Lemma 30.4.4. Let $X$ be a quasi-compact quasi-separated scheme. Let $X = U_1 \cup \ldots \cup U_ n$ be an open covering with each $U_ i$ quasi-compact and separated (for example affine). Set

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

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

Proof. Note that since $X$ is quasi-separated and $U_ i$ quasi-compact the numbers $t(\bigcap _{i \in I} U_ i)$ are finite. Proof using induction on $n$. If $n = 1$ then the result follows from Lemma 30.4.2. If $n > 1$, write $X = U \cup V$ with $U = U_1 \cup \ldots \cup U_{n - 1}$ and $V = U_ n$. We apply the Mayer-Vietoris long exact sequence

$0 \to H^0(X, \mathcal{F}) \to H^0(U, \mathcal{F}) \oplus H^0(V, \mathcal{F}) \to H^0(U \cap V, \mathcal{F}) \to H^1(X, \mathcal{F}) \to \ldots$

see Cohomology, Lemma 20.8.2. To finish the proof for $q \geq d$ we will show that $H^ q(V, \mathcal{F})$, $H^ q(U, \mathcal{F})$, and $H^{q - 1}(U \cap V, \mathcal{F})$ vanish. By the case $n = 1$ we have $H^ q(V, \mathcal{F}) = 0$ for $q \geq t(V) = t(U_ n)$. Since $t(V) \leq d$ this proves what we want. By induction hypothesis we have $H^ q(U, \mathcal{F}) = 0$ for

$q \geq \max \nolimits _{I \subset \{ 1, \ldots , n - 1\} } \left(|I| + t(\bigcap \nolimits _{i \in I} U_ i) - 1\right)$

Since the integer on the right is less than or equal to $d$, this proves what we want. Finally we may use our induction hypothesis for the open $U \cap V = (U_1 \cap U_ n) \cup \ldots \cup (U_{n - 1} \cap U_ n)$ to get the vanishing of $H^ q(U \cap V, \mathcal{F}) = 0$ for

$q \geq \max \nolimits _{I \subset \{ 1, \ldots , n - 1\} } \left(|I| + t(U_ n \cap \bigcap \nolimits _{i \in I} U_ i) - 1\right)$

Since the integer on the right is strictly less than $d$ the lemma 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 $i\geq d$.

Comment #8604 by nkym on

"Since $V$ is affine, we have $H^i(V,¥mathcal{F}) = 0$ for $i¥geq 0$" is not true for $i= 0$.

Comment #8605 by nkym on

Also, V is not necessarily affine in the first place.

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