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.
Best Answer
No such ring exists.
Suppose otherwise. Let $I$ be a non-principal ideal, generated by a collection of elements $f_\alpha$ indexed by the set of ordinals $\alpha<\gamma$ for some $\gamma$. Consider the set $S$ of ordinals $\beta$ with the property that the ideal generated by $f_\alpha$ with $\alpha<\beta$ is not equal to the ideal generated by $f_\alpha$ with $\alpha\leq \beta$.
$I$ is generated by the $f_\beta$ with $\beta \in S$, so if $S$ is finite, then $I$ is finitely generated and thus is principal.
On the other hand, if $S$ is infinite, then take a countable subset $T= \{\beta_1<\beta_2<\dots\}$ of $S$. If the ideal generated by the corresponding set of $f_\beta$'s were principal, its generator would have to be in some $\langle f_{\beta_k} \mid k\leq i \rangle$ for some $i$ (since any element of $\langle f_{\beta}\mid \beta \in T\rangle$ is a finite combination of $f_\beta$'s and therefore lies in some such ideal). Now no $\beta_j$ with $j>i$ could be in $T$.
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The same argument shows that all rings for which any countably generated ideal is finitely generated, have all their ideals finitely generated.
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Corrected thanks to David's questions.