The Stacks project

60.17 Crystals in quasi-coherent modules

In Situation 60.5.1. Set $X = \mathop{\mathrm{Spec}}(C)$ and $S = \mathop{\mathrm{Spec}}(A)$. We are going to classify crystals in quasi-coherent modules on $\text{Cris}(X/S)$. Before we do so we fix some notation.

Choose a polynomial ring $P = A[x_ i]$ over $A$ and a surjection $P \to C$ of $A$-algebras with kernel $J = \mathop{\mathrm{Ker}}(P \to C)$. Set

60.17.0.1
\begin{equation} \label{crystalline-equation-D} D = \mathop{\mathrm{lim}}\nolimits _ e D_{P, \gamma }(J) / p^ eD_{P, \gamma }(J) \end{equation}

for the $p$-adically completed divided power envelope. This ring comes with a divided power ideal $\bar J$ and divided power structure $\bar\gamma $, see Lemma 60.5.5. Set $D_ e = D/p^ eD$ and denote $\bar J_ e$ the image of $\bar J$ in $D_ e$. We will use the short hand

60.17.0.2
\begin{equation} \label{crystalline-equation-omega-D} \Omega _ D = \mathop{\mathrm{lim}}\nolimits _ e \Omega _{D_ e/A, \bar\gamma } = \mathop{\mathrm{lim}}\nolimits _ e \Omega _{D/A, \bar\gamma }/p^ e\Omega _{D/A, \bar\gamma } \end{equation}

for the $p$-adic completion of the module of divided power differentials, see Lemma 60.6.10. It is also the $p$-adic completion of $\Omega _{D_{P, \gamma }(J)/A, \bar\gamma }$ which is free on $\text{d}x_ i$, see Lemma 60.6.6. Hence any element of $\Omega _ D$ can be written uniquely as a sum $\sum f_ i\text{d}x_ i$ with for all $e$ only finitely many $f_ i$ not in $p^ eD$. Moreover, the maps $\text{d}_{D_ e/A, \bar\gamma } : D_ e \to \Omega _{D_ e/A, \bar\gamma }$ fit together to define a divided power $A$-derivation

60.17.0.3
\begin{equation} \label{crystalline-equation-derivation-D} \text{d} : D \longrightarrow \Omega _ D \end{equation}

on $p$-adic completions.

We will also need the “products $\mathop{\mathrm{Spec}}(D(n))$ of $\mathop{\mathrm{Spec}}(D)$”, see Proposition 60.21.1 and its proof for an explanation. Formally these are defined as follows. For $n \geq 0$ let $J(n) = \mathop{\mathrm{Ker}}(P \otimes _ A \ldots \otimes _ A P \to C)$ where the tensor product has $n + 1$ factors. We set

60.17.0.4
\begin{equation} \label{crystalline-equation-Dn} D(n) = \mathop{\mathrm{lim}}\nolimits _ e D_{P \otimes _ A \ldots \otimes _ A P, \gamma }(J(n))/ p^ eD_{P \otimes _ A \ldots \otimes _ A P, \gamma }(J(n)) \end{equation}

equal to the $p$-adic completion of the divided power envelope. We denote $\bar J(n)$ its divided power ideal and $\bar\gamma (n)$ its divided powers. We also introduce $D(n)_ e = D(n)/p^ eD(n)$ as well as the $p$-adically completed module of differentials

60.17.0.5
\begin{equation} \label{crystalline-equation-omega-Dn} \Omega _{D(n)} = \mathop{\mathrm{lim}}\nolimits _ e \Omega _{D(n)_ e/A, \bar\gamma } = \mathop{\mathrm{lim}}\nolimits _ e \Omega _{D(n)/A, \bar\gamma }/p^ e\Omega _{D(n)/A, \bar\gamma } \end{equation}

and derivation

60.17.0.6
\begin{equation} \label{crystalline-equation-derivation-Dn} \text{d} : D(n) \longrightarrow \Omega _{D(n)} \end{equation}

Of course we have $D = D(0)$. Note that the rings $D(0), D(1), D(2), \ldots $ form a cosimplicial object in the category of divided power rings.

