# The Stacks Project

## Tag 01CY

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

Lemma 17.22.10. 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 17.22.5. 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 10.19.1 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 te section $s'$ of the lemma. $\square$

The code snippet corresponding to this tag is a part of the file modules.tex and is located in lines 3847–3863 (see updates for more information).

\begin{lemma}
\label{lemma-s-open}
\begin{slogan}
A (local) trivialisation of a linebundle
is the same as a (local) nonvanishing section.
\end{slogan}
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 $s' (s|_{X_s}) = 1$.
\end{lemma}

\begin{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 \ref{lemma-constructions-invertible}.
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 \ref{algebra-lemma-NAK} 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 te section $s'$ of the lemma.
\end{proof}

Comment #2443 by Junho Won on March 3, 2017 a 2:13 am UTC

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

Comment #2486 by Johan (site) on April 13, 2017 a 10:44 pm UTC

Thanks, fixed here. Did I get your name right?

Comment #2594 by Rogier Brussee on June 4, 2017 a 7:27 pm UTC

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

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