The Stacks project

30.21 Applications of the theorem on formal functions

We will add more here as needed. For the moment we need the following characterization of finite morphisms in the Noetherian case.

Lemma 30.21.1. (For a more general version see More on Morphisms, Lemma 37.40.1.) 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. $\square$

As a consequence we have the following useful result.

slogan

Lemma 30.21.2. (For a more general version see More on Morphisms, Lemma 37.40.2.) Let $f : X \to S$ be a morphism of schemes. Let $s \in S$. Assume

  1. $S$ is locally Noetherian,

  2. $f$ is proper, and

  3. $f^{-1}(\{ s\} )$ is a finite set.

Then there exists an open neighbourhood $V \subset S$ of $s$ such that $f|_{f^{-1}(V)} : f^{-1}(V) \to V$ is finite.

Proof. The morphism $f$ is quasi-finite at all the points of $f^{-1}(\{ s\} )$ by Morphisms, Lemma 29.20.7. By Morphisms, Lemma 29.55.2 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. $\square$

Lemma 30.21.3. Let $f : X \to Y$ be a proper morphism of schemes with $Y$ Noetherian. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Let $y \in Y$ be a point such that $\mathcal{L}_ y$ is ample on $X_ y$. Then there exists a $d_0$ such that for all $d \geq d_0$ we have

\[ R^ pf_*(\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d})_ y = 0 \text{ for }p > 0 \]

and the map

\[ f_*(\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d})_ y \longrightarrow H^0(X_ y, \mathcal{F}_ y \otimes _{\mathcal{O}_{X_ y}} \mathcal{L}_ y^{\otimes d}) \]

is surjective.

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 $\mathcal{B}$ abd 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

\[ H^ p(X_ y, \mathfrak m^ n \mathcal{F}/ \mathfrak m^{n + 1}\mathcal{F} \otimes _{\mathcal{O}_{X_ y}} \mathcal{L}_ y^{\otimes d}) = 0 \]

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

\[ 0 \to \mathfrak m^ n\mathcal{F}/\mathfrak m^{n + 1} \mathcal{F} \to \mathcal{F}/\mathfrak m^{n + 1} \mathcal{F} \to \mathcal{F}/\mathfrak m^ n \mathcal{F} \to 0 \]

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

\[ 0 \to \mathfrak m^ n\mathcal{F}/\mathfrak m^{n + 1} \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d} \to \mathcal{F}/\mathfrak m^{n + 1} \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d} \to \mathcal{F}/\mathfrak m^ n \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes d} \to 0 \]

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. $\square$

Lemma 30.21.4. (For a more general version see More on Morphisms, Lemma 37.46.3.) Let $f : X \to Y$ be a proper morphism of schemes with $Y$ Noetherian. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $y \in Y$ be a point such that $\mathcal{L}_ y$ is ample on $X_ y$. Then there is an open neighbourhood $V \subset Y$ of $y$ such that $\mathcal{L}|_{f^{-1}(V)}$ is ample on $f^{-1}(V)/V$.

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

\[ \varphi _ y = \varphi _{\mathcal{L}_ y^{\otimes d}, (s_{y, 0}, \ldots , s_{y, r})} : X_ y \to \mathbf{P}^ r_{\kappa (y)}. \]

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

\[ \varphi = \varphi _{\mathcal{L}^{\otimes d}, (s_0, \ldots , s_ r)} : X \longrightarrow \mathbf{P}^ r_ Y \]

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. $\square$


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