Domains in which no nonzero prime ideal is countably generated In my answer to Do there exist a non-PIR in which every countably generated prime ideal is principal?, I gave an example of an integral domain in which no nonzero prime ideal is countably generated.  However, my example is rather complicated, and so I have the following question:

What are some other (hopefully simpler) examples of integral domains in which no nonzero prime ideal is countably generated (besides the trivial examples of fields)?

 A: One natural example of such a ring is the ring $R$ of meromorphic functions on $\mathbb{C}$ (or any connected open subset of $\mathbb{C}$) which have only finitely many poles.  This ring is clearly a domain and not a field.  For $f\in R$, we will write $Z(f)$ for the set of zeroes of $f$.  The key fact we use is that if $f,g\in R$, then there exists $h\in (f,g)$ such that $Z(h)=Z(f)\cap Z(g)$ (see A problem about generalization of Bezout equation to entire functions).
Now suppose $I\subset R$ is a countably generated nonzero proper ideal; we will prove $I$ is not prime.  To prove this, let $f\in I$ be nonzero.  Let $C$ be a countable set of generators of $I$.  For each finite subset $F\subseteq C$, let $h_F\in I$ be such that $Z(h_F)=Z(f)\cap\bigcap_{g\in F}Z(g)$.  Since $I$ is a proper ideal, $h_F$ is not a unit, so $Z(h_F)$ is infinite.  Since $f$ is nonzero, $Z(f)$ is countably infinite.  Now since there are only countably many finite subsets $F\subseteq C$, we can diagonalize to partition $Z(f)$ into two subsets $S$ and $T$ such that $S\cap Z(h_F)$ and $T\cap Z(h_F)$ are both infinite for each $F$.  Let $f_0\in R$ be a function whose zero set is $S$ with zeroes of the same multiplicity as $f$ and let $f_1$ be a function whose zero set is $T$ with zeroes of the same multiplicity as $f$.  Then $f_0f_1$ is divisible by $f$, so $f_0f_1\in I$.
However, I claim that neither $f_0$ nor $f_1$ is in $I$, so $I$ is not prime.  Indeed, if $f_0$ were in $I$, it would be in the ideal generated by some finite subset $F\subseteq C$.  But then $f_0$ would be $0$ at all but finitely many points of $\bigcap_{g\in F}Z(g)$ (the finitely many exceptions coming from the fact that elements of $R$ can have finitely many poles to cancel the zeroes of the elements of $F$).  Since $T\cap Z(h_F)\subseteq T\cap \bigcap_{g\in F}Z(g)$ is infinite and $f_0$ has no zeroes on $T$, this is a contradiction.  Thus $f_0\not\in I$, and by a similar argument $f_1\not\in I$ as well.
(More conceptually, what is going on in this proof is that if $I$ were prime, then the collection of zero sets of its elements would give a nonprincipal ultrafilter on $Z(f)$ for any nonzero $f\in I$.  If $I$ were countably generated, this ultrafilter would be countably generated, but a nonprincipal ultrafilter cannot be countably generated.  For some related discussion and another neat place this ring pops up as a counterexample, see my answer to Is the axiom of choice necessary to prove that closed points in the Zariski topology are maximal ideals?.)
