The Stacks Project


Tag 01K2

25.19. Quasi-compact morphisms

A scheme is quasi-compact if its underlying topological space is quasi-compact. There is a relative notion which is defined as follows.

Definition 25.19.1. A morphism of schemes is called quasi-compact if the underlying map of topological spaces is quasi-compact, see Topology, Definition 5.12.1.

Lemma 25.19.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

  1. $f : X \to S$ is quasi-compact,
  2. the inverse image of every affine open is quasi-compact, and
  3. there exists some affine open covering $S = \bigcup_{i \in I} U_i$ such that $f^{-1}(U_i)$ is quasi-compact for all $i$.

Proof. Suppose we are given a covering $S = \bigcup_{i \in I} U_i$ as in (3). First, let $U \subset S$ be any affine open. For any $u \in U$ we can find an index $i(u) \in I$ such that $u \in U_{i(u)}$. As standard opens form a basis for the topology on $U_{i(u)}$ we can find $W_u \subset U \cap U_{i(u)}$ which is standard open in $U_{i(u)}$. By compactness we can find finitely many points $u_1, \ldots, u_n \in U$ such that $U = \bigcup_{j = 1}^n W_{u_j}$. For each $j$ write $f^{-1}U_{i(u_j)} = \bigcup_{k \in K_j} V_{jk}$ as a finite union of affine opens. Since $W_{u_j} \subset U_{i(u_j)}$ is a standard open we see that $f^{-1}(W_{u_j}) \cap V_{jk}$ is a standard open of $V_{jk}$, see Algebra, Lemma 10.16.4. Hence $f^{-1}(W_{u_j}) \cap V_{jk}$ is affine, and so $f^{-1}(W_{u_j})$ is a finite union of affines. This proves that the inverse image of any affine open is a finite union of affine opens.

Next, assume that the inverse image of every affine open is a finite union of affine opens. Let $K \subset S$ be any quasi-compact open. Since $S$ has a basis of the topology consisting of affine opens we see that $K$ is a finite union of affine opens. Hence the inverse image of $K$ is a finite union of affine opens. Hence $f$ is quasi-compact.

Finally, assume that $f$ is quasi-compact. In this case the argument of the previous paragraph shows that the inverse image of any affine is a finite union of affine opens. $\square$

Lemma 25.19.3. Being quasi-compact is a property of morphisms of schemes over a base which is preserved under arbitrary base change.

Proof. Omitted. $\square$

Lemma 25.19.4. The composition of quasi-compact morphisms is quasi-compact.

Proof. This follows from the definitions and Topology, Lemma 5.12.2. $\square$

Lemma 25.19.5. A closed immersion is quasi-compact.

Proof. Follows from the definitions and Topology, Lemma 5.12.3. $\square$

Example 25.19.6. An open immersion is in general not quasi-compact. The standard example of this is the open subspace $U \subset X$, where $X = \mathop{\rm Spec}(k[x_1, x_2, x_3, \ldots])$, where $U$ is $X \setminus \{0\}$, and where $0$ is the point of $X$ corresponding to the maximal ideal $(x_1, x_2, x_3, \ldots)$.

Lemma 25.19.7. Let $f : X \to S$ be a quasi-compact morphism of schemes. The following are equivalent

  1. $f(X) \subset S$ is closed, and
  2. $f(X) \subset S$ is stable under specialization.

Proof. We have (1) $\Rightarrow$ (2) by Topology, Lemma 5.19.2. Assume (2). Let $U \subset S$ be an affine open. It suffices to prove that $f(X) \cap U$ is closed. Since $U \cap f(X)$ is stable under specializations, we have reduced to the case where $S$ is affine. Because $f$ is quasi-compact we deduce that $X = f^{-1}(S)$ is quasi-compact as $S$ is affine. Thus we may write $X = \bigcup_{i = 1}^n U_i$ with $U_i \subset X$ open affine. Say $S = \mathop{\rm Spec}(R)$ and $U_i = \mathop{\rm Spec}(A_i)$ for some $R$-algebra $A_i$. Then $f(X) = \mathop{\rm Im}(\mathop{\rm Spec}(A_1 \times \ldots \times A_n) \to \mathop{\rm Spec}(R))$. Thus the lemma follows from Algebra, Lemma 10.40.5. $\square$

Lemma 25.19.8. Let $f : X \to S$ be a quasi-compact morphism of schemes. Then $f$ is closed if and only if specializations lift along $f$, see Topology, Definition 5.19.3.

