The Stacks Project


Tag 02JY

Chapter 28: Morphisms of Schemes > Section 28.24: Flat morphisms

Lemma 28.24.11. Let $f : X \to Y$ be a quasi-compact, surjective, flat morphism. A subset $T \subset Y$ is open (resp. closed) if and only $f^{-1}(T)$ is open (resp. closed). In other words, $f$ is a submersive morphism.

Proof. The question is local on $Y$, hence we may assume that $Y$ is affine. In this case $X$ is quasi-compact as $f$ is quasi-compact. Write $X = X_1 \cup \ldots \cup X_n$ as a finite union of affine opens. Then $f' : X' = X_1 \amalg \ldots \amalg X_n \to Y$ is a surjective flat morphism of affine schemes. Note that for $T \subset Y$ we have $(f')^{-1}(T) = f^{-1}(T) \cap X_1 \amalg \ldots \amalg f^{-1}(T) \cap X_n$. Hence, $f^{-1}(T)$ is open if and only if $(f')^{-1}(T)$ is open. Thus we may assume both $X$ and $Y$ are affine.

Let $f : \mathop{\rm Spec}(B) \to \mathop{\rm Spec}(A)$ be a surjective morphism of affine schemes corresponding to a flat ring map $A \to B$. Suppose that $f^{-1}(T)$ is closed, say $f^{-1}(T) = V(I)$ for $I \subset A$ an ideal. Then $T = f(f^{-1}(T)) = f(V(I))$ is the image of $\mathop{\rm Spec}(A/I) \to \mathop{\rm Spec}(B)$ (here we use that $f$ is surjective). On the other hand, generalizations lift along $f$ (Lemma 28.24.8). Hence by Topology, Lemma 5.19.5 we see that $Y \setminus T = f(X \setminus f^{-1}(T))$ is stable under generalization. Hence $T$ is stable under specialization (Topology, Lemma 5.19.2). Thus $T$ is closed by Algebra, Lemma 10.40.5. $\square$

    The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 4372–4378 (see updates for more information).

    \begin{lemma}
    \label{lemma-fpqc-quotient-topology}
    Let $f : X \to Y$ be a quasi-compact, surjective, flat morphism.
    A subset $T \subset Y$ is open (resp.\ closed) if and only
    $f^{-1}(T)$ is open (resp.\ closed). In other words, $f$ is
    a submersive morphism.
    \end{lemma}
    
    \begin{proof}
    The question is local on $Y$, hence we may assume that $Y$ is affine.
    In this case $X$ is quasi-compact as $f$ is quasi-compact.
    Write $X = X_1 \cup \ldots \cup X_n$ as a finite union of affine opens.
    Then $f' : X' = X_1 \amalg \ldots \amalg X_n \to Y$ is a surjective
    flat morphism of affine schemes. Note that for $T \subset Y$ we have
    $(f')^{-1}(T) = f^{-1}(T) \cap X_1 \amalg \ldots \amalg f^{-1}(T) \cap X_n$.
    Hence, $f^{-1}(T)$ is open if and only if $(f')^{-1}(T)$ is open.
    Thus we may assume both $X$ and $Y$ are affine.
    
    \medskip\noindent
    Let $f : \Spec(B) \to \Spec(A)$ be a surjective
    morphism of affine schemes corresponding to a flat ring map $A \to B$.
    Suppose that $f^{-1}(T)$ is closed, say $f^{-1}(T) = V(I)$ for $I \subset A$
    an ideal. Then $T = f(f^{-1}(T)) = f(V(I))$ is the image of
    $\Spec(A/I) \to \Spec(B)$ (here we use that $f$
    is surjective). On the other hand, generalizations lift along $f$
    (Lemma \ref{lemma-generalizations-lift-flat}).
    Hence by Topology, Lemma \ref{topology-lemma-lift-specializations-images}
    we see that $Y \setminus T = f(X \setminus f^{-1}(T))$ is stable under
    generalization. Hence $T$ is stable under specialization
    (Topology, Lemma \ref{topology-lemma-open-closed-specialization}).
    Thus $T$ is closed by
    Algebra, Lemma \ref{algebra-lemma-image-stable-specialization-closed}.
    \end{proof}

    Comments (0)

    There are no comments yet for this tag.

    There is also 1 comment on Section 28.24: Morphisms of Schemes.

    Add a comment on tag 02JY

    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?