8

Theorem 7.1 in chapter 2 of Hartshorne's text says that invertible sheaves on a scheme $X$ together with its given global generating sections correspond to morphisms from $X$ to $\mathbb{P}_A^n$ (here $A$ is some fixed ring). I wonder if given any invertible sheaf $\mathcal{L}$ on a scheme $X$, there always exists a finite set of global sections in $\mathcal{L}(X)$ that generate $\mathcal{L}$?

mr.bigproblem
  • 860
  • 4
    Check the definition of very ampleness. – Marci Jul 06 '13 at 05:12
  • 5
    But the sheaf may not have any global sections? Take the sheaf associated with the canonical divisor of the projective line for example. There are no global differentials. Yet the sheaf is invertible (corresponds to negation of the divisor) and the inverse has global section. – Jyrki Lahtonen Jul 06 '13 at 05:13
  • 3
    There is no reason to downvote this question: as Jyrki explains, an invertible sheaf may have no non-zero global section but this is far from evident if one has only just learned the definitions. – Georges Elencwajg Jul 06 '13 at 07:36

1 Answers1

11

It is not true for general schemes $X$ that the global sections $\mathcal L(X)$ of an invertible sheaf $\mathcal L$ on $X$ generate the stalks of $\mathcal L$ because it may happen that $\mathcal L$ has no non-zero section at all: this is the case for Serre's twisting sheaves $\mathcal O_{\mathbb P^n}(r)$ on $\mathbb P^n$ as soon as $r$ is negative.
(Jyrki's example in his comment is the case $n=1, r=-2$)

However your question has an affirmative answer if $X$ is affine: Serre (him again!) has proved that in the affine case any coherent sheaf $\mathcal F$ on $X$ has its stalks generated by the global sections $\mathcal F(X)$ of $\mathcal F$. [This result is known as Serre's Theorem A]
Since any affine scheme is quasi-compact, even finitely many global sections suffice to generate all the stalks of $\mathcal F$.

Georges Elencwajg
  • 150,790