Throughout this paper $R$ denotes an associative ring with identity, $\sigma:R\longrightarrow R$ is a nonzero endomorphism. A ring $R$ is called reduced if it has no nonzero nilpotent elements, and a ring $R$ is called an abelian ring if all its idempotents are central. According to Cohn [4], a ring $R$ is called reversible if $ab=0$ implies $ba=0$ for all $a, b\in R$. Recently, Baser et al. [3] defined a ring $R$ to be right (left) $\alpha$-shifting if whenever $a\alpha(b)=0 (\alpha(a)b=0)$ for $a, b\in R$, $b\alpha(a)=0 (\alpha(b)a=0)$, which is a generalization of revesible rings. Recall that a ring $R$ is semicommutative if $ab=0$ implies $aRb=0$ for all $a, b\in R$. Baser et al. [2] extended the notion of semicommutative rings and called a ring $R$ $\alpha$-semicommutative if $ab=0$ implies $aR\alpha(b)=0$ for all $a, b\in R$. Another generalization of semicommutative rings is the semicommutative $\alpha$-rings. Wang et al. [17] called a ring $R$ right (left) semicommutative $\alpha$-ring if $a\alpha(b)=0$ ($\alpha(a)b=0)$ implies $\alpha(a)Rb=0$ ($aR\alpha(b)=0)$ for all $a, b\in R$, and investigated characterizations of generalized semicommutative rings. According to Lamber [13], a ring $R$ is called symmetric if $abc=0$ implies $acb=0$ for all $a, b, c\in R$. Anderson and Camillo [1] showed that a ring $R$ is symmetric if and only if $r_{1}r_{2}\cdots r_{n}=0$ implies $r_\sigma(1)r_\sigma(2)\cdots r_\sigma(n)=0$ for any permutation $\sigma$ of the set $\{1, 2, \cdots, n\}$ and $r_{i}\in R$. There are many papers to study symmetric rings and their generalization (see [6, 8, 11, 14, 16]). In Kwak [12], an endomorphism $\alpha$ of a ring $R$ is called right (left) symmetric if whenever $abc=0$ for $a, b, c\in R$, $ac\alpha(b)=0 (\alpha(b)ac=0)$. A ring $R$ is called right (left) $\alpha$-symmetric if there exists a right (left) symmetric endomorphism $\alpha$ of R. The notion of an $\alpha$-symmetric ring is a generalization of $\alpha$-rigid rings as well as an extension of symmetric rings. Following [15], a ring $R$ is called a weak symmetric ring if $abc\in{\rm nil}(R)$ implies that $acb\in{\rm nil}(R)$ for all $a, b, c\in R$, where ${\rm nil}(R)$ is the set of all nilpotent elements of $R$. Let $\alpha$ be an endomorphism, and $\delta$ an $\alpha$-derivation of $R$, that is, $\delta$ is an additive map such that $\delta(ab)=\delta(a)b+\alpha(a)\delta(b)$, for $a, b\in R$. When $\alpha=id_{R}$, an $\alpha$-derivation $\delta$ is called a derivation of $R$. A ring $R$ is called a weak $\alpha$-symmetric provided that $abc\in{\rm nil}(R)$ implies $ac\alpha(b)\in{\rm nil}(R)$ for $a, b, c\in R$. Moreover, $R$ is called a weak $\delta$-symmetric if for $a, b, c\in R, abc\in {\rm nil}(R)$ implies that $ac\delta(b)\in{\rm nil}(R)$. If $R$ is both weak $\alpha$-symmetric and weak $\delta$-symmetric, then $R$ is called a weak $(\alpha, \delta)$-symmetric ring. In [15], Ouyang and Chen studed the related properties of weak symmetric rings and weak $(\sigma, \delta)$-symmetric rings.

Motivated by the above, for an endomorphism $\sigma$ of a ring $R$, and a $\sigma$-derivation $\delta$ of the $R$, we introduce in this article the notions of symmetric $\sigma$-ring and weak symmetric $(\sigma, \delta)$-rings to extend symmetric rings and weak symmetric rings respectively, and investigate their properties. First, we discuss the relationship between symmetric $\sigma$-rings and related rings. Next, we investigate the extension properties of weak symmetric $(\sigma, \delta$)-rings. Several known results are obtained as corollaries of our results.

As a generalization of symmetric rings, we now introduce the notion of a symmetric $\sigma$-ring.

Definition 2.1  Let $R$ be a ring, $\sigma$ a nonzero endomorphism of $R$. We say that $R$ is a symmetric $\sigma$-ring, if $ab\sigma(c)=0$ implies $ac\sigma(b)=0$, for any $a, b, c\in R$.

Similarly, a ring $R$ is said to be a left symmetric $\sigma$-ring whenever $\sigma(a)bc=0$ implies $\sigma(b)ac=0$, for $a, b, c\in R$.

Obviously, if $\sigma=id_{R}$, the identity endomorphism of $R$, then a (left) symmetric $\sigma$-ring is a symmetric ring.

The next example shows that if $\sigma\neq id_{R}$, a symmetric $\sigma$-ring need not be symmetric and a symmetric $\sigma$-ring need not be a left symmetric $\sigma$-ring yet. Therefore, the classes of symmetric $\sigma$-ring and left symmetric $\sigma$-ring are non-trivial extension of symmetric rings, and the symmetric $\sigma$-property for a ring is not left-right symmetric, and the concepts of symmetric $\sigma$-rings and that of left symmetric $\sigma$-rings are independent of each other.

Example 2.2  Consider the ring $R=\left\{\left(\begin{array}{cc} a \;\; b \\ 0 \;\; c \end{array} \right)| a, b, c\in \mathbb{Z}\right\}$, where $\mathbb{Z}$ is the ring of integers, the endomorphism $\sigma:R\rightarrow R$, $\sigma\left(\left( \begin{array} {cc} a \;\; b \\ 0 \;\; c \end{array} \right )\right )=\left( \begin{array} {cc} a \;\; 0 \\ 0 \;\; 0 \end{array} \right )$. It is easy to verify that $R$ is not symmetric. Let

$ \textbf{A}=\left( \begin{array} {cc} a_{1} \;\; b_{1} \\ 0 \;\;\; c_{1} \end{array} \right ), \textbf{B}=\left( \begin{array} {cc} a_{2} \;\; b_{2} \\ 0 \;\;\; c_{2} \end{array} \right ), \textbf{C}=\left( \begin{array} {cc} a_{3} \;\; b_{3} \\ 0 \;\;\; c_{3} \end{array} \right )\in R $

with $\textbf{A}\textbf{B}\sigma(\textbf{C})=0$, then $a_{1}a_{2}a_{3}=0$, so we have $a_{1}a_{3}a_{2}=0$ and $\textbf{A}\textbf{C}\sigma(\textbf{B})=0$, concluding that $R$ is a symmetric $\sigma$-ring. For

$ \textbf{A}=\left( \begin{array} {cc} 0 \;\; 1 \\ 0 \;\; 1 \end{array} \right ), \textbf{B}=\left( \begin{array} {cc} 1 \;\; 1 \\ 0 \;\; 0 \end{array} \right ), \textbf{C}=\left( \begin{array} {cc} 1 \;\; 1 \\ 0 \;\; 1 \end{array} \right )\in R, $

we have $\sigma(\textbf{A})\textbf{B}\textbf{C}=0$, but $\sigma(\textbf{B})\textbf{A}\textbf{C}=\left( \begin{array} {cc} 0 \;\; 1 \\ 0 \;\; 0 \end{array} \right )\neq 0$, thus $R$ is not a symmetric $\sigma$-ring.

