Commutative ring

[Grigorios Kostakos]

Let $R$ a ring in which $x^3 = x$, for every $x\in R$. Prove that the ring $R$ is commutative.


P.S. Herstein's Topics has this with one $\ast$.

[Papapetros Vaggelis]

Hi Grigorios.

Let us prove the following :

Suppose that \(\displaystyle{\left(R,+,\cdot\right)}\) is a ring such that

\(\displaystyle{\left(\exists\,n\geq 2\right)\,\left(\forall\,x\in R\right)\,,x^n=x}\). Then,

1. If \(\displaystyle{a\,,b\in R}\) such that \(\displaystyle{a\,b=0}\), then,

\(\displaystyle{b\,a=(b\,a)^n=b\,(a\,b)^{n-1}\,a=0}\).

2. If \(\displaystyle{a^2=k\,a}\) for \(\displaystyle{a\in R}\) and \(\displaystyle{k\in\mathbb{Z}}\), then

\(\displaystyle{k\,a\in Z(R)}\).

Indeed, \(\displaystyle{0=(k\,a)\,x-a^2\,x=a(k\,x-a\,x)\stackrel{1}{=}(k\,x-x\,a)\,a\,,\forall\,x\in R}\), so,

\(\displaystyle{x\,(k\,a)-a\,x\,a=(k\,x-a\,x)\,a=0=(k\,a)\,x-a\,x\,a\,,\forall\,x\in R}\), that is

\(\displaystyle{x\,(k\,a)=(k\,a)\,x\,,\forall\,x\in R\implies k\,a\in Z(R)}\).

and

3.

\(\displaystyle{(a^{n-1})^2=a^{2\,n-2}=a^{n}\,a^{n-2}=a\,a^{n-2}=a^{n-1}\stackrel{(2)}{\implies} a^{n-1}\in Z(R)}\).

For our problem,

\(\displaystyle{(a^2)^2=a^4=a^3\,a=a\,a=a^2\,,\forall\,a\in R}\), so \(\displaystyle{a^2\in Z(R)\,,\forall\,a\in R}\).

Finally, for each \(\displaystyle{x\,,y\in R}\) holds

\(\displaystyle{\begin{aligned} x\,y&=(x\,y)^3\\&=x(y\,x)^2\,y\\&=(y\,x)^2\,x\,y\\&=y\,x\,y\,x^2\,y\\&=y\,x\,x^2\,y\,y\\&=y\,x^3\,y^2\\&=y\,x\,y^2\\&=y\,y^2\,x\\&=y^3\,x\\&=y\,x\end{aligned}}\)

and the ring \(\displaystyle{R}\) is commutative.