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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.

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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? – ulrich Sep 23 '11 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 '11 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. – ulrich Sep 24 '11 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 '11 at 19:44
Yes, that's right. Since that's what you wanted I will undelete the answer. – ulrich Sep 25 '11 at 6:51
up vote 1 down vote accepted

No, this is very far from being true.

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.

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Won't the condition on the Hodge filtration sort this out though? More generally, I thought the statement the OP wants follows from Serre-Tate theory...? – anon Sep 22 '11 at 19:08
I don't understand the relevance of Serre-Tate theory to the question. In any case, according to the OP what I wrote does answer his question. – ulrich Sep 25 '11 at 6:53

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