n - 1} \alpha ({k_s}(y)} \sum\limits_{t = 0}^{n - 1} {{h_t}} (y)){u^{s + t}}. \end{array} $

It follows that $\sum\limits_{s+t=k}\alpha(k_{s}(y))h_{t}(y)=0$, where $k=0, 1, \cdots, n-1$. If $\alpha(k_{0}(y))=0$, take $r=u^{n-1}$. Then we have $0\neq r\in R[u]$, and so

$ \alpha(f(y))r=(\sum\limits_{s=0}^{n-1}(\sum\limits_{i=0}^{p}\alpha(a_{is})y^{i})u^{s})u^{n-1}=(\sum\limits_{i=0}^{p}\alpha(a_{i0})y^{i})u^{n-1}=\alpha(k_{0}(y))u^{n-1}=0. $

If $\alpha(k_{0}(y))\neq 0$, it follows from $g(y)\neq 0$ that there is a minimal $k \in \{0, 1, \cdots n-1 \}$ such that $h_{k}(y)\neq 0$ and $\alpha(k_{0}(y))h_{k}(y)=0$. Since $R$ is right $\alpha$-McCoy, there exists a nonzero element $c\in R$ such that $\alpha(k_{0}(y))c=0$. Let $r'=cu^{n-1}$. Then we have $0\neq r'\in R[u]$ and

$ \alpha(f(y))r'=(\sum\limits_{s=0}^{n-1}(\sum\limits_{i=0}^{p}\alpha(a_{is})y^{i})u^s)cu^{n-1} =(\sum\limits_{i=0}^{p}\alpha(a_{i0})y^{i})cu^{n-1}=0. $

Conversely, suppose that

$ f(y)=\sum\limits_{i=0}^{p}a_{i}y^{i}, g(y)=\sum\limits_{j=0}^{q}b_{j}y^{j} \in R[y] \backslash \{0\} $

such that $\alpha(f(y))g(y)=0$. Since $f(y)$ and $g(y)$ are nonzero polynomials of $R[x]/(x^{n})[y]$ and $R[x]/(x^{n})$ is right $\alpha$-McCoy, it follows that there exists $0\neq r_{1}(x)=\sum\limits_{k=0}^{n-1}c_{k}x^{k} \in R[x]/(x^{n})$ such that $\alpha(f(y))r_{1}(x)=0$. Let $c_{k_{0}}\neq 0$ with $k_{0}$ minimal. Then we obtain $\alpha(f(y))c_{k_{0}}=0$ and so $R$ is right $\alpha$-McCoy.

Corollary 2.11  Let $R$ be a ring and $n$ any positive integer. Then $R$ is right McCoy if and only if $R[x]/(x^{n})$ is right McCoy.

A ring $R$ is called right Ore if given $a, b \in R$ with $b$ regular, there exist $a_{1}, b_{1}\in R$ with $b_{1}$ regular such that $ab_{1}=ba_{1}$. It is well-known that $R$ is a right Ore ring if and only if the classical right quotient ring $Q(R)$ of $R$ exists. Suppose that the classical right quotient ring $Q(R)$ of $R$ exists. Then for an endomorphism $\alpha$ of $R$ and any $ab^{-1} \in Q(R)$ where $a, b \in R$ with $b$ regular, the induced map $\bar{\alpha}:Q(R)\rightarrow Q(R)$ defined by $\bar{\alpha}(ab^{-1})=\alpha(a)\alpha(b)^{-1}$ is also an endomorphism.

Proposition 2.12  Let $R$ be a right Ore ring with $Q(R)$ the classical right quotient ring of $R$. If $\alpha$ is an endomorphism of $R$, then $R$ is right $\alpha$-McCoy if and only if $Q(R)$ is right $\alpha$-McCoy.

Proof  Let $F(x)=\sum\limits_{i=0}^{m}\delta_{i}x^{i}, G(x)=\sum\limits_{j=0}^{n}\beta_{j}x^{j}$ be nonzero polynomials in $Q[x]$ with $\alpha(F(x))G(x)=0$. By [6, Proposition 2.1.16], we may assume that $\delta_{i}=a_{i}u^{-1}$ and $\beta_{j}=b_{j}v^{-1}$ with $a_{i}, b_{j}\in R$ for each $i, j$ and regular elements $u, v\in R$. Moreover, for each $j$, there exists $c_{j}\in R$ and a regular element $w\in R$ such that $\alpha(u)^{-1}b_{j}=c_{j}w^{-1}$ also by [6, Proposition 2.1.16]. Let $f(x)=\sum\limits_{i=0}^{m}a_{i}x^{i}, g(x)=\sum\limits_{j=0}^{n}c_{j}x^{j}$. Then we have

