**Proof.**
A finite morphism is proper according to Morphisms of Spaces, Lemma 67.45.9. A finite morphism is quasi-finite according to Morphisms of Spaces, Lemma 67.45.8. A quasi-finite morphism has discrete fibres $X_ k$, see Morphisms of Spaces, Lemma 67.27.5. Hence a finite morphism is proper and has discrete fibres $X_ k$.

Assume $f$ is proper with discrete fibres $X_ k$. We want to show $f$ is finite. In fact it suffices to prove $f$ is affine. Namely, if $f$ is affine, then it follows that $f$ is integral by Morphisms of Spaces, Lemma 67.45.7 whereupon it follows from Morphisms of Spaces, Lemma 67.45.6 that $f$ is finite.

To show that $f$ is affine we may assume that $Y$ is affine, and our goal is to show that $X$ is affine too. Since $f$ is proper we see that $X$ is separated and quasi-compact. We will show that 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 69.15.1 and 69.5.1. Then it follows that $X$ is affine from Proposition 69.16.7. By Lemma 69.22.8 we conclude that the stalks of $R^1f_*\mathcal{F}$ are zero for all geometric points of $Y$. In other words, $R^1f_*\mathcal{F} = 0$. Hence we see from the Leray Spectral Sequence for $f$ that $H^1(X , \mathcal{F}) = H^1(Y, f_*\mathcal{F})$. Since $Y$ is affine, and $f_*\mathcal{F}$ is quasi-coherent (Morphisms of Spaces, Lemma 67.11.2) we conclude $H^1(Y, f_*\mathcal{F}) = 0$ from Cohomology of Schemes, Lemma 30.2.2. Hence $H^1(X, \mathcal{F}) = 0$ as desired.
$\square$

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