A finitely generated, locally free module over a domain which is not projective?

It is impossible to produce an example of a finitely generated flat $R$-module that is not projective when $R$ is an integral domain. See: Cartier, "Questions de rationalité des diviseurs en géométrie algébrique," here, Appendice, Lemme 5, p. 249. Also see Bourbaki Algèbre Chapitre X (Algèbre Homologique, "AH") X.169 Exercise Sect. 1, No. 13. I also sketch an alternate proof that there are no such examples for $R$ an integral domain below.

Observe that, for finitely generated $R$-modules $M$, being locally free in the weaker sense is equivalent to being flat [Bourbaki, AC II.3.4 Pr. 15, combined with AH X.169 Exercise Sect. 1, No. 14(c).]. ($R$ doesn't have to be noetherian for this, though many books seem to assume it.)

There's a concrete way to interpret projectivity for finitely generated flat modules. We begin by translating Bourbaki's criterion into the language of invariant factors. For any finitely generated flat $R$-module $M$ and any nonnegative integer $n$, the $n$-th invariant factor $I_n(M)$ is the annihilator of the $n$-th exterior power of $M$.

Lemma. (Bourbaki's criterion) A finitely generated flat $R$-module $M$ is projective if and only if, for any nonnegative integer $n$, the set $V(I_n(M))$ is open in $\mathrm{Spec}(R)$.

This openness translates to finite generation.

Proposition. If $M$ is a finitely generated flat $R$-module, then $M$ is projective iff its invariant factors are finitely generated.

Corollary. The following conditions are equivalent for a ring $R$: (1) Every flat cyclic $R$-module is projective. (2) Every finitely generated flat $R$-module is projective.

Corollary. Over an integral domain $R$, every finitely generated flat $R$-module is projective.

Corollary. A flat ideal $I$ of $R$ is projective iff its annihilator is finitely generated.

Example. Let me try to give an example of a principal ideal of a ring $R$ that is locally free in the weak sense but not projective. Of course my point is not the nature of this counterexample itself, but rather the way in which one uses the criteria above to produce it.

Let $S:=\bigoplus_{n=1}^{\infty}\mathbf{F}_2$, and let $R=\mathbf{Z}[S]$. (The elements of $R$ are thus expressions $\ell+s$, where $\ell\in\mathbf{Z}$ and $s=(s_1,s_2,\dots)$ of elements of $\mathbf{F}_2$ that eventually stabilize at $0$.) Consider the ideal $I=(2+0)$.

I first claim that for any prime ideal $\mathfrak{p}\in\mathrm{Spec}(R)$, the $R_{\mathfrak{p}}$-module $I_{\mathfrak{p}}$ is free of rank $0$ or $1$. There are three cases: (1) If $x\notin\mathfrak{p}$, then $I_{\mathfrak{p}}=R_{\mathfrak{p}}$. (2) If $x\in\mathfrak{p}$ and $\mathfrak{p}$ does not contain $S$, then $I_{\mathfrak{p}}=0$. (3) Finally, if both $x\in\mathfrak{p}$ and $S\subset\mathfrak{p}$, then $I_{\mathfrak{p}}$ is a principal ideal of $R_{\mathfrak{p}}$ with trivial annihilator.

It remains to show that $I$ is not projective as an $R$-module. But its annihilator is $S$, which is not finitely generated over $R$.

[This answer was reorganized on the recommendation of Pete Clark.]

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.