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I know that an ideal P in $\mathbb{Z}_n$ is prime if and only if $\mathbb{Z}_n/P$ is an integral domain and an ideal m in $\mathbb{Z}_n$ is maximal if and only if $\mathbb{Z}_n$/m is a field. I think I've figured out that $\mathbb{Z}_p$ where p is a prime that divides n make up the maximal ideals.

I have no idea how to figure out which are prime. Help?

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Hint: Write $m=\Pi_{i=1}^{i=n}p_i^{n_i}$ where $p_i$ is a prime number, the prime and maximal ideals are generated by the class of $p_1^{n_1}..p_i^{n_i-1}..p_n^{i_n}$. The Chinese remainder theorem implies that $\mathbb{Z}_m\simeq \mathbb{Z}/p_1^{n_1}\times...\times \mathbb{Z}/p_1^{n_1}$ and $\mathbb{Z}_m/p_1^{n_1}..p_i^{n_i-1}..p_n^{i_n}\simeq\mathbb{Z}/p_i$.

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  • $\begingroup$ so what are the prime ideals? I don't really follow this $\endgroup$ – John Smith Dec 17 '17 at 2:31
  • $\begingroup$ the prime and maximal are the same $\endgroup$ – Tsemo Aristide Dec 17 '17 at 2:33

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