# Difference between PID and principal ideal ring

All rings are commutative, associative and with 1.

Wikipedia states that the difference between PID and Principal Ideal Ring is that the former has to be integral domain while the latter does not.

It is well known that in PID (in the sense of Wikipedia) every prime is maximal. Can someone point out an example of Principal Ideal Ring $A$ and some prime ideal $I\subset A$ which is not maximal?

• See here, particularly the link to MO. This may or may not be a complete answer to your question, but it seems relevant. – rogerl Sep 10 '14 at 15:44
• The ring $\mathbb{Z} \times \mathbb{Z}$ is a PIR and the ideal generated by $(0,1)$ is prime and not maximal. – Crostul Sep 10 '14 at 15:49

Not quite: every nonzero prime ideal in a PID is a maximal ideal. There are, in fact, PIR's where every prime is maximal. One such example is $\Bbb Z/4\Bbb Z$. A PID with all primes maximal is already a field.
Can someone point out an example of Principal Ideal Ring A and some prime ideal $I\subseteq A$ which is not maximal?
Well, $\Bbb Z$ is already such an example, since $\{0\}$ is prime, but if you want other examples with nonzero nonmaximal primes, you can just take the product ring of any finite number of copies of $\Bbb Z$ (as Crostul already employed in the comments.)
$I=\{0\}\times \Bbb Z$ is a prime ideal of $A=\Bbb Z\times \Bbb Z$ since $A/I\cong \Bbb Z$ is a domain.