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My attempt: if $p$ is prime, then $\mathbb{Z}_p$ is a field and the nonzero prime ideals have the form $(f)$, $f$ irreducible in $\mathbb{Z}_p[x]$ (is there a way to find them all?); else we can write $p=p_1^{e_1}\cdots p_n^{e_n}$ and by the Chinese Remainder Theorem we have $\mathbb{Z}/(p)\simeq \mathbb{Z}/(p_1^{e_1}) \times \dots \times \mathbb{Z}/(p_n^{e_n})$. I was planning to use it somehow...

Is it true that $\mathbb{Z}/(p)[x]\simeq \mathbb{Z}/(p_1^{e_1})[x] \times \dots \times \mathbb{Z}/(p_n^{e_n})[x]$?

If afirmative and I could find the prime ideals of each $\mathbb{Z}/(p_i^{e_i})[x]$ then the prime ideals of $\mathbb{Z}/(p)[x]$ would be $\mathbb{Z}/(p_1^{e_1})[x] \times \dots \times \mathrm{Spec}(\mathbb{Z}/(p_i^{e_i})[x])\times \dots \times \mathbb{Z}/(p_n^{e_n})[x]$. Among these, I really don't know which are maximal.

What I wrote is right?

How can I continue?

How can I find the maximals?

Thanks!

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    $\begingroup$ The first sentence is wrong: a prime ideal in $\;\Bbb F_p[x]\;$ is one that is maximal, and this means that it is generated by an irreducible polynomial over $\;\Bbb F_p\;$ . $\endgroup$ – DonAntonio Sep 30 '16 at 14:43
  • $\begingroup$ Oh, you're right. This is more complicated than I thought... $\endgroup$ – Koto Sep 30 '16 at 15:10
  • $\begingroup$ In fact it is way simpler in some sense, as if $\;F\;$ is a commutative unitary ring, then $\;F[x]\;$ is a PID iff $\;F\;$ is a field. Thus, $\;\Bbb F_p[x]\;$ is PID (in fact, an euclidean domain and also a UFD and etc.), so an element in it is irreducible iff it is prime, and then a non-zero ideal is prime iff it is maximal iff it is generated by an irreducible (or prime) polynomial. Now, yes: this gives us much more prime elements (prime=maximal ideals) as thought before, but in a sense it is also simpler. $\endgroup$ – DonAntonio Sep 30 '16 at 16:04
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    $\begingroup$ $\mathbb Z_p$ or $\mathbb F_p$? $\endgroup$ – Santiago Sep 30 '16 at 21:04
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    $\begingroup$ @LeoMito All have you done is correct. Now notice that a prime ideal in $(\mathbb Z/p^k\mathbb Z)[X]=\mathbb Z[X]/(p^k)$ contains $p$, so it is of the form $(p)$, or $(p,f)$ with $f$ irreducible modulo $p$. $\endgroup$ – user26857 Sep 30 '16 at 21:04

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