Counterexample - modules over non-Noetherian domain

Question

Does anyone know an example of a (necessarily non-Noetherian) domain $A$ and a finitely generated $A$-module $M$ with the property that $M_f$ is not free for any nonzero $f \in A$? This would provide a counterexample to a standard algebraic geometry exercise in the non-Noetherian case.

Answer 1

Actually I don't think there are any counterexamples: the proof seems to go through without the Noetherian hypothesis. Let $K$ be the fraction field of $A$ and choose a $K$-basis for $M \otimes_A K$. By clearing denominators we can choose this basis in the image of $M$, so we get a morphism $\varphi : A^{\oplus r} \to M$ which induces an isomorphism $K^{\oplus r} \cong M \otimes_A K$. By exactness of localization it follows that $\text{ker } \varphi$ and $\text{coker } \varphi$ vanish at the generic point, i.e. they are torsion. Since $A$ is a domain $A^{\oplus r}$ is torsion-free, so $\text{ker } \varphi = 0$ already. On the other hand, $\text{coker } \varphi$ is finitely generated because $M$ is, so we can find a nonzero $f \in A$ which annihilates $\text{coker } \varphi$. This is the desired $f$.