The next example provides that if $\sigma\neq id_{R}$, then there exists a symmetric ring which is not a symmetric $\sigma$-ring.

Example 2.3  Let $\mathbb{Z}_{2}$ be the ring of integers modulo $2$. We consider ring $R=\mathbb{Z}_{2}\bigoplus \mathbb{Z}_{2}$ with the usual addition and multiplication. Then $R$ is a commutative reduced ring, and so $R$ is symmetric. Now let $\sigma: R \longrightarrow R$ given by $\sigma((a, b))= (b, a)$. Then $\sigma$ is an endomorphism of $R$. For $A=(1, 0), B=(0, 1), C=(1, 1)\in R$, we have $AB\sigma(C)=(1, 0)(0, 1)(1, 1)=0$, but $AC\sigma(B)=(1, 0)(1, 1)(1, 0)=(1, 0)\neq 0$. Thus $R$ is not a symmetric $\sigma$-ring.

The next example shows that symmetric $\sigma$-rings need not be $\sigma$-rigid rings.

Example 2.4  Consider the ring $R=\left\{\left(\begin{array}{cc} a \;\; b \\ 0 \;\; a \end{array} \right)| a, b\in \mathbb{Z}\right\}$ and the automorphism $\sigma:R\rightarrow R$,

$ \sigma\left(\left( \begin{array} {cc} a \;\; b \\ 0 \;\; a \end{array} \right )\right )=\left( \begin{array} {cc} a \;\; -b \\ 0 \;\; a \end{array} \right ). $

$R$ is not reduced and hence not $\sigma$-rigid. But $R$ is a symmetric $\sigma$-ring. In fact, for any

$ \textbf{A}=\left(\begin{array}{cc} a \;\; b \\ 0 \;\; a \end{array}\right), \textbf{B}=\left(\begin{array} {cc} c \;\; d \\ 0 \;\; c \end{array}\right), \textbf{C}=\left(\begin{array} {cc} e \;\; f \\ 0 \;\; e \end{array}\right )\in R $

with $\textbf{A}\textbf{B}\sigma(\textbf{C})=0$, we have $ace=0, -acf+ade+bce=0$, it follows that $a=0$ or $c=0$ or $e=0$. If $a=0$, then $acf=ade=bce=0$, and then $aec=-aed+afc+bec=0$, hence

$ \textbf{A}\textbf{C}\sigma(\textbf{B})=\left( \begin{array} {cc} aec \;\; -aed+afc+bec \\ 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; aec \end{array} \right )=0. $

Similarly, for $c=0$ or $e=0$, we have $\textbf{A}\textbf{C}\sigma(\textbf{B})=0$.

Proposition 2.5  For a nonzero endomorphism $\sigma$ of a ring $R$, the following statements are equivalent:

(1) $R$ is a symmtric $\sigma$-ring;

(2) $l_{R}(b\sigma(c))\subseteq l_{R}(c\sigma(b))$ for any $a, b, c \in R$;

(3) $AB\sigma(C)=0 \Longleftrightarrow AC\sigma(B)=0$ for any $A, B, C \subseteq R$

Proof   $(1)\Longleftrightarrow(3)$ Suppose $AC\sigma(B)=0$ for $A, B, C \subseteq R$. Then $ab\sigma(c)=0$ for any $a\in A, b\in B, c\in C$, and hence $ac\sigma(b)=0$. Therefore, $AC\sigma(B)=\{\sum a_{i}c_{i}\sigma(b_{i})|a_{i}\in A, b_{i}\in B, c_{i}\in C \}=0$. The converse is obvious.

$(1)\Longleftrightarrow(2)$ It is clear.

Proposition 2.6  Let $\sigma$ be a nonzero endomorphism of a ring $R$. Then we have the following:

(1) If $\sigma^{2}=id_{R}$, then $R$ is a right (left) $\sigma$-shifting ring if and only if $R$ is a right (left) semicommutative $\sigma$-ring;

(2) If $R$ is a reversible ring, then $R$ is a right (left) $\sigma$-shifting ring if and only if $R$ is a right (left) semicommutative $\sigma$-ring.

Proof  (1) Suppose that $R$ is right $\sigma$-shifting and $a\sigma(b)=0$ for $a, b\in R$. Then we have $b\sigma(a)=0$, $\sigma(b)\sigma^{2}(a)=0$ and $\sigma(b)\sigma^{2}(a)\alpha(R)=0$. It implies that $\sigma(a)R\sigma^{2}(b)=0$ since $R$ is $\sigma$-shifting, and hence $\sigma(a)Rb=0$ by $\sigma^{2}=id_{R}$.

(2) Suppose that $R$ is left $\sigma$-shifting and $\sigma(a)b=0$ for $a, b\in R$. Then $b\sigma(a)=0$ since $R$ is reversible, and hence $b\sigma(a)\sigma(r)=b\sigma(ar)=0$ for all $r\in R$. By the assumption, we have $ar\sigma(b)=0$, including that $R$ is a left semicommutative $\sigma$-ring. Conversely, assume that $R$ is a left semicommutative $\sigma$-ring. If $a, b\in R$ with $a\sigma(b)=0$, then $\sigma(b)a=0$ since $R$ is reversible. So we obtain that $bR\sigma(a)=0$ since $R$ is a left semicommutative $\sigma$-ring, and hence $b\sigma(a)=0$. So $R$ is left $\sigma$-shifting.

Proposition 2.7  Let $\sigma$ be a monomorphism of a ring $R$. If $R$ is a symmetric $\sigma$-ring, then $R$ is semicommutative.

