The Stacks Project


Tag 07PZ

54.25. Pulling back along purely inseparable maps

By an $\alpha_p$-cover we mean a morphism of the form $$ X' = \mathop{\rm Spec}(C[z]/(z^p - c)) \longrightarrow \mathop{\rm Spec}(C) = X $$ where $C$ is an $\mathbf{F}_p$-algebra and $c \in C$. Equivalently, $X'$ is an $\alpha_p$-torsor over $X$. An iterated $\alpha_p$-cover1 is a morphism of schemes in characteristic $p$ which is locally on the target a composition of finitely many $\alpha_p$-covers. In this section we prove that pullback along such a morphism induces a quasi-isomorphism on crystalline cohomology after inverting the prime $p$. In fact, we prove a precise version of this result. We begin with a preliminary lemma whose formulation needs some notation.

Assume we have a ring map $B \to B'$ and quotients $\Omega_B \to \Omega$ and $\Omega_{B'} \to \Omega'$ satisfying the assumptions of Remark 54.6.11. Thus (54.6.11.1) provides a canonical map of complexes $$ c_M^\bullet : M \otimes_B \Omega^\bullet \longrightarrow M \otimes_B (\Omega')^\bullet $$ for all $B$-modules $M$ endowed with integrable connection $\nabla : M \to M \otimes_B \Omega_B$.

Suppose we have $a \in B$, $z \in B'$, and a map $\theta : B' \to B'$ satisfying the following assumptions

  1. $\text{d}(a) = 0$,
  2. $\Omega' = B' \otimes_B \Omega \oplus B'\text{d}z$; we write $\text{d}(f) = \text{d}_1(f) + \partial_z(f) \text{d}z$ with $\text{d}_1(f) \in B' \otimes \Omega$ and $\partial_z(f) \in N'$ for all $f \in B'$,
  3. $\theta : B' \to B'$ is $B$-linear,
  4. $\partial_z \circ \theta = a$,
  5. $B \to B'$ is universally injective (and hence $\Omega \to \Omega'$ is injective),
  6. $af - \theta(\partial_z(f)) \in B$ for all $f \in B'$,
  7. $(\theta \otimes 1)(\text{d}_1(f)) - \text{d}_1(\theta(f)) \in \Omega$ for all $f \in B'$ where $\theta \otimes 1 : B' \otimes \Omega \to B' \otimes \Omega$

These conditions are not logically independent. For example, assumption (4) implies that $\partial_z(af - \theta(\partial_z(f))) = 0$. Hence if the image of $B \to B'$ is the collection of elements annihilated by $\partial_z$, then (6) follows. A similar argument can be made for condition (7).

Lemma 54.25.1. In the situation above there exists a map of complexes $$ e_M^\bullet : M \otimes_B (\Omega')^\bullet \longrightarrow M \otimes_B \Omega^\bullet $$ such that $c_M^\bullet \circ e_M^\bullet$ and $e_M^\bullet \circ c_M^\bullet$ are homotopic to multiplication by $a$.

Proof. In this proof all tensor products are over $B$. Assumption (2) implies that $$ M \otimes (\Omega')^i = (B' \otimes M \otimes \Omega^i) \oplus (B' \text{d}z \otimes M \otimes \Omega^{i - 1}) $$ for all $i \geq 0$. A collection of additive generators for $M \otimes (\Omega')^i$ is formed by elements of the form $f \omega$ and elements of the form $f \text{d}z \wedge \eta$ where $f \in B'$, $\omega \in M \otimes \Omega^i$, and $\eta \in M \otimes \Omega^{i - 1}$.

For $f \in B'$ we write $$ \epsilon(f) = af - \theta(\partial_z(f)) \quad\text{and}\quad \epsilon'(f) = (\theta \otimes 1)(\text{d}_1(f)) - \text{d}_1(\theta(f)) $$ so that $\epsilon(f) \in B$ and $\epsilon'(f) \in \Omega$ by assumptions (6) and (7). We define $e_M^\bullet$ by the rules $e^i_M(f\omega) = \epsilon(f) \omega$ and $e^i_M(f \text{d}z \wedge \eta) = \epsilon'(f) \wedge \eta$. We will see below that the collection of maps $e^i_M$ is a map of complexes.

We define $$ h^i : M \otimes_B (\Omega')^i \longrightarrow M \otimes_B (\Omega')^{i - 1} $$ by the rules $h^i(f \omega) = 0$ and $h^i(f \text{d}z \wedge \eta) = \theta(f) \eta$ for elements as above. We claim that $$ \text{d} \circ h + h \circ \text{d} = a - c_M^\bullet \circ e_M^\bullet $$ Note that multiplication by $a$ is a map of complexes by (1). Hence, since $c_M^\bullet$ is an injective map of complexes by assumption (5), we conclude that $e_M^\bullet$ is a map of complexes. To prove the claim we compute \begin{align*} (\text{d} \circ h + h \circ \text{d})(f \omega) & = h\left(\text{d}(f) \wedge \omega + f \nabla(\omega)\right) \\ & = \theta(\partial_z(f)) \omega \\ & = a f\omega - \epsilon(f)\omega \\ & = a f \omega - c^i_M(e^i_M(f\omega)) \end{align*} The second equality because $\text{d}z$ does not occur in $\nabla(\omega)$ and the third equality by assumption (6). Similarly, we have \begin{align*} (\text{d} \circ h + h \circ \text{d})(f \text{d}z \wedge \eta) & = \text{d}(\theta(f) \eta) + h\left(\text{d}(f) \wedge \text{d}z \wedge \eta - f \text{d}z \wedge \nabla(\eta)\right) \\ & = \text{d}(\theta(f)) \wedge \eta + \theta(f) \nabla(\eta) - (\theta \otimes 1)(\text{d}_1(f)) \wedge \eta - \theta(f) \nabla(\eta) \\ & = \text{d}_1(\theta(f)) \wedge \eta + \partial_z(\theta(f)) \text{d}z \wedge \eta - (\theta \otimes 1)(\text{d}_1(f)) \wedge \eta \\ & = a f \text{d}z \wedge \eta - \epsilon'(f) \wedge \eta \\ & = a f \text{d}z \wedge \eta - c^i_M(e^i_M(f \text{d}z \wedge \eta)) \end{align*} The second equality because $\text{d}(f) \wedge \text{d}z \wedge \eta = - \text{d}z \wedge \text{d}_1(f) \wedge \eta$. The fourth equality by assumption (4). On the other hand it is immediate from the definitions that $e^i_M(c^i_M(\omega)) = \epsilon(1) \omega = a \omega$. This proves the lemma. $\square$

