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.

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