Proposition 31.11.2. Let $f : X \to S$ be a morphism of schemes. Assume that $f$ is surjective and integral, and assume that $X$ is affine. Then $S$ is affine.

** A scheme admitting a surjective integral map from an affine scheme is affine. **

**Proof.**
Since $f$ is surjective and $X$ is quasi-compact we see that $S$ is quasi-compact. Since $X$ is separated and $f$ is surjective and universally closed (Morphisms, Lemma 28.42.7), we see that $S$ is separated (Morphisms, Lemma 28.39.11).

By Lemma 31.7.2 we can write $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$ with $X_ i \to S$ finite. By Lemma 31.4.13 we see that for $i$ sufficiently large the scheme $X_ i$ is affine. Moreover, since $X \to S$ factors through each $X_ i$ we see that $X_ i \to S$ is surjective. Hence we conclude that $S$ is affine by Lemma 31.11.1. $\square$

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## Comments (1)

Comment #3025 by Brian Lawrence on