The Stacks project

Proof. If $n = \ell n'$ for some prime number $\ell $, then we get a short exact sequence $0 \to M[\ell ] \to M \to M' \to 0$ of finite $\Lambda $-modules and $M'$ is annihilated by $n'$. This produces a corresponding short exact sequence of constant sheaves, which in turn gives rise to an exact sequence of cohomology modules

Thus, if we can show the result in case $M$ is annihilated by a prime number, then by induction on $n$ we win.

Let $\ell $ be a prime number such that $\ell $ annihilates $M$. Then we can replace $\Lambda $ by the $\mathbf{F}_\ell $-algebra $\Lambda /\ell \Lambda $. Namely, the cohomology of $\mathcal{F}$ as a sheaf of $\Lambda $-modules is the same as the cohomology of $\mathcal{F}$ as a sheaf of $\Lambda /\ell \Lambda $-modules, for example by Cohomology on Sites, Lemma 21.12.4.

Assume $\ell $ be a prime number such that $\ell $ annihilates $M$ and $\Lambda $. Let us reduce to the case where $M$ is a finite free $\Lambda $-module. Namely, choose a short exact sequence

This determines an exact sequence

By descending induction on $q$ we get the result for $M$ if we know the result for $\Lambda ^{\oplus m}$. Here we use that we know that our cohomology groups vanish in degrees $> 2$ by Theorem 59.83.10.

Let $\ell $ be a prime number and assume that $\ell $ annihilates $\Lambda $. It remains to show that the cohomology groups $H^ q_{\acute{e}tale}(X, \underline{\Lambda })$ are finite $\Lambda $-modules. We will use a trick to show this; the “correct” argument uses a coefficient theorem which we will show later. Choose a basis $\Lambda = \bigoplus _{i \in I} \mathbf{F}_\ell e_ i$ such that $e_0 = 1$ for some $0 \in I$. The choice of this basis determines an isomorphism

of sheaves on $X_{\acute{e}tale}$. Thus we see that

since taking cohomology over $X$ commutes with direct sums by Theorem 59.51.3 (or Lemma 59.51.4 or Lemma 59.52.2). Since we already know that $H^ q_{\acute{e}tale}(X, \underline{\mathbf{F}_\ell })$ is a finite dimensional $\mathbf{F}_\ell $-vector space (by Theorem 59.83.10), we see that $H^ q_{\acute{e}tale}(X, \underline{\Lambda })$ is free over $\Lambda $ of the same rank. Namely, given a basis $\xi _1, \ldots , \xi _ m$ of $H^ q_{\acute{e}tale}(X, \underline{\mathbf{F}_\ell })$ we see that $\xi _1 e_0, \ldots , \xi _ m e_0$ form a $\Lambda $-basis for $H^ q_{\acute{e}tale}(X, \underline{\Lambda })$. $\square$

Proof. Namely, we have $H^ q_{\acute{e}tale}(X, \mathcal{F}) = H^ q_{\acute{e}tale}(Y, f_*\mathcal{F})$ by the vanishing of $R^ qf_*$ for $q > 0$ (Proposition 59.55.2) and the Leray spectral sequence (Cohomology on Sites, Lemma 21.14.6). $\square$

Proof. Since $X'$ is dense, we see that $Z = X \setminus X'$ has dimension $0$ and hence is a finite set $Z = \{ x_1, \ldots , x_ n\} $ of $k$-rational points. Consider the short exact sequence

of Lemma 59.70.8. Observe that $H^ q_{\acute{e}tale}(X, i_*i^{-1}\mathcal{F}) = H^ q_{\acute{e}tale}(Z, i^*\mathcal{F})$. Namely, $i : Z \to X$ is a closed immersion, hence finite, hence we have the vanishing of $R^ qi_*$ for $q > 0$ by Proposition 59.55.2, and hence the equality follows from the Leray spectral sequence (Cohomology on Sites, Lemma 21.14.6). Since $Z$ is a disjoint union of spectra of algebraically closed fields, we conclude that $H^ q_{\acute{e}tale}(Z, i^*\mathcal{F}) = 0$ for $q > 0$ and

