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Let $p$ be a prime. Pick each correct statement from below. Upto isomorphism,

  1. there are exactly two abelian groups of order $p^2$.
  2. there are exactly two groups of order $p^2$.
  3. there are exactly two commutative rings of order $p^2$.
  4. there are exactly one integral domain of order $p^2$.

My try: The number of abelian groups up to isomorphism, of order $p^2$ is equal to the partition of $2$. So 1 is correct. Again any group of order $p^2$ is abelian, hence 2 is also correct. But I am confused with 3 and 4. In case of 4, we know that there is a unique field of order $p^2$, and a field is an integral domain. But will this imply 4? Any hints/ answer will be helpful. Thanks.

Kushal Bhuyan
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    For 4., a finite integral domain is a field. – Watson Sep 02 '16 at 14:26
  • Oh yes, missed that. thanks @Watson. What about 3? – Kushal Bhuyan Sep 02 '16 at 14:27
  • You may want to try your hand on proving that there are exactly three commutative rings with identity having $p^2$ elements (up to isomorphism). – egreg Sep 02 '16 at 15:00
  • hmmm any hints @egreg? I definitely want to prove. – Kushal Bhuyan Sep 02 '16 at 15:36
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    A finite commutative ring is artinian, so it is a product of local rings. Either there is a unique factor or two; in the latter case we get $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$; in the former case, either the ring is a field (and is the unique Galois field of order $p^2$) or it has a nontrivial maximal ideal (and is $\mathbb{Z}/p^2\mathbb{Z}$). – egreg Sep 02 '16 at 15:42

1 Answers1

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For $4.$, a finite integral domain is a field.

For $3.$, think about $\Bbb Z/p^2\Bbb Z$, $\Bbb Z/p\Bbb Z \times \Bbb Z/p\Bbb Z$ and $\Bbb F_{p^2}$.

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Watson
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