Example 54.25.2. A standard example of the situation above occurs when $B' = B\langle z \rangle$ is the divided power polynomial ring over a divided power ring $(B, J, \delta)$ with divided powers $\delta'$ on $J' = B'_{+} + JB' \subset B'$. Namely, we take $\Omega = \Omega_{B, \delta}$ and $\Omega' = \Omega_{B', \delta'}$. In this case we can take $a = 1$ and $$ \theta( \sum b_m z^{[m]} ) = \sum b_m z^{[m + 1]} $$ Note that $$ f - \theta(\partial_z(f)) = f(0) $$ equals the constant term. It follows that in this case Lemma 54.25.1 recovers the crystalline Poincaré lemma (Lemma 54.20.2).

Lemma 54.25.3. In Situation 54.5.1. Assume $D$ and $\Omega_D$ are as in (54.17.0.1) and (54.17.0.2). Let $\lambda \in D$. Let $D'$ be the $p$-adic completion of $$ D[z]\langle \xi \rangle/(\xi - (z^p - \lambda)) $$ and let $\Omega_{D'}$ be the $p$-adic completion of the module of divided power differentials of $D'$ over $A$. For any pair $(M, \nabla)$ over $D$ satisfying (1), (2), (3), and (4) the canonical map of complexes (54.6.11.1) $$ c_M^\bullet : M \otimes_D^\wedge \Omega^\bullet_D \longrightarrow M \otimes_D^\wedge \Omega^\bullet_{D'} $$ has the following property: There exists a map $e_M^\bullet$ in the opposite direction such that both $c_M^\bullet \circ e_M^\bullet$ and $e_M^\bullet \circ c_M^\bullet$ are homotopic to multiplication by $p$.

Proof. We will prove this using Lemma 54.25.1 with $a = p$. Thus we have to find $\theta : D' \to D'$ and prove (1), (2), (3), (4), (5), (6), (7). We first collect some information about the rings $D$ and $D'$ and the modules $\Omega_D$ and $\Omega_{D'}$.

Writing $$ D[z]\langle \xi \rangle/(\xi - (z^p - \lambda)) = D\langle \xi \rangle[z]/(z^p - \xi - \lambda) $$ we see that $D'$ is the $p$-adic completion of the free $D$-module $$ \bigoplus\nolimits_{i = 0, \ldots, p - 1} \bigoplus\nolimits_{n \geq 0} z^i \xi^{[n]} D $$ where $\xi^{[0]} = 1$. It follows that $D \to D'$ has a continuous $D$-linear section, in particular $D \to D'$ is universally injective, i.e., (5) holds. We think of $D'$ as a divided power algebra over $A$ with divided power ideal $\overline{J}' = \overline{J}D' + (\xi)$. Then $D'$ is also the $p$-adic completion of the divided power envelope of the ideal generated by $z^p - \lambda$ in $D$, see Lemma 54.2.4. Hence $$ \Omega_{D'} = \Omega_D \otimes_D^\wedge D' \oplus D'\text{d}z $$ by Lemma 54.6.6. This proves (2). Note that (1) is obvious.

At this point we construct $\theta$. (We wrote a PARI/gp script theta.gp verifying some of the formulas in this proof which can be found in the scripts subdirectory of the Stacks project.) Before we do so we compute the derivative of the elements $z^i \xi^{[n]}$. We have $\text{d}z^i = i z^{i - 1} \text{d}z$. For $n \geq 1$ we have $$ \text{d}\xi^{[n]} = \xi^{[n - 1]} \text{d}\xi = - \xi^{[n - 1]}\text{d}\lambda + p z^{p - 1} \xi^{[n - 1]}\text{d}z $$ because $\xi = z^p - \lambda$. For $0 < i < p$ and $n \geq 1$ we have \begin{align*} \text{d}(z^i\xi^{[n]}) & = iz^{i - 1}\xi^{[n]}\text{d}z + z^i\xi^{[n - 1]}\text{d}\xi \\ & = iz^{i - 1}\xi^{[n]}\text{d}z + z^i\xi^{[n - 1]}\text{d}(z^p - \lambda) \\ & = - z^i\xi^{[n - 1]}\text{d}\lambda + (iz^{i - 1}\xi^{[n]} + pz^{i + p - 1}\xi^{[n - 1]})\text{d}z \\ & = - z^i\xi^{[n - 1]}\text{d}\lambda + (iz^{i - 1}\xi^{[n]} + pz^{i - 1}(\xi + \lambda)\xi^{[n - 1]})\text{d}z \\ & = - z^i\xi^{[n - 1]}\text{d}\lambda + ((i + pn)z^{i - 1}\xi^{[n]} + p\lambda z^{i - 1}\xi^{[n - 1]})\text{d}z \end{align*} the last equality because $\xi \xi^{[n - 1]} = n\xi^{[n]}$. Thus we see that \begin{align*} \partial_z(z^i) & = i z^{i - 1} \\ \partial_z(\xi^{[n]}) & = p z^{p - 1} \xi^{[n - 1]} \\ \partial_z(z^i\xi^{[n]}) & = (i + pn) z^{i - 1} \xi^{[n]} + p \lambda z^{i - 1}\xi^{[n - 1]} \end{align*} Motivated by these formulas we define $\theta$ by the rules $$ \begin{matrix} \theta(z^j) & = & p\frac{z^{j + 1}}{j + 1} & j = 0, \ldots p - 1, \\ \theta(z^{p - 1}\xi^{[m]}) & = & \xi^{[m + 1]} & m \geq 1, \\ \theta(z^j \xi^{[m]}) & = & \frac{p z^{j + 1} \xi^{[m]} - \theta(p\lambda z^j \xi^{[m - 1]})}{(j + 1 + pm)} & 0 \leq j < p - 1, m \geq 1 \end{matrix} $$ where in the last line we use induction on $m$ to define our choice of $\theta$. Working this out we get (for $0 \leq j < p - 1$ and $1 \leq m$) $$ \theta(z^j \xi^{[m]}) = \textstyle{\frac{p z^{j + 1} \xi^{[m]}}{(j + 1 + pm)} - \frac{p^2 \lambda z^{j + 1} \xi^{[m - 1]}}{(j + 1 + pm)(j + 1 + p(m - 1))} + \ldots + \frac{(-1)^m p^{m + 1} \lambda^m z^{j + 1}} {(j + 1 + pm) \ldots (j + 1)}} $$ although we will not use this expression below. It is clear that $\theta$ extends uniquely to a $p$-adically continuous $D$-linear map on $D'$. By construction we have (3) and (4). It remains to prove (6) and (7).

