The Stacks project

Proof. Let $x \in X$ be a closed point. Let $U \subset X$ be an affine open neighbourhood of $x$. Write $U = \mathop{\mathrm{Spec}}(A)$ and let $\mathfrak m \subset A$ be the maximal ideal corresponding to $x$. Set $Z = X \setminus U$ and $Z' = Z \cup \{ x\}$. By Schemes, Lemma 26.12.4 there are quasi-coherent sheaves of ideals $\mathcal{I}$, resp. $\mathcal{I}'$ cutting out the reduced closed subschemes $Z$, resp. $Z'$. Consider the short exact sequence

Since $x$ is a closed point of $X$ and $x \not\in Z$ we see that $\mathcal{I}/\mathcal{I}'$ is supported at $x$. In fact, the restriction of $\mathcal{I}/\mathcal{I'}$ to $U$ corresponds to the $A$-module $A/\mathfrak m$. Hence we see that $\Gamma (X, \mathcal{I}/\mathcal{I'}) = A/\mathfrak m$. Since by assumption $H^1(X, \mathcal{I}') = 0$ we see there exists a global section $f \in \Gamma (X, \mathcal{I})$ which maps to the element $1 \in A/\mathfrak m$ as a section of $\mathcal{I}/\mathcal{I'}$. Clearly we have $x \in X_ f \subset U$. This implies that $X_ f = D(f_ A)$ where $f_ A$ is the image of $f$ in $A = \Gamma (U, \mathcal{O}_ X)$. In particular $X_ f$ is affine.

Consider the union $W = \bigcup X_ f$ over all $f \in \Gamma (X, \mathcal{O}_ X)$ such that $X_ f$ is affine. Obviously $W$ is open in $X$. By the arguments above every closed point of $X$ is contained in $W$. The closed subset $X \setminus W$ of $X$ is also quasi-compact (see Topology, Lemma 5.12.3). Hence it has a closed point if it is nonempty (see Topology, Lemma 5.12.8). This would contradict the fact that all closed points are in $W$. Hence we conclude $X = W$.

Choose finitely many $f_1, \ldots , f_ n \in \Gamma (X, \mathcal{O}_ X)$ such that $X = X_{f_1} \cup \ldots \cup X_{f_ n}$ and such that each $X_{f_ i}$ is affine. This is possible as we've seen above. By Properties, Lemma 28.27.3 to finish the proof it suffices to show that $f_1, \ldots , f_ n$ generate the unit ideal in $\Gamma (X, \mathcal{O}_ X)$. Consider the short exact sequence

The arrow defined by $f_1, \ldots , f_ n$ is surjective since the opens $X_{f_ i}$ cover $X$. We let $\mathcal{F}$ be the kernel of this surjective map. Observe that $\mathcal{F}$ has a filtration

so that each subquotient $\mathcal{F}_ i/\mathcal{F}_{i - 1}$ is isomorphic to a quasi-coherent sheaf of ideals. Namely we can take $\mathcal{F}_ i$ to be the intersection of $\mathcal{F}$ with the first $i$ direct summands of $\mathcal{O}_ X^{\oplus n}$. The assumption of the lemma implies that $H^1(X, \mathcal{F}_ i/\mathcal{F}_{i - 1}) = 0$ for all $i$. This implies that $H^1(X, \mathcal{F}_2) = 0$ because it is sandwiched between $H^1(X, \mathcal{F}_1)$ and $H^1(X, \mathcal{F}_2/\mathcal{F}_1)$. Continuing like this we deduce that $H^1(X, \mathcal{F}) = 0$. Therefore we conclude that the map

is surjective as desired. $\square$