Lemma 59.84.3. Let $\Lambda$ be a Noetherian ring, let $k$ be an algebraically closed field, let $X$ be a separated finite type scheme over $k$ of dimension $\leq 1$, let $\mathcal{F}$ be a constructible sheaf of $\Lambda$-modules on $X_{\acute{e}tale}$, and let $j : X' \to X$ be the inclusion of a dense open subscheme. Then $H^ q_{\acute{e}tale}(X, \mathcal{F})$ is a finite $\Lambda$-module if and only if $H^ q_{\acute{e}tale}(X, j_!j^{-1}\mathcal{F})$ is a finite $\Lambda$-module.

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

$0 \to j_!j^{-1}\mathcal{F} \to \mathcal{F} \to i_*i^{-1}\mathcal{F} \to 0$

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

$H^0_{\acute{e}tale}(Z, i^{-1}\mathcal{F}) = \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{F}_{x_ i}$

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

$0 \to H^0_{\acute{e}tale}(X, j_!j^{-1}\mathcal{F}) \to H^0_{\acute{e}tale}(X, \mathcal{F}) \to H^0_{\acute{e}tale}(Z, i^{-1}\mathcal{F}) \to H^1_{\acute{e}tale}(X, j_!j^{-1}\mathcal{F}) \to H^1_{\acute{e}tale}(X, \mathcal{F}) \to 0$

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. $\square$

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).