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Tag 01PW

Chapter 27: Properties of Schemes > Section 27.17: Sections over principal opens

Lemma 27.17.2. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible sheaf on $X$. Let $s \in \Gamma(X, \mathcal{L})$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_X$-module.

  1. If $X$ is quasi-compact, then (27.17.1.1) is injective, and
  2. if $X$ is quasi-compact and quasi-separated, then (27.17.1.1) is an isomorphism.

In particular, the canonical map $$ \Gamma_*(X, \mathcal{L})_{(s)} \longrightarrow \Gamma(X_s, \mathcal{O}_X),\quad a/s^n \longmapsto a \otimes s^{-n} $$ is an isomorphism if $X$ is quasi-compact and quasi-separated.

Proof. Assume $X$ is quasi-compact. Choose a finite affine open covering $X = U_1 \cup \ldots \cup U_m$ with $U_j$ affine and $\mathcal{L}|_{U_j} \cong \mathcal{O}_{U_j}$. Via this isomorphism, the image $s|_{U_j}$ corresponds to some $f_j \in \Gamma(U_j, \mathcal{O}_{U_j})$. Then $X_s \cap U_j = D(f_j)$.

Proof of (1). Let $t/s^n$ be an element in the kernel of (27.17.1.1). Then $t|_{X_s} = 0$. Hence $(t|_{U_j})|_{D(f_j)} = 0$. By Lemma 27.17.1 we conclude that $f_j^{e_j} t|_{U_j} = 0$ for some $e_j \geq 0$. Let $e = \max(e_j)$. Then we see that $t \otimes s^e$ restricts to zero on $U_j$ for all $j$, hence is zero. Since $t/s^n$ is equal to $t \otimes s^e/s^{n + e}$ in $\Gamma_*(X, \mathcal{L}, \mathcal{F})_{(s)}$ we conclude that $t/s^n = 0$ as desired.

Proof of (2). Assume $X$ is quasi-compact and quasi-separated. Then $U_j \cap U_{j'}$ is quasi-compact for all pairs $j, j'$, see Schemes, Lemma 25.21.7. By part (1) we know (27.17.1.1) is injective. Let $t' \in \Gamma(X_s, \mathcal{F}|_{X_s})$. For every $j$, there exist an integer $n_j \geq 0$ and $t'_j \in \Gamma(U_j, \mathcal{F}|_{U_j})$ such that $t'|_{D(f_j)}$ corresponds to $t'_j/f_j^{e_j}$ via the isomorphism of Lemma 27.17.1. Set $e = \max(e_j)$ and $$ t_j = t'_j \otimes s|_{U_j}^e \in \Gamma(U_j, (\mathcal{F} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes e})|_{U_j}) $$ Then we see that $t_j|_{U_j \cap U_{j'}}$ and $t_{j'}|_{U_j \cap U_{j'}}$ map to the same section of $\mathcal{F}$ over $U_j \cap U_{j'} \cap X_s$. By quasi-compactness of $U_j \cap U_{j'}$ and part (1) there exists an integer $e' \geq 0$ such that $$ t_j|_{U_j \cap U_{j'}} \otimes s^{e'}|_{U_j \cap U_{j'}} = t_{j'}|_{U_j \cap U_{j'}} \otimes s^{e'}|_{U_j \cap U_{j'}} $$ as sections of $\mathcal{F} \otimes \mathcal{L}^{\otimes e + e'}$ over $U_j \cap U_{j'}$. We may choose the same $e'$ to work for all pairs $j, j'$. Then the sheaf conditions implies there is a section $t \in \Gamma(X, \mathcal{F} \otimes \mathcal{L}^{\otimes e + e'})$ whose restriction to $U_j$ is $t_j \otimes s^{e'}|_{U_j}$. A simple computation shows that $t/s^{e + e'}$ maps to $t'$ as desired. $\square$

    The code snippet corresponding to this tag is a part of the file properties.tex and is located in lines 2003–2022 (see updates for more information).