Proof of (6) and (7). As $\theta$ is $D$-linear and continuous it suffices to prove that $p - \theta \circ \partial_z$, resp. $(\theta \otimes 1) \circ \text{d}_1 - \text{d}_1 \circ \theta$ gives an element of $D$, resp. $\Omega_D$ when evaluated on the elements $z^i\xi^{[n]}$2. Set $D_0 = \mathbf{Z}_{(p)}[\lambda]$ and $D_0' = \mathbf{Z}_{(p)}[z, \lambda]\langle \xi \rangle/(\xi - z^p + \lambda)$. Observe that each of the expressions above is an element of $D_0'$ or $\Omega_{D_0'}$. Hence it suffices to prove the result in the case of $D_0 \to D_0'$. Note that $D_0$ and $D_0'$ are torsion free rings and that $D_0 \otimes \mathbf{Q} = \mathbf{Q}[\lambda]$ and $D'_0 \otimes \mathbf{Q} = \mathbf{Q}[z, \lambda]$. Hence $D_0 \subset D'_0$ is the subring of elements annihilated by $\partial_z$ and (6) follows from (4), see the discussion directly preceding Lemma 54.25.1. Similarly, we have $\text{d}_1(f) = \partial_\lambda(f)\text{d}\lambda$ hence $$ \left((\theta \otimes 1) \circ \text{d}_1 - \text{d}_1 \circ \theta\right)(f) = \left(\theta(\partial_\lambda(f)) - \partial_\lambda(\theta(f))\right) \text{d}\lambda $$ Applying $\partial_z$ to the coefficient we obtain \begin{align*} \partial_z\left( \theta(\partial_\lambda(f)) - \partial_\lambda(\theta(f)) \right) & = p \partial_\lambda(f) - \partial_z(\partial_\lambda(\theta(f))) \\ & = p \partial_\lambda(f) - \partial_\lambda(\partial_z(\theta(f))) \\ & = p \partial_\lambda(f) - \partial_\lambda(p f) = 0 \end{align*} whence the coefficient does not depend on $z$ as desired. This finishes the proof of the lemma. $\square$

Note that an iterated $\alpha_p$-cover $X' \to X$ (as defined in the introduction to this section) is finite locally free. Hence if $X$ is connected the degree of $X' \to X$ is constant and is a power of $p$.

Lemma 54.25.4. Let $p$ be a prime number. Let $(S, \mathcal{I}, \gamma)$ be a divided power scheme over $\mathbf{Z}_{(p)}$ with $p \in \mathcal{I}$. We set $S_0 = V(\mathcal{I}) \subset S$. Let $f : X' \to X$ be an iterated $\alpha_p$-cover of schemes over $S_0$ with constant degree $q$. Let $\mathcal{F}$ be any crystal in quasi-coherent sheaves on $X$ and set $\mathcal{F}' = f_{\text{cris}}^*\mathcal{F}$. In the distinguished triangle $$ Ru_{X/S, *}\mathcal{F} \longrightarrow f_*Ru_{X'/S, *}\mathcal{F}' \longrightarrow E \longrightarrow Ru_{X/S, *}\mathcal{F}[1] $$ the object $E$ has cohomology sheaves annihilated by $q$.

Proof. Note that $X' \to X$ is a homeomorphism hence we can identify the underlying topological spaces of $X$ and $X'$. The question is clearly local on $X$, hence we may assume $X$, $X'$, and $S$ affine and $X' \to X$ given as a composition $$ X' = X_n \to X_{n - 1} \to X_{n - 2} \to \ldots \to X_0 = X $$ where each morphism $X_{i + 1} \to X_i$ is an $\alpha_p$-cover. Denote $\mathcal{F}_i$ the pullback of $\mathcal{F}$ to $X_i$. It suffices to prove that each of the maps $$ R\Gamma(\text{Cris}(X_i/S), \mathcal{F}_i) \longrightarrow R\Gamma(\text{Cris}(X_{i + 1}/S), \mathcal{F}_{i + 1}) $$ fits into a triangle whose third member has cohomology groups annihilated by $p$. (This uses axiom TR4 for the triangulated category $D(X)$. Details omitted.)

