The Stacks project

Lemma 60.18.2. In Situation 60.7.5. Assume moreover $X$ and $S$ are affine schemes. Consider the full subcategory $\mathcal{C} \subset \text{Cris}(X/S)$ consisting of divided power thickenings $(X, T, \delta )$ endowed with the chaotic topology (see Sites, Example 7.6.6). For any locally quasi-coherent $\mathcal{O}_{X/S}$-module $\mathcal{F}$ we have

\[ R\Gamma (\mathcal{C}, \mathcal{F}|_\mathcal {C}) = R\Gamma (\text{Cris}(X/S), \mathcal{F}) \]

Proof. Denote $\text{AffineCris}(X/S)$ the fully subcategory of $\text{Cris}(X/S)$ consisting of those objects $(U, T, \delta )$ with $U$ and $T$ affine. We turn this into a site by saying a family of morphisms $\{ (U_ i, T_ i, \delta _ i) \to (U, T, \delta )\} _{i \in I}$ of $\text{AffineCris}(X/S)$ is a covering if and only if it is a covering of $\text{Cris}(X/S)$. With this definition the inclusion functor

\[ \text{AffineCris}(X/S) \longrightarrow \text{Cris}(X/S) \]

is a special cocontinuous functor as defined in Sites, Definition 7.29.2. The proof of this is exactly the same as the proof of Topologies, Lemma 34.3.10. Thus we see that the topos of sheaves on $\text{Cris}(X/S)$ is the same as the topos of sheaves on $\text{AffineCris}(X/S)$ via restriction by the displayed inclusion functor. Therefore we have to prove the corresponding statement for the inclusion $\mathcal{C} \subset \text{AffineCris}(X/S)$.

We will use without further mention that $\mathcal{C}$ and $\text{AffineCris}(X/S)$ have products and fibre products (details omitted, see Lemma 60.8.2). The inclusion functor $u : \mathcal{C} \to \text{AffineCris}(X/S)$ is fully faithful, continuous, and commutes with products and fibre products. We claim it defines a morphism of ringed sites

\[ f : (\text{AffineCris}(X/S), \mathcal{O}_{X/S}) \longrightarrow (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_{X/S}|_\mathcal {C}) \]

To see this we will use Sites, Lemma 7.14.6. Note that $\mathcal{C}$ has fibre products and $u$ commutes with them so the categories $\mathcal{I}^ u_{(U, T, \delta )}$ are disjoint unions of directed categories (by Sites, Lemma 7.5.1 and Categories, Lemma 4.19.8). Hence it suffices to show that $\mathcal{I}^ u_{(U, T, \delta )}$ is connected. Nonempty follows from Lemma 60.5.6: since $U$ and $T$ are affine that lemma says there is at least one object $(X, T', \delta ')$ of $\mathcal{C}$ and a morphism $(U, T, \delta ) \to (X, T', \delta ')$ of divided power thickenings. Connectedness follows from the fact that $\mathcal{C}$ has products and that $u$ commutes with them (compare with the proof of Sites, Lemma 7.5.2).

Note that $f_*\mathcal{F} = \mathcal{F}|_\mathcal {C}$. Hence the lemma follows if $R^ pf_*\mathcal{F} = 0$ for $p > 0$, see Cohomology on Sites, Lemma 21.14.6. By Cohomology on Sites, Lemma 21.7.4 it suffices to show that $H^ p(\text{AffineCris}(X/S)/(X, T, \delta ), \mathcal{F}) = 0$ for all $(X, T, \delta )$. This follows from Lemma 60.18.1 because the topos of the site $\text{AffineCris}(X/S)/(X, T, \delta )$ is equivalent to the topos of the site $\text{Cris}(X/S)/(X, T, \delta )$ used in the lemma. $\square$

Comments (0)

There are also:

  • 4 comment(s) on Section 60.18: General remarks on cohomology

Post a comment

Your email address will not be published. Required fields are marked.

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.

All contributions are licensed under the GNU Free Documentation License.

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 07JK. Beware of the difference between the letter 'O' and the digit '0'.