Lemma 27.13.9. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible sheaf and let $s_0, \ldots , s_ n$ be global sections of $\mathcal{L}$ which generate it. Let $\mathcal{F}$ be the kernel of the induced map $\mathcal{O}_ X^{\oplus n + 1} \to \mathcal{L}$. Then $\mathcal{F} \otimes \mathcal{L}$ is globally generated.
Proof. In fact the result is true if $X$ is any locally ringed space. The sheaf $\mathcal{F}$ is a finite locally free $\mathcal{O}_ X$-module of rank $n$. The elements
with $s_ j$ in the $i$th spot and $-s_ i$ in the $j$th spot map to zero in $\mathcal{L}^{\otimes 2}$. Hence $s_{ij} \in \Gamma (X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L})$. A local computation shows that these sections generate $\mathcal{F} \otimes \mathcal{L}$.
Alternative proof. Consider the morphism $\varphi : X \to \mathbf{P}^ n_\mathbf {Z}$ associated to the pair $(\mathcal{L}, (s_0, \ldots , s_ n))$. Since the pullback of $\mathcal{O}(1)$ is $\mathcal{L}$ and since the pullback of $T_ i$ is $s_ i$, it suffices to prove the lemma in the case of $\mathbf{P}^ n_\mathbf {Z}$. In this case the sheaf $\mathcal{F}$ corresponds to the graded $S = \mathbf{Z}[T_0, \ldots , T_ n]$ module $M$ which fits into the short exact sequence
where the second map is given by $T_0, \ldots , T_ n$. In this case the statement above translates into the statement that the elements
generate the graded module $M(1)$ over $S$. We omit the details. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)