Prove that up-to isomorphism there is exactly one integral domain of order $p^2$ .

Does there exist only two non-commutative rings of order $p^2$ upto isomorphism?

We know that any group of order $p^2$ is abelian and also any abelian group of order $p^2$ is either isomorphic to $\Bbb Z_p\times \Bbb Z_p$ or $\Bbb Z_{p^2}$.

In order for the rings to be isomorphic the corresponding groups should be isomorphic.But I am unable to extend the result for the rings.

Please give some tips.

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    $\begingroup$ Any finite integral domain is a field. The only field of order $p^2$ is $\mathbb{F}_{p^2}.$ Are you sure the question is correct? $\endgroup$ – dhk628 Jul 22 '16 at 2:37
  • $\begingroup$ @dhk628: I believe in the OPs question, integral domains need not be commutative. Your remark demonstrates why the claim is untrue in the commutative case. $\endgroup$ – Alex Wertheim Jul 22 '16 at 2:38
  • $\begingroup$ @AlexWertheim Aren't integral domains defined to be commutative rings with no non-zero zero-divisors? $\endgroup$ – dhk628 Jul 22 '16 at 2:41
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    $\begingroup$ On the other hand, it's bugging me that it doesn't seem possible to me that a ring of order $p^2$ is not commutative. At least if it's unital. Either the ring is $\mathbb{Z}_{p^2}$, or else the image of $\mathbb{Z}$ is $\mathbb{F}_p$, and the ring is a quadratic $\mathbb{F}_p$-algebra, which must be commutative... $\endgroup$ – Ben Blum-Smith Jul 22 '16 at 2:50
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    $\begingroup$ @AlexWertheim apparently we really need to ask what "domain" is supposed to mean here, because even under the very loose interpretation I've given it in my solution, the result still does not work out. $\endgroup$ – rschwieb Jul 22 '16 at 4:19

Let $R$ be a finite nonzero ring, not necessarily commutative, not necessarily having identity, but having the property that $ab=0$ implies that one of $a,b$ is zero.

Let $a$ be nonzero in $R$. Then left multiplication by $a$ is a bijection of $R$ and there exists $b$ such that $ab=a$. It follows that $ab^2=ab$ and $b^2=b\neq0$ so that $b$ is a nonzero idempotent. At this question you can see why $b$ is the identity element of $R$.

By Wedderburn's little theorem, $R$ is a field. It is well-known that there is only one field of order $p^n$ for any given prime $p$ and $n>0$.

So, I don't see how the question could be correct under any interpretation, so far. Please recheck and confirm the source of your question.

  • $\begingroup$ Nice. I was beginning to wonder if the classical commutative result need even hold under these more lax assumptions, but this answer resolves that neatly. The intent/assumptions in the OP's question remains mystifying, however... $\endgroup$ – Alex Wertheim Jul 22 '16 at 4:42
  • $\begingroup$ I have checked the source and have edited the question accordingly.Sorry for the error. $\endgroup$ – Learnmore Jul 22 '16 at 5:40

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