Hence we may assume that $S = \mathop{\rm Spec}(A)$, $X = \mathop{\rm Spec}(C)$, $X' = \mathop{\rm Spec}(C')$ and $C' = C[z]/(z^p - c)$ for some $c \in C$. Choose a polynomial algebra $P$ over $A$ and a surjection $P \to C$. Let $D$ be the $p$-adically completed divided power envelop of $\mathop{\rm Ker}(P \to C)$ in $P$ as in (54.17.0.1). Set $P' = P[z]$ with surjection $P' \to C'$ mapping $z$ to the class of $z$ in $C'$. Choose a lift $\lambda \in D$ of $c \in C$. Then we see that the $p$-adically completed divided power envelope $D'$ of $\mathop{\rm Ker}(P' \to C')$ in $P'$ is isomorphic to the $p$-adic completion of $D[z]\langle \xi \rangle/(\xi - (z^p - \lambda))$, see Lemma 54.25.3 and its proof. Thus we see that the result follows from this lemma by the computation of cohomology of crystals in quasi-coherent modules in Proposition 54.21.3. $\square$

The bound in the following lemma is probably not optimal.

Lemma 54.25.5. With notations and assumptions as in Lemma 54.25.4 the map $$ f^* : H^i(\text{Cris}(X/S), \mathcal{F}) \longrightarrow H^i(\text{Cris}(X'/S), \mathcal{F}') $$ has kernel and cokernel annihilated by $q^{i + 1}$.

Proof. This follows from the fact that $E$ has nonzero cohomology sheaves in degrees $-1$ and up, so that the spectral sequence $H^a(\mathcal{H}^b(E)) \Rightarrow H^{a + b}(E)$ converges. This combined with the long exact cohomology sequence associated to a distinguished triangle gives the bound. $\square$

In Situation 54.7.5 assume that $p \in \mathcal{I}$. Set $$ X^{(1)} = X \times_{S_0, F_{S_0}} S_0. $$ Denote $F_{X/S_0} : X \to X^{(1)}$ the relative Frobenius morphism.

Lemma 54.25.6. In the situation above, assume that $X \to S_0$ is smooth of relative dimension $d$. Then $F_{X/S_0}$ is an iterated $\alpha_p$-cover of degree $p^d$. Hence Lemmas 54.25.4 and 54.25.5 apply to this situation. In particular, for any crystal in quasi-coherent modules $\mathcal{G}$ on $\text{Cris}(X^{(1)}/S)$ the map $$ F_{X/S_0}^* : H^i(\text{Cris}(X^{(1)}/S), \mathcal{G}) \longrightarrow H^i(\text{Cris}(X/S), F_{X/S_0, \text{cris}}^*\mathcal{G}) $$ has kernel and cokernel annihilated by $p^{d(i + 1)}$.

Proof. It suffices to prove the first statement. To see this we may assume that $X$ is étale over $\mathbf{A}^d_{S_0}$, see Morphisms, Lemma 28.34.20. Denote $\varphi : X \to \mathbf{A}^d_{S_0}$ this étale morphism. In this case the relative Frobenius of $X/S_0$ fits into a diagram $$ \xymatrix{ X \ar[d] \ar[r] & X^{(1)} \ar[d] \\ \mathbf{A}^d_{S_0} \ar[r] & \mathbf{A}^d_{S_0} } $$ where the lower horizontal arrow is the relative frobenius morphism of $\mathbf{A}^d_{S_0}$ over $S_0$. This is the morphism which raises all the coordinates to the $p$th power, hence it is an iterated $\alpha_p$-cover. The proof is finished by observing that the diagram is a fibre square, see the proof of Étale Cohomology, Theorem 53.86.3. $\square$

  1. This is nonstandard notation.
  2. This can be done by direct computation: It turns out that $p - \theta \circ \partial_z$ evaluated on $z^i\xi^{[n]}$ gives zero except for $1$ which is mapped to $p$ and $\xi$ which is mapped to $-p\lambda$. It turns out that $(\theta \otimes 1) \circ \text{d}_1 - \text{d}_1 \circ \theta$ evaluated on $z^i\xi^{[n]}$ gives zero except for $z^{p - 1}\xi$ which is mapped to $-\lambda$.

The code snippet corresponding to this tag is a part of the file crystalline.tex and is located in lines 5011–5523 (see updates for more information).

\section{Pulling back along purely inseparable maps}
\label{section-pull-back-along-pth-root}

\noindent
By an $\alpha_p$-cover we mean a morphism of the form
$$
X' = \Spec(C[z]/(z^p - c)) \longrightarrow \Spec(C) = X
$$
where $C$ is an $\mathbf{F}_p$-algebra and $c \in C$. Equivalently,
$X'$ is an $\alpha_p$-torsor over $X$. An {\it iterated
$\alpha_p$-cover}\footnote{This is nonstandard notation.}
is a morphism of schemes in characteristic
$p$ which is locally on the target a composition of finitely many
$\alpha_p$-covers. In this section we prove that pullback along
such a morphism induces a quasi-isomorphism on crystalline cohomology
after inverting the prime $p$. In fact, we prove a precise version
of this result. We begin with a preliminary lemma whose formulation
needs some notation.

\medskip\noindent
Assume we have a ring map $B \to B'$ and quotients $\Omega_B \to \Omega$ and
$\Omega_{B'} \to \Omega'$ satisfying the assumptions of
Remark \ref{remark-base-change-connection}.
Thus (\ref{equation-base-change-map-complexes}) provides a
canonical map of complexes
$$
c_M^\bullet :
M \otimes_B \Omega^\bullet
\longrightarrow
M \otimes_B (\Omega')^\bullet
$$
for all $B$-modules $M$ endowed with integrable connection
$\nabla : M \to M \otimes_B \Omega_B$.

