The Stacks project

Lemma 60.17.3. In the situation above there is a functor

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

Proof. Let $\mathcal{F}$ be a crystal in quasi-coherent modules on $X/S$. Set $T_ e = \mathop{\mathrm{Spec}}(D_ e)$ so that $(X, T_ e, \bar\gamma )$ is an object of $\text{Cris}(X/S)$ for $e \gg 0$. We have morphisms

\[ (X, T_ e, \bar\gamma ) \to (X, T_{e + 1}, \bar\gamma ) \to \ldots \]

which are closed immersions. We set

\[ M = \mathop{\mathrm{lim}}\nolimits _ e \Gamma ((X, T_ e, \bar\gamma ), \mathcal{F}) = \mathop{\mathrm{lim}}\nolimits _ e \Gamma (T_ e, \mathcal{F}_{T_ e}) = \mathop{\mathrm{lim}}\nolimits _ e M_ e \]

Note that since $\mathcal{F}$ is locally quasi-coherent we have $\mathcal{F}_{T_ e} = \widetilde{M_ e}$. Since $\mathcal{F}$ is a crystal we have $M_ e = M_{e + 1}/p^ eM_{e + 1}$. Hence we see that $M_ e = M/p^ eM$ and that $M$ is $p$-adically complete, see Algebra, Lemma 10.98.2.

By Lemma 60.15.1 we know that $\mathcal{F}$ comes endowed with a canonical integrable connection $\nabla : \mathcal{F} \to \mathcal{F} \otimes \Omega _{X/S}$. If we evaluate this connection on the objects $T_ e$ constructed above we obtain a canonical integrable connection

\[ \nabla : M \longrightarrow M \otimes ^\wedge _ D \Omega _ D \]

To see that this is topologically nilpotent we work out what this means.

Now we can do the same procedure for the rings $D(n)$. This produces a $p$-adically complete $D(n)$-module $M(n)$. Again using the crystal property of $\mathcal{F}$ we obtain isomorphisms

\[ M \otimes ^\wedge _{D, p_0} D(1) \rightarrow M(1) \leftarrow M \otimes ^\wedge _{D, p_1} D(1) \]

compare with the proof of Lemma 60.15.1. Denote $c$ the composition from left to right. Pick $m \in M$. Write $\xi _ i = x_ i \otimes 1 - 1 \otimes x_ i$. Using ( we can write uniquely

\[ c(m \otimes 1) = \sum \nolimits _ K \theta _ K(m) \otimes \prod \xi _ i^{[k_ i]} \]

for some $\theta _ K(m) \in M$ where the sum is over multi-indices $K = (k_ i)$ with $k_ i \geq 0$ and $\sum k_ i < \infty $. Set $\theta _ i = \theta _ K$ where $K$ has a $1$ in the $i$th spot and zeros elsewhere. We have

\[ \nabla (m) = \sum \theta _ i(m) \text{d}x_ i. \]

as can be seen by comparing with the definition of $\nabla $. Namely, the defining equation is $p_1^*m = \nabla (m) - c(p_0^*m)$ in Lemma 60.15.1 but the sign works out because in the Stacks project we consistently use $\text{d}f = p_1(f) - p_0(f)$ modulo the ideal of the diagonal squared, and hence $\xi _ i = x_ i \otimes 1 - 1 \otimes x_ i$ maps to $-\text{d}x_ i$ modulo the ideal of the diagonal squared.

Denote $q_ i : D \to D(2)$ and $q_{ij} : D(1) \to D(2)$ the coprojections corresponding to the indices $i, j$. As in the last paragraph of the proof of Lemma 60.15.1 we see that

\[ q_{02}^*c = q_{12}^*c \circ q_{01}^*c. \]

This means that

\[ \sum \nolimits _{K''} \theta _{K''}(m) \otimes \prod {\zeta ''_ i}^{[k''_ i]} = \sum \nolimits _{K', K} \theta _{K'}(\theta _ K(m)) \otimes \prod {\zeta '_ i}^{[k'_ i]} \prod \zeta _ i^{[k_ i]} \]

in $M \otimes ^\wedge _{D, q_2} D(2)$ where

\begin{align*} \zeta _ i & = x_ i \otimes 1 \otimes 1 - 1 \otimes x_ i \otimes 1,\\ \zeta '_ i & = 1 \otimes x_ i \otimes 1 - 1 \otimes 1 \otimes x_ i,\\ \zeta ''_ i & = x_ i \otimes 1 \otimes 1 - 1 \otimes 1 \otimes x_ i. \end{align*}

In particular $\zeta ''_ i = \zeta _ i + \zeta '_ i$ and we have that $D(2)$ is the $p$-adic completion of the divided power polynomial ring in $\zeta _ i, \zeta '_ i$ over $q_2(D)$, see Lemma 60.17.1. Comparing coefficients in the expression above it follows immediately that $\theta _ i \circ \theta _ j = \theta _ j \circ \theta _ i$ (this provides an alternative proof of the integrability of $\nabla $) and that

\[ \theta _ K(m) = (\prod \theta _ i^{k_ i})(m). \]

In particular, as the sum expressing $c(m \otimes 1)$ above has to converge $p$-adically we conclude that for each $i$ and each $m \in M$ only a finite number of $\theta _ i^ k(m)$ are allowed to be nonzero modulo $p$. $\square$

Comments (8)

Comment #3059 by on

In the first paragraph of the proof you said "Hence we see that and that is -adically complete." I could not understand why this is true. Do you have a reference for that? The only reference that I could find in this direction is EGA1, 7.2.9, where Grothendieck proved that if you start with a system of modules satisfying , then after taking the limit you could recover those via taking quotients. But there, is assumed to finitely generated over , and this condition seems to have been used in the argument. If I understand correctly the finiteness condition is used in the citation of the proposition 10.24 in Atiyah-McDonald.

Comment #3063 by Zhang on

@Lei: I think this is because F is a crystal, and is a map in the cris site. By the definition of a crystal, the pullback map is an isomorphism. As M is defined to be the limit, it is by fiat p-adically complete. Right?

Comment #3132 by Lei Zhang on

Sorry, it is still not clear to me. I know that is a crystal so that . But this does not explain why the limit is complete. If one can prove that , then this fine.

Comment #3133 by Lei Zhang on

Sorry, it is still not clear to me. I know that is a crystal so that . But this does not explain why the limit is complete. If one can prove that , then this fine.

Comment #3163 by on

OK, I put in a reference to the relevant algebra lemma which is Lemma 10.98.2. See changes here.

Comment #3175 by Lei Zhang on

@Johan: Thanks, this is very convincing!

Comment #5924 by Matthieu Romagny on

In the statement of the Lemma, I guess that conditions (07JB), (07JC), (07JD), and (07JE) refer to the four conditions listed just before. However, on my browser at least they are displayed as (1) (2) (3) (4). Is there anything that can be done about it ?

Comment #5925 by on

There are two solutions to this problem I think.

It would make sense to let the "tags vs. numbers" toggle switch to tags on numbered lists (provided they have tags assigned to their entries of course, which is not the case for every list). This is a good idea, and I'll try to implement it soon. For now, it's listed at

Moreover, it is in fact possible to link things within MathJax, so that's another possible improvement, listed at That way, you can click on the tag in the equation and go to the precise statement.

For now, you'll have to go the tag listed in the statement of the lemma by typing the tag in the search box in the top-right corner.

Thanks for noticing!

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