The Stacks project

Proof. The implication (1) $\Rightarrow $ (2) follows from Cohomology of Schemes, Lemma 30.16.1 because assumption (1) implies that $X$ is a scheme. The implication (2) $\Rightarrow $ (1) is Cohomology of Spaces, Lemma 69.16.9. $\square$

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 Cohomology of Spaces, Lemma 69.11.2. 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 Cohomology of Spaces, Section 69.22 with $I$ replaced by $\mathfrak m$. Since $X_ y$ is the closed subspace 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 Cohomology of Spaces, Lemma 69.12.8. 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 Cohomology of Schemes, Lemma 30.19.3 part (2) there exists a $d_0$ such that

for all $p > 0$, $d \geq d_0$, and $n \geq 0$. By Cohomology of Spaces, Lemma 69.8.3 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 (Cohomology of Spaces, Theorem 69.22.5) 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 Cohomology of Spaces, Lemma 69.20.3 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 Cohomology of Spaces, Lemma 69.22.4 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 71.15.2 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 subspace 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 $\varphi _ y$ (strictly speaking we need to write out a proof that the construction of morphisms into projective space given in Constructions, Section 27.13 also works to describe morphisms of algebraic spaces into projective space; we omit the details).

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 Cohomology of Spaces, Lemma 69.23.2 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. In particular $X$ is a scheme. 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. $\square$