So I've seen the classic proof that $R[x]$ is a PID if and only if $R$ is a field, using the contrapositive in a way modeled after the proof that $(2, x) \subset \mathbb{Z}[x]$ is not principally generated. The first step seems to be noting that if $R[x]$ is an integral domain then so is $R$. Now, I am wondering if $R[x]$ is not necessarily an integral domain but still is a principal ideal ring, is there still anything interesting we can say about $R$?
Thanks.
Every quotient ring of a principal ideal ring $A$ is a principal ideal ring, since ideals in $A/I$ are $J/I$ where $I \subset J \subset A$. Since $R \cong R[x]/(x)$, if $R[x]$ is a principal ideal ring then so is $R$.
The same argument shows that when $R[x]$ is Noetherian, so is $R$.
This question is a duplicate: see this question and this question.
The best answer, for a commutative ring $R$, is that $R[x]$ is a principal ideal ring if and only if $R$ is isomorphic to a finite product of fields.
