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

1) $M$ is projective;
2) $M$ is max-locally free, meaning that $M_m$ is free for every maximal ideal $m$;
3) $M$ is prime-locally free, meaning that $M_p$ is free for every prime ideal $m$;
4) $M$ 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 it projective?

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!

up vote 8 down vote accepted

Being projective is indeed a local property for the Zariski topology. In fact, it is even local for the fpqc topology --- this is a famous theorem of Raynaud and Gruson (see MR0308104).

A point worth noting: the proof of fpqc descent for projectivity in Raynaud-Gruson is apparently incorrect (as I learned today from Gabber in connection with something else), but the result is nonetheless true.

Here's the deal. RG deduce the result in 3.1.4(1) of part II of the paper, using 3.1.3 of part II and the fact that faithfully flat ring maps satisfy the property they call (C) there. (Briefly, a ring map satisfies property (C) when flat modules over the base ring which satisfy a certain "Mittag-Leffler" condition after the scalar extension actually satisfy the ML condition before the scalar extension. The content of 3.1.3 in part II is that this condition (C) implies descent of projectivity for flat modules. So the problem is to prove an interesting class of maps satisfies property (C).) But RG's proof of (C) for faithfully flat ring maps in 3.1.4(1) of part II rests on another result (2.5.2 in part II) which Gruson has said is incorrect (in his paper "Dimension homologique...."). That's the problem. Gabber says he does not know a counterexample to this 2.5.2 part II result. (I guess Gruson didn't give one when he said it is false.) Anyway, so to make the proof complete, it is necessary to verify that the ring extension of interest (such as faithfully flat in general, or Zariski-covering in case of the question) satisfies the property which RG call (C). Gabber says that this is an easy exercise adapting the method of proof of 3.1.4 in part I of the paper (which is the case of countably presented modules).

I only ever read part I of the paper, never part II (part I was already exhausting enough, and quite spectacular/useful by itself), so in particular I do not know where an error occurs (if Gruson is right) in the proof of 2.5.2 part II. Maybe someone who has read the argument can identify where an error or gap occurs, and hopefully work out Gabber's exercise. (Bhargav?) If so, please let me know.

  • So are Zariski and etale descent still OK? – JBorger Feb 25 '10 at 22:36
  • According to Gabber the whole statement is ok, but the proof in RG seems to be inadequate. However, just because someone else said it is ok doesn't mean one should accept it as true (especially for something like this, for which the proof is ultimately not going to be on the order of difficulty of the decomposition theorem, even if it may require some ingenuity). Care to give it a shot? I am too busy right now to revisit this stuff, but I will be very happy if someone more inspired sorts out an argument. – BCnrd Feb 25 '10 at 22:51
  • Sorry, mate. Too much stuff to do that I actually need. (BTW, my question was whether the RG argument is sufficient to cover the cases of Zariski and etale descent.) – JBorger Feb 26 '10 at 1:01
  • Jim, I'm hoping someone who already read part II will clear up the issue. – BCnrd Feb 26 '10 at 5:00
  • 8
    Alex Perry has given a correct proof of the fpqc descent of projectivity (arXiv:1011.0038v1). He shows (C) (Perry: Lemma 9.2) directly without using the incorrect II.2.5.2. – David Rydh Jan 23 '11 at 19:41

This is not an answer to your question about Zariski-local projectivity, but it is relevant to being locally free and you might be interested.

One can get away with finitely generated rather than finitely presented if one has a little more to work with. In particular, if $M$ is finitely generated and flat over $R$ and either
(i) $S$ is a multiplicative set consisting of non-zero divisors such that $S^{-1}M$ is projective over $S^{-1}R$
(ii) $M/rad(R)M$ is $R/rad(R)$-projective
then $M$ is projective.

This first result is due to Endo and the second is not so hard. More details as well as more of these types of results can be found in Vasconcelos' paper "On Finitely Generated Flat Modules".

Well, your module M is certainly flat, because flatness is a local property (Atiyah-McDonald, Proposition 3.10, but it is also basically easy; first show being 0 is a local property, then show injectivity of morphisms is a local property, now show flatness is).

Note that this implies that any finitely presented such module is projective, since a finitely presented flat module is projective.

I suspect there is a counterexample in general, even with a finitely generated flat module. Try A a commutative von Neumann regular ring, like an infinite product of fields

$A = k_1 \times k_2 \times ... $

Take a finitely generated module that is not projective (but is necessarily flat, since the ring is von Neumann regular), which must exist but which I am having a little trouble writing down at the moment. I bet this will do the job. If no one else fixes this, I will try to do so later.

EDIT: A few minutes thought made this clear. Take I to be the direct sum of all the copies of k_i, and M to be the quotient A/I.

The (prime =) maximal ideals of A are the ideals m_i consisting of the elements which are 0 at every coordinate except the ith one. The local ring is k_i, which is a field. So actually any module M is locally free, but this particular M is not projective (because then the direct sum of the k_i would have to split off the product as a summand).

RE-EDIT: OK, so there are more maximal ideals than the m_i. However, for any commutative ring R, R is von Neumann regular if and only if the localizations of R at its maximal ideals are fields (Lam, A first course in noncommutative rings Ex. 4.15). Thus, any module M is max-locally free (and prime-locally free because primes =maximals for vNR rings). So any non-projective module, such as the one above, is a prime-locally free module that is not projective.

However, I think this module is probably not Zariski-locally free. So it shows that (3) does not imply (1), but says nothing about (4) implying (1).

  • There are more prime ideals than the ones you mentioned. For example I is an ideal so it's containined in a maximal ideal but it's not contained in any of the m_i. – Kevin Buzzard Nov 24 '09 at 21:57
  • Two problems: 1) Those m_i are not all the maximal ideals in your ring... e.g. the ideal I you describe must be in some maximal ideal by Zorn's Lemma, but clearly isn't in any of the m_i. 2) My question is abut a module which is Zariski-locally free, not just max-locally free, a weaker condition. – Andrew Critch Nov 24 '09 at 22:00

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