Let $R$ be a, say, noetherian ring and $M$ an $R$-module. The Wikipedia article on 'locally free sheaf' tells me that the following two statements are equivalent:

  1. The module $M$ is locally free (Edit: this means there is an open cover $\{U_i\}$ of $Spec R$ such that every $\tilde{M}_{|U_i}$ is free as an ${\mathcal{O}_{Spec R}}{|U_i}$-module.)
  2. $M_p$ is a free $R_p$-module for every prime ideal $p$ of $R$.

I see that these two things are equivalent if $M$ is finitely generated but I cannot see this in general, even if $R$ is noetherian. Am I missing something or is there a mistake on Wikipedia?

If the latter case is true, has anybody an example of a (non-finitely generated) $R$-module $M$ over a noetherian $R$ such that $M_p=(R_p)^{n_p}$ for every prime ideal $p$ of $R$ and such that $M$ is not locally free?

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    $\begingroup$ There couldn't be a mistake in Wikipedia. $\endgroup$ – Tom Goodwillie Nov 4 '10 at 16:29
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    $\begingroup$ The current version of <a href="en.wikipedia.org/wiki/Locally_free_sheaf">wikipedia</…> is really wrong. There are examples below with quasi-coherent sheaves. One can also consider the example $X$ equal to the spectrum of a discrete valuation ring $R$, ${\mathcal F}(X)={\mathcal O}_X(X)$ and ${\mathcal F}(U)={\mathcal O_X}(U)^2$ where $U$ is the complementary of the closed point of $X$. Then $\mathcal F$ is not quasi-coherent (so can not be locally free in the correction definition), but its stalks are free. $\endgroup$ – Qing Liu Nov 4 '10 at 21:45

Dear roger123, let $R$ be a commutative ring and $M$ an $R$-module ( which I do not suppose finitely generated). In order to minimize the risk of misunderstandings, allow me to introduce the following terminology:

Locfree The module $M$ is locally free if for every $P \in Spec (R)$ there is an element $f \in Spec(R)$ such that $f \notin P$ and that $M_f$ is a free $R_f$ - module.

Punctfree The module $M$ is punctually free if for every $P \in Spec (R)$ the $R_{P} $ - module $M_P$ is free.

Fact 1 Every locally free module is punctually free. Clear.

Fact 2 Despite Wikipedia's claim, it is false that a punctually free module is locally free.

Fact 3 However if the punctually free $R$- module $M$ is also finitely presented, then it is indeed locally free.

Fact 4 A finitely generated module is locally free if and only it is projective.

Fact 5 A projective module over a local ring is free.This was proved by Kaplansky and is remarkable in that, let me repeat it, the module $M$ is not supposed to be finitely generated.

A family of counterexamples to support Fact 2 Let R be a Von Neumann regular ring. This means that every $r\in R$ can be written $r=r^2s$ for some $s\in R$. For example, every Boolean ring is Von Neumann regular. Take a non-principal ideal $I \subset R$. Then the $R$- module $R/I$ is finitely generated (by one generator: the class of 1 !), all its localizations are free but it is not locally free because it is not projective (cf. Fact 4) .The standard way of manufacturing that kind of examples is to take for R an infinite product of fields $\prod \limits_{j \in J}K_j$ and for $I$ the set of families $(a_j)_{j\in J}$ with $a_j =0$ except for finitely many $j$ 's.

Final irony In the above section on counterexamples I claimed that the $R$ - module $R/I$ is not projective.This is because in all generality a quotient $R/I$ of a ring $R$ by an ideal $I$ can only be $R$ - projective if $I$ is principal . And I learned this fact in...Wikipedia !

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    $\begingroup$ Thank you for your answer Georges! I think that for a finitely presented $R$ module Locfree is the same as Punctfree. If $R$ is noetherian, finitely presented is equivalent to finitely generated. So I think your example only works with a non-noetherian Ring, right? $\endgroup$ – roger123 Nov 5 '10 at 7:43
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    $\begingroup$ Dear roger123: Yes, everything you write is absolutely correct. For finitely presented modules, the equivalence of Locfree and Punctfree follows from Facts 1 and 3. Moreover your remark about the example is extremely interesting. Indeed, R/I can only be non finitely presented if I is non finitely generated. But I only assumed $I$ non-principal. So we are inexorably led to the conclusion that in a Von Neumann regular ring, every finitely generated ideal is principal. I did not know this so I checked five minutes ago in Rotman's book on Homological Algebra (3d ed.): it is stated in Lemma 4.8 ! $\endgroup$ – Georges Elencwajg Nov 5 '10 at 8:46

I'm a bit confused about your question since an $R$-module $M$ is defined to be locally free if $M_{\mathfrak{p}}$ is a free $R_\mathfrak{p}$-module for all primes $\mathfrak{p}$. I'll assume that you are asking for any example where $M_{\mathfrak{p}}$ is a free $R_\mathfrak{p}$-module for all primes $\mathfrak{p}$ but $M$ is not projective. If this is not what you are after, just ignore this answer.

Let $R=\mathbb{Z}$ and let $M$ be the submodule of $\mathbb{Q}$ generated by all $\frac{1}{p}$ where $p$ is a prime number. Then $M$ is not projective/free, but it is locally free since $ M_{(p)}=\mathbb{Z}_{(p)}\frac{1}{p}=R_{(p)} $ and $M_{(0)}=\mathbb{Q}=R_{(0)}$.

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    $\begingroup$ Dear Michael: your defn of "locally free" for finitely generated modules is wrong in the sense that it isn't too useful when $R$ is not noetherian. The right defn for f.gentd modules is local freeness for Zariski topology (i.e., over Zariski-open covering, acquires a basis), and this is equivalent to the stalk condition in the noetherian case. But not otherwise. A counterexample is $M=R/I$ for $R=\prod_{n \ge 0} \mathbf{F}_2$ and $I$ an ideal that isn't finitely generated. This $M$ is not loc. free but each $R_P$ is $\mathbf{F}_2$ (needs some thought) and $M_P$ is 0 or 1-diml. $\endgroup$ – BCnrd Nov 4 '10 at 17:55
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    $\begingroup$ Thank you for the comment. In my case $R$ is noetherian and $M$ is unfortunately not finitely generated. $\endgroup$ – roger123 Nov 4 '10 at 18:36
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    $\begingroup$ Dear roger123: The example in my comment, viewed as an $R$-algebra, is a counterexample to the sufficiency aspect in EGA IV$_4$, 18.4.12(ii) (the mistake is assuming locally finite type rather than locally finite presentation, and the error in the proof occurs when they say "donc $j$ est une immersion ouverte"). This bogus sufficiency claim is invoked in the proof of 18.4.14, but that proof can be easily made OK by verifying the locally finite presentation condition holds in the relevant cases for that argument. $\endgroup$ – BCnrd Nov 6 '10 at 3:21

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