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. $\square$

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