## Rings satisfying x^4 = x are commutative

Posted: April 19, 2012 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , ,

Let $R$ be a ring, which may or may not have $1.$ We proved in here that if $x^3=x$ for all $x \in R,$ then $R$ is commutative.  A similar approach shows that if $x^4=x$ for all $x \in R,$ then $R$ is commutative.

Problem. Prove that if $x^4=x$ for all $x \in R,$ then $R$ is commutative.

Solution. Clearly $R$ is reduced, i.e. $R$ has no nonzero nilpotent element. Note that $2x=0$ for all $x \in R$ because $x=x^4=(-x)^4=-x.$ Hence $x^2+x$ is an idempotent for every $x \in R$ because

$(x^2+x)^2=x^4+2x^3+x^2=x^2+x.$

Thus $x^2+x$ is central for all $x \in R,$ by Remark 3 in this post.  Therefore $(x^2+y)^2+x^2+y$ is central for all $x,y \in R.$ But

$(x^2+y)^2+x^2+y=x^2+x+y^2+y+ x^2y+yx^2$

and hence $x^2y+yx^2$ is central. Thus $(x^2y+yx^2)x^2=x^2(x^2y+yx^2),$ which gives $xy=yx. \ \Box$

1. Esam Kamal says:

But What about the following questin?
If R is a ring such that a^5=a for every a in R. Is it commutative?
Also What about the following question?
If R is a ring such that a^n=a for every a in R where n is a natural number. Is it commutative?

• Yaghoub Sharifi says:
2. Otman says:

The same for rings where x^2=x for every x in the ring.

• Yaghoub says:

Yes, but, as I’m sure you know, the case $x^2=x$ is trivial.