The Stacks project

Lemma 33.35.11. Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$. If $\mathcal{F}$ is $m$-regular, then the multiplication map

\[ H^0(\mathbf{P}^ n_ k, \mathcal{F}(m)) \otimes _ k H^0(\mathbf{P}^ n_ k, \mathcal{O}(1)) \longrightarrow H^0(\mathbf{P}^ n_ k, \mathcal{F}(m + 1)) \]

is surjective.

Proof. Let $k'/k$ be an extension of fields. Let $\mathcal{F}'$ be as in Lemma 33.35.8. By Cohomology of Schemes, Lemma 30.5.2 the base change of the linear map of the lemma to $k'$ is the same linear map for the sheaf $\mathcal{F}'$. Since $k \to k'$ is faithfully flat it suffices to prove the lemma over $k'$, i.e., we may assume $k$ is infinite.

Assume $k$ is infinite. We prove the lemma by induction on $n$. The case $n = 0$ is trivial as $\mathcal{O}(1) \cong \mathcal{O}$ is generated by $T_0$. For $n > 0$ apply Lemma 33.35.3 and tensor the sequence by $\mathcal{O}(m + 1)$ to get

\[ 0 \to \mathcal{F}(m) \xrightarrow {s} \mathcal{F}(m + 1) \to i_*\mathcal{G}(m + 1) \to 0 \]

see Remark 33.35.5. Let $t \in H^0(\mathbf{P}^ n_ k, \mathcal{F}(m + 1))$. By induction the image $\overline{t} \in H^0(H, \mathcal{G}(m + 1))$ is the image of $\sum g_ i \otimes \overline{s}_ i$ with $\overline{s}_ i \in \Gamma (H, \mathcal{O}(1))$ and $g_ i \in H^0(H, \mathcal{G}(m))$. Since $\mathcal{F}$ is $m$-regular we have $H^1(\mathbf{P}^ n_ k, \mathcal{F}(m - 1)) = 0$, hence long exact cohomology sequence associated to the short exact sequence

\[ 0 \to \mathcal{F}(m - 1) \xrightarrow {s} \mathcal{F}(m) \to i_*\mathcal{G}(m) \to 0 \]

shows we can lift $g_ i$ to $f_ i \in H^0(\mathbf{P}^ n_ k, \mathcal{F}(m))$. We can also lift $\overline{s}_ i$ to $s_ i \in H^0(\mathbf{P}^ n_ k, \mathcal{O}(1))$ (see proof of Lemma 33.35.2 for example). After subtracting the image of $\sum f_ i \otimes s_ i$ from $t$ we see that we may assume $\overline{t} = 0$. But this exactly means that $t$ is the image of $f \otimes s$ for some $f \in H^0(\mathbf{P}^ n_ k, \mathcal{F}(m))$ as desired. $\square$

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