\medskip\noindent
Suppose we have $a \in B$, $z \in B'$, and a map $\theta : B' \to B'$
satisfying the following assumptions
\begin{enumerate}
\item
\label{item-d-a-zero}
$\text{d}(a) = 0$,
\item
\label{item-direct-sum}
$\Omega' = B' \otimes_B \Omega \oplus B'\text{d}z$; we write
$\text{d}(f) = \text{d}_1(f) + \partial_z(f) \text{d}z$
with $\text{d}_1(f) \in B' \otimes \Omega$ and $\partial_z(f) \in N'$
for all $f \in B'$,
\item
\label{item-theta-linear}
$\theta : B' \to B'$ is $B$-linear,
\item
\label{item-integrate}
$\partial_z \circ \theta = a$,
\item
\label{item-injective}
$B \to B'$ is universally injective (and hence $\Omega \to \Omega'$
is injective),
\item
\label{item-factor}
$af - \theta(\partial_z(f)) \in B$ for all $f \in B'$,
\item
\label{item-horizontal}
$(\theta \otimes 1)(\text{d}_1(f)) - \text{d}_1(\theta(f)) \in \Omega$
for all $f \in B'$ where
$\theta \otimes 1 : B' \otimes \Omega \to B' \otimes \Omega$
\end{enumerate}
These conditions are not logically independent.
For example, assumption (\ref{item-integrate}) implies
that $\partial_z(af - \theta(\partial_z(f))) = 0$.
Hence if the image of $B \to B'$ is the collection of
elements annihilated by $\partial_z$, then (\ref{item-factor})
follows. A similar argument can be made for condition (\ref{item-horizontal}).

\begin{lemma}
\label{lemma-find-homotopy}
In the situation above there exists a map of complexes
$$
e_M^\bullet :
M \otimes_B (\Omega')^\bullet
\longrightarrow
M \otimes_B \Omega^\bullet
$$
such that $c_M^\bullet \circ e_M^\bullet$
and $e_M^\bullet \circ c_M^\bullet$ are homotopic to
multiplication by $a$.
\end{lemma}

\begin{proof}
In this proof all tensor products are over $B$.
Assumption (\ref{item-direct-sum}) implies that
$$
M \otimes (\Omega')^i =
(B' \otimes M \otimes \Omega^i)
\oplus
(B' \text{d}z \otimes M \otimes \Omega^{i - 1})
$$
for all $i \geq 0$. A collection of additive generators for
$M \otimes (\Omega')^i$ is formed by elements of the form
$f \omega$ and elements of the form $f \text{d}z \wedge \eta$
where $f \in B'$, $\omega \in M \otimes \Omega^i$, and
$\eta \in M \otimes \Omega^{i - 1}$.

\medskip\noindent
For $f \in B'$ we write
$$
\epsilon(f) = af - \theta(\partial_z(f))
\quad\text{and}\quad
\epsilon'(f) = (\theta \otimes 1)(\text{d}_1(f)) - \text{d}_1(\theta(f))
$$
so that $\epsilon(f) \in B$ and $\epsilon'(f) \in \Omega$ by
assumptions (\ref{item-factor}) and (\ref{item-horizontal}).
We define $e_M^\bullet$ by the rules
$e^i_M(f\omega) = \epsilon(f) \omega$ and
$e^i_M(f \text{d}z \wedge \eta) = \epsilon'(f) \wedge \eta$.
We will see below that the collection of maps $e^i_M$ is a map of complexes.

\medskip\noindent
We define
$$
h^i : M \otimes_B (\Omega')^i \longrightarrow M \otimes_B (\Omega')^{i - 1}
$$
by the rules $h^i(f \omega) = 0$ and
$h^i(f \text{d}z \wedge \eta) = \theta(f) \eta$
for elements as above.  We claim that
$$
\text{d} \circ h + h \circ \text{d} = a - c_M^\bullet \circ e_M^\bullet
$$
Note that multiplication by $a$ is a map of complexes
by (\ref{item-d-a-zero}). Hence, since $c_M^\bullet$ is an injective map
of complexes by assumption (\ref{item-injective}), we conclude that
$e_M^\bullet$ is a map of complexes. To prove the claim we compute
\begin{align*}
(\text{d} \circ h + h \circ \text{d})(f \omega)
& =
h\left(\text{d}(f) \wedge \omega + f \nabla(\omega)\right)
\\
& =
\theta(\partial_z(f)) \omega
\\
& =
a f\omega - \epsilon(f)\omega 
\\
& =
a f \omega - c^i_M(e^i_M(f\omega))
\end{align*}
The second equality because $\text{d}z$ does not occur in $\nabla(\omega)$
and the third equality by assumption (6). Similarly, we have
\begin{align*}
(\text{d} \circ h + h \circ \text{d})(f \text{d}z \wedge \eta)
& =
\text{d}(\theta(f) \eta) +
h\left(\text{d}(f) \wedge \text{d}z \wedge \eta -
f \text{d}z \wedge \nabla(\eta)\right)
\\
& =
\text{d}(\theta(f)) \wedge \eta + \theta(f) \nabla(\eta)
- (\theta \otimes 1)(\text{d}_1(f)) \wedge \eta
- \theta(f) \nabla(\eta)
\\
& =
\text{d}_1(\theta(f)) \wedge \eta +
\partial_z(\theta(f)) \text{d}z \wedge \eta -
(\theta \otimes 1)(\text{d}_1(f)) \wedge \eta
\\
& =
a f \text{d}z \wedge \eta - \epsilon'(f) \wedge \eta \\
& = a f \text{d}z \wedge \eta - c^i_M(e^i_M(f \text{d}z \wedge \eta))
\end{align*}
The second equality because
$\text{d}(f) \wedge \text{d}z \wedge \eta =
- \text{d}z \wedge \text{d}_1(f) \wedge \eta$.
The fourth equality by assumption (\ref{item-integrate}).
On the other hand it is immediate from the definitions
that $e^i_M(c^i_M(\omega)) = \epsilon(1) \omega = a \omega$.
This proves the lemma.
\end{proof}

