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

Let $R$ be a ring with identity such that there is a positive integer $n\geq 2$ for which $r^n=r$ for all $r\in R$. Prove $R$ is commutative.

I had proven before that If $n=2$ it is commutative as follows:

$r+s=(r+s)^2=r^2+rs+sr+r^2=r+rs+sr+s\implies 0=rs+sr\implies sr=-rs$

On the other hand $-r=(-1)r=(-1)^2r=r$.

So $sr=-rs=rs$ as desired.

I seem to be stumped even with $n=3$.

• I assume you require $n\ge2$, otherwise the condition is trivial (and the claim is untrue). – Jason Feb 18 '15 at 1:57