Proof. According to Topology, Lemma 5.19.6 if $f$ is closed then specializations lift along $f$. Conversely, suppose that specializations lift along $f$. Let $Z \subset X$ be a closed subset. We may think of $Z$ as a scheme with the reduced induced scheme structure, see Definition 25.12.5. Since $Z \subset X$ is closed the restriction of $f$ to $Z$ is still quasi-compact. Moreover specializations lift along $Z \to S$ as well, see Topology, Lemma 5.19.4. Hence it suffices to prove $f(X)$ is closed if specializations lift along $f$. In particular $f(X)$ is stable under specializations, see Topology, Lemma 5.19.5. Thus $f(X)$ is closed by Lemma 25.19.7. $\square$

    The code snippet corresponding to this tag is a part of the file schemes.tex and is located in lines 3389–3547 (see updates for more information).

    \section{Quasi-compact morphisms}
    \label{section-quasi-compact}
    
    \noindent
    A scheme is {\it quasi-compact} if its underlying topological space is
    quasi-compact. There is a relative notion which is defined as follows.
    
    \begin{definition}
    \label{definition-quasi-compact}
    A morphism of schemes is called {\it quasi-compact}
    if the underlying map of topological spaces is
    quasi-compact, see
    Topology, Definition \ref{topology-definition-quasi-compact}.
    \end{definition}
    
    \begin{lemma}
    \label{lemma-quasi-compact-affine}
    Let $f : X \to S$ be a morphism of schemes.
    The following are equivalent
    \begin{enumerate}
    \item $f : X \to S$ is quasi-compact,
    \item the inverse image of every affine open is quasi-compact, and
    \item there exists some affine open covering $S = \bigcup_{i \in I} U_i$
    such that $f^{-1}(U_i)$ is quasi-compact for all $i$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Suppose we are given a covering $S = \bigcup_{i \in I} U_i$ as in (3).
    First, let $U \subset S$ be any affine open. For any $u \in U$
    we can find an index $i(u) \in I$ such that $u \in U_{i(u)}$.
    As standard opens form a basis for the topology on $U_{i(u)}$ we can find
    $W_u \subset U \cap U_{i(u)}$ which is standard open in $U_{i(u)}$.
    By compactness we can find finitely many points $u_1, \ldots, u_n \in U$
    such that $U = \bigcup_{j = 1}^n W_{u_j}$. For each $j$ write
    $f^{-1}U_{i(u_j)} = \bigcup_{k \in K_j} V_{jk}$ as a finite
    union of affine opens. Since $W_{u_j} \subset U_{i(u_j)}$ is a standard
    open we see that $f^{-1}(W_{u_j}) \cap V_{jk}$ is a standard
    open of $V_{jk}$, see Algebra, Lemma \ref{algebra-lemma-spec-functorial}.
    Hence $f^{-1}(W_{u_j}) \cap V_{jk}$ is affine, and so
    $f^{-1}(W_{u_j})$ is a finite union of affines. This proves that the
    inverse image of any affine open is a finite union of affine opens.
    
    \medskip\noindent
    Next, assume that the inverse image of every affine open is a finite
    union of affine opens.
    Let $K \subset S$ be any quasi-compact open. Since $S$ has a basis
    of the topology consisting of affine opens we see that $K$ is a finite
    union of affine opens. Hence the inverse image of $K$ is a finite
    union of affine opens. Hence $f$ is quasi-compact.
    
    \medskip\noindent
    Finally, assume that $f$ is quasi-compact. In this case the argument
    of the previous paragraph shows that the inverse image of any affine
    is a finite union of affine opens.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-quasi-compact-preserved-base-change}
    Being quasi-compact is a property of morphisms of schemes
    over a base which is preserved under arbitrary base change.
    \end{lemma}
    
    \begin{proof}
    Omitted.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-composition-quasi-compact}
    The composition of quasi-compact morphisms is quasi-compact.
    \end{lemma}
    
    \begin{proof}
    This follows from the definitions and
    Topology, Lemma \ref{topology-lemma-composition-quasi-compact}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-closed-immersion-quasi-compact}
    A closed immersion is quasi-compact.
    \end{lemma}
    
    \begin{proof}
    Follows from the definitions and
    Topology, Lemma \ref{topology-lemma-closed-in-quasi-compact}.
    \end{proof}
    
