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Tag 02NH

Chapter 28: Morphisms of Schemes > Section 28.19: Quasi-finite morphisms

Lemma 28.19.10. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

  1. $f$ is quasi-finite, and
  2. $f$ is locally of finite type, quasi-compact, and has finite fibres.

Proof. Assume $f$ is quasi-finite. In particular $f$ is locally of finite type and quasi-compact (since it is of finite type). Let $s \in S$. Since every $x \in X_s$ is isolated in $X_s$ we see that $X_s = \bigcup_{x \in X_s} \{x\}$ is an open covering. As $f$ is quasi-compact, the fibre $X_s$ is quasi-compact. Hence we see that $X_s$ is finite.

Conversely, assume $f$ is locally of finite type, quasi-compact and has finite fibres. Then it is locally quasi-finite by Lemma 28.19.7. Hence it is quasi-finite by Lemma 28.19.9. $\square$

    The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 3394–3402 (see updates for more information).

    \begin{lemma}
    \label{lemma-quasi-finite}
    Let $f : X \to S$ be a morphism of schemes.
    The following are equivalent:
    \begin{enumerate}
    \item $f$ is quasi-finite, and
    \item $f$ is locally of finite type, quasi-compact, and has finite fibres.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Assume $f$ is quasi-finite. In particular $f$ is locally of finite type
    and quasi-compact (since it is of finite type). Let $s \in S$. Since
    every $x \in X_s$ is isolated in $X_s$ we see that
    $X_s = \bigcup_{x \in X_s} \{x\}$ is an open covering. As $f$
    is quasi-compact, the fibre $X_s$ is quasi-compact. Hence we see
    that $X_s$ is finite.
    
    \medskip\noindent
    Conversely, assume $f$ is locally of finite type, quasi-compact
    and has finite fibres. Then it is locally quasi-finite by
    Lemma \ref{lemma-finite-fibre}. Hence it is quasi-finite by
    Lemma \ref{lemma-quasi-finite-locally-quasi-compact}.
    \end{proof}

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