$ \begin{eqnarray*}0&\;\;=\;\;&\alpha(F(x))G(x)=(\sum\limits_{i=0}^{m}\alpha(\delta_{i})x^{i})(\sum\limits_{j=0}^{n}\beta_{j}x^{j})\\ &=\;\;&\sum\limits_{k=0}^{m+n}(\sum\limits_{i+j=k}\alpha(a_{i})c_{j}(vw)^{-1})x^{k} =\alpha(f(x))g(x)(vw)^{-1}. \end{eqnarray*} $

This implies that $\alpha(f(x))g(x)=0$. Then there exists a nonzero $r\in R$ such that $\alpha(f(x))r=0$ since $R$ is a right $\alpha$-McCoy ring. Then $\alpha(a_{i})r=0$ for each $i$, and hence $\alpha(\delta_{i})(\alpha(u)r)=0$. Now $Q$ being right $\alpha$-McCoy follows from the fact that $\alpha(F(x))(\alpha(u)r)=0$ since $\alpha(u)r$ is a nonzero element of $Q$. On the other hand, note that if

$ m(x)=\sum\limits_{i=0}^{m}a_{i}x^{i}, n(x)=\sum\limits_{j=0}^{n}b_{j}x^{j}\in R[x] $

such that $\alpha(m(x))n(x)=0$. Then there exists a nonzero element $\gamma$ in $Q$ such that $\alpha(m(x))\gamma=0$ since $Q$ is right $\alpha$-McCoy. We may assume $\gamma=d\kappa^{-1}$ with $d$ a nonzero element in $R$ and $\kappa$ a regular element. So we obtain $\alpha(m(x))d\kappa^{-1}=0$, and hence $\alpha(m(x))d=0$. This implies that $R$ is right $\alpha$-McCoy. This completes the proof.

Corollary 2.13  Let $R$ be a right Ore ring and $Q(R)$ be the classical right quotient ring of $R$. Then $R$ is right McCoy if and only if $Q(R)$ is right McCoy.

Comparing with the definition of a weak McCoy ring, we give the following definition of weak $\alpha$-McCoy rings accordingly.

Definition 3.1  An endomorphism $\alpha$ of a ring $R$ is called right (resp., left) weak McCoy, if for each pair of nonzero polynomials $f(x)=\sum\limits_{i=0}^{m}a_{i}x^{i}$ and $g(x)=\sum\limits_{j=0}^{n}b_{j}x^{j}\in R[x]$ with $\alpha(f(x))g(x)=0$ (resp., $f(x)\alpha(g(x))=0$), there exists a nonzero element $r\in R$ such that $\alpha(a_{i})r \in nil(R)$ (resp., $r\alpha(b_{j})\in nil(R)$). A ring $R$ is called right (resp., left) weak $\alpha$-McCoy if there exists a right (resp., left) weak McCoy endomorphism $\alpha$ of $R$. $R$ is a weak $\alpha$-McCoy ring if it is both right and left $\alpha$-McCoy.

It is clear that every right $\alpha$-McCoy ring is right weak $\alpha$-McCoy. It was shown in [10, Theorem 2.1] that $T_{n}(R)$ is not McCoy for $n\geq 2$. The following example shows that there exists a weak $\alpha$-McCoy endomorphism $\alpha$ of a ring $S$ such that $S$ is not an $\alpha$-McCoy ring.

Example 3.2  Let $R$ be a reduced ring. Consider the ring

$ S = \left\{ {\left( {\begin{array}{*{20}c} a&b \\ 0&c \\ \end{array} } \right)\left| {a, b, c \in R} \right.} \right\}. \\ $

Let $\alpha: S\rightarrow S$ be an endomorphism defined by

$ \alpha \left( {\left( {\begin{array}{*{10}c} a&b \\ 0&c \\ \end{array} } \right)} \right) = \left( {\begin{array}{*{20}c} a &-b \\ 0&c \\ \end{array} } \right). $

On the other hand, let

$ f(x)=\left( \begin{array}{cc} 0&-1 \\ 0&0 \\ \end{array} \right)+\left( \begin{array}{cc} 1&0 \\ 0&0 \\ \end{array} \right)x, ~~g(x)=\left( \begin{array}{cc} 0&1 \\ 0&0 \\ \end{array} \right)+\left( \begin{array}{cc} 0&0 \\ 0&-1 \\ \end{array} \right)x $

be elements in $S[x]$. It is straightforward to check that $\alpha(f(x))g(x)=0$, and we can not find a nonzero element $r \in S$ such that $\alpha(f(x)r=0$. This implies that $S$ is not an $\alpha$-McCoy ring. However, $S$ is a right weak $\alpha$-McCoy ring by the following Proposition 3.3.

