The Stacks project

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 $i\geq d$.

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