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.



Do there exist non-PIDs in which every countably generated ideal is principal?
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. 


