Lemma 30.21.1. (For a more general version see More on Morphisms, Lemma 37.43.1.) Let $f : X \to S$ be a morphism of schemes. Assume $S$ is locally Noetherian. The following are equivalent

1. $f$ is finite, and

2. $f$ is proper with finite fibres.

Proof. A finite morphism is proper according to Morphisms, Lemma 29.44.11. A finite morphism is quasi-finite according to Morphisms, Lemma 29.44.10. A quasi-finite morphism has finite fibres, see Morphisms, Lemma 29.20.10. Hence a finite morphism is proper and has finite fibres.

Assume $f$ is proper with finite fibres. 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, Lemma 29.44.7 whereupon it follows from Morphisms, Lemma 29.44.4 that $f$ is finite.

To show that $f$ is affine we may assume that $S$ 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. Hence we may use the criterion of Lemma 30.3.2 to prove that $X$ is affine. To see this let $\mathcal{I} \subset \mathcal{O}_ X$ be a finite type ideal sheaf. In particular $\mathcal{I}$ is a coherent sheaf on $X$. By Lemma 30.20.8 we conclude that $R^1f_*\mathcal{I}_ s = 0$ for all $s \in S$. In other words, $R^1f_*\mathcal{I} = 0$. Hence we see from the Leray Spectral Sequence for $f$ that $H^1(X , \mathcal{I}) = H^1(S, f_*\mathcal{I})$. Since $S$ is affine, and $f_*\mathcal{I}$ is quasi-coherent (Schemes, Lemma 26.24.1) we conclude $H^1(S, f_*\mathcal{I}) = 0$ from Lemma 30.2.2 as desired. Hence $H^1(X, \mathcal{I}) = 0$ as desired. $\square$

Comment #6826 by Fiasco on

sorry, I have a little question about the proof of lemma30.21.1. I think we might need the higher direct image $R^if_*I=0$ for all $i>0$ (not only $i=1$) when we use leray spectral sequence to get $H^1(X,I)=H^1(S,f _﹡I)$, is it right?

Comment #6968 by on

Well, you can use that, but you don't have to. Because a low degree exact sequence associated to the Leray spectral sequence for $f$ and $\mathcal{I}$ reads $0 \to H^1(S, f_*\mathcal{I}) \to H^1(X, \mathcal{I}) \to H^0(S, R^1f_*\mathcal{I})$. OK?

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