# lifting the isomorphisms between abelian schemes over PD thickenings

Assume that X and Y are abelian schemes (or even abelian varieties) over a base T. If $\ S --> T$ is a PD nilpotent thickening (i.e. the ideal of $\ S$ in $\ T$ is a nilpotent divided power ideal) and S is of characteristic $\ p>0$ . If $\ X_{0}$ and $\ Y_{0}$ are reductions of $\ X$ and $\ Y$ to $\ S$. If $\ f_{0} : X_{0} ---> Y_{0}$ is an isomorphism over S, is this true that this isomorphism (if it can be lifted!) lifts to an isomorphism $\ f : X ---> Y$ ? one may also assume that $\ H¹_ {cris} (f_{0})$ preserves the Hodge filtration.

• I deleted my answer since I think I had misunderstood the question. Are you asking whether any lift of $f_0$ to a morphism $f:X \to Y$ is necessarily an isomorphism?
– naf
Sep 23, 2011 at 11:31
• Yes! in fact this is what I ask and I think your answer was correct. and if I remember properly your counterexample also satisfied the condition on Hodge filtration.So it perfectly answered my question. Thank you very much.
– Jack
Sep 23, 2011 at 18:00
• Now I am confused since in my example there was no lift. In fact, I think that if a lift exists then it must indeed be an isomorphism.
– naf
Sep 24, 2011 at 14:47
• As far as I remember, in your answer you took $S = Spec(\mathbb{Z}/p)$ and $T = Spec(\mathbb{Z}/p^2)$ and then using the fact that The versal deformation space of an elliptic curve $E$ over $S$ is isomorphic to $Spec(\mathbb{Z}_p[[x]])$, you rgued that there are lifts of $E$ that are not isomorphims over $T$. So in particular the identitity map of $E$ does not lift to an isomorphism.
– Jack
Sep 24, 2011 at 19:44
• Yes, that's right. Since that's what you wanted I will undelete the answer.
– naf
Sep 25, 2011 at 6:51

For a counterexample, let $S = Spec(\mathbb{Z}/p)$ and $T = Spec(\mathbb{Z}/p^2)$. The versal deformation space of an elliptic curve $E$ over $S$ is isomorphic to $Spec(\mathbb{Z}_p[[x]])$ so lifts of $E$ to $T$ are parametrized by the set of homomorphisms of local algebras $Hom(\mathbb{Z}_p[[x]], \mathbb{Z}/p^2) = p\mathbb{Z}/p^2$. So there do exist lifts for which the identity map of $E$ does not lift to an isomorphism.