A (local) trivialisation of a linebundle is the same as a (local) nonvanishing section.

Lemma 17.24.10. Let $X$ be a ringed space. Assume that each stalk $\mathcal{O}_{X, x}$ is a local ring with maximal ideal $\mathfrak m_ x$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. For any section $s \in \Gamma (X, \mathcal{L})$ the set

$X_ s = \{ x \in X \mid \text{image }s \not\in \mathfrak m_ x\mathcal{L}_ x\}$

is open in $X$. The map $s : \mathcal{O}_{X_ s} \to \mathcal{L}|_{X_ s}$ is an isomorphism, and there exists a section $s'$ of $\mathcal{L}^{\otimes -1}$ over $X_ s$ such that $s' (s|_{X_ s}) = 1$.

Proof. Suppose $x \in X_ s$. We have an isomorphism

$\mathcal{L}_ x \otimes _{\mathcal{O}_{X, x}} (\mathcal{L}^{\otimes -1})_ x \longrightarrow \mathcal{O}_{X, x}$

by Lemma 17.24.5. Both $\mathcal{L}_ x$ and $(\mathcal{L}^{\otimes -1})_ x$ are free $\mathcal{O}_{X, x}$-modules of rank $1$. We conclude from Algebra, Nakayama's Lemma 10.20.1 that $s_ x$ is a basis for $\mathcal{L}_ x$. Hence there exists a basis element $t_ x \in (\mathcal{L}^{\otimes -1})_ x$ such that $s_ x \otimes t_ x$ maps to $1$. Choose an open neighbourhood $U$ of $x$ such that $t_ x$ comes from a section $t$ of $\mathcal{L}^{\otimes -1}$ over $U$ and such that $s \otimes t$ maps to $1 \in \mathcal{O}_ X(U)$. Clearly, for every $x' \in U$ we see that $s$ generates the module $\mathcal{L}_{x'}$. Hence $U \subset X_ s$. This proves that $X_ s$ is open. Moreover, the section $t$ constructed over $U$ above is unique, and hence these glue to give the section $s'$ of the lemma. $\square$

Comment #2443 by Junho Won on

(Typo) I think there should be no $_x$ in $(\mathcal{L}^{\otimes-1})_x$ in the fourth sentence of the proof.

Comment #2594 by Rogier Brussee on

Suggested slogan: A (local) trivialisation of a linebundle is the same as a (local) non vanishing section.

Comment #4898 by Florian Stümpfl on

There is a typo in the last sentence: "... To give the section...".

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