# Grothendieck-Messing theory

Hello,

I would like to work out some examples of deformation of isogenies via Grothendieck-Messing theory. Let's take an easy example: Let A be an abelian variety over $k=\overline{\mathbb{F}_p}$ and $f:A\rightarrow A$ be the multiplication by $p$ map. Let $M=\mathbb{D}(A)$ be the Dieudonné module of $A$. It is a free $W(k)$-module of rank $2 dim(A)$. Finally let $R=\frac{k[t]}{(t^2)}$.

Then lifting $A$ to $R$ boils down to lifting the Hodge filtration $$0 \rightarrow \frac{FM}{pM} \rightarrow \frac{M}{pM} \rightarrow \frac{VM}{pM} \rightarrow 0$$ to a filtration over $R$: $$0 \rightarrow P \rightarrow \frac{M}{pM}\otimes R \rightarrow P' \rightarrow 0$$

Here I have used the fact that what is written sometimes $\mathbb{D}(A)_k$ is actually $\frac{M}{pM}$, and $\mathbb{D}(A)_R=\mathbb{D}(A)_k \otimes R$ (am I right ?). Now, let's move on to deforming the isogeny $f$. Lifting $f$ to $R$ is equivalent to lifting two such Hodge filtrations in a compatible way with $f$. Here is what I don't understand:

The map $f_k$ induced by $f$ on $\mathbb{D}(A)_k=\frac{M}{pM}$ is obviously the zero map. But how can I compute the map induced on $\mathbb{D}(A)_R$ ? Is it simply $f_k\otimes R=0$ ? I doubt it.

Thank you!

Jean-Stefan

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The map induced by $f$ on $\mathbb{D}(A)$ is just multiplication by $p$! Why do you find this so unbelievable? – Keerthi Madapusi Pera Oct 11 '12 at 19:37
Yes, the map induced on $\mathbb{D}(A)$ is multiplication by $p$, but what about the map induced on $\mathbb{D}(A)_R$ ? – Kskvrt Oct 11 '12 at 19:53
You may use the universal property of tensor product to transfer multiplication by $p$ to the right side. – S. Carnahan Oct 12 '12 at 1:04
By $\mathbb{D}(A)$ I meant the Dieudonne crystal of $A$. – Keerthi Madapusi Pera Oct 12 '12 at 11:11
There's something I don't quite understand: To an abelian variety over $k=\overline{\mathbb{F}_p}$, we can associate a Dieudonné module $\mathbb{D}(A)$ which is a free $W(k)$-module. Now, according to Grothendick-Messing theory, for every locally nilpotent divided power extension $R/k$, there is also a free $R$-module $\mathbb{D}(A)_R$. What is the connection between $\mathbb{D}(A)$ and $\mathbb{D}(A)_R$ ? My first thought was that $\mathbb{D}(A)_R=\mathbb{D}(A)\otimes_{W(k)} R$, where $R$ is considered a $W(k)$-module via the map $W(k)\rightarrow k \rightarrow R$. Is this true ? – Kskvrt Oct 12 '12 at 12:18