Lemma 67.17.1. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Assume

$f$ finite,

$f$ surjective,

$Y$ affine, and

$X$ Noetherian.

Then $X$ is affine.

Lemma 67.17.1. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Assume

$f$ finite,

$f$ surjective,

$Y$ affine, and

$X$ Noetherian.

Then $X$ is affine.

**Proof.**
We will prove that under the assumptions of the lemma for any coherent $\mathcal{O}_ X$-module $\mathcal{F}$ we have $H^1(X, \mathcal{F}) = 0$. This implies that $H^1(X, \mathcal{F}) = 0$ for every quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ by Lemmas 67.15.1 and 67.5.1. Then it follows that $X$ is affine from Proposition 67.16.7.

Let $\mathcal{P}$ be the property of coherent sheaves $\mathcal{F}$ on $X$ defined by the rule

\[ \mathcal{P}(\mathcal{F}) \Leftrightarrow H^1(X, \mathcal{F}) = 0. \]

We are going to apply Lemma 67.14.5. Thus we have to verify (1), (2) and (3) of that lemma for $\mathcal{P}$. Property (1) follows from the long exact cohomology sequence associated to a short exact sequence of sheaves. Property (2) follows since $H^1(X, -)$ is an additive functor. To see (3) let $i : Z \to X$ be a reduced closed subspace with $|Z|$ irreducible. Let $W = Z \times _ X Y$ and denote $i' : W \to Y$ the corresponding closed immersion. Denote $f' : W \to Z$ the other projection which is a finite morphism of algebraic spaces. Since $W$ is a closed subscheme of $Y$, it is affine. We claim that $\mathcal{G} = f_*i'_*\mathcal{O}_ W = i_*f'_*\mathcal{O}_ W$ satisfies properties (3)(a) and (3)(b) of Lemma 67.14.5 which will finish the proof. Property (3)(a) is clear as $W \to Z$ is surjective (because $f$ is surjective). To see (3)(b) let $\mathcal{I}$ be a nonzero quasi-coherent sheaf of ideals on $Z$. We simply take $\mathcal{G}' = \mathcal{I} \mathcal{G}$. Namely, we have

\[ \mathcal{I} \mathcal{G} = f'_*(\mathcal{I}') \]

where $\mathcal{I}' = \mathop{\mathrm{Im}}((f')^*\mathcal{I} \to \mathcal{O}_ W)$. This is true because $f'$ is a (representable) affine morphism of algebraic spaces and hence the result can be checked on an étale covering of $Z$ by a scheme in which case the result is Cohomology of Schemes, Lemma 30.13.2. Finally, $f'$ is affine, hence $R^1f'_*\mathcal{I}' = 0$ by Lemma 67.8.2. As $W$ is affine we have $H^1(W, \mathcal{I}') = 0$ hence the Leray spectral sequence (in the form Cohomology on Sites, Lemma 21.14.6) implies that $H^1(Z, f'_*\mathcal{I}') = 0$. Since $i : Z \to X$ is affine we conclude that $R^1i_*f'_*\mathcal{I}' = 0$ hence $H^1(X, i_*f'_*\mathcal{I}') = 0$ by Leray again and we win. $\square$

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