Lemma 59.84.1. Let $\Lambda$ be a Noetherian ring, let $M$ be a finite $\Lambda$-module which is annihilated by an integer $n > 0$, let $k$ be an algebraically closed field, and let $X$ be a separated, finite type scheme of dimension $\leq 1$ over $k$. Then

1. $H^ q_{\acute{e}tale}(X, \underline{M})$ is a finite $\Lambda$-module if $n$ is prime to $\text{char}(k)$,

2. $H^ q_{\acute{e}tale}(X, \underline{M})$ is a finite $\Lambda$-module if $X$ is proper.

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

$H^ q_{\acute{e}tale}(X, \underline{M[n]}) \to H^ q_{\acute{e}tale}(X, \underline{M}) \to H^ q_{\acute{e}tale}(X, \underline{M'})$

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

$0 \to N \to \Lambda ^{\oplus m} \to M \to 0$

This determines an exact sequence

$H^ q_{\acute{e}tale}(X, \underline{\Lambda ^{\oplus m}}) \to H^ q_{\acute{e}tale}(X, \underline{M}) \to H^{q + 1}_{\acute{e}tale}(X, \underline{N})$

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

$\underline{\Lambda } = \bigoplus \underline{\mathbf{F}_\ell } e_ i$

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

$H^ q_{\acute{e}tale}(X, \underline{\Lambda }) = H^ q_{\acute{e}tale}(X, \bigoplus \underline{\mathbf{F}_\ell } e_ i) = \bigoplus H^ q_{\acute{e}tale}(X, \underline{\mathbf{F}_\ell })e_ i$

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 $H^ q_{\acute{e}tale}(X, \underline{\Lambda })$. $\square$

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).