A (local) trivialisation of a linebundle is the same as a (local) nonvanishing section.
Lemma 17.25.10. 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 17.25.5. 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
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