The Stacks project

Lemma 33.48.1. Let $k$ be a field. Let $X$ be a proper scheme over $k$. Let $\mathcal{L}$ be an ample invertible $\mathcal{O}_ X$-module. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. If $\text{Ass}(\mathcal{F})$ does not contain any closed points, then $\Gamma (X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = 0$ for $n \ll 0$.

Proof. For a coherent $\mathcal{O}_ X$-module $\mathcal{F}$ let $\mathcal{P}(\mathcal{F})$ be the property: there exists an $n_0 \in \mathbf{Z}$ such that for $n \leq n_0$ every section $s$ of $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}$ has support consisting only of closed points. Since $\text{Ass}(\mathcal{F}) = \text{Ass}(\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n})$ we see that it suffices to prove $\mathcal{P}$ holds for all coherent modules on $X$. To do this we will prove that conditions (1), (2), and (3) of Cohomology of Schemes, Lemma 30.12.8 are satisfied.

To see condition (1) suppose that

\[ 0 \to \mathcal{F}_1 \to \mathcal{F} \to \mathcal{F}_2 \to 0 \]

is a short exact sequence of coherent $\mathcal{O}_ X$-modules such that we have $\mathcal{P}$ for $\mathcal{F}_ i$, $i = 1, 2$. Let $n_1, n_2$ be the cutoffs we find. Let $\mathcal{F}'_2 \subset \mathcal{F}_2$ be the maximal coherent submodule whose support is a finite set of closed points. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the annihilator of $\mathcal{F}'_2$. Since $\mathcal{L}$ is ample, we can find an $e > 0$ such that $\mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes e}$ is globally generated. Set $n_0 = \min (n_2, n_1 - e)$. Let $n \leq n_0$ and let $t$ be a global section of $\mathcal{F} \otimes \mathcal{L}^{\otimes n}$. The image of $t$ in $\mathcal{F}_2 \otimes \mathcal{L}^{\otimes n}$ falls into $\mathcal{F}'_2 \otimes \mathcal{L}^{\otimes n}$ because $n \leq n_2$. Hence for any $s \in \Gamma (X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes e})$ the product $t \otimes s$ lies in $\mathcal{F}_1 \otimes \mathcal{L}^{\otimes n + e}$. Thus $t \otimes s$ has support contained in the finite set of closed points in $\text{Ass}(\mathcal{F}_1)$ because $n + e \leq n_1$. Since by our choice of $e$ we may choose $s$ invertible in any point not in the support of $\mathcal{F}'_2$ we conclude that the support of $t$ is contained in the union of the finite set of closed points in $\text{Ass}(\mathcal{F}_1)$ and the finite set of closed points in $\text{Ass}(\mathcal{F}_2)$. This finishes the proof of condition (1).

Condition (2) is immediate.

For condition (3) we choose $\mathcal{G} = \mathcal{O}_ Z$. In this case, if $Z$ is a closed point of $X$, then there is nothing the show. If $\dim (Z) > 0$, then we will show that $\Gamma (Z, \mathcal{L}^{\otimes n}|_ Z) = 0$ for $n < 0$. Namely, let $s$ be a nonzero section of a negative power of $\mathcal{L}|_ Z$. Choose a nonzero section $t$ of a positive power of $\mathcal{L}|_ Z$ (this is possible as $\mathcal{L}$ is ample, see Properties, Proposition 28.26.13). Then $s^{\deg (t)} \otimes t^{\deg (s)}$ is a nonzero global section of $\mathcal{O}_ Z$ (because $Z$ is integral) and hence a unit (Lemma 33.9.3). This implies that $t$ is a trivializing section of a positive power of $\mathcal{L}$. Thus the function $n \mapsto \dim _ k \Gamma (X, \mathcal{L}^{\otimes n})$ is bounded on an infinite set of positive integers which contradicts asymptotic Riemann-Roch (Proposition 33.45.13) since $\dim (Z) > 0$. $\square$


Comments (2)

Comment #5454 by Matt Larson on

There is a missing parenthesis in the second paragraph.


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 0FD7. Beware of the difference between the letter 'O' and the digit '0'.