Proof  Assume that $R$ is a symmetric $\sigma$-ring with a monomorphism $\sigma$. Since $1\in R$, $R$ is a right $\sigma$-shifting ring. For $a, b\in R$, if $ab=0$, then $\sigma(a)\sigma(b)=0$, and hence $b\sigma(\sigma(a))=0$. So we have $rb\sigma(\sigma(a))=0$ and $\sigma(a)\sigma(rb)=\sigma(arb)=0$ for any $r\in R$. It shows that $arb=0$ since $\sigma$ is a monomorphism of $R$, entailing that $R$ is semicommutative.

Proposition 2.8  Let $\sigma$ be an endomorphism of a ring $R$ with $\sigma(e)=e$ for any $e^{2}=e\in R$. If $R$ is a symmetric $\sigma$-ring, then $R$, $R[x]$ and $R[x;\sigma]$ are all abelian.

Proof  Assume that $R$ is a symmetric $\sigma$-ring. Then $R$ is a right $\sigma$-shifting ring. For any $r\in R$, we have

$ e\sigma(1-e)\sigma(r)=e\sigma((1-e)r)=0, \\ (1-e)\sigma(e)\sigma(r)=(1-e)\sigma(er)=0. $

Hence $(1-e)r\sigma(e)=0, er\sigma(1-e)=0$ since $R$ is right $\sigma$-shifting. Thus we get $re=ere=er$, proving that $R$ is an abelian ring.

Now, suppose that $f^{2}(x)=f(x)\in R[x;\sigma]$, where $f(x)=\sum\limits^{m}_{i=0}e_{i}x^{i}$. Then we have,

$ \sum\limits^{m}_{k=0}(\sum\limits_{i+j=k}e_{i}\sigma^{i}(e_{j}))x^{k}=\sum\limits^{m}_{i=0}e_{i}x^{i}. $

It follows that the following system of equations:

$ e^{2}_{0}=e_{0}; $

(2.1)

$ e_{0}e_{1}+e_{1}\sigma(e_{0})=e_{1}; $

(2.2)

$ e_{0}e_{2}+e_{2}\sigma^{2}(e_{0})+e_{1}\sigma(e_{1})=e_{2} $

(2.3)

$ \begin{array}{l} \;\;\;\;\;\;\;\;\;\;\; \vdots \\ {e_0}{e_n} + {e_1}\sigma ({e_{n - 1}}) + {e_2}{\sigma ^2}({e_{n - 2}}) + \cdots + {e_n}{e_0} = {e_n}. \end{array} $

(2.4)

From eq. (2.2), we have $2e_{1}e_{0}=e_{1}, 2e_{1}e_{0}(1-e_{0})=e_{1}(1-e_{0})$ and $e_{1}=e_{1}e_{0}, e_{1}=0$ since $\sigma(e_{0})=e_{0}$ is central. Eq. (2.3) yields $2e_{0}e_{2}=e_{2}$ and so $e_{2}=0$ by the same method as above. Continuing this procedure implies $e_{i}=0$ for $i=1, 2, \cdots, m$. Consequently, $f(x)=e_{0}=e_{0}^{2}\in R$ is central.

Let $R_{\gamma}$ be a ring and $\sigma_{\gamma}$ an endomorphism of $R_{\gamma}$ for each $\gamma\in \Gamma$. Then $\sigma:\Pi_{\gamma\in \Gamma}R_{\gamma}\rightarrow \Pi_{\gamma\in \Gamma}R_{\gamma}$, $\sigma((a_{\gamma})_{\gamma\in \Gamma})=(\sigma_{\gamma}(a_{\gamma}))_{\gamma\in \Gamma}$ is an endomorphism of the direct product $\Pi_{\gamma\in \Gamma}R_{\gamma}$ of $R_{ \gamma}, \gamma\in \Gamma$.

The following proposition is a direct verification.

Proposition 2.9  $\Pi_{\gamma\in \Gamma}R_{\gamma}$ is a symmetric $\sigma$-ring if and only if $R_{\gamma}$ is a symmetric $\sigma_{\gamma}$-ring for each $\gamma\in \Gamma$.

Given a ring $R$ and a bimodule ${}_R\!M_{R}$, 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). $

$T(R, M)$ is isomorphic to the ring of all matrices $\left( \begin{array} {cc} r&m \\ 0&r \end{array} \right )$, where $r\in R, m\in M$ and the usual matrix operations are used. For an endomorphism $\sigma$ of a ring $R$, the map $\bar{\sigma}:T(R, R)\rightarrow T(R, R)$ defined by $\bar{\sigma}((a, b))=(\sigma(a), \sigma(b))$ is an endomorphism of $T(R, R)$, where $(a, b)\in T(R, R), a, b\in R$.

Proposition 2.10  Let $R$ be a reduced ring with an endomorphism $\sigma$. If $R$ is a symmetric $\sigma$-ring, then $T(R, R)$ is a symmetric $\bar{\sigma}$-ring.

Proof  Suppose that $R$ is a symmetric $\sigma$-ring. Let $\textbf{A}=\left( \begin{array} {cc} a_{1} \;\; b_{1} \\ 0 \;\; a_{1} \end{array} \right ), \textbf{B}=\left( \begin{array} {cc} a_{2} \;\; b_{2} \\ 0 \;\; a_{2} \end{array} \right ), \textbf{C}=\left( \begin{array} {cc} a_{3} \;\; b_{3} \\ 0 \;\; a_{3} \end{array} \right )\in T(R, R)$ with $\textbf{A}\textbf{B}\bar{\sigma}(\textbf{C})=0$. Then we have

$ a_{1}a_{2}\sigma(a_{3})=0; $

(2.5)

$ a_{1}a_{2}\sigma(b_{3})+a_{1}b_{2}\sigma(a_{3})+b_{1}a_{2}\sigma(a_{3})=0. $

(2.6)

It is known that reduced rings are symmetric rings. Multiplying eq. (2.5) on the right side by $b_{1}$ gives $a_{1}b_{1}a_{2}\sigma(a_{3})=0$. If we multiply eq. (2.6) on the left side by $a_{1}$, then we have

$ a_{1}a_{1}a_{2}\sigma(b_{3})+a_{1}a_{1}b_{2}\sigma(a_{3})=0. $

(2.7)

Multiplying eq. (2.5) on the left side by $a_{1}$ and on the right side by $b_{2}$ gives $a_{1}a_{1}b_{2}\sigma(a_{3})a_{2}=0$. Multiplying eq. (2.7) by $a_{2}$ on the right side gives $0=a_{1}a_{1}a_{2}\sigma(b_{3})a_{2}=a_{1}a_{2}\sigma(b_{3})a_{1}a_{2}\sigma(b_{3})=(a_{1}a_{2}\sigma(b_{3}))^{2}$, so $a_{1}a_{2}\sigma(b_{3})=0$. Thus we have the following equation

