Lemma 30.4.4. Let $X$ be a quasi-compact quasi-separated scheme. Let $X = U_1 \cup \ldots \cup U_ t$ be an affine open covering. Set

$d = \max \nolimits _{I \subset \{ 1, \ldots , t\} } \left(|I| + t(\bigcap \nolimits _{i \in I} U_ i)\right)$

where $t(U)$ is the minimal number of affines needed to cover the scheme $U$. Then $H^ n(X, \mathcal{F}) = 0$ for all $n \geq d$ and all quasi-coherent sheaves $\mathcal{F}$.

Proof. Note that since $X$ is quasi-separated the numbers $t(\bigcap _{i \in I} U_ i)$ are finite. Let $\mathcal{U} : X = \bigcup _{i = 1}^ t U_ i$. By Cohomology, Lemma 20.11.5 there is a spectral sequence

$E_2^{p, q} = \check{H}^ p(\mathcal{U}, \underline{H}^ q(\mathcal{F}))$

converging to $H^{p + q}(U, \mathcal{F})$. By Cohomology, Lemma 20.23.6 we have

$E_2^{p, q} = H^ p(\check{\mathcal{C}}_{alt}^\bullet ( \mathcal{U}, \underline{H}^ q(\mathcal{F}))$

The alternating Čech complex with values in the presheaf $\underline{H}^ q(\mathcal{F})$ vanishes in high degrees by Lemma 30.4.2, more precisely $E_2^{p, q} = 0$ for $p + q \geq d$. Hence the result follows. $\square$

Comment #7726 by Ryo Suzuki on

This lemma can be proved in the same fashion with first proof of lemma 30.4.2. Moreover, it can be slightly strengthened the statement.

First, reformulate the statement as follows: Let X be a quasi-compact quasi-separated scheme. Let $X = U_1\cup \dots U_t$ be an open covering where $U_i$ is quasi-compact separated. Set . Then $H^n(X,F) = 0$ for $n\geq d$ and all quasi-coherent sheaf $F$.

I will prove it by induction on $t$. If $t=1$, the result follows from 30.4.2. If $t>1$, write $X = U\cup V$ with $V=U_t$ quasi-compact separated and $U = U_1\cup \dots \cup U_{t-1}$. Notice that $U\cap V$ can be written as $(U_1\cap V)\cup \dots \cup (U_{t^1}\cup V)$, where $U_i\cap V$ is quasi-compact separated.

Let $a = \max_{I\subset \{1,\dots,t-1\}} (|I|+t(\bigcap_{i\in I} U_i) - 1)$, $b=t(V)$, $c = \max_{I\subset \{1,\dots,t-1\}} (|I|+t(V\cap \bigcap_{i\in I} U_i)-1)$, $d = \max(a,b,c+1) = \max_{I\subset \{1,\dots,t\}} (|I|+(t(\bigcap_{i\in I} U_i)) - 1)$. So, By induction we have $H^i(U,F) =0$ for $i\geq a$, $H^i(V,F)=0$ for $i\geq b$, and $H^i(U\cap V,F) = 0$ for $i\geq c$. By Mayer-Vietoris sequence, $H^i(X,F)=0$ for $i\geq d$.

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