Lemma 30.4.4 (071L)—The Stacks project

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. $\square$

Comments (5)

Comment #7726 by Ryo Suzuki on August 31, 2022 at 04:36

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 $d = \max_{I\subset \{1,\dots,t\}} (|I| + t(\bigcap_{i\in I} U_i) -1)$ . 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 #7976 by Stacks Project on January 13, 2023 at 15:01

Nice! Both easier proof and a stronger statement. Thanks very much. See this commit.

Comment #8604 by nkym on July 31, 2023 at 13:41

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

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