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First off, by the isomorphism (correspondence) theorem, every prime ideal of $\mathbb Z[x]/(9x^2+1)$ corresponds to a prime ideal of Z[x] which contains the ideal $(9x^2+1)$. Now $(9x^2+1)$ is clearly a prime ideal since $\mathbb Z[x]$ is a UFD and in a UFD irreducible elements are prime (hence the ideal generated by the prime $9x^2+1$ is a prime ideal). So the ideal $(9x^2+1)$ is a prime ideal of $\mathbb Z[x]$ which contains the ideal $(9x^2+1)$. But this ideal is not maximal using the fact that principal ideals in $\mathbb Z[x]$ are not maximal ideals. Hence we have found a prime ideal of $\mathbb Z[x]$ which contains $(9x^2+1)$ and is not maximal, so then $\mathbb Z[x]/(9x^2+1)$ also has a prime ideal which is not maximal.

Could someone confirm or point out a mistake in my solution? Thank you.

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    $\begingroup$ How do you know that the principal ideals of $\mathbb Z[X]$ are not maximal? This result is significantly harder to prove than your exercise. $\endgroup$ – user26857 Feb 7 '18 at 7:40
  • $\begingroup$ $(9x^2+1)\subsetneq(7,9x^2+1)$, and the last one is maximal. $\endgroup$ – user26857 Feb 7 '18 at 7:42
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You are entirely right. And the non-maximal prime ideal in $\mathbb Z[x]/(9x^2+1)$, that corresponds to $(9x^2+1)\subseteq \Bbb Z[x]$, is the zero ideal.

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  • $\begingroup$ Thank you so much Arthur. Would there also exist a non zero prime ideal which is not maximal? $\endgroup$ – PaulDavis Feb 3 '18 at 18:53
  • $\begingroup$ @PaulDavis No, there isn't. Any non-zero prime ideal of $\Bbb Z[x]/(9x^2+1)$ is a maximal ideal. $\endgroup$ – Arthur Feb 3 '18 at 21:51
  • $\begingroup$ Thanks, is there an easy way to see why? What is a good strategy for finding out which ideals are prime/maximal for quotient rings in general? $\endgroup$ – PaulDavis Feb 3 '18 at 22:32
  • $\begingroup$ integral scheme has a unique generic point and in the affine case that is precisely the 0 ideal $\endgroup$ – Prince M Feb 7 '18 at 7:52

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