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Tag 01CY

Chapter 17: Sheaves of Modules > Section 17.22: Invertible modules

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 3847–3859 (see updates for more information).

    \begin{lemma}
    \label{lemma-s-open}
    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}

    Comments (2)

    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?

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