Lemma 60.17.1. Let $D$ and $D(n)$ be as in (60.17.0.1) and (60.17.0.4). The coprojection $P \to P \otimes _ A \ldots \otimes _ A P$, $f \mapsto f \otimes 1 \otimes \ldots \otimes 1$ induces an isomorphism

60.17.1.1
\begin{equation} \label{crystalline-equation-structure-Dn} D(n) = \mathop{\mathrm{lim}}\nolimits _ e D\langle \xi _ i(j) \rangle /p^ eD\langle \xi _ i(j) \rangle \end{equation}

of algebras over $D$ with

\[ \xi _ i(j) = x_ i \otimes 1 \otimes \ldots \otimes 1 - 1 \otimes \ldots \otimes 1 \otimes x_ i \otimes 1 \otimes \ldots \otimes 1 \]

for $j = 1, \ldots , n$ where the second $x_ i$ is placed in the $j + 1$st slot; recall that $D(n)$ is constructed starting with the $n + 1$-fold tensor product of $P$ over $A$.

Proof. We have

\[ P \otimes _ A \ldots \otimes _ A P = P[\xi _ i(j)] \]

and $J(n)$ is generated by $J$ and the elements $\xi _ i(j)$. Hence the lemma follows from Lemma 60.2.5. $\square$

Lemma 60.17.2. Let $D$ and $D(n)$ be as in (60.17.0.1) and (60.17.0.4). Then $(D, \bar J, \bar\gamma )$ and $(D(n), \bar J(n), \bar\gamma (n))$ are objects of $\text{Cris}^\wedge (C/A)$, see Remark 60.5.4, and

\[ D(n) = \coprod \nolimits _{j = 0, \ldots , n} D \]

in $\text{Cris}^\wedge (C/A)$.

Proof. The first assertion is clear. For the second, if $(B \to C, \delta )$ is an object of $\text{Cris}^\wedge (C/A)$, then we have

\[ \mathop{\mathrm{Mor}}\nolimits _{\text{Cris}^\wedge (C/A)}(D, B) = \mathop{\mathrm{Hom}}\nolimits _ A((P, J), (B, \mathop{\mathrm{Ker}}(B \to C))) \]

and similarly for $D(n)$ replacing $(P, J)$ by $(P \otimes _ A \ldots \otimes _ A P, J(n))$. The property on coproducts follows as $P \otimes _ A \ldots \otimes _ A P$ is a coproduct. $\square$

In the lemma below we will consider pairs $(M, \nabla )$ satisfying the following conditions

  1. $M$ is a $p$-adically complete $D$-module,

  2. $\nabla : M \to M \otimes ^\wedge _ D \Omega _ D$ is a connection, i.e., $\nabla (fm) = m \otimes \text{d}f + f\nabla (m)$,

  3. $\nabla $ is integrable (see Remark 60.6.8), and

  4. $\nabla $ is topologically quasi-nilpotent: If we write $\nabla (m) = \sum \theta _ i(m)\text{d}x_ i$ for some operators $\theta _ i : M \to M$, then for any $m \in M$ there are only finitely many pairs $(i, k)$ such that $\theta _ i^ k(m) \not\in pM$.

The operators $\theta _ i$ are sometimes denoted $\nabla _{\partial /\partial x_ i}$ in the literature. In the following lemma we construct a functor from crystals in quasi-coherent modules on $\text{Cris}(X/S)$ to the category of such pairs. We will show this functor is an equivalence in Proposition 60.17.4.

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 (60.17.1.1) 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 $p$. $\square$

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. $\square$

Lemma 60.17.5. In Situation 60.5.1. 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 $D_{P', \gamma }(J')$. There are homomorphisms of divided power $A$-algebras

\[ a : D \longrightarrow D',\quad b : D' \longrightarrow D \]

compatible with the maps $D \to C$ and $D' \to C$ such that $a \circ b = \text{id}_{D'}$. These maps induce an equivalence of categories of pairs $(M, \nabla )$ satisfying (1), (2), (3), and (4) over $D$ and pairs $(M', \nabla ')$ satisfying (1), (2), (3), and (4)1 over $D'$. In particular, the equivalence of categories of Proposition 60.17.4 also holds for the corresponding functor towards pairs over $D'$.

Proof. First, suppose that $P' = A[y_1, \ldots , y_ m]$ is a polynomial algebra over $A$. In this case, we can find ring maps $P \to P'$ and $P' \to P$ compatible with the maps to $C$ which induce maps $a : D \to D'$ and $b : D' \to D$ as in the lemma. Using completed base change along $a$ and $b$ we obtain functors between the categories of modules with connection satisfying properties (1), (2), (3), and (4) simply because these these categories are equivalent to the category of quasi-coherent crystals by Proposition 60.17.4 (and this equivalence is compatible with the base change operation as shown in the proof of the proposition).

