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Is the following true:

Let A be a Noetherian ring, and M a not necessarily finitely generated A module. Suppose that Tor_1^A(M,k_p)=0 for the residue fields k_p for all primes p\subset A.

Does this imply that M is flat? NB:if instead of Tor_1 one imposes that all Tor_i are zero, then it's easy to see.

Is the same true without the Noetherian hypothesis?

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Dear unknown, do you actually mean $Tor_1^A(M,A/p)$? If yes, I might know an answer. – Hailong Dao Apr 5 '10 at 19:37
i think (s)he means the field of fractions of A/p, or does the tor computation not change in this situation? – Sean Tilson Apr 5 '10 at 23:45

As far as I understand, this is false. Here is an example (familiar to $D$-module people): $A=k[x,y]$; $M=k[a,b]$ on which $x$ (resp. $y$) acts as $\frac{d}{da}$ (resp. $\frac{d}{db}$). Since the action of both $x$ and $y$ is locally nilpotent, $M$ is supported at the origin of $Spec(A)$. Therefore, the only non-zero Tor's of the kind you consider are $Tor_i(M,k)$, where both $x$ and $y$ act on $k$ by zero. These Tor's are easy to compute (they amount to computing de Rham cohomology of affine plane with coordinates $a$ and $b$), and they are non-zero precisely when $i=2$. (Essentially, the calculation repeats the proof of Kashiwara's Lemma.)

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