which is a finite $\Lambda $-module $\mathcal{F}_{x_ i}$ is finite due to the assumption that $\mathcal{F}$ is a constructible sheaf of $\Lambda $-modules. The long exact cohomology sequence gives an exact sequence

and isomorphisms $H^0_{\acute{e}tale}(X, j_!j^{-1}\mathcal{F}) \to H^0_{\acute{e}tale}(X, \mathcal{F})$ for $q > 1$. The lemma follows easily from this. $\square$

Proof. Since $\dim (X) \leq 1$ there is an open $V \subset X$ which is disjoint from $U$ such that $X' = U \cup V$ is dense open in $X$ (details omitted). If $j' : X' \to X$ denotes the inclusion morphism, then we see that $j_!\underline{M}$ is a direct summand of $j'_!\underline{M}$. Hence it suffices to prove the lemma in case $U$ is open and dense in $X$. This case follows from Lemmas 59.84.3 and 59.84.1. $\square$

Proof. Immediate from the long exact sequence of cohomology. $\square$

Proof. Let $f : V \to U$ be a finite étale morphism of degree prime to $\ell $ as in Lemma 59.66.4. The discussion in Section 59.66 gives maps

whose composition is an isomorphism. Hence it suffices to prove the finiteness of $H^ q_{\acute{e}tale}(X, j_!f_*f^{-1}\mathcal{G})$. By Zariski's Main theorem (More on Morphisms, Lemma 37.43.3) we can choose a diagram

with $\overline{f} : Y \to X$ finite and $j'$ an open immersion with dense image. Since $f$ is finite and $V$ dense in $Y$ we have $V = U \times _ X Y$. By Lemma 59.70.9 we have

By Lemma 59.84.2 it suffices to consider $j'_!f^{-1}\mathcal{G}$. The existence of the filtration given by Lemma 59.66.4, the fact that $j'_!$ is exact, and Lemma 59.84.5 reduces us to the case $\mathcal{F} = j'_!\underline{M}$ for a finite $\Lambda $-module $M$ which is Lemma 59.84.4. $\square$

Proof. without further mention. Write $\mathcal{F} = \mathcal{F}_1 \oplus \ldots \oplus \mathcal{F}_ r$ where $\mathcal{F}_ i$ is annihilated by $\ell _ i^{n_ i}$ for some prime $\ell _ i$ and integer $n_ i > 0$. By Lemma 59.84.5 it suffices to prove the theorem for $\mathcal{F}_ i$. Thus we may and do assume that $\ell ^ n$ kills $\mathcal{F}$ for some prime $\ell $ and integer $n > 0$.

Since $\mathcal{F}$ is constructible as a sheaf of $\Lambda $-modules, there is a dense open $U \subset X$ such that $\mathcal{F}|_ U$ is a finite type, locally constant sheaf of $\Lambda $-modules. Since $\dim (X) \leq 1$ we may assume, after shrinking $U$, that $U = U_1 \amalg \ldots \amalg U_ n$ is a disjoint union of irreducible schemes (just remove the closed points which lie in the intersections of $\geq 2$ components of $U$). By Lemma 59.84.3 we reduce to the case $\mathcal{F} = j_!\mathcal{G}$ where $\mathcal{G}$ is a finite type, locally constant sheaf of $\Lambda $-modules on $U$ (and annihilated by $\ell ^ n$).

Since we chose $U = U_1 \amalg \ldots \amalg U_ n$ with $U_ i$ irreducible we have

where $j_ i : U_ i \to X$ is the inclusion morphism. The case of $j_{i!}(\mathcal{G}|_{U_ i})$ is handled in Lemma 59.84.6. $\square$