Proposition 68.15.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that $f$ is surjective and integral, and assume that $X$ is affine. Then $Y$ is affine.

Proof. We may and do view $f : X \to Y$ as a morphism of algebraic spaces over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ (see Spaces, Definition 63.16.2). Note that integral morphisms are affine and universally closed, see Morphisms of Spaces, Lemma 65.45.7. By Morphisms of Spaces, Lemma 65.9.8 we see that $Y$ is a separated algebraic space. As $f$ is surjective and $X$ is quasi-compact we see that $Y$ is quasi-compact.

Consider the sheaf $\mathcal{A} = f_*\mathcal{O}_ X$. This is a quasi-coherent sheaf of $\mathcal{O}_ Y$-algebras, see Morphisms of Spaces, Lemma 65.11.2. By Lemma 68.9.1 we can write $\mathcal{A} = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i$ as a filtered colimit of finite type $\mathcal{O}_ Y$-modules. Let $\mathcal{A}_ i \subset \mathcal{A}$ be the $\mathcal{O}_ Y$-subalgebra generated by $\mathcal{F}_ i$. Since the map of algebras $\mathcal{O}_ Y \to \mathcal{A}$ is integral, we see that each $\mathcal{A}_ i$ is a finite quasi-coherent $\mathcal{O}_ Y$-algebra. Hence

$X_ i = \underline{\mathop{\mathrm{Spec}}}_ Y(\mathcal{A}_ i) \longrightarrow Y$

is a finite morphism of algebraic spaces. Here $\underline{\mathop{\mathrm{Spec}}}$ is the construction of Morphisms of Spaces, Lemma 65.20.7. It is clear that $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$. Hence by Lemma 68.5.10 we see that for $i$ sufficiently large the scheme $X_ i$ is affine. Moreover, since $X \to Y$ factors through each $X_ i$ we see that $X_ i \to Y$ is surjective. Hence we conclude that $Y$ is affine by Lemma 68.15.1. $\square$

## Comments (2)

Comment #1921 by Matthieu Romagny on

typo in first sentence of proof: a morphism of algebraic spaces

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• 1 comment(s) on Section 68.15: Characterizing affine spaces

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