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I know that with $F$ a field, $F[x]/(f(x))$ is a field iff $f(x)$ is irreducible in $F$. Due to the fact that in a UFD irreducible elements are necessarily prime, we would have that $(f(x))$ is both prime and maximal.

Do there exist any rings $R$ such that $R[x]$ is not a UFD and contains nonzero ideals that are prime but not maximal?

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    $\begingroup$ $F[x]$ already contains an ideal that is prime but not maximal, namely $(0)$. You want to talk about nonzero prime ideals, in which case you want to look at Krull dimension: en.wikipedia.org/wiki/Krull_dimension $\endgroup$ – Qiaochu Yuan May 16 '13 at 5:26
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    $\begingroup$ You don't need the non-UFD condition. For example, take $\,P:=\langle\,x\,\rangle\le\Bbb Z[x]\,$ , which is a prime ideal that is not maximal in a UFD . $\endgroup$ – DonAntonio May 16 '13 at 6:01
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Take $R$ an integral domain which is not a field. Since $R[X]/(X)\simeq R$, we find that $(X)$ is a prime and non-maximal ideal. Furthermore, if $R$ is not a UFD, then $R[X]$ is not a UFD either. (Now take $R=\mathbb Z[\sqrt{-3}]$, for example.)

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  • $\begingroup$ To add to the above answer, it's worth noting that one can iterate this procedure to create a ring with arbitrarily large chains of prime ideals. So for instance, R[X,Y,Z] then has the prime ideals $(X) \subset (X,Y) \subset (X,Y,Z)$ none of which are maximal. $\endgroup$ – Alexander Aug 2 '13 at 12:51

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