Lemma 32.11.1. Let $f : X \to S$ be a morphism of schemes. Assume that $f$ is surjective and finite, and assume that $X$ is affine. Then $S$ is affine.

** A scheme, admitting a finite surjective 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 29.44.7), we see that $S$ is separated (Morphisms, Lemma 29.41.11).

By Lemma 32.9.8 we can write $X = \mathop{\mathrm{lim}}\nolimits _ a X_ a$ with $X_ a \to S$ finite and of finite presentation. By Lemma 32.4.13 we see that $X_ a$ is affine for some $a \in A$. Replacing $X$ by $X_ a$ we may assume that $X \to S$ is surjective, finite, of finite presentation and that $X$ is affine.

By Proposition 32.5.4 we may write $S = \mathop{\mathrm{lim}}\nolimits _{i \in I} S_ i$ as a directed limits of schemes of finite type over $\mathbf{Z}$. By Lemma 32.10.1 we can after shrinking $I$ assume there exist schemes $X_ i \to S_ i$ of finite presentation such that $X_{i'} = X_ i \times _ S S_{i'}$ for $i' \geq i$ and such that $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$. By Lemma 32.8.3 we may assume that $X_ i \to S_ i$ is finite for all $i \in I$ as well. By Lemma 32.4.13 once again we may assume that $X_ i$ is affine for all $i \in I$. Hence the result follows from the Noetherian case, see Cohomology of Schemes, Lemma 30.13.3. $\square$

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