Crystalline cohomology via the syntomic site - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T14:37:17Z http://mathoverflow.net/feeds/question/65865 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65865/crystalline-cohomology-via-the-syntomic-site Crystalline cohomology via the syntomic site unknown 2011-05-24T15:45:23Z 2011-05-25T12:17:15Z <p>Hello,</p> <p>Let $k$ be a field of characteristic $p > 0$, and let $Y$ be a $k$-scheme. Consider the sites $Y_{syn}$ and $(Y/W_n)_{cris}$ (where $W_n$ are the Witt vectors of $k$ of length $n$), of $Y$ with its syntomic topology and its crystalline topology. Then the assignment $\mathcal O_{cris}:Z\mapsto H^0_{cris}(Y/W_n)$ is a sheaf on $Y_{syn}$. It is a fact that <code>$H^*_{syn}(Y,\mathcal O_{cris})$</code> is canonically isomorphic to $H^{*}_{cris}(Y/W_n)$, but I don't see how to prove it.</p> <p>So my question is: How does one prove this fact?</p> <p>Thanks!</p> http://mathoverflow.net/questions/65865/crystalline-cohomology-via-the-syntomic-site/65917#65917 Answer by Nicolas Ojeda Bar for Crystalline cohomology via the syntomic site Nicolas Ojeda Bar 2011-05-25T01:08:11Z 2011-05-25T02:00:05Z <p>A sketch of the proof is as follows:</p> <p>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 $\alpha_{*}(F)(U) = H^0_{syn-cris}(U/W_n,F)$.</p> <p>Lemma. $R^i\beta_*\mathcal O_{Y/W_n} = R^i\alpha_*\mathcal O_{Y/W_n} = 0$ for $i > 0$.</p> <p>This implies the result in your question by a standard application of the Leray spectral sequence.</p> <p>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).</p> <p>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 $A[T]\to A[T^{p^{-n}}]$ is a syntomic cover that has the property that the image of $dT$ is zero.</p>