The Stacks project

Proof. We prove this result by induction on $d$.

If $d = 0$, then $X$ is a profinite space, see Topology, Lemma 5.23.8. Thus (1) holds by Lemma 20.22.3. If $U \subset X$ is quasi-compact open, then $U$ is also closed as a quasi-compact subset of a Hausdorff space. Hence $X = U \amalg (X \setminus U)$ as a topological space and we see that (2) holds. Given $Z$ as in (3) we consider the long exact sequence

Since $X$ and $U = X \setminus Z$ are profinite (namely $U$ is quasi-compact because $Z$ is constructible) and since we have (2) and (1) we obtain the desired vanishing of the cohomology groups with support in $Z$.

Induction step. Assume $d \geq 1$ and assume the proposition is valid for all spectral spaces of dimension $< d$. We first prove part (2) for $X$. Let $U$ be a quasi-compact open. Let $\xi \in H^ d(U, \mathcal{F})$. Set $Z = X \setminus U$. Let $W \subset X$ be the set of points specializing to $Z$. By Lemma 20.22.1 we have

where the colimit is over the quasi-compact open neighbourhoods $V$ of $Z$ in $X$. By Topology, Lemma 5.24.7 we see that $W \setminus Z$ is a spectral space. Since every point of $W$ specializes to a point of $Z$, we see that $W \setminus Z$ is a spectral space of Krull dimension $< d$. By induction hypothesis we see that the image of $\xi $ in $H^ d(W \setminus Z, \mathcal{F}|_{W \setminus Z})$ is zero. By the displayed formula, there exists a $Z \subset V \subset X$ quasi-compact open such that $\xi |_{V \setminus Z} = 0$. Since $V \setminus Z = V \cap U$ we conclude by the Mayer-Vietoris (Lemma 20.8.2) for the covering $X = U \cup V$ that there exists a $\tilde\xi \in H^ d(X, \mathcal{F})$ which restricts to $\xi $ on $U$ and to zero on $V$. In other words, part (2) is true.

Proof of part (1) assuming (2). Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet $. Set

For $U \subset X$ quasi-compact open we have a map of exact sequences as follows

The sheaf $\mathcal{I}^{d - 1}$ is flasque by Lemma 20.12.2 and the fact that $d \geq 1$. By part (2) we see that the right vertical arrow is surjective. We conclude by a diagram chase that the map $\mathcal{G}(X) \to \mathcal{G}(U)$ is surjective. By Lemma 20.12.6 we conclude that $\check{H}^ q(\mathcal{U}, \mathcal{G}) = 0$ for $q > 0$ and any finite covering $\mathcal{U} : U = U_1 \cup \ldots \cup U_ n$ of a quasi-compact open by quasi-compact opens. Applying Lemma 20.11.9 we find that $H^ q(U, \mathcal{G}) = 0$ for all $q > 0$ and all quasi-compact opens $U$ of $X$. By Leray's acyclicity lemma (Derived Categories, Lemma 13.16.7) we conclude that

In particular the cohomology group vanishes if $q > d$.

Proof of (3). Given $Z$ as in (3) we consider the long exact sequence

Since $X$ and $U = X \setminus Z$ are spectral spaces (Topology, Lemma 5.23.5) of dimension $\leq d$ and since we have (2) and (1) we obtain the desired vanishing. $\square$