The Stacks project

Remark 60.24.12 (Perfectness). Let $p$ be a prime number. Let $(A, I, \gamma )$ be a divided power ring with $p$ nilpotent in $A$. Set $S = \mathop{\mathrm{Spec}}(A)$ and $S_0 = \mathop{\mathrm{Spec}}(A/I)$. Let $X$ be a proper smooth scheme over $S_0$. Let $\mathcal{F}$ be a crystal in finite locally free quasi-coherent $\mathcal{O}_{X/S}$-modules. Then $R\Gamma (\text{Cris}(X/S), \mathcal{F})$ is a perfect object of $D(A)$.

Hints: By Remark 60.24.9 we have

\[ R\Gamma (\text{Cris}(X/S), \mathcal{F}) \otimes _ A^\mathbf {L} A/I \cong R\Gamma (\text{Cris}(X/S_0), \mathcal{F}|_{\text{Cris}(X/S_0)}) \]

By Remark 60.24.11 we have

\[ R\Gamma (\text{Cris}(X/S_0), \mathcal{F}|_{\text{Cris}(X/S_0)}) = R\Gamma (X, \mathcal{F}_ X \otimes \Omega ^\bullet _{X/S_0}) \]

Using the stupid filtration on the de Rham complex we see that the last displayed complex is perfect in $D(A/I)$ as soon as the complexes

\[ R\Gamma (X, \mathcal{F}_ X \otimes \Omega ^ q_{X/S_0}) \]

are perfect complexes in $D(A/I)$, see More on Algebra, Lemma 15.74.4. This is true by standard arguments in coherent cohomology using that $\mathcal{F}_ X \otimes \Omega ^ q_{X/S_0}$ is a finite locally free sheaf and $X \to S_0$ is proper and flat (insert future reference here). Applying More on Algebra, Lemma 15.78.4 we see that

\[ R\Gamma (\text{Cris}(X/S), \mathcal{F}) \otimes _ A^\mathbf {L} A/I^ n \]

is a perfect object of $D(A/I^ n)$ for all $n$. This isn't quite enough unless $A$ is Noetherian. Namely, even though $I$ is locally nilpotent by our assumption that $p$ is nilpotent, see Divided Power Algebra, Lemma 23.2.6, we cannot conclude that $I^ n = 0$ for some $n$. A counter example is $\mathbf{F}_ p\langle x \rangle $. To prove it in general when $\mathcal{F} = \mathcal{O}_{X/S}$ the argument of https://math.columbia.edu/~dejong/wordpress/?p=2227 works. When the coefficients $\mathcal{F}$ are non-trivial the argument of [Faltings-very] seems to be as follows. Reduce to the case $pA = 0$ by More on Algebra, Lemma 15.78.4. In this case the Frobenius map $A \to A$, $a \mapsto a^ p$ factors as $A \to A/I \xrightarrow {\varphi } A$ (as $x^ p = 0$ for $x \in I$). Set $X^{(1)} = X \otimes _{A/I, \varphi } A$. The absolute Frobenius morphism of $X$ factors through a morphism $F_ X : X \to X^{(1)}$ (a kind of relative Frobenius). Affine locally if $X = \mathop{\mathrm{Spec}}(C)$ then $X^{(1)} = \mathop{\mathrm{Spec}}( C \otimes _{A/I, \varphi } A)$ and $F_ X$ corresponds to $C \otimes _{A/I, \varphi } A \to C$, $c \otimes a \mapsto c^ pa$. This defines morphisms of ringed topoi

\[ (X/S)_{\text{cris}} \xrightarrow {(F_ X)_{\text{cris}}} (X^{(1)}/S)_{\text{cris}} \xrightarrow {u_{X^{(1)}/S}} \mathop{\mathit{Sh}}\nolimits (X^{(1)}_{Zar}) \]

whose composition is denoted $\text{Frob}_ X$. One then shows that $R\text{Frob}_{X, *}\mathcal{F}$ is representable by a perfect complex of $\mathcal{O}_{X^{(1)}}$-modules(!) by a local calculation.


Comments (0)

There are also:

  • 2 comment(s) on Section 60.24: Some further results

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