Proposition 68.16.7. A quasi-compact and quasi-separated algebraic space is affine if and only if all higher cohomology groups of quasi-coherent sheaves vanish. More precisely, any algebraic space as in Situation 68.16.1 is an affine scheme.

** Serre's criterion for affineness in the setting of algebraic spaces. **

**Proof.**
Choose an affine scheme $U = \mathop{\mathrm{Spec}}(B)$ and a surjective étale morphism $\varphi : U \to X$. Set $R = U \times _ X U$. As $p$ is separated (Lemma 68.16.6) we see that $R$ is a closed subscheme of $U \times _{\mathop{\mathrm{Spec}}(A)} U = \mathop{\mathrm{Spec}}(B \otimes _ A B)$. Hence $R = \mathop{\mathrm{Spec}}(C)$ is affine too and the ring map

is surjective. Let us denote the two maps $s, t : B \to C$ as usual. Pick $g_1, \ldots , g_ m \in B$ such that $s(g_1), \ldots , s(g_ m)$ generate $C$ over $t : B \to C$ (which is possible as $t : B \to C$ is of finite presentation and the displayed map is surjective). Then $g_1, \ldots , g_ m$ give global sections of $\varphi _*\mathcal{O}_ U$ and the map

is surjective: you can check this by restricting to $U$. Namely, $\varphi ^*\varphi _*\mathcal{O}_ U = t_*\mathcal{O}_ R$ (by Lemma 68.11.2) hence you get exactly the condition that $s(g_ i)$ generate $C$ over $t : B \to C$. By the vanishing of $H^1$ of the kernel we see that

is surjective. Thus we conclude that $B$ is a finite type $A$-algebra. Hence $X \to \mathop{\mathrm{Spec}}(A)$ is of finite type and separated. By Lemma 68.16.5 and Morphisms of Spaces, Lemma 66.27.5 it is also locally quasi-finite. Hence $X \to \mathop{\mathrm{Spec}}(A)$ is representable by Morphisms of Spaces, Lemma 66.51.1 and $X$ is a scheme. Finally $X$ is affine, hence equal to $\mathop{\mathrm{Spec}}(A)$, by an application of Cohomology of Schemes, Lemma 30.3.1. $\square$

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## Comments (2)

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