3 added 2 characters in body

A sketch of the proof is as follows:

Consider the site $Y_{syn-cris}$ where the objects are the same as in $Y_{cris}$ but the covering families are surjective syntomic families. Then there are maps of topoi: $\alpha : Sh(Y_{syn-cris})\to Sh(Y_{syn})$ and $\beta : Sh(Y_{syn-cris})\to Sh((Y/W_n)_ {cris})$, defined by $\beta_*(F)(U,T) = F(U, T)$ and $\alpha_{*}(F)(U) = H^0_{syn-cris}(U/W_n,F)$.

Lemma. $R^i\beta_*\mathcal O_{Y/W_n} = R^i\alpha_*\mathcal O_{Y/W_n} = 0$ for $i > 0$.

This implies the result in your question by a standard application of the Leray spectral sequence.

As for the lemma: To prove that $R^i\beta_*\mathcal O_{Y/W_n} = 0$ for $i > 0$, it boils down to checking that if $U$ is an affine in $(Y/W_n)_ {cris}$ then $H^i_{syn}(U,\mathcal O_U) = 0$ for $i > 0$, and this follows the theory of the Cech complex (it is acyclic because of faithful flatness of the syntomic cover).

To prove that $R^i\alpha_*\mathcal O_{Y/W_n} = 0$, we have to check that if $U$ is an open of $Y_{syn}$, and $s\in H^i_{cris}(U/W_n)$ (strictly speaking we have to do the computation with syntomic-cristalline cohomology, but by the previous part, the cohomology groups coincide with the crystalline cohomology groups) then there exists a syntomic cover $U_i\to U$ such that $s\mid U_i = 0\in H^i_{cris}(U_i/W_n)$. Now recall that we can compute this cohomology groups as the hypercohomology groups of the de Rham complex of the divided power envelope of some embedding into a smooth scheme. That means, after shrinking, we can represent $s$ as an $i$-form. We need to find a syntomic cover such that when we restrict $s$ to this cover, it vanishes. To do this, note that $A[T]\to A[T^{p^-1}]$ A[T^{p^{-n}}]$is a syntomic cover that has the property that the image of$dT$is zero. 2 added 60 characters in body A sketch of the proof is as follows: Consider the site$Y_{syn-cris}$where the objects are the same as in$Y_{cris}$but the covering families are surjective syntomic families. Then there are maps of topoi:$\alpha : Sh(Y_{syn-cris})\to Sh(Y_{syn})$and$\beta : Sh(Y_{syn-cris})\to Sh((Y/W_n)_ {cris})$, defined by$\beta_*(F)(U,T) = F(U, T)$and$\alpha_{*}(F)(U) = H^0_{syn-cris}(U/W_n,F)$. Lemma.$R^i\beta_*\mathcal O_{Y/W_n} = R^i\alpha_*\mathcal O_{Y/W_n} = 0$for$i > 0$. This implies the result in your question by a standard application of the Leray spectral sequence. As for the lemma: To prove that$R^i\beta_*\mathcal O_{Y/W_n} = 0$for$i > 0$, it boils down to checking that if$U$is an affine in$(Y/W_n)_ {cris}$then$H^i_{syn}(U,\mathcal O_U) = 0$for$i > 0$, and this follows from computing using the theory of the Cech complex (it is acyclic because of faithful flatness of the syntomic cover). To prove that$R^i\alpha_*\mathcal O_{Y/W_n} = 0$, we have to check that if$U$is an open of$Y_{syn}$, and$s\in H^i_{cris}(U/W_n)$(strictly speaking we have to do the computation with syntomic-cristalline cohomology, but by the previous part, the cohomology groups coincide with the crystalline cohomology groups) then there exists a syntomic cover$U_i\to U$such that$s\mid U_i = 0\in H^i_{cris}(U_i/W_n)$. Now recall that we can compute this cohomology groups as the hypercohomology groups of the de Rham complex of the divided power envelope of some embedding into a smooth scheme. That means, after shrinking, we can represent$s$as an$i$-form. We need to find a syntomic cover such that when we restrict$s$to this cover, it vanishes. To do this, note that$A[T]\to A[T^{p^-1}]$is a syntomic cover that has the property that the image of$dT$is zero. 1 A sketch of the proof is as follows: Consider the site$Y_{syn-cris}$where the objects are the same as in$Y_{cris}$but the covering families are surjective syntomic families. Then there are maps of topoi:$\alpha : Sh(Y_{syn-cris})\to Sh(Y_{syn})$and$\beta : Sh(Y_{syn-cris})\to Sh((Y/W_n)_ {cris})$, defined by$\beta_*(F)(U,T) = F(U, T)$and$\alpha_{*}(F)(U) = H^0_{syn-cris}(U/W_n,F)$. Lemma.$R^i\beta_*\mathcal O_{Y/W_n} = R^i\alpha_*\mathcal O_{Y/W_n} = 0$for$i > 0$. This implies the result in your question by a standard application of the Leray spectral sequence. As for the lemma: To prove that$R^i\beta_*\mathcal O_{Y/W_n} = 0$for$i > 0$, it boils down to checking that if$U$is an affine in$(Y/W_n)_ {cris}$then$H^i_{syn}(U,\mathcal O_U) = 0$for$i > 0$, and this follows from computing using the Cech complex. To prove that$R^i\alpha_*\mathcal O_{Y/W_n} = 0$, we have to check that if$U$is an open of$Y_{syn}$, and$s\in H^i_{cris}(U/W_n)$(strictly speaking we have to do the computation with syntomic-cristalline cohomology, but by the previous part, the cohomology groups coincide with the crystalline cohomology groups) then there exists a syntomic cover$U_i\to U$such that$s\mid U_i = 0\in H^i_{cris}(U_i/W_n)$. Now recall that we can compute this cohomology groups as the hypercohomology groups of the de Rham complex of the divided power envelope of some embedding into a smooth scheme. That means, after shrinking, we can represent$s$as an$i$-form. We need to find a syntomic cover such that when we restrict$s$to this cover, it vanishes. To do this, note that$A[T]\to A[T^{p^-1}]$is a syntomic cover that has the property that the image of$dT\$ is zero.