Let $R$ be a finite ring and suppose that for each $x\in R$ there is some $n(x) > 1$ so that $x^{n(x)} = x.$ Show using the Artin-Wedderburn theorem and Wedderburn's theorem on finite division rings that $R$ is commutative.
The version of Wedderburn's theorem I'm using essentially states that finite division rings are finite fields.
The version of the Artin-Wedderburn theorem I'm using states that if $R$ is an artinian ring that has no nonzero nil two-sided ideals, then $R$ is isomorphic to a direct product of matrix rings $\prod_{i=1}^s M_{n_i}(D_i),$ where the $D_i$ are division rings. A nil two-sided ideal is a two-sided ideal $I$ such that for every $x\in I,$ there exists some $n(x) > 0$ so that $x^{n(x)}= 0.$
I think that the division rings $D_i$ given in the Artin-Wedderburn theorem should be finite and thus by Wedderburn's theorem they must be fields, which are commutative. If I can prove that $R$ is isomorphic to $\prod_{i=1}^s M_{n_i}(D_i)$ then commutativity easily follows, but I'm not sure how to show that $R$ is an artinian ring and has no nonzero nil two-sided ideals. I think it might be useful to assume $R$ is not artinian and get a contradiction and assume $R$ has a nonzero nil two-sided ideal and get a contradiction. For instance, suppose $R$ has a nonzero nil two-sided ideal $I$. Then picking a nonzero element $x$ of $I$, we have that $\exists n > 0$ so that $x^n = 0.$ Let $n$ be minimal so that $x^n = 0.$ Then for every $k\ge n,x^k = x^{n}x^{k-n} = 0.$ Since $x\in R, \exists n_2 > 1$ so that $x^{n_2} = x,$ so we must have $n_2 < n.$ Write $n = qn_2 + r$ for some $q > 0, 0\leq r < n_2.$ Then $x^n = x^{qn_2 + r} = (x^{n_2})^q x^r = x^{q+r} = 0$. But I'm not sure how to get a contradiction from this.
Best Answer
If $R$ is finite, it must be Artinian because each element of a descending chain must have (at least) one element less and therefore it has to terminate.
For the second part of your argument, $q+r < qn_2 + r = n$ (since $n_2 > 1$) and since you've shown $x^{q+r} = 0$ then that contradicts the minimality of $n$.