\begin{example}
\label{example-integrate}
A standard example of the situation above occurs when
$B' = B\langle z \rangle$ is the divided power polynomial ring
over a divided power ring $(B, J, \delta)$ with divided powers
$\delta'$ on $J' = B'_{+} + JB' \subset B'$. Namely, we take
$\Omega = \Omega_{B, \delta}$ and $\Omega' = \Omega_{B', \delta'}$.
In this case we can take $a = 1$ and
$$
\theta( \sum b_m z^{[m]} ) = \sum b_m z^{[m + 1]}
$$
Note that
$$
f - \theta(\partial_z(f)) = f(0)
$$
equals the constant term. It follows that in this case
Lemma \ref{lemma-find-homotopy}
recovers the crystalline Poincar\'e lemma
(Lemma \ref{lemma-relative-poincare}).
\end{example}

\begin{lemma}
\label{lemma-computation}
In Situation \ref{situation-affine}. Assume $D$ and $\Omega_D$ are as in
(\ref{equation-D}) and (\ref{equation-omega-D}).
Let $\lambda \in D$. Let $D'$ be the $p$-adic completion of
$$
D[z]\langle \xi \rangle/(\xi - (z^p - \lambda))
$$
and let $\Omega_{D'}$ be the $p$-adic completion of the module of
divided power differentials of $D'$ over $A$. For any pair $(M, \nabla)$
over $D$ satisfying (\ref{item-complete}), (\ref{item-connection}),
(\ref{item-integrable}), and (\ref{item-topologically-quasi-nilpotent})
the canonical map of complexes (\ref{equation-base-change-map-complexes})
$$
c_M^\bullet : M \otimes_D^\wedge \Omega^\bullet_D
\longrightarrow
M \otimes_D^\wedge \Omega^\bullet_{D'}
$$
has the following property: There exists a map $e_M^\bullet$
in the opposite direction such that both $c_M^\bullet \circ e_M^\bullet$
and $e_M^\bullet \circ c_M^\bullet$ are homotopic to multiplication by $p$.
\end{lemma}

\begin{proof}
We will prove this using Lemma \ref{lemma-find-homotopy} with $a = p$.
Thus we have to find $\theta : D' \to D'$ and prove
(\ref{item-d-a-zero}), (\ref{item-direct-sum}), (\ref{item-theta-linear}),
(\ref{item-integrate}), (\ref{item-injective}), (\ref{item-factor}),
(\ref{item-horizontal}). We first collect some information about the rings
$D$ and $D'$ and the modules $\Omega_D$ and $\Omega_{D'}$.

\medskip\noindent
Writing
$$
D[z]\langle \xi \rangle/(\xi - (z^p - \lambda))
=
D\langle \xi \rangle[z]/(z^p - \xi - \lambda)
$$
we see that $D'$ is the $p$-adic completion of the free $D$-module
$$
\bigoplus\nolimits_{i = 0, \ldots, p - 1}
\bigoplus\nolimits_{n \geq 0}
z^i \xi^{[n]} D
$$
where $\xi^{[0]} = 1$.
It follows that $D \to D'$ has a continuous $D$-linear section, in particular
$D \to D'$ is universally injective, i.e., (\ref{item-injective}) holds.
We think of $D'$ as a divided power algebra
over $A$ with divided power ideal $\overline{J}' = \overline{J}D' + (\xi)$.
Then $D'$ is also the $p$-adic completion of the divided power envelope
of the ideal generated by $z^p - \lambda$ in $D$, see
Lemma \ref{lemma-describe-divided-power-envelope}. Hence
$$
\Omega_{D'} = \Omega_D \otimes_D^\wedge D' \oplus D'\text{d}z
$$
by Lemma \ref{lemma-module-differentials-divided-power-envelope}.
This proves (\ref{item-direct-sum}). Note that (\ref{item-d-a-zero})
is obvious.

