Let $M$ be a filtered module over a filtered algebra $A$, and suppose $gr(M)$ is flat over $gr(A)$, where $gr$ means the associated graded module and algebra, respectively.

What can one say in general about the flatness of $M$ over $A$, or with relevant assumptions (for instance in the above, we should assume both filtrations are complete to avoid dumb counterexamples)? Are there good references for this sort of question? I have played with the various definitions of flatness trying to find an obvious relationship, but I find flatness proofs confusing.

The particular examples I have in mind are comparing $U(\mathfrak{g})$-modules to $S(\mathfrak{g})$-modules, and $D(X)$-modules to $O(T^*X)$-modules for affine varieties $X$, if it helps. I suspect the answer doesn't depend on any of the details though.

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    $\begingroup$ If $A$, $M$ and $N$ are a filtered algebra and two filtered modules, there is a spectral sequence going from $\mathrm{Tor}^{\mathrm{gr}A}(\mathrm{gr}M,\mathrm{gr}N)$ to $\mathrm{Tor}^A(M,N)$ (I don't think you need any hypthesis on anything to get this, and in particular convergence, but one should proably check...). In particular, if $\mathrm{gr}M$ is $\mathrm{gr}A$-flat, then you get that $\mathrm{Tor}^A(M,\mathord-)$ vanishes (in positive degrees) on modules $N$ endowed with filtrations. Maybe that is enough for you? $\endgroup$ – Mariano Suárez-Álvarez Jul 27 '10 at 17:32
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    $\begingroup$ Mariano, surely there have to be hypotheses on the filtrations (satisfied in interesting examples), such as separatedness and/or exhaustiveness, since otherwise one could always use dumb filtrations so that the gr's are all zero (and in such cases there can't possibly be a general spectral sequence which actually converges to Tor). $\endgroup$ – BCnrd Jul 27 '10 at 18:07
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    $\begingroup$ @Mariano: Thanks I will see if this is helpful. I really want M to be flat over A without reservations, in my examples. Maybe this gives a start. @BCnrd: Yes, this is my fault for being incomplete in the setup. The filtration will be increasing (think degree of differential operator), and will be complete, $A=\cup_i A_i$. I'll add that to my question. $\endgroup$ – David Jordan Jul 27 '10 at 18:19
  • $\begingroup$ I thought about using that $gr(A)$ is the $t=0$ fiber of the Rees algebra associated to the filtration, while the other fibers are all isomorphic to $A$. However, this reduces to another problem I don't know how to answer, which is how does the flatness of module $M_t$ over algebra $R_t = \sum t^k A_k$ change, as $t$ varies? I think there are statements like local criterion of flatness, generic flatness that apply when $A$ is commutative, but I don't know how heavily they use commutativity. $\endgroup$ – David Jordan Jul 27 '10 at 18:43
  • $\begingroup$ @BCnrd, sure! I'm too sed to think about those conditions as the reasonable background on top of which you add the hypotheses :) $\endgroup$ – Mariano Suárez-Álvarez Jul 27 '10 at 21:36

Let me suppose, as in your examples, that we have a base field $k$.

It is well known that to check that a right $A$-module $M$ is flat it is enough to show that whenever $I\leq_\ell A$ is a left ideal, the map $M\otimes_AI\to M\otimes_A A$ induced by the inclusion $I\to A$ is injective. This condition can be rewritten: $M$ is flat iff for each left ideal $I\leq_\ell A$ we have $\mathrm{Tor}^A_1(M,A/I)=0$.

So now suppose $A$ and $M$ are (exhaustively, separatedly, increasingly from zero) filtered in such a way that $\mathrm{gr}M$ is a flat $\mathrm{gr}A$-module.

