Lemma 68.16.9. Let $S$ be a scheme. Let $X$ be a Noetherian algebraic space over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume that for every coherent $\mathcal{O}_ X$-module $\mathcal{F}$ there exists an $n \geq 1$ such that $H^1(X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = 0$. Then $X$ is a scheme and $\mathcal{L}$ is ample on $X$.

Proof. Let $s \in H^0(X, \mathcal{L}^{\otimes d})$ be a global section. Let $U \subset X$ be the open subspace over which $s$ is a generator of $\mathcal{L}^{\otimes d}$. In particular we have $\mathcal{L}^{\otimes d}|_ U \cong \mathcal{O}_ U$. We claim that $U$ is affine.

Proof of the claim. We will show that $H^1(U, \mathcal{F}) = 0$ for every quasi-coherent $\mathcal{O}_ U$-module $\mathcal{F}$. This will prove the claim by Proposition 68.16.7. Denote $j : U \to X$ the inclusion morphism. Since étale locally the morphism $j$ is affine (by Morphisms, Lemma 29.11.10) we see that $j$ is affine (Morphisms of Spaces, Lemma 66.20.3). Hence we have

$H^1(U, \mathcal{F}) = H^1(X, j_*\mathcal{F})$

by Lemma 68.8.2 (and Cohomology on Sites, Lemma 21.14.6). Write $j_*\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ as a filtered colimit of coherent $\mathcal{O}_ X$-modules, see Lemma 68.15.1. Then

$H^1(X, j_*\mathcal{F}) = \mathop{\mathrm{colim}}\nolimits H^1(X, \mathcal{F}_ i)$

by Lemma 68.5.1. Thus it suffices to show that $H^1(X, \mathcal{F}_ i)$ maps to zero in $H^1(U, j^*\mathcal{F}_ i)$. By assumption there exists an $n \geq 1$ such that

$H^1(X, \mathcal{F}_ i \otimes _{\mathcal{O}_ X} (\mathcal{O}_ X \oplus \mathcal{L} \oplus \ldots \oplus \mathcal{L}^{\otimes d - 1}) \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = 0$

Hence there exists an $a \geq 0$ such that $H^1(X, \mathcal{F}_ i \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes ad}) = 0$. On the other hand, the map

$s^ a : \mathcal{F}_ i \longrightarrow \mathcal{F}_ i \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes ad}$

is an isomorphism after restriction to $U$. Contemplating the commutative diagram

$\xymatrix{ H^1(X, \mathcal{F}_ i) \ar[r] \ar[d]_{s^ a} & H^1(U, j^*\mathcal{F}_ i) \ar[d]^{\cong } \\ H^1(X, \mathcal{F}_ i \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes ad}) \ar[r] & H^1(U, j^*(\mathcal{F}_ i \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes ad})) }$

we conclude that the map $H^1(X, \mathcal{F}_ i) \to H^1(U, j^*\mathcal{F}_ i)$ is zero and the claim holds.

Let $x \in |X|$ be a closed point. By Decent Spaces, Lemma 67.14.6 we can represent $x$ by a closed immersion $i : \mathop{\mathrm{Spec}}(k) \to X$ (this also uses that a quasi-separated algebraic space is decent, see Decent Spaces, Section 67.6). Thus $\mathcal{O}_ X \to i_*\mathcal{O}_{\mathop{\mathrm{Spec}}(k)}$ is surjective. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the kernel and choose $d \geq 1$ such that $H^1(X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d}) = 0$. Then

$H^0(X, \mathcal{L}^{\otimes d}) \to H^0(X, i_*\mathcal{O}_{\mathop{\mathrm{Spec}}(k)} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d}) = H^0(\mathop{\mathrm{Spec}}(k), i^*\mathcal{L}^{\otimes d}) \cong k$

is surjective by the long exact cohomology sequence. Hence there exists an $s \in H^0(X, \mathcal{L}^{\otimes d})$ such that $x \in U$ where $U$ is the open subspace corresponding to $s$ as above. Thus $x$ is in the schematic locus (see Properties of Spaces, Lemma 65.13.1) of $X$ by our claim.

To conclude that $X$ is a scheme, it suffices to show that any open subset of $|X|$ which contains all the closed points is equal to $|X|$. This follows from the fact that $|X|$ is a Noetherian topological space, see Properties of Spaces, Lemma 65.24.3. Finally, if $X$ is a scheme, then we can apply Cohomology of Schemes, Lemma 30.3.3 to conclude that $\mathcal{L}$ is ample. $\square$

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