The Stacks project

Proof. Using assumption (1) and Lemma 60.18.2 we see that $R\Gamma (\text{Cris}(X/S), \mathcal{F})$ is isomorphic to $R\Gamma (\mathcal{C}, \mathcal{F})$. Note that the categories $\mathcal{C}$ used in Lemmas 60.18.2 and 60.18.3 agree. Let $f : T \to T'$ be a closed immersion as in (2). Surjectivity of $c_ f : f^*\mathcal{F}_{T'} \to \mathcal{F}_ T$ is equivalent to surjectivity of $\mathcal{F}_{T'} \to f_*\mathcal{F}_ T$. Hence, if $\mathcal{F}$ satisfies (1) and (2), then we obtain a short exact sequence

of quasi-coherent $\mathcal{O}_{T'}$-modules on $T'$, see Schemes, Section 26.24 and in particular Lemma 26.24.1. Thus, if $T'$ is affine, then we conclude that the restriction map $\mathcal{F}(U', T', \delta ') \to \mathcal{F}(U, T, \delta )$ is surjective by the vanishing of $H^1(T', \mathcal{K})$, see Cohomology of Schemes, Lemma 30.2.2. Hence the transition maps of the inverse systems in Lemma 60.18.3 are surjective. We conclude that $R^ pg_*(\mathcal{F}|_\mathcal {C}) = 0$ for all $p \geq 1$ where $g$ is as in Lemma 60.18.3. The object $D$ of the category $\mathcal{C}^\wedge $ satisfies the assumption of Lemma 60.18.4 by Lemma 60.5.7 with

in $\mathcal{C}$ because $D(n)$ is the $n + 1$-fold coproduct of $D$ in $\text{Cris}^\wedge (C/A)$, see Lemma 60.17.2. Thus we win. $\square$

Proof. Using Lemma 60.12.6 it follows that $\mathcal{H} = \mathcal{F} \otimes _{\mathcal{O}_{X/S}} \Omega ^ i_{X/S}$ also satisfies assumptions (1) and (2) of Proposition 60.21.1. Write $M(n)_ e = \Gamma ((X, T(n)_ e, \delta (n)), \mathcal{F})$ so that $M(n) = \mathop{\mathrm{lim}}\nolimits _ e M(n)_ e$. Then

By Lemma 60.19.3 the cosimplicial modules

are homotopic to zero. Because the transition maps $M(n)_{e + 1} \to M(n)_ e$ are surjective, we see that the inverse limit of the associated complexes are acyclic¹. Hence the vanishing of cohomology of $\mathcal{H}$ by Proposition 60.21.1. $\square$

Proof. We will prove this using the two spectral sequences associated to the double complex $K^{*, *}$ with terms

What do we know so far? Well, Lemma 60.19.3 tells us that each column $K^{a, *}$, $a > 0$ is acyclic. Proposition 60.21.1 tells us that the first column $K^{0, *}$ is quasi-isomorphic to $R\Gamma (\text{Cris}(X/S), \mathcal{F})$. Hence the first spectral sequence associated to the double complex shows that there is a canonical quasi-isomorphism of $R\Gamma (\text{Cris}(X/S), \mathcal{F})$ with $\text{Tot}(K^{*, *})$.

Next, let's consider the rows $K^{*, b}$. By Lemma 60.17.1 each of the $b + 1$ maps $D \to D(b)$ presents $D(b)$ as the $p$-adic completion of a divided power polynomial algebra over $D$. Hence Lemma 60.20.2 shows that the map

is a quasi-isomorphism. Note that each of these maps defines the same map on cohomology (and even the same map in the derived category) as the inverse is given by the co-diagonal map $D(b) \to D$ (corresponding to the multiplication map $P \otimes _ A \ldots \otimes _ A P \to P$). Hence if we look at the $E_1$ page of the second spectral sequence we obtain

with differentials

as each of these is the alternation sum of the given identifications $H^ a(M \otimes ^\wedge _ D\Omega ^*_ D) = E_1^{a, 0} = E_1^{a, 1} = \ldots $. Thus we see that the $E_2$ page is equal $H^ a(M \otimes ^\wedge _ D\Omega ^*_ D)$ on the first row and zero elsewhere. It follows that the identification of $M \otimes ^\wedge _ D\Omega ^*_ D$ with the first row induces a quasi-isomorphism of $M \otimes ^\wedge _ D\Omega ^*_ D$ with $\text{Tot}(K^{*, *})$. $\square$

