60.25 Pulling back along purely inseparable maps

By an $\alpha _ p$-cover we mean a morphism of the form

$X' = \mathop{\mathrm{Spec}}(C[z]/(z^ p - c)) \longrightarrow \mathop{\mathrm{Spec}}(C) = X$

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 60.6.9. Thus (60.6.9.1) 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 $\partial _ z(f) \in B'$ 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 60.25.1. 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 60.25.2. 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 60.25.1 recovers the crystalline Poincaré lemma (Lemma 60.20.2).

Lemma 60.25.3. In Situation 60.5.1. Assume $D$ and $\Omega _ D$ are as in (60.17.0.1) and (60.17.0.2). 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 (60.6.9.1)

$c_ M^\bullet : M \otimes _ D^\wedge \Omega ^\bullet _ D \longrightarrow M \otimes _ D^\wedge \Omega ^\bullet _{D'}$

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 60.25.1 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 60.2.4. Hence

$\Omega _{D'} = \Omega _ D \otimes _ D^\wedge D' \oplus D'\text{d}z$

by Lemma 60.6.6. 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 60.25.1. 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 60.25.4. 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 $S = \mathop{\mathrm{Spec}}(A)$, $X = \mathop{\mathrm{Spec}}(C)$, $X' = \mathop{\mathrm{Spec}}(C')$ 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 $\mathop{\mathrm{Ker}}(P \to C)$ in $P$ as in (60.17.0.1). 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 $\mathop{\mathrm{Ker}}(P' \to C')$ in $P'$ is isomorphic to the $p$-adic completion of $D[z]\langle \xi \rangle /(\xi - (z^ p - \lambda ))$, see Lemma 60.25.3 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 60.21.3. $\square$

The bound in the following lemma is probably not optimal.

Lemma 60.25.5. With notations and assumptions as in Lemma 60.25.4 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 60.7.5 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 60.25.6. 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 60.25.4 and 60.25.5 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 29.36.20. 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 Étale Morphisms, Lemma 41.14.3. $\square$

[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 $-\lambda$.

Comment #6268 by Mingchen on

Is $N'=B'$ in condition (2)?

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