In this paper we investigate commutativity of rings with unity satisfying any one of the properties: \[ \begin{aligned} &\lbrace 1- g(yx^{m}) \rbrace \ [yx^{m} - x^{r} f (yx^{m}) \ x^s, x] \lbrace 1- h (yx^{m}) \rbrace = 0, \\&\lbrace 1- g(yx^{m}) \rbrace \ [x^{m} y - x^{r} f (yx^{m}) x^{s}, x] \lbrace 1- h (yx^{m}) \rbrace = 0, \\&y^{t} [x,y^{n}] = g (x) [f (x), y] h (x)\ {\mathrm and} \ \ [x,y^{n}] \ y^{t} = g (x) [f (x), y] h (x) \end{aligned} \] for some $f(X)$ in $X^{2} {\mathbb Z}[X]$ and $g(X)$, $ h(X)$ in ${\mathbb Z} [X]$, 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 generalize a number of commutativity theorems established recently.
Suppose that $R$ is an associative ring with identity $1$, $J(R)$ the Jacobson radical of $R$, and $N(R)$ 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 $x,y \in R \setminus J(R)$ and (ii) $[(xy)^m + y^mx^m, x] = 0 = [(yx)^m + x^my^m, x]$, for all $x,y \in R \setminus J(R)$. This result is also valid if (i) and (ii) are replaced by (i)$^{\prime }$ $[x^m,y^m] = 0$ for all $x,y \in R \setminus N(R)$ and (ii)$^{\prime }$ $[(xy)^m + y^m x^m, x] = 0 = [(yx)^m + x^m y^m, x]$ for all $x,y \in R\backslash N(R) $. Other similar commutativity theorems are also discussed.
Let $p$, $ q$ and $r$ be fixed non-negative integers. In this note, it is shown that if $R$ is left (right) $s$-unital ring satisfying $[f(x^py^q) - x^ry, x] = 0$ ($[f(x^py^q) - yx^r, x] = 0$, respectively) where $f(\lambda ) \in {\lambda }^2{\mathbb Z}[\lambda ]$, then $R$ is commutative. Moreover, commutativity of $R$ is also obtained under different sets of constraints on integral exponents. Also, we provide some counterexamples which show that the hypotheses are not altogether superfluous. Thus, many well-known commutativity theorems become corollaries of our results.