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This is a statement in an article that I am studying for my final work in the university, but I don't know why, I would like to have a proof of that: "Since the maximal ideals of Zpn[x] are precisely the ideals (p, f) with f representing an irreducible polynomial in Zp[x]." Thank you.

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    $\begingroup$ What is Zpn[X] ? $\endgroup$ Mar 30, 2011 at 15:13

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Any maximal ideal $I$ is a prime ideal, and your ring has $p^n=0$, hence $p\in I$. Then ideals containing $p$ correspond to ideals of the quotient $\mathbb Z_{p^n}[x]/(p)$, which is just $\mathbb Z_p[x]$. This is a principal ideal domain, and (in a PID) a principal ideal is maximal iff its generator is irreducible.

(Assuming “Zpn[x]” means $\mathbb Z_{p^n}[x]$, which I'd prefer to write as $(\mathbb Z/p^n\mathbb Z)[x]$.)

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    $\begingroup$ What's the reason for the down-vote? $\endgroup$ Mar 30, 2011 at 15:41
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    $\begingroup$ I suspect someone (not me) is chastizing you for answering a question inappropriate for the site. $\endgroup$ Mar 30, 2011 at 15:48
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    $\begingroup$ How do we know that $\mathbb{Z}_{p^n}$ does not denote the valuation ring of the degree $n$ unramified extension of $\mathbb{Q}_p$? $\endgroup$ Mar 30, 2011 at 16:09
  • $\begingroup$ (I mean, I'm guessing it doesn't, but I suppose you take my point...) $\endgroup$ Mar 31, 2011 at 3:09

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