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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!

Gabriel Soranzo
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the L
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4 Answers4

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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.

SvanN
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Pete L. Clark
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    Thank you for the reference. For the intrested reader, I should mention that Bourbaki proves the following much stronger result: If $M$ is a finitely presented module and if for some prime $p$, the module $M_p$ is free, then there exists a neighborhood $D(f)$ of $p$ such that $M_f$ is free over $A_f$, so that "being free at a point" implies "being free at a neigborhood of a point". – the L Jan 08 '11 at 16:51
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    @anonymous: Dear anonymous, more generally, if $M,N$ are finitely presented modules and $M_p, N_p$ are isomorphic, then $M_f \simeq N_f$ for some $f \in A-p$. This is explained in more detail (and generality) in EGA IV-8, where it is proved that (in a certain sense) if ${R_\alpha}$ is an inductive system of rings, then the category of f.p. modules over $\varinjlim R_\alpha$ is the "colimit of the categories of modules over $R_\alpha$," which has many useful applications in reducing questions about arbitrary f.p. modules to questions about finitely generated modules over noetherian rings. – Akhil Mathew Jan 15 '11 at 18:40
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    (contd.) For instance, Grothendieck's generic flatness lemma (if $A$ noetherian integral domain, $B$ finite $A$-algebra, $M$ finite $B$-module, then there is $f \in A$ such that $M_f$ is flat over $A$) works for finitely presented modules because any such module must "descend" to a finitely presented module over a noetherian (e.g. finitely generated) subring. – Akhil Mathew Jan 15 '11 at 18:41
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For future reference, I have written up a constructive and reasonably self-contained, if somewhat dense, proof (one page). The basic idea is to first verify that idempotent matrices over local rings are equivalent to diagonal matrices with entries $1$ and $0$, thus showing that finitely generated projective modules over local rings are free.

Ingo Blechschmidt
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  • Well, I think there are simpler proofs for finitely generated projective modules over local rings are free. Never mind, +1. – user26857 May 08 '15 at 16:40
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    @user26857: I would be interested to hear them! Are they constructive in the technical sense that they avoid the use of the law of excluded middle? I only know of my write-up and the one in Mulvey's gem Intuitionistic algebra and representations of rings. – Ingo Blechschmidt May 08 '15 at 16:54
  • http://math.stackexchange.com/questions/147754/finitely-generated-projective-module – user26857 May 08 '15 at 17:15
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    @user26857: Thanks! The problem is that a proof beginning with "Consider a minimal system of generators" is not constructive, since constructively one cannot show that any inhabited subset of the natural numbers contains a minimal element. In some cases though, the problem can be avoided by a clever rewording; I didn't try that in this case hard enough yet. – Ingo Blechschmidt May 08 '15 at 18:03
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This answer shows that if its stalk $M_\mathfrak{p}$ at $\mathfrak{p}$ is free, then there is an open neighbourhood $\mathfrak{p}\in D(f)$ on which its value $M_f$ is also free.

Proof: Reduce to the case that the natural map $M\to M_\mathfrak{p}$ is an injection by localising some $f_0\in A$. Now use the basis of $M_\mathfrak{p}$ to give a surjection $\alpha:A^n\to M$ whose localisation at $\mathfrak{p}$ is the isomorphism $A^n_\mathfrak{p}\to M_\mathfrak{p}$. Thus $\ker \alpha_\mathfrak{p}=0$, so $\ker \alpha_f=0$ for some $f\in A$. Thus $M_f\simeq A^n_f$ is free, completing the proof.

Thus the quasicompact space $\text{Spec}A$ admits a finite cover by $D(f_i)$'s, on each of which $M_{f_i}$ is free. $\text{Spec}A=\cup D(f_i)$ means precisely that $(f_1,...,f_n)=A$.

Pulcinella
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I have a somehow naïve idea, that $M_{p}=\mbox{colim}_{f\notin p}M_{f}$. Now choose a basis $x_{i}$ for $M_{p}$ and we may assume $x_{i}\in M$, then they are linear independent over $A$. Then we "consider the distance of $M_{f}$ from being free as $f$ goes to infinity", which is the exact sequence

$$0\rightarrow \bigoplus A_{f}x_{i} \rightarrow M_{f} \rightarrow M_{f}/\bigoplus A_{f}x_{i}\rightarrow 0$$

Taking colimit with $f$ and note that localization and taking colimit are exact,

$$\mbox{colim} M_{f}/\bigoplus A_{f}x_{i}=M_{p}/\bigoplus A_{p}x_{i}=0$$

Hence there exist some $f$, such that $M_{f}/\bigoplus A_{f}x_{i}=0$, provided $M$ is finitely generated. Now for every $p$ there is a distinguished open neighborhood, hence the ideal generated by these $f$'s must be trivial.

Y_q
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