$ \begin{align} \ a_{1}b_{2}\sigma(a_{3})+b_{1}a_{2}\sigma(a_{3})=0. \end{align} $

(2.8)

If we multiply eq. (2.5) by $b_{2}$ on the right side, then we get $a_{1}b_{2}\sigma(a_{3})a_{2}=0$. Multiplying eq. (2.8) by $a_{2}$ on the right side gives $0=b_{1}a_{2}\sigma(a_{3})a_{2}=b_{1}a_{2}\sigma(a_{3})a_{2}b_{1}\sigma(a_{3})=(b_{1}a_{2}\sigma(a_{3}))^{2}$. Thus we obtain $b_{1}a_{2}\sigma(a_{3})=0, a_{1}b_{2}\sigma(a_{3})=0$, and hence we have $a_{1}a_{3}\sigma(a_{2})=a_{1}b_{3}\sigma(a_{2})=a_{1}a_{3}\sigma(b_{2})=b_{1}a_{3}\sigma(a_{2})=0$ since $R$ is a symmetric $\sigma$-ring.So $\textbf{A}\textbf{C}\bar{\sigma}(\textbf{B})=0$, proving that $T(R, R)$ is a symmetric $\bar{\sigma}$-ring.

Corollary 2.11  (see [8], Corollary 2.4) Let R be a reduced ring, then $T(R, R)$ is a symmetric ring.

Proposition 2.12  Let $\sigma$ be an endomorphism of an abelian ring $R$ with $\sigma(e)=e$ for any $e^{2}=e\in R$. Then the following statements are equivalent:

(1) $R$ is a symmetric $\sigma$-ring;

(2) $eR$ and $(1-e)R$ are symmetric $\sigma$-rings.

Proof  $(1)\Rightarrow (2)$ Since $\sigma(eR)\subseteq eR, \sigma((1-e)R)\subseteq (1-e)R$, it is obvious by the definition.

$(2)\Rightarrow (1)$ Let $a, b, c\in R$ with $ab\sigma(c)=0$. Then $eab\sigma(c)=0$ and $(1-e)ab\sigma(c)=0$. By the assumption, we get $eab\sigma(c)=e^{3}ab\sigma(c)=eaebe\sigma(c)=eaeb\sigma(ec)=0$ and $(1-e)ab\sigma(c)=(1-e)a(1-e)b\sigma((1-e)c)=0$. Since $eR$ and $(1-e)R$ are symmetric $\sigma$-rings, $eaec\sigma(eb)=eac\sigma(b)=0$ and $(1-e)a(1-e)c\sigma((1-e)b)=(1-e)ac\sigma(b)=0$, hence $ac\sigma(b)=eac\sigma(b)+(1-e)ac\sigma(b)=0$, proving that $R$ is a symmetric $\alpha$-ring.

Corollary 2.13  (see [8], Proposition 3.6(2)) Let $R$ be an abelian ring. Then $R$ is symmetric if and only if $eR$ and $(1-e)R$ are symmetric.

Recall that for a monomorphism $\sigma$ of a ring $R$, an over-ring $A$ of $R$ is a Jordan extension of $R$ if $\sigma$ can be extended to an automorphism of $A$ and $A=\bigcup\limits^{\infty}_{n=0}\sigma^{-n}(R)$ (see [10]).

Proposition 2.14  Let $A$ be the corresponding Jordan extension of a ring $R$ and $\sigma$ be a monomorphism of $R$. Then $R$ is a symmetric $\sigma$-ring if and only if $A$ is a symmetric $\sigma$-ring.

Proof  Since $\sigma (R)\subseteq R$, it suffices to obtain the necessity.

Assume that $R$ is a symmetric $\sigma$-ring and $ab\sigma(c)=0$ for $a, b, c\in A$. By the definition of $A$, there exists $n\geqslant 0$ such that $\sigma^{n}(a), \sigma^{n}(b), \sigma^{n}(c)\in R$. It follows that $\sigma^{n}(a)\sigma^{n}(b)\sigma(\sigma^{n}(c))=\sigma^{n}(ab\sigma(c))=0$. Since $R$ is a symmetric $\sigma$-ring, $\sigma^{n}(a)\sigma^{n}(c)\sigma(\sigma^{n}(b))=\sigma^{n}(ac\sigma(b))=0$. Then, we have $ac\sigma(b)=0$ since $\sigma$ is a monomorphism, and proving that $A$ is a symmetric $\sigma$-ring.

Proposition 2.15  Let $R$ be a ring with an endomorphism $\sigma$, $S$ a ring and $\tau: R\rightarrow S$ a ring isomorphism. Then $R$ is a symmetric $\sigma$-ring if and only if $S$ is a symmetric $\tau \sigma \tau^{-1}$-ring.

Proof  For $a, b, c\in R$, let $a^{'}=\tau(a), b^{'}=\tau(b)$ and $c^{'}=\tau(c)\in S$. Suppose that $R$ is a symmetric $\sigma$-ring and $a^{'}b^{'}\tau \sigma \tau^{-1}(c^{'})=0$ for $a^{'}, b^{'}, c^{'}\in S$. Then we have $\tau(a)\tau(b)\tau \sigma \tau^{-1}(\tau(c))=\tau(ab\sigma(c))=0$, hence $ab\sigma(c)=0$ since $\tau$ is a isomorphism. By the assumption, we get $ac\sigma(b)=0$, so $a^{'}c^{'}\tau \sigma \tau^{-1}(b^{'})=\tau(ac\sigma(b))=0$, including that $S$ is a symmetric $\tau \sigma \tau^{-1}$-ring. On the contrary, assume that $S$ is a symmetric $\tau \sigma \tau^{-1}$-ring and $ab\sigma(c)=0$ for $a, b, c\in R$. Then $a^{'}b^{'}\tau \sigma \tau^{-1}(c^{'})=\tau(ab\sigma(c))=0$. By the assumption, we get $a^{'}c^{'}\tau \sigma \tau^{-1}(b^{'})=\tau(ac\sigma(b))=0$, this implies $ac\sigma(b)=0$. So $R$ is a symmetric $\sigma$-ring.

As a extended weak symmetric rings, we now introduce the notion of a weak symmetric $(\sigma, \delta)$-ring.

Definition 3.1  Let $\sigma$ be an endomorphism and $\delta$ a $\sigma$-derivation of a ring $R$. A ring $R$ is called a weak symmetric $\sigma$-ring if $ab\sigma(c)\in{\rm nil}(R)$ implies $ac\sigma(b)\in{\rm nil}(R)$, for $a, b, c\in R$. Moreover, $R$ is called a weak symmetric $\delta$-ring if for $a, b, c \in R, ab\delta(c) \in{\rm nil}(R)$ implies $ac\delta(b)\in{\rm nil}(R)$. If $R$ is both a weak symmetric $\sigma$-ring and a weak symmetric $\delta$-ring, then $R$ is called a weak symmetric $(\sigma, \delta)$-ring.

