The Stacks Project


Tag 01B1

17.8. Modules locally generated by sections

Let $(X, \mathcal{O}_X)$ be a ringed space. In this and the following section we will often restrict sheaves to open subspaces $U \subset X$, see Sheaves, Section 6.31. In particular, we will often denote the open subspace by $(U, \mathcal{O}_U)$ instead of the more correct notation $(U, \mathcal{O}_X|_U)$, see Sheaves, Definition 6.31.2.

Consider the open immersion $j : U = (0 , \infty) \to \mathbf{R} = X$, and the abelian sheaf $j_!\underline{\mathbf{Z}}_U$. By Sheaves, Section 6.31 the stalk of $j_!\underline{\mathbf{Z}}_U$ at $x = 0$ is $0$. In fact the sections of this sheaf over any open interval containing $0$ are $0$. Thus there is no open neighbourhood of the point $0$ over which the sheaf can be generated by sections.

Definition 17.8.1. Let $(X, \mathcal{O}_X)$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_X$-modules. We say that $\mathcal{F}$ is locally generated by sections if for every $x \in X$ there exists an open neighbourhood $U$ such that $\mathcal{F}|_U$ is globally generated as a sheaf of $\mathcal{O}_U$-modules.

In other words there exists a set $I$ and for each $i$ a section $s_i \in \mathcal{F}(U)$ such that the associated map $$ \bigoplus\nolimits_{i \in I} \mathcal{O}_U \longrightarrow \mathcal{F}|_U $$ is surjective.

Lemma 17.8.2. Let $f : (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ be a morphism of ringed spaces. The pullback $f^*\mathcal{G}$ is locally generated by sections if $\mathcal{G}$ is locally generated by sections.

Proof. Given an open subspace $V$ of $Y$ we may consider the commutative diagram of ringed spaces $$ \xymatrix{ (f^{-1}V, \mathcal{O}_{f^{-1}V}) \ar[r]_{j'} \ar[d]_{f'} & (X, \mathcal{O}_X) \ar[d]^f \\ (V, \mathcal{O}_V) \ar[r]^j & (Y, \mathcal{O}_Y) } $$ We know that $f^*\mathcal{G}|_{f^{-1}V} \cong (f')^*(\mathcal{G}|_V)$, see Sheaves, Lemma 6.26.3. Thus we may assume that $\mathcal{G}$ is globally generated.

We have seen that $f^*$ commutes with all colimits, and is right exact, see Lemma 17.3.3. Thus if we have a surjection $$ \bigoplus\nolimits_{i \in I} \mathcal{O}_Y \to \mathcal{G} \to 0 $$ then upon applying $f^*$ we obtain the surjection $$ \bigoplus\nolimits_{i \in I} \mathcal{O}_X \to f^*\mathcal{G} \to 0. $$ This implies the lemma. $\square$

    The code snippet corresponding to this tag is a part of the file modules.tex and is located in lines 710–811 (see updates for more information).

    \section{Modules locally generated by sections}
    \label{section-locally-generated}
    
    \noindent
    Let $(X, \mathcal{O}_X)$ be a ringed space.
    In this and the following section we will often restrict
    sheaves to open subspaces $U \subset X$, see
    Sheaves, Section \ref{sheaves-section-open-immersions}.
    In particular, we will often denote the open subspace
    by $(U, \mathcal{O}_U)$ instead of the more correct
    notation $(U, \mathcal{O}_X|_U)$, see
    Sheaves, Definition \ref{sheaves-definition-restriction}.
    
    \medskip\noindent
    Consider the open immersion
    $j : U = (0 , \infty) \to \mathbf{R} = X$, and the abelian sheaf
    $j_!\underline{\mathbf{Z}}_U$. By Sheaves, Section
    \ref{sheaves-section-open-immersions} the stalk of
    $j_!\underline{\mathbf{Z}}_U$ at $x = 0$ is $0$. In fact the
    sections of this sheaf over any open interval containing $0$
    are $0$. Thus there is no open neighbourhood of the point
    $0$ over which the sheaf can be generated by sections.
    
    \begin{definition}
    \label{definition-locally-generated}
    Let $(X, \mathcal{O}_X)$ be a ringed space.
    Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_X$-modules.
    We say that $\mathcal{F}$ is {\it locally generated by sections}
    if for every $x \in X$ there exists an open
    neighbourhood $U$ such that $\mathcal{F}|_U$
    is globally generated as a sheaf of $\mathcal{O}_U$-modules.
    \end{definition}
    
    \noindent
    In other words there exists a set $I$ and for
    each $i$ a section $s_i \in \mathcal{F}(U)$ such
    that the associated map
    $$
    \bigoplus\nolimits_{i \in I} \mathcal{O}_U
    \longrightarrow
    \mathcal{F}|_U
    $$
    is surjective.
    
    \begin{lemma}
    \label{lemma-pullback-locally-generated}
    Let $f : (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$
    be a morphism of ringed spaces.
    The pullback $f^*\mathcal{G}$ is locally generated by sections
    if $\mathcal{G}$ is locally generated by sections.
    \end{lemma}
    
    \begin{proof}
    Given an open subspace $V$ of $Y$ we may
    consider the commutative diagram of ringed spaces
    $$
    \xymatrix{
    (f^{-1}V, \mathcal{O}_{f^{-1}V}) \ar[r]_{j'} \ar[d]_{f'} &
    (X, \mathcal{O}_X) \ar[d]^f \\
    (V, \mathcal{O}_V) \ar[r]^j &
    (Y, \mathcal{O}_Y)
    }
    $$
    We know that $f^*\mathcal{G}|_{f^{-1}V} \cong (f')^*(\mathcal{G}|_V)$,
    see Sheaves, Lemma \ref{sheaves-lemma-push-pull-composition-modules}.
    Thus we may assume that $\mathcal{G}$ is globally generated.
    
    \medskip\noindent
    We have seen that $f^*$ commutes with all colimits,
    and is right exact, see Lemma \ref{lemma-exactness-pushforward-pullback}.
    Thus if we have a surjection
    $$
    \bigoplus\nolimits_{i \in I}
    \mathcal{O}_Y
    \to
    \mathcal{G}
    \to
    0
    $$
    then upon applying $f^*$ we obtain the surjection
    $$
    \bigoplus\nolimits_{i \in I}
    \mathcal{O}_X
    \to
    f^*\mathcal{G}
    \to
    0.
    $$
    This implies the lemma.
    \end{proof}

    Comments (2)

    Comment #432 by Herman Rohrbach on January 25, 2014 a 3:13 pm UTC

    There is a small error (typo?) in the first line of the proof of lemma 17.8.2; it says "...an open subspace $V$ of $X$...", which should be "...an open subspace $V$ of $Y$..."

    Comment #433 by Johan (site) on January 25, 2014 a 11:43 pm UTC

    Fixed here. Thanks!

    Add a comment on tag 01B1

    Your email address will not be published. Required fields are marked.

    In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?