Question

Let's say I have a field k and a flat family of k[t]-modules M over the formal disk Spec k[[h]].

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

Answer 1

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.

Answer 2

[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.