The Stacks Project


Tag 056J

Chapter 28: Morphisms of Schemes > Section 28.5: Supports of modules

Lemma 28.5.3. Let $\mathcal{F}$ be a finite type quasi-coherent module on a scheme $X$. Then

  1. The support of $\mathcal{F}$ is closed.
  2. For $x \in X$ we have $$ x \in \text{Supp}(\mathcal{F}) \Leftrightarrow \mathcal{F}_x \not = 0 \Leftrightarrow \mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \kappa(x) \not = 0. $$
  3. For any morphism of schemes $f : Y \to X$ the pullback $f^*\mathcal{F}$ is of finite type as well and we have $\text{Supp}(f^*\mathcal{F}) = f^{-1}(\text{Supp}(\mathcal{F}))$.

Proof. Part (1) is a reformulation of Modules, Lemma 17.9.6. You can also combine Lemma 28.5.1, Properties, Lemma 27.16.1, and Algebra, Lemma 10.39.5 to see this. The first equivalence in (2) is the definition of support, and the second equivalence follows from Nakayama's lemma, see Algebra, Lemma 10.19.1. Let $f : Y \to X$ be a morphism of schemes. Note that $f^*\mathcal{F}$ is of finite type by Modules, Lemma 17.9.2. For the final assertion, let $y \in Y$ with image $x \in X$. Recall that $$ (f^*\mathcal{F})_y = \mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \mathcal{O}_{Y, y}, $$ see Sheaves, Lemma 6.26.4. Hence $(f^*\mathcal{F})_y \otimes \kappa(y)$ is nonzero if and only if $\mathcal{F}_x \otimes \kappa(x)$ is nonzero. By (2) this implies $x \in \text{Supp}(\mathcal{F})$ if and only if $y \in \text{Supp}(f^*\mathcal{F})$, which is the content of assertion (3). $\square$

    The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 645–663 (see updates for more information).

    \begin{lemma}
    \label{lemma-support-finite-type}
    Let $\mathcal{F}$ be a finite type quasi-coherent module
    on a scheme $X$. Then
    \begin{enumerate}
    \item The support of $\mathcal{F}$ is closed.
    \item For $x \in X$ we have
    $$
    x \in \text{Supp}(\mathcal{F})
    \Leftrightarrow
    \mathcal{F}_x \not = 0
    \Leftrightarrow
    \mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \kappa(x) \not = 0.
    $$
    \item For any morphism of schemes $f : Y \to X$ the pullback
    $f^*\mathcal{F}$ is of finite type as well and we have
    $\text{Supp}(f^*\mathcal{F}) = f^{-1}(\text{Supp}(\mathcal{F}))$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Part (1) is a reformulation of
    Modules, Lemma \ref{modules-lemma-support-finite-type-closed}.
    You can also combine
    Lemma \ref{lemma-support-affine-open},
    Properties, Lemma \ref{properties-lemma-finite-type-module},
    and
    Algebra, Lemma \ref{algebra-lemma-support-closed}
    to see this. The first equivalence in (2) is the definition
    of support, and the second equivalence follows from
    Nakayama's lemma, see
    Algebra, Lemma \ref{algebra-lemma-NAK}.
    Let $f : Y \to X$ be a morphism of schemes. Note that
    $f^*\mathcal{F}$ is of finite type by
    Modules, Lemma \ref{modules-lemma-pullback-finite-type}.
    For the final assertion, let $y \in Y$ with image $x \in X$.
    Recall that
    $$
    (f^*\mathcal{F})_y =
    \mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \mathcal{O}_{Y, y},
    $$
    see
    Sheaves, Lemma \ref{sheaves-lemma-stalk-pullback-modules}.
    Hence $(f^*\mathcal{F})_y \otimes \kappa(y)$ is nonzero
    if and only if $\mathcal{F}_x \otimes \kappa(x)$ is nonzero.
    By (2) this implies $x \in \text{Supp}(\mathcal{F})$ if and only
    if $y \in \text{Supp}(f^*\mathcal{F})$, which is the content of
    assertion (3).
    \end{proof}

    Comments (0)

    There are no comments yet for this tag.

    Add a comment on tag 056J

    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?