Question

Let$R$ be a ring such that$x^3=x$ for every$x\in R$ . Prove that R is commutative.

Short answer

It is proved that $R$ is commutative.

Step-by-step solution

Definition of a commutative ring

A commutative ringis a ring $R$ that satisfies the following axiom:

$ab=ba$for every$a,b\in R$[commutative multiplicative]

Show that R   is commutative

Let us demonstrate this by proving that$xy-yx=0$ for every $x,y\in R$.

Use the identity $x^3=x$ to calculate as follows:

role="math" localid="1648450165589" $\begin{array}{rcl}\left(x+y\right)& =& {\left(x+y\right)}^3\\ & =& x^3+x^{2}y+xyz+y{x}^2+x{y}^2+yxy+y^{2}x+y^3.....\left(1\right)\end{array}$

In case of$y=x$ , this entails$2x=8x$ and$6x=0$ for every $x\in R$. It is observed that,

$\begin{array}{rcl}\left(x-y\right)& =& {\left(x-y\right)}^3\\ & =& x^3-x^{2}y-xyz-y{x}^2+x{y}^2+yxy+y^{2}x-y^3....\left(2\right)\end{array}$

Subtract equation (2) from equation (1) as follows:

$\begin{array}{rcl}2y& =& 2\left(x^{2}y+xyx+y{x}^2\right)+2y^3\\ 2y& =& 2\left(x^{2}y+xyx+y{x}^2\right)+2y\\ 0& =& 2\left(x^{2}y+xyx+y{x}^2\right)\end{array}$

Multiply left and right of the above equation and use$x^3=x$ as follows:

$2\left(xy+x^{2}yx+xy{x}^2\right)=0....\left(3\right)$

and$2\left(x^{2}yx+xy{x}^2+yx\right)=0....\left(4\right)$

Subtract equation (4) from equation (3) as follows:

$2\left(xy-yx\right)=0$

Make note of the following:

$\begin{array}{rcl}{z}^2+z& =& {\left(z^2+z\right)}^3\\ & =& z^6+3z^5+3z^4+z^3\\ & =& z^2+3z+3z^2+z\\ & =& 4\left(z^2+z\right)\end{array}$

This follows that $3\left(z^2+z\right)=0$ . With $z=x+y$, then,

$\begin{array}{rcl}3\left({\left(x+y\right)}^2+x+y\right)& =& 3\left(x^2+xy+yx+y^2+x+y\right)\\ & =& 3\left(x^2+x\right)+3\left(y^2+y\right)+3\left(xy+yx\right)\end{array}$

This follows that$3xy+3yx=0$ . It indicates $3\left(xy-yx\right)=0$because$6yx=0$ . As a result,

$\begin{array}{rcl}xy-yx& =& 3\left(xy-yx\right)-2\left(xy-yx\right)\\ & =& 0-0\\ & =& 0\end{array}$

Hence, it is proved that $R$ is commutative.