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Tag 02LS

Chapter 36: More on Morphisms > Section 36.38: Zariski's Main Theorem

Lemma 36.38.4. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

  1. $f$ is finite,
  2. $f$ is proper with finite fibres,
  3. $f$ is proper and locally quasi-finite,
  4. $f$ is universally closed, separated, locally of finite type and has finite fibres.

Proof. We have (1) implies (2) by Morphisms, Lemmas 28.42.10, 28.19.10, and 28.42.9. We have (2) implies (3) by Morphisms, Lemma 28.19.7. We have (3) implies (4) by the definition of proper morphisms and Morphisms, Lemmas 28.19.9 and 28.19.10.

Assume (3). Pick $s \in S$. By Morphisms, Lemma 28.19.7 we see that all the finitely many points of $X_s$ are isolated in $X_s$. Choose an elementary étale neighbourhood $(U, u) \to (S, s)$ and decomposition $X_U = V \amalg W$ as in Lemma 36.36.6. Note that $W_u = \emptyset$ because all points of $X_s$ are isolated. Since $f$ is universally closed we see that the image of $W$ in $U$ is a closed set not containing $u$. After shrinking $U$ we may assume that $W = \emptyset$. In other words we see that $X_U = V$ is finite over $U$. Since $s \in S$ was arbitrary this means there exists a family $\{U_i \to S\}$ of étale morphisms whose images cover $S$ such that the base changes $X_{U_i} \to U_i$ are finite. Note that $\{U_i \to S\}$ is an étale covering, see Topologies, Definition 33.4.1. Hence it is an fpqc covering, see Topologies, Lemma 33.9.6. Hence we conclude $f$ is finite by Descent, Lemma 34.20.23. $\square$

    The code snippet corresponding to this tag is a part of the file more-morphisms.tex and is located in lines 10861–10872 (see updates for more information).

    \begin{lemma}
    \label{lemma-characterize-finite}
    Let $f : X \to S$ be a morphism of schemes.
    The following are equivalent:
    \begin{enumerate}
    \item $f$ is finite,
    \item $f$ is proper with finite fibres,
    \item $f$ is proper and locally quasi-finite,
    \item $f$ is universally closed, separated, locally of finite type
    and has finite fibres.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    We have (1) implies (2) by
    Morphisms, Lemmas \ref{morphisms-lemma-finite-proper},
    \ref{morphisms-lemma-quasi-finite},
    and \ref{morphisms-lemma-finite-quasi-finite}.
    We have (2) implies (3) by Morphisms, Lemma \ref{morphisms-lemma-finite-fibre}.
    We have (3) implies (4) by the definition of proper morphisms and
    Morphisms, Lemmas \ref{morphisms-lemma-quasi-finite-locally-quasi-compact} and
    \ref{morphisms-lemma-quasi-finite}.
    
    \medskip\noindent
    Assume (3). Pick $s \in S$. By
    Morphisms, Lemma \ref{morphisms-lemma-finite-fibre} we
    see that all the finitely many points of $X_s$ are isolated in $X_s$.
    Choose an elementary \'etale neighbourhood $(U, u) \to (S, s)$
    and decomposition $X_U = V \amalg W$ as in
    Lemma \ref{lemma-etale-splits-off-quasi-finite-part}.
    Note that $W_u = \emptyset$ because all points of $X_s$ are isolated.
    Since $f$ is universally closed we see that
    the image of $W$ in $U$ is a closed set not containing $u$.
    After shrinking $U$ we may assume that $W = \emptyset$.
    In other words we see that $X_U = V$ is finite over $U$.
    Since $s \in S$ was arbitrary
    this means there exists a family $\{U_i \to S\}$
    of \'etale morphisms whose images cover $S$ such that
    the base changes $X_{U_i} \to U_i$ are finite.
    Note that $\{U_i \to S\}$ is an \'etale covering,
    see Topologies, Definition \ref{topologies-definition-etale-covering}.
    Hence it is an fpqc covering, see
    Topologies,
    Lemma \ref{topologies-lemma-zariski-etale-smooth-syntomic-fppf-fpqc}.
    Hence we conclude $f$ is finite by
    Descent, Lemma \ref{descent-lemma-descending-property-finite}.
    \end{proof}

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