    \begin{lemma}
    \label{lemma-invert-s-sections}
    Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible sheaf on $X$.
    Let $s \in \Gamma(X, \mathcal{L})$. Let $\mathcal{F}$ be a quasi-coherent
    $\mathcal{O}_X$-module.
    \begin{enumerate}
    \item If $X$ is quasi-compact, then (\ref{equation-module-invert-s})
    is injective, and
    \item if $X$ is quasi-compact and quasi-separated, then
    (\ref{equation-module-invert-s}) is an isomorphism.
    \end{enumerate}
    In particular, the canonical map
    $$
    \Gamma_*(X, \mathcal{L})_{(s)}
    \longrightarrow
    \Gamma(X_s, \mathcal{O}_X),\quad
    a/s^n \longmapsto a \otimes s^{-n}
    $$
    is an isomorphism if $X$ is quasi-compact and quasi-separated.
    \end{lemma}
    
    \begin{proof}
    Assume $X$ is quasi-compact. Choose a finite affine open covering
    $X = U_1 \cup \ldots \cup U_m$ with $U_j$ affine and
    $\mathcal{L}|_{U_j} \cong \mathcal{O}_{U_j}$. Via this isomorphism,
    the image $s|_{U_j}$ corresponds to some
    $f_j \in \Gamma(U_j, \mathcal{O}_{U_j})$. Then
    $X_s \cap U_j = D(f_j)$.
    
    \medskip\noindent
    Proof of (1). Let $t/s^n$ be an element in the kernel of
    (\ref{equation-module-invert-s}). Then $t|_{X_s} = 0$.
    Hence $(t|_{U_j})|_{D(f_j)} = 0$. By
    Lemma \ref{lemma-invert-f-sections} we conclude that
    $f_j^{e_j} t|_{U_j} = 0$ for some
    $e_j \geq 0$. Let $e = \max(e_j)$. Then we see that $t \otimes s^e$
    restricts to zero on $U_j$ for all $j$, hence is zero. Since $t/s^n$
    is equal to $t \otimes s^e/s^{n + e}$ in
    $\Gamma_*(X, \mathcal{L}, \mathcal{F})_{(s)}$ we conclude that $t/s^n = 0$
    as desired.
    
    \medskip\noindent
    Proof of (2). Assume $X$ is quasi-compact and quasi-separated.
    Then $U_j \cap U_{j'}$ is quasi-compact for all pairs $j, j'$, see
    Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated}.
    By part (1) we know (\ref{equation-module-invert-s}) is injective.
    Let $t' \in \Gamma(X_s, \mathcal{F}|_{X_s})$. For every $j$, there exist an
    integer $n_j \geq 0$ and $t'_j \in \Gamma(U_j, \mathcal{F}|_{U_j})$ such that
    $t'|_{D(f_j)}$ corresponds to $t'_j/f_j^{e_j}$
    via the isomorphism of Lemma \ref{lemma-invert-f-sections}.
    Set $e = \max(e_j)$ and
    $$
    t_j = t'_j \otimes s|_{U_j}^e \in
    \Gamma(U_j,
    (\mathcal{F} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes e})|_{U_j})
    $$
    Then we see that $t_j|_{U_j \cap U_{j'}}$ and $t_{j'}|_{U_j \cap U_{j'}}$
    map to the same section of $\mathcal{F}$ over $U_j \cap U_{j'} \cap X_s$.
    By quasi-compactness of $U_j \cap U_{j'}$ and part (1) there exists an
    integer $e' \geq 0$ such that
    $$
    t_j|_{U_j \cap U_{j'}} \otimes s^{e'}|_{U_j \cap U_{j'}} =
    t_{j'}|_{U_j \cap U_{j'}} \otimes s^{e'}|_{U_j \cap U_{j'}}
    $$
    as sections of $\mathcal{F} \otimes \mathcal{L}^{\otimes e + e'}$ over
    $U_j \cap U_{j'}$. We may choose the same $e'$ to work for all pairs
    $j, j'$. Then the sheaf conditions implies there is a section
    $t \in \Gamma(X, \mathcal{F} \otimes \mathcal{L}^{\otimes e + e'})$
    whose restriction to $U_j$ is $t_j \otimes s^{e'}|_{U_j}$.
    A simple computation shows that $t/s^{e + e'}$ maps to $t'$
    as desired.
    \end{proof}

    Comments (2)

    Comment #2595 by Rogier Brussee on June 4, 2017 a 7:38 pm UTC

    Suggested slogan: Multiplication by a section in a line bundle is an isomorphism unless.

    Comment #2623 by Johan (site) on July 7, 2017 a 12:23 pm UTC

    Hmmm... not sure what you wanted to say...

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