The Stacks project

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. $\square$

Proof. The morphism $f$ is quasi-finite at all the points of $f^{-1}(\{ s\} )$ by Morphisms, Lemma 29.20.7. By Morphisms, Lemma 29.56.2 the set of points at which $f$ is quasi-finite is an open $U \subset X$. Let $Z = X \setminus U$. Then $s \not\in f(Z)$. Since $f$ is proper the set $f(Z) \subset S$ is closed. Choose any open neighbourhood $V \subset S$ of $s$ with $Z \cap V = \emptyset $. Then $f^{-1}(V) \to V$ is locally quasi-finite and proper. Hence it is quasi-finite (Morphisms, Lemma 29.20.9), hence has finite fibres (Morphisms, Lemma 29.20.10), hence is finite by Lemma 30.21.1. $\square$

Proof. Note that $\mathcal{O}_{Y, y}$ is a Noetherian local ring. Consider the canonical morphism $c : \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) \to Y$, see Schemes, Equation (26.13.1.1). This is a flat morphism as it identifies local rings. Denote momentarily $f' : X' \to \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$ the base change of $f$ to this local ring. We see that $c^*R^ pf_*\mathcal{F} = R^ pf'_*\mathcal{F}'$ by Lemma 30.5.2. Moreover, the fibres $X_ y$ and $X'_ y$ are identified. Hence we may assume that $Y = \mathop{\mathrm{Spec}}(A)$ is the spectrum of a Noetherian local ring $(A, \mathfrak m, \kappa )$ and $y \in Y$ corresponds to $\mathfrak m$. In this case $R^ pf_*(\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d})_ y = H^ p(X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d})$ for all $p \geq 0$. Denote $f_ y : X_ y \to \mathop{\mathrm{Spec}}(\kappa )$ the projection.

Let $B = \text{Gr}_\mathfrak m(A) = \bigoplus _{n \geq 0} \mathfrak m^ n/\mathfrak m^{n + 1}$. Consider the sheaf $\mathcal{B} = f_ y^*\widetilde{B}$ of quasi-coherent graded $\mathcal{O}_{X_ y}$-algebras. We will use notation as in Section 30.20 with $I$ replaced by $\mathfrak m$. Since $X_ y$ is the closed subscheme of $X$ cut out by $\mathfrak m\mathcal{O}_ X$ we may think of $\mathfrak m^ n\mathcal{F}/\mathfrak m^{n + 1}\mathcal{F}$ as a coherent $\mathcal{O}_{X_ y}$-module, see Lemma 30.9.8. Then $\bigoplus _{n \geq 0} \mathfrak m^ n\mathcal{F}/\mathfrak m^{n + 1}\mathcal{F}$ is a quasi-coherent graded $\mathcal{B}$-module of finite type because it is generated in degree zero over $\mathcal{B}$ and because the degree zero part is $\mathcal{F}_ y = \mathcal{F}/\mathfrak m \mathcal{F}$ which is a coherent $\mathcal{O}_{X_ y}$-module. Hence by Lemma 30.19.3 part (2) we see that

for all $p > 0$, $d \geq d_0$, and $n \geq 0$. By Lemma 30.2.4 this is the same as the statement that $ H^ p(X, \mathfrak m^ n \mathcal{F}/ \mathfrak m^{n + 1}\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d}) = 0 $ for all $p > 0$, $d \geq d_0$, and $n \geq 0$.

Consider the short exact sequences

of coherent $\mathcal{O}_ X$-modules. Tensoring with $\mathcal{L}^{\otimes d}$ is an exact functor and we obtain short exact sequences

Using the long exact cohomology sequence and the vanishing above we conclude (using induction) that

1. $H^ p(X, \mathcal{F}/\mathfrak m^ n \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d}) = 0$ for all $p > 0$, $d \geq d_0$, and $n \geq 0$, and

2. $H^0(X, \mathcal{F}/\mathfrak m^ n \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d}) \to H^0(X_ y, \mathcal{F}_ y \otimes _{\mathcal{O}_{X_ y}} \mathcal{L}_ y^{\otimes d})$ is surjective for all $d \geq d_0$ and $n \geq 1$.

By the theorem on formal functions (Theorem 30.20.5) we find that the $\mathfrak m$-adic completion of $H^ p(X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d})$ is zero for all $d \geq d_0$ and $p > 0$. Since $H^ p(X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d})$ is a finite $A$-module by Lemma 30.19.2 it follows from Nakayama's lemma (Algebra, Lemma 10.20.1) that $H^ p(X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d})$ is zero for all $d \geq d_0$ and $p > 0$. For $p = 0$ we deduce from Lemma 30.20.4 part (3) that $H^0(X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d}) \to H^0(X_ y, \mathcal{F}_ y \otimes _{\mathcal{O}_{X_ y}} \mathcal{L}_ y^{\otimes d})$ is surjective, which gives the final statement of the lemma. $\square$

Proof. Pick $d_0$ as in Lemma 30.21.3 for $\mathcal{F} = \mathcal{O}_ X$. Pick $d \geq d_0$ so that we can find $r \geq 0$ and sections $s_{y, 0}, \ldots , s_{y, r} \in H^0(X_ y, \mathcal{L}_ y^{\otimes d})$ which define a closed immersion

This is possible by Morphisms, Lemma 29.39.4 but we also use Morphisms, Lemma 29.41.7 to see that $\varphi _ y$ is a closed immersion and Constructions, Section 27.13 for the description of morphisms into projective space in terms of invertible sheaves and sections. By our choice of $d_0$, after replacing $Y$ by an open neighbourhood of $y$, we can choose $s_0, \ldots , s_ r \in H^0(X, \mathcal{L}^{\otimes d})$ mapping to $s_{y, 0}, \ldots , s_{y, r}$. Let $X_{s_ i} \subset X$ be the open subset where $s_ i$ is a generator of $\mathcal{L}^{\otimes d}$. Since the $s_{y, i}$ generate $\mathcal{L}_ y^{\otimes d}$ we see that $X_ y \subset U = \bigcup X_{s_ i}$. Since $X \to Y$ is closed, we see that there is an open neighbourhood $y \in V \subset Y$ such that $f^{-1}(V) \subset U$. After replacing $Y$ by $V$ we may assume that the $s_ i$ generate $\mathcal{L}^{\otimes d}$. Thus we obtain a morphism

with $\mathcal{L}^{\otimes d} \cong \varphi ^*\mathcal{O}_{\mathbf{P}^ r_ Y}(1)$ whose base change to $y$ gives $\varphi _ y$.

We will finish the proof by a sleight of hand; the “correct” proof proceeds by directly showing that $\varphi $ is a closed immersion after base changing to an open neighbourhood of $y$. Namely, by Lemma 30.21.2 we see that $\varphi $ is a finite over an open neighbourhood of the fibre $\mathbf{P}^ r_{\kappa (y)}$ of $\mathbf{P}^ r_ Y \to Y$ above $y$. Using that $\mathbf{P}^ r_ Y \to Y$ is closed, after shrinking $Y$ we may assume that $\varphi $ is finite. Then $\mathcal{L}^{\otimes d} \cong \varphi ^*\mathcal{O}_{\mathbf{P}^ r_ Y}(1)$ is ample by the very general Morphisms, Lemma 29.37.7. $\square$