I know that if $A$ is a PID and $I$ an ideal, then $A/I$ need not be a PID, since it's not even a domain unless $I$ is prime.
However, I can't quite seem to find my mistake in the following "proof" that the opposite result holds.
Since $A$ is a PID, it's a Noetherian module over itself, hence $A/I$
is an Noetherian $A$-module as well. This implies every submodule of
$A/I$ is finitely generated, which is the same as saying every ideal
of $A/I$ is finitely generated. Since $A$ is a PID, every such ideal
is generated by a single element.
Where am I going wrong?
Best Answer
That proof only proves that every ideal of $A/I$ is principal. It never says anything about $A/I$ being a domain.
(Also, the conclusion that every ideal of $A/I$ is principal is correct, but the logic used at the end of your proof is very unclear. How does every ideal of $A/I$ being finitely generated and $A$ being a PID mean that every ideal of $A/I$ is generated by a single element? You can instead reach this inclusion by observing that if $J\subseteq A/I$ is an ideal, then it is $J'/I$ for some ideal $J'\subseteq A$, and then a generator of $J'$ gives a generator of $J$.)