Lemma 69.16.9. 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 69.16.7. 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 67.20.3). Hence we have
H^1(U, \mathcal{F}) = H^1(X, j_*\mathcal{F})
by Lemma 69.8.2 (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 69.15.1. Then
H^1(X, j_*\mathcal{F}) = \mathop{\mathrm{colim}}\nolimits H^1(X, \mathcal{F}_ i)
by Lemma 69.5.1. 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 68.14.6 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 68.6). 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 66.13.1) 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 66.24.3. 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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