Similarly, a ring $R$ is said to be a left weak symmetric $(\sigma, \delta)$-ring if $\sigma(a)bc\in {\rm nil}(R)$ then $\sigma(b)ac\in{\rm nil}(R)$, and if $\delta(a)bc\in {\rm nil}(R)$ then $\delta(b)ac\in{\rm nil}(R)$, for $a, b, c\in R$.

It is easy to see that every subring $S$ with $\sigma(S)\subseteq S$, $\delta(S)\subseteq S$ of a (left) weak symmetric $(\sigma, \delta)$-ring is also a (left) weak symmetric $(\sigma, \delta)$-ring.

Consider the $R$ and $\sigma$ in Example 2.3. Taking $\delta=0$, then this example shows that the notions of weak symmetric $(\sigma, \delta)$-rings are not left-right symmetric. Obviously, if $\sigma=id_{R}, \delta=0$, then a (left) weak symmetric $(\sigma, \delta)$-ring is a weak symmetric ring. The next example provides that if $\sigma\neq id_{R}, \delta\neq 0$, then there exists a weak symmetric ring which is not a weak symmetric $(\sigma, \delta)$-ring.

Example 3.2  Let $\mathbb{Z}_{2}$ be the ring of integers modulo $2$, and consider the ring $R=\mathbb{Z}_{2}\bigoplus \mathbb{Z}_{2}$ with the usual addition and multiplication. Then $R$ is a commutative reduced ring, and so $R$ is weak symmetric. Now let $\sigma: R \longrightarrow R$ given by $\sigma((a, b))= (b, a)$ and $\sigma: R \longrightarrow R$ given by $\delta((a, b))=(1, 0)(a, b)-\sigma(a, b)(1, 0)$ for each $(a, b)\in \mathbb{Z}_{2}$. Then $\sigma$ is an endomorphism of $R$ and $\delta$ is a $\sigma$-derivation of $R$. For $A=(1, 0), B=(0, 1), C=(1, 1)\in R$, we have $AB\sigma(C)=(1, 0)(0, 1)(1, 1)=0\in{\rm nil}(R)$, but $AC\sigma(B)=(1, 0)(1, 1)(1, 0)=(1, 0)$ is not in ${\rm nil}(R)$. Thus $R$ is not weak symmetric $(\sigma, \delta)$-ring.

In the following, we always suppose that $\sigma$ is an endomorphism and $\delta$ a $\sigma$-derivation of $R$.

Now we consider the $n$-by-$n$ upper triangular matrix ring $T_{n}(R)$ over $R$ and the ring

$ S_{n}(R)=\left\{\begin{pmatrix} a_{0} \;\; a_{1} \;\; a_{2} \;\; \cdots \;\; a_{n-1}\\ 0 \;\;\; a_{0} \;\;\; a_{1} \;\;\; \cdots \;\;\; a_{n-2}\\ \vdots \;\;\;\;\; \vdots \;\;\;\;\; \vdots \;\;\;\;\; \ddots \;\;\;\;\; \vdots\\ 0 \;\;\;\; 0 \;\;\;\; 0 \;\;\;\; \cdots \;\;\;\; a_{0} \end{pmatrix}|a_{i}\in R, i=0, 1, \cdots, n-1\right\}. $

For an endomorphism $\sigma$ and a $\sigma$-derivation $\delta$ of $R$, the natural extension $\bar{\sigma}:T_{n}(R)\longrightarrow T_{n}(R)$ defined by $\bar{\sigma}((a_{ij})) = (\sigma(a_{ij}))$ is an endomorphism of $T_{n}(R)$ and $\bar{\delta}: T_{n}(R)\longrightarrow T_{n}(R)$ defined by $\delta((a_{ij}))=(\delta(a_{ij}))$ is a $\bar{\sigma}$-derivation of $T_{n}(R)$.

Proposition 3.3  The following statements are equivalent:

(1) $R$ is a weak symmetric $(\sigma, \delta)$-ring;

(2) $T_{n}(R)$ is a weak symmetric $(\bar{\sigma}, \bar{\delta})$-ring;

(3) $S_{n}(R)$ is a weak symmetric $(\bar{\sigma}, \bar{\delta})$-ring.

Proof  $(1)\Longrightarrow(2)$ Let $\textbf{A}=(a_{ij}), \textbf{B}=(b_{ij}), \textbf{C}=(c_{ij})\in T_{n}(R)$, where $a_{ij}=0, b_{ij}=0, c_{ij}=0$, for all $i>j$, with $\textbf{A}\textbf{B}\sigma(\textbf{C})\in{\rm nil}(T_{n}(R))$ and $\textbf{A}\textbf{B}\delta(\textbf{C})\in{\rm nil}(T_{n}(R))$. Then $a_{ii}b_{ii}\sigma(c_{ii})\in{\rm nil}(R)$, $a_{ii}b_{ii}\delta(c_{ii})\in{\rm nil}(R)$ for all $0\leq i\leq n$, and so $a_{ii}c_{ii}\sigma(b_{ii})\in{\rm nil}(R)$, $a_{ii}c_{ii}\delta(b_{ii})\in{\rm nil}(R)$ since $R$ is a weak symmetric $(\sigma, \delta)$-ring. It follows that $\textbf{A}\textbf{C}\bar{\sigma}(\textbf{B})\in{\rm nil}(T_{n}(R))$ and $\textbf{A}\textbf{C}\bar{\delta}(\textbf{B})\in{\rm nil}(T_{n}(R))$. Therefore, $T_{n}(R)$ is a weak symmetric $(\bar{\sigma}, \bar{\delta})$-ring.

$(2)\Longrightarrow(1)$ Suppose that $T_{n}(R)$ is a weak symmetric $(\bar{\sigma}, \bar{\delta})$-ring. For $a, b, c\in R$ with $ab\alpha(c)=0$ and $ab\delta(c)=0$, we have $aEbE\bar{\sigma}(cE)=0$ and $aEbE\bar{\delta}(cE)=0$, and hence $aEcE\bar{\sigma}(bE)=0$ and $aEcE\bar{\delta}(bE)=0$ since $T_{n}(R)$ is a weak symmetric $(\bar{\sigma}, \bar{\delta})$-ring, where $E$ denote the identity matrix. This implies that $ac\sigma(b)=0$ and $ac\delta(b)=0$. So $R$ is a weak symmetric $(\sigma, \delta)$-ring.

$(1)\Longleftrightarrow(3)$ It is similar to $(1)\Longleftrightarrow(2)$.

