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

Question

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?

@OLP: $\mathbb{Z}[x]/\simeq \mathbb{Z}$ which is not a field, so $$ is not maximal. — walkar, Mar 02 '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 02 '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 02 '15 at 23:57

score 3 · Accepted Answer · answered Mar 02 '15 at 23:46

3

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

answered Mar 02 '15 at 23:46

lhf

216,483

Thanks! Would you happen to have suggestions that involve adjoining only one element? I'm not so familiar with adjoins more than one element and our class hasn't touched on it much. I am grateful for the help, in any case. – OLP Mar 02 '15 at 23:48
@OLP, see http://en.wikipedia.org/wiki/Irreducible_element#Example and http://math.stackexchange.com/q/106681/589. – lhf Mar 02 '15 at 23:58

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

1 Answers1

Linked