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

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?

• Why not take $x$ instead of $x^2$? – Ben Mar 2 '15 at 23:49
• I was thinking that the ideal generated by x was maximal. – OLP Mar 2 '15 at 23:51
• @OLP: $\mathbb{Z}[x]/<x>\simeq \mathbb{Z}$ which is not a field, so $<x>$ is not maximal. – walkar Mar 2 '15 at 23:53
• @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. – Robert Cardona Mar 2 '15 at 23:53
• 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. – OLP Mar 2 '15 at 23:57

Try $D=K[x,y]$ and $p=x$, where $K$ is an integral domain (a field, for instance).