Corollary 3.4  The trivial extension $T(R, R)$ of $R$ by $R$ is a weak symmetric $(\bar{\sigma}, \bar{\delta})$-ring if and only if R is a weak symmetric $(\sigma, \delta)$-ring.

Proof  By the isomorphism $T(R, R)\cong T_{2}(R)$, we obtain the proof.

Corollary 3.5  $R[x]/\langle x^n\rangle$ is a weak symmetric $(\bar{\sigma}, \bar{\delta})$-ring if and only if $R$ is a weak symmetric $(\sigma, \delta)$-ring, where $\langle x^n\rangle$ is an ideal of $R$ generated by $x^n$ and $n$ is any positive integer.

Proof  By the isomorphism $R[x]/\langle x^n\rangle \cong S_{n}(R)$, we obtain the proof.

An ring $R$ is said to be an NI ring [9] provided that ${\rm nil}(R)={\rm Nil}^{*}(R)$, where ${\rm Nil}^{*}(R)$ denotes the upper nil radical of $R$.

Proposition 3.6  Let $R$ be an NI ring, $e^{2}=e\in R$ a central idempotent element of $R$. If $\sigma(e)=e, \sigma(1)=1, \delta(e)=\delta(1)=0$, then the following statements are equivalent:

(1) $R$ is a weak symmetric $(\sigma, \delta)$-ring;

(2) $eR$ and $(1-e)R$ are weak symmetric $(\sigma, \delta)$-rings.

Proof  $(1)\Longrightarrow (2)$ Suppose that $ab\sigma(c)\in{\rm nil}(I), ab\delta(c)\in{\rm nil}(I)$ for $a, b, c\in I$, where $I$ denotes $eR$ (resp., $(1-e)R$). Then we have $ac\sigma(b)\in{\rm nil}(R), ac\delta(b)\in{\rm nil}(R)$ since $R$ is a weak symmetric $(\sigma, \delta)$-ring, and hence $ac\sigma(b)\in ({\rm nil}(R)\bigcap I)={\rm nil}(I), ac\delta(b)\in ({\rm nil}(R)\bigcap I)={\rm nil}(I)$.

$(2)\Longrightarrow(1)$ Let $a, b, c\in R$ with $ab\sigma(c)\in {\rm nil}(R), ab\delta(c)\in{\rm nil}(R)$. Then $eaebe\sigma(c)\in {\rm nil}(eR)$ and $(1-e)a(1-e)b(1-e)\sigma(c)\in{\rm nil}((1-e)R)$ since $eR$, $(1-e)R$ are ideals of $R$ and $e\in eR$, $1-e\in (1-e)R$. It follows that $eaece\sigma(b)=eac\sigma(b)\in{\rm nil}(eR)$, and $(1-e)a(1-e)c\sigma((1-e)b)=(1-e)ac\sigma(b)\in {\rm nil}((1-e)R)$ since $eR$ and $(1-e)R$ are weak symmetric $(\sigma, \delta)$-rings. Hence $ac\sigma(b)\in{\rm nil}(R)$ because ${\rm nil}(R)$ is an ideal of $R$. On the other hand, by assumption we have $\delta(ex)=\delta(e)x+\sigma(e)\delta(x)=e\delta(x)$ and $\delta((1-e)x)=(1-e)\delta(x)$ for any $x\in R$. Thus, from $ab\delta(c)\in{\rm nil}(R)$ we have $eaeb\delta(ec)\in{\rm nil}(R), (1-e)a(1-e)b\delta((1-e)c) \in{\rm nil}(R)$. Hence $eaec\sigma(eb)=eac\delta(b)\in{\rm nil}(R)$ and $1-e)a(1-e)c\delta((1-e)b)=(1-e)ac\delta(b)\in{\rm nil}(R)$ since $eR$ and $(1-e)R$ are weak symmetric $(\sigma, \delta)$-rings. This implies that $ac\delta(b)\in{\rm nil}(R)$ since $R$ is an NI ring. Therefore, $R$ is a weak symmetric $(\sigma, \delta)$-ring.

An ideal $I$ of a ring $R$ is said to be $(\sigma, \delta)$-stable if $\sigma(I)\subseteq I$ and $\delta(I)\subseteq I$. If $I$ is a $(\sigma, \delta)$-stable ideal, then $\bar{\sigma}: R/I\longrightarrow R/I$ defined by $\bar{\sigma}(\bar{a})=\bar{\sigma(a)}$ for $\bar{a}\in R/I$ is an endomorphism of the factor ring $R/I$, and $\bar{\delta}: R/I \longrightarrow R/I$ defined by $\delta(\bar{a})=\bar{\delta(a)}$ for $\bar{a}\in R/I$ is an additive map of the ring $R/I$. We can easily see that $\bar{\delta}$ is a $\bar{\sigma}$-derivation of the ring $R/I$.

Theorem 3.7   Let $I$ be a $(\sigma, \delta)$-stable and weak symmetric $(\sigma, \delta)$-ideal of $R$. If $I\subseteq{\rm nil}(R)$, then $R/I$ is a weak symmetric $(\bar{\sigma}, \bar{\delta})$-ring if and only $R$ is a weak symmetric $(\sigma, \delta)$-ring.

Proof   Suppose $\bar{a}\bar{b}\bar{\sigma}(\bar{c})\in{\rm nil}(R/I)$ and $\bar{a}\bar{b}\bar{\delta}(\bar{c})\in{\rm nil}(R/I)$. Then there exist some positive integer $m, n$ such that $(ab\sigma(c))^{n}\in I$, $(ab\delta(c))^{n}\in I$. Thus $ab\sigma(c)\in{\rm nil}(R)$ and $ab\delta(c)\in{\rm nil}(R)$ since $I \subseteq{\rm nil}(R)$. Because $R$ is a weak symmetric $(\sigma, \delta)$-ring, we get $ac\sigma(b)\in{\rm nil}(R)$ and $ac\delta(b)\in{\rm nil}(R)$. It follows that $\bar{a}\bar{c}\bar{\sigma}\bar{(b)}\in{\rm nil}(R/I)$ and $\bar{a}\bar{c}\bar{\delta}\bar{(b)} \in{\rm nil}(R/I)$. Hence $R/I$ is a weak symmetric $(\sigma, \delta)$-ring.

