If $a^2=a\enspace\forall~a\in R$, then the ring $R$ is commutative

Question

I think of the following proof:

Let $a, b\in R$. Then $(ab)^2=ab$ or, $(ab)(ab)=a^2b^2$ or, $a(ba)b=a(ab)b$ or, $ba=ab$ (by left and right cancellation laws). Hence $R$ is commutative.

Is this proof correct? Justify.

Answer 1

Incorrect. You don't have cancellations laws in a general ring.

For example, $\overline{2}.\overline{3} = 0$ in $\mathbb{Z}/6\mathbb{Z}$, yet you can't cancel.