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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.

Bernard
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gete
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    Where is your second equality coming from? Because the easiest proof of it that I can think of is to first prove that $R$ is commutative.

    It's also not generally the case that rings are cancellative: indeed, if such a ring is cancellative, then cancelling $a$ (on either side) in $a^2 = a$ gives $a = 1$, so your ring is trivial.

     – user3482749 Dec 10 '19 at 09:58
  • Compare https://math.stackexchange.com/q/2211767, https://math.stackexchange.com/q/10274. – Martin R Dec 10 '19 at 10:03
  • "Is this proof correct? Justify." sounds like this proof is not yours but rather from an exercise book. In that case what do you think about this proof? – Arnaud Mortier Dec 10 '19 at 10:03

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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.

J. De Ro
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