Is projectiveness a Zariski-local property of modules? (Answered: Yes!) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T14:19:14Z http://mathoverflow.net/feeds/question/6719 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6719/is-projectiveness-a-zariski-local-property-of-modules-answered-yes Is projectiveness a Zariski-local property of modules? (Answered: Yes!) Andrew Critch 2009-11-24T19:24:37Z 2010-02-25T22:09:00Z <p>I know that for a finitely presented $A$-module $M$ ($A$ a commutative ring), TFAE:</p> <p>1) $M$ is <em>projective</em>; <br>2) $M$ is <i>max-locally free</i>, meaning that $M_m$ is free for every maximal ideal $m$; <br>3) $M$ is <i>prime-locally free</i>, meaning that $M_p$ is free for every prime ideal $m$; <br>4) $M$ is <i>Zariski-locally free</i>, meaning that there are some $f_1,\ldots,f_n$ generating the unit ideal in $A$ such that each $M_{f_i}$ is free.</p> <p>(Reference: Eisenbud commutative algebra, p. 136 / end of chapter 4).</p> <p>I know that (1) implies (2) without finite presentation: see <a href="http://www.jstor.org/stable/1970252" rel="nofollow">Kaplansky (1958): <em>Projective Modules</em></a>, 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.</p> <p>So now I'm wondering in particular if you need finite presentation to prove (4) implies (1), and more generally,</p> <blockquote> <p>If $M$ is <em>Zariski-locally projective</em> (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 <em>projective</em>?</p> </blockquote> <p>If so, how can I see this directly / commutative-algebraically?</p> <p><hr /></p> <p><b>Follow up:</b> I checked out Bhargav's reference, <a href="http://www.springerlink.com/content/t0675rw74565u076/" rel="nofollow">Raynaud and Gruson: <em>Critères de platitude et de projectivité</em></a>. 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.</p> <p>Tricky stuff!</p> http://mathoverflow.net/questions/6719/is-projectiveness-a-zariski-local-property-of-modules-answered-yes/6730#6730 Answer by Mark Hovey for Is projectiveness a Zariski-local property of modules? (Answered: Yes!) Mark Hovey 2009-11-24T20:42:23Z 2009-11-24T22:28:13Z <p>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). </p> <p>Note that this implies that any finitely presented such module is projective, since a finitely presented flat module is projective. </p> <p>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</p> <p>$A = k_1 \times k_2 \times ...$</p> <p>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. </p> <p>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.</p> <p>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). </p> <p>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. </p> <p>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). </p> http://mathoverflow.net/questions/6719/is-projectiveness-a-zariski-local-property-of-modules-answered-yes/6736#6736 Answer by Greg Stevenson for Is projectiveness a Zariski-local property of modules? (Answered: Yes!) Greg Stevenson 2009-11-24T21:43:38Z 2009-11-24T22:06:41Z <p>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.</p> <p>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<br /> (i) $S$ is a multiplicative set consisting of non-zero divisors such that $S^{-1}M$ is projective over $S^{-1}R$<br /> or<br /> (ii) $M/rad(R)M$ is $R/rad(R)$-projective<br /> then $M$ is projective.</p> <p>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".</p> http://mathoverflow.net/questions/6719/is-projectiveness-a-zariski-local-property-of-modules-answered-yes/6738#6738 Answer by Bhargav for Is projectiveness a Zariski-local property of modules? (Answered: Yes!) Bhargav 2009-11-24T21:59:48Z 2009-11-24T21:59:48Z <p>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).</p> http://mathoverflow.net/questions/6719/is-projectiveness-a-zariski-local-property-of-modules-answered-yes/16448#16448 Answer by BCnrd for Is projectiveness a Zariski-local property of modules? (Answered: Yes!) BCnrd 2010-02-25T21:03:15Z 2010-02-25T22:09:00Z <p>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. </p> <p>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).</p> <p>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. </p>