Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.