Lemma 33.48.2 (Enriques-Severi-Zariski). 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. Assume that for $x \in X$ closed we have $\text{depth}(\mathcal{F}_ x) \geq 2$. Then $H^1(X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes m}) = 0$ for $m \ll 0$.

**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. $\square$

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