Proposition 3.3  Let $R$ be a right weak $\alpha$-McCoy ring and $\alpha$ an endomorphism of $R$. Then $T_{n}(R)$ is a right weak $\alpha$-McCoy ring.

Proof  Let $f(x)=\sum\limits_{i=0}^{m}A_{i}x^{i}, g(x)=\sum\limits_{j=0}^{n}B_{j}x^{j}$ be nonzero polynomials in $T_{n}(R)[x]$ with $\alpha(f(x))g(x)=0$, where $A_{i}, B_{j} \in T_{n}(R)$ for all $i, j$. If we denote by $E_{ij}$ the usual matrix unit with 1 in the $(i, j)$-coordinate and zero elsewhere, then for each $\alpha(A_{i})$ there exists a nonzero element $C=rE_{1n}$ such that $\alpha(A_{i})C \in nil(T_{n}(R))$, where $0\neq r\in R$.

Given a ring $R$ and a bimodule $_RM_R$, the trivial extension of $R$ by $M$ is the ring $T(R, M)=R\bigoplus M$ with the usual addition and the following multiplication

$ (r_1, m_1)(r_2, m_2)=(r_1r_2, r_1m_2+m_1r_2). $

This is isomorphic to the ring of all matrix $\left(% \begin{array}{cc} r&m \\ 0&r \\ \end{array}% \right), $ where $r\in R$, $m\in M$ and the usual matrix operations are used. For an endomorphism $\alpha$ of a ring $R$ and the trivial extension $T(R, R)$ of $R$, $\overline{\alpha}$: $T(R, R)\rightarrow T(R, R)$ defined by

$ \overline \alpha \left( {\left( {\begin{array}{*{20}c} a \;\; b \\ 0 \;\; a \\ \end{array}} \right)} \right) = \left( {\begin{array}{*{20}c} {\alpha \left( a \right)} \;\; {\alpha \left( b \right)} \\ 0 \;\;\;\;\;\; {\alpha \left( a \right)} \\ \end{array}} \right) $

is an endomorphism of $T(R, R)$. Since $T(R, 0)$ is isomorphic to $R$, we can identify the restriction of $\overline{\alpha}$ by $T(R, 0)$ to $\alpha$.

Corollary 3.4  If $R$ is right weak $\alpha$-McCoy, then the trivial extension $T(R, R)$ is a right weak $\alpha$-McCoy ring.

Based on Proposition 3.3, one may suspect that if $R$ is a weak $\alpha$-McCoy ring, then the $n$-by-$n$ full matrix ring $M_{n}(R)$ is weak $\alpha$-McCoy with $n\geq 2$. But the following example erases the possibility.

Example 3.5  Let $R$ be a reduced ring. Then $R$ is a weak $\alpha$-McCoy ring. Put $S=M_{n}(R)$ and let $\alpha$ be an endomorphism of $S$ defined by

$ \alpha \left( {\left( {\begin{array}{*{10}c} a \;\; b \\ c \;\; d \\ \end{array} } \right)} \right) = \left( {\begin{array}{*{20}c} a \;\; -b \\ -c \;\; d \\ \end{array} } \right). $

We also let

$ f(x)=\left( \begin{array}{cc} 0 \;\; -1 \\ 0 \;\;\;\; 0 \\ \end{array} \right)+\left( \begin{array}{cc} 1 \;\; 0 \\ 0 \;\; 0 \\ \end{array} \right)x, ~~g(x)=\left( \begin{array}{cc} 1 \;\; 1 \\ 0 \;\; 0 \\ \end{array} \right)+\left( \begin{array}{cc} 0 \;\;\;\; 0 \\ -1\;\; -1 \\ \end{array} \right)x $

be polynomials in $S[x]$. Then we have $\alpha(f(x))g(x)=0$, and it is easy to check that $S$ is not a weak $\alpha$-McCoy ring.

Now we consider the case of direct limits of direct systems of right weak $\alpha$-McCoy rings.

Proposition 3.6  The direct limit of a direct system of right weak $\alpha$-McCoy rings is also right weak $\alpha$-McCoy.