\medskip\noindent
At this point we construct $\theta$. (We wrote a PARI/gp script theta.gp
verifying some of the formulas in this proof which can be found in the
scripts subdirectory of the Stacks project.) Before we do so we compute
the derivative of the elements $z^i \xi^{[n]}$. We have
$\text{d}z^i = i z^{i - 1} \text{d}z$. For $n \geq 1$ we have
$$
\text{d}\xi^{[n]} =
\xi^{[n - 1]} \text{d}\xi =
- \xi^{[n - 1]}\text{d}\lambda + p z^{p - 1} \xi^{[n - 1]}\text{d}z
$$
because $\xi = z^p - \lambda$. For $0 < i < p$ and $n \geq 1$ we have
\begin{align*}
\text{d}(z^i\xi^{[n]})
& =
iz^{i - 1}\xi^{[n]}\text{d}z + z^i\xi^{[n - 1]}\text{d}\xi \\
& =
iz^{i - 1}\xi^{[n]}\text{d}z + z^i\xi^{[n - 1]}\text{d}(z^p - \lambda) \\
& =
- z^i\xi^{[n - 1]}\text{d}\lambda +
(iz^{i - 1}\xi^{[n]} + pz^{i + p - 1}\xi^{[n - 1]})\text{d}z \\
& =
- z^i\xi^{[n - 1]}\text{d}\lambda +
(iz^{i - 1}\xi^{[n]} + pz^{i - 1}(\xi + \lambda)\xi^{[n - 1]})\text{d}z \\
& =
- z^i\xi^{[n - 1]}\text{d}\lambda +
((i + pn)z^{i - 1}\xi^{[n]} + p\lambda z^{i - 1}\xi^{[n - 1]})\text{d}z
\end{align*}
the last equality because $\xi \xi^{[n - 1]} = n\xi^{[n]}$.
Thus we see that
\begin{align*}
\partial_z(z^i) & = i z^{i - 1} \\
\partial_z(\xi^{[n]}) & = p z^{p - 1} \xi^{[n - 1]} \\
\partial_z(z^i\xi^{[n]}) & =
(i + pn) z^{i - 1} \xi^{[n]} + p \lambda z^{i - 1}\xi^{[n - 1]}
\end{align*}
Motivated by these formulas we define $\theta$ by the rules
$$
\begin{matrix}
\theta(z^j)
& = & p\frac{z^{j + 1}}{j + 1}
& j = 0, \ldots p - 1, \\
\theta(z^{p - 1}\xi^{[m]})
& = & \xi^{[m + 1]}
& m \geq 1, \\
\theta(z^j \xi^{[m]})
& = &
\frac{p z^{j + 1} \xi^{[m]} - \theta(p\lambda z^j \xi^{[m - 1]})}{(j + 1 + pm)}
& 0 \leq j < p - 1, m \geq 1
\end{matrix}
$$
where in the last line we use induction on $m$ to define our choice of
$\theta$. Working this out we get (for $0 \leq j < p - 1$ and $1 \leq m$)
$$
\theta(z^j \xi^{[m]}) =
\textstyle{\frac{p z^{j + 1} \xi^{[m]}}{(j + 1 + pm)} -
\frac{p^2 \lambda z^{j + 1} \xi^{[m - 1]}}{(j + 1 + pm)(j + 1 + p(m - 1))} +
\ldots +
\frac{(-1)^m p^{m + 1} \lambda^m z^{j + 1}}
{(j + 1 + pm) \ldots (j + 1)}}
$$
although we will not use this expression below. It is clear that $\theta$
extends uniquely to a $p$-adically continuous $D$-linear map on $D'$.
By construction we have (\ref{item-theta-linear}) and (\ref{item-integrate}).
It remains to prove (\ref{item-factor}) and (\ref{item-horizontal}).

\medskip\noindent
Proof of (\ref{item-factor}) and (\ref{item-horizontal}).
As $\theta$ is $D$-linear and continuous it suffices to prove that
$p - \theta \circ \partial_z$,
resp.\ $(\theta \otimes 1) \circ \text{d}_1 - \text{d}_1 \circ \theta$
gives an element of $D$, resp.\ $\Omega_D$ when evaluated on the
elements $z^i\xi^{[n]}$\footnote{This can be done by direct computation:
It turns out that $p - \theta \circ \partial_z$ evaluated on
$z^i\xi^{[n]}$ gives zero except for $1$ which is mapped to $p$ and
$\xi$ which is mapped to $-p\lambda$. It turns out that 
$(\theta \otimes 1) \circ \text{d}_1 - \text{d}_1 \circ \theta$
evaluated on $z^i\xi^{[n]}$ gives zero except for $z^{p - 1}\xi$
which is mapped to $-\lambda$.}.
Set $D_0 = \mathbf{Z}_{(p)}[\lambda]$ and
$D_0' = \mathbf{Z}_{(p)}[z, \lambda]\langle \xi \rangle/(\xi - z^p + \lambda)$.
Observe that each of the expressions above is an element of
$D_0'$ or $\Omega_{D_0'}$. Hence it suffices to prove the result
in the case of $D_0 \to D_0'$. Note that $D_0$ and $D_0'$
are torsion free rings and that $D_0 \otimes \mathbf{Q} = \mathbf{Q}[\lambda]$
and $D'_0 \otimes \mathbf{Q} = \mathbf{Q}[z, \lambda]$.
Hence $D_0 \subset D'_0$ is the subring of elements annihilated
by $\partial_z$ and (\ref{item-factor})
follows from (\ref{item-integrate}), see the discussion directly preceding
Lemma \ref{lemma-find-homotopy}. Similarly, we have
$\text{d}_1(f) = \partial_\lambda(f)\text{d}\lambda$ hence
$$
\left((\theta \otimes 1) \circ \text{d}_1 - \text{d}_1 \circ \theta\right)(f)
=
\left(\theta(\partial_\lambda(f)) - \partial_\lambda(\theta(f))\right)
\text{d}\lambda
$$
Applying $\partial_z$ to the coefficient we obtain
\begin{align*}
\partial_z\left(
\theta(\partial_\lambda(f)) - \partial_\lambda(\theta(f))
\right)
& =
p \partial_\lambda(f) - \partial_z(\partial_\lambda(\theta(f))) \\
& =
p \partial_\lambda(f) - \partial_\lambda(\partial_z(\theta(f))) \\
& =
p \partial_\lambda(f) - \partial_\lambda(p f) = 0
\end{align*}
whence the coefficient does not depend on $z$ as desired.
This finishes the proof of the lemma.
\end{proof}

\noindent
Note that an iterated $\alpha_p$-cover $X' \to X$ (as defined in the
introduction to this section) is finite locally free. Hence if $X$ is
connected the degree of $X' \to X$ is constant and is a power of $p$.

