Let $A$ be a commutative noetherian ring, and let $M$ be a finitely generated projective $A$-module. It is well known and easy to prove that $M$ is locally free in the sense that for every $p \in\operatorname{Spec} A$, the module $M_p$ is a free $A_p$-module.
Is it true that projectives are also locally free in the following (more geometric?) sense:
There are elements $f_1,\dots,f_n \in A$ such that $(f_1,\dots,f_n) = 1$, and such that $M_{f_i}$ is a free $A_{f_i}$-module for all $1\le i \le n$.
Is this true? if so, can you provide a reference or explain how to prove it?
Thanks!
Best Answer
Yes, this is true. See this Math Overflow question for a precise statement and a reference to its proof in Bourbaki's Commutative Algebra.
This result is also stated in my commutative algebra notes, but the proof is not unfortunately not yet written up there. I certainly hope that this will be remedied soon though, as I will be teaching a course out of these notes starting on Monday. When the proof gets written, I will update this answer with a page number.
Added: Here is something in the MO answer that I decided was worth a comment here. For finitely generated modules, this stronger version of local freeness is actually equivalent to projectivity, whereas the weaker "pointwise local freeness" is subtly weaker in general.