The Stacks project

Lemma 63.13.2. Let $f : X \to S$ be a proper smooth morphism of schemes with geometrically connected fibres of dimension $1$. Let $\ell $ be a prime number. Then $R^ qf_*\underline{\mathbf{Z}/\ell \mathbf{Z}}$ is a constructible.

Proof. We may assume $S$ is affine. Say $S = \mathop{\mathrm{Spec}}(A)$. Then, if we write $A = \bigcup A_ i$ as the union of its finite type $\mathbf{Z}$-subalgebras, we can find an $i$ and a morphism $f_ i : X_ i \to S_ i = \mathop{\mathrm{Spec}}(A_ i)$ of finite type whose base change to $S$ is $f : X \to S$, see Limits, Lemma 32.10.1. After increasing $i$ we may assume $f_ i : X_ i \to S_ i$ is smooth, proper, and of relative dimension $1$, see Limits, Lemmas 32.13.1 32.8.9, and 32.18.4. By More on Morphisms, Lemma 37.53.8 we obtain an open subscheme $U_ i \subset S_ i$ such that the fibres of $f_ i : X_ i \to S_ i$ over $U_ i$ are geometrically connected. Then $S \to S_ i$ maps into $U_ i$. We may replace $X \to S$ by $f_ i : f_ i^{-1}(U_ i) \to U_ i$ to reduce to the case discussed in the next paragraph.

Assume $S$ is Noetherian. We may write $S = U \cup Z$ where $U$ is the open subscheme defined by the nonvanishing of $\ell $ and $Z = V(\ell ) \subset S$. Since the formation of $R^ qf_*\underline{\mathbf{Z}/\ell \mathbf{Z}}$ commutes with arbitrary base change (Étale Cohomology, Theorem 59.91.11), it suffices to prove the result over $U$ and over $Z$. Thus we reduce to the following two cases: (a) $\ell $ is invertible on $S$ and (b) $\ell $ is zero on $S$.

Case (a). We claim that in this case the sheaves $R^ qf_*\underline{\mathbf{Z}/\ell \mathbf{Z}}$ are finite locally constant on $S$. First, by proper base change (in the form of Étale Cohomology, Lemma 59.91.13) and by finiteness (Étale Cohomology, Theorem 59.83.10) we see that the stalks of $R^ qf_*\underline{\mathbf{Z}/\ell \mathbf{Z}}$ are finite. By Étale Cohomology, Lemma 59.94.4 all specialization maps are isomorphisms. We conclude the claim holds by Étale Cohomology, Lemma 59.75.6.

Case (b). Here $\ell = p$ is a prime and $S$ is a scheme over $\mathop{\mathrm{Spec}}(\mathbf{F}_ p)$. By the same references as above we already know that the stalks of $R^ qf_*\underline{\mathbf{Z}/p\mathbf{Z}}$ are finite and zero for $q \geq 2$. It follows from Étale Cohomology, Lemma 59.39.3 that $f_*\underline{\mathbf{Z}/p\mathbf{Z}} = \underline{\mathbf{Z}/p\mathbf{Z}}$. It remains to prove that $R^1f_*\underline{\mathbf{Z}/p\mathbf{Z}}$ is constructible. Consider the Artin-Schreyer sequence

\[ 0 \to \underline{\mathbf{Z}/p\mathbf{Z}} \to \mathcal{O}_ X \xrightarrow {F - 1} \mathcal{O}_ X \to 0 \]

See Étale Cohomology, Section 59.63. Recall that $f_*\mathcal{O}_ X = \mathcal{O}_ S$ and $R^1f_*\mathcal{O}_ X$ is a finite locally free $\mathcal{O}_ S$-module of rank equal to the genera of the fibres of $X \to S$, see Algebraic Curves, Lemma 53.20.13. We conclude that we have a short exact sequence

\[ 0 \to \mathop{\mathrm{Coker}}(F - 1 : \mathcal{O}_ S \to \mathcal{O}_ S) \to R^1f_*\underline{\mathbf{Z}/p\mathbf{Z}} \to \mathop{\mathrm{Ker}}(F - 1 : R^1f_*\mathcal{O}_ X \to R^1f_*\mathcal{O}_ X) \to 0 \]

Applying Lemma 63.13.1 we win. $\square$

Comments (0)

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