Proposition 32.11.2. Let $f : X \to S$ be a morphism of schemes. Assume $X$ is affine and that $f$ is surjective and universally closed^{1}. Then $S$ is affine.

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

**Proof.**
By Morphisms, Lemma 29.41.11 the scheme $S$ is separated. Then by Morphisms, Lemma 29.11.11 we find that $f$ is affine. Whereupon by Morphisms, Lemma 29.44.7 we see that $f$ is integral.

By the preceding paragraph, we may assume $f : X \to S$ is surjective and integral, $X$ is affine, and $S$ is separated. Since $f$ is surjective and $X$ is quasi-compact we also deduce that $S$ is quasi-compact.

By Lemma 32.7.3 we can write $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$ with $X_ i \to S$ finite. By Lemma 32.4.13 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 32.11.1. $\square$

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