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Tag 07JH

Proposition 51.17.4. The functor $$\begin{matrix} \text{crystals in quasi-coherent} \\ \mathcal{O}_{X/S}\text{-modules on }\text{Cris}(X/S) \end{matrix} \longrightarrow \begin{matrix} \text{pairs }(M, \nabla)\text{ satisfying} \\ \text{(1), (2), (3), and (4)} \end{matrix}$$ of Lemma 51.17.3 is an equivalence of categories.

Proof. Let $(M, \nabla)$ be given. We are going to construct a crystal in quasi-coherent modules $\mathcal{F}$. Write $\nabla(m) = \sum \theta_i(m)\text{d}x_i$. Then $\theta_i \circ \theta_j = \theta_j \circ \theta_i$ and we can set $\theta_K(m) = (\prod \theta_i^{k_i})(m)$ for any multi-index $K = (k_i)$ with $k_i \geq 0$ and $\sum k_i < \infty$.

Let $(U, T, \delta)$ be any object of $\text{Cris}(X/S)$ with $T$ affine. Say $T = \mathop{\rm Spec}(B)$ and the ideal of $U \to T$ is $J_B \subset B$. By Lemma 51.5.6 there exists an integer $e$ and a morphism $$f : (U, T, \delta) \longrightarrow (X, T_e, \bar\gamma)$$ where $T_e = \mathop{\rm Spec}(D_e)$ as in the proof of Lemma 51.17.3. Choose such an $e$ and $f$; denote $f : D \to B$ also the corresponding divided power $A$-algebra map. We will set $\mathcal{F}_T$ equal to the quasi-coherent sheaf of $\mathcal{O}_T$-modules associated to the $B$-module $$M \otimes_{D, f} B.$$ However, we have to show that this is independent of the choice of $f$. Suppose that $g : D \to B$ is a second such morphism. Since $f$ and $g$ are morphisms in $\text{Cris}(X/S)$ we see that the image of $f - g : D \to B$ is contained in the divided power ideal $J_B$. Write $\xi_i = f(x_i) - g(x_i) \in J_B$. By analogy with the proof of Lemma 51.17.3 we define an isomorphism $$c_{f, g} : M \otimes_{D, f} B \longrightarrow M \otimes_{D, g} B$$ by the formula $$m \otimes 1 \longmapsto \sum\nolimits_K \theta_K(m) \otimes \prod \xi_i^{[k_i]}$$ which makes sense by our remarks above and the fact that $\nabla$ is topologically quasi-nilpotent (so the sum is finite!). A computation shows that $$c_{g, h} \circ c_{f, g} = c_{f, h}$$ if given a third morphism $h : (U, T, \delta) \longrightarrow (X, T_e, \bar\gamma)$. It is also true that $c_{f, f} = 1$. Hence these maps are all isomorphisms and we see that the module $\mathcal{F}_T$ is independent of the choice of $f$.

If $a : (U', T', \delta') \to (U, T, \delta)$ is a morphism of affine objects of $\text{Cris}(X/S)$, then choosing $f' = f \circ a$ it is clear that there exists a canonical isomorphism $a^*\mathcal{F}_T \to \mathcal{F}_{T'}$. We omit the verification that this map is independent of the choice of $f$. Using these maps as the restriction maps it is clear that we obtain a crystal in quasi-coherent modules on the full subcategory of $\text{Cris}(X/S)$ consisting of affine objects. We omit the proof that this extends to a crystal on all of $\text{Cris}(X/S)$. We also omit the proof that this procedure is a functor and that it is quasi-inverse to the functor constructed in Lemma 51.17.3. $\square$

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\begin{proposition}
\label{proposition-crystals-on-affine}
The functor
$$\begin{matrix} \text{crystals in quasi-coherent} \\ \mathcal{O}_{X/S}\text{-modules on }\text{Cris}(X/S) \end{matrix} \longrightarrow \begin{matrix} \text{pairs }(M, \nabla)\text{ satisfying} \\ \text{(\ref{item-complete}), (\ref{item-connection}), (\ref{item-integrable}), and (\ref{item-topologically-quasi-nilpotent})} \end{matrix}$$
of Lemma \ref{lemma-crystals-on-affine}
is an equivalence of categories.
\end{proposition}

\begin{proof}
Let $(M, \nabla)$ be given. We are going to construct
a crystal in quasi-coherent modules $\mathcal{F}$.
Write $\nabla(m) = \sum \theta_i(m)\text{d}x_i$.
Then $\theta_i \circ \theta_j = \theta_j \circ \theta_i$ and we
can set $\theta_K(m) = (\prod \theta_i^{k_i})(m)$ for any multi-index
$K = (k_i)$ with $k_i \geq 0$ and $\sum k_i < \infty$.

\medskip\noindent
Let $(U, T, \delta)$ be any object of $\text{Cris}(X/S)$ with $T$ affine.
Say $T = \Spec(B)$ and the ideal of $U \to T$ is $J_B \subset B$.
By Lemma \ref{lemma-set-generators} there exists an integer $e$ and a morphism
$$f : (U, T, \delta) \longrightarrow (X, T_e, \bar\gamma)$$
where $T_e = \Spec(D_e)$ as in the proof of
Lemma \ref{lemma-crystals-on-affine}.
Choose such an $e$ and $f$; denote $f : D \to B$ also the corresponding
divided power $A$-algebra map. We will set $\mathcal{F}_T$ equal to the
quasi-coherent sheaf of $\mathcal{O}_T$-modules associated to the $B$-module
$$M \otimes_{D, f} B.$$
However, we have to show that this is independent of the choice of $f$.
Suppose that $g : D \to B$ is a second such morphism. Since $f$ and $g$
are morphisms in $\text{Cris}(X/S)$ we see that the image of
$f - g : D \to B$ is contained in the divided power ideal $J_B$.
Write $\xi_i = f(x_i) - g(x_i) \in J_B$. By analogy with the proof
of Lemma \ref{lemma-crystals-on-affine} we define an isomorphism
$$c_{f, g} : M \otimes_{D, f} B \longrightarrow M \otimes_{D, g} B$$
by the formula
$$m \otimes 1 \longmapsto \sum\nolimits_K \theta_K(m) \otimes \prod \xi_i^{[k_i]}$$
which makes sense by our remarks above and the fact that $\nabla$
is topologically quasi-nilpotent (so the sum is finite!).
A computation shows that
$$c_{g, h} \circ c_{f, g} = c_{f, h}$$
if given a third morphism
$h : (U, T, \delta) \longrightarrow (X, T_e, \bar\gamma)$.
It is also true that $c_{f, f} = 1$.
Hence these maps are all isomorphisms and we see that
the module $\mathcal{F}_T$ is independent of the choice of $f$.

\medskip\noindent
If $a : (U', T', \delta') \to (U, T, \delta)$ is a morphism of affine objects
of $\text{Cris}(X/S)$, then choosing $f' = f \circ a$ it is clear
that there exists a canonical isomorphism
$a^*\mathcal{F}_T \to \mathcal{F}_{T'}$. We omit the verification that this
map is independent of the choice of $f$. Using these maps as the restriction
maps it is clear that we obtain a crystal in quasi-coherent modules
on the full subcategory of $\text{Cris}(X/S)$ consisting of affine objects.
We omit the proof that this extends to a crystal on all of
$\text{Cris}(X/S)$. We also omit the proof that this procedure is a functor
and that it is quasi-inverse to the functor constructed in
Lemma \ref{lemma-crystals-on-affine}.
\end{proof}

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