Lemma 62.13.2. 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 32.18.3. 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 arbtrary 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 62.13.1 we win. $\square$

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