Commutative ring

Let $\mathcal{R}$ such that $x^3=x$ forall $x \in \mathcal{R}$. Prove that $\mathcal{R}$ is commutative.

Solution

We call a ring $\mathcal{R}$, J-ring (Jacobson ring), if for any $x \in \mathcal{R}$ there is a natural number $n(x) >1$ such that $x^{n(x)}=x$. (In fact, Jacobson has proven that any J-ring is commutative, for the proof you may take a look at Non-commutative Rings written by Herstein.)

Lemma 1: If $\mathcal{R}$ is a J-ring, then $N(\mathcal{R})= \{0 \}$ where $N(\mathcal{R})$ denotes the nilradical of $\mathcal{R}$.

Proof: Let $0\not= x\in N(\mathcal{R})$. Then there is a smallest natural number greater than $1$ such that $x^m=0$. Since $\mathcal{R}$ is a J-ring, there exists an $n>1$ such that $x^n=x$. Let $m=nq+r$ where $0 \leq r <n$. Therefore,

\begin{align*}
x^m &=x^{nq+r} \\
 &=(x^n)^qx^r \\
 &= x^qx^r\\
 &= x^{q+r} \\
 &=0
\end{align*}

However, $q+r<m$, which is a contradiction, since $m$ was chosen to be the smallest number satisfying $x^m=0$.

Lemma 2: Suppose that in a ring $\mathcal{R}$, $N(\mathcal{R})= \{0 \}$, then any idempotent element $a$ , that is, $a^2=a$, lies in the center $\mathcal{Z}(\mathcal{R})$.

Proof: Suppose that $x \in \mathcal{R}$. Then

\begin{align*}
(axa-ax)^2 &= (axa-ax)(axa-ax)\\
 &=axaaxa-axaxa-axaax+axax \\
 &= axaxa-axaxa-axax+axax\\
 &= 0
\end{align*}

Since $N(\mathcal{R})= \{0 \}$, then we have $axa-ax=0 \Rightarrow axa=ax$. With the same approach and by considering $(axa-xa)^2$ we will obtain $axa=xa$. Hence, $ax=xa$ and since $x$ was an arbitrary element of $\mathcal{R}$ then $a \in \mathcal{Z}(\mathcal{R})$.

Lemma 3: In a J-ring $\mathcal{R}$, we have $x^{n(x)-1} \in \mathcal{Z}(\mathcal{R}).$

Proof:
\begin{align*}
(x^{n(x)-1})^2 &=x^{2n(x)-2} \\
 &=x^{n(x)}x^{n(x)-2} \\
 &= xx^{n(x)-2} \\
 &=x^{n(x)-1}
\end{align*}

 Thus $x^{n(x)-1}$ is an idempotent element of $\mathcal{R}$ and by Lemmata 1 & 2 we get the result.

Returning to the original question we have $n=3$ and $x^2 \in \mathcal{Z}(\mathcal{R})$ for any $x \in \mathcal{R}$. Thus:

\begin{align*}
xy &=(xy)^3 \\
 &=xyxyxy \\
 &= x(yx)^2y\\
 &=(yx)^2xy \\
 &= yxyx^2y \\
 &=yx^3y^2 \\
 &=yxy^2 \\
 &=y^3x \\
 &=yx
\end{align*}

and indeed $\mathcal{R}$ is a commutative ring.