Conversely, assume that $R/I$ is a weak symmetric $(\bar{\sigma}, \bar{\delta})$-ring. Let $ab\sigma(c)\in{\rm nil}(R)$, $ab\delta(c)\in{\rm nil}(R)$ for $a, b, c \in R$. Then $\bar{a}\bar{b}\bar{\sigma(c)}\in{\rm nil}(R/I)$, $\bar{a}\bar{b}\bar{\delta(c)}\in{\rm nil}(R/I)$. Thus we have $\bar{a}\bar{c}\bar{\sigma(b)}=\bar{ac\sigma(b)}\in{\rm nil}(R/I)$, and $\bar{a}\bar{c}\bar{\delta(b)}=\bar{ac\delta(b)}\in{\rm nil}(R/I)$ since $R/I$ is a weak symmetric $(\sigma, \delta)$-ring. So there exist some positive integers $s$ and $t$ such that $(ac\sigma(b))^{s}\in I$ and $(ac\delta(b))^{t}\in I$. Thus $ac\sigma(b)\in{\rm nil}(I)$ and $ac\delta(b)\in{\rm nil}(I)$. Therefore, $R$ is a weak symmetric $(\sigma, \delta)$-ring.

Corollary 3.8  Let $\sigma$ be an endomorphism and $I$ a weak symmetric $\sigma$-ideal of $R$. If $I\subseteq{\rm nil}(R)$, then $R/I$ is a weak symmetric $\bar{\sigma}$-ring if and only $R$ is a weak symmetric $\sigma$-ring.

Corollary 3.9  Let $\delta$ be a derivation and $I$ a weak symmetric $\delta$-ideal of $R$. If $I\subseteq{\rm nil}(R)$, then $R/I$ is a weak symmetric $\bar{\delta}$-ring if and only $R$ is a weak symmetric $\delta$-ring.

Corollary 3.10  Let $I$ be a weak symmetric ideal of $R$. If $I\subseteq{\rm nil}(R)$, then $R/I$ is a weak symmetric ring if and only $R$ is a weak symmetric ring.

According to Chen et al. [5], a ring $R$ is called weakly 2-primal if the set of nilpotent elements in $R$ coincides with its Levitzki radical, that is, $nil(R)=L$-rad($R$). Semicommutative rings, $2$-primal rings [9] and locally $2$-primal rings [6] are weakly 2-primal rings, and weakly $2$-primal rings are NI-ring.

Lemma 3.11  If $R$ is a weakly 2-primal ring and $f(x)=a_{0}+a_{1}x +\cdots+a_{n}x^{n}\in R[x]$. Then $f(x)\in {\rm nil}(R[x])$ if and only if $a_{i}\in{\rm nil}(R)$ for each $0\leq i\leq n$. that is, we have

$ {\rm nil}(R[x])={\rm nil}(R)[x]. $

Proof  Suppose that $f(x)=a_{0}+a_{1}x +\cdots+a_{n}x^{n}\in R[x]\in{\rm nil}(R[x])$. Then by [7], Proposition 1.3, we obtain $a_{i}\in{\rm nil}(R)$ for each $0\leq i\leq n$, and so ${\rm nil}(R[x])\subseteq {\rm nil}(R)[x]$. Now assume that

$ f(x)=a_{0}+a_{1}x +\cdots+a_{n}x^{n}\in R[x]\in{\rm nil}(R)[x]. $

Consider the finite subset $\{a_{0}, a_{1}, \cdots, a_{n}\}$. Since $R$ is weakly 2-primal and hence ${\rm nil}(R)=L$-rad$(R)$. Then the subring $\langle a_{0}, a_{1}, \cdots, a_{n} \rangle$ of $R$ generated by $\{a_{0}, a_{1}, \cdots, a_{n}\}$ is nilpotent, so there exists a positive integer $k$ such that any product of $k$ elements $a_{i1}a_{i2}\cdots a_{ik}$ from $\{a_{0}, a_{1}, \cdots, a_{n}\}$ is zero. Hence we obtain that $f(x)^{k+1}=0$ and so $f(x)\in {\rm nil}(R[x])$. Thus, we have ${\rm nil}(R[x])={\rm nil}(R)[x]$.

Let $\sigma$ be an endomorphism and $\delta$ a $\sigma$-derivation of $R$. Then the map $\bar{\sigma}: R[x]\longrightarrow R[x]$ defined by $\bar{\sigma}(\sum\limits_{i=0}^{m}a_ix^{i})=\sum\limits_{i=0}^{m} \sigma(a_i)x^{i}$ is an endomorphism of the polynomial ring $R[x]$ and clearly this map extends $\sigma$, and the $\sigma$-derivation $\delta$ of $R$ is also extended to $\bar{\delta}: R[x]\longrightarrow R[x]$ defined by $\bar{\delta}(\sum\limits_{i=0}^{m}a_ix^{i})=\sum\limits_{i=0}^{m} \delta(a_i)x^{i}$. We can easily see that $\bar{\delta}$ is a $\bar{\sigma}$-derivation of the ring $R[x]$.

Theorem 3.12  Let $R$ be a weakly $2$-primal ring, $\sigma$ an endomorphism and $\delta$ a $\sigma$-derivation of $R$. Then $R$ is a weak symmetric $(\sigma, \delta)$-ring if and only if $R[x]$ is a weak symmetric $(\bar{\sigma}, \bar{\delta)}$-ring.

Proof  Since any subring $S$ with $\sigma(S)\subseteq S$, $\delta(S)\subseteq S$ of a (left) weak symmetric $(\sigma, \delta)$-ring is also a (left) weak symmetric $(\sigma, \delta)$-ring. Thus it is easy to verify that if $R[x]$ is a weak symmetric $(\bar{\sigma}, \bar{\delta})$-ring, then $R$ is a weak symmetric $(\sigma, \delta)$-ring.

Conversely, assume that $R$ is a weak symmetric $(\sigma, \delta)$-ring. Let $f(x)=a_{0}+a_{1}x +\cdots+a_{n}x^{n}, g(x)=b_{0}+b_{1}x +\cdots+b_{m}x^{m}$, and $h(x)=c_{0}+c_{1}x +\cdots+c_{l}x^{l}$ $\in R[x]$ with $fg\bar{\delta} (h)\in {\rm nil}(R[x])$. Then we have the following equations by Lemma 3.11:

$ a_{0}b_{0}\delta(c_{0})=\Delta_{0}\in{\rm nil}(R); $

(3.1)

$ a_{0}b_{0}\delta(c_{1})+a_{0}b_{1}\delta(c_{0})+a_{1}b_{0}\delta(c_{0})=\Delta_{1}\in {\rm nil}(R); $

(3.2)

$ a_{0}b_{0}\delta(c_{2})+a_{0}b_{1}\delta(c_{1})+a_{0}b_{2}\delta(c_{0})+a_{1}b_{0}\delta(c_{1})+a_{1}b_{1}\delta(c_{0})+a_{2}b_{0}\delta(c_{0})\nonumber\\ =\Delta_{2}\in {\rm nil}(R); $

(3.3)

