TITLE: are deformations of torsion modules always torsion?
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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.