# 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 $n=3$.

Thanks in advance, regards.

PS: This is homework.

• I assume you require $n\ge2$, otherwise the condition is trivial (and the claim is untrue). – Jason Feb 18 '15 at 1:57
• Oh yes, I forgot. thanks for noticing, that is important. – Jorge Fernández Hidalgo Feb 18 '15 at 2:00
• This question has been asked here in various forms, though in full generality I think this is a bit difficult for a homework problem. Perhaps you'll find this link helpful: math.ucla.edu/~ggim/W14-110BH.pdf – Alex Wertheim Feb 18 '15 at 3:02
• See the answer in math.stackexchange.com/questions/360958/… . I have no the book of Herestein. However, in this special case might exist a simple proof. Please, ask your professor to show the solution of this homework, and share the ideas with us. – vesszabo Feb 25 '15 at 17:56
• Here is one more link, namely Herstein has a paper entitled "An elementary proof of a theorem of Jacobson", where he proves it on 4 pages: projecteuclid.org/download/pdf_1/euclid.dmj/1077465581 – Christian Lomp Jun 13 '18 at 9:41