$ \;\;\;\;\;\;\;\;\;\vdots\nonumber\\ a_{0}b_{0}\delta(c_{n-1})+a_{0}b_{1}\delta(c_{n-2})+\cdots+a_{n-2}b_{1}\delta(c_{0})+ a_{n-1}b_{0}\delta(c_{0})=\Delta_{n-1}\in{\rm nil}(R); $

(3.4)

$ a_{0}b_{0}\delta(c_{n})+a_{0}b_{1}\delta(c_{n-1})+\cdots+a_{n-1}b_{1}\delta(c_{0})+a_{n}b_{0}\delta(c_{0})=\Delta_{n}\in {\rm nil}(R). $

(3.5)

Since $R$ is NI, ${\rm nil}(R)$ is an ideal of $R$. eq. (3.1) implies $\delta(c_{0})a_{0}b_{0}\in {\rm nil}(R), b_{0}\delta(c_{0})a_{0}\in{\rm nil}(R)$. If multiply eq. (3.2) on the left side by $b_{0}\delta(c_{0})$, then we have $b_{0}\delta(c_{0})a_{0}b_{0}\delta(c_{1})\in {\rm nil}(R), b_{0}\delta(c_{0})a_{0}b_{1}\delta(c_{0})\in{\rm nil}(R)$. It implies that $b_{0}\delta(c_{0})a_{1}b_{0}\delta(c_{0})\in {\rm nil}(R)$ and $a_{1}b_{0}\delta(c_{0})\in{\rm nil}(R)$. So we obtain that

$ \begin{align} \label{eq14} \ a_{0}b_{0}\delta(c_{1})+a_{0}b_{1}\delta(c_{0})=\Delta_{1}^{'}\in {\rm nil}(R). \end{align} $

(3.6)

If multiply eq. (3.6) on the right side by $a_{0}b_{0}$, then we have $a_{0}b_{1}\delta(c_{0})a_{0}b_{0}\in {\rm nil}(R)$, $a_{0}b_{0}\delta(c_{1})a_{0}b_{0}\in{\rm nil}(R)$, and hence $a_{0}b_{1}\delta(c_{0})\in {\rm nil}(R), a_{0}b_{0}\delta(c_{1})\in{\rm nil}(R)$.

If multiply eq. (3.3) on the right side by $a_{0}b_{0}, a_{0}b_{1}, a_{0}b_{2}$, and $a_{1}b_{0}$, respectively, then we obtain $a_{0}b_{0}\delta(c_{2}), a_{0}b_{1}\delta(c_{1}), a_{0}b_{2}\delta(c_{0}), a_{2}b_{0}\delta(c_{0}), a_{1}b_{0}\delta(c_{1}), a_{1}b_{1}\delta(c_{0})\in {\rm nil}(R)$ in turn.

Inductively assume that $a_{i}b_{j}\delta(c_{k})\in{\rm nil}(R) {\rm for} i+j+k\leq n-1.$ We apply the above method to eq. (3.5). First, If multiply eq. (3.5) on the left side by $b_{0}\delta(c_{0})$, then we have $a_{n}b_{0}\delta(c_{0})\in {\rm nil}(R)$ by the induction hypotheses, and

$ \begin{align} \label{eq15} \ a_{0}b_{1}\delta(c_{n-1})+a_{0}b_{2}\delta(c_{n-2})+\cdots+a_{n-1}b_{1}\delta(c_{0})=\Delta_{n}^{''}\in {\rm nil}(R). \end{align} $

(3.7)

If we multiply eq. $(15)$ on the right side by $a_{0}b_{1}$, it gives $a_{0}b_{1}\delta(c_{n-1})\in{\rm nil}(R)$, and

$ \begin{align} \label{eq16} \ a_{0}b_{2}\delta(c_{n-2})+a_{0}b_{3}\delta(c_{n-3})+\cdots+a_{n-1}b_{1}\delta(c_{0})=\Delta_{n}^{'''}\in {\rm nil}(R). \end{align} $

(3.8)

If multiply eq. (3.8) on the right side by $a_{0}b_{2}$, $a_{0}b_{3}, \cdots, a_{n-1}b_{1}$, respectively, then we obtain $a_{0}b_{2}\delta(c_{n-2})\in {\rm nil}(R)$, $a_{0}b_{3}\delta(c_{n-3})\in {\rm nil}(R), \cdots, a_{n-1}b_{1}\delta(c_{0})\in{\rm nil}(R)$ in turn. By induction, this shows that $a_{i}b_{j}\delta(c_{k})=0$ for all $i, j$ and $k$ with $i+j+k=n$, and hence $a_{i}c_{k}\delta(b_{j})\in{\rm nil}(R)$, for all $\ i, j, k$ with $i+j+k\leq n$ since $R$ is a weak symmetric $(\sigma, \delta)$-ring. Since the coefficients of $fh\bar{\delta}(g)$ can be written as sums $\sum a_{i}c_{k}\delta(b_{j})$ and $nil(R)$ is an ideal of $R$, this yields $fh\bar{\delta}(g)\in{\rm nil}(R)$ by Lemma 3.11.

Similarly, if $f(x)=a_{0}+a_{1}x +\cdots+a_{n}x^{n}, g(x)=b_{0}+b_{1}x +\cdots+b_{m}x^{m}$, and $h(x)=c_{0}+c_{1}x +\cdots+c_{l}x^{l}$ $\in R[x]$ with $fg\bar{\sigma}(h)\in {\rm nil}(R[x])$, by the same method as above, we can obtain $a_{i}c_{k}\sigma(b_{j})\in{\rm nil}(R)$ for all $\ i, j, k$ with $i+j+k\leq n$. This yields $fh\bar{\sigma}(g)\in{\rm nil}(R)$ by Lemma 3.11. Therefore, $R[x]$ is a weak symmetric $(\bar{\sigma}, \bar{\delta})$-ring.

Corollary 3.13  Let $R$ be a weakly $2$-primal ring and $\sigma$ an endomorphism of $R$. Then $R$ is a weak symmetric $\sigma$-ring if and only if $R[x]$ is a weak symmetric $\bar{\sigma}$-ring.

Corollary 3.14  Let $R$ be a weakly $2$-primal ring and $\delta$ a derivation of $R$. Then $R$ is a weak symmetric $\delta$-ring if and only if $R[x]$ is a weak symmetric $\bar{\delta}$-ring.

Corollary 3.15  Let $R$ be a weakly $2$-primal ring. Then $R$ is a weak symmetric ring if and only if $R[x]$ is a weak symmetric ring.

Corollary 3.16(see [15], Corollary 3.10) Let $R$ be a semicommutative ring. Then $R$ is weak symmetric if and only if $R[x]$ is weak symmetric.