Pick a left ideal $I\leq_\ell A$; notice that the filtration on $A$ induces a filtration on the quotient $A/I$. We can compute $\mathrm{Tor}^A_\bullet(M,A/I)$ as the homology of the homologically graded complex $$\cdots\to M\otimes_kA^{\otimes_kp}\otimes_kA/I\to M\otimes_kA^{\otimes_k(p-1)}\otimes_kA/I\to\cdots$$ with certain differentials whose formula does not fit in this margin, coming from the bar resolution. Now the filtrations on $M$, on $A$ and on $A/I$ all collaborate to provide a filtration of our complex. We've gotten ourselves a positively homologicaly graded with a canonically bounded below, increasing, exhaustive and separated filtration. The corresponding spectral sequence then converges, and its limit is $\mathrm{Tor}^A_\bullet(M,A/I)$. Its $E^0$ term is the complex $$\cdots\to\mathrm{gr}M\otimes_k\mathrm{gr}A^{\otimes_kp}\otimes_k\mathrm{gr}(A/I)\to \mathrm{gr}M\otimes_k\mathrm{gr}A^{\otimes_k(p-1)}\otimes_k\mathrm{gr}(A/I)\to\cdots$$ with, again, the bar differential, and its homology, which is the $E^1$ page of the spectral sequence, is then precisely $\mathrm{Tor}^{\mathrm{gr}A}_\bullet(\mathrm{gr}M,\mathrm{gr}(A/I))$. Since we are assuming that $\mathrm{gr}M$ is $\mathrm{gr}A$-flat, this last $\mathrm{Tor}$ vanishes in positive degrees, so the limit of the spectral sequence also vanishes in positive degrees. In particular, $\mathrm{Tor}^A_1(M,A/I)=0$.

NB: As Victor observed above in a comment, Bjork's Rings of differential operators proves in its Proposition 3.12 that $\mathrm{w.dim}_AM\leq\mathrm{w.dim}_{\mathrm{gr}A}\mathrm{gr}M$ (here $\mathrm{w.dim}$ is the flat dimension) from which it follows at once that $M$ is flat as soon as $\mathrm{gr}M$ is; the argument given is essentialy the same one as mine. I am very suprised about not having found this result in McConnell and Robson's!

  • $\begingroup$ Dear Mariano, thanks for this thorough explanation, as well as your helpful comments above. If I could select a second answer, I would =] $\endgroup$ – David Jordan Jul 28 '10 at 0:18
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    $\begingroup$ David, you can always change your selection of the answer to accept if you have a reason (e.g. a new, more relevant answer has been posted) --- it's considered perfectly normal. $\endgroup$ – Victor Protsak Jul 28 '10 at 1:58
  • $\begingroup$ Done. I didn't know that was a feature, though it makes sense. $\endgroup$ – David Jordan Jul 29 '10 at 20:32

Dear David,

See Prop. 1.2 of Algebras of p-adic distributions and admissible representations by Peter Schneider and Jeremy Teitelbaum, for one such result:

Proposition 1.2. Suppose that $gr^{\bullet} R$ and $gr^{\bullet} A$ are left noetherian and that $gr^{\bullet} A$ as a right $gr^{\bullet}R$-module (via $gr^{\bullet}\phi$) is flat; then $A$ is flat as a right $R$-module (via $\phi$).

There are some hypotheses which you can find in the preamble to section 1 of the paper, as well as useful references to literature on this kind of question.

  • $\begingroup$ Dear Emerton, thanks this is perfect! The paper you suggest is way outside my area so I wouldn't have come across it myself in a million years. In fact, I am in just the situation you describe: my module $M$ is itself an algebra over $A$, via a homomorphism $\phi$, and both are Noetherian (I didn't mention this in the question for fear of clouding it with too many details). I will also check out the very useful-seeming textbook Zariskian Filtrations, to which they refer. $\endgroup$ – David Jordan Jul 27 '10 at 19:47
  • $\begingroup$ Great, I'm glad I could help! Cheers, Matt. $\endgroup$ – Emerton Jul 27 '10 at 19:51
  • $\begingroup$ In EGA 0$_{\rm{III}}$, 10.2 there is some discussion of flatness with graded modules, but there I think the gradings are all defined by an ideal of the ring (and everything is commutative, so useless for the intended applications). Matt, thanks for the pointer to the excellent reference; I hope the implicit hypotheses are not too restrictive. $\endgroup$ – BCnrd Jul 27 '10 at 20:49
  • $\begingroup$ The hypothesis of completeness used in that paper is not satisfied in neither of David's examples, though $\endgroup$ – Mariano Suárez-Álvarez Jul 27 '10 at 21:40
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    $\begingroup$ @Emerton: The Schneider-Teitelbaum result seems like it would actually require more input to me (visibly from the extra Noetherian hypotheses) because Mariano's techniques rely on being able to compute Tor via spectral sequence, and in particular on being able to pass the tensor product across the inverse limit coming from the filtration. These methods won't compute $\mathbb{Z}_p \otimes \mathbb{Z}_p$ correctly and so more work seems to be required in the decreasing case. $\endgroup$ – Tyler Lawson Jul 28 '10 at 1:38

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