Rings satisfying x^3 = x are commutative

Posted: December 13, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , , , ,

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!).

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