Let's say I have a field $\mathbb{K}$ and a flat family of $\mathbb{K}[t]$-modules $M$ over the formal disk $Spec \mathbb{K}[[h]]$.

Now, assume that $M/hM$ is torsion as a $\mathbb{K}[t]$-module (but NOT finitely generated). Can I conclude that $M[h^{-1}]$ is torsion as a $\mathbb{K}((h))[t]$-module?

  • $\begingroup$ Oh boy! Our first bounty. $\endgroup$ – Anton Geraschenko Oct 10 '09 at 21:29
  • $\begingroup$ I figure it's worth a shot. I think it's probably true and obvious, but I'm having some kind of mental block coming up with a proof. $\endgroup$ – Ben Webster Oct 10 '09 at 23:49
  • $\begingroup$ Reminder about bounties: If you don't accept an answer within the 1 week period, the highest-voted answer is automatically accepted (half the bounty is awarded to the answerer, and half disappears). If you accept an answer before the week is over, the bounty immediately ends and the decision is binding (you can't change your mind about which answer to accept like you normally can). $\endgroup$ – Anton Geraschenko Oct 11 '09 at 20:02

Let M = k[[h]][x] = \bigoplus_{i=0}^{\infty}{ k[[h]]x^i }. We make this into a flat family of k[t] modules by setting t x^i = h x^{i+1}. Or in other words, p(h,t) \in k[[h]][t] acts by multiplication by p(h,xh) (wrt the natural ring structure on k[[h]][x]). Clearly M/hM = k[x] with the action by t equal to zero. Consider M[h^{-1}] = k((h))[x]. Since k((h))[x] is without zero divisors, and p(h,hx) is nonzero so long as p(h,t) was nonzero, we see that M[h^{-1}] is torsion free as a k((h))[t] module.


[I'm going to work over k[h] as the base instead; I don't think anything changes, but if I'm wrong you should let me know.]

Consider the case M = k[s,t,h]/(st-h^2). Setting h=0 yields M_0 = k[s,t]/(st), which is certainly torsion over k[t]. But inverting h yields M' = k[s,t,h,h^{-1}]/(st-h^2). This is a (Z-)graded k[t]-module, so to show it is torsion-free it suffices to show that there are no homogeneous torsion elements.

Let p be any element of k[t] and suppose that pg = f*(st-h^2) for some f in k[s,t,h,h^{-1}] and some homogeneous element g in k[s,t,h,h^{-1}]. Again by homogeneity, we can assume that both f and p are homogeneous, so in particular p = t^k for some k.

But now, we have g * t^k = f * (st-h^2). The ring k[s,t,h,h^{-1}] is a UFD, so either t|f or t|(st-h^2). The latter is false; every degree-1 element of this ring looks like (as + bt + ch)(p(s/h, t/h)), and clearing denominators, to have a solution to pt = st-h^2 would be to have p(s,t)(as + bt + ch)t = h^j(st-h^2), which can't happen since the left-hand-side doesn't have any terms of degree > 1 in h. Hence t|f, so repeating this argument, t^k | f. Dividing both sides by t^k shows that g is divisible by (st-h^2), so g=0 as an element of M'.

I haven't checked that M is flat over k[h], but the definition of M certainly suggests that this should be the case.

  • $\begingroup$ I'm confused by your answer. Most importantly, why do you think M is flat? It's not clear to me that it is, but I'm going to need some kind of justification before I believe you. $\endgroup$ – Ben Webster Oct 6 '09 at 16:29
  • $\begingroup$ A module over a P.I.D. is flat iff it is torsion-free. So the fact that no polynomial in k[h] divides (st-h^2) in k[s,t,h] immediately implies that M is flat over k[h]. $\endgroup$ – Charley Oct 7 '09 at 4:16
  • $\begingroup$ k[s,t]/(st) is certainly not torsion over t! I knew something about this counterexample smelled fishy! $\endgroup$ – Ben Webster Oct 10 '09 at 21:24
  • $\begingroup$ I see what's going on now. I misinterpreted "torsion" as "not torsion-free." $\endgroup$ – Charley Oct 12 '09 at 15:13

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