Do there exist non-PIDs in which every countably generated ideal is principal? 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.
 A: 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.
A: 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 $T$.   
**
The same argument shows that all rings for which any countably generated ideal is finitely generated, have all their ideals finitely generated.  
**
Corrected thanks to David's questions.  
A: Such rings exist. Here is a lemma which might be useful in proving such rings exist:
Lemma: Given any ordered abelian group $A$, there is a valuation ring $k[[A]]$ with valuation group isomorphic to $A$.
Construction: The ring $k[[A]]$ is the ring of formal sums 
$$\sum_{i_1,i_2, \ldots, i_r=0}^{\infty} k_{i_1 i_2 \cdots i_r} t^{ i_1 a_1 + i_2 a_2+ \cdots i_r a_r}$$
where $a_1$, $a_2$, ..., $a_r$ are allowed to be any elements in $A_{>0}$. The key point to notice is that, if such a sum has nonzero leading term, then its multiplicative inverse is also such a sum. 
This lemma let's you transfer the problem from rings to groups. But my attempted construction of a group with the needed property is wrong. I'm not sure if this is salvageable.
Take $P$ to be a totally ordered set in which every countable subset has a minimal element, but which does not itself have a minimal element. (See here.) Let $A$ be the free abelian group generated by $P$, with lexicographic ordering. That is to say, $\sum_{p \in S \subset P} c(p) p$ is positive if $c(p_0)>0$ for $p_0 = \max S$. 
I think that $A$ also has the property that every countable subset has a minimal element. No, it doesn't. Consider the set $e$, $e-f$, $e-2f$, $e-3f$, etcetera, where $0 < f < e$. 
Oh well, maybe this will give somebody else a better idea.
A: The question is fully settled by Hugh Thomas' anwer, but let me mention this related interesting fact. 
Theorem. There is a ring R and ideal I on R, such that every countable subset
of I is contained in a principal subideal of I, but I is
not principal.
Proof. Let I be the ideal of nonstationary subsets of
ω1, in the power set P(ω1), which is a Boolean
algebra and hence a Boolean ring. That is, I consists of those subsets of ω1 that 
are disjoint from a closed unbounded subset of ω1. It is an elementary set-theoretic fact that the intersection of any countably many closed unbounded subsets of ω1 is still closed and unbounded, and thus the union of countably many
non-stationary sets remains non-stationary. Thus, every
countable subset of I is contained in a principal subideal
of I. But I is not principal, since the complement of any
singleton is stationary. QED
In the previous example, the ideal I is not maximal. If one assumes the existence of a measurable cardinal (a large cardinal notion), however, then the example can be made with I maximal. 
Theorem. If there is a measurable cardinal, then there is a
ring R with a maximal ideal I, such that every countable
subset of I is contained in a principal sub-ideal of I, but
I is not principal.
Proof. Let κ be a measurable cardinal, which means that
there is a nonprincipal κ-complete ultrafilter U on the
power set P(κ), which is a Boolean algebra and thus a
Boolean ring. The ideal I dual to U is also κ-complete,
which means that I is closed under unions of size less than
κ. In particular, since kappa is uncountable, this
means that the union of any countably many elements of I
remains in I, and this union set generates a principal
subideal of I containing the given countable set. The ideal I is maximal since U was an ultrafilter. QED
I'm not sure at the moment whether the situation of this last theorem requires a measurable cardinal or not, but I'll think about it.