\begin{lemma}
\label{lemma-pullback-along-p-power-cover}
Let $p$ be a prime number. Let $(S, \mathcal{I}, \gamma)$ be a divided power
scheme over $\mathbf{Z}_{(p)}$ with $p \in \mathcal{I}$. We set
$S_0 = V(\mathcal{I}) \subset S$. Let $f : X' \to X$ be an iterated
$\alpha_p$-cover of schemes over $S_0$ with constant degree $q$. Let
$\mathcal{F}$ be any crystal in quasi-coherent sheaves on $X$ and set
$\mathcal{F}' = f_{\text{cris}}^*\mathcal{F}$.
In the distinguished triangle
$$
Ru_{X/S, *}\mathcal{F}
\longrightarrow
f_*Ru_{X'/S, *}\mathcal{F}'
\longrightarrow
E
\longrightarrow
Ru_{X/S, *}\mathcal{F}[1]
$$
the object $E$ has cohomology sheaves annihilated by $q$.
\end{lemma}

\begin{proof}
Note that $X' \to X$ is a homeomorphism hence we can identify the underlying
topological spaces of $X$ and $X'$. The question is clearly local on $X$,
hence we may assume $X$, $X'$, and $S$ affine and $X' \to X$ given as a
composition
$$
X' = X_n \to X_{n - 1} \to X_{n - 2} \to \ldots \to X_0 = X
$$
where each morphism $X_{i + 1} \to X_i$ is an $\alpha_p$-cover.
Denote $\mathcal{F}_i$ the pullback of $\mathcal{F}$ to $X_i$.
It suffices to prove that each of the maps
$$
R\Gamma(\text{Cris}(X_i/S), \mathcal{F}_i)
\longrightarrow
R\Gamma(\text{Cris}(X_{i + 1}/S), \mathcal{F}_{i + 1})
$$
fits into a triangle whose third member has cohomology groups annihilated
by $p$. (This uses axiom TR4 for the triangulated category $D(X)$. Details
omitted.)

\medskip\noindent
Hence we may assume that $S = \Spec(A)$, $X = \Spec(C)$, $X' = \Spec(C')$
and $C' = C[z]/(z^p - c)$ for some $c \in C$. Choose a polynomial algebra
$P$ over $A$ and a surjection $P \to C$. Let $D$ be the $p$-adically completed
divided power envelop of $\Ker(P \to C)$ in $P$ as in (\ref{equation-D}).
Set $P' = P[z]$ with surjection $P' \to C'$ mapping $z$ to the class of $z$
in $C'$. Choose a lift $\lambda \in D$ of $c \in C$. Then we see that
the $p$-adically completed divided power envelope $D'$ of
$\Ker(P' \to C')$ in $P'$ is isomorphic to the $p$-adic completion of
$D[z]\langle \xi \rangle/(\xi - (z^p - \lambda))$, see
Lemma \ref{lemma-computation} and its proof.
Thus we see that the result follows from this lemma
by the computation of cohomology of crystals in quasi-coherent modules in
Proposition \ref{proposition-compute-cohomology-crystal}.
\end{proof}

\noindent
The bound in the following lemma is probably not optimal.

\begin{lemma}
\label{lemma-pullback-along-p-power-cover-cohomology}
With notations and assumptions as in
Lemma \ref{lemma-pullback-along-p-power-cover}
the map
$$
f^* :
H^i(\text{Cris}(X/S), \mathcal{F})
\longrightarrow
H^i(\text{Cris}(X'/S), \mathcal{F}')
$$
has kernel and cokernel annihilated by $q^{i + 1}$.
\end{lemma}

\begin{proof}
This follows from the fact that $E$ has nonzero cohomology sheaves in
degrees $-1$ and up, so that the spectral sequence
$H^a(\mathcal{H}^b(E)) \Rightarrow H^{a + b}(E)$ converges.
This combined with the long exact cohomology sequence associated
to a distinguished triangle gives the bound.
\end{proof}

\noindent
In Situation \ref{situation-global} assume that $p \in \mathcal{I}$.
Set
$$
X^{(1)} = X \times_{S_0, F_{S_0}} S_0.
$$
Denote $F_{X/S_0} : X \to X^{(1)}$ the relative Frobenius morphism.

\begin{lemma}
\label{lemma-pullback-relative-frobenius}
In the situation above, assume that $X \to S_0$ is smooth of relative
dimension $d$. Then $F_{X/S_0}$ is an iterated $\alpha_p$-cover
of degree $p^d$. Hence Lemmas \ref{lemma-pullback-along-p-power-cover} and
\ref{lemma-pullback-along-p-power-cover-cohomology} apply to this
situation. In particular, for any crystal in quasi-coherent modules
$\mathcal{G}$ on $\text{Cris}(X^{(1)}/S)$ the map
$$
F_{X/S_0}^* : H^i(\text{Cris}(X^{(1)}/S), \mathcal{G})
\longrightarrow
H^i(\text{Cris}(X/S), F_{X/S_0, \text{cris}}^*\mathcal{G})
$$
has kernel and cokernel annihilated by $p^{d(i + 1)}$.
\end{lemma}

\begin{proof}
It suffices to prove the first statement. To see this we may assume
that $X$ is \'etale over $\mathbf{A}^d_{S_0}$, see
Morphisms, Lemma \ref{morphisms-lemma-smooth-etale-over-affine-space}.
Denote $\varphi : X \to \mathbf{A}^d_{S_0}$ this \'etale morphism.
In this case the relative Frobenius of $X/S_0$ fits into a diagram
$$
\xymatrix{
X \ar[d] \ar[r] & X^{(1)} \ar[d] \\
\mathbf{A}^d_{S_0} \ar[r] & \mathbf{A}^d_{S_0}
}
$$
where the lower horizontal arrow is the relative frobenius morphism
of $\mathbf{A}^d_{S_0}$ over $S_0$. This is the morphism which raises
all the coordinates to the $p$th power, hence it is an iterated
$\alpha_p$-cover. The proof is finished by observing that the diagram
is a fibre square, see the proof of
\'Etale Cohomology, Theorem \ref{etale-cohomology-theorem-baffling}.
\end{proof}

Comments (0)

There are no comments yet for this tag.

Add a comment on tag 07PZ

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 lower-right corner).

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 box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

This captcha seems more appropriate than the usual illegible gibberish, right?