## Tag `07LH`

Chapter 54: Crystalline Cohomology > Section 54.21: Cohomology in the affine case

Lemma 54.21.4. Assumptions as in Proposition 54.21.3. 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 54.17.5. 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 54.17.5. 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 54.21.3. 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 54.20.2 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 54.20.2 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. $\square$

The code snippet corresponding to this tag is a part of the file `crystalline.tex` and is located in lines 4028–4040 (see updates for more information).

```
\begin{lemma}
\label{lemma-compute-cohomology-crystal-smooth}
Assumptions as in Proposition \ref{proposition-compute-cohomology-crystal}.
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 \ref{lemma-crystals-on-affine-smooth}. Then the complex
$$
M' \otimes^\wedge_{D'} \Omega^*_{D'}
$$
computes $R\Gamma(\text{Cris}(X/S), \mathcal{F})$.
\end{lemma}
\begin{proof}
Choose $a : D \to D'$ and $b : D' \to D$ as in
Lemma \ref{lemma-crystals-on-affine-smooth}.
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 \ref{proposition-compute-cohomology-crystal}.
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.
\medskip\noindent
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 {\it 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 \ref{lemma-relative-poincare}
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 \ref{lemma-relative-poincare} 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.
\end{proof}
```

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