Proof. Choose $a : D \to D'$ and $b : D' \to D$ as in Lemma 60.17.5. Note that the base change $M = M' \otimes _{D', b} D$ with its connection $\nabla $ corresponds to $\mathcal{F}$. Hence we know that $M \otimes ^\wedge _ D \Omega _ D^*$ computes the crystalline cohomology of $\mathcal{F}$, see Proposition 60.21.3. Hence it suffices to show that the base change maps (induced by $a$ and $b$)

are quasi-isomorphisms. Since $a \circ b = \text{id}_{D'}$ we see that the composition one way around is the identity on the complex $M' \otimes ^\wedge _{D'} \Omega ^*_{D'}$. Hence it suffices to show that the map

induced by $b \circ a : D \to D$ is a quasi-isomorphism. (Note that we have the same complex on both sides as $M = M' \otimes ^\wedge _{D', b} D$, hence $M \otimes ^\wedge _{D, b \circ a} D = M' \otimes ^\wedge _{D', b \circ a \circ b} D = M' \otimes ^\wedge _{D', b} D = M$.) In fact, we claim that for any divided power $A$-algebra homomorphism $\rho : D \to D$ compatible with the augmentation to $C$ the induced map $M \otimes ^\wedge _ D \Omega ^*_ D \to M \otimes ^\wedge _{D, \rho } \Omega ^*_ D$ is a quasi-isomorphism.

Write $\rho (x_ i) = x_ i + z_ i$. The elements $z_ i$ are in the divided power ideal of $D$ because $\rho $ is compatible with the augmentation to $C$. Hence we can factor the map $\rho $ as a composition

where the first map is given by $x_ i \mapsto x_ i + \xi _ i$ and the second map is the divided power $D$-algebra map which maps $\xi _ i$ to $z_ i$. (This uses the universal properties of polynomial algebra, divided power polynomial algebras, divided power envelopes, and $p$-adic completion.) Note that there exists an automorphism $\alpha $ of $D\langle \xi _ i \rangle ^\wedge $ with $\alpha (x_ i) = x_ i - \xi _ i$ and $\alpha (\xi _ i) = \xi _ i$. Applying Lemma 60.20.2 to $\alpha \circ \sigma $ (which maps $x_ i$ to $x_ i$) and using that $\alpha $ is an isomorphism we conclude that $\sigma $ induces a quasi-isomorphism of $M \otimes ^\wedge _ D \Omega ^*_ D$ with $M \otimes ^\wedge _{D, \sigma } \Omega ^*_{D\langle x_ i \rangle ^\wedge }$. On the other hand the map $\tau $ has as a left inverse the map $D \to D\langle x_ i \rangle ^\wedge $, $x_ i \mapsto x_ i$ and we conclude (using Lemma 60.20.2 once more) that $\tau $ induces a quasi-isomorphism of $M \otimes ^\wedge _{D, \sigma } \Omega ^*_{D\langle x_ i \rangle ^\wedge }$ with $M \otimes ^\wedge _{D, \tau \circ \sigma } \Omega ^*_ D$. Composing these two quasi-isomorphisms we obtain that $\rho $ induces a quasi-isomorphism $M \otimes ^\wedge _ D \Omega ^*_ D \to M \otimes ^\wedge _{D, \rho } \Omega ^*_ D$ as desired. $\square$