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Find an integral domain $D$ containing an irreducible element $p$ such that $D/\langle p \rangle$ is not a field.

I'm working on homework. I think I need to find p such that the ideal generated by $p$ is not maximal. So I think I need an integral domain which is not a PID. If $p$ did not have to be irreducible, I think I could use the ideal genrated by $x^2 \in \mathbb Z[x]$. Any suggestions?

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    $\begingroup$ Why not take $x$ instead of $x^2$? $\endgroup$ – Ben Mar 2 '15 at 23:49
  • $\begingroup$ I was thinking that the ideal generated by x was maximal. $\endgroup$ – OLP Mar 2 '15 at 23:51
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    $\begingroup$ @OLP: $\mathbb{Z}[x]/<x>\simeq \mathbb{Z}$ which is not a field, so $<x>$ is not maximal. $\endgroup$ – walkar Mar 2 '15 at 23:53
  • $\begingroup$ @OLP, it depends on what $k$ is. If $k = \mathbb Z$ and you're looking at $k[x]$, $(x) \subsetneq (2, x)$ and so it's not maximal. $\endgroup$ – Robert Cardona Mar 2 '15 at 23:53
  • $\begingroup$ Thanks, both of you. This stuff is all so new. I, still learning to use the higher level theorems and to stop thinking in terms of manipulating elements. $\endgroup$ – OLP Mar 2 '15 at 23:57
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Try $D=K[x,y]$ and $p=x$, where $K$ is an integral domain (a field, for instance).

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