The Stacks project

Part (1) is the main theorem of [Scheiderer].

Proposition 20.22.4. Let $X$ be a spectral space of Krull dimension $d$. Let $\mathcal{F}$ be an abelian sheaf on $X$.

  1. $H^ q(X, \mathcal{F}) = 0$ for $q > d$,

  2. $H^ d(X, \mathcal{F}) \to H^ d(U, \mathcal{F})$ is surjective for every quasi-compact open $U \subset X$,

  3. $H^ q_ Z(X, \mathcal{F}) = 0$ for $q > d$ and any constructible closed subset $Z \subset X$.

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

In the remark preceding 0A3G, it is said that this result improves Grothendieck's result for Noetherian topological spaces 02UX. This confuses me because Noetherian topological spaces are not in general spectral (sometimes they fail to be sober). Can the remark be made more precise?

Comment #7398 by on

It is not a mathematical statement to say that "result A is an improvement of result B", so I think it is fine to leave the slightly imprecise formulation of that English sentence for now. And I really do think this result is an improvement since sheaves on a space are the same as sheaves on the corresponding sobrification and the sobrification of a Noetherian topological space is Noetherian. Cheers!

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