The Stacks project

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

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 $\dim (Z) > 0$. $\square$

Proof. Choose a closed immersion $i : X \to \mathbf{P}^ n_ k$ such that $i^*\mathcal{O}(1) \cong \mathcal{L}^{\otimes e}$ for some $e > 0$ (see Morphisms, Lemma 29.39.4). Then it suffices to prove the lemma for

on $\mathbf{P}^ n_ k$. Namely, we have

by Cohomology of Schemes, Lemma 30.2.4. Also, if $y \in \mathbf{P}^ n_ k$ is a closed point then $\text{depth}(\mathcal{G}_ y) = \infty $ if $y \not\in i(X)$ and $\text{depth}(\mathcal{G}_ y) = \text{depth}(\mathcal{F}_ x)$ if $y = i(x)$ because in this case $\mathcal{G}_ y \cong \mathcal{F}_ x^{\oplus e}$ as a module over $\mathcal{O}_{\mathbf{P}^ n_ k, x}$ and we can use for example Algebra, Lemma 10.72.11 to get the equality.

Assume $X = \mathbf{P}^ n_ k$ and $\mathcal{L} = \mathcal{O}(1)$ and $k$ is infinite. Choose $s \in H^0(\mathbf{P}^1_ k, \mathcal{O}(1))$ which determines an exact sequence

as in Lemma 33.35.3. Since the map $\mathcal{F}(-1) \to \mathcal{F}$ is affine locally given by multiplying by a nonzerodivisor on $\mathcal{F}$ we see that for $x \in \mathbf{P}^ n_ k$ closed we have $\text{depth}(\mathcal{G}_ x) \geq 1$, see Algebra, Lemma 10.72.7. Hence by Lemma 33.48.1 we have $H^0(\mathcal{G}(m)) = 0$ for $m \ll 0$. Looking at the long exact sequence of cohomology after twisting (see Remark 33.35.5) we find that the sequence of numbers

stabilizes for $m \leq m_0$ for some integer $m_0$. Let $N$ be the common dimension of these spaces for $m \leq m_0$. We have to show $N = 0$.

For $d > 0$ and $m \leq m_0$ consider the bilinear map

By linear algebra, there is a codimension $\leq N^2$ subspace $V_ m \subset H^0(\mathbf{P}^ n_ k, \mathcal{O}(d))$ such that multiplication by $s' \in V_ m$ annihilates $H^1(\mathbf{P}^ n_ k, \mathcal{F}(m - d))$. Observe that for $m' < m \leq m_0$ the diagram

commutes with isomorphisms going vertically. Thus $V_ m = V$ is independent of $m \leq m_0$. For $x \in \text{Ass}(\mathcal{F})$ set $Z = \overline{\{ x\} }$. For $d$ large enough the linear map

has rank $> N^2$ because $\dim (Z) \geq 1$ (for example this follows from asymptotic Riemann-Roch and ampleness $\mathcal{O}(1)$; details omitted). Hence we can find $s' \in V$ such that $s'$ does not vanish in any associated point of $\mathcal{F}$ (use that the set of associated points is finite). Then we obtain

and as before we conclude as before that multiplication by $s'$ on $H^1(\mathbf{P}^ n_ k, \mathcal{F}(m - d))$ is injective for $m \ll 0$. This contradicts the choice of $s'$ unless $N = 0$ as desired.

We still have to treat the case where $k$ is finite. In this case let $K/k$ be any infinite algebraic field extension. Denote $\mathcal{F}_ K$ and $\mathcal{L}_ K$ the pullbacks of $\mathcal{F}$ and $\mathcal{L}$ to $X_ K = \mathop{\mathrm{Spec}}(K) \times _{\mathop{\mathrm{Spec}}(k)} X$. We have

by Cohomology of Schemes, Lemma 30.5.2. On the other hand, a closed point $x_ K$ of $X_ K$ maps to a closed point $x$ of $X$ because $K/k$ is an algebraic extension. The ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{X_ K, x_ K}$ is flat (Lemma 33.5.1). Hence we have

by Algebra, Lemma 10.163.1 (in fact equality holds here but we don't need it). Therefore the result over $k$ follows from the result over the infinite field $K$ and the proof is complete. $\square$

Proof. Since $s$ is a regular section, so is $s^ n \in \Gamma (X, \mathcal{L}^{\otimes n})$ for all $n > 1$. Moreover, the inclusion morphism $Z(s) \to Z(s^ n)$ is a bijection on underlying topological spaces. Hence if $Z(s)$ is disconnected, so is $Z(s^ n)$. Now consider the canonical short exact sequence

Consider the $k$-algebra $R_ n = \Gamma (X, \mathcal{O}_{Z(s^ n)})$. If $Z(s)$ is disconnected, i.e., $Z(s^ n)$ is disconnected, then either $R_ n$ is zero in case $Z(s^ n) = \emptyset $ or $R_ n$ contains a nontrivial idempotent in case $Z(s^ n) = U \amalg V$ with $U, V \subset Z(s^ n)$ open and nonempty (the reader may wish to consult Lemma 33.9.3). Thus the map $\Gamma (X, \mathcal{O}_ X) \to R_ n$ cannot be an isomorphism. It follows that either $H^0(X, \mathcal{L}^{\otimes -n})$ or $H^1(X, \mathcal{L}^{\otimes -n})$ is nonzero for infinitely many positive $n$. This contradicts Lemma 33.48.1 or 33.48.2 and the proof is complete. $\square$