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. $\square$

Proof. By Properties, Lemma 28.22.3 every quasi-coherent sheaf of ideals is a directed colimit of quasi-coherent sheaves of ideals of finite type. By Cohomology, Lemma 20.19.1 taking cohomology on $X$ commutes with directed colimits. Hence we see that $H^1(X, \mathcal{I}) = 0$ for every quasi-coherent sheaf of ideals on $X$. In other words we see that Lemma 30.3.1 applies. $\square$

Proof. This is proved in exactly the same way as Lemma 30.3.1. Let $x \in X$ be a closed point. Let $U \subset X$ be an affine open neighbourhood of $x$ such that $\mathcal{L}|_ U \cong \mathcal{O}_ U$. 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

For every $n \geq 1$ we obtain a short exact sequence

By our assumption we may pick $n$ such that $H^1(X, \mathcal{I}' \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = 0$. 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 $A/\mathfrak m$. Since $\mathcal{L}$ is trivial on $U$ we see that the restriction of $\mathcal{I}/\mathcal{I}' \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}$ to $U$ also corresponds to the $A$-module $A/\mathfrak m$. Hence we see that $\Gamma (X, \mathcal{I}/\mathcal{I'} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = A/\mathfrak m$. By our choice of $n$ we see there exists a global section $s \in \Gamma (X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n})$ which maps to the element $1 \in A/\mathfrak m$. Clearly we have $x \in X_ s \subset U$ because $s$ vanishes at points of $Z$. This implies that $X_ s = D(f)$ where $f \in A$ is the image of $s$ in $A \cong \Gamma (U, \mathcal{L}^{\otimes n})$. In particular $X_ s$ is affine.

Consider the union $W = \bigcup X_ s$ over all $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ for $n \geq 1$ such that $X_ s$ 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 $X = W$. This means that $\mathcal{L}$ is ample by Properties, Definition 28.26.1. $\square$

Proof. Let $V \subset Y$ be an affine open subscheme. We have to show that $U = f^{-1}(V)$ is affine. The inclusion morphism $V \to Y$ is quasi-compact by Schemes, Lemma 26.21.14. Hence the base change $U \to X$ is quasi-compact, see Schemes, Lemma 26.19.3. Thus any quasi-coherent sheaf of ideals $\mathcal{I}$ on $U$ extends to a quasi-coherent sheaf of ideals on $X$, see Properties, Lemma 28.22.1. Since the formation of $R^1f_*$ is local on $Y$ (Cohomology, Section 20.7) we conclude that $R^1(U \to V)_*\mathcal{I} = 0$ by the assumption in the lemma. Hence by the Leray Spectral sequence (Cohomology, Lemma 20.13.4) we conclude that $H^1(U, \mathcal{I}) = H^1(V, (U \to V)_*\mathcal{I})$. Since $(U \to V)_*\mathcal{I}$ is quasi-coherent by Schemes, Lemma 26.24.1, we have $H^1(V, (U \to V)_*\mathcal{I}) = 0$ by Lemma 30.2.2. Thus we find that $U$ is affine by Lemma 30.3.1. $\square$