I know that for a finitely presented $A$-module $M$ ($A$ a commutative ring), TFAE:

- $M$ is
*projective*; - $M$ is
*max-locally free*, meaning that $M_{\mathfrak m}$ is free for every maximal ideal $\mathfrak m$; - $M$ is
*prime-locally free*, meaning that $M_{\mathfrak{p}}$ is free for every prime ideal $\mathfrak p$; - $M$ is
*Zariski-locally free*, meaning that there are some $f_1,\ldots,f_n$ generating the unit ideal in $A$ such that each $M_{f_i}$ is free.

(Reference: Eisenbud commutative algebra, p. 136 / end of chapter 4).

I know that (1) implies (2) without finite presentation: see Kaplansky (1958): *Projective Modules*, p. 374. (He doesn't even assume $A$ is commutative, and uses an awesome lemma that any projective module is a direct sum of countably-generated submodules.) Finite presentation is used to prove (3) implies (4), as is often the case when passing from stalks of a sheaf to actual open sets.

So now I'm wondering in particular if you need finite presentation to prove (4) implies (1), and more generally,

If $M$ is

Zariski-locally projective(meaning there are some $f_1,\ldots,f_n$ generating the unit ideal in $A$ such that each $M_{f_i}$ is projective), is itprojective?

If so, how can I see this directly / commutative-algebraically?

**Follow up:** I checked out Bhargav's reference, Raynaud and Gruson: *Critères de platitude et de projectivité*. It turns out (on p. 81) they actually use the same technique as Kaplansky in the paper I linked above, of writing a module as a transfinite union with countably generated successive quotients, which they call a "Kaplansky division" when these quotients are direct summands. The conclusion that projectiveness is Zariski-local is stated as Example 3.1.4(3) on the bottom of page 82.

Tricky stuff!