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Every finite commutative ring $R$ must be of the form $\frac{ \mathbb{Z} }{n_1 \mathbb{Z}} \times \frac{\mathbb{Z}}{n_2 \mathbb{Z}} \times \dots \frac{\mathbb{Z}}{n_k \mathbb{Z}}$.

I am unable to find a counterexample. I heard that a field of order $4$ works, but I cannot see the reason. Can anyone explain why a field of order $4$ cannot be of the form above.

GraduateStudent
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  • You should just ask for clarification of the comment in your first post, rather than making an entirely new one just for the comment. – rschwieb Oct 12 '20 at 14:09

2 Answers2

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If there are at least two factors (with $n_k\ne1$) in the product, then for sure the product has non-trivial zero divisors, which $\Bbb F_4$ can't have. The only $\Bbb Z/n\Bbb Z$ with four elements is $\Bbb Z/4\Bbb Z$, which is not a field.

  • It might add clarity to note that "if there are at least two factors in the product" only allows the possibility $\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$. – Greg Martin Oct 12 '20 at 05:59
  • @GregMartin In truth, it isn't quite the point that I was making, but yes it's a valid remark. –  Oct 12 '20 at 06:04
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Paraphrasing Gae. S.'s answer:

  • Finite fields are finite commutative rings
  • Finite field do not have zero devisors (because they are fields)
  • Every ring of the form $\frac{\mathbb{Z}}{n_1\mathbb{Z}}\times ...\times \frac{\mathbb{Z}}{n_k\mathbb{Z}}$ has zero devisors unless it is $\frac{\mathbb{Z}}{p\mathbb{Z}}$ for some prime $p$ in which case it has $p$ elements. But it can't be $p=4$.
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