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I encountered a problem:

Every finite integral domain is isomorphic to $ \mathbb{ Z }_{p} $.

I know that finite integral domain is isomorphic to a field, but I have no idea on how to construct a homomorphism to $\mathbb{Z}_{p}$ (or maybe it is wrong, but I haven come up with a counterexample).

user26857
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Peter Liu
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    take $\mathbb{Z_{3}}[i]$ for instance it is a finite integral domain with 9 elements but I don't see it being isomorphic to $\mathbb{Z_{p}}$ for any prime p, assuming p is a prime. – Shreya Dec 02 '15 at 12:43
  • @shrey Oh, yes, you are right. Here p may not be a prime, but your example $\mathbb{Z}_{3}[i]$ still works since it is not a cyclic group. – Peter Liu Dec 02 '15 at 12:46
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    This is not true. There are finite fields of each order $p^n$. These are certainly finite integral domains but they are not isomorphic to $\mathbb Z_p$ if $n>1$. – lhf Dec 02 '15 at 13:00
  • Related: http://math.stackexchange.com/questions/62548 – Watson Jan 19 '17 at 13:03

2 Answers2

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Thanks to @shery and @lhf, the conterexamples come from their comments.

A finite integral domain which actually is a (finite) field will have the order of the form ${p}^{n}$, where $p$ is a prime. Then when $n>1$, it is not isomorphic to $\mathbb{Z}_{{p}^{n}}$ since $\mathbb{Z}_{{p}^{n}}$ is not a field.

user26857
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Peter Liu
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The smallest counter example would be the integral domain $$R = \left\{ \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}, \begin{pmatrix}0 & 1\\ 1 & 1\end{pmatrix}, \begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}, \begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix} \right\}$$ under matrix multiplication and addition.

Siong Thye Goh
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