TITLE: Do there exist non-PIDs in which every countably generated ideal is principal?
QUESTION [32 upvotes]: The title pretty much says it all: suppose $R$ is a commutative integral domain such that every countably generated ideal is principal.  Must $R$ be a principal ideal domain?
More generally: for which pairs of cardinals $\alpha < \beta$ is it the case that: for any commutative domain, if every ideal with a generating set of cardinality at most $\alpha$ is principal, then any ideal with a generating set of cardinality at most $\beta$ is principal?
Examples: Yes if $2 \leq \alpha < \beta < \aleph_0$; no if $\beta = \aleph_0$ and $\alpha < \beta$: 
take any non-Noetherian Bezout domain (e.g. a non-discrete valuation domain).
My guess is that valuation domains in general might be useful to answer the question, although I promise I have not yet worked out an answer on my own.

REPLY [6 votes]: Sorry to dig up an old question, but in case anybody else randomly lands here, here's a quick side note about a way that this can be generalized.
Theorem:  If every countably generated ideal of a ring $R$ is finitely generated, then $R$ is Noetherian.  Hence, if $n < \infty$ and every countably generated ideal is $n$-generated, then every ideal is $n$-generated.
Proof:  By contrapositive.  If $R$ is not Noetherian, then we can make an infinite properly ascending chain $I_1 \subsetneq I_2 \subsetneq \cdots$ of finitely generated ideals.  The union of this chain is a countably generated ideal, and it cannot be finitely generated, because that would cause the chain to terminate at some point.