TITLE: A finitely generated, locally free module over a domain which is not projective?
QUESTION [35 upvotes]: This is a followup to a previous question
What is the right definition of the Picard group of a commutative ring?  
where I was worried about the distinction between invertible modules and rank one projective modules over an arbitrary commutative ring.  I was worrying too much, because of the following theorem [Bourbaki, Commutative Algebra, Section II.5.2, Theorem 1]:
Let $R$ be a commutative ring and $M$ a finitely generated $R$-module.  The following are equivalent:
(i) $M$ is projective.
(ii) $M$ is finitely presented and locally free in the weaker sense: $\forall \mathfrak{p} \in \operatorname{Spec}(R), \ M_{\mathfrak{p}} \cong R_{\mathfrak{p}}^{r(\mathfrak{p})}$.
(iii) $M$ is locally free in the weaker sense and its rank function $\mathfrak{p} \mapsto r(\mathfrak{p})$ is locally constant on $\operatorname{Spec}(R)$.
(iv) $M$ is locally free in the stronger sense: there exist $f_1,\ldots,f_n \in R$, generating the unit ideal, such that for each $i$, $M_{f_i}$ is a free $R_{f_i}$-module.
(v) For every maximal ideal $\mathfrak{m}$ of $R$, there exists $f \in R \setminus \mathfrak{m}$ such that $M_f$ is a free $R_f$-module.  
This answers my previous question, because the rank function of an invertible module is identically one.
In order to really feel like I understand what's going on here, I would like to see an example of a finitely generated locally free [in the weaker sense of (ii) above] module which is not projective.  Thus $R$ must be non-Noetherian.  The wikipedia article on projective modules contains some nice information, in particular sketching an example of such a module over a Boolean ring.  For a Boolean ring though the localization at every prime ideal is simply $\mathbb{Z}/2\mathbb{Z}$, so it is not too surprising that there are more locally free modules than projectives.
I would like to see an example with $R$ an integral domain, if possible.  It would be especially nice if you can give a reference to one of the standard texts on commutative algebra which contains such an example or at least a citation of such an example.

REPLY [14 votes]: After mulling things over for a while, I realize that assuming the results from Bourbaki which I mentioned in the statement of the question, there is a very straightforward answer to my question.  (I mean this as no slight to Clark Barwick's excellent answer, which came instantaneously and contains lots of other valuable information.  Rather, I mean that had I thought more carefully I would not have needed to ask the question at all.)
The key is the following simple result:
Lemma: Let $R$ be a[n always commutative] ring, $\mathfrak{p}_1 \subset \mathfrak{p}_2$ prime ideals of $R$, and $M$ a finitely generated locally free (in the weaker sense) $R$-module.  Then $r(\mathfrak{p}_1) = r(\mathfrak{p}_2)$.
The proof is obvious, once you realize that localizing at $\mathfrak{p}_1$ is the same as localizing at $\mathfrak{p}_2$ and then localizing at (the ideal in $R_{\mathfrak{p}_2}$ naturally corresponding to) $\mathfrak{p}_1$.  (This is the same argument that allows you to see that is enough to require $M_{\mathfrak{p}}$ to be free at every maximal ideal $\mathfrak{p}$.)  Also I am using that the rank of a finitely generated free module over a [commutative!] ring is well-determined, as one sees by tensoring to the quotient field of some maximal ideal.
[I had some kind of psychological block coming from a vague memory that the rank function was merely semicontinuous.  As far as I can see now, semicontinuity does not come up anywhere in the study of the rank function.  I had to see essentially this argument in print -- in Milnor's Introduction to Algebraic K-Theory just this evening -- in order to become unblocked.]
Corollary: Let $R$ be a ring with a unique minimal prime ideal (e.g. an integral domain).  Then any finitely generated locally free $R$-module has constant rank function and therefore (by the Bourbaki result above) is projective.
This also illustrates why the most natural counterexamples come from zero-dimensional rings: the rank function is determined by its behavior on the minimal primes, so you might as well look at the zero-dimensional case.