Proof for general smooth $P'$. By the first paragraph of the proof we may assume $P = A[y_1, \ldots , y_ m]$ which gives us a surjection $P \to P'$ compatible with the map to $C$. Hence we obtain a surjective map $a : D \to D'$ by functoriality of divided power envelopes and completion. Pick $e$ large enough so that $D_ e$ is a divided power thickening of $C$ over $A$. Then $D_ e \to C$ is a surjection whose kernel is locally nilpotent, see Divided Power Algebra, Lemma 23.2.6. Setting $D'_ e = D'/p^ eD'$ we see that the kernel of $D_ e \to D'_ e$ is locally nilpotent. Hence by Algebra, Lemma 10.138.17 we can find a lift $\beta _ e : P' \to D_ e$ of the map $P' \to D'_ e$. Note that $D_{e + i + 1} \to D_{e + i} \times _{D'_{e + i}} D'_{e + i + 1}$ is surjective with square zero kernel for any $i \geq 0$ because $p^{e + i}D \to p^{e + i}D'$ is surjective. Applying the usual lifting property (Algebra, Proposition 10.138.13) successively to the diagrams

\[ \xymatrix{ P' \ar[r] & D_{e + i} \times _{D'_{e + i}} D'_{e + i + 1} \\ A \ar[u] \ar[r] & D_{e + i + 1} \ar[u] } \]

we see that we can find an $A$-algebra map $\beta : P' \to D$ whose composition with $a$ is the given map $P' \to D'$. By the universal property of the divided power envelope we obtain a map $D_{P', \gamma }(J') \to D$. As $D$ is $p$-adically complete we obtain $b : D' \to D$ such that $a \circ b = \text{id}_{D'}$.

Consider the base change functors

\[ F : (M, \nabla ) \longmapsto (M \otimes ^\wedge _{D, a} D', \nabla ') \quad \text{and}\quad G : (M', \nabla ') \longmapsto (M' \otimes ^\wedge _{D', b} D, \nabla ) \]

on modules with connections satisfying (1), (2), and (3). See Remark 60.6.9. Since $a \circ b = \text{id}_{D'}$ we see that $F \circ G$ is the identity functor. Let us say that $(M', \nabla ')$ has property (4) if this is true for $G(M', \nabla ')$. A formal argument now shows that to finish the proof it suffices to show that $G(F(M, \nabla ))$ is isomorphic to $(M, \nabla )$ in the case that $(M, \nabla )$ satisfies all four conditions (1), (2), (3), and (4). For this we use the functorial isomorphism

\[ c_{\text{id}_ D, b \circ a} : M \otimes _{D, \text{id}_ D} D \longrightarrow M \otimes _{D, b \circ a} D \]

of the proof of Proposition 60.17.4 (which requires the topological quasi-nilpotency of $\nabla $ which we have assumed). It remains to prove that this map is horizontal, i.e., compatible with connections, which we omit.

The last statement of the proof now follows. $\square$

Remark 60.17.6. The equivalence of Proposition 60.17.4 holds if we start with a surjection $P \to C$ where $P/A$ satisfies the strong lifting property of Algebra, Lemma 10.138.17. To prove this we can argue as in the proof of Lemma 60.17.5. (Details will be added here if we ever need this.) Presumably there is also a direct proof of this result, but the advantage of using polynomial rings is that the rings $D(n)$ are $p$-adic completions of divided power polynomial rings and the algebra is simplified.

[1] This condition is tricky to formulate for $(M', \nabla ')$ over $D'$. See proof.

Comments (4)

Comment #1446 by Matthieu Romagny on

Typo: just before the statement of 46.17.3, replace "equivalent" by "equivalence"

Comment #7204 by Jefferson Baudin on

There may be a mistake: right after Tag 07J9, you say that the p-adic completion of is freely generated by the , invoking Lemma 07HW.

However, if I understood well, this lemma only applies if , where I call the ideal on which the divided power structure on is defined, and I believe this inclusion may not hold right ? (for example if we already have , then ).

Comment #7205 by on

We are assuming we are in Situation 60.5.1 and hence , see first sentence of this section.


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 07J7. Beware of the difference between the letter 'O' and the digit '0'.