Are quotients of a PID by non-prime ideals ever a PID?

Question

I've proved that any quotient of a PID by a prime ideal is again a PID as an exercise, and got into thinking about quotients by non-prime ideals. Are they ever a PID? More concretely, say $R$ is a PID and $I$ is an ideal of $R$ which is not prime. Is $R/I$ ever a PID?

Testing with $R=\mathbb{Z}$ and $I=4\mathbb{Z}$ this isn't the case, since $\mathbb{Z}/4\mathbb{Z}$ isn't even an integral domain (and by definition cannot be a proper PID).

In rough terms the proof of the earlier statement follows: Let $R$ be a PID and let $P$ be a prime ideal. If $P$ is the zero ideal, then $R/P$ is isomorphic to $R$ and thus a PID. If $P$ is a nonzero prime ideal of $R$ it must be maximal. Thus $R/P$ is a field, which is trivially a PID.

I have a hard time imagining this proof is "reversible", since there should be more cases of $R/P$ being a PID but no field. Instead I got about as far as this:

Suppose $I$ is an non-prime ideal of the PID $R$. Thus there must exists some $(ab)\in I$ such that $a\notin I$ and $b\notin I$. Since $R$ contains no zero divisors, we must have that $a$ and $b$ are nonzero. Thus $R/I$ contains elements $a+I$ and $b+I$. However $$(a+I)(b+I)=(ab)+I=I$$ since $(ab)\in I$. Thus $R/I$ contains zero divisors, and can't be a PID.

Is this correct? How could one think otherwise? Thanks in advance!

Answer 1

True, quotients by non-prime ideals have zero-divisors, so are not domains at all. Nevertheless, quotients of principal ideal rings $R$ by arbitrary ideals $I$ are still principal ideal rings, by the same proof idea: given an ideal $J$ in $R/I$, its inverse image $q^{-1}(J)$ under the quotient map $q:R\to R/I$ is an ideal in $R$, so is of the form $R\cdot r_o$ for some $r_o$. Then $J$ is generated by $q(r_o)$.

Answer 2

As noted in the comments; if $R$ is a PID and $I$ is an ideal of $R$ that is not prime, then $R/I$ is not an integral domain, so certainly $R/I$ is not a PID.