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The following statement is well-known:

$A$ a commutative Noetherian ring, $M$ a finitely generated $A$-module. Than $M$ is flat if and only if $M_{\mathfrak{p}}$ is free for all $\mathfrak{p}$.

My question is: do we need the assumption that $A$ is Noetherian? I have a proof (from Matsumura) which doesn't require that assumption, but the fact that other references (e.g. Atiyah, Wikipedia) are including this assumption makes me rather uneasy.

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you can also look at my criteria which is the most general I found – brunoh Jan 1 at 9:30
up vote 29 down vote accepted

By request, my earlier comments are being upgraded to an answer, as follows. For finitely generated modules over any local ring $A$, flat implies free (i.e., Theorem 7.10 of Matsumura's CRT book is correct: that's what proofs are for). So the answer to the question asked is "no". The CRT book uses the "equational criterion for flatness", which isn't in Atiyah-MacDonald (and so is why the noetherian hypothesis was imposed there). This criterion is in the Wikipedia entry for "flat module", but Wikipedia has many entries on flatness so it's not a surprise that this criterion under "flat module" would not be appropriately invoked in whatever Wikipedia entry was seen by the OP.

An awe-inspiring globalization by Raynaud-Gruson (in their overall awesome paper, really with authors in that order) is given without noetherian hypotheses: if $A$ has finitely many associated primes (e.g., any noetherian ring, or any domain whatsoever) and if $M$ is a finitely generated flat $A$-module then it's finitely presented (so Zariski-locally free!). See 3.4.6 (part I) of Raynaud-Gruson (set $X=S$ there). By 3.4.7(iii) of R-G, the finiteness condition on the set of associated primes cannot be removed, as any absolutely flat ring that isn't a finite product of fields provides a counterexample. (An explicit counterexample is provided by the link at the end of Daniel Litt's answer, namely a finitely generated flat module that is not finitely presented, over everyone's favorite crazy ring $\prod_{n=0}^{\infty} \mathbf{F}_2$.)

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I think the following criterium is more general and much easier to prove than the one in the excellent RG paper – brunoh Jan 1 at 9:28

I believe the non-Noetherian statement is that "flat and finitely presented" implies locally free (i.e., projective). A proof of this can be found, for instance, in Weibel's An Introduction to Homological Algebra.

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This is correct, but here locally free means "Zariski locally free", which (at least a priori) is stronger than just the stalks being free, which is what Kwan's question is about. – Emerton Jul 27 '10 at 15:08
@Akhil: The point being that finitely generated and Noetherian implies finitely presented. – Daniel Litt Jul 27 '10 at 15:10
Also, in the Noetherian setting, stalkwise free is equivalent to Zariski locally free. – Emerton Jul 27 '10 at 15:14
I agree with your statement, but the question is whether we can replace "finitely presented" with "finitely generated". – ashpool Jul 27 '10 at 15:22

This is to expand on Akhil's answer. Locally free implies flat easily, so let's look at the other direction. It suffices to assume $A$ is local with maximal ideal $\mathfrak{m}$.

Pick a basis of $M/\mathfrak{m}M$; by Nakayama, this lifts to a surjective map $A^n\to M$. We want to show this map is injective. If $M$ is finitely presented (or if $A$ is Noetherian) then the kernel is finitely generated. But tensoring with $A/\mathfrak{m}A$ kills the kernel, so by Nakayama again the map is injective.

The finite generation of the kernel is the key point.

UPDATE: Exercise 24.4.F in Ravi Vakil's notes gives a finitely generated, not finitely presented module which is flat but not projective. By BCnrd's comment on Akhil's answer it is, however, stalk-wise free.

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I agree with Akhil's answer wholeheartedly. But the question is whether we can relax the condition of finite presentation with finite generation. – ashpool Jul 27 '10 at 15:26
Aren't you saying that the sequence obtained $0\rightarrow K\rightarrow A^n\rightarrow M\rightarrow 0$ by lifting a basis for $M/\mathfrak{m}M$ remains exact upon tensoring with the residue field? I don't understand why this is the case, maybe 'cause I don't see where you're using the flatness of $M$. Is it true that $M$ is necessarily projective, i.e., does finite flat over a Noetherian ring imply projective? – Keenan Kidwell Jul 27 '10 at 15:32
That should be "Noetherian local ring." – Keenan Kidwell Jul 27 '10 at 15:33
@Keenan: The exactness of that sequence is exactly where I'm using the flatness of $M$. Consider the LES of Tor. And finite flat over a Noetherian local ring implies free, so obviously projective. – Daniel Litt Jul 27 '10 at 15:37
Also, that example in your UPDATE is great. Does anyone know a published reference with an argument? (I now mention it as an aside in Exercise 25.4.E in the July ~21 notes <a href="">here</a>.) – Ravi Vakil Jul 21 '11 at 17:51

This is too long for a comment, so I'm forced to include it as an answer, even though it isn't one. In defense of ashpool's question: until reading Brian Conrad's excellent answer, I'd assumed that finite presentation (not just finite type) hypotheses were necessary (for the statement that flat modules over a local ring are free). Reason 1: there is a (slightly) easier proof in the finite presentation case, and when I proved it myself (something my advisor wisely told me to always do, rather than looking things up), I was led to the finite presentation argument (see 25.4.2 on the June 27 2011 version of the notes here --- it will be fixed in later versions), and didn't think to go further. Reason 2: When sources such as Mumford (see the 2nd edition of the Red Book, p. 215) state the result with finite presentation hypotheses, my prejudices are confirmed. Then when I read Matsumura, I see what I expect to see, and not what is on the page.

Also, I liked Matt's interesting point: you only need Noetherian hypotheses to get from "free stalks" to "locally free sheaf".

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Dear Ravi,I have noticed that much confusion here and on similar questions is caused by the ambiguity of the adverb "locally" in "locally blah", which can mean either that you can cover your scheme by opens on which you have "blah", or more weakly that stalks are "blah". Since you are writing a book which, if there is any justice, should become a standard, you might introduce a terminology which would eradicate these misunderstandings. I think Matt's and Daniel's distinction between "Zariski locally blah" [or just "locally blah"] and "stalk-blah" [or "stalkwise-blah"] is quite satisfactory. – Georges Elencwajg Jul 20 '11 at 20:28
Update . I've just checked Ravi's notes: he has been using the suggested terminology for ages! On the one hand I feel silly about my comment above and on the other I'm happy that these wonderful notes will contribute to the elimination of the confusion alluded to above. – Georges Elencwajg Jul 21 '11 at 19:31
@RaviVakil Perhaps you will find my simple little criteria (which is the most general I think) useful for your extraordinary notes! – brunoh Feb 13 at 16:17

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