The Stacks project

Proposition 60.17.4. The functor

\[ \begin{matrix} \text{crystals in quasi-coherent} \\ \mathcal{O}_{X/S}\text{-modules on }\text{Cris}(X/S) \end{matrix} \longrightarrow \begin{matrix} \text{pairs }(M, \nabla )\text{ satisfying} \\ \text{(07JB), (07JC), (07JD), and (07JE)} \end{matrix} \]

of Lemma 60.17.3 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 $T = \mathop{\mathrm{Spec}}(B)$ and the ideal of $U \to T$ is $J_ B \subset B$. By Lemma 60.5.6 there exists an integer $e$ and a morphism

\[ f : (U, T, \delta ) \longrightarrow (X, T_ e, \bar\gamma ) \]

where $T_ e = \mathop{\mathrm{Spec}}(D_ e)$ as in the proof of Lemma 60.17.3. 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 60.17.3 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 60.17.3. $\square$

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