Lemma 60.21.4. Assumptions as in Proposition 60.21.3. Let $A \to P' \to C$ be ring maps with $A \to P'$ smooth and $P' \to C$ surjective with kernel $J'$. Let $D'$ be the $p$-adic completion of $D_{P', \gamma }(J')$. Let $(M', \nabla ')$ be the pair over $D'$ corresponding to $\mathcal{F}$, see Lemma 60.17.5. Then the complex

$M' \otimes ^\wedge _{D'} \Omega ^*_{D'}$

computes $R\Gamma (\text{Cris}(X/S), \mathcal{F})$.

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 $b$)

$M' \otimes ^\wedge _{D'} \Omega ^*_{D'} \longrightarrow M \otimes ^\wedge _ D \Omega ^*_ D \quad \text{and}\quad M \otimes ^\wedge _ D \Omega ^*_ D \longrightarrow M' \otimes ^\wedge _{D'} \Omega ^*_{D'}$

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

$M \otimes ^\wedge _ D \Omega ^*_ D \longrightarrow M \otimes ^\wedge _ D \Omega ^*_ D$

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

$D \xrightarrow {\sigma } D\langle \xi _ i \rangle ^\wedge \xrightarrow {\tau } D$

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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07LH. Beware of the difference between the letter 'O' and the digit '0'.