The Stacks project

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. $\square$


Comments (1)

Comment #7877 by qyk on

In the proof (2nd paragraph), there is an "abd", which I think should be an "and".


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0D2M. Beware of the difference between the letter 'O' and the digit '0'.