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
H^{q - 1}(X, \mathcal{F}) \to H^{q - 1}(X \setminus Z, \mathcal{F}) \to H^ q_ Z(X, \mathcal{F}) \to H^ q(X, \mathcal{F})
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
H^ d(W \setminus Z, \mathcal{F}|_{W \setminus Z}) = \mathop{\mathrm{colim}}\nolimits _{Z \subset V} H^ d(V \setminus Z, \mathcal{F})
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
\mathcal{G} = \mathop{\mathrm{Im}}(\mathcal{I}^{d - 1} \to \mathcal{I}^ d) = \mathop{\mathrm{Ker}}(\mathcal{I}^ d \to \mathcal{I}^{d + 1})
For U \subset X quasi-compact open we have a map of exact sequences as follows
\xymatrix{ \mathcal{I}^{d - 1}(X) \ar[r] \ar[d] & \mathcal{G}(X) \ar[r] \ar[d] & H^ d(X, \mathcal{F}) \ar[d] \ar[r] & 0 \\ \mathcal{I}^{d - 1}(U) \ar[r] & \mathcal{G}(U) \ar[r] & H^ d(U, \mathcal{F}) \ar[r] & 0 }
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
H^ q(X, \mathcal{F}) = H^ q\left( \Gamma (X, \mathcal{I}^0) \to \ldots \to \Gamma (X, \mathcal{I}^{d - 1}) \to \Gamma (X, \mathcal{G}) \right)
In particular the cohomology group vanishes if q > d.
Proof of (3). Given Z as in (3) we consider the long exact sequence
H^{q - 1}(X, \mathcal{F}) \to H^{q - 1}(X \setminus Z, \mathcal{F}) \to H^ q_ Z(X, \mathcal{F}) \to H^ q(X, \mathcal{F})
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
Comments (2)
Comment #7382 by Badam Baplan on
Comment #7398 by Johan on