Suppose that $R$ is an associative ring with identity $1$, $J\left(R\right)$ the Jacobson radical of $R$, and $N\left(R\right)$ the set of nilpotent elements of $R$. Let $m\ge 1$ be a fixed positive integer and $R$ an $m$-torsion-free ring with identity $1$. The main result of the present paper asserts that $R$ is commutative if $R$ satisfies both the conditions (i) $[x^m,y^m]=0$ for all $x,y\in R\setminus J\left(R\right)$ and (ii) $[{\left(xy\right)}^m+y^m{x}^m,x]=0=[{\left(yx\right)}^m+x^m{y}^m,x]$, for all $x,y\in R\setminus J\left(R\right)$. This result is also valid if (i) and (ii) are replaced by (i)${}^{\text{'}}$ $[x^m,y^m]=0$ for all $x,y\in R\setminus N\left(R\right)$ and (ii)${}^{\text{'}}$ $[{\left(xy\right)}^m+y^m{x}^m,x]=0=[{\left(yx\right)}^m+x^m{y}^m,x]$ for all $x,y\in R\setminus N\left(R\right)$. Other similar...

In this paper we investigate commutativity of rings with unity satisfying any one of the properties: $$\begin{array}{cc}& \{1-g\left(y{x}^m\right)\}[y{x}^m-x^{r}f\left(y{x}^m\right)x^s,x]\{1-h\left(y{x}^m\right)\}=0,\hfill \\ & \{1-g\left(y{x}^m\right)\}[x^{m}y-x^{r}f\left(y{x}^m\right)x^s,x]\{1-h\left(y{x}^m\right)\}=0,\hfill \\ & y^t[x,y^n]=g\left(x\right)[f\left(x\right),y]h\left(x\right)\mathrm{a}nd[x,y^n]y^t=g\left(x\right)[f\left(x\right),y]h\left(x\right)\hfill \end{array}$$ for some $f\left(X\right)$ in $X^2\mathbb{Z}\left[X\right]$ and $g\left(X\right)$, $h\left(X\right)$ in $\mathbb{Z}\left[X\right]$, where $m\ge 0$, $r\ge 0$, $s\ge 0$, $n>0$, $t>0$ are non-negative integers. We also extend these results to the case when integral exponents in the underlying conditions are no longer fixed, rather they depend on the pair of ring elements $x$ and $y$ for their values. Further, under different appropriate constraints on commutators, commutativity of rings has been studied. These results...

Let $m\ge 0,r\ge 0,s\ge 0,q\ge 0$ be fixed integers. Suppose that $R$ is an associative ring with unity $1$ in which for each $x,y\in R$ there exist polynomials $f\left(X\right)\in X^{2}ZZ\left[X\right],g\left(X\right),h\left(X\right)\in XZZ\left[X\right]$ such that $\{1-g\left(y{x}^m\right)\}[x,x^{r}y-x^{s}f\left(y{x}^m\right)x^q]\{1-h\left(y{x}^m\right)\}=0$. Then $R$ is commutative. Further, result is extended to the case when the integral exponents in the above property depend on the choice of $x$ and $y$. Finally, commutativity of one sided s-unital ring is also obtained when $R$ satisfies some related ring properties.