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
s_{ij} = (0, \ldots , 0, s_ j, 0, \ldots , 0, -s_ i, 0, \ldots , 0) \in \Gamma (X, \mathcal{L}^{\oplus n + 1})
with s_ j in the ith spot and -s_ i in the jth 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
0 \to M \to S^{\oplus n + 1} \to S(1) \to 0
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
T_{ij} = (0, \ldots , 0, T_ j, 0, \ldots , 0, -T_ i, 0, \ldots , 0) \in M(1)_0
generate the graded module M(1) over S. We omit the details. \square
Comments (0)