## Rings satisfying x^3 = x are commutative

Posted: December 13, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Throughout $R$ is a ring.

Theorem (Jacobson). If for every $x \in R$ there exists some $n > 1$ such that $x^n=x,$ then $R$ is commutative.

The proof of Jacobson’s theorem can be found in any standard ring theory textbooks. Note that $n,$ in Jacobson’s theorem, doesn’t have to be fixed, i.e. it could depend on $x.$ See this post for the proof of the theorem when $n$ is fixed. Here we only discuss a very special case of the theorem, i.e. when $n=3.$

Definitions. An element $x \in R$ is called idempotent if $x^2=x.$ The center of $R$ is

$Z(R)=\{x \in R: \ xy=yx \ \text{for all} \ y \in R \}.$

It is easy to see that $Z(R)$ is a subring of $R.$ An element $x \in R$ is called central if $x \in Z(R).$ Obviously $R$ is commutative iff $Z(R)=R,$ i.e. every element of $R$ is central.

Problem. Prove that if $x^3=x$ for all $x \in R,$ then $R$ is commutative.

Solution.  First note that $R$ is reduced, i.e. $R$ has no nonzero nilpotent element. For every $x \in R$ we have $(x^2)^2=x^4 = x^2$ and so $x^2$ is idempotent for all $x \in R.$ Hence, by Remark 3 in this post, $x^2$ is central for all $x \in R.$ Now, since

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

we have $2x=(x^2+x)^2-2x^2$ and thus $2x$ is central. Also, since

$x^2+x=(x^2+x)^3=x^6+3x^5+3x^4+x^3=4x^2+4x,$

we have $3x=-3x^2$ and so $3x$ is central. Thus $x = 3x-2x$ is central and so $R$ is commutative.  $\Box$

A similar argument shows that if $x^4=x$ for all $x \in R,$ then $R$ is commutative (see here!).