    \begin{example}
    \label{example-open-immersion-not-quasi-compact}
    An open immersion is in general not quasi-compact.
    The standard example of this is the open subspace
    $U \subset X$, where $X = \Spec(k[x_1, x_2, x_3, \ldots])$,
    where $U$ is $X \setminus \{0\}$, and where $0$ is the point
    of $X$ corresponding to the maximal ideal
    $(x_1, x_2, x_3, \ldots)$.
    \end{example}
    
    \begin{lemma}
    \label{lemma-image-quasi-compact-closed}
    Let $f : X \to S$ be a quasi-compact morphism of schemes.
    The following are equivalent
    \begin{enumerate}
    \item $f(X) \subset S$ is closed, and
    \item $f(X) \subset S$ is stable under specialization.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    We have (1) $\Rightarrow$ (2) by
    Topology, Lemma \ref{topology-lemma-open-closed-specialization}.
    Assume (2). Let $U \subset S$ be an affine open. It suffices to prove
    that $f(X) \cap U$ is closed. Since $U \cap f(X)$ is stable under
    specializations, we have reduced to the case where $S$ is affine.
    Because $f$ is quasi-compact we deduce that $X = f^{-1}(S)$ is
    quasi-compact as $S$ is affine. Thus we may write
    $X = \bigcup_{i = 1}^n U_i$ with $U_i \subset X$ open affine.
    Say $S = \Spec(R)$ and
    $U_i = \Spec(A_i)$ for some $R$-algebra $A_i$.
    Then $f(X) = \Im(\Spec(A_1 \times \ldots \times A_n)
    \to \Spec(R))$. Thus the lemma follows from
    Algebra, Lemma \ref{algebra-lemma-image-stable-specialization-closed}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-quasi-compact-closed}
    Let $f : X \to S$ be a quasi-compact morphism of schemes.
    Then $f$ is closed if and only if specializations lift
    along $f$, see
    Topology, Definition \ref{topology-definition-lift-specializations}.
    \end{lemma}
    
    \begin{proof}
    According to
    Topology, Lemma \ref{topology-lemma-closed-open-map-specialization}
    if $f$ is closed then specializations lift along $f$.
    Conversely, suppose that specializations lift along $f$.
    Let $Z \subset X$ be a closed subset. We may think of $Z$
    as a scheme with the reduced induced scheme structure, see
    Definition \ref{definition-reduced-induced-scheme}.
    Since $Z \subset X$ is closed the restriction
    of $f$ to $Z$ is still quasi-compact. Moreover specializations lift
    along $Z \to S$ as well,
    see Topology, Lemma \ref{topology-lemma-lift-specialization-composition}.
    Hence it suffices to prove $f(X)$ is closed if specializations lift along $f$.
    In particular $f(X)$ is stable under specializations, see
    Topology, Lemma \ref{topology-lemma-lift-specializations-images}.
    Thus $f(X)$ is closed by
    Lemma \ref{lemma-image-quasi-compact-closed}.
    \end{proof}

    Comments (8)

    Comment #278 by Keenan Kidwell on August 9, 2013 a 2:49 am UTC

    In the proof of Lemma 01K4, the assumption that $W_u$ is standard open in $U$ is never used. The only part that matters is that $W_u$ is standard open in $U_{i(u)}$, so the result about the existence of simultaneous standard opens in intersections of affines isn't needed (only that standard opens form a base for the Zariski topology of an affine scheme).

    Comment #279 by Johan (site) on August 21, 2013 a 2:41 pm UTC

    Fixed here. Thanks!

    Comment #1321 by Erik Visse on February 17, 2015 a 2:41 pm UTC

    There is a typo in the first line of the proof of Lemma 25.19.2. X should be S.

    Comment #1344 by Johan (site) on March 12, 2015 a 7:05 pm UTC

    Thanks! This is fixed here.

    Comment #1474 by sdf on May 24, 2015 a 7:53 pm UTC

    The two instances of $X$ in the second paragraph of Lemma 25.19.2. should be S, since $f:X\rightarrow S$.

    Comment #1492 by Johan (site) on June 7, 2015 a 12:46 pm UTC

    Thanks. Fixed here.

    Comment #2462 by Andrea on March 25, 2017 a 4:52 pm UTC

    There is a typo in the sixth line of the proof of lemma 25.19.2: instead of $U_{i(u)}$ there should be $U_{i(u_j)}$, I think.

    Comment #2499 by Johan (site) on April 13, 2017 a 11:26 pm UTC

    Thanks, fixed here.

    Add a comment on tag 01K2

    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?