Proof  Let $D=\{R_{i}, \phi_{ij}\}$ be a direct system of right weak $\alpha$-McCoy rings $R_{i}$ for $i\in I$ and ring homomorphisms $\phi_{ij}:R_{i}\rightarrow R_{j}$ for each $i\leq j$ satisfying $\phi_{ij}(1)=1$, where $I$ is a direct partially ordered set. Let $R=\underrightarrow {\lim }R_i$ be the direct limit of $D$ with $\iota_{i}:R_{i}\rightarrow R$ and $\iota_{j}\phi_{ij}=\iota_{i}$. We shall prove that $R$ is a right weak $\alpha$-McCoy ring. Let $\alpha$ be an endomorphism of $R$ and take $x, y\in R$. It follows that $x=\iota_{i}(x_{i}), y=\iota_{j}(y_{j})$ for some $i, j\in I$ and there is $k\in I$ such that $i\leq k, j\leq k$. Now define $x+y=\iota_{k}(\phi_{ik}(x_{i})+\phi_{jk}(y_{j}))$ and $xy=\iota_{k}(\phi_{ik}(x_{i})\phi_{jk}(y_{j}))$, where $\phi_{ik}(x_{i})$ and $\phi_{jk}(y_{j})$ are in $R_{k}$. It is easy to see that $R$ forms a ring with $0=\iota_{i}(0)$ and $1=\iota_{i}(1)$. Let $\alpha(f(x))g(x)=0$ with $f(x)=\sum\limits_{s=0}^{m}a_{s}x^{s}$ and $g(x)=\sum\limits_{t=0}^{n}b_{t}x^{t}$ in $R[x]$. Then there are $i_{s}, j_{t}, k\in I$ such that $\alpha(a_s)=\iota _{i_s } \left( {a_{i_s } } \right), b_t = \iota _{j_t } \left( {b_{j_t } } \right), i_{s}\leq k, j_{t}\leq k$. So we have

$ \alpha(a_{s})b_{t}=\iota_{k}(\phi_{i_sk}(a_{i_s)}\phi_{j_tk}(b_{j_t})), $

and hence

$ \begin{eqnarray*}\alpha(f(x))g(x)&\;\;=\;\;&(\sum\limits_{s=0}^{m}\iota_{k}(\phi_{i_sk}(a_{i_s}))x^{s}) (\sum\limits_{t=0}^{n}\iota_{k}(\phi_{j_tk}(b_{j_t}))x^{t})\\ &=\;\;& \sum\limits_{d=0}^{m+n}(\sum\limits_{s+t=d}\iota_{k}(\phi_{i_sk}(a_{i_s)}\phi_{j_tk}(b_{j_t})))x^{d}=0\end{eqnarray*} $

in $R_{k}[x]$ since $\alpha(f(x))g(x)=0$. On the other hand, since $R_{k}$ is right weak $\alpha$-McCoy, there exists $s_{k}\in R_{k}\backslash \{0\}$ such that $\iota_{k}(\phi_{i_sk}(a_{i_s}))s_{k}\in nil(R_{k})$ for all $0\leq i\leq m$. Let $s=\iota_{k}(s_{k})$. Then we have $\alpha(a_{s})s\in nil(R)$ and $R$ is right weak $\alpha$-McCoy.

Corollary 3.7  The direct limit of a direct system of right weak McCoy rings is right weak McCoy.

Proposition 3.8  Let $R$ be a ring and $I$ an ideal of $R$ such that $R/I$ is right weak $\alpha$-McCoy. If $I\subseteq nil(R)$, then $R$ is a right weak $\alpha$-McCoy ring.

Proof  Let $f(x)=\sum\limits_{i=0}^{m}a_{i}x^{i}$ and $g(x)=\sum\limits_{j=0}^{n}b_{j}x^{j}$ be polynomials in $R[x]$ with $\alpha(f(x))g(x)=0$. Then we have $\sum\limits_{i=0}^{m}\alpha(\bar{a}_{i})x^{i})(\sum\limits_{j=0}^{n}\bar{b}_{j}x^{j})=0$ in $R/I$. Since $R/I$ is right weak $\alpha$-McCoy, there exists $n_{i} \in \mathbb{N}$ and $s\not\in I$ such that $(\alpha(\bar{a}_{i})\bar{s})^{n_{i}}=0$. It follows that $(\alpha(a_{i})s)^{n_{i}}\in nil(R)$ since $I\subseteq nil(R)$. This completes the proof.