Proving a ring in which $r^n=r$ for all $r$ is commutative.

Solution 1:

The proof is found in the literature but is perhaps too long for an answer here on math.SE. We have long suffered from a lack of a canonical answer for this question, so this is an attempt at an approximation to one.

Even more is true:

If for each $x$ there exists an integer $n(x)>1$ such that $x^{n(x)}-x$ is in the center of $R$, then $R$ is commutative.

There are a few proofs in the literature, ones which are probably too long for a post here.

You should consult your local library to obtain a copy of one of these (probably preferring the newest one you can get):

N. Jacobson, "Structure theory for algebraic algebras of bounded degree", Ann. of Math. 46 (1945), 695–707.

Herstein, I. N. "An elementary proof of a theorem of Jacobson", Duke Mathematical Journal 21.1 (1954): 45-48.

Rogers, Kenneth. "An elementary proof of a theorem of Jacobson", Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg. Vol. 35. No. 3. Springer Berlin/Heidelberg, 1971.

A proof of an even more general nature is also discussed as Theorem 3.2.3 in:

Herstein, Israel Nathan. Noncommutative rings. Vol. 15. American Mathematical Soc., 1994.

There are also useful related questions on this site:

1. $x^n = x$ implies commutativity, a universal algebraic proof?