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## Can one compare integral structures on de Rham and crystalline cohomology?

Suppose $\mathfrak{X}$ is a smooth projective scheme of finite type over $\mathbb{Z}_p$, with generic fibre $X$. Then there are comparison theorems relating de Rham and crystalline cohomology, $H^i_{\mathrm{dR}}(X / \mathbb{Q}_p) \cong H^i_{\mathrm{cris}}(\mathfrak{X}, \mathbb{Q}_p)$.

Does this work integrally, i.e. does this isomorphism match up $H^i_{\mathrm{dR}}(\mathfrak{X} / \mathbb{Z}_p)$ with $H^i_{\mathrm{cris}}(\mathfrak{X}, \mathbb{Z}_p)$?

Similarly: what if one introduces also the etale cohomology $H^i_{\mathrm{et}}(X_{\overline{\mathbb{Q}}_p}, \mathbb{Q}_p)$? This contains a natural lattice $H^i_{\mathrm{et}}(X_{\overline{\mathbb{Q}}_p}, \mathbb{Z}_p)$. Does this match up with $H^i_{\mathrm{cris}}(\mathfrak{X}, \mathbb{Z}_p)$, via Fontaine's functor $\mathbb{D}_{\rm cris}$?

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Won't applying Fontaine's $\mathbb{D}_{\rm cris}$ to $H^i_{\mathrm{et}}(X_{\overline{\mathbb{Q}}_p}, \mathbb{Z}_p)$ forget the integral structure? (since $B_{\rm cris}$ is a $\mathbb{Q}_p$-algebra) – jnewton Nov 18 2010 at 13:12
Probably you mean to use ${\rm{H}}^i(\mathfrak{X}_0/\mathbf{Z}_p)$ (i.e., cohomology of special fiber, relative to $\mathbf{Z}_p$ as a PD-thickening of the residue field). The integral comparison isom is valid if $e < p-1$, which is to say $p > 2$, since then the divided powers are top. nilpotent. The integral comparison morphism underlies the one with $p$ inverted; it is in the book of Berthelot and Ogus on crystalline cohom. I think that the etale-crystalline case (using $A_{\rm{cris}}$, right?) fails for $i = 2 \dim X \ge p$ since $t^p \in p A_{\rm{cris}}$. – BCnrd Nov 18 2010 at 14:57
@jnewton: As BCnrd says, one can define $\mathbb{D}_{\mathrm{cris}}(T)$ for $T$ a finitely-generated $\mathbb{Z}_p$-module with an action of Galois, using Fontaine's ring $\mathbb{A}_{\mathrm{cris}}$. – David Loeffler Nov 18 2010 at 15:25
@BCnrd: Thanks, that is really useful. I am mainly interested in curves, so $2 \mathrm{dim}(X) < p$ is a pretty mild condition :-) – David Loeffler Nov 18 2010 at 15:28
Dear David: if you're just interested in curves then things become a lot more concrete (and simpler to prove, though $(1/2)\infty = \infty$...). Presumably the degree-1 cohomology is what you care about, and so if $J$ denotes the relative Jacobian then the crystalline cohomology is the (contradvariant) Dieudonne module of the $p$-divisible group $G$ of $J_0$ whereas the etale cohomology is the $p$-adic Tate module of the generic fiber of $J$ (or of $G$). So you're really asking about comparison for $p$-divisible groups in the absolutely unramified case, yes? If so, I can dig up a reference. – BCnrd Nov 18 2010 at 16:01