(This is a follow-up to a previous question; as often happens, I didn't include enough hypotheses, so I'm asking a new, more-likely-to-be-true question).

Consider the formal plane $\operatorname{Spec}\mathbb C[[t,h]]$, and let $\mathcal F$ be a quasi-coherent sheaf which is graded for the grading on $\mathbb C[[t,h]]$ where both $t$ and $h$ have degree 1 (EDIT: by which I mean that it is the completion of a graded $\mathbb C[t,h]$-module). Now assume that $\mathcal F$ is flat over infinitely many lines through the origin of the form $t=ah$ for $a\in \mathbb C$ (EDIT: by which I mean that there are infinitely many complex numbers $a\in \mathbb C$ such that the pullback to the subscheme $\operatorname{Spec} \mathbb C[[t,h]]/(t-ah)$ is free as a module over $\mathbb C[[t,h]]/(t-ah)$).

Can we conclude that $\mathcal F$ is flat?

For coherent sheaves, I think I can see an argument by comparing dimensions of residues (if the dimension of the residue at the closed point is the same as at the generic points of infinitely lines, it also must be the same as that at the generic point), but of course, that doesn't make sense for quasi-coherent sheaves.

Is there some better argument, or can the behavior of these modules be more pathological than I realize? I'd really appreciate a pointer toward a reference, as well (or at least guidance on where to look for it).

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    $\begingroup$ What do you mean by saying that ${\mathcal{F}}$ is flat over a line? And what do you mean by saying that it is graded? Usually people consider graded modules over graded rings and a graded ring is a direct SUM of its graded components, while $C[[t,h]]$ is a direct PRODUCT. Does it mean that the module is completion of a graded module over $C[t,h]$? $\endgroup$ – Sasha Dec 16 '10 at 9:18
  • $\begingroup$ Minor nitpick: I think the standard name is "formal polydisk" instead of "formal plane". $\endgroup$ – S. Carnahan Dec 17 '10 at 20:04

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