most recent 30 from http://mathoverflow.net 2013-05-23T16:08:01Z http://mathoverflow.net/feeds/question/20413 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20413/flatness-of-modules-via-tor unknown (google) 2010-04-05T19:23:29Z 2010-04-06T01:56:38Z

<p>Is the following true:</p> <p>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.</p> <p>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.</p> <p>Is the same true without the Noetherian hypothesis?</p>

http://mathoverflow.net/questions/20413/flatness-of-modules-via-tor/20462#20462 t3suji 2010-04-06T01:56:38Z 2010-04-06T01:56:38Z

<p>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.)</p>